On a Class of Inverse Problems for a Heat Equation with Involution Perturbation

On a class of inverse problems for a heat equation with involution perturbation

Nasser Al-Salti∗†, Mokhtar Kirane‡ and Berikbol T. Torebek§

Abstract A class of inverse problems for a heat equation with involution perturbation is considered using four diﬀerent boundary conditions, namely, Dirichlet, Neumann, periodic and anti-periodic boundary conditions. Proved theorems on existence and uniqueness of solutions to these problems are presented. Solutions are obtained in the form of series expansion using a set of appropriate orthogonal basis for each problem. Convergence of the obtained solutions is also discussed.

Keywords: Inverse Problems, Heat Equation, Involution Perturbation. 2000 AMS Classiﬁcation: 35R30; 35K05; 39B52.

1. Introduction Differential equations with modified arguments are equations in which the unknown function and its derivatives are evaluated with modifications of time or space variables; such equations are called in general functional differential equations. Among such equations, one can single out equations with involutions [?]. 1.1. Definition. [?, ?] A function α(x) 6≡ x that maps a set of real numbers, Γ, onto itself and satisfies on Γ the condition α (α(x)) = x, or α−1(x) = α(x) is called an involution on Γ. Equations containing involution are equations with an alternating deviation (at x∗ < x being equations with advanced, and at x∗ > x being equations with delay, where x∗ is a fixed point of the mapping α(x)). Equations with involutions have been studied by many researchers, for example, Ashyralyev [?, ?], Babbage [?], Przewoerska-Rolewicz [?, ?, ?, ?, ?, ?, ?], Aftabizadeh and Col. [?], Andreev [?, ?], Burlutskayaa and Col. [?], Gupta [?, ?, ?], Kirane [?], Watkins [?], and Wiener [?, ?, ?, ?]. Spectral problems and inverse problems for equations with involutions have received a lot of attention as well, see for example, [?, ?, ?, ?, ?], and equations with a delay in the space variable have been the subject of many research papers, see for example, [?, ?].

∗Department of Mathematics and Statistics, Sultan Qaboos University, PO Box 36, Al-Khodh, PC 123, Muscat, Oman, Email: [email protected] †Corresponding Author. ‡LaSIE, Facultédes Sciences et Technologies, Universitéde La Rochelle, Avenue Michel Crépeau, 17000, La Rochelle, France, Email: [email protected] §Institute of Mathematics and Mathematical Modeling, 050010, Pushkin st., 125, Almaty, Kazakhstan, Email: [email protected] 2

Furthermore, for the equations containing transformation of the spatial variable in the diﬀusion term, we can cite the talk of Cabada and Tojo [?], where they gave an example that describes a concrete situation in physics: Consider a metal wire around a thin sheet of insulating material in a way that some parts overlap some others as shown in Figure ??.

Figure 1. An application of heat equation with involution [?]

Assuming that the position y = 0 is the lowest of the wire, and the insulation goes up to the left at −Y and to the right up to Y. For the proximity of two sections of wires they added the third term with modifications on the spatial variable to the right-hand side of the heat equation with respect to the wire: ∂T ∂2T ∂2T (y, t) = a (y, t) + b (−y, t). ∂t ∂y2 ∂y2 Such equations have also a purely theoretical value. For general facts about partial functional differential equations and for properties of equations with involutions in particular, we refer the reader to the books of Skubachevskii [?], Wu [?] and Cabada and Tojo [?]. In this paper, we consider inverse problems for a heat equation with involution using four different boundary conditions. We seek formal solutions to these problems in a form of series expansions using orthogonal basis obtained by separation of variables and we also examine the convergence of the obtained series solutions. The main results on existence and uniqueness are formulated in four theorems in the last section of this paper along with an illustrating example. Concerning inverse problems for heat equations, some recent works have been implemented by Kaliev [?, ?], Sadybekov [?, ?] and Kirane [?, ?].

2. Statements of Problems Consider the heat equation

(2.1) ut (x, t) − uxx (x, t) + εuxx (−x, t) = f (x) , (x, t) ∈ Ω, where, ε is a nonzero real number such that |ε| < 1 and Ω is a rectangular domain given by Ω = {−π < x < π, 0 < t < T }. Our aim is to ﬁnd a regular solution to the following four inverse problems:

IP1: Inverse Problem with Dirichlet Boundary Conditions. Find a pair of functions u (x, t) and f (x) in the domain Ω satisfying equation (??) and the conditions (2.2) u (x, 0) = ϕ (x) , u (x, T ) = ψ (x) , x ∈ [−π, π] , and the homogeneous Dirichlet boundary conditions (2.3) u (−π, t) = 0, u (π, t) = 0, t ∈ [0,T ] , 3 where ϕ (x) and ψ (x) are given, suﬃciently smooth functions.

IP2: Inverse Problem with Neumann Boundary Conditions. Find a pair of functions u (x, t) and f (x) in the domain Ω satisfying equation (??), conditions (??) and the homogeneous Neumann boundary conditions

(2.4) ux (−π, t) = 0, ux (π, t) = 0, t ∈ [0,T ] . IP3: Inverse Problem with Periodic Boundary Conditions. Find a pair of functions u (x, t) and f (x) in the domain Ω satisfying equation (??), conditions (??) and the periodic boundary conditions

(2.5) u (−π, t) = u (π, t) , ux (−π, t) = ux (π, t) , t ∈ [0,T ] . IP4: Inverse Problem with Anti-Periodic Boundary Conditions. Find a pair of functions u (x, t) and f (x) in the domain Ω satisfying equation (??), conditions (??) and the anti-periodic boundary conditions

(2.6) u (−π, t) = −u (π, t) , ux (−π, t) = −ux (π, t) , t ∈ [0,T ] . By a regular solution of problems IP1, IP2, IP3 and IP4, we mean a pair of 2,1 functions u (x, t) and f (x) of the class u (x, t) ∈ Cx,t (Ω) , f (x) ∈ C [−π, π] .

3. Solution Method Here we seek solutions to problems IP1, IP2, IP3 and IP4 in a form of series expansion using a set of functions that form orthogonal basis in L2(−π, π). To ﬁnd the appropriate set of functions for each problem, we shall solve the homogeneous equation corresponding to equation (??) along with the associated boundary conditions using separation of variables.

