Integral theorems

Line and surface integrals, Gauss, Stokes and Green, potentials.

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Integrals and the integral theorems
Potentials
- Helmholtz decomposition and potentials
- The Dirac delta in calculations

Integrals and the integral theorems

Theorem

Line integrals and conservative fields

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Defines the line integral of a vector field along a piecewise $C^1$ curve and proves that path independence, the existence of a single-valued potential and exactness of $\mathbf{F}\cdot d\mathbf{r}$ are the same condition, that each forces $\nabla\times\mathbf{F} = \mathbf{0}$ , and that the converse holds on a star-shaped (more generally simply connected) domain.

Prerequisites: Gradient and directional derivative, Partial derivatives and the chain rule, Connectedness and simple connectivity, Differentiation under the integral sign

Definition(Line integral, conservative field, potential, exact differential)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.1, Eqs. (11.1)–(11.2), pp. 377–379 and §11.1.3, p. 382; Lee (2013), Ch. 11, pp. 288–292

Let $\Omega \subseteq \mathbb{R}^3$ be open and let $\mathbf{F} = (F_1, F_2, F_3)$ be a continuous vector field on $\Omega$ .

A curve segment in $\Omega$ is a continuous map $\gamma : [a,b] \to \Omega$ . It is piecewise $C^1$ if $[a,b]$ can be cut at finitely many points $a = t_0 < t_1 < \dots < t_k = b$ so that $\gamma$ is $C^1$ on each $[t_{i-1}, t_i]$ ; the one-sided velocities at the cuts need not agree. The segment is closed if $\gamma(a) = \gamma(b)$ .
The line integral of $\mathbf{F}$ along a piecewise $C^1$ segment $\gamma$ is $\int_{\gamma} \mathbf{F}\cdot d\mathbf{r} \;=\; \sum_{i=1}^{k} \int_{t_{i-1}}^{t_i} \mathbf{F}\bigl(\gamma(t)\bigr)\cdot \gamma'(t)\, dt .$ In Cartesian components this is $\int_\gamma (F_1\,dx + F_2\,dy + F_3\,dz)$ . The integral of a scalar field $\varphi$ with respect to arc length is $\int_\gamma \varphi\, ds = \int_a^b \varphi(\gamma(t))\,\lvert \gamma'(t)\rvert\, dt$ , and $\int_\gamma \varphi\, d\mathbf{r}$ and $\int_\gamma \mathbf{a}\times d\mathbf{r}$ are defined componentwise in the same way.
$\mathbf{F}$ is conservative on $\Omega$ if $\oint_\gamma \mathbf{F}\cdot d\mathbf{r} = 0$ for every piecewise $C^1$ closed segment $\gamma$ in $\Omega$ .
A potential for $\mathbf{F}$ is a function $\phi$ on $\Omega$ , single-valued and of class $C^1$ , with $\mathbf{F} = \nabla\phi$ . Equivalently, $\mathbf{F}\cdot d\mathbf{r} = d\phi$ is an exact differential: the one-form $F_1\,dx + F_2\,dy + F_3\,dz$ is the total differential of $\phi$ . Physicists usually write $\mathbf{F} = -\nabla\phi$ instead, so that $\phi$ is the work done against $\mathbf{F}$ ; the two conventions differ only in the sign of $\phi$ .
$\Omega$ is simply connected if every closed curve in $\Omega$ can be contracted to a point without leaving $\Omega$ ; it is star-shaped about $\mathbf{c} \in \Omega$ if the straight segment from $\mathbf{c}$ to any point of $\Omega$ lies in $\Omega$ . A convex set is star-shaped, and a star-shaped set is simply connected.

Theorem(Parameter independence)AdaptedAdapted from Lee (2013), Prop. 11.37, p. 290

Let $\gamma : [a,b] \to \Omega$ be piecewise $C^1$ and let $\psi : [c,d] \to [a,b]$ be a $C^1$ bijection with $\psi'$ of one sign. Then

\int_{\gamma \circ \psi} \mathbf{F}\cdot d\mathbf{r} \;=\; \begin{cases} +\displaystyle\int_{\gamma} \mathbf{F}\cdot d\mathbf{r}, & \psi \text{ increasing},\\[2mm] -\displaystyle\int_{\gamma} \mathbf{F}\cdot d\mathbf{r}, & \psi \text{ decreasing}. \end{cases}

So the integral depends only on the image curve and the direction of travel along it, not on the parametrization, and reversing the direction reverses the sign. Splitting a curve at an interior point adds the two pieces' integrals.

Theorem(Path independence, potentials and the curl)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.4, p. 387; Lee (2013), Thm 11.39, p. 291; Lee (2013), Thm 11.42, pp. 292–294; Lee (2013), Prop. 11.44, p. 294; Lee (2013), Thm 11.49, pp. 296–297

Let $\Omega \subseteq \mathbb{R}^3$ be open and connected and let $\mathbf{F}$ be a $C^1$ vector field on $\Omega$ (that is, $\mathbf{F}$ has continuous first partial derivatives there). Consider the four conditions

Path independence. $\int_\gamma \mathbf{F}\cdot d\mathbf{r}$ has the same value for any two piecewise $C^1$ segments in $\Omega$ with the same initial and final point; equivalently, $\mathbf{F}$ is conservative.
Potential. There is a single-valued $\phi \in C^2(\Omega)$ with $\mathbf{F} = \nabla\phi$ .
Exactness. $\mathbf{F}\cdot d\mathbf{r}$ is an exact differential on $\Omega$ .
Vanishing curl. $\nabla\times\mathbf{F} = \mathbf{0}$ at every point of $\Omega$ .

Then:

Conditions 1, 2 and 3 are equivalent, with no hypothesis on the shape of $\Omega$ , and each of them implies condition 4. When they hold, $\phi$ is determined up to one additive constant and $\int_{\gamma} \mathbf{F}\cdot d\mathbf{r} = \phi\bigl(\gamma(b)\bigr) - \phi\bigl(\gamma(a)\bigr)$ for every piecewise $C^1$ segment $\gamma : [a,b] \to \Omega$ .
If $\Omega$ is star-shaped, condition 4 implies the other three, so all four are equivalent. The same conclusion holds whenever $\Omega$ is simply connected.

A field satisfying these conditions is called conservative, and $\phi$ is its scalar potential.

NoteOriginal

Which implications need which hypothesis. Riley, Hobson and Bence state the four conditions as equivalent for a field with continuous partial derivatives on a simply connected region (§11.4, p. 387), so their list carries the topological hypothesis throughout. Lee separates the implications: conservative $\Leftrightarrow$ exact needs no hypothesis on the domain (Thm 11.42), exact $\Rightarrow$ closed needs none (Prop. 11.44), and only the converse of the last one needs the domain to be star-shaped (Thm 11.49) or simply connected (Cor. 16.27). The statement above follows Lee's division, which is the sharper one; the two accounts agree on simply connected domains, which is the case Riley, Hobson and Bence discuss.

Idea of the proofOriginal

Each implication is one short idea. If $\mathbf{F}$ is a gradient, the chain rule turns $\mathbf{F}\cdot\gamma'$ into the $t$ -derivative of $\phi\circ\gamma$ , and the fundamental theorem of calculus leaves only the endpoint values: that is path independence. Conversely, if the integral depends only on the endpoints, then defining $\phi$ at a point as the integral from a fixed base point is unambiguous, and probing it along a short coordinate segment recovers the corresponding component of $\mathbf{F}$ . A gradient has symmetric second derivatives, which is exactly what $\nabla\times\nabla\phi = \mathbf{0}$ says. The last implication is the only one with any geometry in it: on a star-shaped domain we can reach every point by one preferred path, the straight segment from the centre, and integrating along it produces a potential — the vanishing of the curl is what makes the resulting derivative collapse to a total derivative in the segment parameter.

ProofAdaptedAdapted from Lee (2013), Thm 11.39, p. 291; Lee (2013), Thm 11.42, pp. 292–294; Lee (2013), Prop. 11.44, p. 294; Lee (2013), Thm 11.49, pp. 296–297; Riley, Hobson & Bence (2006), §11.4, pp. 387–389

Throughout, $\Omega$ is open and connected, so any two of its points are joined by a piecewise $C^1$ segment in $\Omega$ : the set of points reachable from a fixed $\mathbf{p}_0$ by such a segment is open (append a straight segment inside a ball of $\Omega$ ) and has open complement in $\Omega$ for the same reason, so by connectedness it is all of $\Omega$ (connectedness).

Step 1: condition 2 implies condition 1, with the endpoint formula. Let $\mathbf{F} = \nabla\phi$ and let $\gamma$ be $C^1$ on a subinterval $[t_{i-1},t_i]$ . By the chain rule,

\frac{d}{dt}\,\phi\bigl(\gamma(t)\bigr) = \nabla\phi\bigl(\gamma(t)\bigr)\cdot\gamma'(t) = \mathbf{F}\bigl(\gamma(t)\bigr)\cdot\gamma'(t) ,

so by the fundamental theorem of calculus that piece contributes $\phi(\gamma(t_i)) - \phi(\gamma(t_{i-1}))$ . Summing over the pieces, the intermediate values cancel and

\int_{\gamma} \mathbf{F}\cdot d\mathbf{r} = \phi\bigl(\gamma(b)\bigr) - \phi\bigl(\gamma(a)\bigr) .

What stands on the right involves the endpoints and nothing else — this is where the hypothesis that $\phi$ is single-valued is used — and it vanishes when $\gamma$ is closed.

Step 2: condition 1 implies condition 2. Assume path independence and fix $\mathbf{p}_0 \in \Omega$ . For $\mathbf{q} \in \Omega$ put

\phi(\mathbf{q}) = \int_{\gamma_{\mathbf{q}}} \mathbf{F}\cdot d\mathbf{r},

where $\gamma_{\mathbf{q}}$ is any piecewise $C^1$ segment in $\Omega$ from $\mathbf{p}_0$ to $\mathbf{q}$ . Such a segment exists by the paragraph above, and the value does not depend on which one is used, by hypothesis; so $\phi$ is a single-valued function on $\Omega$ .

Fix $\mathbf{q} \in \Omega$ and $j \in \{1,2,3\}$ . Since $\Omega$ is open there is $\varepsilon > 0$ with $\mathbf{q} + s\,\mathbf{e}_j \in \Omega$ for $\lvert s \rvert < \varepsilon$ . Concatenating a path to $\mathbf{q}$ with the straight segment $\sigma \mapsto \mathbf{q} + \sigma\mathbf{e}_j$ , $0 \le \sigma \le s$ , gives an admissible path to $\mathbf{q} + s\mathbf{e}_j$ , whence

\phi(\mathbf{q} + s\,\mathbf{e}_j) - \phi(\mathbf{q}) = \int_0^s F_j(\mathbf{q} + \sigma\,\mathbf{e}_j)\, d\sigma ,

because along that segment $d\mathbf{r} = \mathbf{e}_j\, d\sigma$ . The integrand is continuous in $\sigma$ , so dividing by $s$ and letting $s \to 0$ gives $\partial\phi/\partial x_j(\mathbf{q}) = F_j(\mathbf{q})$ . Hence $\nabla\phi = \mathbf{F}$ , and $\phi \in C^1$ because its partial derivatives are the continuous functions $F_j$ ; since $\mathbf{F}$ is $C^1$ by hypothesis, in fact $\phi \in C^2$ .

Step 3: conditions 2 and 3 are the same statement. By definition, $\mathbf{F}\cdot d\mathbf{r}$ is exact precisely when there is a single-valued $\phi$ with $d\phi = F_1\,dx + F_2\,dy + F_3\,dz$ , and $d\phi = \nabla\phi\cdot d\mathbf{r}$ identifies the coefficients one by one: $\partial\phi/\partial x = F_1$ , and likewise for $y$ and $z$ . So condition 3 holds if and only if condition 2 does.

Step 4: condition 2 implies condition 4. With $\mathbf{F} = \nabla\phi$ and $\phi \in C^2$ by Step 2, the mixed second partial derivatives of $\phi$ are symmetric (Clairaut's theorem), so

(\nabla\times\mathbf{F})_1 = \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = \frac{\partial^2\phi}{\partial y\,\partial z} - \frac{\partial^2\phi}{\partial z\,\partial y} = 0 ,

and the other two components vanish by the same computation with the indices permuted cyclically.

Step 5: condition 4 implies condition 2 when $\Omega$ is star-shaped. Translate so that $\Omega$ is star-shaped about the origin. For $\mathbf{x} \in \Omega$ the segment $t \mapsto t\mathbf{x}$ , $0 \le t \le 1$ , lies in $\Omega$ , so

\phi(\mathbf{x}) \;=\; \int_0^1 \mathbf{F}(t\mathbf{x})\cdot\mathbf{x}\, dt \;=\; \int_0^1 \sum_{i} F_i(t\mathbf{x})\,x_i \, dt

is defined on all of $\Omega$ ; it is the line integral of $\mathbf{F}$ along that one segment. The integrand and its partial derivatives in $\mathbf{x}$ are continuous on a neighbourhood of $\{t\mathbf{x} : 0 \le t \le 1\}$ because $\mathbf{F}$ is $C^1$ , and the $t$ -interval is compact, so we may differentiate under the integral sign:

\frac{\partial \phi}{\partial x_j}(\mathbf{x}) = \int_0^1 \Bigl[\, t\sum_i \frac{\partial F_i}{\partial x_j}(t\mathbf{x})\, x_i \;+\; F_j(t\mathbf{x}) \,\Bigr]\, dt .

Now the hypothesis $\nabla\times\mathbf{F} = \mathbf{0}$ enters: it says precisely that $\partial F_i/\partial x_j = \partial F_j/\partial x_i$ for every pair $i \ne j$ , and for $i = j$ this is trivially true. Substituting,

\frac{\partial \phi}{\partial x_j}(\mathbf{x}) = \int_0^1 \Bigl[\, t\sum_i x_i\,\frac{\partial F_j}{\partial x_i}(t\mathbf{x}) \;+\; F_j(t\mathbf{x}) \,\Bigr]\, dt = \int_0^1 \frac{d}{dt}\Bigl[\, t\,F_j(t\mathbf{x}) \,\Bigr]\, dt = F_j(\mathbf{x}) ,

the middle equality being the chain rule applied to $t \mapsto t\,F_j(t\mathbf{x})$ and the last one the fundamental theorem of calculus, since the bracket vanishes at $t = 0$ and equals $F_j(\mathbf{x})$ at $t = 1$ . So $\mathbf{F} = \nabla\phi$ .

Step 6: uniqueness of $\phi$ . If $\nabla\phi = \nabla\tilde\phi$ on the connected open set $\Omega$ then $\nabla(\phi - \tilde\phi) = \mathbf{0}$ ; a function with vanishing gradient is constant along every segment inside $\Omega$ by Step 1, hence constant on $\Omega$ by connectedness. $\square$

NoteAdaptedAdapted from Lee (2013), Ch. 16, Thm 16.26 and Cor. 16.27, p. 421

The simply connected case. Step 5 proves the implication $4 \Rightarrow 2$ only for star-shaped domains, which is all that an elementary argument gives: the proof needs a family of paths, one to each point, varying smoothly with the endpoint. For a general simply connected $\Omega$ the implication is still true, but the proof runs through homotopy invariance of line integrals of closed fields — Lee's Thm 16.26 — and is beyond the scope of this unit; see the Poincaré lemma for forms. The ingredients are: a closed field has the same integral along path-homotopic curves; in a simply connected domain every closed curve is path-homotopic to a constant curve; hence every closed circulation vanishes, and Step 2 supplies the potential.

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Simple connectivity, in Step 5 only. On $\Omega = \mathbb{R}^3$ minus the $z$ -axis, the field $\mathbf{F} = \frac{1}{x^2+y^2}\,(-y,\; x,\; 0)$ is $C^\infty$ and has $\nabla\times\mathbf{F} = \mathbf{0}$ everywhere (checked in the second example below), yet its circulation around the unit circle in the $z = 0$ plane is $2\pi \ne 0$ . So condition 4 does not imply conditions 1–3 on this $\Omega$ , which is not simply connected: a loop around the axis cannot be contracted inside $\Omega$ . Restricted to the half-space $x > 0$ , which is convex, the same field is $\nabla\bigl(\arctan(y/x)\bigr)$ .
Openness and connectedness of $\Omega$ . Openness is used in Step 2 to find the probing segment, and in Step 6. Connectedness is used to join points by paths and to pin $\phi$ down to one constant; on a disconnected $\Omega$ the conclusions hold on each component separately, with an independent constant in each.
Continuity of $\mathbf{F}$ versus $\mathbf{F} \in C^1$ . Steps 1–3 and 6 need only that $\mathbf{F}$ is continuous. Step 4 needs the first derivatives of $\mathbf{F}$ to exist and be continuous, since it is a statement about them; Step 5 needs them too, to differentiate under the integral sign.
Piecewise $C^1$ rather than $C^1$ paths. Admitting corners costs nothing in Step 1 and makes Step 2 easy, because the probing segment can simply be appended to an arbitrary path. Lee notes (after Prop. 11.40) that restricting to smooth closed curves would make the equivalence harder to prove.
Single-valuedness of $\phi$ . This is not decoration. The field above has the multivalued "potential" $\varphi = \arctan(y/x)$ , i.e. the azimuthal angle, which increases by $2\pi$ on each circuit; that increase is the circulation. Riley, Hobson and Bence note the same phenomenon for the magnetic scalar potential of a current-carrying wire (p. 389): the results apply as long as the loop encloses no net current.
A field can fail to be conservative for the plainest reason. For $\mathbf{F} = (-y, x, 0)$ on $\mathbb{R}^3$ we get $\nabla\times\mathbf{F} = (0,0,2) \ne \mathbf{0}$ , and Arfken's Example 3.7.1 computes two integrals of this field from $(0,0)$ to $(1,1)$ with different values. Here no topology is involved; condition 4 already fails.

Example(A conservative field: potential and work)Original

On $\Omega = \mathbb{R}^3$ take

\mathbf{F} = \bigl(y^2 z + 2x,\;\; 2xyz + 1,\;\; xy^2 + 3z^2\bigr).

Test condition 4. Componentwise, $\frac{\partial F_3}{\partial y} = 2xy = \frac{\partial F_2}{\partial z},\qquad \frac{\partial F_1}{\partial z} = y^2 = \frac{\partial F_3}{\partial x},\qquad \frac{\partial F_2}{\partial x} = 2yz = \frac{\partial F_1}{\partial y},$ so $\nabla\times\mathbf{F} = \mathbf{0}$ . Since $\mathbb{R}^3$ is convex, hence star-shaped about the origin, the theorem gives a potential.
Build it by the formula of Step 5. With $\mathbf{F}(t\mathbf{x})\cdot\mathbf{x} = 4t^3 x y^2 z + 2t x^2 + y + 3t^2 z^3$ , $\phi(\mathbf{x}) = \int_0^1 \bigl(4t^3 x y^2 z + 2t x^2 + y + 3t^2 z^3\bigr)\, dt = x y^2 z + x^2 + y + z^3 .$ Check: $\partial\phi/\partial x = y^2z + 2x$ , $\partial\phi/\partial y = 2xyz + 1$ , $\partial\phi/\partial z = xy^2 + 3z^2$ , as required. Any other potential differs from this by a constant.
Work between two points. The work done by $\mathbf{F}$ from $(0,0,0)$ to $(1,2,3)$ is $\phi(1,2,3) - \phi(0,0,0) = \bigl(1\cdot 4\cdot 3 + 1 + 2 + 27\bigr) - 0 = 42 ,$ along every path in $\mathbb{R}^3$ joining the two points.
Two explicit paths, as a check. Along the straight segment $\gamma(t) = (t, 2t, 3t)$ , $0 \le t \le 1$ , the three components of $\mathbf{F}$ become $12t^3 + 2t$ , $12t^3 + 1$ and $4t^3 + 27t^2$ , and $\gamma' = (1,2,3)$ , so $\int_{\gamma}\mathbf{F}\cdot d\mathbf{r} = \int_0^1 \bigl(48t^3 + 81t^2 + 2t + 2\bigr)\, dt = 12 + 27 + 1 + 2 = 42 .$ Along the three-leg path $(0,0,0) \to (1,0,0) \to (1,2,0) \to (1,2,3)$ , each leg has only one varying coordinate. On the first, $y = z = 0$ and $F_1 = 2x$ , giving $\int_0^1 2x\, dx = 1$ . On the second, $x = 1$ , $z = 0$ and $F_2 = 1$ , giving $\int_0^2 1\, dy = 2$ . On the third, $x = 1$ , $y = 2$ and $F_3 = 4 + 3z^2$ , giving $\int_0^3 (4 + 3z^2)\, dz = 12 + 27 = 39$ . The total is $1 + 2 + 39 = 42$ , the same value.

Example(A curl-free field that is not conservative)Original

Let $\Omega = \mathbb{R}^3 \setminus \{x = y = 0\}$ , the complement of the $z$ -axis, and

\mathbf{F} = \Bigl(\frac{-y}{x^2+y^2},\;\; \frac{x}{x^2+y^2},\;\; 0\Bigr) .

The curl vanishes. The third component of the curl is $\frac{\partial}{\partial x}\frac{x}{x^2+y^2} - \frac{\partial}{\partial y}\frac{-y}{x^2+y^2} = \frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} + \frac{(x^2+y^2) - 2y^2}{(x^2+y^2)^2} = \frac{2(x^2+y^2) - 2(x^2+y^2)}{(x^2+y^2)^2} = 0 ,$ and the first two components vanish because $F_3 = 0$ and $F_1, F_2$ do not depend on $z$ .
The circulation does not. On the unit circle $\gamma(\theta) = (\cos\theta, \sin\theta, 0)$ , $0 \le \theta \le 2\pi$ , we have $x^2 + y^2 = 1$ and $\mathbf{F}(\gamma)\cdot\gamma' = (-\sin\theta)(-\sin\theta) + (\cos\theta)(\cos\theta) = 1 ,$ so $\oint_\gamma \mathbf{F}\cdot d\mathbf{r} = 2\pi$ .
Reading off the obstruction. By the theorem, $\mathbf{F}$ has no single-valued potential on $\Omega$ , and $\Omega$ is therefore not simply connected — the circulation $2\pi$ is a certificate of that. On any simply connected subdomain the potential exists: on $\{x > 0\}$ it is $\arctan(y/x)$ , on $\{y > 0\}$ it is $-\arctan(x/y)$ , and the two differ by $\pi/2$ on the overlap, which is consistent because each is only determined up to a constant on its own domain.
Circulation counts turns. For the circle of radius $R$ traversed $n$ times the same computation gives $2\pi n$ , independent of $R$ : the integral measures how many times the loop winds about the axis (see the winding number).

ApplicationOriginal

The physical content of the theorem is that work against a conservative force is storable as potential energy. For a force field $\mathbf{F} = -\nabla V$ , the work done by the field along any path from $\mathbf{a}$ to $\mathbf{b}$ is $V(\mathbf{a}) - V(\mathbf{b})$ , so the sum of kinetic and potential energy is constant along a trajectory: gravity, electrostatics and linear springs are all of this kind, and the whole machinery of potentials in Hamiltonian mechanics rests on it.

Line integrals that are not path independent are equally important, and the curl is the diagnostic. The electromotive force $\oint \mathbf{E}\cdot d\mathbf{r}$ around a circuit is nonzero exactly when $\nabla\times\mathbf{E} \ne \mathbf{0}$ , which by Faraday's law means a changing magnetic flux — this is how a generator works. Ampère's law $\oint_C \mathbf{B}\cdot d\mathbf{r} = \mu_0 I$ (Riley, Hobson and Bence, p. 381) is the circulation of $\mathbf{B}$ , and the field of a straight wire is, up to the factor $\mu_0 I/2\pi$ , exactly the curl-free field of the second example: its circulation is nonzero only because the wire has been removed from the domain.

In thermodynamics the same distinction is the difference between a state function and a process quantity. Internal energy has an exact differential $dU$ , so $\oint dU = 0$ around any cycle; heat and work do not, which is why a heat engine can run. In fluid mechanics the circulation $\oint \mathbf{u}\cdot d\mathbf{r}$ of the velocity field around a closed material curve is the quantity conserved by Kelvin's theorem, and an irrotational flow is one whose velocity has a single-valued potential — the starting point of classical aerofoil theory, where the lift is proportional to a circulation made nonzero by cutting the aerofoil out of the domain.

Definition

Surface integrals and flux

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Defines a regular parametrized surface patch, its scalar surface element $dS = \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\,du\,dv$ and its vector element $d\mathbf{S} = \hat{\mathbf{n}}\,dS$ , proves that both are independent of the parametrization, fixes the orientation conventions for closed and open surfaces, and gives the graph and projection formulas used to compute a flux integral.

