Integral Equations and Picard Iteration

Definition Let $f$ be continuous on a region $D$ of the $ty$-plane that contains $(t_0, y_0)$, and let $J$ be an interval containing $t_0$. A continuous function $y$ on $J$ whose graph lies in $D$ satisfies the integral form of the problem if

for every $t$ in $J$. The letter $s$ is the variable of integration and $t$ is the upper limit. The integral form is equivalent to the initial-value problem $y' = f(t, y)$, $y(t_0) = y_0$ on $J$, with one-sided derivatives at an endpoint of $J$. If $y$ is a differentiable solution of the problem, then $y'(t) = f(t, y(t))$ is continuous, and integrating it from $t_0$ to $t$ gives the integral equation. Conversely, if a continuous $y$ satisfies the integral equation, the integrand $s \mapsto f(s, y(s))$ is continuous, so by the fundamental theorem of calculus the right side is differentiable with derivative $f(t, y(t))$. Hence $y' = f(t, y)$, and setting $t = t_0$ makes the integral vanish and gives $y(t_0) = y_0$.

Definition Let $f$ be continuous near $(t_0, y_0)$ and let $J$ be an interval containing $t_0$. For a continuous function $u$ on $J$ whose graph stays where $f$ is defined, the Picard operator $\mathcal{T}$ produces the function

on $J$. Picard iteration chooses a continuous starting function $\varphi_0$ on $J$, often the constant $y_0$, and defines the successive approximations $\varphi_{n+1} = \mathcal{T}\varphi_n$ for $n = 0, 1, 2, \dots$. Every $\varphi_n$ with $n \ge 1$ has the correct value $\varphi_n(t_0) = y_0$. By the integral form, a continuous function $y$ with graph in the region solves the initial-value problem on $J$ exactly when $\mathcal{T}y = y$. The iteration can be carried out only while each new graph stays in the region where $f$ is defined, and a convergence proof has to check that.

Definition Let $J$ be a closed and bounded interval. For a continuous function $u$ on $J$, the sup norm is

which exists by the extreme value theorem. The distance between continuous functions $u$ and $v$ is $\lVert u - v \rVert_\infty$, the largest gap between their graphs over $J$. A sequence $u_n$ converges to $u$ in this norm exactly when it converges to $u$ uniformly on $J$. The continuous functions on $J$ form a complete space under this norm. Every Cauchy sequence, one with $\lVert u_n - u_m \rVert_\infty \to 0$ as $n$ and $m$ grow, converges uniformly to a continuous function on $J$. A set of continuous functions defined by a closed condition, such as $\lvert u(t) - y_0 \rvert \le b$ for every $t$ in $J$, is complete as well, because a uniform limit still satisfies the condition.

Definition Let $a > 0$ and $b > 0$, and let $f$ be continuous on the rectangle $Q$ of points $(t, y)$ with $\lvert t - t_0 \rvert \le a$ and $\lvert y - y_0 \rvert \le b$. Suppose $\lvert f \rvert \le M$ on $Q$ and

for all points $(t, y_1)$ and $(t, y_2)$ of $Q$. Take $0 < h \le a$, let $J = [t_0 - h, t_0 + h]$, and let $X$ be the set of continuous functions $u$ on $J$ with $\lvert u(t) - y_0 \rvert \le b$ for every $t$ in $J$. If $Mh \le b$, then $\mathcal{T}$ maps $X$ into $X$, because

For $u$ and $v$ in $X$ the Lipschitz bound gives $\lvert (\mathcal{T}u)(t) - (\mathcal{T}v)(t) \rvert \le L \lvert t - t_0 \rvert \, \lVert u - v \rVert_\infty$, and therefore

If also $Lh < 1$, the operator $\mathcal{T}$ is a contraction on $X$ with factor $q = Lh$. The same estimates hold on the one-sided interval $[t_0, t_0 + h]$.

Definition A fixed point of an operator $\mathcal{T}$ on a set $X$ is an element $y$ of $X$ with $\mathcal{T}y = y$. The contraction mapping theorem, in the form used here, states the following. Let $X$ be a nonempty set of continuous functions on a closed and bounded interval that is closed under uniform limits, and let $\mathcal{T}$ map $X$ into $X$ with $\lVert \mathcal{T}u - \mathcal{T}v \rVert_\infty \le q \lVert u - v \rVert_\infty$ for all $u$ and $v$ in $X$, where $0 \le q < 1$. Then $\mathcal{T}$ has exactly one fixed point $y$ in $X$, and for every $\varphi_0$ in $X$ the iterates $\varphi_{n+1} = \mathcal{T}\varphi_n$ converge uniformly to $y$ with

The reason is that the successive changes obey $\lVert \varphi_{k+1} - \varphi_k \rVert_\infty \le q^k \lVert \varphi_1 - \varphi_0 \rVert_\infty$, so all changes after step $n$ add up to at most the geometric tail above. The sequence is therefore Cauchy, and completeness gives a limit $y$ in $X$. Since $\lVert \mathcal{T}\varphi_n - \mathcal{T}y \rVert_\infty \le q \lVert \varphi_n - y \rVert_\infty \to 0$, letting $n \to \infty$ in $\varphi_{n+1} = \mathcal{T}\varphi_n$ gives $y = \mathcal{T}y$. Two fixed points $y$ and $z$ would satisfy $\lVert y - z \rVert_\infty \le q \lVert y - z \rVert_\infty$, which forces $y = z$. For the Picard operator under the hypotheses of the contraction estimate, the fixed points in $X$ are exactly the solutions of the initial-value problem on $J$ whose graphs lie in $Q$. So the problem has exactly one such solution, and Picard iteration from any $\varphi_0$ in $X$ converges to it uniformly on $J$. This is a constructive proof of the local part of the existence and uniqueness theorem. The limit is taken in the integral equation, where the Lipschitz bound controls the integrands uniformly, and no derivative of a limit is ever exchanged with a limit of derivatives.

Definition The problem $y' = f(t, y)$, $y(t_0) = y_0$ has local existence if it has a solution on some interval containing $t_0$, however short. It has a global solution on a given interval $I$ containing $t_0$, such as $[t_0, \infty)$, if one solution is defined on all of $I$. The contraction estimate proves local existence on $\lvert t - t_0 \rvert \le h$ whenever $h \le a$, $Mh \le b$ and $Lh < 1$. These conditions are sufficient for that proof, not necessary for convergence. Starting from the constant $\varphi_0 = y_0$, with $h \le a$ and $Mh \le b$, induction gives

for $\lvert t - t_0 \rvert \le h$. These bounds have a finite sum whatever the size of $Lh$, so the iterates converge on that whole interval. One way to guarantee global existence is a bound that holds for all states. If $f$ is continuous for $t$ in an open interval $I$ and all real $y$, and on each closed and bounded subinterval of $I$ it satisfies one Lipschitz bound valid for every $y$, then for every $t_0$ in $I$ and every $y_0$ the solution exists on all of $I$. Linear equations $y' = p(t)y + g(t)$ with $p$ and $g$ continuous on $I$ are of this kind. Without such a bound, local existence at every point does not prevent a solution from blowing up at a finite time.