I am shy and love mathematics, music and sports.

### Concavity maximum principle

Let $u\in C^2(\Omega)\cap C(\bar{\Omega})$ satisfy the elliptic equation

$\displaystyle Lu=a^{ij}(Du)u_{ij}-b(x,u,Du)=0\quad \text{in }\Omega$

Assume $b$ is jointly concave with respect to $(x,u)$ and

$\displaystyle \frac{\partial b}{\partial u}\geq 0$

Then

$\mathcal{C}(y_1,y_3,\lambda)=u(\lambda y_3+(1-\lambda)y_1)-\lambda u(y_3)-(1-\lambda)u(y_1)$

can not achieve positive maximum in the interior of $\Omega\times \Omega$.

See the paper Korevaar 1983

### Mean curvature equations of translators

Suppose ${X:M^n\rightarrow \mathbb{R}}$ is the position vector of ${M^n}$ which flows by mean curvature, that is

$\displaystyle \frac{\partial X}{\partial t}=-HN \ \ \ \ \ (1)$

It is well known that for mean curvature flow the evolution equation of mean curvature is

$\displaystyle \partial_tH=\Delta H+H|A|^2 \ \ \ \ \ (2)$

Translator is a special type solution of mean curvature flow such that it moves by translation. Namely, if ${M_0^n}$ is a translator moves to direction ${V}$, then ${M_0^n+tV}$ satisfies,

$\displaystyle \left(\frac{\partial X}{\partial t}\right)^\perp=-HN. \ \ \ \ \ (3)$

The mean curvature of a translator satisfies ${H=-\langle V,N\rangle}$ and

$\displaystyle \Delta H+H|A|^2+\langle\nabla H,\nabla u\rangle=0 \ \ \ \ \ (4)$

where ${u=\langle X_0,V\rangle}$ is called height function.

How do we connect (2) and (3)? Actually it results from the difference of (1) and (3). Recall that (1) follows from (3) composed with some self diffeomorphism. Let us suppose ${\Phi_t}$ is that diffeomorphism (${\Phi_0=id}$). Then ${X_t=X_0(\Phi_t)+tV}$ satisfies (1). Since ${X_t}$ is also translator, ${H(X_t)=-\langle V,N\rangle(X_t)}$. Using ${\partial_t N=\nabla H}$,

$\displaystyle \partial_t|_{t=0}H(X_t)=- \partial_t|_{t=0}\langle V,N\rangle(X_t)=-\langle V, \nabla H\rangle(X_t)$

Since ${\nabla u=V^T}$, (2) will imply

$\displaystyle \Delta H+H|A|^2=-\langle\nabla u,\nabla H\rangle\quad \text{on } X_t.$

This just means (4) holds.

### Mean curvature of a graph on cylinder

Let ${\Sigma}$ is the cylinder ${\{x: x_1^2+x_2^2=1\}}$ in ${\mathbb{R}^3}$. Suppose ${M}$ is a graph on the cylinder ${\Sigma}$ can be written as

$\displaystyle M=\{u(x)\nu_{\Sigma}(x):x\in \Sigma\},$

where ${\nu_{\Sigma}}$ is the normal vector of ${\Sigma}$, that is ${\nu=(0,0,1)}$. What is the mean curvature of ${M\subset \mathbb{R}^3}$?

We can parametrize the cylinder by ${(\theta,z)}$ where ${x_1=\cos\theta}$, ${x_2=\sin\theta}$, ${x_3=z}$, ${0\leq \theta\leq 2\pi}$. One can write ${M}$ as

$\displaystyle F(\theta,z)=(u(\theta,z)\cos\theta,u(\theta, z)\sin\theta,z)$

Then the induced metric on ${M}$ is

$\displaystyle g=|dx|^2=[u^2+u_\theta^2]d\theta^2+2u_\theta u_zd\theta dz+(u_z^2+1)dz^2$

$\displaystyle g^{-1}=\frac{1}{u^2(1+u_z^2)+u_\theta^2}\begin{bmatrix} 1+u_z^2 & -u_\theta u_z \\ -u_\theta u_z& u^2+u_\theta^2 \end{bmatrix}$

