The Euler-Lagrange equation

Functionals, their first variation and the equations extremals satisfy, with boundary conditions and constraints.

In this chapter 8

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Extremals of functionals
Boundary conditions and constraints

Extremals of functionals

Definition

Functionals and the first variation

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Defines functionals on spaces of curves, the admissible variations and the first variation, proves that the first variation vanishes at an extremum, and proves the fundamental lemma that turns this into a pointwise equation.

Prerequisites: Taylor's theorem

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 1, §§1–3, pp. 1–2, 6–8, 12–13

A functional is a rule $J$ that assigns a real number $J[y]$ to each function $y$ in some class. The class used throughout this chapter is $C^1[a,b]$ , the continuously differentiable real functions on a closed interval, with the norm

\|y\|_1 = \max_{a\le x\le b}|y(x)| + \max_{a\le x\le b}|y'(x)| .

The typical functional is an integral functional

J[y] = \int_a^b F\bigl(x, y(x), y'(x)\bigr)\,dx ,

where the integrand $F(x, y, p)$ is a given function of three real variables. We write $F_y$ and $F_{p}$ for its partial derivatives in the second and third slots; following custom, $F_p$ is also written $F_{y'}$ .

The admissible functions are the $y \in C^1[a,b]$ that satisfy the side conditions of the problem, for instance the fixed end values $y(a) = A$ , $y(b) = B$ . They do not form a vector space: the sum of two admissible functions has the wrong end values.
An admissible variation (or direction) at $y$ is a function $h$ such that $y + \varepsilon h$ is admissible for all small $|\varepsilon|$ . For fixed end values these are exactly the functions in
$C^1_0[a,b] = \{\,h \in C^1[a,b] : h(a) = h(b) = 0\,\}.$
The first variation of $J$ at $y$ in the direction $h$ is
$\delta J[y; h] = \frac{d}{d\varepsilon} J[y + \varepsilon h]\Big|_{\varepsilon = 0},$
whenever this derivative exists. The function $y$ is stationary if $\delta J[y;h] = 0$ for every admissible variation $h$ .
$J$ has a weak local minimum at an admissible $y$ if there is $r > 0$ with $J[\tilde y] \ge J[y]$ for every admissible $\tilde y$ with $\|\tilde y - y\|_1 < r$ ; a weak local maximum is defined in the same way. If the comparison is made with the smaller norm $\max|\tilde y - y|$ instead, the extremum is called strong. Every strong extremum is a weak one, so a necessary condition for weak extrema also holds for strong ones.

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 1, Thm 2, p. 13 and Lemma 1, p. 9; Evans (2010), §8.1.2, pp. 428–429

Let $F$ have continuous first partial derivatives on $[a,b]\times\mathbb{R}\times\mathbb{R}$ , and let $y, h \in C^1[a,b]$ .

First variation of an integral functional. The derivative $\delta J[y;h]$ exists and
$\delta J[y; h] = \int_a^b \Bigl( F_y\bigl(x, y, y'\bigr)\, h + F_{y'}\bigl(x, y, y'\bigr)\, h' \Bigr)\, dx .$
It is linear in $h$ .
Necessary condition. If $J$ has a weak local extremum at an admissible $y$ , then $\delta J[y; h] = 0$ for every admissible variation $h$ .
Fundamental lemma. Let $k \ge 0$ be an integer and $g$ continuous on $[a,b]$ . If $\int_a^b g\,h\,dx = 0$ for every $h \in C^k[a,b]$ with $h(a) = h(b) = 0$ , then $g(x) = 0$ for all $x \in [a,b]$ .

Idea of the proofOriginal

Restricting $J$ to the straight line $\varepsilon \mapsto y + \varepsilon h$ in function space turns it into an ordinary function of one real variable. An extremum of $J$ is then an extremum of that function at $\varepsilon = 0$ , so its derivative, the first variation, must vanish. The fundamental lemma says that a continuous function which integrates to zero against every test function must vanish identically: if it were positive somewhere, a test function concentrated there would detect it.

ProofAdaptedAdapted from Gelfand & Fomin (1963), Lemma 1, p. 9 and Thm 2, p. 13; Evans (2010), pp. 428–429

Part 1. Write $\varphi(\varepsilon) = J[y + \varepsilon h]$ and fix $\varepsilon \ne 0$ with $|\varepsilon| \le 1$ . For each $x$ , the mean value theorem (the case $n = 0$ of Taylor's theorem) applied to $t \mapsto F(x, y + t h, y' + t h')$ on the interval between $0$ and $\varepsilon$ gives a $\theta = \theta(x) \in (0,1)$ with

\frac{F(x, y + \varepsilon h, y' + \varepsilon h') - F(x, y, y')}{\varepsilon} = F_y(x, y + \theta\varepsilon h, y' + \theta\varepsilon h')\,h + F_{y'}(x, y + \theta\varepsilon h, y' + \theta\varepsilon h')\,h' .

All arguments $(x, y + s h, y' + s h')$ with $|s| \le 1$ lie in a compact box $K \subset [a,b]\times\mathbb{R}^2$ , because $y, y', h, h'$ are bounded. $F_y$ and $F_{y'}$ are continuous, hence uniformly continuous on $K$ . The arguments above differ from $(x, y, y')$ by at most $|\varepsilon|(|h| + |h'|) \le |\varepsilon|\,\|h\|_1$ , so the right-hand side converges to $F_y h + F_{y'} h'$ uniformly in $x$ as $\varepsilon \to 0$ . Uniform convergence lets us pass the limit through the integral, which gives the formula. Linearity in $h$ is visible in the formula.

Part 2. Suppose $y$ is a weak local minimum (for a maximum replace $J$ by $-J$ ) and let $h$ be an admissible variation. For small $|\varepsilon|$ the function $y + \varepsilon h$ is admissible, and $\|(y + \varepsilon h) - y\|_1 = |\varepsilon|\,\|h\|_1$ is below the radius $r$ of the definition. Hence $\varphi(\varepsilon) \ge \varphi(0)$ for all small $|\varepsilon|$ : the real function $\varphi$ has an interior local minimum at $0$ . Since $\varphi$ is differentiable at $0$ by part 1, $\varphi'(0) = 0$ , that is, $\delta J[y; h] = 0$ .

Part 3. Suppose, for a contradiction, that $g(x_0) \ne 0$ for some $x_0$ ; replacing $g$ by $-g$ we may take $g(x_0) > 0$ . By continuity there is an interval $[c, d] \subset [a,b]$ with $c < d$ on which $g > g(x_0)/2$ . Define

h(x) = \begin{cases} \bigl((x - c)(d - x)\bigr)^{k+1}, & c \le x \le d,\\ 0, & \text{otherwise.} \end{cases}

At $x = c$ the polynomial branch has a zero of order $k+1$ , so it and its first $k$ derivatives vanish there, matching the zero branch; the same holds at $d$ . So $h \in C^k[a,b]$ , $h(a) = h(b) = 0$ , and $h > 0$ on $(c, d)$ . Then

\int_a^b g\,h\,dx = \int_c^d g\,h\,dx \ge \frac{g(x_0)}{2}\int_c^d h\,dx > 0 ,

contradicting the hypothesis. Hence $g \equiv 0$ . $\square$

RemarkOriginal

Where the hypotheses are used.

Continuity of $F_y$ , $F_{y'}$ is what makes the difference quotients converge uniformly in part 1. It is also why the problem is posed in $C^1$ with the norm $\|\cdot\|_1$ : an integrand that depends on $y'$ needs $y'$ to be controlled. Arc length is an example. A zigzag curve can be uniformly within $\varepsilon$ of a straight segment yet be much longer, so arc length is not continuous for the max norm, although it is continuous for $\|\cdot\|_1$ .
Admissibility of $y + \varepsilon h$ for small $\varepsilon$ of both signs is used in part 2. If only $\varepsilon \ge 0$ were allowed (a one-sided constraint such as $y \ge 0$ that is active somewhere), we would get only $\delta J[y; h] \ge 0$ .
Continuity of $g$ is used in part 3 to make $g$ positive on a whole interval. Without it the lemma fails: the function equal to $1$ at $x = (a+b)/2$ and $0$ elsewhere integrates to zero against every $h$ but is not zero. For merely integrable $g$ the conclusion becomes " $g = 0$ except on a set that does not affect integrals".
The class of test functions can be shrunk to $C^k$ functions vanishing near $a$ and $b$ , or to infinitely differentiable ones, since the $h$ used in part 3 can be replaced by a smooth bump on $(c,d)$ . The vanishing of $\delta J$ is a statement about every admissible direction; it says nothing about second-order behaviour, which decides between minimum, maximum and saddle (see the second variation).

ExampleOriginal

Take $J[y] = \int_0^1 \bigl(y'^2 + 2xy\bigr)\,dx$ on functions with $y(0) = 0$ , $y(1) = 1$ . Is $y_0(x) = x$ a candidate for a minimum?

Here $F = p^2 + 2xy$ , so $F_y = 2x$ and $F_{y'} = 2y'$ . For the admissible variation $h(x) = x(1 - x)$ , which has $h' = 1 - 2x$ , part 1 gives

\delta J[y_0; h] = \int_0^1 \bigl(2x\cdot x(1-x) + 2\cdot 1\cdot(1 - 2x)\bigr)\,dx = 2\Bigl(\frac13 - \frac14\Bigr) + 0 = \frac16 .

As a check, expand directly: $(y_0 + \varepsilon h)' = 1 + \varepsilon(1 - 2x)$ , so

J[y_0 + \varepsilon h] = \int_0^1 \Bigl(1 + 2\varepsilon(1 - 2x) + \varepsilon^2(1 - 2x)^2 + 2x^2 + 2\varepsilon x^2(1 - x)\Bigr)dx = \frac53 + \frac{\varepsilon}{6} + \frac{\varepsilon^2}{3},

whose derivative at $\varepsilon = 0$ is $\frac16$ , as computed. Since $\delta J[y_0; h] \ne 0$ , part 2 says $y_0$ is not an extremum. Indeed $\varepsilon = -\frac14$ gives $J = \frac53 - \frac1{24} + \frac1{48} = \frac53 - \frac1{48} < J[y_0]$ .

ApplicationOriginal

When the first variation can be written as $\delta J[y;h] = \int_a^b G(x)\,h(x)\,dx$ for a continuous $G$ , physicists call $G(x)$ the functional derivative $\delta J/\delta y(x)$ . It plays the role of the gradient. In density-functional theory the ground-state density makes the energy functional stationary under a particle-number constraint. In statistical field theory the mean field solves $\delta \mathcal F/\delta\phi(x) = 0$ for the free-energy functional $\mathcal F$ . In data assimilation for weather prediction, the adjoint model computes the functional derivative of a misfit functional with respect to the initial state. In each case the fundamental lemma is what turns " $\delta J = 0$ for all $h$ " into a pointwise equation.

Theorem

The Euler-Lagrange equation

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A continuously differentiable curve with fixed endpoints that makes an integral functional stationary satisfies the Euler-Lagrange equation, and it is automatically twice differentiable near every point where F_{y'y'} != 0 along it.

Prerequisites: Functionals and the first variation, Implicit and inverse function theorems

TheoremAdaptedAdapted from Gelfand & Fomin (1963), Ch. 1, §4.1, Thm 1, p. 15 and Thm 3, p. 17

Let $F(x, y, p)$ have continuous first partial derivatives on $[a,b]\times\mathbb{R}\times\mathbb{R}$ , and let

J[y] = \int_a^b F\bigl(x, y(x), y'(x)\bigr)\,dx ,\qquad y \in C^1[a,b],\quad y(a) = A,\ y(b) = B .

