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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1


Volume 13 (2015)

Integro-differential systems with variable exponents of nonlinearity

Oleh Buhrii / Nataliya Buhrii
Published Online: 2017-06-22 | DOI: https://doi.org/10.1515/math-2017-0069


Some nonlinear integro-differential equations of fourth order with variable exponents of the nonlinearity are considered. The initial-boundary value problem for these equations is investigated and the existence theorem for the problem is proved.

Keywords: Nonlinear parabolic equation; Integro-differential equation; Generalized Lebesgue space; Generalized Sobolev space; Variable exponents of nonlinearity

MSC 2010: 47G20; 46E35; 35K52; 35K55

1 Introduction

Let n, N ∈ ℕ and T > 0 be fixed numbers, Ω ⊂ ℝn be a bounded domain with the boundary ∂Ω, Q0,T = Ω × (0, T), S0,T = ∂Ω × (0, T). We seek a weak solution u = (u1, . . ., uN) : Q0,T → ℝN of the problem uk,t+αΔ2uki=1naik(x,t)|uxi|p(x)2uk,xixi+Δbk(x,t)|u|γ(x)2uk+(Nu)k==i,j=1nfijk(x,t)xixji=1nfik(x,t)xi+f0k(x,t),(x,t)Q0,T,k=1,N¯,(1) u|S0,T=0,Δu|S0,T=0,u|t=0=u0(x).(2) Here α > 0 is a number, Δ:=2x12+2x22+...+2xn2 is the Laplacian, Δ2 := Δ(Δ), |u| := (|u1|2 + . . . + |uN|)1/2, |uxi|:=(|u1xi|2+...+|uNxi|)1/2,i=1,n¯, (Nu)k(x,t):=(Gu)k(x,t)+(Bu)k(x,t)+ϕk((Eu)k(x,t)),(x,t)Q0,T,(3) (Gu)k(x,t):=gk(x,t)|u(x,t)|q(x)2uk(x,t),(x,t)Q0,T,(4) (Bu)k(x,t):=βk(x,t)(uk(x,t)),(x,t)Q0,T,(5) (Eu)k(x,t):=Ωϵk(x,t,y)u~k(x+y,t)u~k(x,t)dy,(x,t)Q0,T,(6) aik, bk, gk, βk, φk, εk, fijk, fik, f0k, p, γ, q, u0 are some functions, (uk)- := max{-uk, 0}, and ũk is zero extension of uk from Q0,T into (ℝn \ Ω) × (0, T), where i,j=1,n¯,k=1,N¯.

Equation (1) describes, for example, the long-scale evolution of the thin liquid films. The function u(x, t) is a height of the liquid films in the point x at the time t, the fourth-order terms describe capillary force of the liquid surface tension, and the second-order terms describe the evaporation (condensation) process into the liquid (see [1-3] for more details). The investigation of the fourth-order degenerate parabolic equations of the thin liquid films was started in [4] by F. Bernis and A. Friedman (see also [2, 5-8], and the references given there). The Dirichlet problem for the Cahn-Hilliard equation (1) (N = 1, α > 0, ai1 = g1 = ε1 = fij1 = fi1 = f01 = 0, where i,j=1,n¯) was considered in [5] where γ(x) ≡ 2m, m ∈ ℕ, and b1 < 0. The corresponding Neumann problem was studied in [6]. The Neumann problem for equation (1) (N = 1, α > 0, αi1 > 0, p(x) ≡ const > 2, g1 = ε1 = fij1 = fi1 = f01 = 0, where i,j=1,n¯) was considered in [2] if γ(x) ≡ 2 and b1 > 0.

The initial-boundary value problems for the parabolic equations with variable exponents of the nonlinearity and without integral terms in equation were considered for instance in [9-15]. Integral terms (6) arise in many applications (see [16-18]). The second-order parabolic equations with variable exponents of the nonlinearity and integral term (6) were considered in [17, 19].

2 Notation and statement of theorem

Let || · ||B ≡ || ·; B|| be a norm of some Banach space B, BN : = B × . . . × B (N times) be the Cartesian product of the B, B be a dual space for B, and 〈·, ·〉B be a scalar product between B and B. We use the notation XY if the Banach space X is continuously embedded into Y; the notation X ↺̄ Y means the continuous and dense embedding; the notation XKY means the compact embedding.

If wB, z = (z1, . . ., zN) ∈ BN, and υ = (υ1, . . ., υN) ∈ BN, then we set v,w:=v1,wB,...,vN,wBRN,v,z:=k=1Nvk,zkBR,(7) and ||z; BN|| := ||z1; B|| + . . . + ||zN; B||.

Suppose that m, d ∈ ℕ, p ∈ [1, ∞], X is the Banach space, 𝖰 is a measurable set in ℝd, 𝓜(𝖰) is a set of all measurable functions υ : 𝖰 → ℝ (see [20, p. 120]), Lip (𝖰) is a set of all Lipschitz-continuous functions υ : 𝖰 → ℝ (see [21, p. 29]), Cm(𝖰) and C0(Q) are determined from [22, p. 9], Lp(𝖰) is the Lebesgue space (see [22, p. 22, 24]), Wm, p(𝖰) and W0m,p(Q) are Sobolev spaces (see [22, p. 45]), Hm(𝖰) := Wm,2(𝖰), H0m(Q):=W0m,2(Q), C([0, T]; X) and Cm([0, T]; X) are determined from [23, p. 147], Lp(0, T; X) is determined from [23, p. 155], Wm,p(0, T; X) is determined from [24, p. 286], Hm(0, T; X) := Wm,2(0, T; X), and B+(Q):={qL(Q)|essinfyQq(y)>0}. If q ∈ ℬ+(𝖰), then by definition, put q0:=essinfyQq(y),q0:=esssupyQq(y),Sq(s):=max{sq0,sq0},s0,(8) q(y):=q(y)q(y)1fora.e.yQnotethat1q(y)+1q(y)=1andqB+(Q),(9) ρq(υ;Q):=Q|υ(y)|q(y)dy,υM(Q).(10) Assume that q ∈ ℬ+(𝖰), q0 > 1, and m ∈ ℕ. The set Lq(y)(Q):={υM(Q)|ρq(υ;Q)<+} is called a generalized Lebesgue space. It is well known that Lq(y)(𝖰) is a Banach space which is reflexive and separable (see [25, p. 599, 600, 604]) with respect to the Luxemburg norm ||υ;Lq(y)(Q)||:=inf{λ>0|ρq(υ/λ;Q)1}. The set Wm,q(y)(𝖰) := {υLq(y)(𝖰) | DαυLq(y)(𝖰), |α| ≤ m} is called a generalized Sobolev space. It is well known that Wm,q(y)(𝖰) is a Banach space which is reflexive and separable (see [25, p. 604]) with respect to the norm ||υ;Wm,q(y)(Q)||:=|α|m||Dαυ;Lq(y)(Q)||.(11) The closure of C0(Q) with respect to the norm (11) is called a generalized Sobolev space and is denoted by W0m,q(y)(Q).

The generalized Lebesgue space was first introduced in [26]. The properties of the generalized Lebesgue and Sobolev spaces were widely studied in [25, 27-30].

Let us define the set ϒ(Ω) ⊂ 𝓜(Ω) as follows. For every p ∈ ϒ(Ω) there exist numbers m ∈ ℕ, s1,s1,....,sm,smR, and open sets Ω1, . . ., Ωm ⊂ Ω such that the following conditions hold:

  1. Ω1, . . ., Ωm consist of the finite numbers of the components with the Lipschitz boundaries;

  2. mesΩj=1mΩj=0;

  3. 1=s1<s2<s1<s3<s2<...<sm1<sm2<n<sm<sm1<sm=+;

  4. for every j ∈ {1, . . ., m} the inequality sjp(x)sj holds a.e. for x ∈ Ωj;

  5. for every k ∈ {1, . . ., m − 1} the inequality sk<R(sk) holds, where R(q):=nqnqif1q<n,arbitarys>1ifnq.(12)

Note that W1,q(Ω) ↺ L𝖱(q)(Ω), where q ∈ [1, +∞) (see [23, p. 47]).

