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I am trying to understand the stability of the forward Euler method. I read there's a model problem to see the stability.

$$y'(t) = \lambda y(t) \qquad t \in (0, \infty)$$ $$y(0) = 1$$

then the book shows this:

$$u_0 = 1 \qquad \text{does this come from the initial condition?}$$

then

$$u_{n+1}=u_n(1+\lambda h) = (1+\lambda h)^{n+1}, \qquad n ≥ 0$$

I cannot understand how $(1+\lambda h)^{n+1}$ appeared

EDIT

then the analysis says that the only way the $$\lim_{n \rightarrow \infty} u_n = 0$$ is that $$|1+\lambda h| < 1$$ (why does it need to be less than 1?) Which gives (according to the book) $$h < \frac{2}{|\lambda|}$$

for the inequallity $|1+\lambda h| < 1$ the thing I get is $\frac{-2}{\lambda} < h$ but still don't know how they get the absolute value of $\lambda$.

how does the book arrive to this conclusion? I don't expect a very complex answer (which is appreciated) but a simple way to understand the intermediate steps not shown.

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    $\begingroup$ Note that stability of numerical schemes for ODE initial value problems depends on the differential equation that you're solving. Different kinds of stability are associated with different test problems. For example, a method is said to be "A-stable" if it is stable for your example problem for all values of lambda and h. A method is "0-stable" if it's stable for the problem y'(t)=0. $\endgroup$ – Brian Borchers Jan 18 '13 at 14:31
  • $\begingroup$ In the model problem, it's assumed that the parameter $\lambda$ is less than 0. The reason for this is that the exact solution is $y(t)=e^{\lambda t}$ and if $\lambda < 0$, then $y(t)$ goes to 0 as $t$ goes to infinity. If $\lambda > 0$, then $y(t)$ grows. $\endgroup$ – Brian Borchers Jan 18 '13 at 14:34
  • $\begingroup$ Thank you for the feedback. Probably I should have said that this is only for the so called Cauchy problem. $\endgroup$ – BRabbit27 Jan 18 '13 at 15:15
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Ok. So $u_0 = 1$, $u_1 = (1+\lambda h)$, $u_2 = (1+\lambda h)^2$, $u_3 = (1+\lambda h)^3$, ..., $u_n = (1+\lambda h)^n$, $u_{n+1} = (1+\lambda h)^{n+1}$ This will answer one part of your question. I don't understand the other part of your question. For assessing stability, let's assume $\lambda < 0$. You can think about the other possibilities yourself later. The true solution to the differential equation is $u_0 e^{-\lambda t} $ When t goes to infinity, the solution goes to zero. This must be the case for the discrete equation, too. The solution of our discrete equation will go to zero, when $(1 + \lambda h) > -1$. So $h < -2/\lambda = 2/|\lambda|$.

An other hint: When you multiply something smaller than one with itself, you will get something smaller than one. When it's bigger, it will grow.

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  • $\begingroup$ Ok. I'll edit the other part. $\endgroup$ – BRabbit27 Jan 18 '13 at 11:33

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