# Fundamental Solution to 1D Laplace's Equation

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## 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 $T''_f = \delta$.

$\blacksquare$