Weak Solution to Dx u = Heaviside Step Function

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Theorem

Let $H: \R \to \closedint 0 1$ be the Heaviside step function.

Let $u : \R \to \R$ be such that:

$\map u x = \begin{cases}

c & : x < 0 \\ x + c & : x > 0 \end{cases}$

where $c \in \R$.

Let $T_u$ be the distribution associated with $u$.


Then $u$ is a weak solution of:

$u' = H$

That is, in the distributional sense it holds that:

$\dfrac \d {\d x} T_u = T_H$


Proof

$u$ is continuous on $\R$ and continously differentiable on $\R \setminus \set 0$.

For $x < 0$ we have $\map {u'} x = 0$.

For $x > 0$ we have $\map {u'} x = 1$.

That is:

$\map {u'} x = \map H x$

Furthermore:

$\ds \lim_{x \mathop \to 0^-} = 0$
$\ds \lim_{x \mathop \to 0^+} = 1$

By the jump rule:

\(\ds T_u'\) \(=\) \(\ds T_H + 0 \cdot \delta\)
\(\ds \) \(=\) \(\ds T_H\)

$\blacksquare$


Proof of no classical solutions


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Aiming for a contradiction, suppose $u$ is a differentiable function such that $u' = H$, then

$\ds \lim_{x \mathop \to 0-} \map {u'} x = 0$
$\ds \lim_{x \mathop \to 0+} \map {u'} x = 1$

contradicts the Intermediate Value Theorem for Derivatives.


$\Box$

Sources

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