Global solvability of Cauchy–Dirichlet problem for fully nonlinear parabolic systems

J. Math. Anal. Appl. 421 (2015) 1428–1454

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Global solvability of Cauchy–Dirichlet problem for fully nonlinear parabolic systems Luisa Fattorusso a , Antonio Tarsia b,∗ a

Dipartimento di Ingegneria dell’Informazione, delle Infrastrutture e dell’Energia Sostenibile, Università degli Studi Mediterranea di Reggio Calabria, Loc. Feo di Vito, I-89060 Reggio Calabria, Italy b Dipartimento di Matematica, Università di Pisa, Largo B. Pontecorvo, 5, I-56127 Pisa, Italy

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 22 March 2014 Available online 28 July 2014 Submitted by V. Radulescu Keywords: Fully nonlinear parabolic systems Near operators theory

We show existence and uniqueness theorems of global strong solutions of Cauchy– Dirichlet problem, for second order fully nonlinear parabolic systems, that satisfy the Campanato’s condition of ellipticity. We use the Campanato’s near operators theory. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Let Ω be a bounded convex open subset of Rn , n ≥ 2, with C 2,1 boundary ∂Ω and let N be a positive integer. Let u : Ω × [0, T ] −→ RN be a vector function, T > 0. We shall set Du = (D1 u, · · · , Dn u), where 2 ∂ u Di = ∂x and (x, t) = (x1 , · · · , xn , t) ∈ Rn × R, Du is a vector of RnN , D2 u = { ∂x∂i ∂x }i,j=1,···,n , so that i j D2 u is a vector of Rn

2

N

; u (x, t) =

∂u(x,t) ∂t .

Moreover, we set

         H = L2 0, T, H 2 ∩ H01 Ω, RN ∩ C 0 [0, T ], H01 Ω, RN ∩ H 1 0, T, L2 Ω, RN , with the norm uH = uL2 (0,T,H 2 ∩H01 (Ω,RN )) + uC 0 ([0,T ],H01 (Ω,RN )) + uH 1 (0,T,L2 (Ω,RN )) . Let us consider the Dirichlet problem ⎧ ⎪ ⎨ u∈ H  a x, t, u(x, t), Du(x, t), D2 u(x, t) − u (x, t) = 0, ⎪ ⎩ u(x, 0) = 0 a.e. in Ω,

a.e. in Ω × [0, T ],

* Corresponding author. E-mail addresses: [email protected] (L. Fattorusso), [email protected] (A. Tarsia). http://dx.doi.org/10.1016/j.jmaa.2014.07.048 0022-247X/© 2014 Elsevier Inc. All rights reserved.

(1)

(2)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

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2

where a : Ω × [0, T ] × RN × RnN × Rn N −→ RN and a(x, t, 0, 0, 0) ∈ L2 (Ω × [0, T ], RN ). Our purpose is to show that, under suitable assumptions on a, Problem (2) is well posed. In order to make this, there is no loss of generality in writing the problem as follows: ⎧u ∈ H ⎪ ⎪     ⎪ ⎨ F x, t, u(x, t), Du(x, t), D2 u(x, t) − u (x, t) = g x, t, u(x, t), Du(x, t) + f (x, t), ⎪ ⎪ ⎪ ⎩

a.e. in Ω × [0, T ],

(3)

u(x, 0) = 0 a.e. in Ω,

where F (x, t, u, Du, 0) = 0 and g(x, t, 0, 0) = 0.1 In our paper we show global existence of solutions of the Problem (3) without differentiability hypothesis on F with respect to the variable D2 u. Indeed, to study our problem, first we suppose that the principal part of the operator satisfies a condition that S. Campanato, in his paper [2], called first “nonlinear condition of Cordes” and subsequently “Condition A” (see for example [3]).2 Definition 1.1 (Condition A or Campanato condition of ellipticity). The operator F : Ω × [0, T ] × RN × 2 RnN × Rn N −→ RN verifies Condition A if there exist three real constants γ, δ, a with γ, δ ≥ 0, γ + δ < 1, n3 a > 0 such that for any u ∈ Rn , p ∈ RnN , M, Q ∈ SN and for a.e. (x, t) ∈ Ω × [0, T ], we have   n 

   Qii − a F (x, t, u, p, M + Q) − F (x, t, u, p, M)     i=1 N   n     ≤ γQn2 N + δ  Qii  .   i=1

(5)

N

It is clear that a verifies Condition A if and only if the operator F defined by (4) verifies it. This hypothesis is fundamental to obtain our result. Our other hypotheses are essentially of two types: assumptions on the growth in u and p and “compatibility assumptions” between the terms containing second order derivatives and those containing lower order derivatives. In particular, we consider the following assumptions: • F is measurable in (x, t) and continuous in the other variables; • there exists a function ω : RN × RnN −→ R+ bounded and continuous in (u, p), with ω(0, 0) = 0 and 2 such that for any u1 , u2 ∈ RN , p1 , p2 ∈ RnN , ξ ∈ Rn N ,   F (x, t, u1 , p1 , ξ) − F (x, t, u2 , p2 , ξ) ≤ ω(u1 − u2 , p1 − p2 )ξn2 N , N 1

a.e. in Ω × [0, T ];

(6)

Indeed, it is enough to assume       2 2 F x, t, u(x, t), Du(x, t), D u(x, t) = a x, t, u(x, t), Du(x, t), D u(x, t) − a x, t, u(x, t), Du(x, t), 0 ,  

   g x, t, u(x, t), Du(x, t) = − a x, t, u(x, t), Du(x, t), 0 − a(x, t, 0, 0, 0) , f (x, t) = −a(x, t, 0, 0, 0);

(4)

We observe that by these assumptions, we have that a.e. in Ω, one has F (x, t, u, Du, 0) = 0 and g(x, t, 0, 0) = 0. 2 In the case of linear operator this condition is equivalent to Cordes condition (see [18]). The connection between this and others ellipticity conditions are studied in [19]. 3 n p = (p1 , · · · , pn ), pi ∈ RN , if p ∈ RnN . ( | )N and  N are respectively the scalar product and the norm in RN . SN is the 1 N k k vector space of N -ples Q = ({Qij }i,j=1,···,n , · · · , {Qij }i,j=1,···,n ) of n × n matrices with Qij = Qji , i, j = 1, · · · , n, k = 1, · · · , N , equipped with the scalar product: (M|Q)n2 N = n i,j=1 (Mij |Qij )N .

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• there exist three positive constants M , α, β such that for any u ∈ RN , p ∈ RnN , one has    β  g(x, t, u, p) ≤ M uα N + pnN , N

a.e. in Ω × [0, T ],

(7)

where if n = 2,

1 ≤ α,

1 ≤ β < 2,

if n = 3,

1 ≤ α ≤ 2,

1 ≤ β < 2,

if n = 4,

α = 1,

β = 1.

• g is a differentiable function in the variables (u, p) and we assume that there exist M1 , M2 > 0 such that for any u ∈ RN , p ∈ RnN    ∂g(x, t, u, p)     ≤ M1 uα−1 + pβ ,  nN N   ∂u N2    ∂g(x, t, u, p)   β−1  α    2 ≤ M2 uN + pnN ,  ∂p

a.e. in Ω × [0, T ]; a.e. in Ω × [0, T ],

(8)

nN

where

∂g(x,t,u,p) ∂u

= { ∂gi (x,t,u,p) }i,j=1,···,N and ∂uj

∂g(x,t,u,p) ∂p

= { ∂gi (x,t,u,p) }i,j=1,···,N ; k=1,···,n . ∂pk j

Moreover, we suppose that at least one of the following conditions is verified: • for any u, v ∈ H, one has

T

     F x, t, u, Du, D2 u − F x, t, v, Dv, D2 v

0 Ω

 

− u (x, t) − v  (x, t)  g(x, t, u, Du) − g(x, t, v, Dv) N dx dt ≤ 0;

(9)

• for any u, v ∈ H, one has

T





 Δu − Δv − u (x, t) − v  (x, t)  g(x, t, u, Du) − g(x, t, v, Dv) N dx dt ≤ 0.

