Stability of forward euler method

Question

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.

Answer 1

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.