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

I am working on the following academic paper [Symmetry and Related Properties
via the Maximum Principle](https://typeset.io/pdf/symmetry-and-related-properties-via-the-maximum-principle-foyy5f1u5o.pdf), which is a classic by Louis Nirenberg. I am trying to understand the next theorem, which is on page 211 anad 216.
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 to the hiperplane $T_{\lambda_1}$ 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 a equation from the theorem 1.~~
**Theorem 2**. Let $u>0$ be a $C^2$ solution of (1.1) in a ring-shaped domain
\begin{equation}
~~R^{\prime} < |x| \leq R.~~
\end{equation}
Then
\begin{equation}
~~\frac{\partial u}{\partial r} \quad \text{for} \quad \frac{R^{\prime}+R}{2}\leq |x|< R.~~
\end{equation}
**Proof**
~~We may again choose any direction $\gamma$ as positive $x_1$ axis. It follows from Theorem 2.1 that in the corresponding maximal cap $\Sigma_{\gamma}$, $\gamma\cdot\text{grad}u<0$.~~
The union of these maximal caps is the region $(R^{\prime}+R)/2 <|x|<R $. \\
Suppose for some point $y$ with $|y|=(R^{\prime} + R)/2$, $u_{r}(y)=0$. Then with $\gamma=y/|y|$ we conclude from the last assertion of Theorem 2.1 that $\Omega=\Sigma_{\gamma}\cap\Sigma^{\prime}_{\gamma}$ which is impossible. \\
~~The proof also shows that for $|x|>(R^{\prime}+R)/2$, $v\cdot\text{grad}u(x)<0$ for any vector $v$ making an angle less than $(\pi/2 -\theta)$ with the vector $x$.~~

I am working on the following academic paper [Symmetry and Related Properties
via the Maximum Principle](https://typeset.io/pdf/symmetry-and-related-properties-via-the-maximum-principle-foyy5f1u5o.pdf), which is a classic by Louis Nirenberg. I am trying to understand the next theorem, which is on page 211 and 216.
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 to 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 a equation from the theorem 1.
**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}
**Proof**
We 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 $\Omega=\Sigma_{\gamma}\cap\Sigma'_{\gamma}$ 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 $x$.

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