#### Abstract

We study the following semilinear biharmonic equation , in , and , on where is the unit ball in and is the exterior unit normal vector. We prove the existence of such that for there exists a minimal (classical) solution , which satisfies . In the extremal case , we prove the existence of a weak solution which is the unique solution even in a very weak sense. Besides, several new difficulties arise and many problems still remain to be solved. We list those of particular interest in the final section.

#### 1. Introduction and Results

In the last forty years, a great deal has been written about existence and multiplicity of solutions to nonlinear second order elliptic problems in bounded and unbounded domains of . Important achievements on this topic have been made by applying various combinations of analytical techniques, which include the variational and topological methods. For the latter, the fundamental tool which has been widely used is the maximum principle. However, for higher order problems, a possible failure of the maximum principle causes several technical difficulties, which attracted the interest of many researchers. In particular, recently, fourth order equations with a singular nonlinearity have been studied extensively. The motivation for considering these equations stems from a model for the steady states of a simple microelectromechanical system (MEMS) which has the general form (see, e.g., [1]) where denotes the biharmonic operator, is a smooth bounded domain, denotes the outward pointing unit normal to and are physically relevant constants, represents the permittivity profile, and is a constant which is increasing with respect to the applied voltage.

Let and in the above model, we obtain, replacing with . Because of the lack of a “maximum principle,” which plays such a crucial role in developing the theory for the Laplacian, for with Dirichlet boundary condition in general domains (i.e., ), very little is known about . As far as we are aware, only the paper [2] studies this problem for general domains. However, if and is a ball, has recently been studied extensively; see, for example, [3–9] and their references. One of the reasons to study in a ball is that a maximum principle holds in this situation; see [10], and so some tools that are well suited for the corresponding second equation can work for . The second reason is that one can easily find an explicit singular radial solution, denoted by , of for and a suitable parameter which satisfy the first boundary condition but not the second. The singular radial solution, called “ghost” singular solution, plays a fundamental role to characterize the “true” singular solution; see, in particular, [4].

In this paper, we will focus essentially our attention on the case where and the is a ball; namely, The corresponding second order problem, which is related to the general study of singularities of minimal hypersurfaces of Euclidean space, has been studied by Meadows; see [11]. In that case, however, the start point was an explicit singular solution (i.e., ) with parameter . When turning to the biharmonic problem , one cannot find any explicit singular solution even “ghost” singular solution, which causes several technical difficulties. The first purpose of the present paper is to extend some well-known results relative to . The second (and perhaps the most important) purpose of the present paper is to emphasize some striking differences between and .

##### 1.1. Preliminaries

Besides classical solution that is which satisfies , let us introduce the class of weak solutions we will be dealing with. We denote by the usual Sobolev space which can be defined by completion as follows: which is a Hilbert space endowed with the scalar product:

*Definition 1. *One says that is a weak solution of provided almost everywhere, , and

When in (3) the equality is replaced by the inequality (resp. ) and , we say that is a weak supersolution (resp., weak subsolution) of (3) provided the following boundary conditions are satisfied: (resp. ) and (resp. ) on .

*Definition 2. *One calls a solution of minimal if almost everywhere in for any further solution of .

If is a classical solution of , then the linearized operator at : which yields the following notion of stability.

*Definition 3. *A classical solution of is semistable provided
If , we say that is stable.

As far as we are concerned with weak solutions, the linearized operator is no longer well defined; however, we introduce the following weaker notion of stability.

*Definition 4. *A weak solution to is said to be weakly stable if and the following holds:

According to the class of solutions which we consider, let us introduce the following values:

*Remark 5. *Clearly, a classical solution is also a weak solution, so that one has . Moreover, by standard elliptic regularity theory for the biharmonic operator [12], any weak solution of which satisfies turns out to be smooth.

Besides, we give a notion of ()-weak solutions, which is an intermediate class between classical and weak solutions.

*Definition 6. *One says that is a ()-weak solution of (3) if and if
One says that is a ()-weak supersolution (resp. ()-weak subsolution) of if for the equality is replaced with (resp. ) and (resp. ), (resp. ) on .

##### 1.2. Main Results

In order to state our results, we denote by the first eigenvalue of the biharmonic operator on with Dirichlet boundary conditions, which is characterized variationally as follows: It is well known that is simple, isolated and that the corresponding eigenfunctions , spherically symmetric, radially decreasing and do not change sign.

