# 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

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