Suppose thatΔ0υ:=υ,Δ1υ:=Δυ,Δrυ:=Δ(Δr1υ), HΔ2r(Ω):={υH2r(Ω)|υ|Ω=Δυ|Ω=...=Δr1υ|Ω=0},rN.(13) By definition, put Z:=HΔ2(Ω),X:=W01,p(x)(Ω),O:=Lq(x)(Ω),H:=L2(Ω), V:=ZXOH,(14) U(Q0,T):={u:(0,T)VN|Dαu[L2(Q0,T)]N,|α|=2,ux1,...,uxn[Lp(x)(Q0,T)]N,u[Lq(x)(Q0,T)]N[L2(Q0,T)]N},(15) and W(Q0,T):={wU(Q0,T)|wt[U(Q0,T)]}. We will need the following assumptions:

  • (𝐏) : p ∈ ℬ+(Ω), p0 > 1, and one of the following alternatives holds:

                (i) p ∈ ϒ(Ω);            (ii) p0 ≤ 𝖱(p0);      (iii) pC(Ω̄);

  • (Γ): γ ∈ ℬ+(Ω), γ0 > 1;

  • (𝐐): q ∈ ℬ+(Ω), q0 > 1;

  • (𝐙): α > 0, γ0 ≤ 2; s0 := min{2, p0, q0}, s0 := max{2, p0, q0}, r ∈ ℕ, and r12max2,1+n(p02)2p0,n(q02)2q0;

  • (𝐀) : aik ∈ 𝓜(𝖰0,T), 0 < a0aik(x, t) ≤ a0 < +∞ for a.e. (x, t) ∈ 𝖰0,T, where i=1,n¯,k=1,N¯;

  • (𝐁) : bk ∈ 𝓜(𝖰0,T), |bk(x, t)| ≤ b0 < +∞ for a.e. (x, t) ∈ 𝖰0,T, where k=1,N¯;

  • (𝐆): gk ∈ 𝓜(𝖰0,T), 0 < g0gk(x, t) ≤ g0 < +∞ for a.e. (x, t) ∈ 𝖰0,T, where k=1,N¯;

  • (𝐁𝐁): β1, . . ., βN ∈ ℬ+(𝖰0,T);

  • (Φ): φk ∈ Lip (ℝ), |φk(ξ)| ≤ φ0|ξ| for every ξ ∈ ℝ, where φ0 ∈ [0, +∞), k=1,N¯;

  • (𝐄) : εk ∈ 𝓜(𝖰0,T × Ω), |εk(x, t, y)| ≤ ε0 < +∞ for a.e. (x, t, y) ∈ 𝖰0,T × Ω, where k=1,N¯;

  • (𝐅) : fijkL2(𝖰0,T), fikLp′(x)(𝖰0,T), f0kLq′(x)(𝖰0,T), where i,j=1,n¯,k=1,N¯;

  • (𝐔): u0HN.

Let us introduce the following notation. If t ∈ (0, T) and if k ∈ {1, . . ., N}, then we set (Λu)k,wZ:=ΩαΔuk(x)Δw(x)dx,uZN,wZ,(16) (A(t)u)k,wX:=Ωi=1naik(x,t)|uxi(x)|p(x)2uk,xi(x)wxi(x)dx,uXN,wX,(17) (Ψ(t)u)k,wZ:=Ωbk(x,t)|u(x)|γ(x)2uk(x)Δw(x)dx,uZN,wZ,(18) (K(t)u)k,wV:=(Λu)k,wZ+(A(t)u)k,wX+(Ψ(t)u)k,wZ,uVN,wV,(19) Fk(t),wV:=Ω[i,j=1nfijk(x,t)wxixj(x)+i=1nfik(x,t)wxi(x)+f0k(x,t)w(x)]dx,wV.(20)

Using (16) and (7), we define the operator Λ : ZN → [ZN] by the rule Λu:=(Λu)1,...,(Λu)N,Λu,υZN:=k=1N(Λu)k,υkZ,uZN,υ=(υ1,....,υN)ZN. Continuing in the same way, we define the operators A(t) : XN → [XN], Ψ(t) : ZN → [ZN], and 𝒦(t) : VN → [VN], where t ∈ [0, T]. We write: F(t):=(F1(t),...,FN(t)),t[0,T],(Nw)(x,t):=((Nw)1(x,t),...,(Nw)N(x,t)),(x,t)Q0,T,(N(t)w)(x):=(Nw)(x,t),(x,t)Q0,T, where F1, . . ., FN are defined in (20), (𝒩w)1, . . ., (𝒩w)n are defined in (3). Clearly, F(t)[VN],N(t)(ONHN)[ONHN],t[0,T]. Likewise we define the operators G(t) : 𝒪N → [𝒪N], B(t) : HNHN, and E(t) : HNHN, where t ∈ [0, T].

For the sake of convenience we have denoted φ(Eu) = (φ1((Eu)1), . . ., φN((Eu)N)) and φk(Euk(t)) = φk((Eu)k(t)), k=1,N¯. By definition, put (u,υ)Ω:=Ωu(x)υ(x)dxifu:ΩRN,υ:ΩR,Ω(u(x),υ(x))RNdxifu,υ:ΩRN,(21)

Definition 2.1

A real-valued function uW(𝖰0,T) ∩ C([0, T]; HN) is called a weak solution of problem ((1), (2)) if u satisfies (2) and for every υU(𝖰0,T) we have ut,υU(Q0,T)+0TK(t)u(t),υ(t)VN+(N(t)u(t),υ(t))Ωdt=0TF(t),υ(t)VNdt.(22)

Theorem 2.2

Suppose that conditions (𝑷)-(𝑼) and ∂Ω ∈ C2r are satisfied. Then problem ((1), (2)) has a weak solution.

3 Auxiliary facts

3.1 Properties of generalized Lebesgue and Sobolev spaces

The following Propositions are needed for the sequel.

Proposition 3.1

(see [31, p. 31]). If q ∈ ℬ+(𝖰) and q0 > 1, then for every η > 0 there exists a number Yq(η) > 0 such that for every a, b ≥ 0 and for a.e. y ∈ 𝖰 the generalized Young inequality abηaq(y)+Yq(η)bq(y)(23) holds. In addition, Yq(η) depends on q0, q0 and it is independent of y, Y2(η)=14η,Y212=12, Yq(+0) = +∞, and Yq(+∞) = 0.

Proposition 3.2

Assume that q ∈ ℬ+(𝖰) and q0 > 1. Then the following statements are satisfied:

  1. (see [25, p. 600]) if q(y) ≥ r(y) ≥ 1 for a.e. y ∈ 𝖰, then Lq(y)(𝖰) ↺ Lr(y)(𝖰) and ||υ;Lr(y)(Q)||(1+mesQ)||υ;Lq(y)(Q)||,υLq(y)(Q);

  2. (see [30, p. 431]) for every uLq(y)(𝖰) and υLq′(y)(𝖰) we get uυL1(𝖰) and the following generalized Hölder inequality is true Ω|u(y)υ(y)|dy2||u;Lq(y)(Q)||||υ;Lq(y)(Q)||.(24)

Proposition 3.3

(see [32, p. 168]). Suppose that q ∈ ℬ+(𝖰), q0 ≥ 1, Sq is defined by (8), and ρq is defined by (10). Then for every υ ∈ 𝓜(𝖰) the following statements are fulfilled:

  1. ||υ; Lq(y)(𝖰)|| ≤ S1/q(ρq(υ; 𝖰)) if ρq(υ; 𝖰) < +∞;

  2. ρq(υ; 𝖰) ≤ Sq(||υ; Lq(y)(𝖰)||) if ||υ; Lq(y)(𝖰)|| < +∞.

Proposition 3.4

Suppose that p ∈ ℬ+(Ω) and p0 > 1. Then the following statements hold:

  1. (see Theorem 3.10 [25, p. 610] and Theorem 2.7 [30, p. 443]) if either p ∈ ϒ(Ω)or pC(Ω̄), then ||υ;W01,p(x)(Ω)||=i=1n||υxi;Lp(x)(Ω)|| is a equivalent norm of W01,p(x)(Ω);

  2. (see Lemma 5 [13, p. 48] and Theorem 3.1 [27, p. 76]) if ux1, . . ., uxnLp(x)(Ω) and either p ∈ ϒ(Ω) or p0 ≤ 𝖱(p0) (see (12)), then uLp(x)(Ω) and the generalized Poincaré inequality ||u;Lp(x)(Ω)||C1i=1n||uxi;Lp(x)(Ω)||+||u;L1(Ω)||, holds, where C1 > 0 is independent of u;

  3. (see Lemma 2 [13, p. 46] and Theorem 3.2 [27, p. 77]) Lp0(0,T;Lp(x)(Ω))¯Lp(x)(Q0,T)¯Lp0(0,T;Lp(x)(Ω)).(25)

3.2 Auxiliary functional spaces

Let ℒ(X, Y) be a space of bounded linear operators from X into Y (see [33, p. 32]), (·, ·)H be the Cartesian product in the Hilbert space H, and HΔ2r(Ω) is defined in (13), where r ∈ ℕ. It is easy to verify that HΔ2r(Ω) is the Hilbert space such that HΔ2r(Ω)H2r(Ω),HΔ2r(Ω)¯L2(Ω)¯[HΔ2r(Ω)].(26) If ∂Ω ⊂ C1, then the following integration by parts formula is true ΩυΔrudx=ΩuΔrυdx,u,υHΔ2r(Ω).(27) Note that for every r ∈ ℕ the space HΔ2r(Ω) is reflexive.

Let {wj}j∈ℕ be a set of all eigenfunctions of the problem Δwj=λjwjinΩ,wj|Ω=0,jN.(28) Here {λj}j∈ℕ ⊂ ℝ+ is the set of the corresponding eigenvalues. Suppose that {wj}j∈ℕ is an orthonormal set in L2(Ω). It is easy to verify that solutions to problem (28) satisfy the equalities (1)rΔrw=λrw,w|Ω=Δw|Ω=...=Δr1w|Ω=0.(29) The following propositions are needed for the sequel.

Proposition 3.5

(see Theorem 8 [34, p. 230]). If ∂Ω ⊂ C2r, then the set {wj}j∈ℕ of all eigenfunction of the problem, (28) is a basis for the space HΔ2r(Ω).

Proposition 3.6

(see Lemma 3 [34, p. 229]). If ∂Ω ⊂ C2r, then there exists a constant C2 > 0 such that for all υHΔ2r(Ω) we obtain ||υ;H2r(Ω)||C2||Δrυ;L2(Ω)||.(30) Define Wr:=[HΔ2r(Ω)]N,Wr:=[Wr],(31) where r is determined from condition (𝐙). We consider the space VN (see (14) ) with respect to the norm ||υ;VN||:=||Δυ;HN||+i=1n||υxi;[Lp(x)(Ω)]N||+||υ;ON||+||υ;HN||. Since r satisfies (𝐙) and (14) holds, it is easy to verify that Wr¯VN¯HN[HN]¯[VN]¯Wr.(32) The following Lemma is needed for the sequel.

Lemma 3.7

L(0, T; HN) ∩ C([0, T]; [VN]) = C([0, T]; HN).

The proof is omitted (see for comparison Lemma 8.1 [35, p. 307]).

We consider the space U(Q0,T) (see (15) ) with respect to the norm ||u;U(Q0,T)||:=i,j=1n||uxixj;[L2(Q0,T)]N||+i=1n||uxi;[Lp(x)(Q0,T)]N||+||u;[Lq(x)(Q0,T)]N||+||u;[L2(Q0,T)]N||. It is easy to verify that the space U(Q0,T) is reflexive. Taking into account the embedding of type (25) and inequality (30), we obtain Ls0(0,T;VN)¯U(Q0,T)¯Ls0(0,T;VN),(33) where s0 and s0 are determined from condition (𝒁). Whence, Ls0s01(0,T;[VN])¯[U(Q0,T)]¯Ls0s01(0,T;[VN]).(34) Similarly, using (32) we obtain Ls0(0,T;Wr)¯U(Q0,T)¯[L2(Q0,T)]N¯[U(Q0,T)]¯Ls0s01(0,T;Wr).(35) Hence an arbitrary element of the spaces [U(Q0,T)] or U(Q0,T) belongs to D((0, T); [VN]). Therefore, we have distributional derivative of uU(Q0,T) ⊂ D((0, T); [VN]). Together with (34), we conclude that an arbitrary element w ∈ [U(Q0,T)] belongs to Ls0s01(0,T;[VN]). Thus, if uU(Q0,T) belongs to Ls0(0, T; VN), then w,uU(Q0,T)=0Tw(t),υ(t)VNdt. In particular, this equality is true if uC([0, T]; VN).

Lemma 3.8

Suppose that conditions (𝑷) and (𝑸) are satisfied, uU(Q0,T), {wμ}μ∈ℕ is a basis for the space V. Then for every ɛ > 0 there exist a number m ∈ ℕ and functions {φμk}μ=1,k=1m,NC([0,T]) such that ||uψm; U(Q0,T)|| < ɛ, where ψm = (ψm1, . . ., ψmN) and ψmk(x,t)=μ=1mφμk(t)wμ(x),(x,t)Q0,T,k=1,N¯.

The proof is omitted (see for comparison [36, p. 5] and [13, 27]).

3.3 Projection operator

Let ℋ be the Hilbert space and 𝒱 be the reflexive separable Banach space such that V¯HH¯V.(37) Notice that if g ∈ 𝒱 and g ∈ ℋ, then g,υV=(g,υ)H,υV.(37) Suppose {wj}j∈ℕ is a orthonormal basis for the space ℋ, m ∈ ℕ is a fixed number, 𝔐 is a set of all linear combinations of the elements from {w1, . . ., wm}, 𝔐 is a orthogonal complements of 𝔐 (see [37, p. 476]). Then (see [37, p. 526]) 𝔐 is a closed subset of ℋ and ℋ = 𝔐 ⊕ 𝔐.

Define an unique orthogonal projection Pm : ℋ → 𝔐 by the rule (see [37, p. 527]) Pmh:=j=1m(h,wj)Hwj,hH.(38) This is a linear self-adjoint continuous operator (see Theorem 7.3.6 [37, p. 515]) such that ||Pmh||H||h||H,hH.(39) If {wj}j∈ℕ ⊂ 𝒱, then let us define an operator P^m:VV (not necessarily self-adjoint) by the rule P^mυ:=PmυforeveryυV.(40) We shall find a conjugate operator P^m:VV. Take elements υ ∈ 𝒱, z ∈ 𝒱. Then z,PmυV=z,j=1m(υ,wj)HwjV=j=1m(υ,wj)Hz,wjV=υ,j=1mz,wjVwjH. Since υ, w1, . . ., wm ∈ 𝒱, (37) yields that υ,j=1mz,wjVwjH=j=1mz,wjVwj,υH=j=1mz,wjVwj,υV. Thus, z,PmυV=P^mz,υV, where P^mz=j=1mz,wjVwj,zV.(41) In addition, (41) implies that P^m(V)V.

Lemma 3.9

Assume that {wj}j∈ℕ is a orthonormal basis for the spacesuch that {wj}jNV,ψ1m,...,ψmmR are some numbers, and F ∈ 𝒱. Then zm=s=1mψsmwsV satisfies zm,w1V=F,w1V,zm,wmV=F,wmV,(42) iff the following equality holds zm=P^mFinV.(43)


Clearly, (43) implies (42). We shall prove that (42) implies (43). Take υ ∈ 𝒱. There exist numbers α1m,...,αmmRsuchthatPmυ=P^mυ=μ=1mαμmwμ. Multiplying both sides of μ-th equality of (42) by αμm and summing the obtained equalities, we get zm,P^mυV=F,P^mυV.Hence,P^mzm,υV=P^mF,υV for every υ ∈ 𝒱. Thus, P^mzm=P^mFinV.(44) Taking into account (37), the inclusions zm, w1, . . ., wm ∈ 𝒱, and the orthonormality condition for {wj}j∈ℕ ⊂ ℋ, from (41) we obtain P^mzm=j=1mzm,wjVwj=j=1ms=1mψsmws,wjHwj=s,j=1mψsm(ws,wj)Hwj=s=1mψsmws=zm. Therefore, (42) yields (43). ☐

In the sequel, we only consider the case ℋ = L2(Ω), 𝒱 = HΔ2r(Ω) (see (13) ), and {wj}j∈ℕ is determined from problem (28). Then (38) implies that (see (21) ) (Pmu)(x)=j=1m(u,wj)Ωwj(x),xΩ,u:ΩR.(45) This operator Pm : L2(Ω) → L2(Ω) is a linear self-adjoint continuous projection operator such that ||Pm||ℒ(L2(Ω),L2(Ω))= 1.

To prove that m belongs to L(HΔ2r(Ω),HΔ2r(Ω)), we take υHΔ2r(Ω). Then ΔrmυL2(Ω) and Corollary 6.2.10 [38, p. 171] implies that there exists a function hL2(Ω) such that ||h||L2(Ω) = 1 and (h, Δrmυ)L2(Ω) = ||Δrmυ||L2(Ω). By (45), (40), (29), and (27) we obtain ||P^mυ||HΔ2r(Ω)=||ΔrP^mυ||L2(Ω)=(h,ΔrP^mυ)L2(Ω)=h,Δrj=1m(v,wj)ΩwjΩ=h,j=1m(υ,wj)ΩΔrwjΩ=h,j=1m(υ,wj)Ω(1)rλjrwjΩ=h,j=1m(υ,(1)rλjrwj)ΩwjΩ=h,j=1m(υ,Δrwj)ΩwjΩ=j=1m(υ,Δrwj)Ω(h,wj)Ω=υ,j=1m(h,wj)ΩΔrwjΩ=(υ,ΔrP^mh)Ω=(Δrυ,P^mh)Ω=(Δrυ,Pmh)Ω. Using Cauchy-Bunyakowski-Schwarz’s inequality and estimating (39) with ℋ = L2(Ω), we show that |(Δrυ, Pmh)Ω| ≤ ||Δrυ||L2(Ω)||Pmh||L2(Ω) ≤ ||Δrυ||L2(Ω)||h||L2(Ω). Therefore, ||P^mυ||HΔ2r(Ω)||υ||HΔ2r(Ω),υHΔ2r(Ω).(46) Suppose now that fLs(0, T; ℋ), s > 1. If Pm : ℋ → 𝔐 is determined from (38), then Pmf(t) ∈ ℋ for every t ∈ [0, T], Pmf(t)=j=1m(f(t),wj)Hwj,(47) and from (39) we get 0T|Pmf(t)|Hsdt0T|f(t)|Hsdt, i.e. ||Pmf;Ls(0,T;H)||||f;Ls(0,T;H)||,fLs(0,T;H).(48) Finally assume that m : 𝒱 → 𝒱 is determined from (40), ℋ = L2(Ω), and 𝒱 = HΔ2r(Ω). Taking into account (46) and (48), we have that ||P^mu;Ls(0,T;HΔ2r(Ω))||||u;Ls(0,T;HΔ2r(Ω))||,uLs(0,T;HΔ2r(Ω)),s1.(49) Clearly, we can prove (38)-(49) if we replace L2(Ω), HΔ2r(Ω) by [L2(Ω)]N, [HΔ2r(Ω)]N respectively.

3.4 Differentiability of the nonlinear expressions

Take a function σ ∈ 𝓜(Ω) and by definition, put ψσ(x)(s):=sσ(x)ifs>0,0ifs0,xΩ.(50) Similarly to Theorem A.1 [39, p. 47], we obtain that if υW1,p(0, T; Lp(Ω)) (1 ≤ p ≤ ∞), then υ+ := max{u, 0} ∈ W1,p(0, T; Lp(Ω)) and (υ+)t = χ̃(υ)υt almost everywhere in Q0,T, where χ~(s):=1ifs>0,0ifs0.(51) The function υ := max{−u, 0} has a similar property.

