The extremal solution of a boundary reaction problem

We consider 
 
$\Delta u = 0$ in $ \Omega$, $\qquad \frac{\partial u}{\partial \nu} =\lambda f(u)$ on $\Gamma_1, \qquad u = 0$ on $\Gamma_2$ 
 
where $\lambda>0$, $f(u) = e^u$ or $f(u) = (1+u)^p$, 
$\Gamma_1$, $\Gamma_2$ is a partition of $\partial \Omega$ and 
$\Omega\subset \mathbb R^N$. We determine sharp conditions on the 
dimension $N$ and $p>1$ such that the extremal solution is 
bounded, where the extremal solution refers to the one associated 
to the largest $\lambda$ for which a solution exists.


Introduction
We study the semilinear boundary value problem where λ > 0 is a parameter, f (u) is a nonlinear smooth function of u, Ω ⊂ R N is a smooth, bounded domain and Γ 1 , Γ 2 is a partition of ∂Ω into surfaces separated by a smooth interface. We will assume that f is smooth, nondecreasing, convex and f (0) > 0, lim inf t→+∞ f (t)t f (t) > 1. (4) Assumption (4) is not essential, but it simplifies some of the arguments and holds for the examples f (u) = e u , f (u) = (1 + u) p , p > 1.
We call u * the extremal solution of (1).

Remark 1.2.
Under assumption (4) we have u * ∈ H 1 (Ω). The proof is analogous to the argument for (5) in [4], so we skip it. Proposition 1.1 suggests the following natural question : is u * a bounded solution?
In the context of (5), no complete answer has been given yet. For the case f (u) = e u , that is the original Gelfand problem, it was shown by Joseph and Lundgren [19] that if Ω is a ball, then u * is bounded if and only if N < 10. Crandall and Rabinowitz [11] showed that if f (u) = e u and N < 10 then for any smooth and bounded domain, u * is bounded. Brezis and Vázquez [4] provided a different proof of the unboundedness of u * in the case Ω = B 1 and N ≥ 10 : they established in particular that a singular solution which is stable must be the extremal one. In applying this criterion in dimension N ≥ 10 they use Hardy's inequality valid for N ≥ 3 : Other explicit nonlinearities, for instance f (u) = (1+u) p with p > 1, are considered in these references, but in the general case, little is known. In this direction, we mention the result of Nedev [21], which asserts that for any function f satisfying (2) and (3), and any smooth bounded domain in R N , N ≤ 3, u * is bounded. This result has been extended by Cabré to the case N = 4 and Ω strictly convex [5]. Finally, Cabré and Capella [6] showed that if Ω is a ball and N ≤ 9 then for any nonlinearity f satisfying (2),(3) the extremal solution is bounded. Proving that u * is unbounded seems to be much more difficult. Besides the radial case Dávila and Dupaigne [13] have shown that in domains that are small perturbations of a ball and for the nonlinearities e u and (1 + u) p the extremal solution is singular for large dimensions (N ≥ 11 and N > 2 + 4p p−1 + 4 p p−1 respectively).
Returning to (1), we are interested in determining whether the extremal solution u * is bounded or singular in the cases f (u) = e u and f (u) = (1 + u) p , p > 1.
In any dimension N ≥ 10 there exists a domain Ω ⊂ R N and a partition in smooth sets Γ 1 , Γ 2 of ∂Ω such that u * ∈ L ∞ (Ω).
The proof is an adaptation of the argument of Brezis and Vázquez [4], using a stability criterion. In our case the singular solution has the form u 0 (x) = − log |x| for x ∈ ∂R N + and its linearized stability in dimension N ≥ 10 is obtained thanks to : which holds for N ≥ 3 and where the best constant is given by where Γ is the Gamma function. Inequality (9) is known as Kato's inequality and a proof of it was given by Herbst [18].
We will give here a different proof of this result which offers a sharper version, analogous to improvements of (8) found by Brezis and Vázquez [4] or Vázquez and Zuazua [22] (see also [2,4,12,17,22] for other improved versions of the Hardy inequality (8)) : As a converse to Theorem 1.3 we prove : Theorem 1.5. Let f (u) = e u , N ≤ 9 and suppose Ω ⊂ R N + is open, bounded and satisfies: • Ω is symmetric with respect to the hyperplanes x 1 = 0, . . . , x N −1 = 0, and • Ω is convex with respect to all directions x 1 , . . . , x N −1 . Then the extremal solution u * of (1) belongs to L ∞ (Ω). Remark 1.6. In order to prove Theorem 1.5, one is at first tempted to imitate the classical argument of Crandall and Rabinowitz [11]: roughly speaking, one uses the stability inequality (7) and the equation (1) with test functions of the form ϕ = e ju , j ≥ 1. This does not lead to the optimal dimension N = 9 but applies to general domains (see Proposition 1.7 below). We use instead test functions ϕ, which are not functions of u, but which have the expected behavior of e ju near a singular point, assuming it exists.

