Angiuli, Luciana

English

Università del Salento: ESE - Salento University Publishing

Chapter 3Estimates of the derivatives ofsolution of parabolic problemsin L1(Ω)As a consequence of Theorem 2.5.2 and Proposition 1.2.7 we have that (A1, D(A1))is sectorial in L1(Ω), then it generates a bounded analytic semigroup T (t) and T (t)u0 isthe solution of ∂tw −Aw = 0 in (0,∞)× Ωw(0) = u0 in Ω〈ADw, ν〉 = 0 in (0,∞)× ∂Ω.for each u0 ∈ L1(Ω). Moreover there exist ci = ci(Ω, µ,M1), i = 0, 1 such that‖T (t)‖L(L1(Ω)) ≤ c0, t > 0 (3.1)andt‖A1T (t)‖L(L1(Ω)) ≤ c1, t > 0. (3.2)Moreover since D(A1) is dense in L1(Ω) by construction, T (t) is strongly continuous inL1(Ω). Hencelimt→0+‖T (t)u0 − u0‖L1(Ω) = 0 for all u0 ∈ L1(Ω) (3.3)Notice that for every u ∈ L1(Ω) and for every t > 0, T (t)u ∈W 2,1(Ω).3.0.1 Estimates of first order derivativesNow, using the gradient estimate (2.115) of the resolvent operator R(λ,A1), we esta-bilish the following further property of the semigroup T (t).6364Proposition 3.0.4. Let Ω, A and B be as in Section 2.5 and let T (t) be the semigroupgenerated by (A1, D(A1)). Then, there exists c2 depending on Ω, µ,M1 such that fort > 0,t1/2‖DT (t)‖L(L1(Ω)) ≤ c2. (3.4)Proof. Let θ′1 be as in Theorem 2.5.3 and suppose ω′1 = 0 (otherwise we considerA1 − ω′1). Let consider the curveΓ = {λ ∈ C; |arg λ| = θ′1, |λ| ≥ 1} ∪ {λ ∈ C : |arg λ| ≤ θ′1, |λ| = 1}oriented counterclockwise. We know that for t > 0T (t) =12pii∫ΓetλR(λ,A1)dλ.Setting λ′ = λt we getT (t) =12pii∫Γeλ′R(λ′/t, A1)t−1dλ′andDiT (t) =12pii∫Γeλ′t−1DiR(λ′/t, A1)dλ′ i = 1, . . . , ntherefore by (2.115)‖DiT (t)‖L(L1(Ω)) ≤ t−1/2∫ΓeReλ′ |λ′|−1/2d|λ′| ≤ ct−1/2 i = 1, . . . , nand the result is proved.Remark 3.0.5. [Neumann boundary conditions] We have stated Theorem 2.5.2 inthe form we most frequently use, but the estimates hold under more general assumptions.In particular, all non tangential boundary conditions are allowed. We denote by cν aconstant which can be used in the inequalities (3.1)–(3.4), when Neumann boundaryconditions are associated with a general uniformly elliptic operator.Remark 3.0.6. [Assumptions on the coefficients bi] The result of generation in L1and estimates (3.1), (3.2) can be achieved under weaker assumptions on coefficients bi.Assume A,B as in (2.106), (2.110) with coefficients satisfying (2.108), (2.107). Then weknow that (A1, D(A1)) generates an analytic