Section 5.1 Statement of the theorem
For simplicity, we restrict our attention to the semilinear problem
\begin{equation}
\left\{
\begin{alignedat}{2}
\Delta u +f(u) \amp=0 \amp\qquad\amp \ina B,\\
u \amp=0 \amp\qquad\amp \ona \partial B,
\end{alignedat}
\right.\tag{5.1}
\end{equation}
where \(B=B_R(0)\) is a ball centred at the origin. If we were looking for explicit solutions to (5.1), a natural thing to try would be to look for radially symmetric solutions \(u(x)=U(\abs x)\text{,}\) for which (5.1) becomes the ODE
1
See Exercise 5.1.4.
\begin{align}
\left\{
\begin{alignedat}{2}
U''(r) + \frac{N-1} r U'(r) + f(U(r)) \amp= 0 \amp\qquad\amp \fora 0 \lt r \lt R, \\
U(R) = U'(0) \amp= 0.
\end{alignedat}
\right.\tag{5.2}
\end{align}
The main result of this chapter is the following.
Theorem 5.1. Gidas–Ni–Nirenberg.
Let \(f \in C^1(\R)\text{,}\) and suppose that \(u \in C^2(\overline B)\) solves (5.1) and that \(u\gt 0\) in \(B\text{.}\) Then
- \(u(x)=U(\abs x)\) is radially symmetric; and
- \(u\) is strictly monotone decreasing in the sense that \(U'(r) \lt 0\) for \(0 \lt r \lt R\text{.}\)
Exercises Exercises
1. (PS8) Necessity of \(u\gt0\) in Theorem 5.1.
Give an explicit counterexample showing that the conclusion of the Gidas–Ni–Nirenberg theorem can fail when the assumption that \(u \gt 0\) in \(B\) is dropped.
Hint.
Take \(N=1\) and think back to Exercise 1.1.1. If you are a big fan of Bessel functions and the Helmholtz equation, you may enjoy coming up with similar examples in dimensions \(N=2\) and \(N=3\text{.}\)
Solution 1.
Consider \(N=1\text{,}\) \(f(u)=u\text{,}\) and \(R=\pi\text{.}\) Then \(u(x) = \sin(x)\) satisfies all of the hypotheses of the theorem, but is clearly not an even function of \(x\text{.}\)
Solution 2.
Another nice counterexample, which I saw in a solution from a few years ago, is to consider \(f\) with \(f(0)=0\) and the function \(u \equiv 0\text{.}\) Clearly \(u\) satisfies \(\Delta u + f(u) = 0\) in \(B\) and \(u=0\) on \(\partial B\text{.}\) While it is radially symmetric, it fails to satisfy the second conclusion of Gidas–Ni–Nirenberg about monotonicity.
2. (PS8) Proving \(u\gt0\).
Let \(\Omega\) be bounded and connected, and suppose that \(u \in
C^2(\overline\Omega)\) satisfies \(\Delta u + f(u) = 0\) in \(\Omega\) and \(u =
0\) on \(\partial \Omega\text{.}\) If \(f(t) \ge 0\) for all \(t \in \R\text{,}\) show that either \(u \gt 0\) in \(\Omega\) or \(u \equiv 0\text{.}\)
Solution.
Note that \(\Delta u = -f(u) \le 0\text{.}\) If we do not have \(u \gt 0\) in \(\Omega\text{,}\) then \(u\) must achieve \(\inf_\Omega u \le 0\) at some point \(x
\in \Omega\text{.}\) But then the strong maximum principle implies that \(u\) is identically constant, and since \(u = 0\) on \(\partial\Omega\) the only possibility is \(u \equiv 0\text{.}\)
Alternatively we could have appealed to Exercise 3.3.3 with comparison function \(v \equiv 0\text{.}\) Note that in either case we need both the boundedness and the connectedness of \(\Omega\) to make the argument work.
3. (PS8) Very old exam question.
Let \(\Omega\) be bounded and suppose that \(u \in
C^2(\Omega) \cap C^0(\overline\Omega)\) solves \(\Delta u = u^3 - u\) in \(\Omega\) and \(u=0\) on \(\partial\Omega\text{.}\) Show that \(-1 \le u \le 1\) throughout \(\Omega\text{.}\)
Hint.
There are many possible techniques for tackling this problem
- Consider the restriction of \(u\) to the set \(\{x \in \Omega : u \gt 1\}\text{.}\)
- Evaluate the PDE at a maximum \(\gt 1\) or a minimum \(\lt -1\text{,}\) and think about the meaning of the sign of \(\Delta u\)
- Use Lemma 5.4 (or the idea behind its proof) to compare \(u\) to an appropriate constant function.
