First-order equations and existence theory

Elementary methods and the existence, uniqueness and stability of solutions.

In this chapter 3

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Methods and existence

Methods and existence

Method

Solving first-order equations

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Gives the six standard closed-form methods for a first-order first-degree equation — separation, exactness, integrating factors, the linear formula, the substitutions for homogeneous and for Bernoulli equations — each with the hypotheses it needs, the reason it works, and worked examples with checkable numbers.

Prerequisites: Partial derivatives and the chain rule, Differentiation under the integral sign, Implicit and inverse function theorems, Line integrals and conservative fields

Note(What the six methods have in common)Original

There is no procedure that solves every first-order equation in closed form. What there is instead is a short list of shapes that can be recognised by inspection, each of which collapses to an integration. Two ideas do all the work.

Separate. If the $x$ -dependence and the $y$ -dependence can be pushed to opposite sides of the equation, each side is integrated on its own and the solution is the resulting relation between $x$ and $y$ .
Make exact. If the left-hand side of $A\,dx + B\,dy = 0$ is the total differential $dU$ of a single function $U$ , the equation says $dU = 0$ , so the solution curves are the level curves $U = \text{const}$ . When it is not such a differential, multiplying by an integrating factor can make it one.

The other four types are reductions to these two: a linear equation is made exact by an integrating factor that can always be written down, and the homogeneous and Bernoulli substitutions turn their equations into a separable and a linear one respectively. All of them but one involve a division — by a factor $\sigma(y)$ , by $B$ or $A$ , by $F(v) - v$ , by $y^n$ — and each division can discard solutions or shrink the region where the work is valid. The exception is the exact-equation method as given below, which divides by nothing and loses nothing; that is part of why it is the shape the others are reduced to. For the rest, keeping track of what was divided by is part of the method, not an afterthought.

Definition(First-order first-degree equations and their solutions)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2, p. 470; Arfken, Weber & Harris (2013), §7.2, Eq. (7.2), p. 331; Coddington (1989), §5.4, p. 200

A first-order equation contains no derivative of $y$ beyond $dy/dx$ , and it is of first degree when $dy/dx$ appears to the first power only. Such an equation can be written in either of two interchangeable ways,

\frac{dy}{dx} = F(x,y) , \qquad\qquad A(x,y)\,dx + B(x,y)\,dy = 0 ,

related by $F = -A/B$ wherever $B \neq 0$ . The first form is the one to use when $y$ is to be found as a function of $x$ ; the second treats the two variables alike, which is what the exactness methods below need. Riley, Hobson and Bence call these the two standard forms and choose between them according to the type of equation at hand.

A solution on an interval $I$ is a differentiable function $\varphi : I \to \mathbb{R}$ whose graph lies in the region where the data are defined and which satisfies $\varphi'(x) = F(x, \varphi(x))$ at every $x \in I$ . A solution of the initial-value problem $y' = F(x,y)$ , $y(x_0) = y_0$ is a solution on an interval containing $x_0$ with $\varphi(x_0) = y_0$ .

The general solution carries one arbitrary constant. (Riley, Hobson and Bence take the general solution of an $n$ th-order equation to contain $n$ arbitrary parameters, or constants; they say explicitly that they assume this rather than prove it, a fully general proof being difficult, p. 470.) Fixing the constant from an initial condition gives a particular solution.

Most of the methods below deliver the answer not as $y = \varphi(x)$ but as an implicit solution: a relation

U(x,y) = k

together with the statement that every differentiable function defined implicitly by that relation is a solution. Where $\partial U/\partial y \neq 0$ the implicit function theorem converts the relation into a function $y = \varphi(x)$ locally; where $\partial U/\partial y = 0$ the solution curve has a vertical tangent and $y$ cannot be solved for, although $x$ may be solvable as a function of $y$ . This is why the differential form, which is symmetric in the two variables, is often the honest place to stop.

Method(Separable equations)AdaptedAdapted from Coddington (1989), §5.2, Theorem 1, p. 188; Riley, Hobson & Bence (2006), §14.2.1, Eq. (14.3), p. 471

An equation has separated variables when it can be written

h(y)\,\frac{dy}{dx} = g(x) ,

Coddington defines the class by the quotient instead, calling $y' = f(x,y)$ separated when $f$ can be written as $g(x)/h(y)$ (p. 186), and then multiplies up to the displayed form. The two forms pick out the same equations but not the same solutions: where $h$ vanishes or is undefined the displayed form can hold and the quotient form cannot, which is the content of the proviso at the end of this block. Riley, Hobson and Bence write the same class as a product, $dy/dx = f(x)\,\sigma(y)$ , which is the case $h = 1/\sigma$ .

Theorem (Coddington, Theorem 1). Assume $g$ is continuous and real-valued on $a < x < b$ and $h$ is continuous and real-valued on $c \le y \le d$ . Pick antiderivatives: any $G$ with $G' = g$ and any $H$ with $H' = h$ . The solutions are then exactly the functions cut out by the relation

H(y) = G(x) + k , \qquad k \ \text{constant},

in the following two senses, for an interval $I$ contained in $a \le x \le b$ (Coddington's wording: he assumes $g$ continuous on the open interval but allows $I$ inside the closed one).

Suppose that, for one such $k$ , the relation determines implicitly a differentiable real-valued $\varphi$ throughout $I$ . Then $h(y)y' = g(x)$ holds along $\varphi$ on $I$ .
Suppose instead that $\varphi$ is given as a solution of $h(y)y' = g(x)$ on $I$ . Then some constant $k$ makes the relation hold along $\varphi$ throughout $I$ .

For the quotient form of the equation Coddington adds a proviso, since dividing by $h$ is dividing: such a $\varphi$ solves $y' = g(x)/h(y)$ on $I$ as long as $h(\varphi(x))$ is non-zero at each point of $I$ .

Idea of the proof(Separable equations)Original

Both directions are the chain rule, read forwards and backwards. The function $x \mapsto H(\varphi(x)) - G(x)$ has derivative $h(\varphi(x))\varphi'(x) - g(x)$ , which is exactly the left side minus the right side of the equation. So " $\varphi$ solves the equation" and " $H(\varphi) - G$ is constant" are two ways of saying that one function has zero derivative. Writing $h(y)\,dy = g(x)\,dx$ and integrating each side separately is a shorthand for the substitution $u = \varphi(t)$ in the integral of the left-hand side.

Proof(Separable equations)AdaptedAdapted from Coddington (1989), §5.2, pp. 187-188

The relation gives a solution. Suppose $H(y) = G(x) + k$ defines a differentiable $\varphi$ on $I$ , so that $H(\varphi(x)) = G(x) + k$ for all $x \in I$ . Differentiate in $x$ : the left side is differentiable because $H$ is (its derivative $h$ is continuous) and $\varphi$ is, and the chain rule gives
$H'(\varphi(x))\,\varphi'(x) = G'(x), \qquad\text{that is}\qquad h(\varphi(x))\,\varphi'(x) = g(x) ,$
which is the equation $h(y)y' = g(x)$ satisfied by $\varphi$ on $I$ . If moreover $h(\varphi(x)) \neq 0$ throughout $I$ , dividing gives $\varphi' = g/h(\varphi)$ , the equation in the form $y' = g(x)/h(y)$ . This is where Coddington's proviso is used, and it is the only place it is needed.
A solution satisfies the relation. Conversely let $\varphi$ solve $h(\varphi)\varphi' = g$ on $I$ and fix $x_0 \in I$ . By step 1 read backwards, $h(\varphi(t))\varphi'(t)$ is the derivative of $t \mapsto H(\varphi(t))$ , so the equation says that $H \circ \varphi$ and $G$ have the same derivative on $I$ . Integrating from $x_0$ to $x$ and using the fundamental theorem of calculus — legitimate because both derivatives are continuous on $I$ —
$H(\varphi(x)) - H(\varphi(x_0)) = G(x) - G(x_0) .$
Setting $k = H(\varphi(x_0)) - G(x_0)$ gives $H(\varphi(x)) = G(x) + k$ on $I$ .
Why "separating" is the same computation. In step 2 the integral $\int_{x_0}^{x} h(\varphi(t))\varphi'(t)\,dt$ was evaluated by recognising the integrand as a derivative; the substitution $u = \varphi(t)$ , $du = \varphi'(t)\,dt$ turns it into $\int_{\varphi(x_0)}^{\varphi(x)} h(u)\,du$ . That substitution is precisely what the formal manipulation $h(y)\,dy = g(x)\,dx$ records. $\square$

Method(Separating the variables)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.1, p. 471

Try to factor the right-hand side as $f(x)\,\sigma(y)$ . Equations that do not look separable often become so after one factorisation.
Record the zeros of $\sigma$ . Each root $y_\ast$ of $\sigma(y_\ast) = 0$ gives a constant solution $y \equiv y_\ast$ . These are exactly the solutions that the next step may destroy.
Divide by $\sigma(y)$ , multiply by $dx$ , and integrate each side with respect to its own variable, remembering a single constant of integration.
Solve the resulting relation for $y$ if it can be solved; otherwise leave it implicit. Impose the initial condition to fix the constant, and check by substitution.
State the interval on which the answer is valid: it ends where the denominator $\sigma(y(x))$ reaches zero, where an integral diverges, or where $y$ itself ceases to exist.

Remark(Solutions lost by dividing)AdaptedAdapted from Coddington (1989), §5.2, pp. 189-190

Separation can miss solutions, and Coddington makes the point with two examples.

For $y' = y^2$ , separating gives $y = -1/(x + k)$ . The identically zero function also solves the equation and is not of that form for any $k$ .
For $y' = 3y^{2/3}$ , separating gives $y = (x+k)^3$ , and again $y \equiv 0$ is a solution outside the family. This equation has worse behaviour still, and Existence and uniqueness of solutions returns to it.

The remedy is the bookkeeping in step 2 of the method: list the roots of $\sigma$ before dividing. In Example 1 below, the two constant solutions found that way turn out to be the asymptotes that every other solution approaches, so the lost solutions are not a technicality but the main features of the phase picture.

Method(Exact equations and the criterion for exactness)AdaptedAdapted from Coddington (1989), §5.3, Theorems 2 and 3, pp. 193, 195; Riley, Hobson & Bence (2006), §5.3, Eq. (5.9), p. 156, and §14.2.2, Eqs. (14.4)-(14.7), p. 472; Arfken, Weber & Harris (2013), §7.2, Eqs. (7.6)-(7.9), p. 333

The equation $A(x,y) + B(x,y)y' = 0$ , written in differential form as $A\,dx + B\,dy = 0$ , is exact in a region $R$ if there is a function $U$ with continuous first partial derivatives in $R$ such that

\frac{\partial U}{\partial x} = A , \qquad \frac{\partial U}{\partial y} = B \qquad\text{throughout } R .

Equivalently, $A\,dx + B\,dy$ is the total differential $dU$ of a single function — an exact differential in the sense of Riley, Hobson and Bence, §5.3. (Coddington writes $M$ and $N$ where we write $A$ and $B$ , and $F$ where we write $U$ ; the two theorems below are his, in our letters.)

Theorem (solutions of an exact equation; Coddington, Theorem 2). Let $A + By' = 0$ be exact in a rectangle $R$ , and let $U$ be real-valued with $U_x = A$ and $U_y = B$ throughout $R$ . Then every differentiable function defined implicitly by a relation

U(x,y) = k \qquad (k \text{ constant})

is a solution; and conversely, no solution with graph in $R$ is of any other kind. In short, the solutions in $R$ are exactly the differentiable functions that the level relations of $U$ define implicitly.

Theorem (criterion; Coddington, Theorem 3). Let $A$ and $B$ be real-valued functions with continuous first partial derivatives on a rectangle

R : \ \lvert x - x_0\rvert \le a , \quad \lvert y - y_0 \rvert \le b \qquad (a, b > 0) .

Then $A + By' = 0$ is exact in $R$ if and only if

\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x} \qquad \text{in } R .

Remark(Which form of the criterion we follow)AdaptedAdapted from Coddington (1989), §5.3, Theorem 3, p. 195; Riley, Hobson & Bence (2006), §5.3, p. 156; Arfken, Weber & Harris (2013), §7.2, p. 333

The three sources state the criterion with different amounts of hypothesis, and this unit follows Coddington's version, where both the differentiability assumptions and the shape of the region are named.

Coddington asks for continuous first partial derivatives of $A$ and $B$ on a rectangle and gets an "if and only if".
Riley, Hobson and Bence derive $\partial A/\partial y = \partial B/\partial x$ from the equality of mixed partials and then record that the criterion is sufficient as well as necessary (p. 156), without naming a region on which that holds.
Arfken, Weber and Harris are more cautious in the other direction: they draw only the one-way conclusion, that exactness forces the criterion (p. 333), and obtain the potential afterwards by integration.

The region matters. On a region that is not simply connected the criterion can hold without a single-valued $U$ existing: on $\mathbb{R}^2$ minus the origin,

A = \frac{-y}{x^2+y^2} , \qquad B = \frac{x}{x^2+y^2}

satisfy $A_y = B_x = (y^2 - x^2)/(x^2+y^2)^2$ , yet $A\,dx + B\,dy = d\theta$ and the polar angle $\theta$ gains $2\pi$ around a loop enclosing the origin, so no continuous $U$ exists on the whole punctured plane. This is the same phenomenon as a curl-free field that is not a gradient field (Line integrals and conservative fields); a rectangle is convex, so every loop inside it bounds a region also inside it, and the obstruction cannot arise there. Taking the criterion as sufficient without a hypothesis on the region, as the shortest statements of it do, is therefore a genuine restriction on where it may be applied.

Idea of the proof(The criterion for exactness)Original

One direction is just that mixed partial derivatives commute: if $A$ and $B$ are a gradient's two components then $A_y$ and $B_x$ are both the mixed second derivative of $U$ .

The other direction has to produce $U$ out of nothing. Integrate $A$ in $x$ along the horizontal segment that runs from the rectangle's vertical centre line $x = x_0$ out to $(x,y)$ ; that forces $U_x = A$ but leaves an unknown function of $y$ , which is fixed by integrating $B$ along the centre line itself, from the centre $(x_0,y_0)$ up to $(x_0,y)$ . The resulting $U$ then has $U_y = B$ for one reason only: differentiating the horizontal integral under the integral sign produces $\int A_y$ , the hypothesis replaces $A_y$ by $B_x$ , and the fundamental theorem of calculus turns $\int B_x$ back into $B$ . The rectangle is what guarantees that the two segments used lie inside the region.

Proof(The criterion for exactness, and Theorem 2)AdaptedAdapted from Coddington (1989), §5.3, pp. 193-196, Eq. (3.8)

Necessity. Suppose the equation is exact in $R$ , with $U_x = A$ and $U_y = B$ . By the hypothesis of Theorem 3 the two first partials of $U$ are then $C^1$ on $R$ , so $U$ is $C^2$ there, and the equality of mixed partials for a $C^2$ function (Partial derivatives and the chain rule) gives $A_y = U_{xy} = U_{yx} = B_x$ in $R$ .

Sufficiency. Assume $A_y = B_x$ in $R$ and define, for $(x,y) \in R$ ,

U(x,y) = \int_{x_0}^{x} A(s,y)\,ds + \int_{y_0}^{y} B(x_0,t)\,dt .

Both integrals are defined: $R$ is a rectangle centred at $(x_0,y_0)$ , so the horizontal segment from $(x_0,y)$ to $(x,y)$ and the vertical segment from $(x_0,y_0)$ to $(x_0,y)$ both lie in $R$ , where $A$ and $B$ are continuous.

$U_x = A$ . The second integral does not involve $x$ , and $A(\cdot,y)$ is continuous, so the fundamental theorem of calculus gives $\partial U/\partial x = A(x,y)$ at every point of $R$ , the derivative being the one-sided one at the two vertical edges.
$U_y = B$ , first at interior points. Fix $x$ and let $y$ range over the open interval $\lvert y - y_0\rvert < b$ , which is the form in which the prerequisite is stated: its parameter set must be open, whereas our $y$ runs over a closed interval. On that open interval we may differentiate the first integral with respect to $y$ under the integral sign, because $A$ and $A_y$ are continuous on the product of it with the closed $s$ -interval between $x_0$ and $x$ (Differentiation under the integral sign); the second integral is differentiated by the fundamental theorem. Hence, for $\lvert y - y_0\rvert < b$ ,
$\frac{\partial U}{\partial y} = \int_{x_0}^{x} \frac{\partial A}{\partial y}(s,y)\,ds + B(x_0,y) = \int_{x_0}^{x} \frac{\partial B}{\partial x}(s,y)\,ds + B(x_0,y) = B(x,y) ,$
where the middle equality is the hypothesis $A_y = B_x$ — the only step that uses it — and the last is the fundamental theorem applied to $s \mapsto B(s,y)$ .

The two edges. To reach the top and bottom of $R$ , integrate what has just been proved: for $\lvert y - y_0 \rvert < b$ the function $y \mapsto U(x,y)$ has the continuous derivative $B(x,y)$ , so $U(x,y) = U(x,y_0) + \int_{y_0}^{y}B(x,t)\,dt$ there. Both sides are continuous in $y$ on the closed interval $\lvert y - y_0\rvert \le b$ — the left because $A$ and $B$ are continuous on $R$ , the right because $B$ is — so the identity holds at $y = y_0 \pm b$ as well, and differentiating it with the fundamental theorem gives $\partial U/\partial y = B(x,y)$ throughout $R$ , one-sidedly on the two horizontal edges.
Since $U_x = A$ and $U_y = B$ are continuous, $U$ has continuous first partials, so the equation is exact in $R$ .

(Coddington reaches $U_y = B$ differently: he writes down a second integral formula for $U$ , with the roles of the two variables exchanged, and verifies that it agrees with the first by an identity between two iterated integrals. The computation in step 2 does the same work by differentiating the first formula directly.)

Theorem 2. Let $U_x = A$ , $U_y = B$ in $R$ and let $\varphi$ be differentiable on an interval $I$ with graph in $R$ . The chain rule gives

\frac{d}{dx}\,U\bigl(x,\varphi(x)\bigr) = A\bigl(x,\varphi(x)\bigr) + B\bigl(x,\varphi(x)\bigr)\varphi'(x) \qquad (x \in I) .

Read from right to left: if $\varphi$ solves $A + By' = 0$ on $I$ , the right side vanishes identically, so $x \mapsto U(x,\varphi(x))$ has zero derivative on the interval $I$ and is therefore a constant $k$ ; thus $\varphi$ is defined implicitly by $U = k$ . Read from left to right: if $U(x,\varphi(x)) = k$ on $I$ , the left side vanishes, so $\varphi$ solves the equation. $\square$

Method(Solving an exact equation)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.2, Eqs. (14.7)-(14.8), pp. 472-473; Arfken, Weber & Harris (2013), §7.2, Eq. (7.10), p. 333

Write the equation as $A\,dx + B\,dy = 0$ .
Test. Compute $\partial A/\partial y$ and $\partial B/\partial x$ . If they differ, the equation is inexact; go to the integrating-factor method below.
Integrate in $x$ with $y$ held fixed: $U(x,y) = \int A(x,y)\,dx + f(y)$ , the "constant" of integration being allowed to depend on $y$ .
Fix $f$ : differentiate the result with respect to $y$ , set it equal to $B(x,y)$ , and solve for $f'(y)$ . Every $x$ must cancel at this point; if it does not, step 2 was wrong. Integrate $f'$ .
The solution is $U(x,y) = k$ . Fix $k$ from the initial condition and solve for $y$ if possible.

Arfken, Weber and Harris give the result of steps 3 and 4 in closed form,

U(x,y) = \int_{x_0}^{x} A(s,y)\,ds + \int_{y_0}^{y} B(x_0,t)\,dt ,

which is the function constructed in the proof above; their Eq. (7.10). They also note that separability and exactness are independent properties: every separable equation is exact, but an exact equation need not be separable (p. 333).

Method(Integrating factors)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.3, Eqs. (14.9)-(14.12), pp. 473-474; Coddington (1989), §5.3, Exs. 2, 3, 4 and 5, pp. 198-199, and Theorem 3, p. 195

Standing hypotheses. Throughout this block $A$ and $B$ are real-valued with continuous first partial derivatives on a rectangle

R : \ \lvert x - x_0\rvert \le a , \quad \lvert y - y_0 \rvert \le b \qquad (a, b > 0) ,

which are the conditions Coddington imposes for his integrating-factor exercises (Ex. 3, p. 198), and $\mu$ is a function on $R$ with continuous first partial derivatives and nowhere zero there. Call such a $\mu$ an integrating factor on $R$ for $A\,dx + B\,dy = 0$ if

\mu A\,dx + \mu B\,dy = 0 \quad\text{is exact in } R .

The case of interest is the one in which the original equation is inexact, $\partial A/\partial y \neq \partial B/\partial x$ somewhere in $R$ ; nothing in the definition requires it, and Coddington's definition (Ex. 2, p. 198) does not either.

Because $\mu A$ and $\mu B$ again have continuous first partial derivatives on $R$ , the criterion theorem above applies to them, so on this rectangle exactness of the multiplied equation is equivalent to a computable condition:

\mu \ \text{is an integrating factor on } R \iff \frac{\partial (\mu A)}{\partial y} = \frac{\partial(\mu B)}{\partial x} \quad\text{in } R .

(This is Coddington's Ex. 3, p. 198, which writes the right-hand condition out as $\mu(A_y - B_x) = B\mu_x - A\mu_y$ and refers it to his Theorem 3.) The hypothesis that the region is a rectangle is doing real work in this equivalence: on a region that is not of this kind the criterion can hold with no potential existing, as the remark above shows.

Riley, Hobson and Bence assert that the differential $A\,dx + B\,dy$ can always be made exact in this way, but that when $\mu$ depends on both variables there is no general method for finding it; it must then be found by inspection or trial (p. 473). Two cases are systematic.

A factor of $x$ alone. Suppose $B$ is nowhere zero on $R$ , so that $\alpha = \frac{1}{B}\left(\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x}\right)$ is defined and continuous on $R$ , and suppose $\alpha$ is independent of $y$ . Being continuous, $\alpha$ has an antiderivative, and $\mu(x) = \exp\!\left(\int \alpha(x)\,dx\right)$ — the exponential of any one of them — is an integrating factor on $R$ . (Coddington, Ex. 4(b), p. 199, whose hypothesis is exactly that the test quantity be continuous and independent of $y$ .)
A factor of $y$ alone. Suppose $A$ is nowhere zero on $R$ , so that $\beta = \frac{1}{A}\left(\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}\right)$ is defined and continuous on $R$ , and suppose $\beta$ is independent of $x$ . Then, for the same reason, $\mu(y) = \exp\!\left(\int \beta(y)\,dy\right)$ is an integrating factor on $R$ . (Coddington, Ex. 5(b), p. 199.)

(Riley, Hobson and Bence call these two test quantities $f(x)$ and $g(y)$ ; they are renamed here because $f$ , $g$ and $h$ already have other meanings in the separable-equation blocks above.) Computing $\alpha$ and $\beta$ is therefore also the test for whether a one-variable factor exists at all. Coddington's exercises state the converses: under the same standing hypotheses, if the equation has an integrating factor depending on $x$ alone, then $\alpha$ above is a continuous function of $x$ alone (Ex. 4(a), p. 199), and correspondingly for $y$ (Ex. 5(a)).

Idea of the proof(Integrating factors)Original

Write out the exactness condition for $\mu A\,dx + \mu B\,dy$ and all the unknown-function terms collect into one: a single term $\mu_y A$ on the left and a single term $\mu_x B$ on the right. If $\mu$ is allowed to depend on both variables that leaves one partial differential equation for one unknown function, which is underdetermined — hence infinitely many integrating factors and no procedure. If $\mu$ is restricted to one variable, one of the two terms disappears and what is left is an ordinary separable equation for $\mu$ , solvable exactly when its right-hand side happens to involve that one variable only.

Proof(Integrating factors)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.3, pp. 473-474; Coddington (1989), §5.3, Theorem 3, p. 195, and Exs. 3, 4 and 5, pp. 198-199

Throughout, $A$ , $B$ and $R$ satisfy the standing hypotheses of the statement, so that a $\mu$ with continuous first partials and no zero on $R$ is an integrating factor there precisely when $\partial(\mu A)/\partial y = \partial(\mu B)/\partial x$ on $R$ — the equivalence recorded in the statement, which is the criterion theorem applied to $\mu A$ and $\mu B$ .

Case 1: $\mu = \mu(x)$ . Then $\partial(\mu A)/\partial y = \mu A_y$ , because $\mu$ does not depend on $y$ , while the product rule gives $\partial(\mu B)/\partial x = \mu B_x + \mu' B$ . So $\mu$ is an integrating factor exactly when

\mu A_y = \mu B_x + \mu' B , \qquad\text{that is}\qquad \frac{\mu'}{\mu} = \frac{A_y - B_x}{B} = \alpha ,

the division by $B$ being legitimate because $B$ is nowhere zero on $R$ , and the division by $\mu$ because $\mu$ is not either.

The left side of that equation depends on $x$ alone, so it can have a solution $\mu(x)$ only if $\alpha$ too is independent of $y$ — which is the hypothesis of case 1, and which also proves the converse quoted in the statement: for an integrating factor $\mu(x)$ as above, $(A_y - B_x)/B$ equals $\mu'/\mu$ and is therefore a continuous function of $x$ alone. Granted the hypothesis, $\alpha$ is continuous on $R$ ( $A_y$ , $B_x$ and $B$ are continuous there and $B \neq 0$ ), so it has an antiderivative; $\mu'/\mu = \alpha$ is separable, with solution $\ln\lvert\mu\rvert = \int\alpha$ , that is $\mu = \exp\!\left(\int\alpha\right)$ . This $\mu$ is positive, hence nowhere zero, and continuously differentiable, since its logarithmic derivative $\alpha$ is continuous — so it meets the conditions a factor must meet. Reading the display backwards, it does satisfy the criterion, hence makes the equation exact on $R$ . This is Coddington's Ex. 4 in both directions.

Case 2: $\mu = \mu(y)$ . Identical with the roles of the variables exchanged, under the hypothesis that $A$ is nowhere zero on $R$ and $\beta$ is independent of $x$ : now $\partial(\mu B)/\partial x = \mu B_x$ and $\partial(\mu A)/\partial y = \mu A_y + \mu' A$ , so the criterion reads $\mu'/\mu = (B_x - A_y)/A = \beta$ and $\mu = \exp\!\left(\int \beta\right)$ . $\square$

Remark(What an integrating factor can cost)Original

It must not vanish. Multiplying by $\mu$ does not change the solution set as long as $\mu \neq 0$ , since $A + By' = 0$ and $\mu A + \mu B y' = 0$ then have the same solutions pointwise. Where $\mu$ vanishes the second equation is satisfied automatically and can acquire solutions the first does not have; this is why "nowhere zero" is part of the definition.
It is not unique. If $\mu_1$ and $\mu_2$ are both integrating factors with potentials $U_1$ and $U_2$ , then both families of level curves consist of the same solution curves, so $U_2$ is constant along every curve on which $U_1$ is: locally, each potential is a function of the other. The application below exhibits two different integrating factors for the same differential and checks that relation explicitly.
The tests divide. The expression $\alpha = (A_y - B_x)/B$ is undefined where $B = 0$ , and $\beta$ where $A = 0$ . A one-variable factor may exist and still be invisible to the tests on a region meeting such a curve, so apply the tests on a region where the relevant denominator keeps a fixed sign.

Method(The linear equation)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.4, Eqs. (14.14)-(14.17), pp. 474-475; Arfken, Weber & Harris (2013), §7.2, Eqs. (7.11)-(7.17) and the two theorems, pp. 336-337

A linear first-order equation is one of the form

\frac{dy}{dx} + P(x)\,y = Q(x) ,

with $P$ and $Q$ continuous on an interval $I$ . Fix $x_0 \in I$ and set

\mu(x) = \exp\!\left(\int_{x_0}^{x} P(t)\,dt\right) > 0 .

Then:

Multiplying the equation by $\mu$ turns its left-hand side into the derivative of a single function: $\frac{d}{dx}\bigl[\mu(x)y\bigr] = \mu(x)Q(x) .$ This is what calling $\mu$ an integrating factor means here. (Only continuity of $P$ and $Q$ is assumed, so this is one step short of the statement above, whose standing hypotheses ask for $A = Py - Q$ and $B = 1$ to have continuous first partials and hence for $P$ and $Q$ to be $C^1$ ; when they are, that statement applies verbatim on any rectangle lying over $I$ , since $B \equiv 1$ is nowhere zero and the test quantity is $(A_y - B_x)/B = P$ , a function of $x$ alone. This includes the case $P \equiv 0$ , where the original equation is already exact and $\mu \equiv 1$ — an integrating factor by that definition, which does not ask the equation to be inexact.)
The general solution on $I$ is $y(x) = \frac{1}{\mu(x)}\left[C + \int_{x_0}^{x}\mu(t)Q(t)\,dt\right] , \qquad C \text{ an arbitrary constant}.$
Through every point $(x_1, y_1)$ of the strip $x_1 \in I$ , $y_1 \in \mathbb{R}$ there passes exactly one solution, and that solution is defined on the whole of $I$ — not merely near $x_1$ .
(Arfken, Weber and Harris, p. 337.) The general solution of the inhomogeneous equation is any one particular solution plus the general solution $C/\mu(x)$ of the homogeneous equation $y' + Py = 0$ ; equivalently, the solution is unique up to an arbitrary multiple of the homogeneous solution. The homogeneous equation has only one linearly independent solution.

Idea of the proof(The linear equation)Original

In differential form the equation is $\bigl(P(x)y - Q(x)\bigr)dx + dy = 0$ , so $B \equiv 1$ and the test quantity $(A_y - B_x)/B$ equals $P(x)$ — automatically a function of $x$ alone. The linear equation is thus the one case where the one-variable recipe never fails, and its integrating factor can be written down from the coefficient. After multiplying, the whole left side is the derivative of the single product $\mu y$ , so one integration finishes the problem. Because the integrals involved exist across the entire interval of continuity, nothing stops the solution part-way: linear equations have no finite-time breakdown.

Proof(The linear equation)AdaptedAdapted from Arfken, Weber & Harris (2013), §7.2, Eqs. (7.12)-(7.17) and p. 337; Riley, Hobson & Bence (2006), §14.2.4, pp. 474-475

What the factor must satisfy. Let $\mu$ be differentiable and positive on $I$ . By the product rule,
$\frac{d}{dx}\bigl[\mu y\bigr] = \mu y' + \mu' y ,$
while multiplying the equation by $\mu$ gives $\mu y' + \mu P y = \mu Q$ . The two left sides agree for every differentiable $y$ precisely when $\mu' = \mu P$ . So the search for an integrating factor is the search for a solution of that one equation; this is Arfken's route, and it bypasses the general theory above.
Solving for the factor. $\mu' = \mu P$ is separable, and $P$ is continuous on $I$ , so $t \mapsto P(t)$ has an antiderivative there and $\mu(x) = \exp\bigl(\int_{x_0}^{x}P\bigr)$ is a solution with $\mu(x_0) = 1$ . Being an exponential it is positive on all of $I$ , so dividing by it later is always legal. (Any constant multiple of this $\mu$ serves equally well; the constant cancels in step 3, which is why the constant of integration in $\int P$ is conventionally dropped.)
One integration. With that $\mu$ , the equation is equivalent to $\bigl[\mu y\bigr]' = \mu Q$ on $I$ . The right side is continuous, hence integrable, and integrating from $x_0$ to $x$ ,
$\mu(x)y(x) - y(x_0) = \int_{x_0}^{x}\mu(t)Q(t)\,dt ,$
using $\mu(x_0) = 1$ . Writing $C = y(x_0)$ and dividing by $\mu(x) > 0$ gives the formula in part 2. Conversely, differentiating that formula returns $\bigl[\mu y\bigr]' = \mu Q$ , so every member of the family is a solution: the family is exactly the set of solutions, which is what part 1 and part 2 assert.
Existence and uniqueness on the whole interval. Given $(x_1,y_1)$ with $x_1 \in I$ , part 3 shows that a solution with $y(x_1) = y_1$ must have $C = \mu(x_1)y_1 - \int_{x_0}^{x_1}\mu Q$ , one value and no other, and that this choice does produce such a solution. Its formula is defined and differentiable at every point of $I$ , because $\mu$ is positive and continuous there and $\mu Q$ is integrable on every subinterval. Hence exactly one solution, on all of $I$ . For a nonlinear equation neither conclusion holds in general: see Existence and uniqueness of solutions.
The structure statements. If $u$ and $w$ both solve the inhomogeneous equation, subtracting the two equations gives $(u-w)' + P(u-w) = 0$ , so the difference solves the homogeneous equation; and by part 2 with $Q \equiv 0$ the homogeneous solutions are exactly the multiples $C/\mu$ . Two of them, $C_1/\mu$ and $C_2/\mu$ , are proportional, so the homogeneous equation has a one-dimensional solution space: at most one linearly independent solution, Arfken's second theorem. $\square$

Method(Solving a linear equation)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.4, p. 475

Divide through by the coefficient of $y'$ to reach $y' + P(x)y = Q(x)$ , and note the interval $I$ on which $P$ and $Q$ are continuous — in particular, exclude the points where the original coefficient of $y'$ vanishes.
Compute $\mu(x) = \exp\!\left(\int P(x)\,dx\right)$ , dropping the constant of integration.
Multiply by $\mu$ and recognise the left side as $(\mu y)'$ .
Integrate once and divide by $\mu$ : $y = \mu^{-1}\bigl[C + \int \mu Q\bigr]$ . Fix $C$ from the initial condition.

Note(The same theorem without normalising)AdaptedAdapted from Hassani (2013), §14.2, Theorem 14.2.1, Eqs. (14.7)-(14.8), p. 420

Hassani states the solution formula for the unnormalised equation $p_1(x)y' + p_0(x)y = q(x)$ , requiring $p_0$ , $p_1$ and $q$ to be continuous on an open interval (his $(a,b)$ , which has nothing to do with the rectangle dimensions $a$ , $b$ used in the exactness criterion above), and obtains the general solution

y = \frac{1}{\mu(x)p_1(x)}\left[C + \int^{x}_{x_1}\mu(t)q(t)\,dt\right] , \qquad \mu(x) = \frac{1}{p_1(x)}\exp\!\left(\int_{x_0}^{x}\frac{p_0(t)}{p_1(t)}\,dt\right) .