Prerequisites: Multiple integrals and change of variables, Vector algebra and index notation, Partial derivatives and the chain rule

Definition(Regular patch, surface element, orientation, flux)AdaptedAdapted from do Carmo (2016), §2-5, pp. 98–100 and Def. 2, p. 100; Riley, Hobson & Bence (2006), §11.5, pp. 389–390; Arfken, Weber & Harris (2013), §3.7, p. 161; Strauss (2008), §A.3, pp. 421–422

Let $D \subseteq \mathbb{R}^2$ be a connected set that is the closure of a bounded open set, with coordinates $(u,v)$ .

A regular parametrized patch is a map $\mathbf{r} : D \to \mathbb{R}^3$ that is $C^1$ , injective on the interior of $D$ , and satisfies $\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}$ there, where $\mathbf{r}_u = \partial\mathbf{r}/\partial u$ and $\mathbf{r}_v = \partial\mathbf{r}/\partial v$ . Its image is the surface $S = \mathbf{r}(D)$ , and $(u,v)$ are its parameters. A piecewise $C^1$ surface is a finite union of such patches meeting along boundary curves, for example the six faces of a cube.
The two vectors $\mathbf{r}_u, \mathbf{r}_v$ span the tangent plane at $\mathbf{r}(u,v)$ . The first fundamental form of the patch records three numbers attached to that plane, whose classical names are $E$ , $F$ and $G$ : $E = \mathbf{r}_u\cdot\mathbf{r}_u, \qquad F = \mathbf{r}_u\cdot\mathbf{r}_v, \qquad G = \mathbf{r}_v\cdot\mathbf{r}_v .$ By the Lagrange identity $\lvert \mathbf{a}\times\mathbf{b}\rvert^2 + (\mathbf{a}\cdot\mathbf{b})^2 = \lvert\mathbf{a}\rvert^2\lvert\mathbf{b}\rvert^2$ (vector algebra), $\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert = \sqrt{EG - F^2} .$
The scalar surface element is $dS = \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\, du\, dv$ ; it is the area of the parallelogram spanned by $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$ . The area of $S$ and the surface integral of a scalar field $\varphi$ over $S$ are $A(S) = \iint_D \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\, du\, dv , \qquad \int_S \varphi\, dS = \iint_D \varphi\bigl(\mathbf{r}(u,v)\bigr)\, \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\, du\, dv .$
A unit normal at $\mathbf{r}(u,v)$ is one of the two unit vectors orthogonal to the tangent plane; the parametrization picks out $\hat{\mathbf{n}} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\lvert \mathbf{r}_u \times \mathbf{r}_v\rvert} .$ An orientation of a surface is a continuous choice of unit normal on it. A surface admitting one is orientable; equivalently (do Carmo's Def. 1) it can be covered by patches whose changes of parameters all have positive Jacobian determinant. The vector surface element is then $d\mathbf{S} = \hat{\mathbf{n}}\, dS = (\mathbf{r}_u \times \mathbf{r}_v)\, du\, dv$ .
The flux of a continuous vector field $\mathbf{F}$ through an oriented surface $S$ is $\int_S \mathbf{F}\cdot d\mathbf{S} \;=\; \int_S \mathbf{F}\cdot\hat{\mathbf{n}}\, dS \;=\; \iint_D \mathbf{F}\bigl(\mathbf{r}(u,v)\bigr)\cdot\bigl(\mathbf{r}_u\times\mathbf{r}_v\bigr)\, du\, dv .$ The integrals $\int_S \varphi\, d\mathbf{S}$ and $\int_S \mathbf{F}\times d\mathbf{S}$ are defined componentwise in the same way; they are vectors, whereas $\int_S \varphi\, dS$ and the flux are scalars. For a closed surface $S$ the notation $\oint_S$ is used.

Definition(Orientation conventions)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.5, Fig. 11.5 and pp. 389–390; Arfken, Weber & Harris (2013), §3.7, p. 161 with Fig. 3.13

Two conventions fix the sign of a flux, and every statement about flux depends on which one is in force.

Closed surface. A closed surface encloses a volume $V$ ; the positive normal points out of $V$ . A closed surface need not be simply connected — the surface of a torus is closed — but it must bound a volume.
Open surface. The boundary of an open surface is one or more closed curves, and a direction of traversal is chosen on each. Then $\hat{\mathbf{n}}$ is given by the right-hand rule relative to that direction: standing at a boundary point, let $\mathbf{u}$ point along the direction of travel and $\mathbf{v}$ point into the surface, tangent to it; the positive normal is the direction of $\mathbf{u}\times\mathbf{v}$ . Used at one boundary curve of a connected $S$ , the rule fixes $\hat{\mathbf{n}}$ there and so, by continuity of the normal field, everywhere on $S$ — which then fixes the traversal of each of the remaining boundary curves. Arfken insists on this formulation rather than on any appeal to "outward" or "upward", precisely so that it covers a surface that curls round on itself, his own illustration being a spherical surface with a hole cut in it.

An open surface need not be simply connected, but it does have to have two sides. The Möbius strip has one, admits no continuous normal field, and carries no flux at all.

Definition(The surface element does not depend on the parametrization)AdaptedAdapted from do Carmo (2016), §2-3, Prop. 1, p. 72 and §2-5, p. 100

Let $\mathbf{r} : D \to \mathbb{R}^3$ and $\bar{\mathbf{r}} : \bar D \to \mathbb{R}^3$ be regular patches with the same image $S$ , and let $h = \mathbf{r}^{-1}\circ\bar{\mathbf{r}}$ , the change of parameters, carry the interior of $\bar D$ onto the interior of $D$ . That $h$ is a $C^1$ diffeomorphism is not an extra assumption but a theorem: for two regular patches covering the same surface it is automatic, being do Carmo's change-of-parameters proposition (§2-3, Proposition 1, p. 72), which he proves from the inverse function theorem. He states it for his smooth surfaces; the same argument applied to $C^1$ patches returns a $C^1$ diffeomorphism, which is what is used below. Then for every continuous $\varphi$ on $S$ ,

\iint_{\bar D} \varphi\bigl(\bar{\mathbf{r}}\bigr)\,\lvert \bar{\mathbf{r}}_{\bar u} \times \bar{\mathbf{r}}_{\bar v}\rvert\, d\bar u\, d\bar v = \iint_{D} \varphi\bigl(\mathbf{r}\bigr)\,\lvert \mathbf{r}_{u} \times \mathbf{r}_{v}\rvert\, d u\, d v ,

so $A(S)$ and $\int_S \varphi\, dS$ are properties of $S$ alone. The two unit normals agree where $\det Dh > 0$ and are opposite where $\det Dh < 0$ ; hence $\int_S \mathbf{F}\cdot d\mathbf{S}$ is determined by $S$ together with a choice of orientation, and changing the orientation changes its sign.

Definition(Three practical forms of the surface element)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.5.1, Eq. (11.10), pp. 391–392; do Carmo (2016), §2-5, Exercise 5, p. 102; Strauss (2008), §A.3, Thm 5 and p. 422

Graph. For $S : z = h(x,y)$ with $h \in C^1$ on $D \subseteq \mathbb{R}^2$ , the parametrization $\mathbf{r}(x,y) = (x, y, h(x,y))$ gives $\mathbf{r}_x\times\mathbf{r}_y = (-h_x, -h_y, 1)$ , hence $d\mathbf{S} = (-h_x, -h_y, 1)\, dx\, dy, \qquad dS = \sqrt{1 + h_x^2 + h_y^2}\; dx\, dy ,$ with the upward orientation. Every regular patch is locally of this form, with respect to one of the three coordinate planes.
Level set, projected onto a coordinate plane. Let $S$ be part of $\{f = 0\}$ with $f \in C^1$ and $\nabla f \ne \mathbf{0}$ on $S$ , so that $\hat{\mathbf{n}} = \pm\nabla f/\lvert \nabla f\rvert$ . Suppose a piece of $S$ both projects one-to-one onto a region $R$ of the $xy$ -plane and has $f_z \ne 0$ at each of its points, so that it is nowhere vertical. Then on that piece $dS = \frac{dA}{\lvert \hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\rvert} = \frac{\lvert \nabla f\rvert}{\lvert f_z\rvert}\, dA , \qquad dA = dx\, dy ,$ both $\lvert\nabla f\rvert$ and $f_z$ being evaluated on $S$ . The two conditions are independent: $f_z \ne 0$ is a hypothesis here and not a consequence of one-to-one projection (see the remark below). If a line parallel to the $z$ -axis meets $S$ more than once — as happens for every closed surface — the surface must first be cut into pieces on which the projection is one-to-one, and the results added. Projections onto the other two coordinate planes are used in the same way, and a piece that is vertical over the $xy$ -plane is handled by projecting it onto one of those instead.
Sphere of radius $a$ . With $\mathbf{r}(\theta,\phi) = a(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$ on the closed rectangle $D = [0,\pi]\times[0,2\pi]$ , which is a parameter domain of the kind the definition asks for, $\mathbf{r}_\theta\times\mathbf{r}_\phi = a\,\sin\theta\;\mathbf{r}, \qquad dS = a^2\sin\theta\, d\theta\, d\phi, \qquad \hat{\mathbf{n}} = \hat{\mathbf{r}} ,$ the outward normal. On the interior $(0,\pi)\times(0,2\pi)$ the map is regular and injective, as required; what it does on the rest of $D$ is to collapse the edges $\theta = 0$ and $\theta = \pi$ to the two poles, where $\mathbf{r}_\theta\times\mathbf{r}_\phi = \mathbf{0}$ , and to cover the meridian $\phi = 0$ twice, at $\phi = 0$ and again at $\phi = 2\pi$ . Those edges are a set of zero area in $D$ , so deleting them changes no integral.

Idea of the proofOriginal

Both claims come down to the change-of-variables formula for double integrals. A change of parameters multiplies $\mathbf{r}_u$ and $\mathbf{r}_v$ by the Jacobian matrix of $h$ , so their cross product is multiplied by $\det Dh$ ; the modulus of that factor is exactly what the change-of-variables formula contributes when the region of integration is pulled back. The two cancel, which is why the area element is geometric. The sign of $\det Dh$ survives in the direction of the cross product, which is why the normal, and so the flux, is geometric only after an orientation has been chosen.

ProofAdaptedAdapted from do Carmo (2016), §2-5, p. 100

Write $(u,v) = h(\bar u, \bar v)$ , so $\bar{\mathbf{r}} = \mathbf{r}\circ h$ on the interior of $\bar D$ .

Step 1 (how the tangent vectors transform). By the chain rule (chain rule),

\bar{\mathbf{r}}_{\bar u} = \mathbf{r}_u\,\frac{\partial u}{\partial \bar u} + \mathbf{r}_v\,\frac{\partial v}{\partial \bar u}, \qquad \bar{\mathbf{r}}_{\bar v} = \mathbf{r}_u\,\frac{\partial u}{\partial \bar v} + \mathbf{r}_v\,\frac{\partial v}{\partial \bar v} .

Step 2 (how the cross product transforms). Expanding the cross product of these two expressions, the two terms in $\mathbf{r}_u\times\mathbf{r}_u$ and $\mathbf{r}_v\times\mathbf{r}_v$ vanish and the remaining two combine:

\bar{\mathbf{r}}_{\bar u}\times\bar{\mathbf{r}}_{\bar v} = \Bigl(\frac{\partial u}{\partial \bar u}\frac{\partial v}{\partial \bar v} - \frac{\partial u}{\partial \bar v}\frac{\partial v}{\partial \bar u}\Bigr)\,\mathbf{r}_u\times\mathbf{r}_v = \bigl(\det Dh\bigr)\; \mathbf{r}_u\times\mathbf{r}_v .

Taking lengths, $\lvert \bar{\mathbf{r}}_{\bar u}\times\bar{\mathbf{r}}_{\bar v}\rvert = \lvert \det Dh\rvert\,\lvert \mathbf{r}_u\times\mathbf{r}_v\rvert$ . Both cross products are nonzero by regularity, so this identity exhibits again what the change-of-parameters proposition already guarantees: $\det Dh$ vanishes nowhere.

Step 3 (change variables). The integrand on the left is therefore

\varphi\bigl(\mathbf{r}(h(\bar u,\bar v))\bigr)\,\lvert \mathbf{r}_u\times\mathbf{r}_v\rvert\bigl(h(\bar u,\bar v)\bigr)\;\lvert\det Dh\rvert ,

which is precisely the integrand produced by the change-of-variables formula applied to the map $h$ and to the continuous function $\psi = (\varphi\circ\mathbf{r})\,\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert$ on the interior of $D$ . Since $h$ is a $C^1$ diffeomorphism between the interiors and $\psi$ is bounded on a bounded region, the formula applies and the two integrals are equal. The boundaries, being Lebesgue null in the plane, contribute nothing.

Step 4 (normals and flux). By Step 2 the two unit normals satisfy $\bar{\hat{\mathbf{n}}} = \operatorname{sgn}(\det Dh)\,\hat{\mathbf{n}}$ . On a connected parameter domain $\det Dh$ has constant sign, since it is continuous and never zero; so the two parametrizations induce either the same normal field or the opposite one. Applying Steps 1–3 to $\varphi = \mathbf{F}\cdot\hat{\mathbf{n}}$ gives

\iint_{\bar D} \mathbf{F}\cdot\bigl(\bar{\mathbf{r}}_{\bar u}\times\bar{\mathbf{r}}_{\bar v}\bigr)\, d\bar u\, d\bar v = \operatorname{sgn}(\det Dh)\iint_{D} \mathbf{F}\cdot\bigl(\mathbf{r}_{u}\times\mathbf{r}_{v}\bigr)\, du\, dv ,

so the flux is well defined once the orientation is fixed, and reverses sign with it. $\square$

Method(Computing a surface integral)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.5.1, pp. 391–392; Arfken, Weber & Harris (2013), §3.7, p. 161

Describe the surface and decide whether it is closed or open. If it is open, note the direction in which its perimeter is to be traversed; if closed, note which side is the outside.
Choose a parametrization $\mathbf{r}(u,v)$ on a parameter domain $D$ that covers $S$ once: spherical angles on a sphere, $(\rho,\phi)$ or $(\phi,z)$ on a cylinder, $(x,y)$ on a graph. Cut $S$ into pieces if no single patch covers it, which is unavoidable for a closed surface.
Compute $\mathbf{r}_u\times\mathbf{r}_v$ once and for all. Its length gives $dS$ ; its direction gives $\hat{\mathbf{n}}$ .
Fix the sign. Compare the direction of $\mathbf{r}_u\times\mathbf{r}_v$ with the convention of step 1 at one convenient point of the patch, and insert an overall minus sign if they disagree. Doing this at one point suffices, because the direction varies continuously.
Substitute and integrate. Express $\mathbf{F}$ or $\varphi$ in the parameters, form the double integral over $D$ , and evaluate it as an iterated integral (multiple integrals).
Add the pieces. For a cut surface, sum the contributions, each with its own sign fixed by step 4.

How it fails. The commonest errors are a parametrization that covers part of $S$ twice, which doubles that part's contribution; forgetting step 4, which flips the sign of a flux; and projecting a closed surface onto a coordinate plane without cutting it, which silently cancels the two halves. If $\mathbf{r}_u\times\mathbf{r}_v$ vanishes on a curve or at isolated points, those sets have zero area and may be deleted, but a parametrization whose cross product vanishes on a set of positive area is not a regular patch there at all.

RemarkOriginal

Where the hypotheses are used, and conventions that differ.

Regularity, $\mathbf{r}_u\times\mathbf{r}_v \ne \mathbf{0}$ . This is what makes the tangent plane two-dimensional and the normal well defined; it is also what makes the change of parameters a diffeomorphism in Step 2. The spherical patch violates it at $\theta = 0, \pi$ , where $\sin\theta = 0$ , and the cone $\mathbf{r}(u,v) = (u\cos v, u\sin v, u)$ violates it at its apex $u = 0$ . In both cases the bad set has zero area and is simply removed.
Injectivity. Needed for $S$ to be covered once. On $(0,\pi)\times(0,4\pi)$ the spherical patch is regular everywhere but covers the sphere twice, and every integral computed from it is twice the right answer.
Orientability. Steps 1–3 of the proof never mention the normal, so areas and scalar surface integrals are defined on non-orientable surfaces too; it is only the flux that needs a globally continuous $\hat{\mathbf{n}}$ . The Möbius strip is the standard example where none exists.
$f_z \ne 0$ in the projection formula. A one-to-one projection does not supply it. Take $f = z^3 - y$ on the slab $\lvert x\rvert < 1$ , $\lvert z\rvert < 1$ : here $\nabla f = (0, -1, 3z^2)$ never vanishes, so $S = \{y = z^3\}$ is a legitimate level surface, and $S$ projects one-to-one onto the $xy$ -plane, since $y$ determines $z = y^{1/3}$ . Yet $f_z = 3z^2$ vanishes on the whole line $z = 0$ of $S$ , where the normal $(0,-1,0)$ is horizontal, the surface is vertical, and the displayed $dS$ has a zero denominator. What the derivation needs is $\lvert\hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\rvert \ne 0$ , which is $f_z \ne 0$ ; one-to-one projection only prevents the surface from doubling back. (This surface is in fact a graph $y = z^3$ over the $xz$ -plane, so the right move for it is to project the other way, as form 2 says.)
The modulus in the projection formula. Riley, Hobson and Bence write Eq. (11.10) as $dS = \lvert\nabla f\rvert\, dA/(\partial f/\partial z)$ , without a modulus sign in the denominator; as they say, both quantities are evaluated on $S$ . That form is correct where $\partial f/\partial z > 0$ , which holds on the upper piece of a surface such as a sphere when $f$ is chosen with an outward-increasing gradient, and it is the piece they project in their example. For the lower piece $\partial f /\partial z < 0$ and the formula returns a negative $dS$ ; the version with $\lvert f_z\rvert$ above holds on both and is what the derivation $dS = dA/\lvert\hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\rvert$ gives directly, since $dS$ and $dA$ are both areas.
$\sqrt{EG - F^2}$ versus $\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert$ . These are equal, by the Lagrange identity. The first form is the one that survives on an abstract surface with a metric, where there is no ambient space and so no cross product, and is the one used in the first fundamental form; the second is quicker when $\mathbf{r}$ is given explicitly in $\mathbb{R}^3$ .
Notation. Arfken writes the vector element $d\boldsymbol{\sigma}$ and the scalar one $dA$ ; Riley, Hobson and Bence write $d\mathbf{S}$ and $dS$ , which we follow. Nothing but the symbols differs.

Example(Area of a piece of the surface

z = xy

)Original

Let $S$ be the part of the graph $z = xy$ lying over the closed unit disc $D = \{x^2 + y^2 \le 1\}$ .

Surface element. Here $h = xy$ , so $h_x = y$ , $h_y = x$ , and by form 1 above $dS = \sqrt{1 + x^2 + y^2}\, dx\, dy$ .
Change to plane polars (change of variables): $x^2 + y^2 = \rho^2$ and $dx\,dy = \rho\, d\rho\, d\phi$ , so $A(S) = \int_0^{2\pi}\!\!\int_0^1 \sqrt{1 + \rho^2}\;\rho\, d\rho\, d\phi = 2\pi \int_0^1 \rho\sqrt{1+\rho^2}\, d\rho .$
Evaluate. With $s = 1 + \rho^2$ , $ds = 2\rho\, d\rho$ , the inner integral is $\tfrac12\int_1^2 \sqrt{s}\, ds = \tfrac13\bigl(2^{3/2} - 1\bigr)$ , so $A(S) = \frac{2\pi}{3}\bigl(2\sqrt{2} - 1\bigr) = 3.82945 \ \text{(5 d.p.)} .$
Sanity check. The disc itself has area $\pi = 3.14159$ , and the surface lies above it with slope $\lvert\nabla h\rvert = \rho \le 1$ , so its area must exceed $\pi$ but not by more than the factor $\max\sqrt{1+\rho^2} = \sqrt{2}$ , i.e. it lies between $3.1416$ and $4.4429$ . It does.

Example(Flux through a paraboloid cap)Original

Let $S$ be the part of the paraboloid $z = 4 - x^2 - y^2$ with $z \ge 0$ , oriented so that the normal has a positive $z$ -component (upward), and let

\mathbf{F} = (y,\; -x,\; z) .

Parametrize as a graph over the disc $D = \{x^2 + y^2 \le 4\}$ : $h = 4 - x^2 - y^2$ , so $h_x = -2x$ , $h_y = -2y$ and $d\mathbf{S} = (-h_x, -h_y, 1)\, dx\, dy = (2x,\; 2y,\; 1)\, dx\, dy .$ Its third component is $+1 > 0$ , so this is already the required orientation and no sign change is needed (step 4 of the method).
Form the integrand. On $S$ , $\mathbf{F}\cdot(2x, 2y, 1) = 2xy - 2xy + (4 - x^2 - y^2) = 4 - x^2 - y^2 .$ The two circulating components of $\mathbf{F}$ cancel against the normal: they are tangent to the circles $x^2+y^2 = \text{const}$ , which lie in $S$ .
Integrate in plane polars. $\int_S \mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi}\!\!\int_0^2 (4 - \rho^2)\,\rho\, d\rho\, d\phi = 2\pi\Bigl[2\rho^2 - \frac{\rho^4}{4}\Bigr]_0^2 = 2\pi(8 - 4) = 8\pi .$
Check by a second parametrization. Using $(\rho,\phi)$ directly, $\mathbf{r} = (\rho\cos\phi, \rho\sin\phi, 4-\rho^2)$ gives $\mathbf{r}_\rho\times\mathbf{r}_\phi = (2\rho^2\cos\phi,\, 2\rho^2\sin\phi,\, \rho)$ , whose third component is positive, and $\mathbf{F}\cdot(\mathbf{r}_\rho\times\mathbf{r}_\phi) = (4-\rho^2)\rho$ : the same integral, as the parametrization-independence statement requires.
Check the sign convention. Reversing the orientation to point downward changes the answer to $-8\pi$ . Were $S$ closed off by the disc $z = 0$ , $x^2+y^2\le 4$ and read as the boundary of the enclosed solid, the upward normal on the cap would be the outward one, so $8\pi$ is the outward flux through the curved part.

ApplicationOriginal

Flux is how a conservation law is written down before it is turned into a differential equation. The mass of fluid crossing a surface per unit time is $\int_S \rho\mathbf{v}\cdot d\mathbf{S}$ ; the heat crossing it is $-\int_S k\nabla T\cdot d\mathbf{S}$ by Fourier's law; the electric charge enclosed is $\varepsilon_0\oint_S \mathbf{E}\cdot d\mathbf{S}$ by Gauss's law, and the magnetic flux through a circuit is $\int_S \mathbf{B}\cdot d\mathbf{S}$ , whose rate of change is the induced electromotive force. In each case the orientation convention is what makes "in" and "out", or the sign of an induced current, unambiguous — which is why Arfken insists on stating the right-hand rule carefully.

Scalar surface integrals carry the quantities that do not have a direction: the total surface charge $\int_S \sigma\, dS$ , the area of a membrane in a surface-tension energy, the drag integral $\int_S \tau\, dS$ over a wetted hull, the radiating area of a heat sink. The parametrization-independence proved above is what lets an engineer compute such a quantity in whatever coordinates the geometry suggests and still get the same number; and in numerical practice it is what justifies replacing a curved panel by a flat quadrilateral of the same area, the basis of boundary-element and panel methods.

The two integral theorems that follow — the divergence theorem and Stokes' theorem — are statements about exactly these integrals, and are the reason the conventions above have to be fixed once and used consistently.

Theorem

The divergence theorem

Adapted

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States Gauss's theorem — the outward flux of a $C^1$ field through the boundary of a bounded region equals the integral of its divergence over the region — proves it for regions bounded by $C^1$ graphs in each coordinate direction, derives the gradient and curl companion identities, and uses it to turn conservation of mass into the continuity equation.

Prerequisites: Surface integrals and flux, Divergence, curl and the Laplacian, Multiple integrals and change of variables, Differentiation under the integral sign

Definition(Simple region)Original

Call a compact set $V \subseteq \mathbb{R}^3$ $z$ -simple if there are a compact region $D \subseteq \mathbb{R}^2$ whose boundary is a piecewise $C^1$ curve and functions $\psi_1 \le \psi_2$ of class $C^1$ on $D$ , with $\psi_1 < \psi_2$ at every interior point of $D$ , such that

V = \bigl\{ (x,y,z) \;:\; (x,y) \in D, \;\; \psi_1(x,y) \le z \le \psi_2(x,y) \bigr\} .

Define $x$ -simple and $y$ -simple in the same way with the roles of the coordinates permuted, and call $V$ a simple region if it is simple in all three directions. An axis-aligned box is one, and so is the corner tetrahedron $\{x, y, z \ge 0,\; x+y+z \le c\}$ , which in each of the three directions has one affine face as its top and one as its bottom; so is the solid between two $C^1$ graphs when it is also simple the other two ways.

