# Ngô Quốc Anh

## November 1, 2011

### An ODE appearing in the Nirenberg problem

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 20:21

It is well-known that the simplest form of the Nirenberg problem is equivalent to solving the following PDE

$-\Delta u + 2= e^u$

in $\mathbb S^2$. Using stereographic projection, one can see that the above PDE is equivalent to

$-\Delta u = e^u$

in $\mathbb R^2$. If we assume that the solution $u$ has finite energy in the sense that

$\displaystyle \int_{\mathbb R^2} u <+\infty,$

it is well-known that the preceding PDE has unique radial solution. In terms of ODE language, our PDE can be rewritten as

$\displaystyle -u''(r)-\frac{1}{r}u'(r)=e^{u(r)},\quad r\geqslant 0.$

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 $r = e^z$. We then have $u(r)=u(e^z)=v(z)$ which implies that

$\displaystyle u'(r) = e^{-z}v'(z), \quad u''(r) = e^{-2z}(v''(z)-v'(z)).$

Substituting this into the ODE yields

$\displaystyle v''(z) = -e^{v(z)+2z}.$

Let $w(z) = v(z)+2z$, then we immediately have $w''(z) = v''(z)$, thus giving us $w'' = -e^{w}$.

Step 2. In order to solve the latter ODE involving $w$, we multiply both sides by $w'$. Hence, we arrive at

$w' w'' = -e^w w'$

and by integrating, we know that

$\displaystyle \frac{1}{2}\int d((w')^2)=\int w' d(w')=\int w' w'' dz =-\int e^w w' dz=-\int d(e^w).$

Thus,

$\displaystyle \dfrac{1}{2}(w')^2 = -e^w + C_1^2$

since $\frac{1}{2}(w')^2$ is non-negative and $e^w$ is positive. In other words,

$\displaystyle (w')^2 = 2C_1^2 - 2e^w,$

which yields

$\displaystyle\dfrac{dw}{dz} = \sqrt{2C_1^2 - 2e^w}$

We again integrate to get

$\displaystyle\int \dfrac{dw}{\sqrt{C_1^2 - e^w}} = \int \sqrt{2}dz$

which obviously helps us to write down the following

$\displaystyle \frac{1}{C_1}\ln \frac{{{C_1} - \sqrt {C_1^2 - {e^w}} }}{{{C_1} + \sqrt {C_1^2 - {e^w}} }} = z\sqrt 2 + {C_2}.$

Solving for $w$ gives

$\displaystyle w = 2\ln \frac{{2{C_1}{e^{\frac{{\sqrt 2 {C_1}z + {C_1}{C_2}}}{2}}}}}{{1 + {e^{\sqrt 2 {C_1}z + {C_1}{C_2}}}}}.$

Hence,

$\displaystyle u(r) = w(z) - 2\ln r = 2\ln \frac{{2{C_1}{e^{\frac{{\sqrt 2 {C_1}\ln r + {C_1}{C_2}}}{2}}}}}{{1 + {e^{\sqrt 2 {C_1}\ln r + {C_1}{C_2}}}}} - 2\ln r,$

or equivalently,

$\displaystyle u(r) = 2\ln \frac{{2{C_1}{e^{\frac{{{C_1}{C_2}}}{2}}}{r^{\frac{{\sqrt 2 {C_1}}}{2} - 1}}}}{{1 + {e^{{C_1}{C_2}}}{r^{\sqrt 2 {C_1}}}}}.$

Step 3. Now we condition our ODE with initial condition, say, $u'(0)=0$ and $u(0)=C$. Then it immediately yields

$\displaystyle \frac{{\sqrt 2 {C_1}}}{2} - 1 = 0,$

that is $C_1=\sqrt 2$. We now use the condition $u(0)=C$ to find $C_2$. We first have

$\displaystyle u(r) = 2\ln \frac{{2\sqrt 2{e^{\frac{{{C_2}}}{{\sqrt 2 }}}}}}{{1 + {e^{\sqrt 2 {C_2}}}{r^2}}}$

which implies that

$\displaystyle u(r) = 2\ln \frac{{\sqrt 2\lambda }}{{1 + \frac{{{\lambda ^2}}}{4}{r^2}}}, \quad \lambda=2 e^{\frac{sqrt 2 C_2}{2}}.$

Therefore, we can write

$\displaystyle u(r) = 2\ln \frac{{4\sqrt 2\lambda }}{{4 + {\lambda ^2}{r^2}}}=\ln \frac{{32{\lambda ^2}}}{{{{(4 + {\lambda ^2}{r^2})}^2}}}.$

1. 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

2. 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

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