Research article Special Issues

Classical solutions of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system

  • We study solutions of the Dirichlet problem for the Brinkman system and for the Darcy-Forchheimer-Brinkman system in the spaces of functions Ck,α(¯Ω;Rm)×Ck1,α(¯Ω), where ΩRm is a bounded domain.

    Citation: Dagmar Medková. Classical solutions of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system[J]. AIMS Mathematics, 2019, 4(6): 1540-1553. doi: 10.3934/math.2019.6.1540

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  • We study solutions of the Dirichlet problem for the Brinkman system and for the Darcy-Forchheimer-Brinkman system in the spaces of functions Ck,α(¯Ω;Rm)×Ck1,α(¯Ω), where ΩRm is a bounded domain.


    The paper is devoted to classical solutions of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system

    pΔv+λv+a|v|v+b(v)v=f,v=0in Ω (1.1)

    where ΩRm is a bounded domain.

    Boundary value problems for the Darcy-Forchheimer-Brinkman system have been extensively studied in the recent years. This system describes flows through porous media saturated with viscous incompressible fluids, where the inertia of such fluids is not negligible. The constants λ,b>0 are determined by the physical properties of the porous medium. (For further details we refer the reader to the book [1,p.17] and the references therein.)

    M. Kohr et al. studied in [2] the transmission problem, where the Darcy-Forchheimer-Brinkman system is considered in a bounded domain Ω+R3 with connected Lipschitz boundary and the Stokes system is given on its complementary domain Ω. Solutions belong to the space H1(Ω±)×L2(Ω±), where H1(Ω)={uL2loc(Ω,R3);juiL2(Ω),(1+|x|2)1/2uj(x)L2(Ω)}. The paper [3] is concerned with another transmission problem. A bounded domain ΩRm with connected Lipschitz boundary splits into two Lipschitz domains Ω+ and Ω. A solution is found satisfying the homogeneous Darcy-Forchheimer-Brinkman system is in Ω and the homogeneous Navier-Stokes system in Ω+. The transmission condition on the interface ΩΩ+ is accompanied by the Robin condition on Ω. The paper [4] investigates the Robin problem for the Darcy-Forchheimer-Brinkman system (1.1) with b=0 in the space Hs(Ω,Rm)×Hs1(Ω), where 1<s<3/2 and ΩRm is a bounded domain with connected Lipschitz boundary, m{2,3}. The mixed Dirichlet-Robin problem and the mixed Dirichlet-Neumann problem for the Darcy-Forchheimer-Brinkman system (1.1) with b=0 are studied in H3/2(Ω,R3)×H1/2(Ω) (see [4] and [5]). Here ΩR3 is a bounded creased domain with connected Lipschitz boundary. M. Kohr et al. discussed in [4] the problem of Navier's type for the Darcy-Forchheimer-Brinkman system (1.1) with b=0 in H1(Ω,R3)×L2(Ω), where ΩR3 is a bounded domain with connected Lipschitz boundary.

    Now we briefly sketch results concerning the Dirichlet problem for the Darcy-Forchheimer-Brinkman system (1.1). It is supposed that ΩRm is a bounded domain with Lipschitz boundary. For f0 and 2m3 solutions of the problem are looked for in Ws,2(Ω,Rm)×Ws1,2(Ω) with 1s<3/2 (see [6], [7] and [8]). The paper [9] is devoted to similar problems on compact Riemannian manifolds. [10] considers bounded solutions of the problem for b=0 and a domain Ω with Ljapunov boundary.

    This paper begins with the study of classical solutions of the Dirichlet problem for the generalized Brinkman system

    pΔv+λv=f,v=0in Ω.

    If ΩRm is a bounded open set with boundary of class Ck,α we prove the existence of a solution (v,p)Ck,α(¯Ω;Rm)×Ck1,α(¯Ω). Unlike the previous papers we do not suppose that Ω is connected. Using the fixed point theorems give the existence of solutions of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system (1.1) in (v,p)Ck,α(¯Ω;Rm)×Ck1,α(¯Ω). Here λ,a,bCmax(k2,0),α(¯Ω). If k3 then a can be arbitrary. If k>3 then there exists vC(¯Ω;Rm) with v=0 such that |v|vCk2,α(¯Ω;Rm). (See Remark 3.3.) So, for k>3 we must suppose that a0.

