Author Archives: Sun's World

I am shy and love mathematics, music and sports.

Hessian of radial functions

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''

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<i_m}\kappa_{i_1}\cdots\kappa_{i_m}=[D\nu]_m

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) 373–383. 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\}

What is the unit normal to this radial graph?

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.

Principle curvature of translator

Suppose {\Sigma\subset\mathbb{R}^{3}} is a translator in {e_{3}} direction. Denote {V=e_{3}^T} and {A} is the second fundamental form of {\Sigma}. Then it is well know that

\displaystyle \Delta A+\nabla_VA+|A|^2A=0 \ \ \ \ \ (1)

Choose a local orthonormal frame {\{\tau_1,\tau_2\}} such that {A(\tau_1,\tau_1)=\lambda}, {A(\tau_2,\tau_2)=\mu} and {A(\tau_1,\tau_2)=0} in the neighborhood of some point {p}. We want to change (1) to some expression on {\lambda} or {\mu}. To do that, we need to apply both sides of (1) to {\tau_1,\tau_1}, getting

\displaystyle (\nabla_VA)(\tau_1,\tau_1)=\nabla_V \lambda-2A(\nabla_V \tau_1,\tau_1)=\nabla_V\lambda \ \ \ \ \ (2)

where we have used {\nabla_V\tau_1\perp \tau_1}. Similarly

\displaystyle (\nabla_{\tau_k} A)(\tau_1,\tau_1)=\nabla_{\tau_k}\lambda

\displaystyle (\nabla_{\tau_k}A)(\tau_1,\tau_2)=(\lambda-\mu)\langle \nabla_{\tau_k}\tau_1,\tau_2\rangle

Now let us calculate the Laplacian of second fundamental form

\displaystyle (\nabla_{\tau_k}^2A)(\tau_k,\tau_1)=\tau_k(\nabla_{\tau_k}\lambda)-2(\nabla_{\tau_k}A)(\nabla_{\tau_k}\tau_1,\tau_1)-(\nabla_{\tau_k}\tau_k)\lambda

\displaystyle =\tau_1(\nabla_{\tau_1}\lambda)-2\frac{(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}{\lambda-\mu}.

Then

\displaystyle (\Delta A)(\tau_1,\tau_1)=(\nabla_{\tau_k}^2A)(\tau_k,\tau_1)= \Delta \lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}.

Combining all the above estimates to (1), we get

\displaystyle \Delta \lambda+\nabla_V\lambda+|A|^2\lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}=0.

where we write {A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}. Using this equation and the other one on {\mu}, one can derive that if {\Sigma} is mean convex then it is actually convex.

Laplacian on graph

Suppose {\Omega\subset \mathbb{R}^n} is a domain and {\Sigma=graph(F)\subset \mathbb{R}^{n+1}} is a hypersurface, where {F=F(u_1,\cdots,u_n)} is a function on {\Omega}. Define {f=f(u_1,\cdots,u_n)} on {\Omega}. Then {f} also can be considered as a function on {\Sigma}. How do we understand {\Delta_\Sigma f}?

Denote {\partial_i=\frac{\partial}{\partial{u_i}}} for short. If we pull the metric of {\mathbb{R}^{n+1}} back to {\Omega}, denote as {g}, then

\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2

where {W=\sqrt{1+|\nabla F|^2}} and {F_{u_i}=\frac{\partial F}{\partial u_i}}. Then one can use the local coordinate to calculate {\Delta_\Sigma f}

\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)

\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}

also one can see from another definition of Laplacian

\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]

\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f

By using the expression of {g^{ij}} stated above, we can calculate

\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}

It follows from the definition of tangential derivative on {\Sigma}, see, that

\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}

then

\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

where {H} is the mean curvature of the {\Sigma}. Combining all the above calculations,

\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

One example of blowing up corner

Consider {u(x,y)=\sqrt{x^2+y^4}} on {\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}. Notify {\mathbb{R}^2_+} has a corner at the origin and {u} is not smooth at the origin. {u\approx x} when {x\geq y^2} and {u\approx y^2} when {x\leq y^2}. We want to resolve {u} by blowing up the origin through a map {\beta}. After blowing up, {W} looks like the following picture.

 

b_map

Denote {W=[\mathbb{R}^2_+,(0,0)]} and {\beta:W\rightarrow \mathbb{R}^2_+} is the blow down map. On {W\backslash lb}(near A), {\beta} takes the form

\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})

where {\xi_1} is the boundary defining function for ff and {\eta_1} is boundary defining function for rb. Similarly on {W\backslash rb}(near B), {\beta} takes the form

\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)

where {\xi_2} is a bdf for lb and {\eta_2} is a bdf for ff. One can verify that {\beta} is a diffeomorphism {\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}. Let {w=\beta^* u}. Then {w} is a polyhomogeneous conormal function on {W}. Its index can be denoted {(E,F,H)} correspond to lb, ff and rb.

\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}

Suppose {\pi_1} is the projection to {x} coordinate. Consider {f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}, {f} is actually a {b-}fibration. Then the push forward map {f_*} maps {w} to a polyhomogeneous function on {\mathbb{R}^+}.

\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy

In order to make {u} is integrable, let us assume {u} support {x\leq 1} and {y\leq 1}. What is the index for {f_* w} on {\mathbb{R}^+}?

\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy

For the first integral, letting {y^2/x=t}

\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}

For the second integral, letting {x/y^2=t}

\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt

Since the Taylor series

\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1

Consequently

\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots

Combining all the above analysis, the index for {f_*w} is {\{(n,0)\}\cup \{\frac{n}{2},0\}}

From another point of view, the vanishing order of {f} on each boundary hypersurface of {W} are {e_f(lb)=1}, {e_f(\text{ff})=2} and {e_f(rb)=0}. {f} maps lb and ff to the boundary of {\mathbb{R}^+}. Therefore the index of {f_*w} is contained in

\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F

Remark: Daniel Grieser, Basics of {b-}Calculus.

Stereographic projection from center

Suppose we are doing stereographic projection at the center. Namely consider the following map

\displaystyle \phi:\mathbb{S}^2\rightarrow \mathbb{R}^2

\displaystyle (x,y,z)\mapsto\frac{(u_1,u_2,-1)}{\lambda}

where {\lambda=\sqrt{u_1^2+u_2^2+1}}. Then one can see {u_1=\frac{x}{z}}, {u_2=\frac{y}{z}}. Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

new_lens

Let us pull the standard metric of {\mathbb{S}^2} to the {\mathbb{R}^2}. For the following statement, we will always omit {\phi^*} and {\phi_*}. Calculation shows,

\displaystyle dx=\left(\frac{1}{\lambda}-\frac{u_1^2}{\lambda^3}\right)du_1-\frac{u_1u_2}{\lambda^3}du_2=z(-1+x^2)du_1+xyzdu_2

\displaystyle dy=-\frac{u_1u_2}{\lambda^3}du_1+\left(\frac{1}{\lambda}-\frac{u_2^2}{\lambda^3}\right)du_2=xyzdu_1+z(-1+y^2)du_2

\displaystyle dz=\frac{1}{\lambda^3}(u_1du_1+u_2du_2)=z^2(xdu_1+ydu_2)

therefore

\displaystyle dx^2+dy^2+dz^2=z^2(1-x^2)du_1^2-2xyz^2du_1du_2+z^2(1-y^2)du^2_2

\displaystyle =\frac{1}{\lambda^2}(\delta_{ij}-\frac{u_iu_j}{\lambda^2})du_idu_j

here we used the fact that {(x,y,z)\in \mathbb{S}^2}. Suppose {\bar\nabla} is the connection on {\mathbb{S}^2} equipped with the standard metric. We want to calculate {\bar \nabla_{\partial_{u_i}}\partial_{u_j}}. To that end, it is better to use the {x,y,z} coordinates in {\mathbb{R}^3}

\displaystyle \partial_{u_1}=z(x^2-1)\partial_x+xyz\partial_y+xz^2\partial_z

\displaystyle \partial_{u_2}=xyz\partial_x+z(y^2-1)\partial_y+yz^2\partial_z

Since we know

\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=\left(\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}\right)^T=\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}-\langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle\partial_r

where {T} means the tangential part to {\mathbb{S}^2} and {\partial_r=x\partial_x+y\partial_y+z\partial_z} is the unit normal to {\mathbb{S}^2}. Using the connection in {\mathbb{R}^3}, we get

\displaystyle \nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}=z^2\left[3x(x^2-1)\partial_x+y(3x^2-1)\partial_y+z(3x^2-1)\partial_z\right]

\displaystyle \langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle=z^2(x^2-1)

One can verify from the above equalities that

\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=2xz\partial_{u_1}

Similarly

\displaystyle \bar \nabla_{\partial_{u_1}}\partial_{u_2}=yz\partial_{u_1}+xz\partial_{u_2}

\displaystyle \bar\nabla_{\partial u_2}\partial_{u_2}=2yz\partial_{u_2}

Frist and second variation of translating soliton

Suppose {F(x,t):\Sigma\times(-\epsilon,\epsilon)\rightarrow \mathbb{R}^{n+1}} be a variation of {\Sigma} with compact support and fixed boundary. Consider the weighted area functional

\displaystyle \mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}dv

What is the critical point of this area functional? To see that,

\displaystyle \frac{\partial}{\partial t}\mathcal{A}(\Sigma)=\int_\Sigma F_t(x_{n+1})e^{x_{n+1}}dv+e^{x_{n+1}}\partial_t dv

From some basic calculation in minimal surface(see colding minicozzi’s book)

\displaystyle \partial_t dv=-\langle F_t,\mathbf{H}\rangle+div_{\Sigma}F_t^T

Stokes’ theorem implies

\displaystyle \int_\Sigma div_\Sigma F_t^Te^{x_{n+1}}dv=-\int_{\Sigma}\langle F_t^T,\nabla_\Sigma x_{n+1}\rangle e^{x_{n+1}}dv=-\int_\Sigma F_t^T(x_{n+1})e^{x_{n+1}}dv

Combining these above fact, we get

\displaystyle \partial_t\mathcal{A}(\Sigma)=\int_\Sigma F_t^{\perp}(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma \langle F_t^{\perp},N\rangle N(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv

\displaystyle =\int_\Sigma \langle F_t,N\rangle \langle N,e_{n+1}\rangle-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma\langle F_t,\langle N,e_{n+1}\rangle N-\mathbf{H}\rangle dv

where {N} is the unit normal of {\Sigma}. Therefore the critical point of {\mathcal{A}} will satisfy

\displaystyle \mathbf{H}=\langle N, e_{n+1}\rangle N=e_{n+1}^{\perp}

This is so called translating soliton.

Now consider the second variation at a translation soliton,

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+2F_t(x_{n+1})^2dv+2F_t(x_{n+1})\partial_tdv+\partial_{tt}dv\right]

\displaystyle =\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+\partial_{tt}dv\right]

While

\displaystyle \partial_{tt}dv=-|\langle A(\cdot,\cdot),F_t\rangle|^2+|\nabla_\Sigma^NF_t|^2+div_\Sigma(F_{tt})

Suppose {F_t=\eta N}, where {\eta\in C_c^\infty(\Sigma)}, then {|\nabla_\Sigma^NF_t|^2=|\nabla \eta|^2}

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma [|\nabla\eta|^2-|A|^2\eta^2]e^{x_{n+1}}dv