semigroup in L1(Ω).We consider a first order perturbing operator C = ∑ni=1(b˜i − bi)Di with b˜i ∈ L∞(Ω)bi 6= b˜i. Let C1 be the realization of C in L1(Ω) with domain D(C1) = W 1,1(Ω). Theoperator C1 is A1− bounded and more precisely for every ε > 0 there exists c(ε) > 0such that‖C1u‖L1(Ω) ≤ ε‖A1u‖L1(Ω) + c(ε)‖u‖L1(Ω)holds for every u ∈ D(A1). Indeed let u ∈ D(A1), (suppose ω1 = 0, otherwise considerA1 − ω1) then u = R(λ,A1)f for every λ ∈ C with Re λ > 0 and f ∈ L1(Ω). Moreover,by (1.7) we can writeu =∫ ∞0e−λsT (s)fds, Reλ > 0.65Thus, in particular for λ > 0‖Du‖L1(Ω) ≤ c‖f‖L1(Ω)∫ ∞0e−λs√sds =c√λ‖f‖L1(Ω) ≤ c(√λ‖u‖L1(Ω) + 1√λ‖A1u‖L1(Ω))This implies that D(A1) ↪→W 1,1(Ω); moreover, minimizing over λ > 0, we get‖Du‖L1(Ω) ≤ c‖u‖1/2L1(Ω)‖A1u‖1/2L1(Ω) ≤ ε‖A1u‖L1(Ω) +cε‖u‖L1(Ω) (3.5)and by Theorem 1.2.10 we conclude. We point out that the first inequality in (3.5) assertsthat W 1,1(Ω) ∈ J1/2(L1(Ω), D(A1)).3.1 Estimates of second order derivativesIn order to proceed, we also need a precise L1-estimate of the second (spatial) deriva-tives of T (t)u0, for u0 ∈ D(A1). This is proved in Proposition 3.1.3 below. The argumentused here is similar to the one used in [18, Theorem 2.4], where Ω is bounded and dif-ferent boundary conditions are imposed. The scheme is the following: we estimate thesecond order derivatives in Proposition 3.1.1, and then, using this result, we characterizethe interpolation space DA(α, 1) = (L1(Ω), D(A))α,1 as a fractional Sobolev space anduse this to improve estimate (3.6) using the W 1,1 norm of u instead of the L1 norm. Westart with the following result.Proposition 3.1.1. Let Ω,A,B be as in Section 2.5. Assume, in addition, c ∈W 1,∞(Ω);then, there exists c3 depending on n, µ, Ω, M1, ‖c‖W 1,∞(Ω), c0 c1, c2, cν such that forevery t ∈ (0, 1) and u ∈ L1(Ω) we havet‖D2T (t)u‖L1(Ω) ≤ c3‖u‖L1(Ω). (3.6)Proof. We set for σ > 0 uσ = T (σ)u andM2 = max{‖A‖2,∞, ‖B‖2,∞, ‖c‖1,∞}. (3.7)By the regularity of the boundary ∂Ω we can consider a partition of unity {(ηh,Uh)}h∈Nsuch that supp ηh ⊂ Uh,∑∞h=0 ηh(x) = 1 for every x ∈ Ω and 0 ≤ ηh ≤ 1 for every h ∈ N,U0 ⊂ Ω, Uh for h ≥ 1 is a ball such that {Uh}h≥1 is a covering of ∂Ω and {Uh}h∈N is acovering of Ω with bounded overlapping, that is there is κ > 0 such that∑h∈NχUh(x) ≤ κ, ∀x ∈ Ω. (3.8)Moreover we choose ηh in such a way 〈A(x)Dηh(x), ν(x)〉 = 0 for every x ∈ ∂Ω and setM := suph∈N ‖ηh‖2,∞, which is finite by the uniform C2 regularity of ∂Ω. We can alsoconsider coordinate functions ψh : Vh → B(0, 1) such that ψh(Vh ∩ Ω) = B+(0, 1) ={y = (y′, yn) ∈ B(0, 1) : yn > 0}, ψh(Vh ∩ ∂Ω) = {y = (y′, yn) ∈ B(0, 1) : yn = 0},66d(ψh)x(a(x)ν(x)) = −en for every x ∈ ∂Ω where d(ψh)x denotes the differential of ψh atx. Finally we suppose that there is a constant Mψ such thatsuph≥1{‖D2ψh‖2,∞, ‖D2ψ−1h ‖2,∞} ≤Mψ.Notice also that we may assume that for all h ≥ 1 the inclusion Uh ⊂⊂ Vh holds, andthat we can choose a C2 domain E such that ψh(Uh ∩ Ω) ⊂ E ⊂ B+(0, 1). Notice thatuσ ∈W 1,1(Ω) and denote by u(t) = T (t)uσ the solution of the problem∂tw −Aw = 0 in (0,∞)× Ωw(0) = uσ in Ω〈ADw, ν〉 = 0 in (0,∞)× ∂Ω.We want to estimate the L1–norm of tD2u(t) by the L1-norm of u; we shall use estimates(3.1)–(3.4). The functions vh(t) = u(t)ηh solve, for every h ∈ N, the problem∂tw −Aw = Ahu(t) in (0,∞)× Ωw(0) = ηhuσ in Ω〈ADw, ν〉 = 0 in (0,∞)× ∂Ω(3.9)whereAhu(t) = −2〈ADηh, Du(t)〉 − u(t) div(ADηh)− u(t) 〈B,Dηh〉 . (3.10)Notice that the derivative Dkvh(t) satisfies the equation ∂t(Dkvh(t)) − A(Dkvh(t)) =Akhu(t), whereAkhu(t) =div ((DkA)D(u(t)ηh)) + 〈(DkB), D(u(t)ηh)〉+ (Dkc)u(t)ηh +Dk(Ahu(t))=div ((DkA)D(u(t)ηh)) + 〈(DkB), D(u(t)ηh)〉+ (Dkc)u(t)ηh (3.11)+Dk[−2〈ADηh, Du(t)〉 − u(t) div(ADηh)− u(t) 〈B,Dηh〉]For Dkvh(t) we consider the problem∂tw −Aw = Akhu(t) in (0,∞)× Ωw(0) = Dk(ηhuσ) in Ω〈ADw, ν〉 = 0 in (0,∞)× ∂Ω(3.12)whose solution is vhk(t) = T (t)Dk(ηhuσ)+∫ t0T (t− s)Akhu(s)ds. Now we consider h = 0,i.e., we draw our attention to the inner part. Since v0(t) = η0u(t) = 0 in Ω \U0, it turnsout that Dkv0(t) is the solution of (3.12) with h = 0. ThenDkv0(t) = T (t)Dk(η0uσ) +∫ t0T (t− s)Ak0u(s)ds, (3.13)where Ak0 is the operator defined in (3.11). Then, differentiating, we obtainD2lkv0(t) = Dl[T (t)Dk(η0uσ)] +∫ t0Dl[T (t− s)Ak0v(s)]ds.67by which, using (3.4),‖D2lkv0(t)‖L1(Ω) ≤ ‖DlT (t)Dk(η0uσ)‖L1(Ω) +∫ t0‖DlT (t− s)Ak0u(s)‖L1(Ω)ds≤ c2√t‖Dk(η0uσ)‖L1(Ω) +∫ t0c2√t− s‖Ak0u(s)‖L1(Ω)ds≤ c22√t‖η0‖W 1,∞‖uσ‖W 1,1(Ω) +∫ t0c2√t− s‖Ak0u(s)‖L1(Ω)dsFinally, estimating ‖Ak0u(s)‖L1(Ω) by (3.11) we get ‖Ak0u(s)‖L1(Ω) ≤ c ‖u(s)‖W 2,1(Ω)where c = c(M,M2). Summing on l and k, using (A.1) and again (3.1), we get‖D2v0(t)‖L1(Ω) ≤ c( 1√t‖uσ‖W 1,1(Ω) +∫ t01√t− s‖D2u(s)‖L1(Ω)ds)where c = c(M,M2, c2, n). We now consider h ≥ 1, i.e., we consider a ball intersecting∂Ω.Using the transformation