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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 \}

$$

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No matter how I compute it , only $\mu$ I always can’t got. I want an detail answer ,so thanks.

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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.

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