3.1. Spectral Problems. Separation of variables leads to the following spectral problems for IP1, IP2, IP3 and IP4, respectively, (3.1) X00(x) − X00(−x) + λX(x) = 0,X(−π) = X(π) = 0, (3.2) X00(x) − X00(−x) + λX(x) = 0,X0(−π) = X0(π) = 0, (3.3) X00(x) − X00(−x) + λX(x) = 0,X(−π) = X(π),X0(−π) = X0(π), (3.4) X00(x) − X00(−x) + λX(x) = 0,X(−π) = −X(π),X0(−π) = −X0(π). The eigenvalue problems (??)-(??) are self-adjoint and hence they have real eigenvalues and their eigenfunctions form a complete orthogonal basis in L2 (−π, π) [?]. Their eigenvalues are, respectively, given by 12 (7.a) λ = (1 − ε) k + , k ∈ ∪ {0} , λ = (1 + ε) k2, k ∈ , 1k 2 N 2k N 12 (8.a) λ = (1 − ε) k2, λ = (1 + ε) k + , k ∈ ∪ {0} , 1k 2k 2 N 2 2 (9.a) λ1k = (1 − ε) k , k ∈ N ∪ {0} , λ2k = (1 + ε) k , k ∈ N, 12 12 (10.a) λ = (1 + ε) k + , λ = (1 − ε) k + , k ∈ ∪ {0} , 1k 2 2k 2 N 4 and the corresponding eigenfunctions are given by 1 (7.b) X = cos k + x, k ∈ ∪ {0} ,X = sin kx, k ∈ , 1k 2 N 2k N 1 (8.b) X = 1,X = cos kx, k ∈ ,X = sin k + x, k ∈ ∪ {0} , 0 1k N 2k 2 N (9.b) X0 = 1,X1k = cos kx, X2k = sin kx, k ∈ N, 1 1 (10.b) X = sin k + x X = cos k + x, k ∈ ∪ {0} . 1k 2 2k 2 N 3.1. Lemma. The systems of functions (??)-(??) are complete and orthogonal in L2 (−π, π) . Proof. Here we present the proof for the system of functions (??). The orthogonality follows from the direct calculations: Z π X1nX2m dx = 0, n ∈ N ∪ {0} , m ∈ N, −π and Z π XinXim dx = 0, m 6= n, i = 1, 2. −π Hence, it only remains to prove the completeness of the system in L2(−π, π), i.e., we need to show that if Z π 1 (3.5) f(x) cos k + x dx = 0, k ∈ N ∪ {0} , −π 2 and Z π (3.6) f(x) sin kx dx = 0, k ∈ N, −π then f(x) ≡ 0 in (−π, π). To show this, we are going to use the fact that 1 cos k + x and {sin kx} are complete in L2(0, π), see [?] for exam- 2 k∈N∪{0} k∈N ple. Now, suppose that the equation (??) holds. We then have Z π 1 Z π 1 0 = f(x) cos k + x dx = (f(x) + f(−x)) cos k + x dx. −π 2 0 2 1 Hence, by the completeness of the system cos k + x in L2(0, π), we 2 k∈N∪{0} have f(x) = −f(−x), −π < x < π. Similarly, if equation (??) holds, we have Z π Z π 0 = f(x) sin kx dx = (f(x) − f(−x)) sin kx dx. −π 0

Then, by the completeness of the system {sin kx} in L2(0, π), we have f(x) = k∈N f(−x), −π < x < π. Therefore, we must have f(x) ≡ 0 in (−π, π). Completeness and orthogonality of the systems of functions (??)-(??) can be proved similarly. Since each one of the systems of eigenfunctions (??)-(??) is complete and forms a basis in L2 (−π, π), the solution pair u(x, t) and f(x) of each inverse problem can be expressed in a form of series expansion using the appropriate set of eigenfunctions. 5

3.2. Existence of Solutions. Here, we give a full proof of existence of solution to the Inverse Problem IP1. Existence of solutions to the other three problems can be proved similarly. Using the orthogonal system (??), the functions u (x, t) and f (x) can be represented as follows ∞ ∞ X 1 X (3.7) u (x, t) = u (t) cos k + x + u (t) sin kx, 1k 2 2k k=0 k=1 ∞ ∞ X 1 X (3.8) f (x) = f cos k + x + f sin kx, 1k 2 2k k=0 k=1 where the coeﬃcients u1k (t) , u2k (t) , f1k, f2k are unknown. Substituting (??) and (??) into equation (??), we obtain the following equations relating the functions u1k (t) , u2k (t) and the constants f1k, f2k: 12 (3.9) u0 (t) + (1 − ε) k + u (t) = f , 1k 2 1k 1k 0 2 (3.10) u2k (t) + (1 + ε) k u2k (t) = f2k. Solving these equations we obtain

2 f1k −(1−ε)(k+ 1 ) t u1k (t) = + C1ke 2 , 1 2 (1 − ε) k + 2

f 2 u (t) = 2k + C e−(1+ε)k t, 2k (1 + ε) k2 2k where the unknown constants C1k,C2k, f1k, f2k are to be determined using the conditions in (??). Let ϕik, ψik, i = 1, 2 be the coeﬃcients of the series expansions of ϕ (x) and ψ (x), respectively, i.e., π π 1 Z 1 1 Z ϕ = ϕ (x) cos k + x dx, ϕ = ϕ (x) sin kx dx, 1k π 2 2k π −π −π π π 1 Z 1 1 Z ψ = ψ (x) cos k + x dx, ψ = ψ (x) sin kx dx. 1k π 2 2k π −π −π Then, the two conditions in (??) leads to

