I have the following question:
Let us denote $H_2^N: = \{u\in (H^2(0,1))^2: u'(0) = u'(1) = 0\}$. Let an operator $L:H_2^N \to (L^2(0,1))^2$ be given by $Lu = -Du'' + Cu$, where $D$ is a positive definite diagonal matrix of diffusion coefficients and $C$ is some constant hermitian matrix. Can we say that $\lambda_1(L)\geq \lambda_1(C)$?
That is what I have done so far: let $\varphi_1\in H_2^N$ be the eigenelement, which corresponds to $\lambda_1(L)$. Then we write \begin{equation} \lambda_1(L) = \frac{\langle L\varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} = \frac{\langle -D\varphi_1'', \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} + \frac{\langle C\varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} = \frac{\lambda_1(-D\Delta)\langle \varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} + \frac{\langle C\varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} = \frac{\langle C\varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle}\geq \lambda_1(C)\frac{\langle \varphi_1, \varphi_1\rangle}{\langle\varphi_1, \varphi_1\rangle} = \lambda_1(C). \end{equation} The above estimation holds due to the fact that $\lambda_1(-D\Delta) = 0$. I am not sure, whether this is correct and will appreciate your comments and help!