# Wronskian

In mathematics, the Wronskian is a function named after Polish mathematician Josef Hoene-Wronski, especially important in the study of differential equations.

Given a set of n functions f1, ..., fn, the Wronskian W(f1, ..., fn) is given by:

$W(f_{1},\ldots ,f_{n})={\begin{vmatrix}f_{1}&f_{2}&\cdots &f_{n}\\f_{1}'&f_{2}'&\cdots &f_{n}'\\\vdots &\vdots &\ddots &\vdots \\f_{1}^{{(n-1)}}&f_{2}^{{(n-1)}}&\cdots &f_{n}^{{(n-1)}}\end{vmatrix}}$

That is, it is the determinant of the matrix constructed by placing the functions in the first row, the first derivative of each function in the second row, and so on through the n-1 derivative, thus forming a square matrix sometimes called a fundamental matrix.

In a second-order linear differential equation, the Wronskian can be computed more easily by Abel's identity.

## The Wronskian and linear independence

The Wronskian can be used to determine whether a set of differentiable functions is linearly independent on a given interval:

• If the Wronskian is non-zero at some point in an interval, then the associated functions are linearly independent on the interval.

This is useful in many situations. For example, if we wish to verify that two solutions of a second-order differential equation are independent, we may use the Wronskian. Note that if the Wronskian is uniformly zero over the interval, the functions may or may not be linearly independent. A common misconception (unfortunately promulgated in many texts) is that $W=0$ everywhere implies linear dependence - that this is not the case can clearly be seen in the third example below. Rather:

• If a set of functions is linearly dependent on an interval, then the corresponding Wronskian is uniformly zero on the interval.

In fact the two bulleted statements are logically equivalent (by transposition); they are simply alternative statements of the same truth. A proof of the theorem is given below.

## Examples

• Consider the functions $x^{2},$ $x,$ and $1,$ defined for x a real number. Take the Wronskian:
$W={\begin{vmatrix}x^{2}&x&1\\2x&1&0\\2&0&0\end{vmatrix}}=-2.$
We see that $W$ is not uniformly zero, so these functions must be linearly independent.
• Consider the functions $2x^{2}+3$, $x^{2}$, and $1$. These functions are clearly dependent, since $2x^{2}+3=2(x^{2})+3(1).$ Thus, the Wronskian must be zero, which follows by a quick calculation:
$W={\begin{vmatrix}2x^{2}+3&x^{2}&1\\4x&2x&0\\4&2&0\end{vmatrix}}=8x-8x=0.$
• As mentioned above, if the Wronskian is zero, it does not mean in general that the functions involved are linearly dependent. Consider the functions $x^{3}$ and $|x^{3}|$; that is, the absolute value of $x^{3}$. The second function can be written as:
$|x^{3}|=\left\{{\begin{matrix}-x^{3},&{\mathrm {if}}\;x<0\\x^{3},&{\mathrm {if}}\;x\geq 0\end{matrix}}\right.$
One can check that these two functions are linear independent over the set of real numbers; however, their Wronskian is seen to be zero:
$W=\left\{{\begin{matrix}{\begin{vmatrix}x^{3}&-x^{3}\\3x^{2}&-3x^{2}\end{vmatrix}}=-3x^{5}+3x^{5}=0,&{\mathrm {if}}\;x<0\\{\begin{vmatrix}x^{3}&x^{3}\\3x^{2}&3x^{2}\end{vmatrix}}=3x^{5}-3x^{5}=0,&{\mathrm {if}}\;x\geq 0\end{matrix}}\right.$

## Abstract definition

There is a sense in which the Wronskian of an n-th order linear differential equation is its n-th exterior power. For that idea to be implemented one must be working with some formulation in which differential equations are sufficiently like vector spaces: for example in the language of vector bundles carrying a connection.

## Proof: Wronskian and linear independence

The theorem is significantly easier to prove by way of its second instantiation above, namely: If the functions are linearly dependent over the interval, then so are the columns of the associated Wronskian matrix (differentiation is a linear operation); consequently the Wronskian determinant is zero at all points of the interval.