(10)

0 Ω

Without at least one of the assumptions (9) and (10) we can’t state that the problem has solution,4 while to have uniqueness of solution we will assume that in (9) the inequality is strict. More precisely, the main result of this paper is the following: Theorem 1.1. We assume that Condition A holds, and the hypotheses (6), (7), (8) and (9) are verified. If f ∈ L2 (Ω × [0, T ], RN ) then the Cauchy–Dirichlet Problem (3) has at least one solution, that is unique if (9) holds with a strict inequality. 4 We remark that these conditions are necessary in the case of linear elliptic equations since, for example, if λ > 0 is an eigenvalue of Δ, then as everybody knows, the problem



2

1

N

u ∈ H ∩ H0 (Ω, R ) Δu(x) + λu(x) = f (x),

isn’t well posed.

q.o. in Ω,

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There are not so many results of global existence of strong solutions for problems of this type. The reason, we think, is that the techniques used in equations are not generally applicable to systems. For example, the theory of viscosity solutions shows the existence of a viscosity solution for equations and then, under suitable hypotheses on the data, regularizes it to obtain a strong solution; these techniques are generally not valid for the systems because they need a maximum principle that is not valid in this case. Other methods for the equations, that lead to the existence of strong solutions, need the assumption that the principal part F is differentiable with respect to variable M. Moreover, these methods cannot be applicable to systems because they also use the maximum principle of Alexandrov–Bakelman–Pucci (see Notes in Chapter 17 and Epilogue in the book of D. Gilbarg and N.S. Trudinger, [7], and the Chapters 14 and 15 in the book of G.M. Lieberman [9]). In our paper we show global existence without differentiability hypothesis on F with respect to the variables Mij and obviously without maximum principle. These results are possible thanks to near operators theory of Campanato making use of Condition A instead of the assumption that F is differentiable (see [2,4,3]). We follow an analogous method to that used for elliptic systems in one of our previous papers (see L. Fattorusso and A. Tarsia [5]). This confirms the effectiveness of Campanato’s theory. Indeed, the developments of the theory allow to extend its applications to fourth order elliptic nonlinear problems of variational type and also nonstationary (see for example [6]). S. Campanato introduced these methods, not only for the study of fully nonlinear elliptic problems, but also for parabolic ones, see for example [3], where he showed the global existence and uniqueness of the solution for systems like F (x, t, D2 u) − u = f (x, t). Condition A is also used by L.G. Softova (see [14] and [15]) to show the global existence of strong solutions for equation F (x, t, D2 u(x, t)) − u (x, t) = g(x, t, u(x, t), Du(x, t)), on the case n = 2, assuming quadratic growth on g plus another condition in order to bound g from below. In this paper to prove our result we proceeded into a sequence of steps. In the first step (see Section 3) we prove the existence and the uniqueness of the solution for the Cauchy–Dirichlet problem, for the following systems:     F x, t, D2 u(x, t) − u (x, t) = g x, t, u(x, t), Du(x, t) + f (x, t), under assumptions (5), (6), (7), (8) and (9) or (10). We make this by using a particular Implicit Functions theorem, coming from near operators theory (see Theorem 2.2), for which the differentiability hypothesis on the function is not necessary; then we use a generalization of the continuity method (see Theorem 2.3). In the second step we prove the existence for systems written in complete form (see Section 4), making use of Schauder–Tychonov’s fixed point theorem. We deal with problems where the dimension is n = 2, 3, 4, and the growths are between 1 and 2. Up to now we are not able to say whether our methods can solve them for larger n, because for the general form of the system it is very hard to get over the difficulties. 2. Preliminaries For convenience of the reader we will repeat, in this section, the main results used in our exposition. Some of these are a consequence of the so-called “near operators theory”. Definition 2.1. Let B be a Banach space endowed with a norm  ·  and X be a set. Let A, B be two operators, A, B : X −→ B. We say that A is near B if there are two positive constants a, k, with 0 < k < 1, such that for all x1 , x2 ∈ X , we have:   

 B(x1 ) − B(x2 ) − a A(x1 ) − A(x2 )  ≤ k B(x1 ) − B(x2 ).

(11)

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The following theorem underlines some properties of these operators. Theorem 2.1. Let A be near to B, then: (i) (ii) (iii) (iv)

if if if if

B is injective (surjective) then A is injective (surjective) too (see [4]); B(X ) is open in B then A(X ) is open in B (see [16]); B(X ) is dense in B then A(X ) is dense in B (see [16]); B(X ) is compact in B then A(X ) is compact B (see [16]).

These results allow us to show following theorem (see Tarsia [17]) that we will use in next section. Theorem 2.2 (Implicit Functions Theorem). Let X be a topological space, Y a set, Z a Banach space, F : X × Y −→ Z, B : Y −→ Z. Suppose that (x0 , y0 ) ∈ X × Y exists such that F(x0 , y0 ) = 0;

(12)

the map x → F(x, y0 ) is continuous in x = x0 ;

(13)

there exist positive numbers α ˜ , k, with k ∈ (0, 1), and a neighbourhood U (x0 ) ⊂ X of x0 , such that for any y1 , y2 ∈ Y , and for any x ∈ U (x0 ),   

 B(y1 ) − B(y2 ) − α ˜ F(x, y1 ) − F(x, y2 ) Z ≤ k B(y1 ) − B(y2 )Z ,

(14)

B is injective;

(15)

B(Y ) is a neighbourhood of z0 = B(y0 ).

(16)

Then the following is true: there exist a ball S(z0 , ρ) ⊂ B(Y ) and a neighbourhood V (x0 ) ⊂ U (x0 ) of x0 , such that there is exactly one solution y = y(x) : V (x0 ) −→ B −1 (S(z0 , ρ)) of the following problem:    F x, y(x) = 0,

for any x ∈ V (x0 ),

y(x0 ) = y0 .

(17)

Moreover, we need following theorem, that is a “nonlinear method of continuity”; it is also showed by near operators theory (see Fattorusso and Tarsia [6]).5 It is a generalization of the well-known method of continuity for linear operators as well as for nonlinear operators (see for example D. Gilbarg and N.S. Trudinger [7]), because we do not assume the hypothesis that the operator is differentiable. Theorem 2.3. Let B be a Banach space normed with  ·  and let X be a set. Let {Ar }r∈[r1 ,r2 ] be a set of operators, Ar : X −→ B. If there exists a positive constant c such that for any s, r ∈ [r1 , r2 ] and u, v ∈ X , we have   

 Ar (u) − Ar (v) − As (u) − As (v)  ≤ c|r − s|Ar (u) − Ar (v),

(18)

then 1. if Ar is injective (surjective) for some r ∈ [r1 , r2 ], then Ar is injective (surjective) for any r ∈ [r1 , r2 ]; 2. if Ar (X ) is open (closed) for some r ∈ [r1 , r2 ], then Ar (X ) is open (closed) for any r ∈ [r1 , r2 ]; 3. if Ar (X ) is dense for some r ∈ [r1 , r2 ], then Ar (X ) is dense for any r ∈ [r1 , r2 ]. 5

In this paper the theorem is shown on the interval [0, 1], but it is trivial to show it on any interval [r1 , r2 ].

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Moreover, we recall the following results that we will use in Section 3. Lemma 2.1 (Miranda–Talenti inequality). Let Ω be a bounded convex open subset of Rn with a C 2 boundary ∂Ω. Then for any u ∈ H 2 ∩ H01 (Ω), it results

n

(Dij u) dx ≤ 2

Ω i,j=1

(Δu)2 dx.

(19)

Ω

The proof is, for example, in [11]. Lemma 2.2. Let Ω be a bounded convex open subset of Rn with a C 2 boundary ∂Ω. Then for any a > 0 and u ∈ L2 (0, T, H 2 ∩ H01 (Ω)) ∩ H 1 (0, T, L2 (Ω)), such that u(x, 0) = 0 a.e. in Ω, we have

T

   

 Δu(x, t)2 + a2 u (x, t)2 dx dt ≤

0 Ω

T

  Δu(x, t) − au (x, t)2 dx dt.