This is the same result as part 2 above, with a differently scaled integrating factor: his $\mu$ carries an extra factor $1/p_1$ , so his $\mu p_1$ plays the role of our $\mu$ . His theorem is stated with no separate hypothesis that $p_1$ is non-vanishing, although the formulas divide by $p_1$ ; a point where $p_1$ vanishes is a singular point of the equation, and there step 1 of the method above cannot be carried out. We use the normalised form throughout.

Method(Homogeneous equations)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.5, Eqs. (14.18)-(14.19), pp. 475-476; Arfken, Weber & Harris (2013), §7.2, p. 334; Coddington (1989), §5.2, Ex. 4, p. 191

A function $w$ is homogeneous of degree $n$ if $w(\lambda x, \lambda y) = \lambda^{n}w(x,y)$ for all $\lambda$ and all $x,y$ . The equation $y' = F(x,y)$ is called homogeneous when its right-hand side can be written as a quotient

F(x,y) = \frac{N(x,y)}{D(x,y)}

in which $N$ and $D$ are homogeneous of one and the same degree $n$ . (The name clashes with "homogeneous" as applied to a linear equation with zero right-hand side; Arfken, Weber and Harris (p. 334) and Coddington (p. 191) both warn about the collision, which is standard and unavoidable. Riley, Hobson and Bence write the quotient as $A/B$ , with the opposite sign convention to the differential form $A\,dx + B\,dy = 0$ of the definition block above, where $F = -A/B$ ; the neutral letters $N$ and $D$ are used here to keep the two apart.) Riley, Hobson and Bence point out the practical test: in each term of the numerator and of the denominator, the powers of $x$ and $y$ must add up to the same number.

Why the substitution works. With $N$ and $D$ homogeneous of degree $n$ , the quotient $F = N/D$ satisfies $F(\lambda x, \lambda y) = F(x,y)$ : it is homogeneous of degree $0$ . Taking $\lambda = 1/x$ on the half-plane $x > 0$ ,

F(x,y) = F\!\left(1, \frac{y}{x}\right) ,

so the right-hand side depends on $x$ and $y$ only through the ratio $v = y/x$ ; write it $F(v)$ , reusing the letter for the one-variable function. (For $x<0$ take $\lambda = -1/x$ and repeat with $F(-1,\cdot)$ .) Substituting $y = vx$ , so that $y' = v + xv'$ ,

v + x\frac{dv}{dx} = F(v) \qquad\Longrightarrow\qquad \frac{dv}{F(v) - v} = \frac{dx}{x} ,

which has separated variables.

Recipe.

Check that every term of the numerator and of the denominator has the same total degree.
Record the roots of $F(v) - v$ : each root $v_\ast$ gives a straight-line solution $y = v_\ast x$ through the origin, and these are the solutions the next step divides away.
Substitute $y = vx$ , separate as displayed, and integrate both sides.
Replace $v$ by $y/x$ .

Method(Bernoulli's equation)AdaptedAdapted from Riley, Hobson & Bence (2006), §14.2.7, Eq. (14.21), pp. 477-478

Bernoulli's equation is

\frac{dy}{dx} + P(x)\,y = Q(x)\,y^{n} , \qquad n \neq 0, 1 .

The two excluded values are the ones for which it is already linear. The extra factor $y^n$ makes it nonlinear, but the substitution

v = y^{1-n}

removes the nonlinearity. Indeed $v' = (1-n)y^{-n}y'$ , so dividing the equation by $y^{n}$ and substituting,

y^{-n}y' + P\,y^{1-n} = Q \qquad\Longrightarrow\qquad \frac{1}{1-n}\,v' + P\,v = Q \qquad\Longrightarrow\qquad \frac{dv}{dx} + (1-n)P(x)\,v = (1-n)Q(x) ,

a linear equation in $v$ , to be solved by the method above and then inverted through $y = v^{1/(1-n)}$ .

Recipe.

Arrange the equation in the displayed form and read off $n$ .
Note the division by $y^{n}$ : for $n > 0$ the function $y \equiv 0$ solves the original equation and is lost, and for non-integer $n$ the work is confined to $y > 0$ , where $y^{n}$ and $y^{1-n}$ are defined.
Solve the linear equation for $v = y^{1-n}$ .
Return to $y$ and impose the initial condition.

Example(Separation, with the lost solutions as asymptotes)Original

Solve $\displaystyle \frac{dy}{dx} = x\,(y^{2}-1)$ with $y(0) = 0$ .

Step 1: the constant solutions. The right-hand side factors as $f(x)\,\sigma(y)$ with $f(x) = x$ and $\sigma(y) = y^2-1$ , so the equation is separable. The roots of $\sigma$ are $y = 1$ and $y = -1$ , giving the two constant solutions $y \equiv 1$ and $y \equiv -1$ . Neither satisfies the initial condition, but they are recorded before dividing.

Step 2: separate and integrate. On the strip $-1 < y < 1$ , where $\sigma \neq 0$ ,

\frac{dy}{y^{2}-1} = x\,dx , \qquad \frac{1}{y^2-1} = \frac{1}{2}\left(\frac{1}{y-1} - \frac{1}{y+1}\right) ,

so integrating both sides gives $\tfrac12 \ln\left\lvert\dfrac{y-1}{y+1}\right\rvert = \dfrac{x^{2}}{2} + c$ , that is

\frac{y-1}{y+1} = \Lambda\, e^{x^{2}} , \qquad \Lambda = \pm e^{2c} \text{ a non-zero constant}.

Step 3: the initial condition. Putting $x = 0$ , $y = 0$ gives $-1 = \Lambda$ . Hence $y - 1 = -e^{x^2}(y+1)$ , and solving for $y$ ,

y(x) = \frac{1 - e^{x^{2}}}{1 + e^{x^{2}}} = -\tanh\!\left(\frac{x^{2}}{2}\right) ,

the last step by multiplying numerator and denominator by $e^{-x^2/2}$ .

Step 4: check. Differentiating, $y' = -x\,\operatorname{sech}^2(x^2/2)$ , while $x(y^2-1) = x\bigl(\tanh^2(x^2/2) - 1\bigr) = -x\operatorname{sech}^2(x^2/2)$ . They agree, and $y(0) = -\tanh 0 = 0$ . Numerically $y(1) = -\tanh(1/2) = -0.4621171573$ and $y(2) = -\tanh 2 = -0.9640275801$ .

Step 5: where the lost solutions went. The solution exists for all real $x$ and satisfies $y(x) \to -1$ as $x \to \pm\infty$ , approaching the constant solution $y \equiv -1$ discarded in step 2. The other discarded solution, $y \equiv 1$ , is the repelling one: any initial value $y(0) > 1$ gives, by the same formula with $\Lambda > 0$ , a solution that blows up at a finite $x$ . The two constant solutions are therefore the organising features of the picture, which is the practical reason for step 2 of the method.

Example(An inexact equation solved with an integrating factor)Original

Solve $\bigl(2y^{2} + 2x + 1\bigr)\,dx + 2y\,dy = 0$ with $y(0) = 1$ , on the half-plane $y > 0$ .

Step 1: it is inexact. With $A = 2y^2 + 2x + 1$ and $B = 2y$ ,

\frac{\partial A}{\partial y} = 4y , \qquad \frac{\partial B}{\partial x} = 0 ,

which differ wherever $y \neq 0$ , so the equation is not exact on any rectangle in $y > 0$ . Nor is it separable, since $y' = -(2y^2+2x+1)/(2y)$ does not factor, and it is not linear in $y$ .

Step 2: a factor of $x$ alone. Take as the region a rectangle $R$ inside $y > 0$ , where $B = 2y$ is nowhere zero, so that the test quantity of case 1 is defined:

\alpha = \frac{1}{B}\left(\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x}\right) = \frac{4y - 0}{2y} = 2 ,

a constant, hence certainly a continuous function independent of $y$ . Case 1 therefore applies and

\mu(x) = \exp\!\left(\int 2\,dx\right) = e^{2x}

is an integrating factor on $R$ ; it is positive, hence nowhere zero, as required.

Step 3: the multiplied equation is exact. Multiplying through by $e^{2x}$ ,

e^{2x}\bigl(2y^{2} + 2x + 1\bigr)dx + 2y\,e^{2x}\,dy = 0 , \qquad \frac{\partial}{\partial y}\Bigl[e^{2x}(2y^2+2x+1)\Bigr] = 4y\,e^{2x} = \frac{\partial}{\partial x}\bigl(2y\,e^{2x}\bigr) ,

confirming exactness.

Step 4: the potential. Integrating the $dx$ coefficient in $x$ at fixed $y$ , with $\int 2x\,e^{2x}dx = x e^{2x} - \tfrac12 e^{2x}$ ,

U(x,y) = \tfrac12 (2y^2+1)e^{2x} + x e^{2x} - \tfrac12 e^{2x} + h(y) = e^{2x}\bigl(y^{2} + x\bigr) + h(y) .

Differentiating in $y$ gives $2y\,e^{2x} + h'(y)$ , which must equal $2y\,e^{2x}$ ; so $h' = 0$ and $h$ is a constant, absorbed into $k$ . The solution is therefore

e^{2x}\bigl(y^{2}+x\bigr) = C .

Step 5: the initial condition and the answer. At $(0,1)$ the left side is $1$ , so $C = 1$ and

y(x) = \sqrt{e^{-2x} - x} \qquad (y > 0) .

Check: $y(0) = 1$ ; and differentiating, $2yy' = -2e^{-2x} - 1 = -2(y^2 + x) - 1 = -(2y^2+2x+1)$ , which is the equation. Numerically $y(0.2) = \sqrt{e^{-0.4}-0.2} = 0.6857988379$ .

An independent check. The substitution $u = y^2$ turns the equation into $u' + 2u = -(2x+1)$ , a linear equation whose integrating factor is the same $e^{2x}$ ; since $\int (2x+1)e^{2x}dx = x e^{2x}$ , it gives $e^{2x}u = -xe^{2x} + C$ , that is $u = y^2 = Ce^{-2x} - x$ , and $u(0)=1$ fixes $C = 1$ again.

Step 6: the interval. The expression under the root is positive for every $x \le 0$ and vanishes where $e^{-2x} = x$ , that is at $x_\ast = 0.4263027510$ ; so the solution exists on $-\infty < x < x_\ast$ , with $y \to 0$ at the right end and $y \sim e^{-x} \to \infty$ as $x \to -\infty$ . At $x_\ast$ the curve has $dy/dx = -(2y^2+2x+1)/(2y) \to -\infty$ : it meets the line $y = 0$ with a vertical tangent, which is what the differential form predicts, since at $y = 0$ the equation reads $(2x+1)\,dx = 0$ and so forces $dx = 0$ as long as $x \neq -\tfrac12$ . The level curve $e^{2x}(y^2+x) = 1$ continues smoothly into $y < 0$ , but the solution as a function $y = \varphi(x)$ with $\varphi > 0$ ends at $x_\ast$ — which is also where the rectangle of step 2, which had to stay inside $y > 0$ , runs out.

Example(A linear equation: a driven relaxation)Original

A body with a single thermal time constant exchanges heat with surroundings at temperature $T_{\mathrm{out}}(t)$ according to Newton's law of cooling,

\frac{dT}{dt} = -\kappa\bigl(T - T_{\mathrm{out}}(t)\bigr) , \qquad \kappa > 0 .

Take $\kappa = 0.25\ \mathrm{h}^{-1}$ , a daily outdoor cycle $T_{\mathrm{out}}(t) = 10 + 8\cos(\omega t)\ {}^{\circ}\mathrm{C}$ with $\omega = 2\pi/24 = 0.2617993878\ \mathrm{h}^{-1}$ , and an initial temperature $T(0) = 20\,^{\circ}\mathrm{C}$ .

Step 1: put it in linear form. $\dfrac{dT}{dt} + \kappa T = \kappa\bigl(10 + 8\cos\omega t\bigr)$ , so $P \equiv \kappa$ and $Q(t) = \kappa(10 + 8\cos\omega t)$ , both continuous on all of $\mathbb{R}$ . Part 3 of the linear theorem therefore gives a unique solution defined for all $t$ .

Step 2: the integrating factor. $\mu(t) = \exp\!\left(\int \kappa\,dt\right) = e^{\kappa t}$ , and the equation becomes $\bigl(e^{\kappa t}T\bigr)' = \kappa\,e^{\kappa t}\bigl(10 + 8\cos\omega t\bigr)$ .

Step 3: integrate. With $\int e^{\kappa t}\cos\omega t\,dt = e^{\kappa t}\,\dfrac{\kappa\cos\omega t + \omega\sin\omega t}{\kappa^{2}+\omega^{2}}$ (check by differentiating),

e^{\kappa t}T = 10\,e^{\kappa t} + \frac{8\kappa\bigl(\kappa\cos\omega t + \omega \sin\omega t\bigr)}{\kappa^{2}+\omega^{2}}\,e^{\kappa t} + C ,

so, writing the bracket as a single cosine with a phase,

T(t) = 10 + \frac{8\kappa}{\sqrt{\kappa^{2}+\omega^{2}}}\cos(\omega t - \phi) + C e^{-\kappa t} , \qquad \tan\phi = \frac{\omega}{\kappa} .

Step 4: the numbers. Here $\omega/\kappa = 1.047197551$ , so

\phi = \arctan(1.047197551) = 0.8084487926\ \mathrm{rad} , \qquad \frac{\kappa}{\sqrt{\kappa^2+\omega^2}} = 0.6906211226 ,

giving a driven amplitude $8 \times 0.6906211226 = 5.524968980\,^{\circ}\mathrm{C}$ . The initial condition fixes $C = 20 - 10 - 5.524968980\cos\phi = 6.184339721$ .

Step 5: read the answer. After the transient $Ce^{-\kappa t}$ has died away (a few multiples of $1/\kappa = 4$ h), the body oscillates about $10\,^{\circ}\mathrm{C}$ with amplitude $5.52\,^{\circ}\mathrm{C}$ instead of the outdoor $8\,^{\circ}\mathrm{C}$ , and its extremes lag the outdoor ones by

\frac{\phi}{\omega} = \frac{0.8084487926}{0.2617993878} = 3.088046918\ \mathrm{h} .

Checks on the formula: $T(0) = 20.00000000$ as imposed; $T(6) = 15.37566281$ , $T(12) = 6.492239865$ , and after two days $T(48) = 13.81569828$ . Since $48\omega = 4\pi$ , the purely periodic part there is $10 + 5.524968980\cos(-\phi) = 13.81566028$ , and the gap between the two is $3.79979\times10^{-5}$ — exactly the surviving transient $Ce^{-12}$ , as it must be.

Example(A homogeneous equation: a logarithmic spiral)Original

Solve $\displaystyle \frac{dy}{dx} = \frac{x + 2y}{2x - y}$ with $y(1) = 0$ .

Step 1: degrees. Every term of the numerator and of the denominator is of total degree $1$ , so the equation is homogeneous of degree $1$ and the substitution $y = vx$ applies on $x > 0$ .

Step 2: the straight-line solutions. Here $F(v) = (1+2v)/(2-v)$ , and

F(v) - v = \frac{1 + 2v - v(2-v)}{2-v} = \frac{1 + v^{2}}{2 - v} ,

whose numerator never vanishes. So there are no straight-line solutions to lose; the division in the next step costs nothing. (At $v = 2$ , that is on the line $y = 2x$ , the original right-hand side is undefined.)

Step 3: separate and integrate. With $y = vx$ ,

x\frac{dv}{dx} = \frac{1+v^{2}}{2-v} \qquad\Longrightarrow\qquad \frac{2-v}{1+v^{2}}\,dv = \frac{dx}{x} ,

and integrating each side,

2\arctan v - \tfrac12 \ln\bigl(1+v^{2}\bigr) = \ln x + c .

Step 4: the initial condition. At $x = 1$ , $y = 0$ we have $v = 0$ , and both sides vanish, so $c = 0$ .

Step 5: recognise the curve. Put $v = y/x = \tan\theta$ and $r = \sqrt{x^2+y^2}$ , so that $1 + v^2 = \sec^2\theta$ and $x = r\cos\theta$ . Then $\tfrac12\ln(1+v^2) = -\ln\cos\theta$ and the relation becomes

2\theta + \ln\cos\theta = \ln x = \ln r + \ln\cos\theta , \qquad\text{that is}\qquad r = e^{2\theta} :

a logarithmic spiral through $(1,0)$ . Check: on $r = e^{2\theta}$ one has $dr/d\theta = 2r$ , so

\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} = \frac{2\sin\theta+\cos\theta}{2\cos\theta - \sin\theta} = \frac{r\cos\theta + 2r\sin\theta}{2r\cos\theta - r\sin\theta} = \frac{x+2y}{2x-y} ,

as required. At $\theta = \pi/4$ the curve is at $r = e^{\pi/2} = 4.810477381$ , that is $x = y = 3.401521177$ , and $2\arctan(y/x) = \pi/2 = \ln r$ confirms the implicit relation.

Step 6: where $y$ stops being a function of $x$ . The denominator $2x - y$ vanishes when $\tan\theta = 2$ , that is at $\theta = \arctan 2 = 1.107148718$ , where $r = 9.154974892$ and $x$ attains its largest value $4.094229238$ on this arc. The spiral continues happily past that point, but as a solution of the equation in the form $y = \varphi(x)$ the answer ends there, with a vertical tangent — the situation anticipated in the definition block.

Example(A Bernoulli equation: logistic growth)Original

Solve $\displaystyle \frac{dy}{dt} = a y - \frac{a}{K}y^{2}$ with $a = 0.5\ \mathrm{yr}^{-1}$ , $K = 1000$ and $y(0) = 50$ .

Step 1: identify it. Written as $y' - ay = -(a/K)y^{2}$ this is Bernoulli's equation with $P = -a$ , $Q = -a/K$ and $n = 2$ , so $n \neq 0,1$ and the substitution is $v = y^{1-n} = y^{-1}$ . Since $n = 2 > 0$ , the function $y \equiv 0$ solves the original equation and will be lost; it is the extinct state, and the initial value $50$ is not it.

Step 2: linearise. With $v = 1/y$ one has $v' = -y'/y^{2}$ , so dividing the equation by $y^{2}$ and substituting,

\frac{y'}{y^{2}} = \frac{a}{y} - \frac{a}{K} \qquad\Longrightarrow\qquad -v' = av - \frac{a}{K} \qquad\Longrightarrow\qquad \frac{dv}{dt} + a v = \frac{a}{K} ,

which is the general Bernoulli reduction $v' + (1-n)Pv = (1-n)Q$ with $1-n = -1$ .

Step 3: solve the linear equation. The integrating factor is $\mu = e^{at}$ , so $\bigl(e^{at}v\bigr)' = (a/K)e^{at}$ and

e^{at}v = \frac{e^{at}}{K} + C , \qquad v(t) = \frac{1}{K} + C\,e^{-at} .

Step 4: back to $y$ , with the initial condition. $v(0) = 1/50$ gives $C = 1/50 - 1/1000 = 19/1000$ , so

y(t) = \frac{1}{v(t)} = \frac{K}{1 + 19\,e^{-at}} = \frac{1000}{1 + 19\,e^{-t/2}} .

Step 5: check the numbers. Substituting back into $y' = ay(1 - y/K)$ gives zero identically. The values are $y(0) = 50$ , $y(5) = 390.6837592$ , $y(10) = 886.5083241$ and $y(20) = 999.1381448$ ; the population passes half the carrying capacity at $t = \ln(19)/a = 5.888877958$ yr, and $y \to K = 1000$ as $t \to \infty$ — approaching the other constant solution, $y \equiv K$ , which separation would have discarded at the same point in the argument. Note that the equation is separable too; the Bernoulli route is preferable here because it replaces a partial-fraction integration by a linear equation that can be solved in one line.

Application(Integrating factors in thermodynamics: entropy and the adiabatic invariant)Original

The clearest physical appearance of these methods is not as a way of solving an equation but as a way of recognising a state function. For a simple compressible system the first law says that the heat absorbed along a path is

\delta Q = dE + p\,dV ,

where $E$ is the internal energy (written $U$ in most thermodynamics texts, and renamed here because $U$ already denotes the potential of an exact differential). The central fact of thermodynamics is that $\delta Q$ is not the differential of any function of state: heat absorbed depends on the path, not only on its endpoints. In the language above, $\delta Q$ is an inexact differential, and the content of the second law is that a particular integrating factor makes it exact.

The ideal gas, explicitly. Take a fixed amount of ideal gas, $n$ moles, with constant heat capacity $C_V$ , so that $E = C_V T$ and $p = nRT/V$ , and use $(T,V)$ as the two state variables. Then

\delta Q = C_V\,dT + \frac{nRT}{V}\,dV ,

so in the notation of this unit $A = C_V$ and $B = nRT/V$ with $x = T$ , $y = V$ . The exactness test fails:

\frac{\partial A}{\partial V} = 0 , \qquad \frac{\partial B}{\partial T} = \frac{nR}{V} \neq 0 .

So no function of the state $(T,V)$ has $\delta Q$ as its differential, which is the precise sense in which "the heat content of a state" does not exist.

Both one-variable tests succeed. This differential is unusual in that the two tests of the integrating-factor statement both apply. With $x = T$ and $y = V$ ,

\alpha = \frac{1}{B}\left(\frac{\partial A}{\partial V} - \frac{\partial B}{\partial T}\right) = \frac{V}{nRT}\left(0 - \frac{nR}{V}\right) = -\frac{1}{T} , \qquad \beta = \frac{1}{A}\left(\frac{\partial B}{\partial T} - \frac{\partial A}{\partial V}\right) = \frac{nR}{C_V V} ,

the first a function of $T$ alone and the second a function of $V$ alone. Case 1 therefore gives $\mu = \exp\!\left(\int(-1/T)\,dT\right) = 1/T$ and case 2 gives $\mu = \exp\!\left(\int nR/(C_VV)\,dV\right) = V^{nR/C_V}$ . The two factors compute two different things, and both are physically familiar.

First integrating factor: $1/T$ . Multiplying by $1/T$ gives $(C_V/T)\,dT + (nR/V)\,dV$ , and now

\frac{\partial}{\partial V}\left(\frac{C_V}{T}\right) = 0 = \frac{\partial}{\partial T}\left(\frac{nR}{V}\right) ,

so the multiplied differential is exact, with potential

S(T,V) = C_V\ln T + nR\ln V ,

the entropy of the ideal gas, up to the additive constant that no potential of a differential ever fixes; we set that constant to zero, which is the choice used below. One checks directly that $\partial S/\partial T = C_V/T$ and $\partial S/\partial V = nR/V$ .

Second integrating factor: $V^{nR/C_V}$ , and what it computes. Following case 2 with the $\beta$ just computed,

\mu(V) = \exp\!\left(\int \frac{nR}{C_V V}\,dV\right) = V^{nR/C_V} = V^{\gamma-1} ,

using $nR = C_p - C_V$ and $\gamma = C_p/C_V$ , so that $nR/C_V = \gamma - 1$ . Multiplying $\delta Q$ by it gives $C_V V^{\gamma-1}dT + nR\,T\,V^{\gamma-2}dV$ , whose two mixed derivatives are both $nRV^{\gamma-2}$ , and whose potential is

\Phi(T,V) = C_V\,T\,V^{\gamma-1} .

The level curves $\Phi = \text{const}$ are $TV^{\gamma-1} = \text{const}$ : the adiabats of the ideal gas. This is exactly what the theory predicts, in two ways. First, $\delta Q = 0$ is a first-order equation in the differential form $A\,dT + B\,dV = 0$ , and once an integrating factor is found, its potential is constant along the solution curves — so an adiabatic process is a solution curve of $\delta Q = 0$ , and $TV^{\gamma-1}$ is the quantity conserved along it. Second, the remark on non-uniqueness above says that two potentials of the same differential must be functions of one another, and indeed

\Phi = C_V\,T\,V^{\gamma-1} = C_V\exp\!\left(\frac{C_V\ln T + nR\ln V}{C_V}\right) = C_V\,e^{S/C_V} ,

with $S$ normalised as above; a different additive constant $S_0$ in $S$ only replaces this by $\Phi = C_V e^{(S-S_0)/C_V}$ . Either way $\Phi$ is a function of $S$ alone, as it must be: the adiabats are the curves of constant entropy.

What the mathematics does and does not deliver. Riley, Hobson and Bence's assertion that an inexact differential in two variables can always be made exact (p. 473) is what makes a function like $S$ available at all for a system with two independent state variables; it says nothing about which factor to choose, and here it offers two with equal ease. The physical content of the second law is that one of them, $1/T$ with $T$ the absolute temperature, works simultaneously for every system and for systems with more state variables than two, where the mere existence of an integrating factor is no longer automatic. The first-order theory supplies the framework and the bookkeeping; the universality of $1/T$ is physics.

A second, more mundane use of the same machinery is the driven relaxation of the third example: Newton's law of cooling, an $RC$ or $RL$ circuit, a first-order sensor, and the linearised response of any system with a single relaxation time all have the form $y' + y/\tau = u(t)/\tau$ , where $u$ is the input — the ambient temperature, the applied voltage, the true gas temperature. In the notation of the linear-equation statement above this is $P \equiv 1/\tau$ and $Q = u/\tau$ , the whole right-hand side. The integrating factor $e^{t/\tau}$ turns it into one integral, and the answer always has the same two parts — a transient $\propto e^{-t/\tau}$ that remembers the initial condition, and a driven response that does not. For sinusoidal forcing at angular frequency $\omega$ the driven part has its amplitude reduced by the factor $(1 + \omega^2\tau^2)^{-1/2}$ and is delayed by $\arctan(\omega\tau)/\omega$ : a first-order low-pass filter. For a thermocouple with $\tau = 4\ \mathrm{s}$ reading a gas temperature that oscillates at $0.1\ \mathrm{Hz}$ , $\omega\tau = 2.513274123$ , so the recorded swing is only $0.3696978476$ of the true swing and arrives $1.897306001\ \mathrm{s}$ late out of a $10\ \mathrm{s}$ period. Reading those two numbers backwards is how such an instrument is calibrated.

Theorem

Existence and uniqueness of solutions

Adapted

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Proves that an initial-value problem $y' = f(x,y)$ , $y(x_0)=y_0$ has exactly one solution on an explicit interval around $x_0$ when $f$ is continuous and Lipschitz in $y$ , by showing that Picard's successive approximations converge uniformly, and shows by example what fails when the Lipschitz condition is dropped.

Prerequisites: Uniform convergence, Compactness, Solving first-order equations

Note(What is at stake)Original

Almost no differential equation of practical interest can be solved in closed form, so almost everything done with one — drawing its phase portrait, integrating it numerically, expanding it in a small parameter, arguing from its symmetries — presupposes that "the solution" exists and that the phrase refers to one object. That presupposition is a theorem, and it is not free: it needs a hypothesis slightly stronger than continuity, and it delivers less than one might hope, namely a solution on an interval that may be much shorter than the one of interest. This unit gives the hypothesis, the interval, the proof, and the two examples that show why nothing weaker will do.

Definition(The initial-value problem, Picard's approximations, and the Lipschitz condition)AdaptedAdapted from Coddington (1989), §5.4, pp. 200-202, and §5.5, p. 208; Hassani (2013), §14.2, Definition 14.2.3, p. 420

Let $f$ be a real-valued function defined on a rectangle

R : \ \lvert x - x_0 \rvert \le a , \quad \lvert y - y_0 \rvert \le b \qquad (a, b > 0) .

Given an interval $I$ that contains $x_0$ , a solution on $I$ of the initial-value problem

y' = f(x,y) , \qquad y(x_0) = y_0

means a differentiable real-valued $\varphi$ on $I$ with $\varphi(x_0) = y_0$ , with $(x,\varphi(x)) \in R$ for every $x \in I$ , and with $\varphi'(x) = f(x,\varphi(x))$ for every $x \in I$ .

The integral equation attached to the problem is

\varphi(x) = y_0 + \int_{x_0}^{x} f\bigl(t, \varphi(t)\bigr)\,dt ,

a solution of which means a continuous $\varphi$ on $I$ whose graph lies in $R$ and which satisfies the displayed identity at every $x \in I$ . The successive approximations, or Picard iterates, are the functions

\varphi_0(x) = y_0 , \qquad \varphi_{k+1}(x) = y_0 + \int_{x_0}^{x} f\bigl(t, \varphi_k(t)\bigr)\,dt \qquad (k = 0, 1, 2, \dots) .

Each one is obtained by feeding the previous one through the right-hand side of the integral equation; a solution of that equation is exactly a function left unchanged by the process.

Finally, let $f$ be defined on a set $S$ . Then $f$ satisfies a Lipschitz condition on $S$ if there is a constant $K > 0$ with

\bigl\lvert f(x,y_1) - f(x,y_2) \bigr\rvert \le K\,\lvert y_1 - y_2 \rvert \qquad \text{for all } (x,y_1), (x,y_2) \in S ,

and $K$ is called a Lipschitz constant. The condition constrains $f$ in its second argument only, uniformly in the first; it says nothing about how $f$ varies with $x$ beyond the continuity assumed separately.

Theorem(Picard-Lindelöf: local existence, uniqueness and an error bound)AdaptedAdapted from Coddington (1989), §5.4, Theorem 5, p. 203, §5.6, Theorems 7 and 8, pp. 210, 214, and §5.8, Theorem 10 with Corollary 1, pp. 222-223

Let the real-valued $f$ be continuous on the rectangle

R : \ \lvert x - x_0 \rvert \le a , \quad \lvert y - y_0 \rvert \le b \qquad (a, b > 0) ,

let

\lvert f(x,y)\rvert \le M \quad\text{for all } (x,y) \in R , \qquad M > 0 ,

and suppose $f$ satisfies a Lipschitz condition with constant $K$ in $R$ . Put

\alpha = \min\{\,a,\ b/M\,\} , \qquad I : \ \lvert x - x_0 \rvert \le \alpha .

Then:

(Existence; Coddington, Theorems 5 and 7.) The successive approximations $\varphi_k$ all exist as continuous functions on $I$ and their graphs lie in $R$ (his Theorem 5, p. 203), and on $I$ they converge to a limit $\varphi$ which solves $y' = f(x,y)$ , $y(x_0) = y_0$ there (his Theorem 7, p. 210). The convergence is moreover uniform on $I$ : Coddington's statement says only "converge", but his proof and ours both produce a bound on $\lvert \varphi - \varphi_k\rvert$ that is independent of $x$ , and step 4 of the proof below uses the uniformity.
(Uniqueness; Coddington, Theorem 10 and its Corollary 1.) If $\varphi$ and $\psi$ are two solutions of that initial-value problem on an interval $I'$ containing $x_0$ , with graphs contained in $R$ , then $\varphi(x) = \psi(x)$ for every $x \in I'$ . In particular the solution produced in part 1 is the only one on $I$ .
(Error bound; Coddington, Theorem 8.) The $k$ th approximation satisfies $\bigl\lvert \varphi(x) - \varphi_k(x) \bigr\rvert \ \le \ \frac{M}{K}\,\frac{(K\alpha)^{k+1}}{(k+1)!}\,e^{K\alpha} \qquad \text{for all } x \in I .$

The hypothesis $\lvert f \rvert \le M$ is stated separately by Coddington, who notes that it is in fact automatic: a continuous function on the closed bounded rectangle $R$ is bounded there (p. 203 and Compactness), and he allows the reader to take the bound as an extra assumption instead. Where he produces the bound he takes it positive, "there exists a constant $M > 0$ " (p. 203); his Theorem 7 writes the hypothesis as $\lvert f\rvert \le M$ without repeating the positivity, which the quotient $b/M$ nevertheless needs, so we carry $M > 0$ in the hypotheses above.

Idea of the proof(Picard-Lindelöf)Original

Differentiating is awkward to iterate and integrating is not, so the first move is to replace the differential equation by the equivalent integral equation; a solution is then a fixed point of the map that sends $\varphi$ to $y_0 + \int f(t,\varphi)$ . Start with the constant function $y_0$ and apply the map over and over.

Two things must be controlled. The iterates must not wander out of the rectangle where $f$ is defined — this is what the number $b/M$ buys, since the graphs can climb no faster than slope $M$ . And the differences between consecutive iterates must shrink fast enough to sum. The Lipschitz condition converts a difference of $f$ -values into a difference of the iterates themselves, so each application of the map integrates the previous difference; integrating $k$ times produces a factor $\lvert x-x_0\rvert^k/k!$ , and a factorial beats everything. The series of differences is then dominated by a convergent series of constants, so the Weierstrass M-test gives uniform convergence, uniform convergence preserves continuity, and the same Lipschitz estimate lets the limit be taken inside the integral. Uniqueness is the same estimate used once more: the difference of two solutions satisfies an integral inequality that forces it to vanish.

Proof(Picard-Lindelöf)AdaptedAdapted from Coddington (1989), §5.4, Theorems 4 and 5, pp. 200-205, §5.6, pp. 210-214, and §5.8, pp. 224-225

Step 0: the integral equation is equivalent. Let $\varphi$ be continuous on $I$ with graph in $R$ . If $\varphi$ solves the initial-value problem, then $\varphi' (t)= f(t,\varphi(t))$ is continuous on $I$ , being the composition of the continuous $f$ with the continuous map $t \mapsto (t,\varphi(t))$ ; integrating from $x_0$ to $x$ and using $\varphi(x_0) = y_0$ gives the integral equation. Conversely, if $\varphi$ satisfies the integral equation, its integrand is continuous, so the fundamental theorem of calculus makes $\varphi$ differentiable with $\varphi'(x) = f(x,\varphi(x))$ , and setting $x = x_0$ kills the integral and gives $\varphi(x_0) = y_0$ . (Coddington's Theorem 4.) From here on we work with the integral equation.

Step 1: the iterates exist on $I$ and stay in $R$ . We show by induction that each $\varphi_k$ is defined and continuous on $I$ and satisfies

\bigl\lvert \varphi_k(x) - y_0 \bigr\rvert \le M\,\lvert x - x_0 \rvert \qquad (x \in I) .

