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- We didn’t go over the short (eight lines) proof of the theorem given the lemmas. Please read this on your own as part of Problem Sheet 9.
Section 5.4 Proof of the theorem
We now embark on the proof of Theorem 5.1. This proof will take place over the course of three lemmas, which inherit the notation and assumptions of the theorem.
Notation 5.5. Notation.
From this point forward in the proof we fix the unit vector \(e=e_1\) and simply write
\begin{align*}
T_\lambda\amp=\{x \in \R^N : x_1=\lambda \}= T_\lambda(e_1)
\end{align*}
and
\begin{align*}
x^\lambda \amp= x^{\lambda,e_1} \\
\amp= x+2(\lambda-x_1)e_1 \\
\amp= (2\lambda-x_1,x_2,\ldots,x_N).
\end{align*}
We let
\begin{align*}
\Sigma(\lambda)\amp=B \cap \{ x_1 \gt \lambda\}
\end{align*}
be the portion of our ball \(B\) to the right of \(T_\lambda\text{;}\) see Figure 5.4. Finally, we define the reflected function
\begin{gather*}
u^\lambda(x) = u(x^\lambda),
\end{gather*}
see Figure 5.5. Since the Laplacian is invariant under reflections (Exercise 2.6.2), \(u^\lambda\) also satisfies \(\Delta u^\lambda +
f(u^\lambda) = 0\) on its domain of definition.
General strategy.
It helps to keep one eye firmly planted on Figure 5.4 and especially Figure 5.5. The idea is compare the graphs of \(u\) and \(u^\lambda\) on the set \(\Sigma(\lambda)\) as \(\lambda\) decreases from \(R\) to \(0\text{,}\) i.e. as we ‘move the plane’ through which things are being reflected. Lemma 5.6 says that, when \(\lambda \approx R\text{,}\) we will have \(u^\lambda \gt u\) on \(\Sigma(\lambda)\text{.}\) In Lemma 5.8, we then decrease \(\lambda\) the until (roughly speaking) this inequality is violated for the first time, say at \(\lambda=\mu\text{.}\) By continuity this could happen either because \(u \equiv u^\lambda\) on \(\Sigma(\lambda)\text{,}\) in which case we are getting close to proving symmetry in the \(x_1\)-direction, or alternatively if \(u \le
u^\lambda\) but \(u \not\equiv u^\lambda\) as shown in Figure 5.5. This scenario is ruled out, however, by Lemma 5.7, whose proof unsurprisingly involves the strong maximum principle.
Lemma 5.6.
Let \(x^* \in \partial B \cap \{ x_1 \gt0\}\text{.}\) Then, for some \(\delta\gt0\text{,}\) we have \(\partial_1 u
\lt 0\) in \(B \cap B_\delta (x^*)\text{.}\)
As with the Hopf lemma, the non-strict inequality \(\partial_1 u(x^*) \le 0\) is obvious from \(u \gt 0\) in \(B\) and \(u=0\) on \(\partial B\) (Lemma 2.48). The content of the lemma is that we have a strict inequality at nearby points.
Proof.
If the lemma were false then there would be a sequence of points \(x_n \in B\) with \(x_n \to x^*\) and \(\partial_1 u (x_n) \ge 0\) for all \(n\text{.}\) On the other hand since \(u \gt 0\) in \(B\) and \(u = 0\) on \(\partial B\text{,}\) we must have
\begin{equation}
\partial_1 u (x^*) \le 0.\tag{5.5}
\end{equation}
Hence by continuity we find \(\partial_1 u (x^*) = 0\text{.}\) Since \(u = 0\) on \(\partial B\) the tangential derivatives are also zero, and so we conclude that \(\nabla u (x^*) = 0\text{.}\)
We next claim that \(\partial_1^2 u(x^*) = 0\) as well. Since \(u(x^*) = \partial_1 u
(x^*) = 0\) and \(u\gt 0\) in \(B\text{,}\) we must have \(\partial_1^2 u (x^*) \ge 0\text{,}\) so assume for the sake of contradiction that \(\partial_1^2 u (x^*) \gt 0\text{.}\) Since \(u \in C^2(\overline B)\text{,}\) we deduce that \(\partial_1^2 u \gt 0\) in some neighborhood \(N = B \cap B_\delta(x^*)\) of \(x^*\text{.}\)
Consider now the straight line segment \(\Gamma\) in the positive \(e_1\) direction, connecting a point \(x_n\) in our sequence to a point \(y_n\) on \(\partial B\text{;}\) see Figure 5.6. For \(n\) large enough we have \(\Gamma \subseteq N\text{,}\) so that
\begin{gather}
\partial_1 u (y_n) - \partial_1 u (x_n)= \int_\Gamma \partial_1^2 u (x) dx_1 \gt 0.\tag{5.6}
\end{gather}
By assumption we have \(\partial_1 u(x_n) \ge 0\text{,}\) and arguing as above (5.5) we find \(\partial_1 u(y_n) \le 0\) since \(y_n \in \partial B\text{.}\) But then the left hand side of (5.6) is \(\le 0\text{,}\) which is a contradiction. Thus the claim is proved and \(\partial_1^2 u(x^*) = 0\text{.}\)
Case 1: \(f(0) \ge 0\).
