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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

# 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

## Abstract

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.

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 $uk,t+αΔ2uk−∑i=1naik(x,t)|uxi|p(x)−2uk,xixi+Δbk(x,t)|u|γ(x)−2uk+(Nu)k==∑i,j=1nfijk(x,t)xixj−∑i=1nfik(x,t)xi+f0k(x,t),(x,t)∈Q0,T,k=1,N¯,$(1) $u|S0,T=0,Δu|S0,T=0,u|t=0=u0(x).$(2) Here α > 0 is a number, $\mathrm{\Delta }:=\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{x}_{1}^{2}}+\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{x}_{2}^{2}}+...+\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{x}_{n}^{2}}$ is the Laplacian, Δ2 := Δ(Δ), |u| := (|u1|2 + . . . + |uN|)1/2, $|{u}_{{x}_{i}}|\phantom{\rule{thinmathspace}{0ex}}:=\phantom{\rule{thinmathspace}{0ex}}\left(|\frac{\mathrm{\partial }{u}_{1}}{\mathrm{\partial }{x}_{i}}{|}^{2}+...+|\frac{\mathrm{\partial }{u}_{N}}{\mathrm{\partial }{x}_{i}}|{\right)}^{1/2},i=\overline{1,n},$ $(Nu)k(x,t):=(Gu)k(x,t)+(Bu)k(x,t)+ϕk((Eu)k(x,t)),(x,t)∈Q0,T,$(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 $i,j=\overline{1,n},\phantom{\rule{thinmathspace}{0ex}}k=\overline{1,N}.$

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 $i,j=\overline{1,n}$) 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 $i,j=\overline{1,n}$) 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 $X\stackrel{K}{\subset }Y$ means the compact embedding.

If wB, z = (z1, . . ., zN) ∈ BN, and υ = (υ1, . . ., υN) ∈ BN, then we set $v,w:=〈v1,w〉B,...,〈vN,w〉B∈RN,〈v,z〉:=∑k=1N〈vk,zk〉B∈R,$(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 ${C}_{0}^{\mathrm{\infty }}\left(\mathsf{Q}\right)$ are determined from [22, p. 9], Lp(𝖰) is the Lebesgue space (see [22, p. 22, 24]), Wm, p(𝖰) and ${W}_{0}^{m,p}\left(\mathsf{Q}\right)$ are Sobolev spaces (see [22, p. 45]), Hm(𝖰) := Wm,2(𝖰), ${H}_{0}^{m}\left(\mathsf{Q}\right):={W}_{0}^{m,2}\left(\mathsf{Q}\right),$ 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 $B+(Q):={q∈L∞(Q)|essinfy∈Q⁡q(y)>0}.$ If q ∈ ℬ+(𝖰), then by definition, put $q0:=essinfy∈Q⁡q(y),q0:=esssupy∈Q⁡q(y),Sq(s):=max{sq0,sq0},s≥0,$(8) $q′(y):=q(y)q(y)−1fora.e.y∈Qnotethat1q(y)+1q′(y)=1andq′∈B+(Q),$(9) $ρq(υ;Q):=∫Q|υ(y)|q(y)dy,υ∈M(Q).$(10) Assume that q ∈ ℬ+(𝖰), q0 > 1, and m ∈ ℕ. The set $Lq(y)(Q):={υ∈M(Q)|ρq(υ;Q)<+∞}$ 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 ${C}_{0}^{\mathrm{\infty }}\left(\mathsf{Q}\right)$ with respect to the norm (11) is called a generalized Sobolev space and is denoted by ${W}_{0}^{m,q\left(y\right)}\left(\mathsf{Q}\right).$

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, ${s}_{1},\phantom{\rule{thinmathspace}{0ex}}{s}_{1}^{\ast },....,{s}_{m},\phantom{\rule{thinmathspace}{0ex}}{s}_{m}^{\ast }\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}\mathbb{R},$ 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. $\mathrm{m}\mathrm{e}\mathrm{s}\left(\mathrm{\Omega }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\bigcup _{j\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{m}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Omega }}_{j}\phantom{\rule{thinmathspace}{0ex}}\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0;$

3. $1={s}_{1}<{s}_{2}<{s}_{1}^{\ast }<{s}_{3}<{s}_{2}^{\ast }<...<{s}_{m-1}<{s}_{m-2}^{\ast }

4. for every j ∈ {1, . . ., m} the inequality ${s}_{j}\le p\left(x\right)\le {s}_{j}^{\ast }$ holds a.e. for x ∈ Ωj;

5. for every k ∈ {1, . . ., m − 1} the inequality ${s}_{k}^{\ast } holds, where $R(q):=nqn−qif1≤q1ifn≤q.$(12)

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

Suppose that${\mathrm{\Delta }}^{0}\upsilon \phantom{\rule{thinmathspace}{0ex}}:=\phantom{\rule{thinmathspace}{0ex}}\upsilon ,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Delta }}^{1}\upsilon \phantom{\rule{thinmathspace}{0ex}}:=\phantom{\rule{thinmathspace}{0ex}}\mathrm{\Delta }\upsilon ,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\Delta }}^{r}\upsilon \phantom{\rule{thinmathspace}{0ex}}:=\phantom{\rule{thinmathspace}{0ex}}\mathrm{\Delta }\left({\mathrm{\Delta }}^{r-1}\upsilon \right),$ $HΔ2r(Ω):={υ∈H2r(Ω)|υ|∂Ω=Δυ|∂Ω=...=Δr−1υ|∂Ω=0},r∈N.$(13) By definition, put $Z:={H}_{\mathrm{\Delta }}^{2}\left(\mathrm{\Omega }\right),\phantom{\rule{thinmathspace}{0ex}}X:={W}_{0}^{1,p\left(x\right)}\left(\mathrm{\Omega }\right),\phantom{\rule{thinmathspace}{0ex}}\mathcal{O}:={L}^{q\left(x\right)}\left(\mathrm{\Omega }\right),\phantom{\rule{thinmathspace}{0ex}}H:={L}^{2}\left(\mathrm{\Omega }\right),$ $V:=Z∩X∩O∩H,$(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 $W(Q0,T):={w∈U(Q0,T)|wt∈[U(Q0,T)]∗}.$ 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 $r≥12max2,1+n(p0−2)2p0,n(q0−2)2q0;$

• (𝐀) : aik ∈ 𝓜(𝖰0,T), 0 < a0aik(x, t) ≤ a0 < +∞ for a.e. (x, t) ∈ 𝖰0,T, where $i=\overline{1,n},\phantom{\rule{thinmathspace}{0ex}}k=\overline{1,N};$

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

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

• (𝐁𝐁): β1, . . ., βN ∈ ℬ+(𝖰0,T);

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

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

• (𝐅) : fijkL2(𝖰0,T), fikLp′(x)(𝖰0,T), f0kLq′(x)(𝖰0,T), where $i,\phantom{\rule{thinmathspace}{0ex}}j=\overline{1,n},\phantom{\rule{thinmathspace}{0ex}}k=\overline{1,N};$

• (𝐔): u0HN.

