# PDE - Two Dimensional Inhomogeneous?? [closed]

I'm looking at a two dimensional, second order, inhomogeneous equation which has no boundary conditions. I realize that there could be zero or infinite solutions to a problem like this, but I can't think of how one would even get started on a general solution for U(x,y)...?

$U_{xx} + U_{yy} = f(x,y)$ for all $x$, $y$; no boundary conditions.

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If you want to get a useful answer, I suggest specifying a bit more about the problem, such as what sort of function $f$ is, what the domain and range are, what sort of solution you are willing to admit ($C^2$ or $C^\infty$ or analytic or what). –  David Roberts Jan 13 '11 at 5:22
It is valid for all x,y. This means that the domain and range are both positive to negative infinity. The forcing function is a delta function with an exponential coefficient. –  thenickname Jan 13 '11 at 17:11
I'm not sure what you mean by a delta function with an exponential coefficient. Is that something like $\delta(x)\exp(g(x))$? –  David Roberts Jan 13 '11 at 22:25
I was going to write $\delta(\exp(g(x)))$, but decided against it, because this function is identically zero. But then, I suppose I'd better check... –  David Roberts Jan 13 '11 at 22:27
thenickname: You say in a comment below that you're not thinking about physical interpretations; however, it's pretty unavoidable here!! Isn't this just the 2D Poisson equation, so $f$ represents charge distribution, and $U$ tells you the electric field generated? (Not that my physical knowledge is up to much). On the other hand, having no boundary conditions is a bit unphysical; I think physicists normally assume that $U$ goes to zero as you go out to infinity and similar things, and this is what will be discussed in most books on this subject, I expect. –  Zen Harper Jan 14 '11 at 10:00
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## closed as not a real question by David Roberts, Andres Caicedo, Denis Serre, Willie Wong, Andrey RekaloJan 14 '11 at 20:36

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I think you should look into the theory of fundamental solutions, or Greens functions. Greens functions are solutions to the equation

$$G_{xx}+G_{yy}=\delta$$

Which are explicitly know in this case (I think it is $\frac{\log(x^2+y^2)}{4\pi}$, but you have to look this up). Solutions of the original equation are then

$$U=G\star f+U'$$

With $\star$ the convolution and $U'$ a solution to the homogeneous Laplace equation $$U_{xx}'+U_{yy}'=0.$$ Of course this requires some regularity of $f$. If $f$ is a compactly supported distribution this has a distributional solution. If one assumes a certain smoothness as well, this will induce smoothness of the solution.

More on this theory can be found in the chapter on fundamental solutions of

MR2680692 Duistermaat, J. J. ; Kolk, J. A. C. Distributions. Theory and applications. Translated from the Dutch by J. P. van Braam Houckgeest. Cornerstones. Birkhäuser Boston, Inc., Boston, MA, 2010. xvi+445 pp. ISBN: 978-0-8176-4672-1

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"...a Green's function is a type of function used to solve inhomogeneous differential equations subject to specific initial conditions or boundary conditions..." My issue is that this question I'm working on is a PURE math question, (meaning that a physical interpretation is not implied), so I am unable to logically set my own boundary conditions. Given the quote above about Green's function, what do you mean when you imply that I can solve this without boundary conditions? –  thenickname Jan 13 '11 at 19:53
"No boundary conditions" is also a sort of boundary condition, as far as Green's functions are concerned. Typically these are the easiest situations to calculate Green's functions for. –  Andrew Homan Jan 13 '11 at 22:31
Doesn't one Green\s function suffice? Suppose you can solve $$U_{xx}+U_{yy}=0$$ with the boundary condition $U|_{\partial\Omega}=g$, for any (nicely behaved) function $g$, on some well behaved domain $\Omega$. then you can also solve the problem $$U_{xx}+U_{yy}=f(x,y)$$ with $U|_{\partial\Omega}=g$. Just write down a solution U=G\star f+V, where $V$ is a solution to $V_{xx}+V_{yy}=0$, subject to the boundary condition $V|_{\partial\Omega}=g-G\star f|_{\partial\Omega}$ –  Thomas Rot Jan 14 '11 at 19:57
Thomas, your answer works only if $G\star f$ exists. –  Deane Yang Jan 14 '11 at 20:11
Ah, you are right. –  Thomas Rot Jan 14 '11 at 20:22