c 2008 Society for Industrial and Applied Mathematics 

SIAM J. CONTROL OPTIM. Vol. 47, No. 2, pp. 1022–1052

A VISCOSITY SOLUTION APPROACH TO THE INFINITE-DIMENSIONAL HJB EQUATION RELATED TO A BOUNDARY CONTROL PROBLEM IN A TRANSPORT EQUATION∗ GIORGIO FABBRI† Abstract. The paper concerns the infinite-dimensional Hamilton–Jacobi–Bellman equation related to an optimal control problem regulated by a linear transport equation with boundary control. A suitable viscosity solution approach is needed in view of the presence of the unbounded controlrelated term in the state equation in the Hilbert setting. An existence-and-uniqueness result is obtained. Key words. Hamilton–Jacobi–Bellman equation, viscosity solution, boundary control AMS subject classifications. 49J20, 49L25 DOI. 10.1137/050638813

1. Introduction. We study the Hamilton–Jacobi–Bellman (HJB) equation related to the infinite-dimensional formulation of an optimal control problem whose state equation is a PDE of transport type. More precisely we consider the PDE ⎧ ∂ ∂ x(s, r) = −μx(s, r) + α(s, r), (s, r) ∈ (0, +∞) × (0, s¯), ⎨ ∂s x(s, r) + β ∂r (1) x(s, 0) = a(s) if s > 0, ⎩ x(0, r) = x0 (r) if r ∈ [0, s¯], where s¯, β are positive constants, μ ∈ R, the initial data x0 is in L2 (0, s¯), and we consider two controls: A boundary control a is in L2loc ([0, +∞); R) and a distributed control α ∈ L2loc ([0, +∞) × [0, s¯]; R).1 By using the approach and the references described in section 2, the above equation can be written as an ordinary differential equation in the Hilbert space H = L2 (0, s¯) as follows:  d ds x(s) = Ax(s) − μx(s) + α(s) + βδ0 a(s), (2) x(0) = x0 , where A is the generator of a suitable C0 semigroup and δ0 is the Dirac delta in 0. Such an unbounded contribution in the Hilbert formulation comes from the presence in the PDE of a boundary control (see [8]). Besides we consider the problem of minimizing the cost functional  ∞ (3) J(x, α(·), a(·)) = e−ρs L(x(s), α(s), a(s))ds, 0

where ρ > 0 and L is globally bounded and satisfies some Lipschitz-type condition, as better described in section 2. The HJB equation related to the control problem with ∗ Received by the editors August 24, 2005; accepted for publication (in revised form) October 16, 2007; published electronically March 5, 2008. http://www.siam.org/journals/sicon/47-2/63881.html † Dipartimento di Scienze Economiche ed Aziendali, Universit` a LUISS - Guido Carli, Roma, Italy ([email protected]). 1 We write “−μx” instead of “μx” because it is the standard way to write the equation in the economic literature, where −μ has the meaning of a depreciation factor (and only the case μ ≥ 0 is used). Here we consider a generic μ ∈ R.

1022

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

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state equation (2) and target functional (3) is (4) ρu(x) − ∇u(x), Ax − ∇u(x), −μxL2 (0,¯s)   βδ0 (∇u(x)), aR + ∇u(x), αL2 (0,¯s) + L(x, α, a) = 0. − inf (α,a)∈Σ×Γ

The sets Γ and Σ will be introduced in section 2; they are suitable subsets, respectively, of R and H. If we define the value function of the control problem as def

V (x) =

inf

J(x, α(·), a(·)),

(α(·),a(·))∈E×A

we wish to prove that V (·) is the unique solution, in a suitable sense, of the HJB equation. We use the viscosity approach. Our main problem is to write a suitable definition of the viscosity solution, so that an existence and uniqueness theorem can be derived for such a solution. The main difficulties we encounter, with respect to the existing literature, are in dealing with the boundary term and the nonanalyticity of the semigroup. We substantially follow the original idea of Crandall and Lions [14], [15]— with some changes, as the reader will rate in Definitions 2.14 and 2.15—of writing test functions as the sum of a “good part” as it is a regular function with differential in D(A∗ ) and a “bad part” represented by some radial function. The main problems arise from the evaluation of the boundary term on the radial part. In order to write a working definition in our case, some further requirements are needed, such as a C 2 regularity of the test functions, the presence of a “remainder term” in the definition of a sub/supersolution, and the B-Lipschitz continuity (see Definition 2.10) of the solution. This last feature guarantees that the maxima and the minima in the definition of a sub/supersolution remain in D(A∗ ) (see Proposition 4.1). Some other comments on the definition of solution (Definitions 2.14 and 2.15) need some technical details and can be found in Remark 2.17. The technique used cannot be easily extended to treat a general nonlinear problem because we use the explicit form of the PDE that we give in (6). A nontrivial generalization would be also that of replacing the constant μ with a function μ(r) in L∞ (0, s¯) (see Remark 4.9 for details). Nevertheless, the problem remains challenging. A brief summary of the literature. Hamilton–Jacobi equations in infinite dimensions, especially when arising from optimal control problems in Hilbert spaces, were first studied by Barbu and Da Prato [3], [4] with a strong solutions approach. The viscosity method, introduced in the study of finite-dimensional HJ equations in [13], was generalized by the same authors in a series of works: The most important for our approach are [14] and [15]. Moreover new variants of the notion of a viscosity solution for HJB equations in Hilbert spaces were given in [28], [34], [35], [33], [16]. The study of viscosity solutions for HJB equations associated to boundary control in PDE is more recent. In this research field there is not an organic and complete theory but some works that adapt the ideas and the techniques of viscosity solutions to problems for particular PDEs, by exploiting their peculiarities (as we do in this work for the problem regulated by a transport equation). For the first order HJB equations see [9], [12] (see also [10], [11]), where some classes of parabolic equations are treated, and [24], where the authors study the HJB equation related to a twodimensional Navier–Stokes equation (see also [32]). It must be noted that all of these works treat the case of analytic A.

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1024

GIORGIO FABBRI

A HJB equation like (4) was treated, but only in the case of a convex objective functional, with a strong solutions approach by adapting Barbu and Da Prato’s technique of convex regularization in [20], [19]. A motivating economic problem. Transport equations are used to model a large variety of phenomena. They are used, for example, in age-structured population models (see, for instance, [26], [2], [27]), in population economics [23], epidemiologic studies, socioeconomic science, and transport phenomena in physics. Problems such as (1) can be used to describe, in economics, capital accumulation processes where an heterogeneous capital is involved, and this is the reason why the study of the infinite-dimensional control problem is of growing interest in the economic fields. For instance, in the vintage capital models x(t, s) may be regarded as the stock of capital goods differentiated with respect to the time t and the age s. Heterogeneous capital, in both the finite- and infinite-dimensional approaches, is used to study depreciation and obsolescence of physical capital, geographical difference in growth, innovation, and R&D. Regarding problems modeled by a transport equation where an infinite-dimensional setting is used, we cite the following papers: [5], [7] on optimal technology adoption in a vintage capital context (in the case of a quadratic cost functional), [25] on capital accumulation, [6] on optimal advertising, and [19], [21] on the case of a general objective convex functional with a strong solutions approach. See also [22]. Moreover, we mention that the infinite-dimensional approach may apply to problems such as issuance of public debt (see [1] for a description of the problem). In that problem a stochastic setting and simple state-control constraints appear, but hopefully the present work can be a first step in this direction. Plan of the paper. The work is organized as follows: In the first section we recall some results on the state equation, we introduce some preliminary remarks on the main operators involved in the problem, we explain some notations, we define the HJB equation, and we give the definition of solution. The second section regards some properties of the value function (in particular, some regularity properties) that will be used in the third section to prove that it is the unique (viscosity) solution of the HJB equation. 2. Notation and preliminary results. 2.1. State equation. In this subsection we will see some properties of the state equation: We write it in three different (and equivalent) forms that point out different properties of the solution. We will use all three forms in the following proofs. We consider the PDE on [0, +∞) × [0, s¯] given by ⎧ ∂ ∂ x(s, r) = −μx(s, r) + α(s, r), (s, r) ∈ (0, +∞) × (0, s¯), ⎨ ∂s x(s, r) + β ∂r (5) x(s, 0) = a(s) if s > 0, ⎩ x(0, r) = x0 (r) if r ∈ [0, s¯]. Given an initial datum x0 ∈ L2 ((0, s¯); R) (from now on simply L2 (0, s¯)), a boundary control a(·) ∈ L2loc ([0, +∞); R), and a distributed control α(·) ∈ L2loc ([0, +∞) × [0, s¯]; R), (5) has a unique solution in L2loc ([0, +∞) × [0, s¯]; R) given by

s r ∈ [βs, s¯], e−μs x0 (r − βs) + 0 e−μτ α(s − τ, r − βτ )dτ,

r/β −μτ −μ (6) x(s, r) = r β a(s − r/β) + 0 e α(s − τ, r − βτ )dτ, r ∈ [0, βs). e In the following x(s, r) is the function defined in (6).

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

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We can rewrite such an equation in a suitable Hilbert space setting. We take the def Hilbert space H = L2 (0, s¯) and the C0 semigroup S(t) given by  def f (r − βs) for r ∈ [βs, s¯], S(s)f [r] = 0 for r ∈ [0, βs). The generator of S(s) is the operator A given by  D(A) = {f ∈ H 1 [0, s¯] : f (0) = 0}, d A(f )[r] = −β dr f (r) (see [7] for a proof in the case where β = 1; the proof in our case can be obtained by simply taking s = βs). In the following we will use the notation esA instead of S(s). Remark 2.1. To avoid confusion if x ∈ L2 (0, s¯) we will use [·] to denote the pointwise evaluation, so x[r] is the value of x in r ∈ [0, s¯]. On the other hand, x(s) will denote the evolution of the solution of the state equation (in the Hilbert space) at time s (as in (7)). That is, x(s) is an element of H, while x[r] is a real number. We want to write an infinite-dimensional formulation of (5), but in L2 (0, s¯) it should appear like  d ds x(s) = Ax(s) − μx(s) + α(s) + βδ0 a(s), (7) x(0) = x0 , where α(s) ∈ L2 (0, s¯) is the function r → α(s, r). Such an expression does not make sense in L2 (0, s¯) for the presence of the unbounded term βδ0 a(s). We can anyway apply formally the variation of constants method to (7) and obtain a mild form of (7) that is continuous from [0, +∞) to L2 (0, s¯). This is what we do in the next definition. Definition 2.2. Given x0 ∈ L2 (0, s¯), a(·) ∈ L2loc ([0, +∞); R), and α(·) ∈ 2 Lloc ([0, +∞); L2 (0, s¯)), the function in C([0, +∞); L2 (0, s¯)) given by (8) x(s) = e−μs esA x0 −A



s

e−μ(s−τ ) e(s−τ )A (a(τ )ν)dτ +

0



s

e−μ(s−τ ) e(s−τ )A α(τ )dτ,

0

where ν : [0, s¯] → R, μ ν : r → e− β r , is called the mild solution of (7). Remark 2.3. We could include the term −μx in the generator of the semigroup A taking a A˜ = A − μ1 as done in [7]. The problem of this approach is that often we will use, in the estimates, the dissipativity of the generator, and A˜ is dissipative only if μ ≥ 0. Proposition 2.4. Given x(s) the function from R+ to L2 (0, s¯) given by (8) and x(s, r) the function from R+ × [0, s¯] to R given by (6), we have x(s)[r] = x(s, r). Proof. See [7]. Eventually we observe that (7) can be rewritten in a precise way in a larger space in which βδ0 belongs. To this extent, we consider the adjoint operator A∗ , whose explicit expression is given by def s) = 0}, D(A∗ ) = {f ∈ H 1 (0, s¯) : f (¯ d ∗ A (f )[r] = β dr f (r).

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GIORGIO FABBRI

We endow, in all of the paper, D(A∗ ) with the graph norm and the related Hilbert structure. We consider the inclusion i : D(A∗ ) → L2 (0, s¯) and its continuous adjoint i∗ : L2 (0, s¯) → D(A∗ ) , where we have identified L2 with its dual. We can extend A to a generator of a C0 semigroup on D(A∗ ) (the domain of the extension will contain L2 (0, s¯)), and we observe that Dirac’s measure δ0 ∈ D(A∗ ) (see [20, Proposition 4.5, page 60] for details). Proposition 2.5. Given T > 0, x0 ∈ L2 (0, s¯), a(·) ∈ L2 (0, T ), α(·) ∈ L2 ([0, T ]; 2 L (0, s¯)), (8) is the unique solution of  d ∗ ds i x(s) = Ax(s) − μx(s) + α(s) + βδ0 a(s), (9) x(0) = x0 in W 1,2 (0, T ; D(A∗ ) ) ∩ C(0, T, H). Moreover, if a(·) ∈ W 1,2 (0, T ), then such a solution will belong to C 1 (0, T ; D(A∗ ) ) ∩ C(0, T ; H). Proof. See [8, Chapter 3.2] (in particular, Theorem 3.1, page 173). 2.2. The definition of the operator B. In this subsection we give the definition of the operator B that will have a fundamental role.2 Note that A∗ and A are negative operators. We take φ ∈ D(A∗ ), so that φ(¯ s) = 0, and then  s¯ −βφ(0)2 A∗ φ, φ = βφ (r)φ(r)dr = 2 0 and for φ ∈ D(A) (so that φ(0) = 0)  s¯ −βφ(¯ s)2 −βφ (r)φ(r)dr = Aφ, φ = . 2 0 Therefore, given a λ > 0, the operators (A − λI) and (A∗ − λI) are strongly negative: (A − λI)x, x ≤ −λ|x|2H for all x ∈ D(A) and (A∗ − λI)x, x ≤ −λ|x|2H for all x ∈ D(A∗ ). We can also directly prove that (A − λI)−1 : H → D(A) is a continuous negative linear operator whose explicit expression is given by  r λ λ 1 (A − λI)−1 (φ)[r] = −e− β r e β τ φ(τ )dτ . β 0 The continuity can be proved directly with not difficult estimates, and the negativity can be proved directly by using an integration by part argument. 2 We could use an abstract approach, noting that A and A∗ are both generators of C semigroups 0 of contractions, and then both are negative (see [17, page 424]), and the set {λ ∈ C : Re(λ) > 0} is in the resolvent of both A and A∗ (Hille–Yosida theorem; see [29, page 53]). Here we can also follow a more direct approach that allows us to find the explicit form of the operator.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

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In the same way we can prove that (A∗ − λI)−1 : H → D(A∗ ) is a continuous and negative linear operator and that its explicit expression is given by  s¯ λ λ 1 r −β τ ∗ −1 β −e e φ(τ )dτ . (A − λI) (φ)[r] = β r def

Eventually we can define B = (A∗ − λI)−1 (A − λI)−1 = ((A − λI)−1 )∗ (A − λI)−1 that is continuous, positive, and self-adjoint.3 Moreover (A∗ − λI)B = (A − λI)−1 ≤ 0, and so A∗ B = (A − λI)−1 + λB ≤ λB; then A∗ B is continuous, and if we choose λ < 1, we have A∗ B ≤ B.

(10)

Thus B satisfies all requirements of the so-called “weak case” of [14]. Remark 2.6. Note that B 1/2 is a particular case of the operator that Renardy found in more generality in [31], and so B 1/2 : H → D(A∗ ) continuously and in particular R(B 1/2 ) ⊆ D(A∗ ).

Notation 2.7. For every x ∈ H we will indicate with |x|B the B-norm that is Bx, xH . We will write HB for the completion of H with respect to the B-norm. Remark 2.8. Thanks to the definition of A∗ , the graph norm on D(A∗ ) is equivalent to the H 1 (0, s¯)-norm. In particular D(A∗ ) is the completion of K = {f |[0,¯s] : f ∈ Cc∞ (R) with supp(f ) ⊆ (−∞, s¯)} with respect to the H 1 (0, s¯)-norm. So, since H 1 (0, s¯) → C([0, s¯]; R), we can apply βδ0 on the elements of D(A∗ ) and δ0 ∈ D(A∗ ) . Notation 2.9. The notation x, yH will indicate the inner product in the Hilbert space H (for example, H = H ≡ L2 (0, s¯) or H = R or D(A∗ ) . . . ). Otherwise, if Z is a Banach space (possibly a Hilbert space) and Z  its dual, the notation x, yZ×Z  will indicate the duality pairing. Eventually x, y ≡ x, yL2 (0,¯s) . 2.3. The control problem and the HJB equation. In this subsection we describe the optimal control problem, state the hypotheses, define the HJB equation of the system, and give a suitable definition of solution of the HJB equation. We consider the optimal control problem governed by the state equation  d ∗ ds i x(s) = Ax(s) − μx(s) + α(s) + βδ0 a(s), (11) x(0) = x that has a unique solution in the sense described in section 2.1. Given two compact subsets Γ and Λ of R, we consider the set of admissible boundary controls given by def

A = {a : [0, +∞) → Γ ⊆ R : a(·) is measurable} . 3 See

[36, Proposition 2, page 273] for a proof of the equality (A∗ − λI)−1 = ((A − λI)−1 )∗ .

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GIORGIO FABBRI

Moreover we call def

Σ = {γ : [0, s¯] → Λ ⊆ R : γ(·) is measurable} . In view of the compactness of Λ we have Σ ⊆ L2 (0, s¯). We define the set of admissible distributed controls as def

E =



 α : [0, +∞) → Σ ⊆ L2 (0, s¯) : α(·) is measurable .

In view of the compactness of Γ and Λ, we have A ⊆ L2loc ([0, +∞); R) and E ⊆ def

def

def

L2loc ([0, +∞) × [0, s¯]; R). We call Γ = supa∈Γ (|a|), Λ = supb∈Λ (|b|), and Σ = supα∈Σ (|α|H ) (they are bounded thanks to the boundedness of Γ and Λ). We call admissible control a couple (α(·), a(·)) ∈ E × A. The cost functional will be of the form  ∞ J(x, α(·), a(·)) = e−ρs L(x(s), α(s), a(s))ds, 0

where L is uniformly continuous and satisfies the following conditions: There exists a CL ≥ 0 with ((L1))

|L(x, α, a) − L(y, α, a)| ≤ CL B(x − y), (x − y)H×H ∀(α, a) ∈ Σ × Γ, |L| ≤ CL < +∞.

