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}.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: