Semilinear heat equation with singular terms

. The main goal of this paper is to analyze the existence and nonexistence as well as the regularity of positive solutions for the following initial parabolic problem


Introduction
Let Ω be a bounded open subset of R N , N ≥ 3, containing the origin.Set Ω T := Ω × (0, T) where T > 0 is a real constant.In this paper we investigate the existence and regularity as well as the uniqueness of solutions to the following initial parabolic problem where σ ≥ 0 and µ ≥ 0. The source terms f and u 0 satisfy Corresponding author.Email: ahmed.youssfi@usmba.ac.ma; ahmed.youssfi@gmail.comand u 0 ∈ L ∞ (Ω) such that ∀w ⊂⊂ Ω ∃d w > 0 : u 0 ≥ d w in w. (1. 3) It is clear that problem (1.1) is strongly related to the following classical Hardy inequality which asserts that for all u ∈ C ∞ 0 (Ω), where Λ N,2 = ( N−2 2 ) 2 is optimal and not achieved (see for instance [20,50] and [11] when Ω = R N ).The presence of a term with negative exponent generally induces a difficulty in defining the notion of solution for the problem (1.1).
In the literature, singular problems like (1.1) are considered and intensively studied in various situations depending on σ or µ.If σ = 0 and µ > 0, the problem (1.1) is reduced to the following heat equation involving the Hardy potential and is studied first by Baras and Goldstein in their pioneering work [15].When the data 0 ≤ f ∈ L 1 (Ω T ) and u 0 is a positive L 1 -function or a positive Radon measure on Ω are not both identically zero (otherwise the result is false since u ≡ 0 is a solution), Baras and Goldstein [15] have proved that if 0 ≤ µ ≤ Λ N,2 then there exists a positive global solution for the problem (1.5), while if µ > Λ N,2 there is no solution.Problem (1.5) with −diva(x, t, ∇u) instead of −∆ was studied in [45], where the author proved that all the solutions have the same asymptotic behaviour, that is they all tend to the solution of the original problem which satisfies a zero initial condition.In [46] the authors studied the influence of the presence of the Hardy potential and the summability of the datum f on the regularity of the solutions of problem (1.5) with the nonlinear operator −diva(x, t, u, ∇u) in the principal part.
The singular Hardy potential appears in the context of combustion theory (see [50] and references therein) and quantum mechanics (see [15] and [50] and references therein).There is a wide literature about problems involving the Hardy potential where the existence and regularity of solutions as well as nonexistence of solutions are analyzed, for instance, we refer to [2-7, 10, 17, 18, 32, 36-38, 54].
Problems involving singularities (like (1.1) with µ = 0) describe naturally several physical phenomena.Stationary cases include the semilinear equation −∆u = f (x)u −σ , x ∈ Ω ⊂ R N , that can be obtained as a generalization to the higher dimension from a one dimensional ODE (N = 1) by some transformations of boundary layer equations for the class of non-Newtonian fluids called pseudoplastic (see [29,39]).As far as we know, semilinear equations with singularities arise in various contexts of chemical heterogeneous catalysts [9], non-Newtonian fluids as well as heat conduction in electrically conducting materials (the term u σ describes the resistivity of the material), see for instance, [31,39].In view of this physical interpretation various generalizations of this later equation considered in the framework of partial differential equations (N ≥ 2) has been the subject of study in many papers.For the mathematical analysis account, the seminal papers [23,49] constitute the starting point of a wide literature about singular semilinear elliptic equations.Far from being complete we quote the list [8,17,19,21,26,27,33,34,40,42,43,52,53,56].
It is worth recalling that due to the meaning of the unknowns (concentrations, populations,. . .), only the positive solutions are relevant in most cases.
As far as the parabolic setting is concerned for problems as in (1.1) with µ = 0, the literature is not rich enough.For problems like (1.1) with p-Laplacian operator, existence results of nonnegative solutions are obtained in [25] for data with higher summability while in [41] the authors proved the existence of nonnegative distributional solutions for non regular data (L 1 and measure) and the uniqueness is proved for energy solutions.Other related problems with singular terms can be found in [12][13][14].
In the case where σ ̸ = 0 and µ = 0, problem (1.1) with a quite more general diffusion operator including the Laplacian one was studied in [24].The authors considered nonnegative data having suitable Lebesgue-type summabilities and assumed the strict positivity on the initial data inside the parabolic cylinder.They have shown, via Harnack's inequality, that this strict positivity is inherited by the constructed solution to the problem, thus giving a meaning to the notion of solution considered.Some regularity results are obtained according to the regularity of f and the values of σ.
Our main goal in this paper is to study the problem (1.1) in the presence of the two singular terms, that is µ > 0 and σ ≥ 0 extending to the evolution case some results obtained for the elliptic problem (with the ∆ p operator instead of Laplacian one) studied in [1].Abdellaoui and Attar [1] investigated the interplay between the summability of f and σ providing the largest class of the datum f for which the problem admits a solution in the sense of distributions.Uniqueness and regularity results on the distributional solutions are also established.In the same spirit, the parabolic case with µ = 0 was investigated in [24].Our work fits in the context of recent work on equations involving the Hardy potential in the case of nonexistence of solutions.We start by studying first the case µ < Λ N,2 := (N−2) 2 4 distinguishing two cases where σ ≥ 1 and f ∈ L 1 (Ω T ) and the case where Then we investigate the case µ = Λ N,2 and σ = 1 with data f ∈ L 1 (Ω T ).In both cases we prove the existence of a weak solution obtained as limit of approximations that belongs to a suitable Sobolev space.The approach we use consists in approximating the singular equation with a regular problem, where the standard techniques (e.g., fixed point argument) can be applied and then passing to the limit to obtain the weak solution of the original problem.The regularity of weak solutions is analyzed according to the Lebesgue summability of f and σ.Furthermore, we prove the uniqueness of finite energy solutions when the source term f has a compact support by extending the formulation of weak solutions to a large class of test functions.Finally, in the case where µ > Λ N,2 we prove a nonexistence result.
The paper is presented as follows.Section 2 contains all the main results (existence, regularity and uniqueness) and also graphic presentations allowing to better locate the obtained results.In Section 3 we first prove an existence result for approximate regular problems of the problem (1.1) and then we give the proof of all the main results Theorem 2.2, Theorem 2.4, Theorem 2.5, Theorem 2.6, Theorem 2.8 and Theorem 2.10.At the end, some results that are necessary for the accomplishment of the work are given in an appendix to make the paper quite self contained.

