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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.

Klara

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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?2012-11-26
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    @Klara, think about first order equation $\dot y=-y+1$. Do the solutions to this equation tend to zero?2012-11-26
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    @Artem no the solution of the DE goes to 1, does that mean that my limit would go to g_0?2012-11-26
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    This means that the limit is not 0.2012-11-27
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    @ 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?2012-11-27
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    “I saw that the solution is x(t)=exp(tA)x 0 +exp(tA)g 0" It's wrong..2012-11-27
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    @Lee do you know the answer? Can you explain it to me?2012-11-27
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    @Klara 'x(t)=exp(tA)x 0 +∫ t 0 exp((t−s)A)g(s)ds' This is correct. And without a given g(s) we cannot compute anything, yet you can analyze the tendancy when t goes to infinity.2012-11-28