Since

$\displaystyle F_\theta=(-u\sin\theta+u_\theta \cos\theta, u\cos\theta+u_\theta\sin \theta,0)$

$\displaystyle F_z=(u_z\cos\theta,u_z\sin\theta,1)$

The normal vector of ${M}$ is

$\displaystyle \nu_M=\frac{1}{\sqrt{u^2(1+u_z^2)+u_\theta^2}}(u\cos\theta+u_\theta\sin\theta,u\sin \theta-u_\theta\cos \theta,-u\,u_z)$

The second fundamental form is

$\displaystyle A_{\theta\theta}=-F_{\theta\theta}\cdot \nu_{M}=u^2-uu_{\theta\theta}+2u_\theta^2$

$\displaystyle A_{\theta z}=-F_{\theta z}\cdot \nu_{M}=-uu_{\theta z}+u_\theta u_z$

$\displaystyle A_{zz}=-F_{zz}\cdot \nu_{M}=-uu_{zz}$

Then the mean curvature is

$\displaystyle H=\text{tr}A=g^{\theta\theta}A_{\theta\theta}+2g^{\theta z}A_{\theta z}+g^{zz}A_{zz}$

$\displaystyle =\frac{-[(1+u_z^2)u_{\theta\theta}+(u^2+u_\theta^2)u_{zz}]u+(1+u_z^2)u^2+2u_\theta^2+2uu_\theta u_zu_{\theta z}}{[u^2(1+u_z^2)+u_\theta^2]^{3/2}}$

If ${v=u-1}$ and ${|v|_{C^4}\ll 1}$, then

$\displaystyle H\sim 1-v_{\theta\theta}-v_{zz}-v$

### Troyanov’s work on prescrbing curvature on compact surfaces with conical singuarities

I will describe some parts of Troynov’s work on conical surface. For details, check his paper. For simplicity, we consider a closed Riemann surface ${S}$ with a real divisor ${\boldsymbol{\beta}=\sum_{i}\beta_ip_i}$. A conformal metric ${ds^2}$ on ${S}$ is said to represent the divisor ${\boldsymbol{\beta}}$ if ${ds^2}$ is smooth Riemannian metric on ${S\backslash supp(\boldsymbol{\beta})}$ and near each ${p_i}$, there exists a neighborhood ${\mathcal{O}_i}$ of ${p_i}$ and coordinate function ${z_i\mathcal{O}_i:\rightarrow \mathbb{R}^2}$ and ${u:\mathcal{O}_i\rightarrow\mathbb{R}}$ such that ${u\in C^2(\mathcal{O}_i-\{p_i\})}$ and

$\displaystyle ds^2=e^{2u}|z_i-a_i|^{2\beta_i}|dz_i|^2\quad \text{ on }\mathcal{O}_i$

where ${a_i=z_i(p_i)}$. ${\theta_i=2\pi(\beta_i+1)}$ is called the angle of the conical singularity. We will always assume ${\beta_i>-1}$.

For example ${\mathbb{C}}$ equipped with the metric ${|z|^{2\beta}|dz|^2}$ is isometric to an Euclidean cone of total angle ${\theta=2\pi(\beta+1)}$.

Suppose now ${(S,\boldsymbol{\beta},ds_0^2)}$ is a closed Riemann surface with conical metric ${ds_0^2}$. Assume that the Gauss curvature ${K_0}$ extends on ${S}$ as a H\”{o}lder continuous function.

Suppose ${ds^2=e^{2u}ds_0^2}$ is a conformal change of the conical metric. Then necessarily we have ${K=e^{-2u}(\Delta_0u+K_0)}$. In order to prescribe the Gauss curvature ${K}$, we need to solve the equation

$\Delta_0u+K_0=Ke^{2u}$.