Main results
We begin by stating the definition of weak solution and finite energy solution of the problem (1.1) and then we state and comment the main results.

1) By a weak solution of the problem (1.1) we mean a function
2) We call a finite energy solution of the problem (1.1) a weak solution u that satisfies u ∈ L 2 (0, T; In Definition 2.1 above all the integrals make sense.Generated by the singular terms, the only difficulty is raised in the right-hand side.Indeed, by Hardy's inequality the integral Ω T uϕ |x| 2 dxdt is finite while we make use of a comparison result with a solution of a problem in [24,Proposition 2.2], where the hypothesis (1.3) is used, for the integral Ω T f ϕ u σ dxdt to be finite.Thus one has ).Throughout this paper, we will use the two real auxiliary truncation functions T k and G k defined for k > 0 respectively as .
Observe that m 1 ≥ 1 if and only if σ ≤ 1.We will prove the existence of solution for the problem (1.1) under the assumption that the datum f satisfies

The case µ < Λ N,2 : existence of weak solutions
The first existence result is the following.
Observe that 1 ≤ m 1 ≤ 2N N+2 for any 0 ≤ σ ≤ 1.We point out that in the case where σ = 0, which yields m 1 = 2N N+2 , we find the result already established in [46,Theorem 1.2] for data f ∈ L r (0, T; L q (Ω)) with r = q ≥ 2N N+2 .It is worth recalling here that 2N N+2 is the Hölder conjugate exponent of the Sobolev exponent 2N  N−2 and by duality argument, data belonging to the Lebesgue space of exponent 2N  N+2 are in force in the dual space L 2 (0, T; H −1 (Ω)).

2.2
The case µ = Λ N,2 : existence of infinite energy solutions In the following result we deal with the case where µ = Λ N,2 .The weak solutions found do not generally belong to the energy space.

The case µ > Λ N,2 : nonexistence of weak solutions
If we assume µ > Λ N,2 then the problem (1.1) has no weak solution.This is stated in the following theorem.The following Figure 2.1 summarizes the different existence results according to the interactions between the singularities.
No positive weak solution u ∈ L q (0, T; W

Regularity of weak solutions
In the following theorem we give some regularity results for the weak solution u of the problem (1.
In the case where 0 < σ ≤ 1, the regularity results obtained in the previous Theorem 2.6 concerns the weak solutions corresponding to data f ∈ L m (Ω T ), with m ≥ m 1 .When we decrease the summability of the data, that is f ∈ L m (Ω T ) with 1 < m < m 1 , we obtain solutions lying in a bigger space than the energy one.Actually, we have the following result.Theorem 2.8.Let Ω be a bounded open subset of R N , N ≥ 3, containing the origin.Assume that (1.3) holds and f ∈ L m (Ω T ), with 1 < m < m 1 and suppose that 0 , the problem (1.1) has a weak solution u such that u ∈ L q (0, T; W Remark 2.9.We point out that for the particular case σ = 0 we obtain that the solution u belongs to L q (0, T; W N+2−m and γ = m(N+2) N−2m+2 .These are exactly the same exponents as those obtained in nonsingular case in [16,Theorem 1.9] Observe that since for σ = 0 we have m 1 = 2N N+2 < m 3 , the result we prove is a refinement of the one in [16,Theorem 1.9].This is not surprising since the effect of Hardy's potential vanishes for µ < Λ N,2 as it is shown in the proof of Theorem 2.8.Remark that we cannot the consider case where σ = 0 and m = 1, since the test functions we use in order to obtain the regularity stated in Theorem 2.8 cannot be chosen.
The following Figure 2.2 summarizes the previous regularity results considering the interplay between the singularity and the summability of the source term f .

Uniqueness of finite energy solutions
As far as the uniqueness is concerned, we give the following result for the finite energy solutions in the case of data with compact support.
Theorem 2.10.Let Ω be a bounded open subset of R N , N ≥ 3, containing the origin.Suppose that 3 Proofs of the results

Approximate problems
Let us consider the following sequence of approximate initial-boundary value problems where The case σ = 0 leads to the variational framework since m 1 = 2N N+2 is the Hölder conjugate exponent of the Sobolev exponent 2 * := 2N N−2 and then by the Sobolev embedding and a duality argument we obtain and the existence of u n can be found in [30,Theorem 3] on page 356.If 0 < σ ≤ 1, the proof of the existence of a solution u n to the approximate problem (3.1), which is based on the Schauder's fixed point theorem, is now classical.For the convenience of the reader we give it here.
Lemma 3.1.Assume that 0 < σ ≤ 1 and µ ≤ Λ N,2 .For each integer n ∈ N the approximate problem Moreover, u n is such that for every ) the unique weak solution (depending on v and n) of the following problem in Ω.
The existence of w can be found in [30,Theorem 3] on page 356 (see also [35]).Let us consider the map S defined by S(v) = w.Taking w as test function in (3.3) we get Thus, by the Hölder inequality we arrive at so that by the Poincaré inequality one has where and C p is the constant in the Poincaré inequality.Therefore by the Young inequality we obtain we have proved that the map S : B → B is well defined.In order to apply Schauder's fixed point theorem over S to guarantee the existence of a solution for (3.1) in the sense of (3.2), we need to check that the map S is continuous and compact on B.
Let us first prove the continuity of S. In order to do this, let {v k } k ⊂ B be a sequence such that Denote by w k := S(v k ) and w := S(v).Then w k is the solution of the problem in Ω. (3.5) We shall prove that lim Observe that up to a subsequence, we can assume that v k → v a.e. in Ω T .So that one has by the dominated convergence theorem we have Thus, testing by w k − w in the difference equations solved by w k and w and using the fact that w k (x, 0) = w(x, 0) = u 0 and the Hölder inequality, we obtain If µ < Λ N,2 then by the Poincaré inequality we obtain , where C p is the Poincaré constant.While if µ = Λ N,2 then by [50, Theorem 2.1] there exists a constant C(Ω) > 0 such that .
Having in mind (3.6) we conclude that the sequence {w k } k converges to w in L 2 (Ω T ) and so S is continuous.
We turn now to prove that S is compact on B. Let {v k } k∈N be a bounded sequence in B. We shall prove that there exists a subsequence of w k := S(v k ) that converges in norm in L 2 (Ω T ).
Taking w k = S(v k ) as a test function in (3.5) solved by w k and using the Hölder inequality we obtain Thus, from the Poincaré and Young inequalities it follows where C is a positive constant not depending on k.Hence, by (3.7) the sequence {w k } k is uniformly bounded in L 2 (0, T; H 1 0 (Ω)).Now, testing by an arbitrary ϕ ∈ L 2 (0, T; H 1 0 (Ω)) in (3.5) we obtain Then,

