Existence and nonexistence results for a class of linear and

semilinear parabolic equations related to some

Caffarelli-Kohn-Nirenberg inequalities

B. Abdellaoui, E. Colorado∗ & I. Peral∗

Departamento de MatematicasUniversidad Autonoma de Madrid

28049 Madrid, Spain

Dedicated to Professor James Serrin in his 76th birthday.

Abstract

In this work we study the problem

ut − div(|x|−2γ∇u) = λuα

|x|2(γ+1)+ f in Ω× (0, T ),

u ≥ 0 in Ω× (0, T ) and u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(1)

Ω ⊂ IRN (N ≥ 2) is a bounded regular domain such that 0 ∈ Ω, λ > 0, α > 0, −1 < γ <N − 2

2, f

and u0 are positive functions such that f ∈ L1(Ω× (0, T )) and u0 ∈ L1(Ω). The main points underanalysis are: i) spectral instantaneous and complete blow-up related to Harnack inequality in the caseα = 1; ii) the nonexistence of solution if α > 1; iii) an uniqueness result for very weak solution; iv)further results on existence of weak solution in the case 0 < α ≤ 1.

1 Introduction.

We take as starting point for this work the results by Baras and Goldstein in the paper [5]. In particularin [22] and [9] those results are related to a Hardy inequality. With this perspective we can see the resultsby Baras-Goldstein as a kind of spectral instantaneous and complete blow-up, roughly speaking, if weconsider the heat equation with a singular perturbation, ut −∆u = λ|x|−2u, then the initial-boundaryproblem has global solution for some large class of initial data if λ < λN,0 := (N − 2)2/4, and has notpositive solution for λ ≥ λN,0. In fact λN,0 is the optimal constant in the Hardy inequality (19) below,in the case γ = 0. It is well known that λN,0 is not attained. In the papers [4], [16] and [18] has been

∗Partially supported by Project BFM2001-0183 of M.C.Y.T. Spain0Mathematical Classification Subject American Mathematical Society: 35K10, 35K15, 35K65, 35J70, 46E35.

Keyword: Parabolic equations, blow-up, Harnack inequality, Hardy-Sobolev inequalities, uniqueness

1

2

obtained results for some class of parabolic equations related to generalized Hardy-Sobolev inequalities,showing that the spectral instantaneous blow-up is not true, in the sense that, even if the parameter λ islarger than the optimal constant in the corresponding Hardy-Sobolev inequality, there exists a solution.In this paper we study the following problem,

ut − div(|x|−2γ∇u) = λuα

|x|2(γ+1)+ f in Ω× (0, T ),

u ≥ 0 in Ω× (0, T ) and u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(2)

where Ω ⊂ IRN (N ≥ 2) is a bounded regular domain such that 0 ∈ Ω, λ > 0, α > 0, −1 < γ <N − 2

2, f

and u0 are positive functions such that f ∈ L1(Ω×(0, T )) and u0 ∈ L1(Ω). Notice that the case γ = 0 andα = 1 is just the problem solved by Baras and Goldstein in [5]. In a contrary direction, in [16] is proved, inparticular, that if (1+γ) < 0, problem (2) with α = 1 has global weak solution for all λ ∈ IR, namely, thespectral instantaneous and complete blow-up does not occur. However, the corresponding Hardy-Sobolevinequality holds. We will discuss in this paper the case (1 + γ) > 0. The fundamental difference with thecase (1 + γ) ≤ 0, is that if (1 + γ) > 0, the associated homogeneous linear differential equation verifiesthe parabolic Harnack inequality (10), proved by Chiarenza and Serapioni in [15] and by Gutierrez andWheeden in [20] for more general cases of degenerate equations. Moreover the interval of γ is optimal.Precisely, if (1 + γ) ≤ 0 we consider u(x, t) = t|x|ρ where ρ > 0. A direct computation shows that

ut − div (|x|−2γ∇u) = |x|ρ − ρt|x|ρ−2(γ+1)(ρ+N − 2(γ + 1)).

As ρ > 0 and γ + 1 ≤ 0, choosing r0, t0 > 0 small enough, u is a supersolution to problem (9) inthe cylinder Br0(0) × (0, t0). Since u(0, t) = 0, even the weak Harnack inequality, given by (12), is notsatisfied. Then we discover that Harnack inequality, joint to the Hardy-Sobolev one given by (19) are theproperties to have the spectral instantaneous and complete blow-up. The case γ = 0 and α > 1, namelythe heat equation with a perturbation uα|x|−2, was studied in [9] where it is shown that the completeand instantaneous blow-up is independent of λ > 0, namely, problem (2) has not positive solutions evenin distribution sense unless the trivial one .The Sobolev space D1,2

0,γ(Ω) is defined as the completion of C∞0 (Ω) under the norm

‖φ‖2γ =

∫

Ω

|x|−2γ |∇φ|2dx.

It is clear that D1,20,γ(Ω) is a Hilbert space. Notice that D1,2

0,γ(Ω) ⊂ W 1,20 (Ω) for γ > 0 and W 1,2

0 (Ω) ⊂

D1,20,γ(Ω) if γ ≤ 0. This is the natural space to work with the elliptic part and find solutions in the classical

energy sense or, more precisely, looking for solutions in a variational framework.Next we precise the more general senses in which we will consider solutions in the case in which the dataof the problem are no variational.First, following [8] and [25], we formulate the sense of entropy solution. As usual we will consider thetruncature Tk(u) = u− sign(u)(|u| − k)+.

Definition 1.1 We say that u ∈ T 1,20,γ (Ω× (0, T )) if Tk(u) ∈ L2((0, T ),D1,2

0,γ(Ω)) for all k > 0.

3

Definition 1.2 Assume that f and u0 are positive functions such that f ∈ L1(Ω×(0, T )) and u0 ∈ L1(Ω).A function u ∈ C([0, T ], L1(Ω)) is an entropy solution to problem

ut − div(|x|−2γ∇u) = f in Ω× (0, T ),u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(3)

if u ∈ T 1,20,γ (Ω× (0, T )) and

∫

Ω

Θk(u − v)(T )dx+

∫ T

0

∫

Ω

vtTk(u− v)dxdt +

∫ T

0

∫

Ω

∇u∇(Tk(u − v))|x|−2γdxdt

=

∫

Ω

Θk(u0 − v(0))dx +

∫ T

0

∫

Ω

fTk(u− v)dxdt,

(4)

for all k > 0 and for all v ∈ L2((0, T );D1,20,γ(Ω)) ∩ L

∞(Ω× [0, T ]) ∩ C([0, T ];L1(Ω)) where

Θk(s) =

∫ s

0

Tk(t)dt. (5)

Notice that as u ∈ C([0, T ], L1(Ω)),

∫

Ω

utTk(u)dx =d

dt

∫

Ω

Θk(u)dx

or

∫

Ω

Θk(u(t))dx −

∫

Ω

Θk(u(0))dx =

∫ t

0

∫

Ω

utTk(u)dxdt.

Using the same arguments as in [8] and [25] it is easy to prove the following result.

Theorem 1.3 Let Ω be a bounded regular domain, assume that f ∈ L1(Ω× (0, T )) and u0 ∈ L1(Ω) arepositive functions. Then problem (3) has an unique positive entropy solution.

As a consequence we have the following comparison principle that we will use systematically.

Lemma 1.4 Let u, v be entropy solutions of (3) with data f, u0 and g, v0 respectively. Suppose that0 ≤ g ≤ f and 0 ≤ v0 ≤ u0, then 0 ≤ v ≤ u.

On the other hand we have the following definition of weak solution.

Definition 1.5 Let f(x, t) ∈ L1(Ω × (0, T )) and u0 ∈ L1(Ω), u ∈ C(

[0, T ), L1(

(1 + |x|−2γ−1)dx))

, is aweak solution to

ut − div(|x|−2γ∇u) = f(x, t),u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in L

1(Ω),(6)

if for all 0 < s < T we have that∫ s

0

∫

Ω

u(−ψt − div(|x|−2γ∇ψ))−

∫

Ω

u0ψ(0) =

∫ s

0

∫

Ω

fψdxdt, (7)

for all ξ ∈ C2(Ω× [0, s]) with ξ(x, s) = 0 on Ω and ξ = 0 on ∂Ω× [0, s].

4

We write u(x, 0) = u0 meaning that limt→0

∫

Ω

|u(x, t)− u0(x)|dx = 0.

In section 3 we will prove the following result on uniqueness.

Theorem 1.6 Assume that u is a weak solution to problem

ut − div(|x|−2γ∇u) = 0,u = 0 on ∂Ω× (0, T ),u(x, 0) = 0 in L1(Ω) in Ω,

(8)

then u ≡ 0.

