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Properties of positive solutions of an Elliptic Equation with negative exponents

In this paper, we study the existence and non-existence result of positive solutions to a singular elliptic equation with negative power on the bounded smooth domain or in the whole Euclidean space. Our model arises in the study of the steady states of thin films and other applied physics. We can get some useful local gradient estimate and L1 lower bound for positive solutions of the elliptic equation. A uniform positive lower bound for convex positive solutions is also obtained. We show that in lower dimensions, there is no stable positive solutions in the whole space. In the whole space of dimension two, we can show that there is no positive smooth solution with finite Morse index. Symmetry properties of related integral equations are also given.


Introduction
In this paper, we mainly study some property of positive solutions of the elliptic equation in the domain Ω ⊂ R n (1) ∆u = u τ , in Ω, where τ < 0. The elliptic equation comes from applied physics, mathematical biology, and the steady states of thin film model of viscous fluids: (2) u t = −div(f (u)∇∆u) − div(g(u)∇u).
As a consequence of the local gradient estimate in Theorem 1, we have the following improvement of Theorem 7.1 in [22].
Corollary 2. For any sequence (u j ) of positive solutions to the equation (1) with τ = −1 and with boundary data φ j ≤ M, and for every compact sub-domain K of Ω, there is a constant C = C(n, K, Ω) such that (4) |∇u j | ≤ (M + 1)C, on K.
Hence, the limit u of any convergent subsequence of (u j ) is a Lipschitz continuous weak solution to a free boundary problem of the equation For the proof of this corollary, we just note that the solution u is a subharmonic function and it attains its maximum only on the boundary ∂Ω. Then we use the gradient bound to get the conclusion. We remark that the result is optimal in the sense that u is not differentiable at its zero point as noticed in [22]. It is unclear how large the zero level set S(u) = {u = 0} is. However, a general study was made in the paper of Jiang and F.H.Lin [16].
Note that the gradient estimate implies the Harnack inequality for positive solutions with u(x) ≥ 1 for all x ∈ B R (x 0 ). One can use this fact in deriving compactness result.
We shall discuss the existence theory of positive solutions to (1) in section 9.
Using the testing function method, we have the following very useful L1 lower bound result.
Theorem 3. Assume τ ≤ 0 and assume that Ω ⊂ R n is an open subset in R n . Let f : R → R + be a positive function such that s τ τ −1 f (s) 1 1−τ ≥ C 0 , f or s > 0 for some constant C 0 . Let u ∈ C 0 (Ω) be a positive weak solution to the equation in Ω. Then for any R > 0 and x 0 ∈ Ω ( with B R (x 0 ) ⊂ Ω), we have absolute constant C(n, τ ) such that We remark that we shall use Theorem 3 for f (u) = u τ and in this case, we can use the convexity of f (u), the spherical average method, and an ODE comparison lemma to get the L 1 lower bound. However, we shall give a proof which can be used on manifolds.
There are many consequences of Theorem 3, and they will be discussed in section 10.
We now try to set up a global upper bound for positive solutions to (1) in R n . Theorem 4. Assume τ ≤ 0. Let 1 ≤ u ∈ C 2 (R n ) be a positive solution to the equation (1) in R n . Then we have absolute constant C(n) such that for all x ∈ R n .
The important consequence of Theorem 4 is the following possibly well-known result.
Proposition 5. Assume that u > 0 is a positive solution to the equation ∆u = 1, in R n . Then u is a polynomial of the form where a 0 > 0,a j ≥ 0 for j = 1, ...n, and 2 j a j = 1.
Proof. We may assume u(x) ≥ 1 by considering u(x) + 1 if necessary. Using our upper bound estimate in Theorem 4, we know u has at most quadratic growth. Consider w(x) = u(x) − |x| 2 2n . Then w is a harmonic function with quadratic growth. Hence, w and u is a quadratic polynomial, which gives the conclusion.
We shall try to study a Liouville property of positive solutions to the equation (1) in the whole space or in the half space. Theorem 6. Assume that n > 2 and τ ≤ −1 − 2n n−2 . We have the Liouville property that there is no positive convex solution u(x) to (1) both on the whole space R n and on the half space R n + with u(x) ≥ 1 everywhere.
Using this Liouville property, we can prove the following compactness result, which can be considered as a generalization of Proposition 16.
Theorem 7. Assume that n > 2 and τ ≤ −1− 2n n−2 . Let Ω be a bounded or unbounded smooth convex domain. Let u be a positive convex solution to (1) on Ω. Then, we have a uniform constant C = C(Ω) such that Note that the convexity property in Theorem 7 can not be removed since u(x) = A|x| 2 1−τ is a non-negative solution to (1) with From the variational point of view, it is very interesting to discuss positive solutions with finite Morse index to the following equation where τ < 0, with finite Morse index. Assume that u ∈ C 2 is a positive solution to (10). Define where φ ∈ C 2 0 (R n ). By definition, we say the positive solution u to (10) with finite Morse index k if there exist L 2 orthogonal nontrivial functions {φ j } k j=1 ⊂ C 2 0 (R n ) such that we have E(φ) < 0 for φ ∈ W := span{φ j } − {0}, and E(φ) ≥ 0 for φ ⊥ W. If k = 0, we say that the solution u is stable. When n = 2, we can prove there are no finite Morse index solutions. This result should have application to bifurcation theory to the equation (1).
Here is the plan of our paper. In last two sections, we discuss symmetry properties of related integral equations. We prove Theorem 1 in section 2 and prove Theorem 3 in section 3. In section 4, we prove Theorem 4. Theorem 7 is proved in section 5. The Liouville property is proved in section 6. The 7-th section is about some discussions to the existence problem and upper bound for positive solutions to (1). The proof of Theorem 8 and some bound about positive solution with finite Morse index are given in section 8. As we mentioned before, we shall discuss the existence theory of positive solutions to (1) in section 9. Many consequences of Theorem 3 will be discussed in section 10. In the following, we shall use C to denote different constants which depend only on n, τ , µ, and the solution u in varying places.

proof of Theorem 1
In this section, we consider the gradient estimate of positive solutions to the elliptic equation where τ < 0. We shall use the maximum principle method [10] (see also [21] and [23]). Recall the following basic formula. Let v be any smooth function in R n . Then, we have 1 2 ∆|∇v| 2 = (∇∆v, ∇v) + |D 2 v| 2 , which can be proved by an elementary calculation. Let us begin with the gradient estimate for positive solution u to the following general elliptic equation: In our case we shall set f (u) = u τ . Set w = log u.

Note that
and then, for any µ > 0, Hence, Then from (⋆) we have Here We remark that in this case, we have Otherwise, we have −P + φF ≤ −P/2 ≤ 0 and 2 n (−P + φF ) 2 − 1 4n Hence, we have P ≤ 4nA(φ, F ′ ). In conclusion, we have on B 0 (R), which implies the conclusion of Theorem 3.
We remark that our gradient estimate can be extended to other elliptic equation like −∆u = u τ .
Note that the right side is bigger than and the left side is less than where C > 0 is an absolute constant depending only on the dimension n. That implies that Using Holder's inequality to the left side, we get Choose p = −τ + 1. Then by our assumption on f , we have It is elementary to compute that , ω n is the volume of the unit ball B 1 (0). This implies the conclusion of Theorem 3. We remark that our argument above can also be used to smooth positive solutions to the following equation Then 0 < ∆u ≤ 1. Using the Gradient estimate in Theorem 1, we can easily get that for all x ∈ B R (0). Here C is a uniform constant. However, this estimate is too rough.
We now use the mean value property to do better. Fix 0 < r = |x| ≤ R/4. Since u > 0 is subharmonic, on one hand, we have On the other hand, let Then ∆w = ∆u − 1 ≤ 0. Hence w is super-harmonic, and for R = 2r,we have .

Compactness result
In this section, we study the point-wise lower bound of positive solutions to the equation (1).
We now prove Theorem 7, which is the Claim: There is a positive constant C = C(Ω) such that for every positive convex solution to (1) with u ∈ S K , we have For otherwise, we have a sequence of positive convex solutions {u j } and a sequence of points {x j } ⊂ Ω such that , Then it is elementary to see that Let Then Ω = R n and by our Harnack gradient estimate and the standard Lp theory, we can extract a convergent subsequence in C 2 (B r (0)) for any r > 0, still denoted by {v j }, with its limitv being a positive convex function satisfying If λ j d(x j , ∂Ω) ≤ C for some constant C, then we have Ω = R n + and we can get a positive solution v ∈ C 2 (R n + ), i.e., ∆v =v τ , in R n + , andv (x) ≥ 1 =v(0). However, both cases give us a contradiction by our Theorem 6. Then, we have proved Theorem 7.

Liouville property
We prove Theorem 6 in this section. Choose any positive number k > 1 and let Ω = {x ∈ R n ; u(x) ≤ k}. By our assumption, Ω is a bounded convex domain. Recall the Pohozaev identity formally. Let g(u) = −u τ and let Multiplying by x∇u to the equation Note that 1 + τ < 0. By integrating the equation above over Ω, we get Note that by multiplying by u to the equation, we have Hence, using ∂ ν u > 0 on the boundary ∂Ω, we have By this we get a contradiction, and then Theorem 6 is true. So we are done.

Discussions
Let −1 < τ < 0. Given a positive data φ on the bounded smooth domain Ω. Consider the boundary problem of positive solutions to (1) on Ω with the boundary condition u = φ on ∂Ω. Let we can get a non-negative minimizer of J(·) on A. For such a minimizer, one need to handle with how large for its zero set. One may see [24] and [14] for more. We now discuss some regularity result for weak solution to (1). We only need to get upper bound for positive weak solutions to (1) for any τ < 0 by assuming a positive lower bound. As in the proof of Theorem 3, we take R > ρ > 0 and a cut-off function ξ = ξ(|x|) such that |∇ξ| ≤ 4 R−ρ , and ξ = 1 on B ρ . Then using u(x) ≥ 1, we have as before that where A, B are uniform constants and We can also derive some other interesting bound without the pointwise lower bound.
Using integration by part, we know that the left side is Let first consider two cases.
(1). If we choose σ = −τ , then we get (2). If we send σ → 0 in (13), then we get In the following, we do iteration. Let σ = −τ + p in (13). Then we have (14) (−τ +p) Then we have We now in the standard Nash-Moser iteration situation. Hence, for any 0 < q, we have a uniform constant C(q, n) such that Once we have a upper bound, we can use the standard Calderon-Zugmund Lp theory to conclude that u is a smooth solution.

Finite Morse Index solutions
Assume that u is the positive solution to (10) with finite Morse index k. Choose a large ball B R (0) which contains the supports of all φ j 's.
. Let ξ be a smooth cut-off function with compact support in T r . Let φ = u −q ξ. Then we have the following stability condition for any ǫ > 0, Using the weak form of the equation (10) with the testing function ξ 2 u −β ,β = 2q + 1 > 0, we have β u −β−1 ξ 2 |Du| 2 ≤ u −β+τ ξ 2 + 2 u −β ξ|Du||Dξ|, and then we have, using the Cauchy-Schwartz inequality, for any δ > 0, Inserting this into the stability condition we get Choose − 1 2 < q < −τ + √ τ 2 − τ and ǫ, δ small enough depending on q, we can have Hence, for some constant C(τ, q), we have Take q > 0 and replace ξ by ξ q+ 1−τ 2 to get Here C(τ, q) is another constant. Using the Young inequality and choosing ǫ small, we get that and then we can find that for all r > 0. Note that the restriction of q is which implies that Set q such that n = 4q 1−τ + 2 and p = 2q + 1 − τ . Then p > n, and we use the lower bound of u to get that for any r > 0. Note that 1 − n p = 1 − 2 1−τ = − τ +1 1−τ . Using our equation we find that |∆u| ∈ L p (R n ). Using the standard L p estimate we find that Theorem 9. Let u is a positive solution with finite Morse index on R n . We now assume that u(x) ≥ u(0) = 1. Then, 1−τ , and then we have the growth estimate In the case when the solution u is stable, we can take T r = B r (0). In this case, we have the following result: Theorem 10. There is no stable positive solutions to (10) Proof. Assume not. Let ξ be a cut-off function such that ξ = 1 on the ball B R (0), ,and ξ = 0 outside B R 2 (0). Then we get from the estimate (16) that We remark that the special case when τ = −1 has been obtained in [22].
We now prove Theorem 8.
Proof. Using the test function φ = ξ in (15), we obtain that where C is independent of R > 1. In the following, we shall let τ = −p, which is often used in the literature about super-linear elliptic equations. We now perform the following scaling: We first claim (20) v(t, θ) ≥ C for t > 2.
It is easily seen thatk(s) has infinitely many positive zeroes for any µ > 0.) Thus, we can easily deduce that q has infinitely many positive zeros. Our claim holds. We denote the zeroes of k as 0 < r 1 < r 2 < ... < r k < ... where r k → +∞ as k → +∞. Let k 0 be such that (29) p u p+1 ≥ 2µ r 2 , r > r k , k ≥ k 0 We now in the position to complete the proof of Theorem 8. Let N > 0 be fixed and i ≥ k 0 . Let h i be the function defined to be k(|x|) for |x| between the ith and (i + 1)th the zeros of k and to be zero otherwise. Then h i ∈ H 1 (R 2 ), h i are orthogonal (in L 2 (R 2 ) or H 1 (R 2 )) and by multiplying (28) by h i and integrating between these zeros we see that is strictly negative at each h i . Hence the span of h i is an (N − 1)dimensional subspace of C ∞ 0 (R 2 ) such that Q(h) < 0. Since h i has compact support it follows easily that there is an (N − 1)-dimensional subspace of H 1 (R 2 ) such that and hence the Morse index of u must be at least N. Since N is arbitrary, the Morse index of u is infinity, a contradiction to our assumption.

Existence theory
We now consider the existence problem of the problem (1). One can easily to find the radial solutions to (1) in the whole space (see Theorem 3.2 in [22] for the case when τ = −1). In general, we have Proposition 11. Assume (−1 =)τ ≤ 0, and let Ω ⊂ R n be a bounded smooth domain. Given any smooth positive boundary data φ. Assume that u is a sub-solution to (1) with u ≤ φ on ∂Ω. Then there is smooth positive solution to (1) with boundary data φ.
We remark that in the case when τ = −1, an existence result has been discussed in [22] by degree argument. Since the proof of Theorem 11 is simple, we give it here.
Proof. Choose a large constant M such that u = M is a super-solution. Then one can use the standard super-sub solution method to get a positive solution.
We give here the variational method. Let It is easy to see that I(·) is bounded from below on the closed convex set A. Since for any u, w ∈ A, (We get this by mean value theorem in Calculus). Then we can use the Sobolev compactness imbedding theorem to get a minimizer u of the functional I(·) in the set A, which is the solution to (1) with the boundary data φ (see [27]).
The advantage of variational methods is that one may prove that the minimizer on the class A is a stable solution in the usual sense. Since we shall not use this fact, we shall not discuss it. The nature question is how to find a sub-solution on a bounded domain. The usual way is to use one-dimensional (or any lower dimensional) solution or radial solution on the whole space. One can also choose a large ball containing the bounded domain, and solve the equation on the ball to get radial solutions on the ball. Such radial solutions are the sub-solutions to the equation on the original domain if the boundary value of the radial solutions on the domain are less than the given boundary data φ.
In particular, as an application of Theorem 11, we have Corollary 12. Assume (−1 =)τ ≤ 0, and let Ω ⊂ R n be a bounded smooth domain. Given any positive smooth boundary data φ in ∂Ω.
Assume that there is a radial solution u(r) in lower dimensional space R k (k < n) or the whole space R n such that φ(x) > u(|x|) on ∂Ω.
Then there is smooth positive solution to (1) with boundary data φ.
Proof. Here we need only to use u(r) as a sub-solution to (1) on the domain Ω in Proposition 11.
Although the argument in the proof of this Corollary is simple, it can be used to study the existence result of positive solutions for a large class of singular elliptic partial differential equations such as ∆u + au log u = 0 with Dirichlet boundary data. One may see [13] and [21] for related results.

Consequences of Theorem 3
As an easy consequence of Theorem 3, we have Proposition 13. Assume τ ≤ 0 and Ω ⊂ R n . Let f be as in Theorem 3 above. Let u ∈ C 0 (Ω) be a positive weak solution to the equation to (6) in Ω. Then for any R > 0 and x 0 ∈ Ω ( with B R (x 0 ) ⊂ R n ), we have absolute constant C(n, τ ) such that The proof of this is direct by using our L1 lower bound (7) since u is subharmonic and the maximum occurs only at boundary point.
Proposition 13 immediately implies the following such that if the positive boundary data φ ≤ C on ∂Ω, the Dirichlet problem to the equation (6) in Ω with u = φ on the boundary ∂Ω has no nontrivial positive weak C 0 -solution.
In fact, we take a ball B R (x 0 ) in the domain Ω and let C = C(n, α)R 2 1−τ . Then we have sup Ω u = sup ∂Ω φ > C.
Using this result, we can find a sequence (u j ) of positive solutions to (1) in the ball B = B 1 (0) such that minu j → 0+ as j → +∞. Choose C large enough, we can solve (1) to get a unique positive radial solution u with u = C on the boundary ∂B. Then there is a constant M > 0 such that |u| C 3 ≤ M. Let φ be as in the Corollary above and let φ t = tC + (1 − t)φ, where t ∈ [0, 1]. Let p > 1 and let δ > 0. Given a smooth function u > 0 in W 2,p (B) with boundary data φ t . Consider the problem with the boundary data v = φ t on the boundary ∂B. Set T t (u) = v and Here M(δ) is a constant coming from the Lp estimate depending on δ. Then the fixed point of T t in U(δ) is a positive solution to (1) with boundary data u = φ t . Note that T t is a compact operator from U(δ) → W 2,p (B), deg(1 − T 1 , U(δ), 0) = 1 and deg(1 − T 0 , U(δ), 0) = 0. Hence, we have t 0 ∈ (0, 1) and u ∈ ∂U(δ) such that T t 0 (u) = u. Choose δ = δ j → 0. Hence, we have a sequence t j ∈ (0, 1) such that T t j (u j ) = u j and u j ∈ ∂U(δ j ) (which implies that min B u j = δ j → 0).
We can do the same thing to (1) in the domain Ω. Hence, we have An open question is, which domain has the uniqueness property in Corollary (15).
Using Theorem 3, we can easily derive the following: Proposition 16. There is no positive solution to (1) in a cone-like unbounded domain Ω with the bound for every ball B R ⊂ Ω with R ≥ 1, for some constant K > 0 and σ > 0.
Proof. Assume we have a positive solution u. Note that we can choose an arbitrary large ball B R in the domain Ω. Then, using our Theorem 3, we get R σ ≤ K, which is not true by sending R → +∞.
We point out that for solution u to (1), the quantity u p is dimensionless. By this, we mean that if u is a solution to (1) in the ball B R (0), then the function Hence, in this sense, the L 1 lower bound estimate in Theorem 3 is the best one.