By Hölder's inequality we have
, so that since the sequence {w k } k is uniformly bounded in L 2 (0, T; H 1 0 (Ω)) then so is {∂ t w k } k in L 1 (0, T; H −1 (Ω)).Therefore, by [47,Corollary 4] there exists a subsequence of {w k } k∈N which converges in norm in L 2 (Ω T ).So S : B → B is compact.Given these conditions on S, Schauder's fixed point theorem provides the existence of a function u n ∈ B such that u n = S(u n ) that is u n solves (3.1) in the sense of (3.2).In particular we have u n ∈ L 2 (0, T; H 1 0 (Ω)) ∩ L ∞ (Ω T ).The last assertion follows from Lemma A.5 (in Appendix).
We also observe that from Lemma A.6 (in Appendix) the sequence {u n } n is increasing.

Proof of Theorem 2.2
The main argument is to get a priori estimates on {u n } n and then to pass to the limit as n → +∞.We divide the proof in four cases, the case where σ = 1, the case σ < 1, the case σ > 1 and the case σ > 1 with supp( f ) ⊂⊂ Ω T .
Taking u n χ(0, τ)(t) as test function in (3.2), with 0 ≤ τ ≤ T, we get Then, by using (1.4) we obtain Passing to the supremum in τ ∈ [0, T], we obtain This shows that the sequence {u n } n is uniformly bounded in L ∞ (0, T; L 2 (Ω))∩L 2 (0, T; H 1 0 (Ω)).Then, there exist a subsequence of {u n } n still indexed by n and a function u ∈ L ∞ (0, T; L 2 (Ω)) ∩ L 2 (0, T; H 1 0 (Ω)) such that u n ⇀ u weakly in L 2 (0, T; H 1 0 (Ω)).Moreover, the boundedness of {∂ t u n } n in the dual space L 2 (0, T; H −1 (Ω)) implies that the sequence {u n } n is relatively compact in L 1 (Ω T ) (see [47,Corollary 4]) and hence for a subsequence, indexed again by n, we have Notice that since u n ⇀ u weakly in L 2 (0, T; H 1 0 (Ω)), we immediately have As regards the first integral in the right-hand side of (3.8), we know that the sequence {u n } is increasing to its limit u so we have Applying Hölder's and Hardy's inequalities we obtain .
As N ≥ 3 and Ω bounded, a straightforward calculation yields the existence of a positive constant C 1 such that Therefore, the function On the other hand, the support supp(ϕ) of the function ϕ is a compact subset of Ω T and so by Lemma A.5 (in Appendix) there exists a constant C supp(ϕ) > 0 such that u n ≥ C supp(ϕ) in supp(ϕ).Then, So that by the Lebesgue dominated convergence theorem we can get Now passing to the limit as n tends to ∞ in (3.8) we obtain The function u n χ (0,τ) ∈ L 2 (0, T; H 1 0 (Ω)), τ ∈ (0, T), is an admissible test function in (3.2).Taking it so and using Hölder's inequality and (1.4) we arrive at where Then, we have By Sobolev's inequality there exists a positive constant C such that For every real numbers a, b ≥ 0 and for every Let ϵ > 0 be arbitrary.For every positive real numbers a and b, the Young inequality yields where p > 1, q = p p−1 and C ϵ = p−1 p(pϵ) in the first term on the right hand side of (3.10) obtaining Choosing ϵ such that 1 − µ Λ N,2 − ϵ > 0 and passing to the supremum in τ ∈ [0, T] we obtain . Therefore, the sequence {u n } n is uniformly bounded in L 2 (0, T; H 1 0 (Ω)) and L ∞ (0, T; L 2 (Ω)).Thus there exist a subsequence of {u n } n , still labelled by n, and a function u ∈ L 2 (0, T; H 1 0 (Ω)) such that Now we shall prove that u is a weak solution of (1.1).For this, let us insert as a test function As in the first case, we can pass to the limit in the above equality to conclude that u is a finite energy solution of (1.1).
In order to prove that {u n } n is uniformly bounded in L 2 (0, T; Inserting G k (u n )χ (0,τ) , with 0 ≤ τ ≤ T, as a test function in (3.2) we obtain (3.12) Observe that the function G k (u n ) is different from zero only on the set B n,k := (x, t) ∈ Ω τ : u n (x, t) > k , and so we have Since Ω (G k (u n (x, 0))) 2 dx ≤ Ω (u 0 (x)) 2 dx and u n + 1 n ≥ k on B n,k inequality (3.12) becomes Taking into account that µ < Λ N,2 by (1.4) we have (3.13) We shall now estimate the term µk dxdt.Let us fix α such that 1 < α < 2 and set

