A nonlinear parabolic equation with drift term

A nonlinear parabolic equation with drift term

Nonlinear Analysis 177 (2018) 397–412 Contents lists available at ScienceDirect Nonlinear Analysis www.elsevier.com/locate/na A nonlinear parabolic...

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Nonlinear Analysis 177 (2018) 397–412

Contents lists available at ScienceDirect

Nonlinear Analysis www.elsevier.com/locate/na

A nonlinear parabolic equation with drift term Fernando Farroni, Gioconda Moscariello* Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Università degli Studi di Napoli “Federico II”, Via Cintia 80126 Napoli, Italy

article

info

Article history: Received 16 April 2018 Accepted 19 April 2018 Communicated by Enzo Mitidieri

abstract We consider weak solutions to a nonlinear evolution diffusion–convection equation with an unbounded convection field. We provide existence and uniqueness results as well as the asymptotic behaviour of solutions. © 2018 Elsevier Ltd. All rights reserved.

To Carlo on the occasion of his 70th birthday MSC: 35K55 35K61 35K65 35K92 Keywords: Nonlinear parabolic equations Existence Asymptotic behaviour

1. Introduction In this paper we study weak solutions to a nonlinear evolution diffusion–convection equation with an unbounded convection field. The general form of our problem reads as follows ⎧ in ΩT , ⎨ ut − div [A(x, t, ∇u) + B(x, t, u)] = − div F u=0 on ∂Ω × (0, T ), (1.1) ⎩ u(x, 0) = u0 (x) in Ω , where Ω is a regular bounded domain of RN with N ⩾ 3 and T > 0. Hereafter ΩT stands for Ω × (0, T ). For the data of the problem we assume that F ∈ L2 (ΩT )

*

and

u0 ∈ L2 (Ω ).

Corresponding author. E-mail addresses: [email protected] (F. Farroni), [email protected] (G. Moscariello).

https://doi.org/10.1016/j.na.2018.04.021 0362-546X/© 2018 Elsevier Ltd. All rights reserved.

(1.2)

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Here A = A(x, t, ξ) : ΩT × RN → RN is a Carath´eodory function satisfying the following conditions |A(x, t, ξ)| ⩽ β|ξ| + g(x, t)

for some β > 0 and g ∈ L2 (ΩT ), 2

⟨A(x, t, ξ) − A(x, t, η), ξ − η⟩ ⩾ α|ξ − η|

for some α > 0

(1.3) (1.4)

for a.e. (x, t) ∈ ΩT and for any ξ, η ∈ RN . Moreover, we assume that B = B(x, t, s) : ΩT × R → RN is a Carath´eodory function verifying the following properties |B(x, t, s) − B (x, t, s′ )| ⩽ b(x, t) |s − s′ |

(1.5)

B(x, t, 0) = 0

(1.6)

for a.e. x ∈ Ω , for any t ∈ (0, T ), for any s, s′ ∈ R and for some suitable nonnegative function b such that ( ) b ∈ L∞ 0, T ; LN,∞ (Ω ) .

(1.7)

In the linear homogeneous case, equation in (1.1) is also known as the Fokker–Planck equation and describes the evolution of some Brownian motion. If the convection term b = b(x, t) is bounded in ΩT , problem (1.1) has been studied by many authors (see e.g. [3] and the references therein). In many applications, the boundedness of the convective field is not available as underlined in [5] in the case of the diffusion model for semiconductor devices. Many definitions of solutions have been introduced for problem (1.1) under consideration as the renormalized solution (see e.g. [16]) and entropy solutions (see e.g. [2]). Here, we assume that conditions (1.5)–(1.7) hold true and we are interested in the existence and uniqueness of a weak solution, according to the following definition. Definition 1.1. We say that ( ) ( ) u ∈ L2 0, T ; W01,2 (Ω ) ∩ L∞ 0, T ; L2 (Ω ) is a weak solution to problem (1.1) if ∫ ∫ − u ∂t φ dx dt + ΩT

⟨A(x, t, ∇u) + B(x, t, u), ∇φ⟩ dx dt

ΩT

∫ ΩT

(1.8)

∫ ⟨F, ∇φ⟩ dx dt +

=

u0 φ(0) dx Ω



for all φ ∈ C (ΩT ) with supp φ ⊂⊂ Ω × [0, T ). The only assumptions (1.5)–(1.7) do not infer in general the existence of such a solution. Indeed, if Ω is the unit ball of RN with N ⩾ 3 then we can see that the problem ] } ( ) {[ ⎧ ( ) ⎪ x 1 x N −2−δ ⎪ ⎪ ut − ∆u − div δ 2+ 1 − |x| x u = − div ⎪ N −δ ⎪ 2−N +δ ⎨ |x| |x| (1.9) in Ω × (0, T ), ⎪ ⎪ ⎪ ⎪ u=0 on ∂Ω × (0, T ), ⎪ ⎩ u(0) = 0 in Ω , does not admit a solution in the sense of Definition 1.1 if N2 − 1 < δ ⩽ N − 1 with δ ̸= N − 2. For further details see Example A.2 in the Appendix. Nevertheless, we can find a proper subset X(ΩT ) of the space ( ) L∞ 0, T ; LN,∞ (Ω ) such that if the conduction term b = b(x, t) belongs to X(ΩT ) then problem (1.1) is uniquely solvable. More precisely, we prove the following result.