Convex polyhedra in general are not simple regions, because the definition asks for $\psi_1, \psi_2$ of class $C^1$ and a solid whose top is made of more than one face has a crease where two of them meet. The functions are forced by $V$ , so there is nothing to choose: for the regular octahedron $\lvert x\rvert + \lvert y\rvert + \lvert z\rvert \le 1$ they are $D = \{\lvert x\rvert + \lvert y\rvert \le 1\}$ , $\psi_2 = 1 - \lvert x\rvert - \lvert y\rvert$ and $\psi_1 = -\psi_2$ , and $\psi_2$ is not differentiable at interior points of $D$ such as $(0, \tfrac12)$ . So the octahedron is not $z$ -simple, and by its symmetry in the three coordinates not simple in any direction. A cube resting on one vertex fails in the same way, with three faces above and three below; so does a tetrahedron with three upper faces, such as the one on the vertices $(0,0,0)$ , $(1,0,0)$ , $(0,1,0)$ , $(\tfrac13,\tfrac13,1)$ , whose apex lies over the interior of its base triangle. Each of these solids is instead a finite union of simple regions, cut along the offending creases, and that is what Step 5 of the proof below exists to handle.

The boundary $\partial V$ of a $z$ -simple region consists of the top $z = \psi_2$ and the bottom $z = \psi_1$ , both graphs over $D$ , together with the lateral part lying over $\partial D$ , on which the outward unit normal $\hat{\mathbf{n}}$ is horizontal, so that $\hat{\mathbf{n}}\cdot\hat{\mathbf{k}} = 0$ there. (That part is the vertical cylinder $\{(x,y) \in \partial D,\; \psi_1 \le z \le \psi_2\}$ , and it degenerates to a curve over whatever part of $\partial D$ carries $\psi_1 = \psi_2$ , as the whole rim of the disc does for the two hemispheres of a ball.)

A simple region is a closed set. When the theorem below is applied to one, $V$ is to be read as its interior and $\overline V$ as the simple region itself; the boundary has zero volume, so which of the two carries the volume integral makes no difference to it.

Theorem(Divergence theorem (Gauss's theorem))AdaptedAdapted from Strauss (2008), Appendix §A.3, Divergence Theorem 7, p. 423; Evans (2010), Appendix §C.2, Thm 1, p. 667; Riley, Hobson & Bence (2006), §11.8, Eq. (11.18), p. 401

Let $V \subseteq \mathbb{R}^3$ be a bounded open region — Strauss's "bounded spatial domain" — whose boundary surface $S = \partial V$ is piecewise $C^1$ , so that the closed region is $\overline V = V \cup S$ ; let $\hat{\mathbf{n}}$ be the unit outward normal on $S$ , and let $\mathbf{F}$ be a vector field of class $C^1$ on $\overline V$ . Then

\int_V \nabla\cdot\mathbf{F}\; dV \;=\; \oint_S \mathbf{F}\cdot\hat{\mathbf{n}}\; dS \;=\; \oint_S \mathbf{F}\cdot d\mathbf{S} .

This is Strauss's Theorem 7. Evans states the same result for a bounded open $U \subseteq \mathbb{R}^n$ with $\partial U$ of class $C^1$ and $\mathbf{u} \in C^1(\overline U; \mathbb{R}^n)$ , and derives it from the componentwise form

\int_U \frac{\partial u}{\partial x_i}\, dx = \int_{\partial U} u\,\nu_i\, dS \qquad (i = 1,\dots,n), \quad u \in C^1(\overline U),

with $\boldsymbol{\nu}$ the outward normal; taking $u = u_i$ and summing over $i$ gives the divergence form. The theorem holds whether or not $V$ is simply connected: the region between two concentric spheres is admissible, with the normal pointing out of $V$ on both components of $S$ , hence into the inner cavity.

Theorem(Companion identities)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.8.2, Eqs. (11.21)–(11.22) and the worked derivation, p. 403; Evans (2010), Appendix §C.2, Thm 2, p. 667; Arfken, Weber & Harris (2013), §3.8, Eqs. (3.89)–(3.90), p. 166

Under the same hypotheses, with $\varphi$ a $C^1$ scalar field and $\mathbf{b}$ a $C^1$ vector field on $\overline V$ :

\int_V \nabla\varphi\; dV = \oint_S \varphi\, d\mathbf{S}, \qquad\qquad \int_V \nabla\times\mathbf{b}\; dV = \oint_S d\mathbf{S}\times\mathbf{b} ,

both equalities between vectors. Three consequences are used constantly:

$\oint_S d\mathbf{S} = \mathbf{0}$ for every closed surface (take $\varphi \equiv 1$ ).
$\displaystyle \operatorname{vol}(V) = \frac13\oint_S \mathbf{r}\cdot d\mathbf{S}$ , since $\nabla\cdot\mathbf{r} = 3$ .
Integration by parts. For $u, v \in C^1(\overline V)$ , applying the componentwise form to $uv$ gives $\int_V \frac{\partial u}{\partial x_i}\, v\; dV = -\int_V u\,\frac{\partial v}{\partial x_i}\; dV + \oint_S u\,v\,\nu_i\; dS ,$ which is Evans's Theorem 2 and the source of Green's identities.

Idea of the proofOriginal

In one variable the fundamental theorem of calculus says that integrating a derivative over an interval leaves only the values at the two ends. The divergence theorem is the same statement in three variables, done one coordinate at a time. Integrate $\partial F_3/\partial z$ first in $z$ : over a region squeezed between two graphs, the inner integral is the value of $F_3$ on the top minus its value on the bottom. The geometric content is then a single observation about the surface element: on the top and bottom of the region, the horizontal area element $dx\,dy$ is exactly $\pm \hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\, dS$ , and on the sides $\hat{\mathbf{n}}\cdot\hat{\mathbf{k}}$ vanishes. So the two-term difference is precisely the flux of $F_3\hat{\mathbf{k}}$ out of the whole boundary. Repeating with $x$ and $y$ and adding the three gives the theorem. Nothing else is involved; the apparent power of the theorem comes from the fact that cutting a region into pieces makes all the internal faces cancel in pairs, which is why the result extends from blocks to arbitrary regions.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §11.3, pp. 384–385 and §11.8, p. 401; Evans (2010), Appendix §C.2, p. 667

We prove the theorem for a region whose closure is a simple region; the hypotheses on $\mathbf{F}$ are those of the statement. Throughout the proof $V$ denotes that closed region, as in the definition, which costs nothing: $S$ has zero volume, so the volume integrals are the same either way.

Step 1 (one component, in the matching direction). Let $V$ be $z$ -simple, with $D$ , $\psi_1$ , $\psi_2$ as in the definition, and write $F_3$ for the third component of $\mathbf{F}$ . Since $\partial F_3/\partial z$ is continuous on the compact set $V$ , the triple integral may be computed as an iterated integral (iterated integrals), innermost in $z$ :

\int_V \frac{\partial F_3}{\partial z}\; dV = \iint_D \left( \int_{\psi_1(x,y)}^{\psi_2(x,y)} \frac{\partial F_3}{\partial z}(x,y,z)\, dz \right) dx\, dy = \iint_D \Bigl[ F_3\bigl(x,y,\psi_2\bigr) - F_3\bigl(x,y,\psi_1\bigr) \Bigr]\, dx\, dy ,

the inner integral being evaluated by the fundamental theorem of calculus in the variable $z$ alone, with $x$ and $y$ held fixed.

Step 2 (the same difference, read as a flux). The top of $\partial V$ is the graph $z = \psi_2$ over $D$ , whose outward normal points upward; by the graph formula for the surface element (surface integrals),

\hat{\mathbf{n}}\, dS = \bigl(-\partial_x\psi_2,\; -\partial_y\psi_2,\; 1\bigr)\, dx\, dy \qquad\text{so}\qquad \hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\; dS = dx\, dy .

On the bottom $z = \psi_1$ the outward normal points downward, which is the opposite choice, so there $\hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\, dS = -dx\, dy$ . On the lateral part the outward normal is horizontal, so $\hat{\mathbf{n}}\cdot\hat{\mathbf{k}} = 0$ and that part contributes nothing. Hence

\oint_S F_3\,\bigl(\hat{\mathbf{n}}\cdot\hat{\mathbf{k}}\bigr)\, dS = \iint_D F_3(x,y,\psi_2)\, dx\, dy - \iint_D F_3(x,y,\psi_1)\, dx\, dy ,

which is the right-hand side of Step 1. Combining,

\int_V \frac{\partial F_3}{\partial z}\; dV = \oint_S F_3\,n_3\; dS . \tag{$\ast$}

Step 3 (the other two components). Because $V$ is also $x$ -simple and $y$ -simple, the argument of Steps 1 and 2 applies with the coordinates permuted. Each direction brings its own data: being $x$ -simple means that $V$ lies between two $C^1$ graphs $x = \chi_1(y,z) \le \chi_2(y,z)$ over a compact region $D' \subseteq \mathbb{R}^2$ in the $(y,z)$ -plane, and these have nothing to do with the $D$ , $\psi_1$ , $\psi_2$ of Step 1; likewise for $y$ . With that reading, Steps 1 and 2 give

\int_V \frac{\partial F_1}{\partial x}\; dV = \oint_S F_1\,n_1\; dS, \qquad \int_V \frac{\partial F_2}{\partial y}\; dV = \oint_S F_2\,n_2\; dS .

Note that each of the three identities needs $V$ to be simple in its own direction; this is the only place where the shape of $V$ is used, and it is used three times.

Step 4 (add). Summing the three identities and using $\nabla\cdot\mathbf{F} = \partial_xF_1 + \partial_yF_2 + \partial_zF_3$ and $\mathbf{F}\cdot\hat{\mathbf{n}} = F_1n_1 + F_2n_2 + F_3n_3$ gives

\int_V \nabla\cdot\mathbf{F}\; dV = \oint_S \mathbf{F}\cdot\hat{\mathbf{n}}\; dS .

Step 5 (regions assembled from simple pieces). Suppose $V = V_1 \cup \dots \cup V_m$ , where the $V_i$ are simple regions meeting only along pieces of their boundaries. Adding the theorem over the pieces, every face interior to $V$ occurs in exactly two of the surface integrals, carrying the outward normal of one piece and therefore the inward normal of the other. The two contributions are equal and opposite and cancel, while the volume integrals add to $\int_V$ . What survives on the surface side is the integral over the part of $\bigcup_i \partial V_i$ lying in $\partial V$ , with the normal pointing out of $V$ . So the theorem holds for $V$ . $\square$

NoteAdaptedAdapted from Lee (2013), Ch. 16, Thm 16.32, pp. 423–424

The general case. The statement above is for any bounded region with a piecewise $C^1$ boundary, which is more than Steps 1–5 deliver: a ball, for instance, is $z$ -simple with graphs $\psi_{1,2} = \mp\sqrt{a^2 - x^2 - y^2}$ that are continuous on the closed disc but not $C^1$ up to its rim. Two devices close the gap, and neither is proved here. The elementary one is exhaustion: apply the theorem to the region between the two graphs over the disc of radius $a - \varepsilon$ , whose boundary now also has a short cylindrical band of area $4\pi(a-\varepsilon)\sqrt{2a\varepsilon - \varepsilon^2} \to 0$ ; as $\varepsilon \to 0$ the volume integrals converge because $\nabla\cdot\mathbf{F}$ is bounded, the band's flux tends to $0$ for the same reason, and the two cap integrals converge to the integral over the sphere. The structural one is Lee's Theorem 16.32, which obtains the divergence theorem on any oriented Riemannian manifold with boundary as a corollary of Stokes's theorem for differential forms; see Stokes' theorem for forms. Strauss and Evans both state the theorem without proof and refer the reader elsewhere, and we follow them in taking it as given in the stated generality.

RemarkOriginal

Where the hypotheses are used, and what fails without them.

$\mathbf{F}$ of class $C^1$ on the closed region. Step 1 needs $\partial F_3/\partial z$ to exist and be integrable on $V$ ; Step 2 needs $F_3$ to have boundary values on $S$ . A field that is $C^1$ only in the open region can fail: $\mathbf{F} = \hat{\mathbf{r}}/r^2$ on the punctured ball $0 < r \le a$ has $\nabla\cdot\mathbf{F} = 0$ everywhere it is defined, so the left side is $0$ , while the outward flux through $r = a$ is $(1/a^2)(4\pi a^2) = 4\pi$ . The theorem is not contradicted — the hypothesis fails at the origin — and the discrepancy is exactly what the Dirac delta is introduced to bookkeep: writing $\nabla\cdot(\hat{\mathbf{r}}/r^2) = 4\pi\delta^{(3)}(\mathbf{r})$ restores the identity. Riley, Hobson and Bence make the same point with a point source of strength $Q$ in an incompressible fluid (p. 405), where $\nabla\cdot\mathbf{v} = Q\delta(\mathbf{r})$ and the flux is $Q$ or $0$ according to whether $S$ encloses the source.
Boundedness of $V$ and the closed surface. The flux integral is over a closed surface bounding a region of nonzero volume. For an open surface the statement is meaningless; the standard manoeuvre, as in Riley, Hobson and Bence's example on p. 401, is to cap the open surface so that the union is closed, apply the theorem, and subtract the cap's flux.
Simple connectivity is not required. Nothing in the proof uses it, and Riley, Hobson and Bence state the theorem for simply and multiply connected closed surfaces alike, asking only that the surface be closed and enclose a non-zero volume (p. 401). A solid torus is admissible, and so is a spherical shell $a \le r \le b$ , whose boundary has two components, with the outward normal $+\hat{\mathbf{r}}$ on $r = b$ and $-\hat{\mathbf{r}}$ on $r = a$ ; forgetting the sign on the inner sphere is the commonest error in applications. Arfken, cited above for the companion identities, does impose the hypothesis: he states Gauss' theorem for a field continuous with its first derivatives over "a simply connected region of $\mathbb{R}^3$ ", adding that a region containing a hole, like a doughnut, is not one (p. 164). Neither his subdivision argument nor ours uses that restriction, so the two sources differ in strength and not in content; we follow Riley, Hobson and Bence, and the difference is logged in verification/discrepancies.md.
Orientability is automatic here. The boundary of a region in $\mathbb{R}^3$ is always two-sided, so the flux is always defined — unlike the situation for a general open surface.
Where the proof could not be shortened. Step 3 cannot be replaced by "by symmetry" unless $V$ really is simple in all three directions. The solid between the paraboloid $z = x^2+y^2$ and the plane $z = 1$ is $z$ -simple with $C^1$ graphs, but as a region over the $(y,z)$ -plane its graphs $x = \pm\sqrt{z - y^2}$ are not $C^1$ up to the curve $z = y^2$ . The identity $(\ast)$ for the third component holds for it directly; the other two need the general statement or the exhaustion argument.
Dimension. Evans's version holds in $\mathbb{R}^n$ for every $n \ge 1$ . For $n = 2$ it is the divergence form of Green's theorem, $\int_D \nabla\cdot\mathbf{f}\, dA = \oint_C \mathbf{f}\cdot\hat{\mathbf{n}}\, ds$ with $\hat{\mathbf{n}}\, ds = (dy, -dx)$ , which Strauss gives as his Eq. (8); for $n = 1$ it is the fundamental theorem of calculus, the "surface" being two points with normals $\pm 1$ .

Example(A tetrahedron, both sides computed)Original

Let $V$ be the tetrahedron with vertices $(0,0,0)$ , $(2,0,0)$ , $(0,2,0)$ , $(0,0,2)$ , that is $\{x,y,z \ge 0,\; x+y+z \le 2\}$ , and let $\mathbf{F} = (x^2,\, y^2,\, z^2)$ . This $V$ is a simple region, so the proof above applies to it directly.

Volume side. $\nabla\cdot\mathbf{F} = 2(x+y+z)$ , so by the symmetry of $V$ in the three coordinates $\int_V \nabla\cdot\mathbf{F}\; dV = 6\int_V x\; dV = 6\int_0^2 x\,\frac{(2-x)^2}{2}\, dx = 6\cdot\frac{2^4}{24} = 4 ,$ where the cross-section of $V$ at height $x$ is a right triangle of area $(2-x)^2/2$ , and $\int_0^a x(a-x)^2\,dx/2 = a^4/24$ .
Surface side: the three coordinate faces. On $\{x = 0\}$ the outward normal is $-\hat{\mathbf{i}}$ , so $\mathbf{F}\cdot\hat{\mathbf{n}} = -x^2 = 0$ there. The same happens on $\{y=0\}$ and $\{z=0\}$ . These three faces contribute nothing — each component of $\mathbf{F}$ vanishes on the face it is normal to.
Surface side: the oblique face. On $x+y+z = 2$ the outward unit normal is $\hat{\mathbf{n}} = (1,1,1)/\sqrt3$ , and the face projects one-to-one onto the triangle $T = \{x,y\ge 0,\, x+y \le 2\}$ with $dS = \sqrt3\, dx\, dy$ (the graph $z = 2-x-y$ has $\sqrt{1 + 1 + 1} = \sqrt3$ ). Hence $\int_{\text{oblique}} \mathbf{F}\cdot\hat{\mathbf{n}}\, dS = \iint_T \frac{x^2+y^2+z^2}{\sqrt3}\,\sqrt3\; dx\, dy = \iint_T \bigl(x^2 + y^2 + (2-x-y)^2\bigr)\, dx\, dy .$ The three terms are equal by the symmetry of $T$ under permuting $x$ , $y$ and $2-x-y$ , so this is $3\iint_T x^2\, dx\, dy = 3\int_0^2 x^2 (2-x)\, dx = 3\Bigl[\frac{2x^3}{3} - \frac{x^4}{4}\Bigr]_0^2 = 3\Bigl(\frac{16}{3} - 4\Bigr) = 4 .$
Comparison. Both sides equal $4$ . As a dimensional check, the face has area $2\sqrt3 = 3.4641$ and $\mathbf{F}\cdot\hat{\mathbf{n}} = (x^2+y^2+z^2)/\sqrt3$ runs from $4/\sqrt3 = 2.309$ at the vertices to $(3\cdot\frac49)/\sqrt3 = 0.770$ at the centroid, so the flux must lie between $2.67$ and $8.00$ ; it does.

Example(A curved boundary: paraboloid and plane)Original

Let $V$ be the solid bounded below by $z = x^2+y^2$ and above by $z = 1$ , and let $\mathbf{F} = \mathbf{r} = (x,y,z)$ .

Volume side. $\nabla\cdot\mathbf{r} = 3$ , and the horizontal cross-section at height $z$ is a disc of radius $\sqrt z$ , of area $\pi z$ , so $\int_V 3\; dV = 3\int_0^1 \pi z\; dz = \frac{3\pi}{2} .$
Surface side: the lid. On $z = 1$ , $x^2+y^2 \le 1$ , the outward normal is $+\hat{\mathbf{k}}$ and $\mathbf{r}\cdot\hat{\mathbf{k}} = z = 1$ , so the flux is the area of the disc, $\pi$ .
Surface side: the paraboloid. Here the outward normal points away from the axis and downward. Writing the surface as the graph $z = h(x,y) = x^2+y^2$ over the unit disc, the upward element is $(-h_x,-h_y,1)\,dx\,dy = (-2x,-2y,1)\,dx\,dy$ ; the outward one is its negative, $(2x, 2y, -1)\, dx\, dy$ . Then $\mathbf{r}\cdot(2x,2y,-1) = 2x^2 + 2y^2 - z = 2\rho^2 - \rho^2 = \rho^2$ on the surface, and in plane polars $\int_{\text{parab}} \mathbf{r}\cdot d\mathbf{S} = \int_0^{2\pi}\!\!\int_0^1 \rho^2\cdot\rho\, d\rho\, d\phi = \frac{2\pi}{4} = \frac{\pi}{2} .$
Comparison. $\pi + \pi/2 = 3\pi/2$ , matching step 1. Note how step 3 would have come out $-\pi/2$ with the orientation taken carelessly, and the two sides would then have differed by $\pi$ — the usual symptom of a sign slip in a flux.
The volume, for free. The same field gives $\operatorname{vol}(V) = \frac13\oint \mathbf{r}\cdot d\mathbf{S} = \frac13\cdot\frac{3\pi}{2} = \frac{\pi}{2}$ , which agrees with $\int_0^1 \pi z\, dz$ .

ApplicationAdaptedAdapted from Riley, Hobson & Bence (2006), §11.8.3, p. 404

The theorem's main use is to convert a conservation law, which is naturally a statement about a region and its boundary, into a partial differential equation, which is a statement at a point. Consider a fluid of density $\rho(\mathbf{r},t)$ moving with velocity $\mathbf{v}(\mathbf{r},t)$ , with no creation or destruction of matter. For any fixed region $V$ in the fluid, the mass inside is $\int_V \rho\, dV$ , and the only way it can change is by transport across $S = \partial V$ , at the rate $\oint_S \rho\mathbf{v}\cdot d\mathbf{S}$ outwards. So

\frac{d}{dt}\int_V \rho\, dV = -\oint_S \rho\mathbf{v}\cdot d\mathbf{S} .

Taking the derivative inside the integral on the left (legitimate for $\rho$ and $\partial\rho/\partial t$ continuous, differentiation under the integral sign) and the divergence theorem on the right,

\int_V \left( \frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) \right) dV = 0 .

The region $V$ was arbitrary and the integrand is continuous, so the integrand vanishes identically: a continuous function whose integral vanishes over every region is zero, since otherwise it would keep one sign on a small ball around a point where it is nonzero. This is the continuity equation

\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) = 0 ,

which reduces to $\nabla\cdot\mathbf{v} = 0$ for an incompressible fluid. The identical argument with $\rho$ read as charge density and $\rho\mathbf{v}$ as current density gives charge conservation; with $\rho$ read as heat content and the flux given by Fourier's law $-k\nabla T$ , it gives the heat equation; and with a source term $\sigma$ on the right it gives the general transport equation of conservation laws. Gauss's law of electrostatics, $\oint_S \mathbf{E}\cdot d\mathbf{S} = q/\varepsilon_0$ , becomes $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$ by the same route, and the integration-by-parts identity above is the first step in every finite-element formulation of an elliptic boundary-value problem (finite elements).

Theorem

Stokes' theorem and Green's theorem in the plane

Adapted

Add to PDF

Proves Green's theorem in the plane for a region bounded by graphs in both coordinate directions, then obtains Stokes' theorem — the circulation of a $C^1$ field round the perimeter of an oriented surface equals the flux of its curl through the surface — by pulling the circulation back to the parameter domain, and fixes the orientation convention that makes the two sides agree in sign.

Prerequisites: Surface integrals and flux, Line integrals and conservative fields, Divergence, curl and the Laplacian, Vector algebra and index notation, Multiple integrals and change of variables, Partial derivatives and the chain rule

Theorem(Green's theorem in the plane)AdaptedAdapted from Strauss (2008), Appendix §A.3, Green's Thm 6, p. 422; Riley, Hobson & Bence (2006), §11.3, Eq. (11.4), p. 384

Let $R$ be a bounded plane domain whose boundary $C = \partial R$ is a piecewise $C^1$ curve. Parametrize $C$ to go round exactly once, keeping $R$ to the left-hand side; for a region with no holes that is the anticlockwise direction, seen from the positive $z$ -axis. Let $P$ and $Q$ be $C^1$ functions on $\overline R = R \cup C$ . Then

\oint_C \bigl(P\, dx + Q\, dy\bigr) \;=\; \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy .

Riley, Hobson and Bence impose their hypothesis on $P$ , $Q$ and the partial derivatives of both: all are to be single-valued, finite and continuous, on $R$ as well as on its boundary curve, and their $R$ is simply connected. For a multiply connected region the line integral is taken over all the boundary curves, each traversed with $R$ on the left — so the outer boundary anticlockwise and each hole clockwise. Strauss's example is the annulus, whose boundary is a pair of circles traversed in opposite senses.

Two equivalent readings, both used below:

Circulation form. With $\mathbf{f} = (P,Q)$ and $\hat{\boldsymbol{\tau}}$ the unit tangent in the direction of travel, the left side is $\oint_C \mathbf{f}\cdot\hat{\boldsymbol{\tau}}\, ds$ , the circulation of $\mathbf{f}$ round $C$ .
Divergence form. Putting $P = -g$ , $Q = f$ for $\mathbf{f} = (f,g)$ , and using $\hat{\mathbf{n}}\, ds = (dy, -dx)$ for the outward normal, the theorem becomes $\iint_R \nabla\cdot\mathbf{f}\; dA = \oint_C \mathbf{f}\cdot\hat{\mathbf{n}}\, ds$ : the two-dimensional divergence theorem.

