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So I'm reading a book about PDE and I have following question: For Poisson's equation in $\mathbb{R}^2: -\bigtriangleup u=f$

The solution is $u(x)=-\frac{1}{2\pi}\int_{\mathbb{R}^2} log(|x-y|)f(y)dy$

At the beginning of the proof it sais: Following calculation would be wrong:

$\bigtriangleup_xu(x)=-\frac{1}{2\pi}\int_{\mathbb{R}^2}\bigtriangleup_x log(|x-y|)f(y)dy=0$

There is no further explanation why this is wrong, so I tried to find out:

$\frac{d^2}{dx_i^2}log(|x-y|)=\frac{1}{(x_1-y_1)^2+(x_2-y_2)^2}-\frac{2(x_i-y_i)^2}{((x_1-y_1)^2+(x_2-y_2)^2)^2}$

So $\bigtriangleup_x log(|x-y|)=0$. This means the first equation must be wrong and it has to be false to simply pull the laplacian into the integral.

I tried to find the reason on the Internet, but the only comment on this I found was "The second derivative cannot be integrated." But no explanation why.

Can anyone explain to me why this cannot be integrated, please? For me this looks like a "normal" function.

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  • $\begingroup$ The main problem with that calculation is that the function $|x-y|$ is not differentiable when $x=y$. The Laplacian basically becomes a Dirac delta which leads to the correct result. See e.g. math.stackexchange.com/questions/1932451/… $\endgroup$ – Winther Nov 5 '18 at 17:25
  • $\begingroup$ Ok, I now want to understand that proof for $\bigtriangleup ln|x|=2\pi\delta(|x|)$ I already don't understand the first statement: Let $\phi \in C^\infty_0(\mathbb{R}^2)$. You need to show \begin{align} \int_{\mathbb{R}^2} \log|x| \Delta \varphi(x)\ dx = 2\pi \varphi(0). \end{align} Why is this equivalent to the equation I want to proof? And then why is $\phi(0)=\delta(\phi)$ is this proof? Also the reason it cannot be integrated is because $\delta(0)=\infty$ and the integral would diverge there? $\endgroup$ – Kekks Nov 5 '18 at 19:00
  • $\begingroup$ To understand it properly you need to know a bit about distributions. They are generalised functions that is meant to act under an integral sign; the Dirac delta is the prime example (and its not a real function so talking about delta at zero is meaningless). The equation you quote is basically the definition of the delta function: it pulls out the value of the function it's integrated along with at $x=0$. $\endgroup$ – Winther Nov 5 '18 at 19:36

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