This raises the following question
Open Problem 1. Does Theorem 1.5 hold in any smooth bounded domain Ω ⊂ R N where K(x, y) = 2x N N ω N |x − y| −N is the Green's function for the Dirichlet problem in R N + (see e.g. [16]). Clearly, w α > 0 in R N + . Moreover w α is harmonic in R N + and w α extends to a function belonging to It is not difficult to verify that for some constant C(N, α) we have In Section 2 we shall show that A heuristic calculation shows that for (1) with nonlinearity f (u) = (1 + u) p , the expected behavior of a solution u which is singular at 0 ∈ ∂Ω should be u(x) ∼ |x| 1 p−1 . The boundedness of u * is then related to the value of C(N, 1 p−1 ). Observe that C(N, 1 p−1 ) is defined for p > N N −1 . In the sequel, when writing C(N, 1 p−1 ) we will implicitly assume that this condition holds.
where Γ 1 ⊂ ∂R N + and Γ 2 ⊂ R N + and such that the following condition holds • Ω is convex with respect to x and • Π N (Ω) = Γ 1 , where Π N is the projection on ∂R N + . If p C(N, 1 p−1 ) > H N or 1 < p < N N −2 then u * is bounded. In the above, Ω is said to be convex with respect to x if (tx , x N ) + ((1 − t)y , x N ) ∈ Ω whenever t ∈ [0, 1], x = (x , x N ) ∈ Ω and y = (y , Remark 1.9. It is not difficult to verify that the same result holds if • Ω is convex with respect to all directions x 1 , . . . , x N −1 and • Ω is symmetric with respect to the hyperplanes x 1 = 0, . . . , x N −1 = 0. there exists a domain Ω such that u * is singular. Remark 1.11. The condition p C(N, 1 p−1 ) ≤ H N is not enough to guarantee that the extremal solution is singular for some domain. Actually this condition can hold for some values of p in the range N N −1 < p < N N −2 . In this case a singular solution exists in some domains, but it does not correspond to the extremal one. See Theorem 6.2 in [4] for a similar phenomenon.
The organization of the paper is as follows. In Section 2 we derive formula (14) and we prove Theorem 1.4 in Section 3. In Section 4 we analyze the exponential case and give a proof of Theorems 1.3 and 1.5. The proofs of Theorems 1.8 and 1.10 are given in Section 5.

Computation of C(N, α)
We write (12) and a simple change of variables that for all rotations e ∈ O(N − 1). and similarly Differentiating with respect to x N yields and observe that it is independent of x ∈ R N −1 .
Using (15) and the radial symmetry of w in the variables x , there exists a The equation ∆w = 0 is equivalent to The initial condition for v is related to (16) v (0) = −C(N, α).
In addition to these initial conditions we remark that w α is a smooth function in R N + and this together with (17) implies that lim Using the change of variables z = it with i the imaginary unit and defining the new unknown h(z) := v(−iz) equation (18) becomes with initial conditions lim t>0, t→0 On the other hand (19) implies The substitution transforms equation (20) into The general solution to (24) is well known. Indeed, equation (24) belongs to the class of Legendre's equations. Following [1], two linearly independent solutions of (24) are given by the Legendre functions P µ ν (z), Q µ ν (z), which are defined in C \ {−1, 1} and analytic in C \ (−∞, 1] (see [1, Formulas 8.1.2 -8.1.6]). Moreover the limits of P µ ν (z), Q µ ν (z) on both sides of (−1, 1) exist and we shall use the notation and a similar notation for Q µ ν . The solution g of (24) is therefore given by , for appropriate constants c 1 , c 2 . These constants are determined by the initial conditions (21), which imply: In order to evaluate C(N, α), we use also condition (22), which is equivalent to But according to [1,Formulas 8 (23),(29) imply that c 1 = 0 and we obtain from (27),(28) From the properties and formulas in [1] the following values can be deduced: The relations (30),(31),(32) and the values (25) yield formula (14).

Improved Kato inequality
We begin with some remarks on (9). Remark 3.1. a) The singular weight 1 |x| in the right-hand side of (9) is optimal, in the sense that it may not be replaced by 1 |x| α with α > 1. This can be easily seen by in a neighborhood of the origin.
Moreover, the infimum in (10) is not achieved. b) In dimension N = 2 the infimum (10) is zero, see [14]. Nonetheless, if the test-functions ϕ are required to vanish on the half line x 1 > 0 then the infimum has been computed in [14] : c) Using Stirling's formula it is easy to see that can be obtained in a more straightforward way using particular test functions. We give a proof of this at the end of Section 3. Also observe that (34) could be deduced from (35).
Let us explain first informally the idea behind the proof of Theorem 1.4, assuming for a moment that a minimizer w ∈ H 1 (R N + ) of (10) exists. w then satisfies the associated Euler-Lagrange equation: Elementary changes of variables show that given R > 0 and a rotation e ∈ O(N −1), and w e :=w(e(x ), x N ) are also minimizers of (10). Thus it is natural to assume w = w R = w e for all R > 0 and e ∈ O(N − 1). In particular a constant multiple of w solves as defined in (12). Observe that C(N, N −2 2 ) = H N by (16) and hence w is indeed a solution of(36).
Following an idea of Brezis and Vázquez (equation (4.6) on page 453 of [4]), we restate (9) Fix such a ϕ ≡ 0 and let w be the function defined by (12). Notice that, on supp ϕ, w is smooth and bounded from above and from below by some positive constants. Hence v := ϕ w ∈ C ∞ 0 (R N + ) is well defined. Now, ϕ = vw, ∇ϕ = v∇w + w∇v and and by Green's formula But by (16) ∂w ∂ν (x) w(x) = H N |x| for x ∈ ∂R N + and hence, The second term in the right hand side of the above inequality yields the improvement of Kato's inequality when ϕ has support in the unit ball. Now we assume ϕ ∈ C ∞ 0 (R N + \ {0} ∩ B) and, as before, set v = ϕ w . Our aim is to prove that given 1 ≤ q < 2 there exists C > 0 such that In spherical coordinates for some C > 0 and all x ∈ B ∩ R N + . Hence Let us compute the Sobolev norm of ϕ : Define |∇v(rθ)| q |w(rθ)| q dθ dr we have by Hölder's inequality since q < 2.
Using |∇w(x)| ≤ C|x| − N 2 we estimate I 2 : From the classical Hardy inequality and therefore Hölder's inequality yields where we have used q < 2. Gathering (40) and (41) we conclude that (39) holds. Now we pass to the proof of item (d) of Remark 3.1. Proof of (35). We shall first show the inequality One may assume that u = u(r, t) where r = |(x 1 , . . . , x N −1 )| and t = x N . Then where ω N −1 is the volume of the unit ball in R N −1 . But We use now the inequality which is one of the classical version of Hardy's inequality (in dimension N − 1). We obtain To prove we consider, for fixed a > 0 and ε ↓ 0, the functioñ and η ≡ 0 outside of B 2 (0) and a suitable choice of a one obtains (42). We omit the details.

The exponential case
We need the following result that characterizes extremal singular solutions belonging to H 1 (Ω), see [4, Theorem 3.1 ]. The proof is an adaptation of the one in this reference.

Remark 4.2.
We have not shown that there is a unique weak solution of (1) when λ = λ * . A result of Martel [20] guarantees that this is indeed the case for problem (5) and this was used by Brezis and Vázquez in the proof of [4, Theorem 3.1 ]. In our context, we take u * = lim λ λ * u λ as the definition of the extremal solution. Knowing that u * ∈ H 1 (Ω), the proof of [4, Theorem 3.1] shows that λ = λ * and v = u * .
To prove Theorems 1.3 and 1.5 it will be convenient to study the function u 0 defined by where as before K(x, y) = 2x N N ω N |x − y| −N . Then u 0 is harmonic in R N + and u 0 (x) = log 1 |x| for x ∈ ∂R N + , x = 0.
Note that Let r = |x |. Then for some v : [0, ∞) → R such that v(0) = 0. We see that where we let The boundary ∂Ω 0 is not smooth itself but Γ 1 , Γ 2 are, and it can be checked that Proposition 1.1 still holds in this case.
It can be verified that Ω 0 can be written as Proof. We give details for N ≥ 4, the case N = 3 being similar. We need to compute v (0). Calculating ∆u 0 in terms of v (see (45)) we obtain that v satisfies (46) We look at the asymptotics of the two integrals above, as t → ∞. For the second integral, we have And for the first integral, .
Going back to (46), we obtain that Now, recall that for x N > 0, lim r→0 v(x N /r) + log 1 r = u 0 (0, x N ) ∈ R exists and is finite. Hence, we must have .
Proof of Theorem 1.3. We have shown that u 0 defined in (44) is a solution to (1) with Ω = Ω 0 and λ = λ 0,N . This solution satisfies the stability condition (43) if and only if (by scaling) In the Appendix we prove that H N ≥ λ 0,N if and only if N ≥ 10 (47) and this completes the proof of the theorem.
Proof of Theorem 1.5.
We prove the theorem by contradiction, assuming that u * is unbounded. We use an idea of Crandall and Rabinowitz [11], but with different test functions. Let where the constants K φ , K ψ are given by Indeed, since u 0 and φ are harmonic in Ω, Clearly, Γ2 φ ∂u0 ∂ν ≤ C, for some constant C independent of ε. So Similarly, since ψ and w (defined in (12)) are harmonic in Ω, we have As before the boundary terms on Γ 2 are bounded independently of ε so Hence,

