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The paper is devoted to classical solutions of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system
∇p−Δv+λv+a|v|v+b(v⋅∇)v=f,∇⋅v=0in Ω | (1.1) |
where Ω⊂Rm 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)={u∈L2loc(Ω,R3);∂jui∈L2(Ω),(1+|x|2)−1/2uj(x)∈L2(Ω)}. The paper [3] is concerned with another transmission problem. A bounded domain Ω⊂Rm 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 ∂Ω−∩∂Ω+ 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)×Hs−1(Ω), where 1<s<3/2 and Ω⊂Rm 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 Ω⊂R3 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 Ω⊂R3 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 Ω⊂Rm is a bounded domain with Lipschitz boundary. For f≡0 and 2≤m≤3 solutions of the problem are looked for in Ws,2(Ω,Rm)×Ws−1,2(Ω) with 1≤s<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
∇p−Δv+λv=f,∇⋅v=0in Ω. |
If Ω⊂Rm is a bounded open set with boundary of class Ck,α we prove the existence of a solution (v,p)∈Ck,α(¯Ω;Rm)×Ck−1,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)×Ck−1,α(¯Ω). Here λ,a,b∈Cmax(k−2,0),α(¯Ω). If k≤3 then a can be arbitrary. If k>3 then there exists v∈C∞(¯Ω;Rm) with ∇⋅v=0 such that |v|v∉Ck−2,α(¯Ω;Rm). (See Remark 3.3.) So, for k>3 we must suppose that a≡0.
Before we investigate the Dirichlet problem for the Brinkman system we need the following auxiliary lemma.
Lemma 2.1. Let Ω⊂Rm be a bounded open set, 0<α<1 and λ∈C0,α(¯Ω) be non-negative. If f∈C0,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 ¯Ω⊂ω. Then we can suppose that f∈C0,α(¯ω,Rm) and λ∈C0,α(¯ω). (See [11,Theorem 1.8.3] or [12,Chapter Ⅵ,§2].) Choose q such that m/(1−α)<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=f−λv. Since v∈W1,q(ω)↪C0,α(¯ω) by [11,Theorem 5.7.8], we infer that v,F∈C0,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 ¯Ω⊂ω1⊂¯ω1⊂ω2⊂¯ω2⊂ω. Fix φ∈C∞(Rm) such that φ=1 on ω1 and φ=0 on Rm∖ω2. Define ˜F=φF in ω and ˜F=0 in Rm∖ω.
For x∈Rm∖{0} and i,j∈{1,2,…,m} define
Eij(x):=12σm{δij(m−2)|x|m−2+xixj|x|m},m≥3 |
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,i≤m, |
∂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):={σ−12ln|x|,m=2,(2−m)−1σ−1m|x|2−m,m>2 |
the fundamental solution for the Laplace equation. Since Fj∈C0,αloc(Rm), [14,Theorem 3.14.2] gives that hΔ∗˜Fj∈C2,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−˜p∈C2(ω1) by [14,Theorem 2.18.2]. Since ˜p∈C1,α(ω) we infer that p∈C1,αloc(ω1). Thus Δv=∇p−F∈C0,αloc(ω1;Rm). According to [14,Proposition 3.18.1] we obtain that v∈C2,αloc(ω1;Rm). (We can prove that v∈C2,αloc(ω1;Rm) also using results in [15].)
