# Dirac Delta function to solve PDE in the sense of distribution

For each Borel set $A$ and $x\in\mathbb{R}^n$, denote Dirac measure centered at $x$ as \begin{equation} \delta_x(A)=\begin{cases} 1 & \mbox{ if } x\in A \\ 0 & \mbox{ otherwise}. \end{cases} \end{equation} Alternatively, we can think $\delta_x$ in the sense of distribution defined by \begin{equation} <\delta_x,\phi> = \phi(x). \end{equation} I want to show that for $c\in\mathbb{R}^n$, $u(x,t):=\delta_{x-ct}$ solves the PDE \begin{equation} u_t+c\cdot Du = 0 \mbox{ in } \mathbb{R}^{n+1} \end{equation} in the sense of distribution.

I tried \begin{equation} <u_t, \phi> = <D_t\delta_{x-ct}, \phi> = -<\delta_{x-ct}, D_t\phi>=-\phi(x-ct)(-c) \end{equation} I am not sure about the last equality because when $c$ is a vector, $<u_t,\phi>$ is a vector? Should it be a scalar instead?

What should I do? Thank you.

You have to use the chain rule in the last step. Putting in some extra parentheses for emphasis, you have:

$$D_t (\phi(x-ct)) = (\nabla \phi)(x-ct) \cdot (-c)$$

which is a scalar.

• @NateEldredge No, $c$ is a vector (the velocity vector underlying the transport).
– Ian
Apr 25, 2015 at 16:26
• I see. Is it the same thing for $x$-derivative? It looks like I only have $\nabla\phi(x-ct)$
– dh16
Apr 25, 2015 at 16:28
• @dh87 The $x$ derivative is already a vector, and then taking the dot product with $c$ recovers a scalar again. But I think I may have lost track of a minus sign somewhere.
– Ian
Apr 25, 2015 at 16:29
• For the second term, do we have $<c\cdot Du, \phi> = c\cdot <Du, \phi>$?
– dh16
Apr 25, 2015 at 16:31
• @dh87 Yes. Essentially, $u$ is a "scalar-valued" distribution so $Du$ is a "vector-valued" distribution. You may be mixing up the distribution-function pairing notation with the inner product notation.
– Ian
Apr 25, 2015 at 16:36