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Math Help - System of first order ODE

  1. #1
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    System of first order ODE

    Hey everybody

    how can i calculate the Solution \in \mathbb{R} of

    x'=\begin{pmatrix}<br />
a & 0 & b\\ <br />
 0&  b& 0\\ <br />
 -b&  0& a<br />
\end{pmatrix} x, \ x \ \in \mathbb{R}^3 ?

    Thank you very much!
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    I tried to figure out the eigenvalues and eigenvectors but i got stuck here:
    <br />
det(A-\lambda E)=det\begin{pmatrix}<br />
a-\lambda & 0  & b\\ <br />
0 & b-\lambda &0 \\ <br />
 -b& 0 & a-\lambda<br />
\end{pmatrix}=(a^2-2a\lambda+\lambda^2)(b-\lambda)-(-b^3+b^2\lambda)=a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3
    <br />
a^2b-a\lambda-2ab\lambda+2a\lambda^2-\lambda^3+b^3=0
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  4. #4
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    Good approach - to find the eigenvalues and eigenvectors. I think you may have skipped too many steps in taking your determinant. I get

    \det\begin{bmatrix}<br />
a-\lambda & 0 & b\\ <br />
0 & b-\lambda &0 \\ <br />
-b& 0 & a-\lambda<br />
\end{bmatrix}=(a-\lambda)(b-\lambda)(a-\lambda)+b(0-(-b)(b-\lambda))

    =(a-\lambda)^{2}(b-\lambda)+b^{2}(b-\lambda)=(b-\lambda)\left[(a-\lambda)^{2}+b^{2}\right]=0.

    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?
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  5. #5
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    so (b-\lambda)[(a-\lambda)^2+b^2]=0 \Rightarrow \lambda_1=b, \ \<br />
\lambda_2=a-ib, \ \ \lambda_3=a+ib

    since I'm looking for a real solution I take \lambda_1=b which gives me the eigenvector (0,1,0)^T.

    So the solution is: x(t)=c_1e^bt(0,1,0)^T? is this correct?

    If so how do i find linear subspaces for which t \rightarrow \infty or t \rightarrow -\infty converges to x=0?
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  6. #6
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    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?
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  7. #7
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    for \lambda_2=a-ib \Rightarrow(i,0,1)^T and for \lambda_3=a+ib \rightarrow  (-i,0,1)^T

    and now?
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  8. #8
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    So, if you call

    A=\begin{bmatrix}<br />
a & 0 & b\\ <br />
0 & b &0 \\ <br />
-b& 0 & a<br />
\end{bmatrix},

    then what is the diagonalization of A?
    Last edited by Ackbeet; November 19th 2010 at 09:28 AM. Reason: Original matrix.
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  9. #9
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    Is it:

    \begin{bmatrix}<br />
\lambda_1 & 0 & 0\\ <br />
0 & \lambda_2 & 0 \\ <br />
0& 0 & \lambda_3<br />
\end{bmatrix},

    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: x'=ax \Rightarrow x(t)=c_1*e^{at}
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  10. #10
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    Yes, you've correctly identified the diagonal matrix D such that there is an invertible matrix P such that A=PDP^{-1}.

    The reason you need to diagonalize A is because the solution to the system \mathbf{x}'=A\mathbf{x}, as you've hinted at, is the following:

    \mathbf{x}(t)=e^{At}\mathbf{x}_{0}.

    So you have to compute this matrix exponential e^{At}, and the best and easiest way to do that is to diagonalize A. 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,

    A^{2}=(PDP^{-1})(PDP^{-1})=PDDP^{-1}=PD^{2}P^{-1},
    A^{3}=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PDDDP^{-1}=PD^{3}P^{-1}.

    In general, you have

    A^{k}=PD^{k}P^{-1}.

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

    e^{At}=Pe^{Dt}P^{-1}.

    The RHS there is easy to compute: e^{Dt} is just the matrix with the elements (e^{Dt})_{ij}=\delta_{ij}(e^{D_{ij}t}). 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 P, and then compute Pe^{Dt}P^{-1}, and then compute Pe^{Dt}P^{-1}\mathbf{x}_{0}, and you're done.

    Make sense?
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  11. #11
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    Just a comment. As the middle row of the matrix is (0,b,0), it might be a whole lot easier to simply solve

    \dot{x} = ax + bz,

    \dot{y} = by,

    \dot{z} = -b x + az.

    For b \ne 0 then z =\dfrac{\dot{x} - ax}{b} giving

    \ddot{x} - 2 a \dot{x} +(a^2+b^2)x = 0.
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  12. #12
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    So i tried the way Danny sugested it and got the solution:

    x(t)=e^{at}(c_1sin(bt)+c_2cos(bt))
    y(t)=c_3 e^{bt}
    z(t)=-e^{at}(c_4sin(bt)-c_5cos(bt))

    is this correct?
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  13. #13
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    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?
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  14. #14
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    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 t \rightarrow \infty or t \rightarrow -\infty converges to x=0
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  15. #15
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    well, if i didnt make any mistakes the solution should be correct!
    True, but you don't know ahead of time if you made any mistakes or not. I recommend computing the LHS (the derivatives), and then computing the RHS and show they are equal.

    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.
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