((L2))

We define formally the HJB equation of the system as (HJB)

ρu(x) − ∇u(x), Ax − ∇u(x), −μx − H(x, ∇u(x)) = 0,

where H is the Hamiltonian of the system and is defined as H : H × D(A∗ ) → R, def

H(x, p) = inf (α,a)∈Σ×Γ (βδ0 (p), aR + p, αH + L(x, α, a)) . Before introducing a suitable definition of the (viscosity) solution of the HJB equation, we give some preliminary definitions. Definition 2.10. A function v ∈ C(H) is Lipschitz with respect to the B-norm, or B-Lipschitz, if there exists a constant C such that def

|v(x) − v(y)| ≤ C|(x − y)|B = C|B 1/2 (x − y)|H for every choice of x and y in H. In the same way we can give the definition of a locally B-Lipschitz function. Definition 2.11. A function v ∈ C(H) is said to be B-continuous at a point x ∈ H if for every xn ∈ H, with xn x and |B(xn − x)| → 0, it holds that v(xn ) → v(x). In the same way we can define the B-upper/lower semicontinuity. Definition 2.12. We say that a function φ such that φ ∈ C 1 (H) and φ is Blower semicontinuous is a test function of type 1, and we write φ ∈ test1, if ∇φ(x) ∈ D(A∗ ) for all x ∈ H and A∗ ∇φ : H → H is continuous. Definition 2.13. We say that g ∈ C 2 (H) is a test function of type 2, and we write g ∈ test2, if g(x) = g0 (|x|) for some nondecreasing function g0 : R+ → R.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

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Definition 2.14. A function u ∈ C(H) bounded and Lipschitz with respect to the B-norm, is a subsolution of the HJB equation (or simply a “ subsolution”) if for every φ ∈ test1, g ∈ test2, and local maximum point x of u − (φ + g) we have (12) ρu(x) − A∗ ∇φ(x), x − ∇φ(x) + ∇g(x), −μx   − inf βδ0 (∇φ(x), a)R + ∇φ(x) + ∇g(x), αH + L(x, α, a) (α,a)∈Σ×Γ



g0 (|x|) Γ2 β . |x| 2

Definition 2.15. A function v ∈ C(H) bounded and Lipschitz with respect to the B-norm is a supersolution of the HJB equation (or simply a “ supersolution”) if for every φ ∈ test1, g ∈ test2, and local minimum point x of v + (φ + g) we have (13) ρv(x) + A∗ ∇φ(x), x + ∇φ(x) + ∇g(x), −μx   − inf − βδ0 (∇φ(x), a)R − ∇φ(x) + ∇g(x), αH + L(x, α, a) (α,a)∈Σ×Γ

≥−

g0 (|x|) Γ2 β . |x| 2

Definition 2.16. A function v ∈ C(H) bounded and Lipschitz with respect to the B-norm is a solution of the HJB equation if it is at the same time a supersolution and a subsolution. Remark 2.17. In the definition of viscosity solution we have used two kinds of test functions: those in test1 and those in test2, which, as usual in the literature, play a different role. In view of their properties the functions of the first set (test1) represent the “good part.” More difficult is to deal with the functions of the set test2, which have the role of localizing the problem. A difficulty of our case is the following: The trajectory is not Lipschitz with respect to the H-norm, and so, given a function g ∈ test2, the term g(x(s)) − g(x) s

(14)

(where x(s) is a trajectory starting from x) cannot be treated with standard arguments. The idea then is to consider only a B-Lipschitz solution so that the maxima/minima considered in Definitions 2.14 and 2.15 are in D(A∗ ). If the starting point x is in D(A∗ ), there are some advantages in the estimate of (14), but some problems remain: In such a case we will prove in Proposition 4.5 that (if α(·) is continuous in 0)    g(x(s)) − g(x)  g0 (|x|) Γ2  ≤ − ∇g(x), −μx + α(0) β + O(s),   s |x| 2 s→0

where the rest O(s) −−−→ 0 and does not depend on the control. So the most challenging case is the one described in the definition. 3. The value function and its properties. The value function is, as usual, the candidate unique solution of the HJB equation. In this section we define the value function V (·) of the problem, and then we verify that it has the regularity properties

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1030

GIORGIO FABBRI

required to be a solution. Namely, we will check that V (·) is B-Lipschitz (Proposition 3.4). To obtain such a result we prove an approximation result (Proposition 3.1) and then a suitable estimate for the solution of the state equation (Proposition 3.3). The value function of our problem is defined as def

V (x) =

inf

J(x, α(·), a(·)).

(α(·),a(·))∈E×A

We consider the functions



ηn : [0, s¯] → R, def

ηn (r) = [2n − 2n2 r]+ (where [·]+ is the positive part). We then define  ∗ Cn : R → H, Cn∗ : γ → γηn . Such functions are linear and continuous, and their adjoints are  Cn : H → R, (15) Cn : x → x, ηn  . The functions Cn approximate the delta measure in some sense. The approximating state equations we consider are  d ∗ ds xn (s) = Axn (s) − μxn (s) + α(s) + βCn a(s), (16) xn (0) = x. In the following proofs we will use (8) together with the mild form of the approximating state equations (that can be found in [30, page 105, equation (2.3)]): −μs sA

(17) xn (s) = e

e

 x+

s

−(s−τ )μ (s−τ )A

e

e

 α(τ )dτ

0

s

e−(s−τ )μ e(s−τ )A βCn∗ a(τ )dτ.

0

Proposition 3.1. For T > 0 and (α(·), a(·)) ∈ E × A lim

sup |xn (s) − x(s)|H = 0.

n→∞ s∈[0,T ]

Proof. By using (8) and (17) we find (18)

   |x(s) − xn (s)| = −A

0

s

e−(s−τ )μ e(s−τ )A (a(τ )ν)dτ   s  − e−(s−τ )μ e(s−τ )A βCn∗ (a(τ ))dτ  . 0

To estimate such an expression we will use the explicit expression of the two terms (as a two-variable function). We simplify the notation by using the extension a ˜(·) of a(·) on all R given by  0 if s < 0, a ˜(s) = a(s) if s ≥ 0.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

So



def





def

s

−A

y(s, r) = (19) yn (s, r) =

e−(s−τ )μ e(s−τ )A (a(τ )ν)dτ

0 s

−(s−τ )μ (s−τ )A

e

e

0



βCn∗ (a(τ ))dτ

r∧(1/n)

=



1031

μ

[r] = e− β r a ˜(s − r/β),

[r]

μ

e− β (r−θ) [2n − 2n2 θ]+ a ˜

0



θ−r + s dθ. β

Now for all s ∈ [0, T ] we have (20) |y(s, ·) − yn (s, ·)|2H  s¯  2  1/n θ − r  − μβ r  −μ (r−θ) 2 + ≤ a ˜(s − r/β) − e β [2n − 2n θ] a ˜ + s dθ dr e  β 1/n  0 2  1/n   r   −μr μ θ−r e β a + s dθ dr + ˜(s − r/β) − e− β (r−θ) [2n − 2n2 θ]+ a ˜  β 0 0 (for s¯ ≤ T )  (21)

|μ|s





e

0

 −

1/n

T

  −μ( r −s) e β a ˜(s − r/β) 

−μ( r−θ β −s)

e 0



θ−r [2n − 2n θ] a ˜ s+ β 2

+



2   1 |μ|/βT dθ dr + e 2Γ . n

Such an estimate does not depend on s; the integral term goes to zero because it is the convolution of a function in L2 (0, T ) with an approximate unit, and the second goes to zero for n → ∞. Proposition 3.2. Let φ ∈ C 1 (H) be such that ∇φ : H → D(A∗ ) is continuous. Then, for an admissible control (α(·), a(·)), if we call x(·) the trajectory starting from x and subject to the control (α(·), a(·)), we have that, for every s > 0, 

s

(22) φ(x(s)) = φ(x) + 0

[A∗ ∇φ(x(τ )), x(τ ) + βδ0 (∇φ(x(τ ))), a(τ )R + ∇φ(x(τ )), α(τ ) + ∇φ(x(τ )), −μx(τ )] dτ.

