The existence of solution and optimal control problems for the Klein-Gordon hemivariational inequality with strongly elliptic operator

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In this paper, we study the optimal control problems of systems governed by Klein-Gordon hemivariational inequalities. We establish the existence of solutions and the optimal control to these problems.

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Nội dung Text: The existence of solution and optimal control problems for the Klein-Gordon hemivariational inequality with strongly elliptic operator

JOURNAL OF SCIENCE OF HNUE Natural Sci., 2008, Vol. 53, N◦ . 5, pp. 20-30 THE EXISTENCE OF SOLUTION AND OPTIMAL CONTROL PROBLEMS FOR THE KLEIN-GORDON HEMIVARIATIONAL INEQUALITY WITH STRONGLY ELLIPTIC OPERATOR Pham Trieu Duong and Nguyen Thanh Nam Hanoi National University of Education Abstract. In this paper, we study the optimal control problems of systems governed by Klein-Gordon hemivariational inequalities. We establish the existence of solutions and the optimal control to these problems. Keywords and phrases: Optimal control problem, hemivariational inequality, monotone operator, hyperbolic equation. 1. Introduction The background of variational problems is in physics, especially in solid me- chanics, where non-monotone and multi-valued constitutive laws lead to hemivari- ational inequalities. In works [4,5] authors studied some applications of hemivaria- tional inequalities, but there is not much literature dealing with the optimal control problems for hemivariational inequalities. The optimal control problems for operator ∆ have been established in [7] where J.Y Park and J.U Jeong received the results on existence of solutions and optimal control. We study the optimal control problem for general strongly elliptic operator. Let Ω be a bounded domain in Rn (n ≥ 2) with the boundary ∂Ω. We first introduce the following abbreviations Q = Ω × (0, T ), Σ = ∂Ω × (0, T ), k · kk,p = k · kW k,p (Ω) , k · kp = k · kLp (Ω) . For simplicity, we denote k · k = k · kL2 (Ω) . We use the following symbol: (·, ·) is scalar product in L2 (Ω) and will also be used for the notation of duality pairing between dual spaces. For each multi-index p = (p1 , · · · , pn ) ∈ Nn , |p| = p1 + . . . + pn and D p u = |p| ∂ u = uxp11 ...xpnn is the generalized derivative up to order p with respect to p1 ∂ x1 · · · ∂ pxnn x = (x1 , ..., xn ). 20
The existence of solution and optimal control problems... Let L(x, t, D) be the following differential operator: m X L(x, t, D) ≡ D p (apq D q ) (m ≥ 1) |p|,|q|=1 where apq ≡ apq (x, t), |p|, |q| = 1, . . . , m are continuous, real-valued functions on
∂apq
Q, apq = (−1)|p|+|q|aqp and
< a positive constant for all (x, t) ∈ Q; |p|, |q| = ∂t 1, . . . , m. Suppose that m X apq (x, t)ξ p ξ q ≥ γ0 |ξ|2m |p|,|q|=1 for all ξ ∈ Rn and (x, t) ∈ Q where ξ = (ξ1 , . . . , ξn ), ξ p = ξ1p1 ξ2p2 . . . ξnpn and γ0 = constant > 0. We denote Z m X a(t, φ(t), ψ(t)) = (−1)m+|p| apq (x, t)D q φ(t)D p ψ(t)dx Ω |p|,|q|=1 It is easy to show that a is a bi-linear form on W0m,2 (Ω)×W0m,2 (Ω) and a(t, φ(t), ψ(t)) = (L(t)φ, ψ). For all φ(t) ∈ W0m,2 (Ω) we have the Garding inequality a(t, φ(t), φ(t)) + λ2 