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This is PDE Evans, 2nd edition: Chapter 9, Exericse 8(b). This is a continuation of my previous question with part (a). I am printing part (a) below for convenience, but now I need help with only part (b).

  1. (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 $\frac{n+2}{n-2}$.

(b) Check that for each $\lambda > 0$, $$u_{\lambda}(x):= \left(\frac{\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-2}2}$$ is also a solution.

Completing part (a) shows that $c=n(n-2)$. So Yamabe's equation is $$-\Delta u = n(n-2)u^{-\frac{n+2}{n-2}}.$$

Now, I got lost again on the derivation for the expression of $-\Delta u$, and I can't seem to find my first mistake. I am typing my detailed work here, in hopes to fix my derivation.

Given $u_\lambda(x):= (\frac{\lambda}{\lambda^2+|x|^2} )^{\frac{n-2}2}$, we find \begin{align} (u_\lambda)_{x_i}&= \frac{n-2}2 \left(\frac{\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-4}2}\left(\frac{\lambda}{\lambda^2+|x|^2} \right)_{x_i} \\ &= \frac{n-2}2 \left(\frac{\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-4}2}\left(\frac{-\lambda}{(\lambda^2+|x|^2)^2}\right)(\lambda^2+|x|^2)_{x_i} \\ &= \frac{2-n}2 \left(\frac{\lambda}{\lambda^2+|x|^2} \right)^{\frac{n-2}2}\left(\frac{1}{\lambda^2+|x|^2}\right)2|x|x_i \\ &= \frac{2-n}{\lambda^2+|x|^2}u_\lambda|x|x_i. \end{align} Thus, \begin{align} (u_{\lambda})_{x_ix_i}&=\left(\frac{2-n}{\lambda^2+|x|^2}u_\lambda|x|x_i \right)_{x_i} \\ &= \left(\frac{2-n}{\lambda^2+|x|^2} u_{\lambda} \right)_{x_i} |x|x_i + \frac{2-n}{\lambda^2+|x|^2}u_\lambda(|x|x_i)_{x_i} \\ &= \left[\left(\frac{2-n}{\lambda^2+|x|^2} \right)_{x_i}u_\lambda+\frac{2-n}{\lambda^2+|x|^2} (u_\lambda)_{x_i} \right] |x|x_i + \frac{2-n}{\lambda^2+|x|^2} u_\lambda (|x|_{x_i} x_i + |x| (x_i)_{x_i}) \\ &= \left[\frac{2(n-2)}{(\lambda^2+|x|^2)^2} |x|_{x_i} u_\lambda + \frac{(2-n)^2}{(\lambda^2+|x|^2)^2} u_\lambda |x|x_i \right] |x|x_i + \frac{2-n}{\lambda^2+|x|^2} u_\lambda \left(\frac{x_i^2}{|x|} + |x|\right) \\ &= \frac{u_\lambda}{(\lambda^2+|x|^2)^2} \left[2(n-2)\frac{x_i}{|x|}+(2-n)^2|x|x_i \right]|x|x_i + \frac{2-n}{\lambda^2+|x|^2}u_\lambda \left(\frac{x_i^2}{|x|} + |x| \right) \\ &= \frac{u_\lambda}{(\lambda^2+|x|^2)^2} 2(n-2)x_i^2+(2-n)^2|x|^2x_i^2 + \frac{2-n}{\lambda^2+|x|^2}u_\lambda \left(\frac{x_i^2}{|x|} + |x| \right). \end{align} Finally, \begin{align} -\Delta u_\lambda =-\sum_{i=1}^n u_{x_i x_i} = -\left[\frac{u_\lambda}{(\lambda^2+|x|^2)^2} 2(n-2)|x|^2+(2-n)^2|x|^4 + \frac{2-n}{\lambda^2+|x|^2}u_\lambda \left(2|x| \right) \right]. \end{align} Now, just like in my previous question (see revision history of that question), I probably made the expression for $-\Delta u_\lambda$ more complicated than it needed to be. How can I equate $-\Delta u_\lambda$ (LHS of equation) to $n(n-2)u_\lambda^{\frac{n+2}{n-2}}$ (RHS of equation)? The expression for the RHS is rather simple, which is $$-\Delta u_\lambda = n(n-2)u_\lambda^{\frac{n+2}{n-2}}=n(n-2)\left(\frac{\lambda}{\lambda^2+|x|^2} \right)^{\frac 12(n+2)}$$

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1 Answers1

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One can use (a):

$$ u_\lambda(x) = \left( \frac{\lambda}{\lambda^2 + |x|^2} \right)^\frac{n-2}{2} = \left( \frac{1}{\lambda (1 + |\frac{x}{\lambda}|^2)} \right)^\frac{n-2}{2} = \left( \frac{1}{\lambda} \right)^\frac{n-2}{2} u(x/\lambda) $$

implies

$$ \Delta u_\lambda(x) = \left( \frac{1}{\lambda} \right)^\frac{n-2}{2} \left( \frac{1}{\lambda} \right)^2 c u(x/\lambda)^\frac{n+2}{n-2} = \left( \frac{1}{\lambda} \right)^\frac{n+2}{2} c u(x/\lambda)^\frac{n+2}{n-2}, $$

wherein

$$ u(x/\lambda)^\frac{n+2}{n-2} = \left( \left( \frac{\lambda^2}{\lambda^2(1 + |\frac{x}{\lambda}|^2)} \right)^\frac{n-2}{2} \right)^\frac{n+2}{n-2} = \lambda^\frac{n+2}{2} \left( \frac{\lambda}{\lambda^2 + |x|^2} \right)^\frac{n+2}{2}. $$

Combining this gives

$$ \Delta u_\lambda(x) = c \left( \frac{\lambda}{\lambda^2 + |x|^2} \right)^\frac{n+2}{2}. $$

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