Category Archives: Geometry

63411

Evolution equations for some geometric flows

March 13, 2020 – 8:37 pm

Suppose {M^n} is the {n}-dimensional strictly convex hypersurface embedded in {\mathbb{R}^{n+1}}. Denote {\boldsymbol{X}:M^n\times[0,T]\rightarrow \mathbb{R}^{n+1}} be the position vector of the embedding. Suppose {M^n} flows by the following form

\displaystyle \frac{\partial}{\partial t}\boldsymbol{X}(x,t)=-F(\mathscr{W}(x,t),\nu(x,t))\nu(x,t)

where {\nu(x,t):M^n\times[0,T)\rightarrow \mathbb{S}^n} is a unit normal to the hypersurface {\boldsymbol{X}(M^n,t)} at the point {\boldsymbol{X}(x,t)} and {\mathscr{W}} is the Weingarten map.

In the following we want to derive some evolution equations for some geometric quantities. Let us use {g_{ij}} to denote the metric of {M^n}, {h_{ij}} denote the second fundamental form.

Since {g_{ij}=\langle \partial_i \boldsymbol{X}, \partial_j \boldsymbol{X}\rangle} 

\displaystyle \partial_t g_{ij}=-2Fh_{ij}

\displaystyle \partial_t g^{ij}=2Fh^{ij}

Since {h_{ij}=-\partial_i\partial_j \boldsymbol{X}\cdot \nu},

\displaystyle \partial_t h_{ij}=-\partial_i\partial_j\partial_t \boldsymbol{X}\cdot \nu=\partial_i\partial_j F+F\partial_i\nu\partial_j\nu

Write in tensor sense

\displaystyle \partial_t h_{ij}=\nabla_i\nabla_j F-Fh_{ik}h^{k}_j

\displaystyle \partial_t h_i^j=g^{jk}\nabla_i\nabla_k F+Fh^{jk}h_{ki}

Since the Gauss curvature {K=\det (h_i^j)}, we can compute

\displaystyle \partial_t K(x,t)=K(h^{-1})^i_j\partial_t h_i^j=K(h^{-1})^{ik}\nabla_i\nabla_kF+FKH

One can compute the special case of {\alpha}-Gauss curvature flow {F=K^\alpha}.

Consider the flow from the Gauss map parametrization. Let us use {s} denote the support function of {\boldsymbol{X}(M^n,t)} and {\mathfrak{r}[s]= Hess_{\bar \nabla }s+\bar g s}.

\displaystyle \frac{\partial}{\partial t}s(z,t)=\Phi(\mathfrak{r}[s](z,t),t)

where {\Phi(X)=-F(X^{-1})}. 

Suppose we are dealing with Gauss curvature flow {F=K}, then {\Phi=-(\det\mathfrak{r}[s])^{-1}}. 

\displaystyle \partial_t s=-(\det\mathfrak{r}[s])^{-1}

\displaystyle \partial_t\Phi(z,t)=(\det\mathfrak{r}[s])^{-1}[((\mathfrak{r}(s))^{-1})^{ij}(\bar\nabla_i\bar\nabla_j\Phi+\bar g_{ij}\Phi)]=\mathcal{L}\Phi-\Phi^2Hwhere {\mathcal{L}=\dot \Phi^{ij}\bar\nabla_i\bar\nabla_j= (\det\mathfrak{r}[s])^{-1}((\mathfrak{r}[s])^{-1})^{ij}\bar \nabla_i\bar\nabla_j}.

If we use normalized flow 

\displaystyle \partial_\tau \tilde s=\tilde s-(\det\mathfrak{r}[\tilde s])^{-1}=\tilde s-\tilde K

Since 

\displaystyle \mathcal{L}\tilde s=\tilde K((\mathfrak{r}[\tilde s])^{-1})^{ij}\bar \nabla_i\bar\nabla_j\tilde s=\tilde K ((\mathfrak{r}[s])^{-1})^{ij}\mathfrak{r}[\tilde s]_{ij}-\tilde K\tilde H\tilde s=n\tilde K-\tilde K\tilde H\tilde s

then we get

\displaystyle (\partial_\tau-\mathcal{L})\tilde s=-(n+1)\tilde K+\tilde s+\tilde K\tilde H \tilde s

Similarly

\displaystyle \partial_\tau \tilde K=\partial_\tau (\det\mathfrak{r}[\tilde s])^{-1}=-\mathcal{L}(\tilde s-\tilde K)-\tilde K\tilde H(\tilde s-\tilde H)

Using the previous relation on {\mathcal{L}\tilde s}, we get

\displaystyle (\partial_\tau-\mathcal{L})\tilde K=-n\tilde K+\tilde K^2\tilde H.