Multiplying the equation (1) by φ and integrating by parts twice yields
Let η ∈ C ∞ (R N ) be such that η ≡ 1 in B R (0) where R > 0 is small and fixed, and η = 0 on Γ 2 . Using the stability condition (7) with ηψ yields where the constant C does not depend on ε. Since ψ 2 = φ on ∂R N + combining (49) and (50) we obtain Using (48) we arrive at where ω N −1 is the area of the N − 1 dimensional sphere.
Next we claim that for any given 0 < σ < 1 there exists r(σ) > 0 such that Observe first that for all 0 < λ < λ * the minimal solution u λ is symmetric in the variables x 1 , . . . , x N −1 by uniqueness of the minimal solution and it achieves its maximum at the origin by the moving plane method (see Proposition 5.2 in [7]).

The power case
Proof of Theorem 1.8. We shall give here the proof of the case p C(N, 1 p−1 ) > H N . If p < N N −2 , the boundedness of u * follows from standard techniques, using the Sobolev trace embedding theorem H 1 (Ω) → L 2(N −1) Observe that pv p−1 = where the infimum is taken over the functions ϕ ∈ C ∞ 0 (R N + ) that do not vanish identically on ∂R N + . Assume that u * is singular. For R > 0 and 0 < λ < λ * let where x λ denotes a point of maximum of u λ . Observe that since u λ is positive and harmonic in Ω, x λ ∈ Γ 1 .
For 0 < λ < λ * , we choose R such that u R (0) = 1 i.e. such that λ Then u R verifies The moving plane method implies that the distance of the point x λ ∈ Γ 1 to Γ 1 ∩ Γ 2 stays bounded away from zero, see [7] for this method in the context of non-linear Neumann condition. Thus implies that φ(R) → +∞ as R → 0. (57) Step 1. We have Proof. Suppose not. Define It follows that for some m x > 0 (and decreasing if necessary δ x ) Indeed, because of (59) and u R (y, 0) ≤ v(y, 0) for |y| ≤ r 0 we immediately obtain u R (y, t) < v(y, t) for |y − x| 2 + t 2 < δ 2 x , |y| ≤ r 0 , t > 0. If (60) is false, then there are sequences y k → x, t k → 0 with |y k | > r 0 and t k |y k −x| → ∞ such that v(y k , t k ) ≤ u R (y k , t k ). Then by the mean value theorem there exists a point ξ k in the segment from (y k , t k ) to ( . This proves (60). Suppose x ∈ Γ R 1 is such that |x| = r 0 and u R (x, 0) < v(x, 0). The by continuity there is δ x > 0 such that (60) still holds.
Also, u satisfies By (57) and the previous step we deduce We claim that µ = 1. Let indeed Thenũ is harmonic in R N + and agrees with u on ∂R N + . Since u is bounded by 1 and u is bounded, we see thatũ − u must be a constant. But then, sinceũ(x, 0) → 0 and u(x, 0) → 0 as |x| → ∞ we see that u ≡ũ. Thus This implies µ ≤ µ p and since µ = 0, we conclude µ = 1.
Step 3. Observe that the supremum in (66) is not attained. Otherwise v − u would achieve a minimum at a point x ∈ ∂R N + , where the normal derivative would be zero. By Hopf's lemma, we would have u ≡ v, which is impossible since u is bounded and v is not. Let x k ∈ ∂R N + be such that |x k | → ∞ and u(x k ) v(x k ) → 1. Let u k (x) = |x k | 1 p−1 u(|x k |x).
Since v is invariant under the above transformation we have u k ≤ v in ∂R N + . Thus, for a subsequence we have u k → u 0 and u 0 solves (64). Since u k ( x k |x k | ) → v(y) where y = lim x k |x k | , again using Hopf's lemma we see that u 0 ≡ v. But u 0 satisfies the condition (65), contradicting (56).
To apply Lemma 4.1 we need to verify that u ∈ H 1 (Ω). We are assuming that pC(N, 1 p−1 ) ≤ H N and p ≥ N N −2 . Actually we must have p > N N −2 . For this it is convenient to observe that : Property (69) is direct from (16) and we leave (70) to the appendix. From these properties we deduce that p > N N −2 and therefore u ∈ H 1 (Ω). This solution satisfies the stability condition (43) if and only if (by scaling) which is guaranteed by Kato's inequality (9). Thus we may apply Lemma 4.1 and conclude that u is the extremal solution.