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.

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