In the case of {\alpha}-Gauss curvature flow, we have {\Phi=-(\det\mathfrak{r}[s])^{-\alpha}}. 

\displaystyle \partial_t s=-(\det\mathfrak{r}[s])^{-\alpha}

\displaystyle \partial_t \Phi=\alpha (\det\mathfrak{r}[s])^{-\alpha}[((\mathfrak{r}[s])^{-1})^{ij}(\bar\nabla_i\bar\nabla_j\Phi+\bar g_{ij}\Phi)]

The normalized flow is 

\displaystyle \partial_\tau \tilde s=\tilde s-\frac{\tilde K^\alpha}{\eta}

where {\eta=\int_{\mathbb{S}^n}\tilde K^{\alpha-1}d\theta/|\mathbb{S}^n|} depends on {\tau} only. 

Then {\mathcal{L}=\frac{\alpha}{\eta}\tilde K^\alpha((\mathfrak{r}[s])^{-1})^{ij}\bar\nabla_i\bar\nabla_j\ }

\displaystyle \mathcal{L}\tilde s=n\alpha\frac{\tilde K^\alpha}{\eta}-\alpha \tilde H\frac{\tilde K^\alpha}{\eta}\tilde s

Then 

\displaystyle (\partial_\tau-\mathcal{L})\tilde s=-(1+n\alpha)\frac{\tilde K^\alpha}{\eta}+\tilde s+\alpha \tilde H\frac{\tilde K^\alpha}{\eta}\tilde s

\displaystyle (\partial_\tau-\mathcal{L})\tilde K^\alpha=-n\alpha \tilde K^\alpha+\alpha \tilde H \frac{\tilde K^{2\alpha}}{\eta}

By Sun's World | Also posted in Mean curvature flow | Comments (0)

Harnack inequality for convex hypersurface

March 12, 2020 – 6:15 pm

Consider the {\alpha}-Gauss curvature flow of a strictly convex hypersuface {\boldsymbol{X}_0} in {\mathbb{R}^{n+1}}

\displaystyle \partial_t \boldsymbol{X}(p,t)=-K^\alpha(p,t)\boldsymbol{N}

Denote the support function of {\boldsymbol{X}} is {s} and {\mathfrak{r}[s]=Hess(s)+Id\cdot s}. Since {\mathfrak{r}[s]=\mathscr{W}^{-1}}. The above flow can be rewritten using the Gauss map parametrization,

\displaystyle \partial_t s=-\mathfrak{r}[s]^{-\alpha}

If we use the notation in [1], {F(X)=(\det X)^\alpha}, {\Phi(X)=-(\det X)^{-\alpha}=sgn(-n\alpha)B^{-n\alpha}} for every positive definite map {X}. 

\displaystyle \Phi(X)=-(\det X)^{-\alpha}=sgn(-n\alpha)B^{-n\alpha}

where {B(X)=(\det X)^{\frac{1}{n}}} is concave. {\Phi} is called {(-n\alpha)}-concave.

\displaystyle \partial_t \Phi+\frac{-n\alpha\Phi}{(-n\alpha-1)t}\leq 0

that is 

\displaystyle \partial_t(\Phi t^{\frac{n\alpha}{n\alpha+1}})\leq 0

Since {\Phi(z,t)=-K^{\alpha}(z,t)}, then

\displaystyle \partial_t (K^{\alpha}(z,t)t^{\frac{n\alpha}{n\alpha+1}})\geq 0

[1] B. Andrews, Harnack inequalities for evolving hypersurfaces, Math. Zeitschrift. 217 (1994) 179–197. doi:10.1007/BF02571941.

By Sun's World | Also posted in Mean curvature flow | Comments (0)

Some special solutions to curve shortening flow

December 2, 2019 – 2:01 pm

Consider {\boldsymbol{\gamma}(p,t)} is a closed curve in {\mathbb{R}^2} which moves by its curvature. More preciesly, let

\displaystyle \partial_t\boldsymbol{\gamma}(p,t)=\kappa \mathbf{N}

where {\mathbf{N}} is the unit normal pointing to the inside of the region inscribed by the curve. Such a flow is curve shortening flow. We are interested in some particular (convex, graphical) solutions of the flow.

