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I am trying to read Fritz John's article, here:


And for the proof of Thm 1.1 in page 303 he writes that one can easily verify the equation (7) is satisfied for $i=1,k=2$, is it really that easy, perhaps I am missing some symmetry here?

I got so far to:

$$[\frac{\partial^2}{\partial \xi_1 \partial \eta_2 } - \frac{\partial^2}{\partial \xi_2 \partial \eta_1} ] \frac{v(\xi , \eta)}{\sum_i (\xi_i -\eta_i)^2} = $$

$$= \frac {\(\frac{\partial^2 v}{\partial \xi_1 \partial \eta_2} -\frac{\partial^2 v }{\partial \xi_2 \partial \eta_1})|\xi-\eta| +2|\xi-\eta|^2[(\xi_2-\eta_2)v_{\xi_1}-(\xi_1-\eta_1)v_{\xi_2}]}{|\xi-\eta|^4}$$

Now somehow I need to calculate another several derivatives wrt to eqaution (6) in the paper, is there some easy way to calculate that I am missing, he writes that it should be easy... :-D

Thanks in advance, Alan.

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1 Answer 1

OK, I think I understand why this is so.

If I calculate the expression:

$$\frac{\partial^2}{\partial \xi_1 \partial \eta_2 }(\frac{ v}{|\xi-\eta|})$$

I get an expression that doesn't depend on $\xi_1 \& \eta_2$ explicitly thus by interchanging variable $\xi_1 \mapsto \xi_2$ and $\eta_2 \mapsto \eta_1$ I get the same expression.

For the love of god it is easy... :-D

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I believe I also spotted a misprint, on page 4 after Eq. (8) it should be: $$\frac{v(\theta \xi +(1-\theta)\eta,\eta )}{|\theta \xi +(1-\theta)\eta -\eta|} = \frac{1}{\theta}\frac{v(\xi,\eta)}{|\xi-\eta|}$$ They forgot $\theta$ in the LHS of the denominator. –  Alan Jun 11 '13 at 3:07

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