Theorem(Stokes' theorem)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.9, Eq. (11.23), p. 406; Arfken, Weber & Harris (2013), §3.8, Eq. (3.92), p. 167 and Eqs. (3.94)–(3.95), p. 168; Lee (2013), Ch. 16, Thm 16.34, p. 427

Let $S$ be a two-sided (orientable) piecewise $C^1$ surface in $\mathbb{R}^3$ with perimeter curve $C = \partial S$ , piecewise $C^1$ , and let $\mathbf{F}$ be a $C^1$ vector field on an open set containing $S$ . Orient $S$ by a choice of unit normal $\hat{\mathbf{n}}$ and let $C$ be traversed in the sense the right-hand rule attaches to $\hat{\mathbf{n}}$ (the convention of surface integrals: with $\mathbf{u}$ along the direction of travel and $\mathbf{v}$ tangent to $S$ and pointing into it, $\hat{\mathbf{n}}$ is the direction of $\mathbf{u}\times\mathbf{v}$ ; Arfken asks of $\mathbf{v}$ only that it be tangent to the surface and lie on it, leaving the sense to his Fig. 3.13, so the words "pointing into it" are the disambiguation made here and in that unit, not his). Then

\int_S \bigl(\nabla\times\mathbf{F}\bigr)\cdot d\mathbf{S} \;=\; \oint_C \mathbf{F}\cdot d\mathbf{r} .

Riley, Hobson and Bence note that the theorem holds for simply and multiply connected open surfaces alike, provided they are two-sided; for a multiply connected $S$ the right side becomes a sum over all the boundary curves, each taken in the sense attached to $\hat{\mathbf{n}}$ . Two immediate consequences:

Closed surfaces. A closed surface has no perimeter, so $\displaystyle\oint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = 0$ for every closed $S$ : a curl has no net flux through any closed surface. (Consistently, $\nabla\cdot(\nabla\times\mathbf{F}) = 0$ .)
Surface independence. Two surfaces spanning the same curve $C$ , with normals agreeing with the same traversal of $C$ , give the same flux of $\nabla\times\mathbf{F}$ — provided $\mathbf{F}$ is $C^1$ on one open set containing both of them. The hypothesis is not a formality: the last remark below exhibits a field, and two surfaces spanning one circle, for which it fails and the two fluxes are not even both defined.

The companion identities obtained by Arfken with an arbitrary constant vector are

\int_S d\mathbf{S}\times\nabla\varphi = \oint_C \varphi\, d\mathbf{r}, \qquad\qquad \int_S (d\mathbf{S}\times\nabla)\times\mathbf{P} = \oint_C d\mathbf{r}\times\mathbf{P} ,

for $\varphi$ and $\mathbf{P}$ of class $C^1$ on an open set containing $S$ , with $\nabla$ in the second acting on $\mathbf{P}$ alone. Arfken states these without derivation, saying only that they come from the same arbitrary-vector technique that gave him the companion identities of the divergence theorem (his Eqs. (3.89)–(3.90)); the third proof below carries that technique out.

Idea of the proofOriginal

Green's theorem is the fundamental theorem of calculus applied twice, once in each coordinate direction, exactly as in the divergence theorem: integrating $\partial P/\partial y$ across the region in the $y$ -direction leaves the values of $P$ on the top and bottom arcs, and those two arcs are the boundary, traversed in opposite senses — which is where the minus sign in front of $\partial P /\partial y$ comes from. The term in $Q$ is the same computation turned on its side.

Stokes' theorem then needs no new analysis, only a change of variables. Push the circulation integral back to the flat parameter domain: a tangent vector to the boundary curve of $S$ is $\mathbf{r}_u\,du + \mathbf{r}_v\,dv$ , so $\mathbf{F}\cdot d\mathbf{r}$ becomes a plane line integral with $P = \mathbf{F}\cdot\mathbf{r}_u$ and $Q = \mathbf{F}\cdot\mathbf{r}_v$ . Green's theorem converts it to a double integral, and the combination $\partial_u Q - \partial_v P$ turns out to be exactly $(\nabla\times\mathbf{F})\cdot(\mathbf{r}_u\times\mathbf{r}_v)$ : the second derivatives of $\mathbf{r}$ cancel because mixed partials commute, and what remains is an algebraic identity about cross products. So the curl appears not because it was put in by hand but because it is what survives.

Proof(Green's theorem)AdaptedAdapted from Riley, Hobson & Bence (2006), §11.3, pp. 384–385

Assume first that $R$ is describable in both of the two ways

R = \bigl\{(x,y) : a \le x \le b,\; y_1(x) \le y \le y_2(x)\bigr\} = \bigl\{(x,y) : c \le y \le d,\; x_1(y) \le x \le x_2(y)\bigr\},

with $y_1 \le y_2$ and $x_1 \le x_2$ piecewise $C^1$ . (This is the situation of Riley, Hobson and Bence's Fig. 11.3; a disc, a convex region and a rectangle are all of this kind.)

Step 1 (the term in $P$ ). Since $\partial P/\partial y$ is continuous on the compact set $\overline R$ , it is bounded and integrable there, so the double integral may be written as an iterated integral (iterated integrals), innermost in $y$ ; the inner integral is then evaluated by the fundamental theorem of calculus at fixed $x$ :

\iint_R \frac{\partial P}{\partial y}\; dx\, dy = \int_a^b \Bigl[ P\bigl(x, y_2(x)\bigr) - P\bigl(x, y_1(x)\bigr) \Bigr]\, dx .

Step 2 (the same difference as a line integral). Traversing $C$ with $R$ on the left means going along the lower arc $y = y_1(x)$ in the direction of increasing $x$ and back along the upper arc $y = y_2(x)$ in the direction of decreasing $x$ . Any vertical segments of $C$ (at $x = a$ or $x = b$ ) contribute nothing to $\oint P\, dx$ because $dx = 0$ on them. Hence

\oint_C P\, dx = \int_a^b P\bigl(x,y_1(x)\bigr)\, dx + \int_b^a P\bigl(x,y_2(x)\bigr)\, dx = -\iint_R \frac{\partial P}{\partial y}\; dx\, dy ,

using Step 1 in the last equality. The minus sign is produced by the direction of travel, not by any choice of ours.

Step 3 (the term in $Q$ ). Now use the second description of $R$ and integrate innermost in $x$ :

\iint_R \frac{\partial Q}{\partial x}\; dx\, dy = \int_c^d \Bigl[ Q\bigl(x_2(y), y\bigr) - Q\bigl(x_1(y), y\bigr) \Bigr]\, dy .

With $R$ on the left, the right arc $x = x_2(y)$ is traversed with $y$ increasing and the left arc $x = x_1(y)$ with $y$ decreasing, while horizontal segments contribute nothing to $\oint Q\, dy$ . So

\oint_C Q\, dy = \int_c^d Q\bigl(x_2(y),y\bigr)\, dy + \int_d^c Q\bigl(x_1(y),y\bigr)\, dy = \iint_R \frac{\partial Q}{\partial x}\; dx\, dy ,

this time with no sign change: the two orientations work out oppositely in the two directions, which is precisely why the integrand of Green's theorem is a difference.

Step 4 (add). Adding Steps 2 and 3 gives the theorem for a region describable in both ways at once.

Step 5 (regions assembled from finitely many such pieces). Suppose next that $R = R_1\cup\dots\cup R_m$ , where each $R_i$ is a region of the kind Steps 1–3 treat, the $R_i$ meet only along arcs of their boundaries, and each $\partial R_i$ is traversed with $R_i$ on the left. Add the theorem over the pieces. An arc interior to $R$ belongs to exactly two of the pieces, which lie on opposite sides of it, so it is traversed once in each direction and its two contributions to $\oint(P\,dx + Q\,dy)$ cancel; the double integrals add to $\iint_R$ . What survives on the line-integral side is the integral over the part of $\bigcup_i\partial R_i$ that lies in $\partial R$ , each arc still carrying $R$ on its left. So the theorem holds for $R$ . The same bookkeeping covers a multiply connected $R$ , whose boundary is then a family of curves, the outer one anticlockwise and each hole clockwise. $\square$

NoteAdaptedAdapted from Strauss (2008), Appendix §A.3, Green's Thm 6, p. 422; Riley, Hobson & Bence (2006), §11.3, Eq. (11.4) and p. 385

What the proof above covers, and what is taken on authority. Steps 1–5 prove Green's theorem for a plane region that is a finite union of pieces each of which is a graph region in both coordinate directions. The statement at the head of this unit is the stronger one — every bounded plane domain with piecewise $C^1$ boundary — and that gap is not closed here, exactly as the gap between simple regions and general ones is left open in the divergence theorem.

The gap is real, not just unaddressed: a bounded domain with piecewise $C^1$ boundary need not admit a finite decomposition of that kind. Let

f(x) = 2 + x^3\sin(1/x) \quad (0 < x \le 1), \qquad f(0) = 2 ,

which is $C^1$ on $[0,1]$ , since $f'(x) = 3x^2\sin(1/x) - x\cos(1/x) \to 0 = f'(0)$ as $x \to 0^+$ , and let $R$ be the region $\{0 < x < 1,\; 0 < y < f(x)\}$ . Let $I_1, I_2, \dots$ be the maximal open intervals of $(0,1)$ on which $\sin(1/x) > 0$ ; there are infinitely many, they are pairwise disjoint, and $f > 2$ exactly on their union. Put $B_j = \{(x,y) : x \in I_j,\ 2 < y < f(x)\}$ , an infinite family of pairwise disjoint nonempty open subsets of $R$ , any two of which are separated by the line $y = 2$ in the sense that no connected subset of $R \cap \{y > 2\}$ meets two of them.

Now suppose $R = \bigcup_{k=1}^{N} R_k$ were a finite decomposition into two-way graph regions, and write the $y$ -description of the $k$ th piece as $R_k = \{(x,y) : c_k \le y \le d_k,\ a_k(y) \le x \le b_k(y)\}$ with $a_k \le b_k$ continuous on $[c_k, d_k]$ . The part of that piece above the line,

R_k \cap \{y > 2\} = \bigl\{ (x,y) : \max(c_k, 2) < y \le d_k,\ a_k(y) \le x \le b_k(y) \bigr\} ,

is connected: each of its horizontal slices is the single interval $[a_k(y), b_k(y)]$ , each slice meets the left edge $\{(a_k(y), y) : \max(c_k, 2) < y \le d_k\}$ , and that edge is connected, being the image of an interval under a continuous map — and a union of connected sets all meeting one connected set is connected. Since the $B_j$ are separated in the sense just noted, a connected subset of $R \cap \{y > 2\}$ meets at most one of them, so each $R_k$ meets at most one $B_j$ . With only $N$ pieces and infinitely many $B_j$ , some $B_j$ is met by no piece at all; as $B_j$ is a nonempty subset of $R$ , this contradicts $\bigcup_k R_k = R$ .

We therefore take the general statement unproved, from Strauss's Green's Theorem 6 (p. 422), who states it for any bounded plane domain with piecewise $C^1$ boundary and refers the reader elsewhere for the proof, and from Riley, Hobson and Bence's Eq. (11.4) (p. 384), whose own derivation is exactly Steps 1–4 for the single region of their Fig. 11.3, the extension to multiply connected regions being introduced on p. 385 with the words "it may further be shown". The proof of Stokes' theorem below uses Green's theorem only on parameter domains, and is written so as to say which of the two cases it is invoking.

Proof(Stokes' theorem for a

C^2

patch)Original

Let $S = \mathbf{r}(D)$ be a regular patch with $\mathbf{r}$ of class $C^2$ on a compact parameter domain $D$ to which Green's theorem applies — either a domain of the two-way graph form that Steps 1–4 of the previous proof treat directly, or a finite union of those, which its Step 5 covers; the note above says what is taken on authority beyond that. Require $\mathbf{r}$ to be injective on the whole of $D$ , not merely regular, and orient $S$ by $\hat{\mathbf{n}} = (\mathbf{r}_u\times\mathbf{r}_v)/\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert$ . Let $\mathbf{F}$ be $C^1$ on an open set containing $S$ .

Injectivity is what makes the two descriptions of the perimeter agree. A regular injective $\mathbf{r}$ on the compact set $D$ is a homeomorphism of $D$ onto $S$ , so it carries the interior of $D$ onto the points of $S$ with a neighbourhood in $S$ and $\partial D$ onto the rest; that is, $\mathbf{r}(\partial D)$ is precisely the perimeter $C = \partial S$ named in the statement. Without injectivity the image of $\partial D$ need not be the boundary of the image at all: a map $\mathbf{r}$ that wraps $D$ twice over the same patch has $\mathbf{r}(\partial D)$ running round $S$ 's rim twice, and the identity proved below would then compute twice the flux, not the circulation of the statement.

Step 1 (orientation bookkeeping). Traverse $\partial D$ anticlockwise, with $D$ on the left, and let $C = \mathbf{r}(\partial D)$ — the perimeter $\partial S$ of the statement, by the injectivity just assumed — carry the induced direction. This is the right-hand traversal relative to $\hat{\mathbf{n}}$ . To see it, look at a boundary point in parameters where $D$ is locally the half-plane $v \ge 0$ with boundary the $u$ -axis: with $D$ on the left, the direction of travel is $+u$ , so the travel vector on $S$ is $\mathbf{u} = \mathbf{r}_u$ , and the tangent to $S$ pointing into the surface is $\mathbf{v} = \mathbf{r}_v$ . Then $\mathbf{u}\times\mathbf{v} = \mathbf{r}_u\times\mathbf{r}_v$ , which is the chosen normal — exactly the right-hand rule in the form fixed in the statement.

Step 2 (pull the circulation back to the parameters). Let $t \mapsto (u(t),v(t))$ parametrize $\partial D$ . The image curve is $t\mapsto\mathbf{r}(u(t),v(t))$ , with velocity $\mathbf{r}_u u' + \mathbf{r}_v v'$ by the chain rule. Therefore

\oint_C \mathbf{F}\cdot d\mathbf{r} = \oint_{\partial D}\Bigl[ \bigl(\mathbf{F}\cdot\mathbf{r}_u\bigr)\, du + \bigl(\mathbf{F}\cdot\mathbf{r}_v\bigr)\, dv \Bigr] ,

a plane line integral with $P = \mathbf{F}(\mathbf{r})\cdot\mathbf{r}_u$ and $Q = \mathbf{F}(\mathbf{r})\cdot\mathbf{r}_v$ as functions of $(u,v)$ . Both are $C^1$ on $D$ : $\mathbf{F}$ is $C^1$ and $\mathbf{r}$ is $C^2$ , so $\mathbf{r}_u$ and $\mathbf{r}_v$ are $C^1$ .

Step 3 (apply Green's theorem). By the previous proof,

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D \left( \frac{\partial}{\partial u}\bigl(\mathbf{F}\cdot\mathbf{r}_v\bigr) - \frac{\partial}{\partial v}\bigl(\mathbf{F}\cdot\mathbf{r}_u\bigr) \right) du\, dv .

Step 4 (the second derivatives of $\mathbf{r}$ cancel). By the product rule and the chain rule,

\frac{\partial}{\partial u}\bigl(\mathbf{F}\cdot\mathbf{r}_v\bigr) = \bigl[(\mathbf{r}_u\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{r}_v + \mathbf{F}\cdot\mathbf{r}_{vu}, \qquad \frac{\partial}{\partial v}\bigl(\mathbf{F}\cdot\mathbf{r}_u\bigr) = \bigl[(\mathbf{r}_v\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{r}_u + \mathbf{F}\cdot\mathbf{r}_{uv} ,

where $(\mathbf{a}\cdot\nabla)\mathbf{F}$ denotes the vector with components $a_m\,\partial F_n/\partial x_m$ . Here the hypothesis $\mathbf{r} \in C^2$ is used: $\mathbf{r}_{uv} = \mathbf{r}_{vu}$ , so the two terms containing them cancel on subtraction, leaving

\frac{\partial Q}{\partial u} - \frac{\partial P}{\partial v} = \bigl[(\mathbf{r}_u\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{r}_v - \bigl[(\mathbf{r}_v\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{r}_u .

Step 5 (this is the curl, dotted with the normal). For any two vectors $\mathbf{a}, \mathbf{b}$ and any differentiable $\mathbf{F}$ , the epsilon-delta identity $\varepsilon_{kmn}\varepsilon_{kpq} = \delta_{mp}\delta_{nq} - \delta_{mq}\delta_{np}$ (vector algebra) gives

\bigl(\nabla\times\mathbf{F}\bigr)\cdot\bigl(\mathbf{a}\times\mathbf{b}\bigr) = \varepsilon_{kmn}\,\partial_mF_n\;\varepsilon_{kpq}\,a_pb_q = a_m\,\partial_mF_n\,b_n - b_m\,\partial_mF_n\,a_n = \bigl[(\mathbf{a}\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{b} - \bigl[(\mathbf{b}\cdot\nabla)\mathbf{F}\bigr]\cdot\mathbf{a} .

With $\mathbf{a} = \mathbf{r}_u$ , $\mathbf{b} = \mathbf{r}_v$ this is exactly the integrand of Step 4.

Step 6 (conclude). Substituting into Step 3,

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D \bigl(\nabla\times\mathbf{F}\bigr)\cdot\bigl(\mathbf{r}_u\times\mathbf{r}_v\bigr)\, du\, dv = \int_S \bigl(\nabla\times\mathbf{F}\bigr)\cdot d\mathbf{S} ,

the last equality being the definition of a flux integral over the patch. For a surface made of several patches, add: interior edges are traversed twice in opposite directions when the patches are consistently oriented, and cancel. $\square$

Proof(The two companion identities)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.8, Eqs. (3.94)–(3.95), p. 168

Both come from Stokes' theorem applied to a field built out of an arbitrary constant vector $\mathbf{a}$ — the technique Arfken points to but does not carry out. Being constant, $\mathbf{a}$ passes through $\nabla$ and through both integral signs; $(\mathbf{a}\cdot\nabla)\mathbf{P}$ keeps the meaning it was given in the previous proof, the vector with components $a_m\,\partial P_n/\partial x_m$ . Throughout, $S$ and $C$ satisfy the hypotheses of the theorem and $\varphi, \mathbf{P}$ are $C^1$ on an open set containing $S$ .

Step 1 (one cyclic rule, used twice). For any three vectors, $(\mathbf{u}\times\mathbf{a})\cdot\mathbf{w} = \det[\mathbf{u}, \mathbf{a}, \mathbf{w}] = \det[\mathbf{a}, \mathbf{w}, \mathbf{u}] = \mathbf{a}\cdot(\mathbf{w}\times\mathbf{u})$ , the two determinants differing by a cyclic permutation of their rows, which is an even permutation.

Step 2 (the scalar identity). Take $\mathbf{F} = \varphi\,\mathbf{a}$ . Then $\nabla\times(\varphi\mathbf{a}) = \nabla\varphi\times\mathbf{a}$ , since the term $\varphi\,\nabla\times\mathbf{a}$ vanishes for constant $\mathbf{a}$ . Apply Stokes' theorem to this $\mathbf{F}$ ; rewrite its flux side by Step 1 with $\mathbf{u} = \nabla\varphi$ and $\mathbf{w} = \hat{\mathbf{n}}$ , and its circulation side by pulling the scalar $\varphi$ past the dot product. This gives

\mathbf{a}\cdot\int_S \bigl(\hat{\mathbf{n}}\times\nabla\varphi\bigr)\, dS = \int_S \bigl(\nabla\varphi\times\mathbf{a}\bigr)\cdot d\mathbf{S} = \oint_C \varphi\,\mathbf{a}\cdot d\mathbf{r} = \mathbf{a}\cdot\oint_C \varphi\, d\mathbf{r} .

Step 3 (strip the arbitrary vector). The outer two members agree for every constant $\mathbf{a}$ ; taking $\mathbf{a} = \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ in turn equates the three components of the two vectors, which is the first identity, Arfken's Eq. (3.94), since $\hat{\mathbf{n}}\,dS = d\mathbf{S}$ .

Step 4 (the vector identity: the left side of Stokes). Take $\mathbf{F} = \mathbf{P}\times\mathbf{a}$ . With the epsilon-delta identity used in the previous proof,

\bigl[\nabla\times(\mathbf{P}\times\mathbf{a})\bigr]_i = \varepsilon_{ijk}\,\partial_j\bigl(\varepsilon_{klm}P_la_m\bigr) = \bigl(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\bigr)a_m\,\partial_jP_l = \bigl[(\mathbf{a}\cdot\nabla)\mathbf{P}\bigr]_i - a_i\,\nabla\cdot\mathbf{P} ,

using $\varepsilon_{ijk}\varepsilon_{klm} = \varepsilon_{kij}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ . On the circulation side, Step 1 with $\mathbf{u} = \mathbf{P}$ and $\mathbf{w} = d\mathbf{r}$ gives $(\mathbf{P}\times\mathbf{a})\cdot d\mathbf{r} = \mathbf{a}\cdot(d\mathbf{r}\times\mathbf{P})$ . So Stokes' theorem reads

\int_S \Bigl[(\mathbf{a}\cdot\nabla)\mathbf{P} - \mathbf{a}\,\bigl(\nabla\cdot\mathbf{P}\bigr)\Bigr]\cdot d\mathbf{S} = \mathbf{a}\cdot\oint_C d\mathbf{r}\times\mathbf{P} .

Step 5 (recognise the left side as $(d\mathbf{S}\times\nabla)\times\mathbf{P}$ ). Expanding that double cross product with $(\hat{\mathbf{n}}\times\nabla)_j = \varepsilon_{jlm}n_l\partial_m$ , and using $\varepsilon_{ijk}\varepsilon_{jlm} = -\varepsilon_{jik}\varepsilon_{jlm} = -(\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})$ ,

\bigl[(\hat{\mathbf{n}}\times\nabla)\times\mathbf{P}\bigr]_i = \varepsilon_{ijk}\,\varepsilon_{jlm}\,n_l\,\partial_mP_k = -\bigl(n_i\,\partial_kP_k - n_k\,\partial_iP_k\bigr) = n_k\,\partial_iP_k - n_i\,\nabla\cdot\mathbf{P} .

Dotting with $\mathbf{a}$ and collecting, $\mathbf{a}\cdot\bigl[(\hat{\mathbf{n}}\times\nabla)\times\mathbf{P}\bigr] = n_k\,(\mathbf{a}\cdot\nabla)P_k - (\mathbf{a}\cdot\hat{\mathbf{n}})\,\nabla\cdot\mathbf{P} = \bigl[(\mathbf{a}\cdot\nabla)\mathbf{P} - \mathbf{a}(\nabla\cdot\mathbf{P})\bigr]\cdot\hat{\mathbf{n}}$ , which is exactly the integrand of Step 4.

Step 6 (conclude). Therefore $\mathbf{a}\cdot\int_S (d\mathbf{S}\times\nabla)\times\mathbf{P} = \mathbf{a}\cdot\oint_C d\mathbf{r}\times\mathbf{P}$ for every constant $\mathbf{a}$ , and stripping $\mathbf{a}$ as in Step 3 gives the second identity, Arfken's Eq. (3.95). $\square$

NoteAdaptedAdapted from Lee (2013), Ch. 16, Thm 16.34 and Eq. (16.14), pp. 426–427; Riley, Hobson & Bence (2006), §11.9, p. 406; Arfken, Weber & Harris (2013), §3.8, p. 167

Other routes, and the general version. Riley, Hobson and Bence and Arfken both derive Stokes' theorem by dividing $S$ into a mesh of small rectangles, applying the circulation definition of the curl to each, and observing that all interior edges cancel in pairs. That argument shows why the theorem is true and is the one to remember, but turning it into a proof requires an estimate of the error on each cell and a limit; the route through Green's theorem above avoids that. The statement in its full generality — $S$ a compact oriented two-dimensional submanifold with boundary in an oriented Riemannian 3-manifold — is Lee's Theorem 16.34, proved by applying Stokes's theorem for differential forms to the 1-form $\mathbf{F}^\flat$ and identifying $d(\mathbf{F}^\flat)$ with the curl; see Stokes' theorem for forms, which contains Green's theorem, Stokes' theorem and the divergence theorem as the three cases $n = 2$ , and $(2,3)$ and $(3,3)$ of one statement.

