Ngô Quốc Anh

September 11, 2014

Construction of non-radial solutions for the equation Δu=u^q with 0<q<1 in the whole space

Filed under: Uncategorized — Ngô Quốc Anh @ 3:01

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

$\displaystyle \Delta u = u^q$

in the whole space $\mathbb R^N$. The construction is basically due to Louis Dupaigne and mainly depends on the unique radial solution of the equation.

To start our construction, let us recall that there is a unique radial solution, denoted by $u_0$, of the equation $\Delta u = u^q$ such that $u_0 (0)=1$ and $u'_0(0)=0$. Moreover, $u_0$ is globally defined and blows up at infinity at a fixed rate

$\displaystyle \lim_{r \to +\infty} \frac{u_0(r)}{r^\alpha} = L$

where $\alpha = \frac{2}{1-q}$ and $L=[\alpha (\alpha + N-2)]^{-1/(q-1)}>0$, see a paper by Yang and Guo published in J. Partial Diff. Eqns. in 2005.

Notice that

$\displaystyle \Delta u_0 = r^{1-N} (r^{N-1} u'_0)'.$

Hence, integrating both sides of the equation for $u_0$ gives

$\displaystyle\frac{du_0}{dr} = r^{1-N} \int_0^r t^{N-1} u_0^q dt$

from which it follows that

$\displaystyle \lim_{r \to +\infty} r^{1-\alpha}\frac{du_0}{dr}=L\alpha.$

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 $\frac{\partial u_0}{\partial x_1}$. Clearly, $u_1$ solves the linearized equation

$\displaystyle -\Delta u_1 = -qu_0^{q-1}u_1$

in $\mathbb R^N$. Since $r = (x_1^2 + \cdots x_N^2)^{1/2}$, we know that $\frac{\partial r}{\partial x_1} = \frac{x_1}{r}$. Hence, we can conclude that

$\displaystyle u_1 = \frac{\partial u_0}{\partial x_1} = \frac{du_0}{dr} \frac{\partial r}{\partial x_1}=\frac{du_0}{dr} \frac{x_1}{r} = o(u_0)$

as $|x| \to \infty$. Thanks to $\Delta u = \Delta u_0 + \varepsilon \big( \Delta u_1 + \Delta v \big)$, the function $v$ needs to satisfy the following

$\displaystyle - \Delta v + qu_0^{q - 1}v = - \frac{1}{\varepsilon } \big( {({u_0} + \varepsilon ({u_1} + v))^q} - u_0^q - qu_0^{q - 1}\varepsilon ({u_1} + v) \big).$

Using the Taylor expansion, we further obtain

$\displaystyle - \Delta v + qu_0^{q - 1}v = \frac{1}{\varepsilon }\frac{{q(1 - q)}}{2}{({u_0} + t\varepsilon ({u_1} + v))^{q - 2}}{({u_1} + v)^2}$

in $\mathbb R^N$ for some $t = t(u_0, u_1, v, \varepsilon) \in [0,1]$. Hence, our aim is to solve the preceding equation for $v$. To this purpose, we shall use the method of sub-super solutions. Clearly, $\underline v\equiv 0$ is always a sub-solution. A direct calculation shows $\overline v = u_0^q+1$ which solves

$\displaystyle - \Delta \overline v + qu_0^{q - 1}\overline v = q(1 - q)u_0^{q - 2}{\left( {\frac{{d{u_0}}}{{dr}}} \right)^2} + qu_0^{q - 1},$

could be a super-solution provided $\varepsilon>0$ is sufficiently small.

Hence, for fixed $R>0$, we can always find a function $v_R \in [\underline v, \overline v]$ in $B(0, R)$. By elliptic regularity, up to a subsequence, $(v_R)_R$ converges in $C_{\rm loc}^2 (\mathbb R^N)$ to a solution $v$ of the previous PDE. Moreover, there holds

$\displaystyle 0 \leqslant v \leqslant u_0^q +1$

in $\mathbb R^N$.

Observe that, when $x=(x_1, 0)$ and $x_1 \to \pm \infty$, there holds $v =o(u_1)$; hence $u$ cannot be radial about any point in $\mathbb R^N$ since the term $u_1+v$ cannot be radial.