Approriate scalling in Yamabe equation

Suppose {(M,g)} is a Riemannian manifold, and {L_g=\Delta_g -\frac{n-2}{4(n-1)}R_g} is the conformal Laplacian. Assume {u>0} satisfies

\displaystyle L_gu+Ku^p=0

where {K} is some fixed constant, {1<p\leq \frac{n+2}{n-2}}. Suppose near a point {x_0\in M}, there is a coordinates {x^1,x^2,\cdots, x^n}. We want to scale the coordinates to {x^i=\lambda y^i},

\displaystyle g(x)=g_{ij}(x)dx^idx^j=\lambda^2 g_{ij}(\lambda y)dy^idy^j=\lambda^2 \hat{g}(y)

By the conformal invariance of {L}, for any {\phi}, we get

\displaystyle L_{g}(\lambda^{-\frac{n-2}{2}}\phi)=\lambda^{-\frac{n+2}{2}}L_{\hat{g}}(\phi)

We want to choose {\phi(y)=\lambda^{\alpha}u(\lambda y)} such that

\displaystyle L_{\hat{g}}(\phi)+K\phi^p=0

which means

\displaystyle L_{\hat{g}}(\lambda^{\alpha}u(\lambda y))=\lambda^{\frac{n+2}{2}}L_g (\lambda^{\alpha-\frac{n-2}{2}}u(\lambda y))=-K\lambda^{\alpha+2} (u(\lambda y))^p

Letting

\displaystyle \alpha+2=\alpha p

we get {\alpha=\frac{2}{p-1}}.

The above proof may not be right.

Or we should look it more directly

\displaystyle L_g(u(x))=\lambda^2 L_{\hat{g}}(u(\lambda y))=\lambda^2 Ku(\lambda y)^p

then

\displaystyle L_{\hat{g}}(\lambda^\alpha u(\lambda y))=K(\lambda^\alpha u(\lambda y))^p

with {\alpha=\frac{2}{p-1}}.

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