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Let $y'(t)=f(t,y(t))$ and $y(0)=y_0$

The backward euler method together with the center rule is given by:

$y(t_k)=hk$ where $h\in (0,\frac{1}{K})$ is the step size.

Recursion: $y_k=y_{k-1}+hf(t_{k-1}+\frac{h}{2},\mu_{k-1})$ where $\mu_{k-1}=y_{k-1}+\frac{h}{2}f(t_{k-1},y_{k-1})$

Question: I want to use the method now for $y'(t)=\lambda y(t), t>0$ and $y(0)=y_0$ to evaluate for which $\lambda$ the sequence $(y_k)_{k\in\mathbb N}$ converges to $0$

I tried to evaluate some $y_k$ but I couldn' tsee anything.

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1 Answer 1

up vote 2 down vote accepted

This doesn't look like backward Euler method to me; it looks more like a 2nd order Runge-Kutta scheme. A backward Euler method would be $y_{k+1}=y_k+hf(t_{k+1},y_{k+1})$; see here for details.

Assuming you actually want to use the recursion which you have given, first note that $f(t,y)=\lambda y$, so that $$\mu_{k-1}=y_{k-1}+\frac{h}{2}\lambda y_{k-1}=(1+h\lambda/2)y_{k-1}.$$

Then, $$y_{k}=y_{k-1}+h\lambda\mu_{k-1}=y_{k-1}+h\lambda(1+h\lambda/2)y_{k-1}=\frac{1}{2}\left(2+2h\lambda+h^2\lambda^2\right)y_{k-1}.$$

You should now be able to determine which choices of $\lambda$ ensures $\lim\limits_{k\rightarrow\infty}y_k=0$.

Added You can also solve the problem analytically, giving $y(t)=Ce^{\lambda t}$, which approaches $0$ as $t\rightarrow\infty$ when $\operatorname{Re}(\lambda)<0$.

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