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.

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