Toeplitz - Hankel integro - Differential equation

JOURNAL OF SCIENCE OF HNUE Mathematical and Physical Sci., 2014, Vol. 59, No. 7, pp. 21-33 This paper is available online at TOEPLITZ - HANKEL INTEGRO - DIFFERENTIAL EQUATION Nguyen Xuan Thao1, Nguyen Anh Dai2 and Nguyen Minh Phuong1 1School of Applied Mathematics and Informatics, Hanoi University of Science and Technology 2Faculty of Mathematics, Hung Yen University of Technology and Education Abstract. There are many applications of the Toeplitz plus Hankel equation in mathematics, physics and medicine. However, this is still an open problem nowadays. In this paper, we study four cases of the Toeplitz plus Hankel integro - differential equations and general integro - differential equation. These equations can be solved in closed form using new convolutions and well-known convolutions. The obtained results were the expansions of these equation types. A new approach to solved classes of equations of this kind is to use the new and known convolutions to obtain explicit solutions as well as solutions in closed form. Keywords: Toeplitz and Hankel integral equation, integro-differential equation, convolution, Hartley transform, Fourier transform, Laplace transform and Fourier cosine transform. 1. Introduction The integral equation with Toeplitz plus Hankel is of the form [5]. f(x) + ∫ ∞ 0 [k1(x+ y) + k2(x− y)]f(y)dy = g(x), x > 0, (1.1) where g, k1, k2 are given and f is an unknown function. This equation has many useful applications in mathematics, physics and medicine [1, 3]. This integral equation can be solved in closed form only in some particular cases of the Hankel kernel k1 and the Toeplitz kernel k2. However, the solution of equation (1.1) in the general case is still open. In various particular cases of equation (1.1), the Toeplitz kernel k2 is an even function and equation (1.1) takes the form f(x) + ∫ ∞ 0 [k1(x+ y) + k2|x− y|]f(y)dy = g(x), x > 0. (1.2) Received June 24, 2014. Accepted September 10, 2014. Contact Nguyen Xuan Thao, e-mail address: thao.nguyenxuan@hust.edu.vn 21 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong We now recall several particular cases of equation (1.2) when the solution is known in closed form. Theorem 1.1. ([10]) Supposing functions g1, g2, k1, k2 ∈ L1(R+), g = g1 + g2, so that the following conditions hold: 1 + √ π 2 Fc(k1 + k2)(y) ̸= 0, ∀y > 0, and g1 = √ π 2 (((g2 ∗ 1 ℓ)− g2) ∗ 2 (k1 − k2))(x). Then the integral equation (1.2) has the unique solution in L1(R+) in the following form f(x) = g2(x)− (g2 ∗ 1 ℓ)(x), where ℓ is a function in L1(R+), defined by (Fcℓ)(y) = √ π 2 Fc(k1 + k2)(y) 1 + √ π 2 Fc(k1 + k2)(y) . The Fourier cosine (f ∗ 1 g) and sine (f ∗ 2 g) convolutions are of the form (f ∗ 1 g)(x) = 1√ 2π ∞∫ 0 f(u)[g(x+ u) + g(x− u)]du, (f ∗ 