Uniqueness of tempered fundamental solution (with support in half space) for the linear KdV equation Show that $E(t,x) = \frac{\textbf{1}_{(0,+\infty)}(t)}{t^{1/3}}\textbf{Ai}(\frac{x}{t^{1/3}})$ is the unique tempered fundamental solution with support in $\{ (t,x) \mid t\geq 0 \}$ of the partial differential operator $P(D) = \partial_t + \frac{1}{3}\partial_x^3$.

I am puzzled about an exercise problem from an undergraduate course about distributions and Fourier transfrom.
The exercise concerns about the constant-coefficients 1+1 dimensional differential operator 
$$P(D) = \partial_t + \frac{1}{3}\partial_x^3.$$ 
The first part of the exercise is to check that 
$$E(t,x) = \frac{\textbf{1}_{(0,+\infty)}(t)}{t^{1/3}}\textbf{Ai}(\frac{x}{t^{1/3}})$$
is a fundamental solution of $P(D)$, where $\textbf{Ai}(x) = \mathcal{F}^{-1}_{\xi \rightarrow x}(e^{i\xi^3/3})$ is the Airy function (it turns out that $\textbf{Ai}$ is $C^\infty$). This part is straightforward and I could work out.
But the second part of the exercise asks to show that $u(t,x) = E(t,x)$ is the unique solution in the sense that
$$  u \in \mathcal{S}'(\mathbb{R}^2_{t,x}) \\ (\partial_t + \frac{1}{3}\partial_x^3)u = \delta_{(0,0)}  \\  \textbf{supp}~u \subset \mathbb{R}_{t \geq 0} \times \mathbb{R}_x$$
I could not work out. I have consulted Hormander's book 
$$\textit{The Analysis of Linear Partial Differential Operators II}$$
 and found some topics very close to it, mostly in Chap. XII "The Cauchy and Mixed Problems", but it seems that there is no theorem I could directly quote.
$\textbf{My Question}$


*

*Is there any general theorem about the uniqueness of tempered fundamental solutions of linear differential operators with constant coefficients? Is there any necessary or sufficient conditions such that $P(D)u = \delta$ has a unique tempered solution $u$, maybe with some requirement of the support $\textbf{supp}~u$ (for instance $\textbf{supp}~u \subset \text{half space}$ like above.)

*Does the requirement $\textbf{supp}~u \subset \mathbb{R}_{t \geq 0} \times \mathbb{R}_x$  take the most important role so that actually this exercise is very easy?
 A: It suffices to show that if a tempered distribution $u$ supported on $\{t\ge0\}$ satisfies the homogeneous equation $(\partial_t+\frac13\partial_x^3)u=0$, then $u=0$. This can be shown using a duality argument.
More precisely, pick $\phi\in\mathcal S_{t,x}$ and it suffices to show $u(\phi)=0$. By Duhamel's formula,
$$ \hat\Phi(t,\xi)=\int_t^\infty -e^{i(t-s)\xi^3}\hat\phi(s,\xi)ds $$
satisfies $(\partial_t+\frac13\partial_x^3)\Phi=\phi$, $\Phi(t,\cdot)\to0$ in $\mathcal S_x$ as $t\to+\infty$, and $\Phi\in C_t^\infty(\mathcal S_x)$ (this is tedious (but routine) to check).
Now let $\rho\in C^\infty$ be such that $\rho=1$ on $(-1,+\infty)$ and $\rho=0$ on $(-\infty,-2)$. Then $\rho(t)\Phi\in S_{t,x}$ and
$$ u((\partial_t+\frac13\partial_x^3)(\rho(t)\Phi))=-((\partial_t+\frac13\partial_x^3)u)(\rho(t)\Phi)=0. $$
Since $\rho=1$ on $(-1,+\infty)$, we have
$$(\partial_t+\frac13\partial_x^3)(\rho(t)\Phi)=(\partial_t+\frac13\partial_x^3)\Phi=\phi\text{ on }\{t\ge0\}.$$
Since $u$ is supported on $\{t\ge0\}$,
$$ 0=u((\partial_t+\frac13\partial_x^3)(\rho(t)\Phi))=u(\phi). $$
PS: We leave it to the reader (and the OP) to think about its generalizations.
PPS: Similar duality arguments are used in the proof of Holmgren's uniqueness theorem, see Section 3.5 of Fritz John, Partial Differential Equations.
