📂Odinary Differential Equations

Wronskian of Two Solutions of a Second Order Linear Differential Equation

Theorem ¹

Suppose $y_{1}$ and $y_{2}$ are solutions to the second-order linear differential equation $y^{\prime \prime}+p(t)y^{\prime}+q(t)y=0$. Then,

1. The Wronskian of $y_{1}$ and $y_{2}$ is expressed in the form of an exponential function.

   $$W [y_{1}, y_{2}] (t)=c e^{-\int p(t) dt}$$

   Where $c$ is a constant that depends on $y_{1},\ y_{2}$.

2. $W[y_{1},y_{2}] (t)$ is either always $0$ or never $0$ at all points.

Explanation

Also known as Abel’s theorem. Although Abel’s theorem typically refers to Abel’s limit theorem, the Boyce textbook on ordinary differential equations refers to the above theorem as Abel’s theorem, presumably because it was derived by Abel.

The key point of the theorem is 2. Because the Wronskian is always either $0$ or not $0$, finding even a single point where it is not $0$ means that $W[y_{1},y_{2}]\ne 0$ and $y_{1},\ y_{2}$ are independent and constitute a fundamental set of solutions. When calculating the Wronskian of any two solutions and it’s uncertain whether it is $0$ or not, proving it is not $0$ by substituting an arbitrary (easy to calculate) value demonstrates the independence of the two solutions.

Proof

Since $y_{1},\ y_{2}$ is a solution to the given differential equation, the following holds true.

$$y_{1}^{\prime \prime}+p(t)y_{1}^{\prime}+q(t)y_{1}=0 \\[1em] y_{2}^{\prime \prime}+p(t)y_{2}^{\prime}+q(t)y_{2}=0$$

Multiplying the above equation by $-y_{2}$ and multiplying the equation below by $y_{1}$ and adding the two yields

$$\begin{equation} (y_{1}y_{2}^{\prime \prime}-y_{1}^{\prime \prime}y_{2})+p(t)(y_{1}y_{2}^{\prime}-y_{1}^{\prime}y_{2})=0 \end{equation}$$

According to the definition of the Wronskian,

$$W[y_{1},y_{2}] (t)=W=y_{1}y_{2}^{\prime}-y_{1}^{\prime}y_{2} \\[1em] W^{\prime}=y_{1}^{\prime}y_{2}^{\prime}+y_{1}y_{2}^{\prime \prime}-y_{1}^{\prime \prime}y_{2}-y_{1}^{\prime}y_{2}^{\prime}=y_{1}y_{2}^{\prime \prime}-y_{1}^{\prime \prime}y_{2}$$

Expressing $(1)$ in terms of $W, W^{\prime}$ gives

$$W^{\prime}+p(t)W=0$$

This is a simple separable first-order differential equation.

$$\begin{align*} \\ && W^{\prime}+p(t)W =&\ 0 \\ \implies && \dfrac{dW}{dt} =&\ -p(t)W \\ \implies && \dfrac{1}{W} dW =&\ -p(t)dt \\ \implies && \ln W =&\ -\int p(t)dt+C \\ \implies && W =&\ ce^{-\int p(t) dt} \end{align*}$$

Since $W$ is in the form of an exponential function, it can never be $0$ unless it is $c=0$. Therefore, the case of $W=0$ is when it is $c=0$, and in this case, irrespective of $t$, it is always $W=0$ at all points. Likewise, in the case of $c \ne 0$, since $W$ is in the form of an exponential function, irrespective of $t$, it is always $W \ne 0$ at all points.

■