Hey everybody

how can i calculate the Solution of

?

Thank you very much!

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- Nov 19th 2010, 12:57 AMLHeinerSystem of first order ODE
Hey everybody

how can i calculate the Solution of

?

Thank you very much! - Nov 19th 2010, 01:02 AMAckbeet
What ideas have you had so far?

- Nov 19th 2010, 02:52 AMLHeiner
I tried to figure out the eigenvalues and eigenvectors but i got stuck here:

- Nov 19th 2010, 04:49 AMAckbeet
Good approach - to find the eigenvalues and eigenvectors. I think you may have skipped too many steps in taking your determinant. I get

This is not the same thing as what you got, if you multiply it all out (sometimes it pays not to do that!). So, what are the eigenvalues? - Nov 19th 2010, 06:14 AMLHeiner
so

since I'm looking for a real solution I take which gives me the eigenvector .

So the solution is: ? is this correct?

If so how do i find linear subspaces for which or converges to x=0? - Nov 19th 2010, 07:18 AMAckbeet
I don't think you can ignore the complex eigenvalues - don't worry, I think they'll turn into sines and cosines. You have correctly found the eigenvector for the real eigenvalue. What are the eigenvectors for the complex eigenvalues?

- Nov 19th 2010, 08:20 AMLHeiner
for and for

and now? - Nov 19th 2010, 08:28 AMAckbeet
So, if you call

then what is the diagonalization of - Nov 19th 2010, 08:48 AMLHeiner
Is it:

But how does this help me in solving the ODE?

In other cases i simply took the eigenvalues and eigenvectors (if complex, split it into sin and cos) and solved the DE with the exponential function (like in the one dimensional case: - Nov 19th 2010, 09:05 AMAckbeet
Yes, you've correctly identified the diagonal matrix such that there is an invertible matrix such that

The reason you need to diagonalize is because the solution to the system , as you've hinted at, is the following:

So you have to compute this matrix exponential and the best and easiest way to do that is to diagonalize Why? Because it's ridiculously easy to compute arbitrary powers of a diagonal matrix (just raise the elements on the diagonal to the desired power!). Also,

In general, you have

Then you just invoke the Taylor series expansion of the exponential, and you find that

The RHS there is easy to compute: is just the matrix with the elements That is, you just exponentiate the elements on the main diagonal.

So I would say that diagonalization, when it can be done, is at the heart of solving linear systems of ODE's with constant coefficients.

So, what remains to be done is this: construct your invertible matrix and then compute , and then compute and you're done.

Make sense? - Nov 19th 2010, 09:28 AMJester
Just a comment. As the middle row of the matrix is (0,b,0), it might be a whole lot easier to simply solve

.

For then giving

. - Nov 20th 2010, 01:09 AMLHeiner
So i tried the way Danny sugested it and got the solution:

is this correct? - Nov 20th 2010, 02:36 AMAckbeet
Well, why not compute the derivative, and see if it satisfies the original system? Here's one of the really great things about differential equations: as long as you're decent at differentiation (which is quite straight-forward, really), you can check your own answers very easily.

What do you get? - Nov 20th 2010, 05:10 AMLHeiner
well, if i didnt make any mistakes the solution should be correct!

can you give me one last hint how i get the linear subspaces which or converges to x=0 - Nov 20th 2010, 06:14 AMAckbeetQuote:

well, if i didnt make any mistakes the solution should be correct!

As for the linear subspaces, could you please state your question a little more carefully? Perhaps rephrase it a bit? I'm not following what you're asking, though I have an inkling.