Liapunov Stability

For Cauchy Problem, I is a bounded interval, and y(t)\in C^n(I),

y^{(n)}=f(t,y(t)),

where t\in I, I wonder the stability for this problem. Thus I looked up the definition for the stability in the sense of Liapunov(zero stable).

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Note: I didn’t give the Cauchy data for this problem, we had assumed all the initial data were given.

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Definition: The stability means:

for any perturbation (\delta_0,\delta(t)) satisfying

|\delta_0|<\epsilon, |\delta(t)|<\epsilon,\forall t\in I

with \epsilon>0 sufficient small to ensure that the solution z(t) to the perturbed problem does exist., then

\exists C>0 independent of \epsilon such that |y(t)-z(t)|

Claim 1: If our force term f(t,y(t)) is uniformly continuous, then we can assure the stability(Liapunov).

Proof: Let \omega(t)=z(t)-y(t), we have

\omega^{(n)}(t)=f(t,z)-f(t,y)+\delta(t).

Hence,

\displaystyle\omega(t)=\delta_0(t)+\int\int\int\dots\int_{\Omega} [f(s,z(s))-f(s,y(s))]\mathrm{d}s+\int\int\int\dots\int_{\Omega}\delta(s)\mathrm{d}s

Thanks to the uniform-continuity.

\displaystyle|\omega(t)|\le \delta_0+\int\int\int\dots\int_{\Omega}L|z(s)-y(s)|\mathrm{d}s+\int\int\int\dots\int_{\Omega}\delta(s)\mathrm{d}s

Which is,

\displaystyle|\omega(t)|\le\delta_0+\int\int\int\dots\int_{\Omega}L|\omega(s)|\mathrm{d}s+\int\int\int\dots\int_{\Omega}\delta(s)\mathrm{d}s

equivalent to:

\displaystyle|\omega(t)|\le L\int\int\int\dots\int_{\Omega} |\omega(s)|\mathrm{d}s+(1+meas(\Omega))\epsilon

By Gronwall lemma,
\displaystyle|\omega(t)|\le (1+meas(\Omega))\epsilon \exp{(L\cdot meas(\Omega))}

Where \Omega=[0,t]^n.

Here C=(1+K)\exp(LK), K=\max_{t\in I} t^n.

About YiMin

This is just a nerd PhD student of Math@UT Austin.

One thought on “Liapunov Stability

  1. YiMin says:

    Here I have to make some comment: the initial data, we say with pertubation f^{(k)}=\delta_k(t), thus the integral need another term \displaystyle\frac{T^n-1}{T-1}\epsilon, which doesn’t matter much on this problem.

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