Finding a fundamental solution of a linear PDE in the half space

Question

Let $\Omega=\mathbb{R}^n\times \mathbb{R}^+$ and $a_0,\ldots, a_n\in \mathbb C$, I must find a fundamental solution of the PDE $$(*)\qquad\qquad\frac{\partial u}{\partial t}(x, t)+\sum_{k=1}^n a_k\frac{\partial u}{\partial x_k}(x, t)+a_0 u(x, t)=f(x, t),\qquad (x, t)\in \Omega.$$ If we call $$L=\frac{\partial }{\partial t} +\sum_{k=1}^n a_k\frac{\partial }{\partial x_k} +a_0,$$ then it is enough to find a fundamental solution for $L$, say $T$ (i.e., $LT=\delta$, and afterwards we just convolute with $f$)

How should I proceed here? I know that every non-zero linear differential operator with constant coefficients has a fundamental solution (and one can even show an explicit formula by means of one of the constructive proofs of the Malgrange-Ehrenpreis theorem), but this is rather long and cheap.

Is there an easier(direct) way to find such solution? Any help or reference is highly appreciated

Answer 1

For the computation of the fourier transform method, some progress:

$LT=\delta$ as you've written, and then we can find $T$ via Fourier transform. I'm going to rename your equation slightly,

$u_t + \textbf{a}\cdot \nabla_x u(x,t) + bu(x,t) = f(x,t)$

I take the convention that(as in Evan's PDE book)

$$\mathcal{F}[u](\xi)=\frac{1}{(2 \pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n}u(x)e^{-ix \cdot \xi}\,dx$$

$$\mathcal{F}^{-1}[u](x)=\frac{1}{(2 \pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n}u(\xi)e^{i\xi \cdot x}\,d\xi$$

Then,

$\mathcal{F}[\partial_t u]=\hat{u}_t$

$\mathcal{F}[a \cdot \nabla u] = i(a \cdot \xi) \hat{u}$

$\mathcal{F}[\delta(x)] = 1$

Hence, we arrive to the equation

$$\hat{T}_t + (i(a \cdot \xi)+b)\hat{T} = 1$$

so the Fourier transform of the fundamental solution satisfies

$$\hat{T}(\xi,t) = C e^{-t(i(a \cdot \xi)+b)} + (i(a \cdot \xi)+b)^{-1}$$

where $C$ is yet to be found. However from here, I've been having some trouble writing out the inverse transform of each of those functions.