By Young's inequality we can write
µk Having in mind (3.9) we get µk where C 5 = C 1 µk β β .Then the Hölder inequality yields µk where and by (1.4) we obtain µk where For arbitrary ϵ > 0, applying the Young inequality (3.11) with a = where . Choosing ϵ such that 1 − µ Λ N,2 − ϵ > 0 and gathering (3.13) and (3.14), we deduce that where C 9 = C 8 + C 4 .Passing to the supremum in τ ∈ [0, T], we conclude that the sequence {G k (u n )} n∈N is uniformly bounded in L 2 (0, T; H 1 0 (Ω)) ∩ L ∞ (0, T; L 2 (Ω)).We now turn to prove that the sequence {T where we have dropped σ > 1 in the second integral on the left-hand side and written ) in the first integral on the right-hand side of the inequality.As ), the first term on the right-hand side of the above inequality can be estimated as So that by (1.4), (3.9), (3.14) and (3.15) there exists a real constant C 10 > 0 such that Then, it follows that the inequality (3.16) reads as with . On the other hand, let Ω ′ ⊂⊂ Ω.By Lemma A.5 (in Appendix) there exists C Ω ′ > 0 such that for all (x, t) ∈ Ω ′ × [0, T].Thus, by (3.17) and (3.18) we get Passing to the supremum in τ ∈ [0, T], we get that the sequence . Therefore, we conclude that the sequence {u n } n∈N is uniformly bounded in L 2 (0, T; H 1 loc (Ω)) ∩ L ∞ (0, T; L 2 (Ω)).As a consequence, there exist a subsequence of {u n } n∈N , relabelled again by n, and a function u ∈ L 2 (0, T; H 1 loc (Ω)) ∩ L ∞ (0, T; L 2 (Ω)) such that u n ⇀ u weakly in L 2 (0, T; H 1 loc (Ω)).On the other hand, let us assume that 4σ (σ+1) 2 − µ Λ N,2 > 0. Taking u σ n χ (0,τ) (t), 0 ≤ τ ≤ T, as a test function in (3.2) and using the Hardy inequality (1.4) we arrive at This shows that u σ+1 2 n is uniformly bounded in L 2 (0, T; H 1 0 (Ω)) and so by the Poincaré inequality the sequence u n is uniformly bounded in L σ+1 (Ω T ) and hence for a subsequence, labelled again by n, we have u n → u a.e. in Ω T .
Testing by an arbitrary function ϕ We shall now pass of the limit in each term of (3.19).Notice that since u n ⇀ u weakly in For the first integral in the right-hand side of (3.19), we know that the sequence {u n } is increasing to its limit u so we obtain By Hölder's and Hardy's inequalities we get .
Since u ∈ L 2 (0, T; H 1 loc (Ω)), a calculation as in (3.9) allows us conclude that the function in Ω T , so that by the Lebesgue dominated convergence theorem one has As regards the last term in (3.19), by Lemma A.5 (in Appendix) there exists a constant C supp(ϕ) > 0 such that u n ≥ C supp(ϕ) in supp(ϕ).Then, So that by the Lebesgue dominated convergence theorem we get Finally passing to the limit as n tends to ∞ in (3.19) we obtain for all ϕ ∈ C ∞ 0 (Ω T ).Furthermore, by Lemma A.5 there exists a constant which shows that u is a weak solution of (1.1).Now assume that σ > 1 is such that 4σ (σ+1) 2 − µ Λ N,2 > 0. For 0 ≤ τ ≤ T let us use u σ n χ (0,τ) as a test function in (3.2).By the Hardy inequality (1.4) we arrive at . Therefore, we deduce that u

