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Variational method for the solution of an inverse problem
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Variational method for the solution of an inverse problem N. Yildirim Aksoy Department of Mathematics, Faculty of Arts and Sciences, Kafkas University, Kars, 36100, Turkey
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Article history: Received 23 September 2015
This paper presents a variational method for the solution of an inverse problem. An inverse problem of determining the unknown coefficient of nonlinear time-dependent Schrödinger equation is considered. Inverse problem is reformulated as a variational problem by means of an observation functional. The existence and uniqueness of solutions of the constituted variational problem are investigated. The differentiability of observation functional is obtained and a necessary condition for the solution of the variational problem is given. © 2016 Elsevier B.V. All rights reserved.
MSC: 35R30 35J10 49J40 Keywords: Inverse problem Schrödinger equation Variational method
1. Introduction The inverse problems of finding the quantum-mechanical potential for Schrödinger equation which describes how the state of a physical system changes in time have drawn a lot of attention in recent years. Especially, inverse problems of determining the quantum-mechanical potential for nonlinear time dependent Schrödinger equation has a great importance in this area. It is known that the inverse problems for nonlinear time-dependent Schrödinger equation usually arise in the dispersion of light beams (waves) in nonlinear medium [1]. Let us consider the wave process describing by nonlinear Schrödinger equation
∂ 2ψ ∂ψ ∂ψ + a0 2 + ia1 (x) − a2 (x)ψ + v(t )ψ + ia3 |ψ|2 ψ = f ∂t ∂x ∂x ψ(x, 0) = ϕ(x), x ∈ (0, l) ,
i
ψ(0, t ) = ψ(l, t ) = 0,
t ∈ (0, T )
(1.1) (1.2) (1.3)
where ψ = ψ(x, t ) is the wave’s complex amplitude, x and t are variables of space and time, respectively, l > 0, T > 0 are √ given numbers, 0 ≤ x ≤ l, 0 ≤ t ≤ T , i = −1 imaginary unit, a0 , a3 > 0 are given real numbers, a2 (x) is a measurable real-valued function that satisfies the condition 0 < µ0 ≤ a2 (x) ≤ µ1
for almost all x ∈ (0, l) , µ0 , µ1 = const. > 0,
(1.4)
a1 (x) is a measurable real-valued function which satisfies the condition
da1 (x) |a1 (x)| ≤ µ2 , ≤ µ3 for almost all x ∈ (0, l) , µ2 , µ3 = const. > 0, , dx a1 (0) = a1 (l) = 0 the functions f , ϕ are given complex-valued functions. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.cam.2016.01.002 0377-0427/© 2016 Elsevier B.V. All rights reserved.
(1.5)
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When we consider the inverse problem of determining the quantum mechanical potential v(t ) and the function ψ(x, t ) from the conditions (1.1)–(1.3), the conditions (1.1)–(1.3) are not sufficient to find the solution of the inverse problem. Therefore, we use an additional condition and the choosing of an additional condition is very important to find the solution by variational methods of an inverse problem. The solutions by variational methods of the inverse problems for Schrö dinger equation were previously studied in [2–14] under different additional conditions. As the different from the before studies, in the present paper, we consider an inverse problem for a nonlinear timedependent Schrödinger equation in the form (1.1) from additional condition ψ1 (x, t ) = ψ2 (x, t ) such that the functions ψ1 (x, t ) = ψ2 (x, t ) are solutions of the following first and second type boundary value problems
∂ 2 ψ1 ∂ψ1 ∂ψ1 + a0 2 + ia1 (x) − a2 (x)ψ1 + v(t )ψ1 + ia3 |ψ1 |2 ψ1 = f1 ∂t ∂x ∂x ψ1 (x, 0) = ϕ1 (x), x ∈ (0, l) ,
(1.7)
ψ1 (0, t ) = ψ1 (l, t ) = 0,
(1.8)
i
(1.6)
t ∈ (0, T )
and
∂ 2 ψ2 ∂ψ2 ∂ψ2 + a0 2 + ia1 (x) − a2 (x)ψ2 + v(t )ψ2 + ia3 |ψ2 |2 ψ2 = f2 ∂t ∂x ∂x ψ2 (x, 0) = ϕ2 (x), x ∈ (0, l) ,
(1.9)
i
∂ψ2 (l, t ) ∂ψ2 (0, t ) = = 0, ∂x ∂x
(1.10)
t ∈ (0, T )
(1.11)
respectively, where
ϕ1 ∈ W˚ 22 (0, l),
∂ϕ2 (l) ∂ϕ2 (0) = = 0, ∂x ∂x
ϕ2 ∈ W22 (0, l),
0 ,1
fk ∈ W2 (Ω ) for k = 1, 2
(1.12)
and Ωt = (0, l) × (0, t ), Ω = ΩT . Thus, we formulate the inverse problem as follows: is to determine the unknown coefficient v(t ) and the functions ψ1 (x, t ), ψ2 (x, t ) from the additional condition
ψ1 (x, t ) = ψ2 (x, t ),
( x, t ) ∈ Ω .
(1.13)
The quantum mechanical potential v(t ) is investigated on the set
V ≡
dv(t ) ≤ b1 for almost all t ∈ (0, T ) , v : v(t ) ∈ W21 (0, T ), |v(t )| ≤ b0 , dt
called as the set of admissible coefficients, which the constants b0 , b1 > 0 are given numbers. Here the Sobolev space W21 (0, T ) is a Hilbert space consisting of all the elements L2 (0, T ) having square summable generalized derivatives of the ˚ 2 (0, l), W 0,1 (Ω ) are widely defined in [15]. first order on (0, T ). The Sobolev spaces W 2 (0, l), W 2
2
2
∂ψ
Here, it should be noted that the Schrödinger equation considered in this paper contains the term ia1 (x) ∂ x as different from before studies. So, this study is more comprehensive than previous. In this paper, in Section 2, the variational formulation of the considered inverse problem and the existence of the solutions of the boundary value problems (1.6)–(1.8), (1.9)–(1.11) are given. In Section 3, the existence and uniqueness of solutions of the constituted variational problem are shown. Finally, a formula for the first variation of observation functional and a necessary condition for the solution of the variational problem are obtained in Section 4. 2. The variational formulation of the inverse problem In this section, we will present a variational formulation of the above inverse problem. If we want to formulate the above inverse problem as a variational problem, we can write it using the method in [16] as follows: to find the minimum of the observation functional Jα (v) = ∥ψ1 − ψ2 ∥2L2 (Ω ) + α∥v − w∥2W 1 (0,T )
(2.1)
2
on the set V under conditions (1.6)–(1.8) and (1.9)–(1.11), where α ≥ 0 is a given number, w ∈ W21 (0, T ) is a given element. We must firstly prove the existence of the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11) to investigate the solution of the variational problem. For this reason, let us define the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11). Under given conditions, by a solution of the problem (1.6)–(1.8), we mean a function ψ1 (x, t ) in ˚ 2 (0, l)) ∩ C 1 ([0, T ], L2 (0, l)), which satisfies Eq. (1.6) for almost all x ∈ (0, l) and any t ∈ [0, T , ] the space B1 ≡ C 0 ([0, T ], W 2 the initial condition (1.7) for almost all x ∈ (0, l) and the boundary condition (1.8) for almost all t ∈ (0, T ). Similarly, by a solution of the problem (1.9)–(1.11), we mean a function ψ2 (x, t ) in the space B2 ≡ C 0 ([0, T ], W22 (0, l)) ∩ C 1 ([0, T ], L2 (0, l)),
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which satisfies Eq. (1.9) for almost all x ∈ (0, l) and any t ∈ [0, T ], the initial condition (1.10) for almost all x ∈ (0, l) and the boundary condition (1.11) for almost all t ∈ (0, T ), where C k ([0, T ], B) is a Banach space of all B− valued, k ≥ 0 times continuously differentiable functions on [0, T ] with the norm
∥ψ∥C k ([0,T ],B) =
k m=0
m d ψ(t ) max 0≤t ≤T dt m
B
for ψ ∈ C k ([0, T ], B). We prove that the following theorem is valid for the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11): Theorem 2.1. Assume that functions a1 (x), a2 (x) and ϕk (x), fk (x, t ) for k = 1, 2 satisfy the conditions (1.5), (1.4) and (1.12), respectively. Then, boundary value problems (1.6)–(1.8) and (1.9)–(1.11) have a unique solution ψ1 ∈ B1 , ψ2 ∈ B2 for any v ∈ V , respectively, and the following estimations are valid:
∥ψ1 (., t )∥
2 ˚ 2 (0,l) W 2
∂ψ1 2 + ≤ c ∥ϕ1 ∥2W˚ 2 (0,l) + ∥ϕ1 ∥6W˚ 2 (0,l) + ∥ϕ1 ∥6W˚ 1 (0,l) 1 2 2 2 ∂ t L2 (0,l) 18 2 6 + ∥ϕ1 ∥W˚ 1 (0,l) + ∥f1 ∥ 0,1 + ∥f1 ∥ 0,1 , 2
W2
(Ω )
W2
(Ω )
∂ψ2 2 ≤ c ∥ϕ2 ∥2W 2 (0,l) + ∥ϕ2 ∥6W 2 (0,l) + ∥ϕ2 ∥6W 1 (0,l) ∥ψ2 (., t )∥2W 2 (0,l) + 2 ∂t 2 2 2 2 L2 (0,l) 18 2 6 + ∥ϕ2 ∥W 1 (0,l) + ∥f2 ∥ 0,1 + ∥f2 ∥ 0,1 2
W2
(Ω )
W2
(Ω )
(2.2)
(2.3)
for any t ∈ [0, T ], where the constants c1 , c2 > 0 are independent from ϕ1 , f 1 , ϕ2 , f2 and t. The proof of Theorem 2.1 is easily obtained from Galerkin’s method. For simplicity, let us rewrite the boundary value problems (1.6)–(1.8) and (1.9)–(1.11) as the following:
∂ 2 ψk ∂ψk ∂ψk + a0 2 + ia1 (x) − a2 (x)ψk + v(t )ψk + ia3 |ψk |2 ψk = fk for k = 1, 2 ∂t ∂x ∂x ψk (x, 0) = ϕk (x), x ∈ (0, l) for k = 1, 2 i
ψ1 (0, t ) = ψ1 (l, t ) = 0,
∂ψ2 (l, t ) ∂ψ2 (0, t ) = = 0, ∂x ∂x
t ∈ (0, T ) ,
(2.4) (2.5) (2.6)
which include the two boundary value problems (1.6)–(1.8) and (1.9)–(1.11) for k = 1 and k = 2, respectively. Thus, when the conditions (1.6)–(1.8) and (1.9)–(1.11) are satisfied, the problem of finding the minimum of the functional (2.1) on the set V is called the variational problem (2.1), (2.4)–(2.6). 3. Existence and uniqueness theorems The variational problems may not always have a solution. For this, in the present section, we investigate the existence of the solution of the considered variational problem. Firstly, we show that the variational problem (2.1), (2.4)–(2.6) has a unique solution for α > 0 on a dense subset G of the space W21 (0, T ) and secondly, the variational problem has at least one solution for any α ≥ 0 on the space W21 (0, T ). We refer to the following theorem to prove the uniqueness of the solution of the variational problem (2.1), (2.4)–(2.6). Theorem 3.1 (The Corollary of Goebel’s Theorem [17]). Let X be a uniformly convex space, U be a closed bounded set on X , the functional I (v) be lower semicontinuous and lower bounded on U, α > 0 be a given number. Then, there is a dense subset G of the space X such that for any w ∈ G the functional 2 Iα (v) = I (v) + α∥v − w∥ X
takes its minimum value at a unique point on U. Lemma 3.1. The functional J0 (v) = ∥ψ1 − ψ2 ∥2L (Ω ) is continuous on the set V . 2 1 Proof. Let 1v ∈ W∞ (0, T ) be an increment of any element v ∈ V such that v + 1v ∈ V , let the functions ψk = ψk (x, t ) ≡ ψk (x, t ; v) for k = 1, 2 be the solutions of the problem (2.4)–(2.6) corresponding to v ∈ V and 1ψk = 1ψk (x, t ) ≡ ψk (x, t ; v + 1v) − ψk (x, t ; v) = ψk∆ − ψk (x, t ) for k = 1, 2, where the functions ψk∆ = ψk (x, t ; v + 1v) for
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k = 1, 2 are solutions of the problem (2.4)–(2.6) for any v + 1v . From the conditions (2.4)–(2.6) we obtain that the functions 1ψk = 1ψk (x, t ) for k = 1, 2 are solutions of the following boundary value problem: i
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk ∂t ∂ x2 ∂x + ia3 |ψk∆ |2 + |ψk |2 1ψk + ψk∆ ψk 1ψ¯ k = −1v(t )ψk , (x, t ) ∈ Ω , k = 1, 2
1ψk (x, 0) = 0,
(3.1)
x ∈ (0, l), k = 1, 2
(3.2)
∂ 1ψ2 (0, t ) ∂ 1ψ2 (l, t ) = = 0, ∂x ∂x
1ψ1 (0, t ) = 1ψ1 (l, t ) = 0,
t ∈ (0, T ) .
(3.3)
¯ k (x, t ) for To evaluate the solution of the boundary value problem (3.1)–(3.3), let us multiply both sides of Eq. (3.1) by 1ψ k = 1, 2 and integrate over Ωt . Then, integrating by part we get
∂ 1ψk 2 ∂ 1ψk + ia1 (x) ∂ 1ψk 1ψ¯ k − a2 (x) |1ψk |2 + (v(t ) + 1v(t ))|1ψk |2 1ψ¯ k − a0 ∂t ∂x ∂x Ωt 2 2 2 2 ¯ + ia3 |ψk∆ | + |ψk | |1ψk | + ia3 ψk∆ ψk (1ψk ) dxdτ = − 1v(t )ψk 1ψ¯ k dxdτ i
Ωt
for k = 1, 2. If we subtract the complex conjugate of the above equality from itself, we obtain
∂ 1ψ k ∂ ∂ 1ψk 2 |1ψk | dxdτ + 1ψ¯ k + a1 (x) 1ψk dxdτ a1 (x) ∂x ∂x Ωt Ωt ∂ t 2 2 2 + 2a3 |ψk∆ | + |ψk | |1ψk | dxdτ = −2 Im 1vψk 1ψ k dxdτ − 2a3 Ωt
Ωt
for t ∈ [0, T ] and k = 1, 2. If we add the term conditions (1.5), (3.2), we obtain
da1 (x) dx
Ωt
Re ψk∆ ψk 1ψ k
2
dxdτ
|1ψk |2 for k = 1, 2 to both sides of the above equality and use the
∥1ψk (., t )∥ + 2a3 (|ψk∆ | + |ψk | )|1ψk | dxdτ = −2 Im 1vψk 1ψ k dxdτ Ωt Ωt 2 da1 (x) 2 − 2a3 Re ψk∆ ψk 1ψ k dxdτ + |1ψk | dxdτ 2 L2 (0,l)
2
2
Ωt
2
dx
Ωt
and by Young’s inequality and the condition (1.5)
∥1ψk (., t )∥2L2 (0,l) + a3 (|ψk∆ |2 + |ψk |2 )|1ψk |2 dxdτ Ωt |1v| |ψk | |1ψk | dxdτ + µ3 |1ψk |2 dxdτ ≤2 Ωt
Ωt
t
l
|1v|2 |ψk |2 dxdτ +
≤ 0
0
Ωt
|1ψk |2 dxdτ + µ3
t
= 0
|1v|2 ∥ψk (., t )∥2L2 (0,l) dτ + (1 + µ3 )
≤ max ∥ψk (., t )∥2L2 (0,l) 0≤t ≤T
Ωt
|1ψk |2 dxdτ
t
0
∥1ψk (., t )∥2L2 (0,l) dτ
T
|1v|2 dτ + (1 + µ3 ) 0
t
0
∥1ψk (., t )∥2L2 (0,l) dτ
for any t ∈ [0, T ] and k = 1, 2. In the above inequality, if we use the estimations (2.2) and (2.3) we get
∥1ψk (., t )∥2L2 (0,l) + a3
Ωt
(|ψk∆ |2 + |ψk |2 )|1ψk |2 dxdτ ≤ c3 ∥1v∥2L2 (0,T ) + (1 + µ3 )
t
0
∥1ψk (., t )∥2L2 (0,l) dτ .
(3.4)
Thus, ıf we take into account a3 > 0, we have the inequality
∥1ψk (., t )∥2L2 (0,l) ≤ c3 ∥1v∥2L2 (0,T ) + (1 + µ3 )
t
0
∥1ψk (., t )∥2L2 (0,l) dτ ,
for any t ∈ [0, T ] and k = 1, 2, where the constant c3 > 0 is independent from 1v and t. Applying the Gronwall’s inequality to the above inequality, we get
∥1ψk (., t )∥2L2 (0,l) ≤ c4 ∥1v∥2L2 (0,T ) for any t ∈ [0, T ], k = 1, 2
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and thus
∥1ψk (., t )∥2L2 (0,l) ≤ c4 ∥1v∥2L2 (0,T ) ≤ c5 ∥1v∥2W 1 (0,T ) ≤ c6 ∥1v∥2W 1 (0,T ) , ∞
2
k = 1, 2
(3.5)
for any t ∈ [0, T ], where the constants c4 , c5 , c6 > 0 are independent from 1v and t. Now, let us find the increment of the functional J0 (v) for any v ∈ V . If we use the definition of the functional J0 (v), we obtain
|ψ1∆ − ψ2∆ |2 dxdt − |ψ1 − ψ2 |2 dxdt 1J0 (v) = J0 (v + 1v) − J0 (v) = Ω Ω = (ψ1∆ − ψ2∆ ) ψ 1∆ − ψ 2∆ − (ψ1 − ψ2 ) ψ 1 − ψ 2 dxdt Ω = (ψ1 − ψ2 ) 1ψ 1 − 1ψ 2 + (1ψ1 − 1ψ2 ) ψ 1 − ψ 2 + |1ψ1 |2 + |1ψ2 |2 − 2Re 1ψ1 1ψ 2 dxdt Ω =2 Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t ) dxdt Ω
+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) − 2
Re 1ψ1 1ψ 2 dxdt .
Ω
(3.6)
Applying the Cauchy–Schwarz inequality to the equality (3.6), we get
|1J0 (v)| ≤ 2 ∥ψ1 ∥L2 (Ω ) ∥1ψ1 ∥L2 (Ω ) + 2 ∥ψ1 ∥L2 (Ω ) ∥1ψ2 ∥L2 (Ω ) + 2 ∥ψ2 ∥L2 (Ω ) ∥1ψ1 ∥L2 (Ω ) + 2 ∥ψ2 ∥L2 (Ω ) ∥1ψ2 ∥L2 (Ω ) + 2 ∥1ψ1 ∥2L2 (Ω ) + 2 ∥1ψ2 ∥2L2 (Ω ) . Using the estimations (2.2), (2.3), (3.5) in the above inequality, we obtain the following inequality for the increment of the functional J0 (v):
2 |J0 (v + 1v) − J0 (v)| ≤ c7 ∥1v∥W∞ 1 (0,T ) + ∥1v∥ 1 W (0,T ) ∞
for any v ∈ V , where the constant c7 > 0 is independent from 1v . Thus, since 1J0 (v) → 0 for ∥1v∥W 1 (0,T ) → 0 and for ∞ any v ∈ V , we can easily say that the functional J0 (v) is continuous on the set V . Now, let us give the following theorem stated the uniqueness of the solution of the variational problem (2.1), (2.4)–(2.6) on a dense subset G of the space W21 (0, T ). Theorem 3.2. Assume that the functions a1 (x), a2 (x) and ϕk (x), fk (x, t ) for k = 1, 2 satisfies the conditions (1.5), (1.4), (1.12), respectively, and w ∈ W21 (0, T ) is a given function. Then, there is a dense subset G of the space W21 (0, T ) such that the variational problem (2.1), (2.4)–(2.6) has a unique solution for any w ∈ G and α > 0. Proof. Since the functional J0 (v) is continuous on the set V , it is a lower semicontinuous functional. Also, since J0 (v) ≥ 0 for any v ∈ V , the functional J0 (v) is lower bounded. Additionally, the set V is a closed, bounded and convex set of the uniformly convex space W21 (0, T ) [18]. Thus, from Theorem 3.1, it follows that there is a dense subset G of the space W21 (0, T ) such that the variational problem (2.1), (2.4)–(2.6) has a unique solution for any w ∈ G and α > 0. The next theorem states that the variational problem (2.1), (2.4)–(2.6) has at least one solution for any α ≥ 0 on the space W21 (0, T ). Theorem 3.3. Suppose that α ≥ 0 and the conditions of Theorem 2.1 hold and w ∈ W21 (0, T ) is a given function. Then, the variational problem (2.1), (2.4)–(2.6) has at least one solution. We can easily make the proof of Theorem 3.3 by using the method in [13]. 4. Differentiability of the functional In the present section, constituting an adjoint problem we obtain the first variation of the functional Jα (v) by means of the adjoint problem. Also, a necessary condition for the solution of the variational problem (2.1), (2.4)–(2.6) is given. Firstly, we reformulate the variational problem (2.1), (2.4)–(2.6) by using Lagrange multiplier functions as follows: Consider the minimization problem L (v, ψ1 , ψ2 , η1 , η2 ) → inf
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where L (v, ψ1 , ψ2 , η1 , η2 ) = Jα (v)
∂ψ1 ∂ 2 ψ1 ∂ψ1 2 + i + a0 2 + ia1 (x) − a2 (x)ψ1 + v(t )ψ1 + ia3 |ψ1 | ψ1 − f1 η1 dxdt 2 Ω ∂t ∂x ∂x 1 ∂ψ 1 ∂ 2ψ 1 ∂ψ 1 2 + −i + a0 − ia1 (x) − a2 (x)ψ 1 + v(t )ψ 1 − ia3 |ψ1 | ψ 1 − f 1 η1 dxdt 2 Ω ∂t ∂ x2 ∂x 1 ∂ψ2 ∂ 2 ψ2 ∂ψ2 + i + a0 2 + ia1 (x) − a2 (x)ψ2 + v(t )ψ2 + ia3 |ψ2 |2 ψ2 − f2 η2 dxdt 2 Ω ∂t ∂x ∂x ∂ψ 2 ∂ 2ψ 2 1 ∂ψ 2 2 −i + a0 − ia1 (x) − a2 (x)ψ 2 + v(t )ψ 2 − ia3 |ψ2 | ψ 2 − f 2 η2 dxdt , + 2 Ω ∂t ∂ x2 ∂x 1
the functions ηk = ηk (x, t ) = ηk (x, t ; v) for k = 1, 2 are Lagrange multipliers and the functions ψk = ψk (x, t ) for k = 1, 2 are solutions of the boundary value problem (2.4)–(2.6). Thus, from the stationarity condition of the Lagrange functional L (v, ψ1 , ψ2 , η1 , η2 ), we obtain the following adjoint problem: i
∂ ∂ 2 ηk ∂ηk + a0 2 + i (a1 (x)ηk ) − a2 (x)ηk + v(t )ηk − 2ia3 |ψk |2 ηk + ia3 ψk2 η¯ k ∂t ∂x ∂x = 2(−1)k (ψ1 (x, t ) − ψ2 (x, t )) ,
ηk (x, T ) = 0,
(x, t ) ∈ Ω , k = 1, 2,
(4.1)
x ∈ (0, l), k = 1, 2
η1 (0, t ) = η1 (l, t ) = 0,
(4.2)
t ∈ (0, T )
∂η2 ∂η2 (0, t ) = (l, t ) = 0, ∂x ∂x
(4.3)
t ∈ (0, T )
(4.4)
where the functions ψk = ψk (x, t ) ≡ ψk (x, t ; v) for k = 1, 2 are solutions of boundary value problem (2.4)–(2.6) for any v ∈ V . As seen, adjoint problem (4.1)–(4.4) includes the two boundary value problems according to the functions η1 and η2 for k = 1, 2. If we apply the transform τ = T − t to adjoint problem, it is seen that the adjoint problem is an initial boundary value problem in the form of problem (2.4)–(2.6). Under given conditions, by a solution of the adjoint problem, we mean two functions η1 (x, t ) in the space B1 and η2 (x, t ) in the space B2 , which satisfy Eq. (4.1) for almost all x ∈ (0, l) and any t ∈ [0, T ], the condition (4.2) for almost all x ∈ (0, l) and the boundary conditions (4.3) and (4.4) for almost all t ∈ (0, T ), respectively. Therefore, we can write the following theorem for the solution of adjoint problem (4.1)–(4.4): Theorem 4.1. Assume that the conditions of Theorem 2.1 hold. Then, the adjoint problem (4.1)–(4.4) has a unique solution η1 ∈ B1 , η2 ∈ B2 for any v ∈ V and the following estimations are valid for the solution:
∥η1 (., t )∥
2 ˚ 2 (0,l) W 2
∂η1 (., t ) 2 + ∂t
∥η2 (., t )∥
2 W22 (0,l)
∂η2 (., t ) 2 + ≤ c9 ∥ψ1 − ψ2 ∥W 0,1 (Ω ) 2 ∂ t L2 (0,l)
L2 (0,l)
≤ c8 ∥ψ1 − ψ2 ∥W 0,1 (Ω ) ,
(4.5)
2
(4.6)
for any t ∈ [0, T ], where the constants c8 , c9 > 0 are independent from t. We can easily prove Theorem 4.1 as similar to the proof of Theorem 2.1 by using Galerkin’s method. 1 Now, let us find the increment of the functional Jα (v) for ∀v ∈ V . Let the function 1v ∈ W∞ (0, T ) be an increment given to any v ∈ V such that v + 1v ∈ V . Then, using the formula (2.1) and (3.6) we can write the increment of the functional Jα (v) for any v ∈ V as the following:
1Jα (v) = Jα (v + 1v) − Jα (v) = 1J0 (v) + α∥v + 1v − w∥2W 1 (0,T ) − α∥v − w∥2W 1 (0,T ) 2
= 1J0 (v) + 2α
0
=2 Ω
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
0
2
dv dt
−
dw dt
Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t ) dxdt
d(1v) dt
dt + α∥1v∥2W 1 (0,T ) 2
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) − 2 + 2α
Ω
–
7
Re 1ψ1 1ψ 2 dxdt
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
dv(t ) dt
0
0
)
−
dw(t ) dt
d1v(t ) dt
dt + α∥1v∥2W 1 (0,T )
(4.7)
2
where the functions 1ψk = 1ψk (x, t ) ≡ ψk (x, t ; v + 1v) − ψk (x, t ; v) for k = 1, 2 are solutions of the problem (3.1)–(3.3) for v ∈ V . Lemma 4.1.
Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t )
2 Ω
=
dxdt
1v(t )Re(1ψ1 (x, t )η1 (x, t ))dxdt + 1v(t )Re(ψ1 (x, t )η1 (x, t ))dxdt Ω 1v(t )Re(1ψ2 (x, t )η2 (x, t ))dxdt + 1v(t )Re(ψ2 (x, t )η2 (x, t ))dxdt + Ω Ω − a3 (|ψ1∆ |2 − |ψ1 |2 )Im(1ψ1 η¯ 1 )dxdt − a3 (|ψ2∆ |2 − |ψ2 |2 )Im(1ψ2 η¯ 2 )dxdt Ω Ω 2 2 |1ψ2 | Im (ψ2 η¯ 2 ) dxdt . − a3 |1ψ1 | Im (ψ1 η¯ 1 ) dxdt − a3 Ω
Ω
Ω
Proof. It is clear that the functions 1ψk = ψk (x, t ; v + 1v) − ψ(x, t ; v) for k = 1, 2 satisfy integral identities
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk 2 ∂ t ∂ x ∂x Ω 2 2 ¯ 1v(t )ψk φ k (x, t )dxdt + ia3 |ψk∆ | + |ψk | 1ψk + ψk∆ ψk 1ψk φ k (x, t )dxdt = −
i
(4.8)
Ω
for any functions φk = φk (x, t ) ∈ L2 (Ω ), k = 1, 2 and the conditions (3.2), (3.3). Also, the functions ηk = ηk (x, t ) for k = 1, 2 satisfy the identities
∂ 2 ηk ∂ ∂ηk + a0 2 + i (a1 (x)ηk ) − a2 (x)ηk + v(t )ηk − 2ia3 |ψk |2 ηk + ia3 ψk2 ηk ξ k dxdt ∂t ∂x ∂x Ω = 2(−1)k (ψ1 (x, t ) − ψ2 (x, t )) ξ k dxdt , (x, t ) ∈ Ω
i
(4.9)
Ω
for any functions ξk = ξk (x, t ) ∈ L2 (Ω ), k = 1, 2. Let us put the functions 1ψ1 ∈ B1 , 1ψ2 ∈ B2 instead of the test functions ξk for k = 1, 2 in identities (4.9), respectively. Later, if we apply by part integration formula to obtained identities, we get
∂ 1ψ k ∂ 2 1ψ k ∂ 1ψ k 2 −i − ia1 (x) + a0 − a2 (x)1ψ k + v(t )1ψ k − 2ia3 |ψk | 1ψ k ηk dxdt ∂t ∂ x2 ∂x Ω 2 k + ia3 ψk η¯ k 1ψ k dxdt = 2(−1) (ψ1 (x, t ) − ψ2 (x, t )) 1ψ k dxdt Ω
Ω
and its complex conjugate
∂ 1ψk ∂ 2 1ψk ∂ 1ψk 2 ( x ) ( x ) 1 ψ + v( t ) 1 ψ + 2ia |ψ | 1 ψ + a0 + ia − a 1 2 k k 3 k k η k dxdt ∂t ∂ x2 ∂x Ω 2 − ia3 ψ k ηk 1ψk dxdt = 2(−1)k ψ 1 (x, t ) − ψ 2 (x, t ) 1ψk dxdt .
i
Ω
(4.10)
Ω
Also, if we put the functions η1 (x, t ) ∈ B1 , η2 (x, t ) ∈ B2 instead of functions φ1 (x, t ), φ2 (x, t ) in (4.8), respectively, we have
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk 2 ∂t ∂x ∂x Ω + ia3 |ψk∆ |2 + |ψk |2 1ψk + ψk∆ ψk 1ψ¯ k ηk (x, t )dxdt = − 1v(t )ψk ηk (x, t )dxdt .
i
Ω
(4.11)
8
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
)
–
For k = 1 and k = 2, respectively, subtracting the (4.10) from (4.11), we obtain the following equalities
2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) 1ψ1 dxdt =
1v(t )1ψ1 (x, t )η1 (x, t )dxdt + 1v(t )ψ1 (x, t )η1 (x, t )dxdt + ia3 (|ψ1∆ |2 − |ψ1 |2 )1ψ1 η1 dxdt Ω
Ω
Ω
+ ia3
ψ1∆ ψ1 1ψ 1 η1 dxdt + ia3
Ω
Ω
2 ψ 1 1ψ1 η1 dxdt
(4.12)
and
−2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) 1ψ2 dxdt =
1v(t )1ψ2 (x, t )η2 (x, t )dxdt + 1v(t )ψ2 (x, t )η2 (x, t )dxdt + ia3 (|ψ2∆ |2 − |ψ2 |2 )1ψ2 η2 dxdt Ω
Ω
Ω
+ ia3
Ω
ψ2∆ ψ2 1ψ 2 η2 dxdt + ia3
Ω
ψ2
2
1ψ2 η2 dxdt .
(4.13)
Summing (4.13) with (4.12), we get
2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt =
1v(t )ψ1 (x, t )η1 (x, t )dxdt + ia3
+ Ω
Ω
Ω
1v(t )1ψ1 (x, t )η1 (x, t )dxdt
(|ψ1∆ |2 − |ψ1 |2 )1ψ1 η1 dxdt
ψ1
2
1v(t )ψ2 (x, t )η2 (x, t )dxdt + ia3
(|ψ2∆ |2 − |ψ2 |2 )1ψ2 η2 dxdt
ψ2
+ ia3
Ω
+ Ω
+ ia3
Ω
ψ1∆ ψ1 1ψ 1 η1 dxdt + ia3
ψ2∆ ψ2 1ψ 2 η2 dxdt + ia3
Ω
Ω
Ω
2
1ψ1 η1 dxdt +
Ω
1v(t )1ψ2 (x, t )η2 (x, t )dxdt
1ψ2 η2 dxdt .
Summing the above equality with its complex conjugate, we obtain
4
Re
Ω
+2 Ω
+2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt = 2
1v(t )Re ψ1 (x, t )η1 (x, t ) dxdt + 2
Ω
1v(t )Re ψ2 (x, t )η2 (x, t ) dxdt − 2a3
− 2a3
Ω
Ω
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt
1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt
Ω
(|ψ1∆ |2 − |ψ1 |2 )Im 1ψ1 η1 dxdt
(|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt − 2a3
Ω
Im ψ1 η1 |1ψ1 |2 dxdt − 2a3
Im ψ2 η2 |1ψ2 |2 dxdt
Ω
which is equivalent to
Re ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt =
2 Ω
+ Ω
+ Ω
1v(t )Re ψ1 (x, t )η1 (x, t ) dxdt +
Ω
1v(t )Re ψ2 (x, t )η2 (x, t ) dxdt − a3
− a3
Ω
Ω
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt
1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt
Ω
(|ψ1∆ |2 − |ψ1 |2 )Im 1ψ1 η1 dxdt
(|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt − a3
Thus, Lemma 4.1 is proved.
Im ψ1 η1 |1ψ1 |2 dxdt − a3
Ω
Im ψ2 η2 |1ψ2 |2 dxdt .
Ω
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
)
–
9
Using Lemma 4.1 in (4.7), we can write the increment of Jα (v) as the following:
1Jα (v) =
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) 1v(t )dxdt
Ω
T
+ 2α
(v(t ) − w(t )) 1v(t )dt + 2α
T
dt
0
0
dv(t )
−
dw(t )
dt
d1v(t ) dt
dt + R
where
R =
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt + 1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt Ω Ω 2 2 − a3 (|ψ1∆ | − |ψ1 | )Im 1ψ1 η1 dxdt − a3 (|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt Ω Ω − a3 Im ψ1 η1 |1ψ1 |2 dxdt − a3 Im ψ2 η2 |1ψ2 |2 dxdt + ∥1ψ1 ∥2L2 (Ω ) Ω
Ω
+ ∥1ψ ∥
2 2 L2 (Ω )
+ α∥1v∥
2 W21 (0,T )
Re 1ψ1 1ψ 2 dxdt .
−2 Ω
(4.14)
Theorem 4.2. Assume that the conditions of Theorem 4.1 are fulfilled and let w ∈ W21 (0, T ) be a given function. Then, the functional Jα (v) is differentiable on the set V and the following formula is valid for its first variation:
δ Jα (v, h) =
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
Ω
+ 2α
T
(v(t ) − w(t )) h(t )dt + 2α 0
T
0
dv(t ) dt
−
dw(t )
dt
dh(t ) dt
dt
1 for any h ∈ W∞ (0, T ), where the functions ψk = ψk (x, t ) for k = 1, 2 are solutions of the initial boundary value problem (2.4)–(2.6), the functions ηk = ηk (x, t ) for k = 1, 2 are solutions of the adjoint problem (4.1)–(4.4) for any v ∈ V .
Proof. Firstly, it is proved that R = o ∥1v∥W 1 (0,T ) , where o ∥1v∥W 1 (0,T ) represents ‘‘higher-order terms’’ which go to ∞ ∞ 0 faster than ∥1v∥W 1 (0,T ) as ∥1v∥W 1 (0,T ) approaches 0, i.e. ∞
∥1v∥
∞
o ∥1v∥W 1 (0,T ) ∞
lim
1 (0,T ) W∞
∥1v∥W∞ 1 0,T ( )
→0
= 0.
From (4.14), we get
|R| ≤
|ψ1∆ |2 − |ψ1 |2 |1ψ1 | |η1 | dxdt |1ψ1 | |η1 | |1v| dxdt + |1ψ2 | |η2 | |1v| dxdt + a3 Ω Ω Ω 2 2 2 + a3 |ψ2∆ | − |ψ2 | |1ψ2 | |η2 | dxdt + a3 |1ψ1 | |ψ1 | |η1 | dxdt Ω Ω |1ψ1 | |1ψ2 | dxdt + ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) + α∥1v∥2W 1 (0,T ) . + a3 |1ψ2 |2 |ψ2 | |η2 | dxdt + 2 Ω
Ω
2
If we apply Young’s inequality to the above inequality, we obtain
|R| ≤
1 1 1 |1ψ1 |2 dxdt + |η1 |2 |1v|2 dxdt + |1ψ2 |2 dxdt + |η2 |2 |1v|2 dxdt 2 Ω 2 Ω 2 Ω 2 Ω + a3 (|ψ1∆ | + |ψ1 |) |1ψ1 |2 |η1 | dxdt + a3 (|ψ2∆ | + |ψ2 |) |1ψ2 |2 |η2 | dxdt Ω Ω 1 1 1 2 2 2 |1ψ1 | |ψ1 | dxdt + a3 |1ψ1 | |η1 |2 dxdt + a3 |1ψ2 |2 |ψ2 |2 dxdt + a3 2 2 2 Ω Ω Ω 1 |1ψ2 |2 |η2 |2 dxdt + |1ψ1 |2 dxdt + |1ψ2 |2 dxdt + a3 1
2
Ω
Ω
Ω
+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) + α∥1v∥2W 1 (0,T ) 2 5 5 1 1 2 2 |η1 |2 |1v|2 dxdt + |η2 |2 |1v|2 dxdt ≤ ∥1ψ1 ∥L2 (Ω ) + ∥1ψ2 ∥L2 (Ω ) + 2
2
2
Ω
2
Ω
10
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
1
× a3 2 1
1
Ω
(|ψ1∆ | + |ψ1 |)2 |1ψ1 |2 dxdt + a3
2 1
)
–
|1ψ1 |2 |η1 |2 dxdt
Ω
|1ψ2 |2 |η2 |2 dxdt (|ψ2∆ | + |ψ2 |)2 |1ψ2 |2 dxdt + a3 2 2 Ω Ω 1 1 1 2 2 2 |1ψ1 | |ψ1 | dxdt + a3 |1ψ1 | |η1 |2 dxdt + a3 |1ψ2 |2 |ψ2 |2 dxdt + a3 2 2 2 Ω Ω Ω 1 |1ψ2 |2 |η2 |2 dxdt + α∥1v∥2W 1 (0,T ) + a3
+ a3
2
Ω
2
which is equivalent to
|R| ≤
5
T max ∥1ψ ∥
5
1
∥1v∥2L2 (0,T ) + T max ∥1ψ ∥ + l max ∥η1 (., t )∥ 2 0≤t ≤T 2 0≤t ≤T 1 2 2 + l max ∥η2 (., t )∥L∞ (0,l) ∥1v∥L2 (0,T ) + a3 |ψ1∆ |2 + |ψ1 |2 |1ψ1 |2 dxdt 2
2 1 L2 (0,l)
0 ≤t ≤T
2
2 2 L2 (0,l)
0 ≤t ≤T
+ a3
Ω
|ψ2∆ |2 + |ψ2 |2 |1ψ2 |2 dxdt + a3
Ω
T
∥η1 (., t )∥2L∞ (0,l) ∥1ψ1 (., t )∥2L2 (0,l) dt
0
T
0
∥ψ1 (., t )∥2L∞ (0,l) ∥1ψ1 (., t )∥2L2 (0,l) dt
T
∥ψ2 (., t )∥2L∞ (0,l) ∥1ψ2 (., t )∥2L2 (0,l) dt + α∥1v∥2W 1 (0,T ) .
+ a3 2
T
2
0
1
1
∥η2 (., t )∥2L∞ (0,l) ∥1ψ2 (., t )∥2L2 (0,l) dt + a3
+ a3
2 L∞ (0,l)
(4.15)
2
0
According to known inequality in [15], we can write the inequality
1 1 ∂ψ(., t ) 2 ∥ψ(., t )∥L22 (0,l) , β2 = const . > 0 ∥ψ(., t )∥L∞ (0,l) ≤ β2 ∂x L2 (0,l) 2 2 ∂ψ(., t ) β2 2 2 ∥ψ(., t )∥L∞ (0,l) ≤ + ∥ψ(., t )∥L2 (0,l) . 2 ∂ x L2 (0,l)
(4.16)
for any t ∈ [0, T ]. Using the inequalities (3.4), (3.5) for k = 1, 2 and (4.16), the estimations (4.5), (4.6) in the inequality (4.15), we obtain
|R| ≤ c10 ∥1v∥2L2 (0,T ) + α∥1v∥2W 1 (0,T ) 2
≤ c11 ∥1v∥2W 1 (0,T ) 2
≤ c12 ∥1v∥2W 1 (0,T ) ∞
which shows that R = o ∥1v∥2
1 0,T W∞ ( )
, where the constants c10 , c11 , c12 > 0 are independent from 1v and t.
Thus, we obtain the following formula for the increment of the functional Jα (v):
1Jα (v) =
Ω
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) 1v(t )dxdt
+ 2α
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
0
dv(t ) dt
0
−
dw(t )
dt
d1v(t ) dt
dt + o ∥1v∥2W 1 (0,T ) . ∞
(4.17)
1 1 1 If we consider the function θ h ∈ W∞ (0, T ) for any 0 < θ < 1 and h ∈ W∞ (0, T ) instead of the function 1v ∈ W∞ (0, T ) in (4.17), we get
1Jα (v) = Jα (v + θ h) − Jα (v) = Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) θ h(t )dxdt Ω
+ 2α
T
(v(t ) − w(t )) θ h(t )dt + 2α 0
T
0
dv(t ) dt
−
dw(t ) dt
d (θ h(t )) dt
dt + o (θ)
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
)
–
11
which is equivalent to
Jα (v + θ h) − Jα (v) = θ
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
Ω
+ 2α
T
(v(t ) − w(t )) h(t )dt + 2α
T
0
0
dv(t ) dt
−
dw(t ) dt
dh(t ) dt
dt + o (θ )
which shows that the following formula for the first variation δ Jα (v, h) of the functional Jα (v) is valid:
δ Jα (v, h) = lim θ → 0+ + 2α
Jα (v + θ h) − Jα (v)
θ
Ω
T
(v(t ) − w(t )) h(t )dt + 2α
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
=
T
0
0
dv(t ) dt
−
dw(t ) dt
dh(t ) dt
dt .
Theorem 4.3. Suppose that the conditions of Theorem 4.2 are fulfilled and let V∗ ≡
v ∗ : v ∗ ∈ V , Jα (v ∗ ) = inf Jα (v) = Jα∗ v∈V
be the set of solutions of the variational problem (2.1), (2.4)–(2.6). Then, for any v ∗ ∈ V∗ the inequality
T ∗ ∗ ∗ ∗ ∗ ∗ Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) v(t ) − v (t ) dxdt + 2α v (t ) − w(t ) v(t ) − v ∗ (t ) dt Ω 0 T ∗ dw(t ) dv(t ) dv ∗ (t ) dv (t ) − − dt ≥ 0, ∀v ∈ V + 2α 0
dt
dt
dt
dt
is valid, where the functions ψk (x, t ) ≡ ψk (x, t , ; v ) and ηk∗ (x, t ) ≡ ηk (x, t ; v ∗ ) for k = 1, 2 are solutions of the boundary value problem (2.4)–(2.6) and adjoint problem (4.1)–(4.4) for the v ∗ ∈ V , respectively. ∗
∗
Proof. Let v ∈ V be any element, v ∗ be any element of the set V ∗ , i.e., the solution of the variational problem (2.1), (2.4)–(2.6). Since the set V is a convex set it is written that v ∗ + θ (v − v ∗ ) ∈ V for v ∗ ∈ V and any v ∈ V , ∀θ ∈ (0, 1). Therefore, according to known theorem from [19] we can write the inequality d dθ
Jα
v + θ (v − v ) ∗
∗
θ=0
= δ Jα (v ∗ , v − v ∗ ) ≥ 0,
∀v ∈ V
which is equivalent to
T ∗ v(t ) − v ∗ (t ) dxdt + 2α v (t ) − w(t ) v(t ) − v ∗ (t ) dt Ω 0 T ∗ dv(t ) dv (t ) dw(t ) dv ∗ (t ) dt ≥ 0, ∀v ∈ V . + 2α − − Re ψ1∗ (x, t )η∗1 (x, t ) + ψ2∗ (x, t )η∗2 (x, t )
0
dt
dt
dt
dt
References [1] M.A. Vorontsov, V.I. Shmalgauzen, The Principles of Adaptive Optics, Izdatel’stvo Nauka, Moscow, 1985 (in Russian). [2] A.G. Butkovskiy, Yu.I. Samoilenko, Control of Quantum-Mechanical Processes and Systems. Mathematics and its Applications (Soviet Series), Vol. 56, Kluwer Academic Publishers Group, Dordrecht, 1990. [3] L. Cardoulis, M. Cristofol, P. Gaitan, Inverse problem for the Schrödinger operator in an unbounded strip, J. Phys. Conf. Ser. 124 (1) (2008). [4] A.D. Iskenderov, On variational formulations of multidimensional inverse problems of mathematical physics, Dokl. USSR 274 (3) (1984) 531–533. [5] A.D. Iskenderov, Identification problem for the time dependent Schrödinger type equation, Proc. Lankaran State Univ. Nat. Sci. Ser. (2005) 31–53. [6] A.D. Iskenderov, G.Ya. Yagubov, A variational method for solving the inverse problem of determining the quantum-mechanical potential, Sov. Math. Dokl. 38 (1989) 637–641. (English Trans.) American Mathematical Society. [7] A.D. Iskenderov, G.Y. Yagubov, Optimal control problem with unbounded potential for multidimensional, nonlinear and nonstationary Schrödinger equation, Proc. Lankaran State Univ. Nat. Sci. ser. (2007) 3–56. [8] A.D. Iskenderov, G.Ya. Yagubov, Inverse problem of determining the quantum potential in the nonstationary Schrodinger equation with the complex coefficient in the nonlinear parts its equation, in: Inverse Problems: Modeling and Simulation–Abstracts, Fethiye, Turkey, 2008. [9] A.D. Iskenderov, G.Ya. Yagubov, N.S. Ibrahimov, M.A. Musayeva, On a problem of identification of nonlinear time-dependent Schrödinger equation with observed overall region, in: XX. International Conf. of Problems of Decision Making under Uncertainties (PDMU 2012), Taras Shevchenko National University of Kyiv, Ukraine, 2012, pp. 159–161. [10] A.D. Iskenderov, G.Ya. Yagubov, M.A. Musayeva, The Identification of Quantum Mechanics Potentials, Çaşıoğlu, Bakü, 2012. [11] N.M. Mahmudov, On an optimal control problem for the Schrödinger equation with the real coefficient, Izv. VUZOV (11) (2010) 31–40. [12] G.Ya. Yagubov, M.A. Musayeva, On the identification problem for nonlinear Schrödinger equation, Differ. Uravn. 3 (1997) 1691–1698. [13] N. Yildirim Aksoy, The variational formulation of an inverse problem for multidimensional nonlinear time dependent Schrödinger equation, J. Inverse Ill-Posed Probl. (2015) http://dx.doi.org/10.1515/jiip-2015-0029.