Suppose {\boldsymbol{\gamma}} locally can written as a graph, say {\boldsymbol{\gamma}=(x,u(x,t))}. Then the flow equation means

\displaystyle u_t=\frac{u_{xx}}{1+u_x^2}

Consider {u(x,t)=g(\lambda t+f(x))} in particular, where {\lambda} is a constant. Then we can decouple the equation into two differential equations.

\displaystyle f''+\mu f'^2-\lambda=0,

\displaystyle g''-\lambda g'^3-\mu g'=0.

where {\mu} is a constant. Depending on the values of {\lambda} and {\mu}, one can have the following four cases

(1) {\lambda=0}, {\mu=0}. Then {u(x,t)=ax} and {\boldsymbol{\gamma}} is actually a static line.

(2) {\lambda\neq 0}, {\mu=0}. Then one can choose {f=\frac{1}{2}\lambda x^2} and {g(x)=\frac{1}{2}\sqrt{\frac{x}{-2\lambda}}}. In this case

\displaystyle u^2+\frac{1}{4}x^2=-\frac{1}{2}t

{\boldsymbol{\gamma}} is a circle.

(3) {\lambda\mu<0}. Say {\lambda=-1} and {\mu=1}. Then one can solve the ODE to get {f(x)=\log\cos x} and {g(x)=const}, {\sinh^{-1}(e^x)}, {\pm \cosh^{-1}(e^x)}. They corresponds to grim reaper, hair clip and paper clip (Angenent oval) respectively.

u=\log\cos x-t

\sinh u=e^{-t}\cos x,\quad \cosh u=e^{-t}\cos x

(4) {\lambda\mu>0}. This can be reduced to (3) by writing 

\displaystyle y=u(x,t),\quad v(y,t)=f^{-1}(-\lambda t+g^{-1}(y)).

K. Nakayama, T. Iizuka, M. Wadati, Curve Lengthing equation and its solutions, J. Phys. Soc. Japan. 63 (1994) 1311-1321.

By Sun's World | Also posted in Mean curvature flow | Tagged | Comments (0)

Smoothness of principal curvatures and principal vectors of a hypersurface

August 23, 2019 – 7:14 pm

Given a C^\infty hypersurface M^n immersed in \mathbb{R}^{n+1}, one can not find a set of C^\infty vector fields such that the second fundamental form of M is diagonalized near a given point.

The right statement is: the principal curvatures are continuous on all of M and C^\infty just on an open, dense subset of M. In a neighborhood of any point in this open, dense subset, one can find a desired C^\infty principal vector fields.

Check the paper: D.H. Singley, Smoothness theorems for the principal curvatures and principal vectors of a hypersurface, Rocky Mt. J. Math. 5 (1975) 135–144. doi:10.1216/RMJ-1975-5-1-135.

By Sun's World | Also posted in Riemannian Geo, Uncategorized | Comments (0)

Convex function, Legendre transform and support function

July 8, 2019 – 3:40 pm

Suppose {u} is a (smooth) locally strictly convex function on {\mathbb{R}^n}, {\phi} is the Legendre transform of {u}. Namely, suppose {u=u(x)} and {y=Du(x)}, then 

\displaystyle \phi(y)=x\cdot Du(x)-u(x),\quad x=D\phi(y)

It follows that the Hessian {D^2u=(D^2\phi)^{-1}}. One can verify that the Legendre transformation of {\phi} is {u}. 

Now consider the graph of {u} by {\Sigma=(x,u(x))}. The downward unit normal of {\Sigma} is 

\displaystyle \nu(x)=\frac{(Du,-1)}{\sqrt{1+|Du|^2}}=\frac{(y,-1)}{\sqrt{1+|y|^2}}.

One can think of {\nu} as the Gauss map of {\Sigma} to {\mathbb{S}^n}. Denote {z=\nu(x)} the coordinates in {\mathbb{S}^n} and {X=(x,u(x))} is the position vector of {\Sigma}. Then the support function of Gaussian mapping is 

\displaystyle w(z)=X\cdot \nu=\frac{\phi(y)}{\sqrt{1+|y|^2}}.

One can also interpret this relation as the following: treat {\phi} is defined over the hyperplane {(y,-1)\subset \mathbb{R}^{n+1}}. Then the homogeneous degree one extension of {w(z)} restricted to this plane is just {\phi(y)=\sqrt{1+|y|^2}w(z)}.

Relation of three functions

One also recover {\Sigma} by just the support function {w(z)}. Since {\Sigma} has nonvanishing curvature, the Gauss map is invertible. Let {X=\nu^{-1}(z)} be its inverse. Then 

\displaystyle w(z)=z\cdot \nu^{-1}(z).