(20)

0 Ω

Proof. Let u ∈ C ∞ (Ω × [0, T ]) such that u(x, t) = 0 in ∂Ω × [0, T ] and u(x, 0) = 0 in Ω. Fixed a > 0, we can write

T

  Δu(x, t) − au (x, t)2 dx dt

0 Ω

T =

   

 Δu(x, t)2 + a2 u (x, t)2 − 2aΔu(x, t)u (x, t) dx dt

0 Ω

T =

   

 Δu(x, t)2 + a2 u (x, t)2 dx dt + a

0 Ω

T ≥

  Du(x, T )2 dx − a

Ω

  Du(x, 0)2 dx

Ω

   

 Δu(x, t)2 + a2 u (x, t)2 dx dt,

(21)

0 Ω

because u(x, 0) = 0 in Ω implies that Du(x, 0) = 0 in Ω. The thesis follows by density of spaces arguments. 2 Lemma 2.3. Let f ∈ L2 (Ω × [0, T ], RN ) and u ∈ L2 (0, T, H 2 ∩ H01 (Ω)) ∩ C 0 ([0, T ], H01 (Ω)) ∩ H 1 (0, T, L2 (Ω)) be solution of following Cauchy–Dirichlet problem 

Δu(x, t) − au (x, t) = f (x, t),

on Ω × [0, T ],

u(x, 0) = 0,

(22)

where a > 0, the following inequality holds

sup [0,T ]

Ω

  Du(x, t)2 dx ≤ 1 a

T 0 Ω

  f (x, t)2 dx dt

(23)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

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Proof. We multiply the equation by u , in Ω:

Δu(x, t)u (x, t) dx − a Ω

integrate and use that u(x, 0) = 0 in Ω implies that Du(x, 0) = 0   2 u  dx =

Ω

Ω 

−

⇐⇒

1 d 2 dt

Ω

  Du(x, t)2 dx + a

Ω

⇐⇒

f (x, t)u (x, t) dx

Du(x, t)Du (x, t) dx − a

⇐⇒

1 2



  2 u  dx =

Ω

  2 u  dx = −

Ω

  Du(x, t)2 dx + a

t

f (x, t)u (x, t) dx

Ω

f (x, t)u (x, t) dx

Ω

  2 u  dx dt = −

0 Ω

Ω



t

f (x, τ )u (x, τ ) dx dτ.

0 Ω

Then we have, for any t ∈ [0, T ],

  Du(x, t)2 dx + 2a

t

  2 u  dx dτ

0 Ω

Ω

t

≤2

  f (x, τ ) u (x, τ ) dx dτ

0 Ω

 t ≤2

  f (x, τ )2 dx dτ

 12  t

0 Ω

=⇒

 12

0 Ω

  Du(x, t)2 dx ≤ 1 a

Ω

See also Baiocchi [1, Section 5].

   u (x, τ )2 dx dτ

t

  f (x, τ )2 dx dτ.

0 Ω

2

Lemma 2.4. Let Ω be a bounded convex open subset of Rn with a C 2 boundary ∂Ω, and let u ∈ L2 (0, T, H 2 ∩ H01 (Ω)) ∩ H 1 (0, T, L2 (Ω)), such that u(x, 0) = 0 a.e. in Ω, then u ∈ C 0 ([0, T ], H01 (Ω)). Moreover, the following inequality holds   uH ≤ C Δu(x, t) − au (x, t)L2 (Ω,RN ) ,

(24)

where C depends on Ω, T , a, with a > 0. Proof. By hypotheses it follows that Δu ∈ L2 (0, T, L2 (Ω)), u ∈ L2 (0, T, L2 (Ω)), then u is the solution of a Cauchy–Dirichlet problem 

Δu(x, t) − au (x, t) = f (x, t), on Ω × [0, T ],

(25)

u(x, 0) = 0 where f ∈ L2 (0, T, L2 (Ω)), a > 0. By the cited results of Baiocchi (see [1]), we have the thesis. We also use following theorem (see for example [12, Theorem 58.X] or [13]).

2

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Theorem 2.4 (Gagliardo–Nirenberg–Sobolev). Let p1 ≥ 1, 1 ≤ p0 ≤ +∞ and let r be a positive integer such N that r − np−1 with the boundary 1 is not an integer greater or equal to zero. If Ω is an open bounded in R 0,1 C , there exist two positive constants C1 and C2 that depend only on Ω, r, p1 and on the distance of r − np−1 1 by N, such that setting 1 σr,h

  1−a h r 1 + = +a − , n p1 n p0

(26)

it holds, for any u ∈ H r,p1 ∩ Lp0 (Ω),   σ1   α D u(x)σ dx Ω |α|=h

  a p1   (1−a)  p1 p0 p0 p0 p1  α   1 0       D u(x) dx u(x) dx u(x) dx ≤ C1 + C2 Ω |α|=r

where σ = σr,h ,

h r

Ω

(27)

Ω

≤ a ≤ 1. Then, if r − np−1 1 ∈ N, the inequality (27) holds for

h r

≤ a < 1.

We will also use the following Schauder–Tychonov theorem (see [8]). Theorem 2.5. Let C be a nonempty convex subset of a locally convex linear topological space E, and let T : C −→ C be a compact map. Then T has a fixed point. 3. The system with principal part F (x, t, D 2 u) In this section we consider the Cauchy–Dirichlet problem ⎧ u ∈ H, ⎪ ⎨     F x, t, D2 u(x, t) − u (x, t) = g x, u(x, t), Du(x, t) + f (x, t), ⎪ ⎩ u(x, 0) = 0 a.e. in Ω.

a.e. in Ω × [0, T ],

(28)

To show the existence and the uniqueness of the solution of this problem, we will use the above mentioned theorem of nonlinear continuity (Theorem 2.3). In order to make this we need the following theorem, that we show by using Implicit Function Theorem (Theorem 2.2). Theorem 3.1. Let us consider ⎧ u∈H ⎪ ⎨   F x, t, D2 u(x, t) − u (x, t) = sg(x, t, u, Du) + f (x, t), ⎪ ⎩ u(x, 0) = 0 a.e. in Ω.

a.e. in Ω × [0, T ],

(29)

If we assume Condition A, (7), (8), (9) or (10) and f ∈ L2 (Ω × [0, T ], RN ), then there exists r0 > 0 such that for each s ∈ (−r0 , r0 ), Problem (29) has one and only one solution. First we show that operator F is near to operator Δ. Indeed, we have the following lemma. Lemma 3.1. If we assume Condition A on F , then for any u, v ∈ H, such that u(x, 0) = v(x, 0) = 0 in Ω, we have

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T



    Δ(u − v)(x, t) − a F x, t, D2 u(x, t) − F x, t, D2 v(x, t) 2 dx dt N

0 Ω

T ≤ (γ + δ)

2



 Δ(u − v)(x, t) − a u (x, t) − v  (x, t) 2 dx dt N

(30)

0 Ω

(here a > 0 is the constant of Condition A). Proof. By Condition A, we obtain6 :

T



    Δ(u − v)(x, t) − a F x, t, D2 u(x, t) − F x, t, D2 v(x, t) 2 dx N

0 Ω

T ≤ γ(γ + δ) 

  2 D (u − v)(x, t)2 dx dt + δ(γ + δ) N

0 Ω

 by Miranda–Talenti inequality (19)

T

≤ (γ + δ)

2

T

  Δ(u − v)(x, t)2 dx dt N

0 Ω

  Δ(u − v)(x, t)2 dx dt (by Lemma 2.2) N

0 Ω

T ≤ (γ + δ)

2



 Δ(u − v)(x, t) − a u (x, t) − v  (x, t) 2 dx dt. N

2

(31)

0 Ω

Lemma 3.2. If we assume on g hypothesis (7) and (8) then following inequality holds for any u1 , u2 ∈ H

T

     g x, t, u1 (x, t), Du1 (x, t) − g x, t, u2 (x, t), Du2 (x, t) 2 dx dt N

0 Ω

ξ ≤ c(dΩ , n) u1 H + u2 H

T



 Δu1 (x, t) − Δu2 (x, t) − a u1 (x, t) − u2 (x, t) 2 dx dt, N

(32)

0 Ω

where ξ = ξ(α, β, n) > 0. Proof. We begin by setting us = su1 + (1 − s)u2 , with u1 , u2 ∈ H, and observing that

T

     g x, t, u1 (x, t), Du1 (x, t) − g x, t, u2 (x, t), Du2 (x, t) 2 dx dt N

0 Ω

2

T   1 d      = g x, t, us (x, t), Dus (x, t) ds dx dt.   ds  0 Ω

0

N

By this and by assumption (8), for any u1 , u2 ∈ H, we have 6

Here we use: for any a, b ∈ R+ , we have (γa + δb)2 ≤ γ(γ + δ)a2 + δ(γ + δ)b2 .

(33)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

T

1437

     g x, t, u1 (x, t), Du1 (x, t) − g x, t, u2 (x, t), Du2 (x, t) 2 dx dt N

0 Ω

 

T 1  N  2  ∂  2      ≤c  ∂uj gi x, t, us (x, t), Dus (x, t)  u1 (x, t) − u2 (x, t) N dx dt ds 0

i,j=1

0 Ω

 

T 1  N n   ∂ 2 2       +c  ∂pjh gi x, t, us (x, t), Dus (x, t)  Du1 (x, t) − Du2 (x, t) nN dx ds dt 0

i,j=1 h=1

0 Ω

 T  1 ≤ c(M1 , M2 ) 0 Ω

T

 1

+ 0 Ω



        us (x, t)2α−2 + Dus (x, t)2β ds u1 (x, t) − u2 (x, t)2 dx dt N nN N

0



us (x, t)2α N

+

2β−2  Dus (x, t)nN ds

  Du1 (x, t) − Du2 (x, t)2 dx dt nN



0

 T ≤ c(M1 , M2 , α, β)



 2α−2 2α−2 u1 N + u2 N u1 − u2 2N dx dt

0 Ω

T

+



2β  2 Du1 2β nN + Du2 nN u1 − u2 N dx dt



 2α 2 u1 2α N + u2 N Du1 − Du2 nN dx dt

0 Ω

T + 0 Ω

T +



2β−2 Du1 nN

+

2β−2  Du1 Du2 nN

 −

Du2 2nN

dx dt

0 Ω

= c(M1 , M2 , α, β)(I1 + I2 + I3 + I4 ).

(34)

We can estimate the last side of (34), that is, the term (I1 +I2 +I3 +I4 ), by using Schwartz–Hölder inequality and Gagliardo–Nirenberg–Sobolev imbedding theorems. We begin estimating the integral I3 . For any α ≥ 1, we have:

T  I3 ≤ c 0



u1 N + u2 N

2pα

 p1  dx

Ω

Du1 − Du2 2q nN dx

 q1 dt,

(35)

Ω

where p1 + 1q = 1. If n = 2, by Sobolev imbedding theorem,

T  I3 ≤ c(dΩ )

α Du1 2nN + Du2 2nN dx

0

u1 − u2 2H 2 (Ω,RN ) dt

Ω

 ≤ c(dΩ ) sup [0,T ]

α T Du1 2nN

Ω

+

Du2 2nN

u1 − u2 2H 2 (Ω,RN ) dt.

dx 0

(36)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1438

Let be n > 2, taking in (35) q =

T 



I3 ≤ c(dΩ , n) 0

n n−2

n 2,

and p =

u1 N + u2 N

nα N

we have  n2

dx

 u1 − u2 2H 2 (Ω,RN ) dt

if nα ≤

Ω

T 



α Du1 2nN + Du2 2nN dx

≤ c(dΩ , n) 0

2n n−2

u1 − u2 2H 2 (Ω,RN ) dt

Ω

 ≤ c(dΩ , n) sup

α T Du1 2nN

[0,T ]

+

Du2 2nN

u1 − u2 2H 2 (Ω,RN ) dt.

dx

(37)

0

Ω

2 Because in this paper we consider α ≥ 1 and, by the above inequality, α ≤ n−2 , the thesis follows for 2 ≤ n ≤ 4. In particular, if n = 4 we must take α = 1, while if n = 3, 1 ≤ α ≤ 2. For this reason in the next estimates we take always n ≤ 4. Now we consider I1 . If n = 4, the thesis is obvious because, by the above remark, we must take α = 1. 2 Let be n = 3 and 1 < α ≤ 2 (because, by above remark, α ≤ n−2 ); if α = 1 the inequality is obvious.



T I1 ≤

sup u1 − u2 2N Ω

0

2(α−1)

u1 N

2(α−1)

+ u2 N

 dx dt

Ω



T ≤ c(dΩ )

u1 −

u2 2H 2 (Ω,RN )

0

 Du1 2nN + Du2 2nN dx

(α−1) dt

Ω



 Du1 2nN + Du2 2nN dx

≤ c(dΩ ) sup [0,T ]

(α−1) T u1 − u2 2H 2 (Ω,RN ) dt.

(38)

0

Ω

If n = 2, we have for any α ≥ 1 the same inequality. Now we consider I2 ; let be n = 2.

T  I2 ≤ c(β) 0

2pβ Du1 nN + Du2 nN dx

 p1 

Ω

u1 −

u2 2q N

 q1 dx

 1 1 where + = 1 , p q

 dt

Ω

 ≤ c(β, dΩ ) sup [0,T ]

 T  Du1 −

Du2 2nN

dx 0

Ω



Du1 nN + Du2 nN

2pβ

 p1 dx

dt.

(39)

Ω

Here to estimate the last integral we can use Gagliardo–Nirenberg–Sobolev theorem (see Theorem 2.4), 2p0 with h = 1, r = 2, 12 ≤ a ≤ 1, n = 2, p1 = 2, a = 12 , σ = 2βp, and then β = p(2+p . By this β ∈ [1, 2) 0) 2,2 assuming opportune values p0 ≥ 2 and p > 1. So we have for any u1 , u2 ∈ H (Ω, RN ) ∩ Lp0 (Ω, RN ),

T  0



Du1 nN + Du2 nN

2pβ

 p1 dx

dt

Ω

 T  ≤ c(β, p0 , dΩ ) 0

Ω

   2 2  D u1  2 + D2 u2 2 2 dx n N n N

 β2  Ω



 u1 pN0 + u2 pN0 dx

 pβ

0

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

 +

  u1 pN0 + u2 pN0 dx

1439



  2β p 0

dt

Ω





≤ c(p0 , β, dΩ ) sup [0,T ]

 + T sup [0,T ]





Du1 2nN

+

Du2 2nN

Ω

Du1 2nN

+

Du2 2nN





 β2 T  dx

n N

0

β  dx

 2 2 D u1  2

 2  + D2 u2 n2 N dx

 β2

Ω

.

(40)

Ω

By this, the thesis follows. Let be n = 3.

T  I2 ≤ c(β) 0

3β Du1 nN + Du2 nN dx

 23 

Ω

 13 dt

Ω



 T 

≤ c(β, dΩ ) sup [0,T ]

u1 − u2 6N dx

Du1 − Du2 2nN dx 0

Ω



Du1 nN + Du2 nN

3β

 23 dx

dt.

(41)

Ω

To estimate the last integral, we use Theorem 2.4, in which we take n = 3, r = 2, h = 1, p1 = 2, a = p0 = 6. So because 12 ≤ a ≤ 1, it follows that 1 < β ≤ 2. So we can write

3 2

− β1 ,

I2 ≤ sup [0,T ]

Du1 − Du2 2nN dx Ω

 ·

T 

C1

n N

0

+ C2 0

u1 6N

+

u2 6N



 β(1−a) 3 dt

dx

Ω



u1 6N

+

u2 6N





 β3 dx

dt

Ω

Ω

 ·



Du1 − Du2 2nN dx

≤ sup [0,T ]

βa 

Ω

T 

 2  + D2 u2 n2 N dx

 2 2 D u1  2

T 

C3

 2 2 D u1  2

n N

0

βa 

Ω

T  + C4 0

2   + D2 u2 n2 N dx



Du1 2nN

+

Du2 2nN



β (1−a) dx

dt

Ω



Du1 2nN

+

Du2 2nN



β dx

 dt .

(42)

Ω

Taking a ≤ 34 , we have  β I2 ≤ C(T, β)u1 − u2 2H u1 H + u2 H . If n = 4, taking p = q = 2, we have the thesis if β = 1 because

(43)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1440

T  I2 ≤ 0



Du1 4nN

+

Du2 4nN



 12  dx

Ω

[0,T ]

 12 dx

dt

Ω



T Du1 − Du2 2nN dx

≤ sup

u1 −

u2 4nN

 2 2 D u1 

nN

 2  + D2 u2 nN dx dt.

(44)

0 Ω

Ω

Now we consider I4 . Let n = 2. For any u1 , u2 ∈ H 2,2 (Ω, RN ),

T  I4 ≤ 0



Du1 nN + Du2 nN

2p(β−1)

 p1  dx

Ω

Du1 − Du2 2q nN dx

 q1 dt

(45)

Ω

(where p1 + 1q = 1). We apply Gagliardo–Nirenberg–Sobolev theorem to each term on the right side of (45). 1 In the first term we apply it taking a = 12 , p0 ≥ 1, and then 2p(β−1) = 14 + 2p10 , so 

 2p(β−1) 2p(β−1)  + Du2 nN Du1 nN dx

 p1

Ω

 ≤ c(p, p0 , Ω)

 2 2 D u1  2

n N

2   + D2 u2 n2 N dx

  β−1 2

Ω

 +



u1 pN0

+

 u2 pN0 dx

 β−1 p 0

Ω

  u1 pN0 + u2 pN0 dx

  2(β−1) p 0

Ω

 ≤ c(p, p0 , Ω)  +

 2 2 D u1  2

n N

Ω

 Du1 2nN

2   + D2 u2 n2 N dx

  β−1 2

(β−1)   2 . + Du2 nN dx



Du1 2nN

+

Du2 2nN



 β−1 2 dx

Ω

(46)

Ω

Now we consider the second term, where as above we take a = 12 , q0 ≥ 1, and then  Du1 − Du2 2q nN dx Ω



≤ c(q, q0 , Ω)

 2  D u1 − D2 u2 2 2 dx n N

u1 − u2 qN0 dx

+ Ω



≤ c(q, q0 , Ω)  + Ω

Ω

=

1 4

+

1 2q0 ,

so

 q1

Ω



1 2q

 q1 

 14 

u1 − u2 qN0 dx

 2q1

0

Ω

0

 2  D u1 − D2 u2 2 2 dx n N

 12  . Du1 − Du2 2nN dx

 14 

Du1 − Du2 2N n dx

 14

Ω

(47)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1441

(q0 −2) The thesis follows because p1 + 1q = 1, by this equality β = 1 + pq00(p and this implies β ∈ [1, 2). 0 +2) Let be n = 3: we have the thesis from (45) applying Theorem 2.4 taking into account that u ∈ H 2,2 ∩ Lq0 , 0 0 0 with q0 ≥ 1, a = 12 , then q = q2q , p = q2q and β = q2q ∈ [1, 2). 0 +1 0 −1 0 +1 If n = 4, the thesis holds trivially. 2

Lemma 3.3. If we assume Condition A on F and hypotheses (7), (8) on g then following inequality holds for any u1 , u2 ∈ Y

T

     g x, t, u1 (x, t), Du1 (x, t) − g x, t, u2 (x, t), Du2 (x, t) 2 dx dt N

0 Ω



ξ

≤ c(dΩ , n, γ, δ) u1 H + u2 H a

T 2

   F x, t, D2 u1 (x, t) − u1 (x, t)

0 Ω

 2

 − F x, t, D2 u2 (x, t) − u2 (x, t) N dx dt,

(48)

where ξ = ξ(α, β, n) > 0. Proof. We use Lemma 3.1 obtaining:

T



 Δu1 (x, t) − au1 (x, t) − Δu2 (x, t) − au2 (x, t) 2 dx dt N

0 Ω

T ≤ (1 + η)





    Δ u1 (x, t) − u2 (x, t) − a F x, t, D2 u1 (x, t) − F x, t, D2 u2 (x, t) 2 dx dt N

0 Ω

         1  aF x, t, D2 u1 (x, t) − au1 (x, t) − aF x, t, D2 u2 (x, t) − au2 (x, t) 2 dx dt + 1+ N η Ω

T ≤ (1 + η)(γ + δ)

2





 Δ u1 (x, t) − u2 (x, t) − a u1 (x, t) − u2 (x, t) 2 dx dt N

0 Ω

 + 1+



1 2 a η

T

  

   F x, t, D2 u1 (x, t) − u1 (x, t) − F x, t, D2 u2 (x, t) − u2 (x, t) 2 dx dt. N

(49)

0 Ω

By Lemma 3.2 and (49), setting η such that (1 + η)(γ + δ)2 < 1, we have the thesis. 2 Now we can give a proof of Theorem 3.1. Proof of Theorem 3.1. In order to use Theorem 2.2, we set X = R; Y = {u : u ∈ H, with u − u0 H < R}, where r > 0 is fixed and we will find it later in the article; Z = L2 (Ω × [0, T ], RN ); F(s, u) = aF (x, t, D2 u) − au (x, t) − sag(x, t, u, Du) − af (x, t), s ∈ R; (s0 , u0 ) = (0, u0 ) where u0 ∈ H is the solution of the system F (x, t, D2 u0 (x, t)) − u0 (x, t) = f (x, t), a.e. in Ω × [0, T ], so that F(s0 , u0 ) = 0;

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1442

B(u) = aF (x, t, D2 u) − au (x, t). We observe that the assumptions of Theorem 2.2 are satisfied, indeed: F : X × Y −→ Z, because for any s ∈ R and u ∈ H, we have   F(s, u)2 = Z

T

    aF x, t, D2 u − au (x, t) − sag(x, t, u, Du) − af (x, t)2 dx dt N

0 Ω

T ≤ ca

2

      

   F x, t, D2 u 2 + u (x, t)2 + |s|2 g(x, t, u, Du)2 + f (x, t)2 dx dt N N N N

0 Ω

< +∞;

(50)

indeed, by assumption (7) and Sobolev imbedding theorems, for any u ∈ H,

T

  g(x, t, u, Du)2 dx dt ≤ c N

0 Ω

T

T u2α N

0 Ω

Du2β nN dx dt < +∞;

dx dt + c 0 Ω

moreover, by Condition A, we have

T

   aF x, t, D2 u 2 dx dt ≤ 2 N σ2

0 Ω

T

       aF x, t, D2 u(x, t) − Δu(x, t)2 + Δu(x, t)2 dx dt N N

0 Ω

2 ≤ 2 σ

T

2  2    2 γ D u(x, t)n2 N + (δ + 1)Δu(x, t)N dx dt < +∞;

(51)

0 Ω

T  obviously, because u ∈ H, 0 Ω u (x, t)2N dx dt < +∞. The last inequality shows that B : H −→ L2 (Ω × [0, T ], RN ). d Moreover, B is near to Δ − a dt as operator by Y and Z (see Definition 2.1, Lemma 3.1). d Since Δ − a dt is a bijection between H and L2 (Ω × [0, T ], RN ), it results by Theorem 2.1 that B also d is a bijection between H and L2 (Ω × [0, T ], RN ). On the other hand, Δ − a dt is an open map, by Banach open map theorem, then by Theorem 2.1, B also is an open map. In particular, B(Y ) is open in Z. Moreover, the function s −→ F(s, u) is continuous in s = 0. The proof is completed by showing that the assumption (14) of Theorem 2.2 is verified: there exist k ∈ (0, 1) and r1 > 0 such that for any u1 , u2 ∈ Y and s ∈ (−r1 , r1 ), we have

T 0 Ω

  

   aF x, t, D2 u1 − au1 (x, t) − aF x, t, D2 u2 − au2 (x, t)    − aF x, t, D2 u1 − au1 (x, t) − sag(x, t, u1 , Du1 )  2

 − aF x, t, D2 u2 − au2 (x, t) − sa g(x, t, u2 , Du2 ) N dx dt

T

≤ ka

2 0 Ω

  

   F x, t, D2 u1 − u1 (x, t) − F x, t, D2 u2 − u2 (x, t) 2 dx dt, N

(52)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1443

i.e.

T

  g(x, t, u1 , Du1 ) − g(x, t, u2 , Du2 )2 dx dt N

2

s

0 Ω

T

 

    F x, t, D2 u1 − u1 (x, t) − F x, t, D2 u2 (x) − u2 (x, t) 2 dx dt. N

≤ ka

2

(53)

0 Ω

This inequality follows by Lemma 3.3, setting k = s2 c(dΩ , n, γ, δ)(R + u0 H )ξ , because u1 , u2 ∈ Y implies u1 H ≤ u0 H + R, u2 H ≤ u0 H + R. Then we have (52), accordingly |s| ∈ (0, r1 ) with r1 such that r12 c(dΩ , n, γ, δ)(R + u0 H )ξ < 1. So (−r1 , r1 ) is the neighbourhood U (s0 ) of s0 = 0 on which is true inequality (14). In this way we have shown that all assumptions of Theorem 2.2 are verified. By this theorem, we can conclude that there exists a ball S(F (·, D2 u0 ), ρ) in B(Y ) and r0 ∈ (0, r1 ), such that there exists one and only one solution us , with s ∈ (0, r0 ), u0 ∈ B −1 (S(F (·, D2 u0 ), ρ)), solution of7 

F(s, us ) = 0 us |s=0 = u0 .