Any reasonable solution of (0) will satisfy

$\displaystyle \int K e^{2u}d\mu_0=\int K_0d\mu_0:=\gamma$

where ${\gamma=2\pi (\chi(S)+\sum_i\beta_i)}$ follows from the Gauss-Bonnet formula for conical surface. We will use variational method to attack it. To that end, define ${H(ds_0^2)}$ to be Sobolev space of functions with

$\displaystyle \int_S|\nabla u|_0^2+u^2d\mu_0<\infty$

and functionals

$\displaystyle \mathscr{F}(u):=\int_{S}|\nabla u|_0^2+2K_0u^2d\mu_0,\quad\mathscr{G}(u):=\int Ke^{2u}d\mu_0$

It is easy to verify that the minimizer of the following variation problem will be a weak solution of (0)

$\displaystyle m=\inf_{u\in H(ds_0^2)}\{\mathscr{F}(u):\mathscr{G}(u)=\gamma\}$

As one can see, there are two immediate questions we need to answer,

(a) ${m>-\infty}$

(b) minimizer exists

The first question follows from the Trudinger inequality. Namely, define ${\tau(S,\boldsymbol{\beta})=4\pi+4\pi\min\{0,\min_i\beta_i\}}$. Then

Lemma: For any fixed ${b<\tau(S,\beta)}$, there exist constant ${C>0}$ such that,

$\displaystyle \int_S e^{bu^2}d\mu_0\leq C$

for any ${u\in H(ds_0^2)}$ and ${\int_S ud\mu_0=0}$ and ${\int_S |\nabla u|_0^2d\mu_0\leq 1}$.

Lemma: Suppose ${\psi\in L^2(ds_0^2)}$ and ${\int_S\psi d\mu_0=1}$. Let ${p>1}$. Then for any fixed ${b<\tau(S,\boldsymbol{\beta})}$, there exists constants ${C=C(b, p,\psi,ds_0^2)}$ such that

$\displaystyle |\int ve^{2u}d\mu_0|\leq C||v||_{L^p(ds_0^2)}\exp\left\{\frac{q}{b}\int_S |\nabla u|_0^2d\mu_0+\int_S2\psi ud\mu_0\right\}$

for any ${v\in L^p(ds_0^2)}$ and ${u\in H(ds_0^2)}$. Here ${q=p/(p-1)}$.

From this key lemma, one can derive that ${\mathscr{F}(u)}$ is lower bounded on ${\mathscr{G}(u)=\gamma}$ provided ${\gamma<\tau}$ and ${\sup K>0}$. To do that, choose ${p}$ and ${b}$ such that ${\gamma< \frac{b}{q}<\tau}$, then it follows from the above lemma that

$\displaystyle \gamma=\int_S Ke^{2u}d\mu_0\leq C||K||_{L^p(ds_0^2)}\exp\left\{\frac{q}{b}\int_S |\nabla u|_0^2+2\int_S\frac{K_0 u}{\gamma}d\mu_0\right\}$

$\displaystyle \leq C|K|_\infty\exp\left\{\frac{1}{\gamma}\mathscr{F}(u)+(-\frac{1}{\gamma}+\frac{q}{b})\int|\nabla u|_0^2\right\}\leq C|K|_\infty\exp\{\frac{1}{\gamma}\mathscr{F}(u)\}.$

For the second question, we need to prove the embedding ${H(ds_0^2)\rightarrow L^2(ds_0^2)}$ is compact(actually it is true for any ${L^p(ds_0^2)}$ for any ${p<\infty}$). Note that this is true if ${ds_0^2}$ is some smooth metric, however ${ds_0^2}$ is conical one. Therefore, we want to compare ${H(ds_0^2)}$ with ${H(ds_1^2)}$ of some smooth metric.

