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Let \begin{equation} P(\xi)=\xi_{1}^2 \xi_{2}^2 + \xi_{3}^2 + i \xi_{4}, \end{equation} and consider the partial differential operator \begin{equation} P(D)=\left( -i \frac{\partial}{\partial x_1},\dots, - i \frac{\partial}{\partial x_4} \right). \end{equation} Hörmander (The Analysis of Linear Partial Differential Operators, Vol. II, Example 10.2.15) states that \begin{equation} E(x) = \begin{cases} (2 \pi)^{-3} \int \exp[i(x_1 \xi_1+x_2 \xi_2 + x_3 \xi_3)) - x_4 (\xi_{1}^2 \xi_{2}^2 + \xi_{3}^2)] d\xi_{1} d \xi_{2} d \xi_{3} & \textit{ if } x_4 > 0, \\ 0 & \textit{if } x_4 \leq 0 \end{cases} \end{equation} is the unique temperate fundamental solution of the operator $P(D)$.

There are two big flaws for me in this sentence.

(i) Hörmander's statement about the uniqueness is for sure false, since $P$ has real zeros. So for example, if $T$ is a temperate fundamental solution, e.g. $T + c_0 + c_1 x_1$ is too, for any $c_0, c_1 \in \mathbb{C}$.

(ii) The most relevant fact is that $E(x)$ is not even well-defined for $x_4 > 0$. Indeed, we have \begin{equation} \int_{\mathbb{R}^3} \exp \left( - x_4 \left(\xi_{1}^{2} \xi_{2}^{2} + \xi_{3}^{2} \right) \right) d \xi_\ d \xi_2 d \xi_3 = \int_{\mathbb{R}} \frac{\pi}{x_4 |\xi_1|} d \xi_1 = \infty. \end{equation}

It seems very strange to me that such a great mathematician like Hörmander made two big mistakes like these in the same sentence. What do you think about?

Thank you very much for your attention.

NOTE. Hörmander uses this example of partial differential operator to show that there are partial differential operators which do not admit a "regular" temperate fundamental solution (for the definition of "regular solution", see Hörmander, The Analysis of Linear Partial Differential Operators, Vol. II, Definition 10.2.2). Actually, his proof that the differential operator $P(D)$ defined above does not admit a regular temperate fundamental solution is correct and has nothing to do with the statement above.

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Let for simplicity $x=(x_1,x_2,x_3)$, similarly for $\xi$.

Let also $Q(\xi) = \xi_1^2\xi_2^2+\xi_3^2$ and define accodringly $Q(D)$.

Consider now the equation $$P(D)u(x,x_4) = Q(D)u(x,x_4) + \partial_{x_4}u(x,x_4) = 0.$$ Note that $Q(D)$ does not act on $x_4$.

Fourier transfrom with respect to $x$ will give us $$Q(\xi)\hat u(\xi,x_4) +\partial_{x_4}\hat u(\xi,x_4) = 0,$$ with the obvious solution

$$\hat u(\xi,x_4) = C\exp( - x_4 Q(\xi)).$$ for positive $x_4$ this function is bounded, hence temperate.

If you multiply it by $H(x_4)$ (the Heavyside function) and put $C=1$, you obtain a globally bounded function (hence temperate on $\Bbb R^4$) that solves the equation $$ Q(\xi)\hat u(\xi,x_4) +\partial_{x_4}\hat u(\xi,x_4) = \delta_{x_4}. $$ The inverse Fourier transform of the right hand side would give you $\delta_0$, and the inverse FT of $H(x_4) \exp( - x_4 Q(\xi))$ gives the expression for $E(x)$ that you provided. As the inverse FT is map from $S'\to S'$, we conclude that $E\in S'$, even though the provided integral might not not converge in the usual sense.

For the unicity part, maybe he meant "vanishing on infinity"?

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  • $\begingroup$ @Zakreveskiy Thank you very very much fro your help. I would have been lost without it! $\endgroup$ Commented Feb 1, 2017 at 10:21
  • $\begingroup$ @MaurizioBarbato you are welcome=) $\endgroup$ Commented Feb 1, 2017 at 10:27
  • $\begingroup$ @Zakreveskiy There are two places where you wrote $x^4$ instead of $x_4$, maybe you can correct this wording error for future readers, because the system does not allow me to do it. $\endgroup$ Commented Feb 1, 2017 at 10:37
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This is not a separate answer, but only a long comment to Zakreveskiy's answer.

