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
- 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 $2$