Weak Solution to Dx u + 3yu = 0 with Heaviside Step Function Boundary Condition

Theorem
Consider the boundary value problem:


 * $\begin{cases}

\dfrac {\partial u} {\partial x} + 3 y u = 0 & : x \in \R_{>0},~ y \in \R \\ & \\ \map u {0, y} = \map H y & : y \in \R \\ \end{cases}$

Then it has a weak solution of the form:


 * $u = e^{-3 y x} \map H y$

Proof
Let $u = e^{-3 y x} \map H y$

We have that:


 * Heaviside Step Function is Locally Integrable
 * Locally Integrable Function defines Distribution
 * Multiplication of Distribution induced by Locally Integrable Function by Smooth Function

Hence, we can define a distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$.

Then in the distributional sense we have that:

That is:

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

Then:

Hence, in the distributional sense we have that:


 * $\dfrac \partial {\partial x} \map H y = \mathbf 0$

where $\mathbf 0$ is the zero distribution.

Therefore:

Moreover: