It has a solution which is not holder continuous around origin. Note that the above equation is not uniformly elliptic.

See the paper, Trudinger, on the regularity of generalized solutions of linear, non-unformly elliptic equations, 1970.

Another example is and the function satisfy the elliptic equation in . This equation is degenerate elliptic. But is not an weight.

Check the paper Fabes Kenig and Serapioni, the local regularity of solutions of degenerate elliptic equations. 1982.

]]>One can see that maps onto the open ball . It is easy to verify that looks like

We want to pull the metric of to , that is . Denote . We have

Therefore, is conformal to .

Next, for some , we want to pull the solution of

to . That is defining on . We shall derive the equation satisfy on . Note that the equation means

Let us use the following notations

It follows from the covariant property of conformal laplacian that for any ,

Note that . We turn to calculate

It is not hard to see

Therefore

Next, to handle the term , applying ,

That is

It follows from the transformation formula that

To summarize the above calculation, one the one hand,

on the other hand

Then we get the equation of

Notice that where is the distance of to the center of the the ball . The equation that satisfies is rotationally symmetric with respect to the center of .

]]>We want to find the Green’s function for , namely

in distribution sense, or equivalently for any ,

Moreover looks like the inverse operator of . Of course such would only be unique modulo some constants.

To do that, we know on , the Green function is

In a general Riemannian manifold, should look like more or less.

Suppose has injectivity radius and is a smooth radial cut off function on whose support lies in and in . Then is well defined on through local coordinate patches, which we still deonte as . Easy computation shows

Green’s formula reads

for all . This actually means

and

After changing the order of integration in the Green’s formula, we could establish that

for any . One can compare this with the equation satisfy. Denote . Define two operators

Also denote . Then Green’s formula means

To find the Green’s function, it is equivalent to find the inverse operator of . One can apply the idea of parametrix, which says the inverse operator can be constructed by

Therefore if we define

Denote and . Then . It is easy to get

then

Consequently

Now let us see the expansion of and . To do that, we need to formula of laplacian

which works for any local coordinates. Since near and , then and

Remark 1We have .

On the other hand, we have . Consequently and

Then

It follows from the expansion of that

Putting all these back to , it yields

Recall the equation of , it leads to

Since ,

then

Moreover, since , because , then

One can use the (3-5) in \cite{Case2017} and on to get

Remark:

Jeffrey S Case. Some energy inequalities involving fractional GJMS operators. Analysis and PDE , 10(2):253–280, 2017.

Jeffrey S Case and Sun-Yung Alice Chang. On Fractional GJMS Operators. Communications on Pure and Applied Mathematics , 69(6):1017–1061, 2016

Many thanks to the help of Jeffrey S Case. The dimension of is and .

]]>Suppose , for any , which means that is a regular hypersurface of . Define the unit normal vector field

and

Then it is easy to the flow will satisfy our request.

For example let be the Paneitz operator of fourth order. Assume it is a positive operator. Define

Choose our and

then and . Consequently

Then one may have the following flow

where .

P. Baird, A. Fardoun, R. Regbaoui, Calculus of Variations Q-curvature flow on 4-manifolds, 27 (2006) 75–104.

M. Gursky, A. Malchiodi, A strong maximum principle for the Paneitz operator and a non-local flow for the $Q$-curvature, J. Eur. Math. Soc. 17 (2015) 2137–2173.

]]>Here is a Riemannian metric with , is the Schouten tensor and is .

Now let us differentiate the two terms respectively, suppose . From conformal change property of scalar curvature, we get

Then

where we have used the fact . Next, it is easy to know

So

Now assume keep the unit volume infinitesimally, that is , then under this constraint means

**Remark: **M.J. Gursky, F. Hang, Y.-J. Lin, Riemannian Manifolds with Positive Yamabe Invariant and Paneitz Operator, Int. Math. Res. Not.

Assume is jointly concave with respect to and

Then

can not achieve positive maximum in the interior of .

See the paper Korevaar 1983

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

Translator is a special type solution of mean curvature flow such that it moves by translation. Namely, if is a translator moves to direction , then satisfies,

The mean curvature of a translator satisfies and

where 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 is that diffeomorphism (). Then satisfies (1). Since is also translator, . Using ,

Since , (2) will imply

This just means (4) holds.

]]>where is the normal vector of , that is . What is the mean curvature of ?

We can parametrize the cylinder by where , , , . One can write as

Then the induced metric on is

Since

The normal vector of is

The second fundamental form is

Then the mean curvature is

If and , then

]]>

where . is called the angle of the conical singularity. We will always assume .

For example equipped with the metric is isometric to an Euclidean cone of total angle .

Suppose now is a closed Riemann surface with conical metric . **Assume that the Gauss curvature extends on as a H\”{o}lder continuous function.**

Suppose is a conformal change of the conical metric. Then necessarily we have . In order to prescribe the Gauss curvature , we need to solve the equation

.

Any reasonable solution of (0) will satisfy

where follows from the Gauss-Bonnet formula for conical surface. We will use variational method to attack it. To that end, define to be Sobolev space of functions with

and functionals

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

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

(a)

(b) minimizer exists

The first question follows from the Trudinger inequality. Namely, define . Then

**Lemma:** For any fixed , there exist constant such that,

for any and and .

**Lemma:** Suppose and . Let . Then for any fixed , there exists constants such that

for any and . Here .

From this key lemma, one can derive that is lower bounded on provided and . To do that, choose and such that , then it follows from the above lemma that

For the second question, we need to prove the embedding is compact(actually it is true for any for any ). Note that this is true if is some smooth metric, however is conical one. Therefore, we want to compare with of some smooth metric.

Suppose is equipped with smooth metric such that . Here is smooth and positive outside the support of and near . If we use the denote the gradient of with respect to , then and

This is only true in two dimension. Now and have inner product

From the above analysis, we need to examine the difference of and . It follows from the singular behavior of that

for any and some . Then for any ,

Then . On the contrary, we need the following inequality

**Lemma: **For conical metric

Then we have . Therefore is compact for . Furthermore, after some effort, one can prove

is also compact for .

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.

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