Controlled Singular Volterra Integral Equations and Pontryagin Maximum Principle
aa r X i v : . [ m a t h . O C ] D ec Controlled Singular Volterra Integral Equationsand Pontryagin Maximum Principle ∗ Ping Lin † and Jiongmin Yong ‡ Abstract.
This paper is concerned with a class of controlled singular Volterra integral equations, whichcould be used to describe problems involving memories. The well-known fractional order ordinary differentialequations of the Riemann–Liouville or Caputo types are strictly special cases of the equations studied inthis paper. Well-posedness and some regularity results in proper spaces are established for such kind ofquestions. For the associated optimal control problem, by using a Liapounoff’s type theorem and the spikevariation technique, we establish a Pontryagin’s type maximum principle for optimal controls. Differentfrom the existing literature, our method enables us to deal with the problem without assuming regularityconditions on the controls, the convexity condition on the control domain, and some additional unnecessaryconditions on the nonlinear terms of the integral equation and the cost functional.
AMS Mathematics Subject Classification:
Keywords:
Singular Volterra integral equation, Fractional ordinary differential equation, optimal con-trol, Pontryagin’s maximum principle
In this paper, we consider the following controlled Volterra integral equation:(1.1) y ( t ) = η ( t ) + Z t f ( t, s, y ( s ) , u ( s )) ds, t ∈ [0 , T ] . We call the above the state equation , where η ( · ) and f ( · , · , · , · ) are given maps, called the free term and the generator of the state equation, respectively, y ( · ) is called the state trajectory taking values in the Euclideanspace R n , and u ( · ) is called the control taking values in some separable metric space U . To measure theperformance of the control, we introduce the cost functional(1.2) J ( u ( · )) = Z T g ( t, y ( t ) , u ( t )) dt + m X j =1 h j ( y ( t j )) , with the two terms on the right hand representing the running cost and the specific instant costs (at 0 t < t < · · · < t m T ), respectively.Equations like (1.1) can be used to describe some dynamics involving memories. In the classical situationsof optimal control for Volterra integral equations, people usually assume that the map f ( · , · , · , · ) is continu-ous, together with some further smoothness/differentiability conditions. Relevant works can be traced backto those by Vinokurov in the later 1960s [45], followed by the works of Angell [4], Kamien-Muller [27],Medhin [33], Carlson [15], Burnap-Kazemi [12], and some recent works by de la Vega [19], Belbas [6, 7], andBonnans–de la Vega–Dupuis [9]. On the other hand, in the past several decades, fractional differential equa-tions have attracted quite a few researchers’ attention due to some very interesting applications in physics, ∗ This work was partially supported by the National Natural Science Foundation of China under grant 11471070, and NSFGrant DMS-1406776. † School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China. E-mail address:[email protected]. ‡ Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA. E-mail address: [email protected]. s = t , and the free term η ( · ) being possibly discontinuous (blowing up) at t = 0. More precisely, thecorresponding controlled state equation of form (1.1) could have the feature that(1.3) η ( t ) = ct − α (or c ) , f ( t, s, y, u ) = e f ( s, y, u )( t − s ) − α , s < t T, ∀ ( y, u ) , for some map e f ( · , · , · ) and constants α ∈ (0 , c ∈ R . Such kind of singularity makes the optimal controlproblems for fractional differential equations different from the classical optimal control problems for Volterraintegral equations as in the above-mentioned literature.The purpose of this paper is to study an optimal control problem with the state equation (1.1) allowing( t, s ) f ( t, s, y, u ) to have some singularity along t = s and allowing the free term η ( · ) to be (unboundedly)discontinuous. We point out that our state equation (1.1) could cover a much wider class of dynamic systemswith various type memories than the ones described by fractional differential equations (with the conditionslike (1.3)). Let us make a little more comments on our state equation. Since the free term η ( · ) is allowedto have some singularities, a natural class for η ( · ) should be L p functions. Then we expect, under suitableconditions, the state trajectory y ( · ) will also be a function in the same class. On the other hand, in thecost functional, we need y ( t j ) to be defined. Therefore, we need to have certain continuity of the statetrajectory. Then it is necessary to narrow the L p space by adding certain continuity. This will lead tosome difficulties in establishing the well-posedness of the state equation in the correct class of functionsthat the state trajectories will belong to. To overcome the difficulty, we introduce certain weighted functionspaces, and extend some classical results, such as Gronwall’s inequality, etc. to the form that will make ourprocedure works.The rest of the paper is organized as follows. In Section 2, necessary preliminaries will be presented. Someresults are interesting by themselves. Well-posedness of the state equation, together with the continuity ofthe solutions, will be established in Section 3. Section 4 is devoted to a proof of Pontryagin’s type maximumprinciple for our optimal control problem of singular integral equations. As a special case, the maximumprinciples for fractional differential equations in the sense of Riemann–Liouville, and Caputo, will be brieflydescribed. Some concluding remarks will be collected in Section 5. In this section, we will present some preliminary results which will be useful later. First of all, let
T > L p (0 , T ; R n ) = n ϕ : [0 , T ] → R n (cid:12)(cid:12) k ϕ ( · ) k L p (0 ,T ; R n ) ≡ (cid:16) Z T | ϕ ( t ) | p dt (cid:17) p < ∞ o , p < ∞ ,L ∞ (0 , T ; R n ) = n ϕ : [0 , T ] → R n (cid:12)(cid:12) k ϕ ( · ) k L ∞ (0 ,T ; R n ) ≡ esssup t ∈ [0 ,T ] | ϕ ( t ) | < ∞ o ,C ([0 , T ]; R n ) = n ϕ : [0 , T ] → R n (cid:12)(cid:12) ϕ ( · ) is continuous o .
2e denote k ϕ ( · ) k p = k ϕ ( · ) k L p (0 ,T ; R n ) , ∀ ϕ ( · ) ∈ L p (0 , T ; R n ) , p ∈ [1 , ∞ ] , k ϕ ( · ) k C = max t ∈ [0 ,T ] | ϕ ( t ) | = k ϕ ( · ) k ∞ , ∀ ϕ ( · ) ∈ C ([0 , T ]; R n ) . Also, we define L p + (0 , T ; R n ) = [ q>p L q (0 , T ; R n ) , p ∈ [1 , ∞ ) ,L p − (0 , T ; R n ) = \ q
} . If w ( s ) = | s − s | γ for some s ∈ [0 , T ] and γ >
0, then a ( · ) | · − s | α ∈ C w ( · ) ([0 , T ]; R n ) , for any α γ , a ( · ) ∈ C ([0 , T ]; R n ). From the above, we should have some feeling about the space C w ( · ) ([0 , T ]; R n ).We denote(2.1) ∆ = (cid:8) ( t, s ) ∈ [0 , T ] (cid:12)(cid:12) s < t T (cid:9) . Note that the “diagonal line” { ( t, t ) | t ∈ [0 , T ] } is not contained in ∆. Thus if ϕ : ∆ → R n with ( t, s ) ϕ ( t, s )being continuous, then ϕ ( · ) could be unbounded as | t − s | → Lemma 2.1.
Let p, q, r > satisfy p + 1 q = 1 + 1 r . Then for any f ( · ) ∈ L p ( R n ) , g ( · ) ∈ L q ( R n ) , (2.2) k f ∗ g k r k f k p k g k q . We now present several results. Some of them should be standard. However, we will provide the proofsfor readers’ convenience.
Lemma 2.2.
Let β ∈ (0 , and ϕ : ∆ → R n . Define (2.3) ψ ( t ) = Z t ϕ ( t, s )( t − s ) − β ds, t ∈ [0 , T ] . (i) Suppose for some p ∈ [1 , ∞ ) , (2.4) Z T sup t ∈ [ s,T ] | ϕ ( t, s ) | p ds < ∞ . hen (2.5) k ψ ( · ) k p T β β (cid:16) Z T sup t ∈ [ s,T ] | ϕ ( t, s ) | p ds (cid:17) p . (ii) Suppose, in addition, t ϕ ( t, s ) satisfies the following (2.6) | ϕ ( t, s ) − ϕ ( t ′ , s ) | ω ( | t − t ′ | ) , ∀ t, t ′ ∈ [ t − σ, t + σ ] , s ∈ [0 , t ∧ t ′ ) , for some modulus of continuity ω : [0 , ∞ ) → [0 , ∞ ) , and for some q > β , σ > , with ( t − σ, t + σ ) ⊆ [0 , T ] ,the following holds: (2.7) Z t + σt − σ sup t ∈ [ s,t + σ ] | ϕ ( t, s ) | q ds < ∞ . Then ψ ( · ) is continuous at t . Consequently, if t ϕ ( t, s ) is continuous uniformly in s ∈ [0 , T ] and (2.4) holds for some p > β , then t ψ ( t ) is continuous on [0 , T ] . Proof. (i) Let ¯ ϕ ( s ) = sup t ∈ ( s,T ] | ϕ ( t, s ) | , s ∈ [0 , T ] . Then ¯ ϕ ( · ) ∈ L p (0 , T ; R ). Hence, (2.5) follows from Young’s inequality for convolution.(ii) Let q > β which is equivalent to κ ≡ (1 − β ) qq − < b ϕ ( s ) = sup t ∈ ( s,t + σ ] | ϕ ( t, s ) | , s ∈ [0 , t + σ ] . For any t − σm < t < t ′ < t + σm with m > | ψ ( t ) − ψ ( t ′ ) | = (cid:12)(cid:12)(cid:12) Z t ϕ ( t, s )( t − s ) − β ds − Z t ′ ϕ ( t ′ , s )( t ′ − s ) − β ds (cid:12)(cid:12)(cid:12) Z t − σm (cid:12)(cid:12)(cid:12) ϕ ( t, s )( t − s ) − β − ϕ ( t ′ , s )( t ′ − s ) − β (cid:12)(cid:12)(cid:12) ds + Z tt − σm | ϕ ( t, s ) | ( t − s ) − β ds + Z t ′ t − σm | ϕ ( t ′ , s ) | ( t ′ − s ) − β ds Z t − σm | ϕ ( t, s ) | (cid:16) t − s ) − β − t ′ − s ) − β (cid:17) ds + Z t − σm | ϕ ( t, s ) − ϕ ( t ′ , s ) | ( t ′ − s ) − β ds + Z tt − σm b ϕ ( s )( t − s ) − β ds + Z t ′ t − σm b ϕ ( s )( t ′ − s ) − β ds ( t ′ − t ) − β Z t − σm b ϕ ( s )( t − s ) − β ( t ′ − s ) − β ds + ω ( | t − t ′ | ) Z t − σm ds ( t ′ − s ) − β + k b ϕ ( · ) k L q ( t − σm ,t ; R n ) (cid:16) Z tt − σm ds ( t − s ) κ (cid:17) q − q + k b ϕ ( · ) k L q ( t − σm ,t ′ ; R n ) (cid:16) Z t ′ t − σm ds ( t ′ − s ) κ (cid:17) q − q ( t ′ − t ) − β ( σm ) − β ) k b ϕ ( · ) k + ω ( | t − t ′ | ) T β β + k b ϕ ( · ) k L q ( t − σ,t + σ ; R n ) h(cid:16) ( σm ) − κ − κ (cid:17) q − q + (cid:16) ( t ′ − t + σm ) − κ − κ (cid:17) q − q i . Hence, for any ε >
0, we first take m > k b ϕ ( · ) k L q ( t − σ,t + σ ; R n ) h(cid:16) ( σm ) − κ − κ (cid:17) q − q + (cid:16) ( σm ) − κ − κ (cid:17) q − q i < ε . Then let δ ∈ (0 , σ ) be small enough so that δ − β ( σm ) − β ) k b ϕ ( · ) k + ω ( δ ) T β β < ε . ψ ( · ) is continuous at t . The last conclusion follows easily from what wejust proved.The above lemma show that for any p ∈ [1 , ∞ ), under condition (2.4), one has ψ ( · ) ∈ L p (0 , T ; R n ).To guarantee the continuity of ψ ( · ) at t ∈ (0 , T ], we need to assume the continuity of t ϕ ( t, s ) for t ∈ [ t − σ, t + σ ], uniformly in s ∈ [0 , t ), together with L q integrability of s sup t ∈ ( s,t + σ ] | ϕ ( t, s ) | for q > β .The following example shows that continuity of ψ ( · ) might fail ¯ ϕ ( · ) does not have a good enough integrability. Example 2.3.