The following Propositions are needed for the sequel.

Proposition 3.10

(see Theorem 2 [24, p. 286]). If X is a Banach space and 1 ≤ p ≤ ∞, then W1,p(0, T; X) ↺ C([0, T]; X) and the following integration by parts formula holds: sτut(t)dt=u(τ)u(s),0s<τT,uW1,p(0,T;X).(52)

Proposition 3.11

(the Aubin theorem, see [40] and [41, p. 393]). If s, h > 1 are fixed numbers, 𝒲, ℒ, ℬ are the Banach spaces, and WKLB, then {uLs(0,T;W)|utLh(0,T;B)}KLs(0,T;L)C([0,T];B).

Lemma 3.12

Suppose that Ω ⊂ ℝn is a bounded C0,1 -domain. Then the integration by parts formula Qs,τwtzdxdt=Ωtwzdxt=st=τQs,τwztdxdt,0s<τT,(53) holds if one of the following alternatives hold:

  1. wLq(x)(Q0,T), where q ∈ ℬ+(Ω) and q0 > 1, wtL1(Q0,T), zL(Q0,T), ztLq′(x)(Q0,T);

  2. w, wtL1(Q0,T), z, ztL(Q0,T).


(i). Take W := {wLq(x)(Q0,T) | wtL1(Q0,T)}, Z := {zL(Q0,T) | ztLq′(x)(Q0,T)}. If ϕC1([0, T]) and zZ, then φzW1,1(0,T;Lq0q01(Ω)). Using (52) with υ = ϕ(t)z(x, t), we get sτφt(t)z(x,t)dt=φ(τ)z(x,τ)φ(s)z(x,s)sτφ(t)zt(x,t)dt,xΩ.(54) Take a function υC1(Ω). By (54), we obtain that Qs,τφtυzdxdt=Ωtφυzdxt=st=τQs,τφυztdxdt.(55) Clearly, C1([0, T]; C1(Ω̄)) ↺̄ W ↺̄ W1,1(0,T; L1(Ω)). Then the set {i=1mφi(t)υi(x)|mN,φ1,...,φmC1([0,T]),υ1,...,υmC1(Ω¯)} is dense in W and (55) yields (53).

We shall omit the proof of (ii) because it is analogous to the previous one. ☐

Lemma 3.13

Suppose that σ ∈ ℬ+(𝖰), p, q ∈ ℬ+(𝖰), p0, q0 > 1, p(y) ≥ σ(y) and q(y)p(y)σ(y) for a.e. y ∈ 𝖰, and ψσ(y) is determined from (50) if we replace σ(x) by σ(y). Then for every uLp(y)(𝖰) we have that ψσ(y)(u)Lp(y)σ(y)(Q), ρp/σ(ψσ(y)(u);Q)ρp(u;Q),(56) ||ψσ(y)(u);Lq(y)(Q)||C3Sσ/p(ρp(u;Q)),(57) where C3 > 0 is independent of u.


Clearly, p(y)σ(y)1fora.e.yQ,|ψσ(y)(u)|p(y)σ(y)=|u+|p(y)|u|p(y)L1(Q). Then by [42, p. 297], we obtain ψσ(y)(u)Lp(y)σ(y)(Q). Moreover, (56) and ||ψσ(y)(u);Lq(y)(Q)||C4||ψσ(y)(u);Lp(y)σ(y)(Q)||C4Sσ/pρp/σ(ψσ(y)(u);Q) hold. This inequality and (56) imply (57). ☐

Lemma 3.14

Suppose that p ∈ ℬ+(𝖰), p0 > 1, θ ∈ 𝓜(Ω × ℝ), for a.e. x ∈ Ω the function ℝ ∋ ξθ(x, ξ) ∈ ℝ is continuously differentiable, and there exists a number M > 0 such that |θ(x,ζ)θ(x,η)|M|ζη|,|θξ(x,ξ)|M(58) for a.e. x ∈ Ω and for every ζ, η, ξ ∈ ℝ. If u, utLp(x)(Q0,T), then θ(x, u), (θ(x, u))tLp(x)(Q0,T) and (θ(x,u))t=θξ(x,u)ut.(59)


Since u, utLp(x)(Q0,T), there exists a sequence {um}mNC1(Q0,T¯)suchthatummuandutmmut strongly in Lp(x)(Q0,T) and almost everywhere in Q0,T. Clearly, (θ(x,um(x,t)))t=limh0θ(x,um(x,t+h))θ(x,um(x,t))um(x,t+h)um(x,t)um(x,t+h)um(x,t)h=θξ(x,um(x,t))utm(x,t), where (x, t) ∈ Q0,T, m ∈ ℕ. In addition, |θ(x, um) − θ(x, u)| ≤ M|umu|. Hence, θ(x,um)mθ(x,u) strongly in Lp(x)(Q0,T) and so θ(x, u) ∈ Lp(x)(Q0,T).

Clearly, θξ(x,um)utmθξ(x,u)ut=Am+Bm, where Am=θξ(x,um)(utmut),Bm=(θξ(x,um)θξ(x,u))ut. On the other hand, |Am|p(x)Mp(x)|utmut|p(x)m0inL1(Q0,T).ThenAmm0 in Lp(x)(Q0,T). Moreover, |Bm|p(x) ≤ (2M|ut|)p(x)L1(Q0,T), Bmm0 almost everywhere in Q0,T, and Bmm0 in Lp(x)(Q0,T). Therefore, θξ(x,um)utmmθξ(x,u)ut in Lp(x)(Q0,T) and so θξ(x, u)utLp(x)(Q0,T).

Finally let us prove (59). Take a function φC0(Q0,T). Then (59) holds because Q0,Tθξ(x,u)utφdxdt=limmQ0,Tθξ(x,um)utmφdxdt=limmQ0,T(θ(x,um))tφdxdt=limmQ0,Tθ(x,um)φtdxdt=Q0,Tθ(x,u)φtdxdt. ☐ Notice that Lemma 3.14 generalizes the results of Lemma 3 [43, p. 18], where the case θ(x, u) = θ(u) was considered.

Corollary 3.15

Suppose that −∞ < a < b < +∞ and one of the following alternatives holds: (i) I = [a, b]; (ii) I = [a, +∞); (iii) I = (−∞, b]. Assume also that p ∈ ℬ+(Ω), p0 > 1, θ ∈ 𝓜 (Ω × I), a.e. for x ∈ Ω the function Iξθ(x, ξ) ∈ ℝ is continuously differentiable, and there exists a number M > 0 such that a.e. or x ∈ Ω and for every ζ, η, ξI, (58) holds. If u, utLp(x)(Q0,T) and u(x, t) ∈ I a.e. for (x, t) ∈ Q0,T, then θ(x, u), (θ(x, u))tLp(x)(Q0,T) and (59) holds.


For the sake of convenience, only the case I = (−∞, b] is considered (see for comparison [44, p. 98]). Let us extend θ outside I as follows Θ(x,ξ):=θ(x,ξ)ifξb,θξ(x,b)ξ+θ(x,b)θξ(x,b)bifξ>b,xΩ. Then Θ satisfies the conditions of Lemma 3.14 and Θ(x, u(x, t)) = θ(x, u(x, t)) for a.e. (x, t) ∈ Q0,T. This completes the proof. ☐

Lemma 3.16

Suppose that p ∈ ℬ+(Ω), p0 > 1, θ ∈ 𝓜(Ω × ℝ), for a.e. x ∈ Ω the function ℝ ∋ ξθ(x, ξ) ∈ ℝ is continuous and the function ℝ \ {ξ1, . . ., ξN} ∋ ξθ(x, ξ) ∈ ℝ is differentiable, and (58) holds for a.e. x ∈ Ω, where (ζ, η ∈ ℝ, ξ ∈ ℝ \ {ξ1, . . ., ξN}. If u, utLp(x) (Q0,T), then θ(x, u), (θ(x, u))tLp(x)(Q0,T) and (59) holds.


For the sake of convenience, only the case N = 1 and ξ1 = 0 is considered (see for comparison [44, p. 100]). It is easy to verify that θ(x,u):=θ(x,u+)+θ(x,u)θ(x,0).(60) Since u, utLp(x)(Q0,T) ⊂ Lp0(Q0,T), we have that (u±)tLp0 (Q0,T) and (u±)t = ±χ̃(u)ut, where χ̃ is determined from (51). Then by Corollary 3.15, we obtain the formulas of type (59) for every term in (60). Therefore, (59) holds. By (58) and (59), we get (θ(x, u))tLp(x)(Q0,T). ☐

Lemma 3.17

Suppose that β ∈ ℬ+(Ω), ψβ(x) is determined from (50) if we replace σ by γ, and χk(s):=1ifs>1k,0ifs1k,kN.(61) If uC1(Q0,T¯)andυ,υtL1(Q0,T), then limk+Q0,Tχk(u)β(x)ψβ(x)1(u)utυdxdt=Ωtψβ(x)(u)υdx|t=0t=TQ0,Tψβ(x)(u)υtdxdt.(62)