It is easy to show that 

\displaystyle \overline{\nabla}_e w(z)=e\cdot \nu^{-1}(z)

for any tangent vector of {\mathbb{S}^n}. Here {\overline{\nabla}= \nabla^{\mathbb{S}^n}}. This tells us that {\overline{\nabla}w(z)} is the projection of {\nu^{-1}(z)} onto the tangent plane 

\displaystyle X=\nu^{-1}(z)=w(z)z+\overline{\nabla}w(z).

Then {d\nu^{-1}(z)=d\overline{\nabla} w+wI}. Therefore, the Weingarten map of {d\nu=(d\overline{\nabla} w+wI)^{-1}}. The eigenvalues of {d\overline{\nabla} w+wI} are the reciprocals of the principle curvatures of {\Sigma}.

There is another way to verify this fact. Suppose the metric on {\Sigma} induced from {\mathbb{R}^{n+1}} is 

\displaystyle g=g_{ij}dx^idx^j,\quad g_{ij}=(\delta_{ij}+u_iu_j)

where {u_i=\partial_{x_i}u}. Denote {\mu=\sqrt{1+|Du|^2}}. Then {g^{ij}=\delta_{ij}-\mu^{-2}u_iu_j}. The second fundamental form of {\Sigma} is {A_{ij}=\mu^{-1}u_{ij}}. Then the principle curvature {\kappa_1,\cdots,\kappa_n} of {\Sigma} are the eigenvalues of the matrix 

\displaystyle g^{ik}A_{kj}=\frac{g^{ik}u_{kj}}{\mu}

It is equivalent to say {\frac{1}{\kappa_1},\cdots, \frac{1}{\kappa_n}} are the eigenvalues of {\mu \phi_{ik}g_{ij}=\mu\phi_{ik}(1+y_ky_j)}

Consider the Stereographic projection from origin (see here), one can prove that the metric of {\mathbb{S}^n} reads 

\displaystyle \sigma^{ij}={\mu^2}(\delta_{ij}+y_iy_j)

and more importantly 

\displaystyle \phi_{y_iy_j}=\mu(\overline{\nabla}_{ij}w+w\sigma_{ij})

then {\mu \phi_{ik}g_{ij}=\sigma^{ik}\phi_{kj}\mu^{-1}=\overline{\nabla}^i_jw+w\delta^i_j}. This just means {\frac{1}{\kappa_1},\cdots, \frac{1}{\kappa_n}} are the eigenvalues of {d\overline{\nabla}w+wI}.

By Sun's World | Also posted in Riemannian Geo | Comments (0)

Upper half plane and disc

April 16, 2019 – 7:50 pm

Suppose {\mathbb{R}^{n+1}_+=\{(y,t)|y\in \mathbb{R}^n,t>0\}} is the upper half plane. Define the map {\pi: \mathbb{R}^{n+1}_+=\{(y,t)\}\rightarrow \mathbb{R}^{n+1}=\{(x,s)\}} by the following 

\displaystyle x=\frac{y}{|y|^2+(t+1)^2}

\displaystyle s=\frac{t+1}{|y|^2+(t+1)^2}-1

One can see that {\pi} maps {\mathbb{R}_+^n} onto the open ball {B=B_{\frac12}((0,-\frac12))\subset \mathbb{R}^{n+1}}. It is easy to verify that {\pi^{-1}} looks like 

\displaystyle y=\frac{x}{|x|^2+(s+1)^2}

\displaystyle t=\frac{s+1}{|x|^2+(s+1)^2}-1

Upper half plane and disc

We want to pull the metric of {\mathbb{R}^{n+1}_+} to {B}, that is {(\pi^{-1})^*(dy^2+dt^2)}. Denote {A=|x|^2+(s+1)^2}. We have 

\displaystyle (\pi^{-1})^*dy_i=A^{-1}dx_i-A^{-2}x_idA

\displaystyle (\pi^{-1})^*dt=A^{-1}ds-A^{-2}(s+1)dA

\displaystyle (\pi^{-1})^*(dy^2+dt^2)=A^{-2}(dx^2+ds^2)

Therefore, {(\pi^{-1})^*(dy^2+dt^2)} is conformal to {dx^2+ds^2}. 

Next, for some {\alpha\geq 0}, we want to pull the solution {u} of 

\displaystyle -\text{div}(t^\alpha \nabla u)=\alpha(n+\alpha-1)t^{\alpha-1}u^{\frac{n+\alpha+1}{n+\alpha-1}}

to {B}. That is defining {\psi(x,s)=A^{-\frac{n+\alpha-1}{2}}u} on {B}. We shall derive the equation {\psi} satisfy on {B}. Note that the equation means 

\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=-\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}.

