# 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 $f(t)=-t^{-\alpha}$

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

Thus, $u(x_1)\leqslant u(x_0).$

Since $f$ is increasing we deduce that $f(u(x_1))\leqslant f(u(x_0)).$

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

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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