That is, there exists one solution of the Problem (29). Now we show that the solution is unique under hypothesis (9). Indeed, if u, v are two solutions of the Problem (29), we have, for any s ∈ (0, r0 ),

T

  

   aF x, t, D2 u − au (x, t) − aF x, t, D2 v − av  (x, t)

0 Ω

2 − sa g(x, t, u, Du) − g(x, t, v, Dv) N dx dt = 0.

This can be written as it follows

T

  

   F x, t, D2 u − u (x, t) − F x, t, D2 v − v  (x, t) 2 dx dt N

0 Ω

T + s2

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt N

0 Ω

T = 2s

   F x, t, D2 u − u (x, t)

0 Ω

  

 − F x, t, v, D2 v − v  (x, t)  g(x, t, u, Du) − g(x, t, v, Dv) N dx dt. It is absurd, because the right hand side is negative by (9). Now we show that the solution is unique under the hypothesis (10). Indeed, let u and v be two solutions of the Problem (29). 7

Here r0 ≤ r1 , so that (−r0 , r0 ) is the neighbourhood V (x0 ) of Theorem 2.2.

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1444

T



    Δ(u − v) − aF x, t, D2 u(x, t) − aF x, t, D2 v(x, t) 2 dx dt N

0 Ω

T =



 Δ(u − v)(x, t) − a(u − v) − as g(x, t, u, Du) − g(x, t, v, Dv)

0 Ω

   − aF x, t, D2 u(x, t) − au − asg(x, t, u, Du) − af (x, t)  2

 + aF x, t, D2 v(x, t) − av  − asg(x, t, v, Dv) − af (x, t) N dx dt

T =





 Δ(u − v) − a u − v  − as g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

T =



 Δ(u − v) − a u − v  2 dx dt + a2 s2 N

0 Ω

T

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt N

0 Ω

T − 2as



    Δu − Δv − a u − v   g(x, t, u, Du) − g(x, t, v, Dv) N dx dt.

(54)

0 Ω

By this and by Lemma 3.1, we have

T 2as



    Δu − Δv − a u − v   g(x, t, u, Du) − g(x, t, v, Dv) N dx dt

0 Ω



≥ 1 − (γ + δ)

2



T



 Δu − Δv − a u − v  2 dx dt

0 Ω

T 2 2

+a s

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt. N

0 Ω

This is in contradiction with hypothesis (10). 2 Now let us consider the Dirichlet Problem (28) and we show the following theorem Theorem 3.2. Let Condition A, (7), (8), (9), or (10) be satisfied. Then if f ∈ L2 (Ω × [0, T ], RN ), Problem (28) has one and only one solution. Proof. We deduce the thesis applying nonlinear method of continuity (see Theorem 2.3), showing that there exists a positive constant c such that for any s, r ∈ [r0 , 1] (where r0 is established by Theorem 3.1) and for any u, v ∈ H, we have 

 Ar (u) − Ar (v) − As (u) − As (v) 

L2 (Ω×[0,T ],RN )

  ≤ c|r − s|Ar (u) − Ar (v)L2 (Ω×[0,T ],RN ) ,

(55)

where     Ar (u) = F x, t, D2 u(x, t) − u (x, t) − rg x, t, u(x, t), Du(x, t) , and then we have to show

(56)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

T 0 Ω

1445

  

   F x, t, D2 u − u − rg(x, t, u, Du) − F x, t, D2 v − v  − rg(x, t, v, Dv)   

  2 − F x, t, D2 u − u − sg(x, t, u, Du) − F x, t, D2 v − v  − sg(x, t, v, Dv) N dx dt

T

≤ C1 |r − s|

2

   F x, t, D2 u − u − rg(x, t, u, Du)

0 Ω

 2

 − F x, t, D2 v − v  − rg(x, t, v, Dv) N dx dt.

(57)

To make this it is enough to take r ∈ [r0 , 1], and u, v ∈ H for have

T

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt N

0 Ω

T ≤ C1

  

  

 F x, t, D2 u − u − F x, t, D2 v − v  − r g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt. (58) N

0 Ω

We first show (58) under hypothesis (9). Indeed, setting r ≥ r0 and u, v ∈ H, we have

T

  

  

 F x, t, D2 u − u − F x, t, D2 v − v  − r g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

T ≥

  

   F x, t, D2 u − u − F x, t, D2 v − v  2 dx dt N

0 Ω

T +

r02

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt. N

(59)

0 Ω

Now we obtain the thesis of theorem showing (58) with hypothesis (10). To make this, we consider the term in the right side of the same inequality and we estimate it.

T a

2

  

  

 F x, t, D2 u − u − F x, t, D2 v − v  − r g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

= Ω

  

   

 aF x, t, D2 u − au − aF x, t, D2 v − av  − Δ[u − v] − a u − v  2 dx dt N

+ Ω





 Δ[u − v] − a u − v  − ra g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

+2



      aF x, t, D2 u − aF x, t, D2 v − Δ[u − v]  Δ[u − v] − a u − v 

Ω

 − ra g(x, t, u, Du) − g(x, t, v, Dv) N dx dt

T ≥ 0 Ω

      a F x, t, D2 u − F x, t, D2 v − Δ[u − v]2 dx dt N

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1446

T +





 Δ(u − v) − a u − v  − ra g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

−

k ν

T



 Δ(u − v) − a u − v  2 dx dt N

0 Ω

T −ν





 Δ[u − v] − a u − v  − ra g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt, N

(60)

0 Ω

where ν ∈ (0, 1).8 Moreover, to estimate the right hand side of (60), we consider each of its terms and we observe that following inequalities hold

T

   Δ(u − v) − a u − v  2 dx dt N

0 Ω

T ≤ (1 + η)



    Δ(u − v) − a F x, t, D2 u − F x, t, D2 v 2 dx dt N

0 Ω



 T        1 2 F x, t, D2 u − F x, t, D2 v − u − v  2 dx dt + 1+ a N η 0 Ω

(by Condition A, and by Lemma 3.1)

T ≤ k(1 + η)

   Δ(u − v) − a u − v  2 dx dt N

0 Ω



 T        1 2 F x, t, D2 u − F x, t, D2 v − u − v  2 dx dt. + 1+ a N η 0 Ω

8

Indeed,

T 0 Ω



 

  2  2   aF x, t, D u − aF x, t, D v − Δ[u − v]  Δ[u − v] − a u − v

 − ra g(x, t, u, Du) − g(x, t, v, Dv) N dx dt  T

≤

      a F x, t, D 2 u − F x, t, D 2 v − Δ[u − v]2 dx dt N

1

2

0 Ω

 T ·





 Δ[u − v] − a u − v  − ra g(x, u, Du) − g(x, v, Dv) 2 dx dt N

1

2

0 Ω

 T 1 2 

 Δ(u − v) − a u − v  2 dx dt ≤ k N 0 Ω

 T ·





 Δ[u − v] − a u − v  − ra g(x, u, Du) − g(x, v, Dv) 2 dx dt N

0 Ω

The last estimate is obtained by Lemma 3.1 setting k = (γ + δ)2 .

1

2

.