Solution 1. Restricting \(u\) to a subdomain
The hint was to consider the restriction of \(u\) to the open set \(\Omega'
:= \{x \in \Omega : u \gt 1\}\text{.}\) Assume for the sake of contradiction that \(\Omega'\) is nonempty. Since \(u=0\lt1\) on \(\partial\Omega\text{,}\) we can check that \(\partial\Omega' \subset \Omega\text{,}\) which in particular implies that \(u=1\) on this set. Thus we have
\begin{alignat*}{2}
\Delta u \amp= u^3 - u \ge 0 \amp\qquad\amp \ina \Omega',\\
u \amp= 1 \amp\qquad\amp \ona \partial\Omega'.
\end{alignat*}
The weak maximum then implies \(u \le 1\) on all of \(\Omega'\text{.}\) But this contradicts the definition of \(\Omega'\text{.}\) The argument for \(u \ge -1\) is similar.
Solution 2. Thinking about \(\Delta u\) at a maximum or minimum
Suppose for the sake of contradiction that \(M =\sup_\Omega u \gt 1\text{.}\) Since \(\Omega\) is bounded, this supremum must be achieved at some \(x \in \overline
\Omega\text{,}\) and since \(u=0\) on \(\partial \Omega\) we must have \(x \in \Omega\text{.}\) Thus \(x\) is a local maximum of \(u\) and so the Hessian \(D^2 u(x)\) is negative semi-definite. In particular its trace \(\Delta u(x) = \delta_{ij} (D^2
u(x))_{ij} \le 0\) by Lemma 2.31. But from the equation we have
\begin{gather*}
\Delta u(x) = u^3-u = M^3 - M = M(M^2-1) \gt 0,
\end{gather*}
a contradiction. Thus \(\sup_\Omega u \le 1\text{.}\) The argument that \(\inf_\Omega u
\ge 1\) is similar.
Solution 3. Using Lemma 5.4
It is natural to wonder if Lemma 5.4 can be used here, and indeed it can. Suppose for simplicity that \(\Omega\) is connected, although the argument can be generalised to work when \(\Omega\) is disconnected. (Of course on an exam it would be dangerous to make additional assumptions beyond what is given to us in the problem!) First we write the given PDE as \(\Delta u + f(u) = 0\) where \(f \in C^1(\R)\) is given by \(f(z)=z-z^3\text{.}\) Let \(M=\max_{\overline \Omega} u\text{,}\) and suppose for the sake of contradiction that \(M \gt 1\text{.}\) As \(u=0 \lt 1\) on \(\partial\Omega\text{,}\) \(u\) must achieve \(M\) at an interior point of \(\Omega\text{.}\) As \(f(M) \lt 0\) the constant function \(M\) also satisfies
\begin{align*}
\Delta u +f(u) = 0 \amp\gt f(M) = \Delta M + f(M),
\end{align*}
in \(\Omega\text{,}\) as well as \(u \le M\text{.}\) The first part of Lemma 5.4 then implies \(u \equiv M\text{,}\) contradicting the fact that \(u \lt M\) on \(\partial\Omega\text{.}\)
4. (PS8) Laplacian of radially symmetric functions.
Let \(B=B_R(0)\text{,}\) and suppose that \(u \in C^2(B)\) is given by \(u(x)=U(\abs x)\) for some \(C^2\) function \(U\text{.}\) By repeatedly using the chain rule, show that for \(x \in B \without \{0\}\) we have
\begin{equation*}
\Delta u(x) = U''(\abs x) + \frac{N-1}{\abs x}U'(\abs x)\text{.}
\end{equation*}
Hint.
You’ve done a lot of the legwork already in Exercise 2.2.2 and Exercise 2.3.2.
Solution.
As in Exercise 2.3.2 we have
\begin{gather*}
\partial_i u(x)
= \partial_i U(\abs x)
= U'(\abs x)\abs x^{-1} x_i \text{.}
\end{gather*}
Differentiating a second time using the product rule, we find
\begin{align*}
\Delta u(x)
\amp = \partial_i \partial_i u(x)\\
\amp = \partial_i \big[ U'(\abs x)\abs x^{-1} x_i \big]\\
\amp = U''(\abs x)\abs x^{-2} x_i x_i
- U'(\abs x)\abs x^{-3} x_i x_i
+ U'(\abs x) \abs x^{-1} \delta_{ii}\\
\amp = U''(\abs x)
- U'(\abs x)\abs x^{-3} x_i x_i
+ U'(\abs x) \abs x^{-1} \delta_{ii}\\
\amp = U''(\abs x)
+ \frac{N-1}{\abs x} U'(\abs x)
\end{align*}
as desired.
