Solving the "Transport" PDE in the sense of distributions with Dirac Delta Source Let $\delta_0$ be the standard Dirac Delta distribution. I wish to solve the PDE $$u_t+cu_x=\delta_0$$ in the sense of distributions with initial condition $u(x,0)=g(x)$ for some continuous $g$. That is, I wish to find $u(x,t)$ such that
$$-\iint_\mathbb{R} u(x,t)(\phi_t+c\phi_x)dA=\phi(0,0)$$
where $\phi$ is any so-called test function.
Can anyone point in me the right direction? I tried to take a Fourier transform but that didn't seem to do much.
Edit:
To respond to a comment, taking the Fourier transform yields:
$$\mathcal{F}(u)_t+cik\mathcal{F}(u)=1$$
This is equivalent to the ODE $$f'(t)+cikf(t)=1$$
This ODE is solved by
$$\mathcal{F}(u)=f(t)=C e^{-(i kc t)} - i/(kc)$$
I'm unsure of where to go from here, or if this is correct.
 A: The r.h.s. of the partially Fourier-transformed equation in OP is incorrect. Indeed, spatial Fourier transformation of the 2D Dirac $\delta_0 =\delta(x)\delta(t)$ gives $\delta(t)$, not $1$. Moreover, the weak form in OP is incorrect too. Integrating by parts, we have
\begin{aligned}
0 &= \iint_{\Bbb R\times\Bbb R_+} (u_t + cu_x-\delta_0)\phi\,\text d x\,\text d t \\
&= -\int_{\Bbb R} g\phi|_{t=0}\, \text d x - \iint_{\Bbb R\times\Bbb R_+} u(\phi_t + c\phi_x)\,\text d x\,\text d t - \phi(0,0)
\end{aligned}
for any test function $\phi$.
The present problem amounts to the computation of the Green's function for the non-homogeneous advection equation $u_t+cu_x=f$. Fourier transformation in space and time of the PDE yields
$$
\text i(\omega-kc)\, \mathcal{F}_t\mathcal{F}_x u = 1
$$
where $\mathcal{F}_t = \int\text dt\, e^{-\text i\omega t}$ and $\mathcal{F}_x = \int\text dx\, e^{\text ik x}$. Thus, the solution is represented as
\begin{aligned}
u(x,t) &= \frac{1}{(2\pi)^2}\iint \frac{e^{\text i(\omega t-kx)}}{\text i (\omega-kc)}\text dk\,\text d\omega \\
&= \frac{\text{sgn}(t)}2 \left( \frac{1}{2\pi} \int e^{-\text i k(x-ct)}\text dk \right) \\
&= \tfrac12 \text{sgn}(t)\, \delta(x-ct)
\end{aligned}
where the residue theorem was used (singularity at $\omega=kc$ -- see this post). Using the superposition principle, the solution to the initial problem may be expressed as
$$
u(x,t) = g(x-ct)+\tfrac12 \text{sgn}(t) \, \delta(x-ct) \, .
$$
As pointed out in the comments, an alternative consists in using Duhamel's principle, cf. this article.
A: OP's first-order initial value problem (IVP) is
$$ \frac{\partial u(x,t)}{\partial t}+ c\frac{\partial u(x,t)}{\partial x}~=~\delta(t)\delta(x), \qquad u(x,t\!=\!0)~=~g(x).\tag{1}$$
One idea is to transform the IVP (1) into the form
$$ \frac{\partial v(x^{\prime},t^{\prime})}{\partial t^{\prime}}~=~\delta(t^{\prime})\delta(x^{\prime}), \qquad v(x^{\prime},t^{\prime}\!=\!0)~=~g(x^{\prime}),\tag{2}$$
by make a suitable linear coordinate transformation $(x,t)\mapsto (x^{\prime},t^{\prime})$. A bit of thought using the chain rule reveals that the coordinate transformation
$$ x~=~x^{\prime}+ct^{\prime}, \qquad t~=~t^{\prime}, \tag{3}$$
will do the job.
The unique solution to the IVP (2) is evidently
$$ v(x^{\prime},t^{\prime})~=~\frac{1}{2}{\rm sgn}(t^{\prime})\delta(x^{\prime})+ g(x^{\prime}). \tag{4}$$ 
Hence the unique solution to the original IVP (1) is
$$ u(x,t)~=~\frac{1}{2}{\rm sgn}(t)\delta(x\!-\!ct)+ g(x\!-\!ct). \tag{5}$$ 