Suppose ${S}$ is equipped with smooth metric ${ds_1^2}$ such that ${ds_0^2=\rho ds_1^2}$. Here ${\rho}$ is smooth and positive outside the support of ${\boldsymbol{\beta}}$ and ${\rho(z)=O(|z|^{2\beta_i})}$ near ${p_i}$. If we use the ${^i\nabla u}$ denote the gradient of ${u}$ with respect to ${ds_i^2}$, then ${^0\nabla=(1/\rho)^1\nabla}$ and

$\displaystyle \int_{S}|\nabla u|_0^2d\mu_0=\int_{S}|\nabla u|_1^2d\mu_1$

This is only true in two dimension. Now ${H(ds_1^2)}$ and ${H(ds_0^2)}$ have inner product

$\displaystyle (u,v)_1=\int_S \langle\nabla u,\nabla v\rangle_1+uv d\mu_1,\quad (u,v)_0=\int_S \langle\nabla u,\nabla v\rangle_0+uv d\mu_0$

From the above analysis, we need to examine the difference of ${L^2(ds^2_1)}$ and ${L^2(ds_0^2)}$. It follows from the singular behavior of ${\rho }$ that

$\displaystyle L^p(ds_0^2)\subset L^q(ds_1^2),\quad L^p(ds_1^2)\subset L^q(ds_0^2)$

for any ${p\geq 1}$ and some ${q\gg p}$. Then for any ${u\in H(ds_0^2)}$,

$\displaystyle |u|_{L^2(ds_1^2)}\leq C|u|_{L^p(ds_0^2)} \leq C|u|_{H(ds_0^2)}$

Then ${H(ds_0^2)\subset H(ds_1^2)}$. On the contrary, we need the following inequality

Lemma: For conical metric ${ds_0^2}$

$\displaystyle |u|_{L^p(ds_0^2)}\leq |u|_{H(ds_0^2)}$

Then we have ${H(ds_0^2)= H(ds_1^2)}$. Therefore ${H(ds_0^2)\subset L^p(ds_0^2)}$ is compact for ${p<\infty}$. Furthermore, after some effort, one can prove

$\displaystyle H(ds_0^2)\rightarrow L^p(ds_0^2)$

$\displaystyle u\mapsto e^{2u}$

is also compact for ${p<\infty}$.

Remark:

[1] M. Troyanov, Prescribing curvature on compact surfaces with conical singularities, Trans. Am. Math. Soc. 324 (1991) 793–821. doi:10.1090/S0002-9947-1991-1005085-9.

### Surface intersection with a ball

Suppose $M$ is 2-dimensional surface in $\mathbb{R}^3$. If $M$ is simply connected, then $B\cap M$ may not be simply connected, where $B$ is ball in $\mathbb{R}^3$. See the following figure.

But if $M$ is minimal surface, then $M\cap B$ must be simply connected. The reason is $M$ has convex haul property.

Learn this example from Jacob Bernstein.

Suppose ${u=u(r)}$ is a radial function on ${\mathbb{R}^n}$, here ${r=|x|}$.

$\displaystyle u_{x_i}=u'\frac{x_i}{r}$

$\displaystyle u_{x_ix_j}=u''\frac{x_ix_j}{r^2}+u'(\frac{\delta_{ij}}{r}-\frac{x_ix_j}{r^3})=\frac{u'}{r}\delta_{ij}+(\frac{u''}{r^2}-\frac{u'}{r^3})x_ix_j$

therefore

$\displaystyle \det D^2u = \left(\frac{u'}{r}\right)^{n}\det[ \delta_{ij}+\frac{r}{u'}(u''-\frac{u'}{r})\frac{x_ix_j}{r^2}]$

$\displaystyle =\left(\frac{u'}{r}\right)^{n}[1+\frac{r}{u'}(u''-\frac{u'}{r})]=\left(\frac{u'}{r}\right)^{n-1}u''$

If we use the polar coordinates ${(r,\theta_1,\cdots, \theta_{n-1})}$, and ${g=dr^2+r^2\sum_{i=1}^{n-1}d\theta_i^2}$ and the following fact

$\displaystyle \nabla_X\partial_r=\begin{cases}\frac{1}{r}X&\text{if } X\text{ is tangent to }\mathbb{S}^{n-1}\\0 \quad &\text{if } X=\partial_r\end{cases}$

then one can calculate the Hessian of ${u}$ under this coordinates

$\displaystyle Hess (u)(\partial_r,\partial_r)=u''$

$\displaystyle Hess (u)(\partial_{\theta_i},\partial_r)=0$

$\displaystyle Hess (u)(\partial_{\theta_i},\partial_{\theta_j})=ru'\delta_{ij}.$

Then

$\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'}{r}\right)^{n-1}u''$

If the metric is $g=dr^2+\phi^2ds_{n-1}^2$, then we will have

$\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'\phi'}{\phi}\right)^{n-1}u''$

### An identity related to generalized divergence theorem

I am trying to verify one proposition proved in Reilly’s paper. For the notation of this, please consult the paper.

