# Ngô Quốc Anh

## July 11, 2011

### Energy functionals associate to integrals of exponential type

Filed under: PDEs — Ngô Quốc Anh @ 15:05

The purpose of this note is to derive some integral functionals $\mathcal F$ associated to the following

$\displaystyle R(x)e^u$

in the weak form in the sense that each critical point of $\mathcal F$ is a weak solution for equation

$\displaystyle R(x)e^u = 0.$

For simplicitly, we denote

$\displaystyle \mathcal I(u) = \int_M R(x)e^udv_g$

where $M$ is a Riemannian manifold with metric $g$ and $u$ a function sitting in an appropriate Sobolev space. To be exact, we shall find a functional $\mathcal F$ so that its first variation, denoted by $\delta\mathcal F$, equals $\mathcal I$.

Type 1. We shall find $\mathcal F$ of the following form

$\displaystyle \mathcal F(u) = C\int_M R(x)e^udv_g$

for some constant $C$ to be determined later. To this purpose, we try to calculate the first variation of $\mathcal F$ in the direction $v$. Indeed,

$\displaystyle\delta \mathcal{F}(u)(v) = C\frac{d}{{dt}}\int_M R (x){e^{u + tv}}d{v_g}{\bigg|_{t = 0}} = C\int_M R (x){e^u}vd{v_g}.$

Therefore, we may choose $C=1$.

Type 2. We shall find $\mathcal F$ of the following form

$\displaystyle \mathcal F(u) = C\ln\left(\int_M R(x)e^udv_g\right)$

for some constant $C$ to be determined later. The first variation of $\mathcal F$ in the direction $v$ can be computed as below

$\displaystyle\begin{gathered} \delta \mathcal{F}(u)(v) = C\frac{d}{{dt}}\ln \left( {\int_M R (x){e^{u + tv}}d{v_g}} \right){\bigg|_{t = 0}} \hfill \\ \qquad= C\dfrac{{\displaystyle\int_M R (x){e^{u + tv}}vd{v_g}}}{{\displaystyle\int_M R (x){e^{u + tv}}d{v_g}}}{\bigg|_{t = 0}} \hfill \\ \qquad= \dfrac{C}{{\displaystyle\int_M R (x){e^u}d{v_g}}}\int_M R (x){e^u}vd{v_g}. \hfill \\ \end{gathered}$

Thus, in this case, we have to choose

$\displaystyle C = \int_M R (x){e^u}d{v_g}.$