Let us introduce the following notation. If t ∈ (0, T) and if k ∈ {1, . . ., N}, then we set $〈(Λu)k,w〉Z:=∫ΩαΔuk(x)Δw(x)dx,u∈ZN,w∈Z,$(16) $〈(A(t)u)k,w〉X:=∫Ω∑i=1naik(x,t)|uxi(x)|p(x)−2uk,xi(x)wxi(x)dx,u∈XN,w∈X,$(17) $〈(Ψ(t)u)k,w〉Z:=∫Ωbk(x,t)|u(x)|γ(x)−2uk(x)Δw(x)dx,u∈ZN,w∈Z,$(18) $〈(K(t)u)k,w〉V:=〈(Λu)k,w〉Z+〈(A(t)u)k,w〉X+〈(Ψ(t)u)k,w〉Z,u∈VN,w∈V,$(19) $〈Fk(t),w〉V:=∫Ω[∑i,j=1nfijk(x,t)wxixj(x)+∑i=1nfik(x,t)wxi(x)+f0k(x,t)w(x)]dx,w∈V.$(20)

Using (16) and (7), we define the operator Λ : ZN → [ZN] by the rule $Λu:=(Λu)1,...,(Λu)N,〈Λu,υ〉ZN:=∑k=1N〈(Λu)k,υk〉Z,u∈ZN,υ=(υ1,....,υN)∈ZN.$ 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, $F(t)∈[VN]∗,N(t)(ON∩HN)⊂[ON∩HN]∗,t∈[0,T].$ 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)), $k=\overline{1,N}.$ By definition, put $(u,υ)Ω:=∫Ωu(x)υ(x)dxifu:Ω→RN,υ:Ω→R,∫Ω(u(x),υ(x))RNdxifu,υ:Ω→RN,$(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 $〈ut,υ〉U(Q0,T)+∫0T〈K(t)u(t),υ(t)〉VN+(N(t)u(t),υ(t))Ωdt=∫0T〈F(t),υ(t)〉VNdt.$(22)

#### Theorem 2.2

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

## 3.1 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 $ab≤ηaq(y)+Yq(η)bq′(y)$(23) holds. In addition, Yq(η) depends on q0, q0 and it is independent of y, ${Y}_{2}\left(\eta \right)=\frac{1}{4\eta },\phantom{\rule{thinmathspace}{0ex}}{Y}_{2}\left(\frac{1}{2}\right)=\frac{1}{2},$ 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 $||υ;Lr(y)(Q)||≤(1+mesQ)||υ;Lq(y)(Q)||,υ∈Lq(y)(Q);$

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 $∫Ω|u(y)υ(y)|dy≤2||u;Lq(y)(Q)||⋅||υ;Lq′(y)(Q)||.$(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 ${W}_{0}^{1,p\left(x\right)}\left(\mathrm{\Omega }\right);$

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 $||u;Lp(x)(Ω)||≤C1∑i=1n||uxi;Lp(x)(Ω)||+||u;L1(Ω)||,$ 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 ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ is defined in (13), where r ∈ ℕ. It is easy to verify that ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ 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 $∫ΩυΔrudx=∫ΩuΔrυdx,u,υ∈HΔ2r(Ω).$(27) Note that for every r ∈ ℕ the space ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ is reflexive.

Let {wj}j∈ℕ be a set of all eigenfunctions of the problem $−Δwj=λjwjinΩ,wj|∂Ω=0,j∈N.$(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 $(−1)rΔrw=λrw,w|∂Ω=Δw|∂Ω=...=Δr−1w|∂Ω=0.$(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 ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right).$

#### Proposition 3.6

(see Lemma 3 [34, p. 229]). If ∂Ω ⊂ C2r, then there exists a constant C2 > 0 such that for all υ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ 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, $Ls0s0−1(0,T;[VN]∗)↺¯[U(Q0,T)]∗↺¯Ls0s0−1(0,T;[VN]∗).$(34) Similarly, using (32) we obtain $Ls0(0,T;Wr)↺¯U(Q0,T)↺¯[L2(Q0,T)]N↺¯[U(Q0,T)]∗↺¯Ls0s0−1(0,T;Wr∗).$(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 ${L}^{\frac{{s}^{0}}{{s}^{0}-1}}\left(0,\phantom{\rule{thinmathspace}{0ex}}T;\phantom{\rule{thinmathspace}{0ex}}\left[{V}^{N}{\right]}^{\ast }\right).$ Thus, if uU(Q0,T) belongs to Ls0(0, T; VN), then $〈w,\phantom{\rule{thinmathspace}{0ex}}u{〉}_{U\left({Q}_{0,T}\right)}={\int }_{0}^{T}{〈w\left(t\right),\phantom{\rule{thinmathspace}{0ex}}\upsilon \left(t\right)〉}_{{V}^{N}}dt.$ 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 $\left\{{\phi }_{\mu k}{\right\}}_{\mu =1,k=1}^{m,\phantom{\rule{thinmathspace}{0ex}}N}\phantom{\rule{thinmathspace}{0ex}}\subset \phantom{\rule{thinmathspace}{0ex}}{C}^{\mathrm{\infty }}\left(\left[0,\phantom{\rule{thinmathspace}{0ex}}T\right]\right)$ such that ||uψm; U(Q0,T)|| < ɛ, where ψm = (ψm1, . . ., ψmN) and ${\psi }_{mk}\left(x,\phantom{\rule{thinmathspace}{0ex}}t\right)=\sum _{\mu =1}^{m}{\phi }_{\mu k}\left(t\right){w}^{\mu }\left(x\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,t\right)\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}{Q}_{0,T},\phantom{\rule{thinmathspace}{0ex}}k=\phantom{\rule{thinmathspace}{0ex}}\overline{1,N}.$

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 $V↺¯H≅H∗↺¯V∗.$(37) Notice that if g ∈ 𝒱 and g ∈ ℋ, then $〈g,υ〉V=(g,υ)H,υ∈V.$(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]) $Pmh:=∑j=1m(h,wj)Hwj,h∈H.$(38) This is a linear self-adjoint continuous operator (see Theorem 7.3.6 [37, p. 515]) such that $||Pmh||H≤||h||H,h∈H.$(39) If {wj}j∈ℕ ⊂ 𝒱, then let us define an operator ${\stackrel{^}{P}}_{m}:\mathcal{V}\to \mathcal{V}$ (not necessarily self-adjoint) by the rule $P^mυ:=Pmυforeveryυ∈V.$(40) We shall find a conjugate operator ${\stackrel{^}{P}}_{m}^{\ast }:{\mathcal{V}}^{\ast }\to {\mathcal{V}}^{\ast }.$ Take elements υ ∈ 𝒱, z ∈ 𝒱. Then $〈z,Pmυ〉V=z,∑j=1m(υ,wj)HwjV=∑j=1m(υ,wj)H〈z,wj〉V=υ,∑j=1m〈z,wj〉VwjH.$ Since υ, w1, . . ., wm ∈ 𝒱, (37) yields that $υ,∑j=1m〈z,wj〉VwjH=∑j=1m〈z,wj〉Vwj,υH=∑j=1m〈z,wj〉Vwj,υV.$ Thus, $〈z,\phantom{\rule{thinmathspace}{0ex}}{P}_{m}\upsilon {〉}_{\mathcal{V}}=〈{\stackrel{^}{P}}_{m}^{\ast }z,\phantom{\rule{thinmathspace}{0ex}}\upsilon {〉}_{\mathcal{V}},$ where $P^m∗z=∑j=1m〈z,wj〉Vwj,z∈V∗.$(41) In addition, (41) implies that ${\stackrel{^}{P}}_{m}^{\ast }\left({\mathcal{V}}^{\ast }\right)\phantom{\rule{thinmathspace}{0ex}}\subset \phantom{\rule{thinmathspace}{0ex}}\mathcal{V}.$