For $k = 0$ the left side is zero. Assume it for $\varphi_k$ . Then for $x \in I$ we have $\lvert x - x_0 \rvert \le \alpha \le b/M$ , so $\lvert \varphi_k(x) - y_0\rvert \le M\alpha \le b$ , and also $\lvert x - x_0\rvert \le \alpha \le a$ ; hence $(x,\varphi_k(x)) \in R$ and the integrand $t \mapsto f(t,\varphi_k(t))$ is defined and continuous on $I$ . Therefore $\varphi_{k+1}$ is defined and continuous (indeed continuously differentiable) on $I$ , and

\bigl\lvert \varphi_{k+1}(x) - y_0 \bigr\rvert = \left\lvert \int_{x_0}^{x} f\bigl(t,\varphi_k(t)\bigr)\,dt \right\rvert \le M\,\lvert x - x_0\rvert ,

using $\lvert f \rvert \le M$ on $R$ . The absolute value outside the integral covers both $x > x_0$ and $x < x_0$ . Both of $\alpha$ 's two candidates are used here: $a$ keeps $x$ inside the rectangle horizontally, $b/M$ keeps the graph inside it vertically. (Coddington's Theorem 5.)

Step 2: consecutive iterates differ by at most a factorial-sized amount. We claim that for every $p \ge 1$ and $x \in I$ ,

\bigl\lvert \varphi_p(x) - \varphi_{p-1}(x) \bigr\rvert \ \le \ \frac{M K^{p-1}\,\lvert x - x_0\rvert^{p}}{p!} .

For $p = 1$ this is the estimate of step 1 with $k=0$ . Assume it for $p$ . Subtracting the defining formulas for $\varphi_{p+1}$ and $\varphi_p$ ,

\varphi_{p+1}(x) - \varphi_p(x) = \int_{x_0}^{x}\Bigl[f\bigl(t,\varphi_p(t)\bigr) - f\bigl(t,\varphi_{p-1}(t)\bigr)\Bigr]dt .

Both graphs lie in $R$ by step 1, so the Lipschitz condition applies to the integrand and gives $\lvert f(t,\varphi_p) - f(t,\varphi_{p-1})\rvert \le K\lvert \varphi_p(t) - \varphi_{p-1}(t)\rvert$ . Hence, using the inductive hypothesis,

\bigl\lvert \varphi_{p+1}(x) - \varphi_p(x)\bigr\rvert \ \le \ K\left\lvert \int_{x_0}^{x} \frac{MK^{p-1}\lvert t-x_0\rvert^{p}}{p!}\,dt\right\rvert = \frac{MK^{p}\,\lvert x-x_0\rvert^{p+1}}{(p+1)!} ,

which is the claim for $p+1$ . The Lipschitz condition enters here for the first time, and this is what it does: it converts the bound already in hand on $\varphi_p - \varphi_{p-1}$ into a bound on the next difference $\varphi_{p+1} - \varphi_p$ , one power of $\lvert x - x_0\rvert$ and one factor of $K$ further on. It is needed twice more, in step 4 and in step 5.

Step 3: the iterates converge uniformly on $I$ . Telescoping,

\varphi_k(x) = \varphi_0(x) + \sum_{p=1}^{k}\bigl[\varphi_p(x) - \varphi_{p-1}(x)\bigr] ,

so $\varphi_k$ is the $k$ th partial sum of the series $\varphi_0 + \sum_{p \ge 1}(\varphi_p - \varphi_{p-1})$ , and convergence of the sequence is convergence of the series. By step 2, on $I$ ,

\bigl\lvert \varphi_p(x) - \varphi_{p-1}(x)\bigr\rvert \le \frac{M}{K}\,\frac{(K\alpha)^{p}}{p!} =: M_p , \qquad \sum_{p \ge 1} M_p = \frac{M}{K}\bigl(e^{K\alpha} - 1\bigr) < \infty ,

and the constants $M_p$ do not depend on $x$ . By the Weierstrass M-test the series converges uniformly on $I$ ; call its sum $\varphi$ . Each $\varphi_k$ is continuous on $I$ , and a uniform limit of continuous functions is continuous, so $\varphi$ is continuous on $I$ . Letting $k \to \infty$ in the estimate of step 1 gives $\lvert \varphi(x) - y_0\rvert \le M\lvert x - x_0\rvert \le b$ , so the graph of $\varphi$ also lies in $R$ .

Step 4: the limit solves the integral equation. Fix $x \in I$ and subtract:

\left\lvert \int_{x_0}^{x} f\bigl(t,\varphi(t)\bigr)dt - \int_{x_0}^{x} f\bigl(t,\varphi_k(t)\bigr)dt \right\rvert \le K\left\lvert \int_{x_0}^{x}\bigl\lvert \varphi(t) - \varphi_k(t)\bigr\rvert dt\right\rvert \le K\alpha \, \sup_{t \in I}\bigl\lvert \varphi(t) - \varphi_k(t)\bigr\rvert ,

the first inequality by the Lipschitz condition (both graphs are in $R$ ), the second by bounding the integrand by its supremum over an interval of length at most $\alpha$ . The right-hand side tends to $0$ by step 3 — and this is exactly where uniform, as opposed to pointwise, convergence is needed, since the bound must be one number serving all $t$ at once. Letting $k \to \infty$ in

\varphi_{k+1}(x) = y_0 + \int_{x_0}^{x} f\bigl(t,\varphi_k(t)\bigr)\,dt

therefore gives $\varphi(x) = y_0 + \int_{x_0}^{x}f(t,\varphi(t))\,dt$ . By step 0, $\varphi$ solves the initial-value problem on $I$ . This proves part 1.

Step 5: uniqueness. Let $\varphi$ and $\psi$ solve the same initial-value problem on an interval $I'$ containing $x_0$ , with graphs in $R$ . By step 0 both satisfy the integral equation, so subtracting and using the Lipschitz condition, for $x \ge x_0$ in $I'$ ,

\bigl\lvert \varphi(x) - \psi(x)\bigr\rvert \le \int_{x_0}^{x}\bigl\lvert f(t,\varphi(t)) - f(t,\psi(t))\bigr\rvert\,dt \le K\int_{x_0}^{x}\bigl\lvert \varphi(t)-\psi(t)\bigr\rvert\,dt .

Set $E(x) = \int_{x_0}^{x}\lvert \varphi - \psi\rvert\,dt$ . Then $E(x_0) = 0$ , $E \ge 0$ , and $E$ is differentiable with $E' = \lvert \varphi - \psi\rvert$ because the integrand is continuous; the display says precisely $E' \le KE$ . Hence

\frac{d}{dx}\Bigl[e^{-Kx}E(x)\Bigr] = e^{-Kx}\bigl[E'(x) - KE(x)\bigr] \le 0 ,

so $e^{-Kx}E(x)$ is non-increasing and therefore at most its value $0$ at $x_0$ : $E(x) \le 0$ for $x \ge x_0$ . Since also $E \ge 0$ , we get $E \equiv 0$ and so $\lvert \varphi - \psi\rvert = E' \equiv 0$ on $I' \cap [x_0,\infty)$ . For $x \le x_0$ apply the same argument to the functions $\tilde\varphi(x) = \varphi(2x_0 - x)$ and $\tilde\psi(x) = \psi(2x_0-x)$ , which solve $y' = -f(2x_0-x,y)$ with the same initial value and the same Lipschitz constant. Hence $\varphi = \psi$ on all of $I'$ , which is part 2. (This is Coddington's route: he proves the general comparison estimate of his Theorem 10 by exactly this differential inequality and then reads uniqueness off as the special case in which the two equations and the two initial values coincide.)

Step 6: the error bound. Subtracting the series of step 3 from its $k$ th partial sum leaves the tail, so by step 2 and the same constants $M_p$ ,

\bigl\lvert \varphi(x) - \varphi_k(x)\bigr\rvert \le \sum_{p = k+1}^{\infty}\frac{M}{K}\frac{(K\alpha)^{p}}{p!} \le \frac{M}{K}\,\frac{(K\alpha)^{k+1}}{(k+1)!}\sum_{q = 0}^{\infty}\frac{(K\alpha)^{q}}{q!} = \frac{M}{K}\,\frac{(K\alpha)^{k+1}}{(k+1)!}\,e^{K\alpha} ,

where the middle step writes $p = k+1+q$ and uses $(k+1+q)! \ge (k+1)!\,q!$ . That is part 3. Since $(K\alpha)^{k+1}/(k+1)!$ is a term of the convergent series for $e^{K\alpha}$ , it tends to $0$ , so the bound does too. $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Coddington (1989), §5.5, pp. 208-209, §5.6, p. 210, and §5.7, Theorem 9, pp. 217-219; Hassani (2013), §14.2, Theorems 14.2.2 and 14.2.4, pp. 420-421; Hirsch, Smale & Devaney (2013), §7.2, p. 143

Continuity of $f$ is used in step 0 (to differentiate the integral and to integrate the derivative) and in step 1 (to know the iterates are defined). Dropping it can destroy existence outright: Hirsch, Smale and Devaney give the vector field on $\mathbb{R}$ equal to $1$ for $x<0$ and $-1$ for $x \ge 0$ , for which no solution passes through $x = 0$ , since the solution would have to decrease at $0$ and increase just to the left of it (p. 143).
The bound $M$ and the number $b/M$ first appear in step 1, where $b/M$ is what keeps the graphs inside $R$ ; that is where the locality of the conclusion comes from, and if $f$ is large the graph may run out of the rectangle quickly and the guaranteed interval shrinks accordingly. $M$ does not stop there: it is the leading factor in the difference estimate of step 2, in the constants $M_p$ of step 3, and hence in the error bound of step 6 and part 3, all of which are proportional to it.
The Lipschitz condition is used in step 2 (to make the differences contract) and in steps 4 and 5. It is not needed for existence: Hassani states the Peano existence theorem (Theorem 14.2.2), that continuity of $f$ on the rectangle together with $\lvert f \rvert \le M$ already gives at least one solution on $\lvert x - x_0\rvert \le \min(a, b/M)$ — the same interval. Coddington says the same in a note after his Theorem 7 and declines to prove it, the proof needing tools beyond his chapter. What the Lipschitz condition adds is uniqueness, and the method of successive approximations itself: Coddington's Ex. 3, pp. 215-216, exhibits a continuous $f$ with $\lvert f\rvert \le 2$ for which the approximations alternate between $x^2$ and $-x^2$ , so that neither convergent subsequence tends to a solution — although the problem does have one.
How to check the Lipschitz condition. Coddington's Theorem 6: if $S$ is a rectangle $\lvert x-x_0\rvert \le a$ , $\lvert y - y_0\rvert\le b$ or a strip $\lvert x - x_0\rvert \le a$ , $\lvert y\rvert < \infty$ , and $f$ is real-valued on $S$ with $\partial f/\partial y$ existing, continuous on $S$ and satisfying $\lvert \partial f/\partial y\rvert \le K$ there, then $f$ satisfies a Lipschitz condition on $S$ with constant $K$ . The proof is one line — $f(x,y_1) - f(x,y_2) = \int_{y_2}^{y_1}(\partial f/\partial y)(x,t)\,dt$ — and this is the test used in practice. (Coddington's Ex. 4, p. 210, notes that the continuity of $\partial f/\partial y$ is not needed, the mean value theorem serving instead.) The converse of the test fails: $f(x,y) = x^{2}\lvert y\rvert$ is Lipschitz on $\lvert x\rvert \le 1$ , $\lvert y\rvert \le 1$ although $\partial f/\partial y$ does not exist at $(x,0)$ when $x \neq 0$ (Coddington, Ex. 3, pp. 209-210).
Being Lipschitz is a property of the region. $f(x,y) = xy^{2}$ satisfies a Lipschitz condition on the rectangle $\lvert x \rvert \le 1$ , $\lvert y\rvert \le 1$ , where $\lvert \partial f/\partial y\rvert = \lvert 2xy\rvert \le 2$ , but on no strip $\lvert x\rvert \le 1$ , $\lvert y \rvert < \infty$ : there the quotient $\lvert f(x,y_1) - f(x,0)\rvert/\lvert y_1\rvert$ equals $\lvert x\rvert\lvert y_1\rvert$ , which is unbounded in $y_1$ whenever $x \neq 0$ (Coddington, pp. 208-209).
Non-local existence. The locality is a consequence of the rectangle, not of the method. Coddington's Theorem 9: if $f$ is continuous on the strip $\lvert x - x_0\rvert \le a$ , $\lvert y\rvert < \infty$ and satisfies a Lipschitz condition with constant $K$ there, the approximations exist and converge on the whole interval $\lvert x - x_0\rvert \le a$ . The linear equation $y' + g(x)y = h(x)$ with $g$ , $h$ continuous and $\lvert g\rvert \le K$ on $\lvert x - x_0\rvert \le a$ is of this kind, which recovers from this theory the global existence proved directly in Solving first-order equations.
The interval depends on the initial value. Hassani's Example 14.2.5: $dy/dx = e^{y}$ with $y(0) = b$ has the solution $y = -\ln(e^{-b} - x)$ , defined for $-\infty < x < e^{-b}$ , so the right-hand end moves as $b$ is varied.

Note(The same theorem in the other sources' hypotheses)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §7.2, p. 144; Strogatz (1994), §2.5, p. 27; Hassani (2013), §14.2, Theorem 14.2.6, p. 421

The result is stated in this handbook with Coddington's hypotheses. Hassani assumes the same and reaches the same interval; the two dynamical-systems sources assume more and say less about the interval. The three read as follows.

Hassani (Theorem 14.2.6). Same statement as ours: $F(x,y)$ defined and continuous on the rectangle $\lvert y - c\rvert \le K$ , $\lvert x - a \rvert \le N$ and satisfying a Lipschitz condition there, $M = \max\lvert F\rvert$ on the rectangle, gives a unique solution with $f(a)=c$ on $\lvert x - a\rvert \le \min(N, K/M)$ . His letters differ from ours — his $K$ is the half-height of the rectangle and his $L$ is the Lipschitz constant — but the content is identical.
Hirsch, Smale and Devaney (p. 144). For an autonomous system $X' = F(X)$ , $X(t_0) = X_0$ in $\mathbb{R}^n$ with $F : \mathbb{R}^n \to \mathbb{R}^n$ of class $C^1$ , there exist an $a > 0$ and a unique solution $X : (t_0-a, t_0+a) \to \mathbb{R}^n$ . This is stronger in dimension and weaker in precision: $C^1$ implies a local Lipschitz condition by the derivative test above, but the number $a$ is not identified. They also observe (p. 146) that uniqueness is not merely local: if two solutions agree at one point and both are defined on an interval $J$ , they agree on all of $J$ , since the set where they agree is both open, by the local theorem, and closed, by continuity.
Strogatz (p. 27). For $\dot x = f(x)$ , $x(0) = x_0$ with $f$ and $f'$ both continuous on an open $x$ -interval $R$ that contains $x_0$ , there is a solution on some interval $(-\tau,\tau)$ and it is unique. Again $C^1$ in place of Lipschitz, and no formula for $\tau$ .

Two conventions to note. Hirsch, Smale and Devaney and Strogatz both state the theorem for autonomous equations, where $f$ does not depend on the independent variable; Coddington and Hassani allow dependence on both. And where we use $x$ for the independent variable, the dynamical-systems sources use $t$ and reserve $x$ for the dependent one. Nothing is lost by the autonomous restriction in Hirsch, Smale and Devaney's $n$ -dimensional statement, since $y' = f(x,y)$ is the autonomous planar system $x' = 1$ , $y' = f(x,y)$ in disguise; Strogatz's one-dimensional statement, which has no room for that extra variable, is genuinely the narrower one.

There is also a route that bypasses the successive approximations as such: observe that the Picard map $\varphi \mapsto y_0 + \int_{x_0}^{\cdot} f(t,\varphi)\,dt$ moves two functions no more than $K\alpha$ times as far apart as they already are, in the supremum distance; when $K\alpha < 1$ it is therefore a contraction on a complete space of continuous functions and has a unique fixed point by the contraction mapping theorem. The proof above avoids the extra hypothesis $K\alpha < 1$ : the factorial in step 2 makes the iterates contract eventually whatever the value of $K\alpha$ .

Example(Without the Lipschitz condition: infinitely many solutions)AdaptedAdapted from Coddington (1989), §5.2, pp. 189-190, §5.5, p. 209, and §5.8, p. 223; Hirsch, Smale & Devaney (2013), §7.2, p. 143; Strogatz (1994), §2.5, Example 2.5.1, p. 27

Consider

y' = 3y^{2/3} , \qquad y(0) = 0 .

The right-hand side is continuous but not Lipschitz. $f(x,y) = 3y^{2/3}$ is continuous on every rectangle. On a rectangle containing the origin, however, take $y_2 = 0$ and $y_1 = \eta > 0$ :

\frac{\lvert f(x,\eta) - f(x,0)\rvert}{\lvert \eta - 0\rvert} = \frac{3\eta^{2/3}}{\eta} = \frac{3}{\eta^{1/3}} \longrightarrow \infty \qquad (\eta \to 0^{+}) ,

so no constant $K$ can serve: $f$ satisfies a Lipschitz condition on no rectangle containing a piece of the line $y = 0$ . (Coddington makes the same computation for $y^{2/3}$ on $\lvert x\rvert \le 1$ , $\lvert y\rvert\le1$ , where the quotient is $y_1^{-1/3}$ , p. 209; the derivative test fails for the same reason, $\partial f/\partial y = 2y^{-1/3}$ being unbounded near $y=0$ .)