2 g)(x) = 1√ 2π ∞∫ 0 f(u)[g(x+ u)− g(x− u)]du, and satisfy the following factorization equalities Fc(f ∗ 1 g)(y) = (Fcf)(y)(Fcg)(y), ∀y > 0, f, g ∈ L1(R+), (1.3) Fs(f ∗ 2 g)(y) = (Fsf)(y)(Fcg)(y), ∀y > 0, f, g ∈ L1(R+). Theorem 1.2. ([10]) Suppose that functions g1, g2, k1, k2 ∈ L1(R+), g = g1 + g2 satisfy the following condition: 1 + √ π 2 Fc(k1 + k2)(y) ̸= 0, ∀y > 0, and g2 = √ π 2 (((g1 ∗ 2 ℓ) + g1) ∗ 1 (k1 + k2))(x). 22 Toeplitz - Hankel integro - differential equation Then the integral equation (1.2) has the unique solution in L1(R+) as follows: f(x) = g1(x) + (g1 ∗ 2 ℓ)(x), where ℓ is a function in L1(R+) which is defined by (Fcℓ)(y) = √ π 2 . Fc(k1 − k2)(y) 1 + √ π 2 Fc(k1 − k2)(y) . The Hartley transform H of f ∈ L1(R) is defined in [6]. (Hf)(y) = 1√ 2π ∫ ∞ −∞ f(x)cas(xy)dx, (1.4) where casx = cos x+sinx. The Hartley transform is an integral transform closely related to the Fourier transform [7] (Ff)(y) = 1√ 2π ∫ ∞ 0 f(x)e−ixydx, f ∈ L1(R). (1.5) It was proposed as an alternative to the Fourier transform by Hartley in 1942. Compared to the Fourier transform, the Hartley transform has advantages when transforming a real function to a real function and being its own inverse [6]. f(x) = 1√ 2π ∫ ∞ −∞ cas(xy)(Hf)(y)dy. (1.6) Instead of considering the Toeplitz plus Hankel integral equation (1.1) onR+, we consider two integral equations f(x) + ∫ ∞ 0 [k(x+ y) + k(x− y)]f(y)dy = g(x), ∀x > 0, (1.7) f(x) + 1 2π ∫ ∞ 0 [f(x+ y) + f(x− y)]k(y)dy = g(x), ∀x > 0, (1.8) where k, g ∈ L1(R) are given functions and f ∈ L1(R+) is an unknown function. Here after, we present a number of new results related to published equations (1.7) and (1.8). Theorem 1.3. ([8]) Let functions k, g ∈ L1(R), satisfy the following conditions i) 1 + (Hk)(x) ̸= 0 for all x ∈ R. ii) g(x)− (g ∗ 3 ℓ)(x) is an even function, where ℓ(x) = H ( (Hk)(y) 1 + (Hk)(y) ) (x). 23 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong Then equation (1.7) has a unique solution f ∈ L1(R+) of the form f(x) = g(x)− (g ∗ 3 ℓ)(x), x ∈ R+, where the Hartley convolution (g ∗ 3 ℓ) is defined by (g ∗ 3 ℓ)(x) = 1 2 √ 2π ∞∫ −∞ g(t)[ℓ(x+ t) + ℓ(x− t) + ℓ(−x− t) + ℓ(−x+ t)]dt, and satisfies the following factorization equality H(g ∗ 3 ℓ)(y) = (Hg)(−y)(Hℓ)(y). Theorem 1.4. ([8]) Let k ∈ L1(R+), g ∈ L1(R), satisfy the following condition 1 + (Fck)(x) ̸= 0 for all x ∈ R. Then equation (1.8) has an unique solution f ∈ L1(R+) of the form f(x) = g(x)− (ℓ ∗ 3 g)(x), x ∈ R, where ℓ(x) = Fc ( (Fck)(y) 1 + (Fck)(y) ) (x) ∈ L1(R+). Theorem 1.5. ([9]) For two arbitrary functions f(x) and g(x) in L1(R+), the generalized convolutions (f γ∗ 4 g) belong to L1(R+). Moreover,the following norm estimation and factorization identity hold: ||(f γ∗ 4 g)||L1(R+) ≤ ||f ||L1(R+)||g||L1(R+), Fc(f γ∗ 4 g) = e−µy(Fcf)(y)(Lg)(y) ∀y > 0. (1.9) Furthermore, the generalized convolution (f γ∗ 4 g) belongs to C0(R+). Where, convolution (f γ∗ 4 g) is a Fourier cosine-Laplace convolution, defined by [19] (f γ∗ 4 g)(x) = 1 π ∞∫ 0 ∞∫ 0 [ v + µ (v + µ)2 + (x− u)2 + v + µ (v + µ)2 + (x+ u)2 ] f(u)g(v)dudv, (1.10) where x > 0. In this paper, we consider Toeplitz - Hankel integro - differential equations (section 2) and integro - differential equations (section 3) which are solved in closed form by using a number of new convolutions published in [8-10]. 24 Toeplitz - Hankel integro - differential equation 2. Toeplitz - Hankel integro - differential equation In this section, we consider two Toeplitz - Hankel integro-differential equation f ′′ (x) + ∞∫ 0 [k1(x+ y) + k2(|x− y|)]f(y)dy = g(x), x> 0, (2.1) where f(x) is an unknown function and satisfies f(x)→ 0 as x→∞, f ′(x)→ 0 as x→ 0 and f ′(x)→ 0 as x→∞. We also consider the following equation f ′′ (x) + 1 2π ∞∫ 0 f(x+ y) + f(x− y)]k(y)dy = g(x), x ∈ R, (2.2) where g(x), k(x) are known functions and f(x) is an unknown function. Theorem 2.1. Suppose that functions g1, g2, k1, k1, g ∈ L1(R+), g = g1 + g2 satisfy the following conditions y2 − √ π 2 Fc(k1 + k2)(y) ̸= 0, ∀y> 0, (2.3) g1(x) = π 2 (( (g2 ∗ 1 e−x) + (ℓ ∗ 1 (g2 ∗ 1 e−x) ) ∗ 2 (k1 − k2) ) (x). (2.4) Then equation (2.1) has a unique solution in L1(R+) of the form f(x) = − √ π 2 (g2 ∗ 1 e−x)(x)− √ π 2 (ℓ ∗ 1 (g2 ∗ 1 e−x))(x), where ℓ is a function in L1(R+) and is defined by (Fcℓ)(y) = Fc(e −x ∗ 1 (k1 + k2))(y) 1− Fc(e−x ∗ √ π 2 (e−x ∗ 1 (k1 + k2)))(y) . (2.5) Proof. Extending g1 to the whole real line as an odd function, f and g2 are even functions, extending g to all whole real lines by g(x) = g1(x) + g2(x). Then the integral equation (2.1) on the whole real line R has the following form: f ′′ (|x|) + ∞∫ 0 [k1(|x+ y|) + k2(|x− y|)]f(y)dy = g(x), x ∈ R. (2.6) Equation (2.6) can be rewritten in the form 25 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong f ′′ (|x|) + 1 2 ∞∫ 0 [k1(|x+ y|) + k2(|x+ y|) + k1(|x− y|) + k2(|x− y|)] +[k1(|x+ y|)− k2(|x+ y|)− k1(|x− y|) + k2(|x− y|)]}f(y)dy = g(x). (2.7) Applying the Fourier cosine to transform both sides of equation (2.7) we obtain −y2(Fcf)(y)+ √ π 2 (Fcf)(y).Fc(k1+k2)(y)+ i √ π 2 (Fsf)(y).Fc(k1−k2)(y) = (Fg)(y). (2.8) Recalling that g(x) = g1(x) + g2(x) where g1 and g2 are the odd and even components of g, then f is a solution of equation (2.6) if both of the below conditions hold −y2(Fcf)(y) + √ π 2 (Fcf)(y).Fc(k1 + k2)(y) = (Fcg2)(y), (2.9) and i √ π 2 (Fsf)(y).Fc(k1 − k2)(y) = −i(Fsg1)(y). (2.10) On the other hand, equation (2.9) can be rewritten in the form (Fcf)(y) = (Fcg2)(y) −y2 +√π 2 Fc(k1 + k2)(y) . Using the relation Fc(e−x)(y) = √ 2 π . 1 1+y2 from formula (1.4.1, p. 23) in [2], we have (Fcf)(y) = −1 1 + y2 . (Fcg2)(y) 1− 1+ √  2 Fc(k1+k2)(y) 1+y2 = − √ π 2 Fc(e −x)(y). (Fcg2)(y) 1− Fc(e−x)(y)[1+ √ π 2 Fc(k1 + k2)(y)] √ π 2 = − √ π 2 Fc(e −x)(y). (Fcg2)(y) 1−√π 2 Fc(e−x)(y)− π2Fc(e−x ∗1(k1 + k2))(y) = − √ π 2 Fc(e −x)(y). (Fcg2)(y) 1−√π 2 Fc[e−x + √ π 2 (e−x ∗ 1 (k1 + k2))](y) = − √ π 2 Fc(e −x)(y).[1 + √ π 2 Fc[e −x + √ π 2 (e−x ∗ 1 (k1 + k2))](y) 1−√π 2 Fc[e−x + √ π 2 (e−x ∗ 1 (k1 + k2))](y) ](Fcg2)(y). Applying the Wiener - Levy theorem [4], by the given equation (2.6) there exists a unique function ℓ ∈ L1(R+) satisfing (2.5). Then we have (Fcf)(y) = − √ π 2 Fc(e −x)(y)[1 + (Fcℓ)(y)](Fcg2)(y). 26 Toeplitz - Hankel integro - differential equation Therefore, (Fcf)(y) = − √ π 2 Fc(g2 ∗ 1 e−x)(y)− √ π 2 Fc(ℓ ∗ 1 (g2 ∗ 1 e−x))(y). Due to the uniqueness of the Fourier cosine, we have f(x) = − √ π 2 (g2 ∗ 1 e−x)(x)− √ π 2 (ℓ ∗ 1 (g2 ∗ 1 e−x))(x). (2.11) Replacing (2.11) in (2.10), we have π 2 [−Fs(g2 ∗ 1 e−x)(y)− Fs(ℓ ∗ 1 (g2 ∗ 1 e−x))(y)]Fc(k1 − k2)(y) = −(Fsg1)(y). Thus Fs(( π 2 g2 ∗ 1 e−x) ∗ 2 (k1 − k2))(y) + π 2 Fs(ℓ ∗ 1 (g2 ∗ 1 e−x)) ∗ 2 (k1 − k2))(y) = (Fsg1)(y). Due to the uniqueness of the Fourier sine we have g1(x) = π 2 [((g2 ∗ 1 e−x) ∗ 2 (k1 − k2))(x) + (ℓ ∗ 1 (g2 ∗ 1 e−x) ∗ 2 (k1 − k2))(x)]. From Parseval identity and the convolutions for the Fourier cosine and Fourier sine transform, the solution of equation (2.1), which is found above, satisfies the conditions (2.3) and (2.4), and belongs to L1(R+). Theorem 2.1 is proven. Similarly, we receive the following theorem. Theorem 2.2. Supposing functions g1, g2, k1, k2, g ∈ L1(R+), g = g1 + g2, and y2 − √ π 2 Fc(k1 − k2)(y) ̸= 0, ∀y > 0 g2 = π 2 ( (k1 + k2) ∗ 1 ( (g1 ∗ 2 ℓ)− ℓ ∗ 1 (g1 ∗ 2 e−x) )) (x) . Then equation (2.1) has the unique solution in L1(R+) in the form f(x) = −(g1 ∗ 1 e−x)(x)− (ℓ1 ∗ 1 (g1 ∗ 2 e−x)(x). Lemma 2.1. Let function f ∈ L1(R+) and suppose f ′(x) → 0 as |x| → ∞ and f(x), f ′(x) → 0 as |x| → ∞. Then Hartley transforms of the differential of function f ∈ L1(R) are as follow (Hf ′)(y) = y√ 2π (Hf)(y), (Hf ′′)(y) = y2(Hf)(y). 27 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong Theorem 2.3. Let functions k, g ∈ L1(R) be given and y2 + (Fck)(y) ̸= 0,∀y ∈ R then equation (2.2) with conditions: f(x)→ 0 as |x| → ∞ and f ′(x)→ 0 as |x| → ∞, has a solution in L1(R) f(x) = √ π 2 [(e−x∗ 5 g)(x) + (ℓ ∗ 5 (e−x∗ 5 g))(x)] x ∈ R, where ℓ ∈ L1(R+) so that: (Fcℓ)(x) = √ π 2 Fc(e −x + e−x ∗ 5 k)(x) 1−√π 2 Fc(e−x + e−x ∗ 5 k)(x) , while Hartley convolution is defined by [8] (f ∗ 5 g)(x) = 1√ 2π ∞∫ 0 f(u)[g(x+ u) + g(x− u)]du, and the following factorization equality holds H(f ∗ 5 g)(y) = (Fcf)(y)(Hg)(y), f ∈ L1(R+), g ∈ L1(R). Proof. Applying Hartley transform to both sides of (2.2), using lemma 2.1 we have y2(Hf)(y) + (Fck)(y)(Hf)(y) = (Hg)(y), y ∈ R. Therefore, (Hf)(y) = (Hg)(y) y2 + (Fck)(y) = (Hg)(y) 1 + y2 . 1 1− 1−(Fck)(y) 1+y2 , from this and formula 1.4.1 page 23 [2] Fc(e −x)(y) = √ 2 π . 1 1 + y2 , we have (Hf)(y) = Fc(e −x)(y). (Hg)(y) 1−√π 2 [1− (Fck)(y)]Fc(e−x)(y) . √ π 2 . Hence, we have (Hf)(y) = H(e−x ∗ 1 g)(y). 1 1− Fc(e−x − e−x ∗ 1 k)(y) √ π 2 . √ π 2 = H(e−x ∗ 1 g)(y) 1 + √π2Fc(e−x − e−x ∗1 k)(y) 1− Fc(e−x − e−x ∗ 1 k)(y) √ π 2 √π 2 . 28 Toeplitz - Hankel integro - differential equation By the Wiener - Levy theorem [4] we have a unique function ℓ ∈ L1(R+) such that (Fcℓ)(x) = √ π 2 Fc(e −x − e−x ∗ 1 k)(y) 1−√π 2 Fc(e−x − e−x ∗ 1 k)(y) . Then, we have (Hf)(y) = H(e−x ∗ 5 g)(y)[1 + (Fcℓ)(y)] √ π 2 = ( H(e−x ∗ 5 g)(y) +H(ℓ ∗ 5 (e−x ∗ 5 g))(y) )√π 2 . Due to the uniqueness of the Hartley transform we obtain a solution of equation (2.2) in L1(R) f(x) = √ π 2 [ (e−x ∗ 5 g)(x) + (l ∗ 5 (e−x ∗ 5 g)(x) ] . Theorem 2.3 is proven. 3. Integro-differential equations In this section we will consider two integro-differential equations: f ′′(x)− λ2f(x) + ∫ ∞ 0 θ(x, u)f(u)du = h(x), x > 0, λ ̸= 0, (3.1) f ′′(x) + ∫ ∞ 0 θ(x, u)f(u)du = h(x), x > 0. (3.2) where θ(x, u) = 1 π ∫ ∞ 0 g(v) [ v + µ (v + µ)2 + (x− u)2 + v + µ (v + µ)2 + (x+ u)2 ] dv, µ > 0, here g(x), h(x) are given functions and f(x) is an unknown function. Lemma 3.1. If it is supposed that f(x) is a two order differentiable function in L1(R+) such that f(0) = f ′(0) = f ′′(0) = 0 and f ′(x), f ′′(x)→ 0 as x→∞, then the following equalities hold (Fcf ′)(y) = y(Fsf)(y), (Fcf ′′)(y) = −y2(Fcf)(y). Theorem 3.1. Let g(x), h(x) ∈ L1(R+) and −y2 − λ2 + γ(Lg)(y) ̸= 0 ∀y > 0, λ ̸= 0, with the initial condition f(0) = f ′(0) = 0 and f(x), f ′(x) → 0 as x → ∞. Then equation (3.1) has a unique solution in L1(R+) in the form f(x) = − √ π 2 [(e−λx ∗ 1 h) + ((e−λx ∗ 1 h) ∗ 1 ℓ)](x), 29 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong with ℓ(x) ∈ L1(R+) defined by (Fcℓ)(y) = √ π 2 Fc(e −λx γ∗ 4 g)(y) 1−√π 2 Fc(e−λx γ∗ 4 g)(y) . (3.3) Proof. We can rewrite equation (3.1) as f ′′(x)−λ2f(x)+1 π ∫ ∞ 0 ∫ ∞ 0 [ v + µ (v + µ)2 + (x− u)2+ v + µ (v + µ)2 + (x+ u)2 ] g(v)f(u)dvdu = h(x) (x > 0) (3.4) Then, we can rewrite equation (3.4) in the form f ′′(x)− λ2f(x) + (f γ∗ 4 g)(x) = h(x). (3.5) Applying the Fourier cosine transform to both sides of equation (3.5), we get (Fcf ′′)(y)− λ2(Fcf)(y) + Fc(f γ∗ 4 g)(y) = (Fch)(y). By Lemma 3.1 and using the factorization equality (1.9) we obtain −y2(Fcf)(y)− λ2(Fcf)(y) + γ(Fcf)(y)(Lg)(y) = (Fch)(y). Solving the above equation with the condition −y2 − λ2 + γ(Lg)(y) ̸= 0, we have (Fcf)(y) = (Fch)(y) −y2 − λ2 + γ(Lg)(y) = (Fch)(y). −1 y2 + λ2 1 1− γ y2+λ2 (Lg)(y) = (Fch)(y). −λ y2 + λ2 1 1− γλ y2+λ2 (Lg)(y) . Using formula Fc(e−λx)(y) = √ 2 π λ y2+λ2 in [3], we give (Fcf)(y) = −√π 2 Fc(e −λx)(y)(Fch)(y) 1− γ√π 2 Fc(e−λx)(y)(Lg)(y) , since (1.3) and (1.9), we have (Fcf)(y) = −√π 2 Fc(e −λx ∗ 1 h)(y) 1−√π 2 Fc(e−λx γ∗ 4 g)(y) = − √ π 2 Fc(e −λx ∗ 1 h)(y) [ 1 + √ π 2 Fc(e −λx γ∗ 4 g)(y) 1−√π 2 Fc(e−λx γ∗ 4 g)(y) ] . 