fˆ(y) := f(ψ−1h (y)) for a generic f defined in Ω ∩ Vh, andsince vh is the solution of (3.9), we get that for every h ≥ 1 the function vˆh(t, y) =ηh(ψ−1h (y))u(t, ψ−1h (y)) is the solution of the following initial-boundary value problemwith homogeneous Neumann boundary conditions∂tw − Aˆw = Aˆhvˆ in (0,+∞)× Ew(0) = ηˆhuˆσ in E∂w∂ν= 0 in (0,+∞)× ∂E(3.14)where Aˆ is the operator defined on B(0, 1) as followsAˆw := div(AˆDw) + 〈Bˆ,Dw〉+ cˆwwhose coefficients (here we omit the index h to simplify the notations and by analogywith (3.9)) are given byAˆ(y) :=(Dψh)(ψ−1h (y)) ·A(ψ−1h (y)) ·(Dψh)t(ψ−1h (y))(Bˆ(y))l :=Tr[(Dψh)(ψ−1h (y)) ·A(ψ−1h (y)) ·H l(ψ−1h (y)) ·(Dψ−1h)t(y)]+Tr[(Dψh)(ψ−1h (y)) ·Gj(y)](Dψh)tjl(ψ−1h (y))−∂∂yj[aˆjl(y)]+[(Dψh)(ψ−1h (y)) ·B(ψ−1h (y))]lcˆ(y) :=c(ψ−1h (y))where H lki = D2ki(ψh)l and Gjki = Dkaij(ψ−1h (y)) and (see (3.10))Aˆhuˆ(t) = −2〈A(ψ−1h (y))(Dψh)tDηˆh, (Dψh)tDuˆ(t)〉− uˆ(t)[div(AˆDηˆh) + 〈Aˆ,Dηˆh〉] .Now, as done before for h = 0, differentiating the equation (now Dk = ∂∂yk ) we obtainthat Dkvˆh solves ∂t(Dkvˆh(t)) − Aˆ(Dkvˆh(t)) = Aˆkhuˆ(t), where Aˆkhvˆ can be obtained by68taking the corresponding term in (3.11). Associated with this operator, we can considerthe problem ∂tw − Aˆw = Aˆkhuˆ(t) in (0,∞)× Ew(0) = Dk(ηˆhuˆσ)in E∂w∂ν= 0 in (0,∞)× ∂E.The function Dkvˆh satisfies the equation and the initial condition. Notice that if k 6= nalso the boundary condition is satisfied since vˆh = 0 in a neighborhood of ∂E ∩ {y ∈Rn | yn > 0}, in the other part of ∂E the operator Dk is a tangential derivative and∂vˆh∂ynis constant for yn = 0. Denote by S the semigroup which gives the solution of thisproblem and notice that the estimates (3.1)–(3.4) hold for S(t), see Remark 3.0.5. ThenDkvˆh(t) = S(t)Dkvˆh(0) +∫ t0S(t− s)Aˆkhuˆ(s)ds. (3.15)Differentiating (3.15) with respect to Dj for any j, we have then proved that the followingholdsD2kj vˆh(t) = DjS(t)Dkvˆh(0) +∫ t0DjS(t− s)Aˆkhuˆ(s)ds. (3.16)Thus, as for v0(t), we have for (k, j) 6= (n, n)‖D2kj vˆh(t)‖L1(E) ≤c2√t‖Dk(ηˆhuˆσ)‖L1(E) +∫ t0c2√t− s‖Aˆkhuˆ(s)‖L1(E)ds≤ c√tσ‖uˆ‖L1(E) +∫ t0c2√t− s‖Aˆkhuˆ(s)‖L1(E)ds.We now estimate D2nnvˆh(t). SinceaˆnnD2nnvˆh(t) = Aˆvˆh(t)−∑(i,j) 6=(n,n)aˆijD2ij vˆh(t)−n∑i,j=1(Diaˆij)Dj vˆh(t)−n∑i=1bˆiDivˆh(t)− cˆvˆh(t)and since aˆ is uniformly elliptic with ellipticity constant proportional to µ, we can finda constant c (depending only on