# Variation under constraint

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.

#### Solutions Collecting From Web of "Variation under constraint"

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.