Proof. In the approximating state equation (16) the unbounded term βδ0 does not appear (βCn∗ are continuous), and then (see [29, Proposition 5.5, page 67]) for every φ(·) ∈ C 1 (H) such that A∗ ∇φ(·) ∈ C(H) we have  (23) φ(xn (s)) = φ(x) +

s

[A∗ ∇φ(xn (τ )), xn (τ ) + ∇φ(xn (τ )), βCn∗ a(τ )

0

+ ∇φ(xn (τ )), α(τ ) + ∇φ(xn (τ )), −μxn (τ )] dτ  s = φ(x) + [A∗ ∇φ(xn (τ )), xn (τ ) + βCn ∇φ(xn (τ )), a(τ ) 0

+ ∇φ(xn (τ )), α(τ ) + ∇φ(xn (τ )), −μxn (τ )] dτ,

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1032

GIORGIO FABBRI

where we passed to the adjoint in view of the continuity of the operator Cn∗ (see (15) for the explicit form of Cn ). Now we prove that every integral term of (23) converges to the corresponding term of (22). This fact, together with the pointwise convergence of φ(xn (s)) to φ(x(s)) (due to Proposition 3.1), will imply the claim. First we note that, in view of Proposition 3.1 and of the continuity of x, xn (τ ) is bounded uniformly in n and τ ∈ [0, s], and, in view of the continuity of ∇φ, ∇φ(xn (τ )) is bounded uniformly in n and τ ∈ [0, s]. So we can apply the Lebesgue theorem (the pointwise convergence is given by Proposition 3.1 and |α(τ )| ≤ Σ), and we derive that  s (24) [∇φ(xn (τ )), α(τ ) + ∇φ(xn (τ )), −μxn (τ )] dτ 0  s n→∞ −−−−→ [∇φ(x(τ )), α(τ ) + ∇φ(x(τ )), −μx(τ )] dτ. 0

Next we observe that, in view of the continuity of A∗ ∇φ and of Proposition 3.1, the term A∗ ∇φ(xn (τ )) is bounded uniformly in n and τ ∈ [0, s], so the same is true for |A∗ ∇φ(xn (τ )) − A∗ ∇φ(x(τ ))|. Therefore we can use the Lebesgue theorem (the pointwise convergence is given by Proposition 3.1) to conclude that  s  s ∗ A ∇φ(xn (τ )), xn (τ ) dτ → A∗ ∇φ(x(τ )), x(τ ) dτ. 0

0

We now have to prove that  s  (25) βCn ∇φ(xn (τ )), a(τ ) dτ → 0

0

s

βδ0 (∇φ(x(τ ))), a(τ )R dτ.

n→∞

We first note that Cn −−−−→ δ0 in H −1 (0, s¯) and then in D(A∗ ) . Indeed given z ∈ H 1 (0, s¯) we have (26)

  |(Cn − δ0 )z| = 

0



  z[τ ]ηn [τ ]dτ − z[0]    τ  1/n    = z[0] + ∂ω z[r]dr ηn [τ ]dτ − z[0]  0  0

(∂ω z is the weak derivative of z), and integrating by parts      1/n  1/n  (27) =  z[0] + ∂ω z[r]dr ηn [r]dr  0 0  1/n  − ∂ω z[τ ] 0

0

τ

   ηn [r]drdτ − z[0] 

1/n and then, writing ηn in explicit form (note that 0 ηn [r]dr = 1),  s¯    1  ≤ χ[0,1/n] [τ ]|∂ω z[τ ]|dτ  ≤ √ zH 1 (0,¯s) . n 0

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1033

n→∞

In summary, by Proposition 3.1, xn (·) −−−−→ x(·) in C([0, T ]; H), then (by hypothesis n→∞ on φ) ∇φ(xn (·)) −−−−→ ∇φ(x(·)) in C([0, T ]; D(A∗ )), and then, by the last estimate, n→∞ βCn (∇φ(xn (·))) −−−−→ βδ0 (∇φ(x(·))) in C([0, T ]; R). Then (25) follows by Cauchy– Schwartz inequality. Proposition 3.3. Given T > 0 and a control (α(·), a(·)) ∈ E × A, there exists cT such that for every x, y ∈ H sup |xx (s) − xy (s)|2B ≤ cT |x − y|2B , s∈[0,T ]

where xy (·) is the solution of  d ∗ ds i x(s) = Ax(s) + α(s) − μx(s) + βδ0 a(s), x(0) = y and xx (·) the solution with initial data x. Proof. We use Proposition 3.2 with φ(x) = Bx, x so that ∇φ(x) = 2Bx. Moreover xx (·) − xy (·) satisfies the equation  d ∗ ds i (xx (s) − xy (s)) = A(xx (s) − xy (s)) − μ(xx (s) − xy (s)), (xx − xy )(0) = x − y (the one of Proposition 3.2 with control identically 0), and then by (10)  (28) |xx (s) − xy (s)|2B = |x − y|2B + 2

s

A∗ B(xx (r) − xy (r)), (xx (r) − xy (r))

0



− μ B(xx (s) − xy (s)), xx (s) − xy (s) dr s

≤ |x − y|2B + 2(1 + |μ|)

B(xx (r) − xy (r)), (xx (r) − xy (r)) dr. 0

Finally we can use Gronwall’s lemma to obtain the claim. Proposition 3.4. The value function V is Lipschitz with respect to the B-norm. Proof. Assume V (y) > V (x), ε > 0, and (α(·), a(·)) ∈ E × A an ε-optimal control for x. We have  ∞ |V (y) − V (x)| − ε ≤ e−ρt |L(xy (s), α(s), a(s)) − L(xx (s), α(s), a(s))|ds. 0

If we look, the explicit form of xx (·) and xy (·) as two-variable functions depends on the initial data only for s ∈ [0, βs¯ ] and afterwards only on the control. Indeed, for s > βs¯ , xx (s) = xy (s), and the previous integral is equal to 

s¯/β

e−ρt |L(xy (s), α(s), a(s)) − L(xx (s), , α(s), a(s))|ds

0

by (L1) and Proposition 3.3  ≤

s¯/β

e−ρt CL |xy (s) − xx (s)|B ds ≤

0

s¯ cs¯CL |x − y|B . β

By letting ε → 0 we have the thesis.

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1034

GIORGIO FABBRI

4. Existence and uniqueness of a solution. In this section we will prove that the value function is a viscosity solution of the HJB equation (Theorem 4.7) and that the HJB equation admits at most one solution (Theorem 4.8). We remind the reader that we use HB to denote the completion of H in the B-norm. This notation will be used in the next propositions. Proposition 4.1. Let u ∈ C(H) be a locally B-Lipschitz function. Let ψ ∈ C 1 (H), and let x be a local maximum (or a local minimum) of u − ψ. Then ∇ψ(x) ∈ R(B 1/2 ) ⊆ D(A∗ ). Proof. We give the proof only when x is a local maximum, as the case of the minima is similar. We take ω ∈ H, with |ω| = 1 and h ∈ (0, 1). Then for every h small enough u(x) − ψ(x) (u(x − hω) − ψ(x − hω)) ≤ , h h so that ψ(x) − ψ(x − hω) ≤ C|w|B , h and by passing to the limit we have ∇ψ(x), ω ≤ C|ω|B . Likewise (u(x + hω) − ψ(x + hω)) u(x) − ψ(x) ≤ , h h so that ψ(x) − ψ(x + hω) ≤ C|w|B , h and by passing to the limit we have − ∇ψ(x), ω ≤ C|ω|B . The two inequalities together then give | ∇ψ(x), ω | ≤ C|ω|B for all ω ∈ H. Then we can consider the linear extension to HB of the continuous linear functional ω → ∇ψ(x), ω that we denote with Φx . By the Riesz representation theorem, we can find zx ∈ HB such that Φx (ω) = zx , ωHB and note afterwards that   (29) zx , ωHB = B 1/2 (zx ), B 1/2 (ω) H   1/2 1/2 = B (B (zx )), ω

∀ω ∈ HB

(HB ) ×(HB )

  = B 1/2 (mx ), ω

(HB ) ×(HB )

,

def

where mx = (B 1/2 (zx )) ∈ H. Now for ω ∈ H     B 1/2 (mx ), ω = B 1/2 (mx ), ω , (HB ) ×(HB )

H

and therefore ∇ψ(x) = B 1/2 (mx ) ∈ R(B 1/2 ) ⊆ D(A∗ ), where the last inclusion follows from Remark 2.6.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1035

4.1. Existence. We start by proving a lemma and two propositions that will be used to prove the existence theorem (Theorem 4.7). We will use the notation introduced in Remark 2.1 on “x(s)” and “x[r].” Moreover we will continue to use the symbol δ0 in the text so that x[0] = δ0 x if x ∈ D(A∗ ). Lemma 4.2. Let x be a function of H 1 (0, s¯), and then  (30)



(i) lim

s→0+

s



(x[r] − x[r − s])2 dr s

s¯−s

(ii) lim

(31)

s→0+

s



= 0, s] − x2 [0] (x[r + s] − x[r]) x2 [¯ x[r]dr = . s 2

Proof. Part (i): We have  s



(x[r] − x[r − s])2 dr = s





ψs [r]dr, 0

where ψs : [0, s¯] → R is defined in the following way: ψs [r] =

if r ∈ [0, s), if r ∈ [s, s¯].

