System Of ODEs - Wronskian

Question

I am reading about the Wronskian but I find a lot of conflicts in the theory, as you can see in the pictures below.

In some sources Wronskian includes the linear independent solutions and in others there are derivatives in it. What is happening?

Answer 1

I want to say that your slide is using a more general definition than one sees on Wikipedia etc. with the derivatives. That is, you can promote a function $f$ to a vector $$\mathbf f(t) = \begin{bmatrix}f(t)\\ \dot f(t) \\ \ddot f(t) \\ \dots\end{bmatrix}$$ to recover the usual idea of the Wronskian in the typical context; you would then find that you can encode a typical second-order ODE, for example $$\ddot x + 2 \lambda~\dot x + \omega^2 x = \alpha$$ as a first-order matrix ODE, $$\dot{\mathbf x} = \begin{bmatrix}\dot x\\\ddot x\end{bmatrix} = \begin{bmatrix}0&1\\-\omega^2&-2\lambda\end{bmatrix}\begin{bmatrix}x\\\dot x\end{bmatrix} + \begin{bmatrix}0\\\alpha\end{bmatrix} = \mathbf M \mathbf x + \mathbf b.$$

Wronskians of the second form you are talking about typically occur in these higher-order contexts; so you would have two solutions to this equation $x_{1,2}$ and you would consider the combination $W = x_1\dot x_2 - \dot x_1 x_2.$

But it seems likely that one could then prove Abel's identity more generally, that given $n$ functions which all obey this equation one might form the determinant $$W(\mathbf x_1, \mathbf x_2, \dots \mathbf x_n) = \begin{vmatrix} \uparrow&\uparrow&&\uparrow\\ \mathbf x_1 & \mathbf x_2 & \dots & \mathbf x_n\\ \downarrow&\downarrow&&\downarrow\end{vmatrix}$$ and then observe that $$ \dot W = \begin{vmatrix} \uparrow&\uparrow&&\uparrow\\ (\mathbf M \mathbf x_1 + \mathbf b) & \mathbf x_2 & \dots & \mathbf x_n\\\downarrow&\downarrow&&\downarrow\end{vmatrix} + \dots + \begin{vmatrix} \uparrow&\uparrow&&\uparrow\\ \mathbf x_1 & \mathbf x_2 & \dots & (\mathbf M \mathbf x_n + \mathbf b)\\\downarrow&\downarrow&&\downarrow\end{vmatrix}. $$ I do not know the full theory but if we assume that $\mathbf b = 0$ then a natural basis to do this in would be the one where $\mathbf M$ is diagonal, assuming it’s diagonalizable at all, and then it would be natural to solve the above ODE with $\mathbf x_1 \propto \hat e_1 = [1;0;0;\dots],$ and likewise for the rest. In that case it would seem to be the case that you would get something like $$\dot W = W~\operatorname{tr}\mathbf M.$$

In the “inhomogeneous” case where $\mathbf b \ne 0$ one typically finds the entire family of solutions by finding one single solution called a “particular solution” $\mathbf M \mathbf x_\text P = \mathbf b$ and then all other solutions to that equation have $$\mathbf M (\mathbf x - \mathbf x_\text P)= 0,$$ thus a solution-in-general is always the sum of a solution-in-general to the “homogeneous” equation $\mathbf M \mathbf x = 0$ plus one particular solution to the inhomogeneous equation. So you might not need to handle the inhomogeneous case for the purposes of using the Wronskian effectively, or it might not make sense there.

(I actually think that this is one of the few cases where the determinant will cleanly distribute over addition, I just can't see a reason why the sum of all of the $\mathbf b$-terms in the determinant would cancel out or form a nice pattern in the result. I suppose you could maybe say that $\mathbf b = \sum_i \beta_i\mathbf x_i$ and then get $\operatorname{tr}\mathbf M + \sum_i \beta_i$ for the resulting term? But one is usually solving for the Wronskian to try and find $\mathbf x_n$ given $\mathbf x_{1\dots n-1}$ and this approach would seem to leave some unknown $\beta_n$ in the expression for the Wronskian in that case.)