2 f1k f1k −(1−ε)(k+ 1 ) T + C1k = ϕ1k, + C1ke 2 = ψ1k, 1 2 1 2 (1 − ε) k + 2 (1 − ε) k + 2

f f 2 2k + C = ϕ , 2k + C e−(1+ε)k T = ψ . (1 + ε) k2 2k 2k (1 + ε) k2 2k 2k Solving these set of algebraic equations, we get 2 ϕ1k − ψ1k 1 C1k = 2 , f1k = (1 − ε) k + (ϕ1k − C1k) , −(1−ε) k+ 1 T 1 − e ( 2 ) 2 ϕ − ψ C = 2k 2k , f = (1 + ε) k2 (ϕ − C ) . 2k 1 − e−(1+ε)k2T 2k 2k 2k 6

Now, substituting u1k (t) , u2k (t) , f1k, f2k into (??) and (??) we get ∞ 2 X −(1−ε) k+ 1 t 1 u (x, t) = ϕ (x) + C e ( 2 ) − 1 cos k + x 1k 2 k=0 ∞ X −(1+ε)k2t + C2k e − 1 sin kx, k=1 and ∞ 2 X 1 1 f (x) = −ϕ00 (x) + εϕ00 (−x) − (1 − ε) k + C cos k + x 2 1k 2 k=0 ∞ X 2 − (1 + ε) k C2k sin kx. k=1 Note that for f (x) ∈ C [−π, π], it is required that ϕ(x) ∈ C2 [−π, π].

3.3. Convergence of Series. In order to justify that the obtained formal solution is indeed a true solution, we need to show that the series appeared in u(x, t) and f(x) as well as the corresponding series representations of uxx(x, t) and ut(x, t) converge uniformly in Ω. For this purpose, let ϕ(i) (−π) = ϕ(i) (π) = 0, i = 0, 2,

ψ(i) (−π) = ψ(i) (π) = 0, i = 0, 2.

Hence, on integration by parts, C1k and C2k can now be rewritten as (3) (3) (3) (3) ϕ2k − ψ2k ϕ1k − ψ1k C1k = ,C2k = − . 1 2 3 −(1+ε)k2T 3 −(1−ε)(k+ 2 ) T 1 1 − e k 1 − e k + 2 where, π π 1 Z 1 Z 1 ϕ(3) = ϕ000 (x) cos kx dx, ϕ(3) = ϕ000 (x) sin k + x dx, 1k π 2k π 2 −π −π

π π 1 Z 1 Z 1 ψ(3) = ψ000 (x) cos kx dx, ψ(3) = ψ000 (x) sin k + x dx. 1k π 2k π 2 −π −π Hence, the series representation of u (x, t) and f(x) can be expressed as

∞ 2 (3) (3) ! X 1 − e−(1+ε)k t ϕ − ψ u (x, t) = ϕ (x) + 1k 1k sin kx 1 − e−(1+ε)k2T k3 k=1 ∞ 1 2 (3) (3) ! −(1−ε)(k+ 2 ) t X 1 − e ϕ2k − ψ2k 1 − 2 cos k + x, −(1−ε)(k+ 1 ) T 1 3 2 k=0 1 − e 2 k + 2 and 7

∞ (3) (3) ! X 1 + ε ϕ − ψ f (x) = −ϕ00 (x) + εϕ00 (−x) + 1k 1k sin kx k 1 − e−(1+ε)k2T k=1 ∞ (3) (3) ! X (1 − ε) ϕ2k − ψ2k 1 − 2 cos k + x. k + 1 −(1−ε)(k+ 1 ) T 2 k=0 2 1 − e 2 For convergence, we then have the following estimates for u (x, t) and f (x)

