A REMARK ON EXISTENCE AND OPTIMAL SUMMABILITY OF SOLUTIONS OF ELLIPTIC PROBLEMS INVOLVING HARDY POTENTIAL

. We study the eﬀect of a zero order term on existence and optimal summability of solutions to the elliptic problem with respect to the summability of f and the value of the parameter a . Here Ω is a bounded domain in R N containing the origin.


1.
Introduction. Regularity results for solutions of elliptic equations in terms of the summability of a source term are nowadays classic. In the linear case we have two types of results: those by Stampacchia for variational operators with discontinuous coefficients and those by Calderón and Zygmund for general operator with regular coefficients. The results by Stampacchia (see [13]) can be summarized as follows in the simpler case of the Laplacian. Consider the solution to −∆u = f with homogeneous Dirichlet boundary conditions. Then we know that • if f ∈ L m (Ω), m > N 2 , then u ∈ W 1,2 0 (Ω) ∩ L ∞ (Ω); • if f ∈ L m (Ω), 2N N +2 ≤ m ≤ N 2 then u ∈ W 1,2 0 (Ω) ∩ L m * * (Ω), m * * = N m N −2m ; • if f ∈ L m (Ω), 1 < m < 2N N +2 then u ∈ W 1,m * 0 (Ω), m * = N m N −m . In this simple case the results are also a consequence of the W 2,m regularity proved by Calderon and Zygmund in [11].
On the other hand we have the following result due to Hardy.
Notice that the potential |x| −2 represents a borderline case in the sense that it belongs to the Marcinkiewicz space M N 2 (Ω) but not to L N 2 (Ω). The main feature of this paper is to analyze the influence of a zero order perturbation term like a u |x| 2 on the regularity behavior of the solutions of elliptic problems. We anticipate that this regularity depends, surprisingly, also on the value of the parameter a. To be precise, we formulate the problem as follows: Let Ω be a bounded, open subset of R N , N > 2, such that 0 ∈ Ω and M : Ω × R → R N 2 , be a bounded and measurable matrix such that Assume that and consider the boundary value problem , then any solution u ∈ L 1 (Ω) of (5) is unbounded. Indeed, assuming by simplicity that f ≥ 0, by the maximum principle we have u > 0 and then −∆u > 0 .
which is easily seen to be unbounded. In the following example we see a precise behavior in terms of a. Consider the radial equation, We look for a solution in the Sobolev space W 1,2 0 (Ω) such that u(1) = 0. An elementary argument provides that such solution is Remark that α + < 0.

