# 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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