Euler-Lagrange equation. If $y$ is stationary for $J$ , that is, $\delta J[y;h] = 0$ for every $h \in C^1[a,b]$ with $h(a) = h(b) = 0$ (in particular, if $y$ is a weak local extremum), then the function $x \mapsto F_{y'}\bigl(x, y(x), y'(x)\bigr)$ is continuously differentiable on $[a,b]$ and
$\frac{d}{dx}\, F_{y'}\bigl(x, y, y'\bigr) = F_y\bigl(x, y, y'\bigr), \qquad a \le x \le b .$
Regularity. Suppose in addition that $F$ has continuous second partial derivatives, and let $x_0 \in [a,b]$ be a point with $F_{y'y'}\bigl(x_0, y(x_0), y'(x_0)\bigr) \ne 0$ . Then there is a $\delta > 0$ such that $y$ is twice continuously differentiable on $[x_0 - \delta, x_0 + \delta]\cap[a,b]$ (a one-sided interval if $x_0$ is an endpoint), and on that interval the equation takes the second-order form
$F_{y'y'}\,y'' + F_{y'y}\,y' + F_{y'x} - F_y = 0 .$

Solutions of the Euler-Lagrange equation are called extremals, whether or not they give an extremum.

Idea of the proofOriginal

The first variation contains $h$ and $h'$ . The textbook move is to integrate the $h'$ term by parts, but that needs $\frac{d}{dx}F_{y'}$ , which we do not yet know exists. Instead we integrate the $h$ term by parts, so that only $h'$ remains. A continuous function that integrates to zero against every $h'$ must be constant. That constant relation shows that $F_{y'}$ along the curve is an antiderivative of $F_y$ . Differentiability of $F_{y'}$ comes out of the proof rather than going in.

ProofAdaptedAdapted from Gelfand & Fomin (1963), Lemmas 2 and 4, pp. 10–11 and Thm 3, pp. 17–18

Along the given curve put $P(x) = F_y(x, y(x), y'(x))$ and $Q(x) = F_{y'}(x, y(x), y'(x))$ . Both are continuous, because $F_y$ , $F_{y'}$ are continuous and $y$ , $y'$ are continuous.

Step 1 (first variation). By the first-variation formula,

0 = \delta J[y; h] = \int_a^b \bigl(P\,h + Q\,h'\bigr)\,dx \qquad\text{for all } h \in C^1_0[a,b] .

Step 2 (move the derivative onto $h$ ). Let $\Phi(x) = \int_a^x P(t)\,dt$ , a $C^1$ function with $\Phi' = P$ by the fundamental theorem of calculus. Integrating by parts and using $h(a) = h(b) = 0$ ,

\int_a^b P\,h\,dx = \bigl[\Phi h\bigr]_a^b - \int_a^b \Phi\,h'\,dx = -\int_a^b \Phi\,h'\,dx .

So Step 1 becomes $\int_a^b (Q - \Phi)\,h'\,dx = 0$ for all $h \in C^1_0[a,b]$ .

Step 3 (lemma of du Bois-Reymond: such a function is constant). Let $g = Q - \Phi$ , which is continuous, and let $c = \frac{1}{b-a}\int_a^b g\,dx$ be its mean. The function $h_*(x) = \int_a^x \bigl(g(t) - c\bigr)\,dt$ is $C^1$ , has $h_*(a) = 0$ , and has $h_*(b) = 0$ by the choice of $c$ ; so it is an admissible test function, with $h_*' = g - c$ . Using it,

\int_a^b (g - c)^2\,dx = \int_a^b (g - c)\,h_*'\,dx = \int_a^b g\,h_*'\,dx - c\,\bigl(h_*(b) - h_*(a)\bigr) = 0 - 0 .

A continuous non-negative function with zero integral vanishes identically, so $g \equiv c$ .

Step 4 (conclusion of part 1). Hence $Q = \Phi + c$ . The right side is $C^1$ with derivative $P$ , so $Q$ is $C^1$ and $Q' = P$ , which is the Euler-Lagrange equation.

Step 5 (regularity at an interior point). Now let $F$ have continuous second derivatives and $F_{y'y'} \ne 0$ at $(x_0, y(x_0), y'(x_0))$ , and suppose first that $x_0 \in (a,b)$ . Consider

G(x, p) = F_{y'}\bigl(x, y(x), p\bigr) - Q(x), \qquad x \in (a,b),\ p \in \mathbb{R} .

$G$ is $C^1$ in $(x,p)$ on the open set $(a,b)\times\mathbb{R}$ , because $F_{y'}$ is $C^1$ , $y$ is $C^1$ , and $Q$ is $C^1$ by Step 4. We have $G(x_0, y'(x_0)) = 0$ and $\partial G/\partial p = F_{y'y'} \ne 0$ there. By the implicit function theorem, there are open intervals $U \ni x_0$ , $U \subset (a,b)$ , and $V \ni y'(x_0)$ and a $C^1$ function $\psi$ on $U$ such that, for $x \in U$ , the only $p \in V$ with $G(x, p) = 0$ is $p = \psi(x)$ . Since $y'$ is continuous, $y'(x) \in V$ for $x$ near $x_0$ , and $G(x, y'(x)) = 0$ by definition of $Q$ . Hence $y' = \psi$ near $x_0$ , so $y'$ is $C^1$ there. Now that $y''$ exists, the chain rule expands $Q' = P$ into the displayed second-order form, which we may solve for $y''$ :

y'' = \frac{F_y - F_{y'x} - F_{y'y}\,y'}{F_{y'y'}} \tag{$\ast$}

at every interior point where $F_{y'y'} \ne 0$ along the curve.

Step 6 (regularity up to an endpoint). Let $x_0 = a$ ; the case $x_0 = b$ is symmetric. The function $x \mapsto F_{y'y'}(x, y(x), y'(x))$ is continuous and nonzero at $a$ , so it is nonzero on $[a, a+\delta]$ for some $\delta > 0$ with $a + \delta < b$ , and the right-hand side $R(x)$ of $(\ast)$ , built from the continuous functions $F_y$ , $F_{y'x}$ , $F_{y'y}$ , $F_{y'y'}$ along the curve and $y'$ , is continuous there. By Step 5, $y'$ is differentiable on $(a, a+\delta)$ with $y'' = R$ . Since $y'$ is continuous on $[a, a+\delta]$ , the mean value theorem gives, for each $x \in (a, a+\delta)$ , a point $\xi_x \in (a,x)$ with $(y'(x) - y'(a))/(x - a) = y''(\xi_x) = R(\xi_x)$ ; letting $x \to a^+$ and using the continuity of $R$ at $a$ , the right derivative of $y'$ at $a$ exists and equals $R(a)$ . So $y'' = R$ on all of $[a, a+\delta]$ and is continuous there. $\square$

RemarkOriginal

Where the hypotheses are used.

Only $y \in C^1$ is assumed. The usual derivation integrates $Q\,h'$ by parts, which presupposes that $Q$ is differentiable, and in practice that $y \in C^2$ . Steps 2 to 4 avoid this. The extra smoothness is then proved in Steps 5 and 6, but only where $F_{y'y'} \ne 0$ .
$F_{y'y'} \ne 0$ cannot be dropped in Step 5. Take $F = (p^2 - 4x^2)^2$ on $[-1, 1]$ with $y(-1) = -1$ , $y(1) = 1$ . Then $J \ge 0$ , and $y(x) = x\,|x|$ satisfies $y' = 2|x|$ , so $F = 0$ along it and it is a global minimum. Along this curve $F_{y'} = 4y'(y'^2 - 4x^2) = 0$ and $F_y = 0$ , so the Euler-Lagrange equation holds, yet $y''$ jumps from $-2$ to $2$ at $x = 0$ . Consistently, $F_{y'y'} = 4(3y'^2 - 4x^2) = 32x^2$ vanishes exactly there.
Strict convexity in the slope is not enough. It does not imply $F_{y'y'} \ne 0$ . Take $F = \frac14 p^4 - xp$ on $[-1,1]$ with $y(\pm 1) = \frac34$ : $p \mapsto F$ is strictly convex, yet $F_{y'y'} = 3p^2$ vanishes at $p = 0$ . Here $y = \frac34|x|^{4/3}$ has $y' = x^{1/3}$ , so $F_{y'} = y'^3 - x = 0$ and $F_y = 0$ : it is a $C^1$ extremal, and it minimises the integrand pointwise (for fixed $x$ the minimum of $\frac14p^4 - xp$ is at $p = x^{1/3}$ ), hence is the global minimum. But $y' = x^{1/3}$ is not differentiable at $0$ .
Fixed endpoints are used in Step 2 to drop the boundary term $[\Phi h]_a^b$ and in Step 3 to make $h_*$ admissible. If an endpoint is free, an extra boundary condition appears (see natural boundary conditions).
The equation is necessary, not sufficient. For $J[y] = \int_0^{2\pi}(y'^2 - y^2)\,dx$ with $y(0) = y(2\pi) = 0$ , the function $y = 0$ is an extremal. But $J[\sin(x/2)] = -\frac{3\pi}{4} < 0$ and $J[\sin 2x] = 3\pi > 0$ , so it is neither a minimum nor a maximum. Deciding between them is the job of the second variation.

Example(Geodesics in the plane)AdaptedAdapted from Arfken, Weber & Harris (2013), Example 22.1.1, p. 1086

The length of the graph of $y \in C^1[a,b]$ is $J[y] = \int_a^b \sqrt{1 + y'^2}\,dx$ , so $F = \sqrt{1 + p^2}$ , with $F_y = 0$ and $F_{y'} = p/\sqrt{1 + p^2}$ .

Euler-Lagrange equation. It reads $\frac{d}{dx}\bigl(y'/\sqrt{1 + y'^2}\bigr) = 0$ , so $y'/\sqrt{1 + y'^2} = k$ is constant, with $|k| < 1$ .
Solve. The map $p \mapsto p/\sqrt{1+p^2}$ is strictly increasing, so $y' = k/\sqrt{1 - k^2} = m$ is constant and $y = mx + n$ : the extremals are straight lines. With $y(a) = A$ , $y(b) = B$ there is exactly one, with $m = (B - A)/(b - a)$ .
It is a minimum. The function $p \mapsto \sqrt{1 + p^2}$ is convex, so it lies above its tangent at $m$ : $\sqrt{1 + p^2} \ge \sqrt{1 + m^2} + \frac{m}{\sqrt{1 + m^2}}(p - m)$ . Put $p = y'(x)$ for any admissible $y$ and integrate. Since $\int_a^b (y' - m)\,dx = (B - A) - m(b - a) = 0$ , this gives $J[y] \ge (b - a)\sqrt{1 + m^2}$ , which is the length of the segment.
Numbers. From $(0, 1)$ to $(3, 5)$ : $m = \frac43$ and the minimal length is $3\sqrt{1 + 16/9} = 5$ . The admissible competitor $y = 1 + \frac43 x + \frac12 x(3 - x)$ has length $5.3062$ (numerical quadrature), larger as it must be.

ExampleOriginal

Minimise $J[y] = \int_0^1 \bigl(y'^2 + y^2\bigr)\,dx$ with $y(0) = 0$ , $y(1) = 1$ .

$F = p^2 + y^2$ , so $F_{y'} = 2y'$ , $F_y = 2y$ , and $F_{y'y'} = 2 \ne 0$ ; by part 2 every extremal is $C^2$ , and the equation is $y'' = y$ .
The general solution is $y = \alpha\cosh x + \beta\sinh x$ . The conditions give $\alpha = 0$ and $\beta = 1/\sinh 1$ , so $y = \sinh x/\sinh 1$ .
Value. Integrate $y'^2$ by parts and use $y'' = y$ : $\int_0^1 (y'^2 + y^2)\,dx = [y\,y']_0^1 - \int_0^1 y\,(y'' - y)\,dx = y(1)\,y'(1) = \coth 1 \approx 1.3130$ .
Comparison. The straight line $y = x$ gives $J = 1 + \frac13 \approx 1.3333$ , slightly more. The extremal is in fact the global minimum: for admissible $y + h$ , $J[y + h] - J[y] = \int_0^1 (h'^2 + h^2)\,dx \ge 0$ , because the cross term $2\int_0^1 (y'h' + yh)\,dx$ is the first variation, which vanishes.