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Theorem 1.1 (Existence and Uniqueness). Let us assume that conditions (1.2)–(1.7) are fulfilled. There ( ) exists a proper convex subset X(ΩT ) of the space L∞ 0, T ; LN,∞ (Ω ) such that, if b ∈ X(ΩT ) ( ) ( ) then problem (1.1) admits a unique solution u ∈ L2 0, T ; W01,2 (Ω ) ∩ C [0, T ]; L2 (Ω ) . ( ) For the definition of X(ΩT ) see Section 2.2. The set X(ΩT ) is larger than L∞ 0, T ; LN,q (Ω ) whenever N ⩽ q < ∞. We see that b belongs to X(ΩT ) essentially if it can be decomposed as the sum of a bounded ( ) function and a L∞ 0, T ; LN,∞ (Ω ) function having a control on its norm. In the stationary case, a result comparable to Theorem 1.1 can be found in [9,10]. We also deal with the time behaviour of the solution to problem (1.1). Indeed, in Section 4 we give estimates on time of the L2 -norm of the solution in terms of the data. Besides the presence of the lower order term, if T = +∞ the L2 -norm of the solution decays exponentially fast, as estimate (4.8) reveals. In a forthcoming paper, we will consider pointwise estimates of the gradient in the spirit of [11,12]. 2. Notation and preliminary results 2.1. Lorentz spaces Let Ω be a bounded domain of RN . For any p, q ∈ (1, ∞), the Lorentz space Lp,q (Ω ) consists of all measurable functions g defined on Ω for which the quantity )1/q ( ∫ ∞ q q−1 p dh ∥g∥p,q = p |Ωh | h 0

is finite, where Ωh = {x ∈ Ω : |g(x)| > h} for any h > 0. Here and in what follows |E| stands for the Lebesgue measure of a measurable subset E of RN . Note that ∥ · ∥p,q is equivalent to a norm and Lp,q (Ω ) becomes a Banach space when endowed with it (see [15]). For p = q, the Lorentz space Lp,p (Ω ) reduces to the Lebesgue space Lp (Ω ). On the other hand, for q = ∞, the class Lp,∞ (Ω ) consists of all measurable functions g defined on Ω for which the quantity ∫ 1 −1 ∥g∥p,∞ = sup |E| p |g| dx E⊂Ω p,∞

is finite. The class L

E

(Ω ) coincides with the Marcinkiewicz class weak-Lp . Moreover, if we set 1

[[g]]p,∞ = sup h|Ωh | p h>0

one has p−1 1

p1+ p

∥g∥p,∞ ⩽ [[g]]p,∞ ⩽ ∥g∥p,∞

(see Lemma A.2 in [1]). For the Lorentz spaces the following inclusions hold Lr (Ω ) ⊂ Lp,q (Ω ) ⊂ Lp,r (Ω ) ⊂ Lp,∞ (Ω ) ⊂ Lq (Ω ) whenever 1 ⩽ q < p < r ⩽ ∞. Moreover, for 1 < p < ∞, 1 ⩽ q ⩽ ∞ and ′ ′ f ∈ Lp,q (Ω ) and g ∈ Lp ,q (Ω ), we have the H¨ older-type inequality ∫ |f (x)g(x)| dx ⩽ ∥f ∥p,q ∥f g∥p′ ,q′ . Ω

1 p

+

1 p′

= 1,

1 q

+

1 q′

= 1, if

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It is well known that L∞ (Ω ) is not a dense subspace of Lp,∞ (Ω ). The distance to L∞ (Ω ) in Lp,∞ (Ω ) is defined as (see [4]) distLp,∞ (Ω) (f, L∞ (Ω )) =

inf

g∈L∞ (Ω)

∥f − g∥Lp,∞ (Ω) .

We conclude this section by recalling the Sobolev embedding theorem in the setting of Lorentz spaces. Theorem 2.1. Let us assume that 1 < p < N and 1 ⩽ q ⩽ p. Then, any function g ∈ W01,1 (Ω ) satisfying ∗ p and |∇g| ∈ Lp,q (Ω ) belongs to Lp ,q (Ω ) where p∗ = NN−p ∥g∥p∗ ,q ⩽ SN,p ∥∇g∥p,q where SN,p =

−1/N p ωN N −p

(2.1)

and ωN stands for the measure of the unit ball in RN .

2.2. The space L∞ (0, T ; Lp,∞ (Ω )) Let [0, T ] be a bounded interval of R which is provided with the usual Lebesgue measure denoted by dt and let Ω be a bounded open subset of RN . Given p ∈ (1, ∞), the class L∞ (0, T ; Lp,∞ (Ω )) is the space of measurable functions f : Ω × [0, T ] → R such that ∥f ∥L∞ (0,T ;Lp,∞ (Ω)) := ess sup ∥f (·, t)∥Lp,∞ (Ω) 0
is finite. For any given δ ⩾ 0, we consider the subset Xδ (ΩT ) of L∞ (0, T ; Lp,∞ (Ω )) defined as Xδ (ΩT ) := {f ∈ L∞ (0, T ; Lp,∞ (Ω )) : distL∞ (0,T ;Lp,∞ (Ω)) (f, L∞ (0, T ; L∞ (Ω ))) ⩽ δ}. In other words, Xδ (ΩT ) consists of all those functions f ∈ L∞ (0, T ; Lp,∞ (Ω )) such that there exists g ∈ L∞ (0, T ; L∞ (Ω )) such that ∥f − g∥L∞ (0,T ;Lp,∞ (Ω)) ⩽ δ. We remark that X0 (ΩT ) is the closure of L∞ (0, T ; L∞ (Ω )) in L∞ (0, T ; Lp,∞ (Ω )) and L∞ (0, T ; Lp,q (Ω )) ⊂ X(ΩT ) for p ⩽ q < ∞. To find a formula for the distance to L∞ (0, T ; L∞ (Ω )) in L∞ (0, T ; Lp,∞ (Ω )), we consider, for κ > 0, the truncation operator at level ±κ, namely s min{|s|, κ}. Tκ (s) = |s| Then distL∞ (0,T ;L∞ (Ω)) (f, L∞ (0, T ; L∞ (Ω ))) = lim ∥f − Tκ (f )∥L∞ (0,T ;Lp,∞ )(Ω) . κ→∞





Indeed, for every g ∈ L (0, T ; L (Ω )) and for every κ ⩾ ∥g∥L∞ (0,T ;L∞ (Ω)) we have for a.e. (x, t) ∈ ΩT |f (x, t) − g(x, t)| ⩾ |f (x, t) − Tκ f (x, t)|. The following lemma then follows. Lemma 2.2. Let δ ⩾ 0 be given. Then, f ∈ Xδ (ΩT ) if and only if lim ∥f − Tκ (f )∥L∞ (0,T ;Lp,∞ (Ω)) ⩽ δ.

κ→∞

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2.3. Some useful asymptotic estimates We recall an abstract result whose application will be essential in the study of the time behaviour of the solution to (1.1). Proposition 2.3. Let t0 ⩾ 0 and T ∈ (t0 , +∞]. Assume that ϕ = ϕ(t) is a continuous and nonnegative function defined in [t0 , T ) verifying ∫ t2 ∫ t2 ϕ(t2 ) − ϕ(t1 ) + M ϕ(t) dt ⩽ g(t) dt t1

t1

for every t0 ⩽ t1 < t2 < T , where M is a positive constant and g is a nonnegative function belonging to L1 ([t0 , T )). Then, for every t ⩾ t0 we get ∫ t −M (t−t0 ) ϕ(t) ⩽ ϕ(t0 )e + g(s) ds. t0

Moreover, if T = +∞ and g belongs to L1 ([t0 , +∞)) there exists t1 > t0 such that ∫ t M g(s) ds ϕ(t) ⩽ Λe− 2 (t) + t/2

for every t ⩾ t1 , where ∫

+∞

Λ = ϕ(t0 ) +

g(s) ds. t0

The proof of the Proposition above is considered in [14] to study stability results in the spirit of [6,7]. 3. The main results 3.1. Existence and uniqueness of solutions Before we give the proof of Theorem 1.1, we consider the following auxiliary parabolic problem ⎧ vt − div [A(x, t, ∇v) + (1 − θ(x, t)) B(x, t, v)] = − div(F − θ(x, t)B(x, t, w)) ⎪ ⎪ ⎨ in ΩT ≡ Ω × (0, T ), ⎪ v = 0 on ∂Ω × (0, T ), ⎪ ⎩ v(0) = u0 (x) in Ω ,

(3.1)

where θ(x, t) =

TM (b(x, t)) b(x, t)

(3.2)

where TM is the usual truncation operator at level ±M . We simply denote by X(ΩT ) the set Xδ (ΩT ) as in Section 2.2 where ( ) α 1 1/N N 0<δ⩽ = ωN − 1 α. (3.3) 2SN,2 2 2 We assume that b ∈ X(ΩT ).

(3.4)

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We follow the strategy of [8–10]. Precisely, assumptions (3.3) and (3.4) allow us to choose M > 0 is chosen in such a way to ensure that sup ∥b(x, t) − TM (b(x, t))∥LN,∞ (Ω) ⩽

0
α . 2SN,2

(3.5)

We start by proving that the operator A : L2 (0, T ; W01,2 (Ω )) → L2 (0, T ; W −1,2 (Ω )) ) ( defined for u, v ∈ L2 0, T ; W01,2 (Ω ) as follows T

∫ ⟨Au, v⟩ :=

∫ ⟨A(x, t, ∇u) + (1 − θ(x, t)) B(x, t, u), ∇v⟩ dx,

dt Ω

0

is coercive because of condition (3.5). ( ) Lemma 3.1. Assume that conditions (3.3) and (3.4) hold true. Then, for each u, v ∈ L2 0, T ; W01,2 (Ω ) , one has ∫ α T 2 (3.6) ⟨Au − Av, u − v⟩ ⩾ ∥∇u − ∇v∥L2 (Ω) dt. 2 0 In particular, ⟨Au, u⟩ ⩾

α 2

∫ 0

T

2

∥∇u∥L2 (Ω) dt.

(3.7)

Proof . We use (1.5), (1.6) and the definition of θ given by (3.2) to get ⟨Au − Av, u − v⟩ ∫ T ∫ = dt ⟨A(x, t, ∇u) − A(x, t, ∇v) + (1 − θ(x, t)) B(x, t, u) − B(x, t, v), ∇(u − v)⟩ dx Ω

0



T

⩾α





2

|∇u − ∇v| dx −

dt Ω

0

T

|b(x, t) − TM (b(x, t))| · |u − v| · |∇u − ∇v| dx.

dt 0

(3.8)

∫ Ω

Next, we use H¨ older inequality in the setting of Lorentz spaces, Sobolev inequality (2.1) and (3.5) to obtain ⟨Au − Av, u − v⟩ ∫ T ∫ 2 dt |∇u − ∇v| dx ⩾α Ω

0 T

∫ − 0 T

∫ ⩾α

∥b(x, t) − TM (b(x, t)) ∥LN,∞ (Ω) · ∥u − v∥L2∗ ,2 (Ω) · ∥∇u − ∇v∥L2 (Ω) dt ∫

2

|∇u − ∇v| dx

dt Ω

0

∫ − SN,2 0

α ⩾ 2

∫ 0

T

T

∥b(x, t) − TM (b(x, t)) ∥LN,∞ (Ω) · ∥∇u − ∇v∥2L2 (Ω) dt 2

∥∇u − ∇v∥L2 (Ω) dt

which proves (3.6). □

(3.9)

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Proof of Theorem 1.1. First, with the help of Schauder’s fixed point theorem, we establish the existence of a local weak solution of (1.1). Next, we will prove that this solution may be extended to an arbitrary large interval which leads to a global weak solution. Let T∗ ∈ (0, T ] be arbitrary chosen. We set { } ( ) M(T∗ ) := w ∈ L2 0, T∗ ; L2 (Ω ) : ∥w∥L2 (0,T∗ ;L2 (Ω)) ⩽ 1 . Then, appealing to the( classical theory operators, for every w ∈ M(T∗ ) there exists a unique ) of monotone ( ) 1,2 2 weak solution v ∈ L 0, T∗ ; W0 (Ω ) ∩ C [0, T∗ ); L2 (Ω ) to problem (3.1). Then, we define a mapping ( ) K : M(T∗ ) → L2 0, T∗ ; L2 (Ω ) by K(w) =: v is the unique solution to problem (3.1). STEP 1: We claim that ∥K(w)∥2L2 (0,T

)⩽1 ⩽C

2 ∗ ;L (Ω)

) ∥K(w)∥2 2 ( 1,2 L 0,T∗ ;W0 (Ω)

holds for all w ∈ M(T∗ ) and for some suitable constant C > 0. Testing by φ = vχ[0,t) in the equation in (3.1) with t ∈ (0, T∗ ), the integration by parts yields ∫ t ∫ 1 ∥v(t)∥2L2 (Ω) + ds ⟨A(x, s, ∇v) + (1 − θ(x, s)) B(x, s, v), ∇v⟩dx 2 0 Ω ∫ t ∫ 1 = ∥u0 ∥2L2 (Ω) + ds ⟨F − θ(x, s)B(x, s, w), ∇v⟩dx 2 0 Ω

(3.10) (3.11)

(3.12)

for any t ∈ (0, T∗ ). We deduce that ∫ t ∫ 1 ∥v(t)∥2L2 (Ω) + ds ⟨A(x, s, ∇v) − A(x, t, 0) dx 2 0 Ω ∫ t ∫ + ds (1 − θ(x, s)) (B(x, s, v) − B(x, t, 0)) , ∇v⟩ dx Ω

0

∫ t ∫ 1 ds ⟨F − θ(x, s)B(x, s, w), ∇v⟩dx = ∥u0 ∥2L2 (Ω) + 2 0 Ω ∫ t ∫ − ds ⟨A(x, s, 0), ∇v⟩dx. 0

(3.13)



Using (3.6), we have ∫ 1 α t 2 ∥v(t)∥2L2 (Ω) + ∥∇v∥L2 (Ω) ds 2 2 0 ∫ t ∫ 1 2 ⩽ ∥u0 ∥L2 (Ω) + ds ⟨F − θ(x, s)B(x, s, w), ∇v⟩dx 2 0 Ω ∫ t ∫ − ds ⟨A(x, s, 0), ∇v⟩dx. 0

(3.14)



We deduce by (1.3) and by H¨ older’s inequality we obtain ∫ t 1 α 2 ∥v(t)∥2L2 (Ω) + ∥∇v∥L2 (Ω) ds 2 2 0 ∫ t( ) 1 ⩽ ∥u0 ∥2L2 (Ω) + ∥F ∥L2 (Ω) + ∥θ(x, s)B(x, s, w)∥L2 (Ω) + ∥g(x, s)∥L2 (Ω) ∥∇v∥L2 (Ω) ds 2 0

(3.15)

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we use Young’s inequality and we get ∫ α t 1 2 2 ∥∇v∥L2 (Ω) ds ∥v(t)∥L2 (Ω) + 2 2 0 ∫ )2 1 1 t( ⩽ ∥u0 ∥2L2 (Ω) + ∥F ∥L2 (Ω) + ∥θ(x, s)B(x, s, w)∥L2 (Ω) + ∥g(x, s)∥L2 (Ω) ds 2 α 0 ∫ t α + ∥∇v∥2L2 (Ω) ds. 4 0

(3.16)

Now, in (3.16) we are able to reabsorb the latter term by the left-hand side, so we manage (3.16) and we obtain ∫ 1 α t 2 ∥∇v∥L2 (Ω) ds ∥v(t)∥2L2 (Ω) + 2 4 0 (3.17) ∫ ) 1 2 t( 2 2 2 2 ⩽ ∥u0 ∥L2 (Ω) + ∥F ∥L2 (Ω) + ∥θ(x, s)B(x, s, w)∥L2 (Ω) + ∥g(x, s)∥L2 (Ω) ds. 2 α 0 Since ∥θ(x, s)B(x, s, w)∥2L2 (Ω) ⩽ M 2 ∥w∥2L2 (Ω) one has ∥v(t)∥2L2 (Ω) + ⩽

∥u0 ∥2L2 (Ω)

⩽ γ1 +

α 4

t



2 + α

2

∥∇v∥L2 (Ω) ds

0

T∗



(

0

) ∥F ∥2L2 (Ω) + M 2 ∥w∥2L2 (Ω) + ∥g(x, s)∥2L2 (Ω) ds

(3.18)

γ2 ∥w∥2L2 (0,T ;L2 (Ω)) ∗

where γ1 := ∥u0 ∥2L2 (Ω) + γ2 :=

) 2( ∥F ∥2L2 (Ω ) + ∥g∥2L2 (Ω ) T T α

(3.19)

2 2 M α

(3.20)

and ∥v∥2L∞ (0,T



+ ;L2 (Ω))

α 4

∫ 0

T∗

2

∥∇v∥L2 (Ω) dt ⩽ γ1 + γ2 ∥w∥2L2 (0,T

From (3.18) we deduce, integrating over (0, T∗ ), ∫ T∗ ∥v(t)∥2L2 (Ω) dt ⩽ γ1 T∗ + γ2 T∗ ∥w∥2L2 (0,T

)

2 ∗ ;L (Ω)

0

while, from (3.21), we deduce ∫ 0

T∗

2

).

2 ∗ ;L (Ω)

∥∇v∥L2 (Ω) dt ⩽

4γ2 4γ1 + ∥w∥2L2 (0,T ;L2 (Ω)) . ∗ α α

(3.21)

(3.22)

(3.23)

Thus, setting T∗ ⩽

1 min 2

{

1 1 , , 2T γ1 γ2

}

we finally prove that (3.10) and (3.11) hold for any w ∈ M(T∗ ). In particular, K maps M(T∗ ) into itself. STEP 2: Next, we prove that K : M(T∗ ) → M(T∗ ) is continuous and compact. For the distributional (time) derivative vt ≡ v ′ of v we have ( ) v ′ ∈ L2 0, T∗ ; W −1,2 (Ω ) .

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Indeed, 2

∥v ′ ∥2L2 (0,T

∗ ;W −1,2 (Ω)

2

2

2

) ⩽ β ∥∇v∥L2 (ΩT∗ ) + ∥g∥L2 (ΩT∗ ) + ∥b(x, t)∥L∞ (0,T∗ ;) · ∥∇v∥L2 (ΩT∗ ) 2

2

+ ∥F ∥L2 (ΩT ) + M 2 ∥w∥L2 (ΩT ) (3.24) ∗ ∗ ) 4(γ1 + γ2 ) ( 2 2 2 β + ∥b(x, t)∥L∞ (0,T∗ ;) + ∥g∥L2 (ΩT ) + M 2 + ∥F ∥L2 (ΩT ) . ⩽ ∗ ∗ α By (3.10) and (3.11), we are in a position to apply Aubin–Lions Lemma; thus K (M(T∗ )) is relatively compact in L2 (0, T∗ ; L2 (Ω )). Now, we prove that K is continuous. Let {wn }n∈N be a sequence in M(T∗ ) which converges in 2 L (0, T∗ ; L2 (Ω )) to a function w ∈ M(T∗ ) as n → ∞. By the compactness of K(M(T∗ )) and relation (3.11), there exists a subsequence {wnk }n∈N of {wn }n∈N such that in L2 (0, T∗ ; L2 (Ω ))

K(wnk ) → v

2

K(wnk ) ⇀ v

in L

(0, T∗ ; W01,2 (Ω ))

(3.25) (3.26)

as n → ∞. Moreover, if we let vnk := K(wnk ), thanks to the assumptions in L2 (Ω × (0, T∗ ))n

˜ A(·, ∇vnk ) ⇀ A

B(x, t, vnk ) → B(x, t, v)

2

in L2 (Ω × (0, T∗ )).

(3.27) (3.28)

By the definition of K, the functions vnk satisfy the following integral identity ∫ T∗ ∫ ∫ T∗ ∫ − dt vnk ∂t φdx + dt ⟨A(x, t, ∇vnk ) + (1 − θ(x, t))B(x, t, vnk ), ∇φ⟩ dx Ω

0



T∗



dt

=

u0 φ(0)dx + Ω

0



0



T∗

(3.29)

∫ dt

0



⟨F + θ(x, t)B(x, t, wnk ), ∇φ⟩ dx

for any φ ∈ C0∞ (Ω × [0, T∗ )). Then, with the aid of (3.25), (3.26), (3.27) and (3.28), by passing to the limit in (3.29) as k → ∞ we have ∫ T∗ ∫ ∫ T∗ ∫ ˜ + (1 − θ(x, t))B(x, t, v), ∇φ⟩ dx − dt v∂t φdx + dt ⟨A 0 Ω 0 Ω (3.30) ∫ ∫ ∫ ∫ T∗

T∗

dt

= 0

u0 φ(0)dx + Ω

⟨F + θ(x, t)B(x, t, w), ∇φ⟩ dx

dt 0



for any φ ∈ C0∞ (Ω × [0, T∗ )). It is not hard to prove (see Lemma A.1 in the Appendix) that ˜ t) = A(x, t, ∇v) A(x,

a.e. in Ω × (0, T∗ )

(3.31)

then v = K(w) and, noticing that this limit is unique, from (3.25)–(3.26) we get as k → ∞ K(wnk ) → K(w)

in L2 (Ω × (0, T∗ ))

thus K is continuous. STEP 3: Now, applying Schauder’s fixed point theorem one finds a function vT∗ ∈ M(T∗ ) such that K(vT∗ ) = vT∗ , that is to say that vT∗ is a weak solution to problem (1.1) in the cylinder Ω × (0, T∗ ). Testing Eq. (1.1) by vT∗ and integrating over the interval (0, T∗ ), using integration by parts, we get ∫ T∗ ∫ 1 2 ∥vT∗ (T∗ )∥L2 (Ω) + dt ⟨A(x, t, ∇vT∗ ) + (1 − θ(x, t)) B(x, t, vT∗ ), ∇vT∗ ⟩dx 2 0 Ω (3.32) ∫ T∗ ∫ 1 2 = ∥u0 ∥L2 (Ω) + dt ⟨F − θ(x, t)B(x, t, vT∗ ), ∇vT∗ ⟩dx 2 0 Ω

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and so by (3.9) and (3.10) 1 1 ∥vT∗ (T∗ )∥2L2 (Ω) ⩽ ∥u0 ∥2L2 (Ω) + c(α) 2 2

T



(

0

) ∥F ∥2L2 (Ω) + ∥g(x, t)∥2L2 (Ω) dt + c(α)M 2 .