To prove the uniqueness Theorem 1.6 we require a Weyl type result for the associated elliptic equationproved in [3]. For the reader convenience, we sketch the proof in the Appendix.In consequence with this uniqueness result all the arguments will be done by approximation, understoodin a convenient way.The paper is organized as follows. In section 2 we state the Harnack inequality (see [15]) and we givethe weak version which we use in the paper. We also state the Hardy-Sobolev inequality for weights,as a particular case of the Cafarelli-Kohn-Nirenberg inequalities. Finally we formulate in this section aconsequence of a Picone identity, see [23], that will be an important tool in the proofs of blow-up. Section3 deals with the proof of Theorem 1.6. Section 4 is devoted to prove the main results about blow-up. Inthe first subsection we study the spectral complete and instantaneous blow-up in the linear case, Theorems4.1 and 4.5. The second subsection deals with the superlinear case, here without any restriction on λ,we are able to prove the nonexistence result, Theorem 4.6, and the complete blow-up phenomena forany positive data, Theorem 4.9. In section 5 we study the sublinear case: we prove the existence of aglobal solution in Theorem 5.1 if the initial data belongs to L2(Ω) and in Theorem 5.5 for more generalinitial data. Moreover we obtain the asymptotic behavior of the solutions as t→ ∞, if the initial data isordered with the stationary solution. In section 6 we state de existence results proved in [16] for the case(1+ γ) ≤ 0 to compare with the case (1+ γ) > 0. Finally in the Appendix we prove a result of regularity(and then the uniqueness) of Weyl type for the associated elliptic equation.

2 Preliminaries.

The Harnack inequality for the parabolic equation will be an important tool in our discussion. Considerthe equation

ut − div(|x|−2γ∇u) = 0 in R, (9)

where γ > −1. We say that u ∈ L2((0, T );D1,20,γ(Ω)) ∩C([0, T ];L

1(Ω)) is a solution in energy sense to (9)if

∫ T

0

∫

Ω

vtu+

∫ T

0

∫

Ω

|x|−2γ∇u∇v = 0, ∀ v ∈ L2((0, T );D1,20,γ(Ω)) ∩C([0, T ];L

1(Ω)).

In the sequel sup, inf will denote the essential supremum, essential infimum respectively. The concreteresult that we use is the following.

5

Theorem 2.1 (Harnack Inequality) Let u be a positive energy solution to (9). We denote R =Bρ(x0) × (t0 − β, t0 + β) ⊂ Ω × (0, T ), where 0 < β < t0. Then there exists C = C(N, γ, ρ, t0, β)such that

supR−

u ≤ C infR+

u, (10)

where R− = Bρ/2(x0)× (t0 −34β, t0 −

14β) and R+ = Bρ/2(x0)× (t0 +

14β, t0 + β).

For a proof, we refer to [15] and also [20] for some extension. We will use the following weak Harnackinequality for positive supersolution.

Theorem 2.2 Let u ∈ L2((0, T );D1,20,γ(Ω)) ∩ C([0, T ];L

1(Ω)) be a positive supersolution to problem (9),then there exists a positive constant C = C(N, γ, ρ, t0, β) such that

∫

B ρ2(x0)

u(x, t)dx ≤ C infR+

u, for all t ∈ (t0 −3

4β, t0 −

1

4β). (11)

Moreover∫ ∫

R−

u(x, t)dxdt ≤ C infR+

u. (12)

Proof: Since u ∈ C([0, T ];L1(Ω)), there exists t ∈ [t0 −34β, t0 −

14β] such that

supt∈(t0−

34β,t0−

14β)

∫

B ρ2(x0)

u(x, t)dx =

∫

B ρ2(x0)

u(x, t)dx. (13)

Let v be the solution to problem

vt − div(|x|−2γ∇v) = 0, in Ω× (t, t0 + β),v(x, t) = u(x, t), x ∈ Ω.

(14)

By comparison, u ≥ v. Let ξ be the positive solution to the adjoint problem

−ξt − div(|x|−2γ∇ξ) = h, in Ω× (t, t0 + β),ξ(x, t0 + β) = 0, x ∈ Ω,

(15)

where h is a bounded positive function that will be chosen later. By the maximum principle, ξ ≥ 0. Usingξ as a test function in (14), we obtain

−

∫

Ω

ξ(x, t)v(x, t)dx+

∫ t0+β

t

∫

Ω

v(−ξt − div(|x|−2γ∇ξ))dxdt = 0,

therefore

∫ t0+β

t

∫

Ω

v(x, t)h(x, t)dxdt =

∫

Ω

u(x, t)ξ(x, t)dx ≥ minB ρ

2(x0)

ξ(x, t)

∫

B ρ2(x0)

u(x, t)dx.

6

By choosing h = χB ρ2(x0)χ(t0+

14β,t0+β), we conclude that

∫ ∫

R+

v(x, t)dxdt ≥ minB ρ

2(x0)

ξ(x, t)

∫

B ρ2(x0)

u(x, t)dx. (16)

Finally by Theorem 2.1 applied to v and Lemma 1.4, we obtain

supR−

v ≤ c infR+

v ≤ c infR+

u. (17)

From (13), (16) and (17), we conclude,

∫ ∫

R−

u(x, t) ≤ c1

∫

B ρ2(x0)

u(x, t)dx ≤ c2 infR+

u,

where c1, c2 are independent on u.

This last result is an important fact to analyze the blow-up behaviour. Notice that the homogeneousparabolic equation associated to problem (2) verifies the weak Harnack inequality (12), if and only if(1 + γ) > 0, according with the counterexample explained in the introduction, i.e., the function u(x, t) =t|x|ρ where ρ > 0.If f ≥ 0, then we can prove the following extension of the Harnack inequality to entropy solution, preciselywe get the following Corollary.

Corollary 2.3 Let f ∈ L1(Ω × (0, T )) be such that f ≥ 0. Assume that u is a positive entropy solutionto problem (3), then there exists a positive constant C = C(N, γ, ρ, t0, β) such that

supt∈(t0−

34β,t0−

14β)

∫

B ρ2(x0)

u(x, t)dx ≤ C infR+

u. (18)

Notice that as a consequence we get the strong maximum principle for positive entropy solution if f ≥ 0.For u ∈ D1,2

0,γ(Ω) we have the following particular case of the Caffarelli-Kohn-Nirenberg inequalities in[12].

Proposition 2.4 (Caffarelli-Kohn-Nirenberg) Let r, α and σ be real constants such that r > 0 and

1

2+α

N,1

r+σ

N> 0.

Then there exists a positive constant C such that for all u ∈ C∞0 (IRN ) we have

∥

∥

∥|x|σu∥

∥

∥

Lr≤ C

∥

∥

∥|x|α|∇u|∥

∥

∥

L2,

if and only if1

r+σ

N=

1

2+α− 1

Nand 0 ≤ α− σ ≤ 1.

7

In particular we get the following extension of the Hardy-Sobolev inequality,

Lemma 2.5 Let N ≥ 2 and −∞ < γ < N−22 . Then for all u ∈ D1,2

0,γ(Ω), we have

λN,γ

∫

IRN

|u|2

|x|2(γ+1)dx ≤

∫

IRN

|∇u|2

|x|2γdx, λN,γ =

(

N − 2(γ + 1)

2

)2

. (19)

Moreover λN,γ is optimal and it is not achieved, see [13].

The following Sobolev inequality will be used in the last section to prove the behavior of the solutions.

Theorem 2.6 (Sobolev Inequality) There exists a positive constant C = C(N, γ) such that for allu ∈ D1,2

0,γ(Ω),

∫

Ω

|u|2∗

γ |x|−2γdx

1/2∗γ

≤ C

∫

Ω

|∇u|2|x|−2γdx

1/2

(20)

where 2∗γ = 2(N−2γ)N−2(γ+1) .

Finally an important tool to prove the nonexistence is the following theorem which proof can be foundin [2].

Theorem 2.7 (Picone Inequality) If u, v ∈ D1,20,γ(Ω) with v > 0 in Ω, and −div(|x|−2γ∇v) is a positive

bounded Radon measure, then

∫

Ω

|x|−2γ |∇u|2dx ≥

∫

Ω

−div (|x|−2γ∇v)

vu2dx.

3 A result on uniqueness

The main result of this part is the following uniqueness Theorem. The proof is deeply inspired in thepapers by Benilan-Brezis-Crandall [6], by Brezis-Crandall [7] and by Pierre [24]. An important tool in theproof is a regularity result for the associated elliptic equation, with the same philosophy that the classicalresult by Weyl for the Laplace operator, obtained in [3]. (See the Appendix at the end of this work.)

Theorem 3.1 Assume that u is a weak solution (in the sense of Definition (1.5)) to problem

ut − div(|x|−2γ∇u) = 0,u = 0 on ∂Ω× (0, T ),u(x, 0) = 0 in L1(Ω) in Ω,

(21)

then u ≡ 0.