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[14] N. Yildirim Aksoy, B. Yildiz, H. Yetişkin, Variational problem with complex coefficient of a nonlinear Schrödinger equation, Proc. Indian Acad. Sci. Math. Sci. 122 (3) (2012) 469–484. [15] O.A. Ladyzhenskaya, V.A. Solonnikov, N.N. Ural’ceva, Linear and Quasilinear Equations of Parabolic Type, Amer. Math. Soc., Providence, RI, 1968 (English Trans.). [16] A.D. Iskenderov, G.Ya. Yagubov, N.S. Ibragimov, N.Y. Aksoy, Variation formulation of the inverse problem of determining the complex-coefficient of equation of quasioptics, Eurasian J. Math. Comput. Appl. 2 (2) (2014) 102–121. [17] M. Goebel, On existence of optimal control, Math. Nachr. 93 (1) (1979) 67–73. [18] K. Yoshida, Functional Analysis, Springer Verlag, 1967. [19] M. Minu, Mathematical Programming, Nauka, Moscow, 1990 (in Russian).
)
–
Contents lists available at ScienceDirect
Journal of Computational and Applied Mathematics journal homepage: www.elsevier.com/locate/cam
Variational method for the solution of an inverse problem N. Yildirim Aksoy Department of Mathematics, Faculty of Arts and Sciences, Kafkas University, Kars, 36100, Turkey
article
abstract
info
Article history: Received 23 September 2015
This paper presents a variational method for the solution of an inverse problem. An inverse problem of determining the unknown coefficient of nonlinear time-dependent Schrödinger equation is considered. Inverse problem is reformulated as a variational problem by means of an observation functional. The existence and uniqueness of solutions of the constituted variational problem are investigated. The differentiability of observation functional is obtained and a necessary condition for the solution of the variational problem is given. © 2016 Elsevier B.V. All rights reserved.
MSC: 35R30 35J10 49J40 Keywords: Inverse problem Schrödinger equation Variational method
1. Introduction The inverse problems of finding the quantum-mechanical potential for Schrödinger equation which describes how the state of a physical system changes in time have drawn a lot of attention in recent years. Especially, inverse problems of determining the quantum-mechanical potential for nonlinear time dependent Schrödinger equation has a great importance in this area. It is known that the inverse problems for nonlinear time-dependent Schrödinger equation usually arise in the dispersion of light beams (waves) in nonlinear medium [1]. Let us consider the wave process describing by nonlinear Schrödinger equation
∂ 2ψ ∂ψ ∂ψ + a0 2 + ia1 (x) − a2 (x)ψ + v(t )ψ + ia3 |ψ|2 ψ = f ∂t ∂x ∂x ψ(x, 0) = ϕ(x), x ∈ (0, l) ,
i
ψ(0, t ) = ψ(l, t ) = 0,
t ∈ (0, T )
(1.1) (1.2) (1.3)
where ψ = ψ(x, t ) is the wave’s complex amplitude, x and t are variables of space and time, respectively, l > 0, T > 0 are √ given numbers, 0 ≤ x ≤ l, 0 ≤ t ≤ T , i = −1 imaginary unit, a0 , a3 > 0 are given real numbers, a2 (x) is a measurable real-valued function that satisfies the condition 0 < µ0 ≤ a2 (x) ≤ µ1
for almost all x ∈ (0, l) , µ0 , µ1 = const. > 0,
(1.4)
a1 (x) is a measurable real-valued function which satisfies the condition
da1 (x) |a1 (x)| ≤ µ2 , ≤ µ3 for almost all x ∈ (0, l) , µ2 , µ3 = const. > 0, , dx a1 (0) = a1 (l) = 0 the functions f , ϕ are given complex-valued functions. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.cam.2016.01.002 0377-0427/© 2016 Elsevier B.V. All rights reserved.
(1.5)
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When we consider the inverse problem of determining the quantum mechanical potential v(t ) and the function ψ(x, t ) from the conditions (1.1)–(1.3), the conditions (1.1)–(1.3) are not sufficient to find the solution of the inverse problem. Therefore, we use an additional condition and the choosing of an additional condition is very important to find the solution by variational methods of an inverse problem. The solutions by variational methods of the inverse problems for Schrö dinger equation were previously studied in [2–14] under different additional conditions. As the different from the before studies, in the present paper, we consider an inverse problem for a nonlinear timedependent Schrödinger equation in the form (1.1) from additional condition ψ1 (x, t ) = ψ2 (x, t ) such that the functions ψ1 (x, t ) = ψ2 (x, t ) are solutions of the following first and second type boundary value problems
∂ 2 ψ1 ∂ψ1 ∂ψ1 + a0 2 + ia1 (x) − a2 (x)ψ1 + v(t )ψ1 + ia3 |ψ1 |2 ψ1 = f1 ∂t ∂x ∂x ψ1 (x, 0) = ϕ1 (x), x ∈ (0, l) ,
(1.7)
ψ1 (0, t ) = ψ1 (l, t ) = 0,
(1.8)
i
(1.6)
t ∈ (0, T )
and
∂ 2 ψ2 ∂ψ2 ∂ψ2 + a0 2 + ia1 (x) − a2 (x)ψ2 + v(t )ψ2 + ia3 |ψ2 |2 ψ2 = f2 ∂t ∂x ∂x ψ2 (x, 0) = ϕ2 (x), x ∈ (0, l) ,
(1.9)
i
∂ψ2 (l, t ) ∂ψ2 (0, t ) = = 0, ∂x ∂x
(1.10)
t ∈ (0, T )
(1.11)
respectively, where
ϕ1 ∈ W˚ 22 (0, l),
∂ϕ2 (l) ∂ϕ2 (0) = = 0, ∂x ∂x
ϕ2 ∈ W22 (0, l),
0 ,1
fk ∈ W2 (Ω ) for k = 1, 2
(1.12)
and Ωt = (0, l) × (0, t ), Ω = ΩT . Thus, we formulate the inverse problem as follows: is to determine the unknown coefficient v(t ) and the functions ψ1 (x, t ), ψ2 (x, t ) from the additional condition
ψ1 (x, t ) = ψ2 (x, t ),
( x, t ) ∈ Ω .
(1.13)
The quantum mechanical potential v(t ) is investigated on the set
V ≡
dv(t ) ≤ b1 for almost all t ∈ (0, T ) , v : v(t ) ∈ W21 (0, T ), |v(t )| ≤ b0 , dt
called as the set of admissible coefficients, which the constants b0 , b1 > 0 are given numbers. Here the Sobolev space W21 (0, T ) is a Hilbert space consisting of all the elements L2 (0, T ) having square summable generalized derivatives of the ˚ 2 (0, l), W 0,1 (Ω ) are widely defined in [15]. first order on (0, T ). The Sobolev spaces W 2 (0, l), W 2
2
2
∂ψ
Here, it should be noted that the Schrödinger equation considered in this paper contains the term ia1 (x) ∂ x as different from before studies. So, this study is more comprehensive than previous. In this paper, in Section 2, the variational formulation of the considered inverse problem and the existence of the solutions of the boundary value problems (1.6)–(1.8), (1.9)–(1.11) are given. In Section 3, the existence and uniqueness of solutions of the constituted variational problem are shown. Finally, a formula for the first variation of observation functional and a necessary condition for the solution of the variational problem are obtained in Section 4. 2. The variational formulation of the inverse problem In this section, we will present a variational formulation of the above inverse problem. If we want to formulate the above inverse problem as a variational problem, we can write it using the method in [16] as follows: to find the minimum of the observation functional Jα (v) = ∥ψ1 − ψ2 ∥2L2 (Ω ) + α∥v − w∥2W 1 (0,T )
(2.1)
2
on the set V under conditions (1.6)–(1.8) and (1.9)–(1.11), where α ≥ 0 is a given number, w ∈ W21 (0, T ) is a given element. We must firstly prove the existence of the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11) to investigate the solution of the variational problem. For this reason, let us define the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11). Under given conditions, by a solution of the problem (1.6)–(1.8), we mean a function ψ1 (x, t ) in ˚ 2 (0, l)) ∩ C 1 ([0, T ], L2 (0, l)), which satisfies Eq. (1.6) for almost all x ∈ (0, l) and any t ∈ [0, T , ] the space B1 ≡ C 0 ([0, T ], W 2 the initial condition (1.7) for almost all x ∈ (0, l) and the boundary condition (1.8) for almost all t ∈ (0, T ). Similarly, by a solution of the problem (1.9)–(1.11), we mean a function ψ2 (x, t ) in the space B2 ≡ C 0 ([0, T ], W22 (0, l)) ∩ C 1 ([0, T ], L2 (0, l)),
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which satisfies Eq. (1.9) for almost all x ∈ (0, l) and any t ∈ [0, T ], the initial condition (1.10) for almost all x ∈ (0, l) and the boundary condition (1.11) for almost all t ∈ (0, T ), where C k ([0, T ], B) is a Banach space of all B− valued, k ≥ 0 times continuously differentiable functions on [0, T ] with the norm
∥ψ∥C k ([0,T ],B) =
k m=0
m d ψ(t ) max 0≤t ≤T dt m
B
for ψ ∈ C k ([0, T ], B). We prove that the following theorem is valid for the solutions of the boundary value problems (1.6)–(1.8) and (1.9)–(1.11): Theorem 2.1. Assume that functions a1 (x), a2 (x) and ϕk (x), fk (x, t ) for k = 1, 2 satisfy the conditions (1.5), (1.4) and (1.12), respectively. Then, boundary value problems (1.6)–(1.8) and (1.9)–(1.11) have a unique solution ψ1 ∈ B1 , ψ2 ∈ B2 for any v ∈ V , respectively, and the following estimations are valid:
∥ψ1 (., t )∥
2 ˚ 2 (0,l) W 2
∂ψ1 2 + ≤ c ∥ϕ1 ∥2W˚ 2 (0,l) + ∥ϕ1 ∥6W˚ 2 (0,l) + ∥ϕ1 ∥6W˚ 1 (0,l) 1 2 2 2 ∂ t L2 (0,l) 18 2 6 + ∥ϕ1 ∥W˚ 1 (0,l) + ∥f1 ∥ 0,1 + ∥f1 ∥ 0,1 , 2
W2
(Ω )
W2
(Ω )
∂ψ2 2 ≤ c ∥ϕ2 ∥2W 2 (0,l) + ∥ϕ2 ∥6W 2 (0,l) + ∥ϕ2 ∥6W 1 (0,l) ∥ψ2 (., t )∥2W 2 (0,l) + 2 ∂t 2 2 2 2 L2 (0,l) 18 2 6 + ∥ϕ2 ∥W 1 (0,l) + ∥f2 ∥ 0,1 + ∥f2 ∥ 0,1 2
W2
(Ω )
W2
(Ω )
(2.2)
(2.3)
for any t ∈ [0, T ], where the constants c1 , c2 > 0 are independent from ϕ1 , f 1 , ϕ2 , f2 and t. The proof of Theorem 2.1 is easily obtained from Galerkin’s method. For simplicity, let us rewrite the boundary value problems (1.6)–(1.8) and (1.9)–(1.11) as the following:
∂ 2 ψk ∂ψk ∂ψk + a0 2 + ia1 (x) − a2 (x)ψk + v(t )ψk + ia3 |ψk |2 ψk = fk for k = 1, 2 ∂t ∂x ∂x ψk (x, 0) = ϕk (x), x ∈ (0, l) for k = 1, 2 i
ψ1 (0, t ) = ψ1 (l, t ) = 0,
∂ψ2 (l, t ) ∂ψ2 (0, t ) = = 0, ∂x ∂x
t ∈ (0, T ) ,
(2.4) (2.5) (2.6)
which include the two boundary value problems (1.6)–(1.8) and (1.9)–(1.11) for k = 1 and k = 2, respectively. Thus, when the conditions (1.6)–(1.8) and (1.9)–(1.11) are satisfied, the problem of finding the minimum of the functional (2.1) on the set V is called the variational problem (2.1), (2.4)–(2.6). 3. Existence and uniqueness theorems The variational problems may not always have a solution. For this, in the present section, we investigate the existence of the solution of the considered variational problem. Firstly, we show that the variational problem (2.1), (2.4)–(2.6) has a unique solution for α > 0 on a dense subset G of the space W21 (0, T ) and secondly, the variational problem has at least one solution for any α ≥ 0 on the space W21 (0, T ). We refer to the following theorem to prove the uniqueness of the solution of the variational problem (2.1), (2.4)–(2.6). Theorem 3.1 (The Corollary of Goebel’s Theorem [17]). Let X be a uniformly convex space, U be a closed bounded set on X , the functional I (v) be lower semicontinuous and lower bounded on U, α > 0 be a given number. Then, there is a dense subset G of the space X such that for any w ∈ G the functional 2 Iα (v) = I (v) + α∥v − w∥ X
takes its minimum value at a unique point on U. Lemma 3.1. The functional J0 (v) = ∥ψ1 − ψ2 ∥2L (Ω ) is continuous on the set V . 2 1 Proof. Let 1v ∈ W∞ (0, T ) be an increment of any element v ∈ V such that v + 1v ∈ V , let the functions ψk = ψk (x, t ) ≡ ψk (x, t ; v) for k = 1, 2 be the solutions of the problem (2.4)–(2.6) corresponding to v ∈ V and 1ψk = 1ψk (x, t ) ≡ ψk (x, t ; v + 1v) − ψk (x, t ; v) = ψk∆ − ψk (x, t ) for k = 1, 2, where the functions ψk∆ = ψk (x, t ; v + 1v) for
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k = 1, 2 are solutions of the problem (2.4)–(2.6) for any v + 1v . From the conditions (2.4)–(2.6) we obtain that the functions 1ψk = 1ψk (x, t ) for k = 1, 2 are solutions of the following boundary value problem: i
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk ∂t ∂ x2 ∂x + ia3 |ψk∆ |2 + |ψk |2 1ψk + ψk∆ ψk 1ψ¯ k = −1v(t )ψk , (x, t ) ∈ Ω , k = 1, 2
1ψk (x, 0) = 0,
(3.1)
x ∈ (0, l), k = 1, 2
(3.2)
∂ 1ψ2 (0, t ) ∂ 1ψ2 (l, t ) = = 0, ∂x ∂x
1ψ1 (0, t ) = 1ψ1 (l, t ) = 0,
t ∈ (0, T ) .
