Ngô Quốc Anh

September 30, 2013

The Lichnerowicz equation under some variable changes

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 4:54

Let us consider the so-called Lichnerowicz equation

-\Delta_g u + hu = fu^{2^\star-1}+au^{-2^\star-1} \quad u>0

on (M,g), a Riemannian manifold of dimension n \geq 3. Here h, f, and a are smooth function with a \geq 0.

  • We first use the the following variable change

    \displaystyle v=\log u \quad u=e^ v.


    \displaystyle\Delta v = \frac{\Delta u}{u} - \frac{|\nabla u|^2}{u^2}


    \displaystyle |\nabla v|^2 = \frac{|\nabla u|^2}{u^2}.

    Therefore, we can write

    \displaystyle -\Delta v =-\frac{\Delta u}{u} +|\nabla v|^2.

    Using this rule, we can rewrite the equation as follows

    \displaystyle \boxed{-\Delta v = -h+fu^{2^\star-2}+au^{-2^\star-2}+|\nabla v|^2=-h+fe^{(2^\star-2)v}+ae^{-(2^\star+2)v}+|\nabla v|^2. }


September 15, 2013

Some integral identities on manifolds with boundary

In this note, I summary several useful integral identities on Riemannian manifolds with boundary.

  1. Suppose that f is a function and X is a 1-form, then

    \displaystyle\boxed{\int_M {f\text{div}Xd{v_g}} = - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}.}

    To prove this, we write everything in local coordinates as follows

    \begin{array}{lcl} \displaystyle\int_M {f \text{div} Xd{v_g}} &=& \displaystyle\int_M {f{\nabla _i}{X^i}d{v_g}} \hfill \\ &=& \displaystyle - \int_M {{\nabla _i}f{X^i}d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}} \hfill \\ &=& \displaystyle - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}\end{array}

    as claimed.

  2. Using the previous identity, we can prove the following

    \displaystyle \boxed{\int_M {{{\left\langle {X,\nabla (\text{div} X)} \right\rangle }_g}d{v_g}} = - \int_M {|\text{div} X|_g^2d{v_g}} + \int_{\partial M} {\text{div} X{{\left\langle {X,\nu } \right\rangle }_g}d{\sigma _g}}}

    where X is again a vector field on M. To prove this, we simply apply the previous identity with f replaced by \text{div}(X) to get the desired result.

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