Let's consider the following PDE in $\mathbb R^d$ : $$\frac{\partial^d u}{\partial x_1...\partial x_d}=f$$ where $f$ is a tempered distribution with support in $\mathbb R^d_+$. There is a result by Hörmander that states that there exists at least one tempered distribution solution to this PDE (this has to do with the division of tempered distributions by polynomials). I want to know if (because of the particularly simple form of this particular equation) all solutions to this PDE are in fact tempered?
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This is false. For example for $d=2$ you have solutions of the form $F(x,y)+g(x)+h(y)$ where $F$ is the standard tempered solution obtained by integration and $g$ and $h$ are ANY distributions, hence not necessarily tempered.
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$\begingroup$ Thank you! What if I add boundary conditions, for example impose the solution $u$ to have support in $\mathbb R_+^d$? $\endgroup$– ThomasCommented Apr 28, 2014 at 11:15
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$\begingroup$ That doesn't change things. Consider the special case with $f=0$ which is particularly transparent $\endgroup$– janacekCommented Apr 28, 2014 at 11:46
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$\begingroup$ It seems to me that it does. In the case $f=0$, all distributions of the form $u=g\otimes 1+ 1\otimes h $ are solutions where $g$ and $h$ are any distributions. But then the support condition imposes that $g$ and $h$ are in fact $0$, so all these solutions are tempered. $\endgroup$– ThomasCommented Apr 28, 2014 at 12:04
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$\begingroup$ just take $u=H(x)\exp(x)$ and so on. Or are we talking at cross purposes? $\endgroup$– janacekCommented Apr 28, 2014 at 12:48
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$\begingroup$ This particular $u$ does not have support in $\mathbb R_+^2$: if I choose $\varphi, \psi \in \mathcal D(\mathbb R)$ such that $\text{supp} \, \varphi \subset \mathbb R_+$ and $\text{supp} \, \psi \subset \mathbb R_-$ then $\text{supp} \, \varphi \otimes \psi \cap \mathbb R_+^2=\emptyset$ and $\langle u , \varphi \otimes \psi \rangle \neq 0$. $\endgroup$– ThomasCommented Apr 28, 2014 at 13:31