ApplicationOriginal

A string under tension $T$ , with weight $w$ per unit length and small sag $y(x)$ between supports at equal height, has potential energy $\int_0^L \bigl(\tfrac{T}{2}y'^2 + w\,y\bigr)\,dx$ to leading order. The Euler-Lagrange equation $T y'' = w$ shows that the sag profile is a parabola, the familiar shallow-cable approximation used for power lines and suspension-bridge decks. The same equation, read as a statement about the action, gives Newton's law for a particle (see Hamilton's principle). Fermat's principle gives the equation of light rays in a stratified atmosphere, the basis of refraction corrections in astronomy and radio propagation.

Method

First integrals and the Beltrami identity

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When the integrand does not involve the unknown function or the independent variable, the second-order Euler-Lagrange equation reduces to a first-order equation, which solves the brachistochrone and the minimal surface of revolution in closed form.

Prerequisites: The Euler-Lagrange equation

MethodAdaptedAdapted from Gelfand & Fomin (1963), Ch. 1, §4.2, pp. 18–19; Riley, Hobson & Bence (2006), §22.2, pp. 777–779

Let $F(x, y, p)$ have continuous first partial derivatives, and let $y$ be an extremal of $J[y] = \int_a^b F(x, y, y')\,dx$ , that is, $\frac{d}{dx}F_{y'} = F_y$ along $y$ .

No $y$ in the integrand. If $F = F(x, p)$ , then $F_{y'}\bigl(x, y'(x)\bigr) = C$ for a constant $C$ .
No $x$ in the integrand (Beltrami identity). If the extremal $y$ is $C^2$ , then $F - y'F_{y'}$ is differentiable along $y$ and
$\frac{d}{dx}\Bigl(F - y'F_{y'}\Bigr) = F_x - y'\Bigl(\frac{d}{dx}F_{y'} - F_y\Bigr) = F_x ,$
the last step because $y$ is an extremal. Hence if $F = F(y, p)$ carries no explicit $x$ , then $F(y, y') - y'\,F_{y'}(y, y') = C$ for a constant $C$ . Both hypotheses are needed: the identity differentiates $y'$ and $F_{y'}$ along the curve, and for a $C^2$ curve that is not an extremal the derivative $\frac{d}{dx}F_{y'}$ need not exist when $F$ is merely $C^1$ . For instance $F = \tfrac23|p|^{3/2}$ has continuous first partial derivatives, but along $y = x^2/2$ it gives $F_{y'} = \mathrm{sgn}(x)\sqrt{|x|}$ , which is not differentiable at $0$ .
No $y'$ in the integrand. If $F = F(x, y)$ , the Euler-Lagrange equation is the algebraic relation $F_y(x, y) = 0$ , which has no free constants and in general cannot meet two prescribed end values.

The constant $C$ in cases 1 and 2 is called a first integral of the Euler-Lagrange equation.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §22.2.2, p. 779

Case 1: $F_y = 0$ , so the Euler-Lagrange equation says that $F_{y'}$ has zero derivative on $[a,b]$ ; it is therefore constant. Case 3 is the Euler-Lagrange equation with $F_{y'} = 0$ .

Case 2: since $y$ is $C^2$ , the chain rule gives $\frac{d}{dx}F(x, y, y') = F_x + F_y\,y' + F_{y'}\,y''$ . By the product rule, $\frac{d}{dx}\bigl(y'F_{y'}\bigr) = y''F_{y'} + y'\frac{d}{dx}F_{y'}$ ; the last derivative exists because $y$ is an extremal, which by the Euler-Lagrange theorem makes $F_{y'}$ along the curve a $C^1$ function of $x$ with derivative $F_y$ . Subtracting, the $y''F_{y'}$ terms cancel and the identity follows. If $F_x = 0$ and $y$ is an extremal, the right-hand side is zero, so $F - y'F_{y'}$ is constant. $\square$

MethodOriginal

To solve $J[y] = \int_a^b F\,dx \to$ stationary, $y(a) = A$ , $y(b) = B$ :

Check which variable is missing from $F$ . If $y$ is missing, use $F_{y'} = C$ . If $x$ is missing, use $F - y'F_{y'} = C$ . To use the Beltrami identity you need the extremal to be $C^2$ : if $F$ has continuous second partial derivatives and $F_{y'y'} \ne 0$ along the expected solution, the regularity part of the Euler-Lagrange theorem supplies that smoothness.
Solve the first-order equation for $y'$ and separate variables. A substitution adapted to the square roots that appear (a trigonometric or hyperbolic one) usually makes the integral elementary. Parametric solutions $x(\theta)$ , $y(\theta)$ are acceptable.
Fix the remaining constant and $C$ from the two end conditions. Expect zero, one or several solutions; the end conditions are not an initial value problem.
Discard spurious solutions. The Beltrami identity is implied by the Euler-Lagrange equation but not conversely: by the identity, a $C^2$ solution of $F - y'F_{y'} = C$ satisfies $y'\,(F_y - \frac{d}{dx}F_{y'}) = 0$ , so the Euler-Lagrange equation follows at every point where $y' \ne 0$ , while at a point with $y'= 0$ the identity says nothing about it. So a solution of the first-order equation with $y' \ne 0$ throughout is an extremal; curves with $y' = 0$ somewhere — constant solutions in particular — must be checked in the full equation.
Compare the values of $J$ among the surviving extremals and with any limiting configurations (curves that leave the admissible class), since the equation is only a necessary condition.

Example(The brachistochrone)AdaptedAdapted from Riley, Hobson & Bence (2006), pp. 784–785; Gelfand & Fomin (1963), p. 26

A bead slides without friction from rest at the origin to the point $B$ that lies $2$ m to the side and $1$ m below, along a wire $y = y(x)$ . Here $y$ is measured downward and $g = 9.81\ \mathrm{m\,s^{-2}}$ . Which wire is fastest?

Functional. Energy conservation gives speed $v = \sqrt{2gy}$ , and $dt = ds/v$ , so
$T[y] = \int_0^2 \sqrt{\frac{1 + y'^2}{2g\,y}}\;dx ,\qquad y(0) = 0,\ y(2) = 1 .$
The integrand is singular at $y = 0$ . On every subinterval $[\delta, 2]$ where $y > 0$ it is smooth, and a fastest curve is also fastest between any two of its own points, so it is an extremal there.
First integral. $x$ is missing. With $F = \sqrt{(1 + p^2)/(2gy)}$ we have $F_{pp} = (2gy)^{-1/2}(1 + p^2)^{-3/2} > 0$ wherever $y > 0$ , so by step 1 of the method the extremal is $C^2$ there and the Beltrami identity applies. Since $F - pF_p = 1/\sqrt{2gy(1 + p^2)}$ , it gives $y\,(1 + y'^2) = 2R$ for a constant $R > 0$ .
Solve. Substitute $y = R(1 - \cos\theta)$ . Then $1 + y'^2 = 2/(1 - \cos\theta)$ , so $y'^2 = \cot^2(\theta/2)$ . Take $dy/dx = \cot(\theta/2)$ , positive while the bead descends. Then $dx = dy\,\tan(\theta/2) = 2R\sin^2(\theta/2)\,d\theta$ , and with $x = 0$ at $\theta = 0$ ,
$x = R(\theta - \sin\theta), \qquad y = R(1 - \cos\theta):$
an inverted cycloid, generated by a circle of radius $R$ rolling along the horizontal through the start.
End condition. At $B$ the ratio $x/y = (\theta - \sin\theta)/(1 - \cos\theta)$ must equal $2$ . Solving numerically gives $\theta_1 = 3.5084$ , and then $R = 1/(1 - \cos\theta_1) = 0.51720$ m. Since $\theta_1 > \pi$ , the bead passes the cycloid's lowest point, $2R = 1.0344$ m deep at $x = \pi R = 1.6248$ m, and climbs the last $3.4$ cm to $B$ .
Travel time. Along the cycloid $ds = 2R\sin(\theta/2)\,d\theta$ and $v = 2\sqrt{gR}\,\sin(\theta/2)$ , so $dt = \sqrt{R/g}\,d\theta$ and
$T = \theta_1\sqrt{R/g} = 3.5084\sqrt{0.51720/9.81} = 0.8056\ \mathrm{s}.$
The straight chute has length $\sqrt5$ m and acceleration $g/\sqrt5$ , so it takes $\sqrt{2\cdot 5/g} = 1.0096$ s, about $25\%$ longer.

Example(Minimal surface of revolution)AdaptedAdapted from Gelfand & Fomin (1963), Example 2, pp. 20–21; Arfken, Weber & Harris (2013), Example 22.1.3, pp. 1088–1092

A soap film spans two coaxial rings of radius $1$ placed at $x = \pm 0.4$ . Rotating $y(x) > 0$ about the $x$ -axis gives the area

A[y] = 2\pi\int_{-0.4}^{0.4} y\sqrt{1 + y'^2}\,dx ,\qquad y(\pm 0.4) = 1 .

First integral. $F = y\sqrt{1 + p^2}$ has no $x$ , and $F - pF_p = y/\sqrt{1 + p^2}$ . So $y/\sqrt{1 + y'^2} = c$ with $0 < c \le y$ . Also $F_{pp} = y(1 + p^2)^{-3/2} > 0$ , so extremals are $C^2$ .
Solve. $y' = \pm\sqrt{y^2/c^2 - 1}$ . Separating variables with $y = c\cosh u$ gives $y = c\cosh\bigl((x - x_0)/c\bigr)$ , a catenary.
Spurious solution. The constant $y = c$ also satisfies step 1 (the cylinder), but the full Euler-Lagrange equation $\frac{d}{dx}\bigl(yy'/\sqrt{1 + y'^2}\bigr) = \sqrt{1 + y'^2}$ fails there: the left side is $0$ and the right side is $1$ . So the cylinder is not stationary, as step 4 of the method warns.
End conditions. Equal end values force $x_0 = 0$ , and then $c\cosh(0.4/c) = 1$ . This equation has two roots, $c = 0.91074$ and $c = 0.15796$ .
Areas. Since $\sqrt{1 + y'^2} = \cosh(x/c)$ , we have $A = 2\pi c\int_{-0.4}^{0.4}\cosh^2(x/c)\,dx = \pi c\bigl(0.8 + c\sinh(0.8/c)\bigr)$ . The shallow catenary gives $A = 4.8838$ ; the deep one gives $A = 6.6013$ .
Comparison. If the film breaks into two flat discs, its area is $2\pi = 6.2832$ . That configuration is not the graph of a $C^1$ function, but it is a limit of admissible ones. The shallow catenoid beats it; the deep one does not, and it is not even a local minimum. Numerically, $c\cosh(d/c) = 1$ has a solution only if the half-separation is $d \le 0.6627$ . Beyond that there is no catenoid, and the film snaps to the two discs.

RemarkOriginal

Where the hypotheses are used. The Beltrami identity differentiates $F(y, y')$ and $y'F_{y'}$ along the curve, so it needs two things: $y \in C^2$ , and the existence of $\frac{d}{dx}F_{y'}$ along the curve. The second is automatic along an extremal but can fail for a general $C^2$ curve when $F$ has only continuous first partials, as the example in case 2 shows; assuming $F \in C^2$ also removes the difficulty. The first-integral relation itself makes sense for $C^1$ curves, but deriving it this way does not. Case 1 needs only the Euler-Lagrange equation. Each first integral reflects a symmetry: no $y$ means invariance under $y \mapsto y + \text{const}$ , and no $x$ means invariance under $x \mapsto x + \text{const}$ . This is the one-dimensional form of Noether's theorem.

ApplicationOriginal

In mechanics, with time as the independent variable, a Lagrangian with no explicit time dependence has the Beltrami first integral $\dot q\,L_{\dot q} - L$ , the energy. A Lagrangian with no explicit $q$ conserves the momentum $L_{\dot q}$ . In optics a medium stratified in the $x$ -direction, with refractive index $n(x)$ constant on the planes $x = \text{const}$ , gives travel time proportional to $\int n(x)\sqrt{1 + y'^2}\,dx$ , with no $y$ . Case 1 then gives $n(x)\,y'/\sqrt{1 + y'^2} = C$ , that is, $n(x)\sin\phi = C$ , where $\phi$ is the angle between the ray and the normal to the layers. This is Snell's law, used in seismic ray tracing and in atmospheric refraction models.

