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Suppose $(M,g)$ is a Riemannian manifold. Let $E(u)=\displaystyle\int_MuP(u) dV_g$, where $P$ is a linear self-adjoint operator, and $dV_g$ is the volume form with respect to $g$. Given a smooth function $k$ which takes both positive and negative values, let $$F=\{u: \int_Mu dV_g=0\mbox{ and } \int_M ke^{4u}dV_g=0\}.$$

Suppose $u\in F$ is a minimizer: $E(u)=\min_{v\in F}E(v)$. The paper I am reading claimed that by the theory of Lagrange multiplier, we have $$P(u)=\alpha +\beta ke^{4u}$$ for some $\alpha, \beta\in \mathbb{R}$. I don't understand how to derive it. Could anyone give me some help? Thank you very much.

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I assume that should be $\int_M ke^{4u}dV_g=0$ and not $\int_M ke^{4v}dV_g=0$ in definition of $F$? –  Rahul Feb 4 '12 at 10:22
    
@Rahul: Yes. That's a typo. I changed it. Thanks! –  Paul Feb 4 '12 at 10:27
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2 Answers

up vote 5 down vote accepted
+50

Firstly, you need also to assume that $\int u \mathrm{d}V_g$ and $\int k e^{4u}\mathrm{d} V_g$ are $C^1$ functionals on your function space.

Then: let $\mathcal{H}$ be a real Hilbert space. Let $S, K_1, K_2$ be $C^1$ functionals such that $K_1$ and $K_2$ intersect transversely, then the set $\mathcal{N} = \{K_1 = k_1\}\cap\{K_2 = k_2\}$ is a $C^1$ Hilbert manifold, and we can apply the theory of Lagrange Multiplier as follows: if $u\in \mathcal{N}$ is a constrained critical point of $S$, we must have $DS(u) = \lambda_1 DK_1(u) + \lambda_2 DK_2(u)$ where $D$ is the Frechet derivative on $\mathcal{H}$.

Applying to the calculus of variations, you have that Frechet derivatives $DS(u):\mathcal{H}\to\mathbb{R}$ satisfy the Gateaux differentiation condition $$ DS(u) \cdot h = \lim_{t\to 0} \frac{1}{t}[ S(u+th) - S(u)] $$ which using calculus of variations and applying to your case where $\mathcal{H} = L^2(M,\mathrm{d}V_g)$ with $\langle f,g\rangle = \int fg \mathrm{d}V_g$, and $S$ is the energy functional, $K_1$ and $K_2$ are the given constraints gives $$ DS[u] \cdot h = \int h P(u) + u P(h) \mathrm{d}V_g $$ where using that $P$ is self-adjoint gives you $$ = \int 2h P(u) \mathrm{d}V_g ~.$$ Similarly you get $$ DK_1[u] \cdot h = \int h \mathrm{d}V_g $$ and $$ DK_2[u] \cdot h = \int 4 k e^{4u} h \mathrm{d}V_g~. $$

Putting this together tells you that for a constrained critical point of $S$ under constraints $K_1,K_2$, the solution $u$ must satisfies that for all $h\in\mathcal{H}$

$$ \int h\left( 2P(u) - \lambda_1 - 4\lambda_2k e^{4u}\right) \mathrm{d}V_g = 0$$

and hence the term inside the parentheses must be 0.


For the case in the paper you are reading, some further justification probably is involved. (For example, you hadn't told us on what function space is $P$ a self-adjoint operator).) If it is $L^2$ as I inferred, $\int ke^{4u}\mathrm{d}V_g$ is not actually $C^1$ (it is not even bounded). Nevertheless, one can usually apply some renormalisation to replace $K_2$ by another functional that is identical to $K_2$ when $|K_2[u]|\leq \epsilon$ and suitably cut-off outside. Slightly more worrisome is that $P$ is likely to be only densely defined on $L^2$ for most geometric applications, which throws some more caveats into the above.

But on a formal level, the above derivation (and more simply speaking, Rahul's answer) explains what you seek.

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btw, for $K_1$ in my notation to be continuous (and hence $C^1$), it is necessary and sufficient that $M$ has finite total volume. –  Willie Wong Feb 8 '12 at 13:41
    
Thank you very much. –  Paul Feb 9 '12 at 3:08
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Let us denote the inner product $\int_M uv\ \mathrm{d}V_g$ by $\langle u,v \rangle$. We want to minimize $E(u) = \langle u,Pu \rangle$ subject to $\langle u,1 \rangle = 0$ and $\langle k,e^{4u} \rangle = 0$.

I'm no expert on Riemannian manifolds or functional analysis, so I don't know if this is the right terminology, but: what I'd call the "gradients" of the objective and constraint functions are $2Pu$, $1$, and $4ke^{4u}$ respectively. In the sense that if you consider a small variation $\delta u$ in $u$, the change in $E(u)$ to first order is $\langle \delta u, 2Pu \rangle$, and so on.

Then by Lagrange multipliers, we immediately find that the solution must satisfy $2Pu + \lambda \cdot 1 + \mu \cdot 4ke^{4u} = 0$. This is equivalent to your desired result.

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