Ngô Quốc Anh

April 1, 2013

A proof of the uniqueness of solutions of the Lichnerowicz equations in the compact case

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 4:38

In this small note, we aim to derive some uniqueness property of solutions of the following PDE

\displaystyle -a\Delta_g u +\text{Scal}_g u =|\sigma|_g^2 u ^{-2\kappa-3}-b\tau^2 u^{2\kappa+1}

on a compact manifold (M,g) where \kappa=\frac{2}{n-2}, a=2\kappa+4, and b=\frac{n-1}{n}.

We assume that \phi_1 and \phi_2 are solutions of the above PDE. Setting \phi=\frac{\phi_2}{\phi_1}. We wish to prove that \phi=1.

Let us consider the following trick basically due to David Maxwell, see this paper. Let \widetilde g= \phi_1^{2\kappa}g. Then the well-known formula for the Laplace-Beltrami operator \Delta_g, which is

\displaystyle {\Delta _g}u = \frac{1}{{\sqrt {\det g} }}{\partial _i}(\sqrt {\det g} {g^{ij}}{\partial _j}u)

helps us to write

\displaystyle {\Delta _{\widetilde g}}u = \phi _1^{ - 2\kappa } \Big({\Delta _g}u + \frac{2}{{{\phi _1}}}{\nabla _g}{\phi _1}{\nabla _g}u \Big).

Consequently,

\displaystyle {\Delta _g}\phi = \phi _1^{2\kappa }{\Delta _{\widetilde g}}\phi - \frac{2}{{{\phi _1}}}{\nabla _g}{\phi _1}{\nabla _g}\phi .

We now calculate - a{\Delta _g}{\phi _2} + {R_g}{\phi _2} as follows.

\begin{array}{lcl} - a{\Delta _g}{\phi _2} + {R_g}{\phi _2} &=& - a{\Delta _g}({\phi _1}\phi ) + {R_g}{\phi _1}\phi \hfill \\ &=& \displaystyle - a({\Delta _g}{\phi _1})\phi - 2a{\nabla _g}{\phi _1}{\nabla _g}\phi - a({\Delta _g}\phi ){\phi _1} + {R_g}{\phi _1}\phi \hfill \\ &=& \displaystyle (|\sigma |_g^2\phi _1^{ - 2\kappa - 3} - b{\tau ^2}\phi _1^{2\kappa + 1})\phi - 2a{\nabla _g}{\phi _1}{\nabla _g}\phi - a({\Delta _g}\phi ){\phi _1} \end{array}

where we have used the fact that \phi_1 solves the PDE. Making use of the formula for \Delta_g \phi gives

\displaystyle - a{\Delta _g}{\phi _2} + {R_g}{\phi _2} = (|\sigma |_g^2\phi _1^{ - 2\kappa - 3} - b{\tau ^2}\phi _1^{2\kappa + 1})\phi - a\phi _1^{2\kappa + 1}{\Delta _{\widetilde g}}\phi.

Since \phi_2 solves the PDE, we further have

\displaystyle (|\sigma |_g^2\phi _1^{ - 2\kappa - 3} - b{\tau ^2}\phi _1^{2\kappa + 1})\phi - a\phi _1^{2\kappa + 1}{\Delta _{\widetilde g}}\phi = |\sigma |_g^2\phi _2^{ - 2\kappa - 3} - b{\tau ^2}\phi _2^{2\kappa + 1}.

If we denote \sigma_1 = \phi_1^{-2}\sigma, we then have

\displaystyle |{\sigma _1}|_{\widetilde g}^2 = {\widetilde g^{ih}}{\widetilde g^{jk}}{({\sigma _1})_{ij}}{({\sigma _1})_{hk}} = \phi _1^{ - 4\kappa }{g^{ih}}{g^{jk}}\phi _1^{ - 4}{\sigma _{ij}}{\sigma _{hk}} = \phi _1^{ - 4\kappa - 4}|\sigma |_g^2.

Therefore, we can rewrite the resulting identity as follows

\displaystyle (\phi _1^{2\kappa + 1}|{\sigma _1}|_{\widetilde g}^2 - b{\tau ^2}\phi _1^{2\kappa + 1})\phi - a\phi _1^{2\kappa + 1}{\Delta _{\widetilde g}}\phi = \phi _1^{2\kappa + 1}|{\sigma _1}|_{\widetilde g}^2\phi ^{ - 2\kappa - 3} - b{\tau ^2}{\phi ^{2\kappa + 1}}\phi _1^{2\kappa + 1}.

