The critical dimension for a fourth order elliptic problem with singular nonlinearity

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1 The critical dimension for a fourth order elliptic problem with singular nonlinearity Craig COWAN Pierpaolo ESPOSITO Nassif GHOUSSOU September 8, 008 Abstract We study the regularity of the extremal solution of the semilinear biharmonic equation u = ( u), which models a simple Micro-Electromechanical System (MEMS) device on a ball R N, under Dirichlet boundary conditions u = νu = 0 on. We complete here the results of [4,, 4] about the existence of a pull-in voltage > 0 such that for (0, ) there exists a minimal (classical) solution u with 0 < u <, while there is none of any kind when >. Our main result asserts that dimension N = 9 is critical for this problem, meaning that the extremal solution u is regular (sup u < ) provided N 8 while u is singular (sup u = ) for N 9, in which case C 0 x 4/3 u (x) x 4/3 on the unit ball, where C 0 := Introduction 3 and := 8(N 3 )(N 8 3 ) 9. The following model has been proposed for the description of the steady-state of a simple Electrostatic MEMS device: α u = ( β Ω u dx + γ ) u + f(x) ( u) +χ R Ω dx ( u) in Ω 0 < u < in Ω u = α ν u = 0 on Ω, where α, β, γ, χ 0, f C(Ω, [0, ]) are fixed, Ω is a bounded domain in R N and 0 is a varying parameter (see for example ernstein and Pelesko [9]). The function u(x) denotes the height above a point x Ω R N of a dielectric membrane clamped on Ω, once it deflects torwards a ground plate fixed at height z =, whenever a positive voltage proportional to is applied. In studying this problem, one typically makes various simplifying assumptions on the parameters α, β, γ, χ, and the first approximation of () that has been studied extensively so far is the equation u = f(x) ( u) in Ω 0 < u < in Ω (S),f u = 0 on Ω, where we have set α = β = χ = 0 and γ = (see for example [6, 8, 9] and the monograph [7]). This simple model, which lends itself to the vast literature on second order semilinear eigenvalue problems, is already Department of Mathematics, University of ritish Columbia, Vancouver,.C. Canada V6T Z. cowan@math.ubc.ca. Dipartimento di Matematica, Università degli Studi Roma Tre, 0046 Roma, Italy. esposito@mat.uniroma3.it. Research supported by M.U.R.S.T., project Variational methods and nonlinear differential equations. Department of Mathematics, University of ritish Columbia, Vancouver,.C. Canada V6T Z. nassif@math.ubc.ca. Research partially supported by the Natural Science and Engineering Research Council of Canada. ()

2 a rich source of interesting mathematical problems. The case when the permittivity profile f is constant (f = ) on a general domain was studied in [6], following the pioneering work of Joseph and Lundgren [3] who had considered the radially symmetric case. The case for a non constant permittivity profile f was advocated by Pelesko [8], taken up by [], and studied in depth in [6, 8, 9]. The starting point of the analysis is the existence of a pull-in voltage (Ω, f), defined as } (Ω, f) := sup { > 0 : there exists a classical solution of (S),f. It is then shown that for every 0 < <, there exists a smooth minimal (smallest) solution of (S),f, while for > there is no solution even in a weak sense. Moreover, the branch u (x) is increasing for each x Ω, and therefore the function u (x) := lim u (x) can be considered as a generalized solution that corresponds to the pull-in voltage. Now the issue of the regularity of this extremal solution which, by elliptic regularity theory, is equivalent to whether sup Ω u < is an important question for many reasons, not the least of which being the fact that it decides whether the set of solutions stops there, or whether a new branch of solutions emanates from a bifurcation state (u, ). This issue turned out to depend closely on the dimension and on the permittivity profile f. Indeed, it was shown in [9] that u is regular in dimensions N 7, while it is not the case for N 8. In other words, the dimension N = 8 is critical for equation (S) (when f =, we simplify the notation (S), into (S) ). On the other hand, it is shown in [8] that the regularity of u can be restored in any dimension, provided we allow for a power law profile x η with η large enough. The case where β = γ = χ = 0 (and α = ) in the above model, that is when we are dealing with the following fourth order analog of (S) ( u) u = in Ω 0 < u < in Ω (P ) u = ν u = 0 on Ω, was also considered by [4,, 4, 9] but with limited success. One of the reasons is the lack of a maximum principle which plays such a crucial role in developing the theory for the Laplacian. Indeed, it is a well known fact that such a principle does not normally hold for general domains Ω (at least for the clamped boundary conditions u = ν u = 0 on Ω) unless one restricts attention to the unit ball Ω = in R N, where one can exploit a positivity preserving property of due to T. oggio [3]. This is precisely what was done in the references mentioned above, where a theory of the minimal branch associated with (P ) is developed along the same lines as for (S). The second obstacle is the well-known difficulty of extracting energy estimates for solutions of fourth order problems from their stability properties. This means that the methods used to analyze the regularity of the extremal solution for (S) could not carry to the corresponding problem for (P ). This is the question we address in this paper as we eventually show the following result. Theorem.. The unique extremal solution u for (P ) in is regular in dimension N 8, while it is singular (i.e, sup u = ) for N 9. In other words, the critical dimension for (P ) in is N = 9, as opposed to being equal to 8 in (S). We add that our methods are heavily inspired by the recent paper of Davila et al. [5] where it is shown that N = 3 is the critical dimension for the fourth order nonlinear eigenvalue problem { u = e u in u = ν u = 0 on, while the critical dimension for its second order counterpart (i.e., the Gelfand problem) is N = 0. In the sequel of the paper, we will always consider problem (P ) on the unit ball. We start by recalling some of the results from [4] concerning (P ), that will be needed in the sequel. We define } := sup { > 0 : there exists a classical solution of (P ),

