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I have a problem with an induction step that appears to be bothersome for me to comprehend it.

Problem: Show that $f(x_1, \dots ,x_n)=(x_1^2+ \dots + x_n^2)^{1- n/2}$ for $(x_1^2 + \dots + x_n^2) \neq 0$ solves the Laplace equation $$ (\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u=0$$ for $ n \geq 3$

Base case

Let $n=3$ then $f(x_1,x_2,x_3)=(x_1^2+ x_2^2+x_3^2)^{-1/2}$ since computing the partial differentials is merely an exercise in differentiation, I will post my results only: $$\partial_{x_1}^2= \frac{2x_1^2-x_2^2-x_3^2}{(x_1^2+x_2^2+x_3)^{5/2}}, \partial_{x_2}^2= \frac{2x_2^2-x_1^2-x_3^2}{(x_1^2+x_2^2+x_3)^{5/2}}, \partial_{x_3}^2= \frac{2x_3^2-x_1^2-x_2^2}{(x_1^2+x_2^2+x_3)^{5/2}} $$ which would indeed work for the base case.


Now I struggle with the induction step because my result makes less sense to me, I assume that $$f(x_1, \dots , x_n, )=(x_1^2+ \dots + x_{n}^2 ) ^{1-(n/2)}=:u $$ solves $$ (\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u=0$$ and want to show that $n \leadsto n+1$. So could I try to say that $$(\partial_{x_1}^2 + \dots + \partial_{x_n}^2+\partial_{x+1}^2)u=\underbrace{(\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u}_{=0}+\partial_{n+1}^2u=0 \\ \implies \partial_{n+1}^2 u =0 \implies \partial_{n+1}^2 =0, \text{ because $u \neq 0$} $$ My induction step seems very confusing to me, because it suggest to me that the last partial differential vanishes, but my induction base case suggest otherwise, i.e. that the partial differentials cancel each other out.

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It is advantageous to set $r(x)^2:= x_1^2 + \dots + x_n^2$, then $f(x) = r(x)^{2-n}$, $\partial_{x_1}f = (2-n)r^{1-n} \partial_{x_1}r$. You get shorter expressions for derivatives.

Also there is no need for induction. Just compute the partial derivatives, and sum up.

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  • $\begingroup$ With that substitution would the second partial derivative look like this $ \partial_{x_1}^2 f= (2-n)(1-n)r^{-n} \partial_{x_1} r + \partial_{x_1}^2 r (2-n)r^{1-n}$ ? $\endgroup$ – Spaced Apr 10 '14 at 8:23
  • $\begingroup$ No. You have to differentiate $\partial_{x_i}r$ before calculation the second-order derivative $\endgroup$ – daw Apr 10 '14 at 8:29
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    $\begingroup$ So $\partial_{x_1} r_1 = \frac{x_1}{r}$ such that I have $\partial_{x_1}f= (2-n)r(x)^{-n} x_1$ and now I need the product and chain rule to continue. $\endgroup$ – Spaced Apr 10 '14 at 8:46
  • $\begingroup$ Thanks a lot for your help, took me a while to see this approach, but I can fully appreciate it now after looking through it. $\endgroup$ – Spaced Apr 10 '14 at 8:53
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    $\begingroup$ Just curious: If we were to prove via induction, how would we approach the inductive step? $\endgroup$ – Anant Apr 11 '14 at 11:30

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