We consider the comparison function \(v \equiv 0\text{.}\) Since we have the inequalities
\begin{equation}
\Delta u + f(u) = 0 \le f(0) = \Delta v + f(v)\tag{5.7}
\end{equation}
Case 2: \(f(0) \lt 0\).
By Exercise 5.4.1, the Hessian \(D^2 u(x^*)\) has the special form
\begin{gather}
\partial_{ij} u(x^*) = A\frac{x^*_i x^*_j}{R^2}\tag{5.8}
\end{gather}
for some constant \(A\text{.}\) In our earlier claim we showed that \(\partial_{11}
u(x^*) = 0\text{,}\) so (5.8) and \(x^*_1 \gt 0\) imply that this constant \(A\) must in fact be zero. In particular, the Laplacian \(\Delta u(x^*) = 0\text{.}\) But this is a contradiction since
\begin{equation*}
\Delta u(x^*) = -f(u(x^*)) = -f(0) \gt 0\text{.}
\end{equation*}
Having arrived at a contradiction in both cases, we conclude that \(\partial_1 u \lt 0\) must hold in \(B \cap B_\delta(x^*)\) for some sufficiently small \(\delta \gt 0\text{.}\)
Lemma 5.7.
Fix \(\lambda \in [0,R)\text{.}\) If \(u \le u^\lambda\) but \(u \not\equiv u^\lambda\) on \(\Sigma(\lambda)\text{,}\) then \(u \lt
u^\lambda\) on \(\Sigma(\lambda)\text{.}\) Moreover, in this case we have \(\partial_1 u \lt 0\) on \(B \cap
T_\lambda\text{.}\)
Proof.
We apply Lemma 5.4 to \(u\) and \(u^\lambda\text{.}\) The inequality \(u \lt u^\lambda\) is immediate from Lemma 5.4(i). So let \(x \in T_\lambda \cap B \subset \partial\Sigma(\lambda)\text{.}\) Then \(x^\lambda = x\text{,}\) and so \(u^\lambda = u\) at \(x\text{.}\) Moreover, \(\Sigma(\lambda)\) has an interior ball at \(x\text{,}\) with outward pointing normal \(n=-e_1\text{,}\) and so Lemma 5.4(ii) gives \(\partial_1 u^\lambda - \partial_1 u \gt 0\) at \(x\text{.}\) By symmetry we have \(\partial_1 u^\lambda = -\partial_1 u\) along \(T_\lambda\text{,}\) and so this implies \(\partial_1 u(x) \lt 0\) as desired.
Lemma 5.8.
For any \(\lambda \in (0,R)\text{,}\) we have
\begin{gather}
u \lt u^\lambda \text{ and } \partial_1 u \lt 0 \text{ on } \Sigma(\lambda).\tag{5.9}
\end{gather}
Proof.