3 and note that we are not restricting our attention to radial solutions. We will deal also with weak solutions: Definition.. We say u is a weak solution of (P ) if 0 u a.e. in, ( u) u φ = φ ( u), φ C4 ( ) H 0 (). L () and We say u is a weak super-solution (resp. weak sub-solution) of (P ) if the equality is replaced with (resp. ) for φ 0. We introduce the notion of regularity and stability: Definition.. We say a weak solution u of (P ) is regular (resp. singular) if u < (resp. =) and stable (resp. semi-stable) if { µ (u) = inf ( φ) φ } ( u) 3 : φ H 0 (), φ L = is positive (resp. non-negative). The following extension of oggio s principle will be frequently used in the sequel (see [, Lemma 6] and [5, Lemma.4]): Lemma. (oggio s Principle). Let u L (). Then u 0 a.e. in, provided one of the following conditions hold:. u C 4 (), u 0 on, and u = u n = 0 on.. u φ dx 0 for all 0 φ C 4 () H 0 (). 3. u H (), u = 0 and u n 0 on, and u φ 0 for all 0 φ H 0 (). Moreover, either u 0 or u > 0 a.e. in. The following theorem summarizes the main results in [4] that will be needed in the sequel: Theorem.. The following assertions hold:. For each 0 < < there exists a classical minimal solution u of (P ). Moreover u is radial and radially decreasing.. For >, there are no weak solutions of (P ). 3. For each x the map u (x) is strictly increasing on (0, ). 4. satisfies the following bounds: { 3(0N N } ) 8 40N + 7N max, 7 8 where ν denotes the first eigenvalue of in H 0 (). 5. For each 0 < < one has µ (u ) > 0. 4ν 7 Using the stability of u, it can be shown that u is uniformly bounded in H0 () and that u is uniformly bounded in L 3 (). Since now u (x) is increasing, the function u (x) := lim u (x) is well defined (in the pointwise sense), u H0 (), u L 3 () and u is a weak solution of (P ). Moreover u is the unique weak solution of (P ). The second result we list is critical in identifying the extremal solution: Theorem.3. If u H 0 () is a singular weak solution of (P ), then u is semi-stable if and only if (u, ) = (u, ). 3

