21 - Global Error Bound for One-Step Methods

Global Error Bound for One-Step Methods

Let $\mathbf{\Psi }$ be a consistent one-step method

Assume increment function $\psi$ is Lipschitz continuous with respect to $\mathbf{y}$
So exists positive constant $L_{\psi }$ such that

$$||\psi (x,\mathbf{y},h,\mathbf{f})-\psi (x,\mathbf{z},h,\mathbf{f})||\le L_{\psi }||\mathbf{y}-\mathbf{z}||\text{ for }(x,\mathbf{y}),(x,\mathbf{z})\text{ in }R$$

for all $0\le h\le h_0$ and for the same $R$ from Picard’s Theorem

Then assuming $(x_n,{\mathbf{y}}_n)$ remains $R$ then

$$e\le \left(\frac{\exp (L_{\psi }(x_N-x_0))-1}{L_{\psi }}\right)\underset{n=0,\cdots ,N-1}{\max }||\tau (x_n,{\mathbf{y}}_n,h,\mathbf{f})||$$

Proof

For generic $n\in \{1,\cdots ,N-1\}$

$$\begin{array}{rl}{e}_{n+1}& =||\mathbf{y}(x_{n+1})-{\mathbf{y}}_{n+1}||\\ & =||\mathbf{y}(x_{n+1})-\mathbf{\Psi }(x_n,{\mathbf{y}}_n,h,\mathbf{f})||\\ & =||\mathbf{y}(x_{n+1})-\mathbf{\Psi }(x_n,\mathbf{y}(x_n),h,\mathbf{f})+\mathbf{\Psi }(x_n,\mathbf{y}(x_n),h,\mathbf{f})-\mathbf{\Psi }(x_n,{\mathbf{y}}_n,h,\mathbf{f})||\\ & \le ||\mathbf{y}(x_{n+1})-\mathbf{\Psi }(x_n,\mathbf{y}(x_n),h,\mathbf{f})||+||\mathbf{\Psi }(x_n,\mathbf{y}(x_n),h,\mathbf{f})-\mathbf{\Psi }(x_n,{\mathbf{y}}_n,h,\mathbf{f})||\\ & =h||\tau (x_n,\mathbf{y}(x_n),h,\mathbf{f})||+||(\mathbf{y}(x_n)+h\psi (x,\mathbf{y}(x_n),h,\mathbf{f}))-({\mathbf{y}}_n+h\psi (x,{\mathbf{y}}_n,h,\mathbf{f}))||\\ & \le h||\tau (x_n,\mathbf{y}(x_n),h,\mathbf{f})||+||\mathbf{y}(x_n)-{\mathbf{y}}_n||+h||\psi (x,\mathbf{y}(x_n),h,\mathbf{f})-\psi (x,{\mathbf{y}}_n,h,\mathbf{f})||\\ & =h||\tau (x_n,\mathbf{y}(x_n),h,\mathbf{f})||+e_n+h||\psi (x,\mathbf{y}(x_n),h,\mathbf{f})-\psi (x,{\mathbf{y}}_n,h,\mathbf{f})||\\ & \le h||\tau (x_n,\mathbf{y}(x_n),h,\mathbf{f})||+e_n+h{L}_{\psi }||\mathbf{y}(x_n)-{\mathbf{y}}_n||,\\ & =h||\tau (x_n,\mathbf{y}(x_n),h,\mathbf{f})||+(1+h{L}_{\psi })e_n\end{array}$$

Iterating recursively then implying (with $e_0=0$)

$$\begin{array}{rl}{e}_{n+1}& \le (1+h{L}_{\psi }{)}^{n+1}{e}_0+h\sum _{k=0}^n(1+h{L}_{\psi }{)}^k\underset{m=0,\cdots ,n}{\max }||\tau (x_m,\mathbf{y}(x_m),h,\mathbf{f})||\\ & =\frac{(1+h{L}_{\psi }{)}^{n+1}-1}{L_{\psi }}\underset{m=0,\cdot ,sn}{\max }||\tau (x_m,\mathbf{y}(x_m),h,\mathbf{f})||\end{array}$$

And as $1+h{L}_{\psi }\le \exp h{L}_{\psi }$ then result follows