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Math Help - Show problem ill-conditioned

  1. #1
    Junior Member TheProphet's Avatar
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    Show problem ill-conditioned

    I am attempting to show that
     y'' -10y' -11y = 0 , with  y(0) = 1 and  y'(0) = -1 is ill-conditioned.

    I assumed a solution of the form  y(x) = e^{rx} which (of course) lead me to  r^{2} - 10r - 11 = 0 . Solutions to this equation are r_{1} = 11 and  r_{2} = -1 .

    Thus,  y(x) = c_{1} e^{11x} + c_{2} e^{-x} . Now we use the boundary conditions given, leading to

     \begin{pmatrix} 1 & 1 \\ 11 & -1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.

    How to go from here? The determinant of the coefficient matrix isn't particularly small, either.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Show problem ill-conditioned

    Quote Originally Posted by TheProphet View Post
    I am attempting to show that
     y'' -10y' -11y = 0 , with  y(0) = 1 and  y'(0) = -1 is ill-conditioned.

    I assumed a solution of the form  y(x) = e^{rx} which (of course) lead me to  r^{2} - 10r - 11 = 0 . Solutions to this equation are r_{1} = 11 and  r_{2} = -1 .

    Thus,  y(x) = c_{1} e^{11x} + c_{2} e^{-x} . Now we use the boundary conditions given, leading to

     \begin{pmatrix} 1 & 1 \\ 11 & -1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.

    How to go from here? The determinant of the coefficient matrix isn't particularly small, either.
    The matrix of the coefficients is non singular, so that there is one and only one solution for c_{1} and c_{2}... precisely is c_{1}=0 and c_{2}=1...

    Kind regards

    \chi \sigma
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Show problem ill-conditioned

    Yes I know the matrix is nonsingular (I mentioned the determinant was non-zero). But I think the problem is to show that a small change in the data (i.e. boundary conditions) changes the solution by much. However, I don't find this to be the case.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Show problem ill-conditioned

    Quote Originally Posted by TheProphet View Post
    Yes I know the matrix is nonsingular (I mentioned the determinant was non-zero). But I think the problem is to show that a small change in the data (i.e. boundary conditions) changes the solution by much. However, I don't find this to be the case.
    Using the language of system theory we can say that adinamic system defined by the DE...

    y^{''} -10\ y^{'} + 11\ y=0 (1)

    ... with general solution...

    y(x) = c_{1}\ e^{11 x} + c_{2}\ e^{-x} (1)

    ... is unstable and that means that initial condition 'only little different' from other can produce a solution that increases without limits with x...

    Kind regards

    \chi \sigma
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  5. #5
    Junior Member TheProphet's Avatar
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    Re: Show problem ill-conditioned

    Ah, I see. Thank you.
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