The energy of a semilinear ODE

Question

I'm currently reading Caffarelli, Gidas, Spruck's paper "Asymptotic Symmetry and Local Behavior of Semilinear Elliptic Equations with Critical Sobolev Growth". For some background, we consider a non-negative radial solution $u(x) = u(|x|)$ to the equation : $$ -\Delta u = u^\frac{n+2}{n-2}$$ in the punctured ball $ B_{1} -\{0\} $. Let $r = |x|$, $t= -\log(r)$ and $$ \psi(t) = r^\frac{n-2}{2}u(r).$$ We have the ODE for $\psi \geq 0$ : $$ \psi'' -\bigg(\frac{n-2}{2}\bigg)^2\psi + \psi^{\frac{n+2}{n-2}}= 0 , \hspace{0.5 cm} 0 < t < \infty \label{1}\tag{1.5}$$ and thus taking antiderivatives, we find that the energy $$ (1.6) \hspace{1 cm} D \equiv (\psi')^2 +\bigg(\frac{n-2}{n}\bigg)\psi^{\frac{2n}{n-2}} - \bigg(\frac{n-2}{2}\bigg)^2\psi^2 \label{2}\tag{1.6} $$ is a constant. The authors then state

1. that by the maximum principle, $\psi$ cannot vanish for any finite $t$ unless $\psi \equiv 0$, and
2. that this forces the estimate $$ 0 \geq D \geq -\bigg(\frac{2}{n}\bigg)\bigg(\frac{n-2}{n}\bigg)^n. $$

Both the assertions 1 and 2 are rather unclear to me (UPDATE: 1 is clear). Thus far, I've been able to show that if 1 holds, we get $$ D \geq -\bigg(\frac{2}{n}\bigg)\bigg(\frac{n-2}{2}\bigg)^n . $$ I did this by using $$ D \geq \bigg(\frac{n-2}{n}\bigg)\psi^{\frac{2n}{n-2}} - \bigg(\frac{n-2}{2}\bigg)^2\psi^2 $$ and taking the sup over $ \psi $ , but this is a different inequality than that in 2 as here we have a $2$ in the denominator instead of an $n$. The other inequality in 2 that $D$ must be non-positive is similarly unclear to me. Appreciate if someone can take a look.

Answer 1

The lower bound you found is the correct one (there is a typo in the paper). If you plug in the value of the specific constant solution described on page 273, with the value of $k$ found on page 272, you find indeed that the correct threshold energy should have a 2 in the denominator, and not $n$.

For the upper bound this is because if $D > 0$ then the constant energy condition (1.6) will require solutions to change signs in finite time.

This is easiest to see if you look at the plot of $f(\psi) = \left( \frac{n-2}2 \right)^2\psi^2 - \left(\frac{n-2}{n}\right)\psi^{2n/(n-2)}$. Below I include the case $n = 4$ for reference.

When $D$ is below the admissible range, this graph is shifted down so as to lie entirely below the horizontal axis. And therefore it is not possible for it (a negative value) to equal $(\psi')^2 \geq0$.

When $D$ in the admissible range and negative, you see that the shifted graph has two "lobes" that sit above the horizontal axis. They are disconnected, and hence solutions must remain along one of the two lobes. In fact, you see that the corresponding solutions are periodic in time, with $\psi$ bouncing between the two positive roots of $f(\psi) + D = 0$.

When $D = 0$, one has to be careful about the analysis since $f'(0) = 0$. But in the paper they provided an explicit solution in that case.

To rule out $D > 0$, let's look at the picture of $f(\psi) + D$. Remember that a solution must follow $(\psi')^2 = f(\psi) + D$.

At issue here is that a solution, starting with $\psi > 0$, will either have $\psi' < 0$ or $\psi' > 0$. In the latter case, in finite time it will hit the turn around point and then since $\psi'' = f'(\psi) < 0$ there it will start decreasing, entering the first case.

But now you see that $\psi' < 0$ and is bounded away from 0, so in finite time following the trajectory the solution must hit $\psi = 0$. (In fact, when $D > 0$ the solution must change signs infinitely many times, and periodically. So the restriction that $D \leq 0$ is entirely due to the assumption made initially that $u$ is a non-negative solution.)