#### Lemma 3.9

Assume that {wj}j∈ℕ is a orthonormal basis for the spacesuch that $\left\{{w}^{j}{\right\}}_{j\in \mathbb{N}}\subset \mathcal{V},\phantom{\rule{thinmathspace}{0ex}}{\psi }_{1}^{m},...,{\psi }_{m}^{m}\in \mathbb{R}$ are some numbers, and F ∈ 𝒱. Then ${z}^{m}=\sum _{s=1}^{m}{\psi }_{s}^{m}{w}^{s}\in \mathcal{V}$ satisfies $〈zm,w1〉V=〈F,w1〉V,⋮〈zm,wm〉V=〈F,wm〉V,$(42) iff the following equality holds $zm=P^m∗FinV∗.$(43)

#### Proof

Clearly, (43) implies (42). We shall prove that (42) implies (43). Take υ ∈ 𝒱. There exist numbers ${\alpha }_{1}^{m},...,{\alpha }_{m}^{m}\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}\mathbb{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\phantom{\rule{thinmathspace}{0ex}}{P}_{m}\upsilon ={\stackrel{^}{P}}_{m}\upsilon =\sum _{\mu =1}^{m}{\alpha }_{\mu }^{m}{w}^{\mu }.$ Multiplying both sides of μ-th equality of (42) by ${\alpha }_{\mu }^{m}$ and summing the obtained equalities, we get $〈{z}^{m},{\stackrel{^}{P}}_{m}\upsilon {〉}_{\mathcal{V}}=〈F,{\stackrel{^}{P}}_{m}\upsilon {〉}_{\mathcal{V}}.\phantom{\rule{thinmathspace}{0ex}}\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\phantom{\rule{thinmathspace}{0ex}}〈{\stackrel{^}{P}}_{m}^{\ast }{z}^{m},\upsilon {〉}_{\mathcal{V}}=〈{\stackrel{^}{P}}_{m}^{\ast }F,\phantom{\rule{thinmathspace}{0ex}}\upsilon {〉}_{\mathcal{V}}$ for every υ ∈ 𝒱. Thus, $P^m∗zm=P^m∗FinV∗.$(44) Taking into account (37), the inclusions zm, w1, . . ., wm ∈ 𝒱, and the orthonormality condition for {wj}j∈ℕ ⊂ ℋ, from (41) we obtain $P^m∗zm=∑j=1m〈zm,wj〉Vwj=∑j=1m∑s=1mψsmws,wjHwj=∑s,j=1mψsm(ws,wj)Hwj=∑s=1mψsmws=zm.$ Therefore, (42) yields (43). ☐

In the sequel, we only consider the case ℋ = L2(Ω), 𝒱 = ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ (see (13) ), and {wj}j∈ℕ is determined from problem (28). Then (38) implies that (see (21) ) $(Pmu)(x)=∑j=1m(u,wj)Ωwj(x),x∈Ω,u:Ω→R.$(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 $\mathcal{L}\left({H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right),\phantom{\rule{thinmathspace}{0ex}}{H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)\right),$ we take υ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$. 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 $||P^mυ||HΔ2r(Ω)=||ΔrP^mυ||L2(Ω)=(h,ΔrP^mυ)L2(Ω)=h,Δr∑j=1m(v,wj)ΩwjΩ=h,∑j=1m(υ,wj)ΩΔrwjΩ=h,∑j=1m(υ,wj)Ω(−1)rλjrwjΩ=h,∑j=1m(υ,(−1)rλjrwj)ΩwjΩ=h,∑j=1m(υ,Δrwj)ΩwjΩ=∑j=1m(υ,Δrwj)Ω(h,wj)Ω=υ,∑j=1m(h,wj)ΩΔrwjΩ=(υ,ΔrP^mh)Ω=(Δrυ,P^mh)Ω=(Δrυ,Pmh)Ω.$ 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, $||P^mυ||HΔ2r(Ω)≤||υ||HΔ2r(Ω),υ∈HΔ2r(Ω).$(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 ${\int }_{0}^{T}|{P}_{m}f\left(t\right){|}_{\mathcal{H}}^{s}\phantom{\rule{thickmathspace}{0ex}}dt\phantom{\rule{thickmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}{\int }_{0}^{T}|f\left(t\right){|}_{\mathcal{H}}^{s}\phantom{\rule{thickmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}$ i.e. $||Pmf;Ls(0,T;H)||≤||f;Ls(0,T;H)||,f∈Ls(0,T;H).$(48) Finally assume that m : 𝒱 → 𝒱 is determined from (40), ℋ = L2(Ω), and 𝒱 = ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$. Taking into account (46) and (48), we have that $||P^mu;Ls(0,T;HΔ2r(Ω))||≤||u;Ls(0,T;HΔ2r(Ω))||,u∈Ls(0,T;HΔ2r(Ω)),s≥1.$(49) Clearly, we can prove (38)-(49) if we replace L2(Ω), ${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$ by [L2(Ω)]N, [${H}_{\mathrm{\Delta }}^{2r}\left(\mathrm{\Omega }\right)$]N respectively.

## 3.4 Differentiability of the nonlinear expressions

Take a function σ ∈ 𝓜(Ω) and by definition, put $ψσ(x)(s):=sσ(x)ifs>0,0ifs≤0,x∈Ω.$(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 $χ~(s):=1ifs>0,0ifs≤0.$(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: $∫sτut(t)dt=u(τ)−u(s),0≤s<τ≤T,u∈W1,p(0,T;X).$(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 $\mathcal{W}\stackrel{K}{\subset }\mathcal{L}↺\mathcal{B},$ then ${u∈Ls(0,T;W)|ut∈Lh(0,T;B)}⊂KLs(0,T;L)∩C([0,T];B).$

#### Lemma 3.12

Suppose that Ω ⊂ ℝn is a bounded C0,1 -domain. Then the integration by parts formula $∫Qs,τwtzdxdt=∫Ωtwzdxt=st=τ−∫Qs,τwztdxdt,0≤s<τ≤T,$(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).

#### Proof

(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 $\phi z\in {W}^{1,1}\left(0,T;\phantom{\rule{thinmathspace}{0ex}}{L}^{\frac{{q}^{0}}{{q}^{0}-1}}\left(\mathrm{\Omega }\right)\right).$ Using (52) with υ = ϕ(t)z(x, t), we get $∫sτφt(t)z(x,t)dt=φ(τ)z(x,τ)−φ(s)z(x,s)−∫sτφ(t)zt(x,t)dt,x∈Ω.$(54) Take a function υC1(Ω). By (54), we obtain that $∫Qs,τφtυzdxdt=∫Ωtφυzdxt=st=τ−∫Qs,τφυztdxdt.$(55) Clearly, C1([0, T]; C1(Ω̄)) ↺̄ W ↺̄ W1,1(0,T; L1(Ω)). Then the set ${∑i=1mφi(t)υi(x)|m∈N,φ1,...,φm∈C1([0,T]),υ1,...,υm∈C1(Ω¯)}$ 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 $q\left(y\right)\le \frac{p\left(y\right)}{\sigma \left(y\right)}$ 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 ${\psi }_{\sigma \left(y\right)}\left(u\right)\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}{L}^{\frac{p\left(y\right)}{\sigma \left(y\right)}}\left(\mathsf{Q}\right),$ $ρp/σ(ψσ(y)(u);Q)≤ρp(u;Q),$(56) $||ψσ(y)(u);Lq(y)(Q)||≤C3Sσ/p(ρp(u;Q)),$(57) where C3 > 0 is independent of u.

#### Proof

Clearly, $\frac{p\left(y\right)}{\sigma \left(y\right)}\ge 1\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}.\mathrm{e}.\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}\mathsf{Q},\phantom{\rule{thinmathspace}{0ex}}|{\psi }_{\sigma \left(y\right)}\left(u\right){|}^{\frac{p\left(y\right)}{\sigma \left(y\right)}}=|{u}^{+}{|}^{p\left(y\right)}\le |u{|}^{p\left(y\right)}\in {L}^{1}\left(\mathsf{Q}\right).$ Then by [42, p. 297], we obtain ${\psi }_{\sigma \left(y\right)}\left(u\right)\in {L}^{\frac{p\left(y\right)}{\sigma \left(y\right)}}\left(\mathsf{Q}\right).$ 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 $|θ(x,ζ)−θ(x,η)|≤M|ζ−η|,|θξ(x,ξ)|≤M$(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)

#### Proof

Since u, utLp(x)(Q0,T), there exists a sequence $\left\{{u}^{m}{\right\}}_{m\in \mathbb{N}}\subset {C}^{1}\left(\overline{{Q}_{0,T}}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\phantom{\rule{thinmathspace}{0ex}}{u}^{m}\underset{m\to \infty }{\to }u\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}{u}_{t}^{m}\underset{m\to \infty }{\to }{u}_{t}$ strongly in Lp(x)(Q0,T) and almost everywhere in Q0,T. Clearly, $(θ(x,um(x,t)))t=limh→0θ(x,um(x,t+h))−θ(x,um(x,t))um(x,t+h)−um(x,t)um(x,t+h)−um(x,t)h=θξ(x,um(x,t))utm(x,t),$ where (x, t) ∈ Q0,T, m ∈ ℕ. In addition, |θ(x, um) − θ(x, u)| ≤ M|umu|. Hence, $\theta \left(x,{u}^{m}\right)\underset{m\to \infty }{\to }\theta \left(x,u\right)$ strongly in Lp(x)(Q0,T) and so θ(x, u) ∈ Lp(x)(Q0,T).

Clearly, ${\theta }_{\xi }\left(x,{u}^{m}\right){u}_{t}^{m}-{\theta }_{\xi }\left(x,u\right){u}_{t}={A}_{m}+{B}_{m},$ where $Am=θξ(x,um)(utm−ut),Bm=(θξ(x,um)−θξ(x,u))ut.$ On the other hand, $|{A}_{m}{|}^{p\left(x\right)}\le {M}^{p\left(x\right)}|{u}_{t}^{m}-{u}_{t}{|}^{p\left(x\right)}\underset{m\to \infty }{\to }0\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}{L}^{1}\left({Q}_{0,T}\right).\phantom{\rule{thinmathspace}{0ex}}\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}{A}_{m}\underset{m\to \infty }{\to }0$ in Lp(x)(Q0,T). Moreover, |Bm|p(x) ≤ (2M|ut|)p(x)L1(Q0,T), ${B}_{m}\underset{m\to \infty }{\to }0$ almost everywhere in Q0,T, and ${B}_{m}\underset{m\to \infty }{\to }0$ in Lp(x)(Q0,T). Therefore, ${\theta }_{\xi }\left(x,\phantom{\rule{thinmathspace}{0ex}}{u}^{m}\right){u}_{t}^{m}\underset{m\to \infty }{\to }{\theta }_{\xi }\left(x,\phantom{\rule{thinmathspace}{0ex}}u\right){u}_{t}$ in Lp(x)(Q0,T) and so θξ(x, u)utLp(x)(Q0,T).

Finally let us prove (59). Take a function $\phi \in {C}_{0}^{\mathrm{\infty }}\left({Q}_{0,T}\right).$ Then (59) holds because $∫Q0,Tθξ(x,u)utφdxdt=limm→∞⁡∫Q0,Tθξ(x,um)utmφdxdt=limm→∞⁡∫Q0,T(θ(x,um))tφdxdt=−limm→∞⁡∫Q0,Tθ(x,um)φtdxdt=−∫Q0,Tθ(x,u)φtdxdt.$ ☐ 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.

#### Proof

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 $Θ(x,ξ):=θ(x,ξ)ifξ≤b,θξ(x,b)ξ+θ(x,b)−θξ(x,b)bifξ>b,x∈Ω.$ 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.

#### Proof

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 $χk(s):=1ifs>1k,0ifs≤1k,k∈N.$(61) If $u\in {C}^{1}\left(\overline{{Q}_{0,T}}\right)\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\upsilon ,\phantom{\rule{thinmathspace}{0ex}}{\upsilon }_{t}\in {L}^{1}\left({Q}_{0,T}\right),$ then $limk→+∞⁡∫Q0,Tχk(u)β(x)ψβ(x)−1(u)utυdxdt=∫Ωtψβ(x)(u)υdx|t=0t=T−∫Q0,Tψβ(x)(u)υtdxdt.$(62)