Separation finds one family of solutions. For $y \neq 0$ , separating as in Solving first-order equations gives $\tfrac13\int y^{-2/3}\,dy = \int dx$ , so $y^{1/3} = x + c$ and $y = (x+c)^3$ . The initial condition picks $c = 0$ and the solution $y = x^3$ .

But there are infinitely many. The constant function $y \equiv 0$ is also a solution with $y(0)=0$ — the solution lost by dividing by $y^{2/3}$ . Worse, the two can be spliced. For any $\tau \ge 0$ (Coddington splices at any positive $k$ , p. 190, and Hirsch, Smale and Devaney likewise take their parameter positive, p. 143; the member $\tau = 0$ , which vanishes for $x \le 0$ and equals $x^3$ afterwards, is equally a solution and is our addition to their families) define

y_\tau(x) = \begin{cases} 0, & x \le \tau,\\[1mm] (x-\tau)^{3}, & x > \tau .\end{cases}

This is differentiable everywhere, including at $x = \tau$ , where the left derivative is $0$ and the right derivative is $\lim_{h\to0^+} h^3/h = 0$ ; and it satisfies the equation, since for $x > \tau$ we have $y_\tau' = 3(x-\tau)^2 = 3\bigl[(x-\tau)^3\bigr]^{2/3} = 3y_\tau^{2/3}$ , while for $x < \tau$ both sides vanish. Every one of these satisfies $y_\tau(0) = 0$ . So the initial-value problem has a one-parameter family of solutions, together with $y \equiv 0$ and with the mirror-image solutions obtained by splicing on the left.

What this does and does not show. Existence is not the issue: $f$ is continuous, so Peano's theorem applies and solutions exist. The Lipschitz condition is what fails, and with it uniqueness — exactly the division of labour described in the remark above. The picture is of a solution that may rest at $0$ for an arbitrary time and then set off, with nothing in the equation to say when. Hirsch, Smale and Devaney give the same example (p. 143), and Strogatz gives the companion $\dot x = x^{1/3}$ from $x_0 = 0$ , with the two solutions $x \equiv 0$ and $x = (\tfrac23 t)^{3/2}$ (p. 27).

Example(Picard iteration, and how sharp the error bound is)Original

Solve

y' = x^{2} + y , \qquad y(0) = 1

by successive approximations, and compare the result with the error bound of part 3.

Step 1: the data of the theorem. Take the rectangle $R : \lvert x\rvert \le \tfrac12$ , $\lvert y - 1\rvert \le 2$ , so $a = \tfrac12$ and $b = 2$ . On $R$ we have $y \in [-1,3]$ and

\lvert f(x,y)\rvert = \lvert x^2 + y\rvert \le \tfrac14 + 3 = \tfrac{13}{4} =: M ,

while $\partial f/\partial y = 1$ is continuous and bounded by $1$ , so by the derivative test $f$ is Lipschitz on $R$ with $K = 1$ . Hence

\alpha = \min\Bigl\{\tfrac12,\ \tfrac{2}{13/4}\Bigr\} = \min\{0.5,\ 0.6154\} = \tfrac12 ,

and the theorem guarantees a unique solution on $\lvert x\rvert \le \tfrac12$ .

Step 2: iterate. With $\varphi_0 \equiv 1$ and $\varphi_{k+1}(x) = 1 + \int_0^x (t^2 + \varphi_k(t))\,dt$ ,

\begin{aligned} \varphi_1(x) &= 1 + x + \tfrac13 x^{3},\\ \varphi_2(x) &= 1 + x + \tfrac12 x^{2} + \tfrac13 x^{3} + \tfrac1{12}x^{4},\\ \varphi_3(x) &= 1 + x + \tfrac12 x^{2} + \tfrac12 x^{3} + \tfrac1{12}x^{4} + \tfrac1{60}x^{5},\\ \varphi_4(x) &= 1 + x + \tfrac12 x^{2} + \tfrac12 x^{3} + \tfrac18 x^{4} + \tfrac1{60}x^{5} + \tfrac1{360}x^{6},\\ \varphi_5(x) &= 1 + x + \tfrac12 x^{2} + \tfrac12 x^{3} + \tfrac18 x^{4} + \tfrac1{40}x^{5} + \tfrac1{360}x^{6} + \tfrac1{2520}x^{7} . \end{aligned}

Each iterate adds one more correct term and perturbs the tail: the coefficient of $x^4$ , for instance, is wrong in $\varphi_2$ and $\varphi_3$ and settles at $\tfrac18$ from $\varphi_4$ on.

Step 3: the exact solution, for comparison. The equation is linear, $y' - y = x^2$ , with integrating factor $e^{-x}$ ; the method of Solving first-order equations gives the particular solution $-(x^2+2x+2)$ and hence

y(x) = 3e^{x} - x^{2} - 2x - 2 ,

whose value at $x=0$ is $1$ and whose derivative is $3e^x - 2x - 2 = x^2 + y$ , as required. Its Taylor series is $1 + x + \tfrac12x^2 + \tfrac12x^3 + \tfrac18x^4 + \tfrac1{40}x^5 + \cdots$ , which is what the iterates are converging to.

Step 4: numbers at the edge of the guaranteed interval. At $x = \alpha = \tfrac12$ the exact value is $y(\tfrac12) = 3e^{1/2} - \tfrac14 - 1 - 2 = 4.946163812 - 3.25 = 1.696163812$ . The bound of part 3, with $M/K = 3.25$ , $K\alpha = 0.5$ and $e^{K\alpha} = 1.648721271$ , is $3.25\,(0.5)^{k+1}e^{0.5}/(k+1)!$ .

$k$	$\varphi_k(1/2)$	actual error	bound of part 3
1	$1.541666667$	$1.54\times10^{-1}$	$6.70\times10^{-1}$
2	$1.671875000$	$2.43\times10^{-2}$	$1.12\times10^{-1}$
3	$1.693229167$	$2.93\times10^{-3}$	$1.40\times10^{-2}$
4	$1.695876736$	$2.87\times10^{-4}$	$1.40\times10^{-3}$
5	$1.696140253$	$2.36\times10^{-5}$	$1.16\times10^{-4}$

Every error is below its bound, by a factor of about five throughout, and both columns fall by roughly an order of magnitude per iteration — the factorial of step 2 at work. The bound is therefore pessimistic by a constant factor but correct in its rate, which is the useful property: it predicts in advance how many iterations are needed for a prescribed accuracy, before any of them is carried out.

Step 5: a caution about the interval. Nothing in the construction says the solution stops at $x = \tfrac12$ ; here it is in fact an entire function. The rectangle was a device, and a different rectangle gives a different $\alpha$ . Enlarging $b$ raises $M$ , so increasing $b$ eventually stops helping: with $a = \tfrac12$ and half-height $b$ , $M = \tfrac14 + 1 + b$ and $b/M = b/(b + \tfrac54)$ , which increases to $1$ as $b \to \infty$ , so $\alpha = \tfrac12$ for every $b \ge \tfrac54$ and the horizontal constraint $a$ is what binds. For this equation the strip theorem quoted in the remark does better: $f$ is Lipschitz with $K=1$ on every strip $\lvert x\rvert \le a$ , so the solution in fact exists for all $x$ .

Remark(Why the theorem has to be local: blow-up)AdaptedAdapted from Strogatz (1994), §2.5, Example 2.5.2, p. 28; Hirsch, Smale & Devaney (2013), §7.2, p. 146

The previous example might suggest that $\alpha$ is always an underestimate of no consequence. It is not. Take

y' = 1 + y^{2} , \qquad y(0) = 0 ,

whose right-hand side is a polynomial, hence continuous with continuous $y$ -derivative everywhere. Separating gives $\arctan y = x$ , so $y = \tan x$ , which exists only for $-\pi/2 < x < \pi/2$ and runs to $\pm\infty$ at the ends: the solution reaches infinity at a finite value of the independent variable. Strogatz calls this blow-up and notes its relevance to models of combustion and other runaway processes (p. 28); Hirsch, Smale and Devaney make the same point with $x' = 1+x^2$ and its solutions $\tan(t+c)$ (p. 146).

What does the theorem predict here? On $R : \lvert x \rvert \le a$ , $\lvert y\rvert \le b$ we may take $M = 1 + b^2$ , so

\alpha = \min\Bigl\{a, \ \frac{b}{1+b^{2}}\Bigr\} \le \max_{b>0}\frac{b}{1+b^{2}} = \frac12 ,

the maximum occurring at $b = 1$ . So no choice of rectangle gives a guaranteed interval longer than $\lvert x \rvert \le \tfrac12$ , against a true interval of half-length $\pi/2 = 1.5708$ . The theorem is pessimistic by a factor of about three — but it is pessimistic about something real, since beyond $\pi/2$ there is nothing to find. No local theorem can distinguish this equation from one whose solutions last forever, because the distinction is not visible in a neighbourhood of the initial point.

Application(Determinism, and where it fails)Original

Trajectories do not cross. Newton's equations for a system with coordinates $q$ become a first-order system for the state $(q,\dot q)$ in phase space: $\dot q = v$ , $\dot v = F(q,v)/m$ . If $F$ is continuously differentiable, the theorem applies in the form quoted from Hirsch, Smale and Devaney, and through each phase point there passes exactly one trajectory. (A force that also depends on $t$ is covered by adjoining $t$ as an extra state variable with $\dot t = 1$ , which restores the autonomous form their statement assumes.) This is the mathematical content of classical determinism, and it is also the licence for every phase-portrait argument: the curves drawn in a phase plane cannot cross, because a crossing point would be an initial condition with two solutions. It is what makes the one-dimensional reasoning of Flows on the line valid.

A typical consequence, used constantly and rarely proved: an equilibrium cannot be reached in finite time. Let $\dot x = f(x)$ be autonomous with a fixed point $x_\ast$ , so that the constant function $x(t) \equiv x_\ast$ is a solution, and assume — this is the hypothesis the argument needs, and it is exactly the one the theorem needs at $x_\ast$ — that $f$ satisfies a Lipschitz condition on some interval around $x_\ast$ , for instance because $f$ is $C^1$ there. Suppose some other solution satisfied $x(t_\ast) = x_\ast$ at a finite time $t_\ast$ . Both would then solve the initial-value problem at $t_\ast$ , so by uniqueness they would be the same solution — and the other one is not constant. Hence a trajectory approaching an equilibrium takes infinitely long to arrive. That is why the separatrix of a pendulum takes forever to reach the inverted position, and why a linear stability analysis predicting $\lvert x - x_\ast\rvert \sim e^{-\lambda t}$ never predicts arrival.

A mechanical system with no unique future. That argument is worth no more than its hypothesis, and the hypothesis can fail in a mechanical system. Consider a bead sliding without friction on a curve $z = h(r)$ with an apex at $r = 0$ , where $r$ is horizontal distance from the apex. For small slopes the equation of motion along the surface reduces to $\ddot r \approx -g\,h'(r)$ . Choose the shape

h(r) = -\frac{2}{3g}\,r^{3/2} , \qquad\text{so that}\qquad -g\,h'(r) = \sqrt{r} \quad (r \ge 0) ,

and take the idealised model $\ddot r = \sqrt r$ , with the bead at rest exactly at the apex: $r(0) = 0$ , $\dot r(0) = 0$ . As a first-order system this is $\dot r = v$ , $\dot v = \sqrt r$ , whose right-hand side is continuous but, near $r = 0$ , not Lipschitz in the state, for the reason seen in the example above. And indeed the problem has infinitely many solutions: $r \equiv 0$ , and for every $T \ge 0$

r_T(t) = \begin{cases} 0, & t \le T, \\[1mm] \dfrac{(t-T)^{4}}{144}, & t > T, \end{cases}

since $\ddot r_T = 12(t-T)^2/144 = (t-T)^2/12$ while $\sqrt{r_T} = (t-T)^2/12$ for $t > T$ , and both sides vanish for $t<T$ ; the two one-sided limits of $r_T$ , $\dot r_T$ and $\ddot r_T$ agree at $t = T$ . The bead may therefore sit motionless at the apex for any length of time and then slide off, with the equations of motion saying nothing about when. This construction is known in the philosophy-of-physics literature as Norton's dome, and it is the cleanest demonstration that classical determinism is a theorem about Lipschitz vector fields rather than an axiom about Newtonian mechanics. The reassurance is that the shape has unbounded curvature at the apex: $h''(r) = -\,r^{-1/2}/(2g) \to -\infty$ as $r \to 0^{+}$ , so the dome is not a smooth surface there, and smooth force laws do not produce this behaviour.

Why numerical work needs the theorem twice. Any integrator for $y' = f(x,y)$ (Integrators for ODEs) computes a sequence of approximate values and is silent about whether there is anything to approximate. Both halves of the theorem are therefore used before the first step is taken: existence, so that the computed sequence has a target, and uniqueness, so that two runs with different step sizes are approximating the same function. The Lipschitz constant is not merely a hypothesis here but an output of the analysis: the classical global error bound for a one-step method of order $p$ and step $h$ on an interval of length $L$ has the form $C h^{p}\bigl(e^{KL}-1\bigr)/K$ , with $K$ the Lipschitz constant of $f$ , so the same number that makes the solution unique also measures how fast errors — numerical or physical — can be amplified. Finally, blow-up has a numerical signature worth recognising: integrating $y' = 1+y^2$ from $y(0)=0$ , an adaptive solver drives its step size towards zero as $x$ approaches $\pi/2 = 1.5708$ and stalls, which is not a bug but the solver detecting that the solution has ceased to exist.

Theorem · advanced

Dependence on initial data and Gronwall's inequality

Adapted

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Gronwall's inequality turns the self-referential integral bound satisfied by the gap between two solutions into an explicit exponential, giving the separation estimate $|Y(t)-Z(t)| \le |Y(t_0)-Z(t_0)|e^{K(t-t_0)}$ , continuity of the flow in the initial point and in parameters, an error bound for a perturbed vector field, and the fact that a solution on a bounded maximal interval leaves every compact subset of its domain.

Prerequisites: Existence and uniqueness of solutions, Compactness

Definition(Lipschitz constants, maximal intervals, and the flow)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.2, p. 394, §17.4, pp. 397-398, and §17.5, p. 402

Throughout, $|\cdot|$ is the Euclidean norm on $\mathbb{R}^n$ and $O$ is an open subset of $\mathbb{R}^n$ (of $\mathbb{R}\times\mathbb{R}^n$ where a nonautonomous field is meant).

A map $F : O \to \mathbb{R}^n$ has Lipschitz constant $K$ on $O$ if

\bigl| F(X) - F(Y) \bigr| \le K\,|X - Y| \qquad \text{for all } X, Y \in O ,

and is locally Lipschitz if every point of $O$ has a neighbourhood on which $F$ has some Lipschitz constant. A $C^1$ map is locally Lipschitz. For a nonautonomous field $F : O \to \mathbb{R}^n$ with $O \subseteq \mathbb{R}\times\mathbb{R}^n$ , " $F$ is Lipschitz in $X$ with constant $K$ " means

\bigl| F(t, X_1) - F(t, X_2) \bigr| \le K\,|X_1 - X_2| \qquad \text{for all } (t, X_1), (t, X_2) \in O ,

with no requirement on how $F$ varies with $t$ beyond continuity.

Let $F : O \to \mathbb{R}^n$ be $C^1$ and fix $X_0 \in O$ . Two solutions of $X' = F(X)$ that agree at one time agree on the whole intersection of their domains: the set where they agree is nonempty, closed by continuity, and open by the local uniqueness theorem of Existence and uniqueness of solutions, hence is all of the (connected) common interval. Consequently the solutions with $X(0) = X_0$ can be glued into one, and there is a largest open interval

J(X_0) = \bigl(\alpha(X_0),\ \beta(X_0)\bigr) \ni 0

carrying a solution with that initial value; this is the maximal interval of existence, and the solution on it is the maximal solution. The flow of the system is

\phi(t, X) = \text{the value at time } t \text{ of the maximal solution with } \phi(0, X) = X ,

defined for $X \in O$ and $t \in J(X)$ .

Theorem(Gronwall's inequality)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.3, p. 395

Let $u : [0, \alpha] \to \mathbb{R}$ be continuous and nonnegative, and let $C \ge 0$ and $K \ge 0$ be constants such that

u(t) \ \le \ C + \int_0^{t} K\,u(s)\,ds \qquad \text{for all } t \in [0, \alpha] .

Then

u(t) \ \le \ C e^{Kt} \qquad \text{for all } t \in [0, \alpha] .

Theorem(Dependence on initial data, on the field, and the shape of the maximal interval)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.3, pp. 394, 397, §17.4, pp. 398, 400, and §17.5, p. 402

Let $O \subseteq \mathbb{R}^n$ be open.

