We introduce the following functional to study Yamabe problem with boundary.


$$
Q\_g(\varphi)=\frac{\int\_M\left(|\nabla \varphi|\_g^2+\frac{n-2}{4(n-1)} R\_g \varphi^2\right) d v+\frac{n-2}{2} \int\_{\partial M} h\_g \varphi^2 d \sigma}{\left(\int\_M|\varphi|^{2 n /(n-2)} d v\right)^{(n-2) / 2}}
$$


$$
Q(M)=\inf \left\{Q\_g(\varphi): \varphi \in C^1(\bar{M}), \varphi \neq 0\right\}
$$


In Escobar's The Yamabe Problem On Manifolds With Boundary,he says if $Q(M)<Q(S^{+}\_{n})$,the functional has a critical point which is the solution of Yamabe Problem.


He also gives the formula of $Q(S\_{n}^{+})$


$$
Q\left(S\_{+}^n\right)=\frac{\int\_{\mathbb{R}\_{+}^n}\left|\nabla u\_{\varepsilon}\right|^2}{\left(\int\_{\mathbb{R}\_{+}^n} u\_{\varepsilon}^{2 n /(n-2)}\right)^{(n-2) / n}}=n(n-2)\left(\int\_{\mathbb{R}\_{+}^n} u\_{\varepsilon}^{\frac{2 n}{n-2}}\right)^{2 / n},
$$
$$
u\_{\varepsilon}\left(x, x\_n\right)=\left(\frac{\varepsilon}{\varepsilon^2+|x|^2+x\_n^2}\right)^{(n-2) / 2}
$$


I wonder how this is derived.


When studying the Yamabe problem on compact manifolds,we use conformal change to calculate $Q(S\_{n})$.Since the functional is conformally invariant, we use conformal change to transfer the problem to $R^{n}$,which becomes the problem of best Sobolev constant.


How can we do this when studying the problem on manifolds with boundary?Any help will be thanked.



How to calculate the infimum of Yamabe functional on upper hemisphere
The first observation is that $Q\_g(\phi)$ is conformally invariant.
Define
\begin{align\*}
 L\_gu & := -\Delta u + \frac{n-2}{4(n-1)}Ru , \\
 B\_gu & := \partial\_\nu u + \frac{n-2}{2}Hu ,
\end{align\*}
where $\nu$ is the outward-pointing unit normal and everything is defined with respect to the metric $g$.
On the one hand, integration by parts gives
\begin{equation\*}
 Q\_g(u) = \frac{\int\_M u\,Lu\,dv\_g + \oint\_{\partial M} u\,Bu\,d\sigma\_g}{\left( \int\_M \lvert u \rvert^{\frac{2n}{n-2}} \, dv\_g \right)^{\frac{n-2}{n}}} .
\end{equation\*}
On the other hand, if $\hat g = w^{\frac{4}{n-2}}g$, then
\begin{align\*}
 L\_{\hat g}u & = w^{-\frac{n+2}{n-2}}L\_g(wu) , \\
 B\_{\hat g}u & = w^{-\frac{n}{n-2}}B\_g(wu) .
\end{align\*}
Therefore
\begin{equation\*}
 Q\_{\hat g}(u) = Q\_g(wu) .
\end{equation\*}
It follows from stereographic projection that $Q(S\_+^n) = Q(\mathbb{R}\_+^n)$, where $\mathbb{R}\_+^n$ is the upper half plane $\{ x\_n \geq 0 \}$.


The next observation is that $u\_\varepsilon$ really minimizes $Q(\mathbb{R}\_+^n)$.
One argument goes as follows:
By the [Concentration Compactness Principle](https://doi.org/10.4171/RMI/12), a minimizer $u$ exists.
Moreover, the usual argument using elliptic regularity and the maximum principle implies that a minimizer is smooth.
By an [adaptation of the Obata argument](https://doi.org/10.1002/cpa.3160430703), if $u$ is a minimizer, then $u^{\frac{4n}{n-2}}dx^2$ is an Einstein metric on $S\_+^n$ with respect to which the boundary is minimal.
This requirement implies that $u$ takes the form $u\_\varepsilon$.
Note that the latter argument proceeds by using stereographic projection again to study the minimizer $u$ on $S\_+^n$.


