Ngô Quốc Anh

May 24, 2013

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

Filed under: PDEs — Ngô Quốc Anh @ 1:39

In this note, a method introduced by David Maxwell was considered. More precisely, we proved that

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

on a compact manifold $(M,g)$ without boundary where $\kappa=\frac{2}{n-2}$, $a=2\kappa+4$, and $b=\frac{n-1}{n}$ admits at most one solution.

In this note, we consider the case when $(M,g)$ has boundary $\partial M$ and together with our PDE above, we have the following Neumann boundary condition

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

where $\nu$ is the outward normal vector field on $\partial M$.

As before, 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 at least $\phi$ is constant. Recall that the following holds

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

where $\widehat g=\phi_1^\frac{4}{n-2}g$ is our conformal change. For the detailed calculation, we refer the reader to the previous notes.

Following the previous notes, we first have

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

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,

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

Note that under the conformal change $\widehat g=\phi_1^\frac{4}{n-2}g$, there holds $\widehat \nu=\phi_1^\frac{2}{n-2}\nu$. Therefore,

$\begin{array}{rcl}\displaystyle \nabla_{\widehat g} \phi \cdot \widehat \nu &=& \displaystyle\phi_1^\frac{2}{n-2} \nabla_g\phi \cdot \nu\\&=& \displaystyle\frac{n-2}{4}\Theta_-\phi_1^\frac{4}{n-2}(\phi-\phi^\frac{n}{n-2}).\end{array}$

Thus, we have proved that

$\displaystyle \nabla_{\widehat g} (\phi -1) \cdot \widehat \nu=\frac{n-2}{4}\Theta_-\phi_1^\frac{4}{n-2}(\phi-\phi^\frac{n}{n-2}).$

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

$\begin{array}{lcl} \displaystyle - \int_M {a{\Delta _{\widehat g}}(\phi - 1){{(\phi - 1)}^ + }d{v_{\widehat g}}} &=& \displaystyle a\int_M {{\nabla _{\widehat g}}(\phi - 1) \cdot {\nabla _{\widehat g}}{{(\phi - 1)}^ + }d{v_{\widehat g}}} - a \int_{\partial M} (\phi-1)^+ \nabla_{\widehat g} (\phi -1) \cdot \widehat \nu ds_{\widehat g}\hfill \\ &=& \displaystyle a\int_{\phi > 1} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} - a \int_{\partial M \cap \{ \phi > 1\}} (\phi-1)\nabla_{\widehat g} (\phi -1) \cdot \widehat \nu ds_{\widehat g}.\end{array}$

Hence,

$\begin{array}{lcl} \displaystyle - \int_M {a{\Delta _{\widehat g}}(\phi - 1){{(\phi - 1)}^ + }d{v_{\widehat g}}}&=&\displaystyle a\int_{\phi > 1} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} \\&&\displaystyle - \frac{(n-2)a}{4} \int_{\partial M \cap \{ \phi > 1\}} \Theta_- (\phi-1) \phi_1^\frac{4}{n-2}(\phi-\phi^\frac{n}{n-2}) ds_{\widehat g}\\&\geqslant & 0\end{array}$

provided $\Theta_- \geqslant 0$ along the boundary $\partial M$. Clearly, if $\phi>1$ then the right hand side of the preceding equality is non-positive. A similar argument using $(\phi-1)^-$ as a test function gives us the same thing. Hence $\phi$ is constant since

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

To prove $\phi=1$, we need to take care the sign of coefficients. An easy way to handle this is to assume that $\Theta_->0$ on $\partial M$.

To conclude our notes, let us consider the following Neumann boundary condition

$\displaystyle \partial_\nu u+\frac{1}{\kappa}H_g u=f$

where $f$ is a scalar function. We then have

$\begin{array}{lcl}\displaystyle f={\partial _\nu }{\phi _2} + \frac{1}{\kappa }H_g{\phi _2} &=& \displaystyle {\partial _\nu }({\phi _1}\phi ) + \frac{1}{\kappa }H_g{\phi _1}\phi \hfill \\ &=& \displaystyle ({\partial _\nu }{\phi _1} + \frac{1}{\kappa }H_g{\phi _1})\phi + ({\partial _\nu }\phi ){\phi _1} \hfill \\ &=& \displaystyle f\phi + ({\partial _\nu }\phi ){\phi _1}. \end{array}$

In other words,

$\displaystyle \partial_\nu \phi = \frac{1-\phi}{\phi_1} f.$

Using our new conformal change, we then have

$\displaystyle \partial_{\widehat\nu} \phi = (1-\phi ) \phi_1^\frac{4-n}{n-2} f.$

Hence,

$\displaystyle\begin{array}{lcl} \displaystyle - \int_M {a{\Delta _{\widehat g}}(\phi - 1){{(\phi - 1)}^ + }d{v_{\widehat g}}}&=&\displaystyle a\int_{\phi > 1} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} \\&&\displaystyle + a\int_{\partial M \cap \{ \phi > 1\}} (\phi-1)^2 \phi_1^\frac{4-n}{n-2} f ds_{\widehat g}\\&\geqslant & 0\end{array}$

provided $f\geqslant 0$ along the boundary $\partial M$.

On the other hand, there holds

$\displaystyle \begin{array}{lcl} \displaystyle - \int_M {a{\Delta _{\widehat g}}(\phi - 1){{(\phi - 1)}^ + }d{v_{\widehat g}}}&=& \int_{\phi > 1} { |{\sigma _1}|_{\widetilde g}^2({\phi ^{ - 2\kappa - 3}} - \phi )(\phi - 1) + b{\tau ^2})(\phi - {\phi ^{2\kappa + 1}})(\phi - 1) d{v_{\widetilde g}}} \\&\leqslant& 0\end{array}.$

Consequently,

$\displaystyle \int_{\phi > 1} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} = 0 =\int_{\partial M \cap \{ \phi > 1\}} (\phi-1)^2 \phi_1^\frac{4-n}{n-2} f ds_{\widehat g}.$

Using the test function $(\phi-1)^-$, we also obtain

$\displaystyle \int_{\phi < 1} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} = 0 =\int_{\partial M \cap \{ \phi < 1\}} (\phi-1)^2 \phi_1^\frac{4-n}{n-2} f ds_{\widehat g}.$

Thus,

$\displaystyle \int_{M} {|{\nabla _{\widehat g}}(\phi - 1){|^2}d{v_{\widehat g}}} = 0 =\int_{\partial M} (\phi-1)^2 \phi_1^\frac{4-n}{n-2} f ds_{\widehat g}.$

We eventually obtain that $\phi$ is constant. If we assume that $f>0$ on $\partial M$, then it must hold that $\phi=1$.