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Orientation is not a convention one may drop. Both sides change sign when the traversal of $C$ is reversed, and both change sign when $\hat{\mathbf{n}}$ is reversed; the theorem is the statement that they change together. Arfken makes the point that this is why no sign ambiguity arises in Faraday's law. Fixing one of the two and reading off the other, as in Step 1, is the only safe procedure.
Two-sidedness. A Möbius strip has a perimeter and a well-defined circulation round it, but no continuous normal field, so the left-hand side is not even defined. This is the one hypothesis with no analogue in the divergence theorem, where the boundary of a solid region is automatically two-sided.
$\mathbf{r} \in C^2$ , used twice. In Step 2 it makes the coefficients $P, Q$ of the pulled-back line integral of class $C^1$ , which is what lets Green's theorem be applied to them in Step 3; in Step 4 it gives $\mathbf{r}_{uv} = \mathbf{r}_{vu}$ , so that the two mixed terms cancel. Our proof therefore covers the $C^2$ case and no more. The statement above is nevertheless given for two-sided piecewise $C^1$ surfaces, on the authority of Riley, Hobson and Bence (§11.9, p. 406), who state the theorem for any two-sided open surface, simply or multiply connected, and attach no smoothness hypothesis to it at all; Arfken's mesh derivation (p. 167) is the same argument, with the same silence. The merely $C^1$ case is proved nowhere in this handbook. It needs $S$ to be approximated by smoother surfaces, and the nearest result in our library, Lee's Theorem 16.34 quoted in the note above, assumes a smooth submanifold and so does not contain it either. What is unproved here is the smoothness of the surface and nothing else: the extension from one patch to a surface made of several is the last line of the proof.
$\mathbf{F} \in C^1$ on a neighbourhood of $S$ , not just on $S$ . The curl involves derivatives of $\mathbf{F}$ in the direction off the surface as well as along it. A field defined only on $S$ has no curl.
$\mathbf{F}$ must be $C^1$ at every point of $S$ , including points not on $C$ . Take $\mathbf{F} = (-y, x, 0)/(x^2+y^2)$ on the complement of the $z$ -axis, for which $\nabla\times\mathbf{F} = \mathbf{0}$ (see line integrals), and let $C$ be the unit circle in the plane $z = 0$ . The circulation is $2\pi$ , but every surface spanning $C$ meets the $z$ -axis, where $\mathbf{F}$ is undefined; so the flux side is not available, and no contradiction arises. If instead $S$ is a disc spanning a circle that does not encircle the axis, both sides are $0$ .
Why the two readings of Green's theorem are the same theorem. In the plane, replacing $\mathbf{f} = (f,g)$ by $(-g, f)$ rotates the field by a quarter turn, which exchanges its tangential and normal components on $C$ and exchanges $\nabla\cdot\mathbf{f}$ with the scalar curl $\partial_xg - \partial_yf$ . Strauss records the substitution $P = -g$ , $Q = f$ explicitly; the two forms carry exactly the same information.
Relation to the plane case. Green's theorem is Stokes' theorem for a flat surface in the plane $z = 0$ with $\hat{\mathbf{n}} = \hat{\mathbf{k}}$ : for $\mathbf{F} = (P, Q, 0)$ the only surviving component of the curl is $(\nabla\times\mathbf{F})_3 = \partial_xQ - \partial_yP$ . Our proof runs in that direction — Green first, then Stokes — so it cannot be reversed without circularity.

Example(Green's theorem: the area inside an astroid)Original

Putting $P = -y$ , $Q = x$ in Green's theorem makes the integrand $\partial_xQ - \partial_yP = 2$ , so for any region $R$ with piecewise $C^1$ boundary

\operatorname{area}(R) = \frac12\oint_C \bigl(x\, dy - y\, dx\bigr) ,

which turns an area into a one-dimensional integral along the boundary. Take the astroid

x = a\cos^3 t, \qquad y = a\sin^3 t, \qquad 0 \le t \le 2\pi \quad (a > 0),

traversed anticlockwise, which encloses a region $R$ with four cusps.

Differentials. $dx = -3a\cos^2 t\,\sin t\; dt$ and $dy = 3a\sin^2 t\,\cos t\; dt$ .
Integrand.
$x\, dy - y\, dx = 3a^2\bigl(\cos^4t\,\sin^2t + \sin^4t\,\cos^2t\bigr)\, dt = 3a^2\cos^2t\,\sin^2t\,\bigl(\cos^2t+\sin^2t\bigr)\, dt = \tfrac34 a^2\sin^2 2t\; dt ,$
using $\cos t\sin t = \tfrac12\sin 2t$ .
Integrate. Since $\int_0^{2\pi}\sin^2 2t\, dt = \pi$ ,
$\operatorname{area}(R) = \frac12\cdot\frac{3a^2}{4}\cdot\pi = \frac{3\pi a^2}{8} .$
For $a = 2$ this is $3\pi/2 = 4.71239$ .
Check, by trapping $R$ between two shapes whose areas are known. On the curve,
$\lvert x\rvert + \lvert y\rvert = a\bigl(\lvert\cos t\rvert^3 + \lvert\sin t\rvert^3\bigr) \le a\bigl(\cos^2 t + \sin^2 t\bigr) = a ,$
since $\lvert\cos t\rvert \le 1$ and $\lvert\sin t\rvert \le 1$ . So the boundary lies in the diamond $\lvert x\rvert + \lvert y\rvert \le a$ , which is convex and therefore contains $R$ as well; the diamond's area is $2a^2$ , so $\operatorname{area}(R) \le 2a^2$ . In the other direction, on the curve
$r^2 = x^2 + y^2 = a^2\bigl(\cos^6 t + \sin^6 t\bigr) = a^2\Bigl(1 - \tfrac34\sin^2 2t\Bigr) \ge \frac{a^2}{4} ,$
so every boundary point is at distance at least $a/2$ from the origin; and $R$ is star-shaped about the origin, each ray from it meeting the curve exactly once, so $R$ contains the disc of radius $a/2$ and $\operatorname{area}(R) \ge \pi a^2/4$ . Hence
$\frac{\pi}{4} = 0.78540 \;\le\; \frac{3\pi}{8} = 1.17810 \;\le\; 2 ,$
in units of $a^2$ : the computed area does lie between the two bounds, nearer the lower one, as the four concave sides should make it.

The cusps at $t = 0, \pi/2, \pi, 3\pi/2$ , where $(x', y') = \mathbf{0}$ , do no harm, but not because the parametrization is smooth. What Green's theorem requires is that the boundary be a piecewise $C^1$ curve, and it is: the four open arcs between consecutive cusps are regular $C^1$ arcs, and the cusps are the finitely many corner points at which the pieces join — exactly as for the corners of a square. The line integral is then computed from the given $C^\infty$ parametrization, whose velocity vanishing at four isolated parameter values affects neither the integrand nor the value of the integral.

Example(Stokes' theorem on a triangle, both sides computed)Original

Let $S$ be the flat triangle with vertices $A = (1,0,0)$ , $B = (0,1,0)$ , $C_0 = (0,0,1)$ and perimeter traversed $A \to B \to C_0 \to A$ , and let

\mathbf{F} = (z,\; x,\; y) .

Orientation. Two edge vectors in the direction of travel are $B - A = (-1,1,0)$ and $C_0 - B = (0,-1,1)$ , and
$(-1,1,0)\times(0,-1,1) = (1,\,1,\,1) ,$
so the right-hand traversal corresponds to $\hat{\mathbf{n}} = (1,1,1)/\sqrt3$ , the normal pointing away from the origin. The triangle is equilateral with side $\lvert B-A\rvert = \sqrt2$ , so its area is $\tfrac{\sqrt3}{4}(\sqrt2)^2 = \tfrac{\sqrt3}{2}$ .
Flux of the curl.
$\nabla\times\mathbf{F} = \Bigl(\frac{\partial y}{\partial y} - \frac{\partial x}{\partial z},\; \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x},\; \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y}\Bigr) = (1,1,1) ,$
a constant field, so
$\int_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = (1,1,1)\cdot\frac{(1,1,1)}{\sqrt3}\;\cdot\frac{\sqrt3}{2} = \sqrt3\cdot\frac{\sqrt3}{2} = \frac{3}{2} .$
Circulation, edge by edge. Each edge is a straight segment $\mathbf{r}(t) = \mathbf{p} + t(\mathbf{q}-\mathbf{p})$ , $0 \le t \le 1$ .
- $A \to B$ : $\mathbf{r} = (1-t,\, t,\, 0)$ , $\mathbf{r}' = (-1,1,0)$ , and $\mathbf{F}(\mathbf{r}) = (z,x,y) = (0,\, 1-t,\, t)$ , so $\mathbf{F}\cdot\mathbf{r}' = 1-t$ and the contribution is $\int_0^1 (1-t)\,dt = \tfrac12$ .
- $B \to C_0$ : $\mathbf{r} = (0,\, 1-t,\, t)$ , $\mathbf{r}' = (0,-1,1)$ , $\mathbf{F}(\mathbf{r}) = (t,\, 0,\, 1-t)$ , so $\mathbf{F}\cdot\mathbf{r}' = 1-t$ : again $\tfrac12$ .
- $C_0 \to A$ : $\mathbf{r} = (t,\, 0,\, 1-t)$ , $\mathbf{r}' = (1,0,-1)$ , $\mathbf{F}(\mathbf{r}) = (1-t,\, t,\, 0)$ , so $\mathbf{F}\cdot\mathbf{r}' = 1-t$ : again $\tfrac12$ .
Hence $\displaystyle\oint_C \mathbf{F}\cdot d\mathbf{r} = 3\cdot\tfrac12 = \tfrac32$ .
Comparison. Both sides equal $\tfrac32$ .
Surface independence, checked. Replace $S$ by the three right triangles $OAB$ , $OBC_0$ , $OC_0A$ with $O$ the origin; together they span the same perimeter, because the three segments $OA$ , $OB$ , $OC_0$ each belong to two of them and cancel. Consistency forces the boundary traversals $A\to B\to O\to A$ , $B\to C_0\to O\to B$ and $C_0\to A\to O\to C_0$ , whose right-hand normals are, by the same cross-product computation as in step 1,
$(-1,1,0)\times(0,-1,0) = \hat{\mathbf{k}}, \qquad (0,-1,1)\times(0,0,-1) = \hat{\mathbf{i}}, \qquad (1,0,-1)\times(-1,0,0) = \hat{\mathbf{j}} .$
Each triangle has area $\tfrac12$ , so each contributes $(1,1,1)\cdot\hat{\mathbf{e}}\cdot\tfrac12 = \tfrac12$ , and the total is again $\tfrac32$ . Note that these three normals point into the tetrahedron $OABC_0$ ; the orientation is fixed by the perimeter, not by any region.

ApplicationAdaptedAdapted from Arfken, Weber & Harris (2013), §3.8, Example 3.8.2, p. 168

Stokes' theorem is what converts the two curl equations of Maxwell's set between their differential and circuit forms. Integrating $\nabla\times\mathbf{B} = \mu_0\mathbf{J}$ over a disc $S$ pierced by a current-carrying wire and applying the theorem gives

\oint_{\partial S}\mathbf{B}\cdot d\mathbf{r} = \mu_0 \int_S \mathbf{J}\cdot d\mathbf{S} = \mu_0 I ,

which is Ampère's law in the form Arfken calls Oersted's law; the surface-independence property is what makes "the current enclosed by the loop" well defined. Integrating $\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t$ over a circuit of area $S$ gives

\oint_{\partial S}\mathbf{E}\cdot d\mathbf{r} = -\frac{d\Phi}{dt}, \qquad \Phi = \int_S\mathbf{B}\cdot d\mathbf{S} ,

Faraday's law: the voltage induced in a loop of wire is minus the rate at which magnetic flux through the loop changes. Two separate things are often run together here. Reversing the sense of the loop reverses $d\mathbf{r}$ and therefore also $d\mathbf{S}$ and the flux $\Phi$ , so both sides change sign together and the prediction does not depend on which way round the loop was labelled: that is Arfken's "no sign ambiguity" remark (p. 168), a statement about orientation. Lenz's law is the content of the minus sign itself — the induced current flows so as to oppose the change in flux — and it is what makes a transformer or an alternator work rather than run away.

In fluid mechanics the circulation $\Gamma = \oint_C\mathbf{u}\cdot d\mathbf{r}$ round a closed curve equals the flux of the vorticity $\boldsymbol{\omega} = \nabla\times\mathbf{u}$ through any surface spanning it; this is the identity behind Kelvin's circulation theorem, the Kutta–Joukowski lift formula, and the statement that a vortex tube's strength is constant along its length. Green's theorem in its own right is the planimeter's mathematics: an instrument that traces the boundary of a shape and reads off $\tfrac12\oint(x\,dy - y\,dx)$ returns its area, and the same formula is how a geographic information system computes the area of a polygon from its vertex list.

Theorem

Green's identities

Adapted

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Derives Green's first and second identities from the divergence theorem and uses the first to prove that the Dirichlet problem for Poisson's equation has at most one solution, that the Neumann problem has one up to an additive constant and only when its data satisfy the compatibility condition $\oint_S h\,dS = \int_V f\,dV$ , and that the Robin problem with positive coefficient has at most one.

Prerequisites: The divergence theorem, Divergence, curl and the Laplacian

Theorem(Green's first and second identities)AdaptedAdapted from Strauss (2008), §7.1, Eq. (G1), p. 180 and §7.2, Eq. (G2), p. 185; Evans (2010), Appendix §C.2, Thm 3, p. 667; Riley, Hobson & Bence (2006), §11.8.1, Eqs. (11.19)–(11.20), p. 403

Let $V \subseteq \mathbb{R}^3$ be a bounded domain whose boundary $S = \partial V$ is piecewise $C^1$ , with unit outward normal $\hat{\mathbf{n}}$ , and write

\frac{\partial u}{\partial n} = \hat{\mathbf{n}}\cdot\nabla u

for the outward normal derivative on $S$ . Then:

Green's first identity. For $u \in C^2(\overline V)$ and $v \in C^1(\overline V)$ , $\oint_S v\,\frac{\partial u}{\partial n}\; dS \;=\; \int_V \nabla v\cdot\nabla u\; dV \;+\; \int_V v\,\nabla^2 u\; dV . \tag{G1}$
The case $v \equiv 1$ . For $u \in C^2(\overline V)$ , $\oint_S \frac{\partial u}{\partial n}\; dS = \int_V \nabla^2 u\; dV .$
Green's second identity. For $u, v \in C^2(\overline V)$ , $\int_V \bigl( u\,\nabla^2 v - v\,\nabla^2 u \bigr)\; dV \;=\; \oint_S \left( u\,\frac{\partial v}{\partial n} - v\,\frac{\partial u}{\partial n} \right) dS . \tag{G2}$

Evans states all three in $\mathbb{R}^n$ for a bounded open $U$ with $C^1$ boundary and $u, v \in C^2(\overline U)$ ; nothing in the statements or the proof is special to three dimensions. Strauss presents (G1) and (G2) without attaching smoothness conditions to $u$ and $v$ — his claim is that they hold whatever the solid region and whatever the two functions — because he obtains them formally from the product rule and the divergence theorem; the hypotheses above are what that derivation actually needs, and are Evans's.

Idea of the proofOriginal

Both identities are the three-dimensional integration-by-parts formula, and they come from one line of algebra plus the divergence theorem. The product rule for the divergence of $v\nabla u$ splits it into a term with no derivative on $v$ and a term with one on each; integrating and converting the left side to a boundary integral gives (G1), with the boundary term playing the part of the $[\,uv\,]_a^b$ of one-variable integration by parts. The second identity is obtained for free: in (G1) the middle term $\int \nabla u\cdot\nabla v$ is symmetric in $u$ and $v$ , so writing (G1) twice with the two functions exchanged and subtracting annihilates it, leaving a relation between the Laplacians inside and the data on the boundary.

That relation is what makes uniqueness proofs work. Setting $v = u$ in (G1) turns the boundary data of a harmonic function into the integral of $\lvert\nabla u\rvert^2$ — a quantity that is non-negative and vanishes only for a constant. So vanishing boundary data forces the solution to be constant, and the proof is three lines long.

ProofAdaptedAdapted from Strauss (2008), §7.1, pp. 179–180 and §7.2, p. 185; Riley, Hobson & Bence (2006), §11.8.1, p. 403; Evans (2010), Appendix §C.2, p. 667

Step 1 (the product rule). For $u \in C^2$ and $v \in C^1$ , differentiating $v\,\partial u/\partial x_i$ and summing over $i$ ,

\nabla\cdot\bigl(v\,\nabla u\bigr) = \sum_i \frac{\partial}{\partial x_i}\left( v\,\frac{\partial u}{\partial x_i} \right) = \sum_i \frac{\partial v}{\partial x_i}\frac{\partial u}{\partial x_i} + v\sum_i\frac{\partial^2 u}{\partial x_i^2} = \nabla v\cdot\nabla u + v\,\nabla^2 u .

Step 2 (apply the divergence theorem). The field $\mathbf{F} = v\nabla u$ is $C^1$ on $\overline V$ , because $v \in C^1(\overline V)$ and $\nabla u \in C^1(\overline V)$ ; so the divergence theorem applies to it:

\int_V \nabla\cdot\bigl(v\nabla u\bigr)\; dV = \oint_S v\,\nabla u\cdot\hat{\mathbf{n}}\; dS = \oint_S v\,\frac{\partial u}{\partial n}\; dS ,

the last equality being the definition of the normal derivative. Substituting Step 1 on the left gives (G1).

Step 3 (the case $v \equiv 1$ ). A constant is $C^1$ with $\nabla v = \mathbf{0}$ , so the first term on the right of (G1) drops out and part 2 follows.

Step 4 (antisymmetrize). Let now $u, v \in C^2(\overline V)$ , so that (G1) may be written both ways round:

\oint_S v\,\frac{\partial u}{\partial n}\, dS = \int_V \nabla v\cdot\nabla u\; dV + \int_V v\,\nabla^2 u\; dV , \qquad \oint_S u\,\frac{\partial v}{\partial n}\, dS = \int_V \nabla u\cdot\nabla v\; dV + \int_V u\,\nabla^2 v\; dV .

The first term on each right-hand side is the same number, since the dot product is symmetric. Subtracting the first equation from the second therefore gives

\oint_S \left( u\,\frac{\partial v}{\partial n} - v\,\frac{\partial u}{\partial n} \right) dS = \int_V \bigl( u\,\nabla^2 v - v\,\nabla^2 u \bigr)\; dV ,

which is (G2). $\square$

Theorem(Uniqueness for Laplace's and Poisson's equations)AdaptedAdapted from Strauss (2008), §7.1, Eq. (4), p. 180 and Eq. (7), p. 182, with Exercises 2 and 3, p. 184

Let $V$ be a bounded connected domain with piecewise $C^1$ boundary $S$ , and let $f$ on $\overline V$ and $h$ on $S$ be given continuous functions. All solutions below are required to be of class $C^2(\overline V)$ .

Dirichlet problem. The problem $\nabla^2 u = f$ in $V$ , $u = h$ on $S$ , has at most one solution.
Neumann problem. For $\nabla^2 u = f$ $\nabla^{2} u = f$ in $V$ $V$ , $\partial u/\partial n = h$ $\partial u / \partial n = h$ on $S$ $S$ :
- a solution exists only if the data satisfy the compatibility condition $\oint_S h\; dS = \int_V f\; dV ;$
- and any two solutions differ by a constant. So the Neumann problem is not well-posed in the strict sense: existence fails for most data, and uniqueness fails by one constant. Strauss notes that when the compatibility condition does hold, a solution can be shown to exist.
Robin problem. If $a$ is continuous with $a > 0$ everywhere on $S$ , the problem $\nabla^2 u = f$ in $V$ , $\partial u/\partial n + a\,u = h$ on $S$ , has at most one solution.

ProofAdaptedAdapted from Strauss (2008), §7.1, p. 180, Eq. (4) and p. 182, Eq. (7)

Each part uses the same device: subtract two solutions, so that the difference $u$ solves the homogeneous problem, then put $v = u$ in (G1).

Step 0 (the energy identity). For $u \in C^2(\overline V)$ with $\nabla^2 u = 0$ in $V$ , taking $v = u$ in (G1) and dropping the last term gives

\oint_S u\,\frac{\partial u}{\partial n}\; dS = \int_V \lvert\nabla u\rvert^2\; dV . \tag{E}

Step 1 (a vanishing lemma). If $\int_V \lvert\nabla u\rvert^2\, dV = 0$ then $\nabla u \equiv \mathbf{0}$ on $V$ : the integrand is continuous and non-negative, so were it positive at a point it would exceed half that value on a ball around the point and the integral would be positive. Since $V$ is connected, a function with vanishing gradient on $V$ is constant there (integrate along a path, line integrals), hence on $\overline V$ by continuity.

Step 2 (Dirichlet). Let $u_1, u_2$ both solve part 1 and put $u = u_1 - u_2$ . Then $\nabla^2u = f - f = 0$ in $V$ and $u = h - h = 0$ on $S$ . The left side of (E) vanishes because $u = 0$ at every point of $S$ , so $\int_V\lvert\nabla u\rvert^2 = 0$ ; by Step 1, $u$ is a constant, and that constant is $0$ because $u$ vanishes on $S$ . Hence $u_1 = u_2$ .

Step 3 (Neumann: the compatibility condition). Suppose $u$ solves part 2. Part 2 of the identities, with $v \equiv 1$ , gives directly

\oint_S h\; dS = \oint_S \frac{\partial u}{\partial n}\; dS = \int_V \nabla^2 u\; dV = \int_V f\; dV .

So data violating this equality admit no solution at all. (For $f \equiv 0$ the condition reads $\oint_S h\, dS = 0$ : a steady temperature field cannot have a net heat flux through its boundary.)

Step 4 (Neumann: uniqueness up to a constant). With $u = u_1 - u_2$ as before, now $\nabla^2u = 0$ in $V$ and $\partial u/\partial n = 0$ on $S$ , so the left side of (E) vanishes because the normal derivative vanishes on $S$ . By Step 1, $u$ is constant; nothing fixes its value, and indeed $u_2 + c$ solves the same problem as $u_2$ for every constant $c$ .

Step 5 (Robin). With $u = u_1 - u_2$ we have $\nabla^2u = 0$ in $V$ and $\partial u/\partial n = -a\,u$ on $S$ . Substituting into (E),

\int_V \lvert\nabla u\rvert^2\; dV = \oint_S u\,\frac{\partial u}{\partial n}\; dS = -\oint_S a\,u^2\; dS .

The left side is $\ge 0$ and the right side is $\le 0$ , since $a > 0$ and $u^2 \ge 0$ ; so both are $0$ . From the left, $\nabla u \equiv \mathbf{0}$ and $u$ is a constant $c$ by Step 1. From the right, $\oint_S a\,c^2\, dS = 0$ with $a > 0$ on a surface of positive area forces $c = 0$ . Hence $u_1 = u_2$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Connectedness of $V$ , in Step 1 only. On a disconnected $V$ the Dirichlet conclusion still holds, since $u$ vanishes on the boundary of each component; but the Neumann solution is then unique only up to an independent constant in each component.
$a > 0$ in the Robin problem, not $a \ge 0$ . With $a \equiv 0$ the Robin problem is the Neumann problem, and uniqueness genuinely fails by a constant: Step 5 breaks at its last sentence. With $a < 0$ somewhere, the two sides of Step 5 can both be nonzero and uniqueness can fail outright. Strauss poses the case $a > 0$ as Exercise 3.
$u \in C^2$ up to the boundary. Both the divergence theorem in Step 2 of the first proof and the evaluation of $u$ or $\partial u/\partial n$ on $S$ require the solution and its first derivatives to extend continuously to $S$ . Dropping this admits spurious "solutions": the harmonic function $u = \operatorname{Re}\bigl((1+z)/(1-z)\bigr)$ on the unit disc in the plane vanishes at almost every boundary point yet is not identically zero — it is unbounded near $z = 1$ and so not in $C^2(\overline V)$ . The energy method does not see such functions, and the uniqueness statements are statements about the class $C^2(\overline V)$ .
Only $v \in C^1$ is needed in (G1). Step 2 of the first proof applies the divergence theorem to $v\nabla u$ , which is $C^1$ as soon as $v$ is $C^1$ and $u$ is $C^2$ . (G2), by contrast, differentiates both functions twice and needs both in $C^2$ .
Nothing here needs $V$ simply connected. The underlying divergence theorem does not, so the identities hold on a spherical shell, with the boundary integral taken over both spheres and $\hat{\mathbf{n}}$ pointing out of $V$ on each.
Symmetry of the boundary condition. Strauss observes that each of the three classical boundary conditions makes the right-hand side of (G2) vanish for every pair $u, v$ satisfying it — he calls such a condition symmetric for $\nabla^2$ . For the Dirichlet condition both terms vanish separately; for the Neumann condition likewise; for the Robin condition $\partial u/\partial n = -au$ the two terms become $-auv$ and $+avu$ and cancel. This is exactly the property that makes $\nabla^2$ self-adjoint on the corresponding function space, and hence the starting point for eigenfunction expansions.
Notation. Strauss writes the Laplacian $\Delta$ and warns against the physicists' $\nabla^2$ ; we use $\nabla^2$ , as do Riley, Hobson and Bence and Arfken. Evans writes $D u$ for the gradient and $\nu$ for the outward normal. Only symbols differ.

NoteOriginal

Moments on the unit sphere and ball, used in the examples. Aligning the polar axis with the $x$ -axis so that $x = r\cos\theta$ , and using $\oint_{S^2} g(\cos\theta)\, dS = 2\pi\int_{-1}^{1} g(t)\, dt$ on the unit sphere,

\oint_{S^2} x^{2m}\, dS = 2\pi\int_{-1}^{1} t^{2m}\, dt = \frac{4\pi}{2m+1}, \qquad \int_{B} x^{2m}\, dV = \left(\int_0^1 r^{2m+2}\, dr\right)\frac{4\pi}{2m+1} = \frac{4\pi}{(2m+1)(2m+3)} ,

where $B$ is the unit ball. In particular $\oint x^2\,dS = \tfrac{4\pi}{3}$ , $\oint x^4\,dS = \tfrac{4\pi}{5}$ , $\oint x^6\,dS = \tfrac{4\pi}{7}$ , and $\int_B x^2\,dV = \tfrac{4\pi}{15}$ , $\int_B x^4\,dV = \tfrac{4\pi}{35}$ . Also, expanding $\oint (x^2+y^2+z^2)^2 dS = \oint 1\, dS = 4\pi$ into $3\oint x^4\, dS + 6\oint x^2y^2\, dS$ gives $\oint x^2y^2\, dS = \tfrac{4\pi}{15}$ , and by symmetry the same for $\oint x^2z^2\, dS$ .

Example(Green's first identity on the unit ball)Original

Let $V = B$ be the unit ball, $S$ the unit sphere, and

u = x^2 + y^2, \qquad v = x^2 .

Both are polynomials, hence $C^\infty$ on $\overline V$ , so (G1) applies.

Ingredients. $\nabla u = (2x, 2y, 0)$ , $\nabla^2 u = 4$ ; $\nabla v = (2x, 0, 0)$ ; and $\nabla u\cdot\nabla v = 4x^2$ . On $S$ the outward normal is $\hat{\mathbf{n}} = (x,y,z)$ , so $\frac{\partial u}{\partial n} = (2x,2y,0)\cdot(x,y,z) = 2(x^2+y^2) = 2(1 - z^2)$ using $x^2+y^2+z^2 = 1$ on $S$ .
Left side. Using the moments above, $\oint_S v\,\frac{\partial u}{\partial n}\, dS = \oint_S 2x^2\bigl(1 - z^2\bigr)\, dS = 2\oint_S x^2\, dS - 2\oint_S x^2z^2\, dS = 2\cdot\frac{4\pi}{3} - 2\cdot\frac{4\pi}{15} = \frac{32\pi}{15} .$
Right side, first term. $\displaystyle\int_B \nabla u\cdot\nabla v\; dV = 4\int_B x^2\, dV = 4\cdot\frac{4\pi}{15} = \frac{16\pi}{15}$ .
Right side, second term. $\displaystyle\int_B v\,\nabla^2u\; dV = 4\int_B x^2\, dV = \frac{16\pi}{15}$ .
Comparison. $\tfrac{16\pi}{15} + \tfrac{16\pi}{15} = \tfrac{32\pi}{15}$ , which is step 2. Numerically both sides are $6.70206$ to five decimal places. Note that all three terms are nonzero here, so this is a genuine test of the identity and not of a symmetry.

Example(Green's second identity on the unit ball)Original

On the same ball take

u = x^2, \qquad v = x^4 ,

so that $\nabla^2 u = 2$ and $\nabla^2 v = 12x^2$ .

Volume side. $\int_B \bigl(u\,\nabla^2v - v\,\nabla^2u\bigr)\, dV = \int_B \bigl(12x^4 - 2x^4\bigr)\, dV = 10\int_B x^4\, dV = 10\cdot\frac{4\pi}{35} = \frac{8\pi}{7} .$
Normal derivatives on $S$ . With $\hat{\mathbf{n}} = (x,y,z)$ on the unit sphere, $\frac{\partial v}{\partial n} = (4x^3,0,0)\cdot(x,y,z) = 4x^4, \qquad \frac{\partial u}{\partial n} = (2x,0,0)\cdot(x,y,z) = 2x^2 .$
Surface side. $\oint_S \left( u\,\frac{\partial v}{\partial n} - v\,\frac{\partial u}{\partial n} \right) dS = \oint_S \bigl(4x^6 - 2x^6\bigr)\, dS = 2\oint_S x^6\, dS = 2\cdot\frac{4\pi}{7} = \frac{8\pi}{7} .$
Comparison. Both sides are $8\pi/7 = 3.59039$ .
A contrast with (G1). Had we chosen $u = x^2$ , $v = y^2$ here, both sides would have come out $0$ by the symmetry of the ball under exchanging $x$ and $y$ — true, but no test of anything. Breaking the symmetry, as above, is what makes the check informative; this is why (G2) is harder to test numerically than (G1).

ApplicationOriginal

Green's second identity is the engine of potential theory. Choosing $v$ to be the fundamental solution $-1/(4\pi\lvert\mathbf{x}-\mathbf{x}_0\rvert)$ of the Laplacian and cutting out a small ball around its singularity turns (G2) into a formula expressing the value of a harmonic function at an interior point as an integral of $u$ and $\partial u/\partial n$ over the boundary — Strauss's representation formula (§7.2). Replacing that fundamental solution by the Green's function of the region, which also vanishes on the boundary, removes the unknown $\partial u/\partial n$ and gives the solution of the Dirichlet problem purely in terms of its data; this is the subject of Green's functions for Laplace's equation and the Poisson integral formula. Riley, Hobson and Bence single out the second identity for exactly this reason.

Green's first identity is the engine of the variational and numerical theory. Read from right to left it says that, for $u$ vanishing on the boundary, $\int v\nabla^2 u = -\int \nabla v\cdot\nabla u$ : a second-order operator has been traded for a symmetric first-order bilinear form. That trade defines the weak formulation of an elliptic problem, makes $-\nabla^2$ manifestly non-negative, and produces the stiffness matrix of the finite-element method; it is also what Strauss's Dirichlet principle rests on, which identifies the harmonic function with the minimizer of $\tfrac12\int\lvert\nabla w\rvert^2$ among functions with the given boundary values (compare the Rayleigh–Ritz method).

In physics the same identity is energy accounting. With $u = v = \phi$ the electrostatic potential and $\nabla^2\phi = -\rho/\varepsilon_0$ , (G1) gives $\int \lvert\mathbf{E}\rvert^2 dV = \int \rho\phi\, dV/\varepsilon_0$ for a potential decaying at infinity, equating field energy with the work of assembling the charges — Arfken's Exercise 3.8.4. The Neumann compatibility condition $\oint_S h\, dS = \int_V f\, dV$ is the statement that a body in steady state cannot absorb heat on balance, and in a numerical solver a violated compatibility condition is the standard reason a Neumann solve fails to converge.

Potentials

Theorem

Helmholtz decomposition and potentials

Adapted

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States and proves the two theorems of potential theory as Arfken gives them — a field is determined by its divergence, its curl and its boundary normal component on a simply connected region, and a field whose source and circulation densities vanish at infinity splits into an irrotational part $-\nabla\varphi$ and a solenoidal part $\nabla\times\mathbf{A}$ with the two Poisson integrals as potentials — and describes the gauge freedom in $\varphi$ and $\mathbf{A}$ .

Prerequisites: The divergence theorem, Stokes' theorem and Green's theorem in the plane, Line integrals and conservative fields, Green's identities, The Dirac delta in calculations, Differentiation under the integral sign

Theorem(Scalar and vector potentials)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, Eqs. (3.96)–(3.100), pp. 171–172 and Eq. (3.102), p. 172, verified on p. 173; Riley, Hobson & Bence (2006), §10.8.2, Eqs. (10.37)–(10.38), p. 356 and §11.4, p. 389

Two one-line identities govern everything in this unit: for $\phi \in C^2$ and $\mathbf{a} \in C^2$ ,

\nabla\times\nabla\phi = \mathbf{0}, \qquad\qquad \nabla\cdot(\nabla\times\mathbf{a}) = 0 .

So a gradient is irrotational and a curl is solenoidal. The converses are the content of potential theory:

Scalar potential. On a simply connected region, a $C^1$ field $\mathbf{F}$ with $\nabla\times\mathbf{F} = \mathbf{0}$ can be written $\mathbf{F} = -\nabla\varphi$ ; Arfken calls $\varphi$ a scalar potential, and the three conditions $\mathbf{F} = -\nabla\varphi$ , $\oint\mathbf{F}\cdot d\mathbf{r} = 0$ and $\nabla\times\mathbf{F} = \mathbf{0}$ are equivalent there. The step from the vanishing circulation to the vanishing curl is Stokes' theorem applied to small loops, which is how Arfken takes it (p. 172); the rest is the theorem on conservative fields, with the physicists' sign. $\varphi$ is determined up to one additive constant.

Why simple connectivity is needed. Arfken imposes it too: the section opens "over a given simply connected region of space (one with no holes)" (p. 171), and the equivalence of his three conditions, Eqs. (3.96), (3.99) and (3.100), is stated within that setting (p. 172). The hypothesis cannot be dropped. On $\mathbb{R}^3$ minus the $z$ -axis the field $(-y, x, 0)/(x^2+y^2)$ has zero curl everywhere yet circulation $2\pi$ around any loop encircling the axis (line integrals), so a curl-free field there need not be a gradient.
Vector potential. A $C^1$ field $\mathbf{B}$ with $\nabla\cdot\mathbf{B} = 0$ admits a field $\mathbf{A}$ with $\mathbf{B} = \nabla\times\mathbf{A}$ . Arfken's Eq. (3.102), p. 172, exhibits one explicitly,
$\mathbf{A} = \hat{\mathbf{e}}_y\!\int_{x_0}^{x}\!\! B_z(x,y,z)\, dx \;+\; \hat{\mathbf{e}}_z\!\left[\int_{y_0}^{y}\!\! B_x(x_0,y,z)\, dy - \int_{x_0}^{x}\!\! B_y(x,y,z)\, dx\right] ,$
and the text on p. 173 verifies $\nabla\times\mathbf{A} = \mathbf{B}$ for it, the $x$ -component being the one that uses $\nabla\cdot\mathbf{B} = 0$ . Riley, Hobson and Bence state the existence of such an $\mathbf{A}$ without proof (p. 389).
Gauge freedom. $\mathbf{A}$ is far from unique — Arfken's own phrase, in the paragraph on p. 173 that follows the verification: for any scalar $\psi$ and any constant vector $\mathbf{k}$ (Riley, Hobson and Bence call this constant vector $\mathbf{c}$ ; that letter is reserved here for Arfken's circulation density in Helmholtz's theorem below),
$\mathbf{A}' = \mathbf{A} + \nabla\psi + \mathbf{k} \qquad\text{also satisfies}\qquad \nabla\times\mathbf{A}' = \mathbf{B} ,$
by the first identity above. Arfken's formula also depends on the arbitrary base point $(x_0, y_0)$ , and a second formula with $x$ and $y$ interchanged (his Eq. (3.103)) gives yet another $\mathbf{A}$ . Replacing $\mathbf{A}$ by $\mathbf{A} + \nabla\psi$ is called a gauge transformation.

Theorem(Uniqueness from divergence, curl and boundary data)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, pp. 177–178

Let $V$ be a bounded simply connected region with piecewise $C^1$ boundary, and let $\mathbf{P}$ and $\mathbf{P}'$ be $C^1$ fields on the closed region $\overline V$ that agree in all three of

\nabla\cdot\mathbf{P} = \nabla\cdot\mathbf{P}', \qquad \nabla\times\mathbf{P} = \nabla\times\mathbf{P}', \qquad \mathbf{P}\cdot\hat{\mathbf{n}} = \mathbf{P}'\cdot\hat{\mathbf{n}} \ \text{ on } \partial V .

Then $\mathbf{P} = \mathbf{P}'$ everywhere in $V$ . Read as a recipe: three pieces of data pin a field down on such a region — how much it spreads out, how much it circulates, and how much of it crosses the boundary.

Boundedness is our hypothesis, not Arfken's. He asks only for a simply connected region, and his proof, which is the one given below, runs through Green's first identity; that identity is available on a bounded region with piecewise $C^1$ boundary and not otherwise (Green's identities). The hypothesis is load-bearing rather than decorative: on $V = \mathbb{R}^3$ , which is simply connected and has no boundary at all, the boundary condition would be vacuous and the conclusion false, since every constant field has zero divergence and zero curl. Arfken applies the theorem to all of space in his proof of Helmholtz's theorem; Step 5 of our second proof below therefore replaces that appeal by a limit over balls, and the difference from the source is recorded in verification/discrepancies.md.

A caveat of Arfken's applies here and to Helmholtz's theorem below. Suppose the divergence or the curl of the field makes sense at some point only as a delta function; that point still belongs to the region, and is not to be excised from it. A point charge sitting inside $V$ is therefore covered, with $\nabla\cdot\mathbf{P}$ understood as a multiple of $\delta^{(3)}$ (the Dirac delta) — whereas removing the point would replace $V$ by a region that is not simply connected in the same way, and the conclusion could fail.

Theorem(Helmholtz's theorem)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, Eqs. (3.122)–(3.125) and the closing paragraph straddling pp. 179–180, pp. 178–180

Suppose a field $\mathbf{P}$ has a source density and a circulation density that both die away at infinity. Then $\mathbf{P}$ splits into two summands, the first of them irrotational and the second solenoidal. Explicitly, with

s = \nabla\cdot\mathbf{P} \quad\text{(the source density)}, \qquad \mathbf{c} = \nabla\times\mathbf{P} \quad\text{(the circulation density)},

the decomposition is

\mathbf{P} = -\nabla\varphi + \nabla\times\mathbf{A}, \qquad \varphi(\mathbf{r}_1) = \frac{1}{4\pi}\int \frac{s(\mathbf{r}_2)}{r_{12}}\; d\tau_2 , \qquad \mathbf{A}(\mathbf{r}_1) = \frac{1}{4\pi}\int \frac{\mathbf{c}(\mathbf{r}_2)}{r_{12}}\; d\tau_2 ,

where $r_{12} = \lvert\mathbf{r}_1 - \mathbf{r}_2\rvert$ and the integrals are over all space. The first part is irrotational because $\nabla\times\nabla\varphi = \mathbf{0}$ and the second is solenoidal because $\nabla\cdot(\nabla\times\mathbf{A}) = 0$ , so the two pieces carry the whole of the divergence and the whole of the curl respectively.

In electromagnetic language, $s$ is the charge density divided by $\varepsilon_0$ and $\mathbf{c}$ is the current density times $\mu_0$ ; Arfken says the theorem is what legitimizes splitting the quantities of electromagnetic theory into an irrotational $\mathbf{E}$ and a solenoidal $\mathbf{B}$ , each with its own potential.

NoteOriginal

On the hypothesis. The hypothesis is stated above exactly as Arfken states it: the source and circulation densities vanish at infinity. He also uses, in appealing to the uniqueness theorem, that the field itself vanishes at infinity. Mere vanishing is not by itself enough for the two integrals to converge — for $\varphi(\mathbf{r}_1)$ to converge one needs $s$ to decay faster than $1/r^2$ , which holds in the case of chief interest, where $s$ and $\mathbf{c}$ have bounded support, as in the first example below. Convergence is not the only thing mere vanishing fails to give: the integration by parts in Step 3 of the second proof below needs $\mathbf{c} = o(1/r)$ , and the identification of $\mathbf{P}^{\star}$ with $\mathbf{P}$ in Step 5 needs a decay rate for the field, neither of which follows from vanishing alone. The remark below sets out all three conditions and which step needs which. We follow the source and state the hypothesis as given; a reader applying the theorem to a slowly decaying source should check those conditions first.

Idea of the proofOriginal

The uniqueness theorem is the energy argument of Green's identities in disguise. The difference of two candidate fields has zero divergence, zero curl and zero normal component. Zero curl on a simply connected region means it is a gradient; zero divergence then makes its potential harmonic; and Green's first identity converts the vanishing boundary data of a harmonic function into the vanishing of $\int\lvert\nabla\varphi\rvert^2$ , which forces the field itself to vanish.

Helmholtz's theorem is then proved without any further analysis, by checking rather than constructing. The two candidate potentials are built out of $1/r_{12}$ for one reason only: $\nabla^2(1/r) = -4\pi\delta^{(3)}$ , so applying the Laplacian to either integral sifts the source back out. That gives the proposed field the right divergence and the right curl — the curl computation needing one extra observation, that a circulation density is itself divergence-free, which kills the unwanted $\nabla(\nabla\cdot\mathbf{A})$ term. Having the right divergence and curl, it is the right field — not by the uniqueness theorem itself, which speaks only of bounded regions, but by running that theorem's energy argument on the ball $r < R$ and letting $R$ grow, the boundary term on the sphere being what the decay of the field kills.

Proof(The uniqueness theorem)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, p. 178; Strauss (2008), §7.1, Eq. (G1), p. 180

Step 1 (reduce to the homogeneous case). Let $\mathbf{Q} = \mathbf{P} - \mathbf{P}'$ . By linearity of $\nabla\cdot$ and $\nabla\times$ and of the normal component,

\nabla\cdot\mathbf{Q} = 0 \ \text{ in } V, \qquad \nabla\times\mathbf{Q} = \mathbf{0} \ \text{ in } V, \qquad \mathbf{Q}\cdot\hat{\mathbf{n}} = 0 \ \text{ on } \partial V .

It suffices to show $\mathbf{Q} \equiv \mathbf{0}$ .

Step 2 (a potential for $\mathbf{Q}$ ). $\mathbf{Q}$ is $C^1$ and irrotational on the simply connected region $V$ , so by part 1 of the first statement there is $\phi$ with $\mathbf{Q} = -\nabla\phi$ (conservative fields). Since $\mathbf{Q}$ is $C^1$ on $\overline V$ , so is $\nabla\phi$ , and $\phi \in C^2(\overline V)$ — which is what Step 4 will need. Here, and only here, simple connectivity is used.

Step 3 ( $\phi$ is harmonic, and its normal derivative vanishes). Taking the divergence, $0 = \nabla\cdot\mathbf{Q} = -\nabla^2\phi$ , so $\nabla^2\phi = 0$ in $V$ . On $\partial V$ , $\partial\phi/\partial n = \hat{\mathbf{n}}\cdot\nabla\phi = -\mathbf{Q}\cdot\hat{\mathbf{n}} = 0$ .

Step 4 (Green's first identity). $V$ is bounded with piecewise $C^1$ boundary and $\phi \in C^2(\overline V)$ , so Green's first identity is available (Green's identities) — this is where boundedness is used. Apply (G1) with $u = v = \phi$ :

\oint_{\partial V} \phi\,\frac{\partial\phi}{\partial n}\; dS = \int_V \lvert\nabla\phi\rvert^2\; dV + \int_V \phi\,\nabla^2\phi\; dV .

The left side vanishes because $\partial\phi/\partial n = 0$ on $\partial V$ (Step 3) — this is where the boundary hypothesis is used — and the last term vanishes because $\phi$ is harmonic. Hence

\int_V \lvert\nabla\phi\rvert^2\; dV = \int_V \mathbf{Q}\cdot\mathbf{Q}\; dV = 0 .

Step 5 (conclude). The integrand $\lvert\mathbf{Q}\rvert^2$ is continuous and non-negative with zero integral, so it vanishes identically: were it positive at a point it would exceed half that value on a ball around the point, making the integral positive. Thus $\mathbf{Q} \equiv \mathbf{0}$ and $\mathbf{P} = \mathbf{P}'$ . $\square$

Proof(Helmholtz's theorem)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, pp. 178–179; Riley, Hobson & Bence (2006), §10.8.2, Eq. (10.41), p. 356 and Eq. (10.42), p. 357

The case proved. The statement above is Arfken's, with his hypothesis that the two densities vanish at infinity. What is proved here is the case his integral formulas are really written for, and it is declared in full before the argument starts rather than being assembled as it goes:

$\mathbf{P}$ is $C^2$ on $\mathbb{R}^3$ , so that $s = \nabla\cdot\mathbf{P}$ is $C^1$ and $\mathbf{c} = \nabla\times\mathbf{P}$ is $C^1$ with $\nabla\cdot\mathbf{c} = 0$ ;
$s$ and $\mathbf{c}$ vanish outside some ball $r \le a$ (compact support);
$\mathbf{P} = O(r^{-2})$ as $r \to \infty$ .

Compact support does three things. It makes the two integrals converge, since each is an integral of a bounded density against the locally integrable kernel $1/r_{12}$ over a bounded set. It makes the single differentiation of Step 3 legitimate under the integral sign, because $\lvert\nabla_1(1/r_{12})\rvert = r_{12}^{-2}$ is locally integrable too and the domain is a fixed compact set (differentiation under the integral sign); the second derivatives are never taken naively, the two Laplacians being read through the identity $\nabla^2(1/r) = -4\pi\delta^{(3)}$ of the Dirac delta, which is itself a statement about integrals of exactly this shape. And it makes the surface term discarded in Step 3 vanish identically for large spheres rather than merely in a limit. The decay $\mathbf{P} = O(r^{-2})$ is the quantitative form of Arfken's "the vector involved vanishes at infinity", and Step 5 is where it is used. What each hypothesis can be weakened to is set out in the remark below.

Write $\mathbf{P}^{\star} = -\nabla\varphi + \nabla\times\mathbf{A}$ for the proposed field, with $\varphi$ and $\mathbf{A}$ the two integrals of the statement. Subscripts on $\nabla$ record which variable is being differentiated.

Step 1 (the divergence comes out right). Since $\nabla\cdot(\nabla\times\mathbf{A}) = 0$ ,

\nabla_1\cdot\mathbf{P}^{\star} = -\nabla_1^2\varphi(\mathbf{r}_1) = -\frac{1}{4\pi}\int \left[\nabla_1^2\frac{1}{r_{12}}\right] s(\mathbf{r}_2)\; d\tau_2 = -\frac{1}{4\pi}\int \bigl[-4\pi\,\delta^{(3)}(\mathbf{r}_1-\mathbf{r}_2)\bigr]\, s(\mathbf{r}_2)\; d\tau_2 = s(\mathbf{r}_1) ,

using $\nabla^2(1/r) = -4\pi\delta^{(3)}$ and then the sifting property. So $\mathbf{P}^{\star}$ has the same divergence as $\mathbf{P}$ .

Step 2 (prepare the curl). Since $\nabla\times\nabla\varphi = \mathbf{0}$ , we need $\nabla\times(\nabla\times\mathbf{A}) = \mathbf{c}$ . Use the identity

\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} ,

which in Cartesian components reads $[\nabla\times(\nabla\times\mathbf{A})]_j = [\nabla(\nabla\cdot\mathbf{A})]_j - \nabla^2A_j$ . Riley, Hobson and Bence's Eq. (10.41) is the vector identity, and they attach to the componentwise step, their Eq. (10.42), the warning that it is available only when the basis vectors are constant in direction, as in Cartesian coordinates, and not in cylindrical or spherical polars (§10.8.2, p. 357). We work in Cartesian coordinates throughout.

Step 3 (the unwanted term vanishes). Differentiating under the integral sign and noting that $s$ and $\mathbf{c}$ do not depend on $\mathbf{r}_1$ ,

\nabla_1\cdot\mathbf{A}(\mathbf{r}_1) = \frac{1}{4\pi}\int \mathbf{c}(\mathbf{r}_2)\cdot\nabla_1\frac{1}{r_{12}}\; d\tau_2 = -\frac{1}{4\pi}\int \mathbf{c}(\mathbf{r}_2)\cdot\nabla_2\frac{1}{r_{12}}\; d\tau_2 ,

the sign change because $1/r_{12}$ depends on $\mathbf{r}_1-\mathbf{r}_2$ , so that differentiating with respect to $\mathbf{r}_2$ instead of $\mathbf{r}_1$ reverses the result. Now integrate by parts in $\mathbf{r}_2$ . Fix $\mathbf{r}_1$ and take $R > \max(a, 2r_1)$ , so that the ball $V_R = \{r_2 < R\}$ contains the whole support of $\mathbf{c}$ in its interior. Write $g = 1/r_{12}$ ; away from $\mathbf{r}_2 = \mathbf{r}_1$ the product rule gives $\nabla_2\cdot(g\,\mathbf{c}) = \mathbf{c}\cdot\nabla_2 g + g\,\nabla_2\cdot\mathbf{c}$ . Apply the divergence theorem on $V_R$ with the ball $r_{12} < \varepsilon$ removed, where $g\,\mathbf{c}$ is $C^1$ . The inner sphere contributes at most $\varepsilon^{-1}\cdot 4\pi\varepsilon^2\cdot\max\lvert\mathbf{c}\rvert \to 0$ , while both volume integrands are absolutely integrable near $\mathbf{r}_1$ ( $g = r_{12}^{-1}$ and $\lvert\nabla_2 g\rvert = r_{12}^{-2}$ , against $d\tau_2 \sim r_{12}^2\,dr_{12}$ ). Letting $\varepsilon \to 0$ therefore gives

\int_{V_R} \mathbf{c}\cdot\nabla_2 g\; d\tau_2 = \oint_{r_2 = R} g\,\mathbf{c}\cdot d\mathbf{S}_2 \;-\; \int_{V_R} g\,\bigl(\nabla_2\cdot\mathbf{c}\bigr)\, d\tau_2 .