(Separation bound.) Suppose $F : O \to \mathbb{R}^n$ has Lipschitz constant $K$ on $O$ . Suppose further that each of $Y$ and $Z$ solves $X' = F(X)$ throughout the interval $[t_0, t_1]$ , taking its values in $O$ . Then
$\bigl| Y(t) - Z(t) \bigr| \ \le \ \bigl| Y(t_0) - Z(t_0) \bigr|\, e^{K(t - t_0)} \qquad \text{for all } t \in [t_0, t_1] .$
(A bounded maximal interval forces escape.) Suppose $F : O \to \mathbb{R}^n$ is $C^1$ , and let $Y$ solve $X' = F(X)$ on its maximal interval $J = (\alpha, \beta)$ , whose right endpoint is assumed finite, $\beta < \infty$ . Then $Y$ leaves every compact subset of $O$ , and does so arbitrarily late: for every compact $C \subset O$ and every $\gamma \in (\alpha, \beta)$ there is a $t \in [\gamma, \beta)$ with $Y(t) \notin C$ . In particular no compact subset of $O$ holds the whole track.
(Continuous dependence on the initial point, uniformly on a closed interval.) Suppose $F : O \to \mathbb{R}^n$ is $C^1$ , and suppose $Y$ solves $X' = F(X)$ at every point of a closed interval $[t_0, t_1]$ , with $Y(t_0) = Y_0$ . Some neighbourhood $U \subseteq \mathbb{R}^n$ of $Y_0$ and some constant $K$ then have this property: each $Z_0 \in U$ is the initial value, at time $t_0$ , of a unique solution $Z$ that also survives to $t_1$ , and
$\bigl| Y(t) - Z(t) \bigr| \ \le \ \bigl| Y_0 - Z_0 \bigr|\, e^{K(t - t_0)} \qquad \text{for all } t \in [t_0, t_1] .$
In particular the flow $\phi(t, X)$ of a $C^1$ field is a continuous function of $X$ .
(Dependence on the field and on parameters.) Let $O \subseteq \mathbb{R}\times\mathbb{R}^n$ be open and contain $(0, X_0)$ , and let $F, G : O \to \mathbb{R}^n$ be $C^1$ in $X$ and continuous in $t$ , with
$\bigl| F(t, X) - G(t, X) \bigr| < \varepsilon \qquad \text{for all } (t, X) \in O .$
Write $K > 0$ for a constant with respect to which $F$ is Lipschitz in its second argument. Suppose $X(t)$ solves $X' = F(t, X)$ and $Y(t)$ solves $Y' = G(t, Y)$ , both throughout an interval $J$ that contains the origin, and both passing through $X_0$ there: $X(0) = X_0 = Y(0)$ . Then
$\bigl| X(t) - Y(t) \bigr| \ \le \ \frac{\varepsilon}{K}\Bigl( e^{K|t|} - 1 \Bigr) \qquad \text{for all } t \in J .$
Separately, if $X' = F_a(X)$ is a system whose field is continuously differentiable in $X$ and in the parameter $a$ jointly, then its flow is continuous in $a$ as well as in $X$ .

Hirsch, Smale and Devaney state part 4 with the factor $\varepsilon/K$ and do not say that $K$ is positive; since that quotient needs it, we carry $K > 0$ in the hypotheses. They also state part 1 with $\exp(K(t - t_0))$ , that is, forward in time from $t_0$ ; see the remark below for the two-sided form.

Part 2 is printed there in the weaker form of its last sentence: given any compact $C \subset O$ , some $t \in (\alpha, \beta)$ has $Y(t) \notin C$ , with no $\gamma$ in the statement. We state the one-sided version because that is what the proof below — which is their proof — actually delivers, the track being used only on $[\gamma, \beta)$ , and because the weaker version is not enough for part 3, whose proof needs the escape to happen after $t_0$ . Hirsch, Smale and Devaney's own proof of part 3 invokes the theorem in the same one-sided way; the gap between their statement and their use of it is logged in verification/discrepancies.md.

Idea of the proofOriginal

The gap between two solutions satisfies no equation anyone can solve, but it does satisfy an inequality about itself: subtracting the two integral equations and using the Lipschitz condition shows that the gap at time $t$ is at most its initial size plus $K$ times the integral of the gap up to $t$ . Gronwall's inequality says that a nonnegative function trapped by such a self-referential bound can do no better than the solution of the corresponding equation, which is the exponential $Ce^{Kt}$ . That single step yields every quantitative statement here.

The rest is bookkeeping. A perturbation of the field contributes an extra term $\varepsilon t$ , which is absorbed by applying Gronwall to the shifted function $u + \varepsilon/K$ instead of to $u$ . A parameter becomes an extra coordinate once the equation $a' = 0$ is appended, so dependence on parameters is dependence on initial data for a larger system. And the escape statement is a compactness argument with no inequalities in it at all: a solution confined to a compact set over the last stretch of its interval moves at bounded speed there, so it has a limit at the end of a finite maximal interval, and that limit is an interior point from which the solution can be continued — which contradicts maximality.

Proof(Gronwall's inequality)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.3, pp. 395-396

The case $C > 0$ . Put

U(t) = C + \int_0^{t} K\,u(s)\,ds \qquad (0 \le t \le \alpha) .

Since $u$ is continuous, the fundamental theorem of calculus makes $U$ differentiable with $U'(t) = K u(t)$ , and $U' \ge 0$ because $K \ge 0$ and $u \ge 0$ . Hence $U$ is nondecreasing, so $U(t) \ge U(0) = C > 0$ on the whole interval; in particular $U$ never vanishes and $\log U$ is defined. The hypothesis is exactly $u \le U$ .

Divide: for every $t$ ,

\frac{U'(t)}{U(t)} = \frac{K\,u(t)}{U(t)} \le K ,

where the inequality uses $u(t) \le U(t)$ together with $K \ge 0$ and $U(t) > 0$ . The left-hand side is $(\log U)'(t)$ , so integrating from $0$ to $t$ gives $\log U(t) - \log C \le Kt$ . Exponentiating, $U(t) \le Ce^{Kt}$ , and therefore $u(t) \le U(t) \le Ce^{Kt}$ .

The case $C = 0$ . For any $c > 0$ the hypothesis still holds with $C$ replaced by $c$ , because $0 \le c$ . The case already proved gives $u(t) \le c e^{Kt}$ for every $c > 0$ ; letting $c \downarrow 0$ gives $u(t) \le 0$ , hence $u \equiv 0$ , which is the asserted bound. $\square$

Proof(Part 1: the separation bound)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.3, p. 396

Write $v(t) = |Y(t) - Z(t)|$ , a continuous nonnegative function on $[t_0, t_1]$ . Both solutions satisfy the integral form of the equation (Existence and uniqueness of solutions, step 0 of its proof), so subtracting,

Y(t) - Z(t) = Y(t_0) - Z(t_0) + \int_{t_0}^{t} \Bigl[ F\bigl(Y(s)\bigr) - F\bigl(Z(s)\bigr) \Bigr]\, ds .

Take norms. The triangle inequality for integrals and then the Lipschitz hypothesis on $F$ — legitimate because $Y(s)$ and $Z(s)$ lie in $O$ for every $s$ , which is where $K$ is a Lipschitz constant — give

v(t) \ \le \ v(t_0) + \int_{t_0}^{t} K\,v(s)\,ds \qquad (t_0 \le t \le t_1) .

Now translate the clock so that Gronwall's inequality applies in the form stated. Set $u(\tau) = v(t_0 + \tau)$ for $\tau \in [0, t_1 - t_0]$ ; substituting $s = t_0 + \sigma$ in the integral turns the display into

u(\tau) \ \le \ v(t_0) + \int_0^{\tau} K\,u(\sigma)\,d\sigma .

This is the hypothesis of Gronwall's inequality with $C = v(t_0) \ge 0$ , so $u(\tau) \le v(t_0)e^{K\tau}$ . Reading $\tau = t - t_0$ gives the claim. $\square$

Proof(Part 2: a bounded maximal interval forces escape)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.4, pp. 398-399

Suppose the conclusion fails for some compact $C \subset O$ and some $\gamma \in (\alpha, \beta)$ : that is, $Y(t) \in C$ for every $t \in [\gamma, \beta)$ . We derive a contradiction with the maximality of $J = (\alpha, \beta)$ . Only the late part of the track is assumed confined; what $Y$ does on $(\alpha, \gamma)$ plays no role below.

The solution has bounded speed. $F$ is continuous and $C$ is compact, so $M = \max_{X \in C} |F(X)|$ is finite and attained (Compactness). For $\gamma \le s \le t < \beta$ the track lies in $C$ , so

\bigl| Y(t) - Y(s) \bigr| = \left| \int_s^{t} F\bigl(Y(\sigma)\bigr)\, d\sigma \right| \le M\,(t - s) .

So $Y$ is Lipschitz, hence uniformly continuous, on $[\gamma, \beta)$ . Here the finiteness of $\beta$ matters: the interval has an endpoint to approach.

It therefore has a limit at $\beta$ . Given $\eta > 0$ , the display shows $|Y(t) - Y(s)| < \eta$ whenever $s, t$ are within $\eta/(M+1)$ of $\beta$ , so $Y(t)$ satisfies the Cauchy criterion as $t \to \beta^-$ . As $\mathbb{R}^n$ is complete the limit exists; call it $Y(\beta)$ . Since $C$ is closed and contains $Y(t)$ for every $t \in [\gamma, \beta)$ , the limit lies in $C$ , hence in $O$ .

The extension still solves the equation. On $[\gamma, \beta)$ the solution satisfies

Y(t) = Y(\gamma) + \int_{\gamma}^{t} F\bigl(Y(s)\bigr)\, ds .

The integrand is continuous on $[\gamma, \beta)$ and bounded there by $M$ , so the integral converges as $t \to \beta^-$ and the identity passes to the limit: $Y(\beta) = Y(\gamma) + \int_{\gamma}^{\beta} F(Y(s))\,ds$ . The extended $Y : [\gamma, \beta] \to C$ is thus continuous and satisfies the integral equation on the closed interval, so by the fundamental theorem of calculus it is differentiable with $Y'(t) = F(Y(t))$ there, the derivative at $\beta$ being the left-hand one.

Maximality is contradicted. $Y(\beta)$ is a point of the open set $O$ , so the local existence theorem (Existence and uniqueness of solutions) supplies a solution $W$ on some $[\beta, \delta]$ with $\delta > \beta$ and $W(\beta) = Y(\beta)$ . Splice: the function equal to $Y$ on $(\alpha, \beta]$ and to $W$ on $[\beta, \delta]$ is continuous, and at $\beta$ its left and right derivatives both equal $F(Y(\beta))$ , so it is differentiable there and solves the equation on $(\alpha, \delta]$ . That is a solution on an interval strictly larger than $J$ , contradicting the assumption that $J$ is maximal. $\square$

Proof(Part 3: continuous dependence on the initial point)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.4, pp. 400-401

A lemma: a locally Lipschitz field is Lipschitz on a compact set. Let $F$ be locally Lipschitz on $O$ and let $C \subset O$ be compact. If no Lipschitz constant worked on $C$ , then for each $m \in \mathbb{N}$ there would be points $X_m, Y_m \in C$ with

\bigl| F(X_m) - F(Y_m) \bigr| > m\,|X_m - Y_m| .

A compact tube around the given solution. The track $\Gamma = \{\,Y(t) : t_0 \le t \le t_1\,\}$ is the continuous image of a compact interval, hence compact, and $\Gamma \subset O$ . Some $r > 0$ has the property that every point within distance $r$ of $\Gamma$ lies in $O$ : were there no such $r$ , we could pick $X_m \notin O$ with $\operatorname{dist}(X_m, \Gamma) \le 1/m$ ; these lie in a bounded set, so a subsequence converges, necessarily to a point of the closed set $\Gamma$ . But the complement of $O$ is closed, so the limit would lie outside $O$ while lying in $\Gamma \subset O$ — impossible. Fix such an $r$ , put $\epsilon = r/2$ , and set

C = \bigl\{\, X \in \mathbb{R}^n : \operatorname{dist}(X, \Gamma) \le 2\epsilon \,\bigr\} \subseteq O , \qquad W = \bigl\{\, X \in \mathbb{R}^n : \operatorname{dist}(X, \Gamma) < 2\epsilon \,\bigr\} .

The set $C$ is closed and bounded, hence compact, and $W$ is open because $X \mapsto \operatorname{dist}(X, \Gamma)$ is continuous; thus $\Gamma \subset W \subseteq C \subseteq O$ . Since $F$ is $C^1$ it is locally Lipschitz, so by the lemma $F$ has a Lipschitz constant $K$ on $C$ , hence also on the open set $W$ — which is the shape part 1 asks for, since part 1 is stated for an open set. This $K$ depends on the whole track, not only on $Y_0$ .

Choice of the neighbourhood. Pick $\delta > 0$ with

\delta \le \tfrac{1}{2}\epsilon \qquad\text{and}\qquad \delta\, e^{K(t_1 - t_0)} \le \tfrac{1}{2}\epsilon ,

and take for $U$ the open ball centred at $Y_0$ whose radius is $\delta$ ; since $Y_0 \in \Gamma$ and $\delta < 2\epsilon$ , we have $U \subseteq W \subseteq O$ . Fix $Z_0 \in U$ and let $Z$ be the maximal solution with $Z(t_0) = Z_0$ , defined on $[t_0, \beta)$ to the right, where $(\alpha_Z, \beta)$ is its maximal interval.

Bootstrap: $Z$ stays in the tube. Write $I = [t_0, t_1] \cap [t_0, \beta)$ for the part of $[t_0, t_1]$ on which both solutions are defined; it is $[t_0, t_1]$ if $\beta > t_1$ , and $[t_0, \beta)$ otherwise. Let

S = \bigl\{\, \tau \in I \ :\ |Z(t) - Y(t)| \le \epsilon \text{ for all } t \in [t_0, \tau] \,\bigr\} , \qquad T = \sup S .

The set $S$ contains $t_0$ , because $|Z(t_0) - Y(t_0)| = |Z_0 - Y_0| < \delta \le \epsilon$ , so $T$ is well defined; and $S$ is an interval with left endpoint $t_0$ , since its defining condition at $\tau$ implies the same condition at every smaller $\tau$ .

Let $t \in [t_0, T)$ . Some $\tau \in S$ satisfies $t \le \tau$ , so the gap is at most $\epsilon$ at every point of $[t_0, t]$ . Then both solutions take their values in $W$ throughout $[t_0, t]$ : $Y(s) \in \Gamma$ , and $\operatorname{dist}(Z(s), \Gamma) \le |Z(s) - Y(s)| \le \epsilon < 2\epsilon$ . So part 1 applies on the closed interval $[t_0, t]$ with the open set $W$ in place of $O$ and with the Lipschitz constant $K$ , and it gives

\bigl| Z(s) - Y(s) \bigr| \le |Z_0 - Y_0|\, e^{K(s - t_0)} < \delta\, e^{K(t_1 - t_0)} \le \tfrac{1}{2}\epsilon \qquad (t_0 \le s \le t) ,

the middle step using $|Z_0 - Y_0| < \delta$ together with $s - t_0 \le t_1 - t_0$ and $K \ge 0$ .

The gap is therefore at most $\tfrac{1}{2}\epsilon$ on $[t_0, T)$ , and also at $t_0$ itself, where it equals $|Z_0 - Y_0| < \delta \le \tfrac{1}{2}\epsilon$ (this covers the case $T = t_0$ , where the previous display says nothing). By continuity it is then at most $\tfrac{1}{2}\epsilon$ at $T$ as well whenever $T \in I$ . Suppose $T$ were not the right endpoint of $I$ . Then $T \in I$ , the gap at $T$ is at most $\tfrac{1}{2}\epsilon < \epsilon$ , and by continuity the gap stays below $\epsilon$ on $[t_0, T + h]$ for some small $h > 0$ with $T + h \in I$ ; that puts $T + h$ in $S$ and contradicts $T = \sup S$ . So $S$ exhausts $I$ , and

\bigl| Z(t) - Y(t) \bigr| \le |Z_0 - Y_0|\, e^{K(t - t_0)} \le \tfrac{1}{2}\epsilon , \qquad Z(t) \in W \subseteq C ,

at every $t \in I$ .

$Z$ survives to $t_1$ . Suppose $\beta \le t_1$ . Then $\beta < \infty$ , $I = [t_0, \beta)$ , and the bootstrap has just put $Z(t)$ in the compact set $C \subset O$ for every $t \in [t_0, \beta)$ . Now $t_0$ lies in the maximal interval $(\alpha_Z, \beta)$ of $Z$ , so part 2, applied to $Z$ with this compact set and with $\gamma = t_0$ , produces a $t \in [t_0, \beta)$ with $Z(t) \notin C$ — a contradiction. This is the step that needs the one-sided form of part 2: the bootstrap controls $Z$ only after $t_0$ , and nothing at all is known about its track before $t_0$ , which may well leave every compact subset of $O$ backwards in time. Therefore $\beta > t_1$ , so $Z$ is defined on all of $[t_0, t_1]$ , it is unique there by Existence and uniqueness of solutions, and the estimate above holds on $[t_0, t_1]$ in the stated form $|Y(t) - Z(t)| \le |Y_0 - Z_0|e^{K(t-t_0)}$ .