30 Toeplitz - Hankel integro - differential equation Applying the Wiener-Levy theorem (see [4, p.63]), there exists a unique function ℓ ∈ L1(R+) which satisfies (3.3), we obtain (Fcf)(y) = − √ π 2 Fc(e −λx ∗ 1 h)(y)(1 + (Fcℓ)(y)) = − √ π 2 [Fc(e −λx ∗ 1 h)(y) + Fc(e −λx ∗ 1 h)(y)(Fcℓ)(y)] = − √ π 2 [Fc(e −λx ∗ 1 h)(y) + Fc((e −λx ∗ 1 h) ∗ 1 ℓ)(y)]. Due to the uniqueness of the Fourier cosine transform, we have f(x) = − √ π 2 [(e−λx ∗ 1 h)(x) + ((e−λx ∗ 1 h) ∗ 1 ℓ)(x)]. Theorem 3.1 is proven. Theorem 3.2. Let h(x), g(x) be functions in L1(R+) such that −y2 + γ(Lg)(y) ̸= 0 for all y > 0, with the initial condition f(0) = f ′(0) = 0 and f(x), f ′(x) → 0 as x → ∞. Then equation (3.2) has a unique solution f(x) in L1(R+) in the form f(x) = − √ π 2 (e−x ∗ 1 h+ (e−x ∗ 1 h) ∗ 1 ℓ)(x), with ℓ(x) ∈ L1(R+) defined by: (Fcℓ)(y) = √ π 2 Fc(e −x + e−x γ∗ 4 g)(y) 1−√π 2 Fc(e−x + e−x γ∗ 4 g)(y) . (3.6) Proof. We can rewrite equation (3.2) as f ′′(x)+ 1 π ∫ ∞ 0 ∫ ∞ 0 [ v + µ (v + µ)2 + (x− u)2+ v + µ (v + µ)2 + (x+ u)2 ] g(v)f(u)dvdu = h(x) (x > 0) (3.7) Then we can rewrite equation (3.7) in the form f ′′(x) + (f γ∗ 4 g)(x) = h(x). (3.8) Applying the Fourier cosine transform to both sides of the equation (3.8), we get (Fcf ′′)(y) + Fc(f γ∗ 4 g)(y) = (Fch)(y). 31 Nguyen Xuan Thao, Nguyen Anh Dai and Nguyen Minh Phuong Since lemma 3.1 and by using factorization property (1.9) we get −y2(Fcf)(y) + γ(Fcf)(y)(Lg)(y) = (Fch)(y). We have (Fcf)(y) = (Fch)(y) −y2 + γ(Lg)(y) = (Fch)(y) −1− y2 + 1 + γ(Lg)(y) = (Fch)(y). −1 1 + y2 . 1 1− 1 1+y2 − γ 1+y2 (Lg)(y) . Then using formula Fc(e−x)(y) = √ 2 π 1 1+y2 in [3], we have (Fcf)(y) = −√π 2 (Fch)(y).Fc(e −x)(y) 1−√π 2 Fc(e−x)(y)− γ √ π 2 Fc(e−x)(y)(Lg)(y) . Thus, from (1.3) and (1.9) we give (Fcf)(y) = −√π 2 Fc(e −x ∗ 1 h)(y) 1−√π 2 Fc(e−x)(y)− √ π 2 Fc(e−x γ∗ 4 g)(y) = −√π 2 Fc(e −x ∗ 1 h)(y) 1−√π 2 Fc(e−x + e−x γ∗ 4 g)(y) = − √ π 2 Fc(e −x ∗ 1 h)(y) [ 1 + √ π 2 Fc(e −x + e−x γ∗ 4 g)(y) 1−√π 2 Fc(e−x + e−x γ∗ 4 g)(y) ] . Due to the Wiener-Levy theorem (in [4, p.63]) there exists a function ℓ in L1(R+) satisfying (3.6), so we have (Fcf)(y) = − √ π 2 Fc(e −x ∗ 1 h)(y)[1 + (Fcℓ)(y)] = − √ π 2 [Fc(e −x ∗ 1 h)(y) + Fc(e −x ∗ 1 h)(y)(Fcℓ)(y)] = − √ π 2 [Fc(e −x ∗ 1 h)(y) + Fc((e −x ∗ 1 h) ∗ 1 ℓ))(y)] = − √ π 2 [Fc(e −x ∗ 1 h+ (e−x ∗ 1 h) ∗ 1 ℓ)(y)] 32 Toeplitz - Hankel integro - differential equation Due to the uniqueness of the Fourier cosine transform, equation (3.2) has a unique solution in L1(R) as follows f(x) = − √ π 2 [e−x ∗ 1 h+ (e−x ∗ 1 h) ∗ 1 ℓ](x). Theorem 3.2 is proven. REFERENCES [1] A. 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