0 (x[r]−x[r−s])2 s

We prove the thesis by means of the Lebesgue theorem. First we show that ψs converges a.e. to zero. For r > s:

ψs [r] ≤

 

  r  r−s ∂ω x[τ ]dτ  s

|x[r] − x[r − s]| ,

where ∂ω x is the weak derivative of x. Now almost every r is a Lebesgue point, which implies that 

  r   r−s ∂ω x(τ )dτ  s

s→0+

−−−−→ |∂ω x[r]|

a.e. in r ∈ (0, s¯],

while the term |x[r] − x[r − s]| goes uniformly to 0. In order to dominate the convergence we note that by Morrey’s theorem ([18, Theorem 4, page 266]) every x ∈ H 1 (0, s¯) is 1/2-Holder continuous, and then there exists a positive constant C such that for every s ∈ (0, s¯] and every r ∈ [s, s¯] we have |x[r] − x[r − s]| √ ≤ C. s Then |ψs [r]| ≤

|x[r] − x[r − s]|2 ≤ C 2, s

and the proof of part (i) is complete.

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1036

GIORGIO FABBRI

Now we prove part (ii): def



s¯−s

(32) I(s) =

s



(x[r + s] − x[r]) x[r]dr s s¯−s

= s s¯−2s

 =− s

(x[r + s]x[r]) dr − s



s¯−2s

(x[r + s]x[r + s]) dr s 0  s¯−s (x[r + s] − x[r]) (x[r + s]x[r]) dr x[r + s]dr + s s s¯−2s  s (x[r + s])2 def dr = −I1 (s) + I2 (s) + I3 (s). + − s 0

By the continuity of x we see that s→0+

I2 (s) −−−−→ x2 [¯ s] and s→0+

I3 (s) −−−−→ −x2 [0]. Moreover, by using arguments similar to those in (i), we find that  lim+ (I(s) − I1 (s)) = lim+

(33)

s→0

s→0

s

s¯−2s

(x[r + s] − x[r])2 dr s  s¯−s (x[r + s] − x[r]) x[r]dr = 0, + lim s s→0+ s¯−2s



so, since I(s) + I1 (s) = I2 (s) + I3 (s), the limit lims→0+ I(s) exists if and only if there , and in such a case they have the same value. As exists the limit lims→0+ I1 (s)+I(s) 2 I2 (s) + I3 (s) s→0+ x2 [¯ I1 (s) + I(s) s] − x2 [0] = , −−−−→ 2 2 2 then  lim

s→0+

s

s¯−s

(x[r + s] − x[r]) x[r]dr s

=

s] − x2 [0] x2 [¯ . 2

Notation 4.3. We will call from now on O(s) a generic function O(·) : [0, +∞) → s→0+

[0, +∞) such that O(s) −−−−→ 0 and O(0) = 0. In what follows such notation will be used to express in particular the estimates that do not depend on the control. Lemma 4.4. Given x ∈ D(A∗ ), there exists an O(s), independent of the control, such that, for every control (α(·), a(·)) ∈ E × A, we have that |x(s) − x| ≤ O(s) (where we called x(s) the trajectory that starts from x and subject to the control (α(·), a(·))).

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1037

Proof. We consider s ∈ (0, 1]. By means of (6) we have (34) x(s) − x2H=  s¯    −μs  = x[r − βs] + e βs

s

−μτ

e

0

2  α(s − τ, r − βτ )dτ − x[r] dr

 2  r/β  μ   −βr  −μτ + a(s − r/β) + e α(s − τ, r − βτ )dτ − x[r] dr e   0 0 2  s¯  s¯  s   −μs 2   e ≤2 x[r − βs] − x[r] dr + 2 e|μ| Λdτ  dr  βs βs 0 2  βs   r/β   |μ|/β  + Γ + e|μ| Λdτ + |x|L∞ (0,¯s)  dr, e   0 0 

βs

where we have used that x ∈ D(A∗ ) ⊆ W 1,2 (0, s¯) ⊆ L∞ (0, s¯)  (35)

≤2

 2  e−μs x[(r − βs) ∧ 0] − x[r]2 dr + 2s2 s¯ e|μ| Λ

s¯ 

0

 2 + sβ e|μ|/β Γ + |x|L∞ + se|μ| Λ .

Observe that in this estimate the control (α(·), a(·)) does not appear. The second and the third terms go to zero for s → 0, while for the first we can use the Lebesgue theorem by observing that   −μs e x[(r − βs) ∧ 0] − x[r] ≤ e|μ| |x|L∞ + |x|L∞ ∀(s, r) ∈ (0, 1] × [0, s¯] s→0

and that |e−μs x[(r − βs) ∧ 0] − x[r]| −−−→ 0 pointwise. Proposition 4.5. Given x ∈ D(A∗ ) and g ∈ test2, there exists an O(s), independent of the control, such that, for every control (α(·), a(·)) ∈ E × A, with a(·) continuous, we have that    g(x(s)) − g(x) s ∇g(x), α(r)  g  (|x|) Γ2   0 − − ∇g(x), −μx ≤ 0 β + O(s)    s s |x| 2 (where we called x(s) the trajectory that starts from x and subject to the control (α(·), a(·))). Proof. First we observe (36)

s ∇g(x), α(r) g(x(s)) − g(x) − ∇g(x), −μx − 0 s s

s ∇g(x), α(r) g(x(s)) − g(y(s)) + g(y(s)) − g(x) − ∇g(x), −μx − 0 , = s s

where y(·) is the mild solution of  y(s) ˙ = Ay(s) + βδ0 a(s), (37) y(0) = x

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1038

GIORGIO FABBRI

corresponding to (11) with μ = 0 and α(·) = 0. The difference x(s) − y(s) can be expressed in the mild form as  s x(s) − y(s) = e(s−τ )A (α(τ ) − μx(τ ))dτ. 0

Now we come back to (36), and we have     g(x(s)) − g(x) s ∇g(x), α(r) dr   0 − − ∇g(x), −μx (38)    s s   

s  g(x(s)) − g(y(s))   g(y(s)) − g(x)  ∇g(x), α(r) dr   0 . ≤ − − ∇g(x), −μx +     s s s In order to estimate the first addendum we use the Taylor expansion as follows: (39)

g(x(s)) − g(y(s)) = s



 x(s) − y(s) ∇g(y(s)), s   x(s) − y(s) + ∇g(ξ(s)) − ∇g(y(s)), , s

where ξ(s) is a point between x(s) and y(s)  (40)

=

∇g(y(s)),

s 0

e(s−τ )A (α(τ ) − μx(τ ))dτ s  +



∇g(ξ(s)) − ∇g(y(s)), s→0

s 0

e(s−τ )A (α(τ ) − μx(τ ))dτ s

 .

s→0

We know by Lemma 4.4 that x(s) −−−→ x and y(s) −−−→ x uniformly in the control s→0 (α(·), a(·)), and then ∇g(y(s)) −−−→ ∇g(x) uniformly in the control and |∇g(y(s)) − s→0 ∇g(ξ(s))| −−−→ 0 uniformly in the control. Moreover, since in addition the control is s→0 bounded and x(s) −−−→ x uniformly in the control, we infer that the term   s  e(s−τ )A (α(τ ) − μx(τ ))dτ    0     s H

is bounded uniformly in the control α(·) and in s. From this we conclude that the second term is (40) goes to zero uniformly in (α(·), a(·)) and that    g(x(s)) − g(y(s)) s ∇g(x), α(r) dr    − 0 (41) − ∇g(x), −μx ≤ O(s)    s s for some function O(s) independent of the control. Now we estimate the second term of (38). We first note that ∇g(x) = g0 (|x|)

x |x|

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

and D2 g(x) = g0 (|x|)

x x ⊗ + g0 (|x|) |x| |x|



I x⊗x − |x| |x|3

1039

.

We consider the Taylor’s expansion of g at x: (42)

g(y(s)) − g(x) ∇g(x), y(s) − x 1 (y(s) − x)T (D2 g(x))(y(s) − x) = + s s 2 s o(|y(s) − x|2 ) +  s  y(s) − x 1 y(s) − x, y(s) − x g0 (|x|) x, + = |x| s 2 s  2  1 g0 (|x|) g0 (|x|) x, y(s) − x o(|y(s) − x|2 ) + + − 2 3 2 |x| |x| s s def

= P 1 + P 2 + P 3.

First we prove that P 2 and P 3 go to zero uniformly in (α(·), a(·)), and then we estimate P 1. We proceed in three steps. Step 1. There exists a constant C such that for every admissible control (α(·), a(·)) ∈ E × A with a(·) continuous, and every s ∈ (0, 1]4    x, y(s) − x   ≤ C.    s We observe first that the explicit solution of y(s)[r] can be found by taking μ = 0 and α = 0 in (6). We have  x[r − βs], r ∈ [βs, s¯], y(s, r) = a(s − r/β), r ∈ [0, βs), so

βs  s¯ x[r](a(s − r/β) − x[r])dr (x[r − βs] − x[r]) x, y(s) − x x[r] (43) = dr + 0 s s s βs



s ¯  s¯−βs βs −x2 [r]dr x[r]x[r + βs]dr (x[r + βs] − x[r]) s¯−βs = + 0 x[r] dr + s s s βs

βs 2

βs x[r]a(s − r/β)dr x [r]dr 0 − 0 + . s s When s → 0, the third and the fifth terms have opposite limits, while the second goes to zero as x is continuous and x(¯ s) = 0. The first term goes to − β2 x2 [0] = A∗ x, x in view of Lemma 4.2. The control appears only in the fourth term that we estimate as follows:  βs  βs   |x[r]|Γdr  0 x[r]a(s − r/β)dr  ≤ β max |x[r]|Γ.  ≤ 0   s s r∈[0,¯ s] 4 In the expression of y(·) the distributed control α(·) does not appear, so we will speak from now on only of the boundary control a(·)

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1040

GIORGIO FABBRI

Step 2. There exists a constant C such that for every admissible control a(·) ∈ A, with a(·) continuous, and every s ∈ (0, 1] |y(s) − x|2 ≤ C. s Indeed (44)

     y(s) − x, y(s) − x   s¯ (x[r − βs] − x[r])2  =   dr     s s βs  βs    2  0 (a(s − r/β) − x[r]) dr  + ,   s

since x ∈ D(A∗ ) ⊆ H 1 (0, s¯) and Lemma 4.2 holds; then the first term goes to zero. Moreover the second term is less than or equal to

βs (45)

0

Γ2 dr + s

βs 0

2|x[r]|Γdr + s

βs 0

|x[r]|2 dr ≤ C. s

This completes Step 2. From Step 2 it follows that o(|y(s) − x|2 ) |y(s) − x|2 s→0+ o(|y(s) − x|2 ) = −−−−→ 0 s |y(s) − x|2 s s→0

uniformly in a(·). Thus |P 3| −−−→ 0 uniformly in a(·). Moreover 2

x, y(s) − x | x, y(s) − x | ≤ |x||y(s) − x|, s s s→0

and then, from Step 1 and Lemma 4.4, |P 2| −−−→ 0 uniformly in a(·). Step 3. (Conclusion.) We now estimate P 1. We can write a more explicit form of P 1 as in the proofs of Steps 1 and 2 ((43), (44), and (45)), and by using the same arguments we can see that there exists a function O(s) (depending only on x and independent of the control) such that for every continuous control a(·) g  (|x|) (46) P 1 = 0 |x|



βs x[s]a(s − r/β)dr 1 0 (a(s − r/β))2 dr + A x, x + s 2 s 

βs 2

βs 1 0 x [r]dr 1 0 −2x[r]a(s − r/β)dr + + O(s). + 2 s 2 s

βs



0

2

The fourth term above does not depend on the control and converges to β x[0] that 2 is the opposite of the first term. The second and the fifth terms are opposite. Then we have that  βs  1 g0 (|x|) g0 (|x|) 1 0 (a(s − r/β))2 dr ≤ O(s) + P 1 = O(s) + βΓ2 . |x| 2 s 2 |x|

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1041

Now, by using the estimates on P 1, P 2, and P 3, we see that     g(y(s)) − g(x)   ≤ O(s) + 1 g0 (|x|) βΓ2 .    s 2 |x| By using this fact and (41) in (38), we have proved the proposition. Proposition 4.6. If x ∈ D(A∗ ) and φ ∈ test1, then there exists an O(s), independent of the control, such that, for every control (α(·), a(·)) ∈ E × A, with a(·) continuous, we have that

(47)

  φ(x(s)) − φ(x) s ∇φ(x), α(r) dr  − 0 − ∇φ(x), −μx   s s 

s βδ0 (∇φ(x)), a(r)R dr  ∗ 0 − A ∇φ(x), x −  ≤ O(s)  s

(where we called x(s) the trajectory that starts from x and subject to the control (α(·), a(·))). Proof. We proceed as in the proof of Proposition 4.5 by observing that φ(x(s)) − φ(x) φ(x(s)) − φ(y(s)) φ(y(s)) − φ(x) = + , s s s where y(·) is the solution of (37). It is possible to prove, by using exactly the same arguments in the proof of Proposition 4.5, that  

s  φ(x(s)) − φ(y(s)) ∇φ(x), α(r) dr   0 − ∇φ(x), −μx −   ≤ O(s),   s s where O(s) does not depend on the control. What is left to show is  

s  φ(y(s)) − φ(x) β δ0 ∇φ(x), a(r)R dr   ∗ 0 − A ∇φ(x), x −   ≤ O(s),   s s where O(s) does not depend on the control. We write (48)

φ(y(s)) − φ(x) def = I0 + I1 = s

 ∇φ(x),

 y(s) − x s   y(s) − x + ∇φ(ξ(s)) − ∇φ(x), , s s→0

where ξ(s) is a point between x and y(s). In view of Lemma 4.4, |y(s) − x| −−−→ 0 s→0 uniformly in the control, so that |ξ(s) − x| −−−→ 0 uniformly in a(·). By hypothesis ∗ ∇φ : H → D(A ) is continuous so that s→0

|∇φ(ξ(s)) − ∇φ(x)|D(A∗ ) −−−→ 0

(49) uniformly in a(·).

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1042

GIORGIO FABBRI

If we read (37) in D(A∗ ) it appears as an equation of the form  ¯ u(t) ˙ = Au(t) + f (t), u(0) = x, where f (t) is a bounded measurable function, with |f (t)|D(A∗ ) ≤ β|δ0 |D(A∗ ) Γ, and A¯ is an extension of A that generates of a C0 -semigroup on D(A∗ ) . So5 we can choose a constant C that depends on x such that, for all admissible continuous control a(·) and all s ∈ (0, 1], |y(s) − x|D(A∗ ) ≤ C. s

(50)

s→0

Thus from (49) and (50), we can say that |I1 | −−−→ 0 uniformly in a(·). Therefore    φ(y(s)) − φ(x) ∇φ(x), y(s) − x  s→0  −−−→ 0  −   s s uniformly in a(·). We now write ∇φ(x), y(s) − x = s

(51)





∇φ(x)[r] βs

βs

(x[r − βs] − x[r]) dr s

∇φ(x)[r](a(s − r/β) − x[r])dr s  s¯  s¯−βs (−∇φ(x)[r]x[r]) ∇φ(x)[r + βs] − ∇φ(x)[r] dr + dr = x[r] s s βs s¯−βs

βs

βs (∇φ(x)[r + βs]x[r])dr ∇φ(x)[r]a(s − r/β)dr 0 + + 0 s s

βs −∇φ(x)[r]x[r]dr . + 0 s +

0

The third and the fifth terms, which do not depend on the control, have opposite limits, the second goes to zero because ∇φ(x) and x are in D(A∗ ), and then x[¯ s] = 0 = ∇φ(x)[¯ s]. The first term tends to A∗ ∇φ(x), x. Finally we observe that the only term that depends on the control is the fourth and  βs 

s  ∇φ(x)[0]a(s − r )dr  s→0  0 ∇φ(x)[r]a(s − r/β)dr 0 −β   −−−→ 0   s s uniformly in a(·) and, since φ(x)[0] is a constant,

s

s ∇φ(x)[0]a(s − r)dr βδ0 ∇φ(x), a(r)R dr 0 = 0 . β s s This complete the proof. We can now prove that the value function is a solution of the HJB equation. Theorem 4.7. The value function V is bounded and B-Lipschitz, and it is a solution of the HJB equation. 5 In

¯ ⊆ D(A∗ ) ; see [20] for a proof. view of the fact that x is in H ⊆ D(A)

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1043

Proof. The boundedness of V follows from the boundedness of L (assumption (L2)). The B-Lipschitz property is the result of Proposition 3.4. It remains to verify that V is a solution of the HJB equation. V is a subsolution: Let x be a local maximum of V − (φ + g) for φ ∈ test1 and g ∈ test2. Due to Proposition 4.1 we know that ∇(φ + g)(x) ∈ D(A∗ ). Moreover we x ∈ know that ∇φ(x) ∈ D(A∗ ) for the definition of the set test1. So ∇g(x) = g0 (|x|) |x| D(A∗ ), which implies that x ∈ D(A∗ ). We can assume that V (x) − (φ + g)(x) = 0. We consider the constant control (α(·), a(·)) ≡ (α, a) ∈ Σ × Γ and x(s) the trajectory starting from x and subject to (α, a). Then for s small enough V (x(s)) − (φ + g)(x(s)) ≤ V (x) − (φ + g)(x), and thanks to Bellman’s principle of optimality we know that  s V (x) ≤ e−ρs V (x(s)) + e−ρr L(x(r), α, a)dr, 0

so that (52)

φ(x(s)) − φ(x) g(x(s)) − g(x) 1 − e−ρs V (x(s)) − − s s s



s −ρr e L(x(r), α, a)dr 0

s

≤ 0.