n,M2, µ, ∂Ω) such that‖D2nnvˆh(t)‖L1(E) =∥∥∥ 1aˆnn(Aˆvˆh(t)−∑(i,j) 6=(n,n)aˆijD2ij vˆh(t) +−n∑i,j=1(Diaˆij)Dj vˆh(t)−n∑i=1bˆiDivˆh(t)− cˆvˆh(t))∥∥∥L1(E)≤ c[ ∑(i,j) 6=(n,n)‖D2ij vˆh(t)‖L1(E) + ‖Aˆvˆh(t)‖L1(E) + ‖Dvˆh(t)‖L1(E) + ‖vˆh(t)‖L1(E)].69Summing up, we may argue in the same way as for h = 0, and get‖D2vˆh(t)‖L1(E) ≤ c′[ 1√t‖uσ ◦ ψ−1h ‖W 1,1(E) +∫ t01√t− s‖D2uˆ(s)‖L1(E)ds+ ‖Aˆvˆh(t)‖L1(E)]≤ c′[ 1√tσ‖u ◦ ψ−1h ‖L1(E) +∫ t01√t− s‖D2uˆ(s)‖L1(E)ds+ ‖Aˆvˆh(t)‖L1(E)]where c′ = c(M,M2,Mψ, n, c2, cν). Coming back to Ω ∩ Uh we obtain‖D2vh(t)‖L1(Ω∩Uh) ≤ c′′[ 1√t‖uσ‖W 1,1(Ω∩Uh) +∫ t01√t− s‖D2u(s)‖L1(Ω∩Uh)ds+ ‖Avh(t)‖L1(Ω∩Uh)](3.17)≤ c′′[ 1√tσ‖u‖L1(Ω∩Uh) +∫ t01√t− s‖D2u(s)‖L1(Ω∩Uh)ds+ ‖Avh(t)‖L1(Ω∩Uh)]where c′′ depends on M,M2,Mψ, n, c2, cν . Now, using (3.1), (3.2) and (3.8), we have‖D2u(t)‖L1(Ω) = ‖D2( ∞∑h=0vh(t))‖L1(Ω) = ‖ ∞∑h=0D2vh(t)‖L1(Ω) (3.18)≤ κc′′[ 1√t‖uσ‖W 1,1(Ω) +∫ t01√t− s‖D2u(s)‖L1(Ω)ds+ ‖Au(t)‖L1(Ω)]≤ c′′′[ 1√tσ‖u‖L1(Ω) +∫ t01√t− s‖D2u(s)‖L1(Ω)ds+ 1√tσ‖u‖L1(Ω)],where c′′′ depends on κ, c′′, c0, c1. Now using Gronwall’s generalized inequality (seeLemma 1.5.7), we get‖D2u(t)‖L1(Ω) ≤ c√tσ‖u‖L1(Ω). (3.19)Then, by taking σ = t, we get ‖D2u(t)‖L1(Ω) ≤ c3t−1‖u‖L1(Ω) for every t ∈ (0, 1).3.1.1 Characterization of interpolation spaces between D(A1) andL1(Ω)We can use Proposition 3.1.1 to characterize some interpolation spaces betweenD(A1)and L1(Ω).Theorem 3.1.2. Let A1 be as in Proposition 3.1.1; then for every α ∈ (0, 1/2) we have(L1(Ω), D(A1))α,1 =W 2α,1(Ω)where W 2α,1 denotes the Sobolev space of fractional order (see Section A.2.1 for details).Proof. It is sufficient to prove that(L1(Ω), D(A1))α,1 = (L1(Ω),W 2,1(Ω) ∩W 1,1A,ν(Ω))α,170in fact using Theorem A.2.7 we complete the proof.First of all, let us observe thatW 2,1(Ω)∩W 1,1A,ν(Ω) ↪→ D(A1). Therefore, using DefinitionA.2.2, we obtain(L1(Ω),W 2,1(Ω) ∩W 1,1A,ν(Ω))α,1 ↪→ (L1(Ω), D(A1))α,1.Conversely, let u0 ∈ (L1(Ω), D(A1))α,1 and set for t ∈ [0, 1]u0 = u0 − T (t)u0 + T (t)u0 = −∫ t0A1T (s)u0ds+ T (t)u0 = v1 + v2.We have‖v1‖L1(Ω) ≤∫ t0‖A1T (s)u0‖L1(Ω)dsand since v2 ∈W 2,1(Ω) ∩W 1,1A,ν(Ω), using (A.1), (3.1) and Proposition 3.1.1, we have‖v2‖W 2,1(Ω) = ‖T (t)u0‖L1(Ω) +n∑i,j=1‖Dij [T (t)u0 − T (1)u0 + T (1)u0]‖L1(Ω)≤ c0‖u0‖L1(Ω) +n∑i,j=1‖Dij∫ 1tT (s/2)A1T (s/2)u0ds‖L1(Ω) + c3‖u0‖L1(Ω)≤ c{‖u0‖L1(Ω) +∫ 1ts−1‖A1T (s/2)u0‖L1(Ω)ds}Therefore for t ∈ [0, 1], setting K(t, u0) := K(t, u0, L1(Ω),W 