Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent

Theorem

Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:

$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

on a closed interval $\closedint a b$.

Then:

$y_1$ and $y_2$ are linearly dependent

if and only if:

the Wronskian $\map W {y_1, y_2}$ of $y_1$ and $y_2$ is zero everywhere on $\closedint a b$.

Proof

Sufficient Condition

Let $y_1$ and $y_2$ are linearly dependent.

Suppose either $y_1$ or $y_2$ is zero everywhere on $\closedint a b$.

Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\closedint a b$.

Thus:

$y_1 {y_2}' - y_2 {y_1}' = 0$

and so by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.

Without loss of generality, it follows by definition of linearly dependent that:

$y_2 = C y_1$

for some $C \in \R$.

Thus by Derivative of Constant Multiple:

${y_2}' = C {y_1}'$

Hence:

\(\ds C\)

\(=\)

\(\ds \frac {y_2} {y_1}\)

\(\ds \leadsto \ \ \)

\(\ds {y_2}'\)

\(=\)

\(\ds \frac {y_2} {y_1} {y_1}'\)

\(\ds \leadsto \ \ \)

\(\ds y_1 {y_2}'\)

\(=\)

\(\ds y_2 {y_1}'\)

\(\ds \leadsto \ \ \)

\(\ds y_1 {y_2}' - y_2 {y_1}'\)

\(=\)

\(\ds 0\)

Hence by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

$\Box$

Necessary Condition

Let $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\closedint a b$.

Without loss of generality, suppose $y_1$ is zero everywhere on $\closedint a b$.

Then by Real Function is Linearly Dependent with Zero Function, $y_1$ and $y_2$ are linearly dependent.

Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.

As $y_1$ is continuous, there is some closed interval:

$\closedint c d \subseteq \closedint a b$

on which:

$\forall x \in \closedint c d: \map {y_1} x \ne 0$

Since $\map W {y_1, y_2} = 0$:

$\dfrac {y_1 {y_2}' - y_2 {y_1}'} { {y_1}^2} = 0$

on $\closedint c d$.

By Quotient Rule for Derivatives, this is the same as:

$\paren {\dfrac {y_2} {y_1} }' = 0$

Integrating with respect to $x$, this gives:

$\dfrac {y_2} {y_1} = k$

Explicitly:

$\forall x \in \closedint c d: \map {y_2} x = k \, \map {y_1} x$

As $\map {y_2} x$ and $k \, \map {y_1} x$ have equal values on $\closedint c d$, their derivatives are equal there as well.

From Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions, it follows that:

$\forall x \in \closedint a b: \map {y_2} x = k \, \map {y_1} x$.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.15$: The General Solution of the Homogeneous Equation: Lemma $2$