Let ϕ ( t, s ) = ¯ ϕ ( s ) = 1 | s − s | , β = 12 , with s ∈ (0 , T ). Then ¯ ϕ ( · ) ∈ L q (0 , T ; R ) with q < < β . Note that ψ ( t ) = Z t ϕ ( t, s )( t − s ) − β ds = Z t ds | s − s | ( t − s ) , t > . By Young’s inequality, we know that the above ψ ( · ) ∈ L p (0 , T ; R ) for some p >
1. Also, one sees thatlim t ↑ s ψ ( t ) = Z s ds | s − s | = ∞ . Thus, ψ ( · ) is discontinuous at s .It is natural to ask if we relax the L q -integrability of ϕ ( · ), what can we say about the continuity of ψ ( · )defined by (2.3)? Let us make it more precise now. Let α i , β ∈ (0 , i ℓ and 0 s < s < · · ·
Lemma 2.4.
Let α i , β ∈ (0 , , i ℓ , and w ( · ) be defined by (2.8) with s < s < · · · < s ℓ T .Let e ϕ : ∆ → R n satisfy (2.11) | e ϕ ( t, s ) − e ϕ ( t ′ , s ) | ω ( | t − t ′ | ) , ∀ ( t, s ) , ( t ′ , s ) ∈ ∆ , for some modulus of continuity ω : [0 , ∞ ) → [0 , ∞ ) , and (2.12) Z T max t ∈ ( s,T ] | e ϕ ( t, s ) | q ds < ∞ , with some (2.13) q > β ∨ α i , i ℓ. efine e ψ ( · ) by (2.9) . Then (2.14) e ψ ( · ) ∈ L ∞ ¯ w ( · ) (0 , T ; R n ) \ (cid:16) ℓ \ i =1 C (cid:0) ( s i − , s i ); R n (cid:1)(cid:17) , where (2.15) ¯ w ( s ) = ℓ Y i =0 | s − s i | (1+ q − α i − β ) + . Consequently, for any ε > , (2.16) e ψ ( · ) ∈ C ¯ w ε ( · ) ([0 , T ]; R n ) , where (2.17) ¯ w ε ( s ) = ℓ Y i =0 | s − s i | (1+ q − α i − β ) + + ε . Further, if (2.18) α i + β > q , then e ψ ( · ) is continuous at s i . Consequently, if (2.19) min i ℓ α i + β > q , then e ψ ( · ) ∈ C ([0 , T ]; R n ) . Proof.
Denote ¯ ϕ ( s ) = max t ∈ [ s,T ] | e ϕ ( t, s ) | , s ∈ [0 , T ] . Define w i ( s ) = Y j = i | s − s j | − α j , s ∈ [0 , T ] , i ℓ, and(2.20) δ = min i ℓ − s i +1 − s i > , ¯ s i = s i + s i +1 , i ℓ − . Then(2.21) s i + δ ¯ s i s i +1 − δ , i = 0 , , · · · , ℓ − . We establish some estimates for e ψ ( · ) on some time intervals, more precisely, on [0 , s ), ( s , ¯ s ], (0 , ¯ s ], [¯ s , s ),( s , ¯ s ], [¯ s , s ). Then induction will apply.(i) For t ∈ [0 , s ), one has | e ψ ( t ) | = (cid:12)(cid:12)(cid:12) Z t e ϕ ( t, s ) w ( s )( s − s ) − α ( t − s ) − β ds (cid:12)(cid:12)(cid:12) K Z t ¯ ϕ ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z t ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z t ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − (cid:17) q − q K (cid:16) Z t ¯ ϕ ( s ) q ds (cid:17) q ( s − t ) α + β − (cid:16) Z t ds (1 + t − ss − t ) (1 − α ) qq − ( t − ss − t ) (1 − β ) qq − (cid:17) q − q . K is a positive constant, which may be different when appears at differentplaces. Now, we let τ = t − ss − t . Then s = t − ( s − t ) τ , ds = ( t − s ) dτ , and | e ψ ( t ) | K (cid:16) Z t ¯ ϕ ( s ) q ds (cid:17) q ( s − t ) α + β − − q (cid:16) Z ts − t dτ (1 + τ ) (1 − α ) qq − τ (1 − β ) qq − (cid:17) q − q ≤ K ( s − t ) α + β − − q h(cid:16) Z dττ (1 − β ) qq − (cid:17) q − q + (cid:16) Z s s − t dττ (2 − α − β ) qq − (cid:17) q − q i K ( s − t ) α + β − − q h (cid:16) s s − t (cid:17) [1 − (2 − α − β ) qq − ] q − q i K ( s − t ) α + β − − q h (cid:16) s − t (cid:17) α + β − − q i K + K ( s − t ) α + β − − q , t ∈ [0 , s ) . Hence,(2.22) ( s − t ) (1 − α − β + q ) + | e ψ ( t ) | K, t ∈ [0 , s ) . (ii) For t ∈ ( s , ¯ s ], one has | e ψ ( t ) | Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds + Z ts ¯ ϕ ( s ) w ( s )( t − s ) − β ds ≡ I + I . For I , we have I K (cid:16) Z s ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − (cid:17) q − q K ( t − s ) α + β − (cid:16) Z s ds ( s − st − s ) (1 − α ) qq − (1 + s − st − s ) (1 − β ) qq − (cid:17) q − q . Now, we let τ = s − st − s . Then s = s − ( t − s ) τ , ds = ( s − t ) dτ , and I K ( t − s ) α + β − − q (cid:16) Z s t − s dττ (1 − α ) qq − (1 + τ ) (1 − β ) qq − (cid:17) q − q K ( t − s ) α + β − − q h(cid:16) Z dττ (1 − α ) qq − (cid:17) q − q + (cid:16) Z s t − s dττ (2 − α − β ) qq − (cid:17) q − q i K ( t − s ) α + β − − q h (cid:16) s t − s (cid:17) [1 − (2 − α − β ) qq − ] q − q i K ( t − s ) α + β − − q h (cid:16) t − s (cid:17) α + β − − q i K + K ( t − s ) α + β − − q , t ∈ ( s , ¯ s ] . Now, we look at I , noting s + δ ¯ s s − δ , I = Z ts ¯ ϕ ( s ) w ( s )( t − s ) − β ds = Z ts ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z ts ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z ts ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − (cid:17) q − q K ( t − s ) α + β − (cid:16) Z ts ds ( s − s t − s ) (1 − α ) qq − (1 − s − s t − s ) (1 − β ) qq − (cid:17) q − q . Note that by (2.13), we have (1 − α ) qq − < , (1 − β ) qq − < , − (1 − α ) qq − α q − q − > , − (1 − β ) qq − βq − q − > . Let τ = s − s t − s . Then s = s + ( t − s ) τ , ds = ( t − s ) dτ , and I K ( t − s ) α + β − − q (cid:16) Z dττ − α q − q − (1 − τ ) − βq − q − (cid:17) q − q = K ( t − s ) α + β − − q B (cid:16) α q − q − , βq − q − (cid:17) q − q . Here, ( a, b ) B ( a, b ) is the Beta function. Hence,(2.23) ( t − s ) (1 − α − β + q ) + | e ψ ( t ) | K, t ∈ ( s , ¯ s ] . (iii) For t ∈ [¯ s , s ), we have | e ψ ( t ) | (cid:12)(cid:12)(cid:12) Z ¯ s e ϕ ( t, s ) w ( s )( t − s ) − β ds (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) Z t ¯ s e ϕ ( t, s ) w ( s )( t − s ) − β ds (cid:12)(cid:12)(cid:12) ≡ I + I . For I , we have I Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds + Z ¯ s s ¯ ϕ ( s ) w ( s )( t − s ) − β ds Z s ¯ ϕ ( s )( s − s ) − α ds + Z ¯ s s ¯ ϕ ( s )( s − s ) − α (¯ s − s ) − β ds K (cid:16) Z s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z s ds ( s − s ) (1 − α ) qq − (cid:17) q − q + K (cid:16) Z ¯ s s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z ¯ s s ds ( s − s ) (1 − α ) qq − (¯ s − s ) (1 − β ) qq − (cid:17) q − q K + K (¯ s − s ) α + β − (cid:16) Z ¯ s s ds ( s − s ¯ s − s ) (1 − α ) qq − (1 − s − s ¯ s − s ) (1 − β ) qq − (cid:17) q − q . Let τ = s − s ¯ s − s . Then s = s + (¯ s − s ) τ , ds = (¯ s − s ) dτ , and I K + K (¯ s − s ) α + β − − q (cid:16) Z dττ − α q − q − (1 − τ ) − βq − q − (cid:17) q − q = K + K (¯ s − s ) α + β − − q B (cid:16) α q − q − , βq − q − (cid:17) q − q K. For I , we have I Z t ¯ s ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z t ¯ s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z t ¯ s ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − ds (cid:17) q − q K ( s − t ) α + β − (cid:16) Z t ¯ s ds (1 + t − ss − t ) (1 − α ) qq − ( t − ss − t ) (1 − β ) qq − ds (cid:17) q − q . τ = t − ss − t . Then s = t − ( s − t ) τ , ds = − ( s − t ) dτ , and I K ( s − t ) α + β − − q (cid:16) Z t − ¯ s s − t dτ (1 + τ ) (1 − α ) qq − τ (1 − β ) qq − (cid:17) q − q K ( s − t ) α + β − − q h(cid:16) Z dττ (1 − β ) qq − (cid:17) q − q + (cid:16) Z s − ¯ s s − t dττ (2 − α − β ) qq − (cid:17) q − q i K ( s − t ) α + β − − q h (cid:16) s − ¯ s s − t (cid:17) [1 − (2 − α − β ) qq − ] q − q i K ( s − t ) α + β − − q h (cid:16) s − t (cid:17) α + β − − q i K ( s − t ) α + β − − q , t ∈ [¯ s , s ) . Then we have(2.24) ( s − t ) (1 − α − β + q ) + | e ψ ( t ) | K, t ∈ [¯ s , s ) . (iv) For t ∈ ( s , ¯ s ], we have | e ψ ( t ) | Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds + Z ts ¯ ϕ ( s ) w ( s )( t − s ) − β ds ≡ I + I . Note that ¯ s s − δ , I = Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds = Z ¯ s ¯ ϕ ( s ) w ( s ) | s − s | − α ( t − s ) − β ds + Z s ¯ s ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K Z ¯ s ¯ ϕ ( s ) | s − s | − α ds + K Z s ¯ s ¯ ϕ ( s )( s − s ) − α [( t − s ) + ( s − s )] − β ds K (cid:16) Z ¯ s ¯ ϕ ( s ) q ds (cid:17) q h(cid:16) Z ¯ s ds | s − s | (1 − α ) qq − (cid:17) q − q +( t − s ) α + β − (cid:16) Z s ¯ s ds ( s − st − s ) (1 − α ) qq − (1 + s − st − s ) (1 − β ) qq − (cid:17) q − q i . Let τ = s − st − s . Then s = s − ( t − s ) τ , ds = − ( t − s ) dτ , and I K h t − s ) α + β − (cid:16) Z s ¯ s ds ( s − st − s ) (1 − α ) qq − (1 + s − st − s ) (1 − β ) qq − (cid:17) q − q i K h t − s ) α + β − − q (cid:16) Z s − ¯ s t − s dττ (1 − α ) qq − (1 + τ ) (1 − β ) qq − (cid:17) q − q i K h t − s ) α + β − − q (cid:16) Z dττ (1 − α ) qq − + Z s − ¯ s t − s dττ (2 − α − β ) qq − (cid:17) q − q i K h t − s ) α + β − − q + ( t − s ) α + β − − q (cid:16) s − ¯ s t − s (cid:17) [1 − (2 − α − β ) qq − ] q − q i K + K ( t − s ) α + β − − q . Now, we look at I , noting s + δ ¯ s s − δ , I = Z ts ¯ ϕ ( s ) w ( s )( t − s ) − β ds = Z ts ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z ts ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z ts ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − (cid:17) q − q K ( t − s ) α + β − (cid:16) Z ts ds ( s − s t − s ) (1 − α ) qq − (1 − s − s t − s ) (1 − β ) qq − (cid:17) q − q . τ = s − s t − s . Then s = s + ( t − s ) τ , ds = ( t − s ) dτ , and I K ( t − s ) α + β − − q (cid:16) Z dττ − α q − q − (1 − τ ) − βq − q − (cid:17) q − q = K ( t − s ) α + β − − q B (cid:16) α q − q − , βq − q − (cid:17) q − q . Hence,(2.25) ( t − s ) (1+ q − α − β ) + | e ψ ( t ) | K, t ∈ ( s , ¯ s ] . (v) For t ∈ [¯ s , s ), | e ψ ( t ) | Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds + Z ¯ s s ¯ ϕ ( s ) w ( s )( t − s ) − β ds + Z t ¯ s ¯ ϕ ( s ) w ( s )( t − s ) − β ds ≡ I + I + I . We look at the three terms one-by-one. Since ¯ s > s + δ , one has I = Z s ¯ ϕ ( s ) w ( s )( t − s ) − β ds K (cid:16) Z ¯ s ¯ ϕ ( s ) | s − s | − α ( t − s ) − β ds + Z s ¯ s ¯ ϕ ( s )( s − s ) − α ( t − s ) − β ds (cid:17) K (cid:16) Z ¯ s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z ¯ s ds | s − s | (1 − α ) qq − (cid:17) q − q + K (cid:16) Z s ¯ s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z s ¯ s ds ( s − s ) (1 − α ) qq − (cid:17) q − q K. For I , since ( t − s ) > ( s − s ), one has I = Z ¯ s s ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z ¯ s s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z ¯ s s ds ( s − s ) (1 − α ) qq − (¯ s − s ) (1 − β ) qq − (cid:17) q − q K (¯ s − s ) α + β − (cid:16) Z ¯ s s ds ( s − s ¯ s − s ) (1 − α ) qq − ( ¯ s − s ¯ s − s ) (1 − β ) qq − (cid:17) q − q . Let τ = s − s ¯ s − s . Then s = s + (¯ s − s ) τ , ds = (¯ s − s ) dτ , and I K (¯ s − s ) α + β − − q (cid:16) Z dττ − α q − q − (1 − τ ) − βq − q − (cid:17) q − q K. Finally, for I , one has I = Z t ¯ s ¯ ϕ ( s ) w ( s )( s − s ) − α ( t − s ) − β ds K (cid:16) Z t ¯ s ¯ ϕ ( s ) q ds (cid:17) q (cid:16) Z t ¯ s ds ( s − s ) (1 − α ) qq − ( t − s ) (1 − β ) qq − (cid:17) q − q K ( s − t ) α + β − (cid:16) Z t ¯ s ds (1 + t − ss − t ) (1 − α ) qq − ( t − ss − t ) (1 − β ) qq − (cid:17) q − q . Let τ = t − ss − t . Then s = t − ( s − t ) τ , ds = − ( s − t ) dτ , and I = Z t ¯ s ¯ ϕ ( s ) w ( s )( t − s ) − β ds K ( s − t ) α + β − − q (cid:16) Z t − ¯ s s − t dτ (1 + τ ) (1 − α ) qq − τ (1 − β ) qq − (cid:17) q − q K ( s − t ) α + β − − q (cid:16) Z dsτ (1 − β ) qq − ds + Z s − ¯ s s − t dsτ (2 − α − β ) qq − (cid:17) q − q K ( s − t ) α + β − − q h (cid:16) s − ¯ s s − t (cid:17) [1 − (2 − α − β ) qq − ] q − q i K + K ( s − t ) α + β − − q . Hence,(2.26) ( s − t ) (1 − α − β + q ) + | e ψ ( t ) | K, t ∈ [¯ s , s ) .
10y induction, we can obtain the following:(2.27) ( s − t ) (1+ q − α − β ) + | e ψ ( t ) | K, t ∈ (0 , s ) , ( t − s i ) (1+ q − α i − β ) + | e ψ ( t ) | K, t ∈ ( s i , ¯ s i ] , i ℓ − , ( s i +1 − t ) (1+ q − α i +1 − β ) + | e ψ ( t ) | K, t ∈ [¯ s i , s i +1 ) , i ℓ − , ( t − s l ) (1+ q − α l − β ) + | e ψ ( t ) | K, t ∈ ( s l , T ] . Hence, for ¯ w ( · ), we have ¯ w ( t ) | e ψ ( t ) | K, ∀ t ∈ [0 , T ] \ { s , s , · · · , s ℓ } . Then (2.16) follows easily. On the other hand, if s = 0 and s l = T , then for any t ∈ [0 , T ] \{ s , s , s , · · · , s ℓ } ,one can find a σ > i = 0 , , , · · · , ℓ −
1, [ t − σ, t + σ ] ⊆ ( s i , s i +1 ). Then Z t + σt − σ sup t ∈ [ s,t + σ ] (cid:12)(cid:12)(cid:12) e ϕ ( t, s ) w ( s ) (cid:12)(cid:12)(cid:12) q ds K Z t + σt − σ ¯ ϕ ( s ) q ds < ∞ . Hence, by the similar argument of Lemma 2.2, e ψ ( · ) is continuous at any such a t . If 0 < s or s ℓ < T , wecan use the similar argument to show the continuity of e ψ ( · ) at t ∈ [0 , T ] \ { s , s , s , · · · , s ℓ } .Finally, if (2.18) holds, then for σ > r < q , with 1 + q − α i < r < β , Z s i + σs i − σ ¯ ϕ ( s ) r w ( s ) r ds K (cid:16) Z s i + σs i − σ ¯ ϕ ( s ) q ds (cid:17) rq (cid:16) Z s i + σs i − σ ds | s − s i | (1 − α i ) rqq − r (cid:17) q − rq < ∞ , since 1 + 1 q − α i < r ⇐⇒ (1 − α i ) rqq − r < . Thus, by the similar argument of Lemma 2.2, we have the continuity of e ψ ( · ) at s i . The last conclusion isclear.From the above, we see that if α i + β > i ℓ and(2.28) ¯ ϕ ( · ) ∈ L αi + β − + (0 , T ; R ) , i ℓ, then e ψ ( · ) ∈ C ([0 , T ]; R n ). On the other hand, if α i + β <
1, and ¯ ϕ ( · ) is essentially non-zero near s i , then e ψ ( · ) will be blow-up near s i , and roughly it will grow no more than | t − s i | α i + β − − q .The following result is a kind of Gronwall’s inequality with a singular kernel. Lemma 2.5.
Let β ∈ (0 , and q > β . Let L ( · ) , a ( · ) , y ( · ) be nonnegative functions with L ( · ) ∈ L q (0 , T ; R ) , a ( · ) , y ( · ) ∈ L qq − (0 , T ; R ) . Suppose (2.29) y ( t ) a ( t ) + Z t L ( s ) y ( s )( t − s ) − β ds, a.e. t ∈ [0 , T ] . Then (2.30) y ( t ) a ( t ) + k − X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c k Z t L ( s ) a ( s ) ds, a.e. t ∈ [0 , T ] , for some constants c i > and β i ∈ (0 , defined by β i = β + i (cid:16) β − q (cid:17) , i k − , with k being the smallest integer satisfying β + k (cid:16) β − q (cid:17) > . roof. First of all, since L ( · ) ∈ L q (0 , T ; R ) and y ( · ) ∈ L q ′ (0 , T ; R ), q ′ = qq − , we know that L ( · ) y ( · ) ∈ L (0 , T ; R ). Hence, the integral on the right hand side of (2.29) is well-defined, as a function in L (0 , T ; R ).Now, we observe the following: y ( t ) a ( t ) + Z t L ( s ) y ( s )( t − s ) − β ds a ( t ) + Z t L ( s ) a ( s )( t − s ) − β ds + Z t L ( s )( t − s ) − β Z s L ( τ ) y ( τ )( s − τ ) − β dτ ds a ( t ) + Z t L ( s ) a ( s )( t − s ) − β ds + Z t L ( τ ) h Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds i y ( τ ) dτ. Let r = s − τt − τ . Then s = τ + ( t − τ ) r and ds = ( t − τ ) dr . Thus, Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds (cid:16) Z tτ L ( s ) q ds (cid:17) q (cid:16) Z tτ ds ( t − s ) (1 − β ) q ′ ( s − τ ) (1 − β ) q ′ ds (cid:17) q ′ k L ( · ) k q (cid:16) Z ( t − τ ) dr [( t − τ )(1 − r )] (1 − β ) q ′ [( t − τ ) r ] (1 − β ) q ′ (cid:17) q ′ = k L ( · ) k q t − τ ) − β ) − q ′ (cid:16) Z dr (1 − r ) (1 − β ) q ′ r (1 − β ) q ′ (cid:17) q ′ . Since q > β which is equivalent to(2.31) 0 < − (1 − β ) q ′ = 1 − (1 − β ) qq − q − − q + βqq − βq − q − , we obtain Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds k L ( · ) k q ( t − τ ) − β ) − q ′ B (cid:16) βq − q − , βq − q − (cid:17) q ′ ≡ c ( t − τ ) − β , with B ( · , · ) being the Beta function and c = k L ( · ) k q B (cid:16) βq − q − , βq − q − (cid:17) q ′ ,β = 1 − (cid:16) − β ) − q ′ (cid:17) = 2 β + 1 q ′ − β + (cid:16) β − q (cid:17) > β. Consequently, y ( t ) a ( t ) + Z t L ( s ) a ( s )( t − s ) − β ds + c Z t L ( s ) y ( s )( t − s ) − β ds a ( t ) + Z t L ( s ) a ( s )( t − s ) − β ds + c Z t L ( s ) a ( s )( t − s ) − β ds + c Z t L ( s )( t − s ) − β Z s L ( τ ) y ( τ )( s − τ ) − β dτ ds = a ( t ) + X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c Z t L ( τ ) h Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds i y ( τ ) dτ, with c = 1 and β = β . Let r = s − τt − τ . Then s = τ + ( t − τ ) r and ds = ( t − τ ) dr . Thus, Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds (cid:16) Z tτ L ( s ) q ds (cid:17) q (cid:16) Z tτ ds ( t − s ) (1 − β ) q ′ ( s − τ ) (1 − β ) q ′ ds (cid:17) q ′ k L ( · ) k q (cid:16) Z ( t − τ ) dr [( t − τ )(1 − r )] (1 − β ) q ′ [( t − τ ) r ] (1 − β ) q ′ (cid:17) q ′ = k L ( · ) k q t − τ ) − β − β − q ′ (cid:16) Z dr (1 − r ) (1 − β ) q ′ r (1 − β ) q ′ (cid:17) q ′ . q > β > β , we have(2.32) 0 < − (1 − β ) q ′ = 1 − (1 − β ) qq − q − − q + β qq − β q − q − . Hence, we obtain c Z tτ L ( s )( t − s ) − β ( s − τ ) − β ds c k L ( · ) k q ( t − τ ) − β − β − q ′ B (cid:16) βq − q − , β q − q − (cid:17) q ′ ≡ c ( t − τ ) − β , with c = c k L ( · ) k q B (cid:16) βq − q − , β q − q − (cid:17) q ′ ,β = 1 − (cid:16) − β − β − q ′ (cid:17) = β + β + 1 q ′ − β + 2 (cid:16) β − q (cid:17) > β. Consequently, y ( t ) a ( t ) + X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c Z t L ( s ) y ( s )( t − s ) − β ds. By induction, we are able to show that y ( t ) a ( t ) + k − X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c k Z t L ( s ) y ( s )( t − s ) − β k ds, with β i = β + i (cid:16) β − q (cid:17) , i k, and recursively defined c i > c i = c i − k L ( · ) k q B (cid:16) βq − q − , β i − q − q − (cid:17) q ′ , i k. We let k > β k >
1. Then the above implies y ( t ) a ( t ) + k − X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c k T β k − Z t L ( s ) y ( s ) ds, a.e. t ∈ [0 , T ] . Now, let z ( t ) = Z t L ( s ) y ( s ) ds, t ∈ [0 , T ] . Then ˙ z ( t ) = L ( t ) y ( t ) L ( t ) a ( t ) + k − X i =0 c i L ( t ) Z t L ( s ) a ( s )( t − s ) − β i ds + c k T β k − L ( t ) z ( t ) . Hence, z ( t ) Z t e c k T βk − R ts L ( τ ) dτ L ( s ) a ( s ) ds + k − X i =0 c i Z t e c k T βk − R ts L ( τ ) dτ L ( s ) Z s L ( τ ) a ( τ )( s − τ ) − β i dτ ds Z t e c k T βk − R ts L ( τ ) dτ L ( s ) a ( s ) ds + k − X i =0 c i K Z t L ( τ ) a ( τ ) dτ Z tτ L ( s )( s − τ ) − β i ds Z t e c k T βk − R ts L ( τ ) dτ L ( s ) a ( s ) ds + k − X i =0 c i K (cid:16) Z tτ L ( s ) q ds (cid:17) q (cid:16) Z tτ s − τ ) (1 − β i ) qq − ds (cid:17) q − q Z t L ( s ) a ( s ) ds c k Z t L ( s ) a ( s ) ds, c k >
0. Hence, y ( t ) a ( t ) + k − X i =0 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c k Z t L ( s ) a ( s ) ds, a.e. t ∈ [0 , T ] , proving our conclusion.Comparing with the Gronwall type inequality appearing in literature on fractional differential equations(see [26], for example), our inequality only involves a finite sum, instead of an infinite series. In this section, we discuss our state equation (1.1), together with the cost functional (1.2). In what follows, U will be a separable metric space with the metric ρ , which could be a non-empty bounded or unboundedset in R m with the metric induced by the usual Euclidean norm. Let u ∈ U be fixed. For any p >
1, wedefine U p [0 , T ] = (cid:8) u : [0 , T ] → U (cid:12)(cid:12) u ( · ) is measurable , ρ ( u ( · ) , u ) ∈ L p (0 , T ; R ) (cid:9) . L p space We introduce the following assumptions for the generator f ( · , · , · , · ) of our state equation. (H1) Let the map f : ∆ × R n × U → R n be measurable. There are nonnegative functions L ( · ) , L ( · ) with(3.1) L ( · ) ∈ L ( β ∨ pp − )+ (0 , T ; R ) , L ( · ) ∈ L ( p βp ∨ (0 , T ; R ) , for some p > = ∞ ) and β ∈ (0 ,
1) such that(3.2) | f ( t, s, y , u ) − f ( t, s, y , u ) | L ( s ) | y − y | ( t − s ) − β , ∀ ( t, s, u ) ∈ ∆ × U, y , y ∈ R n , | f ( t, s, , u ) | L ( s ) ρ ( u, u ) + L ( s )( t − s ) − β , ∀ ( t, s, u ) ∈ ∆ × U. Note that the larger the β ∈ (0 , f ( · , · , · , · ). Also, (3.2)imply(3.3) | f ( t, s, y, u ) | L ( s ) (cid:2) | y | + ρ ( u, u ) (cid:3) + L ( s )( t − s ) − β , ∀ ( t, s, y, u ) ∈ ∆ × R n × U. We now present the well-posedness of the state equation (1.1) in L p spaces. Theorem 3.1.
Let (H1) hold with some p > and β ∈ (0 , . Then for any η ( · ) ∈ L p (0 , T ; R n ) and u ( · ) ∈ U p [0 , T ] , (1.1) admits a unique solution y ( · ) ≡ y ( · ; η ( · ) , u ( · )) ∈ L p (0 , T ; R n ) , and the followingestimates hold (3.4) k y ( · ) k p k η ( · ) k p + K (cid:16) k ρ ( u ( · ) , u ) k p (cid:17) . If ( η ( · ) , u ( · )) , ( η ( · ) , u ( · )) ∈ L p (0 , T ; R n ) × U p [0 , T ] and y ( · ) , y ( · ) are the solutions of (1.1) correspondingto ( η ( · ) , u ( · )) and ( η ( · ) , u ( · )) , respectively, then (3.5) k y ( · ) − y ( · ) k p K n k η ( · ) − η ( · ) k p + h Z T (cid:16) Z t | f ( t, s, y ( s ) , u ( s )) − f ( t, s, y ( s ) , u ( s )) | ds (cid:17) p dt i p o . roof. Fix any η ( · ) ∈ L p (0 , T ; R n ) and u ( · ) ∈ U p [0 , T ]. For any z ( · ) ∈ L p (0 , T ; R n ), define T [ z ( · )]( t ) = η ( t ) + Z t f ( t, s, z ( s ) , u ( s )) ds, t ∈ [0 , T ] . Denote θ ( t ) = t β − I (0 , ∞ ) ( t ), where I (0 , ∞ ) is the characteristic function of (0 , ∞ ). Then(3.6) k T [ z ( · )] k p k η ( · ) k p + (cid:16) Z T (cid:12)(cid:12)(cid:12) Z t f ( t, s, z ( s ) , u ( s )) ds (cid:12)(cid:12)(cid:12) p dt (cid:17) p k η ( · ) k p + h Z T (cid:16) Z t L ( s ) (cid:2) | z ( s ) | + ρ ( u ( s ) , u ) (cid:3) + L ( s )( t − s ) − β ds (cid:17) p dt i p k η ( · ) k p + k θ ( · ) ∗ (cid:8) L ( · ) (cid:2) ρ ( u ( · ) , u ) + | z ( · ) | (cid:3)(cid:9) k p + k θ ( · ) ∗ L ( · ) k p ≡ k η ( · ) k p + I + I . Now, we split the proof into three cases.
