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.

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