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

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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:

\(\ds \dfrac {\partial u}{\partial x}\) \(=\) \(\ds \dfrac {\partial}{\partial x} \paren {e^{-3 y x} \map H y}\)
\(\ds \) \(=\) \(\ds e^{-3 y x} \paren {-3y} \map H y + e^{-3 y x} \dfrac {\partial \map H y}{\partial x}\)

That is:

\(\ds \dfrac {\partial T_u}{\partial x}\) \(=\) \(\ds \dfrac {\partial}{\partial x} T_{e^{-3 y x} \map H y}\)
\(\ds \) \(=\) \(\ds \dfrac {\partial}{\partial x} \paren {e^{-3 y x} T_{\map H y} }\) Multiplication of Distribution induced by Locally Integrable Function by Smooth Function
\(\ds \) \(=\) \(\ds e^{-3 y x} \paren {-3y} T_{\map H y} + e^{-3 y x} \dfrac {\partial}{\partial x} T_{\map H y}\) Product Rule for Distributional Derivatives of Distributions multiplied by Smooth Functions


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

Then:

\(\ds \dfrac {\partial}{\partial x} \map {T_{\map H y} } \phi\) \(=\) \(\ds - \map {T_{\map H y} } {\dfrac {\partial \phi}{\partial x} }\)
\(\ds \) \(=\) \(\ds - \int_{-\infty}^\infty \int_{-\infty}^\infty \map H y \dfrac {\partial \phi}{\partial x} \rd x \rd y\)
\(\ds \) \(=\) \(\ds - \int_0^\infty \int_{-\infty}^\infty \dfrac {\partial \phi}{\partial x} \rd x \rd y\)
\(\ds \) \(=\) \(\ds - \int_0^\infty \paren {\bigintlimits {\map \phi {x, y} } {x \mathop = - \infty} {x \mathop = \infty} } \rd y\)
\(\ds \) \(=\) \(\ds - \int_0^\infty 0 \rd y\) Definition of Test Function
\(\ds \) \(=\) \(\ds 0\)

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:

\(\ds \dfrac {\partial u} {\partial x} + 3 y u\) \(=\) \(\ds e^{-3xy} \paren {-3y} \map H y + 3y e^{-3yx} \map H y\)
\(\ds \) \(=\) \(\ds \mathbf 0\)

Moreover:

\(\ds \map u {0, y}\) \(=\) \(\ds e^{-0} \map H y\)
\(\ds \) \(=\) \(\ds \map H y\)

$\blacksquare$


Sources

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