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The following question was part of a problem sheet in numerical maths:

Given the scalar equation $\ddot{y} = - \lambda y$, $\lambda > 0$

show that for $h > h_{critical}$, the approximate solution goes to infinity as $t$ goes to infinity. This behaviour is called blow-up.

The Stormer-Verlet-method is $y_{k+1} = -y_{k-1} + 2y_k - h^2 \lambda y_k$, which can be rewritten as

$$ \left( \begin{array}{c} y_{k+1} \\ y_k \end{array} \right) = \left( \begin{array}{cc} 2 - h^2 \lambda & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} y_{k} \\ y_{k-1} \end{array} \right)$$

Then in the solution they write that there is no blow-up if and only if the eigenvalues $\mu_i$ of the matrix $$ B = \left( \begin{array}{cc} 2 - h^2 \lambda & -1 \\ 1 & 0 \end{array} \right)$$ are all $\leq 1$.

My question is: how do they get to this claim about the relationship between eigenvalues and blow-up?

I read that the 2 norm $|| . ||_2$ is the same as the spectral norm if the matrices are square. Then I was wondering if the reason could be something like $$ || Av||_2 \leq || A ||_2 || v ||_2$$ and then using that this is the spectral norm: $$ || Av||_2 \leq || A ||_2 || v ||_2 \leq ( max_i \mu_i ) || v ||_2$$ where the max is taken over the eigenvalues of $A$.

The problem with this is, that I need the other direction: $$ ( max_i \mu_i ) || v ||_2 \leq || Av||_2 $$

Then, obviously, if the biggest eigenvalue is bigger than 1, the approximation after $k$ steps is $A^k y_0$ and using the above I get a blow-up.

Many thanks for your help.

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1 Answer 1

up vote 4 down vote accepted

Your solution will be $(y_{n+1},y_n)^T=B^n(y_1,y_0)^T$ so if you expand $(y_1,y_0)^T$ in terms of eigenvectors of $B$, any one which corresponds to an eigenvalue greater than 1 in absolute value will blow up.

Added in response to the comment: You find the eignevectors of $B$. As long as $B$ is well-behaved, they will span the space, so form a basis. So if the eigenvectors are $\vec{x_1}$ and $\vec{x_2}$ corresponding to $\lambda_1$ and $\lambda_2$ respectively,you would write $(y_1,y_0)^T=a_1\vec{x_1}+a_2\vec{x_2}$ and get$(y_{n+1},y_n)^T=a_1\lambda_1^n\vec{x_1}+a_2\lambda_2^n\vec{x_2}$

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Thanks for your help. What does it mean to expand a vector in terms of eigenvectors of a matrix? How do I write $(y_0 , y_1 )^T$ in terms of $B$? –  Rudy the Reindeer Jan 24 '11 at 17:54

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