# Show problem ill-conditioned

• Aug 29th 2011, 02:24 AM
TheProphet
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.
• Aug 29th 2011, 03:21 AM
chisigma
Re: Show problem ill-conditioned
Quote:

Originally Posted by TheProphet
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

$\chi$ $\sigma$
• Aug 29th 2011, 04:06 AM
TheProphet
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.
• Aug 29th 2011, 04:19 AM
chisigma
Re: Show problem ill-conditioned
Quote:

Originally Posted by TheProphet
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

$\chi$ $\sigma$
• Aug 29th 2011, 04:48 AM
TheProphet
Re: Show problem ill-conditioned
Ah, I see. Thank you.