Theorem

Several functions and higher derivatives

Adapted

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A functional of several unknown functions gives one Euler-Lagrange equation per function, and an integrand containing derivatives up to order m gives the Euler-Poisson equation of order 2m.

Prerequisites: Functionals and the first variation, The Euler-Lagrange equation

TheoremAdaptedAdapted from Gelfand & Fomin (1963), §9 Theorem, p. 35 and §11, Eq. (22), p. 42

(a) Several unknown functions. Let $F(x, y_1, \dots, y_n, p_1, \dots, p_n)$ have continuous first partial derivatives, and let $y = (y_1, \dots, y_n)$ , with each $y_i \in C^1[a,b]$ and prescribed values $y_i(a)$ , $y_i(b)$ , be stationary for

J[y] = \int_a^b F\bigl(x, y_1, \dots, y_n, y_1', \dots, y_n'\bigr)\,dx ,

that is, $\delta J[y; h] = 0$ for all $h = (h_1, \dots, h_n)$ with $h_i \in C^1[a,b]$ and $h_i(a) = h_i(b) = 0$ . Then each $F_{y_i'}$ is $C^1$ along $y$ and

\frac{d}{dx}F_{y_i'} = F_{y_i}, \qquad i = 1, \dots, n .

(b) Higher derivatives (Euler-Poisson equation). Let $m \ge 1$ , let $F(x, y, p_1, \dots, p_m)$ have continuous partial derivatives of order up to $m + 1$ , and write $F_{y^{(k)}}$ for $\partial F/\partial p_k$ (with $F_{y^{(0)}} = F_y$ ). Let $y \in C^{2m}[a,b]$ , with $y, y', \dots, y^{(m-1)}$ prescribed at both ends, be stationary for

J[y] = \int_a^b F\bigl(x, y, y', \dots, y^{(m)}\bigr)\,dx

under all variations $h \in C^m[a,b]$ with $h^{(j)}(a) = h^{(j)}(b) = 0$ for $0 \le j \le m - 1$ , that is, $\delta J[y; h] = 0$ for all such $h$ . Note the two classes: the curve $y$ is assumed to lie in the smaller class $C^{2m}[a,b]$ , while it is tested against the wider class of competitors $y + \varepsilon h$ with $h \in C^m[a,b]$ , which are only $C^m$ . Then

\sum_{k=0}^{m} (-1)^k \frac{d^k}{dx^k} F_{y^{(k)}}\bigl(x, y, \dots, y^{(m)}\bigr) = 0 , \qquad a \le x \le b .

Idea of the proofOriginal

For several functions, vary one component at a time: with all other components frozen, the problem is a one-function problem, and the single Euler-Lagrange equation applies to it. For higher derivatives, the first variation contains $h, h', \dots, h^{(m)}$ . Integrating the term with $h^{(k)}$ by parts $k$ times moves every derivative onto the coefficient. The boundary terms vanish because the variation and its first $m - 1$ derivatives vanish at the ends, and the fundamental lemma finishes the argument.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), p. 1097; Gelfand & Fomin (1963), pp. 35, 41–42

(a) Fix $i$ and take $h$ with $h_j = 0$ for $j \ne i$ . Then $\varepsilon \mapsto J[y + \varepsilon h]$ involves only the $i$ -th component, and the computation of the first variation (whose proof works verbatim with more arguments) gives

0 = \delta J[y; h] = \int_a^b \bigl(F_{y_i}\,h_i + F_{y_i'}\,h_i'\bigr)\,dx \qquad\text{for all } h_i \in C^1_0[a,b] ,

where $F_{y_i}$ , $F_{y_i'}$ are evaluated along $y$ and are continuous functions of $x$ . This is exactly the relation from which Steps 2 to 4 of the proof of the Euler-Lagrange theorem deduce that $F_{y_i'}$ is $C^1$ with derivative $F_{y_i}$ . That argument used nothing about where the two continuous coefficients came from.

(b) Step 1 (first variation). As in the one-derivative case, the mean value theorem and uniform continuity on a compact set give

\delta J[y; h] = \int_a^b \sum_{k=0}^{m} F_{y^{(k)}}\,h^{(k)}\,dx = 0 .

Step 2 (regularity of the coefficients). Put $c_k(x) = F_{y^{(k)}}(x, y(x), \dots, y^{(m)}(x))$ . Because $F$ has continuous derivatives of order $m + 1$ and $y \in C^{2m}$ , the chain rule shows that $c_k$ can be differentiated $k$ times with continuous result: each differentiation raises the order of the derivatives of $y$ that appear by one (at most to $m + k \le 2m$ ) and the order of the derivatives of $F$ by one (at most to $k + 1 \le m + 1$ ).

Step 3 (integrate by parts $k$ times). For $1 \le k \le m$ ,

\int_a^b c_k\,h^{(k)}\,dx = \Bigl[\sum_{j=0}^{k-1} (-1)^j c_k^{(j)}\,h^{(k-1-j)}\Bigr]_a^b + (-1)^k\int_a^b c_k^{(k)}\,h\,dx .

Every boundary term contains a derivative $h^{(k-1-j)}$ of order at most $k - 1 \le m - 1$ , which vanishes at $a$ and $b$ . Hence

\int_a^b \Bigl(\sum_{k=0}^m (-1)^k c_k^{(k)}\Bigr)\,h\,dx = 0 .

Step 4 (fundamental lemma). The bracket is continuous. The test functions $h \in C^m$ with $h^{(j)}(a) = h^{(j)}(b) = 0$ for $j < m$ include the bumps $\bigl((x - c)(d - x)\bigr)^{m+1}$ used in the fundamental lemma, since these vanish to order $m + 1$ at $c$ and $d$ . That proof therefore applies and the bracket vanishes identically. $\square$

RemarkAdaptedAdapted from Gelfand & Fomin (1963), §9 Remarks 1–2, p. 36; Gelfand & Fomin (1963), §11 Remark, p. 42

Where the hypotheses are used.

In (a), the components $h_i$ can be varied independently. This fails when the $y_i$ are tied together by a constraint, and then multipliers are needed (see pointwise constraints).
In (b), $y \in C^{2m}$ and $F \in C^{m+1}$ are used only in Step 2, to make the repeated integration by parts legitimate. As in the one-derivative case, a generalised du Bois-Reymond lemma shows that a stationary $y \in C^m$ automatically has the needed smoothness of the combinations $c_k$ , but we do not prove that here.
The $2m$ end conditions on $y, \dots, y^{(m-1)}$ are used in Step 3. If some of them are dropped, the corresponding boundary terms survive and become natural boundary conditions (see free endpoints).
Different integrands, same equations. If $\Phi(x, y_1, \dots, y_n)$ is $C^2$ , then adding $\frac{d}{dx}\Phi = \Phi_x + \sum_i \Phi_{y_i}y_i'$ to $F$ changes $J$ only by $\Phi(b, y(b)) - \Phi(a, y(a))$ , which is the same for all admissible curves. So the Euler-Lagrange equations do not change. Such total derivatives are called null Lagrangians. The same holds when $F$ is multiplied by a nonzero constant.

Example(A coupled system)Original

Find the extremal of $J[y, z] = \int_0^1 \bigl(y'^2 + z'^2 + 2yz\bigr)\,dx$ with $y(0) = 0$ , $y(1) = 1$ , $z(0) = z(1) = 0$ .

Equations. $F_{y'} = 2y'$ , $F_y = 2z$ , $F_{z'} = 2z'$ , $F_z = 2y$ . Part (a) gives $y'' = z$ and $z'' = y$ .
Decouple. With $u = y + z$ and $v = y - z$ we get $u'' = u$ , $v'' = -v$ , with $u(0) = v(0) = 0$ and $u(1) = v(1) = 1$ . So $u = \sinh x/\sinh 1$ , $v = \sin x/\sin 1$ , and
$y = \frac12\Bigl(\frac{\sinh x}{\sinh 1} + \frac{\sin x}{\sin 1}\Bigr), \qquad z = \frac12\Bigl(\frac{\sinh x}{\sinh 1} - \frac{\sin x}{\sin 1}\Bigr).$
Values. $y(\tfrac12) = 0.50658$ and $z(\tfrac12) = -0.06317$ .
Value of $J$ . Integrating $y'^2 + z'^2$ by parts and using the equations, the integral reduces to the boundary term $[\,yy' + zz'\,]_0^1 = y'(1) = \frac12(\coth 1 + \cot 1) = 0.97756$ . The competitor $y = x$ , $z = 0$ gives $J = 1$ .

Example(Clamped beam)Original

A beam of length $L = 2$ m and bending stiffness $EI = 1000\ \mathrm{N\,m^2}$ , clamped horizontally at both ends, carries a uniform load $q = 600\ \mathrm{N/m}$ . Its deflection $w(x)$ (positive downward) minimises the potential energy

J[w] = \int_0^L \Bigl(\frac{EI}{2}\,w''^2 - q\,w\Bigr)\,dx , \qquad w(0) = w'(0) = w(L) = w'(L) = 0 .

Euler-Poisson equation ( $m = 2$ ): $F_w = -q$ , $F_{w'} = 0$ , $F_{w''} = EI\,w''$ , so $-q + \frac{d^2}{dx^2}\bigl(EI\,w''\bigr) = 0$ , that is, $EI\,w'''' = q$ .
Solve. Integrating four times and imposing the four clamping conditions gives $w = \dfrac{q}{24\,EI}\,x^2(L - x)^2$ . Check: $x^2(L - x)^2 = x^4 - 2Lx^3 + L^2x^2$ has fourth derivative $24$ , and $w$ and $w'$ vanish at both ends because of the double roots.
Number. The largest deflection is at mid-span: $w(1) = \dfrac{600}{24\cdot 1000}\cdot 1 \cdot 1 = 0.025$ m, which agrees with the textbook formula $qL^4/(384\,EI)$ .

ApplicationOriginal

Part (a) is the mathematical core of Lagrangian mechanics with several degrees of freedom (see Hamilton's principle). It also gives ray equations in three-dimensional inhomogeneous media, where $y(x)$ and $z(x)$ describe the ray. Part (b) governs every energy that penalises curvature: Euler-Bernoulli beams and plates in structural engineering, smoothing splines in data analysis (minimising $\int f''^2$ plus a misfit gives piecewise cubics), and the bending energy of lipid membranes and DNA in biophysics.

Theorem

Several independent variables

Adapted

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A function of several variables that makes a multiple-integral functional stationary under fixed boundary values satisfies a second-order partial differential equation in divergence form, which gives the minimal-surface equation and the field equations of a Lagrangian density.

Prerequisites: The Euler-Lagrange equation, The divergence theorem

TheoremAdaptedAdapted from Gelfand & Fomin (1963), §5, pp. 22–24; Evans (2010), §8.1.2, Eq. (9), p. 429

Let $\Omega \subset \mathbb{R}^n$ be a bounded open set whose boundary is regular enough for the divergence theorem (for example piecewise $C^1$ ). Let $F(x, u, p)$ , with $x \in \overline\Omega$ , $u \in \mathbb{R}$ , $p = (p_1, \dots, p_n) \in \mathbb{R}^n$ , have continuous second partial derivatives, and consider

J[u] = \int_\Omega F\bigl(x, u(x), \nabla u(x)\bigr)\,dx , \qquad u = g \ \text{on } \partial\Omega .

If $u \in C^2(\overline\Omega)$ is stationary, that is, $\frac{d}{d\varepsilon}J[u + \varepsilon h]\big|_{\varepsilon=0} = 0$ for every $h \in C^1(\overline\Omega)$ with $h = 0$ on $\partial\Omega$ , then

F_u\bigl(x, u, \nabla u\bigr) - \sum_{i=1}^n \frac{\partial}{\partial x_i}\Bigl[F_{p_i}\bigl(x, u, \nabla u\bigr)\Bigr] = 0 \qquad\text{in } \Omega .