By using Propositions 4.5 and 4.6 and letting s → 0, we obtain (53) ρV (x) − ∇φ(x), −μx − ∇g(x), −μx ∗ − A ∇φ(x), x + βδ0 (∇φ(x)), aR + ∇φ(x), α + ∇g(x), α + L(x, α, a) ≤

g0 (|x|) Γ2 β . |x| 2

By taking the inf (α,a)∈Σ×Γ we obtain the subsolution inequality. V is a supersolution: Let φ ∈ test1 and g ∈ test2 and x be a minimum for V + (φ + g) and such that V + (φ + g)(x) = 0. Then as observed above x ∈ D(A∗ ). For some ε > 0 let (αε (·), aε (·)) be an ε2 -optimal strategy. With no loss of generality we can assume aε (·) continuous. We call x(s) the trajectory starting from x and subject to (αε (·), aε (·)). Now for s small enough V (x(s)) + (φ + g)(x(s)) ≥ V (x) + (φ + g)(x), and due to the ε2 -optimality and Bellman’s principle we know that  s e−ρr L(x(r), αε (r), aε (r))dr. V (x) + ε2 ≥ e−ρs V (x(s)) + 0

Then for s = ε we have (54)

1 − e−ρε φ(x(ε)) − φ(x) g(x(ε)) − g(x) V (x(ε)) + + ε ε

ε −ρr ε e L(x(r), αε (r), aε (r))dr ε2 + − 0 ≥ 0. ε ε

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1044

GIORGIO FABBRI

In view of Propositions 4.5 and 4.6 we can choose, independently of the control (αε (·), aε (·)), an O(s) such that: (55) ρV (x) + A∗ ∇φ(x), x + ∇φ(x) + ∇g(x), −μx  ε −βδ0 (∇φ(x), aε (r)R + e−ρr L(x, αε (r), aε (r))dr 0 − ε 

ε ∇φ(x) + ∇g(x), αε (r) dr g  (|x|) Γ2 0 ≥ O(ε) − 0 β . − ε |x| 2 Next we take the infimum as α and a varying in Σ × Γ inside the integral and let ε → 0 to obtain that (56) ρV (x) + A∗ ∇φ(x), x + ∇φ(x) + ∇g(x), −μx − inf − βδ0 (∇φ(x)), aR + L(x, α, a) − ∇φ(x) + ∇g(x), α (α,a)∈Σ×Γ

≥−

g0 (|x|) Γ2 β |x| 2

ε→0

(we observe again that the fact that O(s) −−−→ 0 uniformly in the control is essential). Therefore V is a solution of the HJB equation. 4.2. Uniqueness. Now we prove a uniqueness result in the case μ = 0. The proof of the case μ = 0 is similar with minor changes. Theorem 4.8. Given a supersolution v of the HJB equation and a subsolution u we have u(x) ≤ v(x) f or every x ∈ H. In particular there exists at most one solution of the HJB equation. Proof. We proceed by contradiction. Assume that u is a subsolution of the HJB equation and v a supersolution, and suppose that there exists x ˇ ∈ H and γ > 0 such that (u(ˇ x) − v(ˇ x)) >

3γ > 0. ρ

We choose γ < 1. Given ϑ > 0 small enough we have 2γ > 0. ρ

u(ˇ x) − v(ˇ x) − ϑ|ˇ x|2 >

(57)

We consider ε > 0 and ψ : H × H → R given by def

ψ(x, y) = u(x) − v(y) −

ϑ ϑ 1 1/2 |B (x − y)|2 − |x|2 − |y|2 . 2ε 2 2

Thanks to the boundedness of u and v, chosen ϑ > 0, there exist Rϑ > 0 such that   (58)

ψ(0, 0) ≥

sup

(ψ(x, y))

+ 1.

(|x|≥Rϑ ) or (|y|≥Rϑ )

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1045

We set S = {(x, y) ∈ H × H : |x| ≤ Rϑ and |y| ≤ Rϑ } . If we choose Rϑ big enough x ˇ ∈ S. By standard techniques (see [29, page 252]), given σ > 0, we can find p and q in H, with |p| < σ and |q| < σ, and such that (x, y) → ψ(x, y) − Bp, x − Bq, y attains a maximum (¯ x, y¯) in S. If we choose σ small enough (for example, such that σBRϑ < 14 γρ ), we know from (58) that such a maximum is in the interior of S and, thanks to (57), that ψ(¯ x, y¯) − Bp, x ¯ − Bq, y¯ >

3γ . 2ρ

Moreover (59)

ψ(¯ x, y¯) >

γ ρ

u(¯ x) − v(¯ y) >

and so

γ . ρ

In order to find an estimate in contradiction with (59) we prove some preliminary estimates. Estimates 1 (on ε): We observe that

M : (0, 1] → R,

 M : ε → sup(x,y)∈H×H u(x) − v(y) −

1 2ε

 1/2   B (x − y)2

is nonincreasing and bounded and so it admits a limit for ε → 0+ . So there exists a ε¯ > 0 such that for every ε1 , ε2 ∈ (0, ε¯] we have that |M (ε1 ) − M (ε2 )| <

(60)

γ 16(1 + |μ|)

2 .

We choose now ε that will be fixed in the proof:   1 (61) ε := min ε¯, 32CL2 (CL is the constant introduced in hypotheses (L1) and (L2)). This concludes the estimates on ε. Now we state and prove a claim that we will use in the estimate on σ and ϑ. Claim. If x ˜ ∈ H and y˜ ∈ H satisfy 2 2 1  1/2 γ  u(˜ x) − v(˜ y) − x − y˜) ≥ M (ε) − , B (˜ 2ε 16(1 + |μ|)

(62) then (63)

2 2 γ 1  1/2 1  x − y˜) ≤ . B (˜ ε 32 (1 + |μ|)

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1046

GIORGIO FABBRI

Proof of the claim. (We follow the idea used in Lemma 3.2 of [16]) 2 1  1/2  x − y˜) B (˜ 4ε 2 2 1  1/2 1  1/2   = u(˜ x) − y(˜ y) − x − y˜) + x − y˜) B (˜ B (˜ 2ε 4ε 2 2 γ 1  1/2  ≥ M (ε) − + x − y˜) , B (˜ 16(1 + |μ|) 4ε

(64) M (ε/2) ≥ u(˜ x) − y(˜ y) −

so that (65)

2 2 1  1/2 γ  x − y˜) ≤ M (ε/2) − M (ε) + B (˜ 4ε 16(1 + |μ|) 2 2 2 γ γ γ ≤ + =2 , 16(1 + |μ|) 16(1 + |μ|) 16(1 + |μ|)

γ )2 follows from the definition of ε where the inequality M (ε/2) − M (ε) < ( 16(1+|μ|) (61) that implies ε ≤ ε¯ and then (60). The claim follows. Note that from (61) we have

√ 1 √ ≥ 4 2CL , ε and then, if x ˜, y˜ satisfy the hypothesis (62) of the claim, we have x − y˜|B ≤ CL |˜

(66)

γ . 32(1 + |μ|)

Estimates 2 (on σ): We have already imposed σ < the following   γ ϑ , (67) σ = min , ϑ, 8ρBRϑ Rϑ

γ/ρ 4 B Rϑ

we set here and in

so that ϑ→0

σ −−−→ 0

(68) and

ϑ→0

σRϑ −−−→ 0.

(69)

We recall that we have already fixed ε in (61). From the choice of σ (67) follows that (70)

ϑ→0

|Bp, x ¯| ≤ BσRϑ −−−→ 0,

ϑ→0

|Bq, y¯| ≤ BσRϑ −−−→ 0.

Moreover, in view of the continuity of the linear operator A∗ B : H → H with norm A∗ B, we have (71)

|A∗ Bp, x ¯| ≤ A∗ BσRϑ −−−→ 0, ϑ→0

|A∗ Bq, y¯| ≤ BσRϑ −−−→ 0. ϑ→0

This concludes the estimates on σ.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1047

Estimates 3 (on ϑ): One can prove that with fixed ε we have 2 ϑ→0

2 ϑ→0

ϑ |¯ x| −−−→ 0,

(72)

ϑ |¯ y | −−−→ 0

(it is a quite standard fact; see, for example, [16]). So (73)

lim (ψ(¯ x, y¯) − Bp, x ¯ − Bq, y¯)

ϑ→0

=

x(x) − v(y) −

sup (x,y)∈H×H

2 1  1/2 γ  (x − y) B  >2 2ε ρ

(where the last inequality follows from (57)). In (67) we chose σ as a function of ϑ, and now we fix ϑ. We begin by taking ϑ<

γ 64βΓ2

so that βϑΓ2 <

(74)

γ . 64

We know from (70) and (71) that, if we choose ϑ small enough, we have |μ| |Bp, x ¯| <

γ , 16

|μ| |Bq, y¯| <

γ , 16

|A∗ Bp, x ¯| <

γ , 16

|A∗ Bq, y¯| <

γ . 16

(75)

From (72) we know that, if we choose ϑ small enough, we have 2

|μ|ϑ |¯ x| <

(76)

γ , 32

2

|μ|ϑ |¯ y| <

γ . 32

Moreover (72) implies also that ϑ→0

ϑ |¯ x| −−−→ 0,

ϑ→0

ϑ |¯ y | −−−→ 0,

and then, if we choose ϑ small enough, we have ϑΣ (|¯ x| + |¯ y |) <

(77)

