Fundamental Solution to 1D Laplace's Equation
Theorem
Let $\ds \map f x = \frac {\size x} 2$ where $\size x$ is the absolute value function.
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Let $T_f \in \map {\DD'} \R$ be the distribution associated with $f$.
Then $T_f$ is the fundamental solution to the $1$-dimensional Laplace's equation.
That is:
- $T_f = \delta$
or in the distributional sense:
- $\dfrac {\d^2} {\d x^2} f = \delta$
Proof
$\ds x \stackrel f {\longrightarrow} \frac {\size x} 2$ 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:
- $\ds x < 0 \implies \paren {\frac {\size x} 2}' = -\frac 1 2$
- $\ds x > 0 \implies \paren {\frac {\size x} 2}' = \frac 1 2$
Altogether:
- $\ds \forall x \in \R \setminus \set 0 : \paren {\frac {\size x} 2}' = \map g x$
where:
- $\map g x = \begin{cases}
\ds \frac 1 2 & : x > 0 \\ \ds - \frac 1 2 & : x < 0 \end{cases}$
Furthermore:
- $\ds \map f {0^+} - \map f {0^-} = 0$
By the Jump Rule:
- $T'_f = T_g$
$x \stackrel g {\longrightarrow} \map g x$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By similar arguments:
- $\ds \forall x \in \R \setminus \set 0 : \map {g'} x = 0$
Furthermore:
- $\ds \map g {0^+} - \map g {0^-} = 1$
By the Jump Rule:
- $T'_g = \delta$
By the definition of distributional derivative:
\(\ds T_f\) | \(=\) | \(\ds \paren {T'_f}'\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds T'_{f'}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds T'_g\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \delta\) |
Hence, if $\ds \map f x = \frac {\size x} 2$, then $T_f = \delta$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense