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I am reading this paper by Liniger and Willoughby on Integrating Stiff ODE. I am hoping that I am providing enough information for an answer. We are looking at an system,

$\dot{x} = -\lambda x$

The paper breaks down the integration as follows, $x(t+h) - x(t) - h [(1 - \mu) \dot{x}(t+h) + \mu \dot{x} (t)] = e_1(t)$,

where

$e_1 = - h^2 \int_0^1(\theta - \mu) \ddot{x}(t + \theta h) d\theta$

I have no idea how this came about, if someone could clue me in I would much appreciate it. I tried both the integral and Legrange remainder theorems, not sure what I am doing wrong but I could not arrive at the result.

Ramesh Kadambi
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    This is similar to https://math.stackexchange.com/questions/3381289/error-in-theta-method and the questions linked there. Does that enlighten the background of the task? – Lutz Lehmann Jun 22 '20 at 18:29
  • @LutzLehmann I am aware and know how to show the error bound based on Lipschitz criteria. The issue I am faced with is how do I go from there to the error form given here. I can show that the error is bounded by $e^{b - a} [ |e_0| + \frac{\tau}{K} ]$ where $K$ is the lipshitz constant for the $x(t)$. Anyway, you may have given me an idea. Thank you. – Ramesh Kadambi Jun 22 '20 at 18:46
  • @LutzLehmann You were right, that is exactly what it is. It is just written as an integral, I am not sure why it threw me off. – Ramesh Kadambi Jun 23 '20 at 11:00
  • @LutzLehmann Looking at my notes, I know what threw me off. I would rather have written the error as $e_1 = -h^2 \int_0^1 (\theta - \mu) \ddot{x}(t + \theta^\prime h) d\theta$. When I integrated it by parts, I assumed $\ddot{x}(t + \theta h)$ is a function of $\theta$. It really is a constant. Am I right? Writing it the way I have it gives me the truncation error $-h^2 (\frac{1}{2} - \mu) \ddot{x}(t + \theta^\prime h)$ when integrated. – Ramesh Kadambi Jun 23 '20 at 11:32

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Consider for $t$ and $h$ fixed the function $$ g(θ)=x(t+θh)-h(θ-μ)\dot x(t+θh) $$ Then \begin{align} e_1(t)=g(1)-g(0)&=\int_0^1g'(θ)\,dθ\\ &=\int_0^1\left[h\dot x(t+θh) -h\dot x(t+θh)-h^2(θ-μ)\ddot x(t+θh)\right]\,dθ\\ &=-h^2\int_0^1(θ-μ)\ddot x(t+θh) \,dθ \end{align} as claimed.

Lutz Lehmann
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