#### Proof

By definition, set $ψβ(x),k(s):=kβ(x)ifs≥k,sβ(x)if1k k ∈ ℕ, k ≥ 2, x ∈ Ω. Clearly, ${\psi }_{\beta \left(x\right),k}\left(s\right)\underset{k\to \infty }{\to }{\psi }_{\beta \left(x\right)}\left(s\right),$ 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 $s=\frac{1}{k}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}s=k.\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{r},\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{\partial }}{\mathrm{\partial }s}{\psi }_{\beta \left(x\right),k}\left(s\right)={\stackrel{~}{\xi }}_{{}_{\beta \left(x\right),k}}\left(s\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}s\ne \frac{1}{k}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}s\ne k.$ 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. $∫Q0,T(ψβ(x),k(u))tυdxdt=∫Ωψβ(x),k(u)υdx|t=0t=T−∫Q0,Tψβ(x),k(u)υtdxdt.$(64) Let $M:=\underset{\left(x,t\right)\in \overline{{Q}_{0,T}}}{max}|u\left(x,t\right)|,\phantom{\rule{thinmathspace}{0ex}}{k}_{0}\in \mathbb{N},\phantom{\rule{thinmathspace}{0ex}}{k}_{0}\ge max\left\{2,\phantom{\rule{thinmathspace}{0ex}}M\right\}.$ 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 $|{\psi }_{\beta \left(x\right),k}\left(u\left(x,t\right)\right)|\le {M}^{\beta \left(x\right)}\mathrm{\forall }\left(x,t\right)\in \overline{{Q}_{0,T}}$ 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 $u\in {C}^{1}\left(\overline{{Q}_{0,T}}\right),$ 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 $p\in {L}_{+}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ and p(x) ≥ σ(x) for a.e. x ∈ Ω, then ${\psi }_{\sigma \left(x\right)}\left(u\right),\phantom{\rule{thinmathspace}{0ex}}\left({\psi }_{\sigma \left(x\right)}\left(u\right){\right)}_{t}\in {L}^{\frac{p\left(x\right)}{\sigma \left(x\right)}}\left({Q}_{0,T}\right),$ equality (65) is true, and the estimate $ρp/σ(ψσ(x)(u))t;Q0,T≤C5S1/σ′ρp(u;Q0,T)S1/σρp(ut;Q0,T)$(66) holds, where C5 > 0 is independent of u.

#### Proof

First let us prove Case 1. Take a function $u\in {C}^{1}\left(\overline{{Q}_{0,T}}\right).\phantom{\rule{thinmathspace}{0ex}}\mathrm{I}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}\upsilon ,\phantom{\rule{thinmathspace}{0ex}}{\upsilon }_{t}\in C\left(\overline{{Q}_{0,T}}\right),{\chi }_{k}$ 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 $∫Q0,Tσ(x)ψσ(x)−1(u)utυdxdt=∫Ωtψσ(x)(u)υdx|t=0t=T−∫Q0,Tψσ(x)(u)υtdxdt.$(67) Taking in (67) the function $\upsilon \in {C}_{0}^{\mathrm{\infty }}\left({Q}_{0,T}\right),$ 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 $\left\{{u}^{m}{\right\}}_{m\in \mathbb{N}}\subset {C}^{1}\left(\overline{{Q}_{0,T}}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\phantom{\rule{thinmathspace}{0ex}}{u}^{m}\underset{m\to \infty }{\to }u\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}{u}_{t}^{m}\underset{m\to \infty }{\to }{u}_{t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{y}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}{L}^{p\left(x\right)}\left({Q}_{0,T}\right),\phantom{\rule{thinmathspace}{0ex}}{u}^{m}\underset{m\to \infty }{\to }u\phantom{\rule{thinmathspace}{0ex}}$ in C([0, T]; Lp(x)(Ω)).

Assume that $\upsilon ,\phantom{\rule{thinmathspace}{0ex}}{\upsilon }_{t}\in \phantom{\rule{thinmathspace}{0ex}}C\left(\overline{{Q}_{0,T}}\right).$ By (67), for every m ∈ ℕ we obtain $∫Q0,Tσ(x)ψσ(x)−1(um)utmυdxdt=∫Ωψσ(x)(um)υdxt=0t=T−∫Q0,Tψσ(x)(um)υtdxdt.$(68) Since 1 < σ(x) ≤ p(x), we get $\frac{p\left(x\right)}{\sigma \left(x\right)-1}>1$ for a.e. x ∈ Ω. Therefore, $ψσ(x)−1(um)⟶m→∞ψσ(x)−1(u)stronglyinLp(x)σ(x)−1(Q0,T).$ Clearly, $\left[{L}^{\frac{p\left(x\right)}{\sigma \left(x\right)-1}}\left({Q}_{0,T}\right){\right]}^{\ast }\cong {L}^{\frac{p\left(x\right)}{p\left(x\right)-\left(\sigma \left(x\right)-1\right)}}\left({Q}_{0,T}\right).$ Since p(x) ≥ (σ(x) − 1) + 1, we have that$p\left(x\right)\ge \frac{p\left(x\right)}{p\left(x\right)-\left(\sigma \left(x\right)-1\right)}$ for a.e. x ∈ Ω. Therefore, $utm⟶m→∞utstronglyinLp(x)p(x)−(σ(x)−1)(Q0,T).$ By Lemma 5.2 [23, p. 19], we obtain $∫Q0,Tσ(x)ψσ(x)−1(um)utmυdxdt⟶m→∞∫Q0,Tσ(x)ψσ(x)−1(u)utυdxdt$(69) and σψσ(x)−1(u)utL1(Q0,T). It is easy to verify that $ψσ(x)(um(t))⟶m→∞ψσ(x)(u(t))stronglyinLp(x)σ(x)(Ω)fort=0andt=T,$(70) $ψσ(x)(um)⟶m→∞ψσ(x)(u)stronglyinLp(x)σ(x)(Q0,T).$(71) Letting m → ∞ in (68) and using (69)-(71), we get (67) and (65).

By Lemma 3.13, we get ${\psi }_{\sigma \left(x\right)}\left(u\right)\in {L}^{\frac{p\left(x\right)}{\sigma \left(x\right)}}\left({Q}_{0,T}\right).$ 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, $\left({\psi }_{\sigma \left(x\right)}\left(u\right){\right)}_{t}\in {L}^{\frac{p\left(x\right)}{\sigma \left(x\right)}}\left({Q}_{0,T}\right).$

By (65) and the generalized Hölder’s inequality, we obtain that $∫Q0,T|(ψσ(x)(u))t|p(x)σ(x)dxdt≤C8|||u|p(x)σ′(x);Lσ′(x)(Q0,T)||⋅|||ut|p(x)σ(x);Lσ(x)(Q0,T)||≤C9S1/σ′∫Q0,T|u|p(x)dxdtS1/σ∫Q0,T|ut|p(x)dxdt.$ 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. $u\in {C}^{1}\left(\overline{{Q}_{0,T}}\right)$ (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 $|u{|}^{r\left(x\right)},\left(|u{|}^{r\left(x\right)}{\right)}_{t}\in {L}^{\frac{p\left(x\right)}{r\left(x\right)}}\left({Q}_{0,T}\right)$).

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. $u\in {C}^{1}\left(\overline{{Q}_{0,T}}\right)$ (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 $|u{|}^{r\left(x\right)-2}u,\phantom{\rule{thinmathspace}{0ex}}\left(|u{|}^{r\left(x\right)-2}u{\right)}_{t}\in {L}^{\frac{p\left(x\right)}{r\left(x\right)-1}}\left({Q}_{0,T}\right)$).

