Euler's Method | Sahithyan's Notes

Euler’s method approximates the solution of a well-posed IVP at equally spaced mesh points, using the first-order Taylor expansion.

Mesh Points

Choose a positive integer $N$ . The step size is $h = \dfrac{b - a}{N}$ . The mesh points are

t_i = a + ih, \quad i = 0, 1, \ldots, N

with $t_0 = a$ , $t_N = b$ , and $t_{i+1} - t_i = h$ .

Approximations are produced only at the mesh points. Values between them come from interpolation.

Difference Equation

First-order Taylor expansion of $y$ about $t_i$ , assuming $y$ is twice differentiable

y(t_{i+1}) = y(t_i) + h f(t_i, y(t_i)) + \frac{h^2}{2} y''(\xi_i), \quad \xi_i \in (t_i, t_{i+1})

Dropping the second-order term and writing $w_i \approx y(t_i)$

\begin{aligned} w_0 &= \alpha \\ w_{i+1} &= w_i + h f(t_i, w_i), \quad i = 0, 1, \ldots, N-1 \end{aligned}

Theoretical Error Bound

If $f$ satisfies a Lipschitz condition with constant $L > 0$ and $|y''(t)| \leq M$ for some $M > 0$ , then for each $i = 0, 1, \ldots, N$

|y(t_i) - w_i| \leq \frac{hM}{2L}\left( e^{L(t_i - a)} - 1 \right)

$M$ : bound on $|y''|$ over $[a, b]$
$L$ : Lipschitz constant of $f$

The bound is proportional to $h$ . Euler’s method has order 1.

Computational Error Bound

With round-off, each step carries an error bounded by $\delta$ . The computed values $u_i$ satisfy

\begin{aligned} u_0 &= \alpha + \delta_0 \\ u_{i+1} &= u_i + h f(t_i, u_i) + \delta_{i+1}, \quad |\delta_i| < \delta \end{aligned}

Then for each $i = 0, 1, \ldots, N$

|y(t_i) - u_i| \leq \frac{1}{L}\left( \frac{hM}{2} + \frac{\delta}{h} \right)\left( e^{L(t_i - a)} - 1 \right) + |\delta_0|\, e^{L(t_i - a)}

As $h \to 0$ the factor $\dfrac{hM}{2} + \dfrac{\delta}{h} \to \infty$ . Error grows with decreasing $h$ once $h$ is small enough. The minimizing step size is

h = \sqrt{\frac{2\delta}{M}}

Worked Example

IVP $y'(t) = y - t^2 + 1$ , $0 \leq t \leq 2$ , $y(0) = 0.5$ , with $N = 4$ , so $h = 0.5$ and $f(t, y) = y - t^2 + 1$ .

\begin{aligned} w_0 &= 0.5 \\ w_1 &= w_0 + 0.5\,(w_0 - 0.0^2 + 1) = 1.25 \\ w_2 &= w_1 + 0.5\,(w_1 - 0.5^2 + 1) = 2.25 \\ w_3 &= w_2 + 0.5\,(w_2 - 1.0^2 + 1) = 3.375 \\ w_4 &= w_3 + 0.5\,(w_3 - 1.5^2 + 1) = 4.4375 \end{aligned}

Hence $y(2) \approx w_4 = 4.4375$ . The exact solution is $y(t) = (t+1)^2 - 0.5 e^t$ , giving $y(2) \approx 5.3055$ .