Dividing both sides by \phi _1^{2\kappa + 1} gives

\displaystyle - a{\Delta _{\widetilde g}}\phi + (|{\sigma _1}|_{\widetilde g}^2 - b{\tau ^2})\phi = |{\sigma _1}|_{\widetilde g}^2\phi ^{ - 2\kappa - 3} - b{\tau ^2}{\phi ^{2\kappa + 1}},

or equivalently,

\displaystyle - a{\Delta _{\widetilde g}}(\phi - 1) = |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi ) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}}).

We now use the test functions (\phi-1)^\pm. Indeed, we first have

\begin{array}{lcl} \displaystyle - \int_M {a{\Delta _{\widetilde g}}(\phi - 1){{(\phi - 1)}^ + }d{v_{\widetilde g}}} &=& \displaystyle a\int_M {{\nabla _{\widetilde g}}(\phi - 1) \cdot {\nabla _{\widetilde g}}{{(\phi - 1)}^ + }d{v_{\widetilde g}}} \hfill \\ &=& \displaystyle a\int_{\phi > 1} {|{\nabla _{\widetilde g}}(\phi - 1){|^2}d{v_{\widetilde g}}} \hfill \\ &\geqslant & 0.\end{array}

Hence

\displaystyle a\int_{\phi > 1} {|{\nabla _{\widetilde g}}(\phi - 1){|^2}d{v_{\widetilde g}}} = \int_{\phi > 1} { |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi )(\phi - 1) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})(\phi - 1) d{v_{\widetilde g}}} .

Clearly, if \phi>1 then the right hand side of the preceding equality is non-positive. Therefore, we must have

\displaystyle\int_{\phi > 1} {|{\nabla _{\widetilde g}}(\phi - 1){|^2}d{v_{\widetilde g}}} = 0

and

\displaystyle\int_{\phi > 1} { |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi )(\phi - 1) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})(\phi - 1) d{v_{\widetilde g}}} = 0.

A similar argument using (\phi-1)^- as a test function shows

\displaystyle\int_{\phi < 1} {|{\nabla _{\widetilde g}}(\phi - 1){|^2}d{v_{\widetilde g}}} = 0

and

\displaystyle\int_{\phi < 1} { |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi )(\phi - 1) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})(\phi - 1) d{v_{\widetilde g}}} = 0.

Hence, in particular,

\displaystyle\int_M {|{\nabla _{\widetilde g}}(\phi - 1){|^2}d{v_{\widetilde g}}} = 0

which says that \phi is constant. If \phi \ne 1 then it follows from

\displaystyle\int_M { |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi )(\phi - 1) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})(\phi - 1) d{v_{\widetilde g}}} = 0

that |\sigma_1|_{\widetilde g}=0 and \tau=0.

Assume that we have some boundary condition, say

\displaystyle \partial_\nu u+\frac{1}{\kappa}Hu=-\frac{1}{2\kappa}\Theta_-u^{\kappa+1}

where \nu is the outward normal vector field on \partial M. Then we have

\begin{array}{lcl}\displaystyle {\partial _\nu }{\phi _2} + \frac{1}{\kappa }H{\phi _2} &=& \displaystyle {\partial _\nu }({\phi _1}\phi ) + \frac{1}{\kappa }H{\phi _1}\phi \hfill \\ &=& \displaystyle ({\partial _\nu }{\phi _1} + \frac{1}{\kappa }H{\phi _1})\phi + ({\partial _\nu }\phi ){\phi _1} \hfill \\ &=& \displaystyle - \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^{\kappa + 1}\phi + ({\partial _\nu }\phi ){\phi _1}. \end{array}

Using the equation for \phi_2, we conclude that

\displaystyle -\frac{1}{{2\kappa }}{\Theta _ - }\phi _2^{\kappa + 1} = - \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^{\kappa + 1}\phi + ({\partial _\nu }\phi ){\phi _1},

that is,

\displaystyle - \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^{\kappa + 1}{\phi ^{\kappa + 1}} = - \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^{\kappa + 1}\phi + ({\partial _\nu }\phi ){\phi _1}.

In particular,

\displaystyle {\partial _\nu }\phi = \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^\kappa \phi - \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^\kappa {\phi ^{\kappa + 1}} = \frac{1}{{2\kappa }}{\Theta _ - }\phi _1^\kappa (\phi - {\phi ^{\kappa + 1}}).

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