Therefore, we are able to extend vT∗ to a weak solution of problem (1.1) in Ω × [0, T ]. STEP 4: uniqueness of the solution. ) ( ( ) 2 Let u, v ∈ L 0, T ; W01,2 (Ω ) ∩ C [0, T ]; L2 (Ω ) be solutions to problem (1.1). Then, by using w = u − v as a test function in the equation appearing in (1.1) and subtracting we get ∫ ∫ 1 α t 2 ds ∥w(t)∥2L2 (Ω) + |∇w| dx 2 2 0 Ω ∫ t ∫ ds θ(x, s)⟨B(x, s, v) − B(x, s, u), ∇w⟩ dx ⩽ ∫

t

∫ |TM (b(x, s))| · |u − v| · |∇w| dx.

ds

+ 0

(3.33)



0



Now, by (1.5), (1.6) and Young’s inequality ∫ ∫ 1 α t 2 ∥w(t)∥2L2 (Ω) + ds |∇w| dx 2 4 0 Ω ∫ t ∫ 2 2 ⩽ C(α) ds |TM (b(x, t))| · |w| dx

(3.34)



0

and so 1 ∥w(t)∥2L2 (Ω) ⩽ CM 2 2



t

(3.35)

∥w(s)∥L2 (Ω) ds. 0

Then, by the Gronwall inequality, since w(0) = 0 ∥w(t)∥L2 (Ω) = 0

for a.e. t ∈ (0, T )

and therefore u = v a.e. in ΩT . □

4. Time behaviour of the solution We begin with the following lemma (see also [2,13]). Lemma 4.1. The solution u of (1.1) satisfies sup |{x ∈ Ω : |u(x, t)| > k}| ⩽ c(α)

∥u0 ∥2L2 (Ω) + ∥F ∥2L2 (Ω

+ ∥g∥2L2 (Ω

log2 (1 + k)

t∈(0,T )

for each k > 0. Proof . We fix T∗ ∈ (0, T ) and we choose φ=

T)

u χ(0,T∗ ) 1 + |u|

T)

+ ∥b∥2L2 (Ω

T)

(4.1)

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as a test function for Eq. (1.1). We observe that ∇φ =

∇u χ(0,T∗ ) . (1 + |u|)2

Therefore we get ∫

T∗





dt

T∗

ut φ dx + Ω

0



[A(x, t, ∇u)∇φ + B(x, t, u)∇φ] dx

dt Ω

0

T∗

(4.2)

∫ F ∇φ dx.

dt

=





0

We rearrange the equation above to get ] ∫ [∫ u(x,T∗ ) ∫ T∗ ∫ r dr dx + dt [A(x, t, ∇u) − A(x, t, 0)] ∇φ dx 1 + |r| Ω 0 0 Ω ] ∫ T∗ ∫ ∫ [∫ u0 (x) r dr dx = dt F ∇φ dx + 1 + |r| 0 Ω Ω 0 ∫ T∗ ∫ − dt [A(x, t, 0)∇φ + B(x, t, u)∇φ] dx. 0

(4.3)



A direct computation gives us ∫ 0

s

r dr = |s| − log(1 + |s|) 1 + |r|

and, since ew − 1 − w ⩾ 12 w2 for any w ⩾ 0, we have ∫ s r 1 dr ⩾ log2 (1 + |s|). 2 0 1 + |r| From (4.3) we have 1 2

T∗

2

|∇u| dx (1 + |u|)2 Ω 0 Ω )1/2 (∫ )1/2 ∫ T∗ (∫ 2 2 |∇u| |F | ⩽ dx dx dt + ∥u0 ∥22 2 2 0 Ω (1 + |u|) Ω (1 + |u|) )1/2 (∫ )1/2 ∫ T∗ (∫ 2 2 |g| |∇u| + dx dx dt 2 2 0 Ω (1 + |u|) Ω (1 + |u|) ∫ T∗ ∫ dx + b(x, t)|u||∇u| dt. (1 + |u|)2 0 Ω



log2 (1 + |u(x, T∗ )|) dx + α





dt

We observe that, since N > 2, we have b(·, t) ∈ L2 (Ω ); from the fact that inequality gives us T∗

|u| 1+|u|

⩽ 1 and

(4.4)

1 1+|u|

2

|∇u| dx (1 + |u|)2 Ω 0 Ω ∫ T∗ ∫ ∫ 2 ( ) α T∗ 3 |∇u| ⩽ ∥F ∥22 + ∥g∥22 + ∥b(x, t)∥22 dt + dt dx + ∥u0 ∥22 2 2α 0 2 0 (1 + |u|) Ω

1 2



log2 (1 + |u(x, T∗ )|) dx + α





⩽ 1, Young’s

dt

(4.5)

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therefore we have 1 2

T∗

2

|∇u| dx 2 Ω 0 Ω (1 + |u|) [∫ ] ∫ T∗ ∫ T∗ T∗ 2 2 2 2 ⩽ c2 (α) ∥F ∥2 dt + ∥u0 ∥2 + ∥g∥2 dt + ∥b(x, t)∥2 dt .



log2 (1 + |u(x, T∗ )|) dx + c1 (α)

0





dt

0

(4.6)