The surface term is exactly zero, not merely small: $\mathbf{c}$ vanishes on the sphere $r_2 = R$ , because $R > a$ and $\mathbf{c}$ is supported in $r_2 \le a$ . This is where the compact support declared in the preamble is used, and it is the one place where Arfken's hypothesis that $\mathbf{c}$ merely vanishes at infinity would not suffice. For a $\mathbf{c}$ without compact support the term is bounded only by $4\pi R^2\cdot\max_{r_2=R}\lvert\mathbf{c}\rvert/(R-r_1)$ , hence by $8\pi R\max_{r_2=R}\lvert\mathbf{c}\rvert$ once $R > 2r_1$ , because the sphere's area grows like $R^2$ while the kernel decays only like $R^{-1}$ ; so it dies as $R \to \infty$ only if $\mathbf{c} = o(1/r)$ , and for $\lvert\mathbf{c}\rvert\sim 1/r$ — a circulation density that does vanish at infinity — the bound does not tend to zero at all. Since $\mathbf{c}$ is supported in $r_2 \le a$ , the remaining volume integral is already the integral over all of space, and no limit in $R$ is needed:

\nabla_1\cdot\mathbf{A}(\mathbf{r}_1) = \frac{1}{4\pi}\int \bigl[\nabla_2\cdot\mathbf{c}(\mathbf{r}_2)\bigr]\,\frac{1}{r_{12}}\; d\tau_2 = 0 ,

because $\mathbf{c} = \nabla\times\mathbf{P}$ is a curl and $\nabla\cdot(\nabla\times\mathbf{P}) = 0$ (here $\mathbf{P} \in C^2$ is used). Hence the term $\nabla(\nabla\cdot\mathbf{A})$ in Step 2 is zero.

Step 4 (the curl comes out right). It remains to show $-\nabla^2\mathbf{A} = \mathbf{c}$ . The vector Laplacian acts componentwise in Cartesian coordinates, so for each $j$ , exactly as in Step 1,

-\nabla_1^2 A_j(\mathbf{r}_1) = -\frac{1}{4\pi}\int c_j(\mathbf{r}_2)\left[\nabla_1^2\frac{1}{r_{12}}\right] d\tau_2 = -\frac{1}{4\pi}\int c_j(\mathbf{r}_2)\bigl[-4\pi\,\delta^{(3)}(\mathbf{r}_1-\mathbf{r}_2)\bigr]\, d\tau_2 = c_j(\mathbf{r}_1) .

Combining with Steps 2 and 3, $\nabla\times\mathbf{P}^{\star} = \mathbf{c} = \nabla\times\mathbf{P}$ .

Step 5 (identify $\mathbf{P}^{\star}$ with $\mathbf{P}$ ). Put $\mathbf{Q} = \mathbf{P} - \mathbf{P}^{\star}$ . By Steps 1 and 4, $\nabla\cdot\mathbf{Q} = 0$ and $\nabla\times\mathbf{Q} = \mathbf{0}$ on the whole of $\mathbb{R}^3$ . The uniqueness theorem cannot be applied with $V = \mathbb{R}^3$ : that region is unbounded and has no boundary, so its boundary hypothesis would be empty, and the conclusion false, as any constant field shows. We run the argument of that theorem on the balls $V_R = \{r < R\}$ instead and let $R \to \infty$ .

Since $\mathbb{R}^3$ is simply connected and $\mathbf{Q}$ is $C^1$ and irrotational, $\mathbf{Q} = -\nabla\chi$ with $\chi \in C^2(\mathbb{R}^3)$ , and $\nabla^2\chi = -\nabla\cdot\mathbf{Q} = 0$ . Each $V_R$ is bounded with a smooth boundary, so Green's first identity applies there with $u = v = \chi$ ; the term in $\nabla^2\chi$ drops out, and $\partial\chi/\partial n = \hat{\mathbf{r}}\cdot\nabla\chi = -\mathbf{Q}\cdot\hat{\mathbf{r}}$ on $r = R$ , so

\int_{V_R} \lvert\mathbf{Q}\rvert^2\, dV \;=\; -\oint_{r=R} \chi\,\bigl(\mathbf{Q}\cdot\hat{\mathbf{r}}\bigr)\, dS .

The constant in $\chi$ may be used to sharpen this. Since $\nabla\cdot\mathbf{Q} = 0$ in $V_R$ , the divergence theorem gives $\oint_{r=R}\mathbf{Q}\cdot\hat{\mathbf{r}}\, dS = 0$ , so $\chi$ may be replaced by $\chi - k$ for any constant $k$ . Choose $k$ to be the value of $\chi$ at one point of the sphere $r = R$ . Any two points of that sphere are joined by a great-circle arc of length at most $\pi R$ , along which $\chi$ changes at a rate at most $\lvert\nabla\chi\rvert = \lvert\mathbf{Q}\rvert$ , so $\lvert \chi - k\rvert \le \pi R\, M(R)$ on the sphere, where $M(R) = \max_{r=R}\lvert\mathbf{Q}\rvert$ . With the sphere's area $4\pi R^2$ ,

\int_{V_R} \lvert\mathbf{Q}\rvert^2\, dV \;\le\; 4\pi R^2 \cdot \pi R\,M(R)\cdot M(R) \;=\; 4\pi^2 R^3 M(R)^2 .

The left side increases with $R$ , so if the right side tends to $0$ along some sequence $R \to \infty$ then $\int_{V_R}\lvert\mathbf{Q}\rvert^2\, dV = 0$ for every $R$ ; the integrand is continuous and non-negative, so $\mathbf{Q} \equiv \mathbf{0}$ , exactly as in Step 5 of the first proof. The theorem therefore follows as soon as $R^3M(R)^2 \to 0$ along a sequence — that is, as soon as $\lvert\mathbf{Q}\rvert$ decays faster than $r^{-3/2}$ .

It does so under the hypotheses declared in the preamble, which is where the compact support of $s$ and $\mathbf{c}$ and the decay $\mathbf{P} = O(r^{-2})$ are used. Indeed for $r_1 = \lvert\mathbf{r}_1\rvert \ge 2a$ every source point has $r_{12} \ge r_1 - a \ge r_1/2$ , so

\lvert\varphi(\mathbf{r}_1)\rvert \le \frac{1}{4\pi}\cdot\frac{2}{r_1}\int\lvert s\rvert\, d\tau_2 , \qquad \lvert\nabla_1\varphi(\mathbf{r}_1)\rvert \le \frac{1}{4\pi}\int \frac{\lvert s\rvert}{r_{12}^2}\, d\tau_2 \le \frac{1}{\pi r_1^2}\int\lvert s\rvert\, d\tau_2 ,

using $\lvert\nabla_1(1/r_{12})\rvert = 1/r_{12}^2$ , and the same bounds hold for $\mathbf{A}$ and its derivatives with $\lvert\mathbf{c}\rvert$ in place of $\lvert s\rvert$ . Hence $\mathbf{P}^{\star} = -\nabla\varphi + \nabla\times\mathbf{A}$ is $O(r^{-2})$ , so $M(R) = O(R^{-2})$ and $R^3M(R)^2 = O(R^{-1}) \to 0$ . Therefore $\mathbf{P}^{\star} = \mathbf{P}$ , that is,

\mathbf{P} = -\nabla\varphi + \nabla\times\mathbf{A} ,

with $-\nabla\varphi$ irrotational and $\nabla\times\mathbf{A}$ solenoidal. $\square$

RemarkOriginal

Where the hypotheses are used, and what the freedom is.

Simple connectivity, in Step 2 of the first proof only. Drop it, keeping every other hypothesis, and uniqueness fails. Take the capped cylindrical shell $V = \{1 \le \rho \le 2,\; 0 \le z \le h\}$ , which is bounded and has piecewise $C^1$ boundary but is not simply connected, and on it the field $\mathbf{Q} = (-y, x, 0)/(x^2+y^2)$ , which is $C^1$ on $\overline V$ because $\rho \ge 1$ there. Then $\nabla\cdot\mathbf{Q} = 0$ , $\nabla\times\mathbf{Q} = \mathbf{0}$ , and $\mathbf{Q}\cdot\hat{\mathbf{n}} = 0$ on the whole of $\partial V$ : on the two cylindrical walls the normal is $\pm\hat{\boldsymbol{\rho}}$ while $\mathbf{Q}$ is azimuthal, and on the two annular caps $z = 0$ and $z = h$ the normal is $\mp\hat{\mathbf{k}}$ while $\mathbf{Q}$ has no $z$ -component. Yet $\mathbf{Q} \ne \mathbf{0}$ , so $\mathbf{P}$ and $\mathbf{P} + \mathbf{Q}$ share all three data. What fails is exactly Step 2: $\mathbf{Q}$ has no single-valued potential on $V$ , its circulation round a circle $\rho = \text{const}$ being $2\pi$ (see the second example of line integrals). The caps matter. The region between two infinite coaxial cylinders falls outside the theorem for a second and independent reason, that it is unbounded, and so isolates nothing; closing it off at $z = 0$ and $z = h$ costs nothing, since $\mathbf{Q}\cdot\hat{\mathbf{n}}$ vanishes on the new faces too.
The normal component, not the tangential one. Specifying $\mathbf{P}\cdot\hat{\mathbf{n}}$ is what makes the left side of Step 4 vanish. Prescribing instead the tangential components would not close the argument in this form.
Decay at infinity, in Steps 3 and 5 of the second proof — and how much of it each step needs. Neither step is served by Arfken's hypothesis that the densities merely vanish at infinity, and the two need different things, so the proof above declares compact support in its preamble and both steps then have what they require outright.

Step 3 discards the surface term $\oint_{r_2=R}(1/r_{12})\,\mathbf{c}\cdot d\mathbf{S}_2$ . That term is bounded by $4\pi R^2\max_{r_2=R}\lvert\mathbf{c}\rvert/(R-r_1)$ , hence by $8\pi R\max_{r_2=R}\lvert\mathbf{c}\rvert$ once $R > 2r_1$ : the sphere's area grows like $R^2$ while the kernel decays only like $R^{-1}$ . So it tends to zero if and only if $\mathbf{c} = o(1/r)$ , and for $\lvert\mathbf{c}\rvert\sim 1/r$ — a field that does vanish at infinity — it does not, leaving $\nabla\cdot\mathbf{A} = 0$ , on which Steps 2 and 4 both rest, unestablished. The correct condition for Step 3 in general is therefore $\mathbf{c} = o(1/r)$ , and it is logged with the convergence condition in verification/discrepancies.md. Under compact support the term is identically zero for every $R > a$ and no limit is needed.

Step 5 needs more again. Its bound $4\pi^2R^3M(R)^2$ tends to zero only if $\lvert\mathbf{Q}\rvert$ decays faster than $r^{-3/2}$ , and that rate is got there from the compact support of $s$ and $\mathbf{c}$ together with $\mathbf{P} = O(r^{-2})$ . Arfken does not need it, because he applies his uniqueness theorem directly on all of space, which the statement above does not permit. Our Step 5 is the limiting argument that replaces that appeal, and the difference from the source is likewise recorded in verification/discrepancies.md.

Convergence of the two integrals is a third, separate condition: as the note above says, $\varphi$ needs $s$ to decay faster than $1/r^2$ . Compact support gives all three conditions at once, which is why the proof is written for that case.
What is unique and what is not. The decomposition into irrotational plus solenoidal is pinned down by the theorem, because the two pieces are reconstructed from $s$ and $\mathbf{c}$ . The potentials are not: $\varphi$ may be shifted by a constant, and $\mathbf{A}$ by any $\nabla\psi$ plus a constant vector. Arfken's Example 3.9.2 exhibits three different vector potentials for a constant field $\mathbf{B} = B_z\hat{\mathbf{e}}_z$ , namely $B_zx\,\hat{\mathbf{e}}_y$ , $-B_zy\,\hat{\mathbf{e}}_x$ and $\tfrac12(\mathbf{B}\times\mathbf{r})$ , and reconciles them by adding $\nabla(Cxy)$ . The particular $\mathbf{A}$ produced by the Helmholtz integral has $\nabla\cdot\mathbf{A} = 0$ , as Step 3 shows; in electromagnetism that choice is the Coulomb gauge.
A field can be decomposed without the integral formulas applying. The second example below splits a field that grows linearly at infinity. The theorem's integral formulas say nothing about it, but the decomposition exists, found by inspection. Conversely, a field decaying at infinity whose source decays only like $1/r^2$ may defeat the formulas, as the note above says.
Arfken's delta caveat. The two theorems are stated to include points where $\nabla\cdot\mathbf{P}$ or $\nabla\times\mathbf{P}$ exists only as a delta. This matters: the field of a point charge is the main application, and it is covered only because the singular point is kept inside the region, with $s$ read as $q\delta^{(3)}/\varepsilon_0$ .
Three dimensions, and what replaces this in general. The construction uses $1/r$ , the fundamental solution of the Laplacian in $\mathbb{R}^3$ , and the curl, which exists only in three dimensions. The statement that survives in any dimension and on curved spaces is the Hodge decomposition of differential forms, of which this is the case of a 1-form in $\mathbb{R}^3$ .

Example(A field with source but no circulation: a ball whose source grows with the radius)Original

Let $a > 0$ and take the source density

s(\mathbf{r}) = \begin{cases} 12\,r/a, & r < a,\\ 0, & r > a,\end{cases} \qquad \mathbf{c} = \mathbf{0} ,

a source that vanishes at the centre, grows linearly out to the value $12$ just inside $r = a$ , and is switched off beyond. Both densities have bounded support, so the Helmholtz integrals converge, and $\mathbf{A} = \mathbf{0}$ because $\mathbf{c} = \mathbf{0}$ . We compute $\varphi$ and recover $\mathbf{P} = -\nabla\varphi$ . The decay condition of the proof above will hold too; its smoothness condition holds everywhere except on the single sphere $r = a$ , where $s$ jumps, and step 4 below comes back to that.

The spherical average of $1/r_{12}$ . For $\lvert\mathbf{r}_1\rvert = r$ fixed, average $1/\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ over the sphere $\lvert\mathbf{r}_2\rvert = t$ . With $\mu$ the cosine of the angle between the two vectors, $r_{12} = \sqrt{r^2+t^2-2rt\mu}$ and
$\frac12\int_{-1}^{1}\frac{d\mu}{\sqrt{r^2+t^2-2rt\mu}} = \frac{1}{2rt}\Bigl[\sqrt{r^2+t^2+2rt} - \sqrt{r^2+t^2-2rt}\Bigr] = \frac{(r+t) - \lvert r-t\rvert}{2rt} = \frac{1}{\max(r,t)} .$
The potential. Since $s$ is radial, slice the ball into shells of radius $t$ and area $4\pi t^2$ and use step 1 on each shell. The factor $4\pi$ cancels against the $1/4\pi$ in front of the Helmholtz integral, leaving
$\varphi(r) = \int_0^a \frac{s(t)\,t^2}{\max(r,t)}\, dt = \frac{12}{a}\int_0^a \frac{t^3}{\max(r,t)}\, dt .$
For $r \le a$ split the range at $t = r$ :
$\varphi(r) = \frac{12}{a}\left[\frac1r\int_0^r t^3\, dt + \int_r^a t^2\, dt\right] = \frac{12}{a}\left[\frac{r^3}{4} + \frac{a^3 - r^3}{3}\right] = 4a^2 - \frac{r^3}{a} ,$
and for $r \ge a$ the whole ball has $\max(r,t) = r$ , so $\varphi(r) = (12/a)\cdot a^4/(4r) = 3a^3/r$ . Hence
$\varphi(r) = \begin{cases} 4a^2 - \dfrac{r^3}{a}, & r \le a,\\[2mm] \dfrac{3a^3}{r}, & r \ge a .\end{cases}$
Both expressions give $\varphi = 3a^2$ at $r = a$ , so $\varphi$ is continuous.
The field. $\mathbf{P} = -\nabla\varphi = -\varphi'(r)\,\hat{\mathbf{r}}$ , so
$\mathbf{P} = \frac{3r^2}{a}\,\hat{\mathbf{r}} \quad (r \le a), \qquad \mathbf{P} = \frac{3a^3}{r^2}\,\hat{\mathbf{r}} \quad (r \ge a) ,$
continuous at $r = a$ with magnitude $3a$ on both sides: the field grows as $r^2$ out to the surface and then falls as $r^{-2}$ .
Check the data, on each side of $r = a$ separately. By the radial formula $\nabla\cdot(f(r)\hat{\mathbf{r}}) = r^{-2}(r^2f)'$ : inside, $r^{-2}(3r^4/a)' = 12r/a = s$ ; outside, $r^{-2}(3a^3)' = 0 = s$ . And $\nabla\times(f(r)\hat{\mathbf{r}}) = \mathbf{0}$ for any radial $f$ , so $\mathbf{c} = \mathbf{0}$ as required. The field decays like $r^{-2}$ , so it vanishes at infinity at the rate the proof's Step 5 asks for.

The check has to be made piecewise, and it is worth saying why. The density $s$ jumps from $12$ to $0$ across $r = a$ , and since $\nabla^2\varphi = -s$ , the second derivative of $\varphi$ must jump by the same amount: $\varphi'' = -6r/a \to -6$ from inside and $\varphi'' = 6a^3/r^3 \to +6$ from outside. So $\varphi$ is $C^1$ but not $C^2$ at $r = a$ , and $\mathbf{P} = -\nabla\varphi$ is continuous there but not differentiable — $\lvert\mathbf{P}\rvert$ has a kink, with slope $+6$ approaching the sphere from inside and $-6$ from outside. The divergence $\nabla\cdot\mathbf{P}$ therefore does not exist on the sphere $r = a$ at all, which is exactly why the verification above is carried out separately on $r < a$ and on $r > a$ .

That this costs nothing is a statement about measure, not about smoothness: $r = a$ is a single surface, of zero volume, so it contributes nothing to either Helmholtz integral and nothing to any flux balance, and $\nabla\cdot\mathbf{P} = s$ holds at every point off it. The example is thus marginally outside the letter of the proof's hypothesis $\mathbf{P}\in C^2(\mathbb{R}^3)$ , on a zero-volume set; a density with a jump across a surface is the normal situation in electrostatics, and the honest reading is that the formulas are being checked wherever the derivatives they contain exist. The second example below, whose field is a polynomial, has no such surface anywhere.
A number. Take $a = 2$ , so $s = 6r$ inside the ball of radius $2$ . Then $\varphi(0) = 16$ and $\varphi(2) = 12$ , while $\lvert\mathbf{P}\rvert = 1.5$ at $r = 1$ , rises to $6$ at $r = 2$ , and is back to $1.5$ at $r = 4$ . The flux out of the sphere $r = 4$ is $1.5\times 4\pi(16) = 96\pi$ , and
$\int s\, d\tau = \int_0^2 6t\cdot 4\pi t^2\, dt = 24\pi\left[\frac{t^4}{4}\right]_0^2 = 96\pi :$
the divergence theorem is satisfied, as it must be.
What it is. This is the electrostatic field of a ball whose charge density grows in proportion to the distance from the centre, with $s = \rho/\varepsilon_0$ . It is worth contrasting with the uniform ball, whose field rises linearly inside; here the enclosed charge grows like $r^4$ rather than $r^3$ , so the field rises as $r^2$ instead. Outside the ball, both look alike — only the total charge is visible, which is the content of the $3a^3/r$ tail.

Example(A field with both parts, split by inspection)Original

Take

\mathbf{F} = (x + y,\;\; y - x,\;\; 2z) \qquad\text{on all of } \mathbb{R}^3 .

Its data. $\nabla\cdot\mathbf{F} = 1 + 1 + 2 = 4$ and $\nabla\times\mathbf{F} = \bigl(\partial_y(2z) - \partial_z(y-x),\; \partial_z(x+y) - \partial_x(2z),\; \partial_x(y-x) - \partial_y(x+y)\bigr) = (0,0,-2) .$ Neither vanishes, so both parts of the decomposition are present.
Guess the pieces. Try $\varphi = -\left(\frac{x^2+y^2}{2} + z^2\right), \qquad \mathbf{A} = \Bigl(0,\, 0,\, \frac{x^2+y^2}{2}\Bigr) .$ Then $-\nabla\varphi = (x, y, 2z)$ and $\nabla\times\mathbf{A} = (\partial_yA_3,\; -\partial_xA_3,\; 0) = (y, -x, 0)$ .
Check. Their sum is $(x+y,\; y-x,\; 2z) = \mathbf{F}$ . Moreover $\nabla\cdot(-\nabla\varphi) = -\nabla^2\varphi = 4$ and $\nabla\times(\nabla\times\mathbf{A}) = \nabla\times(y,-x,0) = (0,0,-2)$ , so the irrotational piece carries all the divergence and the solenoidal piece all the curl, as Helmholtz's theorem says they must.
Gauge freedom, made explicit. Replace $\mathbf{A}$ by $\mathbf{A}' = \mathbf{A} + \nabla(Kxy) = \bigl(Ky,\; Kx,\; \tfrac12(x^2+y^2)\bigr)$ for any constant $K$ . Then $\nabla\times\mathbf{A}' = \bigl(\partial_yA'_3 - \partial_zA'_2,\; \partial_zA'_1 - \partial_xA'_3,\; \partial_xA'_2 - \partial_yA'_1\bigr) = (y,\; -x,\; K - K) = (y,-x,0) ,$ unchanged. Note that $\nabla\cdot\mathbf{A} = 0$ while $\nabla\cdot\mathbf{A}' = 0$ too here, so this family stays in the Coulomb gauge; a $\psi$ with $\nabla^2\psi \ne 0$ would leave it.
Why the integral formulas do not apply. $\mathbf{F}$ grows linearly with $\lvert\mathbf{r}\rvert$ , and $s = 4$ , $\mathbf{c} = (0,0,-2)$ are constants: they do not vanish at infinity, and the integrals $\int s/r_{12}\, d\tau_2$ and $\int\mathbf{c}/r_{12}\, d\tau_2$ both diverge. The decomposition nevertheless exists. The theorem gives a sufficient condition and a formula, not a characterization of the fields that split.

ApplicationAdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, Example 3.9.3 and Eqs. (3.108)–(3.111), pp. 174–175

Electromagnetism is built on this decomposition. Because $\nabla\cdot\mathbf{B} = 0$ always, $\mathbf{B}$ has a vector potential, $\mathbf{B} = \nabla\times\mathbf{A}$ ; substituting into Faraday's law shows that $\mathbf{E} + \partial\mathbf{A}/\partial t$ is irrotational, hence a gradient, which defines the scalar potential:

\mathbf{B} = \nabla\times\mathbf{A}, \qquad \mathbf{E} = -\nabla\varphi - \frac{\partial\mathbf{A}}{\partial t} .

Six field components have been replaced by four potential components, and the gauge freedom $\mathbf{A}\mapsto\mathbf{A}+\nabla\psi$ remains, which amounts to a free choice of $\nabla\cdot\mathbf{A}$ . Arfken's Example 3.9.3 makes the choice $c^{-2}\partial_t\varphi + \nabla\cdot\mathbf{A} = 0$ , the Lorentz gauge, which decouples the equations into $\nabla^2\varphi - c^{-2}\partial_t^2\varphi = -\rho/\varepsilon_0$ and $c^{-2}\partial_t^2\mathbf{A} - \nabla^2\mathbf{A} = \mu_0\mathbf{J}$ : two wave equations, one with the charge density as source and one with the current density. The invariance of all predictions under gauge transformations is, as Arfken remarks, the precursor of gauge theory in modern physics.

In fluid mechanics the same split separates a flow into a potential part and a vortical part. Removing the irrotational part is what every incompressible Navier–Stokes solver does at each time step: given a provisional velocity field, project out its gradient part by solving a Poisson equation for the pressure, leaving a divergence-free field. That projection is the Helmholtz decomposition, and in the discrete setting its uniqueness is what makes the pressure well determined up to a constant. In magnetostatics the vector potential of a current loop is the Helmholtz integral $\mathbf{A} = (\mu_0/4\pi)\int \mathbf{J}/r_{12}\, d\tau_2$ , from which the Biot–Savart law follows by taking the curl; and in image processing and computer graphics the decomposition of a vector field into curl-free and divergence-free parts is a standard tool for smoothing and for detecting sources and vortices in measured flow fields.

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The Dirac delta in calculations

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Introduces the delta as the limit of a normalized family of peaked functions, proves that the Gaussian family sifts the value of any bounded function continuous at the point, collects the working rules (evenness, rescaling, $\delta(g(x))$ , the derivative, $\theta' = \delta$ ), defines the three-dimensional delta, and proves $\nabla^2(1/r) = -4\pi\delta^{(3)}(\mathbf{r})$ from Green's second identity.