For sure Zakreveskiy's answer shows how the solution $E$ can be found "formally", but it is cleary not a proof of the fact that $E$ is a temperate fundamental solution of $P(D)$, since we cannot take Fourier transform of distributions "only with respect to some variables". But now I understood what Hörmander meant: the rigorous definition of $E$ is the following: \begin{equation} E(\phi)= \int_{\mathbb{R}^4} H(x_4) \exp(-x_4 Q(\xi)) \left( \int_{\mathbb{R}^3} \exp(i \xi \cdot x) \phi(x,x_4) dx \right) d \xi d x_4 \quad (\phi \in \mathcal{S}(\mathbb{R}^4)). \end{equation}

By using simple inequalities (see e.g. the proof of Theorem (10.2) in Blanchard and Brüning, Mathematical Methods in Physics) we see that the map \begin{equation} \phi(x,x_4) \in \mathcal{S}(\mathbb{R}^4) \mapsto \psi(\xi,x_4)=\int_{\mathbb{R}^3} \exp(i \xi \cdot x) \phi(x,x_4) d x \end{equation} is a linear continuous map from $\mathcal{S}(\mathbb{R}^4)$ into $\mathcal{S}(\mathbb{R}^4)$. So the map $E$ defined above is actually a tempered distribution. By using the definition of $E$ and the properties of Fourier transforms on $\mathcal{S}(\mathbb{R}^3)$ it is a plain exercise to give a rigorous proof that $P(D)E=\delta_{0}$. Indeed we have for each $\phi \in \mathcal{S}(\mathbb{R}^4)$ (note that all the monomials of $Q$ have even degree):

\begin{multline} (P(D)E)(\phi)= E \left[Q(D)\phi - \frac{\partial \phi}{\partial x_4} \right] = \\ = \int_{\mathbb{R}^4} H(x_4)\exp(-x_4 Q(\xi)) \left( \int_{\mathbb{R}^3} \exp(i \xi \cdot x) \left[(Q(D)\phi)(x,x_4) - \frac{\partial \phi}{\partial x_4} (x,x_4) \right] dx \right) d \xi d x_4 = \\ = \int_{\mathbb{R}^4} H(x_4)\exp(-x_4 Q(\xi)) \left[ Q(\xi) \psi(\xi,x_4)- \frac{\partial \psi}{\partial x_4}(\xi,x_4) \right] d \xi d x_4 = \\ = \int_{\mathbb{R}^3} \left( \int_{0}^{\infty} \exp(-x_4 Q(\xi)) \left[ Q(\xi) \psi(\xi,x_4)- \frac{\partial \psi}{\partial x_4}(\xi,x_4) \right] d x_4 \right)d \xi = \\ = \int_{\mathbb{R}^3} \left( \int_{0}^{\infty} \left[ - \frac{\partial (\exp(-x_4 Q(\xi))}{\partial x_4} \psi(\xi,x_4) - \exp(-x_4 Q(\xi)) \frac{\partial \psi}{\partial x_4}(\xi,x_4) \right] d x_4 \right) d \xi = \\ = \int_{\mathbb{R}^3} \left( \int_{0}^{\infty} - \frac{\partial [\exp(-x_4 Q(\xi))\psi]}{\partial x_4} d x_4 \right) d \xi = \\ = \int_{\mathbb{R}^3} \psi(\xi,0) d \xi = \int_{\mathbb{R}^3} e^{-i 0 \cdot \xi} \psi(\xi,0) d \xi = \phi(0), \end{multline} where the last passage derives from the Inversion Theorem.

As for the "uniqueness" issue, I don't think Hörmander meant something about the behavior at infinity, otherwise he would have explicitly stated what he precisely meant (hi style was incredibly precise). I think this was only a slip of tongue and the reason why I think so is the following. In the notes to Chapter X of The Analysis of Linear Partial Differential Operators, Vol. II, he says that Example 10.2.15 is taken from On Fundamental Solutions.

Actually Enqvist at p.29 gives an example of a partial differential operator $P(D)$ which has no real zeros, so that there is only one temperate fundamental solution of the operator $P(D)$ and he proves that this solution is not "regular". But this is not the example 10.2.15 of Hörmander! It is very likely that Hörmander included Enqvist's example in a first draft of his masterpiece, but then for some reason he changed his mind and decided to give another example, while the sentence "unique tempered fundamental solution" remain unchanged by a simple distraction.

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