4 The effect of boundary conditions on the pull-in voltage As in [5], we are led to examine problem (P ) with non-homogeneous boundary conditions such as where α, β are given. ( u) u = in α < u < in (P ),α,β u = α, ν u = β on, First, let us notice that natural restrictions on α and β are necessary. Indeed, let Φ(x) := (α β ) + β x denote the unique solution of { Φ = 0 in () Φ = α, ν Φ = β on. y lemma. the function u Φ is positive in which yields to sup Φ < sup u. To insure that α is a classical sub-solution of (P ),α,β, we impose α and β 0, and condition sup Φ < rewrites as α β <. We will then say that the pair (α, β), with β 0, is admissible if α β <. This section will be devoted to obtaining analogous results for (P ),α,β (when (α, β) is an admissible pair) which were valid for (P ). To cut down on notations we won t always indicate the α and β. For example in this section u and u will denote the minimal and extremal solution of (P ),α,β. We first introduce a notion of weak solution: Definition.. We say u is a weak solution of (P ),α,β if α u a.e. in, ( u) L () and (u Φ) φ = φ ( u), φ C4 ( ) H 0 (), where Φ is given in (). We say u is a weak super-solution (resp. weak sub-solution) of (P ),α,β equality is replaced with (resp. ) for φ 0. if the We now define as before := sup{ > 0 : (P ),α,β has a classical solution} and := sup{ > 0 : (P ),α,β has a weak solution}. Observe that by the Implicit Function Theorem, we can classically solve (P ),α,β for small s. Therefore, (and also ) is well defined. Let now U be a weak super-solution of (P ),α,β. Recall the following existence result: Theorem. ([]). For every 0 f L () there exists a unique 0 u L () which satisfies u φ = fφ for all φ C 4 ( ) H 0 (). 4

5 We can introduce the following weak iterative scheme: u 0 = U and (inductively) let u n, n, be the solution of (u n Φ) φ φ = ( u n ) φ C 4 ( ) H0 () given by theorem.. Since 0 is a sub-solution of (P ),α,β, inductively it is easily shown by lemma. that α u n+ u n U for every n 0. Since by Lebesgue theorem the function u = there holds ( u n ) ( U) L (), lim u n is a weak solution of (P ),α,β so that α u U. Namely, n + Theorem.. Assume the existence of a weak super-solution U of (P ),α,β. solution u of (P ),α,β so that α u U a.e. in. Then there exists a weak In particular, for every (0, ) we find a weak solution of (P ),α,β. In the same range of s this is still true for regular weak solutions as shown by: Theorem.3. Let (α, β) be an admissible pair and u be a weak solution of (P ),α,β. 0 < µ < there is a regular solution. Then for every Proof. Let ε (0, ) be given and let ū = ( ε)u + εφ, where Φ is given in (). We have that sup ū ( ε) + ε sup Φ <, inf ū ( ε)α + ε inf Φ = α and for every 0 φ C 4 ( ) H0 () there holds: (ū Φ) φ = ( ε) (u Φ) φ φ = ( ε) ( u) = ( ε) 3 φ ( ū + ε(φ )) ( ε)3 φ ( ū). Note that 0 ( ε)( u) = ū+ε(φ ) < ū. So ū is a weak super-solution of (P ) ( ε)3,α,β so that sup ū <. y theorem. we get the existence of a weak solution w of (P ) ( ε)3,α,β so that α w ū. In particular, sup w < and w is a regular weak solution. Since ε (0, ) is arbitrarily chosen, the proof is done. Theorem.3 implies the existence of a regular weak solution U for every (0, ). Introduce now a classical iterative scheme: u 0 = 0 and (inductively) u n = v n + Φ, n, where v n H0 () is the (radial) solution of v n = (u n Φ) = ( u n ) in. (3) Since v n H0 (), u n is also a weak solution of (3), and by lemma. we know that α u n u n+ U for every n 0. Since sup u n sup U < for n 0, we get that ( u n ) L () and the existence of v n is guaranteed. Since v n is easily seen to be uniformly bounded in H0 (), we have that u := lim n does hold pointwise and weakly in H (). y Lebesgue theorem, we have that u is a radial n + weak solution of (P ),α,β so that sup u sup U <. y elliptic regularity theory [] u C ( ) and u Φ = ν (u Φ) = 0 on. So we can integrate by parts to get u φ = (u Φ)φ = 5 (u Φ) φ = φ ( u )

6 for every φ C 4 ( ) H0 (). Hence, u is a radial classical solution of (P ),α,β showing that =. Moreover, since Φ and v := u Φ are radially decreasing in view of [0], we get that u is radially decreasing too. Since the argument above shows that u < U for any other classical solution U of (P ) µ,α,β with µ, we have that u is exactly the minimal solution and u is strictly increasing as. In particular, we can define u in the usual way: u (x) = lim u (x). Finally, we have that Theorem.4. Let (α, β) be an admissible pair. Then < +. Proof. Let u be a classical solution of (P ),α,β and let (ψ, ν ) denote the first eigenpair of in H0 () with ψ > 0. Now, let C be such that (β ψ α ν ψ) = C ψ. Multiplying (P ),α,β by ψ and then integrating by parts one arrives at ( ) ( u) ν u C ψ = 0. Since ψ > 0 there must exist a point x where Since α < u( x) <, hence one can conclude that which shows that < +. In conclusion, we have shown that ( u( x)) ν u( x) C 0. sup (ν u + C)( u) α<u< Theorem.5. Let (α, β) be an admissible pair. Then (0, + ) and we have that:. For each 0 < < there exists a classical, minimal solution u of (P ),α,β. Moreover u is radial and radially decreasing.. For each x the map u (x) is strictly increasing on (0, ). 3. For > there are no weak solutions of (P ),α,β.. Stability of the minimal branch of solutions This section is devoted to the proof of the following stability result for minimal solutions: Theorem.6. Suppose (α, β) is an admissible pair.. The minimal solution u is stable: µ (u ) > 0, and is the unique semi-stable H () weak solution of (P ),α,β.. The function u := lim u is a well-defined semi-stable H () weak solution of (P ),α,β. 3. When u is classical solution, then µ (u ) = 0 and u is the unique H () weak solution of (P ),α,β. 4. If v is a singular, semi-stable H () weak solution of (P ),α,β, then v = u and = 6

7 In theorem.6 we use the notion of H () weak solutions, which is an intermediate class between classical and weak solutions: Definition.. We say u is a H () weak solution of (P ),α,β if u Φ H0 (), α u a.e. in, ( u) L () and φ u φ = ( u), φ C4 ( ) H0 (), where Φ is given in (). We say u is a H () weak super-solution (resp. H () weak sub-solution) of (P ),α,β if for φ 0 the equality is replaced with (resp. ) and u α (resp. ), ν u β (resp. ) on. The crucial tool is a comparison result which is valid exactly in this class: Lemma.. Let (α, β) be an admissible pair and u be a semi-stable H () weak solution of (P ),α,β. Assume U is a H () weak super-solution of (P ),α,β so that U Φ H 0 (). Then. u U a.e. in ;. If u is a classical solution and µ (u) = 0 then U = u. Proof. (i) Define w := u U. Then by means of the Moreau decomposition [7] for the biharmonic operator there exist w, w H0 (), with w = w + w, w 0 a.e., w 0 in the H () weak sense and w w = 0. y lemma. w 0 a.e. in. Given 0 φ Cc () we have w φ (f(u) f(u))φ, where f(u) := ( u). Since u is semi-stable one has f (u)w ( w ) = w w Since w w one has which re-arranged gives f (u)ww (f(u) f(u))w, fw 0, (f(u) f(u))w. where f(u) = f(u) f(u) f (u)(u U). The strict convexity of f gives f 0 and f < 0 whenever u U. Since w 0 a.e. in one sees that w 0 a.e. in. The inequality u U a.e. in is then established. (ii) Since u is a classical solution, it is easy to see that the infimum in µ (u) is attained at some φ. The function φ is then the first eigenfunction of in H0 (). Now we show that φ is of fixed sign. ( u) 3 Using the above decomposition, one has φ = φ + φ where φ i H0 () for i =,, φ 0, φ φ = 0 and φ 0 in the H0 () weak sense. If φ changes sign, then φ 0 and φ < 0 in (recall that either φ < 0 or φ = 0 a.e. in ). We can write now: 0 = µ (u) ( (φ φ )) f (u)(φ φ ) (φ φ ) < ( φ) f (u)φ φ = µ (u) in view of φ φ < φ φ in a set of positive measure, leading to a contradiction. So we can assume φ 0, and by the oggi s principle we have φ > 0 in. For 0 t define g(t) = [tu + ( t)u] φ f(tu + ( t)u)φ, 7

8 where φ is the above first eigenfunction. Since f is convex one sees that g(t) [tf(u) + ( t)f(u) f(tu + ( t)u)] φ 0 for every t 0. Since g(0) = 0 and g (0) = (U u) φ f (u)(u u)φ = 0, we get that g (0) = f (u)(u u) φ 0. Since f (u)φ > 0 in, we finally get that U = u a.e. in. ased on lemma.(3) a more general version of lemma. is available: Lemma.. Let (α, β) be an admissible pair and β 0. Let u be a semi-stable H () weak sub-solution of (P ),α,β with u = α, ν u = β β on. Assume that U is a H () weak super-solution of (P ),α,β with U = α, ν U = β on. Then U u a.e. in. Proof. Let ũ H0 () denote a weak solution to ũ = (u U) in. Since ũ u + U = 0 and ν (ũ u + U) 0 on, by lemma. one has that ũ u U a.e. in. y means of the Moreau decomposition [7] we write ũ as ũ = w + v, where w, v H0 (), w 0 a.e. in, v 0 in a H () weak sense and w v = 0. Then for 0 φ C4 ( ) H0 () one has ũ φ = (u U) φ (f(u) f(u))φ. In particular, we have that Since by semi-stability of u we get that ũ w (f(u) f(u))w. f (u)w ( w) = ũ w, f (u)w (f(u) f(u))w. y lemma. we have v 0 and then w ũ u U a.e. in. So we see that 0 (f(u) f(u) f (u)(u U)) w. The strict convexity of f implies as in lemma. that U u a.e. in. We need also some a-priori estimates along the minimal branch u : Lemma.3. Let (α, β) be an admissible pair. Then one has (u Φ) ( u ) 3 u Φ ( u ), where Φ is given in (). In particular, there is a constant C > 0 so that ( u ) + ( u ) 3 C (4) for every (0, ). 8

9 Proof. Testing (P ),α,β on u Φ C 4 ( ) H 0 (), we see that u Φ ( u ) = u (u Φ) = ( (u Φ)) (u Φ) ( u ) 3 in view of Φ = 0. In particular, for δ > 0 small we have that { u Φ δ} ( u ) 3 δ { u Φ δ} δ by means of Young s inequality. Since for δ small for some C > 0, we get that { u Φ δ} for some C > 0 and for every (0, ). Since ( u ) = u Φ + { u Φ δ} (u Φ) ( u ) 3 δ ( u ) 3 + C δ ( u ) 3 C ( u ) 3 C ( u Φ ( u ) δ ( u ) + C δ + C in view of Young s and Hölder s inequalities, estimate (4) is finally established. ( u ) ) 3 ( u ) 3 We conclude now with Proof (of theorem.6): () Since u <, the infimum defining µ (u ) is achieved at a first eigenfunction for every (0, ). Since u (x) is increasing for every x, it is easily seen that µ (u ) is an increasing, continuous function on (0, ). Define := sup{0 < < : µ (u ) > 0}. We have that =. Indeed, otherwise we would have that µ (u ) = 0, and for every µ (, ) u µ would be a classical super-solution of (P ),α,β. A contradiction arises since lemma. implies u µ = u. Finally, lemma. guarantees uniqueness in the class of semi-stable H () weak solutions. () y (4) it follows that u u in a pointwise sense and weakly in H (), and u L 3 (). In particular, u is a H () weak solution of (P ),α,β which is also semi-stable as limiting function of the semi-stable solutions {u }. (3) Whenever u <, the function u is a classical solution, and by the Implicit Function Theorem we have that µ (u ) = 0 to prevent the continuation of the minimal branch beyond. y lemma. u is then the unique H () weak solution of (P ),α,β. An alternative approach which we do not pursue here based on the very definition of the extremal solution u is available in [4] when α = β = 0 (see also [5]) to show that u is the unique weak solution of (P ), regardless of whether u is regular or not. (4) If <, by uniqueness v = u. So v is not singular and a contradiction arises. y theorem.5(3) we have that =. Since v is a semi-stable H () weak solution of (P ),α,β and u is a H () weak super-solution of (P ),α,β, we can apply lemma. to get v u a.e. in. Since u is a semi-stable solution too, we can reverse the roles of v and u in lemma. to see that v u a.e. in. So equality v = u holds and the proof is done. 9