Propsotion 2.4 Let ${{D}}$ be a domain in ${\mathbb{R}^n}$. ${f}$ is a smooth function on ${{D}}$. Then

$\displaystyle \int_{{D}}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\left(\tilde S_q(z)z_n-\sum_{\alpha\beta}\tilde T_{q-1}(z)^{\alpha\beta}z_{\alpha}z_{n,\beta}\right)dA$

Proof: Take an orthonormal frame field ${\{e_\alpha,e_n\}}$ such that ${\{e_\alpha\}}$ is tangent to ${\partial D}$. Notice

$\displaystyle D^2(f)(X,Y)=X(Yf)-(\nabla_{X}Y)f$

$\displaystyle D^2(f)=\left(\begin{matrix} z_{,\alpha\beta}-z_nA_{\alpha\beta} & z_{n,\alpha}\\ z_{n,\beta} & f_{nn} \end{matrix} \right)$

where ${f_{nn}=D^2(f)(e_n,e_n)}$. It follows from Remark 2.3 that

$\displaystyle \int_{D}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA$

where ${t=(t_1,\cdots,t_n)}$ is the outward unit normal to ${\partial D}$. Changing the coordinates to ${e_\alpha}$ and ${e_n}$, we can get

$\displaystyle \int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA=\int_{\partial D}T_q(f)^{\alpha n}z_{\alpha}+T_{q}(f)^{nn}z_ndA$

It is easy to see

$\displaystyle T_q(f)^{nn}=\tilde S_{q}(z)$

and

$\displaystyle T_q(f)^{\alpha n}z_\alpha=\sum\delta\binom{i_1,i_2\cdots,n,\alpha}{j_1,j_2\cdots,\beta, n}z_\alpha$

$\displaystyle =-\sum\delta\binom{\alpha_1,\alpha_2\cdots,\alpha}{\beta_1,\beta_2\cdots,\beta}f_{n\beta}z_\alpha=-\tilde T_{q-1}(z)^{\alpha\beta}z_\alpha z_{n,\beta}$

Therefore the proposition is established.

Remark: Robert Reilly, On the hessian of a function and the curvature of its graph

### Some identities related to mean curvature of order m

For any ${n\times n}$ (not necessarily symmetric) matrix ${\mathcal{A}}$, we let ${[\mathcal{A}]_m}$ denote the sum of its ${m\times m}$ principle minors. For any hypersurface which is a graph of ${u\in C^2(\Omega)}$, where ${\Omega\subset \mathbb{R}^n}$. We have its downward unit normal is

$\displaystyle (\nu,\nu_{n+1})=\left(\frac{Du}{\sqrt{1+|Du|^2}},\frac{-1}{\sqrt{1+|Du|^2}}\right).$

The principle curvatures are taken from the eigenvalues of the Jacobian matrix ${[D\nu]}$. One can define its ${m}$ mean curvature using the notation of above

$\displaystyle H_m=\sum_{i_1<\cdots

Now let us consider general matrix ${\mathcal{A}}$,

$\displaystyle A_m=[\mathcal{A}]_m=\frac{1}{m!}\sum \delta\binom{i_1,\cdots,i_m}{j_1,\cdots,j_m}a_{i_1j_1}\cdots a_{i_mj_m}$

where ${\delta}$ is the generalized Kronecker delta.