(3.3)
¯ k (x, t ) for To evaluate the solution of the boundary value problem (3.1)–(3.3), let us multiply both sides of Eq. (3.1) by 1ψ k = 1, 2 and integrate over Ωt . Then, integrating by part we get
∂ 1ψk 2 ∂ 1ψk + ia1 (x) ∂ 1ψk 1ψ¯ k − a2 (x) |1ψk |2 + (v(t ) + 1v(t ))|1ψk |2 1ψ¯ k − a0 ∂t ∂x ∂x Ωt 2 2 2 2 ¯ + ia3 |ψk∆ | + |ψk | |1ψk | + ia3 ψk∆ ψk (1ψk ) dxdτ = − 1v(t )ψk 1ψ¯ k dxdτ i
Ωt
for k = 1, 2. If we subtract the complex conjugate of the above equality from itself, we obtain
∂ 1ψ k ∂ ∂ 1ψk 2 |1ψk | dxdτ + 1ψ¯ k + a1 (x) 1ψk dxdτ a1 (x) ∂x ∂x Ωt Ωt ∂ t 2 2 2 + 2a3 |ψk∆ | + |ψk | |1ψk | dxdτ = −2 Im 1vψk 1ψ k dxdτ − 2a3 Ωt
Ωt
for t ∈ [0, T ] and k = 1, 2. If we add the term conditions (1.5), (3.2), we obtain
da1 (x) dx
Ωt
Re ψk∆ ψk 1ψ k
2
dxdτ
|1ψk |2 for k = 1, 2 to both sides of the above equality and use the
∥1ψk (., t )∥ + 2a3 (|ψk∆ | + |ψk | )|1ψk | dxdτ = −2 Im 1vψk 1ψ k dxdτ Ωt Ωt 2 da1 (x) 2 − 2a3 Re ψk∆ ψk 1ψ k dxdτ + |1ψk | dxdτ 2 L2 (0,l)
2
2
Ωt
2
dx
Ωt
and by Young’s inequality and the condition (1.5)
∥1ψk (., t )∥2L2 (0,l) + a3 (|ψk∆ |2 + |ψk |2 )|1ψk |2 dxdτ Ωt |1v| |ψk | |1ψk | dxdτ + µ3 |1ψk |2 dxdτ ≤2 Ωt
Ωt
t
l
|1v|2 |ψk |2 dxdτ +
≤ 0
0
Ωt
|1ψk |2 dxdτ + µ3
t
= 0
|1v|2 ∥ψk (., t )∥2L2 (0,l) dτ + (1 + µ3 )
≤ max ∥ψk (., t )∥2L2 (0,l) 0≤t ≤T
Ωt
|1ψk |2 dxdτ
t
0
∥1ψk (., t )∥2L2 (0,l) dτ
T
|1v|2 dτ + (1 + µ3 ) 0
t
0
∥1ψk (., t )∥2L2 (0,l) dτ
for any t ∈ [0, T ] and k = 1, 2. In the above inequality, if we use the estimations (2.2) and (2.3) we get
∥1ψk (., t )∥2L2 (0,l) + a3
Ωt
(|ψk∆ |2 + |ψk |2 )|1ψk |2 dxdτ ≤ c3 ∥1v∥2L2 (0,T ) + (1 + µ3 )
t
0
∥1ψk (., t )∥2L2 (0,l) dτ .
(3.4)
Thus, ıf we take into account a3 > 0, we have the inequality
∥1ψk (., t )∥2L2 (0,l) ≤ c3 ∥1v∥2L2 (0,T ) + (1 + µ3 )
t
0
∥1ψk (., t )∥2L2 (0,l) dτ ,
for any t ∈ [0, T ] and k = 1, 2, where the constant c3 > 0 is independent from 1v and t. Applying the Gronwall’s inequality to the above inequality, we get
∥1ψk (., t )∥2L2 (0,l) ≤ c4 ∥1v∥2L2 (0,T ) for any t ∈ [0, T ], k = 1, 2
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and thus
∥1ψk (., t )∥2L2 (0,l) ≤ c4 ∥1v∥2L2 (0,T ) ≤ c5 ∥1v∥2W 1 (0,T ) ≤ c6 ∥1v∥2W 1 (0,T ) , ∞
2
k = 1, 2
(3.5)
for any t ∈ [0, T ], where the constants c4 , c5 , c6 > 0 are independent from 1v and t. Now, let us find the increment of the functional J0 (v) for any v ∈ V . If we use the definition of the functional J0 (v), we obtain
|ψ1∆ − ψ2∆ |2 dxdt − |ψ1 − ψ2 |2 dxdt 1J0 (v) = J0 (v + 1v) − J0 (v) = Ω Ω = (ψ1∆ − ψ2∆ ) ψ 1∆ − ψ 2∆ − (ψ1 − ψ2 ) ψ 1 − ψ 2 dxdt Ω = (ψ1 − ψ2 ) 1ψ 1 − 1ψ 2 + (1ψ1 − 1ψ2 ) ψ 1 − ψ 2 + |1ψ1 |2 + |1ψ2 |2 − 2Re 1ψ1 1ψ 2 dxdt Ω =2 Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t ) dxdt Ω
+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) − 2
Re 1ψ1 1ψ 2 dxdt .
Ω
(3.6)
Applying the Cauchy–Schwarz inequality to the equality (3.6), we get
|1J0 (v)| ≤ 2 ∥ψ1 ∥L2 (Ω ) ∥1ψ1 ∥L2 (Ω ) + 2 ∥ψ1 ∥L2 (Ω ) ∥1ψ2 ∥L2 (Ω ) + 2 ∥ψ2 ∥L2 (Ω ) ∥1ψ1 ∥L2 (Ω ) + 2 ∥ψ2 ∥L2 (Ω ) ∥1ψ2 ∥L2 (Ω ) + 2 ∥1ψ1 ∥2L2 (Ω ) + 2 ∥1ψ2 ∥2L2 (Ω ) . Using the estimations (2.2), (2.3), (3.5) in the above inequality, we obtain the following inequality for the increment of the functional J0 (v):
2 |J0 (v + 1v) − J0 (v)| ≤ c7 ∥1v∥W∞ 1 (0,T ) + ∥1v∥ 1 W (0,T ) ∞
for any v ∈ V , where the constant c7 > 0 is independent from 1v . Thus, since 1J0 (v) → 0 for ∥1v∥W 1 (0,T ) → 0 and for ∞ any v ∈ V , we can easily say that the functional J0 (v) is continuous on the set V . Now, let us give the following theorem stated the uniqueness of the solution of the variational problem (2.1), (2.4)–(2.6) on a dense subset G of the space W21 (0, T ). Theorem 3.2. Assume that the functions a1 (x), a2 (x) and ϕk (x), fk (x, t ) for k = 1, 2 satisfies the conditions (1.5), (1.4), (1.12), respectively, and w ∈ W21 (0, T ) is a given function. Then, there is a dense subset G of the space W21 (0, T ) such that the variational problem (2.1), (2.4)–(2.6) has a unique solution for any w ∈ G and α > 0. Proof. Since the functional J0 (v) is continuous on the set V , it is a lower semicontinuous functional. Also, since J0 (v) ≥ 0 for any v ∈ V , the functional J0 (v) is lower bounded. Additionally, the set V is a closed, bounded and convex set of the uniformly convex space W21 (0, T ) [18]. Thus, from Theorem 3.1, it follows that there is a dense subset G of the space W21 (0, T ) such that the variational problem (2.1), (2.4)–(2.6) has a unique solution for any w ∈ G and α > 0. The next theorem states that the variational problem (2.1), (2.4)–(2.6) has at least one solution for any α ≥ 0 on the space W21 (0, T ). Theorem 3.3. Suppose that α ≥ 0 and the conditions of Theorem 2.1 hold and w ∈ W21 (0, T ) is a given function. Then, the variational problem (2.1), (2.4)–(2.6) has at least one solution. We can easily make the proof of Theorem 3.3 by using the method in [13]. 4. Differentiability of the functional In the present section, constituting an adjoint problem we obtain the first variation of the functional Jα (v) by means of the adjoint problem. Also, a necessary condition for the solution of the variational problem (2.1), (2.4)–(2.6) is given. Firstly, we reformulate the variational problem (2.1), (2.4)–(2.6) by using Lagrange multiplier functions as follows: Consider the minimization problem L (v, ψ1 , ψ2 , η1 , η2 ) → inf
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where L (v, ψ1 , ψ2 , η1 , η2 ) = Jα (v)
∂ψ1 ∂ 2 ψ1 ∂ψ1 2 + i + a0 2 + ia1 (x) − a2 (x)ψ1 + v(t )ψ1 + ia3 |ψ1 | ψ1 − f1 η1 dxdt 2 Ω ∂t ∂x ∂x 1 ∂ψ 1 ∂ 2ψ 1 ∂ψ 1 2 + −i + a0 − ia1 (x) − a2 (x)ψ 1 + v(t )ψ 1 − ia3 |ψ1 | ψ 1 − f 1 η1 dxdt 2 Ω ∂t ∂ x2 ∂x 1 ∂ψ2 ∂ 2 ψ2 ∂ψ2 + i + a0 2 + ia1 (x) − a2 (x)ψ2 + v(t )ψ2 + ia3 |ψ2 |2 ψ2 − f2 η2 dxdt 2 Ω ∂t ∂x ∂x ∂ψ 2 ∂ 2ψ 2 1 ∂ψ 2 2 −i + a0 − ia1 (x) − a2 (x)ψ 2 + v(t )ψ 2 − ia3 |ψ2 | ψ 2 − f 2 η2 dxdt , + 2 Ω ∂t ∂ x2 ∂x 1
the functions ηk = ηk (x, t ) = ηk (x, t ; v) for k = 1, 2 are Lagrange multipliers and the functions ψk = ψk (x, t ) for k = 1, 2 are solutions of the boundary value problem (2.4)–(2.6). Thus, from the stationarity condition of the Lagrange functional L (v, ψ1 , ψ2 , η1 , η2 ), we obtain the following adjoint problem: i
∂ ∂ 2 ηk ∂ηk + a0 2 + i (a1 (x)ηk ) − a2 (x)ηk + v(t )ηk − 2ia3 |ψk |2 ηk + ia3 ψk2 η¯ k ∂t ∂x ∂x = 2(−1)k (ψ1 (x, t ) − ψ2 (x, t )) ,
ηk (x, T ) = 0,
(x, t ) ∈ Ω , k = 1, 2,
(4.1)
x ∈ (0, l), k = 1, 2
η1 (0, t ) = η1 (l, t ) = 0,
(4.2)
t ∈ (0, T )
∂η2 ∂η2 (0, t ) = (l, t ) = 0, ∂x ∂x
(4.3)
t ∈ (0, T )
(4.4)
where the functions ψk = ψk (x, t ) ≡ ψk (x, t ; v) for k = 1, 2 are solutions of boundary value problem (2.4)–(2.6) for any v ∈ V . As seen, adjoint problem (4.1)–(4.4) includes the two boundary value problems according to the functions η1 and η2 for k = 1, 2. If we apply the transform τ = T − t to adjoint problem, it is seen that the adjoint problem is an initial boundary value problem in the form of problem (2.4)–(2.6). Under given conditions, by a solution of the adjoint problem, we mean two functions η1 (x, t ) in the space B1 and η2 (x, t ) in the space B2 , which satisfy Eq. (4.1) for almost all x ∈ (0, l) and any t ∈ [0, T ], the condition (4.2) for almost all x ∈ (0, l) and the boundary conditions (4.3) and (4.4) for almost all t ∈ (0, T ), respectively. Therefore, we can write the following theorem for the solution of adjoint problem (4.1)–(4.4): Theorem 4.1. Assume that the conditions of Theorem 2.1 hold. Then, the adjoint problem (4.1)–(4.4) has a unique solution η1 ∈ B1 , η2 ∈ B2 for any v ∈ V and the following estimations are valid for the solution:
∥η1 (., t )∥
2 ˚ 2 (0,l) W 2
∂η1 (., t ) 2 + ∂t
∥η2 (., t )∥
2 W22 (0,l)
∂η2 (., t ) 2 + ≤ c9 ∥ψ1 − ψ2 ∥W 0,1 (Ω ) 2 ∂ t L2 (0,l)
L2 (0,l)
≤ c8 ∥ψ1 − ψ2 ∥W 0,1 (Ω ) ,
(4.5)
2
(4.6)
for any t ∈ [0, T ], where the constants c8 , c9 > 0 are independent from t. We can easily prove Theorem 4.1 as similar to the proof of Theorem 2.1 by using Galerkin’s method. 1 Now, let us find the increment of the functional Jα (v) for ∀v ∈ V . Let the function 1v ∈ W∞ (0, T ) be an increment given to any v ∈ V such that v + 1v ∈ V . Then, using the formula (2.1) and (3.6) we can write the increment of the functional Jα (v) for any v ∈ V as the following:
1Jα (v) = Jα (v + 1v) − Jα (v) = 1J0 (v) + α∥v + 1v − w∥2W 1 (0,T ) − α∥v − w∥2W 1 (0,T ) 2
= 1J0 (v) + 2α
0
=2 Ω
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
0
2
dv dt
−
dw dt
Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t ) dxdt
d(1v) dt
dt + α∥1v∥2W 1 (0,T ) 2
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+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) − 2 + 2α
Ω
–
7
Re 1ψ1 1ψ 2 dxdt
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
dv(t ) dt
0
0
)
−
dw(t ) dt
d1v(t ) dt
dt + α∥1v∥2W 1 (0,T )
(4.7)
2
where the functions 1ψk = 1ψk (x, t ) ≡ ψk (x, t ; v + 1v) − ψk (x, t ; v) for k = 1, 2 are solutions of the problem (3.1)–(3.3) for v ∈ V . Lemma 4.1.
Re (ψ1 (x, t ) − ψ2 (x, t )) 1ψ 1 (x, t ) − 1ψ 2 (x, t )
2 Ω
=
dxdt
1v(t )Re(1ψ1 (x, t )η1 (x, t ))dxdt + 1v(t )Re(ψ1 (x, t )η1 (x, t ))dxdt Ω 1v(t )Re(1ψ2 (x, t )η2 (x, t ))dxdt + 1v(t )Re(ψ2 (x, t )η2 (x, t ))dxdt + Ω Ω − a3 (|ψ1∆ |2 − |ψ1 |2 )Im(1ψ1 η¯ 1 )dxdt − a3 (|ψ2∆ |2 − |ψ2 |2 )Im(1ψ2 η¯ 2 )dxdt Ω Ω 2 2 |1ψ2 | Im (ψ2 η¯ 2 ) dxdt . − a3 |1ψ1 | Im (ψ1 η¯ 1 ) dxdt − a3 Ω
Ω
Ω
Proof. It is clear that the functions 1ψk = ψk (x, t ; v + 1v) − ψ(x, t ; v) for k = 1, 2 satisfy integral identities
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk 2 ∂ t ∂ x ∂x Ω 2 2 ¯ 1v(t )ψk φ k (x, t )dxdt + ia3 |ψk∆ | + |ψk | 1ψk + ψk∆ ψk 1ψk φ k (x, t )dxdt = −
i
(4.8)
Ω
for any functions φk = φk (x, t ) ∈ L2 (Ω ), k = 1, 2 and the conditions (3.2), (3.3). Also, the functions ηk = ηk (x, t ) for k = 1, 2 satisfy the identities
∂ 2 ηk ∂ ∂ηk + a0 2 + i (a1 (x)ηk ) − a2 (x)ηk + v(t )ηk − 2ia3 |ψk |2 ηk + ia3 ψk2 ηk ξ k dxdt ∂t ∂x ∂x Ω = 2(−1)k (ψ1 (x, t ) − ψ2 (x, t )) ξ k dxdt , (x, t ) ∈ Ω
i
(4.9)
Ω
for any functions ξk = ξk (x, t ) ∈ L2 (Ω ), k = 1, 2. Let us put the functions 1ψ1 ∈ B1 , 1ψ2 ∈ B2 instead of the test functions ξk for k = 1, 2 in identities (4.9), respectively. Later, if we apply by part integration formula to obtained identities, we get
∂ 1ψ k ∂ 2 1ψ k ∂ 1ψ k 2 −i − ia1 (x) + a0 − a2 (x)1ψ k + v(t )1ψ k − 2ia3 |ψk | 1ψ k ηk dxdt ∂t ∂ x2 ∂x Ω 2 k + ia3 ψk η¯ k 1ψ k dxdt = 2(−1) (ψ1 (x, t ) − ψ2 (x, t )) 1ψ k dxdt Ω
Ω
and its complex conjugate
∂ 1ψk ∂ 2 1ψk ∂ 1ψk 2 ( x ) ( x ) 1 ψ + v( t ) 1 ψ + 2ia |ψ | 1 ψ + a0 + ia − a 1 2 k k 3 k k η k dxdt ∂t ∂ x2 ∂x Ω 2 − ia3 ψ k ηk 1ψk dxdt = 2(−1)k ψ 1 (x, t ) − ψ 2 (x, t ) 1ψk dxdt .
i
Ω
(4.10)
Ω
Also, if we put the functions η1 (x, t ) ∈ B1 , η2 (x, t ) ∈ B2 instead of functions φ1 (x, t ), φ2 (x, t ) in (4.8), respectively, we have
∂ 1ψk ∂ 2 1ψk ∂ 1ψk + a0 + ia1 (x) − a2 (x)1ψk + (v + 1v)1ψk 2 ∂t ∂x ∂x Ω + ia3 |ψk∆ |2 + |ψk |2 1ψk + ψk∆ ψk 1ψ¯ k ηk (x, t )dxdt = − 1v(t )ψk ηk (x, t )dxdt .
i
Ω
(4.11)
8
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)
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For k = 1 and k = 2, respectively, subtracting the (4.10) from (4.11), we obtain the following equalities
2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) 1ψ1 dxdt =
1v(t )1ψ1 (x, t )η1 (x, t )dxdt + 1v(t )ψ1 (x, t )η1 (x, t )dxdt + ia3 (|ψ1∆ |2 − |ψ1 |2 )1ψ1 η1 dxdt Ω
Ω
Ω
+ ia3
ψ1∆ ψ1 1ψ 1 η1 dxdt + ia3
Ω
Ω
2 ψ 1 1ψ1 η1 dxdt
(4.12)
and
−2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) 1ψ2 dxdt =
1v(t )1ψ2 (x, t )η2 (x, t )dxdt + 1v(t )ψ2 (x, t )η2 (x, t )dxdt + ia3 (|ψ2∆ |2 − |ψ2 |2 )1ψ2 η2 dxdt Ω
Ω
Ω
+ ia3
Ω
ψ2∆ ψ2 1ψ 2 η2 dxdt + ia3
Ω
ψ2
2
1ψ2 η2 dxdt .
(4.13)
Summing (4.13) with (4.12), we get
2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt =
1v(t )ψ1 (x, t )η1 (x, t )dxdt + ia3
+ Ω
Ω
Ω
1v(t )1ψ1 (x, t )η1 (x, t )dxdt
(|ψ1∆ |2 − |ψ1 |2 )1ψ1 η1 dxdt
ψ1
2
1v(t )ψ2 (x, t )η2 (x, t )dxdt + ia3
(|ψ2∆ |2 − |ψ2 |2 )1ψ2 η2 dxdt
ψ2
+ ia3
Ω
+ Ω
+ ia3
Ω
ψ1∆ ψ1 1ψ 1 η1 dxdt + ia3
ψ2∆ ψ2 1ψ 2 η2 dxdt + ia3
Ω
Ω
Ω
2
1ψ1 η1 dxdt +
Ω
1v(t )1ψ2 (x, t )η2 (x, t )dxdt
1ψ2 η2 dxdt .
Summing the above equality with its complex conjugate, we obtain
4
Re
Ω
+2 Ω
+2 Ω
ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt = 2
1v(t )Re ψ1 (x, t )η1 (x, t ) dxdt + 2
Ω
1v(t )Re ψ2 (x, t )η2 (x, t ) dxdt − 2a3
− 2a3
Ω
Ω
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt
1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt
Ω
(|ψ1∆ |2 − |ψ1 |2 )Im 1ψ1 η1 dxdt
(|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt − 2a3
Ω
Im ψ1 η1 |1ψ1 |2 dxdt − 2a3
Im ψ2 η2 |1ψ2 |2 dxdt
Ω
which is equivalent to
Re ψ 1 (x, t ) − ψ 2 (x, t ) (1ψ1 − 1ψ2 ) dxdt =
2 Ω
+ Ω
+ Ω
1v(t )Re ψ1 (x, t )η1 (x, t ) dxdt +
Ω
1v(t )Re ψ2 (x, t )η2 (x, t ) dxdt − a3
− a3
Ω
Ω
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt
1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt
Ω
(|ψ1∆ |2 − |ψ1 |2 )Im 1ψ1 η1 dxdt
(|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt − a3
Thus, Lemma 4.1 is proved.