The derivative $\partial/\partial x_i$ is the total one: it acts on $x_i$ both directly and through $u$ and $\nabla u$ . For several unknown fields $u_1, \dots, u_N$ and $F(x, u_1, \dots, u_N, \nabla u_1, \dots, \nabla u_N)$ , the same holds for each field:

\frac{\partial F}{\partial u_\alpha} - \sum_{i=1}^n \frac{\partial}{\partial x_i}\frac{\partial F}{\partial(\partial_i u_\alpha)} = 0, \qquad \alpha = 1, \dots, N .

Idea of the proofOriginal

The argument is the one-variable argument with the divergence theorem in place of integration by parts. The first variation is $\int (F_u h + F_p\cdot\nabla h)$ . The second term is the integral of $\operatorname{div}(h F_p)$ minus $h\operatorname{div}F_p$ , and the divergence part becomes a boundary integral that vanishes because $h = 0$ there. What remains is $\int(F_u - \operatorname{div}F_p)\,h$ , and a ball-shaped bump function shows that the bracket must be zero.

ProofAdaptedAdapted from Evans (2010), pp. 428–429; Gelfand & Fomin (1963), pp. 22–24

Step 1 (first variation). Fix an admissible $h$ . On the compact set $\overline\Omega \times [-1, 1]$ the function $\varepsilon \mapsto F(x, u + \varepsilon h, \nabla u + \varepsilon\nabla h)$ has a derivative in $\varepsilon$ that is continuous in $(x, \varepsilon)$ . The argument of the first variation (mean value theorem plus uniform continuity) therefore gives

0 = \delta J[u; h] = \int_\Omega \Bigl(F_u\,h + \sum_{i=1}^n F_{p_i}\,\frac{\partial h}{\partial x_i}\Bigr)\,dx ,

with $F_u$ , $F_{p_i}$ evaluated at $(x, u(x), \nabla u(x))$ .

Step 2 (divergence theorem). Let $V(x)$ be the vector field with components $V_i = F_{p_i}(x, u, \nabla u)$ . Because $F$ is $C^2$ and $u$ is $C^2$ , $V$ is $C^1$ on $\overline\Omega$ . The product rule gives $\operatorname{div}(hV) = \nabla h\cdot V + h\operatorname{div}V$ , and the divergence theorem gives $\int_\Omega \operatorname{div}(hV)\,dx = \int_{\partial\Omega} h\,V\cdot\nu\,dS = 0$ , since $h = 0$ on $\partial\Omega$ . Hence

\int_\Omega \nabla h\cdot V\,dx = -\int_\Omega h\operatorname{div}V\,dx, \qquad\text{so}\qquad \int_\Omega \bigl(F_u - \operatorname{div}V\bigr)\,h\,dx = 0 .

Step 3 (fundamental lemma in $\Omega$ ). The function $G = F_u - \operatorname{div}V$ is continuous. Suppose $G(x_0) > 0$ at some $x_0 \in \Omega$ (otherwise replace $G$ by $-G$ ). Then $G > 0$ on a closed ball $\overline{B} = \{|x - x_0| \le r\} \subset \Omega$ . Let $h(x) = (r^2 - |x - x_0|^2)^2$ on $\overline B$ and $h = 0$ elsewhere. This $h$ is $C^1$ , because both $h$ and its gradient $-4(r^2 - |x - x_0|^2)(x - x_0)$ vanish on the sphere $|x - x_0| = r$ . It also vanishes on $\partial\Omega$ and is positive inside $B$ . Then $\int_\Omega G h\,dx > 0$ , a contradiction. So $G \equiv 0$ in $\Omega$ , which is the stated equation.

Several fields. Vary one field $u_\alpha$ at a time, keeping the others fixed, and repeat Steps 1 to 3. $\square$

RemarkOriginal

Where the hypotheses are used.

$u \in C^2$ and $F \in C^2$ are used in Step 2, to make $V = F_p(x, u, \nabla u)$ continuously differentiable so that the divergence theorem applies. In several variables there is no du Bois-Reymond shortcut. Proving that a minimiser in a larger class (a Sobolev space) is smooth is a deep matter. For uniformly convex $F$ it is the De Giorgi-Nash-Moser theory, and it can fail for systems.
The boundary condition $h = 0$ on $\partial\Omega$ kills the boundary integral in Step 2. If $u$ is not prescribed on part of the boundary, stationarity also forces $V\cdot\nu = F_p\cdot\nu = 0$ there, the multi-dimensional natural boundary condition. For the Dirichlet integral below it is the Neumann condition $\partial u/\partial\nu = 0$ .
Stationary, not minimal. The theorem uses only $\delta J = 0$ , so it applies equally to actions such as $\iint(\rho u_t^2 - T u_x^2)\,dx\,dt$ , whose solutions are saddle points. The Euler-Lagrange equation of that action is the wave equation $\rho u_{tt} = T u_{xx}$ .

Example(Minimal surfaces)AdaptedAdapted from Gelfand & Fomin (1963), Example, p. 24; Evans (2010), Example 4, pp. 430–431

A soap film spanning a wire whose projection bounds $\Omega \subset \mathbb{R}^2$ , written as a height function $u(x,y)$ , has area $\int_\Omega \sqrt{1 + u_x^2 + u_y^2}\,dx\,dy$ . With $F = \sqrt{1 + |p|^2}$ we have $F_u = 0$ and $F_{p} = p/\sqrt{1 + |p|^2}$ . The stationarity condition, called the minimal surface equation, therefore reads

\operatorname{div}\frac{\nabla u}{\sqrt{1 + |\nabla u|^2}} = 0 \quad\Longleftrightarrow\quad (1 + u_y^2)\,u_{xx} - 2u_xu_y\,u_{xy} + (1 + u_x^2)\,u_{yy} = 0 .

The second form follows from the first by carrying out the divergence and clearing the factor $(1 + |\nabla u|^2)^{-3/2}$ . Geometrically, its left side equals $2(1 + |\nabla u|^2)^{3/2}H$ , where $H$ is the mean curvature; a stationary film is therefore a surface with $H = 0$ everywhere (see surface curvature).

Check on Scherk's surface $u = \ln(\cos y/\cos x)$ on $|x|, |y| < \pi/2$ . We have $u_x = \tan x$ , $u_y = -\tan y$ , $u_{xx} = \sec^2 x$ , $u_{yy} = -\sec^2 y$ and $u_{xy} = 0$ . Then

(1 + \tan^2 y)\sec^2 x + (1 + \tan^2 x)(-\sec^2 y) = \sec^2 y\sec^2 x - \sec^2 x\sec^2 y = 0 ,

so the equation holds exactly.

Example(Poisson's equation on the disc)Original

Minimise $J[u] = \int_D \bigl(\tfrac12|\nabla u|^2 - u\bigr)\,dx\,dy$ over the unit disc $D$ with $u = 0$ on the circle.

Equation. $F = \tfrac12|p|^2 - u$ , so $F_u = -1$ and $F_p = p$ . The Euler-Lagrange equation $-1 - \operatorname{div}\nabla u = 0$ is $-\Delta u = 1$ .
Solution. The radial function $u = (1 - r^2)/4$ has $\Delta u = u_{rr} + u_r/r = -\tfrac12 - \tfrac12 = -1$ and vanishes at $r = 1$ .
Value. Multiply $-\Delta u = 1$ by $u$ and use Green's identity (the divergence theorem with $u = 0$ on the boundary): $\int_D |\nabla u|^2 = \int_D u$ . So $J[u] = -\tfrac12\int_D u = -\tfrac12\cdot 2\pi\int_0^1 \tfrac{1 - r^2}{4}\,r\,dr = -\tfrac{\pi}{16} \approx -0.19635$ .
A trial function. Among $v = c\cos(\pi r/2)$ , the energy is a quadratic in $c$ . Its minimum is at $c = 32(\pi - 2)/\bigl(\pi^2(\pi^2 + 4)\bigr) \approx 0.26687$ (compare $u(0) = 0.25$ ), with value $J \approx -0.19395$ . This lies above $-\pi/16$ , as it must, since $u$ is the minimiser. Indeed $J[u + h] - J[u] = \tfrac12\int_D|\nabla h|^2 \ge 0$ .

ApplicationOriginal

In field theory a Lagrangian density $\mathcal L(\phi, \partial_\mu\phi)$ on space-time gives the field equations $\partial\mathcal L/\partial\phi - \sum_\mu \partial_\mu\bigl(\partial\mathcal L/\partial(\partial_\mu\phi)\bigr) = 0$ . For $\mathcal L = \tfrac12\phi_t^2 - \tfrac12|\nabla\phi|^2 - \tfrac12 m^2\phi^2$ this is the Klein-Gordon equation $\phi_{tt} - \Delta\phi + m^2\phi = 0$ . Maxwell's equations and the equations of linear elasticity arise the same way from their densities. In engineering, writing a boundary value problem as the minimisation of an energy is the starting point of the finite element method, and the equation in this unit is what that energy minimisation solves. In image processing, total-variation denoising minimises $\int\bigl(|\nabla u| + \tfrac{\mu}{2}(u - f)^2\bigr)$ , whose Euler-Lagrange equation is a nonlinear diffusion equation.

Boundary conditions and constraints

Theorem

Free endpoints and natural boundary conditions

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When an endpoint value is not prescribed, a stationary curve must satisfy an extra boundary condition, F_{y'} = 0 at a free end and the transversality condition when the endpoint slides along a given curve, so the problem still has as many conditions as constants.

Prerequisites: The Euler-Lagrange equation, Implicit and inverse function theorems, Differentiation under the integral sign

TheoremAdaptedAdapted from Gelfand & Fomin (1963), §6, p. 26 and §14, p. 60

Let $F(x, y, p)$ have continuous first partial derivatives and let $J[y] = \int F(x, y, y')\,dx$ .

Free end values (natural boundary conditions). Let $y \in C^1[a,b]$ , and let the admissible variations be
$\mathcal{H} = \bigl\{\,h \in C^1[a,b] : h(a) = 0\,\bigr\}$
when $y(a) = A$ is prescribed and $y(b)$ is free, and $\mathcal{H} = C^1[a,b]$ when both end values are free. If $\delta J[y; h] = 0$ for every $h \in \mathcal{H}$ , then $y$ satisfies the Euler-Lagrange equation on $[a,b]$ and
$F_{y'}\bigl(b, y(b), y'(b)\bigr) = 0 ;$
in the second case also $F_{y'}\bigl(a, y(a), y'(a)\bigr) = 0$ .
End sliding on a curve (transversality condition). Let $x_1 > a$ and $\delta > 0$ , let $F$ have continuous first partial derivatives on $[a, x_1 + \delta]\times\mathbb{R}\times\mathbb{R}$ , and let $\psi$ be $C^1$ on a neighbourhood of $x_1$ . Let $y \in C^1[a, x_1]$ satisfy $y(a) = A$ and $y(x_1) = \psi(x_1)$ , let $\tilde y \in C^1[a, x_1 + \delta]$ be an extension of $y$ , and assume $y'(x_1) \ne \psi'(x_1)$ , so the curve is not tangent to the target curve $Y = \psi(X)$ . For $h \in C^1[a, x_1+\delta]$ with $h(a) = 0$ and $|\varepsilon|$ small, the varied curve $\tilde y + \varepsilon h$ meets the target at a unique abscissa $s(\varepsilon)$ near $x_1$ , with $s(0) = x_1$ (Step 1 of the proof), so its restriction to $[a, s(\varepsilon)]$ is an admissible competitor; write
$\varphi_h(\varepsilon) = \int_a^{s(\varepsilon)} F\bigl(x, \tilde y + \varepsilon h, \tilde y' + \varepsilon h'\bigr)\,dx .$
Suppose $y$ is stationary in the sense that $\varphi_h'(0) = 0$ for every such $h$ . Then $y$ is an extremal on $[a, x_1]$ and
$\Bigl[F + \bigl(\psi' - y'\bigr)F_{y'}\Bigr]_{x = x_1} = 0 .$
For integrands $F = f(x, y)\sqrt{1 + p^2}$ with $f \ne 0$ , this reduces to $y'(x_1)\,\psi'(x_1) = -1$ : the extremal meets the target curve at a right angle.

