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. $\endgroup$ Aug 23, 2017 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). $\endgroup$ Aug 23, 2017 at 15:41

1 Answer 1


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, 2017 at 11:10
  • $\begingroup$ Are you sure that the Wronskian of $f_1, f_3$ does not depend on $x$? $\endgroup$ Aug 25, 2017 at 11:58
  • $\begingroup$ Could you please elaborate on what is meant by dependent on $x$? $\endgroup$
    – Daniyal
    Aug 25, 2017 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. $\endgroup$ Aug 25, 2017 at 14:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .