Consider the eigenvalue problem
\begin{cases}
-\Delta u = \lambda u ~~~~~~~~x\in \Omega \\
u|_{\partial \Omega}=0
\end{cases}
when $\Omega$ is a circle, I see a picture from this as following
which is the picture of first eigenfunction on unit circle. Denote the first eigenfunction as $\varphi$, then ,seemly ,the $|\nabla \varphi|$ has a low boundary. In fact, I always think $|\nabla \varphi|\rightarrow 0$ when $x\rightarrow \partial \Omega$. But seemly, I am wrong. So, I want some reference about the low bound of $|\nabla \varphi|$. But only upper bound I can find it. Who can help me ,thanks.
The Hopf lemma (which you can find for instance in Evan's book) prevents this. Because we know that the first eigenfunction is positive, we have $$ -\Delta u \geq 0, x \in \Omega$$ $$ u = 0 , x \in \partial \Omega$$
So if the boundary of $\Omega$ is $C^2$ (which is the case for the circle), Hopf lemma tels you that $$\frac{\partial u(x)}{\partial n} < 0 , \forall x \in \partial \Omega$$ where $n$ is the outer normal.
Because of the Dirichlet boundary conditions, the tangent derivative vanishes at the boundary so $$ | \nabla u(x) | = |\frac{\partial u(x)}{\partial n}| > 0 , \forall x \in \partial \Omega $$
In the particular case of the circle, you can compute the solution explicitly: $$u(x) = a J_0(\frac{\nu_0}{R} |x| )$$ where a is an arbitrary non-zero constant and $\nu_0$ is the first zero of the Bessel function $J_0$. then $$\frac{\partial u(x)}{\partial n} = \frac{a\nu_0}{R} J_0'(\nu_0) \not = 0$$