10 3 Regularity of the extremal solution for N 8 We now return to the issue of the regularity of the extremal solution in problem (P ). Unless stated otherwise, u and u refer to the minimal and extremal solutions of (P ). We shall show that the extremal solution u is regular provided N 8. We first begin by showing that it is indeed the case in small dimensions: Theorem 3.. u is regular in dimensions N 4. Proof. As already observed, (4) implies that f(u ) = ( u ) L 3 (). Since u is radial and radially decreasing, we need to show that u (0) < to get the regularity of u. The integrability of f(u ) along with elliptic regularity theory shows that u W 4, 3 (). y the Sobolev imbedding theorem we get that u is a Lipschitz function in. Now suppose u (0) = and N 3. Since u C x in for some C > 0, one sees that = C 3 x 3 ( u ) 3 <. A contradiction arises and hence u is regular for N 3. For N = 4 we need to be more careful and observe that u C, 3 ( ) by the Sobolev imbedding theorem. If u (0) =, then u (0) = 0 and u C in x 4 3 for some C > 0. We now obtain a contradiction exactly as above. We now tackle the regularity of u for 5 N 8. We start with the following crucial result: Theorem 3.. Let N 5 and (u, ) be the extremal pair of (P ). When u is singular, then ( where C 0 := ) 3 and := 8(N 3 )(N 8 3 ) 9. u (x) C 0 x 4 3 in, Proof. First note that theorem.(4) gives the lower bound: For δ > 0, we define u δ (x) := C δ x 4 3 with C δ := u δ H loc (RN ), u δ 8 40N + 7N =. (5) 8 ( ) + δ 3 >. Note that since N 5, we have that L 3 loc (RN ) and u δ is a H weak solution of u δ = + δ ( u δ ) in R N. We claim that u δ u in and, letting δ 0, the proof would be done. Assume by contradiction that the set Γ := {r (0, ) : u δ (r) > u (r)} is non-empty and let r = sup Γ. Since u δ () = C δ < 0 = u () 0

11 in view of C δ >, we have that 0 < r < and one infers that α := u (r ) = u δ (r ), β := (u ) (r ) u δ(r ). Setting u δ,r (r) = r 4 3 (u δ (r r) )+, we easily see that u δ,r is a H () weak super-solution of (P ) +δ,α,β, where α := r 4 3 (α ) +, β := r 3 β. Similarly, let us define u r (r) = r 4 3 (u (r r) )+. The dilation map w w r with w r (r) = r 4 3 is a correspondence between solutions of (P ) on and of (P ) 4 on, r 3 r,0 (defined in the obvi- H integrability. In particular, (u r, ) is the extremal pair of (P ) 4 on, r 3 r,0 ous way). Moreover, u r is a singular semi-stable H () weak solution of (P ),α,β. (w(r r) )+ which preserves the Since u is radially decreasing, we have that β 0. Define the function w as w(x) := (α β β )+ x +γ(x), where γ is a solution of γ = in with γ = ν γ = 0 on. Then w is a classical solution of { w = in w = α, ν w = β on. Since ( u r ), by lemma. we have u r w a.e. in. Since w(0) = α β + γ(0) and γ(0) > 0, the bound u r a.e. in yields to α β <. Namely, (α, β ) is an admissible pair and by theorem.6(4) we get that (u r, ) coincides with the extremal pair of (P ),α,β in. Since (α, β ) is an admissible pair and u δ,r is a H () weak super-solution of (P ) +δ,α,β, by theorem. we get the existence of a weak solution of (P ) +δ,α,β. Since + δ >, we contradict the fact that is the extremal parameter of (P ),α,β. Thanks to this lower estimate on u, we get Theorem 3.3. Let 5 N 8. The extremal solution u of (P ) is regular. Proof. Assume that u is singular. For ε > 0 set ψ(x) := x 4 N +ε and note that ( ψ) = (H N + O(ε)) x N+ε, where H N := N (N 4). 6 Given η C0 (), and since N 5, we can use the test function ηψ H0 () into the stability inequality to obtain ψ ( u ) 3 ( ψ) + O(), where O() is a bounded function as ε 0. y theorem 3. we find that and then Computing the integrals one arrives at ψ x 4 ( ψ) + O(), x N+ε (H N + O(ε)) x N+ε + O(). H N + O(ε). As ε 0 finally we obtain H N. Graphing this relation one sees that N 9.

