Show problem ill-conditioned

I am attempting to show that

$\displaystyle y'' -10y' -11y = 0 $, with $\displaystyle y(0) = 1 $ and $\displaystyle y'(0) = -1 $ is ill-conditioned.

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

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

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

How to go from here? The determinant of the coefficient matrix isn't particularly small, either.

Re: Show problem ill-conditioned

Quote:

Originally Posted by

**TheProphet** I am attempting to show that

$\displaystyle y'' -10y' -11y = 0 $, with $\displaystyle y(0) = 1 $ and $\displaystyle y'(0) = -1 $ is ill-conditioned.

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

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

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

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 $\displaystyle c_{1}$ and $\displaystyle c_{2}$... precisely is $\displaystyle c_{1}=0$ and $\displaystyle c_{2}=1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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.

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...

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

... with general solution...

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

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Show problem ill-conditioned