The Wronskian

The Wronskian

Given an inhomogeneous linear second-order ode:

$$\ddot{x} + p(t) \dot{x} + q(t)x = 0$$

Suppose that $x = X_1(t)$ and $x = X_2(t)$ are two solutions to the ode.

According to the principle of superposition, we can write the general solution to the ode as:

$$x = c_1 X_1(t) + c_2 X_2(t)$$

To fulfill a given initial condition

$$x(t_0) = x_0, \quad \dot{x}(t_0) = u_0$$

We need to solve a system of linear equations:

$$\begin{aligned} c_1 X_1(t_0) + c_2 X_2(t_0) &= x_0 \quad (1)\\ c_1 \dot{X_1}(t_0) + c_2 \dot{X_2}(t_0) &= u_0 \quad (2) \end{aligned}$$

To solve (1) and (2), we need to find the RREF of:

$$\begin{bmatrix} X_1(t_0) & X_2(t_0) & x_0 \\ \dot{X_1}(t_0) & \dot{X_2}(t_0) & u_0 \end{bmatrix}$$

Thus, we need to find the inverse of:

$$\begin{bmatrix} X_1(t_0) & X_2(t_0) \\ \dot{X_1}(t_0) & \dot{X_2}(t_0) \end{bmatrix}$$

The inverse of the matrix exist when:

$$\begin{vmatrix} X_1(t_0) & X_2(t_0) \\ \dot{X_1}(t_0) & \dot{X_2}(t_0) \end{vmatrix} \ne 0$$

The Wronskian W is given by:

$$W = \begin{vmatrix} X_1(t_0) & X_2(t_0) \\ \dot{X_1}(t_0) & \dot{X_2}(t_0) \end{vmatrix} = X_1(t_0) \dot{X_2}(t_0) - \dot{X_1}(t_0)X_2(t_0) \ne 0$$

Example

$$\begin{aligned} \dot{X_1}(t) &= \alpha e^{\alpha t} \\ \dot{X_2}(t) &= \beta e^{\beta t} \\ W &= \begin{vmatrix} e^{\alpha t} & e^{\beta t} \\ \alpha e^{\alpha t} & \beta e^{\beta t} \end{vmatrix} \\ &= \beta e^{\alpha t} e^{\beta t} - \alpha e^{\alpha t} e^{\beta t} \end{aligned}$$

Thus, $W \ne 0$ when $\alpha \ne \beta$.