∞ ϕ(3) + ψ(3) ∞ ϕ(3) + ψ(3) X 1k 1k X 2k 2k |u (x, t)| ≤ |ϕ (x)| + c + c k3 1 3 k=1 k=0 k + 2 and

∞ 2 2 X (3) (3) 2 |f (x)| ≤ |ϕ00 (x)| + |ϕ00 (−x)| + c ϕ + ψ + 1k 1k k2 k=1 ∞ 2 2 ! X (3) (3) 2 + c ϕ + ψ + , 2k 2k 1 2 k=0 k + 2 for some positive constant c. Here, for the estimate of f(x), we have used the inequality 2ab ≤ a2 + b2. The convergence of the series in the estimate of u(x, t) (3) (3) is clearly achieved if ϕik , ψik , i = 1, 2 are ﬁnite. This can be ensured by assum- 000 000 ing that ϕ (x) and ψ (x) ∈ L2(−π, π). Furthermore, by Bessel inequality for trigonometric series, the following series converge:

∞ 2 X (3) 2 ϕ ≤C kϕ000 (x)k , i = 1, 2, ik L2(−π,π) k=1

∞ 2 X (3) 2 ψ ≤C kψ000 (x)k , i = 1, 2. ik L2(−π,π) k=1 Therefore, by the Weierstrass M-test (see [?]), the series representations of u(x, t) and f(x) converge absolutely and uniformly in the region Ω. The convergence of the series representations of uxx(x, t) and ut(x, t) which are obtained by term-wise diﬀerentiation of the series representation of u(x, t) can be shown is a similar way.

3.4. Uniqueness of Solution. Suppose that there are two solution sets {u1 (x, t) , f1 (x)} and {u2 (x, t) , f2 (x)} to the Inverse Problem IP1. Denote

u (x, t) = u1 (x, t) − u2 (x, t) , and

f (x) = f1 (x) − f2 (x) . Then, the functions u (x, t) and f (x) clearly satisfy equation (??), the boundary conditions in (??) and the homogeneous conditions (3.11) u (x, 0) = 0, u (x, T ) = 0, x ∈ [−π, π] 8

Let us now introduce the following

π 1 Z 1 (3.12) u (t) = u (x, t) cos k + x dx, k ∈ ∪ {0} , 1k π 2 N −π

π 1 Z (3.13) u (t) = u (x, t) sin kx dx, k ∈ , 2k π N −π

π 1 Z 1 (3.14) f = f (x) cos k + x dx, k ∈ ∪ {0} , 1k π 2 N −π

π 1 Z (3.15) f = f (x) sin kx dx, k ∈ . 2k π N −π Note that the homogeneous conditions in (??) lead to

(3.16) uik(0) = uik(T ) = 0, i = 1, 2.

Diﬀerentiating equation (??) gives

π 1 Z 1 u0 (t) = (u (x, t) − εu (−x, t)) cos k + x dx + f , 1k π xx xx 2 1k −π which on integrating by parts and using the conditions in (??) reduces to

12 u0 (t) = (ε − 1) k + u + f . 1k 2 1k 1k

One can then easily show that this equation together with the conditions u1k(0) = u1k(T ) = 0 imply that

f1k = 0, u1k (t) ≡ 0.

Similarly, for u2k and f2k as given in (??) and (??), respectively, one can show that

f2k = 0, u2k (t) ≡ 0.

Therefore, due to the completeness of the system of eigenfunctions (??) in L2 (−π, π), we must have f (t) ≡ 0, u (x, t) ≡ 0, (x, t) ∈ Ω¯. This ends the proof of uniqueness of solution to the Inverse Problem IP1. Unique- ness of solutions to the Inverse Problems IP2, IP3 and IP4 can be proved in a similar way. 9

4. Main Results and Example Solution 4.1. Main Results. The main results for the Inverse Problems IP1, IP2, IP3 and IP4 can be summarized in the following theorems:

2 000 000 4.1. Theorem. Let ϕ (x) , ψ (x) ∈ C [−π, π], ϕ (x), ψ (x) ∈ L2(−π, π) and ϕ(i) (±π) = ψ(i) (±π) = 0, i = 0, 2. Then, a unique solution to the Inverse Problem IP1 exists and it can be written in the form

∞ 2 (3) (3) ! X 1 − e−(1+ε)k t ϕ − ψ u (x, t) = ϕ (x) + 1k 1k sin kx 1 − e−(1+ε)k2T k3 k=1 ∞ 1 2 (3) (3) ! −(1−ε)(k+ 2 ) t X 1 − e ϕ2k − ψ2k 1 − 2 cos k + x, −(1−ε)(k+ 1 ) T 1 3 2 k=0 1 − e 2 k + 2

∞ (3) (3) ! X 1 + ε ϕ − ψ f (x) = −ϕ00 (x) + εϕ00 (−x) + 1k 1k sin kx k 1 − e−(1+ε)k2T k=1 ∞ (3) (3) ! X (1 − ε) ϕ2k − ψ2k 1 − 2 cos k + x, k + 1 −(1−ε)(k+ 1 ) T 2 k=0 2 1 − e 2 where π π 1 Z 1 Z 1 ϕ(3) = ϕ000 (x) cos kx dx, ϕ(3) = ϕ000 (x) sin k + x dx, 1k π 2k π 2 −π −π

π π 1 Z 1 Z 1 ψ(3) = ψ000 (x) cos kx dx, ψ(3) = ψ000 (x) sin k + x dx. 1k π 2k π 2 −π −π

2 000 000 4.2. Theorem. Let ϕ (x) , ψ (x) ∈ C [−π, π], ϕ (x), ψ (x) ∈ L2(−π, π) and ϕ0 (±π) = ψ0 (±π) = 0. Then a unique solution to the Inverse Problem IP2 exists and it can be written in the form

2 ∞ 1 − e−(1−ε)k t ψ(3) − ϕ(3) t X 2k 2k u (x, t) = ϕ (x) + (ψ − ϕ ) + cos kx T 0 0 1 − e−(1−ε)k2T k3 k=1 2 −(1+ε) k+ 1 t (3) (3) ∞ 1 − e ( 2 ) ψ − ϕ X 1k 1k 1 − sin k + x, 1 2 3 −(1+ε)(k+ 2 ) T 1 2 k=0 1 − e k + 2

∞ (1 − ε) ϕ(3) − ψ(3) ψ0 − ϕ0 X 2k 2k f (x) = −ϕ00 (x) + εϕ00 (−x) + − cos kx T k 1 − e−(1−ε)k2T k=1 ∞ (1 + ε) ϕ(3) − ψ(3) X 1k 1k 1 + sin k + x, 1 2 1 −(1+ε)(k+ 2 ) T 2 k=0 k + 2 1 − e 10 where π π 1 Z 1 Z ϕ = ϕ (x) dx, ψ = ψ (x) dx, 0 2π 0 2π −π −π

π π 1 Z 1 1 Z ϕ(3) = ϕ000 (x) cos k + x dx, ϕ(3) = ϕ000 (x) sin kx dx, 1k π 2 2k π −π −π

π π 1 Z 1 1 Z ψ(3) = ψ000 (x) cos k + x dx, ψ(3) = ψ000 (x) sin kx dx. 1k π 2 2k π −π −π

2 000 000 4.3. Theorem. Let ϕ (x) , ψ (x) ∈ C [−π, π], ϕ (x), ψ (x) ∈ L2(−π, π) and ϕ(i) (−π) = ϕ(i) (π) , ψ(i) (−π) = ψ(i) (π) , i = 0, 1, 2. Then, a unique solution to the Inverse Problem IP3 exists and it can be written in the form

