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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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closed as not a real question by David Roberts, Andrés E. Caicedo, Denis Serre, Willie Wong, Andrey Rekalo Jan 14 '11 at 20:36

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

\begin{equation} G_{xx}+G_{yy}=\delta \end{equation}

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

\begin{equation} U=G\star f+U' \end{equation}

With $\star$ the convolution and $U'$ a solution to the homogeneous Laplace equation \begin{equation} U_{xx}'+U_{yy}'=0. \end{equation} 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 \begin{equation} U_{xx}+U_{yy}=0 \end{equation} 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 \begin{equation} U_{xx}+U_{yy}=f(x,y) \end{equation} 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

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