Im ψ1 η1 |1ψ1 |2 dxdt − a3
Ω
Im ψ2 η2 |1ψ2 |2 dxdt .
Ω
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
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Using Lemma 4.1 in (4.7), we can write the increment of Jα (v) as the following:
1Jα (v) =
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) 1v(t )dxdt
Ω
T
+ 2α
(v(t ) − w(t )) 1v(t )dt + 2α
T
dt
0
0
dv(t )
−
dw(t )
dt
d1v(t ) dt
dt + R
where
R =
1v(t )Re 1ψ1 (x, t )η1 (x, t ) dxdt + 1v(t )Re 1ψ2 (x, t )η2 (x, t ) dxdt Ω Ω 2 2 − a3 (|ψ1∆ | − |ψ1 | )Im 1ψ1 η1 dxdt − a3 (|ψ2∆ |2 − |ψ2 |2 )Im 1ψ2 η2 dxdt Ω Ω − a3 Im ψ1 η1 |1ψ1 |2 dxdt − a3 Im ψ2 η2 |1ψ2 |2 dxdt + ∥1ψ1 ∥2L2 (Ω ) Ω
Ω
+ ∥1ψ ∥
2 2 L2 (Ω )
+ α∥1v∥
2 W21 (0,T )
Re 1ψ1 1ψ 2 dxdt .
−2 Ω
(4.14)
Theorem 4.2. Assume that the conditions of Theorem 4.1 are fulfilled and let w ∈ W21 (0, T ) be a given function. Then, the functional Jα (v) is differentiable on the set V and the following formula is valid for its first variation:
δ Jα (v, h) =
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
Ω
+ 2α
T
(v(t ) − w(t )) h(t )dt + 2α 0
T
0
dv(t ) dt
−
dw(t )
dt
dh(t ) dt
dt
1 for any h ∈ W∞ (0, T ), where the functions ψk = ψk (x, t ) for k = 1, 2 are solutions of the initial boundary value problem (2.4)–(2.6), the functions ηk = ηk (x, t ) for k = 1, 2 are solutions of the adjoint problem (4.1)–(4.4) for any v ∈ V .
Proof. Firstly, it is proved that R = o ∥1v∥W 1 (0,T ) , where o ∥1v∥W 1 (0,T ) represents ‘‘higher-order terms’’ which go to ∞ ∞ 0 faster than ∥1v∥W 1 (0,T ) as ∥1v∥W 1 (0,T ) approaches 0, i.e. ∞
∥1v∥
∞
o ∥1v∥W 1 (0,T ) ∞
lim
1 (0,T ) W∞
∥1v∥W∞ 1 0,T ( )
→0
= 0.
From (4.14), we get
|R| ≤
|ψ1∆ |2 − |ψ1 |2 |1ψ1 | |η1 | dxdt |1ψ1 | |η1 | |1v| dxdt + |1ψ2 | |η2 | |1v| dxdt + a3 Ω Ω Ω 2 2 2 + a3 |ψ2∆ | − |ψ2 | |1ψ2 | |η2 | dxdt + a3 |1ψ1 | |ψ1 | |η1 | dxdt Ω Ω |1ψ1 | |1ψ2 | dxdt + ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) + α∥1v∥2W 1 (0,T ) . + a3 |1ψ2 |2 |ψ2 | |η2 | dxdt + 2 Ω
Ω
2
If we apply Young’s inequality to the above inequality, we obtain
|R| ≤
1 1 1 |1ψ1 |2 dxdt + |η1 |2 |1v|2 dxdt + |1ψ2 |2 dxdt + |η2 |2 |1v|2 dxdt 2 Ω 2 Ω 2 Ω 2 Ω + a3 (|ψ1∆ | + |ψ1 |) |1ψ1 |2 |η1 | dxdt + a3 (|ψ2∆ | + |ψ2 |) |1ψ2 |2 |η2 | dxdt Ω Ω 1 1 1 2 2 2 |1ψ1 | |ψ1 | dxdt + a3 |1ψ1 | |η1 |2 dxdt + a3 |1ψ2 |2 |ψ2 |2 dxdt + a3 2 2 2 Ω Ω Ω 1 |1ψ2 |2 |η2 |2 dxdt + |1ψ1 |2 dxdt + |1ψ2 |2 dxdt + a3 1
2
Ω
Ω
Ω
+ ∥1ψ1 ∥2L2 (Ω ) + ∥1ψ2 ∥2L2 (Ω ) + α∥1v∥2W 1 (0,T ) 2 5 5 1 1 2 2 |η1 |2 |1v|2 dxdt + |η2 |2 |1v|2 dxdt ≤ ∥1ψ1 ∥L2 (Ω ) + ∥1ψ2 ∥L2 (Ω ) + 2
2
2
Ω
2
Ω
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N.Y. Aksoy / Journal of Computational and Applied Mathematics (
1
× a3 2 1
1
Ω
(|ψ1∆ | + |ψ1 |)2 |1ψ1 |2 dxdt + a3
2 1
)
–
|1ψ1 |2 |η1 |2 dxdt
Ω
|1ψ2 |2 |η2 |2 dxdt (|ψ2∆ | + |ψ2 |)2 |1ψ2 |2 dxdt + a3 2 2 Ω Ω 1 1 1 2 2 2 |1ψ1 | |ψ1 | dxdt + a3 |1ψ1 | |η1 |2 dxdt + a3 |1ψ2 |2 |ψ2 |2 dxdt + a3 2 2 2 Ω Ω Ω 1 |1ψ2 |2 |η2 |2 dxdt + α∥1v∥2W 1 (0,T ) + a3
+ a3
2
Ω
2
which is equivalent to
|R| ≤
5
T max ∥1ψ ∥
5
1
∥1v∥2L2 (0,T ) + T max ∥1ψ ∥ + l max ∥η1 (., t )∥ 2 0≤t ≤T 2 0≤t ≤T 1 2 2 + l max ∥η2 (., t )∥L∞ (0,l) ∥1v∥L2 (0,T ) + a3 |ψ1∆ |2 + |ψ1 |2 |1ψ1 |2 dxdt 2
2 1 L2 (0,l)
0 ≤t ≤T
2
2 2 L2 (0,l)
0 ≤t ≤T
+ a3
Ω
|ψ2∆ |2 + |ψ2 |2 |1ψ2 |2 dxdt + a3
Ω
T
∥η1 (., t )∥2L∞ (0,l) ∥1ψ1 (., t )∥2L2 (0,l) dt
0
T
0
∥ψ1 (., t )∥2L∞ (0,l) ∥1ψ1 (., t )∥2L2 (0,l) dt
T
∥ψ2 (., t )∥2L∞ (0,l) ∥1ψ2 (., t )∥2L2 (0,l) dt + α∥1v∥2W 1 (0,T ) .
+ a3 2
T
2
0
1
1
∥η2 (., t )∥2L∞ (0,l) ∥1ψ2 (., t )∥2L2 (0,l) dt + a3
+ a3
2 L∞ (0,l)
(4.15)
2
0
According to known inequality in [15], we can write the inequality
1 1 ∂ψ(., t ) 2 ∥ψ(., t )∥L22 (0,l) , β2 = const . > 0 ∥ψ(., t )∥L∞ (0,l) ≤ β2 ∂x L2 (0,l) 2 2 ∂ψ(., t ) β2 2 2 ∥ψ(., t )∥L∞ (0,l) ≤ + ∥ψ(., t )∥L2 (0,l) . 2 ∂ x L2 (0,l)
(4.16)
for any t ∈ [0, T ]. Using the inequalities (3.4), (3.5) for k = 1, 2 and (4.16), the estimations (4.5), (4.6) in the inequality (4.15), we obtain
|R| ≤ c10 ∥1v∥2L2 (0,T ) + α∥1v∥2W 1 (0,T ) 2
≤ c11 ∥1v∥2W 1 (0,T ) 2
≤ c12 ∥1v∥2W 1 (0,T ) ∞
which shows that R = o ∥1v∥2
1 0,T W∞ ( )
, where the constants c10 , c11 , c12 > 0 are independent from 1v and t.
Thus, we obtain the following formula for the increment of the functional Jα (v):
1Jα (v) =
Ω
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) 1v(t )dxdt
+ 2α
T
(v(t ) − w(t )) 1v(t )dt + 2α
T
0
dv(t ) dt
0
−
dw(t )
dt
d1v(t ) dt
dt + o ∥1v∥2W 1 (0,T ) . ∞
(4.17)
1 1 1 If we consider the function θ h ∈ W∞ (0, T ) for any 0 < θ < 1 and h ∈ W∞ (0, T ) instead of the function 1v ∈ W∞ (0, T ) in (4.17), we get
1Jα (v) = Jα (v + θ h) − Jα (v) = Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) θ h(t )dxdt Ω
+ 2α
T
(v(t ) − w(t )) θ h(t )dt + 2α 0
T
0
dv(t ) dt
−
dw(t ) dt
d (θ h(t )) dt
dt + o (θ)
N.Y. Aksoy / Journal of Computational and Applied Mathematics (
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which is equivalent to
Jα (v + θ h) − Jα (v) = θ
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
Ω
+ 2α
T
(v(t ) − w(t )) h(t )dt + 2α
T
0
0
dv(t ) dt
−
dw(t ) dt
dh(t ) dt
dt + o (θ )
which shows that the following formula for the first variation δ Jα (v, h) of the functional Jα (v) is valid:
δ Jα (v, h) = lim θ → 0+ + 2α
Jα (v + θ h) − Jα (v)
θ
Ω
T
(v(t ) − w(t )) h(t )dt + 2α
Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) h(t )dxdt
=
T
0
0
dv(t ) dt
−
dw(t ) dt
dh(t ) dt
dt .
Theorem 4.3. Suppose that the conditions of Theorem 4.2 are fulfilled and let V∗ ≡
v ∗ : v ∗ ∈ V , Jα (v ∗ ) = inf Jα (v) = Jα∗ v∈V
be the set of solutions of the variational problem (2.1), (2.4)–(2.6). Then, for any v ∗ ∈ V∗ the inequality
T ∗ ∗ ∗ ∗ ∗ ∗ Re ψ1 (x, t )η1 (x, t ) + ψ2 (x, t )η2 (x, t ) v(t ) − v (t ) dxdt + 2α v (t ) − w(t ) v(t ) − v ∗ (t ) dt Ω 0 T ∗ dw(t ) dv(t ) dv ∗ (t ) dv (t ) − − dt ≥ 0, ∀v ∈ V + 2α 0
dt
dt
dt
dt
is valid, where the functions ψk (x, t ) ≡ ψk (x, t , ; v ) and ηk∗ (x, t ) ≡ ηk (x, t ; v ∗ ) for k = 1, 2 are solutions of the boundary value problem (2.4)–(2.6) and adjoint problem (4.1)–(4.4) for the v ∗ ∈ V , respectively. ∗
∗
Proof. Let v ∈ V be any element, v ∗ be any element of the set V ∗ , i.e., the solution of the variational problem (2.1), (2.4)–(2.6). Since the set V is a convex set it is written that v ∗ + θ (v − v ∗ ) ∈ V for v ∗ ∈ V and any v ∈ V , ∀θ ∈ (0, 1). Therefore, according to known theorem from [19] we can write the inequality d dθ
Jα
v + θ (v − v ) ∗
∗
θ=0
= δ Jα (v ∗ , v − v ∗ ) ≥ 0,
∀v ∈ V
which is equivalent to
T ∗ v(t ) − v ∗ (t ) dxdt + 2α v (t ) − w(t ) v(t ) − v ∗ (t ) dt Ω 0 T ∗ dv(t ) dv (t ) dw(t ) dv ∗ (t ) dt ≥ 0, ∀v ∈ V . + 2α − − Re ψ1∗ (x, t )η∗1 (x, t ) + ψ2∗ (x, t )η∗2 (x, t )
0
dt
dt
dt
dt
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