I always can't compute right.$u=u(x),R=R(x)$ and $\tau$ is constant, and $M$ is compact manifold.If $u$ is the minimizer of
$$
\inf\{\int_M [\tau(4|\nabla u|^2+Ru^2)-u^2\ln u^2-nu^2](4\pi\tau)^{-n/2dV}
~~|~~\int_M u^2(4\pi\tau)^{-n/2}dV =1 \}
$$
,how to compute the Euler-Lagrange of
$$
\int_M [\tau(4|\nabla u|^2+Ru^2)-u^2\ln u^2-nu^2](4\pi\tau)^{-n/2dV}
$$
under the constraint
$$~~\int_M u^2(4\pi\tau)^{-n/2}dV =1$$
In my book ,the answer is $$ \tau(-4\Delta u+Ru)-2u\ln u-nu=\mu u $$
$\mu$ is defined as below. $$ \mu=\inf\{\int_M [\tau(4|\nabla u|^2+Ru^2)-u^2\ln u^2-nu^2](4\pi\tau)^{-n/2dV} ~~|~~\int_M u^2(4\pi\tau)^{-n/2}dV =1 \} $$
No matter how I compute it , only $\mu$ I always can't got. I want an detail answer ,so thanks.
That's Lagrange multiplier. Whenever you want to maximize $F$ subject to $G = 1$, then at the minimum $u$ there is $\mu$ so that $\nabla F = \mu \nabla G$. In your answer $$ \tag{1} \tau(-4\Delta u+Ru)-2u\ln u-nu=\mu u, $$ the left hand side is $\frac 12 \nabla F$, where $$F(u) = \int_M [\tau(4|\nabla u|^2+Ru^2)-u^2\ln u^2-nu^2]dV$$ and the right hand side is $\frac 12 \nabla G$, where $$G(u) = \int_M u^2dV. $$ Since you have assumed that $u$ is the minimum, by multiplying $(1)$ by $(4\pi\tau)^{-n/2}u$ and then integrate over $M$ (and use integration by part), you have $$\begin{split} \mu&= \int_M\mu u^2 (4\pi\tau)^{-n/2}dV \\ &= \int_M [\tau(4|\nabla u|^2+Ru^2)-u^2\ln u^2-nu^2](4\pi\tau)^{-n/2dV} \end{split}$$ which is the same as the minimum you want.