Idea of the proofOriginal

Allowing more competitors gives more information. The variations that vanish at both ends still give the Euler-Lagrange equation. Once that equation holds, the first variation reduces to the boundary terms. A variation that does not vanish at a free end then forces the coefficient of $h$ there to vanish. When the end slides along a curve, moving the end changes both the integration range and the end value. The two first-order effects must cancel, and the resulting condition is transversality.

ProofAdaptedAdapted from Gelfand & Fomin (1963), §6, pp. 25–26 and §§13–14, pp. 55–60; Riley, Hobson & Bence (2006), pp. 782–784

Part 1. In both cases the variations with $h(a) = h(b) = 0$ belong to $\mathcal{H}$ , so by the Euler-Lagrange theorem the function $Q(x) = F_{y'}(x, y, y')$ is $C^1$ and $Q' = F_y$ . For any $h \in \mathcal{H}$ we may now integrate by parts and use $Q' = F_y$ :

0 = \delta J[y;h] = \int_a^b \bigl(F_y\,h + Q\,h'\bigr)\,dx = \int_a^b\bigl(F_y - Q'\bigr)\,h\,dx + Q(b)\,h(b) - Q(a)\,h(a) = Q(b)\,h(b) - Q(a)\,h(a) .

If $y(a) = A$ is prescribed, every $h \in \mathcal{H}$ has $h(a) = 0$ , and the choice $h(x) = (x - a)/(b - a)$ , which lies in $\mathcal{H}$ and has $h(b) = 1$ , gives $Q(b) = 0$ . If both end values are free, then $\mathcal{H} = C^1[a,b]$ : the same $h$ still gives $Q(b) = 0$ , and $h(x) = (b - x)/(b - a)$ , now admissible because no condition at $a$ is imposed, gives $Q(a) = 0$ .

Part 2. Fix the extension $\tilde y \in C^1[a, x_1+\delta]$ of the hypothesis.

Step 1 (a family of admissible curves). Take any $h \in C^1[a, x_1 + \delta]$ with $h(a) = 0$ . We want to know where the curve $\tilde y + \varepsilon h$ meets the target. Let $G(s, \varepsilon) = \tilde y(s) + \varepsilon h(s) - \psi(s)$ . Then $G(x_1, 0) = 0$ and $\partial G/\partial s\,(x_1, 0) = y'(x_1) - \psi'(x_1) \ne 0$ by the non-tangency assumption. The implicit function theorem gives a $C^1$ function $s(\varepsilon)$ with $s(0) = x_1$ and $G(s(\varepsilon), \varepsilon) = 0$ , unique near $x_1$ , and

s'(0) = -\frac{\partial G/\partial\varepsilon}{\partial G/\partial s}\Big|_{(x_1, 0)} = \frac{h(x_1)}{\psi'(x_1) - y'(x_1)} .

So $\tilde y + \varepsilon h$ restricted to $[a, s(\varepsilon)]$ is admissible, and it tends to $y$ as $\varepsilon \to 0$ .

Step 2 (differentiate). The integrand of $\varphi_h$ is $C^1$ in $(x, \varepsilon)$ , because $F$ is $C^1$ on $[a, x_1+\delta]\times\mathbb{R}\times\mathbb{R}$ , and the upper limit $s$ is $C^1$ by Step 1. So by the Leibniz rule with a variable upper limit $\varphi_h$ is differentiable at $0$ , and the stationarity hypothesis gives

0 = \varphi_h'(0) = F\bigl(x_1, y, y'\bigr)\,s'(0) + \int_a^{x_1}\bigl(F_y\,h + F_{y'}\,h'\bigr)\,dx .

Step 3 (Euler-Lagrange equation). Let $k \in C^1[a, x_1]$ with $k(a) = k(x_1) = 0$ , and extend it to $h \in C^1[a, x_1 + \delta]$ by its tangent line at $x_1$ , $h(x) = k'(x_1)(x - x_1)$ for $x > x_1$ ; this $h$ is $C^1$ because $k(x_1) = 0$ , and $h(a) = 0$ . Since $h(x_1) = 0$ we have $s'(0) = 0$ by the formula of Step 1, so Step 2 leaves $\int_a^{x_1}(F_y\,k + F_{y'}\,k')\,dx = 0$ . These are exactly the variations with fixed ends on $[a, x_1]$ , so by the Euler-Lagrange theorem $y$ is an extremal on $[a, x_1]$ and $Q = F_{y'}(x, y, y')$ is $C^1$ there with $Q' = F_y$ .

Step 4 (boundary terms). For general $h$ , integrate by parts as in Part 1. The integral reduces to $F_{y'}(x_1)\,h(x_1)$ , and Step 2 becomes

0 = h(x_1)\Bigl[\frac{F}{\psi' - y'} + F_{y'}\Bigr]_{x = x_1} .

Choose $h$ with $h(x_1) \ne 0$ and multiply by $\psi' - y' \ne 0$ to get the transversality condition.

Step 5 (orthogonality). For $F = f\sqrt{1 + p^2}$ we have $F_{y'} = f\,p/\sqrt{1 + p^2}$ , so

F + (\psi' - p)F_{y'} = \frac{f}{\sqrt{1 + p^2}}\Bigl(1 + p^2 + \psi' p - p^2\Bigr) = \frac{f\,(1 + p\,\psi')}{\sqrt{1 + p^2}} .

Since $f \ne 0$ , this vanishes exactly when $p\,\psi' = -1$ . $\square$

RemarkOriginal

Where the hypotheses are used.

Counting conditions. A second-order Euler-Lagrange equation has two free constants. In Part 1 the lost condition $y(b) = B$ is replaced by $F_{y'} = 0$ . In Part 2 there is a third unknown, $x_1$ , and two conditions at that end, $y(x_1) = \psi(x_1)$ and transversality. Conditions imposed on the admissible class, such as $y(a) = A$ , are called essential. Conditions that come out of stationarity, such as $F_{y'}(b) = 0$ , are called natural. A minimiser satisfies the natural conditions automatically, which is why finite-element codes never impose them explicitly.
Non-tangency $y'(x_1) \ne \psi'(x_1)$ is used in Step 1 of Part 2 to apply the implicit function theorem. Without it, the end of a varied curve need not move smoothly along the target.
Stationarity, not a weak extremum, in Part 2. The competitors there are defined on intervals $[a, s(\varepsilon)]$ with different right endpoints, so the $C^1$ -norm notion of a weak local extremum, which compares functions on one fixed interval, does not apply to them. Gelfand and Fomin measure the distance between two curves with different end abscissae by the $C^1$ distance of extensions to a common interval plus the Euclidean distances between the two pairs of endpoints (Gelfand & Fomin (1963), §13, Eq. (3), p. 54). In that distance each family of Step 1 converges to $y$ , because $s(\varepsilon) \to x_1$ and $\tilde y + \varepsilon h \to \tilde y$ in $C^1[a, x_1 + \delta]$ . Hence a curve that minimises or maximises $J$ among the admissible curves within some such distance of it makes every $\varphi_h$ have a local extremum at $\varepsilon = 0$ , and $\varphi_h'(0) = 0$ follows because $\varphi_h$ is differentiable there (Step 2). Part 2 assumes only this stationarity, which is all the proof uses.
Special targets. For a vertical line $x = b$ the endpoint cannot move horizontally, and Part 1 applies with $F_{y'} = 0$ . For a horizontal target $\psi = \text{const}$ the condition reads $F - y'F_{y'} = 0$ .
The same bookkeeping gives the natural conditions for higher derivatives and for several independent variables. In the latter case $F_p\cdot\nu = 0$ on the free part of the boundary.

Example(A free right end)Original

Minimise $J[u] = \int_0^1\bigl(\tfrac12 u'^2 - u\bigr)\,dx$ with $u(0) = 0$ and $u(1)$ free.

Euler-Lagrange equation: $F_u = -1$ and $F_{u'} = u'$ , so $u'' = -1$ .
Natural condition: $F_{u'}(1) = u'(1) = 0$ .
Solve: $u = -\tfrac12 x^2 + \alpha x + \beta$ , with $\beta = 0$ from $u(0) = 0$ and $\alpha = 1$ from $u'(1) = 0$ . So $u = x - \tfrac12 x^2$ , with $u(1) = \tfrac12$ .
Value: $J[u] = \int_0^1\bigl(\tfrac12(1 - x)^2 - x + \tfrac12 x^2\bigr)dx = \tfrac16 - \tfrac12 + \tfrac16 = -\tfrac16$ . The best straight line $u = ax$ has $J = a^2/2 - a/2$ , minimised at $a = \tfrac12$ with $J = -\tfrac18 > -\tfrac16$ , consistent with $u$ being the minimiser.

Physically, $u$ is the deflection of a string fixed at $x = 0$ , attached at $x = 1$ to a ring sliding freely on a vertical rod, and loaded uniformly. At the free end the string must leave the rod horizontally.

Example(Shortest distance to a parabola)Original

Find the shortest graph from $P = (0, 3)$ to the parabola $Y = X^2$ .

Functional and extremals. $J = \int_0^{x_1}\sqrt{1 + y'^2}\,dx$ with $y(0) = 3$ and $y(x_1) = x_1^2$ . As shown in the Euler-Lagrange unit, the extremals are straight lines.
Transversality. Here $f = 1$ , so the line must meet the parabola at a right angle. A line from $(0, 3)$ to $(t, t^2)$ has direction $(t, t^2 - 3)$ and the parabola has tangent $(1, 2t)$ . Orthogonality gives $t + 2t(t^2 - 3) = t\,(2t^2 - 5) = 0$ .
Candidates. The root $t = 0$ gives a vertical segment, which is not a graph $y(x)$ , so the theorem does not apply to it; it is in fact the longest of the stationary segments. The root $t = \sqrt{5/2}$ gives the foot point $(1.5811, 2.5)$ ; the root $t = -\sqrt{5/2}$ has $x_1 < 0 = a$ , so it falls outside the set-up $J = \int_a^{x_1}$ with $x_1 > a$ , and it is the mirror image of the first solution, found by running the same computation to the left. At the foot point the line has slope $-0.5/1.5811 = -0.31623$ and the parabola has slope $3.1623$ , whose product is $-1$ .
Distance: $\sqrt{5/2 + (5/2 - 3)^2} = \sqrt{11}/2 \approx 1.6583$ , shorter than the vertical distance $3$ .

Example(Freeing an end lowers the minimum)Original

Minimise $J[y] = \int_0^1\bigl(y'^2 + y^2\bigr)\,dx$ with $y(0) = 1$ , first with $y(1)$ free and then with $y(1) = 0$ imposed.

Equations. $F_y = 2y$ and $F_{y'} = 2y'$ , so the Euler-Lagrange equation is $y'' = y$ in both problems; $F_{y'y'} = 2 \ne 0$ , so the extremals are $C^2$ .
Free end. The natural condition is $F_{y'}(1) = 2y'(1) = 0$ . With $y = \alpha\cosh x + \beta\sinh x$ , the conditions $y(0) = 1$ and $y'(1) = 0$ give $y = \cosh(1 - x)/\cosh 1$ .
Its value. Integrating by parts and using $y'' = y$ , $J[y] = [y\,y']_0^1 = y(1)y'(1) - y(0)y'(0) = \tanh 1 = 0.76159$ . It is the global minimum: for admissible $y + h$ (so $h(0) = 0$ , $h(1)$ unrestricted) the cross term is $2[y'h]_0^1 - 2\int_0^1(y'' - y)h\,dx = 0$ , because $y'(1) = 0$ and $h(0) = 0$ , so $J[y+h] - J[y] = \int_0^1(h'^2 + h^2)\,dx \ge 0$ .
Fixed end. Imposing $y(1) = 0$ instead gives $y = \sinh(1 - x)/\sinh 1$ and, by the same boundary-term computation, $J = -y(0)y'(0) = \coth 1 = 1.31304$ .
Comparison. The free-end minimum is the smaller, as it must be: its competitors include all the curves of the fixed-end problem. The natural condition $y'(1) = 0$ is what the free-end minimiser does of its own accord instead of meeting a prescribed value.

