I have to prove that, if $A(t)$ is a real symmetric $N\times N$ matrix whose eigenvalues are all less than $-1$ for all $t$, then if we consider $u$ to be the solution of $\dot u(t)=A(t)u(t),$ Then $\lim_{t\to\infty}|u(t)|^2=0.$ The only thing I was able to do was to write down the solution, but nothing more.. can you help me? Many thanks..
Prove that the solution tends to $0$ as $t$ goes to infinity
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ordinary-differential-equations
1 Answers
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Note that $\frac{\mathrm d}{\mathrm dt}|u(t)|^2=2\langle u(t),u'(t)\rangle=2\langle u(t),A(t)u(t)\rangle\leqslant-2\langle u(t),u(t)\rangle=-2|u(t)|^2, $ hence $ |u(t)|^2\leqslant|u(0)|^2\,\mathrm e^{-2t}. $
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0Excellent.. th$a$nk you did.. – 2012-08-04