I am working on a physics problem which leads to an inhomogeneous PDE $$ \left[ \frac{\partial}{\partial t} + \vec{v} \cdot \nabla_\vec{r} \right] f(\vec{r},\vec{p},t) = g(\vec{r},\vec{p},t) $$ which I want to solve using a Green´s function, which leads to $$ \left[ \frac{\partial}{\partial t} + \vec{v} \cdot \nabla_\vec{r} \right] G(\vec{r}-\vec{r}^\prime;t-t^\prime) = \delta(\vec{r}-\vec{r}^\prime) \delta(t-t^\prime)$$ such that $$ f(\vec{r},\vec{p},t) = \int G(\vec{r}-\vec{r}^\prime;t-t^\prime) g(\vec{r},\vec{p},t) \ \mathrm{d}\vec{r}^\prime \mathrm{d}t.$$ Here $\vec{v}$ is a constant and the function $g(\vec{r},\vec{p},t)$ does not need to be specified for my problem (I think). $\vec{r} = (x,y,z)$ is the usual coordinate vector.
My problem is that I don´t know how to solve the PDE for $G(\vec{r}-\vec{r}^\prime;t-t^\prime)$ since I don´t have much experience with PDEs or Green´s functions. At first I thought I could solve it by separation but this didn´t work, so I am stuck right now. Help would be very much appreciated, thanks in advance!
Edit: After Sal´s comment I tried solving it by Fourier transform, which leads to $$ G(\vec{k},\omega) = \frac{-i e^{-i(\vec{k} \cdot \vec{r}^\prime + \omega t^\prime)}}{\omega+\vec{v} \cdot \vec{k}}. $$ The Green´s function can then be obtained by inverse FT $$ G(\vec{r}-\vec{r}^\prime;t-t^\prime) = \frac{1}{(2\pi)^4} \int G(\vec{k},\omega) e^{i \vec{k} \cdot \vec{r}} e^{i \omega t}\ \mathrm{d}\vec{k} \mathrm{d}\omega. $$
Unfortunately this integral diverges. So Fourier transform doen´t lead to a solution. Of couse it could be possible that I have made a mistake somewhere.
Best Answer
This is the 3D inhomogeneous linear transport equation (advection). OK, so let's make things a bit simpler by solving $$ (\partial_t + \vec v\cdot \nabla_{\vec r})\, G = \delta(\vec r)\delta(t) , \qquad \vec r = (x,y,z) $$ as we can always shift the Dirac deltas later on. We take the Fourier transform $\mathcal{F}[\cdot] = \iint (\cdot)\, \text{e}^{-\text i (\omega t - \vec k\cdot \vec r)}\text d \vec r\, \text d t$ of the PDE to get $$ \mathcal{F}[G] = \mathcal{F}_{\vec r} \mathcal{F}_t [G] = \frac{1}{\text{i}(\omega - \vec k\cdot\vec v)} \, . $$ The next step is the computation of the inverse Fourier transform w.r.t. $\omega$. The integrand has a pole at $\omega = \vec k\cdot\vec v$, which can be assumed positive without loss of generality. Here, we might want to use the residue theorem, and integrate $\mathcal{F}[G]\, \text{e}^{\text i \omega t}$ over a closed path $\mathcal{C}$ of the complex plane $\omega \in \Bbb C$, with positive orientation. Here we consider a loop that includes a half-circle of radius $R$ in the upper half-plane, a half-circle of radius $\epsilon$ in the lower half-plane around the pole, and a path along the real line. Thus, \begin{aligned} \oint_\mathcal{C} \mathcal{F}[G]\, \text{e}^{\text i \omega t} &= 2\pi\, \text{e}^{\text i \vec k\cdot\vec v t} \\ &= \int_{0}^{\pi} \frac{\exp(\text i R \text{e}^{\text i\theta} t) \text{i}R \text{e}^{\text i\theta}}{\text{i}(R \text{e}^{\text i\theta} - \vec k\cdot\vec v)} \text d \theta + \int_{-\pi}^0 \text{e}^{\text i (\vec k\cdot\vec v + \epsilon \exp(\text i\theta)) t} \text d \theta \\ &\quad + \left(\int_{-R}^{\vec k\cdot\vec v-\epsilon}+\int_{\vec k\cdot\vec v+\epsilon}^{R}\right) \frac{\text{e}^{\text i \omega t} \text d \omega}{\text{i}(\omega - \vec k\cdot\vec v)} \, . \end{aligned} The first two terms corresponding to the circles have limit $\pi\, \text{e}^{\text i \vec k\cdot\vec v t}$ as $R\to +\infty$, $\epsilon \to 0$ for positive times. The last term gives us $2\pi\, \mathcal{F}_{\vec r}[G]$. Thus, we end up with $\mathcal{F}_{\vec r}[G] = \tfrac12 \text{e}^{\text i \vec k\cdot\vec v t}$, whose inverse Fourier transform w.r.t. $\vec k$ equals $G = \tfrac12 \delta(\vec r - \vec v t)$. We can proceed similarly for negative times, giving us $$G = \tfrac12\, \text{sign}(t)\, \delta(\vec r - \vec v t)\, .$$ See also this post for the scalar case, and other answers.