kφ(t)k2 ≥ λ1 kφ(t)k2m,2 where λ1 , λ2 are constants, not independent of φ, x, t. We consider in Q the problem: Minimize J(y, u) with all pair (y, u) satisfy Klein-Gordon hemivariational inequality (denoted by (∗)) y 00 + αy 0 + (−1)m βL(x, t, D)y + δ|y|γ y + Ξ = f + Bu, (1.1) ∂j y = 0, j = 0, . . . , m − 1, (1.2) ∂ν j y(0, x) = y0 (x); y 0(0, x) = y1 (x) ∀x ∈ Ω, (1.3) Ξ(x, t) ∈ φ(y(x, t)) for a.e (x, t) ∈ Q (1.4) where ν is outer unit normal to S ; α, β, δ are positive constants (independent of t); δ ≥ 1. Function φ is discontinuous and nonlinear multi-valued mapping by filling in jumps of locally bounded b. u denotes the control variable in the real space Hilbert 21
Pham Trieu Duong and Nguyen Thanh Nam U , B is a bounded linear operator and f is a given function. The cost function J(y, u) is given by Z T J(y, u) = [g(y(t)) + h(u(t))] dt 0 where g, h are convex functions. In this paper we consider the existence of solution and optimal control of problem (1.1) − (1.4). 2. Main results Definition 2.1. Function y is said to be a weak solution of (∗) if y ∈ L (0, T ; W0m,2(Ω)∩ ∞ Lγ+2 (Ω)), y 0 ∈ L∞ (0, T ; L2(Ω)) there exists Ξ ∈ L∞ (0, T ; L2(Ω)) and the following relations hold: Z T Z T Z T Z T 00 0 (y (t), η(t)w) dt+α (y (t), η(t)w) dt+β a(t, y(t), η(t)w) dt+ (Ξ, η(t)w) dt 0 0 0 0 Z T Z T Z T γ +δ (|y(t)| y(t), η(t)w) dt = (f (t), η(t)w) dt + (Bu, η(t)w) dt 0 0 0 for all w ∈ W0m,2 (Ω) ∩Lγ+2 (Ω) and η ∈ D(0, T ) y(0, x) = y0 ; y 0(0, x) = y1 , on Ω Ξ(x, t) ∈ φ(y(x, t)) for a.e (x, t) ∈ Q. We will use the evolutional triple 0 W0m,2 (Ω) ∩ Lγ+2 (Ω) ⊂ L2 (Ω) ⊂ W0−m,2 (Ω) + L(γ+2) (Ω) which means that embedding W0m,2 (Ω)∩Lγ+2 (Ω) ,→ L2 (Ω) and L2 (Ω) ,→ W0−m,2 (Ω)+ L(γ+2) (Ω) are continuous, dense and compact. (W0m,2 (Ω) ∩ Lγ+2 (Ω) is defined with 0 norm k · km,2 + k · kγ+2 ). Scalar function k , given by k(s) = |s|γ s. It is easy to show that |k 0 (s)| ≤ (γ + 1)|s|γ . N We recall the Sobolev embedding: if m < we have continuous embedding W0m,2 (Ω) ,→ 2 2N N N N N L N − 2m (Ω). We suppose that ≤m< and 1 ≤ γ ≤ (or m = 4 2 2(N − 2m) 2 and γ ≥ 1) then the nonlinear operator k : W0m,2 (Ω) → L2 (Ω), y 7−→ k ◦ y 22
The existence of solution and optimal control problems... is well defined, especially it is locally Lipschitz. We have the following well-known lemma which is proved in [5]: Lemma 2.1. The operator k : W m,2 0 → L2 (Ω); y 7−→ k ◦ y is locally Lipschitz. (Ω) That is, there exists a constant l > 0 such that: kk(ψ) − k(ρ)k ≤ l(kψkm,2 + kρkm,2 )γ kψ − ρkm,2 . We assume that the following conditions hold (Hb ) b : Q × R is a locally bounded function satisfying the following conditions: (i) (x, t) −→ b(x, t, ξ) is continuous on Q for all ξ ∈ R, (ii) (x, t, ξ) −→ b(x, t, ξ) is measurable in Q × R, (iii) |b(x, t, ξ)| ≤ µ(1 + |ξ|) ∀(x, t, ξ) ∈ Q × R (µ = constant > 0), (iv) b(x, t, ξ)ξ ≥ θξ 2 ∀ξ ∈ R (θ = constant > 0). (HB ) B : L2 (0, T ; U) −→ L2 (0, T ; L2(Ω)) is bounded linear operator, where U is a real Hilbert space. (HU ) Uad is a closed convex, bounded subset of L2 (0, T ; U). (Hg ) g : L2 (Ω) −→ (−∞, +∞] is convex and lower semi-continuous. Moreover there exists r1 > 0 and r2 ∈ R such that g(y) ≥ r1 kyk2 + r2 ∀y ∈ L2 (Ω). (Hh ) h : U −→ (−∞, +∞] is convex lower semi-continuous. Moreover, there exists r3 > 0 and r4 ∈ R such that h(u) ≥ r3 kukU + r4 , ∀u ∈ U . The multi-valued function φ : Q × R −→ 2R is obtained by filling in jumps of a function b(x, t, ·) : R −→ R by means of the functions b , b , b, b : R −→ R as follows: b (x, t, ξ) = ess inf b(x, t, s); b (x, t, ξ) = ess sup b(x, t, s), |s−ξ|≤ |s−ξ|≤ b(x, t, ξ) = lim+ b (x, t, ξ); b(x, t, ξ) = lim+ b (x, t, ξ), →0 →0 φ(x, t, ξ) = [b(x, t, ξ), b(x, t, ξ)]. We shall need a regularization of b defined by: Z ∞ n b (x, t, s) = n b(x, t, s − τ )ρ(nτ )dτ, −∞ 23
Pham Trieu Duong and Nguyen Thanh Nam R1 where ρ ∈ C0∞ ((−1, 1)), ρ ≥ 0 and ρ(τ )dτ = 1. −1 It is easy to show that Z 1 n n |b (x, t, s)| ≤ µ(2 + |s|)n ρ(nτ )dτ = µ(2 + |s|). 1 −n Theorem 2.1. Under hypotheses (H ), (H (HL ), and (y0 , y1 , f ) ∈ (W0m,2 (Ω) ∩ b B ), Lγ+2 (Ω)) × L2 (Ω) × L2 (0, T ; L2(Ω)). Then (∗) has at least one weak solution for every u ∈ L2 (0, T ; U). Theorem 2.2. Assume that the conditions of Theorem 2.1; (H U ), (Hg ), (Hh ) hold. Then the optimal control problem (P ) has at least one solution. 3. Existence of solution for the Klein - Gordon hemivari- ational inequality Proof of Theorem 2.1. We represent by {w } a basis in W0m,2 (Ω) ∩ Lγ+2 (Ω) j j≥0 which is orthogonal in L2 (Ω). Let Vn = Span{w1 , . . . , wn }. We denote n X yn (t) = gjn (t)wj (3.1) j=1 to be solution to the approximate equation: 0 (yn00 (t), wj ) + α(yn (t), wj ) + βa(t, yn (t), wj ) + δ(|yn (t)|γ yn (t), wj )+ + (bn (yn (t)), wj ) = (f (t), wj ) + (Bu, wj ) ∀j = 1, . . . , n (3.2) yn (0) = y0n → y0 in W0m,2 (Ω) ∩ Lγ+2 (Ω) yn0 (t) = y1n → y1 in L2 (Ω) as n → ∞. Because the function k(yn ) = |yn |γ yn is locally Lipschitz, by standard methods of ODE we can prove the existence of a solution to the above equation on some interval [0, tn ). Step 1: We will prove kyn (t)km,2 + kyn (t)kγ+2 ≤ constant (independent of n, t). From that, we get tn = T . Indeed: 1 By (3.1), in equation (3.2) we can replace wj by yn0 (t). Putting En (t) = kyn0 (t)k2 + 2 δ γ+2 kyn (t)kγ+2 we obtain γ+2 d 0 En (t) + αkyn (t)k2 + βa(t, yn (t), yn0 (t)) dt (3.3) = (f (t), yn0 (t)) + (Bu, yn0 (t)) − (bn (yn (t)), yn0 (t)). 24
The existence of solution and optimal control problems... By apq = (−1)|p|+|q|aqp , we have m X 2 (−1)m+|p| apq D q yn (t)D p yn0 (t) |p|,|q|=1 m d X = [ (−1)m+|p| apq D q yn (t)D p yn (t)] dt |p|,|q|=1 Xm ∂apq q − (−1)m+|p| D yn (t)D p yn (t). ∂t |p|,|q|=1 Integrating (3.3) over (0, t) with t ≤ tn , then adding side by side with Z t 2 βλ2 kyn (t)k = β(2λ2 (yn (s), yn0 (s))ds + λ2 ky0n k2 ), 0 (where λ2 is coefficient in the Garding inequality), we come to Z t 2En (t)+2α kyn0 (s)k2 ds + β[a(t, yn (t), yn (t)) + λ2 kyn (t)k2 ] 0 Z t Z t Z t 0 0 = 2 (f (s), yn (s)) ds + 2 (Bu, yn (s)) ds − 2 (bn (yn (s)), yn0 (s)) ds 0 0 0 Z t 2 + 2En (0) + βa(0, y0n , y0n ) + βλ2 ky0n k + 2βλ2 (yn (s), yn0 (s)) ds 0 Z t m X ∂apq q + (−1)m+|p| D yn (s)D p yn (s) dxds. (3.4) 0 |p|,|q|=1 ∂s We will provide some estimates of terms of (3.4) by using the Gronwall inequality. It is easy to show that 2En (0) + βa(0, y0n , y0n ) ≤ C (3.5) where C is denoted a general constant which is positive and independent of n, t. By continuous embedding W0m,2 (Ω) ,→ L2 (Ω) we have Z t Z tZ Z t n 2 2 2 kb (yn (s))k ds ≤ 2µ (4 + |yn (x, s)| ) dxds ≤ C + C kyn (s)k2m,2 ds. 0 0 Ω 0 (3.6) Combining Schwartz inequality and Cauchy inequality we have: Z t Z Z t n 0 1 t n 2 | (b (yn (s)), yn (s)) ds| ≤ kb (yn (s))k ds + kyn0 (s)k2 ds 0 2 0 0 Z t Z t ≤C +C 2 kyn (s)km,2 ds + C kyn0 (s)k2 ds. (3.7) 0 0 25
