the first eigenfunction of Dirichlet problem Let $\Omega$ be a bounded planar domain which has a axis of symmetry and $T:\Bbb{R}^2\longrightarrow\Bbb{R}^2$ symmetry with respect to this axis. Let $u_{1}(x)$ be the first eigenfunction of Dirichlet problem $$-\Delta u=\lambda u~~in~~ \Omega\\
u=0~~in ~~\partial\Omega$$ How can I show that $u_1(x)=u_1(Tx)$?
Thanks for help.

Let $v(x)=u_1(Tx)$. You should check that this function has the same properties as $u_1$ itself:


*

*$-\Delta v=\lambda_1 v$ in $\Omega$ 

*$v=0$ on $\partial \Omega$


Then appeal to the fact that the first eigenvalue of the Dirichlet Laplacian is simple: its eigenspace is one-dimensional. It follows that  $v=cu_1$ for some constant $c$. The constant must be $1$, for otherwise $u_1$ would have to vanish on the axis of symmetry, which can't happen (the first eigenfunction has constant sign).  
