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  • Let $\operatorname{g}\left(x,t\right)=\alpha t + x$ and $\alpha < 0$ is a parameter s.t. $\left\vert\alpha\right\vert$ is large.
  • To numerically solve the equation $Y'= \operatorname{g}\left(x,Y\right)$, the Backward Euler Method is used.
  • Find how small $\omega$ should be so the iteration $$ y_{m + 1}^{\left(i + 1\right)} = y_{m} + \omega\operatorname{g}\left(\,{x_{m + 1},y_{m + 1}^{\left(i\right)}}\,\right),\quad i = 0,1,2,\ldots $$ will converge to $y_{m + 1}$.

The first thought I got to solve this problem is to use the error formula, but I'm not sure how to continue from there.

Felix Marin
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rudytheduck
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  • You have a fixed-point iteration. Apply convergence criteria for these. – Lutz Lehmann Dec 06 '22 at 23:48
  • I'm not sure what the convergence criteria are. The only thing I see in my textbook is that if $ \left| h \cdot \frac{\partial f(x_{n+1},y_{n+1})}{\partial z}\right| <1$ then the error will converge to $0$ as long as the initial guess $y_{n+1}^{(0)}$ is good enough – rudytheduck Dec 07 '22 at 01:25
  • So does it converge when $\left| \omega \alpha \right|<1$? Which implies $\frac{-1}{\alpha}<\omega < \frac{1}{\alpha}$ – rudytheduck Dec 07 '22 at 01:42
  • Yes, exactly that. Usually you want to integrate forwards, so $0<ω$. – Lutz Lehmann Dec 07 '22 at 05:43

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