Weak Solution to Dx u = u

Theorem

Let $H$ be the Heaviside step function.

Let $\map u {x, t} = \map H t e^x$

Then $u$ is a weak solution of the partial differential equation $\ds \dfrac {\partial u} {\partial x} = u$.

That is, for the distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$ in the distributional sense it holds that:

$\ds \dfrac {\partial T_u} {\partial x} = T_u$

Proof

Let $\phi \in \map \DD {\R^2}$ be a test function.

Then in the distributional sense we have that:

 $\ds \paren {\dfrac \partial {\partial x} - 1} \map {T_u} \phi$ $=$ $\ds -\map {T_u} {\dfrac {\partial \phi} {\partial x} } - \map {T_u} \phi$ Definition of Distributional Partial Derivative $\ds$ $=$ $\ds -\iint_{\R^2} \paren {\map H t e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + \map H t e^x \map \phi {x, t} }\rd x \rd t$ Definition of Distribution $\ds$ $=$ $\ds -\int_0^\infty \int_{-\infty}^\infty \paren {e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + e^x \map \phi {x, t} }\rd x \rd t$ Definition of Heaviside Step Function $\ds$ $=$ $\ds -\int_0^\infty \int_{-\infty}^\infty \dfrac \partial {\partial x} \paren {e^x \map \phi {x, t} }\rd x \rd t$ Product Rule for Derivatives $\ds$ $=$ $\ds -\int_0^\infty \bigintlimits {e^x \map \phi {x, t} } {x \mathop = -\infty} {x \mathop = \infty} \rd t$ Fundamental Theorem of Calculus $\ds$ $=$ $\ds -\int_0^\infty 0 \rd t$ Definition of Test Function $\ds$ $=$ $\ds 0$

$\blacksquare$