    Before we investigate the Dirichlet problem for the Brinkman system we need the following auxiliary lemma.

    Lemma 2.1. Let ΩRm be a bounded open set, 0<α<1 and λC0,α(¯Ω) be non-negative. If fC0,α(¯Ω,Rm) then there exists a solution (v,p)C2,α(¯Ω;Rm)×C1,α(¯Ω) of

    Δv+λv+p=f,v=0in Ω. (2.1)

    Proof. Choose a bounded domain ω with smooth boundary such that ¯Ωω. Then we can suppose that fC0,α(¯ω,Rm) and λC0,α(¯ω). (See [11,Theorem 1.8.3] or [12,Chapter Ⅵ,§2].) Choose q such that m/(1α)<q<. According to Lemma 3.6 in the Appendix there exists a solution (v,p)W1,q(ω,Rm)×Lq(ω) of

    Δv+λv+p=f,v=0in ω,v=0on ω.

    Put F=fλv. Since vW1,q(ω)C0,α(¯ω) by [11,Theorem 5.7.8], we infer that v,FC0,α(¯ω) by Lemma 3.7 in the Appendix. Then

    Δv+p=F,v=0in ω.

    Choose bounded open sets ω1 and ω2 such that ¯Ωω1¯ω1ω2¯ω2ω. Fix φC(Rm) such that φ=1 on ω1 and φ=0 on Rmω2. Define ˜F=φF in ω and ˜F=0 in Rmω.

    For xRm{0} and i,j{1,2,,m} define

    Eij(x):=12σm{δij(m2)|x|m2+xixj|x|m},m3
    Eij(x):=12σ2{δijln1|x|+xixj|x|2},m=2,
    Qj(x):=1σmxj|x|m

    where σm is the area of the unit sphere in Rm. Then E={Eij}, Q=(Q1,,Qm) form a fundamental tensor of the Stokes system, i.e.,

    ΔEij+λEij+iQj=δ0δij,im,
    1E1j++mEmj=0,

    where δ0 is the Dirac measure. (See for example [13].) Define ˜v:=E˜F and ˜p:=Q˜F. Then

    Δ˜v+˜p=˜F,˜v=0in Rm.

    Define

    hΔ(x):={σ12ln|x|,m=2,(2m)1σ1m|x|2m,m>2

    the fundamental solution for the Laplace equation. Since FjC0,αloc(Rm), [14,Theorem 3.14.2] gives that hΔ˜FjC2,αloc(Rm) for j=1,,m. Since Qj=jhΔ, we infer

    ˜p=1(hΔ˜F1)++m(hΔ˜Fm)C1,αloc(Rm).

    Since

    Δ(v˜v)+(p˜p)=F˜F=0,(v˜v)=0in ω1,

    we infer that

    Δ(p˜p)=(p˜p)=Δ(v˜v)=Δ[(v˜v)]=0in ω1

    in the sense of distributions. Thus p˜pC2(ω1) by [14,Theorem 2.18.2]. Since ˜pC1,α(ω) we infer that pC1,αloc(ω1). Thus Δv=pFC0,αloc(ω1;Rm). According to [14,Proposition 3.18.1] we obtain that vC2,αloc(ω1;Rm). (We can prove that vC2,αloc(ω1;Rm) also using results in [15].)

    Theorem 2.2. Let 0<β<α<1. Suppose that ΩRm is a bounded domain with boundary of class C1,α and λC0,β(¯Ω) is non-negative. Let fC0,β(¯Ω,Rm), gC1,β(Ω,Rm). Then there exist vC1,β(¯Ω;Rm)C2(Ω;Rm) and pC0,β(¯Ω)C1(Ω) solving

    Δv+λv+p=f,v=0in Ω,v=gon Ω (2.2)

    if and only if

    Ωng dσ=0. (2.3)

    A velocity v is unique and a pressure p is unique up to an additive constant. Moreover,

    vC1,β(¯Ω)+pC0,β(¯Ω)C(fC0,β(Ω)+gC1,β(Ω)+|Ωp dx|).

