Initial Value Problem | Sahithyan's Notes

Also known as IVP. A first-order ODE together with the solution value at the left endpoint.

y'(t) = f(t, y), \quad a \leq t \leq b, \quad y(a) = \alpha

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

|f(t, y_1) - f(t, y_2)| \leq L|y_1 - y_2|

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 $a \leq t \leq b$ .

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

z'(t) = f(t, z) + \delta(t), \quad a \leq t \leq b, \quad z(a) = \alpha + \delta_0

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:

t \in [a, b]

Here:

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

Well-Posed Problem

The IVP is well-posed iff:

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

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.

If the Lipschitz condition fails, check the definition directly.

Worked Examples

Example 1

Find a Lipschitz constant in $y$ for $f(t,y) = t^2 - 3y$ on $D = [0,2] \times \mathbb{R}$ .

Solution Reveal Hide

\frac{\partial f}{\partial y}(t,y) = -3

\left| \frac{\partial f}{\partial y}(t,y) \right| = 3

The partial derivative test gives $L = 3$ .

Example 2

Show that the IVP $y' = 1 + t\sin(ty)$ , $0 \leq t \leq 2$ , $y(0) = 0$ has a unique solution.

Solution Reveal Hide

\left| \dfrac{\partial f}{\partial y}(t, y) \right| = |t^2 \cos(ty)| \leq 4

on $[0, 2] \times \mathbb{R}$ . Since $f(t, y) = 1 + t\sin(ty)$ is continuous, $f$ is Lipschitz in $y$ with $L = 4$ . The IVP has a unique solution on $[0, 2]$ .

Example 3

Use the Lipschitz condition to decide whether $f(t,y) = \sin(t) + 4y$ has a unique solution for every initial value on $[0,1]$ .

Solution Reveal Hide

$f$ is continuous on $[0,1] \times \mathbb{R}$ .

\frac{\partial f}{\partial y}(t,y) = 4

\left| \frac{\partial f}{\partial y}(t,y) \right| = 4

$f$ is Lipschitz in $y$ with $L = 4$ . Every IVP with this right-hand side has a unique solution on $[0,1]$ .

Example 4

Show that $f(t,y) = \sqrt{|y|}$ does not satisfy a Lipschitz condition in $y$ on $D = [0,1] \times [0,1]$ .

Solution Reveal Hide

Take $y_1 = h^2$ and $y_2 = 0$ , where $0 < h \leq 1$ . Then

|f(t,y_1) - f(t,y_2)| = |h - 0| = h

|y_1-y_2| = h^2

If a Lipschitz constant $L$ existed, then $h \leq Lh^2$ . Hence $1 \leq Lh$ .

This fails when $h < \dfrac{1}{L}$ . No Lipschitz constant exists on $D$ .

Example 5

Show the IVP $y'(t) = y - t^2 + 1$ , $0 \leq t \leq 2$ , $y(0) = 0.5$ is well-posed.

Solution Reveal Hide

$f(t, y) = y - t^2 + 1$ 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 $\delta_0$

z'(t) = z - t^2 + 1 + \delta, \quad z(0) = 0.5 + \delta_0

Both problems have closed-form solutions

y(t) = (t+1)^2 - 0.5 e^t, \qquad z(t) = (t+1)^2 - \delta + (\delta_0 + \delta - 0.5)\, e^t

|z(t) - y(t)| = \left| (\delta_0 + \delta)\, e^t - \delta \right| \leq (2 e^2 + 1)\, \varepsilon

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 $\varepsilon > 0$ .

Example 6

Show the IVP $y'(t) = t e^{3t} - 2y$ , $0 \leq t \leq 1$ , $y(0) = 0$ is well-posed.

Solution Reveal Hide

$f(t, y) = t e^{3t} - 2y$ is continuous on $D = [0, 1] \times \mathbb{R}$ .

\left| \frac{\partial f}{\partial y} \right| = |-2| = 2

$f$ is Lipschitz in $y$ with $L = 2$ , so the IVP is well-posed.