(61)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1447

Taking η > 0 such that k(1 + η) < 1, by this it follows [1 − k(1 + η)]η 1+η

T

   Δ[u − v] − a u − v  2 dx dt N

0 Ω

T

≤ a2

       F x, t, D2 u − F x, t, D2 v − u − v  2 dx dt. N

(62)

0 Ω

Besides, to estimate the first term in the right side of (60), we observe that, setting ε ∈ (0, 1) and taking into account (62), we have

T

      aF x, t, D2 u − aF x, t, D2 v − Δ(u − v)2 dx dt N

0 Ω

T

      2 a2 F x, t, D2 u(x, t) − F x, t, D2 v(x, t) − u − v  N

≥ 0 Ω

  2 + Δ[u − v] − a u − v  N dx dt

T −ε

      2 a2 F x, t, D2 u − F x, t, D2 v − u − v  N dx dt

0 Ω

−

1 ε

T

   Δ[u − v] − a u − v  2 dx dt N

0 Ω

T = (1 − ε)a

2

       F x, t, D2 u − F x, t, D2 v − u − v  2 dx N

0 Ω



1 + 1− ε

 T

   Δ[u − v] − a u − v  2 dx dt N

0 Ω

 ≥

  T    1 η[1 − k(1 + η)] Δ[u − v] − a u − v  2 dx dt. + 1− (1 − ε) N 1+η ε

(63)

0 Ω

Now we examine the second and the fourth term in the right side of (60), by using (10) we can write:

T





 Δ[u − v] − a u − v  − ra g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

T ≥ 0 Ω



 Δ[u − v] − a u − v  2 dx dt + r2 a2 N

T

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt. N

0 Ω

By (60), taking into account (63), (64) and Lemma 2.2 we obtain

(64)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1448

T

    

 

 a F x, t, D2 u − u − a F x, t, D2 v − v  − ra g(x, t, u, Du) − g(x, t, v, Dv) 2 dx dt N

0 Ω

≥

    T  

 1 k η[1 − k(1 + η)] Δ[u − v] − a u − v  2 dx dt + 1− + 1−ν− (1 − ε) N 1+η ε ν 0 Ω

T

+ (1 − ν)r2 a2

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt. N

(65)

0 Ω

From this inequality, it follows the thesis with r ≥ r0 , since it is possible to verify that there exist ε ∈ (0, 1), ν ∈ (0, 1) and η small enough such that      η[1 − k(1 + η)] 1 k (1 − ε) + 1− + 1−ν− > 0. 1+η ε ν To show this inequality, first we set τ =

η[1−k(1+η)] 1+η

(66)

so that (66) becomes

  ϕ(ε) = −ε2 τ ν + ε ντ + 2ν − ν 2 − k − ν > 0, −k 9 then we verify that ϕ has a positive maximum in the point ε0 = ντ +2ν−ν . 2τ ν 2 2 2 Indeed, ϕ(ε0 ) > 0 if [ν τ + 2ν − ν − k] > 4ν τ . This is true if we take τ small enough, i.e. η small √ enough. In fact if τ = 0, we have [2ν − (ν 2 + k)]2 > 0, for ν = 1 ± 1 − k. 2 2

4. Existence and uniqueness of solution for the system in complete form Let us consider the Dirichlet problem in its complete form (see (3)). We prove the following theorem. Theorem 4.1. Let F satisfy Condition A and the hypothesis (6), while g satisfies hypotheses (7), (8). Moreover, we also assume (9). Then if f ∈ L2 (Ω × [0, T ], RN ), Problem (3) has one solution, that it is unique if (9) holds with a strict inequality. To prove this theorem, we will use Theorem 3.1 showing the existence of a solution of the system     F x, t, u(x, t), Du(x, t), D2 u(x, t) − u (x, t) = s g x, t, u(x, t), Du(x, t) + f (x, t),

(67)

with s small enough, and then we will apply the nonlinear continuity method (Theorem 2.3). The first step of the proof is the following lemma. Lemma 4.1. Under the assumptions of Theorem 4.1, there exists r0 > 0 such that for any s ∈ [−r0 , r0 ] system (67) has one solution in H. Proof. In order to apply the Schauder–Tychonov theorem, we consider 9

It is easy to show ε0 ∈ (0, 1), because we can consider the following system 

ν 2 − ν(τ + 2) + k < 0, ν 2 + ν(τ − 2) + k > 0,

and we observe that it has solutions ν such that 0 < ν1 < ν < ν2 < 1 where ν1 =

2−τ +



(τ −2)2 −4k , 2

ν2 =

2+τ +

 (2+τ )2 −4k . 2

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1449

   T : L2 0, T, H01 Ω, RN −→ H,

(68)

such that to any w ∈ L2 (0, T, H01 (Ω, RN )), it maps u = T (w) that is the solution of problem ⎧ u ∈ H, ⎪ ⎨     F x, t, w(x, t), Dw(x, t), D2 u(x, t) − u (x, t) = sg x, t, u(x, t), Du(x, t) + f (x, t), ⎪ ⎩ a.e. in Ω × [0, T ].

(69)

By Theorem 3.1, if s is small enough, we know that Problem (69) has one and only one solution. Moreover, we consider the imbeddings          J1 : L2 0, T, H 2 ∩ H01 Ω, RN ∩ H 1 0, T, L2 Ω, RN −→ H θ 0, T, H 2(1−θ) Ω, RN ,

θ ∈ [0, 1]

and10       J2 : H θ 0, T, H 2(1−θ) Ω, RN −→ L2 0, T, H01 Ω, RN ,

θ∈

  1 . 0, 2

J1 is continuous, while J2 is compact by Rellich theorem, and we set T1 = J2 ◦ J1 ◦ T .

(70)

We deduce the thesis by the Schauder–Tychonov theorem applied to the map T1 , showing that    the set of the solutions of (69), for any w ∈ L2 0, T, H01 Ω, RN , is bounded;

(71)

T is continuous.

(72)

We begin showing (71). If u ∈ H is the solution of Problem (69), we can write

T

  Δu − au 2 dx dt N

0 Ω

T =



   Δu − au − a F x, t, w, Dw, D2 u − u − sg(x, t, u, Du) − f (x, t) 2 dx dt N

0 Ω

T ≤ (1 + ε)

  Δu − aF (x, t, w, Dw, D2 u)2 dx dt N

0 Ω

  T 2  1 + 1+ a2 sg(x, t, u, Du) − f (x, t)N dx dt (by Lemma 3.1) ε 0 Ω

T ≤ (1 + ε)(γ + δ)

2

  Δu − au 2 dx dt N

0 Ω

T 2

+ c(ε, a)s

  g(x, t, u, Du)2 dx dt + c(ε, a) N

0 Ω

10

See, for example, [10, Chap. 1, Proposition 2.3 and Remark 9.5].

T 0 Ω

  f (x, t)2 dx dt. N

(73)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1450

By hypothesis (7) on the growth of g and by Sobolev imbedding theorems, we can write

T

  g(x, t, u, Du)2 dx dt N

0 Ω

 T ≤c

 2  D u(x, t)2 dx dt + N

0 Ω

T

≤c

T

  Du(x, t)2 dx dt + N

0 Ω

T

  u(x, t)2 dx dt N



0 Ω

  Δu(x, t)2 dx dt. N

0 Ω

Then, by this and (73), we have

T

  Δu(x, t) − au (x, t)2 dx dt N

0 Ω

 

T

T 2  2c(ε, a) 2  2 Δu(x, t) − au (x, t) dx dt + ≤ cs f N dx dt = R1 . N 1 − (1 + ε)(γ + δ)2 0 Ω

(74)

0 Ω

If s is small enough, by Lemma 2.2, we obtain following inequality that complete the proof of (71):

T u2H

≤ c(a, γ, δ, s)

  f (x)2 dx dt. N

(75)

0 Ω

Now we show (72), i.e. T is continuous. h→+∞

Let be vh , v ∈ L2 (0, T, H01 (Ω, RN )) such that vh −→ v in L2 (0, T, H01 (Ω, RN )), setting uh = T vh and h→+∞

u = T v, we will prove that to say uh −→ u in H is as to say

T





 Δ uh (x) − u(x) − a uh − v  2 dx dt h→+∞ −→ 0. N

0 Ω

Taking into account that uh and u are the solutions of (69), we can write

T



 Δ[uh − u] − a uh − u 2 dx dt N

0 Ω

T =

  

  Δ[uh − u] − a uh − u − a F x, t, vh , Dvh , D2 uh − uh

0 Ω

 

  − sg(x, t, uh , Duh ) − f − F x, t, vh , Dvh , D2 u − a F x, t, vh , Dvh , D2 u   2 − F x, t, v, Dv, D2 u + u + s g(x, t, u, Du) + f N dx dt

T

≤ (1 + ε) 0 Ω

  

  Δ[uh − u] − a F x, t, vh , Dvh , D2 uh − F x, t, vh , Dvh , D2 u 2 dx dt N

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

T + c(a, ε)