ApplicationOriginal

Natural boundary conditions are the free-edge conditions of continuum mechanics. A cantilever beam with energy $\int\bigl(\tfrac{EI}{2}w''^2 - qw\bigr)$ is clamped at one end ( $w = w' = 0$ , essential) and free at the other. There stationarity yields $w'' = 0$ (no bending moment) and $w''' = 0$ (no shear force). For a membrane or a heat-conducting body with a free or insulated edge, the natural condition is $\partial u/\partial\nu = 0$ . In optimal control and in optics, transversality conditions fix the free terminal time or position: a ray leaving a point and ending on a mirror surface of its choice must meet that surface at right angles.

Method

Isoperimetric problems

Adapted

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To make one integral stationary while another integral is held fixed, add a constant Lagrange multiplier times the constraint integrand and solve the Euler-Lagrange equation of the sum; this gives the circular arc of Dido's problem and the catenary of a hanging chain.

Prerequisites: The Euler-Lagrange equation, Lagrange multipliers

MethodAdaptedAdapted from Gelfand & Fomin (1963), Ch. 2, §12.1, Thm 1, p. 43

Let $F(x, y, p)$ and $G(x, y, p)$ have continuous first partial derivatives. Among $y \in C^1[a,b]$ with $y(a) = A$ , $y(b) = B$ and

K[y] = \int_a^b G(x, y, y')\,dx = \ell ,

suppose $y$ gives a weak local extremum of $J[y] = \int_a^b F(x, y, y')\,dx$ . Assume that $y$ is not an extremal of $K$ , that is, $\delta K[y; h_2] \ne 0$ for some $h_2 \in C^1_0[a,b]$ . Then there is a constant $\lambda$ such that $y$ is an extremal of $J + \lambda K$ :

\frac{d}{dx}\bigl(F + \lambda G\bigr)_{y'} = \bigl(F + \lambda G\bigr)_y \qquad\text{on } [a,b] .

With several constraints $K_j[y] = \ell_j$ ( $j = 1, \dots, k$ ), the same conclusion holds for $F + \sum_j\lambda_j G_j$ , provided the linear map $h \mapsto \bigl(\delta K_1[y;h], \dots, \delta K_k[y;h]\bigr)$ from $C^1_0[a,b]$ onto $\mathbb{R}^k$ is surjective.

Idea of the proofOriginal

A variation in an arbitrary direction $h_1$ would usually spoil the constraint. We repair it with a second, fixed direction $h_2$ that does change $K$ . In the two-parameter family $y + s h_1 + t h_2$ , the condition $K = \ell$ cuts out a curve through the origin of the $(s,t)$ -plane. Along that curve $J$ has an extremum at the origin. This is an ordinary two-variable problem with one constraint, and the finite-dimensional Lagrange multiplier rule gives a multiplier. The multiplier comes out as a ratio computed from $h_2$ alone, so it is the same for every $h_1$ .

ProofAdaptedAdapted from Gelfand & Fomin (1963), proof of Thm 1, pp. 43–45

Step 1 (two-parameter family). Fix $h_2$ as in the hypothesis and let $h_1 \in C^1_0[a,b]$ be arbitrary. Define

\Phi(s,t) = J[y + s h_1 + t h_2], \qquad \Psi(s,t) = K[y + s h_1 + t h_2] .

The argument of the first variation, applied along each coordinate direction, shows that $\Phi$ and $\Psi$ have partial derivatives in $s$ and $t$ given by integrals with continuous integrands; these depend continuously on $(s, t)$ , so $\Phi, \Psi$ are $C^1$ near $(0,0)$ . At the origin,

\Phi_s = \delta J[y; h_1],\quad \Phi_t = \delta J[y; h_2],\quad \Psi_s = \delta K[y; h_1],\quad \Psi_t = \delta K[y; h_2] \ne 0 .

Step 2 (finite-dimensional extremum). For $(s, t)$ near the origin with $\Psi(s,t) = \ell$ , the function $y + s h_1 + t h_2$ has the right end values, satisfies the constraint and lies within $|s|\,\|h_1\|_1 + |t|\,\|h_2\|_1$ of $y$ in the $C^1$ norm. So $\Phi$ restricted to the level set $\{\Psi = \ell\}$ has a local extremum at $(0, 0)$ , and $\nabla\Psi(0,0) \ne 0$ because $\Psi_t \ne 0$ .

Step 3 (multiplier). By the Lagrange multiplier rule there is $\mu$ with $\nabla\Phi(0,0) = \mu\,\nabla\Psi(0,0)$ . The $t$ -component gives $\mu = \delta J[y;h_2]/\delta K[y;h_2]$ , which does not depend on $h_1$ . Put $\lambda = -\mu$ . The $s$ -component then reads

\delta J[y; h_1] + \lambda\,\delta K[y; h_1] = 0 \qquad\text{for every } h_1 \in C^1_0[a,b] .

Step 4 (Euler-Lagrange equation). The left side is the first variation of the functional with integrand $F + \lambda G$ , which has continuous first derivatives. So $y$ is stationary for it, and the Euler-Lagrange theorem gives the equation. $\square$

For $k$ constraints the same proof works with $k$ repairing directions $h_2, \dots, h_{k+1}$ chosen so that the $k \times k$ matrix $\bigl(\delta K_j[y; h_{i+1}]\bigr)$ is invertible, which is possible by the surjectivity hypothesis. We do not write it out.

MethodAdaptedAdapted from Riley, Hobson & Bence (2006), §22.4, pp. 785–786; Gelfand & Fomin (1963), p. 45

To make $J[y] = \int_a^b F\,dx$ stationary subject to $y(a) = A$ , $y(b) = B$ , $\int_a^b G\,dx = \ell$ :

Form $F^* = F + \lambda G$ with an unknown constant $\lambda$ .
Write the Euler-Lagrange equation of $F^*$ . If $F^*$ has no $x$ , use the Beltrami first integral $F^* - y'F^*_{y'} = C$ .
Solve. The general solution contains two constants of integration and $\lambda$ .
Determine all three from the two end conditions and the constraint $\int G\,dx = \ell$ .
Check the non-degeneracy hypothesis (the solution must not also be an extremal of $K$ ), and compare values of $J$ among the candidates.

The multiplier usually has a physical meaning: with the convention $F + \lambda G$ , the number $-\lambda$ is the rate of change of the optimal value of $J$ with respect to $\ell$ . In the examples below it is a radius and a tension.

Example(Dido's problem)AdaptedAdapted from Gelfand & Fomin (1963), Example 1, p. 49

A rope of length $\ell = 2.5$ is tied to the points $(\pm 1, 0)$ on a straight shore. Which curve $y(x) \ge 0$ encloses the largest area with the shore?

Set-up. Maximise $J = \int_{-1}^1 y\,dx$ subject to $K = \int_{-1}^1\sqrt{1 + y'^2}\,dx = 2.5$ and $y(\pm 1) = 0$ . So $F^* = y + \lambda\sqrt{1 + y'^2}$ .
Equation. $F^*_y = 1$ and $F^*_{y'} = \lambda y'/\sqrt{1 + y'^2}$ , so $\frac{d}{dx}\bigl(\lambda y'/\sqrt{1 + y'^2}\bigr) = 1$ . Integrating, $\lambda y'/\sqrt{1 + y'^2} = x - x_0$ .
Solve. Solving for $y'$ and integrating once more gives $(x - x_0)^2 + (y - y_0)^2 = \lambda^2$ : an arc of a circle of radius $|\lambda|$ . Symmetry of the end conditions gives $x_0 = 0$ .
Constants. Let the arc subtend the angle $2\alpha$ at its centre $(0, y_0)$ , with radius $r$ . The chord gives $r\sin\alpha = 1$ and the length gives $2r\alpha = 2.5$ . Hence $\sin\alpha/\alpha = 0.8$ , so $\alpha = 1.13110$ and $r = 1.10512$ . The centre lies at depth $r\cos\alpha = 0.47041$ below the shore, and the arc rises $0.63471$ above it.
Area. A circular segment has area $\tfrac12 r^2(2\alpha - \sin 2\alpha) = 0.91099$ . For comparison, the parabola $y = k(1 - x^2)$ of the same length has $k = 0.67406$ and area $\tfrac43 k = 0.89875$ , which is smaller.
Non-degeneracy. The arc is not a straight segment, so it is not an extremal of length, and the theorem applies. For $\ell > \pi$ the optimal arc would exceed a semicircle and could not be written as a graph over the shore; a parametric formulation is needed then.

Example(The hanging chain)AdaptedAdapted from Riley, Hobson & Bence (2006), pp. 786–787

A uniform chain of length $3$ hangs from the points $(\pm 1, 0)$ , with $y$ measured upward. Its shape minimises the potential energy, proportional to $\int y\,ds$ , at fixed length.

Set-up. $J = \int_{-1}^1 y\sqrt{1 + y'^2}\,dx$ , $K = \int_{-1}^1\sqrt{1 + y'^2}\,dx = 3$ , $y(\pm 1) = 0$ . So $F^* = (y + \lambda)\sqrt{1 + y'^2}$ .
First integral. $F^*$ has no $x$ , and $F^* - y'F^*_{y'} = (y + \lambda)/\sqrt{1 + y'^2} = c$ .
Solve. This is the equation of the minimal surface of revolution for $Y = y + \lambda$ , so $y + \lambda = c\cosh\bigl((x - x_0)/c\bigr)$ . Symmetry gives $x_0 = 0$ .
Constants. The length is $\int_{-1}^1\cosh(x/c)\,dx = 2c\sinh(1/c) = 3$ , so $c = 0.61647$ . The end condition gives $\lambda = c\cosh(1/c) = 1.62174$ .
Shape. The lowest point is $y(0) = c - \lambda = -1.00527$ , so the chain sags $1.0053$ below its supports. The multiplier $\lambda$ is the height of the supports above the "directrix" $y = -\lambda$ . Physically, the tension at a point at height $y$ equals the weight per unit length times $y + \lambda$ .

RemarkOriginal

Where the hypotheses are used. The non-degeneracy assumption $\delta K[y; h_2] \ne 0$ is used in Step 2 to make $\nabla\Psi \ne 0$ , so that the finite-dimensional multiplier rule applies. It cannot be dropped. Take $K[y] = \int_0^1\sqrt{1 + y'^2}\,dx = 1$ with $y(0) = y(1) = 0$ . The only admissible curve is $y = 0$ , which is therefore an extremum of $J = \int_0^1 y\,dx$ . But the Euler-Lagrange equation of $y + \lambda\sqrt{1 + y'^2}$ along $y = 0$ reads $0 = 1$ , false for every $\lambda$ . Here $y = 0$ is an extremal of $K$ (a straight line), so the theorem does not apply. In such degenerate cases one can only assert the weaker rule that $\lambda_0 F + \lambda G$ is stationary for some $(\lambda_0, \lambda) \ne (0, 0)$ .

ApplicationOriginal

In statistical mechanics and information theory, maximising the entropy $-\int f\ln f\,dx$ of a probability density subject to $\int f\,dx = 1$ and $\int x^2 f\,dx = \sigma^2$ gives $-\ln f - 1 + \lambda_0 + \lambda_1 x^2 = 0$ (no derivatives appear, so the Euler-Lagrange equation is algebraic). Hence $f$ is a Gaussian, with $\lambda_1 = -1/(2\sigma^2)$ and $\lambda_0 - 1$ the logarithm of the normalising constant. With an energy constraint in place of the variance this gives the Boltzmann distribution, where the multiplier is $-1/(k_BT)$ . Minimising $\int p\,y'^2 + q\,y^2$ subject to $\int w\,y^2 = 1$ gives a Sturm-Liouville equation whose multiplier is an eigenvalue (see variational eigenvalues). This is also how a normalised quantum ground state is computed.

Method

Pointwise constraints

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To make an integral stationary over curves forced to lie on a surface g = 0, add to the integrand g times a multiplier that is a function of the independent variable; for arc length this shows that geodesics are the curves whose acceleration is normal to the surface, such as great circles on a sphere and helices on a cylinder.

