1 Sectorial operators

Transcription

1 1 1 Sectorial operators Definition 1.1 Let X and A : D(A) X X be a Banach space and a linear closed operator, respectively. If the relationships i) ρ(a) Σ φ = {λ C : arg λ < φ}, where φ (π/2, π); ii) R(λ, A) L(X) M λ, λ Σ φ; are satisfied, then A is called a sectorial operator. iη 2 Imλ iη 1 σ(a) µ λ ε 1 ε 2 λ -i η1 -iη 2 Figure 1: solvent set and spectrum of a sectorial operator. mark 1.1 As one can see in the previous figure 12.1, the resolvent set of a sectorial operator contains at least a concave angle. However, definition 1.1 makes no assumption on the nature of the origin with respect to the spectrum of A. Definition 1.2 For any η (π/2, φ) and ε >, the family of linear bounded operators {T (t)} t is defined by T (t) = 1 e tλ R(λ, A) dλ, t R +, (1.1) T () =I, integration being performed along the unbounded curve γ(ε, η), oriented from e iη e iη, which is defined by to γ(ε, η) = {re iη C : r ε} {εe iθ C : θ [ η, η]} {re iη C : r ε} := γ 1 γ 2 γ 3. The integral appearing in (1.1) is called the Dunford integral.

2 2 mark 1.2 The curve related to the Dunford integral does not pass through the origin that, as we have already noticed, may be a singular point for the integrand. Definition 1.3 The Laplace transform of a function u : R + X is defined by ũ(λ) = + e tλ u(t) dt for all those values of λ for which t e tλ u(t) is Riemann-integrable over R + generalized sense. in a mark 1.3 For any λ C such that t e tλ u(t) is absolutely Riemann-integrable over R +, t e tλ u(t) is absolutely Riemann-integrable over R + for any λ C such that λ > λ. mark 1.4 Note that by a formal integration by parts the Laplace transform of u can be easily expressed in terms of ũ and u() (provided we suppose that e tλ u(t) in X as t + ). Indeed, we have (u ) (λ) = + e tλ u (t) dt = e tλ u(t) + +λ + Consider now the Cauchy problem { u (t) Au(t) =, u() = x. e tλ u(t) dt = u() + λũ(λ). Performing the Laplace transformation on both sides of the differential equation, we easily obtain λũ(λ) Aũ(λ) = x, which implies ũ(λ) = R(λ, A)x. If the function λ R(λ, A)x turns out to be integrable over the line λ = ξ for some ξ R, then the theory of the Laplace transformation yields the formula for the preimage u(t) = 1 ξ+i ξ i e tλ R(λ, A)x dλ, t R +. However such a property need not to hold for a generic x X. mark 1.5 Observe that the Dunford integral of the holomorphic function λ e tλ R(λ, A) is well defined since the following inequalities hold for any λ γ 1 (ε, η) γ 3 (ε, η): e tλ R(λ, A) L(X) e t λ R(λ, A) L(X) Met λ e t λ cos η M λ λ. (1.2)

3 3 mark 1.6 For any t R + we have e tλ R(λ, A) dλ = γ(εt 1,η) e tλ R(λ, A) dλ, since the difference of the two integrals is a curvilinear integral along a closed bounded path of a holomorphic function in a simply connected region containing the image of γ, and hence vanishes because of Cauchy s theorem [RU2]. Lemma 1.1 T (t) L(X) for any t R + and T (t) L(X) Mc(ε, η). mark 1.7 In other words, the family of linear bounded operators {T (t)} t is equibounded. Proof. From the equation T (t) = 1 3 j=1 γ j (εt 1,η) e tλ R(λ, A) dλ and from the following inequality, which takes advantage of theorem : e tλ R(λ, A) dλ e tλ R(λ, A) L(X) dλ we easily derive the following relationships: i) if λ γ 3 (εt 1, η), λ = r, then λ = r cos η. This implies ii) iii) γ 3 (εt 1,η) + e tλ R(λ, A) dλ M εt 1 + = M ρ=rt likewise, if λ γ 1 (εt 1, η), we get γ 1 (εt 1,η) ε tr cos η dr e r ρ cos η dρ e ρ := Mc 1(ε, η), t R + ; e tλ R(λ, A) dλ Mc 1(ε, η), t R + ; if λ γ 2 (εt 1, η), then λ = εt 1 and λ = εt 1 cos θ. As a consequence γ 2 (εt 1,η) η e tλ R(λ, A) dλ M η e ε cos θ 1 εt 1 εt 1 dθ M η η e ε dθ := Mc 2 (ε, η), t R +.

4 4 Therefore T (t) L(X) and T (t) L(X) 2Mc 1 (ε, η) + Mc 2 (ε, η) := Mc(ε, η), t R +. mark 1.8 The constant c(ε, η) may be explicitly computed. Lemma 1.2 T C (R + ; L(X)). Moreover, Dt k T (t) = 1 λ k e tλ R(λ, A) dλ, t R +, and D k t T (t) L(X) c k (ε, η)t k, t R +, k N. mark 1.9 The derivatives of T are not bounded on a right neighbourhood of t =, More precisely, they become more and more unbounded as n increases. In contrast to this, they get more regular in a neighbourhood of +. Proof. For any triplet (λ, t, t) C R R such that λ ϵ and < t t 2t the following inequalities hold e tλ e t λ = e t λ [e (t t )λ 1] = e t λ e (t t )λ 1 = e t λ t t λe λs ds et λ λ t t e s λ ds t t e t λ λ e tε ds = e t λ+t ε λ t t. Hence e tλ e t λ t t et λ+t ε λ, < t t 2t, λ ε. Interchanging t and t and considering the cases λ and λ ε separately, we get e tλ e t λ t t et λ+tε λ e t ( λ+3ε)/2 λ, < t /2 t t, λ ε. Combining the bounds found so far we easily deduce the inequality { e tλ e t λ e t ( λ+2ε)/2 t t λ, < t /2 t t, λ ε, e t ( λ+ε) λ, < t t 2t, λ ε. (1.3)

5 5 Observe that the function appearing on the right hand side of (1.3) is integrable along the curve γ(ε, η). Taking the limit as t t on both sides of the identity T (t) T (t ) t t = 1 e tλ e t λ t t R(λ, A) dλ, t, t R +, t t, and applying the dominated convergence theorem to the integral on the right hand side, we deduce that T (t ) exists in L(X) and that T (t ) = 1 λe tλ R(λ, A) dλ, t R +. Therefore, T is continuous and differentiable on R +. Proceeding as above we can first show that T C(R + ; L(X)), and then by induction that T C (R + ; L(X)) and the following formulae hold: T (k) (t) = 1 Furthermore we obtain λ k e λt R(λ, A) dλ, t R +, k N. Dt k T (t) = 1 λ k e λt R(λ, A) dλ γ(εt 1,η) 1 = ξ k t k 1 e ξ R(ξt 1, A) dξ. λ=ξt 1 Finally, assumption (1.2) implies the following equations, where t R + and k N: Dt k T (t) L(X) 1 ξ k ξ k 1 Me t d ξ 2π ξ t 1 = 1 2π Mt k ξ k 1 e ξ dξ := c k (ε, η)t k. Lemma 1.3 T (t) L(X; D(A)) and AT (k) (t) = T (k+1) (t) in L(X) for any t R +. Moreover AT (k) (t) L(X) Mc k+1 (ε, η)t k 1, t R +, k N. Proof. First we show that T (t)x D(A), x X, and AT (t)x = T (t)x, t R +, x X. From the definition of the resolvent we deduce e tλ R(λ, A)x D(A), t R +, λ γ(ε, η), x X.

6 6 Moreover AR(λ, A) = I + λr(λ, A) L(X), λ Σ φ. (1.4) Hence the function λ e tλ [ I + λr(λ, A)] is integrable along γ(ε, η). By theorem?? we deduce that T (t) : X D(A) and the following equation holds AT (t) = 1 e tλ AR(λ, A) dλ, t R +. Further, from (1.4) it follows that for any x X and any t R + we have AT (t)x = 1 e tλ AR(λ, A)x dλ = 1 ( ) e tλ dλ x + 1 λe tλ R(λ, A)x dλ, = T (t)x. Notice that, by Cauchy s theorem, the second integral vanishes since the exponential tends to in the sector Σ φ = {λ C : φ arg λ 2π φ}. Therefore, we conclude that the equation AT (t) = T (t) holds for any t R +. Proceeding by induction, for any t R + and any k N we get that the following equalities hold in L(X): AT (k) (t) = 1 λ k+1 e tλ R(λ, A) dλ = T (k+1) (t). Finally, from the estimates for T (k) we can derive those for AT (k) : AT (k) (t) L(X) Mc k+1 (ε, η)t k 1, t R +, k N. Lemma 1.4 The function t T (t)x is continuous in [, + ) if and only if x D(A). mark 1.1 At the origin the semigroup is only strongly continuous (cf. the related concept of pointwise convergence), but not (generally speaking) uniformly continuous, i.e. T (t) need not converge to I in L(X) as t +. Proof. First assume that x D(A), and note that the equalities λr(λ, A)x = x + AR(λ, A)x = x + R(λ, A)Ax imply T (t)x = 1 e tλ λr(λ, A)x dλ λ

7 7 = 1 ( e tλ dλ λ ) x + 1 e tλ R(λ, A)Ax dλ λ. By the residue theorem [RU2], the first integral in the last member equals 1. Hence T (t)x x = 1 e tλ R(λ, A)Ax dλ λ. Since the integrand is holomorphic, Cauchy s theorem implies T (t)x x = 1 e λt R(λ, A)Ax dλ λ. γ(εt 1,η) From this relationship and definition we deduce the estimates T (t)x x 1 e t λ R(λ, A) L(X) Ax dλ 2π λ γ(εt 1,η) M 2π Ax = λ=t 1 µ γ(εt 1,η) M 2π Ax t e t λ dλ e λ 2 µ dµ 2, as t +. µ Now let x D(A). By definition there exists a sequence {x n } D(A) such that x n x in X. Consider then the estimates (cf. lemma 1.1) T (t)x x T (t)(x x n ) + T (t)x n x n + x n x [c (ε, η)m + 1] x x n + T (t)x n x n. As x n D(A) for any n N, the second term in the last side of the previous inequality tends to as t +. This implies the relationship lim sup t + T (t)x x [c (ε, η) M + 1] x x n, n N. Taking the limit as n +, we obtain the chain of inequalities which imply lim inf t + T (t)x x lim sup T (t)x x, t + lim T (t)x x =. t + This relationship is equivalent to saying that T (t)x x for t + and for any x D(A). Finally, assume that x X and t T (t)x is continuous in t =. Since T (t)x D(A) for any t R +, it is immediate to deduce that This fulfills the proof. x = lim t + T (t)x D(A).

8 8 mark 1.11 From lemmas it immediately follows that, for any x D(A), the Cauchy problem { u (t) = Au(t), t R + u() = x admits the solution u(t) = T (t)x. However, we do not yet know whether such a solution is unique. This property is true, but will be proved in the following (cfr. theorem??). Lemma 1.5 T (t) satisfies the semigroup property T (t 1 + t 2 ) = T (t 1 )T (t 2 ), t 1, t 2 [, + ). (1.5) Proof. First observe that property (1.5) is immediately satisfied if t 1 = or t 2 = since T () = I. Assume now t 1, t 2 R + and consider the two curves γ(ε 1, η 1 ) and γ(ε 2, η 2 ), drawn in picture 12.2, with < ε 1 < ε 2 and π/2 < η 2 < η 1 < π Finally, observe that the following iη 2 Im λ iη 1 σ(a) µ λ ε 1 ε 2 γ(ε1,η1) γ(ε,η ) 2 2 λ -i η1 -iη 2 Figure 2: Integration path

9 9 chain of identities prove the assertion: T (t 1 )T (t 2 ) = 1 ( )( 4π γ(ε ) e t1λ R(λ, A) dλ e t2µ R(µ, A) dµ 2 2,η 2 ) (change the product of integrals into iterated integrals) = 1 e t1λ dλ e t2µ R(λ, A)R(µ, A) dµ 4π 2 (use the resolvent equation, cf. lemma??) = 1 e t1λ dλ e t 1 2µ R(λ, A) dµ 4π 2 λ µ 1 4π 2 e t1λ dλ (use theorem??) = 1 e t1λ R(λ, A) dλ 4π 2 1 e t2µ R(µ, A) dµ 4π 2 e t 2µ R(µ, A) dµ λ µ e t 2µ λ µ dµ e t 1λ λ µ dλ (use Cauchy s theorem) = 1 e t1µ e t2µ R(µ, A) dµ = T (t 1 + t 2 ). As a consequence of the previous lemmata we immediately deduce the following results: Theorem 1.1 {T (t)} t is an equibounded semigroup in L(X). Theorem 1.2 For any x D(A) and any t R + t T (s)x ds D(A) and A t t mark 1.12 The mapping x T (s)x ds = [T (t) I]x. T (s)xds is regularizing in the sense that, for any t R +, it maps an element x D(A) into an element of D(A). Proof. First observe that for any ε (, 1), t R +, ξ ρ(a) and x D(A) the

10 1 following equalities hold: t εt T (s)x ds = t εt (ξi A)R(ξ, A)T (s)xds = ξr(ξ, A) = ξr(ξ, A) = ξr(ξ, A) t εt t εt t εt T (s)x ds T (s)x ds t εt t εt R(ξ, A)AT (s)x ds D s [R(ξ, A)T (s)x]ds T (s)x ds R(ξ, A)T (t)x + R(ξ, A)T (εt)x. Making use of the assumption x D(A) and letting ε +, we obtain the identity t { T (s)xds = R(ξ, A) ξ t } T (s)xds T (t)x + x, t R +. (1.6) Since the resolvent sends any vector in X into the domain of A, we get t T (t)xds D(A), t R +, x D(A). Applying (ξi A) to both sides of (1.6), we obtain i.e. (ξi A) A t t T (s)xds = ξ t T (s)xds = [ T (t) + I] x, T (s)xds T (t)x + x, t R +, t R +, x D(A) and so the assertion is proved. mark 1.13 If s AT (s)x = T (s)x R loc (R + ; X), then we easily derive the formula ferences t AT (s)xds = T (t)x x, t R +. [DE] [GR] [HA] [HP] Denisov, A. M. (2). Introduction to Inverse Problems. VSP, Utrecht, The Netherlands. Groetsch, C. W. (1993). Inverse Problems in the Mathematical Sciences. Vieweg, Braunschweig Wiesbaden. Halmos, P. (197). Topology. Academic Press. Hille E. and Phillips R. S. (1957). Functional Analysis and Semi-Groups. American Mathematical Society, Colloquium Publications, vol. 31.

11 11 [NE] Nečas, J. (1964). Problèms directes en théorie des équations elliptiques. Dunod, Paris. [RU1] Rudin, W. (1966). al and Complex Analysis. McGraw Hill, New York London. [RU2] Rudin, W. (1991). Functional Analysis. McGraw Hill, Singapore. [SC] Schwartz, L. (1967). Course d Analyse. Hermann, Paris.