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Question: Consider $\Phi (x,h)= (1-2xh+h^2)^{-1/2}$. Show that $$(1-x^2){\partial^2 \Phi\over\partial x^2} -2x{\partial\Phi\over\partial x} +h{\partial^2\over\partial h^2} (h\Phi)=0$$

This is what I’m able to come up with (again, now third time): \begin{align} \Phi_x&=h\Phi^3\\ \Phi_{xx}&=3h^2\Phi^5\\ \ (h \Phi)_{hh}&=\Phi^3(2x-3h)+3h\Phi^5(x-h)^2 \end{align}

But this makes the LHS of the equation in the question to evaluate to $3h^2\Phi^3(\Phi^4-1)$ which is clearly not zero.

What went wrong?

Got that: I was writing wrongly that $\Phi^2(1-2xh+h^2)$ was equal to $\Phi^4-1$. It is $0$, in fact.

Atom
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  • None of the expressions involve a power of $\Phi$ greater than $5$, but you say it is equal to a polymial of degree $7$ in $\Phi$? – Paul Sinclair Nov 15 '19 at 01:18
  • I get the same derivatives, but when I plug them into the equation, I do not get the same simplification you do. The expression I get in place of $\Phi^4 - 1$ does indeed reduce to $0$. – Paul Sinclair Nov 15 '19 at 01:30
  • @PaulSinclair Yep, I also am now getting zero. Thanks! – Atom Nov 15 '19 at 03:08

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