#### Proof

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 $|s|r(x)=ψr(x)(s)+ψr(x)(−s),|s|r(x)−2s=ψr(x)−1(s)−ψr(x)−1(−s),x∈Ω,s∈R.$

## 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 ${\phi }^{0}=\left({\phi }_{1}^{0},...,{\phi }_{\ell }^{0}\right)$ ∈ 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 ξ$\overline{{D}_{R}}$ := {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 $y(τ)≤(C+∫0τB(t)e−∫0tA(s)dsdt)e∫0τA(t)dt,τ∈[0,T].$(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 $(L(t,ξ),ξ)Rℓ≥−α(t)|ξ|2−β(t)$(78) holds, then problem (74) has a global weak solution ϕW1,p (0, T; ℝ).

#### Proof

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 $φ(τ)=φ0+∫0τM(t)dt−∫0τL(t,φ(t))dt$(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 $R:={(|φ0|2+∫0T[2β(t)+|M(t)|2]dt)e∫0T[2α(t)+1]dt}1/2,$(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 ξ$\overline{{D}_{R}}$ := {y ∈ ℝ ||y| ≤ R} inequality (75) holds.

Taking into account (see (78)) the following inequalities $(L(t,φ(t)),φ(t))Rℓ≥−α(t)|φ(t)|2−β(t),(M(t),φ(t))Rℓ≤|M(t)|⋅|φ(t)|≤12|M(t)|2+12|φ(t)|2,$ from (74) we get $(φ′(t),φ(t))Rℓ−α(t)|φ(t)|2−β(t)≤12|M(t)|2+12|φ(t)|2,t∈[0,b¯).$ Hence, $∫0τ(φ′(t),φ(t))Rℓdt≤∫0τ[(α(t)+12)|φ(t)|2+β(t)+12|M(t)|2]dt,τ∈|(0,b¯).$(81) Since ϕW1,p(0, τ; ℝ) and p ≥ 2, we obtain $|φ|2∈W1,p2(0,τ),(|φ(t)|2)′=2(φ′(t),φ(t))Rℓ,t∈(0,τ),$ (see Case 1.ii of Theorem 3.19). Hence Proposition 3.10 implies that $∫0τ(φ′(t),φ(t))Rℓdt=12|φ(τ)|2−12|φ(0)|2.$ 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 $y(τ)≤(C+∫0τB(t)e−∫0tA(s)dsdt)e∫0τA(t)dt≤(C+∫0τB(t)dt)e∫0τA(t)dt≤R2,$ 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 $|φ(t2)−φ(t1)|=∫t1t2L(t,φ(t))dt≤∫t1t2hR(t)dt⟶t1,t2→b¯−00.$ Finally we have an existence of the finite limit $\underset{t\to \overline{b}-0}{lim}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phi \left(t\right)$. Then solution to problem (74) can be extended to [0, ] by the rule $\phi \left(\overline{b}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}:=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to \overline{b}-0}{lim}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phi \left(t\right)<\mathrm{\infty }$. 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 $w\left(x,\xi \right)=\sum _{l=1}^{m}{\xi }_{l}{w}^{l}\left(x\right)$. Then the function $I(t,ξ):=∫Ωg(x,t)|w(x,ξ)|q(x)−2w(x,ξ)z(x)dx,t∈(0,T),ξ∈Rm,$(82) satisfies the L -Carathéodory condition.

#### Proof

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 ${\xi }_{1}^{0}\in \mathbb{R}$. Take $\xi =\left({\xi }_{1},\phantom{\rule{thinmathspace}{0ex}}{\xi }_{2},...,{\xi }_{m}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\xi }^{0}=\left({\xi }_{1}^{0},{\xi }_{2},...,{\xi }_{m}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}|\xi -{\xi }^{0}|\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1$.

By Theorem 2.1 [52, p. 2], we get $||η1|q(x)−2η1−|η2|q(x)−2η2|≤C10(|η1|+|η2|)q(x)−1−β(x)|η1−η2|β(x),$(83) where 0 < β(x) ≤ min{1, q(x) − 1}, η1, η2 ∈ ℝ, C10 > 0 is independent of η1, η2, x. Hence, $|I(t,ξ)−I(t,ξ0)|=|∫Ωg|w(x,ξ)|q(x)−2w(x,ξ)−|w(x,ξ0)|q(x)−2w(x,ξ0)zdx|≤C11∫Ω(|w(x,ξ)|+|w(x,ξ0)|)q(x)−1−β(x)|w(x,ξ)−w(x,ξ0)|β(x)|z|dx=C11(I1+I2),$(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 $h(x,ξ,ξ0)=(|w(x,ξ)|+|w(x,ξ0)|)q(x)−1−β(x)|w(x,ξ)−w(x,ξ0)|β(x)|z(x)|,x∈Ω.$ By taking β(x) = q(x) − 1, where x ∈ Ω1, we obtain $I1=∫Ω1|w(x,ξ)−w(x,ξ0)|q(x)−1|z(x)|dx=∫Ω1|ξ1−ξ10|q(x)−1|w1(x)|q(x)−1|z(x)|dx≤|ξ1−ξ10|q0−1∫Ω1|w1(x)|q(x)−1|z(x)|dx=C12|ξ1−ξ10|q0−1→ξ1→ξ100.$ By taking β(x) = 1, where x ∈ Ω2, we obtain $I2=∫Ω2(|w(x,ξ)−w(x,ξ0)|)q(x)−2|w(x,ξ)−w(x,ξ0)|⋅|z(x)|dx=|ξ1−ξ10|∫Ω2(|w(x,ξ)|+|w(x,ξ0)|)q(x)−2|w1(x)|⋅|z(x)|dx≤C13(ξ10)|ξ1−ξ10|→ξ1→ξ100.$ Therefore, by (84), we obtain that $|I\left(t,\xi \right)-I\left(t,{\xi }^{0}\right)|\underset{{\xi }_{1}\to {\xi }_{1}^{0}}{\to }0$. 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 $||Eu;Ls(Q0,τ)||≤C14||u;Ls(Q0,τ)||,u∈Ls(Q0,T),τ∈(0,T],$(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), $w\left(x,\xi \right)=\sum _{l=1}^{m}{\xi }_{l}{w}^{l}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in \mathrm{\Omega }$, and the operator E is determined from (85). Then the function $J(t,ξ):=∫Ωϕ(Ew(⋅,ξ))(x,t)z(x)dx,t∈(0,T),ξ∈Rm,$(87) satisfies the L -Carathéodory condition.

#### Proof

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 ${\xi }_{1}^{0}\in \mathbb{R}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{T}\mathrm{a}\mathrm{k}\mathrm{e}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\xi =\left({\xi }_{1},{\xi }_{2},...,{\xi }_{m}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\xi }^{0}=\left({\xi }_{1}^{0},{\xi }_{2},...,{\xi }_{m}\right)$. Then $|J(t,ξ)−J(t,ξ0)|≤∫Ωϕ((Ew(⋅,ξ))(x,t))−ϕ((Ew(⋅,ξ0))(x,t))⋅z(x)|dx≤C15∫Ω(Ew(⋅,ξ))(x,t)−(Ew(⋅,ξ0))(x,t)⋅|z(x)|dx=C15∫Ω|∫Ωε(x,t,y)(w(x+y,ξ)−w(x,ξ)−(w(x+y,ξ0)−w(x,ξ0))dy|⋅|z(x)|dx≤C16|ξ1−ξ10|∫Ω∫Ω|w1(x+y)|+w1(x)⋅|z(x)|dxdy=C17|ξ1−ξ10|⟶ξ1→ξ100.$ 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, $Ψ(t)u,υ≤C19S1/γ′(Sγ(||u;HN||))||υ;ZN||,u,υ∈ZN,t∈(0,T),$(89) where S1/γ′ and Sγ are defined by (8), C19 > 0 is independent of u, υ and t.

#### Proof

Similar to [54, p. 159], we use the generalized Hölder inequality, Proposition 3.3 with q = γ, and notation (7). We get the estimate $|Ψ(t)u,υ|=∫Ω∑k=1Nbk(x,t)|u|γ(x)−2ukΔυkdx≤b0∫Ω|u|γ(x)−1|Δυ|dx≤2b0|||u|γ(x)−1;Lγ′(x)(Ω)||⋅|||Δυ|;Lγ(x)(Ω)||≤2b0S1/γ′(∫Ω|u|(γ(x)−1)γ′(x)dx)×|||Δυ|;Lγ(x)(Ω)||≤C20S1/γ′(Sγ(||u;[Lγ(x)(Ω)]N||))⋅||Δυ;[Lγ(x)(Ω)]N||.$ 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 $ΨV(t)u,υV:=Ψ(t)u,υ,u,υ∈V,t∈[0,T].$ By (89), we obtain $ΨV(t)u,υV≤C21S1/γ′(Sγ(||u;V||))||u;V||,u,υ∈V,t∈(0,T),$(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 ${\mathcal{W}}_{r}^{\ast }$.

#### 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 $zk(x)=∑ℓ=1mξℓkwℓ(x),x∈Ω,k=1,N.$ Then $(L(t,ξ),ξ)RmN≥∫Ωα2|Δz|2+a0∑i=1n|zxi|p(x)+g0|z|q(x)−C22|z|2dx−C23,t∈(0,T),$(92) where C22, C23 > 0 are independent of z, ξ and t.

#### Proof

Clearly, $(L(t,ξ),ξ)RmN=K(t)z,z+(N(t)z,z)Ω=∑k=1N∫Ωα|Δzk|2+∑i=1naik(t)|zxi|p(x)−2|zk,xi|2+bk(t)|z|γ(x)−2zkΔzk+gk(t)|z|q(x)−2|zk|2+βk(t)|(zk)−|2dx+(ϕ(Ez(t)),z)Ω.$(93) Taking into account (A), (G), and (BB), we obtain $∑k=1Nα|Δzk|2+∑i=1naik(t)|zxi|p(x)−2|zk,xi|2+gk(t)|z|q(x)−2|zk|2+βk(t)|(zk)−|2≥α|Δz|2+a0|zxi|p(x)+g0|z|q(x).$(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 $(ϕ(Ez(t)),z)Ω≤ϕ0∫Ω|Ez(t)|⋅|z|dx≤C26||Ez(t);[L2(Ω)]N||⋅||z;[L2(Ω)]N||≤C18||z;[L2(Ω)]N||⋅||z;[L2(Ω)]N||≤C27∫Ω|z|2dx,$(96) where C27 > 0 is independent of z, t and m.

Using (94)-(96) and choosing ${\chi }_{1}=\frac{\alpha }{2}$ 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), $MmN:={x↦(∑μ=1mαμ1mwμ(x),...,∑μ=1mαμNmwμ(x))|αμkm∈R,k=1,N,¯μ=1,m¯},m∈N,$ r is determined from condition (Z), 𝒲r and ${\mathcal{W}}_{r}^{\ast }$ are defined by (31), and V is defined by (14). Taking into account Proposition 3.5 and (32), we obtain that ${\mathfrak{M}}^{N}:=\bigcup _{m\in \mathbb{N}}{\mathfrak{M}}_{m}^{N}$ is dense in 𝒲r and VN.

Take m ∈ ℕ and um := $\left({u}_{1}^{m},...,{u}_{N}^{m}\right)$, where $ukm(x,t):=∑μ=1mφμkm(t)wμ(x),(x,t)∈Q0,T,k=1,N¯,$ ${\phi }^{m}:=\left({\phi }_{11}^{m},{\phi }_{21}^{m},...,{\phi }_{m1}^{m},...,{\phi }_{1N}^{m},{\phi }_{2N}^{m},...,{\phi }_{mN}^{m}\right)$ 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 ${u}_{0}^{m}:=\left({u}_{01}^{m},...,{u}_{0N}^{m}\right)$ satisfies the condition $u0m⟶m→∞u0stronglyinHN,$ and ${u}_{0k}^{m}\left(x\right):=\sum _{\mu =1}^{m}{\beta }_{\mu k}^{m}{w}^{\mu }\left(x\right),x\in \mathrm{\Omega },\phantom{\rule{thinmathspace}{0ex}}k=\overline{1,N}$. 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 $(L(t,φm),φm)RmN≥−C28∫Ω|um|2dx−C29≥−C30(m)∫Ω∑k=1N∑μ=1m|φμkm|2|wμ(x)|2dx−C29=−C31(m)|φm|2−C29,$(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, $um∈H1(0,T;Wr)⊂H1(0,T;[H2r(Ω)]N)⊂[H1(Q0,T)]N.$(102)

Step 2. Multiplying both sides of the corresponding equality (97) by ${\phi }_{\mu k}^{m}\left(t\right)$, summing the obtained equalities, and integrating in t ∈ (0, τ) ⊂ (0, T), we get $∫Q0,τ(utm,um)dxdt+∫0τ(L(t,φm(t))RmNdt=∫Q0,τ∑i,j=1n(fij,uxixjm)+∑i=1n(fi,uxim)+(f0,um)dxdt,τ∈(0,T].$(103) By (102), similar to Case 1.ii of Theorem 3.19 (with p(x) = r(x) ≡ 2), we obtain $|um|2∈W1,1(0,T;L1(Ω)),(|um|2)t=2(utm,um).$ Then, the integration by parts formula and (99) yield that $∫Q0,τ(utm,um)dxdt=12∫Ω|um(x,τ)|2dx−12∫Ω|u0m(x)|2dx.$ By (92), we get $∫0τ(L(t,φm(t)),φm(t)RmNdt≥∫Q0,τα2|Δum|2+a0∑i=1n|uxim|p(x)+g0|um|q(x)−C32|um|2dx−C33,$ where C32, C33 > 0 are independent of m and τ. In addition, Young’s inequality, the condition ∂Ω ∈ C2, and estimate (30) yield that $|∫Q0,τ∑i,j=1n(fij,uxixjm)dxdt|≤∫Q0,τ∑i,j=1nχ1|uxixjm|2+14χ1|fij|2dxdt≤∫Q0,τχ1C34|Δum|2+14χ1∑i,j=1n|fij|2dxdt,$ where 𝜒1 > 0, the constant C34 > 0 is independent of m and 𝜒1. By (23), we get $|∑i=1n(fi,uxim)+(f0,um)|≤χ2∑i=1n|uxim|p(x)+Yp(χ2)∑i=1n|fi|p′(x)+χ3|um|q(x)+Yq(χ3)|f0|q′(x).$ According to the above remarks, from (103) we have the following inequality $12∫Qτ|um|2dx+∫Q0,τα2−χ1C34|Δum|2+(a0−χ2)∑i=1n|uxim|p(x)+(g0−χ3)|um|q(x)dxdt≤12∫Ω|u0m|2dx+C35(χ1,χ2,χ3)(1+∫Q0,τ∑i,j=1n|fij|2+∑i=1n|fi|p′(x)+|f0|q′(x)dxdt+∫Q0,τ|um|2dxdt),τ∈(0,T],$(104) where C35 > 0 is independent of m and τ.