$\displaystyle \delta\binom{i_1 \dots i_p }{j_1 \dots j_p} = \begin{cases} +1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an even permutation of } j_1 \dots j_p \\ -1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an odd permutation of } j_1 \dots j_p \\ \;\;0 & \quad \text{in all other cases}.\end{cases}$

Then we define the Newton tensor

$\displaystyle T^{ij}_m=\frac{\partial A_m}{\partial a_{ij}}=\frac{1}{(m-1)!}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}a_{i_2j_2}\cdots a_{i_m j_m}.$

For any vector field ${X}$ on ${\mathbb{R}^n}$, ${DX}$ is a matrix, where ${D=(D_{1},\cdots,D_n)}$ and ${|X|\neq 0}$, denote ${\tilde X=X/|X|}$, we have

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}X_{j_2}]\cdots [D_{i_m}X_{j_m}].$

Since for any ${1\leq p,k,l\leq n}$

$\displaystyle D_k\tilde X_l=\frac{D_kX_{l}}{|X|}-\frac{\sum_{p=1}^nX_pD_kX_pX_l}{|X|^3}$

$\displaystyle \sum_{i,j,i_2,j_2}\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_jX_pX_{j_2}D_{i_2}X_p=0$

because $\delta$ is skew-symmetric in $j,j_2$. Then

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=|X|^{m-1}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}\tilde X_{j_2}]\cdots [D_{i_m}\tilde X_{j_m}].$

$=(m-1)!|X|^{m-1}T_m^{ij}(D\tilde X)X_iX_j=(m-1)!|X|^{m+1}T_m^{ij}(D\tilde X)\tilde X_i\tilde X_j.$

Applying the formula $(T^{ij}_m(\mathcal{A}))=[\mathcal{A}]_{m-1}I-T_{m-1}(\mathcal{A})\cdot \mathcal{A}$(Check [1] Propsition 1.2) and $(D\tilde X)\tilde X=0$, we get

$T^{ij}_m(DX)X_iX_j=|X|^{m+1}[D\tilde X]_{m-1}$

It follows from the result of Reilly, Remark 2.3(a), that

$\displaystyle mA_m[DX]=D_i[T^{ij}_mX_j]$

Suppose ${X}$ is a vector field normal to ${\partial \Omega}$, then

$\displaystyle m\int_\Omega [DX]_m=\int_{\partial \Omega} T^{ij}_m X_j\gamma_i=\int_{\partial \Omega}(X\cdot \gamma)^m[D\gamma]_{m-1}=\int_{\partial \Omega} (X\cdot \gamma)^m H_{m-1}(\partial \Omega)$

where ${\gamma}$ is the outer ward unit normal to ${\partial \Omega}$.

Remark: [1] R.C. Reilly, On the Hessian of a function and the curvatures of its graph., Michigan Math. J. 20 (1974) 373â€“383. doi:10.1307/mmj/1029001155.

[2] N. Trudinger, Apriori bounds for graphs with prescribed curvature.

We probably have to assume $DX$ is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

### Sherman-Morrison Formula

Suppose $\eta\in \mathbb{R}^n$ is a column vector and $M_{n\times n}$ is an invertible matrix.  Set $A=M+\eta \eta^T$, then

$A^{-1}=M^{-1}-\frac{M^{-1}\eta\eta^TM^{-1}}{1+\eta^T M^{-1}\eta}$

This formula has more general forms.

### Unit normal to a radial graph over sphere

Consider $\Omega\subset \mathbb{S}^n$ is a domain in the sphere. $S$ is a radial graph over $\Omega$.

$\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}\subset \mathbb{R}^{n+1}$

What is the unit normal to this radial graph in $\mathbb{R}^{n+1}$?

Suppose $\{e_1,\cdots,e_n\}$ is a smooth local frame on $\Omega$. Let $\nabla$ be the covariant derivative on $\mathbb{S}^n$. Tangent space of $S$ consists of $\{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n$ which are

$\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x$

In order to get the unit normal, we need some simplification. Let us assume $\{e_i\}$ are orthonormal basis of the tangent space of $\Omega$ and $\nabla v=e_1(v)e_1$. Then

$\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2$

Then we obtain an orthonormal basis of the tangent of $S$

$\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}$

We are able to get the normal by projecting $x$ to this subspace

$\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.$

After normalization, the (outer)unit normal can be written

$\frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}$

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.