It is well-known that the simplest form of the Nirenberg problem is equivalent to solving the following PDE
in . Using stereographic projection, one can see that the above PDE is equivalent to
in . If we assume that the solution has finite energy in the sense that
it is well-known that the preceding PDE has unique radial solution. In terms of ODE language, our PDE can be rewritten as
The purpose of this note is to find solutions to the above ODE. Our approach consists of several steps as shown below.
Step 1. Let . We then have which implies that
Substituting this into the ODE yields
Let , then we immediately have , thus giving us .
Step 2. In order to solve the latter ODE involving , we multiply both sides by . Hence, we arrive at
and by integrating, we know that
Thus,
since is non-negative and is positive. In other words,
which yields
We again integrate to get
which obviously helps us to write down the following
Solving for gives
Hence,
or equivalently,
Step 3. Now we condition our ODE with initial condition, say, and . Then it immediately yields
that is . We now use the condition to find . We first have
which implies that
Therefore, we can write
Hi Ngo, it’s true, often many PDE problems are reduced to an Ode problems, for insance think about the application of the moving plane method to proof the existence of radial solution as in the classical Gidas-Ni-Nirenberg work.
Comment by Fab — November 7, 2011 @ 19:40
If you have time, you can read several posts of mine at https://anhngq.wordpress.com/tag/moving-planes/ and at https://anhngq.wordpress.com/tag/moving-spheres/. The method of moving planes is nothing to do with the existence of radial symmetric solutions to equations. It is just to claim that under certain conditions, it is necessary to have such a symmetric result.
Comment by Ngô Quốc Anh — November 7, 2011 @ 19:43
hi, where do you find the proof?
Comment by prop — June 11, 2012 @ 22:08
I did by myself cause I have not found any proof for that.
Comment by Ngô Quốc Anh — June 12, 2012 @ 1:46