Characteristic roots problem

May 2010
10
0
baltimore, md
I am almost completely lost with this problem.

I say almost completely because i have a vague understanding of the principle but my professor rushed through this lecture and our text book doesnt cover this.(Crying)

Here is the problem:

Prove that every solution of the equation:

y'' + a1y' + a2y = 0 (a1 = const; a + 2 = const)

is bounded on [0, infinty) if and only if the real parts of the characteristic roots are non-positive and the roots with zero real part have multiplicity 1.

Any guidance on this would be appreciated.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I am almost completely lost with this problem.

I say almost completely because i have a vague understanding of the principle but my professor rushed through this lecture and our text book doesnt cover this.(Crying)

Here is the problem:

Prove that every solution of the equation:

y'' + a1y' + a2y = 0 (a1 = const; a + 2 = const)

is bounded on [0, infinty) if and only if the real parts of the characteristic roots are non-positive and the roots with zero real part have multiplicity 1.

Any guidance on this would be appreciated.
Since this is "if and only if", you have to do it both ways- but they are pretty similar.

Suppose that "real parts of the characteristic roots are non-positive and the roots with zero real part have multiplicity 1". Then look at the solutions corresponding to each characteristic root. If a+ ib is a root, then solutions are of the form \(\displaystyle Ce^{ax}cos(bx)\) and \(\displaystyle Ce^{ax}sin(bx)\) for some constant C. Obviously cos(bx) and sin(bx) are bounded. What about \(\displaystyle e^{ax}\) with a negative and x positive?

If a+ bi is a multiple root, you can have solutions of the form \(\displaystyle Cx^ne^{ax}cos(bx)\) and \(\displaystyle e^{ax}sin(bx))\). You will need to show that \(\displaystyle x^ne^{-ax}\), again with a negative and x positive, is bounded for all n.

If the real part is 0, then the root is simply bi and the solution is C(cos(bx)+ i sin(bx)) which is clearly bounded.

If bi is a multiple root, we could have solutions of the form \(\displaystyle Cx^n(cos(bx)\) and \(\displaystyle Cx^nsin(bx)\) which are not bounded. That's why you need "roots with zero real part have multiplicity 1".
 
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May 2010
10
0
baltimore, md
Thank you for the input. Ill have to look over the few notes i have on this and do some internet search to see if i can understand it better.

ill probably be back later with a follow up question.