Theorem 2.2. Let 0<β<α<1. Suppose that Ω⊂Rm is a bounded domain with boundary of class C1,α and λ∈C0,β(¯Ω) is non-negative. Let f∈C0,β(¯Ω,Rm), g∈C1,β(∂Ω,Rm). Then there exist v∈C1,β(¯Ω;Rm)∩C2(Ω;Rm) and p∈C0,β(¯Ω)∩C1(Ω) solving
−Δv+λv+∇p=f,∇⋅v=0in Ω,v=gon ∂Ω | (2.2) |
if and only if
∫∂Ωn⋅g dσ=0. | (2.3) |
A velocity v is unique and a pressure p is unique up to an additive constant. Moreover,
‖v‖C1,β(¯Ω)+‖p‖C0,β(¯Ω)≤C(‖f‖C0,β(Ω)+‖g‖C1,β(∂Ω)+|∫Ωp dx|). |
Proof. If v∈C1,β(¯Ω;Rm)∩C2(Ω;Rm), p∈C0,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 ˆg∈C1,β(∂Ω,Rm). Choose q such that m/(1−β)<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 λ≡0 then ˆv∈C1,β(¯Ω;Rm), ˆp∈C0,β(¯Ω) by [16,Theorem 5.2]. Moreover, ˆv∈C∞(Ω;Rm), ˆp∈C∞(Ω). (See for example [17,§1.2].) Thus v∈C1,β(¯Ω;Rm)∩C2(Ω;Rm), p∈C0,β(¯Ω)∩C1(Ω).
Let now λ be general. Since v∈W1,q(Ω)↪C0,β(¯Ω) by [11,Theorem 5.7.8], Lemma 3.7 in the Appendix gives that λv∈C0,β(¯Ω). Therefore
−Δv+∇p=(f−λv)∈C0,β(¯Ω;Rm). |
We have proved that v∈C1,β(¯Ω;Rm)∩C2(Ω;Rm), p∈C0,β(¯Ω)∩C1(Ω).
Denote by Y the set of all g∈W1−1/q,q(∂Ω;Rm) satisfying (2.3). Define X:={v∈W1,q(Ω;Rm);∇⋅v=0 in Ω,v|∂Ω∈Y},
Uλ(v,p):=[−Δv+λv+∇p,v,∫Ωp dx]. |
Then Uλ:X×Lq(Ω)→W−1,q(Ω;Rm)×Y×R1 is an isomorphism by Lemma 3.6. Denote Z=[X∩C1,β(¯Ω;Rm)]×C0,β(¯Ω), W=C0,β(¯Ω,Rm)×[Y∩C1,β(∂Ω,Rm)]×R1. We have proved that U−1λ(W)⊂Z. Since U−1λ:W→Z is closed, it is continuous by the Closed graph theorem.
Theorem 2.3. Let 0<α<1 and k∈N0. Suppose that Ω⊂Rm is a bounded domain with boundary of class Ck+2,α and λ∈Ck,α(¯Ω) is non-negative. Let f∈Ck,α(¯Ω,Rm), g∈Ck+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,
‖v‖Ck+2,α(¯Ω)+‖p‖Ck+1,α(¯Ω)≤C(‖f‖Ck,α(¯Ω)+‖g‖Ck+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 g∈Ck+2,α(∂Ω;Rm) satisfying (2.3). Define X:={v∈Ck+2,α(¯Ω;Rm);∇⋅v=0 in Ω,v|∂Ω∈Y},
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 u↦λu 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 Uλ−U0: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 Ω⊂Rm be open, 0<α≤1 and k∈N. Put l=max(k−2,0). Let b∈Cl,α(¯Ω;Rm). Define
Lb(u,v):=b(u⋅∇)v. | (3.1) |
Then there exists C1∈(0,∞) such that if u,v∈Ck,α(¯Ω;Rm), then Lb(u,v)∈Cl,α(¯Ω;Rm) and
‖Lb(u,v)‖Cl,α(¯Ω)≤C1‖u‖Ck,α(¯Ω)‖v‖Ck,α(¯Ω), | (3.2) |
‖Lb(u,u)−Lb(v,v)‖Cl,α(¯Ω)≤C1‖u−v‖Ck,α(¯Ω)(‖v‖Ck,α(¯Ω)+‖u‖Ck,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(u−v,u)+Lb(v,u−v), |
the estimate (3.3) is a consequence of the estimate (3.2).
Lemma 3.2. Let Ω⊂Rm be a bounded domain with boundary of class C1,α and 0<β≤α<1. Suppose that a∈C0,β(¯Ω). For v∈C(¯Ω;Rm) define
Aav:=a|v|v. | (3.4) |
1. Then there exists a constant C1 such that for u,v∈C1,0(¯Ω;Rm) it holds
‖Aav‖C0,β(¯Ω)≤C1‖v‖2C1,0(¯Ω), | (3.5) |
‖Aav−Aau‖C0,β(¯Ω)≤C1‖v−u‖C1,0(¯Ω)[‖v‖C1,0(¯Ω)+‖u‖C1,0(¯Ω)]. | (3.6) |
2. If a∈C1,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
‖Aav‖C1,β(¯Ω;Rm)≤C2‖v‖2C2,0(¯Ω;Rm). | (3.7) |
Proof. Easy calculation yields that
‖ |v| ‖C0,β(¯Ω)≤‖v‖C0,β(¯Ω;Rm). |
So, according to Lemma 3.7 in the Appendix
‖Aav‖C0,β(¯Ω;Rm)≤4‖a‖C0,β(¯Ω)‖v‖2C0,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
‖Aav−Aau‖C0(¯Ω)≤c1‖v−u‖C0(¯Ω)(‖v‖C0(¯Ω)+‖u‖C0(¯Ω)). | (3.8) |
We now calculate derivatives of A1v. If x∈Ω 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)| ‖v‖C1,0(¯Ω). | (3.10) |
Let now x∈Ω and v(x)=0. Denote ej=(δ1j,…,δmj). Then
∂j[|v(x)|vk(x)]=limt→0|v(x+tej)|[vk(x+tej)−vk(x)]t=|v(x)|∂jvk(x)=0 |
and (3.10) holds too. Suppose now that x∈∂Ω. 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
‖A1v‖C1,0(¯Ω)≤c2‖v‖2C1,0(¯Ω). | (3.11) |
Suppose that u,v∈C1,0(¯Ω;Rm). Let x∈Ω. 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)| ‖v‖C1,0(¯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)|. |
≤6‖u−v‖C1,0(¯Ω)(‖v‖C1,0(¯Ω)+‖u‖C1,0(¯Ω)). |
This inequality, (3.12) and (3.8) give that there exists a constant c3 such that
‖A1u−A1v‖C1,0(¯Ω)≤c3‖u−v‖C1,0(¯Ω)(‖v‖C1,0(¯Ω)+‖u‖C1,0(¯Ω)). |
Since C1,0(¯Ω;Rm)↪C0,β(¯Ω;Rm) by [19,Lemma 6.36], there exists a constant c4 such that
‖A1u−A1v‖C0,β(¯Ω)≤c4‖u−v‖C1,0(¯Ω)(‖v‖C1,0(¯Ω)+‖u‖C1,0(¯Ω)). |
Since Aav=aA1v, Lemma 3.7 gives that there exists a constant C1 such that (3.6) holds.
Let now v∈C2,0(¯Ω;Rm). We are going to calculate ∇2A1v. If x∈Ω and v(x)≠0, then we obtain from (3.9)
∂l∂j[|v(x)|vk(x)]=|v(x)|∂l∂jvk(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)⋅∂l∂jv(x)]|v(x)|−vk(x)[v(x)⋅∂jv(x)][v(x)⋅∂lv(x)]|v(x)|3. | (3.13) |
So,
|∂l∂j[|v(x)|vk(x)]|≤6‖v‖2C2,0(¯Ω). | (3.14) |
Now we calculate ∂l∂j[|v(x)|vk(x)] in the sense of distributions. For ϵ≥0 denote Ω(ϵ):={x∈Ω;|v(x)|>ϵ}, V(ϵ):=Ω∖¯Ω(ϵ). Suppose that φ∈C∞(Rm) has compact support in Ω. We have proved that if v(x)=0, then ∂j[|v(x)|vk(x)]=0. Thus
⟨∂l∂j[|v|vk],φ⟩=−∫Ω[∂lφ(x)]∂j[|v(x)|vk(x)] dx |
=−limϵ↓0∫Ω(ϵ)[∂lφ(x)]∂j[|v(x)|vk(x)] dx. |
According to the Green formula
⟨∂l∂j[|v|vk],φ⟩=limϵ↓0[∫Ω(ϵ)φ∂l∂j[|v|vk] dx−∫∂Ω(ϵ)φnl∂j[|v|vk] dσ]. |
If x∈∂Ω(ϵ)∩Ω then |v(x)|=ϵ. If x∈∂Ω(ϵ)∖Ω then φ(x)=0. Thus we obtain by (3.14) and (3.9)
⟨∂l∂j[|v|vk],φ⟩=∫Ω(0)φ∂l∂j[|v|vk]−limϵ↓0∫∂Ω(ϵ)φnl[ϵ∂jvk+vkv⋅∂jvϵ]. | (3.15) |
According to the Green formula
|limϵ↓0∫∂Ω(ϵ)φnl[ϵ∂jvk+vkv⋅∂jvϵ]|≤limϵ↓0|ϵ∫∂Ω(ϵ)φnl∂jvk dσ| |
+limϵ↓0|∫∂V(ϵ)φnlvkv⋅∂jvϵ dσ|=limϵ↓0ϵ|∫Ω(ϵ)∂l[φ∂jvk] dx| |
+limϵ↓0|∫V(ϵ)1ϵ[∂lφvkv⋅∂jv+φ∂lvkv⋅∂jv+φvk∂lv⋅∂jv+φvkv⋅∂l∂jv] dx| |
≤0+limϵ↓0∫V(ϵ)∖V(0)‖φ‖C1(¯Ω)‖v‖2C2(¯Ω) dx=0. |
This and (3.15) give
⟨∂l∂j[|v|vk],φ⟩=∫Ω(0)φ∂l∂j[|v|vk]. |
So, if we define ∂l∂j[|v|vk](x) by (3.13) for v(x)≠0, and ∂l∂j[|v|vk](x)=0 for v(x)=0, then this function is ∂l∂j[|v|vk] in sense of distributions. (3.14) forces A1v∈W2,∞(Ω;Rm) and
‖A1v‖W2,∞(Ω)≤c5‖v‖2C2,0(¯Ω). |
W2,∞(Ω)↪C1,β(¯Ω) compactly by [20,Theorem 6.3]. So,
‖A1v‖C1,β(¯Ω)≤c6‖v‖2C2,0(¯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 vk→v in C2,0(¯Ω;Rm) and ‖A1vk−A1v‖C1,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 w∈C1,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 a∈C∞(¯Ω). We cannot show that Aa:Ck,β(¯Ω;Rm)→Ck−2,β(¯Ω;Rm) for k≥4 as shows the following example. Fix z∈Ω and define v(x):=(x2−z2,0,…,0). Then v∈C∞(Rm;Rm) but |v|v∉C2(Ω;Rm).
Theorem 3.4. Let 0<β≤α<1 and k∈N. If k=1 suppose that β<α. Put l=max(k−2,0). Let Ω⊂Rm be a bounded domain with boundary of class Ck,α. Let a,b,λ∈Cl,β(¯Ω) and λ≥0. If k≥3 suppose that a≡0. Then there exist δ,ϵ,C∈(0,∞) such that the following holds: If g∈Ck,β(∂Ω;Rm) satisfying (2.3), F∈Cl,β(¯Ω;Rm) and
‖g‖Ck,β(∂Ω)+‖F‖Cl,β(¯Ω)<δ, | (3.17) |
then there exists a unique solution (u,p)∈[Ck,β(¯Ω;Rm)∩C2(Ω;Rm)]×[Ck−1,β(¯Ω)∩C1(Ω)] of the Dirichlet problem for the Darcy-Forchheimer-Brinkman system
∇p−Δu+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
‖u‖Ck,β(¯Ω)<ϵ. | (3.20) |
Moreover,
‖u‖Ck,β(¯Ω)+‖p‖Ck−1,β(¯Ω)≤C(‖g‖Ck,β(∂Ω)+‖F‖Cl,β(¯Ω)). | (3.21) |
If ˜g∈Ck,β(∂Ω;Rm), ˜F∈Cl,β(¯Ω;Rm), ˜u∈Ck,β(¯Ω;Rm)∩C2(Ω;Rm) and ˜p∈Ck−1,β(¯Ω)∩C1(Ω),
∇˜p−Δ˜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 ‖˜u‖Ck,β(¯Ω)<ϵ, then
‖u−˜u‖Ck,β(¯Ω)+‖p−˜p‖Ck−1,β(¯Ω)≤C(‖g−˜g‖Ck,β(∂Ω)+‖F−˜F‖Cl,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
‖Dabv‖Cl,β(¯Ω)≤C1‖v‖2Ck,β(¯Ω), | (3.24) |
‖Dabv−Dabu‖Cl,β(¯Ω)≤C1‖v−u‖Ck,β(¯Ω)[‖v‖Ck,β(¯Ω)+‖u‖Ck,β(¯Ω)]. | (3.25) |
By Theorem 2.2 and Theorem 2.3 there exists a constant C2 such that for each g∈Ck,β(∂Ω;Rm) satisfying (2.3) and f∈Cl,β(¯Ω;Rm) there exists a unique solution (u,p)∈[Ck,β(¯Ω;Rm)∩C2(Ω;Rm)]×[Ck−1,β(¯Ω)∩C1(Ω)] of the Dirichlet problem (2.2), (3.19). Moreover,
‖u‖Ck,β(¯Ω)+‖p‖Ck−1,β(¯Ω)≤C2(‖g‖Ck,β(∂Ω)+‖f‖Cl,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=F−Dabu. Put
ϵ:=14(C1+1)(C2+1),δ:=ϵ2(C2+1). |
If (u,p),(˜u,˜p)∈[Ck,β(¯Ω;Rm)∩C2(Ω;Rm)]×[Ck−1,β(¯Ω)∩C1(Ω)] are solutions of (3.18), (3.19) and (3.22) with (3.20) and ‖˜u‖Ck,β(¯Ω)<ϵ, then
‖u−˜u‖Ck,β(¯Ω)+‖p−˜p‖Ck−1,β(¯Ω)≤C2[‖g−˜g‖Ck,β(∂Ω)+‖F−˜F‖Cl,β(¯Ω) |
+‖Dab(u,u)−Dab(˜u,˜u)‖Cl,β(¯Ω)]≤C2[‖g−˜g‖Ck,β(∂Ω) |
+‖F−˜F‖Cl,β(¯Ω)+2ϵC1‖u−˜u‖Ck,β(¯Ω)]. |
Since 2C1C2ϵ<1/2 we get subtracting 2ϵC1C2‖u−˜u‖Ck,β(¯Ω) from the both sides
‖u−˜u‖Ck,β(¯Ω)+‖p−˜p‖Ck−1,β(¯Ω)≤2C2[‖g−˜g‖Ck,β(∂Ω)+‖F−˜F‖Cl,β(¯Ω)]. |
Therefore a solution of (3.18) satisfying (3.19) and (3.20) is unique. Putting ˜p≡0, ˜u≡0, ˜F≡0 and ˜g≡0, we obtain (3.21) with C=2C2.
