Angiuli, Luciana

English

Università del Salento: ESE - Salento University Publishing

Appendix BHeat kernel estimates ondomainsIn this section we collect some upper and lower estimates for the integral kernel ofthe semigroup associated with the parabolic problem∂tw −Aw = 0 in (0,∞)× Ωw(0) = u0 in Ω〈ADw, ν〉 = 0 in (0,∞)× ∂Ω.(B.1)under the hypotheses summarized at the beginning of Chapter 5. Since we shall deal withseveral semigroups, the exponential notation seems to us to be clearer, as it emphasizesthe relevant elliptic generator. In fact we considerA0 = div(A ·D), A′ = div(A ·D) +B ·D and A = div(A ·D) +B ·D + cand the related semigroups e−tA0 , e−tA′and e−tA whose kernels p0, p′ and p are suchthat, e.g.,e−tAf(x) =∫Ωp(t, x, y)f(y) dyand the analogous expressions for e−tA0 and e−tA′hold.We first recall upper estimates directly for p, that are well-known. On the contrary, lowerestimates are known in the symmetric case, i.e., for p0. After observing that there is nodifficulty in passing from p′ to p, we shall deduce lower estimates for p′, deducing themfrom those on p0 via a perturbation argument. The proofs in Section B.2.2 are due to G.Metafune, E.M. Ouhabaz and D. Pallara whom we thank for communicating the aboveresults and allowing us to reproduce them here.125126B.1 Gaussian upper bounds for heat kernelsWe collect the known Gaussian upper bound results in the following statement andwe refer to [45, Theorem 5.7] for the proof.Theorem B.1.1. (Kernel estimates)Let Ω be an open set of Rn uniformly regular of class C2. Let A, B be as in (2.3)–(2.7)and let (T (t))t≥0 be the analytic semigroup generated by the realization of A in L1(Ω)with homogeneous boundary conditions Bu = 0; for the kernel p : (0,+∞)×Ω×Ω→ R ofthe semigroup (T (t))t≥0 the following estimates hold: there exist b, c1 > 0, a real numberω such that for |α|, |β| < 2, x, y ∈ Ω and t > 0|DαxDβy p(t, x, y)| ≤c1tn+|α|+|β|2e−b|x−y|2t eωt. (B.2)B.1.1 Some norm estimatesImmediate consequences of the Gaussian upper bound are the following L1 −Lp andLp − L∞ estimates.Proposition B.1.2. Let p ≥ 1 and let e−tA be the semigroup generated by A. Thenthere exist c2, c3 > 0 such that‖e−tA‖L(L1,Lp) ≤ c2t−n2 (1− 1p ) 0 < t < 1, (B.3)and‖e−tA‖L(Lp,L∞) ≤ c3t−n2p 0 < t < 1. (B.4)Proof. Let f ∈ L1(Ω); then, using (B.2) we get‖e−tAf‖pLp(Ω) =∫Ω∣∣ ∫Ωp(t, x, y)f(y) dy∣∣p dx≤ ‖f‖pL1(Ω)∫Ω‖p(t, x, ·)‖pL∞(Ω) dx≤ c1t−np/2‖f‖pL1(Ω)∫Ωe−b|x−y|2t dx= c′1t−n2 (p−1)‖f‖pL1(Ω)Thus‖e−tAf‖Lp(Ω) ≤ c2t−n2 (1− 1p )‖f‖L1(Ω)127and (B.3) is proved. Similarly, let f ∈ Lp, and p′ = p/(p− 