    Proof. If vC1,β(¯Ω;Rm)C2(Ω;Rm), pC0,β(¯Ω)C1(Ω) solve (2.2) then (2.3) holds by the Green formula.

    Suppose now that (2.3) holds. According to Lemma 2.1 there exists (˜v,˜p)C2,β(¯Ω;Rm)×C1,β(¯Ω) such that

    Δ˜v+λ˜v+˜p=f,˜v=0in Ω.

    Put ˆg:=g˜v. Then ˆgC1,β(Ω,Rm). Choose q such that m/(1β)<q<. According to Lemma 3.6 there exists a solution (ˆv,ˆp)W1,q(Ω,Rm)×Lq(Ω) of

    Δˆv+λˆv+ˆp=0,ˆv=0in Ω,ˆv=ˆgon Ω.

    A velocity ˆv is unique and a pressure ˆp is unique up to an additive constant. Define v:=˜v+ˆv, p:=˜p+ˆp. Then (v,p) is a solution of (2.2).

    If λ0 then ˆvC1,β(¯Ω;Rm), ˆpC0,β(¯Ω) by [16,Theorem 5.2]. Moreover, ˆvC(Ω;Rm), ˆpC(Ω). (See for example [17,§1.2].) Thus vC1,β(¯Ω;Rm)C2(Ω;Rm), pC0,β(¯Ω)C1(Ω).

    Let now λ be general. Since vW1,q(Ω)C0,β(¯Ω) by [11,Theorem 5.7.8], Lemma 3.7 in the Appendix gives that λvC0,β(¯Ω). Therefore

    Δv+p=(fλv)C0,β(¯Ω;Rm).

    We have proved that vC1,β(¯Ω;Rm)C2(Ω;Rm), pC0,β(¯Ω)C1(Ω).

    Denote by Y the set of all gW11/q,q(Ω;Rm) satisfying (2.3). Define X:={vW1,q(Ω;Rm);v=0 in Ω,v|ΩY},

    Uλ(v,p):=[Δv+λv+p,v,Ωp dx].

    Then Uλ:X×Lq(Ω)W1,q(Ω;Rm)×Y×R1 is an isomorphism by Lemma 3.6. Denote Z=[XC1,β(¯Ω;Rm)]×C0,β(¯Ω), W=C0,β(¯Ω,Rm)×[YC1,β(Ω,Rm)]×R1. We have proved that U1λ(W)Z. Since U1λ:WZ is closed, it is continuous by the Closed graph theorem.

    Theorem 2.3. Let 0<α<1 and kN0. Suppose that ΩRm is a bounded domain with boundary of class Ck+2,α and λCk,α(¯Ω) is non-negative. Let fCk,α(¯Ω,Rm), gCk+2,α(Ω,Rm). Then there exists a solution (v,p)Ck+2,α(¯Ω;Rm)×Ck+1,α(¯Ω) of (2.2) if and only if (2.3) holds. A velocity v is unique and a pressure p is unique up to an additive constant. Moreover,

    vCk+2,α(¯Ω)+pCk+1,α(¯Ω)C(fCk,α(¯Ω)+gCk+2,α(Ω)+|Ωp dx|).

    Proof. (2.3) is a necessary condition for the solvability of the problem (2.2) by Theorem 2.2.

    Denote by Y the set of all gCk+2,α(Ω;Rm) satisfying (2.3). Define X:={vCk+2,α(¯Ω;Rm);v=0 in Ω,v|ΩY},

    Uλ(v,p):=[Δv+λv+p,v,Ωp dσ].

    Then U0:X×Ck+1,α(¯Ω)Ck,α(¯Ω;Rm)×Y×R1 is an isomorphism by Theorem 2.2 and [18,Theorem IV.7.1]. Since uλu is a bounded operator on Ck,α(¯Ω,Rm) by Lemma 3.7 and Ck+2,α(¯Ω,Rm)Ck,α(¯Ω,Rm) is compact by [19,Lemma 6.36], the operator UλU0:X×Ck+1,α(¯Ω)Ck,α(¯Ω;Rm)×Y×R1 is compact. Hence the operator Uλ:X×Ck+1,α(¯Ω)Ck,α(¯Ω;Rm)×Y×R1 is Fredholm with index 0. The kernel of Uλ is trivial by Theorem 2.2. Therefore, Uλ:X×Ck+1,α(¯Ω)Ck,α(¯Ω;Rm)×Y×R1 is an isomorphism.

    We prove the existence of a classical solution of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system using fixed point theorems. The following two lemmas are crucial for it.

    Lemma 3.1. Let ΩRm be open, 0<α1 and kN. Put l=max(k2,0). Let bCl,α(¯Ω;Rm). Define

    Lb(u,v):=b(u)v. (3.1)

    Then there exists C1(0,) such that if u,vCk,α(¯Ω;Rm), then Lb(u,v)Cl,α(¯Ω;Rm) and

    Lb(u,v)Cl,α(¯Ω)C1uCk,α(¯Ω)vCk,α(¯Ω), (3.2)
    Lb(u,u)Lb(v,v)Cl,α(¯Ω)C1uvCk,α(¯Ω)(vCk,α(¯Ω)+uCk,α(¯Ω)). (3.3)

    Proof. Lemma 3.7 in the Appendix forces that Lb(u,v)Cl,α(¯Ω;Rm) and the estimate (3.2) holds true. Since

    Lb(u,u)Lb(v,v)=Lb(uv,u)+Lb(v,uv),

    the estimate (3.3) is a consequence of the estimate (3.2).

    Lemma 3.2. Let ΩRm be a bounded domain with boundary of class C1,α and 0<βα<1. Suppose that aC0,β(¯Ω). For vC(¯Ω;Rm) define

    Aav:=a|v|v. (3.4)

    1. Then there exists a constant C1 such that for u,vC1,0(¯Ω;Rm) it holds

    AavC0,β(¯Ω)C1v2C1,0(¯Ω), (3.5)
    AavAauC0,β(¯Ω)C1vuC1,0(¯Ω)[vC1,0(¯Ω)+uC1,0(¯Ω)]. (3.6)

    2. If aC1,β(¯Ω), then there exists a positive constant C2 such that Aa:C2,0(¯Ω;Rm)C1,β(¯Ω;Rm) is a compact continuous mapping and

    AavC1,β(¯Ω;Rm)C2v2C2,0(¯Ω;Rm). (3.7)

    Proof. Easy calculation yields that

     |v| C0,β(¯Ω)vC0,β(¯Ω;Rm).

    So, according to Lemma 3.7 in the Appendix

    AavC0,β(¯Ω;Rm)4aC0,β(¯Ω)v2C0,β(¯Ω;Rm).

    Since C1,0(¯Ω;Rm)C0,β(¯Ω;Rm) by [19,Lemma 6.36], we obtain the estimate (3.5).

    Clearly

    |Aav(x)Aav(x)||a(x)| ||v(x)||u(x)|| |v(x)|+|a(x)||u(x)||v(x)u(x)|.

    Hence there exists a constant c1 such that

    AavAauC0(¯Ω)c1vuC0(¯Ω)(vC0(¯Ω)+uC0(¯Ω)). (3.8)

    We now calculate derivatives of A1v. If xΩ and v(x)0 then

    j[|v(x)|vk(x)]=|v(x)|jvk(x)+vk(x)[v(x)jv(x)]|v(x)|. (3.9)

    Hence

    |A1v(x)|(m+1)2|v(x)| vC1,0(¯Ω). (3.10)

    Let now xΩ and v(x)=0. Denote ej=(δ1j,,δmj). Then

    j[|v(x)|vk(x)]=limt0|v(x+tej)|[vk(x+tej)vk(x)]t=|v(x)|jvk(x)=0

    and (3.10) holds too. Suppose now that xΩ. If v(x)0 then j[|v(x)|vk(x)] can be continuously extended to x by (3.9). If v(x)=0 then (3.10) gives that A1v can be continuously extended to x by A1v(x)=0. The estimates (3.5) and (3.10) give that there exists a constant c2 such that

    A1vC1,0(¯Ω)c2v2C1,0(¯Ω). (3.11)

    Suppose that u,vC1,0(¯Ω;Rm). Let xΩ. Suppose first that u(x)=0. Since A1u(x)=0, (3.10) gives

    |A1v(x)A1u(x)|=|A1v(x)|(m+1)2|v(x)u(x)| vC1,0(¯Ω). (3.12)

    Let now |v(x)||u(x)|>0. According to (3.9)

    |j[|v(x)|vk(x)]j[|u(x)|uk(x)]||v(x)u(x)||jvk(x)|+|u(x)||jvk(x)juk(x)|
    +|u(x)||vk(x)uk(x)||v(x)jv(x)|+|u(x)||uk(x)||v(x)u(x)||jv(x)||v(x)| |u(x)|
    +|uk(x)||u(x)|2|jv(x)ju(x)|+|u(x)v(x)||uk(x)||u(x)ju(x)||v(x)| |u(x)|.
    6uvC1,0(¯Ω)(vC1,0(¯Ω)+uC1,0(¯Ω)).

    This inequality, (3.12) and (3.8) give that there exists a constant c3 such that

    A1uA1vC1,0(¯Ω)c3uvC1,0(¯Ω)(vC1,0(¯Ω)+uC1,0(¯Ω)).

    Since C1,0(¯Ω;Rm)C0,β(¯Ω;Rm) by [19,Lemma 6.36], there exists a constant c4 such that

    A1uA1vC0,β(¯Ω)c4uvC1,0(¯Ω)(vC1,0(¯Ω)+uC1,0(¯Ω)).

    Since Aav=aA1v, Lemma 3.7 gives that there exists a constant C1 such that (3.6) holds.

    Let now vC2,0(¯Ω;Rm). We are going to calculate 2A1v. If xΩ and v(x)0, then we obtain from (3.9)

    lj[|v(x)|vk(x)]=|v(x)|ljvk(x)+jvk(x)[v(x)lv(x)]|v(x)|+lvk(x)[v(x)jv(x)]+vk(x)[lv(x)jv(x)]+vk(x)[v(x)ljv(x)]|v(x)|vk(x)[v(x)jv(x)][v(x)lv(x)]|v(x)|3. (3.13)

    So,

    |lj[|v(x)|vk(x)]|6v2C2,0(¯Ω). (3.14)

    Now we calculate lj[|v(x)|vk(x)] in the sense of distributions. For ϵ0 denote Ω(ϵ):={xΩ;|v(x)|>ϵ}, V(ϵ):=Ω¯Ω(ϵ). Suppose that φC(Rm) has compact support in Ω. We have proved that if v(x)=0, then j[|v(x)|vk(x)]=0. Thus

    lj[|v|vk],φ=Ω[lφ(x)]j[|v(x)|vk(x)] dx
    =limϵ0Ω(ϵ)[lφ(x)]j[|v(x)|vk(x)] dx.

    According to the Green formula

    lj[|v|vk],φ=limϵ0[Ω(ϵ)φlj[|v|vk] dxΩ(ϵ)φnlj[|v|vk] dσ].

    If xΩ(ϵ)Ω then |v(x)|=ϵ. If xΩ(ϵ)Ω then φ(x)=0. Thus we obtain by (3.14) and (3.9)

    lj[|v|vk],φ=Ω(0)φlj[|v|vk]limϵ0Ω(ϵ)φnl[ϵjvk+vkvjvϵ]. (3.15)

    According to the Green formula

    |limϵ0Ω(ϵ)φnl[ϵjvk+vkvjvϵ]|limϵ0|ϵΩ(ϵ)φnljvk dσ|
    +limϵ0|V(ϵ)φnlvkvjvϵ dσ|=limϵ0ϵ|Ω(ϵ)l[φjvk] dx|
    +limϵ0|V(ϵ)1ϵ[lφvkvjv+φlvkvjv+φvklvjv+φvkvljv] dx|
    0+limϵ0V(ϵ)V(0)φC1(¯Ω)v2C2(¯Ω) dx=0.