Symmetry about Related equations
We also consider the integral equation with n ≥ 2, 0 < µ < n, h(x) is a positive smooth function, and τ < 0. This integral equation is closely related to the elliptic differential equation the elliptic equation: It is clearly that the positive solution to (32) is bounded from above by h. We shall use the following notation. Given any hyperplane π in R n . For any point x ∈ R n , let x π be the reflection of x about the plane π and let π(x) ∈ π be projection of x into π. Define u π (x) = u(x π ) for any function u : R n → R.
In the famous paper of Gidas and Spruck [11], they proved that for n > 2 and µ = 2, and 1 ≤ τ < n+2 n−2 , the only non-negative solution to the equation −∆u = u τ , in R n , is zero. However, the negative index τ case has not been treated before X.W.Xu's recent work [29] . So, the following result can be considered as a generalization of their result to the equation (32).
Assume that for some plane π, we have h(x) = h(π(x)). Then u(x) is symmetric to the plane π.
Using the Kelvin transformation, we have another kind of result.
The proof of Theorem 18 will be given in the last section. In recent years, there are important progress in the study of symmetry properties of non-negative solutions to Yamabe type equations. In particular, X.Xu [29] has obtained some related results to ours. His equation is the following which corresponds to the elliptic equation: One should be caution about the negative sign before the Laplacian operator, which makes the equation can not have any positive solution on the whole space. We will use a symmetry method (see also [5] and [7]) to prove our results. This symmetry method is powerful in our case since we can use the behavior at infinity of the solution.
We now give a proof of Theorem 17: After using a rotation, we may assume that the hyperplane π is orthonormal to x 1 axis at the origin. So we may assume that h(x) = h is a constant in the following argument.
Proof of Theorem 17.
We want to show that for sufficiently positive values of λ, Σ − λ must be empty.
Then by Lemma 19, for any (y)dy. It follows first from inequality (37) and the Holder inequality that, for any q > n/(n − µ), . By condition (34), we can choose N sufficiently large, such that for λ > N, we have

Now (39) implies that
u λ − u L q (Σ − λ ) = 0, and therefore Σ − λ must be measure zero, and hence empty. Then using the standard moving plane trick (see ( [5]) and [7]), we know that the solution u is symmetric in the variable x 1 .
So we complete the proof of Theorem 17. Finally, we remark that in some cases, using the analysis of ODE, we see that there is no radially symmetric positive solution to (32) with the condition (34). However, we shall not do this here.

proof of theorem 18
Let us now introduce the Kelvin type transform of u as follow v(x) = |x| µ−n u( x |x| 2 ) for any x = 0. Then by elementary calculations, one can see that (32) is transformed into the following: where α = (n + µ) − (n − µ)τ > 0. Obviously, v(x) has a singularity at origin. Since u is locally bounded, it is easy to see that v(x) has no singularity at infinity, i.e., for any domain Ω that is a positive distance away from the origin, Ω v β (y)dy < ∞.
For the second equality, we have made the transform y = z/|z| 2 . Since Ω is a positive distance away from the origin, Ω ⋆ , the image of Ω under this transform, is bounded. Also note that β(n − µ) − 2n > 0. Then we get the estimate (41).
Then it is easy to see that Here we have used the fact that |x − y λ | = |x λ − y|. Substituting x by x λ , we get This implies (42).

Proof of Theorem 2. Define
We want to show that for sufficiently negative values of λ, Σ − λ and Σ + λ must be empty. Whenever x, y ∈ Σ λ , we have that |x − y| ≤ |x λ − y|. Moreover, since λ < 0, |y λ | ≥ |y| for any y ∈ Σ λ . Then by Lemma 20, for any x ∈ Σ − λ , It follows first from inequality (43) and then the Holder inequality that, for any q > n/(n−µ), which will be used below in the form that τ < µ, . By condition (41), we can choose N sufficiently large, such that for λ > N, we have

Now (45) implies that
v λ − v L q (Σ − λ ) = 0, and therefore Σ − λ must be measure zero. Then we can use the moving plane trick as before to get that v is symmetric at zero with respect to the x − 1 direction. This completes the proof of Theorem 18.