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:

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

$\blacksquare$

See also

• Solution to Linear First Order ODE with Constant Coefficients

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions