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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

on a closed interval $\left[{a \,.\,.\, b}\right]$.


Then:

$y_1$ and $y_2$ are linearly dependent

if and only if:

the Wronskian $W \left({y_1, y_2}\right)$ of $y_1$ and $y_2$ is zero everywhere on $\left[{a \,.\,.\, b}\right]$.


Proof

Sufficient Condition

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

Suppose either $y_1$ or $y_2$ is zero everywhere on $\left[{a \,.\,.\, b}\right]$.

Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\left[{a \,.\,.\, b}\right]$.

Thus:

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

and so by definition $W \left({y_1, y_2}\right) = 0$ everywhere on $\left[{a \,.\,.\, b}\right]$.


Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\left[{a \,.\,.\, b}\right]$.

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:

\(\displaystyle C\) \(=\) \(\displaystyle \frac {y_2} {y_1}\)
\(\displaystyle \implies \ \ \) \(\displaystyle {y_2}'\) \(=\) \(\displaystyle \frac {y_2} {y_1}{y_1}'\)
\(\displaystyle \implies \ \ \) \(\displaystyle y_1 {y_2}'\) \(=\) \(\displaystyle y_2 {y_1}'\)
\(\displaystyle \implies \ \ \) \(\displaystyle y_1 {y_2}' - y_2 {y_1}'\) \(=\) \(\displaystyle 0\)

Hence by definition $W \left({y_1, y_2}\right) = 0$ everywhere on $\left[{a \,.\,.\, b}\right]$.

$\Box$


Necessary Condition

Let $W \left({y_1, y_2}\right) = 0$ everywhere on $\left[{a \,.\,.\, b}\right]$.

One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\left[{a \,.\,.\, b}\right]$.

Without loss of generality, suppose $y_1$ is zero everywhere on $\left[{a \,.\,.\, b}\right]$.

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 $\left[{a \,.\,.\, b}\right]$.

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

$\left[{c \,.\,.\, d}\right] \subseteq \left[{a \,.\,.\, b}\right]$

on which:

$\forall x \in \left[{c \,.\,.\, d}\right]: y_1 \left({x}\right) \ne 0$


Since $W \left({y_1, y_2}\right) = 0$:

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

on $\left[{c \,.\,.\, d}\right]$.

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

$\left({\dfrac {y_2} {y_1} }\right)' = 0$

Integrating with respect to $x$, this gives:

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

Explicitly:

$\forall x \in \left[{c \,.\,.\, d}\right]: y_2 \left({x}\right) = k y_1 \left({x}\right)$

As $y_2 \left({x}\right)$ and $k y_1 \left({x}\right)$ have equal values on $\left[{c \,.\,.\, d}\right]$, 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:

$\forall x \in \left[{a \,.\,.\, b}\right]: y_2 \left({x}\right) = k y_1 \left({x}\right)$.

$\blacksquare$


Sources