Wrońskian | Topic Playlist

Given $n$ functions $y_{1}(x)$ to $y_{n}(x)$, the Wrońskian is defined as,

\begin{equation} W(y_{1},y_{2},\ldots,y_{n}) = \text{det} \begin{bmatrix} y_{1} & y_{2} & \ldots & y_{n} \\ y_{1}' & y_{2}' & \ldots & y_{n}' \\ \vdots & \vdots & & \vdots \\ y_{1}^{(n-1)} & y_{2}^{(n-1)} & \ldots & y_{n}^{(n-1)} \end{bmatrix} \end{equation}

For just two functions, the Wrońskian evaluates to,

\begin{equation} W(y_{1},y_{2}) = \text{det} \begin{bmatrix} y_{1} & y_{2} \\ y_{1}' & y_{2}' \end{bmatrix} = y_{1}y_{2}' - y_{1}'y_{2} \end{equation}

When solving an IVP of the form,

\begin{equation} y'' +p(x)y' +q(x)y = 0 \qquad y(x_{0}) = y_{0} \qquad y'(x_{0}) = y'_{0} \end{equation}

We'd like to verify that we can solve for any initial condition and that the two homogeneous solutions $y_{1}(x)$ and $y_{2}(x)$ we found are fundamentally different from each other. By writing the equations for the initial conditions as a matrix equation, we can use the tools of linear algebra to figure out.

\begin{equation} y(x) = c_{1}y_{1}(x)+c_{2}y_{2}(x) \end{equation} \begin{equation} y(x_{0}) = c_{1}y_{1}(x_{0})+c_{2}y_{2}(x_{0}) = y_{0} \end{equation} \begin{equation} y'(x_{0}) = c_{1}y_{1}'(x_{0})+c_{2}y_{2}'(x_{0}) = y'_{0} \end{equation} \begin{equation} \begin{bmatrix} y_{1}(x_{0}) & y_{2}(x_{0}) \\ y_{1}'(x_{0}) & y_{2}'(x_{0}) \end{bmatrix}\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} y_{0} \\ y'_{0} \end{bmatrix} \end{equation}

The determinant of the matrix is the Wrońskian $W(y_{1},y_{2})$ evaluated at $x=x_{0}$, and the matrix equation is guaranteed to have a solution if and only if the determinant of the matrix is nonzero. Another name for this is the two homogenous solutions $y_{1}(x)$ and $y_{2}(x)$ are linearly independent functions and form a fundamental set of solutions to the differential equation.

Given $y_{1}(x),y_{2}(x),\ldots,y_{n}(x)$ solve an IVP with ODE

\begin{equation} y^{(n)} + p_{n-1}(x)y^{(n-1)} + p_{n-2}(x)y^{(n-2)} + \cdots + p_{0}(x)y = 0 \end{equation}

If $W(y_{1},y_{2},\ldots,y_{n}) \neq 0$ except at isolated points in an interval $I$, then the set of functions $y_{1},y_{2},\ldots,y_{n}$ form a fundamental set of solutions on the interval $I$.

Reading this very carefully, the theorem doesn't say the functions $y_{1}(x),y_{2}(x),\ldots,y_{n}(x)$ will be linearly dependent if the $W(y_{1},y_{2},\ldots,y_{n}) = 0$ in an interval $I$ instead of just isolated points. In fact, the set of functions can be linearly independent if $W(y_{1},y_{2},\ldots,y_{n}) = 0$ in an interval $I$. The only thing the theorem guarantees is $y_{1}(x),y_{2}(x),\ldots,y_{n}(x)$ are linearly independent if $W(y_{1},y_{2},\ldots,y_{n}) \neq 0$ except at isolated points in an interval $I$. If $W(y_{1},y_{2},\ldots,y_{n}) = 0$ in an interval $I$, the set of functions $y_{1}(x),y_{2}(x),\ldots,y_{n}(x)$ may or may not be linearly independent in the interval $I$.

For the IVP, when the $W(y_{1},y_{2},\ldots,y_{n}) \neq 0$ at $x = x_{0}$, then the IVP will have a solution. If the $W(y_{1},y_{2},\ldots,y_{n}) = 0$ at $x = x_{0}$, the IVP may or may not have a solution.