We may now state the following theorem.

Theorem 7. *There exists such that, for , possesses a minimal classical solution, denoted by , which is positive and stable. Moreover, satisfies the following bounds:
*

It is remarkable that at there is an immediate switch from existence of regular minimal solutions to nonexistence of any (even singular) solution. The only possibly singular minimal solution corresponds to . This result is known from [13] for the corresponding second order problems, but the method used there may not be carried over to fourth order problems. Nevertheless, the result extends to biharmonic case in the following theorem.

Theorem 8. *The following holds:
**
In particular, for there are no solutions, even in the weak sense. Furthermore, for almost every , there exists
**
and is a weakly stable -weak solution of , which is called the extremal solution.**If , then the extremal solution of is smooth; that is, exists in the topology of . It is the unique regular solution to .*

From the above theorem, we note that the function exists in any dimension and does solve in the weak sense and it is a classical solution in dimensions . This will allow us to start another branch of nonminimal (unstable) solutions. Besides, inspired by [3, 14, 15], we get the following uniqueness of the extremal solution of , which gives Theorem 9.

Theorem 9. *Let be a weak supersolution of with parameter . Then, ; in particular, has a unique weak solution.*

From this theorem, we know that there are no strict supersolutions to .

Corollary 10. *Let be a weak solution of such that . Then, is weakly stable if and only if and .*

We may also characterize the uniform convergence to 0 of as by giving the precise rate of its extinction.

Theorem 11. *For all , let be the minimal solution of and let
**
Then, for all and all , and
*

##### 1.3. Key Ingredients

Now, we give some comparison principles which will be used throughout the paper.

Lemma 12 (Boggio's principle, [10]). *If satisfies
**
then in .*

Lemma 13. *Let and suppose that
**
for all such that in , . Then, in . Moreover, or almost everywhere, in .*

For a proof, see Lemma 17 in [16]. From this lemma, we know that any solution of is necessarily positive almost everywhere inside the ball.

Lemma 14. *If is radial, in in the weak sense, that is,
**
and , then in .*

*Proof. *The proof is standard; see [15]; we give a proof here for the sake of completeness. We only deal with the case for simplicity. Solve
in the sense that and for all . Then, in by Lemma 24.

Let so that in . Define . Then, in and since is radial we find that is a constant. It follows that . Using the boundary conditions, we deduce and , which imply .

Lemma 15. *Let almost everywhere. Then, there exists a unique such that and
**
Moreover, there exists which does not depend on such that .*

*Proof. *The proof is standard; see [16]; we give a proof here for the sake of completeness. The uniqueness is clear. Indeed, let and be two solutions of (19). Then, satisfies
Given any , let be the solution of
It follows that
Since is arbitrary, we deduce that .

For the existence, given an integer , we set , so that as in . Let be the solution of
The sequence is clearly monotone nondecreasing. It is also a Cauchy sequence in since
where is defined by
Hence,
Passing to the limit in (23) (after multiplication by ), we obtain (19) and according to Lemma 13. Finally, taking in (19), we obtain
and the proof is completed.

Proposition 16. *Assume the existence of a weak supersolution of . Then, there exists a weak solution of so that almost everywhere in .*

*Proof. *By means of a standard monotone iteration argument, set and define recursively as the unique solution of
then we have
and Lemma 24 yields almost everywhere for all . Since
the claim follows from the Lebesgue convergence theorem.

We complete these preliminary results by proving a key lemma which provides a comparison principle.

Lemma 17. *Assume is a weakly stable ()-weak subsolution of and is ()-weak supersolution of . Then,*(1)* almost everywhere in ;*(2)*if is a classical solution such that and is any classical supersolution of , then .*

*Proof. *(1) Define . Then, by the Moreau decomposition [9] for the biharmonic operator, there exists , with , almost everywhere, in the weak sense, and
By Lemma 12, we have that almost everywhere in .

Given now , we have that
where . Since is stable, one has
Since , one also has
which once rearranged gives
where . The strict convexity of gives and whenever . Since almost everywhere in , one sees that almost everywhere in . The inequality almost everywhere in is then established.

(2) Let be the first eigenfunction of in ; we, now, for , define
where is the above first eigenfunction. Since is convex, one sees that
for every . Since and
we get that
Since in , we finally get that almost everywhere in .

#### 2. Existence Results: Proofs of Theorems 7 and 8

##### 2.1. The Branch of Minimal Solutions

Let us define

Proposition 18. *For all , there exists a minimal classical solution of which is smooth and stable. Moreover, *(i)*the map *,
* for **, is differentiable and strictly increasing;*(ii)*the map ** is decreasing on **;*(iii)*let ** be a regular solution of ** for **,; if ** is not the minimal solution, then **.*

*Proof. *First, we show that does not consist of just . To this end, let be the first eigenfunction of the biharmonic operator subject to Dirichlet boundary conditions on which we normalize by and let be the corresponding eigenvalue. Next, we are going to prove that for the function is a supersolution of as long as is sufficiently small. We have
moreover,
provided that

Notice that
and that on . Thus, looking at the function , for , it is easily seen that we can choose sufficiently small such that
Since is a subsolution of , the classical subsuper solution theorem provides a classical solution to . With such function , we can use the Boggio principle to show straightforwardly that the iterative scheme
gives rise to a monotone sequence satisfying
for all . Therefore, the minimal solution is obtained as the increasing limit
Again from the Boggio positivity preserving property (Lemma 12), we obtain ; in particular, from standard elliptic regularity theory for the biharmonic operator, it follows that is smooth. In order to prove stability, let us argue as follows: set
clearly . Now, suppose by contradiction that and let sufficiently small such that and is the corresponding minimal solution. By the definition and left continuity of the map , we have necessarily . Since is a supersolution of , by Lemma 17, we get and thus , a contradiction.

Since each is stable, then by setting , we get that is invertible for . It then follows from Implicit Function Theorem that is differentiable with respect to .

Now, we prove that the map is strictly increasing on . Consider and their corresponding minimal positive solutions and , and let be a solution for . The same as the above iterative scheme, we have
and in particular in . Therefore, for all .

Finally, by differentiating with respect to and since is nondecreasing, we get
Applying the strong maximum principle, we conclude that on for all .

That is decreasing follows easily from the variational characterization of , the monotonicity of , and the monotonicity of with respect to , and the proof of (ii) is completed.

Now, we give the proof of (iii). Let be the minimal solution for so that . If the linearization around had nonnegative first eigenvalue, then Lemma 17 would also yield so that and necessarily coincide, a contradiction.

##### 2.2. Weak Solutions versus Classical Solutions

Lemma 19. *Let be a weak solution of with . Then, for sufficiently small, the problem possesses a classical solution.*

*Proof. *Let be the unique solution of
provided by Lemma 15. By hypothesis, we have
By uniqueness, we get
whereas Lemma 13 yields almost everywhere in and hence we may assume
Therefore,
thus is a weak supersolution of and Proposition 16 yields a weak solution of which satisfies
and then classical by Remark 5.

*Remark 20. *From this lemma, we know that ; in what follows, we always denote by the largest possible value of such that has a solution, unless otherwise stated.

Proposition 21. *Up to a subsequence, the convergence
**
holds in and the extremal solution satisfies
**
In particular, the extremal solution is weakly stable, and if , then .*

*Proof. *Since is stable, we have
Next, it is easy to check that the following elementary inequality holds: there exists a constant such that
which, used in (60), yields
where is independent of . From the above inequality, we get
Therefore, we may assume in and, by monotone convergence theorem (59), holds after integration by parts. Since for all , in particular we have
and passing to the limit as , we obtain that is weakly stable. Finally, if and hence is a classical solution of , the linearized operator at
is well defined on the space . If , then the Implicit Function Theorem applied to the function
would yield a solution for contradicting the definition of ; thus .

Corollary 22. *There exists a constant independent of such that, for each , the minimal solution satisfies .*

*Proof. *From Proposition 21, we have
So
From this, we easily obtain , and the proof is completed.

Corollary 23. *For dimensions , the extremal solution is regular; that is, exists in the topology of .*

*Proof. *Since is radial and radially decreasing, we need just to show that to get the regularity of . Since according to Corollary 22, we have that by the standard elliptic regularity theory. And then by the Sobolev imbedding theorem, we have . So if , one can easily see that . As , we get
hence
A contradiction arises, so is regular for .

##### 2.3. The Upper and Lower Bounds for

Lemma 24. *Consider
**
where is the first eigenvalue of in .*

*Proof. *Let be a solution of and let denote the first eigenpair of in with ; then
and this implies
Since , there must exist a point where
And one can conclude that .

The lower bound for is obtained by finding a suitable supersolution. For example, if for some parameter there exists a supersolution, then by Proposition 18.

Lemma 25. *For , one has
*

*Proof. *For any and , let . Then, by direct calculation, we find the following facts:
So we have
Now, let and ; we have

Also for any take ; one concludes from (78) that
Set ; one can obtain that

Choosing , , one concludes that is a supersolution of and according to Proposition 18. Besides, we consider the function
which satisfies for and
Now, the idea is to obtain from a supersolution of , for a suitable choice of and for in a suitable range of the form . For simple calculation, we have
and thus
from which we deduce that
and the proof is completed.

We complete this section by giving proofs of Theorems 7 and 8.

*Proofs of Theorems 7 and 8.* The proof of Theorem 7 follows from Proposition 18 and Lemmas 24 and 25. For the proof of Theorem 8, we only need to prove the uniqueness of the regular extremal solution ; the other parts of Theorem 8 follow from Lemma 19 and Corollary 23. Indeed, if the extremal solution is regular, we can easily check that by Implicit Function Theorem, since, otherwise, we can continue the minimal branch beyond . And then the uniqueness follows from (ii) of the Lemma 17.

#### 3. Uniqueness of the Extremal Solution: Proof of Theorem 9

*Proof of Theorem 9. *Suppose that satisfies
and . Notice that the construction of minimal solutions in Proposition 18 for carries over to , but just in the weak sense; precisely, we may assume that for there exists a minimal weak solution. In other words, it is legitimate to assume
The idea of the proof is as follows: first, we prove that the function
is a supersolution to the following perturbation of problem :
for a standard cut-off function and to be suitably chosen; besides, a solution is understood in weak sense unless otherwise stated. Second, we construct, for some , a supersolution to by using a solution of (89) and this will enable us to build up a weak solution of for and thus necessarily .

Indeed, we first observe that for and for some
To prove this, we recall Green's function for with Dirichlet boundary conditions
where is the Dirac mass at . Boggio gave an explicit formula for which was used in [17] to prove that in dimension
where
Formula (92) yields
for some , and this in turn implies that, for smooth functions and such that and ,
Using a standard approximation procedure, we conclude that
Since , , we deduce (92).

Let . Then, by Taylor's theorem,
for some and
for some . Adding (97) and (98) yields
and in turn we obtain
Thus, is a weak supersolution of (89) with and the cut-off with support in . Now, reasoning as in Lemma 19, we may assume, for sufficiently small, that (89) possesses a classical solution with parameter replaced by . Set and let be the unique classical solution of the following:
We also, by the Boggio principle, have that there exists sufficiently large such that . Next, let and set
Choosing sufficiently small, we obtain ; moreover, from
we have again by the Boggio principle that and eventually that . Finally, we have
since . Thus, it is enough to choose to provide a classical solution to for , which is a contradiction; this completes the proof of Theorem 9.

#### 4. Behavior of the Minimal Solutions as : Proof of Theorem 11

*Proof of Theorem 11. *We first show that
Since this standard, we just briefly sketch its proof. By Theorem 7, we know that
Then, by multiplying by and by integrating by parts, we obtain that remains bounded. Hence, up to a subsequence, converges in the weak topology to 0, which is the unique solution of . By convergence of the norms, we infer that the convergence is in the norm topology.

Next, note that satisfies
Therefore, ; one concludes that by Lemma 13.

In order to prove the last statement of Theorem 11, note that from (106) we know that for all there exists such that .

So, fix and let . Then,
This shows that for all , and the proof is completed according to the arbitrariness of .

#### 5. Further Results and Open Problems

First, we give the following result which is the main tool to guarantee that is singular. At the same time, it gives a precise estimate for . The proof of this result is based on an upper estimate of by a stable singular subsolution.

Proposition 26. *Suppose there exist , and a singular radial function with such that
**
If , then and is singular.*

*Proof. *First, note that (111) and yield . Equation (110) implies that is a weak subsolution of . If now , then by Lemma 17 would necessarily be below the minimal solution , which is a contradiction since is singular while is regular. In the following, we will prove that is singular.

Now, let in such a way that
Setting , we claim that
Note that by the choice of we have , and therefore to prove (113) it suffices to show that, for , we have in . Indeed, fix such and note that
Assume that