γ . 32

Moreover in view of (73) we know that, if we choose ϑ small enough, then x ¯ and y¯ satisfy the hypothesis (62) of the claim, and then, from (63), we have 2 2 1  1/2 γ γ 1 1 γ  (78) (¯ x − y ¯ ) ≤ ≤ ≤ B  ε 32 (1 + |μ|) 32 (1 + |μ|) 32 (we took 0 < γ < 1 and then γ 2 < γ). From (63) in the same way we obtain 2 |μ|  1/2 γ  (79) x − y¯) ≤ B (¯ ε 32 and, from (66), (80)

  γ   CL B 1/2 (¯ . x − y¯) ≤ 32

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1048

GIORGIO FABBRI

Eventually, if we choose ϑ small enough in (68), we have 2B|p|Σ ≤ 2BσΣ ≤

(81)

γ 64

and 2σβδ0 ◦ BΓ ≤

(82)

γ , 32

where δ0 ◦ B is the norm of the linear continuous functional δ0 ◦ B : H → R. Finally we fix ϑ > 0 small enough to satisfy (75), (76), (77), (78), (79) (80), (81), and (82). Now we proceed with the proof of the theorem. The map x → u(x) −

1 1/2 ϑ |B (x − y¯)|2 + |x|2 + Bp, x 2ε 2

attains a maximum at x ¯, and y → v(y) +

1 1/2 ϑ |B (¯ x − y)|2 + |y|2 + Bq, y 2ε 2

attains a minimum at y¯. Note that, due to Proposition 4.1, x ¯ and y¯ are in D(A∗ ). We can now use the definition of sub- and supersolution to obtain 1 ∗ 1 x − y¯), x ¯ − B(¯ ¯ A B(¯ x − y¯), −μ¯ x − A∗ Bp, x ε ε 1 − Bp, −μ¯ x − ϑ ¯ x, −μ¯ x − inf x − y¯)), aR + βδ0 (Bp), aR βδ0 (B(¯ (α,a)∈Σ×Γ ε ϑβΓ2 1 x − y¯), α + Bp, α + L(¯ x, α, a) ≤ + ϑ ¯ x, α + B(¯ ε 2

(83) ρu(¯ x) −

and 1 ∗ 1 A B(¯ x − y¯), −μ¯ y  + A∗ Bq, y¯ x − y¯), y¯ − B(¯ ε ε 1 βδ0 (B(¯ x − y¯)), aR − βδ0 (Bq), aR + Bq, −μ¯ y  + ϑ ¯ y , −μ¯ y − inf (α,a)∈Σ×Γ ε ϑβΓ2 1 x − y¯), α − Bq, α + L(¯ y , α, a) ≥ − . − ϑ ¯ y , α + B(¯ ε 2

(84) ρv(¯ y) −

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1049

INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

The two estimates together give (85) ρu(¯ x) − ρv(¯ y) − −

1 ∗ A B(¯ x − y¯), (¯ x − y¯) ε

1 B(¯ x − y¯), −μ(¯ x − y¯) − A∗ Bp, x ¯ − A∗ Bq, y¯ ε

− Bp, −μ¯ x − Bq, −μ¯ y  − ϑ ¯ x, −μ¯ x − ϑ ¯ y , −μ¯ y 1 βδ0 (B(¯ x − y¯)), aR + βδ0 (Bp), aR − inf (α,a)∈Σ×Γ ε 1 + ϑ ¯ x, α + B(¯ x − y¯), α + Bp, α + L(¯ x, α, a) ε 1 βδ0 (B(¯ x − y¯)), aR − βδ0 (Bq), aR + inf (α,a)∈Σ×Γ ε − ϑ ¯ y , α +

1 B(¯ x − y¯), α − Bq, α + L(¯ y , α, a) ≤ βϑΓ2 . ε

We now note that from (10) A∗ B ≤ B, and then we have (86)

1 1 1 x − y¯), (¯ x − y¯) = − |¯ x − y¯|2B . − A∗ B(¯ x − y¯), (¯ x − y¯) ≥ − B(¯ ε ε ε

Moreover we observe that 1 βδ0 (B(¯ (87) − inf x − y¯)), aR + βδ0 (Bp), aR (α,a)∈Σ×Γ ε

1 B(¯ x − y¯), α + Bp, α + L(¯ x, α, a) ε 1 + inf βδ0 (B(¯ x − y¯)), aR − βδ0 (Bq), aR (α,a)∈Σ×Γ ε 1 x − y¯), α − Bq, α + L(¯ y , α, a) − ϑ ¯ y , α + B(¯ ε inf − βδ0 (Bp), aR − βδ0 (Bq), aR + L(¯ y , α, a) − L(¯ x, α, a) + ϑ ¯ x, α +



(α,a)∈Σ×Γ

− ϑ ¯ y , α − ϑ ¯ x, α − Bq, α − Bp, α ≥

 inf (α,a)∈Σ×Γ





sup (α,a)∈Σ×Γ



 sup (α,a)∈Σ×Γ

 L(¯ y , α, a) − L(¯ x, α, a)

βδ0 (Bp), aR + βδ0 (Bq), aR

 ϑ ¯ y , α + ϑ ¯ x, α −

 sup



Bq, α + Bp, α



(α,a)∈Σ×Γ

≥ −CL |¯ x − y¯|B − 2σβδ0 ◦ BΓ − Σϑ(|¯ x| + |¯ y |) − 2BσΣ.

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1050

GIORGIO FABBRI

Thus by using (86) and (87) in (85) we derive   1 x − y¯|2B (88) ρ u(¯ x) − v(¯ y ) − |¯ ε μ x − y¯), −(¯ x − y¯) − A∗ Bp, x − B(¯ ¯ − A∗ Bq, y¯ ε − Bp, −μ¯ x − Bq, −μ¯ y  − ϑ ¯ x, −μ¯ x − ϑ ¯ y , −μ¯ y x − y¯|B − 2σβδ0 ◦ BΓ − Σϑ(|¯ x| + |¯ y |) − 2BσΣ − βϑΓ2 ≤ 0, − CL |¯ and from the preceding and from (78), (79), (75), (76), (80), (82), (77), (81), and (74) we obtain γ γ γ γ γ γ γ γ (89) ρ(u(¯ x) − v(¯ y )) − 2 −4 −2 − − − − − ≤ 0; 32 16 32 32 32 32 64 64 that is, 1 ρ(u(¯ x) − v(¯ y )) − γ ≤ 0. 2   Now we recall that from (59) we have ρ u(¯ x) − v(¯ y ) > γ, and then we obtain from (90) (90)

1 1 1 γ = γ − γ < ρ(u(¯ x) − v(¯ y )) − γ ≤ 0, 2 2 2 which yields a contradiction because γ > 0. The theorem is proved. Remark 4.9. Now we can better explain the remark in the introduction saying that it is difficult to treat the case of a nonconstant coefficient. We can estimate 2 the term 1ε B(¯ x − y¯), −μ(¯ x − y¯) because we use the term 1ε |x − y|B to penalize the doubling of the variables with respect to the B-norm. If μ is a function of r, such a term is replaced by 1ε B(¯ x − y¯), −μ(·)(¯ x − y¯), where −μ(·)(¯ x − y¯) is the pointwise product of the L∞ (0, s¯) function μ(·) and the L2 (0, s¯) function (¯ x − y¯), which cannot be estimated by using similar arguments. ´ ech for his hospitality, Acknowledgments. The author thanks Prof. Andrzej Swi¸ his great kindness, many useful suggestions, and stimulating conversations. Thanks to Silvia Faggian for the invaluable advice. REFERENCES [1] M. Adamo, A. L. Amadori, M. Bernaschi, C. La Chioma, A. Marigo, B. Piccoli, S. Sbaraglia, A. Uboldi, D. Vergni, P. Fabbri, D. Iacovoni, F. Natale, S. Scalera, L. Spilotro, and A. Valletta, Optimal strategies for the issuances of public debt securities, Int. J. Theor. Appl. Finance, 7 (2004), pp. 805–822. [2] S. Anit ¸ a, Analysis and Control of Age-Dependent Population Dynamics, Math. Model. Theory Appl. 11, Kluwer Academic Publishers, Dordrecht, 2000. [3] V. Barbu and G. Da Prato, Hamilton-Jacobi Equations in Hilbert Spaces, Res. Notes Math. 86, Pitman (Advanced Publishing Program), Boston, MA, 1983. [4] V. Barbu, G. Da Prato, and C. Popa, Existence and uniqueness of the dynamic programming equation in Hilbert space, Nonlinear Anal., 7 (1983), pp. 283–299. [5] E. Barucci and F. Gozzi, Optimal investment in a vintage capital model, Res. Econ., (1998), pp. 159–188. [6] E. Barucci and F. Gozzi, Optimal advertising with a continuum of goods, Ann. Oper. Res., 88 (1999), pp. 15–29.

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INFINITE-DIMENSIONAL HJB ARISING FROM TRANSPORT PDE

1051

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For instance, in the vintage capital models x(t, s) may be regarded ... Heterogeneous capital, in both the finite- and infinite-dimensional approaches, is used ...... Finally we fix ϑ > 0 small enough to satisfy (75), (76), (77), (78), (79) (80), (81), and (82). Now we proceed with the proof of the theorem. The map x ↦→ u(x) −.

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