Proof of Theorem 2.4
Let 0 ≤ τ ≤ T. Taking u n χ (0,τ) (t) as a test function in (3.2), we get Passing to the supremum in τ ∈ [0, T] and using Theorem A.1 (in Appendix) we conclude that the sequence {u n } n is uniformly bounded in L q (0, T; W ), for all q < 2. As a consequence, there exist a subsequence of {u n } n , still indexed by n, and a function u ∈ L q (0, T; W Arguing in a similar way as in the case 1, we conclude that u is a weak solution of the problem (1.1).

Proof of Theorem 2.5
Suppose that µ > Λ N,2 .Arguing by contradiction, assume that (1.1) admits a positive weak solution u.Thus u is also a weak solution to the problem in ∂Ω × (0, T).
By virtue of Lemma A.3 (in Appendix) we have for any small enough parabolic cylinder B r 1 (0) × (t 1 , t 2 ) ⊂⊂ Ω T where α 1 is defined in (A.1).As in our equation λ = Λ N,2 we have α 1 = N−2 2 .Since u > 0 and f ≥ 0 we have in particular On the other hand, since Gathering (3.20) and (3.21) we obtain which is a contradiction.Therefore, if µ > Λ N,2 the problem (1.1) has no positive weak solution.

Proof of Theorem 2.6
The proofs of (i) and (ii) are similar.We only give the proof of (i).
• Proof of (a) -We shall establish an a priori L ∞ -estimate for the solution u n of (3.2).To do so, we use standard ideas that can be found in several nonsingular cases as for instance in [22,28,48,51,55,57].Despite being classic, we give the proof for the convenience of the reader.
Observe that since G k (u n ) is different from zero only on the set A k,n and according to the choice of k, one has

Note that the Hölder inequality implies
.
Taking into account that on the subset A k,n one has ∇G k (u n ) = ∇u n a.e. in Ω, so that Hardy's inequality yields Then passing to the supremum in τ ∈ (0, T) we obtain On the other hand, since . Therefore, by [28, Proposition 3.1] there exists a positive constant c such that , we obtain using (3.22) Observe that both integrals are on the subset A k,n .Using Hölder's inequality in the right-hand side term with exponents 2N+4 N and 2N+4 N+4 , we get , from which it follows , we use again Hölder's inequality obtaining Since m > N 2 + 1 we have β > 1 and then we can apply the first item of [48,Lemma 4.1] to conclude that there exists a constant • Proof of (b) -Using u γ n χ (0,τ) , 0 < τ < T, as a test function in (3.2) and applying the Hölder's inequality and (1.4) we arrive at . Since we have supposed that γ 2 γ+1 2 − µ Λ N,2 > 0, we discuss the two cases σ = γ and σ < γ.Thus, if σ = γ we immediately have While If σ < γ, we compute (γ − σ)m ′ = (γ + 1) N+2 N < (γ + 1) N N−2 .Therefore, by (3.23) there exists a positive constant C such that Using the Sobolev inequality in the first term on the right hand side of the above inequality, we conclude that there exists a positive constant C 1 such that Finally we choose ϵ such that γ ⇀ v weakly in L 2 (0, T; H 1 0 (Ω)).Now according to the proof of the second item of Theorem 2.2, we know that u n ⇀ u weakly in L 2 (0, T; H 1 loc (Ω)) so that identifying almost everywhere the limits one has v = u γ+1 2 ∈ L 2 (0, T; H 1 0 (Ω)).

