Proving Lyapunov Stability for \(\dot{x} = Ax + B(t)x\)

  • #1
motherh
27
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A question I am doing hints that the solution (y,[itex]\dot{y}[/itex]) = (0,0) of [itex]\ddot{y}[/itex] - [itex]\frac{2}{t}[/itex][itex]\dot{y}[/itex] + y = 0 is unstable. I believe (although I am not 100% sure) that is true however I am struggling to prove it.

I can rewrite the equation as a system of equations in matrix form to get

[itex]\dot{x}[/itex] = Ax + B(t)x,

where A = [{0,1},{-1,0}], B(t) = [{0,0},{0,[itex]\frac{2}{t}[/itex]}].

This the form of all the theorems I appear to have. But all my theorems require me to find an eigenvalue of A with positive real part - which I can't here.

So basically have I made a mistake already or is there another theorem anybody knows of that can tell me the trivial solution is unstable?
 
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  • #2
Anybody?
 
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