Let $y\left(t\right):={\int }_{\mathrm{\Omega }}|{u}^{m}\left(x,t\right){|}^{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}dx,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\in \left[0,T\right]$. Choosing 𝜒1, 𝜒2, 𝜒3 > 0 sufficiently small, from (104) we can obtain that $y\left(\tau \right)\le {C}_{36}+{C}_{37}{\int }_{0}^{\tau }y\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\tau \in \left(0,T\right]$. Then the Gronwall-Bellman Lemma yields that $∫Ω|um(x,τ)|2dx≤C38,τ∈(0,T],$(105) and so $∫Q0,τ|um|2dxdt≤C38T,τ∈(0,T].$(106) Using (104), (106), and choosing 𝜒1, 𝜒2, 𝜒3 > 0 sufficiently small, we get $∫Q0,τ|Δum|2+∑i=1n|uxim|p(x)+|um|q(x)dxdt≤C39,τ∈(0,T].$(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 $umj⟶j→∞u∗−weaklyinL∞(0,T;HN)andweaklyinU(Q0,T).$(108)

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

Using (24), (86), (106) and (107), we get $Aum,υU(Q0,T)=∫Q0,T∑k=1NαΔukmΔυk+∑i=1naik|uxim|p(x)−2uk,ximυk,xi+bk|um|γ(x)−2ukmΔυk+gk|um|q(x)−2ukmυk−βk(ukm)−υk+ϕk(Eukm)υkdxdt≤C40∫Q0,T|Δum|⋅|Δυ|+∑i=1n|uxim|p(x)−1|υxi|+|um|γ(x)−1|Δυ|+|um|q(x)−1|υ|+|um|⋅|υ|+|Eum|⋅|υ|dxdt≤C40|||Δum|;L2(Q0,T)||⋅|||Δυ|;L2(Q0,T)||+2∑i=1n|||uxim|p(x)−1;Lp′(x)(Q0,T)||×|||υxi|;Lp(x)(Q0,T)||+2|||um|γ(x)−1;Lγ′(x)(Q0,T)||⋅|||Δυ|;Lγ(x)(Q0,T)||+2|||um|q(x)−1;Lq′(x)(Q0,T)||⋅|||υ|;Lq(x)(Q0,T)||+|||um|;L2(Q0,T)||⋅|||υ|;L2(Q0,T)||+|||Eum|;L2(Q0,T)||⋅|||υ|;L2(Q0,T)||≤C41||υ;U(Q0,T)||,$ where C41 > 0 is independent of m, υ. Then $||Aum;[U(Q0,T)∗]||≤C41$(111) and so $Aumj⟶j→∞χweaklyin[U(Q0,T)]∗.$(112)

Step 3. Suppose that the numbers r and s0 are determined from condition (Z), the spaces 𝒲r and ${\mathcal{W}}_{r}^{\ast }$ 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 $utm=P^m∗(F−Aum).$(113) By (49), we get $||P^mf;Ls0(0,T);Wr)||≤||f;Ls0(0,T;Wr)||,f∈Ls0(0,T;Wr).$(114) Since ||D*||ℒ(B*, A*) = ||D||ℒ(A, B) for every D ∈ ℒ(A, B) (see [42, p. 231]), using (114), we have $||P^m∗h;Ls0s0−1(0,T;Wr∗)||≤||h;Ls0s0−1(0,T;Wr∗)||,h∈Ls0s0−1(0,T;Wr∗).$(115) Taking into account (115), (35), and (109), we obtain $||P^m∗F;Ls0s0−1(0,T;Wr∗)||≤||F;Ls0s0−1(0,T;Wr∗)||≤C42||F;[U(Q0,T)]∗||≤C43.$(116) By (115), (110), (111), and (35), we get $||P^m∗Aum;Ls0s0−1(0,T;Wr∗)||≤||Aum;Ls0s0−1(0,T;Wr∗)||≤C44||Aum;[U(Q0,T)]∗||≤C45.$(117) Using (113), (116), and (117) (see for comparison [54, 55]), we obtain $||utm;Ls0s0−1(0,T;Wr∗)||≤C46.$(118) Here C43,...,C46 > 0 are independent of m. Therefore, $utmj⟶j→∞utweaklyinLs0s0−1(0,T;Wr∗).$(119)

Step 4. Suppose the numbers r and so are determined from condition (Z). Then (32) implies that ${V}^{N}\stackrel{K}{\subset }{H}^{N}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}↺\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathcal{W}}_{r}^{\ast }$. 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 $umj⟶j→∞ustronglyinL2(0,T;HN)andinC([0,T];Wr∗),$(121) $umj⟶j→∞ualmosteverywhereinQ0,T.$(122) Clearly, ${V}^{N}\stackrel{K}{\subset }\left[{H}_{0}^{1}\left(\mathrm{\Omega }\right){\right]}^{N}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}↺\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathcal{W}}_{r}^{\ast }$. Then (120), (118), and the Aubin theorem yield that $umj⟶j→∞ustronglyinL2(0,T;[H01(Ω)]N).$(123) Hence for every i ∈ {1,...,n} we have $∫Q0,T|uximj−uxi|2dxdt≤||umj−u;L2(0,T;[H01(Ω)]N||2⟶j→∞0.$ Thus ${u}_{{x}_{i}}^{{m}_{j}}\underset{j\to \infty }{\to }{u}_{{x}_{i}}$ strongly in [L2(Q0,T)]N and so Lemma 1.18 [23, p. 39] implies that $uximj⟶j→∞uxialmosteverywhereinQ0,T,i=1,n¯.$(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 $\phi \in {C}_{0}^{\mathrm{\infty }}\left(\left(0,T\right)\right),\phantom{\rule{thinmathspace}{0ex}}w\in {\mathfrak{M}}_{k}^{N},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{N},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k\le {m}_{j},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j\in \mathbb{N}$. 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 ${u}_{t}\in {L}^{\frac{{s}^{0}}{{s}^{0}-1}}\left(0,T;\left[{V}^{N}{\right]}^{\ast }\right)$ 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

Accepted: 2017-05-10

Published Online: 2017-06-22

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 859–883, ISSN (Online) 2391-5455,

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