By definition, set ψβ(x),k(s):=kβ(x)ifsk,sβ(x)if1k<s<k,1kβ(x)ifs1k,ξ~β(x),k(s):=β(x)sβ(x)1if1k<s<k,0ifs1kandsk, k ∈ ℕ, k ≥ 2, x ∈ Ω. Clearly, ψβ(x),k(s)kψβ(x)(s), where s ∈ ℝ, x ∈ Ω. In addition, for k ∈ ℕ (k ≥ 2) and x ∈ Ω the function sψβ(x),k(s) has the Lipschitz property in ℝ and it is not differentiable only in the point s=1kands=k.Moreover,sψβ(x),k(s)=ξ~β(x),k(s)ifs1kandsk. Whence, by Lemma 3.16, we obtain (ψβ(x),k(u))t=ξ~β(x),k(u)utalmosteverywhereinQ0,T.(63) Thus, ψβ(x),k(u), (ψβ(x),k(u))tL(Q0,T). Using case (ii) of Lemma 3.12 with z = ψβ(x),k(u) and w = υ, we get (53), i.e. Q0,T(ψβ(x),k(u))tυdxdt=Ωψβ(x),k(u)υdx|t=0t=TQ0,Tψβ(x),k(u)υtdxdt.(64) Let M:=max(x,t)Q0,T¯|u(x,t)|,k0N,k0max{2,M}. Since |u| ≤ Mk0k, from (63) we have (ψβ(x),k(u))t=ξ~β(x),k(u)ut=χt(u)β(x)ψβ(x)1(u)ut, where kk0. By |ψβ(x),k(u(x,t))|Mβ(x)(x,t)Q0,T¯ and Lebesgue’s Dominate Convergence Theorem (see [33, p. 90]), we obtain limk+Ωtψβ(x),k(u)υdx=Ωtψβ(x)(u)υdxift=0andt=T,limk+Q0,Tψβ(x),k(u)υtdxdt=Q0,Tψβ(x)(u)υtdxdt. Therefore, (62) follows from (64). ☐

Theorem 3.18

Suppose that σ ∈ ℬ+(Ω), σ0 > 1, and the function ψσ(x) is determined from (50). Then the following statements are satisfied:

  1. if uC1(Q0,T¯), then ψσ(x)(u), (ψσ(x)(u))tL(Q0,T) and (ψσ(x)(u))t=σ(x)ψσ(x)1(u)ut;(65)

  2. if u, utLp(x)(Q0,T), where pL+(Ω) and p(x) ≥ σ(x) for a.e. x ∈ Ω, then ψσ(x)(u),(ψσ(x)(u))tLp(x)σ(x)(Q0,T), equality (65) is true, and the estimate ρp/σ(ψσ(x)(u))t;Q0,TC5S1/σρp(u;Q0,T)S1/σρp(ut;Q0,T)(66) holds, where C5 > 0 is independent of u.


First let us prove Case 1. Take a function uC1(Q0,T¯).Ifυ,υtC(Q0,T¯),χk is determined from (61), and k ∈ ℕ, then |χk(u)σ(x)ψσ(x)−1(u)utυ| ≤ C6, where C6 > 0 is independent of k, x, t. Hence, Lebesgue’s Dominate Convergence Theorem (see [33, p. 90]) yields that limk+Q0,Tχk(u)σ(x)ψσ(x)1(u)utυdxdt=Q0,Tσ(x)ψσ(x)1(u)utυdxdt. Using (62) with β = σ > 1, we obtain Q0,Tσ(x)ψσ(x)1(u)utυdxdt=Ωtψσ(x)(u)υdx|t=0t=TQ0,Tψσ(x)(u)υtdxdt.(67) Taking in (67) the function υC0(Q0,T), we get Q0,Tσ(x)ψσ(x)1(u)utυdxdt=Q0,Tψσ(x)(u)υtdxdt (notice that σψσ(x)−1(u)utL(Q0,T) becauseσ > 1). Therefore, (65) holds.

Since σ0 > 1, from (50) we have ψσ(x)L(Q0,T) and from (65) we have (ψσ(x)(u))tL(Q0,T).

Now let us prove Case 2. Suppose uU, where U := {uLp(x)(Q0,T) | utLp(x)(Q0,T)}.

Clearly, C1([0, T]; C1(Ω̅)) ↺̄ (5 W1,p0(0, T; Lp(x)(Ω)) ↺̄ U ↺̄ W1,p0(0, T; Lp(x)(Ω)). Then there exists a sequence {um}mNC1(Q0,T¯)suchthatummuandutmmutstronglyinLp(x)(Q0,T),ummu in C([0, T]; Lp(x)(Ω)).

Assume that υ,υtC(Q0,T¯). By (67), for every m ∈ ℕ we obtain Q0,Tσ(x)ψσ(x)1(um)utmυdxdt=Ωψσ(x)(um)υdxt=0t=TQ0,Tψσ(x)(um)υtdxdt.(68) Since 1 < σ(x) ≤ p(x), we get p(x)σ(x)1>1 for a.e. x ∈ Ω. Therefore, ψσ(x)1(um)mψσ(x)1(u)stronglyinLp(x)σ(x)1(Q0,T). Clearly, [Lp(x)σ(x)1(Q0,T)]Lp(x)p(x)(σ(x)1)(Q0,T). Since p(x) ≥ (σ(x) − 1) + 1, we have thatp(x)p(x)p(x)(σ(x)1) for a.e. x ∈ Ω. Therefore, utmmutstronglyinLp(x)p(x)(σ(x)1)(Q0,T). By Lemma 5.2 [23, p. 19], we obtain Q0,Tσ(x)ψσ(x)1(um)utmυdxdtmQ0,Tσ(x)ψσ(x)1(u)utυdxdt(69) and σψσ(x)−1(u)utL1(Q0,T). It is easy to verify that ψσ(x)(um(t))mψσ(x)(u(t))stronglyinLp(x)σ(x)(Ω)fort=0andt=T,(70) ψσ(x)(um)mψσ(x)(u)stronglyinLp(x)σ(x)(Q0,T).(71) Letting m → ∞ in (68) and using (69)-(71), we get (67) and (65).

By Lemma 3.13, we get ψσ(x)(u)Lp(x)σ(x)(Q0,T). By (65) and generalized Young’s inequality, we obtain |(ψσ(x)(u))t|p(x)σ(x)σ(x)p(x)σ(x)|u|p(x)σ(x)|ut|p(x)σ(x)C7(|u|p(x)+|ut|p(x))L1(Q0,T). Thus, (ψσ(x)(u))tLp(x)σ(x)(Q0,T).

By (65) and the generalized Hölder’s inequality, we obtain that Q0,T|(ψσ(x)(u))t|p(x)σ(x)dxdtC8|||u|p(x)σ(x);Lσ(x)(Q0,T)|||||ut|p(x)σ(x);Lσ(x)(Q0,T)||C9S1/σQ0,T|u|p(x)dxdtS1/σQ0,T|ut|p(x)dxdt. This implies (66) and completes the proof of Theorem 3.18. ☐

Note that the case σ(x) ≡ σ ∈ (0, 1] is considered in [45].

Theorem 3.19

Suppose that r ∈ ℬ+(Ω). Then the following statements are satisfied:

  1. If r0 > 1, then the equality (|u|r(x))t=r(x)|u|r(x)2uut(72) is true if one of the following alternatives holds:

    1. uC1(Q0,T¯) (here we have |u|r(x), (|u|r(x))tL(Q0,T));

    2. u, utLp(x)(Q0,T) and p(x) ≥ r(x) for a.e. x ∈ Ω (here we have |u|r(x),(|u|r(x))tLp(x)r(x)(Q0,T)).

  2. If r0 > 2, then the equality (|u|r(x)2u)t=(r(x)1)|u|r(x)2ut(73) is true if one of the following alternatives hold:

    1. uC1(Q0,T¯) (here we have |u|r(x)-2u, (|u|r(x)-2u)tL(Q0,T));

    2. u, utLp(x)(Q0,T) and p(x) ≥ r(x) − 1 for a.e. x ∈ Ω (here |u|r(x)2u,(|u|r(x)2u)tLp(x)r(x)1(Q0,T)).


Suppose that ψr(x)−2 is determined from (50) if we replace σ by r − 2. Then the proof follows from Theorem 3.18 since |s|r(x)=ψr(x)(s)+ψr(x)(s),|s|r(x)2s=ψr(x)1(s)ψr(x)1(s),xΩ,sR.

3.5 Cauchy’s problem for system of ordinary differential equations

Take Q = (0, T) × ℝ, where ℓ ∈ ℕ. In this section, we seek a weak solution ϕ : [0, T] → ℝ of the problem φ(t)+L(t,φ(t))=M(t),t[0,T],φ(0)=φ0,(74) where M : [0, T] → ℝ, L : Q → ℝ are some functions (for the sake of convenience we have assumed that L(t, 0) = 0 for every t ∈ [0, T]) and φ0=(φ10,...,φ0) ∈ ℝ.

The following Definitions are needed for the sequel.

Definition 3.20

A real-valued function ϕW1,1 (0, T; ℝk) is called a weak solution of problem (74) if u satisfies the initial value condition and satisfies the equation almost everywhere.

Definition 3.21

We shall say that a function L : Q → ℝ satisfies the Carathéodory condition if for every ξ ∈ ℝ the function (0, T) ∋ tL(t, ξ) ∈ ℝ is measurable and if for a.e. t ∈ (0, T) the functionξL(t, ξ) ∈ ℝ is continuous.

Definition 3.22

(see [46, p. 241]). We shall say that a function L : Q → ℝ satisfies the Lp-Carathéodory condition if L satisfies the Carathéodory condition and for every R > 0 there exists a function hRLp (0, T) such that |L(t,ξ)|hR(t)(75) a.e. for t ∈ (0, T) and for every ξDR¯ := {y ∈ ℝ | |y| ≤ R}.

Proposition 3.23

(Gronwall-Bellman’s Lemma [47, p. 25]). Suppose that A, BL1 (0, T) and yC([0, T]) are nonnegative functions. If for every τ ∈ [0, T] we have y(τ)C+0τ[A(t)y(t)+B(t)]dt,(76) where C is a nonnegative number, then the following inequality is true y(τ)(C+0τB(t)e0tA(s)dsdt)e0τA(t)dt,τ[0,T].(77) We will need the following Theorem.

Theorem 3.24

(Carathéodory-LaSalle’s Theorem). Suppose that p ≥ 2, function L : Q → ℝ satisfies Lp-Carathéodory condition, MLp (0, T; ℝ), and ϕ0 ∈ ℝ. If there exists a nonnegative functions α, βL1 (0, T) such that for every ξ ∈ ℝ and for a.e. t ∈ [0, T] the inequality (L(t,ξ),ξ)Rα(t)|ξ|2β(t)(78) holds, then problem (74) has a global weak solution ϕW1,p (0, T; ℝ).


We modify the method employed in the proof of Theorem 3 [48, p. 240]. According to the Carathéodory Theorem [49, p. 17], we have a local weak solution ϕW1,p(0, b; ℝ) (b ∈ (0, T]) to the Cauchy problem (74) such that for every τ ∈ [0, b] the equality φ(τ)=φ0+0τM(t)dt0τL(t,φ(t))dt(79) holds. If b = T, then Theorem 3.24 is proved. If b < T, then we take ϕ1 := ϕ(b) and consider the equation from (74) with new initial value condition ϕ(b) = ϕ1. Using the Carathéodory Theorem and (79), we extend solution to problem (74) into [b, b1], where b1T etc. Thus, similarly to [50, p. 22-24], we have one of the following possibility:

  1. solution to problem (74) can be extended into [0, T];

  2. there exists a weak solution to problem (74) which is defined on right maximal interval of existence [0, ), where T.

We shall prove that Case 2 is impossible. Assume the converse. Then for every τ ∈ (0, ) this local weak solution ϕ belongs to W1,p (0, τ; ℝ). Define R:={(|φ0|2+0T[2β(t)+|M(t)|2]dt)e0T[2α(t)+1]dt}1/2,(80) where α and β are determined from (78). Since L satisfies the Lp-Carathéodory condition and R is determined from (80), there exists a function hRLp(0, T) such that for a.e. t ∈ (0, T) and for every ξDR¯ := {y ∈ ℝ ||y| ≤ R} inequality (75) holds.

Taking into account (see (78)) the following inequalities (L(t,φ(t)),φ(t))Rα(t)|φ(t)|2β(t),(M(t),φ(t))R|M(t)||φ(t)|12|M(t)|2+12|φ(t)|2, from (74) we get (φ(t),φ(t))Rα(t)|φ(t)|2β(t)12|M(t)|2+12|φ(t)|2,t[0,b¯). Hence, 0τ(φ(t),φ(t))Rdt0τ[(α(t)+12)|φ(t)|2+β(t)+12|M(t)|2]dt,τ|(0,b¯).(81) Since ϕW1,p(0, τ; ℝ) and p ≥ 2, we obtain |φ|2W1,p2(0,τ),(|φ(t)|2)=2(φ(t),φ(t))R,t(0,τ), (see Case 1.ii of Theorem 3.19). Hence Proposition 3.10 implies that 0τ(φ(t),φ(t))Rdt=12|φ(τ)|212|φ(0)|2. Whence (81) has a form (76), where C = |ϕ0|2, y(t)=|φ(t)|2,A(t)=2α(t)+1,B(t)=2β(t)+|M(t)|2,t(0,τ). Therefore, from (77) we get y(τ)(C+0τB(t)e0tA(s)dsdt)e0τA(t)dt(C+0τB(t)dt)e0τA(t)dtR2, where R is determined from (80). Thus |ϕ(τ)| ≤ R, τ ∈ (0, ), i.e. the point ϕ(t) belongs to DR, where t ∈ (0, ). By (75), we have that |L(t, ϕ(t))| ≤ hR(t), where t ∈ (0, ). Therefore, (79) yields that |φ(t2)φ(t1)|=t1t2L(t,φ(t))dtt1t2hR(t)dtt1,t2b¯00. Finally we have an existence of the finite limit limtb¯0φ(t). Then solution to problem (74) can be extended to [0, ] by the rule φ(b¯):=limtb¯0φ(t)<. This contradiction completes the proof Theorem 3.24. ☐

If L is slowly continuous with respect to the ϕ, then Theorem 3.24 follows from Theorem 3 [48, p. 240]. If M ≡ 0 and L is continuous, then Theorem 3.24 coincides with Lemma 4 [51, p. 67].

3.6 Some integral expressions

The following lemmas will be needed in the sequel.

Lemma 3.25

(see for comparison Lemma 2.3 [31, p. 26]). Suppose that condition (Q) is satisfied, gL(Q0, T), zLq(x)(Ω), m ∈ ℕ, ξ = (ξ1,..., ξm) ∈ ℝm, w1,..., wmLq(x)(Ω), and w(x,ξ)=l=1mξlwl(x). Then the function I(t,ξ):=Ωg(x,t)|w(x,ξ)|q(x)2w(x,ξ)z(x)dx,t(0,T),ξRm,(82) satisfies the L -Carathéodory condition.


Step 1. The Fubini Theorem [33, p. 91] yields that I(·, ξ) ∈ L1(0, T). Then the function [0, T] ∋ tI(t, ξ) ∈ ℝ is measurable.

Step 2. We prove that the function ℝ ∋ ξ1I(t, ξ1,...,ξm) ∈ ℝ is continuous at the point ξ10R. Take ξ=(ξ1,ξ2,...,ξm),ξ0=(ξ10,ξ2,...,ξm),where|ξξ0|1.

By Theorem 2.1 [52, p. 2], we get ||η1|q(x)2η1|η2|q(x)2η2|C10(|η1|+|η2|)q(x)1β(x)|η1η2|β(x),(83) where 0 < β(x) ≤ min{1, q(x) − 1}, η1, η2 ∈ ℝ, C10 > 0 is independent of η1, η2, x. Hence, |I(t,ξ)I(t,ξ0)|=|Ωg|w(x,ξ)|q(x)2w(x,ξ)|w(x,ξ0)|q(x)2w(x,ξ0)zdx|C11Ω(|w(x,ξ)|+|w(x,ξ0)|)q(x)1β(x)|w(x,ξ)w(x,ξ0)|β(x)|z|dx=C11(I1+I2),(84) where I1=Ω1h(x,ξ,ξ0)dx,I2=Ω2h(x,ξ,ξ0)dx, Ω1 = {x ∈ Ω | q(x) ≤ 2}, Ω2 = {x ∈ Ω | q(x) > 2}, and h(x,ξ,ξ0)=(|w(x,ξ)|+|w(x,ξ0)|)q(x)1β(x)|w(x,ξ)w(x,ξ0)|β(x)|z(x)|,xΩ. By taking β(x) = q(x) − 1, where x ∈ Ω1, we obtain I1=Ω1|w(x,ξ)w(x,ξ0)|q(x)1|z(x)|dx=Ω1|ξ1ξ10|q(x)1|w1(x)|q(x)1|z(x)|dx|ξ1ξ10|q01Ω1|w1(x)|q(x)1|z(x)|dx=C12|ξ1ξ10|q01ξ1ξ100. By taking β(x) = 1, where x ∈ Ω2, we obtain I2=Ω2(|w(x,ξ)w(x,ξ0)|)q(x)2|w(x,ξ)w(x,ξ0)||z(x)|dx=|ξ1ξ10|Ω2(|w(x,ξ)|+|w(x,ξ0)|)q(x)2|w1(x)||z(x)|dxC13(ξ10)|ξ1ξ10|ξ1ξ100. Therefore, by (84), we obtain that |I(t,ξ)I(t,ξ0)|ξ1ξ100. Continuing in the same way, we see that I is continuous with respect to ξ2,..., ξm.

Step 3. Taking into account the results of Step 1 and Step 2, we obtain that the function I satisfies the Carathéeodory condition. Since gL(Q0, T), the L-Carathéodory condition holds. ☐

Lemma 3.26

Suppose that condition (E) is satisfied, (Eu)(x,t):=Ωε(x,t,y)u~(x+y,t)u~(x,t)dy,(x,t)Q0,T,(85) where uL1(Q0, T), ũ is the zero extension of u from Q0, T into (ℝn \ Ω) × (0, T). Then for every s > 1 the operator E : Ls (Q0, T) → Ls(Q0, T) is linear bounded continuous and ||Eu;Ls(Q0,τ)||C14||u;Ls(Q0,τ)||,uLs(Q0,T),τ(0,T],(86) where C14 > 0 is independent of u and τ.

The proof is trivial.

Lemma 3.27

Suppose that φ ∈ Lip(ℝ), ε ∈ L(Q0, T × Ω), zL2(Ω), m ∈ ℕ, ξ = (ξ1,..., ξm) ∈ ℝm, w1,..., wmL2(Ω), w(x,ξ)=l=1mξlwl(x),xΩ, and the operator E is determined from (85). Then the function J(t,ξ):=Ωϕ(Ew(,ξ))(x,t)z(x)dx,t(0,T),ξRm,(87) satisfies the L -Carathéodory condition.


Step 1. Lemma 3.26 implies that EwL2(Q0, T) if ξ ∈ ℝm. Hence φ(Ew) ∈ L2(Q0, T) ⊂ L1(Q0, T). The Fubini Theorem [33, p. 91] yields that J(·, ξ) ∈ L1(0, T). Then the function [0, T] ∋ tJ(t, ξ) ∈ ℝ is measurable.

Step 2. Take a point t ∈ (0, T). We prove that the function ℝ ∋ ξ1I(t, ξ1,...,ξm) ∈ ℝ is continuous at the point ξ10R.Takeξ=(ξ1,ξ2,...,ξm),ξ0=(ξ10,ξ2,...,ξm). Then |J(t,ξ)J(t,ξ0)|Ωϕ((Ew(,ξ))(x,t))ϕ((Ew(,ξ0))(x,t))z(x)|dxC15Ω(Ew(,ξ))(x,t)(Ew(,ξ0))(x,t)|z(x)|dx=C15Ω|Ωε(x,t,y)(w(x+y,ξ)w(x,ξ)(w(x+y,ξ0)w(x,ξ0))dy||z(x)|dxC16|ξ1ξ10|ΩΩ|w1(x+y)|+w1(x)|z(x)|dxdy=C17|ξ1ξ10|ξ1ξ100. Continuing in the same way, we see that J is continuous with respect to ξ2,...,ξm.

Step 3. Taking into account the results of Step 1 and Step 2, we obtain that the function J satisfies the Carathéodory condition. Since ε ∈ L(Q0, T × Ω), the L-Carathéodory condition holds. ☐

Clearly, the operator Λ(t) : ZN → [ZN]* (see (16)) is linear, bounded, continuous and monotone. Similarly as in Theorem 3.4 [53, p. 454], we prove that A(t) : XN → [XN]* (see (17)) is bounded, semicontinuous and monotone if p ∈ ℬ +(Ω), p0 > 1, and condition (A) is satisfied. The operator G(t) : 𝒪N → [𝒪N]* (see (4)) is bounded, semicontinuous and monotone. Similarly to (86), we get the estimate ||(Ew)(t);[Ls(Ω)]N||C18||w;[Ls(Ω)]N||,w[Ls(Ω)]N,t[0,T],(88) where s > 1 and C18 > 0 is independent of w and t. Using condition (Φ), we get that the operator [Ls(Q0, T)]Nuφ(Eu) ∈ [Ls(Q0, T)]N is bounded and continuous.

Lemma 3.28

Suppose that conditions (Γ), (B), and (Z) are satisfied, the operator ψ is determined from (18). Then ψ(t) : ZN → [ZN]* is bounded and semicontinuous. Moreover, Ψ(t)u,υC19S1/γ(Sγ(||u;HN||))||υ;ZN||,u,υZN,t(0,T),(89) where S1/γ′ and Sγ are defined by (8), C19 > 0 is independent of u, υ and t.


Similar to [54, p. 159], we use the generalized Hölder inequality, Proposition 3.3 with q = γ, and notation (7). We get the estimate |Ψ(t)u,υ|=Ωk=1Nbk(x,t)|u|γ(x)2ukΔυkdxb0Ω|u|γ(x)1|Δυ|dx2b0|||u|γ(x)1;Lγ(x)(Ω)|||||Δυ|;Lγ(x)(Ω)||2b0S1/γ(Ω|u|(γ(x)1)γ(x)dx)×|||Δυ|;Lγ(x)(Ω)||C20S1/γ(Sγ(||u;[Lγ(x)(Ω)]N||))||Δυ;[Lγ(x)(Ω)]N||. Since γ0 ≤ 2, we obtain that (89) holds and the operator Ψ is bounded. We omit the proof that Ψ is semicontinuous (it is similar to the proof of Lemma 3.25). ☐

Let us consider the Banach space 𝒱 such that 𝒱 ↺ ZN. Let us define the family of operators ψ𝒱(t) : 𝒱 → 𝒱* by the rule ΨV(t)u,υV:=Ψ(t)u,υ,u,υV,t[0,T]. By (89), we obtain ΨV(t)u,υVC21S1/γ(Sγ(||u;V||))||u;V||,u,υV,t(0,T),(90) where C21 > 0 is independent of u, υ and t. Then ψ𝒱: 𝒱 → 𝒱 is bounded. We will replace this space 𝒱 by VN and 𝒲r. For the sake of convenience we have replaced ψVN and ψ𝒲r by ψ and we have replaced 〈·, ·〉VN and 〈·, ·〉𝒲r by 〈·, ·〉. The same notation we need for Λ(t), A(t), and 𝒦(t), t ∈ (0, T). According to the above remarks, we have that the operator 𝒦(t) (see (19) ) is bounded and semicontinuous from VN into [VN]* and is bounded from 𝒲r into Wr.

Lemma 3.29

Suppose that (Γ), (Q), (A)-(E), (7), and (21) hold. Assume also that α > 0, p ∈ ℬ +(Ω), p0 > 1, {wj}j∈ℕV, m ∈ ℕ, L = (L11, L21,...,Lm1,..., L1N, L2N,..., LmN), where Lμk(t,ξ)=(K(t)z)k,wμ+((N(t)z)k,wμ)Ω,k=1,N¯,μ=1,m¯,t(0,T),(91) ξ = (ξ11, ξ21,...,ξm1,...,ξ1N, ξ2N,...,ξmN), z = (z1,...,zN), and zk(x)==1mξkw(x),xΩ,k=1,N. Then (L(t,ξ),ξ)RmNΩα2|Δz|2+a0i=1n|zxi|p(x)+g0|z|q(x)C22|z|2dxC23,t(0,T),(92) where C22, C23 > 0 are independent of z, ξ and t.


Clearly, (L(t,ξ),ξ)RmN=K(t)z,z+(N(t)z,z)Ω=k=1NΩα|Δzk|2+i=1naik(t)|zxi|p(x)2|zk,xi|2+bk(t)|z|γ(x)2zkΔzk+gk(t)|z|q(x)2|zk|2+βk(t)|(zk)|2dx+(ϕ(Ez(t)),z)Ω.(93) Taking into account (A), (G), and (BB), we obtain k=1Nα|Δzk|2+i=1naik(t)|zxi|p(x)2|zk,xi|2+gk(t)|z|q(x)2|zk|2+βk(t)|(zk)|2α|Δz|2+a0|zxi|p(x)+g0|z|q(x).(94) Using the generalized Young inequality, we get k=1N|bk|z|γ(x)2zkΔzk|=b0|z|γ(x)1|Δz|C24(χ1)|z|γ(x)+χ1|Δz|γ(x)χ1|Δz|2+C25(χ1)(1+|z|2),(95) where χ1 > 0, C25(χ1) > 0 is independent of x, t, k and m.

Taking into account condition (Φ), Cauchy-Bunyakowski-Schwarz’s inequality, and (88), we obtain (ϕ(Ez(t)),z)Ωϕ0Ω|Ez(t)||z|dxC26||Ez(t);[L2(Ω)]N||||z;[L2(Ω)]N||C18||z;[L2(Ω)]N||||z;[L2(Ω)]N||C27Ω|z|2dx,(96) where C27 > 0 is independent of z, t and m.

Using (94)-(96) and choosing χ1=α2 we can show that (93) yields (92). ☐

4 Proof of main Theorem

The solution will be constructed via Faedo-Galerkin’s method.

Step 1. Let {wj}j∈ℕ be a set of all eigenfunctions of the problem (28) which are an orthonormal in L2(Ω), MmN:={x(μ=1mαμ1mwμ(x),...,μ=1mαμNmwμ(x))|αμkmR,k=1,N,¯μ=1,m¯},mN, r is determined from condition (Z), 𝒲r and Wr are defined by (31), and V is defined by (14). Taking into account Proposition 3.5 and (32), we obtain that MN:=mNMmN is dense in 𝒲r and VN.

Take m ∈ ℕ and um := (u1m,...,uNm), where ukm(x,t):=μ=1mφμkm(t)wμ(x),(x,t)Q0,T,k=1,N¯, φm:=(φ11m,φ21m,...,φm1m,...,φ1Nm,φ2Nm,...,φmNm) is a solution to the problem utm(t),wμ+K(t)um(t),wμ+(N(t)um(t),wμ)Ω=F(t),wμ,t(0,T),(97) φμkm(0)=βμkm,k=1,N¯,μ=1,m¯(98) (see (3), (19), and (20) for definition of the elements of 𝒩, 𝒦, and F), the functions u0m:=(u01m,...,u0Nm) satisfies the condition u0mmu0stronglyinHN, and u0km(x):=μ=1mβμkmwμ(x),xΩ,k=1,N¯. Clearly, um(0)=μ=1mφμ1m(0)wμ(x),...,μ=1mφμNm(0)wμ(x)=u0m.(99) The problem ((97), (98)) coincides with (74) if ℓ = mN, φ0=(β11m,β21m,...,βm1m,...,β1Nm,β2Nm,...,βmNm),M=(M11,M21,...,Mm1,...,M1N,M2N,...,MmN),Mμk(t)=Fk(t),wμ,L=(L11,L21,...,Lm1,...,L1N,L2N,...,LmN),Lμk(t,φm)=(K(t)um(t))k,wμ+(N(t)um(t))k,wμΩ,k=1,N¯,μ=1,m¯,t(0,T).(100) By (F), we have ML2(0, T; ℝmN). Taking into account the lemmas such as Lemmas 3.27 and 3.25, we see that L satisfies the L-Carathéodory condition. From (92) we obtain (L(t,φm),φm)RmNC28Ω|um|2dxC29C30(m)Ωk=1Nμ=1m|φμkm|2|wμ(x)|2dxC29=C31(m)|φm|2C29,(101) where C29, C31 > 0 are independent of t, ϕm. Then Carathéodory-LaSalle’s Theorem 3.24 implies that there exists a solution ϕmH1 (0, T; ℝmN) to problem (97), (98). If we combine the condition ∂Ω ∈ C2r with Proposition 3.5 and embedding (26), we get {wj}j∈ℕ ⊂ 𝒲r ⊂ [H2r (Ω)]N. Thus, umH1(0,T;Wr)H1(0,T;[H2r(Ω)]N)[H1(Q0,T)]N.(102)

