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Let $M := C_0^{\infty}(\mathbb{R}^n)$ denote the smooth maps with compact support. Then we have a map

$\Delta:M\rightarrow M,\,\, f\mapsto \Delta f$,

where $\Delta f = \sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}f$ is the Laplacian. I am wondering if $\Delta$ is surjective, i.e. if for any $f\in M$ there exists an $F\in M$ with $\Delta F = f$. Is that true?

Thanks for your help!

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    $\begingroup$ Laplacian is the divergence of gradient. By the divergence theorem the integral of Laplacian must be zero. $\endgroup$ – user31373 Sep 13 '12 at 11:14
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    $\begingroup$ @LKV: I don't understand how this helps me. Could you please elaborate? Thanks. $\endgroup$ – Sh4pe Sep 13 '12 at 13:02
  • $\begingroup$ re LVK's comment , start with $\mathbb R^1$ $\endgroup$ – mike Sep 13 '12 at 14:30
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It is very far from being surjective. Note that if $f\in C^\infty_0$ and $u$ is any harmonic function in the entire space, then $\int (\Delta f)u=\int f(\Delta u)=0$ (integration by parts or Green). This imposes infinitely many independent restrictions on the functions that can be represented as Laplacians of smooth compactly supported functions in every dimension above $1$ (in dimension $1$ the only harmonic functions are linear).

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  • $\begingroup$ Could you please specify what exactly you mean by "infinitely many independent restrictions"? How can I construct an $f$ from you comment that cannot be written as $\Delta F$? $\endgroup$ – Sh4pe Sep 13 '12 at 12:56
  • $\begingroup$ Ah - yeah. Got it now. Thank you! $\endgroup$ – Sh4pe Sep 13 '12 at 14:30
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This can not be true. If it would, it would imply that the fundamental solution of laplace equation is unique, which is obviously false. Igor.

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  • $\begingroup$ I came up with this question when I was asking myself wheather the fundamental solution of the laplace equation is unique. So it is not? Why is this obvious? Any hints to literature maybe? Thank you for your answer. $\endgroup$ – Sh4pe Sep 13 '12 at 13:56
  • $\begingroup$ Damn, I'm stupid: if you add to any fundamental solution an harmonic function, you get another one... Well, I'm a beginner in this topic, so please be patient with me ;) $\endgroup$ – Sh4pe Sep 13 '12 at 14:58
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    $\begingroup$ Very good literature: amazon.de/Dostojewski/lm/3T09G5CNUPF7K $\endgroup$ – AlexisZorbas Sep 13 '12 at 14:58
  • $\begingroup$ See also: amazon.de/Numerical-Schemes-Conservation-Advances-Mathematics/… $\endgroup$ – AlexisZorbas Sep 13 '12 at 15:05

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