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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0Where did you used that the matrix is simmetric? Was it just a redundant information? – 2012-08-04
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0Why $\leqslant-1$ and not $\lt0$? – 2012-08-04
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0To ensure that $\langle v,Av\rangle\leqslant-\langle v,v\rangle$. – 2012-08-04
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0I don't know.. form your proof $<0$ seems to suffice indeed.. i copied the text as i found it.. – 2012-08-04
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0I don't get your point about symmetry.. can you please provide more details? – 2012-08-04
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0Yes: real symmetric matrices are diagonalizable by orthogonal matrices hence $A(t)=P(t)^*D(t)P(t)$ with $P(t)$ orthogonal and $D(t)$ diagonal with diagonal entries the eigenvalues of $A(t)$. Thus, $\langle v,A(t)v\rangle=\langle w,D(t)w\rangle\leqslant-2|w|^2=-2|v|^2$ with $w=P(t)v$. – 2012-08-04
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0Excellent.. thank you did.. – 2012-08-04