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The solution to the initial value problem $$x'(t)=Ax(t)+g(t)\quad\text{with}\quad x(0)=x_0$$ is $$x(t)=\exp(tA)x_0+\int_0^t \exp((t-s)A)g(s)\,\mathrm{d}s$$

Suppose that all eigenvalues of $A$, satisfy $\mathrm{Re}(\alpha_j)<0$. I have to find $$\lim_{t\to\infty} x(t)$$ when $$\lim_{t\to\infty} |g(t)|=g_0$$

I saw that the solution is $$x(t)=\exp(tA)x_0+\exp(tA)g_0$$ I get the same answer that it goes to zero just like in the case when $$\lim_{t\to\infty} |g(t)|=0$$ Is this right? If the first term goes to zero why doesn't the second as well?

This question is not answered in the dublicate. Thank you for your help.


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marked as duplicate by robjohn Nov 24 '12 at 9:53

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1 Answer 1

You just check it with the explicit formula of the solution you give above.. The first part goes to 0 obviously, the second part needs one more comment that both exp(At) and g(t) is bounded.

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could you please help me with my updated version? –  Klara Nov 26 '12 at 16:29
@Klara, think about first order equation $\dot y=-y+1$. Do the solutions to this equation tend to zero? –  Artem Nov 26 '12 at 18:28
@Artem no the solution of the DE goes to 1, does that mean that my limit would go to g_0? –  Klara Nov 26 '12 at 22:29
This means that the limit is not 0. –  Artem Nov 27 '12 at 0:04
@ Artem Is it right that x(t)=\exp(tA)x_0+\exp(tA)g_0. If yes, then how is the second term any different from the first? –  Klara Nov 27 '12 at 5:42

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