Initial Value Problems

u'(t) = f(t, u(t)), \qquad a \le t \le b, \qquad u(a) = u_0

If the equation is linear, $u'(t) = g(t) + u\,h(t)$ , then with $\rho(t) = \exp\big(-\int_a^t h(s)\, ds\big)$ ,

\rho(t)u(t) = u_0 + \int_a^t \rho(s)g(s)\, ds

(see integrating factor).

If $\big|\frac{\partial f}{\partial u}\big| \le L$ for all $a \le t \le b$ and all $u$ , the solution exists and is unique, and the solution $u(t; u_0 + \delta)$ with perturbed initial value $u(a) = u_0 + \delta$ satisfies

\| u(t; u_0 + \delta) - u(t; u_0) \|_\infty \le |\delta| e^{L(b - a)}

Higher order An equation of order $m$ becomes a first-order system with $u_1 = y$ , $u_2 = y'$ , $u_3 = y'', \ldots$ , e.g. for $y'' = g(y, y')$ ,

u_1' = u_2, \qquad u_2' = g(u_1, u_2)

One-Step Methods

$t_i = a + ih$ , $h = \frac{b - a}{n}$ , $u_i \approx \hat{u}(t_i)$ with $\hat{u}$ the exact solution, and

u_{i+1} = u_i + h\,\phi(t_i, u_i, h), \qquad i = 0, \ldots, n - 1

The local truncation error substitutes the exact solution into the method and divides by $h$ ,

\tau_{i+1}(h) = \frac{\hat{u}(t_{i+1}) - \hat{u}(t_i)}{h} - \phi(t_i, \hat{u}(t_i), h)

and the method is consistent if $\tau \to 0$ as $h \to 0$ . If $|\tau_{i+1}(h)| \le Ch^p$ and $\big|\frac{\partial \phi}{\partial u}\big| \le L$ for all $a \le t \le b$ , $u$ and $h > 0$ , then the global error satisfies

|\hat{u}(t_i) - u_i| \le \frac{Ch^p}{L}\big[e^{L(t_i - a)} - 1\big] = O(h^p)

as $h \to 0$ , but not as $t \to \infty$ . So the global error has the same order of accuracy as the LTE.

Euler's Method

$\phi = f$ ,

u_{i+1} = u_i + hf(t_i, u_i), \qquad i = 0, \ldots, n - 1

By Taylor expansion,

\hat{u}(t_{i+1}) - [\hat{u}(t_i) + hf(t_i, \hat{u}(t_i))] = [\hat{u}(t_i) + h\hat{u}'(t_i) + \tfrac{1}{2}h^2\hat{u}''(t_i) + \cdots] - [\hat{u}(t_i) + h\hat{u}'(t_i)] = \tfrac{1}{2}h^2\hat{u}''(t_i) + O(h^3)

so the LTE is $O(h)$ and Euler is consistent and first order: an $O(h^2)$ error per step over $O(h^{-1}) = O(n)$ steps gives a global error of $O(h)$ .