What is the intuition behind the Wronskian? I do not want an explanation that involves a rigorous proof. I just want to learn how to intuitively think about it.

Edit: And since it is tied to the weight function in the theory of differential equations, it would also be great to have an intuitive connection between the two included in the answer!


2 Answers 2


Every homogeneous linear ordinary differential equation $L^{(n)}f=0$ of order $n$ can be equivalently put into a form of linear system $\vec{f}\,'(z)=A(z)\vec f(z)$ with $A(z)\in\mathrm{Mat}_{n\times n}$, for example by setting $$\vec f =(\begin{array}{cccc} f& f' & \ldots & f^{(n-1)}\end{array})^T.$$ Let us take $n$ vector solutions $\vec f_1,\ldots, \vec f_n$ and combine them into a matrix $F=(\begin{array}{ccc}\vec f_1 &\ldots &\vec f_n \end{array})$. Clearly, $F(z)\in\mathrm{Mat}_{n\times n}$ satisfies the same equation $$F'(z)=A(z)F(z).\tag{1}$$ While it is difficult and usually impossible to solve the latter system explicitly (this is equivalent to the original problem), one can easily show that the specific combination $W(z)=\operatorname{det} F(z)$ satisfies a scalar linear ODE of 1st order, namely, $$W'(z)=W(z)\operatorname{Tr}A(z),$$ which implies that $W(z)=\mathrm{const}\cdot \exp \int \operatorname{Tr}A(z)\,dz $. This specific combination is called wronskian. Its merits are :

  • Its form can be explicitly computed as shown above ; this is because the only unknown piece is the value of the constant which can be calculated even when the solutions themselves are e.g. complicated special functions (it suffices to keep only their asymptotics).

  • Vanishing of $W(z)$ is clearly equivalent to linear dependence of solutions.

  • $\begingroup$ "Vanishing of W(z) is clearly equivalent to linear dependence of solutions." — Could you please elaborate on this? I don't exactly see why this is the case. $\endgroup$
    – 1110101001
    Mar 14, 2017 at 2:39
  • 3
    $\begingroup$ @1110101001 Vanishing of $W(z)$ at some point implies its vanishing for all $z$ ($\mathrm{const}$ above is $0$). Thus $\mathrm{det}\, F(z)$ is identically $0$. $\endgroup$ Mar 14, 2017 at 12:39

To me, the key formula in understanding the Wronskian of a linear homogeneous first order differential system is the following: $$ \text{if}\quad \frac{d A}{dt}(t)=B(t)A(t)\quad \text{then}\quad \frac{d}{dt}\left(\det A(t)\right) = \mathrm{tr}\, B(t)\det A(t).$$

Here $A, B$ are square matrices and $\mathrm{tr}$ stands for "trace" and this formula is often called "Liouville's formula".

You apply it as follows. Consider the system of $n$ differential equations $$ \frac{d\boldsymbol{x}}{dt} = B(t) \boldsymbol x(t).$$ (Boldface denotes elements of $\mathbb R^n$, considered as column vectors). Taking $n$ solutions $\boldsymbol{x}_1\ldots \boldsymbol{x}_n$ one forms the matrix $$ A(t)=\begin{bmatrix} \boldsymbol{x}_1,\ \ldots\, , \boldsymbol{x}_n\end{bmatrix}\in\mathbb{R}^{n\times n}.$$

The Wronskian determinant $\det A(t)$ is geometrically interpreted as the oriented volume of the parallelepiped spanned by $\boldsymbol{x}_1\ldots \boldsymbol{x}_n$. Liouville's formula says that this volume's evolution is governed by the trace of $B(t)$.

For a single differential equation of order $n$ $$ x^{(n)}(t)+ b_{n-1}(t)x^{(n-1)}(t)+\ldots +b_0(t) x(t) = 0$$ one introduces the vector $$ \boldsymbol{x}(t)= \begin{bmatrix} x(t) \\ x'(t) \\ \vdots \\ x^{(n-1)}(t)\end{bmatrix}$$ and the matrix ("companion matrix") $$ B(t)=\begin{bmatrix} 0& 1& 0& \ldots& \ldots& \ldots \\ 0 & 0 & 1 & 0 & \ldots&\ldots \\ & & &\vdots & & \\ 0& \ldots& \ldots& \ldots& 0 & 1 \\ -b_0(t) & -b_1(t)& \ldots &\ldots & -b_{n-2}(t) & -b_{n-1}(t)\end{bmatrix}$$ so that the equation is equivalent to $\frac{d\boldsymbol{x}}{dt}=B(t)\boldsymbol{x}(t)$. Often, one says that $\boldsymbol{x}$ belongs to the phase space of the original equation. (This is especially true when $n=2$, in which case the phase space is typically interpreted the Cartesian product of position and velocity.)

In the case of a linear homogeneous differential equation of order $n$, the Wronskian is interpreted as an oriented volume in phase space. Liouville's formula says that the evolution of the Wronskian is governed by the term $\mathrm{tr}\, (B(t))=-b_{n-1}(t)$.

Further development. This interpretation fits well in the frame of Liouville's theorem for Hamiltonian systems. See also here.

  • $\begingroup$ Hello, this seems really interesting. Can this be related to the probability current? Since the probability current in quantum mechanics can be written as a determinant of a Wronskian: $W\left(\psi,\psi^\ast\right)=\left(\begin{matrix}\psi&\psi^\ast\\\psi_x&\psi_x^\ast\\\end{matrix}\right) \rightarrow j=\frac{\hbar\left|W\right|}{2mi}$ $\endgroup$ Feb 17 at 16:03
  • $\begingroup$ @bananenheld: as you said, "probability current" is the physicist's name for a particular Wronskian, so yes, the concepts are related. $\endgroup$ Feb 25 at 16:42
  • $\begingroup$ Thank you for your reaction. Does this mean that this particular wronskian suggests that the probability current is geometrically interpreted as the parallepiped (in this case parallellogram) spanned by $\psi$ and $\psi^*$? $\endgroup$ Mar 4 at 7:28

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