1), then again by (B.2)‖e−tAf‖L∞(Ω) = supx∈Ω∣∣∣ ∫Ωp(t, x, y)f(y) dy∣∣∣≤ ‖f‖Lp(Ω) supx∈Ω‖p(t, x, ·)‖Lp′ (Ω)≤ c1‖f‖Lp(Ω) supx∈Ω(t−n2 p′∫Ωe−bp′ |x−y|2t dy)1/p′= c3t−n(p′−1)2p′ ‖f‖Lp(Ω) = c3t−n2p ‖f‖Lp(Ω).B.2 Gaussian lower boundsThis section is devoted to obtain Gaussian lower bounds for p(t, x, y). Such lowerbounds in the symmetric case can be deduced from Gaussian upper bounds and Ho¨ldercontinuity of the kernel.Remark B.2.1. One can easily observe that if some Gaussian lower bounds are esta-blished for p′, the same hold for p, more precisely p(t, x, y) ≥ e−ωtp′(t, x, y). Indeed,since c ∈ L∞(Ω), then there exists ω > 0 such that −ω ≤ c(x) ≤ ω a.e. x ∈ Ω. Let f ≥ 0in Ω and consider u and v solutions respectively of the problems∂tu = div(A ·Du) +B ·Du in (0,+∞)× Ωu(0, x) = f(x) in Ω〈ADu, ν〉 = 0 in (0,+∞)× ∂Ω.(B.5)and ∂tv = div(A ·Dv) +B ·Dv + cv in (0,+∞)× Ωv(0, x) = f(x) in Ω〈ADv, ν〉 = 0 in (0,+∞)× ∂Ω.(B.6)By the maximum principle we deduce that u ≥ 0. We want to prove that v ≥ e−ωtu,hence p(t, x, y) ≥ e−ωtp′(t, x, y) as announced. The problem satisfied by z = v−w, withw = e−ωtu, is ∂tz −Az = (c+ ω)w ≥ 0 in (0,+∞)× Ωz(0, x) = 0 in Ω〈ADz, ν〉 = 0 in (0,+∞)× ∂Ω.(B.7)Thus applying again the maximum principle we deduce z ≥ 0, i.e.p(t, x, y) ≥ p′(t, x, y)e−ωt.As a consequence of Remark B.2.1 we can restrict the study to the operator A′ = A− cand our aim will be to deduce Gaussian lower bound for p′.128B.2.1 The symmetric caseWe first consider the symmetric case and show lower bounds for p0 (more details arecontained in [34]). Under our assumptions on the coefficients, p0 is Ho¨lder continuous,that is|p0(t, x, y)− p0(t, x′, y)| ≤ kt−n/2−γ/2|x− x′|γ , for all x, x′, y ∈ Ω (B.8)for some γ > 0 and k > 0 independent on y. Moreover it satisfies the Gaussian upperbound in Theorem B.1.1 and the conservation property holds:∫Ωp0(t, x, y) dy = 1 forall t > 0 and x ∈ Ω.The first step shows that an on-diagonal lower bound can be deduced from a Gaussianupper bound and the conservation property.Proposition B.2.2. There exists a constant C > 0 such that for all t > 0 and a.e.x ∈ Ωp0(t, x, x) ≥ Ct−n/2 (B.9)Proof. Fix δ > 0; we have∫Ω\B(x,δ√t)p0(t, x, y) dy ≤ c1t−n/2∫Ω\B(x,δ√t)e−b2|x−y|2t e−b2|x−y|2t dy≤ c1t−n/2e− b2 δ2∫Rne−b2|x−y|2t dy≤ ke− b2 δ2 .Now, for δ large enough, ke−b2 δ2 ≤ 12 , thus a.e. x ∈ Ω∫Ω∩B(x,δ√t)p0(t, x, y) dy = 1−∫Ω\B(x,δ√t)p0(t, x, y) dy≥ 12.It follows by the semigroup property and the symmetry of p0 thatp0(t, x, x) =∫Ωp0(t/2, x, y)p0(t/2, y, x) dy=∫Ω|p0(t/2, x, y)|2 dy≥∫Ω∩B(x,δ√t)|p0(t/2, x, y)|2 dy≥ 1|Ω ∩B(x, δ√t)|(∫Ω∩B(x,δ√t)p0(t/2, x, y) dy)2≥ 14|B(x, δ√t)| ≥ Ct−n/2129for some positive constant C.The following step consists in deducing an off-diagonal Gaussian lower bound fromthe on-diagonal one, by exploiting the Ho¨lder continuity of p0.Proposition B.2.3. There exist positive constants C and η such thatp0(t, x, y) ≥ Ct−n/2 (B.10)for all x, y ∈ Ω and t > 0, sufficiently small such that |x− y| ≤ η√t.Proof. Since by (B.8)|p0(t, x, y)− p0(t, x′, y)| ≤ kt−n2−γ2 |x− x′|γfor all x, x′, y ∈ Ω we havep0(t, x, y) ≥ p0(t, y, y)− kt−n/2−γ/2|x− y|γThus, using estimate (B.9),p0(t, x, y) ≥ Ct−n/2 − kt−n/2( |x− y|√t)γ= Ct−n/2(1−( |x− y|√t)γ)≥ Ct−n/2for |x− y| ≤ 12√t, which shows (B.10).Let us now extend the previous estimate to arbitrary x, y in Ω.Theorem B.2.4. Let p0(t, x, y) be the heat kernel of A0. There exist constants c0, C0 > 0such thatp0(t, x, y) ≥ C0t−n/2e−c0|x−y|2t (B.11)for all x, y ∈ Ω and t > 0.Proof. Let x, y ∈ Ω. Fix N ∈ N and consider a finite sequence of points xi,0 ≤ i ≤ N in Ω such that x0 = x, xN = y, [xi, xi+1] ⊂ Ω and |xi − xi+1| ≤ K |x−y|N =: rfor all i = 0, . . . , N − 1. Then by the semigroup property and the positivity of p0(t, x, y),we havep0(t,x, y) =∫Ω. . .∫Ωp0(tN, x, z1)p0(tN, z1, z2) . . . p0(tN, zN−1, y) dz1 . . . dzN−1≥∫B(x1,r)∩Ω. . .∫B(xN−1,r)∩Ωp0(tN, x, z1)p0(tN, z1, z2) . . . p0(tN, zN−1, y) dz1 . . . dzN−1Let us observe that if zi ∈ B(xi, r) and zi+1 ∈ B(xi+1, r) (where we have set z0 = x andzN = y), then it holds that|zi − zi+1| ≤ |xi − xi+1|+ 2r ≤ (K + 2)r i = 0, . . . , N − 1.130If (K + 2)|x − y| ≤ η√t (η as in Proposition B.2.3) then |x − y| ≤ η√t. In this case(B.11) follows from (B.9) and Proposition B.2.3.If (K + 2)|x− y| > η√t, we choose N ≥ 2 to be the smallest integer such that(K + 2)|x− y|√N≤ η√t.this yields that |zi − zi+1| ≤ (K + 2) |x−y|N ≤ η√tN for i = 0, . . . , N − 1. Then usingProposition B.2.3 in the above integrals we getp0(t, x, y) ≥∫B(x1,r)∩Ω. . .∫B(xN−1,r)∩Ωp0(tN, x, z1) . . . p0(tN, zN−1, y) dz1 . . . dzN−1≥ CN[( tN)−n2 ]N ∫B(x1,r)∩Ω. . .∫B(xN−1,r)∩Ωdz1 . . . dzN−1≥ k(n,Ω)CN[( tN)−n2 ]N[( tN)n2]N−1≥ k(n,Ω)e−C′N( tN)−n2, (B.12)where we have used the regularity of Ω in order to say that there exists a constantk(n,Ω) > 0 such that for all x ∈ Ω, |Ω∩B(x, r)| ≥ k(n,Ω)|B(x, r)|. Finally, by definitionof N , we have N − 1 ≤ Kγ |x−y|2t , thus from (B.12)p0(t, x, y) ≥ C0t−n/2e−c0|x−y|2t .This concludes the proof.B.2.2 The non-symmetric caseNotice that in the proof of Proposition B.2.3 symmetry has not been used. Thereforeif p′(t, ·, y) is Ho¨lder continuous and p′(t, x, x) ≥ ct−n/2, using an argument similar toProposition B.2.3 and Theorem B.2.4, we get Gaussian lower bound for p′(t, x, y), too.Moreover Theorem B.2.4 holds also without assumptions of symmetry and Ho¨lder con-tinuity. Its proof uses only estimate (B.10).Let us show that the L1 → L∞ norm of the difference e−tA′−e−tA0 is relatively small.Now, we prove a result which allows us to conclude without assuming Ho¨lder continuityfor p′.Proposition B.2.5. There exists C > 0 such that‖e−tA′ − e−tA0‖L(L1,L∞) ≤ Ct−n2+ 12 (B.13)Proof. The integral representation of the solution gives thate−tA′ − e−tA0 =∫ t0e−(t−s)A′B ·De−sA0ds=∫ t/20e−(t−s)A′B ·De−sA0ds+∫ tt/2e−(t−s)A′B ·De−sA0ds131Now, by using (B.3), (B.4), the fact that De−sA0 ∈ L(Lp) for 1 < p ≤ 2 and that‖De− s2A0‖L(Lp) ≤ c4s−1/2, we get for p > 1 (close to 1) the following estimate∥∥∥∫ t/20e−(t−s)A′B ·De−sA0fds∥∥∥L∞(Ω)≤ ‖B‖∞‖f‖L1(Ω)∫ t/20‖e−(t−s)A′‖L(Lp,L∞)‖De− s2A0‖L(Lp)‖e− s2A0‖L(L1,Lp) ds≤ C‖f‖L1(Ω)∫ t/20(t− s)−n/2ps−1/2s−n2 (1− 1p ) ds≤ Ct−n/2p‖f‖L1(Ω)∫ t/20s−1/2s−n2 (1− 1p ) ds= Ct−n2+12 ‖f‖L1(Ω) (B.14)where C = C(c2, c3, c4, ‖B‖∞). Moreover from (B.2) we have that ‖De− s2A0‖L(Lq,L∞) ≤c5s− 12− 1q . Thus, using an exponent q close to ∞ we get∥∥∥∫ tt/2e−(t−s)A′B ·De−sA0fds∥∥∥L∞(Ω)≤ ‖B‖∞‖f‖L1(Ω)∫ tt/2‖De− s2A0‖L(Lq,L∞)‖e− s2A0‖L(L1,Lq)ds≤ C‖f‖L1(Ω)∫ tt/2s−1/2s−n/2 ds≤ Ct−n2+ 12 ‖f‖L1(Ω) (B.15)where C = C(c2, c5, ‖B‖∞). Summing up (B.14) and (B.15) we get the claim.As an immediate consequence we deduce a Gaussian lower bound for p(t, x, y).Theorem B.2.6. Let p′(t, x, y) be the fundamental solution of ∂t−A′. Then there existpositive constants C1, c1 such thatp′(t, x, y) ≥ C1t−n/2e−c1|x−y|2tfor all x, y ∈ Ω and t > 0, sufficiently small.Proof. Since‖e−tA′ − e−tA0‖L1→L∞ ≤ Ct−n2+ 12 (B.16)by the Dunford-Pettis theorem (see [7] for a proof) we havesupx,y∈Ω|p′(t, x, y)− p0(t, x, y)| ≤ Ct−n2+ 12whence, for |x− y| ≤ η√t (η as in Proposition B.2.3) we getp′(t, x, y) ≥ p0(t, x, y)− Ct−n2+ 12≥ Ct−n2 (1−√t)≥ Ct−n2132for t ≤ δ0 independent of x, y. Thus Proposition B.2.3 is true also for p′(t, x, y) andproceeding as before we deduce (B.11) also for p′(t, x, y).From Remark B.2.1 we finally deduce the following.Corollary B.2.7. Let p(t, x, y) be the heat kernel of ∂t −A. Then there exist constantsc1, C1 > 0 such thatp(t, x, y) ≥ C1t−n/2e−c1|x−y|2t e−ωt (B.17)for all x, y ∈ Ω and t > 0 small.

Appendix B: heat kernel estimates on domains

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Appendix B: heat kernel estimates on domains

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