Weak Solution to Dx u = u

Theorem

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

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:

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

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

Then in the distributional sense we have that:

$\ds \paren {\dfrac \partial {\partial x} - 1} \map {T_u} \phi$

$=$

$\ds -\map {T_u} {\dfrac {\partial \phi} {\partial x} } - \map {T_u} \phi$

$\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 $

$=$

$\ds -\int_0^\infty \int_{-\infty}^\infty \paren {e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + e^x \map \phi {x, t} }\rd x \rd t$

$\ds $

$=$

$\ds -\int_0^\infty \int_{-\infty}^\infty \dfrac \partial {\partial x} \paren {e^x \map \phi {x, t} }\rd x \rd t$

$\ds $

$=$

$\ds -\int_0^\infty \bigintlimits {e^x \map \phi {x, t} } {x \mathop = -\infty} {x \mathop = \infty} \rd t$

$\ds $

$=$

$\ds -\int_0^\infty 0 \rd t$

Definition of Test Function

$\ds $

$=$

$\ds 0$

$\blacksquare$

2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.3$: A glimpse of distribution theory. Weak solutions