Wronskian Theorem

The Wronskian provides a powerful test for linear independence of solutions and plays a central role in the theory of linear differential equations.

DefinitionWronskian

For two functions $y_1(x)$ and $y_2(x)$ , the Wronskian is defined as:

$W(y_1, y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_1'y_2$

TheoremWronskian Test for Linear Independence

Let $y_1$ and $y_2$ be solutions of:

$y'' + p(x)y' + q(x)y = 0$

on an interval $I$ where $p$ and $q$ are continuous. Then:

(a) If $W(y_1, y_2)(x_0) \neq 0$ for some $x_0 \in I$ , then $y_1$ and $y_2$ are linearly independent on $I$ .

(b) If $y_1$ and $y_2$ are linearly independent on $I$ , then $W(y_1, y_2)(x) \neq 0$ for all $x \in I$ .

(c) If $y_1$ and $y_2$ are linearly dependent, then $W(y_1, y_2)(x) = 0$ for all $x \in I$ .

In other words, for solutions of a linear ODE, the Wronskian is either identically zero (dependent) or never zero (independent).

TheoremAbel's Formula

If $y_1$ and $y_2$ are solutions of $y'' + p(x)y' + q(x)y = 0$ , then their Wronskian satisfies:

$W(y_1, y_2)(x) = Ce^{-\int p(x)dx}$

where $C$ is a constant. This formula shows that the Wronskian never changes sign and depends only on the coefficient $p(x)$ .

Proof of Abel's Formula

Differentiate the Wronskian:

$W' = (y_1y_2')' - (y_1'y_2)' = y_1'y_2' + y_1y_2'' - y_1''y_2 - y_1'y_2' = y_1y_2'' - y_1''y_2$

Since $y_1$ and $y_2$ satisfy the ODE:

$y_1'' = -p(x)y_1' - q(x)y_1$ $y_2'' = -p(x)y_2' - q(x)y_2$

Substituting:

$W' = y_1[-p(x)y_2' - q(x)y_2] - [-p(x)y_1' - q(x)y_1]y_2$

$= -p(x)[y_1y_2' - y_1'y_2] = -p(x)W$

This is a first-order linear ODE for $W$ , with solution:

$W(x) = Ce^{-\int p(x)dx}$

ExampleComputing Wronskian

For $y'' + 4y = 0$ , the solutions are $y_1 = \cos 2x$ and $y_2 = \sin 2x$ .

Direct computation:

$W = \cos 2x \cdot 2\cos 2x - (-2\sin 2x) \cdot \sin 2x = 2\cos^2 2x + 2\sin^2 2x = 2$

Using Abel's formula with $p(x) = 0$ :

$W = Ce^{0} = C$

Computing at $x = 0$ : $W(0) = \cos 0 \cdot 2\cos 0 - 0 = 2$ , so $C = 2$ .

Both methods give $W(x) = 2 \neq 0$ , confirming linear independence.

ExampleDependent Functions

Consider $y_1 = x^2$ and $y_2 = 3x^2$ (linearly dependent since $y_2 = 3y_1$ ).

$W = x^2 \cdot 6x - 2x \cdot 3x^2 = 6x^3 - 6x^3 = 0$

The Wronskian is identically zero, as expected for dependent functions.

Remark

Important Warning: If two functions are NOT solutions of the same linear ODE, then $W = 0$ does NOT imply linear dependence. For example, $f(x) = x^2$ and $g(x) = |x|^3$ have $W = 0$ at $x = 0$ but are linearly independent on any interval containing zero.

The Wronskian test for linear independence only works reliably when the functions are solutions of a linear ODE.

ExampleUsing Wronskian in Variation of Parameters

The Wronskian appears in the variation of parameters formulas:

$u_1' = -\frac{y_2f}{W}, \quad u_2' = \frac{y_1f}{W}$

The condition $W \neq 0$ ensures these formulas are well-defined, which is guaranteed when $y_1$ and $y_2$ are linearly independent solutions.

The Wronskian theorem provides both a theoretical tool for understanding solution structure and a practical computational method for verifying linear independence.