Prerequisites: Several functions and higher derivatives, Lagrange multipliers, Implicit and inverse function theorems

MethodAdaptedAdapted from Gelfand & Fomin (1963), Ch. 2, §12.2, Thm 2, p. 46

Let $q = (q_1, \dots, q_n)$ be the unknown functions of $t \in [t_0, t_1]$ , with $n \ge 2$ . Let $F(t, q, p)$ be $C^2$ on an open set $U \subset \mathbb{R}\times\mathbb{R}^n\times\mathbb{R}^n$ , and let $g(t, q)$ be $C^2$ . Consider

J[q] = \int_{t_0}^{t_1} F\bigl(t, q, q'\bigr)\,dt

over $C^1$ curves with $(t, q, q') \in U$ , prescribed $q(t_0)$ , $q(t_1)$ , and the pointwise (holonomic) constraint

g\bigl(t, q(t)\bigr) = 0 \qquad\text{for every } t \in [t_0, t_1] .

Suppose $q \in C^2[t_0, t_1]$ is a weak local extremum and that the gradient $\nabla_q g = (g_{q_1}, \dots, g_{q_n})$ does not vanish at any point $(t, q(t))$ . Then there is a continuous function $\lambda(t)$ such that

\frac{d}{dt}F_{q_i'} - F_{q_i} = \lambda(t)\,g_{q_i}\bigl(t, q(t)\bigr), \qquad i = 1, \dots, n ,

that is, $q$ is an extremal of $\int\bigl(F + \lambda(t)\,g\bigr)\,dt$ , treated as a functional of $q$ with $\lambda$ fixed. Explicitly, $\lambda = -\bigl(E\cdot\nabla_q g\bigr)/|\nabla_q g|^2$ , where $E_i = F_{q_i} - \frac{d}{dt}F_{q_i'}$ .

Idea of the proofOriginal

Near any point of the curve, the constraint can be solved for one coordinate in terms of the others. That turns the problem locally into an unconstrained one in the remaining coordinates. Varying those coordinates freely shows that the vector $E$ of Euler-Lagrange expressions has no component along the surface: it is parallel to $\nabla_q g$ at every $t$ . The proportionality factor is the multiplier. Unlike the isoperimetric case, there is one constraint for every $t$ , so the multiplier is a function of $t$ rather than a single number.

ProofAdaptedAdapted from Gelfand & Fomin (1963), proof of Thm 2, pp. 46–48

Write $g_i = g_{q_i}(t, q(t))$ and $E_i(t) = F_{q_i} - \frac{d}{dt}F_{q_i'}$ , evaluated along the curve. Because $F$ is $C^2$ and $q$ is $C^2$ , each $F_{q_i'}(t, q, q')$ is $C^1$ , so $E_i$ is continuous. It suffices to show that $E$ is parallel to $\nabla_q g$ at every $t$ . Then $E = -\lambda\nabla_q g$ with $\lambda = -(E\cdot\nabla_q g)/|\nabla_q g|^2$ , which is continuous because $\nabla_q g \ne 0$ , and this is the claim.

Step 1 (local solution of the constraint). Fix $t_* \in [t_0, t_1]$ and choose $k$ with $g_k(t_*) \ne 0$ . Write $\hat q$ for the coordinates other than $q_k$ . By the implicit function theorem, there are a neighbourhood $N$ of $(t_*, \hat q(t_*))$ , an interval $W \ni q_k(t_*)$ and a $C^2$ function $\phi$ on $N$ such that, for $(t, \hat q) \in N$ and $q_k \in W$ , $g(t, q) = 0$ holds exactly when $q_k = \phi(t, \hat q)$ . Moreover $\phi_{q_i} = -g_{q_i}/g_{q_k}$ for $i \ne k$ . By continuity there is an interval $I \ni t_*$ , open relative to $[t_0, t_1]$ , on which $(t, \hat q(t)) \in N$ and $q_k(t) \in W$ . On $I$ , therefore, $q_k(t) = \phi(t, \hat q(t))$ .

Step 2 (admissible variations). Fix $i \ne k$ and a $C^2$ function $\eta$ whose support is a compact subset of $I \cap (t_0, t_1)$ . Define the varied curve to agree with $q$ outside $I$ , and on $I$ to have $\hat q + \varepsilon\eta\,e_i$ in the $\hat q$ -slots and $\phi(t, \hat q + \varepsilon\eta\,e_i)$ in the $k$ -th slot. For small $|\varepsilon|$ this curve is $C^2$ , lies on the surface, has the same endpoints, and tends to $q$ in the $C^1$ norm as $\varepsilon \to 0$ . Its $k$ -th component changes at first order by

\zeta(t) = \phi_{q_i}\bigl(t, \hat q(t)\bigr)\,\eta(t) = -\frac{g_i}{g_k}\,\eta(t),

a $C^1$ function with the same support as $\eta$ .

Step 3 (first variation). The derivative of the integrand in $\varepsilon$ is continuous in $(t, \varepsilon)$ , since $\phi$ is $C^2$ . Differentiating under the integral sign, as in the case of several functions, gives

0 = \int\Bigl(F_{q_i}\,\eta + F_{q_i'}\,\eta' + F_{q_k}\,\zeta + F_{q_k'}\,\zeta'\Bigr)\,dt = \int\bigl(E_i\,\eta + E_k\,\zeta\bigr)\,dt = \int\Bigl(E_i - E_k\frac{g_i}{g_k}\Bigr)\eta\,dt .

Here we integrated the $\eta'$ and $\zeta'$ terms by parts; the boundary terms vanish because $\eta$ and $\zeta$ vanish near the ends of their support. The bracket is continuous on $I$ . By the fundamental lemma on subintervals of $I$ , $E_i\,g_k = E_k\,g_i$ on $I$ , for every $i \ne k$ , and trivially for $i = k$ .

Step 4 (conclusion). On $I$ , then, $E = (E_k/g_k)\,\nabla_q g$ is parallel to $\nabla_q g$ . Since $t_*$ was arbitrary, this holds on all of $[t_0, t_1]$ . $\square$

MethodOriginal

To make $\int F(t, q, q')\,dt$ stationary over curves on $g(t, q) = 0$ with fixed ends:

Form $F + \lambda(t)\,g(t, q)$ with an unknown function $\lambda$ .
Write the $n$ Euler-Lagrange equations in $q$ (no equation for $\lambda$ ). Together with $g = 0$ these are $n + 1$ equations for $n + 1$ unknown functions.
Eliminate $\lambda$ . Either take the components of the equations tangent to the surface, or solve one equation for $\lambda$ and substitute.
Fix the $2n$ constants of integration by the end conditions. Check that $\nabla_q g \ne 0$ along the solution.
Alternatively, parametrise the surface and solve an unconstrained problem in the parameters. That avoids $\lambda$ but loses its interpretation as a constraint force.

With several constraints $g_1 = \dots = g_m = 0$ ( $m < n$ ) one uses $F + \sum_j\lambda_j(t)\,g_j$ . The analogous theorem needs the gradients $\nabla_q g_j$ to be linearly independent along the curve; we state it without proof.

Example(Geodesics on a sphere)AdaptedAdapted from Gelfand & Fomin (1963), Example 2, pp. 49–50

Find the shortest curve on the unit sphere $g = x^2 + y^2 + z^2 - 1 = 0$ from $P = (1, 0, 0)$ to $Q = (\tfrac12, \tfrac12, \tfrac1{\sqrt2})$ .

Functional. Length is $\int|r'(t)|\,dt$ , with $F = |p|$ , which is $C^2$ away from $p = 0$ ; so we work with regular curves. Length does not depend on the parametrisation, so we may parametrise the extremal by arc length $s$ , making $|r'| = 1$ .
Equations. $F_q = 0$ and $F_{p} = p/|p|$ , so with $|r'| = 1$ the equations read $\frac{d}{ds}r' = \lambda\nabla g$ , that is, $r'' = 2\lambda\,r$ : the acceleration is normal to the sphere.
Solve. $\frac{d}{ds}(r\times r') = r'\times r' + r\times r'' = 0$ , so $r\times r' = N$ is a constant vector. It is nonzero, since $r \perp r'$ (differentiate $|r|^2 = 1$ ) and $|r| = |r'| = 1$ . Then $r\cdot N = 0$ : the curve lies in the plane through the origin with normal $N$ , so it is an arc of a great circle.
Multiplier. Differentiating $r\cdot r' = 0$ gives $1 + r\cdot r'' = 0$ , and $r\cdot r'' = 2\lambda$ , so $\lambda = -\tfrac12$ is constant.
Numbers. $P\cdot Q = \tfrac12$ , so the shorter great-circle arc has length $\arccos\tfrac12 = \pi/3 \approx 1.0472$ . Explicitly $r(s) = \cos s\,P + \sin s\,u$ with $u = (0, 1/\sqrt3, \sqrt2/\sqrt3)$ , and $r(\pi/3) = Q$ . The complementary arc, of length $5\pi/3 \approx 5.2360$ , satisfies the same equations but is not the shortest: the method yields necessary conditions only.

Example(Geodesics on a cylinder)AdaptedAdapted from do Carmo (2016), Example 4, pp. 249–250

On $g = x^2 + y^2 - 1 = 0$ , write a unit-speed curve as $r(s) = (\cos\theta(s), \sin\theta(s), z(s))$ .

Equations. As above, $r'' = \lambda\nabla g = 2\lambda(s)\,(\cos\theta, \sin\theta, 0)$ . Computing directly, $r'' = \bigl(-\theta''\sin\theta - \theta'^2\cos\theta,\ \theta''\cos\theta - \theta'^2\sin\theta,\ z''\bigr)$ .
Tangential components. The $z$ -component gives $z'' = 0$ . Dotting with $(-\sin\theta, \cos\theta, 0)$ , which is tangent to the cylinder and orthogonal to the right side, gives $\theta'' = 0$ . So $\theta$ and $z$ are linear in $s$ : the geodesics are helices, including horizontal circles and vertical lines as limiting cases. The normal component gives $-\theta'^2 = 2\lambda$ , so $\lambda = -\theta'^2/2$ .
Numbers. From $(1, 0, 0)$ to $(-1, 0, \pi)$ , $\theta$ may run from $0$ to $\pi$ , to $-\pi$ , to $3\pi$ , and so on, while $z$ runs from $0$ to $\pi$ . The lengths are $\sqrt{(\Delta\theta)^2 + \pi^2}$ : $\pi\sqrt2 \approx 4.4429$ for $\Delta\theta = \pm\pi$ (two shortest helices) and $\pi\sqrt{10} \approx 9.9346$ for $\Delta\theta = 3\pi$ . On the shortest ones, $\theta' = \pi/(\pi\sqrt2) = 1/\sqrt2$ , so $\lambda = -\tfrac14$ .

RemarkOriginal

Where the hypotheses are used.

$\nabla_q g \ne 0$ is used in Step 1 to solve the constraint locally, and again to define $\lambda$ . It cannot be dropped. The function $\tilde g = (x^2 + y^2 - 1)^2$ describes the same cylinder, but $\nabla\tilde g = 0$ on it. With $\tilde g$ , the multiplier equations would say $r'' = 0$ , which the shortest helices violate.
$C^2$ smoothness of $F$ , $g$ and $q$ makes $E_i$ continuous and $\phi$ twice differentiable, as needed for differentiating under the integral and for the fundamental lemma.
Reparametrisation in the examples relies on the length integrand being positively homogeneous of degree one in $p$ . For other integrands the parameter cannot be changed freely.
The conclusion for arc length, that the acceleration of a unit-speed geodesic is normal to the surface, is the standard characterisation of geodesics in differential geometry (see geodesics).

ApplicationOriginal

In mechanics, Hamilton's principle with a holonomic constraint gives equations of motion of the form $m\,\ddot r = -\nabla V + \lambda(t)\nabla g$ (see Hamilton's principle). The term $\lambda\nabla g$ is the force of constraint: the tension in a pendulum rod, or the normal force of a surface on a sliding particle. Computing $\lambda$ tells an engineer when a particle leaves a surface or how strong a rod must be. Constrained molecular-dynamics algorithms such as SHAKE and RATTLE fix bond lengths in exactly this way, by solving for the multipliers at every time step. In general relativity, geodesics of space-time are the extremals of the analogous length or action integral.