Determinant

Wronskian Determinants of Two Functions

We are going to look more into second order linear homogenous differential equations, but before we do, we need to first learn about a type of determinant known as a Wronskian Determinant which we define below.

Definition: Let $f$ and $g$ be two differentiable functions. Then the Wronskian Determinant of $f$ and $g$ is the $2\times 2$ determinant $W(f,g)={\begin{vmatrix}f(x)&g(x)\\f'(x)&g'(x)\end{vmatrix}}=f(x)g'(x)+f'(x)g(x)$.

Sometimes the term "Wronskian" by itself is used to mean the same thing as "Wronskian Determinant". Furthermore, sometimes we can just write "$W$", or "$W(x)$" instead of $W(f,g)$ to represent the Wronskian of $f$ and $g$.

Let's look at some examples of computing the Wronskian determinant of two differentiable functions.

Example 1

Determine the Wronskian of the functions $f(x)=x^{2}$ and $g(x)=3x^{2}$. For what values of $x$ is the Wronskian equal to zero?

We note that $f$ and $g$ are both differentiable functions and that $f'(x)=2x$ and $g'(x)=6x$. Therefore the Wronskian of $f$ and $g$ is:

${\begin{aligned}\quad {\begin{vmatrix}x^{2}&3x^{2}\\2x&6x\end{vmatrix}}=x^{2}\cdot 6x-3x^{2}\cdot 2x=6x^{3}-6x^{3}=0\end{aligned}}$

Therefore the Wronskian of $f$ and $g$ is equal to zero for all $x\in \mathbb {R}$.

Example 2

Determine the Wronskian of the functions $f(x)=e^{x}\sin t$ and $g(x)=e^{x}\cos x$. For what values of $x$ is the Wronskian equal to zero?

We note that $f$ and $g$ are both differentiable functions and that $f'(x)=e^{x}\sin x+e^{x}\cos x$ and $g'(x)=e^{x}\cos x-e^{x}\sin x$. Therefore the Wronskian of $f$ and $g$ is:

${\begin{aligned}\quad {\begin{vmatrix}e^{x}\sin x&e^{x}\cos x\\e^{x}\sin x+e^{x}\cos x&e^{x}\cos x-e^{x}\sin x\end{vmatrix}}=\left(e^{x}\sin x\right)\cdot \left(e^{x}\cos x-e^{x}\sin x\right)-\left(e^{x}\cos x\right)\cdot \left(e^{x}\sin x+e^{x}\cos x\right)\\\quad =e^{2x}\sin x\cos x-e^{2x}\sin ^{2}x-e^{2x}\sin x\cos x-e^{2x}\cos ^{2}x=-e^{2x}(\cos ^{2}x+\sin ^{2}x)=-e^{2x}\end{aligned}}$

Note that $-e^{2x}<0$ for all $x\in \mathbb {R}$ so the Wronskian of $f$ and $g$ is zero nowhere.

Wronskian Determinants and Linear Homogenous Differential Equations

Recall that if we have the second order linear homogenous differential equation ${\frac {d^{2}y}{dt^{2}}}+p(t){\frac {dy}{dt}}+q(t)y=0$ and $y=y_{1}(t)$ and $y=y_{2}(t)$ are solutions to this differential equation, then for constants $C$ and $D$, the linear combination $y=Cy_{1}(t)+Dy_{2}(t)$ is also a solution to this differential equation. One question to ask if whether or not all of the solutions to this differential equation are in this form as we do not want to miss any other potential solutions.

Suppose that we are given an initial value problem to a second order linear homogenous differential equation in this form alongside the initial conditions $y(t_{0})=y_{0}$ and $y'(t_{0})=y'_{0}$. Applying the first initial condition $y(t_{0})=y_{0}$ to our solution $y=Cy_{1}(t)+Dy_{2}(t)$ and we have that:

${\begin{aligned}\quad y_{0}=Cy_{1}(t_{0})+Dy_{2}(t_{0})\end{aligned}}$

Now the derivative of $y=Cy_{1}(t)+Dy_{2}(t)$ is $y'=Cy_{1}'(t)+Dy_{2}'(t)$. Applying the second initial condition $y'(t_{0})=y'_{0}$ and we have that:

${\begin{aligned}\quad y'_{0}=Cy_{1}'(t_{0})+Dy_{2}'(t_{0})\end{aligned}}$

We need the unknown constants $C$ and $D$ to satisfy both of these equations, that is, we want to solve the system $\left\{{\begin{matrix}Cy_{1}(t_{0})+Dy_{2}(t_{0})=y_{0}\\Cy_{1}'(t_{0})+Dy_{2}'(t_{0})=y'_{0}\end{matrix}}\right.$ for the constants $C$ and $D$. We note that we have a system of two equations with two unknowns and thus, a unique solution exists if the determinant of the augmented matrix for this system is nonzero, that is:

${\begin{aligned}W={\begin{vmatrix}y_{1}(t_{0})&y_{2}(t_{0})\\y_{1}'(t_{0})&y_{2}'(t_{0})\end{vmatrix}}=y_{1}(t_{0})y_{2}'(t_{0})-y_{1}'(t_{0})y_{2}(t_{0})\neq 0\end{aligned}}$

Notice though that this determinant $W$ is simple the Wronskian of the functions $y_{1}$ and $y_{2}$ evaluated at $t_{0}$. If $W\neq 0$, then we can apply Cramer's Rule from linear algebra to find the values of the constants $C$ and $D$. We have that:

${\begin{aligned}\quad C={\frac {\begin{vmatrix}y_{0}&y_{2}(t_{0})\\y'_{0})&y_{2}'(t_{0})\end{vmatrix}}{\begin{vmatrix}y_{1}(t_{0})&y_{2}(t_{0})\\y_{1}'(t_{0})&y_{2}'(t_{0})\end{vmatrix}}}={\frac {y_{0}y_{2}'(t_{0})-y'_{0}y_{2}(t_{0})}{y_{1}(t_{0})y_{2}'(t_{0})-y_{1}'(t_{0})y_{2}(t_{0})}}\quad \quad D={\frac {\begin{vmatrix}y_{1}(t_{0})&y_{0}\\y_{1}'(t_{0})&y'_{0}\end{vmatrix}}{\begin{vmatrix}y_{1}(t_{0})&y_{2}(t_{0})\\y_{1}'(t_{0})&y_{2}'(t_{0})\end{vmatrix}}}={\frac {y'_{0}y_{1}(t_{0})-y_{0}y_{1}'(t_{0})}{y_{1}(t_{0})y_{2}'(t_{0})-y_{1}'(t_{0})y_{2}(t_{0})}}\end{aligned}}$

If this determinant $W$ equals zero, then either no solutions for $C$ and $D$ exist, or infinitely many solutions for the values of $C$ and $D$ may exist. The following theorem summarizes what we have just found.

Theorem 1: Let ${\frac {d^{2}y}{dt^{2}}}+p(t){\frac {dy}{dt}}+q(t)y=0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous functions on an open interval $I$ such that $t_{0}\in I$ and with the initial conditions $y(t_{0})=y_{0}$ and $y'(t_{0})=y'_{0}$. If $y=y_{1}(t)$ and $y=y_{2}(t)$ are solutions to this differential equation then there exists constants $C$ and $D$ for which $y=Cy_{1}(t)+Dy_{2}(t)$ is a solution to the initial value problem if and only if the Wronskian at $t_{0}$ is nonzero, that is $W(y_{1},y_{2}){\bigg |}_{t_{0}}=y_{1}(t_{0})y_{2}'(t_{0})-y_{1}'(t_{0})y_{2}(t_{0})\neq 0$.