Let us use the following notations 

\displaystyle \beta=n+\alpha-1.

It follows from the covariant property of conformal laplacian that for any {u=u(y,t)}, 

\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{-\frac{n-1}{2}}u).

Note that {A^{-\frac{n-1}{2}}u=A^{\frac{\alpha}{2}}\psi}. We turn to calculate 

\displaystyle \Delta_{(x,t)}(A^{\frac{\alpha}{2}}\psi)=(\Delta A^{\frac{\alpha}{2}})\psi+A^{\frac{\alpha}{2}}\Delta\psi+2\nabla A^{\frac{\alpha}{2}}\nabla \psi.

It is not hard to see 

\displaystyle \Delta A^{\frac{\alpha}{2}}=\alpha\beta A^{\frac{\alpha}{2}-1},\quad 2\nabla A^{\frac{\alpha}{2}}\nabla \psi=2\alpha A^{\frac{\alpha}{2}-1}\nabla\psi\cdot(x,s+1).

Therefore 

\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{\frac{\alpha}{2}}\psi)=A^{\frac{\beta+4}{2}}\Delta\psi+2\alpha A^{\frac{\beta+2}{2}}\nabla \psi\cdot(x,s+1)+\alpha\beta A^{\frac{\beta+2}{2}}\psi.

Next, to handle the term {\partial_t u}, applying {\partial_t A=-2(1+t)A^{2}=-2(1+s)A^{3}}, 

\displaystyle \partial_t u=\partial_t(A^{\frac{\beta}{2}}\psi)=-\beta(1+t)A^{\frac{\beta+2}{2}}\psi+A^{\frac{\beta}{2}}\partial_x \psi[-2y(1+t)A^2]+

\displaystyle A^{\frac{\beta}{2}}\partial_s\psi[A-2(1+t)^2A^2].

That is 

\displaystyle \partial_t u=A^{\frac{\beta}{2}}(1+s)[-\beta\psi-2x\partial_x\psi-2(1+s)\partial_s\psi]+A^{\frac{\beta+2}{2}}\partial_s\psi

\displaystyle =-A^{\frac{\beta}{2}}(1+s)[\beta\psi+2\nabla\psi\cdot(x,s+\frac12)]+A^{\frac{\beta}{2}}[A-(1+s)]\partial_s\psi.

It follows from the transformation formula that 

\displaystyle \frac{\alpha}{t}\partial_t u=\frac{\alpha A}{s+1-A}\partial_t u

To summarize the above calculation, one the one hand, 

\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=A^{\frac{\beta+4}{2}}\Delta\psi-\frac{2\alpha}{s+1-A}A^{\frac{\beta+4}{2}}\nabla\psi\cdot(x,s+\frac12)-\frac{\alpha\beta}{s+1-A}A^\frac{\beta+4}{2}\psi

on the other hand 

\displaystyle -\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}=-\alpha\beta\frac{A}{s+1-A}A^{\frac{n+\alpha+1}{2}}\psi^{\frac{n+\alpha+1}{n+\alpha-1}}.

Then we get the equation of {\psi} 

\displaystyle \Delta \psi-\frac{2\alpha \nabla\psi\cdot(x,s+\frac12)}{s+1-A}=\alpha\beta\frac{\psi-\psi^{\frac{n+\alpha+1}{n+\alpha-1}}}{s+1-A}.

Notice that {s+1-A=\frac14-r^2} where {r=\sqrt{|x|^2+(s+\frac12)^2}} is the distance of {(x,s)} to the center {(0,-\frac12)} of the the ball {B}. The equation that {\psi} satisfies is rotationally symmetric with respect to the center of {B}. 

By Sun's World | Also posted in Elliptic PDE, Riemannian Geo | Tagged | Comments (0)

Mean curvature equations of translators

January 11, 2019 – 2:51 am

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.

By Sun's World | Also posted in Mean curvature flow, Uncategorized | Comments (0)

Mean curvature of a graph on cylinder

November 12, 2018 – 8:12 pm

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

 

By Sun's World | Also posted in Mean curvature flow | Comments (0)

Surface intersection with a ball

October 23, 2018 – 7:47 pm

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.

nose

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.

 

 

By Sun's World | Also posted in Riemannian Geo | Comments (1)

Hessian of radial functions

September 23, 2018 – 2:15 am

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''

By Sun's World | Also posted in Riemannian Geo | Tagged | Comments (0)
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