8

Proof. To simplify the notation we set Lγv = div(|x|−2γ∇v). Let u be a weak solution to (8), we claimthat

u(t)− u(s) = Lγ

∫ t

s

u(σ)dσ in D′(Ω).

To prove the claim we set X(t) =

∫

Ω

u(x, t)φ(x)dx where t ∈ (0, T ) and φ ∈ C∞0 (Ω). Notice that X ∈

L1(0, T ). Let α ∈ C∞0 (0, T ), then we have

〈X ′, α〉 ≡ −〈X,α′〉

= −

∫ T

0

∫

Ω

u(x, t)φ(x)α′(t)dxdt =

∫ T

0

∫

Ω

u(x, t)α(t)Lγ(φ(x))dxdt.

Hence we conclude that

X ′(t) =

∫

Ω

u(x, t)Lγ(φ(x))dx in D′(0, T ). (22)

Since

∫

Ω

u(x, t)Lγ(φ(x))dx ∈ L1(0, T ), integrating (22) we conclude the claim.

As

∫

Ω

|u(x, s)|dx→ 0 as s→ 0 we obtain that

u(t) = Lγ

∫ t

0

u(σ)dσ in D′(Ω). (23)

Let v(t) be the weak solution to the elliptic problem

−Lγv(t) = u(t), v(t)∣

∣

∂Ω= 0. (24)

By the uniqueness result, Corollary 7.4 in the Appendix, since u(t) ∈ L1(Ω), therefore v coincide withthe entropy solution to the elliptic problem (24).

Hence we conclude that v(t) = −∫ t

0u(σ)dσ. Then v is a differentiable function in (0, T ) and

∂v

∂t(t) = u(t) ∈ L1

dµ(Ω× (0, T )), dµ = (|x|−2γ−1 + 1)dxdt.

Therefore we obtain that v is a solution to problem

−Lγv(t) = u(t) = −∂v

∂t, v(t)

∣

∣

∂Ω= 0. (25)

For v we can use Tk(v) as a test function in (25). Hence we get

∫

Ω

|∇Tk(v)|2|x|−2γdx = −

∫

Ω

∂v

∂tTk(v)dx = −

d

dt

∫

Ω

Θk(v)dx

,

9

where Θk is defined in (5). By integration we obtain

∫ T

0

∫

Ω

|∇Tk(v)|2|x|−2γdx+

∫ T

0

∫

Ω

Θk(v(T ))dx = 0.

Since Θk(s) ≥ 0 we conclude that |∇Tk(v)| ≡ 0 for all k > 0 and then v ≡ 0, that implies u ≡ 0.

4 Nonexistence results: Blow-up.

Since we have the uniqueness Theorem 3.1, it is sufficient to work with entropy solutions, that meansthat we will have in mind the techniques and results for this kind of solutions.

4.1 The linear case, α = 1 and λ > λN,γ, spectral instantaneous and complete

blow-up.

This subsection deals with the following problem

ut − div(|x|−2γ∇u) = λu

|x|2(γ+1), u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ) and u(x, 0) = u0(x),(26)

where u0 is a positive function such that u0 ∈ L1(Ω) and λ > λN,γ .The main result about problem (26) is the following Theorem.

Theorem 4.1 Let u0 ∈ L1(Ω) be a positive function and assume that λ > λN,γ. Then if u0 6= 0, problem(26) has not positive weak solution.

Proof. Let u ≥ 0 be a positive weak solution to (26). By Corollary 2.3 we obtain that u > 0.We set

an(x) =1

|x|2(γ+1) + 1n

, (27)

andgn(s) = minn, s. (28)

Let un be the unique solution to problem

(un)t − div(|x|−2γ∇un) = λan(x)gn(u) in Ω× (0, T ),un = 0 on ∂Ω× (0, T ), and un(x, 0) = u0(x) in Ω.

(29)

Notice that un ∈ L∞(Ω× (0, T )). Since u0 6= 0, then using the Harnack inequality we obtain that un > 0in Ω× (0, T ) and un converge to u in C([0, T ];L1(Ω)). Since an(x)gn(u) is increasing with respect ton, then we obtain that un is also increasing.Using again the Harnack inequality we get

u1 ≥ ε a.e. in Bη(0)× (τ, T )

10

for some 0 < τ < T . Since u ≥ u1, then u ≥ ε a.e. in Bη(0)× (τ, T ). For ε > 0 we can choose a constant

c = c(η, p, ε) such that v(x) = c(log1

|x|− log

1

|η|) satisfies

vt − div(|x|−2γ∇v) =ε

|x|2(γ+1)in Bη(0)× (τ, T ),

v|∂Bη(0) = 0.(30)

We define w(x, t) = (t− τ)v(x), for t ∈ (τ, T ) which verifies

wt − div(|x|−2γ∇w) = v(x) + (t− τ)ε

|x|2(γ+1)in Bη(0)× (τ, T ),

w|∂Bη(0) = 0,(31)

and limt→τ

∫

Bη(0)

w(x, t)φ(x)dx = 0, for all φ ∈ C∞0 (Bη(0)). Hence w(x, t) is an energy and then the weak

solution to (31).

Since wt − div(|x|−2γ∇w) ≤ v(x) + Tε

|x|2(γ+1)and lim

|x|→0|x|2(γ+1)v(x) = 0, one can chose ε and η such

that wt − div(|x|−2γ∇w) ≤ε

|x|2(γ+1)in Bη(0)× (τ, T ).

Let wn be the unique positive solution to problem

(wn)t − div(|x|−2γ∇wn) = εan(x) in Br(0)× (τ, T ),(wn)|∂Br(0) = 0,

(32)

then by comparison,w(x, t) ≤ w1(x, t) = lim

n→∞wn(x, t).

Since un(x, t) ≥ u1(x, t) ≥ ε, as above by the weak comparison principle, un(x, t) ≥ wn(x, t) for all(x, t) ∈ Br(0)× (τ, T ).Taking into account that lim

|x|→0w(x, t) = ∞ uniformly in t ∈ [τ1, T1] ⊂ (τ, T ), we obtain

lim|x|→0

u(x, t) = lim|x|→0

limn→∞

un(x, t) = ∞ ∀t ∈ [τ1, T1].

Then for all c >> 1 there exists n0 ∈ IN and η1 > 0 such that for all n ≥ n0,un(x, t) ≥ c in Bη1(0) anduniformly on t ∈ [τ1, T1].Let Bη(0) ⊂⊂ Bη1(0) and consider ψ ∈ C∞

0 (Bη(0)). By using Theorem 2.7, we get

∫

Bη(0)

|∇ψ|2|x|−2γdx ≥

∫

Bη(0)

|ψ|2−div(|x|−2γ∇un)

undx, for all t ∈ [τ1, T1].

Hence∫

Bη(0)

|∇ψ|2|x|−2γdx ≥ λ

∫

Bη(0)

|ψ|2an(x)gn(u)

undx−

∫

Bη(0)

|ψ|2(un)tun

dx (33)

11

and by integration we get∫ T1

τ1

∫

Bη(0)

|∇ψ|2|x|−2γdxdt ≥ λ

∫ T1

τ1

∫

Bη(0)

|ψ|2an(x)gn(u)

undxdt −

∫ T1

τ1

∫

Bη(0)

|ψ|2(un)tun

dxdt.

We estimate directly the last integral on (τ, T ) as follows,

(T1 − τ1)

∫

Bη(0)

|∇ψ|2|x|−2γdxdt ≥ λ

T1∫

τ1

∫

Bη(0)

|ψ|2an(x)gn(u)

udxdt −

∫

Bη(0)

|ψ|2 log(un(x, T1))dx, (34)

because if x ∈ Bη(0) then un(x, T1) >> 1. As gn(u)an(x) րu

|x|2(γ+1)∈ L1(Bη(0)) and

|ψ|2

u∈

L∞(Bη(0)), then using the Monotone Convergence Theorem we conclude that

λ

∫ T1

τ1

∫

Bη(0)

|ψ|2an(x)gn(u)

udxdt → λ(T1 − τ1)

∫

Bη(0)

|ψ|2

|x|2(γ+1)dx.

We set ζ(x) = 1T1−τ1

log(u(x, T1)). Since u(·, T1) ∈ L1(Bη(0)) we obtain that ζ ∈ Lp(Bη(0)) for all p ≥ 1.In particular ζ ∈ Lp(Bη(0)) where p > N/2. We set

λ(η) = infφ∈C∞

0 (Bη(0)):φ 6=0

∫

Bη(0)

|∇φ|2dx

∫

Bη(0)

ζφ2dx

.

Then we have that the eigenvalue, λ(η) → ∞ as η → 0. Hence for all ε > 0 we get the existence of ηε in

such a way that λ(ηε) ≥1

ε. Let ε be such that

λ

1 + ε> λN,γ . We set η = ηε, then from (34) we obtain

that∫

Bη(0)

|∇ψ|2|x|−2γdx ≥λ

1 + ε

∫

Bη(0)

|ψ|2

|x|2(γ+1)dx.

Sinceλ

1 + ε> λN,γ we reach a contradiction with the optimality of λN,γ .

As a consequence we get the following corollaries.

Corollary 4.2 If u0 ≡ 0 then the unique nonnegative weak solution to problem (26) is u ≡ 0.

Corollary 4.3 Suppose f ∈ L1(Ω × (0, T )) and u0 ∈ L1(Ω) are positive functions such that (f, u0) 6=(0, 0). Then, problem

ut − div(|x|−2γ∇u) = λu

|x|2(γ+1)+ f, u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(35)

where λ > λN,γ has not weak positive solution.

12

Notice that in the proof of Theorem 4.1 no explicit representation formula are needed, and then it is analternative proof to the one given by Baras-Goldstein, [5], in the case γ = 0. See also Cabre-Martel, [11].Moreover we prove the following blow-up result.

Lemma 4.4 Let un be the minimal solution to problem

∂un∂t

− div(|x|−2γ∇un) = λan(x)gn(un) in Ω× (0, T ),

un ≥ 0 in Ω,un|∂Ω = 0, and un(x, 0) = u0(x),

(36)

where an(x), gn(s) are defined in (27), (28). Let u0 ∈ L∞(Ω) be such that u0 ≥ 0 and u0 6≡ 0. Then forall r > 0 such that B4r(0) ⊂ Ω and all t ∈ (0, T ) we have

∫

Br(0)

un(x, t)dx → ∞ as n→ ∞. (37)

Proof. We argue by contradiction. Let τ > 0, r0 > 0 be such that∫

Br0(0)

un(x, τ)dx ≤ C0 for all n.

Since un are minimal then un is a non decreasing sequence. Therefore, there exists u(x, τ) ∈ L1(Br0(0))such that un(x, τ) ր u(x, τ) for all x ∈ Br0(0). Without loss of generality we can assume thatB4r0(0) ⊂ Ω,then using weak Harnack inequality for un we get the existence of C1 ≡ C1(τ,N, γ, r0) such that for allt ∈ (τ/4, τ/3) we have∫

Br0(0)

un(x, t)dx ≤ C1 infun(x) |x ∈ Br0(0)× (τ/2,3

2τ) ≤ C1

1

|Br0(0)|

∫

Br0(0)

un(x, τ)dx ≤ C2 (38)

for all n ∈ IN .Fixed τ in a such way that (τ/5, τ/4) ⊂ (τ/4, τ/3), then we can define u(x, t) for all (x, t) ∈ Br0(0) ×(τ/5, τ/4) and

∫

Br0(0)u(x, t)dx < ∞, uniformly in t ∈ (τ/5, τ/4). Moreover, using Theorem 2.2, we get

the existence of ε1 > 0 such that un ≥ u1 > ε1 in Br0(0)× (τ/5, τ/4) ⊂ (0, T ).Let µ = λ inf

Br0(0)×(τ/5,τ/4)

g1(u1(x, t)), and ε = minε1, µ. Since un(x, t) ≥ wn(x, t) where wn is the

solution to problem

(wn)t − div(|x|−2γ∇wn) = εan(x) in Br0(0)× (τ/5, τ/4),(wn)|∂Br0

(0) = 0, wn(x, τ/5) = ε in Br0(0)(39)

and using the fact that wn ↑ w where lim|x|→0

w(x, t) = +∞ uniformly in (τ2, τ3) where τ/5 < τ2 < τ3 < τ/4

(see the proof of Theorem 4.1), so we conclude that limn→∞

lim|x|→0

un(x, t) = +∞ uniformly in (τ2, τ3).

Let Bη(0) ⊂⊂ Ω be such that un(x, t) >> 1 for all x ∈ Bη(0) × (t1, t2) with τ2 < t1 < t2 < τ3, andψ ∈ C∞

0 (Bη(0)). Using Theorem 2.7 we obtain∫

Bη(0)

|∇ψ|2|x|−2γdx ≥ λ

∫

Bη(0)

|ψ|2an(x)gn(un)

undx−

∫

Bη(0)

|ψ|2(un)tun

dx. (40)

13

By integration we get∫ t2

t1

∫

Bη(0)

|∇ψ|2|x|−2γdxdt ≥ λ

∫ t2

t1

∫

Bη(0)

|ψ|2an(x)gn(un)

undxdt−

∫ t2

t1

∫

Bη(0)

|ψ|2(un)tun

dxdt.

Since un ↑ u in L1(Bη(0)× (t1, t2)), then using the same argument as in the proof of Theorem 4.1 we get∫

Bη(0)

|∇ψ|2|x|−2γdx ≥λ

1 + ε

∫

Bη(0)

|ψ|2

|x|2(γ+1)dx,

where ε can be chosen such thatλ

1 + ε> λN,γ . Then we reach a contradiction with the optimality of

λN,γ .

We are now able to formulate the main blow-up result.

Theorem 4.5 Let un be the minimal positive solution to problem (36), then for all (x0, t0) ∈ Ω× (0, T )we have lim

n→∞un(x0, t0) = ∞.

Proof. Let (x0, t0) ∈ Ω × (0, T ), if x0 ∈ Br(0), such that B4r(0) ⊂ Ω, then using Harnack inequalitywe get

un(x0, t0) ≥ c

∫

Br(0)

un(x, t0 − ε)dx,

where c = c(N, γ, r) and ε > 0 is such that t0 − ε > 0.By Lemma 4.4 we find that

∫

Br(0)un(x, t0 − ε)dx→ ∞, hence lim

n→∞un(x0, t0) = ∞.

If x0 /∈ Br(0), we first suppose that Br(0) ∩ Br(x0) 6= ∅, then there exists y ∈ Ω such that Bη(y) ⊂Br(0) ∩Br(x0). Therefore using again the Harnack inequality and Lemma 4.4 we obtain

un(x0, t0) ≥ c

∫

Br(x0)

un(x, t0 − ε)dx ≥ c

∫

Bη(y)

un(x, t0 − ε)dx

≥ c infx∈Bη(y)

un(x, t0 − ε) ≥ cun(x1, t0 − ε),

where c is a positive constant independent of un. Since x1 ∈ Br(0), then un(x1, t0−ε) → ∞, and we obtainthat un(x0, t0) → ∞ for n → ∞. In the general case, if x0 ∈ Ω arbitrary, then since d(x,Br(0)) < ∞,using an iteration argument, in a finite number of steps we conclude.

4.2 The super-linear case: α > 1.

In this subsection we deal with the following problem,

ut − div(|x|−2γ∇u) =uα

|x|2(γ+1)in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω

(41)

where α > 1. Here the behaviour of the problem is quite different to the linear case, in particular it isindependent of λ. So we formulate the result with λ = 1.

14

Theorem 4.6 Let u0 be a positive function such that u0 ∈ L∞(Ω) and u0 6≡ 0, then problem (41) hasnot weak positive solution. In the case where u0 ≡ 0 then the unique nonnegative solution is u ≡ 0.

Proof. Assume that (41) has a positive weak solution u. Let an be defined in (27). Define gn(s) =minn, sα and fn(x, t) = an(x)gn(u(x, t)); in this way fn(x, t) ∈ L∞(Ω×(0, T )) and fn(x, t) ր f(x, t) =uα(x, t)

|x|2(γ+1)in L1(Ω × (0, T )). Since |x|2γfn ∈ Lr(Ω, |x|−2γdx) with r >

N − 2γ

2, we have the required

integrability to obtain L∞ estimates. (See [14]). Consider un the unique positive global solution to problem

(un)t − div(|x|−2γ∇un) = fn(x, t) in Ω× (0, T ),un = 0 on ∂Ω× (0, T ),un(x, 0) = u0(x) in Ω.

(42)

Then, un ∈ L∞(Ω× (0, T )) and, since u0 6≡ 0, weak Harnack inequality implies that un > 0 in Ω× (0, T ).Notice that fn is increasing with respect to n, and then we obtain that un is also increasing andconverges to u in C([0, T ];L1(Ω)). Again Harnack inequality gives us that

u ≥ u1 ≥ ε a.e. in Bη(0)× (τ, T )

for some 0 < τ < T and Bη(0) ⊂⊂ Ω. By the same arguments as in the proof of Theorem 4.1 we findthat

lim|x|→0

u(x, t) = lim|x|→0

limn→∞

un(x, t) = ∞ uniformly in t ∈ (τ, T ).

Then for all c > 0 there exists n0 ∈ IN and η1 > 0 such that un(x, t) ≥ c for all x ∈ Bη1(0) and for allt ∈ (τ, T ). We chose c > sup1, α− 1 such that

(α − 1)cα−1 − 4(α− 1)log(c)

T − τ> λN,γ , (43)

where λN,γ is the optimal constant in the Hardy inequality given in Proposition 2.4. Moreover there existsη, 0 < η << 1 and n0 ∈ IN such that if n ≥ n0 then un ≥ 2c in Bη(0)× (τ, T ).Next we use a kind of arguments as in [9]. Define the function,

φ(s) =

1

α− 1

(

1

cα−1 −1

sα−1

)

if s > c,

1

cα(s− c) if s ≤ c.

(44)

Notice that φ ∈ C1(IR), is a concave function and φ′(s) =1

sαfor all s > c, in particular 0 < φ′(s) <

1

cα.

Since φ(s) is a regular function for s > c, and un > c for all n ≥ n0, then

−div(|x|−2γ∇φ(un)) = φ′(un)(−div(|x|−2γ∇un)− φ′′(un)|x|−2γ |∇un|

2.

It is clear that φ(un) ≤1

α−11

cα−1 . Since φ′′ < 0 and un(x) > c > 1 in Bη(0), we obtain

(φ(un))t − div(|x|−2γ∇φ(un)) ≥ φ′(un)(

(un)t − div(|x|−2γ∇un))

= φ′(un)fn.

15

Consider ψ ∈ C∞0 (Bη(0)). Using Theorem 2.7, we get

∫

Bη(0)

|∇ψ|2|x|−2γdx ≥

∫

Bη(0)

|ψ|2−div(|x|−2γ∇φ(un))

φ(un)dx,

hence∫

Bη(0)

|∇ψ|2|x|−2γdx ≥

∫

Bη(0)

|ψ|2fnφ′(un)

φ(un)dx−

∫

Bη(0)

|ψ|2(un)t(φ

′(un))

φ(un)dx. (45)

By integration over (τ, T ) we obtain

∫ T

τ

∫

Bη(0)

|∇ψ|2|x|−2γdxdt ≥

∫ T

τ

∫

Bη(0)

|ψ|2fnφ′(un)

φ(un)dxdt −

∫ T

τ

∫

Bη(0)

|ψ|2(un)t(φ

′(un))

φ(un)dxdt. (46)

Since x ∈ Bη(0), then un(x, t) > 2c. Therefore we get

1

φ(un)≥ (α− 1)cα−1.

On the other hand φ′(un)fn → φ′(u)uα

|x|2(γ+1)a.e. in Ω. Since φ′(un) ≤ a, for all n ≥ n0 and x ∈ Bη(0),

we obtain that φ′(un)fn ≤ af . Hence by the dominated convergence theorem we conclude that

φ′(un)fn → φ′(u)uα

|x|2(γ+1)=

1

|x|2(γ+1)in L1(Bη(0)× (τ, T )).

Therefore we obtain∫ T

τ

∫

Bη(0)

|ψ|2fnφ′(un)

φ(un)dxdt ≥ (T − τ)(α − 1)cα−1

∫

Bη(0)

|ψ|2

|x|2(γ+1)dx+ o(1).

To estimate the term involving the t derivative, we proceed as follows. Let we denote,

I =∣

∣

∣

∫

Bη(0)

∫ T

τ

(un)t(φ′(un))

φ(un)dtdx

∣

∣

∣ =∣

∣

∣

∫

Bη(0)

|ψ|2 (logφ(un(T ))− logφ(un(τ))) dx∣

∣

∣.

Since φ(un) ≤ φ(u) we get

I ≤ 2

∫

Bη(0)

|ψ|2| logφ(u(T ))|dx ≤ 4(α− 1) log(c)

∫

Bη(0)

|ψ|2dx,

hence, by substitution in (46) and integrating in time, we reach the inequality,∫

Bη(0)

|∇ψ|2|x|−2γdx ≥

(

(α− 1)cα−1 − 4(α− 1)log(c)

T − τ

)∫

Bη(0)

|ψ|p

|x|2(γ+1)dx := Λ

∫

Bη(0)

|ψ|2

|x|2(γ+1)dx,

where

Λ = (α− 1)cα−1 − 4(α− 1)log(c)

T − τ> λN,γ

by (43), which is a contradiction with the optimality of λN,γ .

16

Corollary 4.7 Suppose f ∈ L1(Ω × (0, T )) and u0 ∈ L1(Ω), are positive functions such that (f, u0) 6=(0, 0). Then, problem

ut − div(|x|−2γ∇u) =uα

|x|2(γ+1)+ f, u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(47)

has not positive weak solution.

As a consequence we have the following blow up result.

Lemma 4.8 Let un be the minimal solution to problem

∂un∂t

− div(|x|−2γ∇un) = an(x)gn(un) in Ω× (0, T ),

un ≥ 0 in Ω, and un|∂Ω = 0,un(0, x) = u0(x).

(48)

Then for all r > 0 such that B4r(0) ⊂ Ω, and for all t ∈ (0, T ) we have∫

Br(0)

un(x, t)dx → ∞ as n→ ∞. (49)

Proof. We proced as in the proof of Lemma 4.4. Assume the existence of τ > 0, r0 > 0 such that∫

Br0(0)

un(x, τ)dx ≤ C for all n.

Since un are increasing, there exists u(x, τ) ∈ L1(Br0(0)) such that un(x, τ) ր u(x, τ) for all x ∈ Br0(0).Without loss of generality we can assume that B4r0(0) ⊂ Ω, then using weak Harnack inequality asin the proof of Lemma 4.4 we get the existence of τ > 0 such that

∫

Br0(0) un(x, t)dx ≤ C(r0, τ) for

all 0 < τ/2 ≤ t < τ . Then we obtain that u(·, t) ∈ L1(Br0(0)) for all t ∈ (τ1/4, τ1) ⊂ (τ/2, τ) and∫

Br0(0)u(x, t)dx < C, where C is independent of t. Moreover as in Lemma 4.4 one can prove that

lim|x|→0

limn→∞

un(x, t) = ∞ uniformly for all t ∈ (τ2, τ3) where τ1/4 < τ2 < τ3 < τ1.

Fixed Br0(0) ⊂⊂ Ω such that un >> c in Br0(0)× (τ2, τ3). Let ψ ∈ C∞0 (Br0(0)), by Theorem 2.7 we get

∫

Br0(0)

|∇ψ|2|x|−2γdx ≥

∫

Br0(0)

−div(|x|−2γ∇un)

unψ2dx.

Therefore as in the proof of Theorem 4.6 we obtain∫

Bη(0)

|∇ψ|2|x|−2γdx ≥

(

(α− 1)cα−1 − 4(α− 1)log(c)

τ3 − τ2

)∫

Bη(0)

|ψ|p

|x|2(γ+1)dx.

Since we can chose c such that(

(α− 1)cα−1 − 4(α− 1) log(c)τ3−τ2

)

> λN,γ , then we reach a contradiction

with the optimality of λN,γ .

As a consequence of the weak Harnack inequality we obtain the main blow-up result.

17

Theorem 4.9 Assume that un is the positive minimal solution to problem (48), then for all (x0, t0) ∈Ω× (0, T ) we have lim

n→∞un(x0, t0) = ∞.

5 Existence and non-uniqueness results for the sub-linear case:

0 < α < 1.

We study in this section the sublinear case, 0 < α < 1.

Theorem 5.1 Let u0 ∈ L2(Ω) be a positive function. Assume that 0 < α < 1, then problem

ut − div(|x|−2γ∇u) =uα

|x|2(γ+1), u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω,

(50)

has a global solution, u ∈ L2((0, T );D1,20,γ(Ω)) ∩ C([0, T ];L

2(Ω)).

Proof. The existence of solution for t small can be easily proved. Moreover using u as a test functionin (50) we obtain that

∫

Ω

u2(x, T )dx+

∫ T

0

∫

Ω

|∇u|2|x|−2γdxdt =

∫

Ω

u20dx+

∫ T

0

∫

Ω

uα+1

|x|2(γ+1)dxdt.

By Holder and Young inequalities we get

∫

Ω

un(x, T )2dx +

∫ T

0

∫

Ω

|∇un|2|x|−2γdxdt

≤

∫

Ω

u20dx+

∫ T

0

∫

Ω

dx

|x|2(γ+1)

1−α2

∫

Ω

u2dx

|x|2(γ+1)

α+12

dt

≤

∫

Ω

u20dx+α+ 1

2ε

α+12

∫ T

0

∫

Ω

u2

|x|2(γ+1)dxdt+ cε−

1−α2 T.

By Proposition 2.4 we obtain

∫

Ω

u2n(x, T )dx+ (1− λ−1N,γ

α+ 1

2ε

α+12 )

∫ T

0

∫

Ω

|∇un|2|x|−2γdxdt ≤ cε−

1−α2 T +

∫

Ω

u20dx

and choosing ε small enough we get that the solution is defined in [0, T ] for arbitrary T .

Consider now the stationary problem

−div(|x|−2γ∇w) =wα

|x|2(γ+1), w|∂Ω = 0. (51)

18

In [1] it is proved that there exists an unique solution w ≥ 0 to (51), that is a stationary solution toproblem (50) with u0 ≡ w.Since α < 1, using a construction by H. Fujita in [17] for the heat equation, we obtain the followingnonuniqueness result.

Theorem 5.2 Let w be the unique positive solution to problem (51). Then problem

ut − div(|x|−2γ∇u) =uα

|x|2(γ+1), u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = 0 in Ω,

(52)

has a positive maximal solution u 6≡ 0 and, moreover, uր w in D1,20,γ(Ω) as t→ ∞.

To prove the above Theorem we will use the following result.

Lemma 5.3 Let v ∈ L2((0, T );D1,20,γ(Ω))∩C([0, T ];L

1(Ω)) be a positive subsolution to problem (52), thend

dt

∫

Ω

(v − w)2+dx ≤ 0, namely

∫

Ω

(v − w)2+dx ≤

∫

Ω

(v(0)− w)2+dx for all t ≥ 0.

Proof. Since v is a subsolution we get

(vt − wt)− div(|x|−2γ∇(v − w)) ≤vα − wα

|x|2(γ+1).

Using (v − w)+ as test function we obtain

d

dt

∫

Ω

(v − w)2+dx+

∫

Ω

|∇(v − w)+|2|x|−2γdx ≤

∫

Ω

(vα − wα)(v − w)+|x|2(γ+1)

dx ≤ α

∫

Ω

wα−1 (v − w)2+|x|2(γ+1)

dx.

By Theorem 2.7 we get∫

Ω

|∇(v − w)+|2|x|−2γdx ≥

∫

Ω

−div(|x|−2γ∇w)

w(v − w)2+dx ≥

∫

Ω

wα−1

|x|2(γ+1)(v − w)2+dx.

Hence we conclude thatd

dt

∫

Ω

(v − w)2+dx ≤ 0.

Proof of Theorem 5.2. Let w be the unique positive solution to problem (51). Since u0 ≡ 0, then w is asupersolution to problem (52). We set u(0) = w and we define un+1 as the unique solution to problem

∂un+1

∂t− div(|x|−2γ∇un+1) =

uαn|x|2(γ+1)

, un+1 ≥ 0 in Ω× (0, t0),

un+1 = 0 on ∂Ω× (0, t0),un+1(x, 0) = 0 in Ω.

(53)

19

Since un is a decreasing sequence and un ≤ u0 for all n, then we get the existence of u solution toproblem (52). Then we have just to prove that u 6= 0. We set wε(x, t) = ε1−α((1− α)t)1−αφ where ε > 0will be chosen later and φ is the first eigenfunction of the following problem

−div(|x|−2γ∇φ) = λ1φ

|x|2γ, φ|∂Ω = 0,

such that ||φ||L∞ = 1. Then we get

∂wε∂t

− div(|x|−2γ∇wε) =(

εα+1t+ ε1−α((1− α)t)1−α))

φ,

wε ≥ 0 in Ω× (0, t0), wε = 0 on ∂Ω× (0, t0) and wε(x, 0) = 0 in Ω.

Choosing ε such that wε ≤ w, then we can prove by induction that un ≥ wε for all n. Therefore weconclude that u ≥ wε > 0.Let v be another solution to (52), then using Lemma 5.3 we conclude that v ≤ w = u(0). Therefore usingan iteration argument one can prove that un ≥ v for all n. Hence u ≥ v and the maximality of u follows.Let s > 0, then one can prove that u(x, t + s) ≥ u(x, t) for all (x, t) ∈ Ω × (0, T ), hence u is increasingwith respect to t for all x ∈ Ω. Taking u as a test function in (52) and using the increasing property, weget

∫

Ω

|∇u(x, t)|2|x|−2γdx ≤

∫

Ω

uα

|x|2(γ+1)dx.

By Holder and Hardy-Sobolev inequalities (see (19)), we obtain that for all t > 0

∫

Ω

|∇u(x, t)|2|x|−2γdx ≤ C(N, γ).

Hence we get the existence of a subsequence tk ↑ ∞ and w ∈ D1,20,γ(Ω) such that u(·, tk) w as tk → ∞

and u(·, tk) → w a.e. in Ω and in Lr(|x|−2γdx,Ω) for all 1 ≤ r < 2∗γ . Since u is increasing in t, we get theconvergence for all subsequence. Then one can prove easily that w is a weak solution to problem (51),and as a consequence w = w.

In the case 0 ≤ u(x, 0) = u0 ≤ w, we can prove the following result.

Lemma 5.4 Let u be a solution to problem (50) with 0 ≤ u(x, 0) ≤ w. Then limt→∞

u(x, t) = w(x) a.e.

Proof: By Lemma 5.3 u(x, t) ≤ w(x) for all (x, t) ∈ Ω× (0,∞). Using w(x) as a test function in problem(50) and taking into account that 0 < α < 1, uαw ≥ uwα, therefore we obtain

d

dt

∫

Ω

u(x, t)w(x)dx +

∫

Ω

uwα

|x|2(γ+1)dx =

∫

Ω

uαw

|x|2(γ+1)wdx ≥

∫

Ω

uwα

|x|2(γ+1)dx.

Then∫

Ω

u(x, t)w(x)dx ≥

∫

Ω

u(x, 0)w(x)dx > 0.

20

As a consequence u(x, t) 6→ 0 as t→ ∞. Since 0 ≤ u(x, t) ≤ w(x) , u(x, t) → w1(x) as t→ ∞, a solutionto the stationary problem, then by uniqueness, w1 ≡ w.

Let ψ(x) ≡ |x|−κc where λ < c < λN,γ and

κc =N − 2(γ + 1)−

√

(N − 2(γ + 1))2 − 4c

2,

then if Ω ≡ B1(0), ψ is a solution to problem

−div(|x|−2γ∇ψ) = cψ

|x|2(γ+1), ψ ≥ 0 and ψ|∂Ω

= 1.

Therefore we have the following Theorem.

Theorem 5.5 Let u0 ∈ L1(Ω) be a nonnegative function such that u0 6= 0. Suppose that one of thefollowing conditions holds

(1) α = 1, λ < λN,γ and

∫

Ω

u0(x)|x|−κcdx <∞,

(2) γ + 1 > 0, 0 < α < α1 = 1− γ+1N−(γ+1) and λ > 0,

(3) γ + 1 ≤ 0, 0 < α < 1 and λ > 0,

then problem

ut − div(|x|−2γ∇u) = λuα

|x|2(γ+1), u ≥ 0 in Ω× (0, T ),

u = 0 on ∂Ω× (0, T ),u(x, 0) = u1(x) in Ω,

(54)

has a weak positive minimal solution.

Proof. The first case follows by using a similar argument as in [5]. For the reader convenience we sketchthe proof here. Let un be the minimal solution to the truncated problem,

dund t

− div(|x|−2γ∇un) = λun

|x|2(γ+1) + 1/n, un ≥ 0 in Ω× (0, T ),

un = 0 on ∂Ω× (0, T ) and un(x, 0) = Tn(u0(x)) in Ω.(55)

Let p ∈ C2(IR) be a convex function satisfying p(0) = p′(0) = 0 which approximate | · |, taking p′(un)ψas a test function in (55) we obtain that

∫

Ω

un(t)ψ(x)dx + (c− λ)

∫ T

0

∫

Ω

unψ

|x|2(γ+1) + 1/ndxdt ≤

∫

Ω

Tn(u0)ψdx ≤

∫

Ω

u0(x)ψ(x)dx. (56)

21

Since un is an increasing sequence and using the fact that ψ > 1, we get the existence of u(x, t) suchthat

∫

Ω

un(t)ψdxր

∫

Ω

u(x, t)ψdx <∞ and

∫ T

0

∫

Ω

un|x|2(γ+1) + 1/n

dxdt ր

∫ T

0

∫

Ω

u(x, t)

|x|2(γ+1)dxdt <∞.

By setting f(x, t) =u

|x|2(γ+1)∈ L1(Ω × (0, T )), then using the approximation procedure we get the

desired result.To prove the other cases we set u0n(x) = Tn(u0(x)) and we assume that λ = 1. Let un be the minimalpositive solution to problem

dundt

− div(|x|−2γ∇un) =uαn

|x|2(γ+1), un ≥ 0 in Ω× (0, T ),

un = 0 on ∂Ω× (0, T ),un(x, 0) = u0n(x) in Ω.

