# Ngô Quốc Anh

## October 12, 2014

### Construction of non-radial solutions for a Lichnerowicz type equation in the whole space

Filed under: Uncategorized — Ngô Quốc Anh @ 1:23

Given $q \in (0,1)$ and $N \geqslant 2$, in this note, we are interested in construction of non-radial solutions for the following Lichnerowicz type equation

$\displaystyle -\Delta u = -u^q + u^{-q-2}$

in the whole space $\mathbb R^N$.

In the previous post, we showed how to construct non-radial solutions of the following equation

$\displaystyle -\Delta u = -u^q.$

Clearly, this equation comes from the Lichnerowicz type equation by writing off the term with a negative exponent.

To start our construction and for simplicity, let us denote by $f$ the following

$\displaystyle f(t) = t^q - t^{-q-2},$

then a simple calculation shows $f'(t)=qt^{q-1} + (q+2)t^{-q-3}$ and $f''(t)=q(q-1)t^{q-2} - (q+2)(q+3)t^{-q-4}$. Hence, the function $f$ is monotone increasing in $[0,+\infty)$. Moreover, there exists a real number $a>0$ sufficiently large such that $f>0$ and $f$ is concave in $[a,+\infty)$. In addition, we can choose the number $a$ even large in such a way that

$\displaystyle \frac 1C f''(t) \leqslant f''(2t) \leqslant C f''(t)$

for some constant $C>0$.

As always, let us start with a radial solution, denoted by $u_0$, of the equation $\Delta u = f(u)$ such that $u_0 (0)=b>a$ and $u'_0(0)=0$. It is well-known that $u_0$ is globally defined and blows up at infinity. Moreover, the following estimate holds

$\displaystyle \frac 1N \sqrt{2F(u_0)} \leqslant \frac{du_0}{dr} \leqslant \sqrt{2F(u_0)}$

where $F(t) = \int_b^t f(s)ds$. Again, a non-radial solution $u$ that we are going to construct is of the following form

$\displaystyle u = u_0 + \varepsilon \big( v + \frac{\partial u_0}{\partial x_1} \big).$

For the sake of simplicity, we denote by $u_1$ the function $\partial u_0/\partial x_1$. That is, we seek for $v$ solving

$\displaystyle -\Delta v + f'(u_0)v = -\frac 1\varepsilon \big[ f(u_0+\varepsilon (u_1 + v)) -f(u_0) - f'(u_0) \varepsilon (u_1 +v)\big]$

which is equivalent to solving

$\displaystyle -\Delta v + f'(u_0)v = -\frac \varepsilon2 f''(u_0+t\varepsilon (u_1 + v)) (u_1 +v)^2$

in $\mathbb R^N$ where $t=t(u_0, u_1, v, \varepsilon) \in [0,1]$.

Since $f$ is concave, $\underline v \equiv 0$ is always a subsolution of the preceding equation. Let $\overline v = f(u_0)+1$. Then a direct calculation shows

$\displaystyle -\Delta \overline v + f'(u_0) \overline v = -f''(u_0) \left( \frac{du_0}{dr} \right)^2 + f'(u_0).$

Using the l’Hopital rule,

$\displaystyle\lim_{t \to +\infty} \frac{f^2(t)}{2F(t)} = \lim_{t \to +\infty} f'(t) =l \in [0,+\infty).$

In view of the inequalities for $du_0/dr$, we obtain

$\displaystyle(u_1 + \overline v)^2 \leqslant C \left( \frac{du_0}{dr} \right)^2$

for all $|x| \geqslant 1$. Using the l’Hopital rule again, we deduce that

$\displaystyle\lim_{t \to +\infty} \frac{2F(t)}{t^2} = \lim_{t \to +\infty} f'(t) = l \in [0, +\infty).$

Hence, $du_0/dr \leqslant Cu_0$. Hence, we have shown that

$\displaystyle -f''(u_0+t\varepsilon (u_1 +\overline v)) \leqslant -Cf''(u_0).$

Thus, $\overline v$ is a super-solution provided $\varepsilon >0$ is chosen sufficiently small; hence showing the existence of a solution $u = u_0+\varepsilon (u_1 + v)$ of our PDE with

$\displaystyle 0\leqslant v \leqslant f(u_0)+1.$

In case $l=0$, clearly $v=o(u_1)$ when $x=(x_1, 0)$ and $x_1 \to \pm \infty$. In case $l>0$, we may assume that $l<1/N^2$, hence upon a change of variable $y=\lambda x$, if necessary, we know that

$\displaystyle \limsup_{|x_1| \to \infty} \frac {v(x_1,0)}{|u_1 (x_1,0)|}<1.$

In conclusion, $u$ cannot be radial about any point in $\mathbb R^N$.