4.3: The Wronskian

View tutorial on YouTube

Suppose that having determined that two solutions of (4.2.1) are $x = X_1(t)$ and $x = X_2(t)$, we attempt to write the general solution to (4.2.1) as (4.2.2). We must then ask whether this general solution will be able to satisfy the two initial conditions given by $$\label{eq:1}x(t_0)=x_0,\quad\overset{.}{x}(t_0)=u_0.$$

Applying these initial conditions to (4.2.2), we obtain $$\begin{align}c_1X_1(t_0)+c_2X_2(t_0)&=x_0,\nonumber \\ c_1\overset{.}{X}_1(t_0)+c_2\overset{.}{X}_2(t_0)&=u_0,\label{eq:2}\end{align}$$ ) which is observed to be a system of two linear equations for the two unknowns $c_1$ and $c_2$. Solution of $\eqref{eq:2}$ by standard methods results in $$c_1=\frac{x_0\overset{.}{X}_2(t_0)-u_0X_2(t_0)}{W},\quad c_2=\frac{u_0X_1(t_0)-x_0\overset{.}{X}_1(t_0)}{W},\nonumber$$ where $W$ is called the Wronskian and is given by $$\label{eq:3}W=X_1(t_0)\overset{.}{X}_2(t_0)-\overset{.}{X}_1(t_0)X_2(t_0).$$

Evidently, the Wronskian must not be equal to zero $(W\neq 0)$ for a solution to exist.

For examples, the two solutions $$X_1(t)=A\sin\omega t,\quad X_2(t)=B\sin\omega t,\nonumber$$ have a zero Wronskian at $t=t_0$, as can be shown by computing $$\begin{aligned} W&=(A\sin\omega t_0)(B\omega\cos\omega t_0)-(A\omega\cos\omega t_0)(B\sin\omega t_0) \\ &=0;\end{aligned}$$ while the two solutions $$X_1(t)=\sin\omega t,\quad X_2(t)=\cos\omega t,\nonumber$$ with $\omega\neq 0$, have a nonzero Wronskian at $t=t_0$, $$\begin{aligned}W&=(\sin\omega t_0)(-\omega\sin\omega t_0)-(\omega\cos\omega t_0)(\cos\omega t_0) \\ &=-\omega.\end{aligned}$$

When the Wronskian is not equal to zero, we say that the two solutions $X_1(t)$ and $X_2(t)$ are linearly independent. The concept of linear independence is borrowed from linear algebra, and indeed, the set of all functions that satisfy (4.2.1) can be shown to form a two-dimensional vector space.