(57)

Assume that γ + 1 > 0. Since α < α1 we get the existence of β > 0 such that

0 < β < 2(γ + 1) and 2(γ + 1)−αβ

2< N(1− α).

Let φ be the positive solution to problem

−div(|x|−2γ∇φ) =1

|x|β, φ|∂Ω = 0,

notice that φ ∈ L∞(Ω). Using φ as a test function in (57) we obtain

d

dt

∫

Ω

unφdx +

∫

Ω

un|x|β

dx =

∫

Ω

uαnφ

|x|2(γ+1). (58)

Using Hardy-Sobolev, Holder and Young inequalities we get

d

dt

∫

Ω

unφdx +

∫

Ω

un|x|β

dx ≤ ε

∫

Ω

unφdx + ε

∫

Ω

un|x|β

dx+ C(N, γ, ε),

where ε can be chosen small. Therefore we conclude that

d

dt

∫

Ω

unφdx ≤ ε

∫

Ω

unφdx+ C.

Using Gronwall Lemma we find that

∫

Ω

un(x, t)φ(x)dx ≤ c1t+ c2 and

∫

Ω

un|x|β

dx < c2 <∞,

22

where c1 and c2 depend only on n, γ and∫

Ω

u0(x)dx.

Therefore we get the existence of a measurable function u ≥ 0 such that un(x, t) ↑ u(x, t) for all (x, t) ∈

Ω× (0, T ),

∫

Ω

u(x, t)φ(x)dx <∞ and

∫ T

0

∫

Ω

u(x, t)

|x|βdxdt <∞.

We set fn(x, t) =uαn(x, t)

|x|2(γ+1), by Holder and Young inequalities we obtain

∫ T

0

∫

Ω

fn(x, t)dxdt ≤ C(N, γ)T +

∫ T

0

∫

Ω

un(x, t)

|x|βdxdt ≤ C(N, γ)T +

∫ T

0

∫

Ω

u(x, t)

|x|βdxdt <∞.

Then using the dominated convergence theorem we conclude that fn(x, t) ↑ f(x, t) =uα

|x|2(γ + 1)∈

L1(Ω× (0, T )). The result now follows by applying the same arguments as in [8] and [25].If γ + 1 ≤ 0, then using Tk(un) as a test function in (54) we get

d

dt

∫

Ω

Θk(un)dx+

∫

Ω

|∇Tk(un)|2|x|−2γdx =

∫

Ω

Tk(un)uαn

|x|2(γ+1)dx ≤

∫

Ω

Tk(un)uαndx ≤

k1−α|Ω|1−α(

∫

Ω

Tk(un)undx)α

.

Since Θk(s) ≥12Tk(s)s, then we obtain

d

dt

∫

Ω

Θk(un)dx+

∫

Ω

|∇Tk(un)|2|x|−2γdx ≤ 2αk1−α|Ω|1−α

(

∫

Ω

Θk(un)dx)α

.

Therefore following the argument above we conclude.

6 Some remarks on existence of solutions in the case (1 + γ) ≤ 0

Consider the problem

ut − div(|x|−2γ∇u) = λu

|x|2(γ+1)in Ω× (0, T ),

u ≥ 0 in Ω× (0, T ) and u = 0 on ∂Ω× (0, T ),u(x, 0) = u0(x) in Ω, u0 ∈ L2(Ω)

(59)

where we assume that (1 + γ) ≤ 0 and λ > 0. In this case we have the following energy estimate

∫

Ω

u2(x, t)dx +

∫ t

0

∫

Ω

|∇u|2|x|−2γdxdτ = λ

∫ t

0

∫

Ω

u2

|x|2(γ+1)dxdτ +

∫

Ω

u20(x)dx.

23

Since (γ + 1) ≤ 0 we have that

λ

∫ t

0

∫

Ω

u2

|x|2(γ+1)dxdτ ≤ c(Ω)λ

∫ t

0

∫

Ω

u2dxdτ,

then using Gronwall inequality and classical arguments to pass to the limit we are able to obtain thefollowing result.

Theorem 6.1 If (1 + γ) ≤ 0, λ > 0 and u0 ∈ L2(Ω), then problem (59) has an unique solution u ∈L2((0, T );D1,2

0,γ(Ω)) ∩C([0, T ];L2(Ω))

This result is a particular case of the existence results in [16]. Details can be seen in such work.

Remark 6.2 Notice that the result is independent of the relation between λ and λN,γ. As we have analyzedabove, the behavior of problem (59) if (1+γ) > 0. In this last case the existence is obtained only if λ ≤ λN,γ.We have proved the nonexistence result in the previous section 4. The existence result for λ ≤ λN,γ isobtained by using the corresponding Hardy-Sobolev inequality which provide an energy estimate. We referto [16] for details and extensions to the quasilinear framework.

7 Appendix: Uniqueness result for the elliptic case

We give a sketch of the proof to the uniqueness result for the elliptic case. We refer to [3] for more details.The results which extend the classical regularity result by Weyl for the Laplace equation are been obtainedfor several authors in particular by F. E. Browder in [10] and L. Nirenberg in [21], among others, in thecase of strongly elliptic equations with regular coefficients. On the other hand there are results by J. Serrinin [26] on non-uniqueness for non-continuous coefficients. In our problems the differential operator is eitherdegenerate or with singular coefficients at 0. Then we can see the uniqueness result as an extension tothese cases of the results quoted above about uniqueness (regularity). One of the main tools in the proofis the following result of Fourier Analysis.

Lemma 7.1 Assume that φ is 2k-times continuously differentiable function in B with k ≥ ([N/4] + 1),

where [·] is the integer part function. Let ci(r) =

∫

SN−1

φ(r, σ)ψi(σ)dσ.

We set Qn(r, σ) =∑

χ(ψi)=n

ci(r)ψi(σ), where χ(ψi) = n means that the degree of ψi is n. Then we have

|Qn(r, σ)|2 ≤ µ(N)||∆k

0φ||2 nN−4k−2 ≤ C(N, k, φ)nN−4k−2,

where ∆k0 denotes the k-times iterated Laplace-Beltrami operator, ||∆k

0φ||2 =

∫

SN−1

|(∆k0φ)(r, σ)|

2dσ| and

µ(N) is a positive constant depending on N . Moreover

φ(r, σ) =

∞∑

n=0

Qn(r, σ) =

∞∑

n=0

∑

χ(ψi)=n

ci(r)ψi(σ)

=∞∑

i=0

ci(r)ψi(σ)

24

and the convergence is uniform in B.

We refer to [19] page 127, Proposition 3.6.2, for a proof.Let Ω be a bounded domain such that 0 ∈ Ω. We begin by the following definition.

Definition 7.2 Let f ∈ L1(Ω) and u ∈ L1(|x|−2γ−1dx,Ω). We say that u is a weak solution to problem

−Lγ(u) ≡ −div(|x|2γ∇u) = f, (60)

if for all φ ∈ C∞0 (Ω) we have

∫

Ω

u(−Lγ(φ))dx =

∫

Ω

fφdx. (61)

The main result in this appendix is the following Theorem.

Theorem 7.3 Let u ∈ L1(|x|−2γ−1dx,Ω) be such that

div(|x|−2γ∇u) = 0, (62)

in the sense given in Definition 7.2. Then u ∈ Cε(Ω) ∩ D1,2γ, loc where ε depends on N and γ.

Proof. Since |x|−2γ ∈ C∞(Ω\0), then u ∈ C∞(Ω\0) by Weyl lemma for regular coefficients. See[10] and [21]. Hence we have just to prove the regularity of u at 0. Without loss of generality we canassume that B1(0) ⊂ Ω. We divide the proof in several steps.Step 1. Since u is regular for r > 0, then expanding it in terms of ψi, the eigenfunction of the Laplace-Beltrami operator −∆σ, in S

N−1, we obtain that

u(r, σ) =∑

i≥0

fi(r)ψi(σ) where fi(r) =

∫

SN−1

u|∂Br(r, σ)φi(σ)dσ.

Since u is classical solution of (62) for r ∈ [ε, 1], then we get

∑

i≥0

f ′′i (r) +

N − 1− 2γ

rf ′1(r)−

cir2fi

ψi(σ) = 0.

Hence

f ′′i (r) +

N − 1− 2γ

rf ′1(r) −

cir2fi = 0 for all i.

