9.1 The theory of initial-value problems
Before approximating a solution we should know that there is one, that it is unique, and that it does not change wildly when the data change slightly (since the data, and every computation, contain errors).
Definition 9.1.1. A function \(f(t,y)\) satisfies a Lipschitz condition in \(y\) on a set \(D\subseteq \mathbb R^2\) if there is a constant \(L>0\), the Lipschitz constant, with \[|f(t,y_1)-f(t,y_2)|\leq L|y_1-y_2|\qquad \text {for all }(t,y_1),(t,y_2)\in D .\]
If \(D\) is convex in \(y\) and \(\left |\frac {\partial f}{\partial y}\right |\leq L\) on \(D\), then \(f\) is Lipschitz with constant \(L\), by the Mean Value Theorem: \(f(t,y_1)-f(t,y_2) = \frac {\partial f}{\partial y}(t,\eta )(y_1-y_2)\).
Theorem 9.1.2 (Existence and uniqueness). Let \(D = \{(t,y): a\leq t\leq b,\ -\infty <y<\infty \}\) and let \(f\) be continuous on \(D\) and Lipschitz in \(y\). Then the initial-value problem (9.1) has a unique solution \(y(t)\) for \(a\leq t\leq b\).
The proof constructs the solution by Picard iteration, \(y_{k+1}(t) = \alpha +\int _a^tf(s,y_k(s))\,ds\), which is fixed-point iteration in a space of functions; the Lipschitz condition makes the map a contraction (compare the Fixed-Point Theorem 2.3.7). The details are given in courses on differential equations.
Definition 9.1.3. Problem (9.1) is well posed if it has a unique solution and if, for every \(\varepsilon >0\), there is \(k>0\) such that any perturbed problem \(\frac {dz}{dt} = f(t,z)+\delta (t)\), \(z(a) = \alpha +\varepsilon _0\), with \(|\varepsilon _0|<\varepsilon \) and \(|\delta (t)|<\varepsilon \), has a unique solution with \(|z(t)-y(t)|<k\varepsilon \) for all \(t\).
Under the hypotheses of Theorem 9.1.2 the problem is also well posed. Numerical methods always solve a perturbed problem, so well-posedness is what makes their results meaningful.
Example 9.1.4. Show that \(f(t,y) = y-t^2+1\) satisfies a Lipschitz condition on \(D = \{0\leq t\leq 2,\ y\in \mathbb R\}\), so that \[y' = y-t^2+1,\qquad 0\leq t\leq 2,\qquad y(0) = 0.5\] is well posed, and verify that its solution is \(y(t) = (t+1)^2-0.5e^t\).
Solution. \(\frac {\partial f}{\partial y} = 1\), so \(L = 1\) will do, and \(f\) is continuous; Theorem 9.1.2 applies. For the solution, \(y' = 2(t+1)-0.5e^t\) and \(y-t^2+1 = t^2+2t+1-0.5e^t-t^2+1 = 2t+2-0.5e^t\): they agree, and \(y(0) = 1-0.5 = 0.5\). This problem is used as the test problem for every method of this section.
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