Case 1. p > − β . In this case, 1 β > pp − , p βp > . For any ε ∈ (0 , β − β ), which is equivalent to (1 − β )(1 + ε ) <
1, define q through the following:1 q = 1 p + 1 −
11 + ε < p + 1 −
11 + β − β = 1 p + β < . The last inequality in the above follows from p > − β . Thus, 1 < q < p and q ց p βp > , as ε ր β − β . Since L ( · ) ∈ L p βp + (0 , T ; R ), we may assume that L ( · ) ∈ L q (0 , T ; R ) (for an ε being close enough to β − β ).Hence, by Young’s inequality, I ≡ k θ ( · ) ∗ L ( · ) k p k θ ( · ) k ε k L ( · ) k q . Also, 1 q − p = 1 −
11 + ε < β.
Thus, p − qpq ր β, as ε ր β − β , which is equivalent to pqp − q ց β , as ε ր β − β . Hence, by L ( · ) ∈ L β + (0 , T ; R ), we could find ε which is close enough to β − β so that L ( · ) ∈ L pqp − q (0 , T ; R ).Then I ≡ k θ ( · ) ∗ (cid:8) L ( · ) (cid:2) | z ( · ) | + ρ ( u ( · ) , u ) (cid:3)(cid:9) k p k θ k ε k L ( · ) (cid:2) | z ( · ) | + ρ ( u ( · ) , u ) (cid:3) k q k θ ( · ) k ε k L ( · ) k pqp − q k | z ( · ) | + ρ ( u ( · ) , u ) k p . Then we see that T : L p (0 , T ; R n ) → L p (0 , T ; R n ). Next, let z ( · ) , z ( · ) ∈ L p (0 , T ; R n ), we look at thefollowing: k T [ z ( · )] − T [ z ( · )] k L p (0 ,δ ; R n ) ≡ (cid:16) Z δ | T [ z ( · )]( t ) − T [ z ( · )]( t ) | p dt (cid:17) p h Z δ (cid:12)(cid:12)(cid:12) Z t (cid:16) f ( t, s, z ( s ) , u ( s )) − f ( t, s, z ( s ) , u ( s )) (cid:17) ds (cid:12)(cid:12)(cid:12) p dt i p h Z δ (cid:16) Z t L ( s ) | z ( s ) − z ( s ) | ( t − s ) − β ds (cid:17) p dt i p ≤ k θ ( · ) ∗ (cid:2) L ( · ) | z ( · ) − z ( · ) | (cid:3) k L p (0 ,δ ; R ) k θ ( · ) k L ε (0 ,δ ; R ) k L ( · ) | z ( · ) − z ( · ) | k L q (0 ,δ ; R ) (cid:16) δ β − (1 − β ) ε β − (1 − β ) ε (cid:17) ε k L ( · ) k L pqp − q (0 ,δ ; R ) k z ( · ) − z ( · ) k L p (0 ,δ ; R n ) . δ > T : L p (0 , δ ; R n ) → L p (0 , δ ; R n ) is a contraction. Hence, it admits a uniquefixed point on L p (0 , δ ; R n ), which is the unique solution of the state equation (1.1) on [0 , δ ].Next, we look (1.1) on [ δ, T ], which can be written as y ( t ) = η ( t ) + Z δ f ( t, s, y ( s ) , u ( s )) ds + Z tδ f ( t, s, y ( s ) , u ( s )) ds, t ∈ [ δ, T ] . Since (similar to (3.6)) (cid:13)(cid:13)(cid:13) η ( · ) + Z δ f ( · , s, y ( s ) , u ( s )) ds (cid:13)(cid:13)(cid:13) L p ( δ,T ; R n ) = h Z Tδ (cid:12)(cid:12)(cid:12) η ( t ) + Z δ f ( t, s, y ( s ) , u ( s )) ds (cid:12)(cid:12)(cid:12) p dt i p (cid:16) Z Tδ | η ( t ) | p dt (cid:17) p + h Z Tδ (cid:16) Z δ | f ( t, s, y ( s ) , u ( s )) | ds (cid:17) p dt i p k η ( · ) k L p ( δ,T ; R n ) + h Z Tδ (cid:16) Z δ L ( s ) (cid:2) | y ( s ) | + ρ ( u ( s ) , u ) (cid:3) + L ( s )( t − s ) − β ds (cid:17) p dt i p k η ( · ) k L p ( δ,T ; R n ) + h Z Tδ (cid:16) Z δ L ( s ) (cid:2) | y ( s ) | + ρ ( u ( s ) , u ) (cid:3) ( t − s ) − β ds (cid:17) p dt i p + h Z Tδ (cid:16) Z δ L ( s )( t − s ) − β ds (cid:17) p dt i p k η ( · ) k L p ( δ,T ; R n ) + K (cid:16) k y ( · ) k L p (0 ,δ ; R n ) + k ρ ( u ( · ) , u ) k L p (0 ,δ ; R ) + 1 (cid:17) . Then using the same argument as above, we obtain the existence and uniqueness of the solution to the stateequation on [0 , δ ]. By induction, we could get the solvability of the state equation on [0 , T ].Now, let ( η ( · ) , u ( · )) , ( η ( · ) , u ( · )) ∈ L p (0 , T ; R n ) × U p [0 , T ] and y ( · ), y ( · ) be the correspondingsolutions. Then | y ( t ) − y ( t ) | | η ( t ) − η ( t ) | + Z t | f ( t, s, y ( s ) , u ( s )) − f ( t, s, y ( s ) , u ( s )) | ds + Z t L ( s ) | y ( s ) − y ( s ) | ( t − s ) − β ds ≡ a ( t ) + Z t L ( s ) | y ( s ) − y ( s ) | ( t − s ) − β ds. Hence, by Lemma 2.5,(3.7) | y ( t ) − y ( t ) | a ( t ) + k − X i =1 c i Z t L ( s ) a ( s )( t − s ) − β i ds + c k Z t L ( s ) a ( s ) ds, a.e. t ∈ [0 , T ] , for some constants c i > β i ∈ [ β, k y ( · ) − y ( · ) k p K (cid:16) Z T a ( t ) p dt (cid:17) p K n k η ( · ) − η ( · ) k p + h Z T (cid:16) Z t | f ( t, s, y ( s ) , u ( s )) − f ( t, s, y ( s ) , u ( s )) | ds (cid:17) p dt i p o , proving the stability estimate. We can use the similar argument to prove this estimate to get (3.4). Case 2. < p − β . In this case, 1 β pp − , p βp . Also, since 1 − β p <
1, for any ε ∈ (0 , p − − β p <
11 + ε . − β )(1 + ε ) <
1. Define q through the following:1 p < q = 1 p + 1 −
11 + ε ր , as ε ր p − . Then 1 q − p = 1 −
11 + ε ր − p as ε ր p − . Thus, pqp − q ց pp − , as ε ր p − . Consequently, by choosing ε > p −
1, we have q > pqp − close enoughto pp − . Hence, k θ ( · ) ∗ L ( · ) k p k θ ( · ) k ε k L ( · ) k q , and k θ ( · ) ∗ (cid:8) L ( · ) (cid:2) | z ( · ) | + ρ ( u ( · ) , u ) (cid:3)(cid:9) k p k θ ( · ) k ε k L ( · ) k pqp − q k | z ( · ) | + ρ ( u ( · ) , u ) k p . The rest of the proof is similar to that of Case 1.
Case 3. p = 1. In this case, the condition reads L ( · ) ∈ L ∞ (0 , T ; R ) and L ( · ) ∈ L (0 , T ; R ). Then k θ ( · ) ∗ L ( · ) k k θ ( · ) k k L ( · ) k , and k θ ( · ) ∗ (cid:8) L ( · ) (cid:2) | z ( · ) | + ρ ( u ( · ) , u ) (cid:3)(cid:9) k k θ ( · ) k k L ( · ) k ∞ k | z ( · ) | + ρ ( u ( · ) , u ) k . The rest of the proof is similar to that of Case 1.Let us make some comments and observations on the above theorem. First of all, the above theoremgives some sufficient conditions under which for ( η ( · ) , u ( · )) ∈ L p (0 , T ; R n ) × U p [0 , T ], equation (1.1) admitsa unique solution y ( · ) ∈ L p (0 , T ; R n ). The conditions we imposed in (H1) are compatibility conditions ofthe integrability for the free term η ( · ), the control u ( · ), and the coefficients L ( · ) and L ( · ). From the above,we see that if ( η ( · ) , u ( · )) ∈ L p (0 , T ; R ) × U p [0 , T ] with p > − β , then by assuming L ( · ) ∈ L β + (0 , T ; R ) and L ( · ) ∈ L β − (0 , T ; R ) (note that p βp < β ), the equation has a unique solution y ( · ) ∈ L p (0 , T ; R n ). Thisis the case, in particular, if η ( · ) ∈ L ∞ (0 , T ; R n ) and U is bounded (under the metric ρ ). We will comeback to this later. On the other hand, if 1 p < − β , that is, say, the free term and/or the control haveweaker integrability, then we need to strengthen the integrability condition for L ( · ) from L β + to L pp − + (inthe current case, pp − > β ) to get L p solution y ( · ). But, the integrability of L ( · ) is only required to be L (0 , T ; R ). Finally, since we have used the contraction mapping theorem to establish the well-posedness ofthe state equation, one can see that the solution to the state equation can be obtained by a Picard iteration.Let us present an example from which we could get some feeling about the above result. Example 3.2.
Consider the following Volterra integral equation(3.8) y ( t ) = 1 | t − | − γ + Z t p ( s − δ − + y ( s ) | s − | − α ( t − s ) − β ds, a.e. t ∈ [0 , T ] , for some α, β ∈ (0 , γ, δ ∈ (0 , T >
1. In this case, we have/can take η ( t ) = 1 | t − | − γ , L ( s ) = 1 | s − | − α , L ( s ) = 1 | s − | − α − δ . In order η ( · ) ∈ L p (0 , T ; R ), we need p (1 − γ ) < ⇐⇒ p < − γ ( 10 , ∞ ) .