By Lemma 5.6, (5.9) holds for \(\lambda\) sufficiently close to \(R\text{.}\) Let \(\mu\) be the critical value such that (5.9) holds for \(\lambda \gt \mu\text{,}\) but not for \(\lambda \lt \mu\) with \(\lambda\approx \mu\text{.}\) If \(\mu = 0\) then we are done, so assume for the sake of contradiction that \(\mu \gt 0\text{.}\)
We first claim (5.9) holds for \(\lambda=\mu\text{.}\) Continuity implies that \(\partial_1 u \le 0\) and \(u \le u^\mu\) in \(\Sigma(\mu)\text{,}\) which puts us in a position to apply Lemma 5.7, provided we can show that \(u \not\equiv u^\mu\) in \(\Sigma(\mu)\text{.}\) To see this, note that any \(x_0 \in \partial\Sigma(\mu) \without T_\mu\text{,}\) we have \(u(x_0) = 0\) since \(x_0 \in \partial B\) but \(u^\mu(x_0) \gt 0\) since \(x_0^\mu \in B\text{.}\) Thus we can apply Lemma 5.7 to conclude that \(u \lt u^\mu\) in \(\Sigma(\mu)\text{,}\) and moreover that \(\partial_1 u \lt 0\) on \(B \cap T_\mu\text{.}\) By the choice of \(\mu\text{,}\) we have \(\partial_1 u \lt 0\) on \(\Sigma(\lambda)\) for all \(\lambda \gt \mu\text{,}\) and so the claim is proved.
Thus \(\partial_1 u \lt 0\) on \(B \cap T_\mu\text{.}\) Moreover, for any \(x \in \partial B \cap T_\mu\) we can use Lemma 5.6 to obtain \(\partial_1 u \lt 0\) in some neighborhood \(B_\delta(x) \cap B\text{.}\) Appealing to the compactness of \(\overline B \cap
T_\mu\text{,}\) we deduce that \(\partial_1 u \lt 0\) in \(\Sigma(\mu-\varepsilon)\) for \(\varepsilon \gt 0\) sufficiently small. This means that the second inequality in (5.9) holds for \(\lambda\) slightly less than \(\mu\text{,}\) and so it must be that the first inequality fails.
1
By popular request, this argument is written out in full in Section D.1.
Since the first inequality in (5.9) fails for \(\lambda \lt
\mu\) with \(\lambda \approx \mu\text{,}\) we can find sequences \(\lambda_j \gt 0\) and \(x_j \in \Sigma(\lambda_j)\) such that \(\lambda_j \nearrow \mu\) as \(j \to \infty\text{,}\) and \(u(x_j) \ge u^\lambda(x_j)\) for all \(j\text{.}\) Taking a subsequence, we can assume that \(x_j\) converges to a limit \(x \in \overline{\Sigma(\mu)}\text{,}\) in which case continuity gives \(u(x) \ge u^\lambda(x)\text{.}\) Since (5.9) holds for \(\lambda=\mu\text{,}\) we must have \(x \in
\partial\Sigma(\mu)\text{.}\) As we saw above, \(u = 0 \le u^\mu\) on \(\partial\Sigma(\mu) \without T_\mu\text{,}\) and so in fact \(x \in T_\mu \cap B\text{.}\) Thus \(x_j^{\lambda_j} \to x^\mu = x\) as well, and in particular the straight line segments \(\Gamma_j\) connecting \(x_j\) to \(x_j^{\lambda_j}\) lie in \(B\) for \(j\) sufficiently large. Applying the mean value theorem, we see that for each \(j\) there must be \(y_j \in \Gamma_j\) with \(\partial_1 u(y_j) \ge 0\text{.}\) But \(y_j \to x\text{,}\) and so by continuity this gives \(\partial_1 u(x) \ge 0\text{,}\) contradicting our earlier conclusion that \(\partial_1 u \lt 0\) on \(B \cap T_\mu\text{.}\) Thus \(\mu = 0\) as desired.
We can now complete the proof of the Gidas–Ni–Nirenberg theorem.
Proof of Theorem 5.1.
By continuity and Lemma 5.8 we have \(u(x) \le u(x^0)\) in \(\Sigma(0) = B \cap \{x_1 \gt 0\}\text{,}\) and moreover that \(\partial_1 u \lt 0\) in \(B \cap
\{ x_1 \gt 0 \}\text{.}\)
Repeating the same argument with \(x_1\) replaced by \(-x_1\text{,}\) we find that \(u(x) \ge u(x^0)\) in \(\Sigma(0)\text{.}\) Hence \(u(x) = u(x^0)\) and \(u\) is symmetric across the plane \(x_1 = 0\text{.}\)
By the same argument \(u\) is symmetric with respect to any plane through the origin, and hence by Lemma 5.3 it is radially symmetric. The condition \(\partial_1 u \lt 0\text{,}\) suitably rotated, gives the radial monotonicity property.
Exercises Exercises
1. A calculation needed in Lemma 5.6.
Let \(B = B_R(0) \subseteq \R^2\text{,}\) and consider the point \(x^* = Re_2 \in \partial B\text{.}\) Note that, near \(x^*\text{,}\) \(\partial B\) can be parametrised as a graph,
\begin{equation*}
(x_1,x_2) = \phi(x_1) := (x_1,\sqrt{R^2-x^2})\text{.}
\end{equation*}
(a)
Suppose that \(f \in C^1(\overline B)\) achieves \(\max_{\partial B}
f\) at \(x^*\text{.}\) By considering the composition \(f \circ \phi\text{,}\) show that \(\partial_1 f=0\) at \(x^*\text{.}\)
Hint.
The summation convention is not particularly useful here. You will need to calculate \(\partial_1 \phi_1\) and \(\partial_1 \phi_2\) at \(x_1=0\text{.}\)
Solution.
The composition \(f \circ \phi\) has a local maximum at \(x_1 = 0\text{.}\) Thus
\begin{equation}
0 = \partial_1 (f \circ \phi)
=
\partial_1 f(\phi) \partial_1 \phi_1
+
\partial_2 f(\phi) \partial_1 \phi_2 \text{.}\tag{5.10}
\end{equation}
Now at \(x_1 = 0\) we have
\begin{equation*}
\phi = x^*,
\quad
\partial_1 \phi_1 = 1,
\quad
\partial_1 \phi_2 = 0,
\end{equation*}
and so (5.10) simplifies to \(\partial_1 f(x^*)=0\) as desired.
(b)
Suppose that \(u \in C^2(\overline B)\) has \(u = 0\) on \(\partial B\text{.}\) By repeatedly differentiating the identity \(u \circ \phi = 0\text{,}\) show that \(\partial_1 u = 0\) and \(\partial_{11} u = R^{-1}\partial_2 u\) at \(x^*\text{.}\)
Hint.
The summation convention is not particularly useful here. You will need to calculate \(\partial_{11} \phi_1\) and \(\partial_{11} \phi_2\) at \(x_1=0\text{.}\)
Solution.
Since \(u = 0\) on \(\partial B\text{,}\) we indeed have \(u \circ \phi = 0\) for \(x_1\) near \(0\text{.}\) Differentiating this identity once we have as above that
\begin{equation*}
0 = \partial_1 u(\phi) \partial_1 \phi_1 + \partial_1 u(\phi) \partial_1 \phi_2,
\end{equation*}
which yields \(\partial_1 u(x^*)=0\) upon setting \(x_1=0\text{.}\) Differentiating a second time, we find
\begin{align*}
0
\amp =
\partial_1(\partial_1 u(\phi) \partial_1 \phi_1 + \partial_1 u(\phi) \partial_1 \phi_2)\\
\amp =
(\partial_{11} u(\phi)\partial_1 \phi_1 + \partial_{12} u(\phi) \partial_1 \phi_2) \partial_1 \phi_1 \\
\amp \qquad
+ \partial_1 u(\phi) \partial_{11} \phi_1 \\
\amp \qquad
+ (\partial_{21} u(\phi)\partial_1 \phi_1 + \partial_{22} u(\phi) \partial_1 \phi_2) \partial_1 \phi_2 \\
\amp \qquad
+ \partial_2 u(\phi) \partial_{11} \phi_2 \text{.}
\end{align*}
Now at \(x_1=0\text{,}\) we have
\begin{equation*}
\phi = x^*,
\quad
\partial_1 \phi_1 = 1,
\quad
\partial_{11} \phi_2 = -R^{-1},
\quad
\partial_1 \phi_2
=
\partial_{11} \phi_1
=
\partial_1 u
= 0\text{,}
\end{equation*}
and so the above formula simplifies dramatically to
\begin{align*}
0
\amp =
\partial_{11} u(x^*)
- R^{-1} \partial_2 u(x^*)\text{,}
\end{align*}
which is the desired identity.
(c)
In the setting of Part b, suppose in addition that \(u \gt 0\) in \(B\) and that \(\partial_2 u =0\) at \(x^*\text{.}\) Show that \(x \cdot \nabla u \le 0\) on \(\partial B\text{.}\) By applying Part a to \(f=x\cdot \nabla u\text{,}\) deduce that \(\partial_{12} u = 0\) at \(x^*\text{,}\) and hence
\begin{equation*}
D^2 u(x^*) =
\begin{pmatrix}
0 \amp 0 \\
0 \amp \partial_{22} u(x^*)
\end{pmatrix}\text{.}
\end{equation*}
Solution.
For \(x \in \partial B\text{,}\) the outward pointing normal vector is \(n=x/R\text{.}\) Since \(u\) achieves its minimum over \(\overline B\) at \(x\text{,}\) we therefore have
\begin{equation*}
x \cdot \nabla u = Rn \cdot \nabla u = R\frac{\partial u}{\partial n} \le 0\text{.}
\end{equation*}
Since \(x \cdot \nabla u = \partial_2 u = 0\) at \(x^*\text{,}\) we conclude that \(f = x
\cdot \nabla u\) achieves \(\max_{\partial B} f\) at \(x^*\text{.}\) Thus by Part a we have \(\partial_1 f = 0\) at \(x^*\text{.}\) Using the product rule we have
\begin{align*}
\partial_1 f
\amp
= \partial_1 (x_1 \partial_1 u + x_2 \partial_2 u) \\
\amp
=
\partial_1 u
+ x_1 \partial_{11} u
+ x_2 \partial_{12} u \text{.}
\end{align*}
Plugging in \(x=x^*\) and simplifying, we conclude that
\begin{equation*}
0 = \partial_1 f(x^*) = R \partial_{12} u(x^*)\text{,}
\end{equation*}
and hence \(\partial_{12} u(x^*)=0\) as desired.
(d)
(Optional) Generalise the result in Part c, first to the case where \(B = B_R(0) \subset \R^N\) and \(x^*=Re_N\text{,}\) and then to the case of general \(x^* \in \partial B\text{,}\) establishing the identity (5.8) needed in the proof of Lemma 5.6.
Hint.
This is not particularly easy; it’s optional for a reason.
Solution.
First consider the case where \(B_R(0) \subset \R^N\) and \(x^* = Re_N\text{.}\) We can now parametrise \(\partial B\) near \(x^*\) using
\begin{equation*}
x = \phi(x_1,\ldots,x_{N-1}) = \Big(x_1,\ldots,x_{N-1},\sqrt{R^2-x_1^2-\cdots-x_{N-1}^2}\Big)\text{.}
\end{equation*}
At \(x_1=\cdots=x_{N-1}=0\text{,}\) we calculate
\begin{align*}
\partial_i \phi_k \amp=
\begin{cases}
1 \amp i=k \lt N \\
0 \amp \text{otherwise}
\end{cases},
\\
\partial_{ij} \phi_k \amp=
\begin{cases}
-R^{-1} \amp i=j \lt N \text{ and } k=N \\
0 \amp \text{otherwise}
\end{cases}\text{.}
\end{align*}
If \(u \in C^2(\overline B)\) satisfies \(u=0\) on \(\partial B\text{,}\) \(u \gt
0\) in \(B\text{,}\) and \(\partial_N u = 0\) at \(x^*\text{,}\) calculations similar to those in the previous two parts then yield \(\nabla u(x^*) = 0\) and \(\partial_{ij} u(x^*) = 0\) unless \(i=j=N\text{,}\) i.e.
