I am having problem understanding the definition of absolute stability. One definition of absolute stability I have heard is
"A numerical solution $ w_n $ to a problem is absolute stable for a given h > 0 if $ \lim_{n\to\infty} w_n = \lim_{t\to\infty} y(t) = 0 $ given that the problem is y'(t) = $\lambda y $ and $y(0) = 1 $.
This works well with the Euler forward method, for example.
$ w_{n+1} = w_n + h\lambda w_n = \dots = (1 + h\lambda)^{n+1}$. From here one can see that $ | 1 + h\lambda| < 1 $ in order for $ \lim_{n\to\infty} w_n = 0 $ which means absolute stability according to the definition I mentioned earlier.
However, I have also heard absolute stability being defined as something like: "If small perturbation in the initial value of an IVP does not affect the behavior of the function after long time, than it is absolute stable".
Again, one can prove Euler forward is absolute stable by showing that a small deviation $ \delta $ of the initial value does not affect the function as t approaches infinity. This is done by showing the error approaches 0 as t approaches infinity.
Are the two definitions the same, just written in other words? Have I done something wrong?
For example, consider the IVP (without perturbation) $y' = \lambda y + f(t), y(0) = y_0 $ and the same IVP but this time with a perturbation: $ u' = \lambda u + f(t), u(0) = y_0 + \delta $.
If I were to prove the first IVP as absolute stable, I could do it using definition 2, since the f(t) will cancel out when I calculate the error $e = u_{n+1} - y_{n+1} $. But I don't know how I could prove it using definition 1, since f(t) will make it more difficult.
To summarize. If I can prove small changes in the initial value doesn't matter after long time, does this mean my method is absolute stable? If not, then how can I prove the latest example is absolute stable?