Yamabe's equation

Question

This is PDE Evans, 2nd edition: Chapter 9, Exericse 8(a).

8. (a) Assume $n \ge 3$. Find a constant $c$ such that $$u(x) := (1+|x|^2)^{\frac{2-n}2}$$ solves Yamabe's equation $$-\Delta u = cu^{\frac{n+2}{n-2}} \quad \text{in }\mathbb{R}^n.$$ Note the appearance of the critical exponent $\frac{n+2}{n-2}$.

My work so far:

Given $u(x):= (1+|x|^2)^{\frac{2-n}2}$, we find that $$u^{\frac{n+2}{n-2}}=(1+|x|^2)^{-\frac{n+2}2}.$$

Also, I differentiate $u(x)$ to find $$u_{x_i}=(2-n)(1+|x|^2)^{-\frac n2} x_i.$$ I differentiate again to find \begin{align} u_{x_i x_i} &= (2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} x_i^2 + (1+x^2)^{-\frac n2}\right] \\ &= (2-n)[-nx_i^2 +(1+x^2)](1+|x|^2)^{-\frac{n+2}2}. \end{align}

Therefore, the negative Laplacian is \begin{align} -\Delta u = -\sum_{i=1}^n u_{x_i x_i} &= (n-2)[-n|x|^2 +n(1+x^2)](1+|x|^2)^{-\frac{n+2}2} \\ &= (n-2)n(1+|x|^2)^{-\frac{n+2}2} \end{align}

Putting these altogether, Yamabe's PDE $-\Delta u=cu^{\frac{n+2}{n-2}}$ becomes $$(n-2)n(1+|x|^2)^{-\frac{n+2}2}=c(1+|x|^2)^{-\frac{n+2}2}.$$

This would mean that $$c=n(n-2).$$

I had fixed my work acordingly (see my revision history), so this is actually the answer. I do have a follow-up question here, though.

Answer 1

You should have $$u_{x_i x_i} = (2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} x_i^2 + (1+|x|^2)^{-\frac n2}\right].$$ That is, not $x_i$, but $x_i^2$ in the squred bracket. Therefore, we have $$-\Delta u=-\sum_{i=1}^nu_{x_i x_i} = -(2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} |x|^2 + n(1+|x|^2)^{-\frac n2}\right]=n(n-2)(1+|x|^2)^{-\frac{n+2}2}$$ where the second equality follows from $|x|^2=\sum_{i=1}^nx_i^2$. As you have calculated, $u^{\frac{n+2}{n-2}}=(1+|x|^2)^{-\frac{n+2}2}$. We have $$-\Delta u=n(n-2)u^{\frac{n+2}{n-2}}.$$

Answer 2

I think While calculating $u_{x_ix_i}$ we have missed the summation for $(1+x^2)$.If we take that into account, the constant would be $n(n-2)$.