    This and (3.15) give

    lj[|v|vk],φ=Ω(0)φlj[|v|vk].

    So, if we define lj[|v|vk](x) by (3.13) for v(x)0, and lj[|v|vk](x)=0 for v(x)=0, then this function is lj[|v|vk] in sense of distributions. (3.14) forces A1vW2,(Ω;Rm) and

    A1vW2,(Ω)c5v2C2,0(¯Ω).

    W2,(Ω)C1,β(¯Ω) compactly by [20,Theorem 6.3]. So,

    A1vC1,β(¯Ω)c6v2C2,0(¯Ω). (3.16)

    Since A1 maps bounded subsets of C2,0(¯Ω;Rm) to bounded subsets of W2,(Ω;Rm) and W2,(Ω)C1,β(¯Ω) compactly, the mapping A1:C2,0(¯Ω;Rm)C1,β(¯Ω;Rm) is compact. We now show that the mapping A1:C2,0(¯Ω;Rm)C1,β(¯Ω;Rm) is continuous. Suppose the opposite. Then there exist {vk}C2,0(¯Ω;Rm) and ϵ>0 such that vkv in C2,0(¯Ω;Rm) and A1vkA1vC1,β(¯Ω)>ϵ. Since A1:C2,0(¯Ω;Rm)C1,β(¯Ω;Rm) is compact, there exists a sub-sequence {vk(n)} of {vk} and wC1,β(¯Ω) such that A1vk(n)w in C1,β(¯Ω). Since A1:C2,0(¯Ω;Rm)C0,β(¯Ω;Rm) is continuous, it must be w=A1v. That is a contradiction.

    Since Aav=aA1v, Lemma 3.7 and (3.16) give that Aa:C2,0(¯Ω;Rm)C1,β(¯Ω;Rm) is a compact continuous mapping and (3.7) holds.

    Remark 3.3. Let Aa be from Lemma 3.2 with aC(¯Ω). We cannot show that Aa:Ck,β(¯Ω;Rm)Ck2,β(¯Ω;Rm) for k4 as shows the following example. Fix zΩ and define v(x):=(x2z2,0,,0). Then vC(Rm;Rm) but |v|vC2(Ω;Rm).

    Theorem 3.4. Let 0<βα<1 and kN. If k=1 suppose that β<α. Put l=max(k2,0). Let ΩRm be a bounded domain with boundary of class Ck,α. Let a,b,λCl,β(¯Ω) and λ0. If k3 suppose that a0. Then there exist δ,ϵ,C(0,) such that the following holds: If gCk,β(Ω;Rm) satisfying (2.3), FCl,β(¯Ω;Rm) and

    gCk,β(Ω)+FCl,β(¯Ω)<δ, (3.17)

    then there exists a unique solution (u,p)[Ck,β(¯Ω;Rm)C2(Ω;Rm)]×[Ck1,β(¯Ω)C1(Ω)] of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system

    pΔu+λu+a|u|u+b(u)u=F,u=0in Ω, (3.18a)
    u=gon Ω (3.18b)

    such that

    Ωp dx=0 (3.19)

    and

    uCk,β(¯Ω)<ϵ. (3.20)

    Moreover,

    uCk,β(¯Ω)+pCk1,β(¯Ω)C(gCk,β(Ω)+FCl,β(¯Ω)). (3.21)

    If ˜gCk,β(Ω;Rm), ˜FCl,β(¯Ω;Rm), ˜uCk,β(¯Ω;Rm)C2(Ω;Rm) and ˜pCk1,β(¯Ω)C1(Ω),

    ˜pΔ˜u+a|˜u|˜u+λ˜u+b(˜u)˜u=˜F,˜u=0in Ω, (3.22a)
    ˜u=˜gon Ω,Ω˜p dx=0 (3.22b)

    and ˜uCk,β(¯Ω)<ϵ, then

    u˜uCk,β(¯Ω)+p˜pCk1,β(¯Ω)C(g˜gCk,β(Ω)+F˜FCl,β(¯Ω)). (3.23)

