Skip to main content
1 of 4
Elio Li
  • 809
  • 4
  • 13

A problem about using the moving plane method to prove radial symmetry of the $C^{2}$ global solution of a elliptic PDE in $R^{2}$

Recently I'm learning using the moving plane method to prove radial symmetry of the $C^{2}$ global solution of a PDE in $R^{2}$.Using maximum principle I have proved that $$w \leqslant 0 \ in \ \Sigma(\lambda) \ for \ \lambda \leqslant 0$$ and $$w>0 \ in \ \Sigma(\lambda) \ for \ \lambda>0$$ using Hopf lemma we have $$w_{1}(0,0)>0 \ for \ y \ \in \mathbb{R}$$ where $\Sigma(\lambda)=\left\{(x, y) \in \mathbb{R}^{2} \mid x<\lambda\right\}$ and $v=u(2 \lambda-x, y)$, $w(x,\lambda)=v-u$.

The paper I'm reading is Monotonicity and symmetry of solutions of fully nonlinear elliptic equations on unbounded domains written by Congming Li.

What I'm confused is that in this paper, Dr.Li said we should use the maximality of 0, trying to follow his work, I set a sequence such that $$\lambda^{k} \searrow 0, x^{k} \in \sum\left(\lambda^{k}\right), w\left(x^{k}, \lambda^{k}\right)>0$$ $$w\left(x^{k}, \lambda^{k}\right)=\max _{x \in \Sigma\left(\lambda^{k}\right) \atop 0\leq\lambda \leq \lambda^{k}} w(x, \lambda)>0$$ $$\nabla_{x} w\left(x^{k}, \lambda^{k}\right)=0,\left\{w_{i j}\left(x^{k}, \lambda^{k}\right)\right\} \leq 0$$ The paper said that we can 'assume' $x^{k} \longrightarrow 0 \in \overline{\sum(0)}$, and taking the limit we have $$\nabla_{x} w(0,0)=0$$ this contradicts to the results I get using Hopf lemma that $$w_{1}(0,0)>0 \ for \ y \ \in \ \mathbb{R}$$ Actually I'm so confused about the construction of this sequence, how to prove that $x^{k} \longrightarrow 0 \in \overline{\sum(0)}$.

Elio Li
  • 809
  • 4
  • 13