Therefore we obtain that

f0(r) = B0 +A0r2(γ+1)−N , fi(r) = Air

αi +Birβi ,

where

αi = −N − 2(γ + 1) +

√

(N − 2(γ + 1))2 + 4ci2

, βi =

√

(N − 2(γ + 1))2 + 4ci − (N − 2(γ + 1))

2,

25

ci = i(N + i− 2), for i ≥ 1; and

A0(ε) =εN−2(γ+1)(f0(ε)− f0(1))

1− εN−2(γ+1)

B0(ε) =f0(1)− εN−2(γ+1)f0(ε)

1− εN−2(γ+1),

and

Ai(ε) =ε−αifi(ε)− εβi−αifi(1)

1− εβi−αi

Bi(ε) =fi(1)− ε−αifi(ε)

1− εβi−αi,

for i ≥ 1, where for all ε > 0 we denote

fi(ε) =

∫

SN−1

u|∂Bε(ε, σ)φi(σ)dσ.

Therefore we conclude that for r ∈ (ε, 1) we have

u(r, σ) =f0(1)− εN−2(γ+1)f0(ε)

1− εN−2(γ+1)+εN−2(γ+1)(f0(ε)− f0(1))

1− εN−2(γ+1)r2(γ+1)−N +

∑

i≥1

ε−αifi(ε)− εβi−αifi(1)

1− εβi−αirαi +

fi(1)− ε−αifi(ε)

1− εβi−αirβi

ψi(σ).

Step 2. We claim that |εN−2(γ+1)fi(ε)| → 0. Notice that, as a consequence, we obtain that |ε−αifi(ε)| → 0as ε→ 0 for all i ≥ 0.We first assume that i ≥ 1, let φ ∈ C∞

0 (B), since u is a solution to (62) and a classical solution for |x| > 0,we obtain

0 =

∫

u(Lγ(φ))dx = limε→0

∫

|x|>ε

uLγ(φ)dx

= limε→0

∫

|x|>ε

φLγ(u)−

∫

|x|=ε

|x|−2γu〈∇φ,x

|x|〉dσ +

∫

|x|=ε

|x|−2γφ〈∇u,x

|x|〉dσ

.

Since Lγu(x) = 0 for |x| ≥ ε, and by setting φ(x) = g(r)ψi0 (σ) where g ∈ C∞[0, a), a < 1, g(0) = 1,

with supp g ⊂ [0, a) and using the fact that for x = rσ we have x∇v = r∂v

∂rand by the orthogonality of

the spherical harmonics, we obtain that

0 = limε→0

− εN−1−2γ

∫

SN−1

(Ai0(ε)εαi0 +Bi0(ε)ε

βi0 )g′(ε)ψ2i0dσ +

εN−2(γ+1)

∫

SN−1

(αi0Ai0(ε)εαi0 + βi0Bi0(ε)ε

βi0 )g(ε)ψ2i0dσ

.

Hence we conclude that

limε→0

εN−1−2γ(Ai0(ε)εαi0 +Bi0(ε)ε

βi0 )g′(ε)− εN−2(γ+1)(αi0Ai0(ε)εαi0 + βi0Bi0 (ε)ε

βi0 )g(ε)

= 0.

26

Therefore we get

0 = limε→0

εN−1−2γ(

[ε−αi0 fi0(ε)− εβi0−αi0 fi0(1)]ε

αi0 + [fi0(1)− ε−αi0 fi0(ε)]εβi0

)

g′(ε)−

εN−2(γ+1)(

αi0 [ε−αi0 fi0(ε)− εβi0

−αi0 fi0(1)]εαi0 + βi0 [fi0(1)− ε−αi0fi0(ε)]ε

βi0

)

g(ε)

.

Hence

limε→0

εN−1−2γ(fi0(ε)− εβi0−αi0 fi0(ε))g

′(ε)− εN−2(γ+1)(αi0fi0(ε)− βi0εβi0

−αi0 fi0(ε))g(ε)

= 0,

finally we obtain that

limε→0

εN−2(γ+1)fi0(ε)[

(ε− εβi0−αi0

+1)g′(ε)− (αi0 − βi0εβi0

−αi0 )g(ε)]

= 0. (63)

As[

(ε− εβi0−αi0

+1)g′(ε)− (αi0 − βi0εβi0

−αi0 )g(ε)]

→ −αi0 ,

then we conclude the claim in the case i ≥ 1.If i = 0, then by setting φ(x) = g(r) and using the same argument as above we obtain that

limε→0

εN−2(γ+1)f0(ε)[

(ε− εN−2(γ+1)+1)g′(ε) + (N − 2(γ + 1))g(ε)]

= 0. (64)

Then we conclude that limε→0

|εN−2(γ+1)f0(ε)| = 0 and the claim is proved.

Step 3. We set

v(r, σ) =∑

i≥0

fi(1)rβiψi(σ),

which is the unique solution in the space Cε(Ω) ∩ D1,2γ, loc to the Dirichlet problem with boundary data

v = u in the unitary sphere. We will prove that u = v a.e. in B. Let φ ∈ C∞0 (Ω) be such that φ(r, σ) =

gi0(r)ψi0 (σ) where gi0 ∈ C∞0 [0, 1) and i0 ≥ 1. Therefore we get

∫

B

v(x)φ(x)|x|−2γ−1dx = fi0(1)

∫ 1

0

gi0(r)rN−2(γ+1)+βi0 dr.

On the other hand we have

Ii0 ≡

∫

B

u(x)φ(x)|x|−2γ−1dx = limε→0

∫

|x|>2ε

u(x)φ(x)|x|−2γ−1dx

= limε→0

∫ 1

2ε

∫

SN−1

∞∑

i=0

(

Ai(ε)rαi +Bi(ε)r

bi)

ψi(σ)φ(r, σ)dσ

= limε→0

∞∑

i=0

∫ 1

2ε

(

Ai(ε)rαi +Bi(ε)r

bi)

gi0(r)rN−2(γ+1)dr

∫

SN−1

ψi(σ)ψi0 (σ)dσ

= limε→0

∫ 1

2ε

(

Ai(ε)rαi +Bi(ε)r

bi)

gi0(r)rN−2(γ+1)dr

= limε→0

∫ 1

2ε

ε−αi0fi0(ε)− εβi0−αi0 fi0(1)

1− εβi0−αi0

rαi0 +fi0(1)− ε−αi0 fi0(ε)

1− εβi0−αi0

rβi0

gi0(r)rN−2(γ+1)dr

.

27

Since by step 2, |εN−2(γ+1)fi0(ε)| → 0 as ε→ 0 we conclude that

Ii0 = fi0(1)

∫ 1

0

gi0(r)rN−2(γ+1)+βi0 dr.

If i0 = 0, then choosing φ(x) = g0(r) ∈ C∞0 [0, 1) we obtain that

I0 = limε→0

∫ 1

2ε

(f0(1)− εN−2(γ+1)f0(ε)

1− εN−2(γ+1)+εN−2(γ+1)(f0(ε)− f0(1))

1− εN−2(γ+1)r2(γ+1)−N

)

g0(r)rN−2(γ+1)dr.

Using the same argument as above we obtain that

I0 = f0(1)

∫ 1

0

g0(r)rN−2(γ+1)dr.

Therefore we conclude that for all φ ∈ C∞0 (B1(0)) such that φ(x) = gi(r)ψi(σ) we have

∫

B

u(x)φ(x)|x|−2γ−1dx =

∫

B

vφ(x)|x|−2γ−1dx. (65)

Notice that (65) holds for all φ ∈ C(B) such that φ(r, σ) = g(r)ψi(σ) where g ∈ C[0, 1].We set

F =

n∑

k=0

gk(r)ψk(σ), such that n ∈ IN and gk ∈ C∞([0, 1]) for all 0 ≤ k ≤ n

.

Then for all φ ∈ F we have∫

B

u(x)φ(x)|x|−2γ−1dx =

∫

B

vφ(x)|x|−2γ−1dx.

Therefore using Lemma 7.1 we get the density of F in C∞0 (B1(0)) with respect to the uniform convergence

norm, hence∫

B

u(x)φ(x)|x|−2γ−1dx =

∫

B

vφ(x)|x|−2γ−1dx,

for all φ ∈ C∞0 (B1(0)). Since u(x)|x|−2γ−1, v(x)|x|−2γ−1 ∈ L1(B1(0)) we obtain that

u(x) = v(x) a.e. x ∈ B1(0).

As a consequence (denoting dµ = |x|−2γ−1dx) we get the next result.

Corollary 7.4 Let f ∈ L1(Ω) then problem

−Lγ(u) = f, u|∂Ω = 0 (66)

has an unique weak solution u ∈ L1(dµ,Ω) such that

||u||L1(dµ,Ω) ≤ C||f ||L1(Ω), (67)

where C depends on N, γ and Ω. Moreover, if γ + 1 > 0 we have

||u||L1(Ω) ≤ C||f ||L1(Ω).

28

Acknowledgments.- The third author wants to thank Professor Haım Brezis for his encouragement toprove the uniqueness results in this work and Professor Cristian E. Gutierrez for his helpful commentsabout parabolic Harnack inequality for degenerate and singular linear parabolic equations.

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