Euler's Method Global Error: How to calculate $C_1$ if $error = C_1 h$

Question

My textbook claims that, for small step size $h$, Euler's method has a global error which is at most proportional to $h$ such that error $= C_1h$. It is then claimed that $C_1$ depends on the initial value problem, but no explanation is given as to how one finds $C_1$.

So if I know $h$, then how can I deduce $C_1$ from the IVP?

Answer 1

Given an IVP: $$\frac{dy}{dt}=f(t,y), y(a)=y_0, t\in [a,b].$$ Here is a Theorem from Numerical Analysis by Sauer:

Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|\le \frac{Mh}{2L}(e^{L(t_i-a)}-1).$$

The proof is based on the following lemma:

Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]\times [\alpha,\beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|\le e^{L(t-a)}|Y(a)-Z(a)|.$$

Sketch of proof of the first theorem:

Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP: $$z_i'=f(t,z_i),z_i(t_i)=w_i, t\in [t_i,t_{i+1}].$$

Then $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|\le |w_i-z_i(t)|+|z_i(t)-y_i|\\ \le e_i+e^{Lh}g_{i-1}\\ \le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})\le \cdots\\ \le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+\cdots +e^{(i-1)Lh}e_1.$$ Since each $e_i\le \frac{h^2M}{2}$, we have $$g_i\le \frac{h^2M}{2}(1+e^{Lh}+\cdots+e^{(i-1)Lh})=\frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}\le \frac{Mh}{2L}(e^{L(t_i-a)}-1).$$

Hope this helps.