Can we show that the normal derivative of the first eigenfunction of $-\Delta$ is negative?

Question

Let $(\lambda_1,e_1)$ denote the first eigenpair of $-\Delta$ on a ball $\Omega:=B_r(0)$ in $\mathbb R^d$, i.e. $\lambda_1>0$ and $e_1\in C^2(\overline\Omega)$ with \begin{align}-\Delta\left. e_1\right|_\Omega&=\lambda_1\left.e_1\right|_\Omega\tag1\\\left.e_1\right|_\Omega&>0\tag2\\\left.e_1\right|_{\partial\Omega}&=0.\tag3\end{align}

How can we show that the derivative $\partial_\nu e_1$ of $e_1$ in the direction of the unit outer normal field $\nu$ on $\partial\Omega$ is negative?

We can show that $e_1$ is rotational invariant and satisfies $$\lambda_1e_1(s)+\frac{n-1}se_1'(s)+e_1''(s)=0\tag4\;\;\;\text{for all }s\in(0,r).$$ Moreover, since $$\langle x,\nabla e_1(x)\rangle=e_1'(\|x\|)\;\;\;\text{for all }x\in\overline\Omega\tag5,$$ we've got $$\partial_\nu e_1(x)=\langle\nu(x),\nabla e_1(x)\rangle=\frac1r\langle x,\nabla e_1(x)\rangle=\frac1re_1'(r)\tag6.$$

Lastly, since $e_1$ is positive in $\Omega$ and $e_1(R)=0$, it should hold $e_1'(R)\le0$.

Can we somehow show that $e_1'(R)=0$ is impossible and use this to conclude the desired claim?

Answer 1

There are several ways to prove it.

1. Just use Hopf lemma.

2. Note that $e_1$ satisfies (4) and $e_1(R)=0$. If $e_1'(R) = 0$ also, then $e_1(s) = 0$ for all $s>0$ by the uniqueness theorem of ODEs.

3. Write $e_1' = f$. Then (4) implies $$\tag{1'} f' = -\frac{n-1}{s} f - \lambda_1 e_1. $$ This implies \begin{align} (s^{n-1} f)' &= (n-1)s^{n-2} f + s^{n-1} f' \\ &= (n-1)s^{n-2} f + s^{n-1} \left(-\frac{n-1}{s} f - \lambda_1 e_1 \right) \\ &= - s^{n-1}\lambda_1 e_1. \end{align} Integrating from $0$ to $R$ and use $f(0) = 0$ gives $$R^{n-1} f(R) = -\int_0^R \lambda_1 s^{n-1} e_1(s) ds. $$ Thus $f(R)$ is clearly negative since $\lambda_1>0$, $s^{n-1} e_1\ge 0$ and is not identically zero.

I guess one can also find $e_1$ explicitly and argue directly. Indeed all eigenvalues and eigenfunctions are known in this case. See here.