2 ∞ 1 − e−(1−ε)k t ϕ(3) − ψ(3) t X 2k 2k u (x, t) = ϕ (x) + (ψ − ϕ ) − cos kx T 0 0 1 − e−(1−ε)k2T k3 k=1 2 ∞ 1 − e−(1+ε)k t ϕ(3) − ψ(3) X 1k 1k + sin kx, 1 − e−(1+ε)k2T k3 k=1

∞ (1 − ε) ϕ(3) − ψ(3) ψ0 − ϕ0 X 2k 2k f (x) = −ϕ00 (x) + εϕ00 (−x) + − cos kx T 1 − e−(1−ε)k2T k k=1 ∞ (1 + ε) ϕ(3) − ψ(3) X 1k 1k + sin kx, 1 − e−(1+ε)k2T k k=1 where π π 1 Z 1 Z ϕ = ϕ (x) dx, ψ = ψ (x) dx, 0 2π 0 2π −π −π

π π 1 Z 1 Z ϕ(3) = ϕ000 (x) cos kx dx, ϕ(3) = ϕ000 (x) sin kx dx, 1k π 2k π −π −π

π π 1 Z 1 Z ψ(3) = ψ000 (x) cos kx dx, ψ(3) = ψ000 (x) sin kx dx. 1k π 2k π −π −π

2 000 000 4.4. Theorem. Let ϕ (x) , ψ (x) ∈ C [−π, π], ϕ (x), ψ (x) ∈ L2(−π, π) and ϕ(i) (−π) = −ϕ(i) (π) , ψ(i) (−π) = −ψ(i) (π) , i = 0, 1, 2. Then, a unique solution 11

Figure 2. Graphs of u(x, t) at diﬀerent times (left) and f(x) (right) for = 0.1 and T = 1.

Figure 3. Graphs of u(x, t) at t = 0.5 (left) and f(x) (right) for diﬀerent values of and for T = 1. to the Inverse Problem IP4 exists and it can be written in the form

2 −(1−ε) k+ 1 t (3) (3) ∞ 1 − e ( 2 ) ϕ − ψ X 2k 2k 1 u (x, t) = ϕ (x) − cos k + x 1 2 3 −(1−ε)(k+ 2 ) T 1 2 k=0 1 − e k + 2 2 −(1+ε) k+ 1 t (3) (3) ∞ 1 − e ( 2 ) ϕ − ψ X 1k 1k 1 + sin k + x, 1 2 3 −(1+ε)(k+ 2 ) T 1 2 k=0 1 − e k + 2

∞ (1 − ε) ϕ(3) − ψ(3) X 2k 2k 1 f (x) = −ϕ00 (x) + εϕ00 (−x) − cos k + x 1 2 1 −(1−ε)(k+ 2 ) T 2 k=0 k + 2 1 − e ∞ (1 + ε) ϕ(3) − ψ(3) X 1k 1k 1 + sin k + x, 1 2 1 −(1+ε)(k+ 2 ) T 2 k=0 k + 2 1 − e where π π 1 Z 1 1 Z 1 ϕ(2) = ϕ000 (x) cos k + x dx, ϕ(2) = ϕ000 (x) sin k + x dx, 1k π 2 2k π 2 −π −π

π π 1 Z 1 1 Z 1 ψ(2) = ψ000 (x) cos k + x dx, ψ(2) = ψ000 (x) sin k + x dx. 1k π 2 2k π 2 −π −π 4.2. Example Solution. For the sake of illustration, we present here a simple example solution for the Inverse Problem IP1. For this purpose, we consider the following choice of conditions (??): u (x, 0) = 0, u (x, T ) = sin x, x ∈ [−π, π] , i.e., we have ϕ (x) = 0 and ψ (x) = sin x. Calculating the coeﬃcients of the series solutions as given in Theorem ??, we get 1 − e−(1+ε)t 1 + u(x, t) = sin x, and f(x) = sin x. 1 − e−(1+ε)T 1 − e−(1+ε)T These solutions are illustrated in Figures ?? and ??. 12

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