    Proof. Let Lb be defined by (3.1), Aa be given by (3.4). Put Dabu:=Lb(u,u)+Aau. According to Lemma 3.1 and Lemma 3.2 there exists a constant C1 such that

    DabvCl,β(¯Ω)C1v2Ck,β(¯Ω), (3.24)
    DabvDabuCl,β(¯Ω)C1vuCk,β(¯Ω)[vCk,β(¯Ω)+uCk,β(¯Ω)]. (3.25)

    By Theorem 2.2 and Theorem 2.3 there exists a constant C2 such that for each gCk,β(Ω;Rm) satisfying (2.3) and fCl,β(¯Ω;Rm) there exists a unique solution (u,p)[Ck,β(¯Ω;Rm)C2(Ω;Rm)]×[Ck1,β(¯Ω)C1(Ω)] of the Dirichlet problem (2.2), (3.19). Moreover,

    uCk,β(¯Ω)+pCk1,β(¯Ω)C2(gCk,β(Ω)+fCl,β(¯Ω)). (3.26)

    Remark that (u,p) is a solution of (3.18) if (u,p) is a solution of (2.2) with f=FDabu. Put

    ϵ:=14(C1+1)(C2+1),δ:=ϵ2(C2+1).

    If (u,p),(˜u,˜p)[Ck,β(¯Ω;Rm)C2(Ω;Rm)]×[Ck1,β(¯Ω)C1(Ω)] are solutions of (3.18), (3.19) and (3.22) with (3.20) and ˜uCk,β(¯Ω)<ϵ, then

    u˜uCk,β(¯Ω)+p˜pCk1,β(¯Ω)C2[g˜gCk,β(Ω)+F˜FCl,β(¯Ω)
    +Dab(u,u)Dab(˜u,˜u)Cl,β(¯Ω)]C2[g˜gCk,β(Ω)
    +F˜FCl,β(¯Ω)+2ϵC1u˜uCk,β(¯Ω)].

    Since 2C1C2ϵ<1/2 we get subtracting 2ϵC1C2u˜uCk,β(¯Ω) from the both sides

    u˜uCk,β(¯Ω)+p˜pCk1,β(¯Ω)2C2[g˜gCk,β(Ω)+F˜FCl,β(¯Ω)].

    Therefore a solution of (3.18) satisfying (3.19) and (3.20) is unique. Putting ˜p0, ˜u0, ˜F0 and ˜g0, we obtain (3.21) with C=2C2.

    Denote X:={vCk,β(¯Ω,Rm);vCk,β(¯Ω)ϵ}. Fix gCk,β(Ω;Rm) and FCl,β(¯Ω;Rm) satisfying (2.3) and (3.17). For vX there exists a unique solution (uv,pv)[Ck,β(¯Ω;Rm)C2(Ω;Rm)]×[Ck1,β(¯Ω)C1(Ω)] of the Dirichlet problem (2.2), (3.19) with f=FDabv. Remember that (uv,pv) is a solution of (3.18) if and only if uv=v. According to (3.26), (3.17) and (3.24)

    uvCk,β(¯Ω)C2[gCk,β(Ω)+FCl,β(¯Ω)+DabvCl,β(¯Ω)]C2δ+C2C1ϵ2.

    As C2δ+C2C1ϵ2<ϵ, we infer uvX. If wX then

    uvuwCk,β(¯Ω)C2DabvDabwCl,β(¯Ω)C2C12ϵwvCk,β(¯Ω)

    by (3.26) and (3.25). Since C2C12ϵ<1, the Fixed point theorem ([21, Satz 1.24]) gives that there exists vX such that uv=v. So, (uv,pv) is a solution of (3.18), (3.19) in [Ck,β(¯Ω;Rm)C2(Ω;Rm)]×[Ck1,β(¯Ω)C1(Ω)] satisfying (3.20).

    In Theorem 3.4 we suppose that a0 for k3. This assumption cannot be removed for k>3 as Remark 3.3 shows. The following theorem is devoted to the case k=3.

    Theorem 3.5. Let 0<βα<1 and ΩRm be a bounded domain with boundary of class C3,α. Let a,b,λC1,β(¯Ω) and λ0. Then there exist δ,ϵ(0,) such that the following holds: If gC3,β(Ω;Rm) satisfies (2.3), FC1,β(¯Ω;Rm) and

    gC3,β(Ω)+FC1,β(¯Ω)<δ, (3.27)

    then there exists a unique solution (u,p)C3,β(¯Ω;Rm)×C2,β(¯Ω) of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system (3.18), (3.19) such that

    uC3,β(¯Ω)<ϵ. (3.28)

    Proof. Let Lb be defined by (3.1), Aa be given by (3.4). We conclude from Lemma 3.1 and Lemma 3.2 that there exists a constant C1 such that

    Lb(v,v)C1,β(¯Ω)+AavC1,β(¯Ω)C1v2C3,β(¯Ω), (3.29)
    Lb(v,v)Lb(u,u)C1,β(¯Ω)C1vuC3,β(¯Ω)[vC3,β(¯Ω)+uC3,β(¯Ω)]. (3.30)

    According to Theorem 2.3 there exists a constant C2 such that for each gC3,β(Ω;Rm) satisfying (2.3) and fC1,β(¯Ω;Rm) there exists a unique solution (u,p)C3,β(¯Ω;Rm)×C2,β(¯Ω) of the Dirichlet problem (2.2), (3.19). Moreover,

    uC3,β(¯Ω)+pC2,β(¯Ω)C2(gC3,β(Ω)+fC1,β(¯Ω)). (3.31)

    Suppose now that

    0<ϵ<14(C1+1)(C2+1),0<δ<ϵ2(C2+1). (3.32)

    Put Xϵ:={vC3,β(¯Ω,Rm);vC3,β(¯Ω)ϵ}. Fix gC3,β(Ω;Rm) and FC1,β(¯Ω;Rm) satisfying (2.3) and (3.27). For vXϵ there exists a unique solution (uv,pv)C3,β(¯Ω;Rm)×C2,β(¯Ω) of the Dirichlet problem (2.2), (3.19) with f=FLb(v,v). Moreover, there is a unique solution (˜uv,˜pv)C3,β(¯Ω;Rm)×C2,β(¯Ω) of

    ˜pvΔ˜uv+λ˜uv=a|v|v,˜uv=0in Ω,
    ˜uv=0on Ω,Ω˜pv dx=0.

    If uv+˜uv=v then (v,pv+˜pv) is a solution of the problem (3.18), (3.19). If wXϵ then

    uvC3,β(¯Ω)+˜uwC3,β(¯Ω)C2(gC3,β(Ω)+FC1,β(¯Ω)+Lb(v,v)C1,β(¯Ω)
    +AawC1,β(¯Ω))C2(δ+C1v2C3,β(¯Ω)+C1w2C3,β(¯Ω))<ϵ

    by (3.31), (3.27), (3.29) and (3.32). So, uv+˜uwXϵ. According to (3.31), (3.30) and (3.32)

    uvuwC3,β(¯Ω)C2Lb(v,v)Lb(w,w)C1,β(¯Ω)
    C1C2vwC3,β(¯Ω)[vC3,β(¯Ω)+wC3,β(¯Ω)]<12vwC3,β(¯Ω).

    So vuv is a contractive mapping on Xϵ. Since Aa:C3,β(¯Ω;Rm)C1,β(¯Ω;Rm) is a compact continuous mapping by Lemma 3.2, the mapping v˜uv is a compact continuous mapping on Xϵ. Lemma 3.8 forces that there exists vXϵ such that uv+˜uv=v. Hence (v,pv+˜pv) is a solution of the problem (3.18), (3.19), (3.28) in C3,β(¯Ω;Rm)×C1,β(¯Ω). By Theorem 3.4, for sufficiently small ϵ and δ there exists at most one solution of the problem (3.18), (3.19), (3.28) in C3,β(¯Ω;Rm)×C1,β(¯Ω).

    Supported by RVO: 67985840 and GAČR grant GA17-01747S.

    The author declares no conflict of interest in this paper.



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