Mixed Partial Derivative of Heaviside Step Function

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Theorem

Let $\tuple {x, y} \stackrel u {\longrightarrow} \map u {x, y}: \R^2 \to \R$ be the Heaviside step function.

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

Let $\delta_{\tuple {0, 0} } \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.


Then in the distributional sense:

$\dfrac {\partial^2 u} {\partial x \partial y} = \delta_{\tuple {0, 0}}$


Proof

Let $\phi \in \map \DD {\R^2}$ be a test function with support on $\openint 0 a^2 := \openint 0 a \times \openint 0 a$ where $\times$ is the Cartesian product and $a > 0$.

Then:

\(\ds \map {\dfrac {\partial^2 u}{\partial x \partial y} } \phi\) \(=\) \(\ds - \map {\dfrac {\partial u}{\partial y} } {\dfrac {\partial \phi}{\partial x} }\) Definition of Distributional Partial Derivative
\(\ds \) \(=\) \(\ds \map u {\dfrac {\partial^2 \phi}{\partial y \partial x} }\) Definition of Distributional Partial Derivative
\(\ds \) \(=\) \(\ds \iint_{\R^2} u \dfrac {\partial^2 \phi}{\partial y \partial x} \rd x \rd y\) Definition of Distribution
\(\ds \) \(=\) \(\ds \iint_{\R^2} u \dfrac {\partial^2 \phi}{\partial x \partial y} \rd x \rd y\) Clairaut's Theorem, Definition of Test Function
\(\ds \) \(=\) \(\ds \int_0^a \int_0^a \dfrac {\partial^2 \phi}{\partial x \partial y} \rd x \rd y\) Definition of Heaviside Step Function, $\phi$ is supported on $\openint 0 a^2$
\(\ds \) \(=\) \(\ds \int_0^a \paren {\map {\dfrac {\partial \phi} {\partial y} } {a, y} - \map {\dfrac {\partial \phi} {\partial y} } {0, y} } \rd y\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds - \int_0^a \map {\dfrac {\partial \phi} {\partial y} } {0, y} \rd y\) $\phi$ is supported on $\openint 0 a^2$
\(\ds \) \(=\) \(\ds \map \phi {0, 0} - \map \phi {0, a}\)
\(\ds \) \(=\) \(\ds \map \phi {0, 0}\) $\phi$ is supported on $\openint 0 a^2$
\(\ds \) \(=\) \(\ds \map {\delta_{\tuple {0, 0} } } \phi\) Definition of Dirac Delta Distribution

$\blacksquare$


Sources

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