Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE

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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 $\closedint a b$.


Let $y_1$ and $y_2$ be linearly independent.


Then their Wronskian is either never zero, or zero everywhere on $\closedint a b$.


Proof

\(\displaystyle \map W {y_1, y_2}\) \(=\) \(\displaystyle y_1 {y_2}' - y_2 {y_1}'\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {W'} {y_1, y_2}\) \(=\) \(\displaystyle \paren {y_1 {y_2}'' + {y_1}' {y_2}'} - \paren {y_2 {y_1}'' + {y_2}' {y_1}'}\) Product Rule
\(\displaystyle \) \(=\) \(\displaystyle y_1 {y_2}'' - y_2 {y_1}''\)


Because $y_1$ and $y_2$ are both particular solutions of $(1)$:

\((2):\quad\) \(\displaystyle {y_1}'' + \map P x {y_1}' + \map Q x y_1\) \(=\) \(\displaystyle 0\)
\((3):\quad\) \(\displaystyle {y_2}'' + \map P x {y_2}' + \map Q x y_2\) \(=\) \(\displaystyle 0\)
\((4):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle y_2 {y_1}'' + \map P x y_2 {y_1}' + \map Q x y_2 y_1\) \(=\) \(\displaystyle 0\) $(2)$ multiplied by $y_2$
\((5):\quad\) \(\displaystyle y_1 {y_2}'' + \map P x y_1 {y_2}' + \map Q x y_1 y_2\) \(=\) \(\displaystyle 0\) $(3)$ multiplied by $y_1$
\((6):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {y_1 {y_2}'' - y_2 {y_1}''} + \map P x \paren {y_1 {y_2}' - y_2 {y_1}'}\) \(=\) \(\displaystyle 0\) $(5)$ subtracted from $(6)$

That is:

$\dfrac {\d P} {\d W} + P W = 0$

This is a linear first order ODE.

From Solution to Linear First Order Ordinary Differential Equation:

$W = C e^{-\int P \rd x}$

The exponential function is never zero:

Therefore:

$W = 0 \iff C = 0$

and the result follows.

$\blacksquare$


Sources