Fundamental Solution to y'' + w y

Theorem
Let $H$ be the Heaviside step function.

Let $\omega \in \R : \omega \ne 0$.

Let $\ds \map f x = \map H x \frac {\map \sin {\omega x}} \omega$.

Let $T_f$ be the distribution associated with $f$.

Then, in the distributional sense, $T_f$ is the fundamental solution of


 * $\paren {\dfrac {\d^2}{\d x^2} + \omega^2} T_f = \delta$

Proof
$x \stackrel f {\longrightarrow} \map H x \frac {\map \sin {\omega x}} \omega$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.

By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.

Moreover:


 * $\ds x < 0 \implies \paren {\map H x \frac {\map \sin {\omega x}}{\omega}}' = 0$.


 * $\ds x > 0 \implies \paren {\map H x \frac {\map \sin {\omega x}}{\omega}}' = \map \cos {\omega x}$.

Altogether:


 * $\forall x \in \R \setminus \set 0 : \paren {{\map H x} \frac {\map \sin {\omega x}} \omega}' = \map H x \map \cos {\omega x}$

Furthermore:


 * $\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \frac {\map \sin {\omega \cdot 0^+}} \omega - \map H {0^-} \frac {\map \sin {\omega \cdot 0^-}} \omega = 0$

By the Jump Rule:


 * $T'_{\map H x \frac {\map \sin {\omega x}} \omega} = T_{\map H x \map \cos {\omega x}}$

$x \stackrel g {\longrightarrow} \map H x \map \cos {\omega x}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.

By Differentiable Function as Distribution we have that $T'_g = T_{g'}$.

Moreover:


 * $\ds x < 0 \implies \paren {\map H x \map \cos \omega}' = 0$.


 * $\ds x > 0 \implies \paren {\map H x \map \cos \omega}' = - \omega \map \sin {\omega x}$.

Altogether:


 * $\forall x \in \R \setminus \set 0 : \paren {\map H x \map \cos \omega}' = - \map H x \omega \map \sin {\omega x}$

Furthermore:


 * $\ds \map g {0^+} - \map g {0^-} = \map H {0^+} \map \cos {\omega \cdot 0^+} - \map H {0^-} \map \cos {\omega \cdot 0^-} = 1$

By the Jump Rule:


 * $T'_{\map H x \map \cos {\omega x}} = T_{- \omega \map H x \map \sin {\omega x}} + \delta$

Hence:


 * $T''_{\map H x \frac {\map \sin {\omega x}} \omega} = T_{- \omega^2 \map H x \frac {\map \sin {\omega x}} \omega} + \delta$

Rearranging the terms and using the linearity of distribution gives:


 * $T''_{\map H x \frac {\map \sin {\omega x}} \omega} + \omega^2 T_{\map H x \frac {\map \sin {\omega x}} \omega} = \delta$