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Let $x(t)$ be the solution to a differential equation: $$ x'(t) = f(t, x(t)) $$ where $f:[t_0, T] \to \mathbb{R}^n$ is a $C^1$ function.

We consider the problem of approximating $x$ using a partition $t_i = t_0 + ih$ for $i = 0 \dots N$ using the $\theta$-method succession $${x_{i+1}} = x_i + h ((1-\theta) f(t_i, x_i) + \theta f(t_{i+1}, x_{i+1}))$$

I want to prove that the truncation error: $$ \tau(h) = \max_t \left( \frac{x(t + h) - x(t)}{h} - ((1-\theta) x'(t) + \theta x'(t+h) \right)$$ can be bound by: $$ \tau(h) \leq \tfrac{1}{2} cMh$$ where $M = \max_{t \in [t_0, T]} ||x''||_{\infty}$ and $c$ satisfies $||x|| \leq c ||x||_{\infty}$ for all $x$.

What I have tried I have been able to prove the weaker bound: $$ \tau(h) \leq (\frac{1}{2} + \theta)cMh$$ using the Taylor expansion. We can write: $$x(t + h) = x(t) + hx'(t) + \frac{h^2}{2} x''(\xi)$$ and $$x'(t + h) = x'(t) + hx''(\eta)$$ for some $\xi, \eta \in (t, t+h)$.

This implies: $$ ||\tau(h)|| = h \left| \left| \frac{1}{2} x''(\xi) - \theta x''(\eta) \right| \right| $$ and we can find the previous bound by bounding: $$ \left| \left| \frac{1}{2} x''(\xi) - \theta x''(\eta) \right| \right| \leq (\frac{1}{2} + \theta) cM $$

Vivianne
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  • See for instance https://math.stackexchange.com/questions/3381289/error-in-theta-method, https://math.stackexchange.com/questions/1213614/numerical-estimates-for-the-convergence-order-of-trapezoidal-like-runge-kutta-me – Lutz Lehmann Mar 03 '23 at 22:32
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    @LutzLehmann I looked at both before posting. If I am not wrong, both of them assume stronger regularity conditions ($C^2$) on $f$, so they are not applicable here. – Vivianne Mar 03 '23 at 22:36

1 Answers1

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Following the cited post, a more convenient midpoint is $t+(1-\theta)h$ so that \begin{align} x(t+(1−θ)h)&=x(t)+(1−θ)hx'(t)+\frac{(1−θ)^2h^2}2x''(\xi_1)\\ &=x(t+h)-θhx'(t+h)+\frac{θ^2h^2}2x''(\xi_2) \end{align} so that \begin{align} \tau(t,h)&=\frac{h}2(θ^2x''(\xi_2)-(1−θ)^2x''(\xi_1))\\ \tau(h)&\le\frac{Mh}4(1+(1-2θ)^2) \end{align} and the last factor is at worst $2$.

Lutz Lehmann
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