12 We can now slightly improve the lower bound (5): Corollary 3.. In any dimension N, we have > = 8(N 3 )(N 8 3 ). (6) 9 Proof. The function ū := x 4 3 is a H () weak solution of (P ),0, 4 3. If by contradiction =, then ū is a H () weak super-solution of (P ) for every (0, ). y lemma. we get that u ū for all <, and then u ū a.e. in. If N 8, u is then regular by theorems 3. and 3.3. y theorem.6(3) there holds µ (u ) = 0. Lemma. then yields that u = ū, which is a contradiction since then u will not satisfy the boundary conditions. If now N 9 and =, then C 0 = in theorem 3., and we then have u ū. It means again that u = ū, a contradiction that completes the proof. 4 The extremal solution is singular for N 9 In this section we will need the following improved Hardy-Rellich inequality, which is valid for N 5 (see [0] and references therein): ( ψ) H N ψ x 4 + C ψ ψ H 0 () where H N := N (N 4) 6 is optimal and C > 0. As in the previous section (u, ) denotes the extremal pair of (P ). We first show the following upper bound on u. Lemma 4.. Let N 9. Then u x 4 3 in. Proof. Let us recall that := 8(N 3 )(N 8 3 ) 9. If =, by the proof of corollary 3. we know that u ū. Now suppose <. We claim that u ū for all (, ). Fix and assume by contradiction that R := inf{0 R : u < ū in (R, )} > 0. y the boundary conditions one has that u (r) < ū(r) as r. Hence, 0 < R < and α := u (R ) = ū(r ) and β := u (R ) ū (R ). Introduce as in the proof of theorem 3. (u ) R and (ū) R. We have that (u ) R is a classical super-solution of (P ),α,β, where α := R 4 3 (α ) +, β := R 3 β. Note that (ū) R is a H () weak sub-solution of (P ),α,β which is also semi-stable in view of H N and the Hardy inequality. y lemma. we deduce that (u ) R (ū) R in. Note that, arguing as in the proof of theorem 3., (α, β ) is an admissible pair. We have shown that u ū in R and a contradiction arises in view of lim ū(x) = and u <. x 0 Hence, u ū in for every (, ), and in particular u ū in is still true. In large dimensions, we have the following upper bound on. Proposition 4.. H N for all N 3.

13 Proof. Let N 3 and set φ 0 := ( x 4 3 x ). Define u := ū φ 0 H0 () and note that u L3 (), 0 u in and u 7 ( u) in \ {0}. So u is H () weak sub-solution of (P ) µ, µ = 7. Since 7 H N, by φ 0 0 we get that ψ µ ( u) 3 H ψ N ( x φ 0 ) H ψ 3 N x 4 ( ψ) for all ψ H 0 (). Hence u is also semi-stable. If 7 <, by lemma. one has u u 7 which contradicts the fact that u is singular. Hence 7 and so H N. The approach for showing that u is singular for N 9, will depend on the sign of H N. Theorem 4.. Let N 9 and H N. Then the extremal solution u of (P ) is singular. Proof. Let ψ Cc has () with ψ =. Then by lemma 4. and the improved Hardy-Rellich inequality one ( ψ) ψ ( u ) 3 ( ψ) ψ H N x 4 C. Hence µ (u ) > 0 and u must be singular: otherwise, one could use the Implicit Function Theorem to continue the minimal branch past. y proposition 4. and theorem 4. we know that u is singular whenever N 3. For 9 N 30 we argue as follows. Proposition 4.. Let 9 N 30 and define φ 0 := 4 3β ( x α x α+β ). Then there exists α, β > 0 such that x 8 3 φ 0 H N ( x φ 0 ) in. (7) Proof. Using Maple one can check that the result holds true for each 9 N 30 where one takes α := 5 and β := 0. Theorem 4.. N 30 and > H N. Then the extremal solution u is singular. Proof. Let φ 0 be as in proposition 4. and note that < H N. Now define u := ū φ 0 H0 (). Observe that 0 u in, u L3 () and by proposition 4. u H N ( u) in. Since by φ 0 0 and the Hardy inequality ψ H N ( u) 3 = H ψ N ( x φ 0 ) H 3 N ψ x 4 ( ψ) for all ψ H0 (), we get that u is a semi-stable H () weak sub-solution of (P ) H N. Since > H N, we have that u is a H () weak super-solution of (P ) H N. y lemma. one sees that u u a.e. in and hence u is singular. 3

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