Proof of Theorem 2.8
The ideas we use are standard and we follow the lines of [24, Theorem 4.1, (i)-(b)].Let us choose u 2δ−1 n χ (0,τ) , 0 < τ < T, as a test function in (3.2) where δ is a positive real constant satisfying This choice made possible by the fact that µ < Λ N,2 implies 1  2 < δ and 2δ−1 δ 2 − µ Λ N,2 > 0 that will be chosen after few lines.We get Passing to the supremum in τ ∈ (0, T) and applying Hardy's inequality (1.4) and then Hölder's inequality, we obtain . Thus, by [28,Proposition 3.1] there exists a positive constant c such that Then, using (3.24) we obtain where To check the upper bound on δ, we notice that δ < 1 is equivalent to m < 2N+4 N(1+σ)+4 .Such an inequality is always satisfied since for σ ≤ 1 we have m < m 1 ≤ 2N+4 N(1+σ)+4 .Therefore, with this choice of δ we obtain Since m < N 2 + 1 we have and so by virtue of Young's inequality the sequence {u n } n is uniformly bounded in Now we shall obtain an estimation on ∇u n .Notice that from (3.24) we get and since {u n } n is uniformly bounded in L γ (Ω T ), we deduce the existence of a positive constant C, not depending on n, such that Let now q ≥ 1 be such that q < 2.An application of Hölder's inequality with exponents 2 q and 2 2−q yields . Now we impose the condition γ = (1−δ)2q 2−q that gives q = m(N+2)(σ+1) N+2−m(1−σ) .Observe that q ≥ m(σ + 1) > 1 and since σ ≤ 1 we have m < m 1 ≤ 2N+4 N(1+σ)+4 which implies q < 2. Thus, the sequence {u n } n is uniformly bounded in L q (0, T; W 1,q 0 (Ω)) ∩ L γ (Ω T ).Therefore, there exist a subsequence of {u n } n , still indexed by n, and a function u ∈ L q (0, T; W Notice that since u n ⇀ u weakly in L q (0, T; W 1,q 0 (Ω)), we immediately have As regards the first integral in the right-hand side of (3.25), we know that the sequence {u n } is increasing to its limit u so we have Applying Hölder's and Hardy's inequalities with exponents 2δ and 2δ 2δ−1 we obtain .
From (3.9) and (3.24) we deduce that the sequence {u δ n } is uniformly bounded in L 2 (0, T; H 1 0 (Ω)) and thus there exist a subsequence of {u δ n }, still indexed by n, and a function v ∈ L 2 (0, T; H 1 0 (Ω)) such that u δ n ⇀ v weakly in L 2 (0, T; H 1 0 (Ω)) and u δ n → v a.e. in Ω T .But we also have u δ n ⇀ v weakly in L q (0, T; W 1,q 0 (Ω)) and hence follows v = u δ ∈ L 2 (0, T; H 1 0 (Ω)).Which shows that the function |x| 2 a.e. in Ω T , the Lebesgue dominated convergence theorem gives On the other hand, the support supp(ϕ) of the function ϕ is a compact subset of Ω T and so by Lemma A.5 (in Appendix) there exists a constant C supp(ϕ) > 0 such that u n ≥ C supp(ϕ) in supp(ϕ).Then, So that by the Lebesgue dominated convergence theorem we get We point out that we also have u ≥ C supp(ϕ) in supp(ϕ).Now passing to the limit as n tends to ∞ in (3.25) we obtain Namely u is a finite energy solution of the problem (1.1).

Proof of Theorem 2.10
Let u, v ∈ L 2 (0, T; H 1 0 (Ω)) be two energy solutions of the problem (1.1) corresponding to the same data u 0 satisfying (1.3) ) is an admissible test function in the formulation of solution (A.8) in Lemma A.7 (in Appendix).Taking it so in the difference of formulations (A.8) solved by u and v, we obtain dν and dropping the negative term, we get Using Ω Θ k ((u − v) + (x, T))dx ≥ 0, the fact that u(x, 0) = v(x, 0) = u 0 (x), Hardy's inequality (1.4) and Hölder's inequality, we arrive at .
Having in mind (3.9) and using again (1.4) we reach that + kµT .
Passing now to the limit as k tends to 0 we obtain

A Appendix
We give here some important lemmas that are necessary for the accomplishment of the proofs of the previous results.The following lemma provides a local comparison result with this radial solution.
Lemma A.6.Assume that µ ≤ Λ N,2 and let u n be a solution of (3.1).The sequence {u n } n∈N is nonnegative and increasing with respect to n ∈ N.
Proof.Writing (3.2) with u n and u n+1 and then subtracting the two corresponding equations, we obtain for every ϕ ∈ L 2 (0, T; H 1 0 (Ω)).Inserting (u n − u n+1 ) + ∈ L 2 (0, T; H 1 0 (Ω)) as a test function in (A.7) and using the fact that T n+1 is a 1-Lipschitzian function, we get Dropping the non-negative parabolic term and using the fact that (u n − u n+1 ) + 1 we obtain