0

This estimate in turn implies (4.1). □ We are in position to state a result concerning the L2 -norm of the solution to problem (1.1). Theorem 4.2 (Behaviour of the Solution). Let all the assumptions of Theorem 1.1 be fulfilled. There exist positive constants M and C0 such that the following estimate ∫ t ∥u(t)∥2L2 (Ω) ⩽ ∥u0 ∥2L2 (Ω) e−M t + C0 h(s) ds (4.7) 0

holds for every t > 0. Here h(t) = ∥F (t)∥2L2 (Ω) + ∥g(t)∥2L2 (Ω) + ∥b(t)∥2L2 (Ω) . Moreover, if h ∈ L1 ((t0 , +∞)) for some t0 ⩾ 0, then the following estimate ∫ ( ) M ∥u(t)∥2L2 (Ω) ⩽ ∥u0 ∥2L2 (Ω) + ∥h∥L1 ((t0 ,+∞)) e− 2 t + C0

t

h(s) ds

(4.8)

t/2

holds for every t ⩾ t0 . Proof . First, we rewrite equation (1.1) as follows ut − div [A(x, t, ∇u) + (1 − θ(x, t))B(x, t, u)] = − div(F − θ(x, t))B(x, t, u) and we use the solution u itself as a test function. Arguing as before, for every t1 , t2 ∈ (0, T ) with t1 < t2 we obtain ∫ t2 ∫ t2 ( ) 1 1 ∥u(t2 )∥22 − ∥u(t1 )∥22 + c1 (α) ∥∇u∥22 dt ⩽ c2 (α) ∥F ∥22 + ∥g∥22 dt 2 2 t1 t1 (4.9) ∫ t2 ∫ + dt TM (b(x, t)) |u| · |∇u| dx. t1



The latter term can be estimated as follows ∫ ∫ ∫ TM (b(x, t)) |u| · |∇u| dx = TM (b(x, t)) |u| · |∇u| dx + |u|⩽k



∫ ⩽k

TM (b(x, t)) |u| · |∇u| dx

|u|>k

∫ |u| · |∇u| dx

b(x, t)|∇u| dx + M |u|⩽k

|u|>k

(4.10) c1 (α) 1/N 2 ∥∇u∥2 + M |Ek | ∥u∥2∗ ,2 ∥∇u∥2 ⩽ + 4 c1 (α) 1/N ⩽ c5 (α, k)∥b∥22 + ∥∇u∥22 + M |Ek | ∥∇u∥22 4 where we have defined Ek := {|u| > k}. We use (4.1) to observe that |Ek | → 0 as k → ∞. Therefore, if 1/N k > 0 is sufficiently large, we have M |Ek | ⩽ c2 4(α) , which means that we can reabsorb by the left-hand side the latter two terms in (4.9). We finally get ∫ t2 ∫ t2 ( ) 1 1 ∥u(t2 )∥22 − ∥u(t1 )∥22 + c6 ∥∇u∥22 dt ⩽ c7 ∥F ∥22 + ∥g∥22 + ∥b∥22 dt. (4.11) 2 2 t1 t1 c5 (α, k)∥b∥22

F. Farroni, G. Moscariello / Nonlinear Analysis 177 (2018) 397–412

409

Finally, by means of Poincar´e inequality, we have 1 1 ∥u(t2 )∥22 − ∥u(t1 )∥22 + c8 2 2



t2

∥u∥22

t2



(

dt ⩽ c7

t1

) ∥F ∥22 + ∥g∥22 + ∥b∥22 dt.

(4.12)

t1

The assertion is a consequence of Proposition 2.3 whenever ϕ(t) = ∥u(t)∥2L2 (Ω) . □ Acknowledgement The Authors are members of the Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). Appendix ˜ be as in (3.27). Then (3.31) holds Lemma A.1. Let all assumptions of Theorem 1.1 be fulfilled and let A true. Proof . Since (3.27) and (3.28) hold, we are allowed to pass to the limit in (3.29) and we get T∗

∫ −





dt

v∂t φdx + Ω

0 T∗



˜ t) + (1 − θ(x, t))B(x, t, v), ∇φ⟩ dx ⟨A(x,

dt Ω

∫ u0 φ(0)dx +



0



0

∫ dt

=

T∗

T∗

(A.1)

∫ ⟨F + θ(x, t)B(x, t, w), ∇φ⟩ dx

dt 0



for any φ ∈ C0∞ (Ω ×[0, T∗ )). In particular, (A.1) holds if we pick v in place of φ as a test function. Therefore, we get ∫ T∗ ∫ 1 ˜ t) + (1 − θ(x, t))B(x, t, v), ∇v⟩ dx dt ⟨A(x, − ∥v(T∗ )∥22 + 2 0 Ω ∫ T∗ ∫ = dt ⟨F + θ(x, t)B(x, t, w), ∇v⟩ dx.

(A.2)



0

On the other hand, (3.29) holds if we pick vnk in place of φ as a test function. Therefore, we get ∫ T∗ ∫ 1 − ∥vnk (T∗ )∥22 + dt ⟨A(x, t, ∇vnk ) + (1 − θ(x, t))B(x, t, vnk ), ∇vnk ⟩ dx 2 0 Ω ∫ T∗ ∫ = dt ⟨F + θ(x, t)B(x, t, wnk ), ∇vnk ⟩ dx.

(A.3)



0

We pass to the limit in the latter relation and, with the aid of (A.2) we obtain ∫

T∗

∫ dt

lim

k→∞



0

∫ ⟨A(x, t, ∇vnk ), ∇vnk ⟩ dx =

T∗

∫ ˜ t), ∇v⟩ dx. ⟨A(x,

dt 0

(A.4)



( ) For our purposes, it will be sufficient to prove that for each ψ ∈ L2 0, T∗ ; W01,2 (Ω ) one has ∫ ˜ t) − A(x, t, ∇ψ), ∇v − ∇ψ⟩ dx dt ⩾ 0. ⟨A(x, Ω×(0,T∗ )

(A.5)

F. Farroni, G. Moscariello / Nonlinear Analysis 177 (2018) 397–412

410

If we set for simplicity vk = vnk , we have ∫ ˜ t) − A(x, t, ∇ψ), ∇v − ∇ψ⟩ dx dt ⟨A(x, Ω×(0,T∗ )

∫ ˜ t) − A(x, t, ∇vk ), ∇v − ∇ψ⟩ dx dt ⟨A(x,

= Ω×(0,T∗ )



∫ ⟨A(x, t, ∇vk ), ∇v⟩ dx dt −

+ Ω×(0,T∗ )

⟨A(x, t, ∇vk ), ∇vk ⟩ dx dt

(A.6)

Ω×(0,T∗ )

∫ −

⟨A(x, t, ∇ψ), ∇v − ∇vk ⟩ dx dt Ω×(0,T∗ )

∫ ⟨A(x, t, ∇vk ) − A(x, t, ∇ψ), ∇vk − ∇ψ⟩ dx dt.

