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[Cross posting https://math.stackexchange.com/questions/1374384/fundamental-solution-for-a-parabolic-pde-with-costant-coefficents ]

I don't know if this question is more appropriate in Mathematics and not here, in this case I will delete it.

As it is well known, the fundamental solution of the heat equation is the function

$G(t,x)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{4t}}$,

for all $t>0,x\in\mathbb{R}^n$.

I wonder if exists (and if you have same references) a similar explicit formula for the fundamental solution for a parabolic PDE with constant coefficents. It is possible that it can be found in "Linear and quasilinear equation of parabolic type" by Ladyzenskaja, Solonnikov and Ural'ceva, but as I cant' consulte the book by now, I don't know.

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  • $\begingroup$ Could you write out the precise form of the PDE you want to solve? Is it $u_t = a^{ij}\partial_i\partial_ju$? $\endgroup$
    – Deane Yang
    Commented Jul 26, 2015 at 14:57
  • $\begingroup$ Yes, it is like you wrote. first I'm interested in second order equation, then in higher order $\endgroup$
    – foo90
    Commented Jul 26, 2015 at 21:47

3 Answers 3

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Yes, it can be written out explicitly. Namely, for the operator $$ \partial_t-\sum_{i,j=1}^na_{ij}\partial_{ij}-\sum_{i=1}^nb_i\partial_i-c, $$ where $A=(a_{ij})>0$, $b=(b_1,\ldots,b_n)$, $$ G(t,x)=\frac{1}{(4\pi t)^{n/2}|A|^{1/2}}\exp\left\{-\frac{(A^{-1}(x-bt),x-bt)}{4t}+ct\right\}. $$ It can be obtained either via the Fourier transform wrt $x$ as in the case of the heat equation or making an affine change of variables in $\mathbb R^{n}$, reducing again to the heat equation.

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  • $\begingroup$ And for an higher order parabolic equation there exists a similar formula? $\endgroup$
    – foo90
    Commented Jul 26, 2015 at 21:54
  • $\begingroup$ Thank you very much. Do you have any reference? $\endgroup$
    – foo90
    Commented Jul 27, 2015 at 9:47
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    $\begingroup$ @foo90 Not off top of my head. $\endgroup$
    – Andrew
    Commented Jul 27, 2015 at 14:16
  • $\begingroup$ @foo90 For the equation $u_t+(-1)^m\Delta^{m}u=0$ the fundamental solution is the inverse Fourier transform of $e^{-|y|^{2m}t}$. For $m>1$ it can be expressed through hypergeometric functions. Say, for $n=1$ and $m=2$ Mathematica gives $$ \frac1{\sqrt{2\pi}}\left(\frac{2 \Gamma \left(\frac{5}{4}\right) \, _0F_2\left(;\frac{1}{2},\frac{3}{4};\frac{x ^4}{256 t}\right)}{\sqrt[4]{t}}-\frac{\Gamma \left(\frac{3}{4}\right) x ^2 \, _0F_2\left(;\frac{5}{4},\frac{3}{2};\frac{x ^4}{256 t}\right)}{4 t^{3/4}}\right). $$ $\endgroup$
    – Andrew
    Commented Jul 28, 2015 at 6:33
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For an operator of the form $L=\partial_t-\sum A_{jk}\partial_j\partial_k$, the fundamental solution is computed in Section 3.3 of Volume I of Hormander's treatise (The Analysis of Linear PDOs). I think one may try to extend the formula to include lower order terms by playing with it a little; for instance if $u$ solves $Lu=0$, then $v = e^{-at}u$ solves $Lv+av=0$, and similarly $v=(a_1x_1+\dots+a_nx_n)u$ solves an equation with additional first order terms.

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  • $\begingroup$ Thank you. And for higher orde parabolic pde do you have some reference? $\endgroup$
    – foo90
    Commented Jul 26, 2015 at 21:59
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    $\begingroup$ Sorry, I have no reference for the higher order case. I guess it should be possible to do something along the lines of Hormander's computations $\endgroup$
    – Piero D'Ancona
    Commented Jul 27, 2015 at 20:57
  • $\begingroup$ I have tried your approach. For the zero order term it works, for the first order terms I haven't succeded, can you write something more? $\endgroup$
    – foo90
    Commented Aug 17, 2015 at 16:31
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    $\begingroup$ You are right, I think that part of my answer is not useful (it leads to first order terms but with a special form). However you have the full formula in Andrew's answer, no? from that answer one might guess that the right transformation to use is $v(x,t)=u(x-bt,t)$ $\endgroup$
    – Piero D'Ancona
    Commented Aug 18, 2015 at 21:17
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A good place to start is the series of articles by P. Wagner and N. Ortner. In addition to their constructive proof of the Malgrange-Ehrenpreis theorem on the existence of fundamental solutions for pde's with constant coefficients (AMS 116---MR 2510844), they have, in a series of previous papers, computed explicit solutions for many concrete examples.

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  • $\begingroup$ I could be wrong since I'm relying on distant memories, but I believe that the theorems of Malgrange and Ehrenpreis apply only to PDE's where the top order term is nondegenerate (i.e., PDE's whose real analytics solutions can be obtained via Cauchy-Kovalevski). Without this assumption there might not be any solutions at all. The heat equation has a degenerate top order term, so I'm not sure that the theorems you cite apply. $\endgroup$
    – Deane Yang
    Commented Jul 26, 2015 at 14:57
  • $\begingroup$ The theorem of Malgrange-Ehrenpreis applies to all linear pde's with constant coefficients. I had assumed that the OP was referring to this situaton (implicitly implied by the phrase "with constant coefficients"---I'm not sure what this phrase can mean when applied to non-linear pde's). If not, then this is, of course, a completely new ball game. $\endgroup$
    – priel
    Commented Jul 26, 2015 at 15:13
  • $\begingroup$ No, you're right. I stand corrected. The Malgrange-Ehrenpreis theorem does not make any assumptions about the top order term at all. $\endgroup$
    – Deane Yang
    Commented Jul 26, 2015 at 15:29

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