2,1(Ω)∩W 1,1A,ν(Ω)) we obtainK(t, u0) = infu0=u10+u20(‖u10‖L1(Ω) + t‖u20‖W 2,1(Ω))≤ ‖v1‖L1(Ω) + t‖v2‖W 2,1(Ω)≤ c(∫ t0‖A1T (s)u0‖L1(Ω)ds+ t‖u0‖L1(Ω)+ t∫ 1ts−1‖A1T (s/2)u0‖L1(Ω)ds)On the other hand, choosing u10 = u0 and u20 = 0 we getK(t, u0) ≤ ‖u0‖L1(Ω).ThereforeK(t, u0) ≤ c(min(1, t)‖u0‖L1(Ω) +∫ t0‖A1T (s)u0‖L1(Ω)ds+ t∫ 1ts−1‖A1T (s/2)u0‖L1(Ω)ds).71Therefore for each α ∈ (0, 1) we get∫ ∞0t−(1+α)K(t, u0)dt ≤ c{‖u0‖L1(Ω)∫ ∞0t−(1+α)min(1, t)dt+∫ ∞0(t−(1+α)∫ t0‖A1T (s)u0‖L1(Ω)ds)dt+∫ ∞0(t−α∫ ∞ts−1‖A1T (s)u0‖L1(Ω)ds)dt}so that using Hardy inequalities stated in Theorem 1.5.6, we get∫ ∞0t−(1+α)K(t, u0)dt ≤ c{‖u0‖L1(Ω) +∫ ∞0s−α‖A1T (s)u0‖L1(Ω)ds}and hence from Theorem 1.3.2 we get(L1(Ω), D(A1))α,1 ↪→ (L1(Ω),W 2,1(Ω) ∩W 1,1A,ν(Ω))α,1so, the result is proved.Using Theorem 3.1.2 we can improve the estimate of Proposition 3.1.1, under addi-tional assumption on the initial datum; in fact, we have the following.Proposition 3.1.3. Let Ω,A,B be as in Section 2.5. Assume, in addition, c ∈W 1,∞(Ω);then, there exist δ ∈ (1/2, 1) and c4 depending on n, µ, Ω, M2, c0, c1, c2, c3 cν such thatfor every t ∈ (0, 1) and u ∈ D(A1) we havetδ‖D2T (t)u‖L1(Ω) ≤ c4‖u‖W 1,1(Ω). (3.20)Proof. We can repeat the proof of Proposition 3.1.1 until the first inequality in(3.18), with σ > 0, so that we have‖D2u(t)‖L1(Ω) ≤κc′′[ 1√t‖uσ‖W 1,1(Ω)+∫ t01√t− s‖D2u(s)‖L1(Ω)ds+ ‖Au(t)‖L1(Ω)](3.21)Using (1.10), we get that for any α, β ∈ (0, 1) there is C such thatt1−α+β‖AT (t)u‖DA(β,1) ≤ C‖u‖DA(α,1).By definition of interpolation, DA(β, 1) is continuously embedded in L1(Ω) for any β ∈(0, 1). Using the fact that DA(α, 1) is the fractional Sobolev space W 2α,1(Ω) for α < 1/2and that W 1,1(Ω) embeds in W 2α,1(Ω) for such α, we obtain, with constants C that maychange from a line to the other,‖AT (t)u‖L1(Ω) ≤C‖AT (t)u‖DA(β,1) ≤Ct1−α+β‖u‖DA(α,1)=Ct1−α+β‖u‖W 2α,1(Ω) ≤ Ct1−α+β‖u‖W 1,1(Ω)72We choose then α ∈ (0, 1/2) and β ∈ (0, 1) is such a way that δ = 1− α + β ∈ (1/2, 1),and (3.21) becomes‖D2u(t)‖L1(Ω) ≤ Ctδ‖uσ‖W 1,1(Ω) +∫ t0C ′√t− s‖D2u(s)‖L1(Ω)ds.Therefore applying the Gronwall’s lemma and passing to the limit as σ → 0 we get(3.20).

Estimates of the derivatives of solution of parabolic problems in L1(Ω)

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Estimates of the derivatives of solution of parabolic problems in L1(Ω)

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