I've got a question regarding the intuition of a Wronskian, in the following sense:

The intuition for the determinant of a square $n \times n $-matrix is that it represents the area/(hyper-)volume between vectors. But what is the intuition behind the Wronskian of let's say two linear functions $f_1 = 0.5x+4$ and $f_2 = -2x+4$ which is $-10$? What is the (geometric/graphical? )intuition of the value $-10$? Is there any intuition in such a sense possible?
Thank you in advance for hints.
Best regards

  • $\begingroup$ Why the need for intuition? It's a quantity used in variation of parameters. $\endgroup$ – Sean Roberson Aug 23 '17 at 15:40
  • 8
    $\begingroup$ It is a signed volume in phase space, that is, in the space $(x, \dot x)$ of pairs (position, velocity). $\endgroup$ – Giuseppe Negro Aug 23 '17 at 15:41

This is best seen for dynamical systems, that is, systems of first order autonomous differential equations: $$ \dot{\mathbf x}(t)= F(\mathbf x(t)),$$ where $\mathbf x\in \mathbb R^n$ is considered as a column vector. In this case, the Wronskian of $n$ solutions $\mathbf x_1 \ldots \mathbf x_n$ is $$ W(t)=\det \begin{bmatrix} \mathbf x_1(t)\ldots \mathbf x_n(t)\end{bmatrix}.$$ In this case it is clear that $W(t)$ is the signed volume of the parallelepiped spanned by $\mathbf x_1(t),\ldots, \mathbf x_n(t)$.

For the higher-order equation $$\frac{d^{n} x}{dt^{n}} = G\left( \frac{d^{n-1} x}{dt^{n-1}} ,\ldots, \frac{dx}{dt}, x\right)$$ we define the Wronskian of $n$ solutions $x_1\ldots x_n$ as follows: $$ W(t)=\det\begin{bmatrix} x_1 & \ldots & x_n \\ \frac{dx_1}{dt} & \ldots & \frac{dx_n}{dt} \\ \ldots & \ldots & \ldots \\ \frac{d^{n-1}x_1}{dt^{n-1}} & \ldots & \frac{d^{n-1}x_n}{dt^{n-1}}\end{bmatrix}.$$ This is the same as the Wronskian of the dynamical system obtained with the substitutions $$ \mathbf x(t) =\begin{bmatrix} x \\ \frac{dx}{dt} \\ \vdots \\ \frac{d^{n-1}x}{dt^{n-1}}\end{bmatrix}, \qquad F(\mathbf x)=\begin{bmatrix} x_2 \\ x_3 \\\vdots \\ G(x_n\ldots x_1)\end{bmatrix}. $$

This makes it apparent that the Wronskian is a signed volume in the phase space of $n$-uples $(x, \dot x\ldots \ x^{(n-1)})$.

  • $\begingroup$ Thank you for elaborating on this issue. I have another question related to this topic. Suppose we are given three functions $f_1(x) = 2x+4$, $f_2(x) = 2x+5$ and $f_3(x) = 1.5x+4$. Computing the pairwise wronskian of $(f_1(x),f_2(x))$ and $(f_1(x),f_3(x))$ yields a value of -2 for both tuples. How does it come, or what is the intuition behind the observation that both tuples lead to the same volume in phase space? $f_2$ has a larger intercept (larger by 1 compared to $f_1$) while $f_3$ has a smaller slope (smaller by 0.5 compared to $f_1$). $\endgroup$ – Daniyal Aug 25 '17 at 11:10
  • $\begingroup$ Are you sure that the Wronskian of $f_1, f_3$ does not depend on $x$? $\endgroup$ – Giuseppe Negro Aug 25 '17 at 11:58
  • $\begingroup$ Could you please elaborate on what is meant by dependent on $x$? $\endgroup$ – Daniyal Aug 25 '17 at 13:13
  • $\begingroup$ Sorry, forget about that. (I was asking whether $W(f_1, f_3) $ was a function of $x$ and it is not). Anyway, I do not see anything mysterious about those two Wronskians. Both equal $-2$ even if they come from different parallelepipeds, but that's nothing new: to make a simple-minded example, the square of side $1$ has the same area as the rectangle with sides $2$ and $1/2$. Something similar happens here. $\endgroup$ – Giuseppe Negro Aug 25 '17 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.