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We have the initial value problem $$\dot{\mathbf{y}} = f(\mathbf{y}), \mathbf{y}(0) = \mathbf{y_0},$$ with $f(\mathbf{y})$ continuously differentiable. There exists a $T > 0$ such that $\mathbf{y_0} = \mathbf{\Phi}^T \mathbf{y_0}$ and $\mathbf{y_0} \neq \mathbf{\Phi}^t \mathbf{y_0}$ for all $0 < t < T$. I already know that $\mathbf{\Phi}^{t+T} \mathbf{y_0} = \mathbf{\Phi}^t \mathbf{y_0}$. Now I want to show that the number $1$ is an eigenvalue of the Wronski matrix $W(T;0,\mathbf{y_0})$. In our lecture about numerical mathematics, we defined the Wronski matrix as follows:

$$W(t;t_0,\mathbf{z}) := \left.\frac{\partial}{\partial \mathbf{y}} \mathbf{\Phi}^{t_0,t} \mathbf{y} \right|_{\mathbf{y} = \mathbf{z}} \in \mathbb{R}^{d,d}$$

The standard solution suggests to write down $\mathbf{\Phi}^t \mathbf{y_0} = \mathbf{\Phi}^{t+T} \mathbf{y_0} = \mathbf{\Phi}^T \mathbf{\Phi}^t \mathbf{y_0}$ which holds for all $t \in \mathbb{R}$. Then I am supposed to differentiate both sides with respect to $t$ and plugging in $t = 0$ should lead me to $$\mathbf{f}(\mathbf{y_0}) = W(T;0,\mathbf{y_0}) \mathbf{f}(\mathbf{y_0}).$$

I don't really understand how to differentiate given equation and how to apply product or chain rule there and how that should give me above equation. After all, they suggest to differentiate with respect to $t$ whereas the Wronski matrix is defined with a partial derivative with respect to $\mathbf{y}$? How do they obtain the Wronski matrix out of it? Could someone just write down every step taken to obtain the desired result? Thanks a lot in advance.

Huy
  • 6,674

1 Answers1

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We have \begin{align*} \frac{d}{dt} \Phi^t y_0 &= f(\Phi^t y_0) \end{align*} as $t \mapsto \Phi^t y_0$ is the solution of the given IVP. Now $t \mapsto \Phi^T(\Phi^t y_0)$ is a composition of $t \mapsto \Phi^t y_0$ with $z \mapsto \Phi^Tz$. The chain rule gives $$ \frac d{dt} \Phi^T\Phi^t y_0 = D\Phi^T(\Phi^t y_0)f(\Phi^t y_0) \tag 1$$ Now, by your definition, the derivative of $\Phi^T = \Phi^{0,T}$ is represented by the matrix $W(0, T; \cdot)$, that is (in the notation of (1)) $D\Phi^T(z) = W(0, T; z)$ for each $z$. So (1) reads $$ \frac d{dt} \Phi^T\Phi^t y_0 = W(0,T; \Phi^t y_0)f(\Phi^t y_0). $$ So, for each $t$ $$ W(0, T; \Phi^t y_0)f(\Phi^t y_0) = f(\Phi^t y_0) $$ with $t = 0$ we have $$ W(0, T; y_0)f(y_0) = f(y_0) $$

martini
  • 84,101