The typical derivation of the fundamental solution of Laplace's equation is to look for a radially symmetric solution because the Laplace equation has radial symmetry, and a similar heuristic can be used to derive the fundamental solution of the heat equation. On page $72$ of Terrence Tao's book (http://www.math.ucla.edu/~tao/preprints/chapter.pdf) he writes that one can use symmmetry methods to derive the fundamental solution of the heat equation as well, and I have also seen similar claims elsewhere. But is this more than a heuristic? Sure we can derive the fundamental solutions using these methods, but is there any a priori reasons to believe that the solutions found should be fundamental solutions, or that fundamental solutions should possess all of the symmetries of the equation?
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I cannot understand why this question has been downvoted. – Giuseppe Negro May 31 '14 at 17:17
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@GiuseppeNegro Neither can I, it appears that a few days ago someone went and downvoted a bunch of my questions just to be malicious. – JLA May 31 '14 at 18:26
1 Answers
Are there a priori reasons to believe that the solutions found should be fundamental solutions?
Yes. The a priori reason is that they are the fundamental solutions for the symmetry-reduced equations (usually ODEs). Of course one still needs to check that the solutions are in fact the correct ones, but once the explicit form of the fundamental solution is available the checking part is often routine.
fundamental solutions should possess all of the symmetries of the equation?
Obviously not. The Laplace equation on $\mathbb{R}^n$ is translation symmetric. The fundamental solution $\propto r^{2-n}$ is not. One has to be a bit more careful about what you mean by symmetry.
Suppose $L$ is a linear operator on smooth functions (perhaps with compact support) defined over some domain $\Omega$. And let $\Phi:\Omega\to\Omega$ be a symmetry of $L$, in the sense that for any smooth function $f$, $(Lf)\circ\Phi = L(f\circ\Phi)$.
Further suppose that $L$ is injective, in the sense that $Lf = 0$ on the whole of $\Omega$ implies that $f = 0$.
Suppose you are looking to solve $$ Lf = g $$ and you know that $g\circ \Phi = g$. Then by the previous discussion you know that if $f$ is a solution, then so is $f\circ\Phi$. But we then have $$ L(f - f\circ\Phi) = 0 \implies f = f\circ\Phi $$
The same argument essentially goes through for distributions instead of $C^\infty_c$, and in the case of the fundamental solution you take $g$ to be the $\delta$-distribution. On $\mathbb{R}^n$, $\delta$ is spherically symmetric, but not translation symmetric. Hence the fundamental solution should also be spherically symmetric, and not translation symmetric.
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Thanks for the nice answer. However what if $L$ isn't injective, like the Laplacian? I guess in this case if you demand the function is $0$ at infinity it is injective... – JLA May 14 '14 at 16:20
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If you consider $\triangle: C^\infty_0(\mathbb{R}^n) \to C^\infty_0(\mathbb{R}^n)$ it is injective. If you allow functions that grow near infinity, and so $L$ is not injective, the notion of a "fundamental solution" itself doesn't make sense. – Willie Wong May 15 '14 at 07:25
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@JLA: what about? Are you worried that it grows a bit? What turns out to be important is the action on $C^\infty_c$. How much distribution theory do you know? (Roughly, you have $G_1, G_2$ two fundamental solutions, then you have $L (G_1 - G_2)* f = 0$ for any $f$. Since the convolution of a distribution against a $C^\infty_c$ function is again $C^\infty$, we have that necessarily $(G_1 - G_2) * f = 0$ for any $f$. But this requires $G_1 - G_2 = 0$. ) – Willie Wong May 16 '14 at 15:47
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@WilleWong I know about distributions, I was just commenting since you said if $L$ is not injective the concept of fundamental solutions doesn't make sense, but it seems you can add any harmonic function to the fundamental solution of the Laplace equation in $2$-D and get something with similar properties, and you can even add harmonic polynomials and it will still be a tempered distribution. – JLA May 16 '14 at 16:18
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@JLA: but if you take a harmonic polynomial $P$ and consider the convolution $P* f$ for $f\in C^\infty_0$ you end up generally outside $C^\infty_0$. That's what my previous comment is getting at. – Willie Wong May 16 '14 at 16:47