1451

     F x, t, vh , Dvh , D2 u − F x, t, v, Dv, D2 u 2 dx dt N

0 Ω

T

  g(x, t, uh , Duh ) − g(x, t, u, Du)2 dx dt. N

2

+ c(a, ε)|s|

(76)

0 Ω

By this, Condition A and assumptions (6), by taking ε small enough such that (1 + ε)(γ + δ)2 < 1, it follows

T



 Δ[uh − u] − a uh − u  dx dt N

0 Ω

 T

  2 2  D u ω(vh − v, Dvh − Dv)2 dx dt N

≤ c(a, ε, γ, δ) 0 Ω

T + |s|

2

   g(x, t, uh , Duh ) − g(x, t, u, Du)2 dx dt . N

(77)

0 Ω

Moreover, by Lemma 3.2, we have by taking as u1 = uh , u2 = u and Y the ball centered in u with radius less than 2R1 (see inequality (74))

T



 Δ[uh − u] − a uh − u  dx dt N

0 Ω

T ≤ c(a, γ, δ, ε)

  2 2  D u ω(vh − v, Dvh − Dv)2 dx dt N

0 Ω

 ξ + c(dΩ , n, a) uh H 2 (Ω,RN ) + uH 2 (Ω,RN )

≤ c(a, γ, δ, ε)

T

|s|2

0

   Δuh − Δu − a uh − v  2 dx dt N

Ω

 2 2   D u ω(vh − v, Dvh − Dv)2 dx N

Ω

+ c(dΩ , n, a)Rξ |s|2



 Δuh − Δu − a uh − u 2 dx, N

Ω

by this, if s is small enough, follows

T



 Δ[uh − u] − a uh − u  dx dt

0 Ω

c(dΩ , a, γ, δ) ≤ 1 − c(n, dΩ , a)s2 Rξ

 T 0 Ω

 2  2 2  D u ω(vh − v, Dvh − Dv) dx dt . N

(78)

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1452

n→+∞

h→+∞

h→+∞

Now we observe that if vh −→ v in H01 (Ω, RN ) then vh −→ v and Dvh −→ Dv in measure on Ω. By a well-known measure theory theorem, we have that h→+∞

ω(vh − v, Dvh − Dv) −→ 0 in measure on Ω. Because ω is a bounded function in RN ×RnN , by Lebesgue bounded convergence theorem (in measure) it results that right hand side of (78) converges to zero. 2 Now we can give the proof of Theorem 4.1. We apply the nonlinear method of continuity to the set of operators   Ar (u) = F x, t, u, Du, D2 u − u − rg(x, t, u, Du), by taking X = H, B = L2 (Ω × [0, T ], RN ) so we have that Ar : X −→ B. Indeed,

T

    F x, t, u, Du, D2 u − u − r g(x, t, u, Du)2 dx dt N

0 Ω

c ≤ 2 a

 T



   Δu − a F x, u, Du, D2 u − F (x, u, Du, 0) 2 dx dt N

0 Ω

T +

  Δu − au 2 dx dt + a2 N

0 Ω

T

  g(x, t, u, Du)2 dx dt N



0 Ω

(see proof of Theorem 3.1)

T ≤ c(γ, δ, a)

  Δu − au 2 dx dt + c(M, a) N

0 Ω

T

2β  u2α N dx + DunN dx dt < +∞,

(79)

0 Ω

for any u ∈ L2 (0, T, H 2 ∩ H01 (Ω)), by (7) and by Sobolev imbedding theorems. Moreover, we have also to show (18); to this purpose we observe that, by inequality (9), we have for any r ≥ r1 , and for any r1 > 0.   Ar (u) − Ar (v)2 2

L (Ω×[0,T ],RN )

T ≥r

2

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt N

0 Ω

− 2r

   F x, t, u, Du, D2 u − u

Ω

 

  − F x, t, v, Dv, D2 v − v   g(x, t, u, Du) − g(x, t, v, Dv) N dx dt

T ≥

r12

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt, N

0 Ω

so for any u, v ∈ H and s, r ∈ [r1 , 1], and for any r1 ∈ (0, 1), we have:

L. Fattorusso, A. Tarsia / J. Math. Anal. Appl. 421 (2015) 1428–1454

1453



 Ar (u) − Ar (v) − As (u) − As (v) 2 2

L (Ω×[0,T ],RN )

T = |r − s|

2

  g(x, t, u, Du) − g(x, t, v, Dv)2 dx dt N

0 Ω

≤

2 |r − s|  Ar (u) − Ar (v)L2 (Ω×[0,T ],RN ) . 2 r1 2

Then we have proved the hypotheses of Theorem 2.3 so we can apply the statement (1) of it. Indeed, we know, by Lemma 4.1, that there exists r0 > 0, such that for any s ∈ [0, r0 ], the operator As is surjective; then taking r1 ≤ r0 , we can conclude that As is surjective for any s ∈ [r1 , 1] and, in particular, for s = 1. This show the existence of solution of Problem (3). Now, if hypothesis (9) strictly holds, we show the uniqueness. On the contrary if we suppose that the Problem (3) has two different solutions u and v, so we can write the following equality    

 F x, t, u(x, t), Du(x, t), D2 u(x, t) − F x, t, v(x, t), Dv(x, t), D2 v(x, t) − u (x, t) − v  (x, t)     = g x, t, u(x, t), Du(x, t) − g x, t, u(x, t), Dv(x, t) , a.e. in Ω × [0, T ],

(80)

in this multiplying by g(x, t, u(x, t), Du(x, t)) − g(x, t, u(x, t), Dv(x, t)) and integrating:

T

  

    F x, t, u, Du, D2 u − F x, t, v, Dv, D2 v − u − v   g(x, t, u, Du) − g(x, t, u, Dv) N dx dt

0 Ω

T =

  g(x, t, u, Du) − g(x, t, u, Dv)2 dx dt, N

(81)

0 Ω

by which we have an absurd, because the left hand side of (81) is negative by assumption (9), while the right hand side is positive. References [1] C. Baiocchi, Teoremi di regolarità per le soluzioni di un’equazione differenziale astratta, Symp. Math., vol. VII, Academic Press, London, New York, 1971. [2] S. Campanato, A Cordes type condition for nonlinear variational systems, Rend. Accad. Naz. Sci. XL Mem. Mat. Appl. (5) 13 (1) (1989) 307–321. [3] S. Campanato, Nonvariational basic parabolic systems of second order, Rend. Accad. Naz. Lincei, Mat. Appl. 2 (1991) 129–136. [4] S. Campanato, On the condition of nearness between operators, Ann. Mat. Pura Appl. 167 (1994) 243–256. [5] L. Fattorusso, A. Tarsia, Global solvability of Dirichlet problem for fully nonlinear elliptic systems, Numer. Funct. Anal. Optim. 35 (7–9) (2014) 1043–1065. [6] L. Fattorusso, A. Tarsia, Von Kármán equations in Lp spaces, Appl. Anal. 92 (11) (2013) 2375–2391. [7] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, reprint of the 1998 edition, Classics Math., Springer-Verlag, Berlin, 2001, xiv+517 pp. [8] A. Granas, J. Dugundji, Fixed Point Theory, Springer Monogr. Math., Springer-Verlag, New York, 2003, xvi+690 pp. [9] G.M. Lieberman, Second Order Parabolic Differential Equations, World Scientific Publishing Co., Inc., River Edge, NJ, 1996, xii+439 pp. [10] E. Magenes, J.L. Lions, Problèmes aux limites non homogènes et applications, vol. 1, Dunod, Paris, 1968. [11] A. Maugeri, D.K. Palagachev, L.G. Softova, Elliptic and Parabolic Equations with Discontinuous Coefficients, Math. Res., vol. 109, Wiley–VCH, 2002. [12] M. Miranda, Istituzioni di Analisi Funzionale Lineare, vol. 1, Unione Matematica Italiana, 1978. [13] L. Nirenberg, An extended interpolation inequality, Ann. Sc. Norm. Super. Pisa Cl. Sci. (3) 20 (1966) 733–737. [14] L.G. Softova, Strong solvability for a class of nonlinear parabolic equations, Le Matematiche 52 (1997) 59–70. [15] L.G. Softova, Strong solvability of the Cauchy–Neumann problem for nonlinear parabolic equations, Inv. Lect., in: Eight Int. Coll. Differ. Equations, Plovdiv ’97, vol. I, Academic Publications, 1998, pp. 111–116.

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