Initial Value Problems

An initial value problem (IVP) is a first-order ODE together with the solution value at the left endpoint.

Most IVPs have no analytical solution. Numerical methods approximate $y$ at discrete points of $[a, b]$.

Lipschitz Condition

$f(t, y)$ satisfies a Lipschitz condition in $y$ on $D \subseteq \mathbb{R}^2$ iff a constant $L > 0$ exists with

for all $(t, y_1), (t, y_2) \in D$. $L$ is a Lipschitz constant for $f$.

Partial Derivative Test

If $D$ is convex and a constant $L > 0$ exists with $\left| \dfrac{\partial f}{\partial y}(t, y) \right| \leq L$ for all $(t, y) \in D$, then $f$ satisfies a Lipschitz condition in $y$ on $D$ with Lipschitz constant $L$.

Existence and Uniqueness

Let $D = [a, b] \times \mathbb{R}$. If $f$ is continuous on $D$ and satisfies a Lipschitz condition in $y$ on $D$, then the IVP has a unique solution $y(t)$ for $a \leq t \leq b$.

For example, consider: $y' = 1 + t\sin(ty)$, $0 \leq t \leq 2$, $y(0) = 0$.

Here $\left| \dfrac{\partial f}{\partial y} \right| = |t^2 \cos(ty)| \leq 4$ on $[0, 2] \times \mathbb{R}$, so a unique solution exists.

Stable Solution

Let $\delta(t)$ be a continuous perturbation of the equation and $\delta_0$ a perturbation of the initial value. The perturbed problem is

The solution $y(t)$ is stable iff constants $\varepsilon_0 > 0$ and $k > 0$ exist such that for every $\varepsilon \in (0, \varepsilon_0)$:

• whenever $|\delta(t)| < \varepsilon$ for all $t \in [a, b]$ and $|\delta_0| < \varepsilon$
• the perturbed problem has a unique solution $z(t)$
• $|z(t) - y(t)| < k\varepsilon$ for all $t \in [a, b]$

Here:

• $k$: stability constant, independent of $\varepsilon$

The perturbed solution stays within $k\varepsilon$ of the original. The change in the solution is bounded proportionally to the size of the perturbations.

Well-Posed Problem

The IVP is well-posed iff:

• a unique solution $y(t)$ exists
• the solution is stable

Check well-posedness before applying any numerical method.

Well-Posedness from Lipschitz

Let $D = [a, b] \times \mathbb{R}$. If $f$ is continuous and satisfies a Lipschitz condition in $y$ on $D$, then the IVP is well-posed.

A well-posed IVP may fail the Lipschitz condition. If Lipschitz fails, check the well-posed definition directly.

Worked Example

$y'(t) = y - t^2 + 1$, $0 \leq t \leq 2$, $y(0) = 0.5$.

$f(t, y) = y - t^2 + 1$ is continuous on $D = [0, 2] \times \mathbb{R}$, and $\left| \dfrac{\partial f}{\partial y} \right| = 1$. So $f$ is Lipschitz in $y$ with $L = 1$, and the IVP is well-posed.

To see the stability bound directly, take a constant equation perturbation $\delta$ and an initial perturbation $\delta_0$

Both problems have closed-form solutions

for $|\delta| < \varepsilon$ and $|\delta_0| < \varepsilon$, using $e^t \leq e^2$ on $[0, 2]$. The bound holds with $k = 2 e^2 + 1$ for every $\varepsilon > 0$.