+ Ω×(0,T∗ )

According to (1.4) we have ∫ ⟨A(x, t, ∇vk ) − A(x, t, ∇ψ), ∇vk − ∇ψ⟩ dx dt ⩾ 0

for each k ⩾ 1

(A.7)

Ω×(0,T∗ )

and, according to (3.27) and (3.26) respectively we have ∫ ˜ t) − A(x, t, ∇vk ), ∇v − ∇ψ⟩ dx dt → 0 ⟨A(x,

as k → ∞

Ω×(0,T∗ )

∫ ⟨A(x, t, ∇ψ), ∇v − ∇vk ⟩ dx dt → 0

as k → ∞.

Ω×(0,T∗ )

On the other hand, according to (3.27) and (A.4) we have ∫ ∫ ⟨A(x, t, ∇vk ), ∇v⟩ dx dt − ⟨A(x, t, ∇vk ), ∇vk ⟩ dx dt → 0 Ω×(0,T∗ )

as k → ∞.

Ω×(0,T∗ )

Thus, taking into account (A.6) and (A.7) and passing to the limit as k → ∞, we conclude that (A.5) holds true. □ Example A.2. Let Ω be the unit ball of RN (which will be also denoted by B1 ) with N ⩾ 3. We fix T > 0 and N − 1 < δ ⩽ N − 1. 2

(A.8)

Let us consider for moment the further assumption that δ ̸= N − 2. We claim that the problem ⎧ {[ ] } ( ) ( ) ⎪ x 1 x ⎪ N −2−δ ⎪ u − ∆u − div δ 2+ 1 − |x| x u = − div ⎪ ⎪ N −δ ⎨ t 2−N +δ |x| |x| in Ω × (0, T ), ⎪ ⎪ ⎪ u = 0 on ∂Ω × (0, T ), ⎪ ⎪ ⎩ u(0) = 0 in Ω , does not admit a solution in the sense of Definition 1.1.

(A.9)

F. Farroni, G. Moscariello / Nonlinear Analysis 177 (2018) 397–412

As a preliminary step, let us consider the function v defined as follows ⎧ ( ) 1 2−N +δ ⎨ 1 − |x| for δ ̸= N − 2 2−N +δ v(x) = ⎩ − log|x| for δ = N − 2.

411

(A.10)

A direct calculation gives us ∇v(x) = −

x N −δ

|x|

therefore, condition (A.8) on the parameter δ gives that v actually belongs to W01,2 (Ω ). We also emphasize that v is a nontrivial solution to problem ⎧ ⎨ −∆v + δ x · ∇v = 0 in Ω , 2 (A.11) |x| ⎩ v=0 on ∂Ω . Suppose that a solution to (A.9) exists. If it is the case, we may test equation (A.9) with the following function φ(x, t) = v(x)e−t χ(t1 ,t2 ) (t) where 0 < t1 < t2 < T . Appealing to the fact that v is a solution to (A.11) we immediately get ∫ t2 ∫ 2 dt e−t |∇v| dx = 0 t1

(A.12)

(A.13)



which leads to a contradiction. Following the same lines as before, one can show that, if δ =N −2 then the problem ⎧ {[ ] } ( ) ⎪ x x x ⎪ ⎪ u − ∆u − div (N − 2) 2 + G(x) u = − div ⎪ 2 ⎪ ⎨ t |x| |x| |x| in Ω × (0, T ), ⎪ ⎪ ⎪ u = 0 on ∂Ω × (0, T ), ⎪ ⎪ ⎩ u(0) = 0 in Ω , does not admit a solution in the sense of Definition 1.1, where { |x| log|x| for x ∈ Ω \ {0}, G(x) = 0 for x = 0

(A.14)

(A.15)

which is still an L∞ function in the unit ball. □ Remark A.3. A rigorous argument to prove that (A.9) does not admit a solution can be obtained by testing with φk (x, t) = v(x)e−t ηk (t)

(A.16)

in place of the function in (A.12). Here v is the same as in (A.10) while, for k ⩾ 1 large enough, we denote by ηk a cut-off function with the following properties ηk ∈ C0∞ (R) 0 ⩽ ηk (t) ⩽ 1 ηk (t) = 1 ηk (t) = 0

for every t ∈ R for every t ∈ [t1 , t2 ] ] [ 1 1 . for every t ∈ R \ t1 − , t2 + k k

F. Farroni, G. Moscariello / Nonlinear Analysis 177 (2018) 397–412

412

So, if we test equation in (A.9) with φk we obtain ∫ T ∫ ∫ 2 −t ηk (t)e dt |∇v| dx = 0



ηk′ (t)e−t u(x, t)v(x) dxdt

ΩT

then, passing to the limit as k → ∞ we obtain the same contradiction as in (A.13). Remark A.4. As a matter of fact, the conductive term b = b(x, t) of both examples can be decomposed as follows b(x, t) =

δ + G(x) |x|

where G ∈ L∞ and bδ (x) :=

δ ∈ LN,∞ (B1 ) |x|

1/N

with ∥bδ ∥LN,∞ (B1 ) = distLN,∞ (B1 ) (bδ , L∞ (B1 )) = δωN .

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