Wronskian determinant

Given functions $f_{1},f_{2},\ldots,f_{n}$, then the Wronskian determinant (or simply the Wronskian) $W(f_{1},f_{2},f_{3},\ldots,f_{n})$ is the determinant of the square matrix

 $W(f_{1},f_{2},f_{3},\ldots,f_{n})=\left\lvert\begin{array}[]{@{}ccccc@{}}f_{1}% &f_{2}&f_{3}&\cdots&f_{n}\\ f_{1}^{\prime}&f_{2}^{\prime}&f_{3}^{\prime}&\cdots&f_{n}^{\prime}\\ f_{1}^{\prime\prime}&f_{2}^{\prime\prime}&f_{3}^{\prime\prime}&\cdots&f_{n}^{% \prime\prime}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ f_{1}^{(n-1)}&f_{2}^{(n-1)}&f_{3}^{(n-1)}&\cdots&f_{n}^{(n-1)}\\ \end{array}\right\rvert$

where $f^{(k)}$ indicates the $k$th derivative of $f$ (not exponentiation).

The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent when there exists a non-trivial linear relation between them, a set of functions $\{f_{1},f_{2},f_{3},\ldots,f_{n}\}$ is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e.,

 $a_{1}f_{1}(t)+a_{2}f_{2}(t)+\cdots+a_{n}f_{n}(t)=0$

for some $a_{1},a_{2},\ldots,a_{n}$, not all zero, at any $t\in I$.

Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian.

Examples

Consider the functions $x^{2}$, $x$, and $1$. Take the Wronskian:

 $W=\left\lvert\begin{array}[]{@{}ccc@{}}x^{2}&x&1\\ 2x&1&0\\ 2&0&0\\ \end{array}\right\rvert=-2$

Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^{2}$ and $x$:

 $W=\left\lvert\begin{array}[]{@{}cc@{}}x^{2}&x\\ 2x&1\\ \end{array}\right\rvert=x^{2}-2x^{2}=-x^{2}$

Here $W=0$ only when $x=0$. Therefore $x^{2}$ and $x$ are independent except at $x=0$.

Consider $2x^{2}+3$, $x^{2}$, and $1$:

 $W=\left\lvert\begin{array}[]{@{}ccc@{}}2x^{2}+3&x^{2}&1\\ 4x&2x&0\\ 4&2&0\\ \end{array}\right\rvert=8x-8x=0$

Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since

 $2x^{2}+3=2(x^{2})+3(1)$

Given $n$ linearly independant functions $f_{1},f_{2},\ldots,f_{n}$, we can use the Wronskian to construct a linear differential equation whose solution space is exactly the span of these functions. Namely, if $g$ satisfies the equation

 $W(f_{1},f_{2},f_{3},\ldots,f_{n},g)=0,$

then $g=a_{1}f_{1}(t)+a_{2}f_{2}(t)+\cdots+a_{n}f_{n}(t)$ for some choice of $a_{1},a_{2},\ldots,a_{n}$.

As a simple illustration of this, let us consider polynomials of at most second order. Such a polynomial is a linear combination of $1$, $x$, and $x^{2}$. We have

 $W(1,x,x^{2},g(x))=\left|\begin{matrix}1&x&x^{2}&g(x)\\ 0&1&2x&g^{\prime}(x)\\ 0&0&2&g^{\prime\prime}(x)\\ 0&0&0&g^{\prime\prime\prime}(x)\end{matrix}\right|=2g^{\prime\prime\prime}(x).$

Hence, the equation is $g^{\prime\prime\prime}(x)=0$ which indeed has exactly polynomials of degree at most two as solutions.

Title Wronskian determinant WronskianDeterminant 2013-03-22 12:22:59 2013-03-22 12:22:59 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Definition msc 34-00 Wronskian GrammianDeterminant