\begin{equation*}
\partial_{ij} u(x^*) = \delta_{iN} \delta_{jN} \partial_{NN} u(x^*) \text{.}
\end{equation*}
Now consider a general \(x^* \in \partial B\text{,}\) let \(A \in \R^{N \times N}\) be an orthogonal matrix with \(x^* = A(Re_N)\text{,}\) and define a function \(v \in C^2(\overline B)\) by \(v(x)=u(Ax)\text{.}\) Assuming that \(u=0\) on \(\partial B\text{,}\) \(u \gt 0\) in \(B\text{,}\) and \(\nabla u =
0\) at \(x^*\text{,}\) we can apply the above argument to \(v\) to get
\begin{equation*}
\partial_{ij} v(Re_N) = \delta_{iN} \delta_{jN} \partial_{NN} v(Re_N) \text{.}
\end{equation*}
Differentiating \(u(x)=v(A^{-1}x)\) twice as in Exercise 2.6.2, we find that
\begin{align*}
\partial_{ij} u(x^*)
\amp
= A_{ki}^{-1} A_{\ell j}^{-1} \partial_{k\ell} v(Re_N)\\
\amp
=
A_{ki}^{-1} A_{\ell j}^{-1}
\delta_{kN} \delta_{\ell N} \partial_{NN} v(Re_N) \\
\amp
=
A_{Ni}^{-1} A_{Nj}^{-1}
\partial_{NN} v(Re_N) \\
\amp
=
A_{iN} A_{jN}
\partial_{NN} v(Re_N) \\
\amp
=
\frac{x^*_i x^*_j}{R^2}
\partial_{NN} v(Re_N) \text{,}
\end{align*}
which is of the desired form in (5.8).
2. (PS9) Monotonicity in an annulus.
Consider the setting of Theorem 5.1, but with the ball \(B_R(0)\) replaced by an annular domain \(\{ R_1 \lt \abs x \lt R_2\}\text{.}\) There is no longer any guarantee that \(u\) is symmetric. Nevertheless, by following the proof of the Theorem 5.1, convince yourself that the radial derivative
\begin{equation*}
\partial_r u := \frac{x}{\abs x} \cdot \nabla u \lt 0
\end{equation*}
for \((R_1+R_2)/2 \lt \abs x \lt R_2\text{.}\)
Hint.
There is no need to rewrite the entire proof of the theorem and all of the lemmas! This question is primarily here to get you to go back and work through all of the steps in the proof, seeing where the modifications need to be made for this different geometry.
Solution.
The proof of Lemma 5.6 only involves a neighborhood of the boundary \(\abs x = R\text{,}\) and so is unchanged for the outer boundary \(\abs x = R_2\) of our annular set. The proofs of Lemmas 5.7–5.8 also continue to work provided \(\lambda \gt (R_1+R_2)/2\text{,}\) since for these values they cannot ‘see’ the hole in the domain. Thus we conclude that (5.9) holds for any \(\lambda \gt (R_1+R_2)/2\text{.}\) This implies \(\partial_1 u \lt 0\) for \(x\) in our annulus with \(x_1 \gt (R_1+R_2)/2\text{,}\) and hence in particular that \(\partial_r u = \partial_1 u \lt 0\) along the line segment \((x_1,0,\ldots,0)\text{,}\) \((R_1+R_2)/2 \lt x_1 \lt R\text{.}\) Rotating coordinates, we obtain \(\partial_r u \lt 0\) for \((R_1+R_2)/2 \lt \abs x \lt R\) as desired.
3. (PS9) Gidas–Ni–Nirenberg for an ellipse.
Consider the setting of Theorem 5.1, but with the ball \(B_R(0)\) replaced by the interior of an ellipse
\begin{equation*}
E = \Big\{(x_1,x_2) \in \R^2 : \frac{x_1^2}{a_1^2} + \frac{x_2^2}{a_2^2} \lt 1\Big\}
\end{equation*}
with \(a_1 \gt a_2 \gt 0\text{.}\) What can you say about \(u\text{?}\)
Hint.
As with Exercise 5.4.2, do not worry too much about writing things out in complete detail. This exercise is mostly meant to get you thinking.
Solution.
Going through the proofs of Lemma 5.6, Lemma 5.7 and Lemma 5.8, everything continues to work. We conclude that \(u\) is even in \(x_1\) with \(\partial_1 u \lt 0\) in \(E \cap \{x_1 \gt 0\}\text{.}\) Rotating coordinates, we get a similar statement with \(x_1\) replaced by \(x_2\text{.}\)
Here we have used the fact that the ellipse is symmetric under reflections in either \(x_1\) or \(x_2\text{.}\) Unlike the ball, it is not symmetric under other reflections, and for this reason we can no longer argue that \(u\) is radially symmetric.
