Wronskian determinant

Given functions $f_1,f_2,\mathrm{\dots },f_n$, then the Wronskian determinant (or simply the Wronskian) $W(f_1,f_2,f_3,\mathrm{\dots },f_n)$ is the determinant of the square matrix

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 $\{f_1,f_2,f_3,\mathrm{\dots },f_n\}$ is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e.,

$$a_1{f}_1(t)+a_2{f}_2(t)+\mathrm{\cdots }+a_n{f}_n(t)=0$$

for some $a_1,a_2,\mathrm{\dots },a_n$, 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:

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

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 $1$:

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 $f_1,f_2,\mathrm{\dots },f_n$, we can usethe Wronskian to construct a linear differential equation whose solution spaceis exactly the span of these functions. Namely, if $g$ satisfies the equation

$$W(f_1,f_2,f_3,\mathrm{\dots },f_n,g)=0,$$

then $g=a_1{f}_1(t)+a_2{f}_2(t)+\mathrm{\cdots }+a_n{f}_n(t)$ for some choice of$a_1,a_2,\mathrm{\dots },a_n$.

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 $x^2$. Wehave

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