Step 2. Multiplying both sides of the corresponding equality (97) by φμkm(t), summing the obtained equalities, and integrating in t ∈ (0, τ) ⊂ (0, T), we get Q0,τ(utm,um)dxdt+0τ(L(t,φm(t))RmNdt=Q0,τi,j=1n(fij,uxixjm)+i=1n(fi,uxim)+(f0,um)dxdt,τ(0,T].(103) By (102), similar to Case 1.ii of Theorem 3.19 (with p(x) = r(x) ≡ 2), we obtain |um|2W1,1(0,T;L1(Ω)),(|um|2)t=2(utm,um). Then, the integration by parts formula and (99) yield that Q0,τ(utm,um)dxdt=12Ω|um(x,τ)|2dx12Ω|u0m(x)|2dx. By (92), we get 0τ(L(t,φm(t)),φm(t)RmNdtQ0,τα2|Δum|2+a0i=1n|uxim|p(x)+g0|um|q(x)C32|um|2dxC33, where C32, C33 > 0 are independent of m and τ. In addition, Young’s inequality, the condition ∂Ω ∈ C2, and estimate (30) yield that |Q0,τi,j=1n(fij,uxixjm)dxdt|Q0,τi,j=1nχ1|uxixjm|2+14χ1|fij|2dxdtQ0,τχ1C34|Δum|2+14χ1i,j=1n|fij|2dxdt, where 𝜒1 > 0, the constant C34 > 0 is independent of m and 𝜒1. By (23), we get |i=1n(fi,uxim)+(f0,um)|χ2i=1n|uxim|p(x)+Yp(χ2)i=1n|fi|p(x)+χ3|um|q(x)+Yq(χ3)|f0|q(x). According to the above remarks, from (103) we have the following inequality 12Qτ|um|2dx+Q0,τα2χ1C34|Δum|2+(a0χ2)i=1n|uxim|p(x)+(g0χ3)|um|q(x)dxdt12Ω|u0m|2dx+C35(χ1,χ2,χ3)(1+Q0,τi,j=1n|fij|2+i=1n|fi|p(x)+|f0|q(x)dxdt+Q0,τ|um|2dxdt),τ(0,T],(104) where C35 > 0 is independent of m and τ.

Let y(t):=Ω|um(x,t)|2dx,t[0,T]. Choosing 𝜒1, 𝜒2, 𝜒3 > 0 sufficiently small, from (104) we can obtain that y(τ)C36+C370τy(t)dt,τ(0,T]. Then the Gronwall-Bellman Lemma yields that Ω|um(x,τ)|2dxC38,τ(0,T],(105) and so Q0,τ|um|2dxdtC38T,τ(0,T].(106) Using (104), (106), and choosing 𝜒1, 𝜒2, 𝜒3 > 0 sufficiently small, we get Q0,τ|Δum|2+i=1n|uxim|p(x)+|um|q(x)dxdtC39,τ(0,T].(107) Here C38, C39 > 0 are independent of m and τ.

By (105)-(107), we have that there exists a sequence {umj}j∈ℕ ⊂ {um}m∈ℕ such that umjjuweaklyinL(0,T;HN)andweaklyinU(Q0,T).(108)

Step 3. We define the element ℱ ∈ [U(Q0, T)]* and the operator 𝒜 : U(Q0, T) → [U(Q0, T)]* by the rules F,υU(Q0,T):=0TF(t),υ(t)dt,υU(Q0,T),(109) Au,υU(Q0,T):=0TK(t)u(t),υ(t)+(N(t)u(t),υ(t))Ωdt,u,υU(Q0,T).(110)

Using (24), (86), (106) and (107), we get Aum,υU(Q0,T)=Q0,Tk=1NαΔukmΔυk+i=1naik|uxim|p(x)2uk,ximυk,xi+bk|um|γ(x)2ukmΔυk+gk|um|q(x)2ukmυkβk(ukm)υk+ϕk(Eukm)υkdxdtC40Q0,T|Δum||Δυ|+i=1n|uxim|p(x)1|υxi|+|um|γ(x)1|Δυ|+|um|q(x)1|υ|+|um||υ|+|Eum||υ|dxdtC40|||Δum|;L2(Q0,T)|||||Δυ|;L2(Q0,T)||+2i=1n|||uxim|p(x)1;Lp(x)(Q0,T)||×|||υxi|;Lp(x)(Q0,T)||+2|||um|γ(x)1;Lγ(x)(Q0,T)|||||Δυ|;Lγ(x)(Q0,T)||+2|||um|q(x)1;Lq(x)(Q0,T)|||||υ|;Lq(x)(Q0,T)||+|||um|;L2(Q0,T)|||||υ|;L2(Q0,T)||+|||Eum|;L2(Q0,T)|||||υ|;L2(Q0,T)||C41||υ;U(Q0,T)||, where C41 > 0 is independent of m, υ. Then ||Aum;[U(Q0,T)]||C41(111) and so Aumjjχweaklyin[U(Q0,T)].(112)

Step 3. Suppose that the numbers r and s0 are determined from condition (Z), the spaces 𝒲r and Wr are defined by (31), Pm : HNHN is the projection operator from (45) (see also (21)), m is defined by (40), where ℋ = HN and 𝒱 = 𝒲r. Similarly to [54, p. 77] and [55, p. 62-63], using Lemma 3.9, notation (109) and (110), we rewrite (97) as utm=P^m(FAum).(113) By (49), we get ||P^mf;Ls0(0,T);Wr)||||f;Ls0(0,T;Wr)||,fLs0(0,T;Wr).(114) Since ||D*||ℒ(B*, A*) = ||D||ℒ(A, B) for every D ∈ ℒ(A, B) (see [42, p. 231]), using (114), we have ||P^mh;Ls0s01(0,T;Wr)||||h;Ls0s01(0,T;Wr)||,hLs0s01(0,T;Wr).(115) Taking into account (115), (35), and (109), we obtain ||P^mF;Ls0s01(0,T;Wr)||||F;Ls0s01(0,T;Wr)||C42||F;[U(Q0,T)]||C43.(116) By (115), (110), (111), and (35), we get ||P^mAum;Ls0s01(0,T;Wr)||||Aum;Ls0s01(0,T;Wr)||C44||Aum;[U(Q0,T)]||C45.(117) Using (113), (116), and (117) (see for comparison [54, 55]), we obtain ||utm;Ls0s01(0,T;Wr)||C46.(118) Here C43,...,C46 > 0 are independent of m. Therefore, utmjjutweaklyinLs0s01(0,T;Wr).(119)

Step 4. Suppose the numbers r and so are determined from condition (Z). Then (32) implies that VNKHNWr. By (33), (106), and (107), we get ||um;Ls0(0,T;VN)||C47||um;U(Q0,T)||C48,(120) where C48 > 0 is independent of m.

Taking into account (120), (118), the Aubin theorem (see Proposition 3.11), and Lemma 1.18 [23, p. 39], we obtain umjjustronglyinL2(0,T;HN)andinC([0,T];Wr),(121) umjjualmosteverywhereinQ0,T.(122) Clearly, VNK[H01(Ω)]NWr. Then (120), (118), and the Aubin theorem yield that umjjustronglyinL2(0,T;[H01(Ω)]N).(123) Hence for every i ∈ {1,...,n} we have Q0,T|uximjuxi|2dxdt||umju;L2(0,T;[H01(Ω)]N||2j0. Thus uximjjuxi strongly in [L2(Q0,T)]N and so Lemma 1.18 [23, p. 39] implies that uximjjuxialmosteverywhereinQ0,T,i=1,n¯.(124) By (122) and (124), we obtain the equality χ = 𝒜u.

Step 5. Using (97) and (102), we obtain 0T(umj(t),w)Ωφ(t)dt+Aumj,wφU(Q0,T)=F,wφU(Q0,T),(125) where φC0((0,T)),wMkN,kN,kmj,jN. Letting j → + ∞ and using Lemma 3.8, we get the equality ut + 𝒜u = ℱ. Whence, ut = ℱ − 𝒜u ∈ [U(Q0, T)]*, uW(Q0, T), and (22) holds. Moreover, we obtain the inclusion utLs0s01(0,T;[VN]) because (34) is true. Hence, uC([0, T]; [VN]*). By (108), we have that uL(0, T; HN). Thus, Lemma 3.7 yields that uC([0, T]; HN) and so u is a weak solution to initial-boundary value problem (1), (2). ☐


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About the article

Received: 2016-10-18

Accepted: 2017-05-10

Published Online: 2017-06-22

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 859–883, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0069.

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