# Wronski matrix of a discrete evolution and condition

To compute the condition

$$\kappa_{\mathrm{abs}}^\infty := \max\{\| W(t,t_0, y_0) \|: t_0 \leq t \leq T \}$$

of an initial value problem on $[t_0, T]$, I need to compute the Wronski matrix $W$.

The approximation of the solution is denoted by the discrete evolution operator $\Psi^{t_0, t}$ and the exact solution is denoted $\Phi^{t_0, t}$ and the Wronski matrix is given as $W(t,t_0, y) := \frac{d}{dy} \Phi^{t_0, t}y \Big |_{y=z}$

My first question: is the condition really a function of the exact solution? According to my understanding numerical analysis deals with finding good approximation functions so the condition should be a measure of how good (stable?) a given approximation is, therefore the Wronski matrix should be a function of $\Psi$, not of $\Phi$.

My second question: how do I explicitly calculate the Wronski matrix for a given problem? Say, I'm using the explicit Euler method $y_{k+1} = y(t_k) + h*f(t_k, y(t_k))$ to solve the ODE $\dot{y} = (\alpha - \beta y)y$ or the system $\dot{u} = (\alpha - \beta v)u, \dot{v} = (\delta u - \gamma)v$.

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Second question: One idea for computing the condition number is to use that the numerical evolution operator $\Psi$ approximates the exact evolution operator $\Phi$. Thus, to compute the condition number, replace $\Phi$ in your definition by $\Psi$. You can compute the derivative $\frac{\partial}{\partial y} \Psi^{t_0,t}(y)$ by finite differences or by evaluating the derivative by hand.
Many thanks. I like your answer to my first question, it's very clear. As for my second question: I'm not entirely satisfied. For example: $\Psi^h y = y + hf(y)$ then trying to solve $\dot{y} = f(y) = (\alpha - \beta y)y$ yields $\frac{\partial}{\partial y} \Psi^h y = 1 + h \alpha - 2h\beta y$ which is a scalar so the maximal determinant is just the maximum value of the function $1 + h\alpha - 2h \beta y(t)$, for $t$ in the domain. Is this correct? – Rudy the Reindeer Jan 26 '11 at 19:41
Second example: using $\Psi^h y = y + h f(y)$ again, this time to solve $\dot{u} = (\alpha - \beta v)u$, $\dot{v} = (\delta u - \gamma)v$, I get $$\Psi^h y = \left( \begin{array}{c} u + h(\alpha - \beta v) u \\\ v + h(\delta u - \gamma)v \end{array} \right)$$ but I don't know what to do now. Can you tell me how to proceed from there? – Rudy the Reindeer Jan 26 '11 at 19:41