Prerequisites: The divergence theorem, Green's identities, Gaussian integrals

Definition(Delta sequence, the delta, the sifting property)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.11, Eqs. (1.149)–(1.157), pp. 75–78; Hassani (2013), §7.3, Eq. (7.20) and Examples 7.3.1–7.3.3, pp. 229–232; Stakgold & Holst (2011), §1.2, pp. 62–63

What one wants is an object $\delta$ concentrated at a single point with unit total weight:

\delta(x) = 0 \ \ (x \ne 0), \qquad \int_a^b f(x)\,\delta(x)\, dx = f(0) \ \ \text{whenever } 0 \in (a,b), \qquad \int_{-\infty}^{\infty}\delta(x)\, dx = 1 .

These are Arfken's Eqs. (1.149)–(1.151). No function has them: a function vanishing off a single point has integral zero. Stakgold and Holst make the same point for the concentrated unit source. So the symbol is given meaning in two steps.

A delta family (or delta sequence) is a family $\{\delta_\varepsilon\}_{\varepsilon > 0}$ of integrable functions with $\int_{-\infty}^{\infty}\delta_\varepsilon = 1$ for every $\varepsilon$ , becoming taller and narrower as $\varepsilon \to 0$ . The standard three are $\delta_\varepsilon(x) = \frac{1}{\sqrt{\varepsilon\pi}}\,e^{-x^2/\varepsilon} \quad\text{(Gaussian)}, \qquad \delta_\varepsilon(x) = \begin{cases} 1/2\varepsilon, & \lvert x\rvert < \varepsilon \\ 0 & \text{otherwise}\end{cases} \quad\text{(rectangular)}, \qquad D_T(x) = \frac{\sin Tx}{\pi x} \quad (T = 1/\varepsilon \to \infty).$
$\delta$ is defined by what it does inside an integral. For a given class of test functions $f$ , $\int f(x)\,\delta(x)\, dx \;:=\; \lim_{\varepsilon\to 0}\int f(x)\,\delta_\varepsilon(x)\, dx ,$ which is Arfken's Eq. (1.157). The limit $\lim_{\varepsilon\to 0}\delta_\varepsilon(x)$ itself does not exist, and the integral on the left is not a Riemann integral; $\delta$ is a distribution, not a function. Every statement below is an abbreviation for a statement about such limits.

The identity $\int f(x)\,\delta(x - x_0)\, dx = f(x_0)$ is called the sifting property; $\delta(x - x_0)$ is the delta translated to $x_0$ . In three dimensions,

\delta^{(3)}(\mathbf{r}) = \delta(x)\,\delta(y)\,\delta(z), \qquad \int_V f(\mathbf{r})\,\delta^{(3)}(\mathbf{r} - \mathbf{a})\, dV = \begin{cases} f(\mathbf{a}), & \mathbf{a} \in V, \\ 0, & \mathbf{a} \notin V,\end{cases}

which is the form Riley, Hobson and Bence use for a point source and Hassani's Eq. (7.24) in $n$ dimensions. The object $\delta^{(3)}$ is localized at a point and carries unit integrated weight whatever coordinate system is in use; in spherical polars this reads $\int f(r,\theta,\phi)\,\delta^{(3)}(\mathbf{r}-\mathbf{r}_1)\, r^2\sin\theta\, dr\, d\theta\, d\phi = f(\mathbf{r}_1)$ , Arfken's Eq. (1.162).

Definition(The Gaussian family sifts)Original

Let $\delta_\varepsilon(x) = (\varepsilon\pi)^{-1/2}e^{-x^2/\varepsilon}$ for $\varepsilon > 0$ , and let $f : \mathbb{R}\to\mathbb{R}$ be bounded and continuous at $x = 0$ . Then

\int_{-\infty}^{\infty}\delta_\varepsilon(x)\, dx = 1 \quad\text{for every } \varepsilon > 0, \qquad\text{and}\qquad \lim_{\varepsilon\to 0^+}\int_{-\infty}^{\infty} f(x)\,\delta_\varepsilon(x)\, dx = f(0) .

Continuity at the one point $x = 0$ is all that is needed; $f$ may be discontinuous elsewhere. Arfken's Eq. (1.150) states the sifting property for "any well-behaved function", and Stakgold and Holst state their Eq. (1.2.6) for $\phi$ continuous at the point in question, which is the hypothesis used here.

Definition(Working rules)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.11, Eqs. (1.158)–(1.162), pp. 78–79; Hassani (2013), §7.3, Eqs. (7.22)–(7.23), pp. 231–232

Each identity below means that the two sides give the same value when integrated against a test function.

Evenness. $\delta(-x) = \delta(x)$ .
Rescaling. $\delta(ax) = \delta(x)/\lvert a\rvert$ for real $a \ne 0$ . (Arfken states Eq. (1.158) for $a>0$ and records the modulus for $a<0$ immediately after.)
Shift. $\int f(x)\,\delta(x - x_0)\, dx = f(x_0)$ .
Composition with a function. If $g$ is $C^1$ with simple zeros $a_i$ on the real axis, so that $g'(a_i) \ne 0$ , then $\delta\bigl(g(x)\bigr) = \sum_i \frac{\delta(x - a_i)}{\lvert g'(a_i)\rvert}, \qquad\text{hence}\qquad \int f(x)\,\delta\bigl(g(x)\bigr)\, dx = \sum_i \frac{f(a_i)}{\lvert g'(a_i)\rvert} .$
Derivative. $\displaystyle\int f(x)\,\delta'(x - x_0)\, dx = -f'(x_0)$ , which may be taken as the definition of $\delta'$ ; more generally the $n$ -th derivative gives $(-1)^nf^{(n)}(x_0)$ .
Step function. With $\theta(x - x_0) = 0$ for $x < x_0$ and $1$ for $x > x_0$ , $\ \dfrac{d}{dx}\theta(x - x_0) = \delta(x - x_0)$ (Hassani's Eq. (7.23)).
Fourier representation. $\displaystyle\delta(x - x') = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(x-x')t}\, dt$ , meaningful only under a further integral sign (Hassani's Eq. (7.22), Arfken's Eq. (1.163)).

Definition(The Laplacian of

1/r

)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.9, Eqs. (3.119)–(3.121), p. 177

Let $r = \lvert\mathbf{r}\rvert$ . Then $\nabla^2(1/r) = 0$ at every point with $r > 0$ , while

\nabla^2\frac{1}{r} = -4\pi\,\delta^{(3)}(\mathbf{r}) ,

meaning precisely that

\int_{\mathbb{R}^3} \frac{1}{r}\,\nabla^2 f(\mathbf{r})\; dV = -4\pi\, f(\mathbf{0}) \qquad\text{for every } f \in C^2(\mathbb{R}^3) \text{ of compact support.}

In the two-point form, $\nabla_1^2\bigl(1/r_{12}\bigr) = -4\pi\,\delta^{(3)}(\mathbf{r}_1-\mathbf{r}_2)$ with $r_{12} = \lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ , the subscript on $\nabla_1$ indicating that the derivatives act on $\mathbf{r}_1$ . Arfken stresses that this identity sidesteps the non-existence of the derivatives of $1/r$ at the origin and is meaningful only inside an integral; it is what makes the point-charge potential $q/4\pi\varepsilon_0 r$ a solution of Poisson's equation $\nabla^2\varphi = -q\,\delta^{(3)}(\mathbf{r})/\varepsilon_0$ .

Idea of the proofOriginal

The sifting property is a statement that a narrowing bump cannot see anything but the value of $f$ at its centre. Split the integral at a small radius $\rho$ : inside, $f$ differs from $f(0)$ by as little as we like, by continuity, and the bump's total mass there is at most $1$ ; outside, $f$ is merely bounded, and the bump's mass out there tends to $0$ because the mass is normalized and the bump is concentrating. Choosing $\rho$ first and then $\varepsilon$ makes both errors small.

For $\nabla^2(1/r)$ the mechanism is geometric. Away from the origin $1/r$ is harmonic, so all the content sits at the one bad point; and the flux of $\nabla(1/r) = -\hat{\mathbf{r}}/r^2$ through a sphere of radius $a$ is $-(1/a^2)(4\pi a^2) = -4\pi$ , the same for every $a$ , because the field weakens exactly as fast as the sphere grows. So the divergence theorem assigns a total "divergence" of $-4\pi$ to any region containing the origin and $0$ to any region avoiding it: that is what $-4\pi\delta^{(3)}$ records. The proof below converts this into a statement about integrals against test functions, by excising a small ball and using Green's second identity on the shell that remains.

Proof(The Gaussian family sifts)Original

Step 1 (normalization). Substituting $x = \sqrt\varepsilon\, s$ and using $\int_{-\infty}^\infty e^{-s^2}\, ds = \sqrt\pi$ (Gaussian integrals),

\int_{-\infty}^{\infty}\delta_\varepsilon(x)\, dx = \frac{1}{\sqrt{\varepsilon\pi}}\int_{-\infty}^{\infty} e^{-x^2/\varepsilon}\, dx = \frac{1}{\sqrt{\varepsilon\pi}}\cdot\sqrt{\varepsilon}\cdot\sqrt{\pi} = 1 .

Step 2 (what has to be estimated). Since $\delta_\varepsilon$ integrates to $1$ ,

\int f\,\delta_\varepsilon\, dx - f(0) = \int_{-\infty}^{\infty}\bigl[f(x) - f(0)\bigr]\,\delta_\varepsilon(x)\, dx ,

and the integral converges absolutely because $f$ is bounded and $\delta_\varepsilon$ is integrable. Let $M = \sup\lvert f\rvert$ and fix $\eta > 0$ .

Step 3 (the near part, using continuity at $0$ ). By continuity of $f$ at $0$ there is $\rho > 0$ with $\lvert f(x) - f(0)\rvert < \eta/2$ for $\lvert x\rvert \le \rho$ . Since $\delta_\varepsilon \ge 0$ and its total integral is $1$ ,

\left\lvert \int_{\lvert x\rvert \le \rho}\bigl[f - f(0)\bigr]\delta_\varepsilon\, dx \right\rvert \;\le\; \frac{\eta}{2}\int_{\lvert x\rvert\le\rho}\delta_\varepsilon\, dx \;\le\; \frac{\eta}{2} .

Step 4 (the far part, using boundedness and concentration). On $\lvert x\rvert > \rho$ we only know $\lvert f - f(0)\rvert \le 2M$ . Substituting $x = \sqrt\varepsilon s$ again,

\int_{\lvert x\rvert > \rho}\delta_\varepsilon(x)\, dx = \frac{2}{\sqrt\pi}\int_{\rho/\sqrt\varepsilon}^{\infty} e^{-s^2}\, ds \;\le\; \frac{2}{\sqrt{\pi}}\;\frac{e^{-\rho^2/\varepsilon}}{2\rho/\sqrt{\varepsilon}} ,

using $\int_T^\infty e^{-s^2}ds \le \int_T^\infty (s/T)e^{-s^2}ds = e^{-T^2}/2T$ for $T > 0$ . The right-hand side tends to $0$ as $\varepsilon \to 0^+$ with $\rho$ held fixed, so there is $\varepsilon_0 > 0$ such that for $0 < \varepsilon < \varepsilon_0$ the far part contributes less than $\eta/2$ in absolute value — here the hypothesis that $f$ is bounded is used, and nothing more.

Step 5 (conclude). For $0 < \varepsilon < \varepsilon_0$ the two estimates give $\bigl\lvert\int f\delta_\varepsilon - f(0)\bigr\rvert < \eta$ . As $\eta > 0$ was arbitrary, the limit is $f(0)$ . $\square$

Proof(The Laplacian of

1/r

)Original

Step 1 ( $1/r$ is harmonic off the origin). For a function of $r$ alone, $\nabla^2F = F'' + (2/r)F'$ (the radial Laplacian). With $F = 1/r$ we have $F' = -1/r^2$ and $F'' = 2/r^3$ , so

\nabla^2\frac{1}{r} = \frac{2}{r^3} - \frac{2}{r}\cdot\frac{1}{r^2} = 0 \qquad (r > 0) .

Step 2 (excise the singularity). Let $f \in C^2(\mathbb{R}^3)$ vanish outside the ball $r < R$ , and for $0 < \varepsilon < R$ let $A_\varepsilon = \{\varepsilon < r < R\}$ , a spherical shell on which $u = 1/r$ is $C^\infty$ . By Green's second identity applied to $u = 1/r$ and $v = f$ on $A_\varepsilon$ , and using $\nabla^2u = 0$ there (Step 1),

\int_{A_\varepsilon}\frac{1}{r}\,\nabla^2 f\; dV = \oint_{\partial A_\varepsilon}\left( \frac{1}{r}\frac{\partial f}{\partial n} - f\,\frac{\partial}{\partial n}\frac{1}{r} \right) dS .

Step 3 (the outer sphere contributes nothing). $\partial A_\varepsilon$ has two components. On $r = R$ both $f$ and $\nabla f$ vanish, because $f \equiv 0$ on a neighbourhood of that sphere, so this component contributes $0$ .

Step 4 (the inner sphere). On $r = \varepsilon$ the normal pointing out of $A_\varepsilon$ points towards the origin, so $\partial/\partial n = -\partial/\partial r$ there, and $-\partial(1/r)/\partial n = +\partial(1/r)/\partial r = -1/\varepsilon^2$ . The inner contribution is therefore

\oint_{r=\varepsilon}\left( -\frac{1}{\varepsilon}\frac{\partial f}{\partial r} - \frac{f}{\varepsilon^2} \right) dS .

The first piece is bounded in modulus by $\varepsilon^{-1}\max\lvert\nabla f\rvert\cdot 4\pi\varepsilon^2 = 4\pi\varepsilon\max\lvert\nabla f\rvert$ , which tends to $0$ ; here $f \in C^1$ with bounded gradient is used. The second piece is exactly $-4\pi$ times the average of $f$ over the sphere of radius $\varepsilon$ , since the sphere has area $4\pi\varepsilon^2$ ; as $\varepsilon\to 0$ that average tends to $f(\mathbf{0})$ by continuity. So the inner contribution tends to $-4\pi f(\mathbf{0})$ .

Step 5 (the left side converges to the full integral). Near the origin $\lvert\nabla^2 f\rvert \le C$ , so

\int_{r<\varepsilon}\left\lvert\frac{\nabla^2f}{r}\right\rvert\, dV \;\le\; C\int_0^{\varepsilon}\frac{1}{r}\,4\pi r^2\, dr = 2\pi C\varepsilon^2 \longrightarrow 0 ,

so $(1/r)\nabla^2f$ is absolutely integrable on $r < R$ and $\int_{A_\varepsilon} \to \int_{r<R} = \int_{\mathbb{R}^3}$ .

Step 6 (conclude). Letting $\varepsilon \to 0$ in Step 2 and using Steps 3–5,

\int_{\mathbb{R}^3}\frac{1}{r}\,\nabla^2 f\; dV = -4\pi f(\mathbf{0}) ,

which is the assertion. Since the distributional Laplacian of $u$ is defined by making $\int (\nabla^2 u)f$ mean $\int u\,\nabla^2 f$ — the two integrations by parts of Green's second identity, with no boundary terms for compactly supported $f$ — this is the statement $\nabla^2(1/r) = -4\pi\delta^{(3)}(\mathbf{r})$ . Translating by $\mathbf{r}_2$ gives the two-point form. $\square$

RemarkOriginal

What the symbols are allowed to mean.

$\delta$ is not a function, and $\delta(0)$ is not a number. Hassani's remark that " $\delta(0) = \infty$ " is a description of the shape of the approximating family, not an equation. Products such as $\delta(x)^2$ and compositions such as $\delta(\delta(x))$ have no meaning at all; the rules of the working list are exactly the operations that survive the limit.
Which family is used does not matter, but the test class does. The Gaussian, rectangular and $\sin Tx/\pi x$ families all sift. The last is not non-negative and not integrable in absolute value on $\mathbb{R}$ , so Steps 3 and 4 above do not apply to it verbatim and it requires more of $f$ — which is why the Fourier representation of rule 7 is the most delicate of the list and is stated by both Arfken and Hassani as meaningful only under a further integral.
The simple-zero hypothesis in rule 4. If $g'(a) = 0$ the rule fails: for $g(x) = x^2$ the right-hand side is undefined, and indeed $\int f(x)\,\delta(x^2)\, dx$ has no finite value, since the rectangular family gives $\int_{-\sqrt\varepsilon}^{\sqrt\varepsilon} f/2\varepsilon \sim f(0)/\sqrt{\varepsilon} \to \infty$ . Arfken's derivation of rule 4 replaces $g$ near each zero by the first term of its Taylor series, which is legitimate only when that term is nonzero.
Where the delta is forced on vector calculus. The field $\mathbf{v} = Q\hat{\mathbf{r}}/4\pi r^2$ of a point source has $\nabla\cdot\mathbf{v} = 0$ wherever it is defined, yet flux $Q$ out of every surface enclosing the origin: the divergence theorem appears to fail. Riley, Hobson and Bence resolve this by writing $\nabla\cdot\mathbf{v} = Q\,\delta^{(3)}(\mathbf{r})$ , after which the theorem holds for every closed surface, enclosing the source or not. This is the same identity as $\nabla^2(1/r) = -4\pi\delta^{(3)}$ , since $\mathbf{v} = -(Q/4\pi)\nabla(1/r)$ .
Dimensions. $\delta(x)$ has the dimensions of $1/x$ , and $\delta^{(3)}(\mathbf{r})$ those of $1/\text{volume}$ — which is why a point charge $q$ has charge density $q\,\delta^{(3)}(\mathbf{r})$ and not $q\,\delta(r)$ . Confusing $\delta^{(3)}(\mathbf{r})$ with the radial $\delta(r)$ is the commonest error in spherical coordinates: for a radially symmetric test function, $\delta^{(3)}(\mathbf{r})$ pairs with $f(0)$ while $\delta(r - a)$ pairs with $4\pi a^2 f(a)$ when the volume element is included.
What is deferred. The systematic theory — which test functions, what topology, why the limits exist, how far the calculus extends — belongs to distributions and tempered distributions. Arfken's §1.11 and Stakgold and Holst's §1.2 both say explicitly that their treatment is provisional in exactly this sense, and this unit follows them: it is the practitioner's rulebook, with the two proofs that make the rules trustworthy.

Example(The Gaussian family sifting a cosine, exactly)Original

Take $f(x) = \cos x$ , which is bounded and continuous, so the theorem applies. Here the approximating integrals can be computed in closed form:

\int_{-\infty}^{\infty}\frac{\cos x}{\sqrt{\varepsilon\pi}}\;e^{-x^2/\varepsilon}\, dx = e^{-\varepsilon/4} .

Derivation. Write $\cos x = \tfrac12(e^{ix} + e^{-ix})$ and complete the square: $-x^2/\varepsilon \pm ix = -\bigl(x/\sqrt\varepsilon \mp i\sqrt\varepsilon/2\bigr)^2 - \varepsilon/4$ . Each of the two Gaussian integrals (Gaussian integrals) contributes $\tfrac12\sqrt{\varepsilon\pi}\,e^{-\varepsilon/4}$ , and dividing by $\sqrt{\varepsilon\pi}$ leaves $e^{-\varepsilon/4}$ .
The limit. $e^{-\varepsilon/4}\to 1 = \cos 0$ as $\varepsilon\to 0$ , as the theorem promises.
Numbers. $\varepsilon = 0.05$ gives $0.987578$ ; $\varepsilon = 0.01$ gives $0.997503$ . The error is $\varepsilon/4$ to leading order, so the approximation converges, but only linearly in the width parameter — one reason numerical work with narrow bumps is delicate.
A function that is bounded but not continuous at $0$ . For $f = \theta$ , the unit step, the same family gives $\int\theta\,\delta_\varepsilon = \tfrac12$ for every $\varepsilon$ , by the evenness of $\delta_\varepsilon$ . The limit exists but is the average of the two one-sided limits, not a value of $f$ : with continuity at the point dropped, the conclusion changes.

Example(A delta composed with a quadratic)Original

Evaluate $\displaystyle I = \int_{-\infty}^{\infty}\bigl(x^4 + 1\bigr)\,\delta\bigl(x^2 - 4\bigr)\, dx$ .

Zeros and slopes. $g(x) = x^2 - 4$ vanishes at $a_{1,2} = \pm 2$ , and $g'(x) = 2x$ , so $\lvert g'(\pm 2)\rvert = 4 \ne 0$ : both zeros are simple and rule 4 applies.
Apply the rule. $I = \frac{f(2)}{\lvert g'(2)\rvert} + \frac{f(-2)}{\lvert g'(-2)\rvert} = \frac{17}{4} + \frac{17}{4} = \frac{17}{2} = 8.5 .$
Check it directly from a family. Using the rectangular family, $\delta_\varepsilon(x^2-4)$ is $1/2\varepsilon$ on the two intervals where $\lvert x^2 - 4\rvert < \varepsilon$ , that is, $\lvert x \mp 2\rvert < \varepsilon/4 + O(\varepsilon^2)$ . Each interval has length $\varepsilon/2$ to leading order, so each contributes $f(\pm 2)\cdot(\varepsilon/2)/(2\varepsilon) = f(\pm2)/4$ , confirming the rule and the factor $\lvert g'\rvert^{-1}$ .
Where it would break. For $\delta(x^2)$ the same computation gives intervals of length $2\sqrt\varepsilon$ and a contribution $f(0)\sqrt{\varepsilon}/\varepsilon\to\infty$ : the double zero makes the expression meaningless, as the remark above says.

Example(

\nabla^2(1/r) = -4\pi\delta^{(3)}

, tested on a Gaussian)Original

Take $f(\mathbf{r}) = e^{-r^2}$ . This is not of compact support, but it and all its derivatives decay faster than any power, so the proof's estimates apply with the outer sphere pushed to infinity, where the boundary terms vanish.

Its Laplacian. For a radial function $\nabla^2f = f'' + (2/r)f'$ with $f' = -2re^{-r^2}$ and $f'' = (4r^2-2)e^{-r^2}$ , so $\nabla^2 f = \bigl(4r^2 - 2 - 4\bigr)e^{-r^2} = \bigl(4r^2 - 6\bigr)e^{-r^2} .$
The integral. With $dV = 4\pi r^2\, dr$ for a radial integrand, $\int_{\mathbb{R}^3}\frac{\nabla^2 f}{r}\; dV = 4\pi\int_0^{\infty}\bigl(4r^3 - 6r\bigr)e^{-r^2}\, dr .$
Evaluate. $\int_0^\infty re^{-r^2}dr = \tfrac12$ and $\int_0^\infty r^3e^{-r^2}dr = \tfrac12$ (substitute $s = r^2$ in each), so the bracket integrates to $4\cdot\tfrac12 - 6\cdot\tfrac12 = -1$ and $\int_{\mathbb{R}^3}\frac{1}{r}\,\nabla^2 f\; dV = -4\pi .$
Comparison. The identity predicts $-4\pi f(\mathbf{0}) = -4\pi\cdot 1 = -4\pi$ . They agree.
A scaled version. For $f = e^{-\lambda r^2}$ the same computation gives $4\pi\int_0^\infty(4\lambda^2r^3 - 6\lambda r)e^{-\lambda r^2}dr = 4\pi(2 - 3) = -4\pi$ , independent of $\lambda$ — as it must be, since $f(\mathbf{0}) = 1$ for every $\lambda$ . The integral is insensitive to the width of the test function, which is the signature of a delta.

ApplicationOriginal

The delta is how a point-like source is written as a density, so that one formula covers both continuous and concentrated sources. A point charge $q$ at $\mathbf{r}_0$ has $\rho = q\,\delta^{(3)}(\mathbf{r}-\mathbf{r}_0)$ , and Poisson's equation $\nabla^2\varphi = -\rho/\varepsilon_0$ then has the solution $\varphi = q/4\pi\varepsilon_0\lvert\mathbf{r}-\mathbf{r}_0\rvert$ because of the identity proved above; superposing gives the Coulomb integral for any charge distribution, and the same algebra with $\rho$ read as mass density gives Newtonian gravity. A point mass in mechanics, a concentrated load on a beam, and an instantaneous injection of tracer into a flow are the same device.

That observation is the whole idea of a Green's function: solve the equation once with a delta on the right, and the solution for arbitrary data follows by superposition, since $f(\mathbf{r}) = \int f(\mathbf{r}')\delta^{(3)}(\mathbf{r}-\mathbf{r}')\, dV'$ expresses any source as a superposition of point sources. In signal processing the same function is the impulse response, and rule 7 — the Fourier representation — is what connects it to the transfer function (convolution). Rule 6, $\theta' = \delta$ , is how a jump in a physical quantity becomes a singular term in its derivative: the surface charge on a conductor, the shock in a compressible flow, the bending moment at a point load.

In probability the delta is the density of a degenerate distribution, so that a mixture of discrete and continuous outcomes has a single density; in quantum mechanics it is the position eigenstate normalization $\langle \mathbf{r}'|\mathbf{r}\rangle = \delta^{(3)}(\mathbf{r}-\mathbf{r}')$ , which is what makes the continuous-index formalism of Hassani's §7.3 work; and rule 4 is used constantly to impose energy or momentum conservation inside a phase-space integral, as in $\delta(E_f - E_i)$ in Fermi's golden rule.