Fundamental Solution to Reduced Linear First Order ODE with Constant Coefficients

Theorem
Let $H$ be the Heaviside step function.

Let $\lambda \in \R$.

Let $\map f x = \map H x \map \exp {\lambda x}$.

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

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


 * $\paren {\dfrac {\d}{\d x} - \lambda} T_f = \delta$

Proof
$x \stackrel f {\longrightarrow} \map H x \map \exp {\lambda x}$ 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 $T'_f = T_{f'}$.

Moreover:


 * $x < 0 \implies \paren {\map H x \map \exp {\lambda x}}' = 0$.


 * $x > 0 \implies \paren {{\map H x} \map \exp { \lambda x}}' = \lambda \map \exp {\lambda x}$.

Altogether:


 * $\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \sin x}' = \lambda \map H x \map \exp {\lambda x}$

Furthermore:


 * $\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \exp {0^+} - \map H {0^-} \map \exp {0^-} = 1$

By the Jump Rule:


 * $T'_{\map H x \map \exp {\lambda x}} = T_{\lambda \map H x \map \exp {\lambda x}} + \delta$

Rearranging the terms and using the linearity of distribution gives:


 * $T'_{\map H x \map \exp {\lambda x}} - \lambda T_{\map H x \map \exp {\lambda x}} = \delta$