# Why $\gamma\cdot\operatorname{grad}u<0$ in the Theorem? (Nirenberg academic paper)

I am working on the following academic paper Symmetry and Related Properties via the Maximum Principle, which is a classic by Louis Nirenberg. I am trying to understand the next theorem, which is on page 211 and 216:

Theorem 2. Let $u>0$ be a $C^2$ solution of (1.1) in a ring-shaped domain \begin{equation} R' < |x| \leq R. \end{equation} Then \begin{equation} \frac{\partial u}{\partial r} < 0 \quad \text{for} \quad \frac{R'+R}{2} \leq |x| < R. \end{equation}ProofWe may again choose any direction $\gamma$ as positive $x_1$ axis. It follows from Theorem 2.1 that in the corresponding maximal cap $\Sigma_{\gamma}$, $\gamma\cdot\operatorname{grad}u < 0$. The union of these maximal caps is the region $(R'+R)/2 < |x| < R$.

Suppose for some point $y$ with $|y|=(R' + R)/2$, $u_{r}(y)=0$. Then with $\gamma=y/|y|$ we conclude from the last assertion of Theorem 2.1 that $\Omega=\Sigma_{\gamma}\cup\Sigma'_{\gamma}$ which is impossible.

The proof also shows that for $|x| > (R'+R)/2$, $v\cdot\operatorname{grad}u(x) < 0$ for any vector $v$ making an angle less than $(\pi/2 - \theta)$ with the vector $x$.

I don't understand why the multiplication of the normal vector of the hyperplane by the gradient of $u$ is less than zero. I have doubts about the final conclusion. My interpretation is that if you are at a point on the border of $\Omega$, the cap is always zero or close to zero so its reflection will be zero too. Hence it is impossible that the join of cap and its reflection is equal to domain, $\Omega$. Is it true?

In the proof, it uses the theorem 2.1 that you can see in the link, $\Sigma=\Sigma_{\gamma}=\Sigma(\lambda_{1})$ is the maximal cap associated with the hyperplane $T_{\lambda_1}$ and \begin{equation} \tag{1.1} \Delta u + f(u)=0 \quad \text{with} \quad u=0 \quad \text{on} \quad |x|=R. \end{equation} is an equation from the theorem 1.

## 1 answer

As the proof of **Theorem 2** suggests, this follows immediately from **Theorem 2.1**. While it's a bit ambiguously worded, to apply **Theorem 2.1**, we need $b_1(x)=0$ in $\Delta u + b_1(x)u_{x_1} + f(u) = 0$, $u > 0$ in $\Omega$, $u = 0$ on a part of $\partial\Omega$, and some basic continuity conditions. The assumptions of **Theorem 2** directly satisfies all of these.

Switching/modernizing notation a bit, I'll write $\nabla$ for the vector derivative which is the gradient when applied to a scalar function, i.e. $\nabla\phi = \operatorname{grad}\phi$. The directional derivative in the direction $\mathbf v$, where $\mathbf v$ is a vector, can be written as $\mathbf v \cdot \nabla$. For a coordinate, $x_i$, the corresponding vector $\mathbf e_i = \nabla x^i$ where $x^i(x_1,\dots, x_n) = x_i$. In other words, $\frac{\partial}{\partial x_i} = \mathbf e_i\cdot\nabla$. For completeness, though it won't be relevant, $\Delta = \nabla\cdot\nabla = \nabla^2$.

In the proof of **Theorem 2**, we are aligning the $x_1$ axis with the vector $\gamma$. That is, $\frac{\partial}{\partial x_1} = \gamma \cdot \nabla$, the directional derivative in the $\gamma$ direction. Therefore, $\gamma \cdot \operatorname{grad} u = \gamma \cdot \nabla u = \frac{\partial u}{\partial x_1} = u_{x_1}$. One of the first consequences of **Theorem 2.1** is that $u_{x_1} < 0$ in $\Sigma=\Sigma_\gamma$. **Theorem 2.1** also has $\Omega = \Sigma \cup \Sigma' \cup (T_{\lambda_1}\cap \Omega)$ if $\gamma \cdot \nabla u = 0$ at some point of $\Omega \cap T_{\lambda_1}$ as a conclusion. For the annulus, the $T_{\lambda_1}$ for varying $\gamma$ will be the lines tangent to the circle halfway into the annulus. Clearly the (closure of the) maximal cap unioned with its reflection is not all of the annulus which is what **Theorem 2.1** implies in this case, thus it can't be the case that $\gamma\cdot\nabla u = 0$ even when $|x| = (R'+R)/2$ where before we only knew this for $|x| > (R'+R)/2$.

The proof of **Theorem 2** is assuming that **Theorem 2.1** holds. The proof of **Theorem 2.1** doesn't come until later in the paper, and it's moderately involved.

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