Pham Trieu Duong and Nguyen Thanh Nam Z t By a similar way we have similar inequalities for (f (s), yn0 (s)) ds and Z 0 t (Bu(s), yn0 (s)) ds. 0 m X m mX 2
∂apq
Using inequality ai bj ≤ (ai + b2j ), and hypotheses
≤ C for i,j=1 2 i,j=1 ∂s all (x, s) ∈ Q we get Z m Z t
t X m+|p| ∂apq
(−1) q p
D yn (s)D yn (s) ds ≤ C kyn (s)k2m,2 ds. (3.8) 0 ∂s 0 |p|,|q|=1 Moreover we have |(yn (s), yn0 (s))| ≤ kyn (s)k.kyn0 (s)k ≤ 1 ≤ (kyn (s)k2 + kyn0 (s)k2 ) ≤ Ckyn0 (s)k2 + Ckyn (s)k2m,2 , 2 thus Z t Z t Z t | (yn (s), yn0 (s)) ds| ≤ C kyn0 (s)k2 ds + C kyn (s)k2m,2 ds. (3.9) 0 0 0 Using the Garding inequality we have a(t, yn (t), yn (t)) + λ2 kyn (t)k2 ≥ λ1 kyn (t)k2m,2 . (3.10) By estimates (3.4), (3.7) − (3.10) we get Z t Z t 2 0 2 0 2 kyn (t)km,2 + kyn (t)k ≤ C + C kyn (t)k ds + C kyn (s)k2m,2 ds. 0 0 From that, using the Gronwall inequality we deduce kyn0 (t)k + kyn (t)km,2 + kyn (t)kγ+2 ≤ constant (independent of n, t) (3.11) (the inequality holds for almost every t ∈ (0, tn ]), hence tn = T . Combining (3.6) and (3.11) we get kbn (yn (t))k ≤ constant (independent of n, t). (3.12) Step 2: Passage to the limit Combining (3.6) and (3.12) we have that relabelling if necessary ∗ yn * y in L∞ (0, T ; W0m,2(Ω) ∩ Lγ+2 (Ω)), ∗ yn0 * y 0 in L∞ (0, T ; L2 (Ω)), ∗ (3.13) bn (yn ) * Ξ in L∞ (0, T ; L2 (Ω)), ∗ D p yn * D p y in L∞ (0, T ; L2 (Ω)). 26
The existence of solution and optimal control problems... From (3.13), using the Aubin-Lion compactness theorem we get yn → y in L2 (0, T ; L2(Ω)). Hence, there exists a subsequence of yn (which is denoted again by yn ) such that a.e yn −→ y in Q. Moreover from k|yn (t)|γ yn (t)k(γ+2)0 = kyn (t)kγ+2 we deduce that ∗ 0 |yn |γ yn * |y|γ y in L∞ (0, T ; L(γ+2) (Ω)). (3.14) Using the estimates (3.13) and (3.14), in (3.2) letting n → ∞ we obtain (y 00(t), w) + α(y 0(t), w) + βa(t, y(t), w) + δ(|y(t)|γ y(t), w)+ + (Ξ, w) = (f (t), w) + (Bu, w) in D0 (0, T ) for all w ∈ W0m,2 (Ω) ∩ Lγ+2 (Ω). (3.15) Step 3: We will prove that (y, Ξ) is a weak solution of (∗). By y(t) ∈ W the m,2 0 (Ω) boundary conditions are satisfied. By a similar way in Section 1.4 of Chapter 1 in [3] we have y 0(0) = y1 and y(0) = y0 , hence the initial conditions are also satisfied. Now we will check the conditions of Ξ. a.e Because yn −→ y in Q then according to Lusin theorem, for each η > 0 we can choose W ⊂ Q such that mease(W ) < η and yn → y uniformly in Q\W . For each 2 > 0 ∃n0 > such that |yn (x, t) − y(x, t)| < ∀(x, t) ∈ Q\W, n > n0 . 2 1 Then, if |yn (x, t) − s| < we have |y(x, t) − s| < for all n > n0 and (x, t) ∈ Q\W n It follows that b (y(x, t)) ≤ bn (yn (x, t)) ≤ b (y(x, t)). For φ ∈ L2 (Q) and φ ≥ 0 then Z Z b (y(x, t))φ(x, t) dxdt ≤ bn (yn (x, t))φ(x, t) dxdt Q\W Q\W Z ≤ b (y(x, t))φ(x, t) dxdt. Q\W Letting n → ∞, then letting → 0 by using Lebesgue monotonous convergence theorem we obtain Z Z b(y(x, t))φ(x, t) dxdt ≤ Ξ(x, t)φ(x, t) dxdt Q\W Q\W Z ≤ b(y(x, t))φ(x, t) dxdt. Q\W 27