ELLIPTIC PROBLEMS INVOLVING HARDY POTENTIAL 515
The fact that the problem is linear and the previous remark allows to show, by duality, that in general there is no solution for data in L 1 (Ω). The meaning of solution here is the following: given f ∈ L 1 (Ω) we say that a function u ∈ L 1 (Ω) such that u |x| 2 ∈ L 1 (Ω) is a weak solution to equation if and only if We will say that the problem is well posed in L 1 (Ω) if there exists a constant C > 0 such that We prove that problem (6) is not well posed in L 1 (Ω). Assume that for each f ∈ L 1 (Ω) we can find u ∈ L 1 (Ω) weak solution to problem (6).
Then, if the problem is well posed in L 1 (Ω) But then, taking the infimum on f ∈ L 1 (Ω) with ||f || 1 ≤ 1, we find that which is false in general as previously shown.
Notation and definitions. Recall the definition of Sobolev conjugate exponent and, for m ≥ 1, Consider the approximate Dirichlet problems.
2. Summability of "variational" solutions. In this section, even though the right hand side belongs to W −1,2 (Ω), a surprising phenomenon appears: a restriction on a has to be assumed in order to have the same results as for the Laplacian.
Precisely we have the next result.
Proof. Since, for any n ∈ N, the function u n belongs to L ∞ (Ω), we can use , as test function in (7) (see also [4]). The following Notice that in order to have an a priori estimate we need that is i.e., if (8) holds. Remark that the definition of γ gives Then, the use of Sobolev inequality implies (S is the Sobolev constant) On the other hand, choosing γ = 2, one can also prove that u n is bounded in W 1,2 0 (Ω), so that, passing to the limit as n tends to infinity yields the result since u n converges to the solution u of (5). The condition a < α λ(m) is optimal as the following example shows.
By elementary integration we find a family of solutions to (11): Notice that if B = 0 the solutions are not in W 1,2 0 (Ω). Therefore we have to choose B = 0, i.e., As ν = N m , the coefficient becomes , then 2 − ν > α + , so that the summability of u is the summability of r α+ and r α+ does not belong to L m * * (Ω).
3. Existence of infinite energy solutions. As in the previous section we start with a result of existence under a restriction on a. Proof. Define for > 0 the test Use v as test function in (7) to obtain Using Hardy and Sobolev inequalities we obtain Now for fixed n and going to zero we have That is, Thus we proved the following a priori estimate: We now follow [4] in order to show that the sequence {u n } is bounded in W 1,m * 0 (Ω). Fix = R in (13) and use (14). Then we have C(a, m, N, α, S, R) = M , so that (use Hölder inequality with exponents 2 m * and 2 2−m * ) Thus we can deduce which implies the boundedness of the sequence {u n } in W 1,m * 0 (Ω). Therefore, as m > 1, there exists u such that, up to a subsequence, u n u, weakly in W 1,m * 0 (Ω), and almost everywhere, u n |x| 2 → u |x| 2 strongly in L 1 (Ω). Hence we can pass to the limit in (7) to prove that u is a distributional solution to (5).
The above result is optimal as the following result shows.
and let u n be the solution of Then an estimate of the type for some nonnegative c 0 (independent on n) cannot hold.
Proof. Let p > 1 be such that 1 p + 1 m * * = 1, let g be in L p (Ω) and let {g n } be a sequence of smooth functions converging to g in L p (Ω). Consider z n ∈ W 1,2 0 (Ω) : −∆z n = a z n |x| 2 + 1 n + g n . Now let f n = |z n | m −2 z n ; hence, choosing u n as test function in the equation for z n and choosing z n as test function in (7), we obtain, using (15): Then Hence, up to subsequences, z n converges weakly in L m (Ω) (and also in W 1,2 0 (Ω)) to z, the unique weak solution of Due to the choice of p, it turns out that z belongs to L p * * (Ω), and this contradicts Example 2.2. Indeed, choosing a ≥ λ(p), we have a > λ(p + ) for every > 0. Hence, the solution in W 1,2 0 (Ω) given by Example 2.2 corresponding to the L p (Ω) function g(x) = 1 |x| ν , with ν = N p+ belongs exactly to M N/α + , and not to L N/α + (Ω). Since N α + ≤ p * * being a ≥ λ(p), we thus have a contradiction. What happens if the datum belongs to L 1 (Ω)? In this case, no estimate can be obtained, as the following example shows. Therefore, by the maximum principle, u n ≥ w n , where w n is the unique solution in W 1,2 0 (Ω) of −∆w n = a zn |x| 2 + f n in Ω , w n = 0 on ∂Ω .
Supposing that f n is radially symmetric, so will be u n , z n and w n . Passing to radial coordinates, we have after straightforward calculations: and therefore Choose now f (x) = 1 |x| N (− ln(|x|)) α , with 1 < α < 2, so that f belongs to L 1 (Ω), and let f n = T n (f ) as usual. Since f n increases to f , Hence, for every |x| = ρ in Ω, Thus the limit of u n is everywhere +∞, and so no estimate can be obtained if the data are bounded in just L 1 (Ω).

Uniqueness.
In this section, we discuss the uniqueness of the solutions obtained as limit of solutions of the regular problems (7) (see [7]). Proof. Consider the boundary value problems (7) and where {g n } is any sequence converging to f in L m (Ω). Therefore, as m > 1, there exists z such that, up to a subsequence, z n converges weakly to z and z is a distributional solution to (5). Then Here, we can repeat the calculation of previous theorems in order to obtain which implies that u − z L m * * (Ω) = 0 , and so uniqueness.

5.1.
A different way to obtain the critical value for a. Consider our problem with the Laplace operator: Then v solves problem Notice that by definition, We are able to show the existence of solution of the original problem by studying the existence of entropy solutions of the transformed problem. To do that, we define the weighted Sobolev space D 1,2 0,γ (Ω) as the completion of C ∞ (Ω) with respect to the norm By using the Caffarelli-Kohn-Nirenberg inequalities of [10] we have in particular the Sobolev inequality stated in the next theorem.
has a unique entropy solution. If f ≥ 0, then v ≥ 0.
Proof. See [2]. It is worth pointing out that in the original problem the result above can be translated as follows. First at all, to have f = g|x| −γ ∈ L 1 (Ω), by Hölder inequality, where g is a nonnegative function in L 1 (Ω). We follow the arguments in [1].
Definition 3. We say that u is a weak solution of problem (23) if u belongs to L 1 (Ω), u |x| 2 belongs to L 1 (Ω), and for all φ ∈ C 2 (Ω) such that φ = 0 on ∂Ω, we have Notice that if u is a weak solution, then u satisfies the equation in the entropy sense (see [2]). Proof. All details can be found in [1].
Notice that, translating the result above to the original problem, we obtain that there exist solution if and only if g ∈ L m (Ω) for m > 2N N +2 and moreover this solution is in an energy space greater than W 1,2 0 (Ω) and given, in a natural way, by the improved Hardy inequality (see [1]).