Ngô Quốc Anh

December 11, 2016

Why the equation Δu+1/u^α=0 in R^n has no positive solution?

Filed under: Uncategorized — Ngô Quốc Anh @ 23:50

Following the idea due to Brezis in his note that we had already mentioned once, it seems that we can show that the following equation

\displaystyle \Delta u + \frac 1{u^\alpha}=0

in \mathbb R^n with n\geqslant 2 has no positive C^2-solution. Here we require \alpha>0.

Indeed, to see this, we first denote


with t>0. Clearly the function f is monotone increasing in its domain. Using this, we rewrite our equation as follows

\Delta u =f(u).

Assume by contradiction that there exists some u solving the equation. Fix any point x_0 \in \mathbb R^n and consider the function

u_\varepsilon(x)=u(x)+\varepsilon |x-x_0|^2, \quad \varepsilon>0, x \in \mathbb R^n.

Since u_\varepsilon(x) \to \infty as |x|\to \infty, the value \min_{\mathbb R^n}u_\varepsilon(x) is achieved at some x_1. Then we have

0\leqslant \Delta u_\varepsilon(x_1)=\Delta u(x_1)+2\varepsilon n=f(u(x_1))+2\varepsilon n.

By definition,

u(x_1)+\varepsilon |x_1-x_0|^2=u_\varepsilon(x_1) \leqslant u_\varepsilon(x_0)=u(x_0).


u(x_1)\leqslant u(x_0).

Since f is increasing we deduce that

f(u(x_1))\leqslant f(u(x_0)).


0\leqslant \Delta u_\varepsilon(x_1) = f(u(x_1))+2\varepsilon n \leqslant f(u(x_0))+2\varepsilon n.

By sending \varepsilon \to 0 we deduce that -u(x_0)^{-\alpha} = f(u(x_0)) \geqslant 0. However this is a contradiction since u>0.

This non-existence result confirms the reason why the authors did study solutions of

\displaystyle \Delta u =\frac 1{u^\alpha}

in \mathbb R^n in this paper.


1 Comment »

  1. Nice proof 🙂

    Comment by Van Hoang Nguyen — December 12, 2016 @ 5:35

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