17n order L ( · ) ∈ L ( β ∨ pp − )+ (0 , T ; R ), one needs1 − αβ < ⇐⇒ α + β > , and (1 − α ) pp − < ⇐⇒ − α < − p ⇐⇒ p > α . Finally, in order L ( · ) ∈ L ( p βp ∨ (0 , T ; R ), one needs(2 − α − δ ) p βp < ⇐⇒ − α − β − δ < p ⇐⇒ p < − α − β − δ ) + , and 2 − α − δ < ⇐⇒ α + δ > . Hence, equation (3.8) has a unique solution y ( · ) ∈ L p (0 , T ; R n ) for any p ∈ [ α , − γ ) ∨ (2 − α − β − δ ) + ), provided(3.9) α + β > , α + δ > . We point out that in general, the solution y ( · ) of the equation (3.8) is not necessarily continuous, even ifthe free term η ( · ) is continuous. In fact, let γ = 1. Then η ( t ) ≡ y ( · ) is positive (which can be seen from a Picard iteration). Consequently,lim t → y ( t ) > t → Z t ds | s − | − α − δ ( t − s ) − β = Z ds (1 − s ) − α − β − δ = ∞ , provided(3.10) 3 − α − β − δ > ⇐⇒ α + β + δ < . This will be the case if we take α = 23 , β = δ = 12 . In this case, the solution y ( · ) ∈ L p (0 , T ; R ) exists with p ∈ ( ,
3) and it is discontinuous at t = 1.Note that in the above example, the solution y ( · ) is discontinuous at t = 1 only, which is the singularityof L ( · ) and L ( · ). It is natural to ask what will be the result for the general situation? Such a question hasits own interest. And also since the values y ( t i ) of y ( · ) are needed in the cost functional (1.2), we would liketo locate the discontinuity points of the solution y ( · ) a priori based on the information of L ( · ) and L ( · ).This leads to the following subsection. In this subsection, we would like to explore the continuity of the solution y ( · ) to the state equation (1.1). Letus begin with some observations. Suppose y ( · ) ∈ L p (0 , T ; R n ) is the unique solution to the state equation(1.1) which is rewritten here:(3.11) y ( t ) = η ( t ) + Z t f ( t, s, y ( s ) , u ( s )) ds, t ∈ [0 , T ] . Then the continuity of y ( · ) is determined by that of η ( · ) and ψ ( · ) ≡ Z · f ( · , s, y ( s ) , u ( s )) ds. The continuity of η ( · ) should be given a priori. Thus, we need to look at the continuity of the above-definedfunction ψ ( · ). Hence, the preliminary results presented in Section 2 will play an interesting role here. Tomake it precise, we introduce the following hypothesis.18 H2)
Let w ( · ) be given by (2.8) with α i ∈ (0 , i ℓ and 0 ≤ s < s < · · · < s ℓ T . Let f : ∆ × R n × U → R n be given by the following:(3.12) f ( t, s, y, u ) = f ( t, s, y, u ) w ( s )( t − s ) − β , ( t, s, y, u ) ∈ ∆ × R n × U, with β ∈ (0 ,
1) and f : ∆ × R n × U → R n being measurable such that(3.13) | f ( t, s, y, u ) − f ( t ′ , s, y, u ) | ω ( | t − t ′ | ) , ∀ ( t, s ) , ( t ′ , s ) ∈ ∆ , ( y, u ) ∈ R n × U, for some modulus of continuity ω : [0 , ∞ ) → [0 , ∞ ), and(3.14) | f ( t, s, y, u ) | ¯ ϕ ( s ) , ∀ ( t, s, y, u ) ∈ ∆ × R n × U, and(3.15) | f ( t, s, y , u ) − f ( t, s, y , u ) | ¯ L ( s )( | y − y | ) , ( t, s, u ) ∈ ∆ × U, y , y ∈ R n , for some measurable functions ¯ ϕ, ¯ L : [0 , T ] → [0 , ∞ ).Note that under (H2), we will have (H1) if one takes the following: L ( s ) = ¯ ϕ ( s ) w ( s ) , L ( s ) = ¯ L ( s ) w ( s ) , s ∈ [0 , T ] . Thus, according to Theorem 3.1, state equation (1.1) admits a unique solution in L p (0 , T ; R n ), under (H2),for any η ( · ) ∈ L p (0 , T ; R n ) with some p ∈ [1 , ∞ ), if(3.16) ¯ ϕ ( · ) w ( · ) ∈ L ( p βp ∨ (0 , T ; R ) , ¯ L ( · ) w ( · ) ∈ L ( β ∨ pp − )+ (0 , T ; R ) . Let us make a simple observation on the above condition. Recall the definition of δ and ¯ s i from (2.20). Itis not hard to see that (3.16) holds if (1 − α i )( β ∨ pp − ) <
1, 0 i ℓ and for some ε ∈ (0 , δ ), the followingholds: ¯ ϕ ( · ) ∈ L ( p βp ∨ ([0 , T ] \ ℓ [ i =0 (( s i − ε ) ∨ , ( s i + ε ) ∧ T ); R ) , ¯ L ( · ) ∈ L ( β ∨ pp − )+ ([0 , T ] \ ℓ [ i =0 (( s i − ε ) ∨ , ( s i + ε ) ∧ T ); R ) , ¯ ϕ ( · ) , ¯ L ( · ) ∈ L ∞ (( s i − ε ) ∨ , ( s i + ε ) ∧ T ; R ) , i ℓ. Namely, due to the special structure of w ( · ), it suffices to have boundedness of ¯ ϕ ( · ) and ¯ L ( · ) near s i (0 i ℓ )and proper integrability of these functions away from the points s i . Therefore, the condition (3.16) is verymild.We have the following result which is a direct consequence of Lemma 2.4. Proposition 3.3.
Let (H2) hold with (3.16) for some p ∈ [1 , ∞ ) , and ¯ ϕ ( · ) ∈ L q (0 , T ; R ) , q > β ∨ α i ,for all i = 0 , , · · · , ℓ . Then for any η ( · ) ∈ L p (0 , T ; R n ) and any u ( · ) ∈ U p [0 , T ] , state equation (1.1) admitsa unique solution y ( · ) ∈ L p (0 , T ; R n ) such that (3.17) y ( · ) − η ( · ) ∈ L ∞ ¯ w ( · ) ((0 , T ); R n ) \ C ¯ w ε ( · ) ([0 , T ]; R n ) , where ¯ w ( · ) and ¯ w ε ( · ) are given in (2.15) and (2.17) . .3 Special cases In this subsection, we look at some special cases.
1. Linear Volterra integral equations.
Consider the following equation:(3.18) y ( t ) = η ( t ) + Z t A ( t, s ) y ( s ) w ( s )( t − s ) − β ds, t ∈ [0 , T ] , where β ∈ (0 , w ( · ) is a weight function defined by (2.8) and A : ∆ → R n satisfies(3.19) | A ( t, s ) | ¯ L ( s ) , ∀ ( t, s ) ∈ ∆ , for some measurable function ¯ L ( · ) satisfying(3.20) ¯ L ( · ) w ( · ) ∈ L ( β ∨ pp − )+ (0 , T ; R ) , with some p ∈ [1 , ∞ ). Then, by Theorem 3.1, for any η ( · ) ∈ L p (0 , T ; R n ), equation (3.18) admits a uniquesolution y ( · ) ∈ L p (0 , T ; R n ). Moreover, if we define operator A by A [ y ( · )]( t ) = Z t A ( t, s ) y ( s ) w ( s )( t − s ) − β ds, t ∈ [0 , T ] , then, thanks to (3.20), by the proof of Theorem 3.1, we see that A : L p (0 , T ; R n ) → L p (0 , T ; R n ) is a linearbounded operator. Our linear integral equation (3.18) reads y ( · ) = η ( · ) + A [ y ( · )] . Therefore, the unique solution y ( · ) admits the following (abstract) representation: y ( · ) = ( I − A ) − η ( · ) = ∞ X k =0 A k η ( · ) . Now, let ( t, τ ) Φ( t, τ ) be the unique solution to the following equation:(3.21) Φ( t, τ ) = A ( t, τ ) w ( τ )( t − τ ) − β + Z tτ A ( t, s )Φ( s, τ ) w ( s )( t − s ) − β ds, τ < t T. Then one has(3.22) y ( t ) = η ( t ) + Z t Φ( t, s ) η ( s ) ds, a.e. t ∈ [0 , T ] . This is called the variation of constant formula .
2. Fractional differential equations.
Let us first recall some basic notions of fractional integrals andderivatives. For α ∈ (0 , I α f ( · )]( t ) = 1Γ( α ) Z t f ( s )( t − s ) − α ds, t > , and as long as the right hand side is well-defined, where Γ( · ) is the Gamma function. We call I α the α -thorder integral operator. Let(3.24) [ D α y ( · )]( t ) = ddt [ I − α y ( · )]( t ) ≡ − α ) ddt Z t y ( s )( t − s ) α ds, and(3.25) [ D α ∗ y ( · )]( t ) = [ D α ( y ( · ) − y (0))]( t ) = [ D α ( y ( · )]( t ) − y (0)Γ(1 − α ) t − α .
20n particular, when y ( · ) ∈ AC ([0 , T ]; R ), the set of all absolutely continuous functions defined on [0 , T ], onehas(3.26) [ D α ∗ y ( · )]( t ) = [ I − α y ′ ( · )]( t ) ≡ − α ) Z t y ′ ( s )( t − s ) α ds, We call D α and D α ∗ the α -th order Riemann-Liouville and
Caputo differential operators , respectively. Wehave the following standard result (see [31], Lemmas 2.5 and 2.22).
Proposition 3.4.
Let α ∈ (0 , . Then for any y ( · ) ∈ L (0 , T ; R ) with [ I − α y ( · )]( · ) ∈ AC ([0 , T ]; R ) . (3.27) I α { D α [ y ( · )] } ( t ) = y ( t ) − I − α [ y ( · )](0)Γ( α ) t − α , a.e. t ∈ (0 , T ]; and for y ( · ) ∈ AC ([0 , T ]; R ) , (3.28) I α { D α ∗ [ y ( · )] } ( t ) = y ( t ) − y (0) . Now, let us consider the following fractional differential equation of Riemann-Liouville type:(3.29) D α [ y ( · )]( t ) = f ( t, y ( t ) , u ( t )) , t ∈ [0 , T ] . Applying the operator I α to the above, we obtain(3.30) y ( t ) = I − α [ y ( · )](0)Γ( α ) t − α + 1Γ( α ) Z t f ( s, y ( s ) , u ( s ))( t − s ) − α ds, t ∈ [0 , T ] . We refer the readers to Theorem 3.1 in [31] for the equivalence of (3.29) and (3.30).Likewise, if we consider the following fractional differential equation of Caputo type:(3.31) D α ∗ [ y ( · )]( t ) = f ( t, y ( t ) , u ( t )) , t ∈ [0 , T ] , applying the operator I α to the above, we obtain(3.32) y ( t ) = y (0) + 1Γ( α ) Z t f ( s, y ( s ) , u ( s ))( t − s ) − α ds, t ∈ [0 , T ] . We refer the readers to Theorem 3.24 in [31] for the equivalence of (3.31) and (3.32).From the above, we see that fractional differential equations of Riemann-Liouville and Caputo types arespecial cases of (1.1).
In this subsection, we consider the following linear backward Volterra integral equation:(3.33) ψ ( t ) = ξ ( t ) + Z Tt A ( s, t ) ⊤ ψ ( s ) w ( t )( s − t ) − β ds, t ∈ [0 , T ] , where A : ∆ → R n × n satisfies (3.19)–(3.20). Such an equation will play an important role in the next section.Let 1 < p < − β . We claim that for any ξ ( · ) ∈ L pp − (0 , T ; R n ), the above equation admits a unique solution ψ ( · ) ∈ L pp − (0 , T ; R n ). In fact, by condition (3.20), we can find an r > β ∨ pp − such that ¯ L ( · ) w ( · ) ∈ L r (0 , T ; R ).By r > β , we can find an ε > ε = 1 − r > − β ⇒ (1 + ε )(1 − β ) < . ψ ( · ) ∈ L pp − (0 , T ; R n ), we have (denoting p ′ = pp − ) (cid:13)(cid:13)(cid:13)Z T · A ( s, · ) ⊤ ψ ( s ) w ( · )( s − · ) − β ds (cid:13)(cid:13)(cid:13) p ′ = nZ T (cid:12)(cid:12)(cid:12)Z Tt A ( s, t ) ⊤ ψ ( s ) w ( t )( s − t ) − β ds (cid:12)(cid:12)(cid:12) p ′ dt o p ′ nZ T (cid:16) ¯ L ( t ) w ( t ) Z Tt | ψ ( s ) | ( s − t ) − β ds (cid:17) p ′ dt o p ′ h Z T (cid:16) ¯ L ( t ) w ( t ) (cid:17) r dt i r h Z T (cid:16) Z Tt | ψ ( s ) | ( s − t ) − β ds (cid:17) p ′ rr − p ′ dt i r − p ′ p ′ r (cid:13)(cid:13)(cid:13) ¯ L ( · ) w ( · ) (cid:13)(cid:13)(cid:13) r k θ ( · ) ∗ | ψ ( · ) | k p ′ rr − p ′ (cid:13)(cid:13)(cid:13) ¯ L ( · ) w ( · ) (cid:13)(cid:13)(cid:13) r k θ ( · ) k ε k ψ ( · ) k p ′ . Here, the Young’s inequality for convolution is used with11 + ε + 1 p ′ = r − p ′ p ′ r + 1 = 1 p ′ − r + 1 ⇐⇒
11 + ε = 1 − r . By a similar argument used in the proof of Theorem 3.1, we get the well-posedness of equation (3.33).
In this section, we discuss the optimal control problem for equation (1.1) with cost functional (1.2). To beginwith, let us introduce the following assumptions. The conditions assumed are more than sufficient. But forthe simplicity of presentation, we prefer to use these stronger conditions. (H3)
Let h j : R n → R , j = 1 , , · · · , m be continuously differentiable, and g : [0 , T ] × R n × U → R be measurable with y g ( t, y, u ) being continuously differentiable. There exist a constant L > ω : [0 , + ∞ ) → [0 , + ∞ ) such that | g ( t, y , u ) − g ( t, y , u ) | L | y − y | + ω (cid:0) ρ ( u , u ) (cid:1) , ∀ ( t, y , u ) , ( t, y , u ) ∈ [0 , T ] × R n × U, | g ( t, , u ) | L, ∀ ( t, u ) ∈ [0 , T ] × U, | g y ( t, y , u ) − g y ( t, y , u ) | ω ( | y − y | + ρ ( u , u )) , ∀ ( t, y , u ) , ( t, y , u ) ∈ [0 , T ] × R n × U. Suppose 0 s < s < · · · < s ℓ T are given as in (H2), and 0 < t < t < · · · < t m T such that(4.1) t j / ∈ { s , s , · · · , s ℓ } , ∀ j = 1 , , · · · , m. Clearly, under (H2)–(H3), our cost functional (1.2) is well-defined. Hence, we can formulate the followingoptimal control problem.
Problem (P)
Find a u ∗ ( · ) ∈ U p [0 , T ] such that(4.2) J ( u ∗ ( · )) = inf u ( · ) ∈ U p [0 ,T ] J ( u ( · )) . Any u ∗ ( · ) satisfying (4.2) is called an optimal control of Problem ( P ), the corresponding state y ∗ ( · ) is calledan optimal state and ( y ∗ ( · ) , u ∗ ( · )) is called an optimal pair .In this section, we shall first give a set of necessary conditions for optimal pairs of Problem (P). Usually,such a result is referred to as a Pontryagin’s maximum principle . Then, we shall show some examples.
In establishing the Pontryagin’s maximum principle for the case that U is not assumed to be convex, weneed the following Liapunoff type theorem (see, Corollary 3.8 of Chapter 4 in [32]). Lemma 4.1.
Let X be a Banach space. For any δ > , let E δ = (cid:8) E ∈ [0 , T ] (cid:12)(cid:12) | E | = δT (cid:9) , where | E | stands for the Lebesgue measure of E . Then for any h ( · ) ∈ C ([0 , T ]; L (0 , T ; X )) , (4.3) inf E ∈ E δ (cid:13)(cid:13)(cid:13) Z T (cid:16) δ E ( s ) − (cid:17) h ( · , s ) ds (cid:13)(cid:13)(cid:13) C ([0 ,T ]; X ) = 0 . P ). Theorem 4.2.
Let (H2)–(H3) hold for some p > , ¯ ϕ ( · ) ∈ L q (0 , T ; R ) , q > β ∨ α i , for all i = 0 , , · · · , ℓ and (3.16) holds. Let f : ∆ × R n × U → R be measurable with y f ( t, s, y, u ) being differentiable. Let η ( · ) ∈ L p (0 , T ; R n ) and η ( · ) be continuous at t j , j = 1 , , · · · , m . Suppose ( y ∗ ( · ) , u ∗ ( · )) is an optimal pair ofProblem (P) . Then there exists a solution ψ ( · ) ∈ L pp − (0 , T ; R n ) of the following adjoint equation (4.4) ψ ( t ) = − g y ( t, y ∗ ( t ) , u ∗ ( t )) ⊤ − m X j =1 [0 ,t j ) ( t ) f y ( t j , t, y ∗ ( t ) , u ∗ ( t )) ⊤ h jy (cid:0) ( y ∗ ( t j ) (cid:1) ⊤ + Z Tt f y ( s, t, y ∗ ( t ) , u ∗ ( t )) ⊤ ψ ( s ) ds, t ∈ [0 , T ] , such that the following maximum condition holds: (4.5) Z Ts ψ ( t ) ⊤ f ( t, s, y ∗ ( s ) , u ∗ ( s )) dt − g ( s, y ∗ ( s ) , u ∗ ( s )) − m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f ( t j , s, y ∗ ( s ) , u ∗ ( s ))= min u ∈ U h Z Ts ψ ( t ) ⊤ f ( t, s, y ∗ ( s ) , u ) dt − g ( s, y ∗ ( s ) , u ) − m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f ( t j , s, y ∗ ( s ) , u ) i , a.e. s ∈ [0 , T ] . Proof.
We split the proof into several steps.
Step 1. A variational inequality.
Let ( y ∗ ( · ) , u ∗ ( · )) be an optimal pair of Problem (P). Fix any u ( · ) ∈ U p [0 , T ]. Denote(4.6) u δ ( t ) = ( u ∗ ( t ) , t ∈ [0 , T ] \ E δ ,u ( t ) , t ∈ E δ , with E δ ⊆ [0 , T ] being measurable and undetermined (see Step 2 ). It is obvious that the control u δ ( · ) is in U p [0 , T ]. Let y δ ( · ) = y ( · ; η ( · ) , u δ ( · )) be the corresponding solution, and let Y δ ( t ) = y δ ( t ) − y ∗ ( t ) δ , t ∈ [0 , T ] . Then, Y δ ( · ) satisfies Y δ ( t ) = 1 δ n Z t h f ( t, s, y δ ( s ) , u δ ( s )) − f ( t, s, y ∗ ( s ) , u δ ( s )) i ds + Z t h f ( t, s, y ∗ ( s ) , u δ ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) i ds o = Z t h Z f y ( t, s, y ∗ ( s ) + τ δY δ ( s ) , u δ ( s )) dτ i Y δ ( s ) ds + 1 δ Z t E δ ( s ) h f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) i ds ≡ Z t f δy ( t, s ) Y δ ( s ) ds + 1 δ Z t E δ ( s ) b f ( t, s ) ds, with(4.7) f δy ( t, s ) = Z f y ( t, s, y ∗ ( s ) + τ δY δ ( s ) , u δ ( s )) dτ, b f ( t, s ) = f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) . ( t, s ) ∈ ∆ .
23y the optimality of ( y ∗ ( · ) , u ∗ ( · )), one has the following variational inequality:(4.8) 0 J ( u δ ( · )) − J ( u ∗ ( · )) δ = 1 δ n Z T h g ( t, y δ ( t ) , u δ ( t )) − g ( t, y ∗ ( t ) , u ∗ ( t )) i ds + m X j =1 h h j ( y δ ( t j )) − h j ( y ∗ ( t j )) io = Z T h Z g y ( t, y ∗ ( t ) + τ δY δ ( t ) , u δ ( t )) dτ i Y δ ( t ) dt + 1 δ Z E δ (cid:2) g ( t, y ∗ ( t ) , u ( t )) − g ( t, y ∗ ( t ) , u ∗ ( t )) (cid:3) dt + m X j =1 h Z h jy ( y ∗ ( t j ) + τ δY δ ( t j )) dτ i Y δ ( t j ) . Step 2. Convergence of Y δ ( · ) and so on. We introduce the following integral equation:(4.9) Y ( t ) = Z t h f y ( t, s, y ∗ ( s ) , u ∗ ( s )) Y ( s ) + b f ( t, s ) i ds, t ∈ [0 , T ] , where b f ( · , · ) is given by (4.7). Under our conditions, the above admits a unique solution Y ( · ) such that Y ( · ) ∈ L p (0 , T ; R n ) \ (cid:16) ℓ \ i =1 C (cid:0) ( s i − , s i ); R n (cid:1)(cid:17) , and Y ( · ) ∈ L ∞ ¯ w ( · ) (0 , T ; R n ) \ C ¯ w ε ( · ) ([0 , T ]; R n ) . We now show that for a suitable choice of E δ , the following holds:lim δ → k Y δ ( · ) − Y ( · ) k p = 0 , lim δ → Y δ ( t j ) = Y ( t j ) , j m. Note that Z t E δ ( s ) h f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) i ds = Z t E δ ( s ) f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) w ( s )( t − s ) − β ds ≡ Z t E δ ( s ) b f ( t, s ) w ( s )( t − s ) − β ds, with b f ( t, s ) = f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) , ( t, s ) ∈ ∆ . By (3.14), we have | b f ( t, s ) | ϕ ( s ) , ( t, s ) ∈ ∆ , with ¯ ϕ ( · ) ∈ L q (0 , T ; R ), where q satisfies q > β ∨ α i , for all i = 0 , , · · · , ℓ . Let h ( t, s ) = [0 ,t ) ( s ) b f ( t, s ) w ( s )( t − s ) − β , ( t, s ) ∈ [0 , T ] . Then for any ¯ p > h Z T (cid:16) Z T (cid:12)(cid:12) h ( t, s ) | ds (cid:17) ¯ p dt i p h Z T (cid:16) Z t ϕ ( s ) w ( s )( t − s ) − β ds (cid:17) ¯ p dt i p ≡ (cid:13)(cid:13)(cid:13) θ ( · ) ∗ ϕ ( · ) w ( · ) (cid:13)(cid:13)(cid:13) ¯ p k θ ( · ) k ε (cid:13)(cid:13)(cid:13) ¯ ϕ ( · ) w ( · ) (cid:13)(cid:13)(cid:13) q < ∞ , ε > ε + 1 q = 1¯ p + 1 , (1 + ε )(1 − β ) < . Then 1¯ p = 1 q + 11 + ε − − (cid:16) β − q (cid:17) + 1 − (1 + ε )(1 − β )1 + ε . Note that on the right hand side of the above, the first term is valued in ( − β,
0) since q > β ; and the secondterm is positive with the range n − (1 + ε )(1 − β )1 + ε (cid:12)(cid:12) ε ∈ [0 , β − β ] o = [0 , β ] . Thus, by suitably choosing ε >
0, we may make ¯ p > p ≥
1, by choosing ε > p > p . Hence, h Z T (cid:16) Z T (cid:12)(cid:12) h ( t, s ) | ds (cid:17) p dt i p < ∞ . Clearly, there exists a sequence of continuous functions h k ( · , · ) such that h Z T (cid:16) Z T | h ( t, s ) − h k ( t, s ) | ds (cid:17) p dt i p < k , ∀ k > . Now, for each h k ( · , · ), applying Lemma 4.1, we have that for any fixed δ >
0, there exists some E k ⊆ E δ such that sup t ∈ [0 ,T ] (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E k ( s ) − (cid:17) h k ( t, s ) ds (cid:12)(cid:12)(cid:12) < k . Then n Z T (cid:12)(cid:12)(cid:12) Z t (cid:16) δ E k ( s ) − (cid:17) b f ( t, s ) w ( s )( t − s ) − β ds (cid:12)(cid:12)(cid:12) p dt o p = n Z T (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E k ( s ) − (cid:17) h ( t, s ) ds (cid:12)(cid:12)(cid:12) p dt o p n Z T (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E k ( s ) − (cid:17) h k ( t, s ) ds (cid:12)(cid:12)(cid:12) p dt o p + n Z T (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E k ( s ) − (cid:17)(cid:2) h ( t, s ) − h k ( t, s ) (cid:3) ds (cid:12)(cid:12)(cid:12) p dt o p k (cid:16) δ + 2 (cid:17) . Hence, for any fixed δ > E ∈ E δ n Z T (cid:12)(cid:12)(cid:12) Z t (cid:16) δ E ( s ) − (cid:17) b f ( t, s ) w ( s )( t − s ) − β ds (cid:12)(cid:12)(cid:12) p dt o p = 0 . Next, for any t j , let us observe Z T [0 ,t j ] ( s ) | b f ( t j , s ) | w ( s )( t j − s ) − β ds = Z t j | b f ( t j , s ) | w ( s )( t j − s ) − β ds Z t j ϕ ( s ) w ( s )( t j − s ) − β ds < ∞ . Hence, inf E ∈ E δ (cid:12)(cid:12)(cid:12) Z t j (cid:16) δ E ( s ) − (cid:17) b f ( t j , s ) w ( s )( t j − s ) − β ds (cid:12)(cid:12)(cid:12) = 0 , j m. Likewise, we also haveinf E ∈ E δ (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E ( s ) − (cid:17)(cid:2) g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s )) (cid:3) ds (cid:12)(cid:12)(cid:12) = 0 . H ( t, s ) = b f ( t, s ) w ( s )( t − s ) − β [0 ,t ) ( s ) b f ( t , s ) w ( s )( t − s ) − β [0 ,t ) ( s )... b f ( t m , s ) w ( s )( t m − s ) − β [0 ,t m ) ( s ) g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s )) Then applying Lemma 4.1 to the above function in a proper product space, we obtain that for any δ > E δ ∈ E δ such that(4.10) n Z T (cid:12)(cid:12)(cid:12) Z t (cid:16) δ E δ ( s ) − (cid:17) b f ( t, s ) w ( s )( t − s ) − β ds (cid:12)(cid:12)(cid:12) p dt o p = o (1) , (cid:12)(cid:12)(cid:12) Z t j (cid:16) δ E δ ( s ) − (cid:17) b f ( t j , s ) w ( s )( t j − s ) − β ds (cid:12)(cid:12)(cid:12) = o (1) , j m, (cid:12)(cid:12)(cid:12) Z T (cid:16) δ E δ ( s ) − (cid:17)(cid:2) g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s )) (cid:3) ds (cid:12)(cid:12)(cid:12) = o (1) . By choosing such a family of E δ , δ >
0, we see that the following convergence hold: lim δ → k Y δ ( · ) − Y ( · ) k p = 0 , lim δ → Y δ ( t j ) = Y ( t j ) , j m, lim δ → Z T (cid:16) δ E δ ( s ) − (cid:17)(cid:2) g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s )) (cid:3) ds = 0 . Hence, we end up with the following variational inequality0 Z T (cid:16) g y ( t, y ∗ ( t ) , u ∗ ( t )) Y ( t ) + g ( t, y ∗ ( t ) , u ( t )) − g ( t, y ∗ ( t ) , u ∗ ( t )) (cid:17) dt + m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) Y ( t j )= Z T (cid:16) g y ( s, y ∗ ( s ) , u ∗ ( s )) Y ( s ) + g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s )) (cid:17) ds + Z T m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f y ( t j , s, y ∗ ( s ) , u ∗ ( s )) Y ( s ) + f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17) ds = Z T (cid:16) g y ( s, y ∗ ( s ) , u ∗ ( s )) + m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f y ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17) Y ( s ) ds + Z T h g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds, with Y ( · ) being the solution to the variational equation (4.9).26 tep 3. Duality. Let ψ ( · ) be the solution to the adjoint equation (4.4). Then we have0 Z T (cid:16) g y ( s, y ∗ ( s ) , u ∗ ( s )) + m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f y ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17) Y ( s ) ds + Z T h g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds = Z T (cid:16) − ψ ( s ) + Z Ts f y ( t, s, y ∗ ( s ) , u ∗ ( s )) ⊤ ψ ( t ) dt (cid:17) ⊤ Y ( s ) ds + Z T h g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds = Z T ψ ( t ) ⊤ (cid:16) − Y ( t ) + Z t f y ( t, s, y ∗ ( s ) , u ∗ ( s )) Y ( s ) ds (cid:17) dt + Z T h g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds = Z T h − ψ ( t ) ⊤ Z t (cid:16) f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) (cid:17) ds i dt + Z T h g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds = Z T h Z Ts − ψ ( t ) ⊤ (cid:16) f ( t, s, y ∗ ( s ) , u ( s )) − f ( t, s, y ∗ ( s ) , u ∗ ( s )) (cid:17) dt + g ( s, y ∗ ( s ) , u ( s )) − g ( s, y ∗ ( s ) , u ∗ ( s ))+ m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) (cid:16) f ( t j , s, y ∗ ( s ) , u ( s )) − f ( t j , s, y ∗ ( s ) , u ∗ ( s )) (cid:17)i ds. Hence, using the Lebesgue point theorem for integrable functions, we reach the following: Z Ts ψ ( t ) ⊤ f ( t, s, y ∗ ( s ) , u ∗ ( s )) dt − g ( s, y ∗ ( s ) , u ∗ ( s )) − m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f ( t j , s, y ∗ ( s ) , u ∗ ( s )) > h Z Ts ψ ( t ) ⊤ f ( t, s, y ∗ ( s ) , u ) dt − g ( s, y ∗ ( s ) , u ) − m X j =1 h jy (cid:0) y ∗ ( t j ) (cid:1) [0 ,t j ] ( s ) f ( t j , s, y ∗ ( s ) , u ) i , a.e. s ∈ [0 , T ] . This gives the maximum condition (4.5).
In recent years, optimal control problems for fractional differential equations have attracted the attention ofsome researchers. However, most of the works on maximum principles for fractional differential equationswere established by convex perturbation technique. See, for instance, Agrawal [1], Agrawal–Defterli–Baleanu273], Frederico–Torres [23] and Kamocki [29] in the sense of Riemann-Liouville case, and Agrawal [2], Bourdin[11] and Hasan–Tangpong–Agrawal [25] in the sense of Caputo case.Let us take a look a recent work [29], in which Kamocki considered the fractional differential equation ofRiemann-Liouville type (3.29) with α ∈ (0 ,
1) and some convex assumptions. The control u ( · ) takes value ina compact set U in R m and f satisfies | f ( t, y , u ) − f ( t, y , u ) | N | y − y | , ∀ y , y ∈ R n , t ∈ [0 , T ] , u ∈ U, | f ( t, , u ) | r ( t ) + γ | u | , ∀ ( t, u ) ∈ [0 , T ] × U, where N > γ > r ( · ) ∈ L p (0 , T ; R ). The corresponding solution belongsto L p (0 , T ; R n ) for some p >
1. When p > I − α [ y ( · )](0) = 0, a Pontryagin’s maximum principlefor Problem (P) was proved. For the case I − α [ y ( · )](0) = 0, maximum principle was obtained only for1 < p < − α .It is easy to check that all the above-mentioned results for fractional differential equations are the specialcases of what we presented in the previous subsection. This paper presented some analysis of singular Volterra integral equations, and established a Pontryagintype maximum principle for an optimal control of such kind of equations. Here are some remarks in order. • As we have indicated, the fractional differential equations of Riemann-Liouville or Caputo types oforder no more than one are fully covered by our results. For fractional differential equations of higher order,similar results can be obtained by properly modifying our approach. • It is easy to see that all the results that we presented will remain true for non-singular Volterra integralequations. • We have allowed to have very general singularity in the free term and the generator. Therefore, ourresults can apply to a much wider class of problems than those covered by fractional differential equationsand non-singular Volterra integral equations.
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