compare of eigenvalues $\lambda_1(D_a)$ and $\lambda_1(D_c)$. Let $f(x)$ be a smooth function on $[-1,1]$, such that $f(x)>0$ for all $x\in(-1,1)$,$f(-1)=f(1)=0$. consider $\gamma\subset\Bbb{R}^2$ the graph of the $f(x)$. Let $T_a$ the symmetry with respect to axis $x$ and $T_c$ the central symmetry with respect to origin. Now consider two domains $D_a$:bounded by the curves $\gamma$ and $T_a(\gamma)$, and $D_c$: bounded by the curves $\gamma$ and $T_c(\gamma)$. Let $\lambda_1(D_a)$ and $\lambda_1(D_c)$ are the eigenvalues of Dirichlet problem on $D_a$ and $D_c$ resp. How is compare of $\lambda_1(D_a)$ and $\lambda_1(D_c)$? ($\lambda_1(D_c)\geq\lambda_1(D_a)$ or $\lambda_1(D_a)\geq \lambda_1(D_c)$)
There exist one theorem in Strauss's book which:
If $\Omega_1\subset\Omega_2$ then $\lambda_1(\Omega_1)\geq\lambda_1(\Omega_2)$.
Thanks.

Fact 1. In the special case 
$$f(-x)\le f(x),\qquad 0\le x\le 1\tag0$$ the inequality $\lambda_1(D_a)\le \lambda_1(D_c)$ holds. 
Proof. Let $u$ be the first eigenfunction for $D_c$. Extend it to $\mathbb R^2$ by zero outside of $D_c$. For $(x,y)\in\mathbb R^2$ define 
$$v(x,y)=\begin{cases} 
\max(u(x,y),u(-x,y))\quad & x\ge 0 \\
\min(u(x,y),u(-x,y))\quad & x\le 0
\end{cases}$$
(This is called the polarization of $u$ with respect to the $y$-axis.) Then the following hold: 
$$\int_{\mathbb R^2} v^2 = \int_{\mathbb R^2} u^2 \tag1$$ $$\int_{\mathbb R^2} |\nabla v|^2 = \int_{\mathbb R^2} |\nabla u|^2 \tag2$$ $$v=0\quad \text{ on } \partial D_c\tag3$$
Here (1) and (3) are relatively easy (you need assumption (0) to prove (3)). The equality (2) is not very straightforward unless you know about something about Sobolev spaces. The paper An approach to symmetrization
via polarization should give you an idea of what is going on here. 
Since $v$ is an eligible function in the variational definition of $\lambda_1$, we have 
$$\lambda_1(D_a)\le \frac{\int |\nabla v|^2}{\int v^2}=\lambda_1(D_c) \tag4$$
as claimed. 
Fact  2. There exist functions $f$ for which $\lambda_1(D_a)<\lambda_1(D_c)$.  
Let $f=\chi_{[1/2,2/5]}$ (I know it's neither continuous nor positive). Then $D_a$ is a rectangle of dimensions $2\times (1/5)$ while $D_c$ is the union of two rectangles of dimensions $1\times (1/5)$. The fundamental frequency of larger rectangle is smaller. Therefore, $\lambda_1(D_a)<\lambda_1(D_c)$ in this case. 
The function $f$ can be approximated by smooth positive functions. The first eigenvalue depends continuously on the domain in various senses (you'll have to dig in the literature). Conclusion: $\lambda_1(D_a)<\lambda_1(D_c)$ holds for some domains. 
An alternative proof of Fact 2 can be given by following the proof of Fact 1 and observing that (4) is a strict inequality  in general. Indeed, $v$ is in general not differentiable, while eigenfunctions are; therefore $v$ is not an eigenfunction.  