Denote X:={v∈Ck,β(¯Ω,Rm);‖v‖Ck,β(¯Ω)≤ϵ}. Fix g∈Ck,β(∂Ω;Rm) and F∈Cl,β(¯Ω;Rm) satisfying (2.3) and (3.17). For v∈X there exists a unique solution (uv,pv)∈[Ck,β(¯Ω;Rm)∩C2(Ω;Rm)]×[Ck−1,β(¯Ω)∩C1(Ω)] of the Dirichlet problem (2.2), (3.19) with f=F−Dabv. 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)
‖uv‖Ck,β(¯Ω)≤C2[‖g‖Ck,β(∂Ω)+‖F‖Cl,β(¯Ω)+‖Dabv‖Cl,β(¯Ω)]≤C2δ+C2C1ϵ2. |
As C2δ+C2C1ϵ2<ϵ, we infer uv∈X. If w∈X then
‖uv−uw‖Ck,β(¯Ω)≤C2‖Dabv−Dabw‖Cl,β(¯Ω)≤C2C12ϵ‖w−v‖Ck,β(¯Ω) |
by (3.26) and (3.25). Since C2C12ϵ<1, the Fixed point theorem ([21, Satz 1.24]) gives that there exists v∈X such that uv=v. So, (uv,pv) is a solution of (3.18), (3.19) in [Ck,β(¯Ω;Rm)∩C2(Ω;Rm)]×[Ck−1,β(¯Ω)∩C1(Ω)] satisfying (3.20).
In Theorem 3.4 we suppose that a≡0 for k≥3. 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<β≤α<1 and Ω⊂Rm be a bounded domain with boundary of class C3,α. Let a,b,λ∈C1,β(¯Ω) and λ≥0. Then there exist δ,ϵ∈(0,∞) such that the following holds: If g∈C3,β(∂Ω;Rm) satisfies (2.3), F∈C1,β(¯Ω;Rm) and
‖g‖C3,β(∂Ω)+‖F‖C1,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
‖u‖C3,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)+‖Aav‖C1,β(¯Ω)≤C1‖v‖2C3,β(¯Ω), | (3.29) |
‖Lb(v,v)−Lb(u,u)‖C1,β(¯Ω)≤C1‖v−u‖C3,β(¯Ω)[‖v‖C3,β(¯Ω)+‖u‖C3,β(¯Ω)]. | (3.30) |
According to Theorem 2.3 there exists a constant C2 such that for each g∈C3,β(∂Ω;Rm) satisfying (2.3) and f∈C1,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,
‖u‖C3,β(¯Ω)+‖p‖C2,β(¯Ω)≤C2(‖g‖C3,β(∂Ω)+‖f‖C1,β(¯Ω)). | (3.31) |
Suppose now that
0<ϵ<14(C1+1)(C2+1),0<δ<ϵ2(C2+1). | (3.32) |
Put Xϵ:={v∈C3,β(¯Ω,Rm);‖v‖C3,β(¯Ω)≤ϵ}. Fix g∈C3,β(∂Ω;Rm) and F∈C1,β(¯Ω;Rm) satisfying (2.3) and (3.27). For v∈Xϵ there exists a unique solution (uv,pv)∈C3,β(¯Ω;Rm)×C2,β(¯Ω) of the Dirichlet problem (2.2), (3.19) with f=F−Lb(v,v). Moreover, there is a unique solution (˜uv,˜pv)∈C3,β(¯Ω;Rm)×C2,β(¯Ω) of
∇˜pv−Δ˜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 w∈Xϵ then
‖uv‖C3,β(¯Ω)+‖˜uw‖C3,β(¯Ω)≤C2(‖g‖C3,β(∂Ω)+‖F‖C1,β(¯Ω)+‖Lb(v,v)‖C1,β(¯Ω) |
+‖Aaw‖C1,β(¯Ω))≤C2(δ+C1‖v‖2C3,β(¯Ω)+C1‖w‖2C3,β(¯Ω))<ϵ |
by (3.31), (3.27), (3.29) and (3.32). So, uv+˜uw∈Xϵ. According to (3.31), (3.30) and (3.32)
‖uv−uw‖C3,β(¯Ω)≤C2‖Lb(v,v)−Lb(w,w)‖C1,β(¯Ω) |
≤C1C2‖v−w‖C3,β(¯Ω)[‖v‖C3,β(¯Ω)+‖w‖C3,β(¯Ω)]<12‖v−w‖C3,β(¯Ω). |
So v↦uv 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 v∈Xϵ 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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