Continuity of the flow in $X$ . Given $X$ and $t \in J(X)$ , apply what has just been proved to the solution $Y = \phi(\cdot, X)$ on the closed interval with endpoints $0$ and $t$ . For $Z_0 \in U$ we get $\phi(t, Z_0)$ defined and

\bigl| \phi(t, Z_0) - \phi(t, X) \bigr| \le |Z_0 - X|\, e^{K|t|} ,

whose right-hand side tends to $0$ as $Z_0 \to X$ . So $\phi$ is continuous in its second argument. $\square$

Proof(Part 4: dependence on the field and on parameters)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §17.3, p. 397, and §17.5, p. 402

Forward time. Take $t \in J$ with $t \ge 0$ . Both solutions satisfy their integral equations and start at the same point $X_0$ , so that term cancels on subtraction:

X(t) - Y(t) = \int_0^{t} \Bigl[ F\bigl(s, X(s)\bigr) - G\bigl(s, Y(s)\bigr) \Bigr] ds .

Insert and remove $F(s, Y(s))$ inside the bracket, so that one difference compares $F$ with itself at two points and the other compares the two fields at one point:

\bigl| X(t) - Y(t) \bigr| \ \le \ \int_0^{t} \bigl| F(s, X(s)) - F(s, Y(s)) \bigr| ds \ + \ \int_0^{t} \bigl| F(s, Y(s)) - G(s, Y(s)) \bigr| ds .

w(t) \ \le \ \varepsilon t + \int_0^{t} K\,w(s)\,ds .

The inhomogeneous term $\varepsilon t$ is not constant, so Gronwall's inequality does not apply to $w$ directly. Apply it instead to the shifted function $u = w + \varepsilon/K$ , which is continuous and nonnegative. Substituting $w = u - \varepsilon/K$ on both sides,

u(t) - \frac{\varepsilon}{K} \ \le \ \varepsilon t + \int_0^{t} K\,u(s)\,ds - \int_0^{t} \varepsilon\,ds = \varepsilon t + \int_0^{t} K\,u(s)\,ds - \varepsilon t ,

and the two copies of $\varepsilon t$ cancel. This is exactly the hypothesis of Gronwall's inequality for $u$ with $C = \varepsilon/K$ , so $u(t) \le (\varepsilon/K)e^{Kt}$ , that is

w(t) \ \le \ \frac{\varepsilon}{K}\bigl(e^{Kt} - 1\bigr) \qquad (t \ge 0) .

The cancellation is where $K > 0$ is used: the shift $\varepsilon/K$ has to be a finite number.

Backward time. Reverse the clock. Put $\tilde X(\tau) = X(-\tau)$ , $\tilde Y(\tau) = Y(-\tau)$ , $\tilde F(\tau, W) = -F(-\tau, W)$ and $\tilde G(\tau, W) = -G(-\tau, W)$ . By the chain rule $\tilde X' = \tilde F(\tau, \tilde X)$ and $\tilde Y' = \tilde G(\tau, \tilde Y)$ , the two new fields again differ by less than $\varepsilon$ , and $\tilde F$ again has Lipschitz constant $K$ in its second argument, since negation changes neither a difference's norm nor the fields' separation. The forward case applied to the tilded system gives the bound for $t \le 0$ with $|t|$ in place of $t$ , which is the statement.

Parameters. Let $F(X, a)$ be $C^1$ in $X$ and $a$ jointly, with $a$ ranging over an open $A \subseteq \mathbb{R}^m$ , and consider on the open set $O \times A \subseteq \mathbb{R}^{n+m}$ the augmented autonomous system

X' = F(X, a) , \qquad a' = 0 .

Its right-hand side is $C^1$ , so part 3 applies to it. Its solutions are precisely the pairs $t \mapsto (\phi_a(t, X), a)$ , where $\phi_a$ is the flow of the original system at parameter value $a$ : the second equation freezes $a$ , and the first is then the original equation. The initial point of such a solution is $(X, a)$ . Hence part 3 says that $(X, a) \mapsto (\phi_a(t, X), a)$ is continuous, that is, $\phi_a(t, X)$ is continuous in $X$ and $a$ together. A parameter has been turned into an initial condition. $\square$

Remark(Where the hypotheses are used)Original

The Lipschitz condition carries part 1 alone. It is the only property of $F$ used, and nothing weaker will do. For $F(x) = 3x^{2/3}$ on $\mathbb{R}$ , which is continuous but not Lipschitz near $0$ , the initial-value problem $x' = F(x)$ , $x(0) = 0$ has both $x \equiv 0$ and $x(t) = t^3$ as solutions (Existence and uniqueness of solutions). Their initial separation is $0$ and their separation at $t = 1$ is $1$ , so no bound of the form $|Y_0 - Z_0|e^{Kt}$ can hold for any $K$ . Continuous dependence is not a bonus attached to existence; it is attached to uniqueness.
$u \ge 0$ and $K \ge 0$ in Gronwall. Nonnegativity of $u$ gives $U' \ge 0$ , hence $U \ge C > 0$ , which is what licenses dividing by $U$ and taking the logarithm. Nonnegativity of $K$ is what turns $u \le U$ into $Ku/U \le K$ . With $K < 0$ the conclusion is false: take $\alpha = 1$ , $u \equiv 1$ , $C = 2$ and $K = -1$ . The hypothesis reads $1 \le 2 - t$ , true on $[0,1]$ , while the asserted conclusion $1 \le 2e^{-t}$ fails for $t > \log 2 \approx 0.693$ .
$\beta < \infty$ in part 2. Drop it and the conclusion fails outright. The equation $x' = 0$ on $O = \mathbb{R}$ has the solution $x \equiv 1$ on the maximal interval $(-\infty, \infty)$ , whose entire track $\{1\}$ sits inside the compact set $[0, 2] \subset O$ : this solution leaves no compact set at any time, late or early. Finiteness of $\beta$ enters the proof at the bounded-speed step, where it is what turns the estimate $|Y(t) - Y(s)| \le M(t-s)$ into the Cauchy criterion as $t \to \beta^-$ , and so produces the limit point from which the solution is continued.
Compactness, not boundedness, in part 2. "Leaves every compact subset of $O$ " is weaker than "blows up". On $O = (0, \infty)$ the equation $x' = -1$ with $x(0) = 1$ has maximal interval $(-\infty, 1)$ and the solution $x(t) = 1 - t$ , which stays bounded as $t \to \beta^- = 1^-$ (on the whole maximal interval it is unbounded, since $x \to +\infty$ as $t \to -\infty$ ; what matters is the approach to the finite endpoint); it escapes every compact $C \subset O$ by running into the boundary point $0$ , which is missing from $O$ . Only when $O = \mathbb{R}^n$ does part 2 force $|Y(t)| \to \infty$ along a sequence.
The closed interval in part 3. The interval $[t_0, t_1]$ must lie inside the maximal interval of the given solution $Y$ , endpoints included. If $t_1$ is the blow-up time itself, no neighbourhood $U$ works: in the example below, $Y_0 = 1$ for $y' = y^2$ blows up at $t = 1$ , and every neighbourhood of $1$ contains initial values $z_0 > 1$ whose solutions blow up at $1/z_0 < 1$ .
$K$ is global in part 1 but manufactured in part 3. Part 1 assumes a Lipschitz constant on all of $O$ , which few interesting fields have. Part 3 assumes only $C^1$ and builds $K$ on a compact tube around the track of $Y$ ; this is why both $U$ and $K$ there depend on the entire solution $Y|_{[t_0,t_1]}$ , not just on $Y_0$ , and why the estimate degrades as $t_1$ grows.

Remark(Two-sided form, and the scalar statement)AdaptedAdapted from Hassani (2013), §14.2, Theorem 14.2.4, pp. 420-421

Part 1 as stated runs forward from $t_0$ . The same bound holds backward, by the time reversal used in the proof of part 4: $X \mapsto -F(X)$ has the same Lipschitz constant. Hassani states the scalar case in that two-sided form at once: if $f$ and $g$ both solve $y' = F(x, y)$ in a domain $D$ on which $F$ satisfies a Lipschitz condition with constant $L$ , then

\bigl| f(x) - g(x) \bigr| \le e^{L|x - a|}\,\bigl| f(a) - g(a) \bigr| ,

from which he reads off that at most one solution curve passes through a given point of $D$ — uniqueness as the special case $f(a) = g(a)$ of continuous dependence. Our part 1 and his theorem are the same statement, written for a system and for a scalar equation respectively.

Hassani also notes, with the example $dy/dx = e^y$ , that under continuity alone the interval on which a solution is defined moves with the initial value; that is the phenomenon parts 2 and 3 are built to handle, and it persists under the Lipschitz hypothesis, as the worked example here shows.

Method(Turning a Lipschitz constant into a stability estimate)Original

To bound how far two solutions of nearby problems can drift apart over a time $T$ :

Write both as integral equations and subtract. The initial values appear as a constant term, the fields inside an integral.
Name a region both solutions are known to stay in over $[t_0, t_0 + T]$ : a tube of radius $\epsilon$ around a track you already know, an invariant region supplied by the physics, or a sublevel set of a conserved quantity. Everything after this is conditional on that confinement, which the bootstrap in part 3 is the general way to justify.
Compute a Lipschitz constant $K$ on that region. For $C^1$ fields on a convex region the mean value inequality gives $K = \sup \lVert DF \rVert$ , the supremum of the operator norm of the Jacobian. Use the smallest region that works: $K$ sits in an exponent, so a loose $K$ is expensive.
Bound the mismatch of the data. Collect the initial-value gap $|Y_0 - Z_0|$ and, if the fields differ, a uniform bound $\varepsilon$ on $|F - G|$ over the region.
Apply Gronwall. With $v = |Y - Z|$ you will have $v(t) \le v(t_0) + \varepsilon (t - t_0) + \int_{t_0}^{t} K v$ . If $\varepsilon = 0$ , read off $v(t) \le v(t_0)e^{K(t-t_0)}$ . Otherwise apply Gronwall to $v + \varepsilon/K$ as in the proof of part 4, giving $v(t) \ \le \ \Bigl( v(t_0) + \frac{\varepsilon}{K} \Bigr) e^{K(t - t_0)} - \frac{\varepsilon}{K} .$
Invert for the design question. To hold $v \le \eta$ out to time $T$ with $\varepsilon = 0$ , start within $\eta e^{-KT}$ ; to hold it with $v(t_0) = 0$ , keep the field error below $\eta K/(e^{KT} - 1)$ .

Example(A blow-up equation: how much slack is in the bound)Original

(a) Two initial values. Take $n = 1$ , $O = \mathbb{R}$ and $F(y) = y^2$ , which is $C^1$ and so locally Lipschitz, but has no Lipschitz constant on all of $\mathbb{R}$ . Separating variables gives the maximal solutions

Y(t) = \frac{1}{1 - t} \quad \bigl(Y(0) = 1\bigr) , \qquad Z(t) = \frac{11}{10 - 11\,t} \quad \Bigl(Z(0) = \tfrac{11}{10}\Bigr) .

Both check directly: $Y' = (1-t)^{-2} = Y^2$ and $Z' = 121(10 - 11t)^{-2} = Z^2$ . Their maximal intervals are $(-\infty, 1)$ and $(-\infty, \tfrac{10}{11})$ , so the right endpoint $\beta(y_0) = 1/y_0$ moves with the initial value, and each solution leaves every compact subset of $\mathbb{R}$ as $t$ approaches it — part 2 in the simplest possible instance.

Fix the window $[t_0, t_1] = [0, \tfrac{2}{5}]$ and the open interval $O' = (-2, 2)$ , on which part 1 will be applied in place of $O$ . Both solutions increase on this window, and

Y\bigl(\tfrac{2}{5}\bigr) = \frac{1}{1 - 2/5} = \frac{5}{3} , \qquad Z\bigl(\tfrac{2}{5}\bigr) = \frac{11}{10 - 22/5} = \frac{55}{28} ,

both smaller than $2$ , so both tracks lie in $[1, \tfrac{55}{28}] \subset O'$ — the containment part 1 needs, and the reason for taking $O'$ open rather than the closed interval $[-2,2]$ . On $O'$ the field has Lipschitz constant $4$ : for $y, z \in O'$ ,

\bigl| y^2 - z^2 \bigr| = |y + z|\,|y - z| \le 4\,|y - z| , \qquad\text{so}\qquad K = 4 .

Part 1 therefore predicts, with $|Y(0) - Z(0)| = \tfrac{1}{10}$ ,

\bigl| Y(t) - Z(t) \bigr| \ \le \ \tfrac{1}{10}\,e^{4t} , \qquad\text{so at } t = \tfrac{2}{5}: \quad \tfrac{1}{10}\,e^{8/5} = 0.495303\ldots

The true gap is

Z\bigl(\tfrac{2}{5}\bigr) - Y\bigl(\tfrac{2}{5}\bigr) = \frac{55}{28} - \frac{5}{3} = \frac{165 - 140}{84} = \frac{25}{84} = 0.297619\ldots

The bound holds, and is about $1.66$ times the truth: the worst case allowed by the Lipschitz constant is attained only by a field that pushes the two solutions apart at the maximal rate everywhere, which $y^2$ does only at the top of the range.

(b) A perturbed field. Now take $F(x) = -x$ and $G(x) = -x + \tfrac{1}{100}$ on $O = \mathbb{R}\times\mathbb{R}$ , both starting from $X(0) = Y(0) = 1$ . Here $F$ is Lipschitz in $x$ with $K = 1$ , and $|F - G| = \tfrac{1}{100}$ identically. Part 4 asks for the strict bound $|F - G| < \varepsilon$ , so apply it with $\varepsilon = \tfrac{1}{100} + \eta$ for arbitrary $\eta > 0$ and let $\eta \downarrow 0$ ; the conclusion is continuous in $\varepsilon$ , so it holds with $\varepsilon = \tfrac{1}{100}$ . The exact solutions are

X(t) = e^{-t} , \qquad Y(t) = \tfrac{1}{100} + \tfrac{99}{100}\,e^{-t} ,

the second because $Y' = -\tfrac{99}{100}e^{-t} = -Y + \tfrac{1}{100}$ and $Y(0) = 1$ . Their gap is $Y(t) - X(t) = \tfrac{1}{100}(1 - e^{-t})$ , so at $t = 1$

\bigl| X(1) - Y(1) \bigr| = \tfrac{1}{100}\bigl(1 - e^{-1}\bigr) = 0.0063212\ldots ,

while part 4 allows

\frac{\varepsilon}{K}\bigl( e^{K} - 1 \bigr) = \tfrac{1}{100}(e - 1) = 0.0171828\ldots

The ratio of the two is $(1 - e^{-1})/(e - 1) = e^{-1} = 0.367879\ldots$ exactly. The slack has a clear cause: the bound assumes the perturbation always pushes away from the unperturbed solution, whereas here the field $-x$ is contracting and pulls the two trajectories back together.

Application(Predictability horizons and the cost of a long integration)AdaptedAdapted from Strogatz (1994), §9.3, pp. 320-322

Gronwall's inequality is the reason "the solution depends continuously on the data" is a usable engineering statement rather than a reassurance. It says that the damage done by an uncertain initial state is at worst exponential in time with rate the Lipschitz constant, and it says so with a computable constant.

Reading the estimate as a forecast horizon. Suppose the state of a system is known to within $\delta$ and a prediction is useful while the error stays below $\eta$ . Part 1 guarantees usefulness for a time at least

T = \frac{1}{K}\,\log\frac{\eta}{\delta} .

The dependence on $\delta$ is logarithmic. Reducing the initial uncertainty by a factor of $1000$ — a very expensive improvement in instrumentation — buys only $\log(1000)/K$ extra units of time. This is the quantitative content of the complaint that better measurements do not buy proportionally better forecasts.

When the bound is sharp. For most systems the estimate is pessimistic, as in part (a) of the example above. Chaotic systems are exactly the ones where something like it is attained. On the Lorenz attractor, numerical experiment shows that a separation vector of initial length $\lVert \delta_0 \rVert$ grows as $\lVert \delta(t) \rVert \approx \lVert \delta_0 \rVert e^{\lambda t}$ with $\lambda \approx 0.9$ Strogatz, §9.3 (Strogatz (1994), pp. 320-322), the rate now measured rather than bounded. Taking $\lambda$ in place of $K$ and asking for $10^{6}$ -fold initial accuracy to be spent down to order-one error gives a horizon of only $\log(10^6)/0.9 \approx 15.4$ time units. Strogatz is careful that the growth is not exactly exponential and must saturate once the separation reaches the diameter of the attractor, so the exponential law describes an initial transient; see Lyapunov exponents and sensitive dependence and The Lorenz equations.

Model error and numerical error. Part 4 is the same statement for an uncertainty in the equations rather than in the state. A solver integrates not $X' = F(t, X)$ but a perturbed field $G$ differing from $F$ by rounding, by a truncated force model, or by interpolation of tabulated coefficients; if $|F - G| \le \varepsilon$ over the region visited, the trajectories can differ by no more than $(\varepsilon/K)(e^{KT} - 1)$ over a time $T$ . That is the characteristic shape of a global error bound for a one-step method, where $\varepsilon$ plays the role of the local error per unit step (Integrators for ODEs). Its exponential growth in $T$ , with no way to improve the exponent without improving $K$ , is why long integrations of chaotic systems are reported as statistics over an ensemble rather than as single trajectories.

Why parameter continuity matters separately. Part 4's second half says that a model's output varies continuously with its parameters. Without it, fitting a parameter to data would be meaningless: the fitting procedure assumes that a small change in the parameter produces a small change in the predicted trajectory over the observation window, which is exactly the augmented-system statement proved above.