Cosine transform on time scales

HNUE JOURNAL OF SCIENCE DOI: 10.18173/2354-1059.2018-0063 Natural Sciences, 2018, Volume 63, Issue 11, pp. 3-10 This paper is available online at COSINE TRANSFORM ON TIME SCALES Nguyen Xuan Thao and Tran Kim Huong Faculty of Mathematics, Hanoi University of Science and Technology Abstract. In this article, we construct and study the cosine (Tc) integral transform on the time scales Th. Some operator properties are obtained. Convolution for the (Tc) transform on the time scales and its application to solve the difference equations are considered. Keyworks: Cosine transform on the time scales, convolution, the difference equation. 1. Introduction Time scales theory emerged in 1988 in Stefan Hilger’s PhD thesis. In 2001-2002, it was given in document [1]. A number of theoretical foundations were introduced in the literature and applied to the dynamics equations on the time scales in documents [1-7]. Calculations on the current time scales are concerned by domestic and foreign mathematicians. A number of articles which cover the application of time scales for studying optimal mathematical problems, diffrential calculus and macroeconomic models applied to problem solving game. In 2008, the research team Robert J. Marks IIa, Ian A. Gravagnea, and John M. Davisb built the Fourier transform on the time scales in [8] as defined by the expression: X (u) = F {x} (u) := ∫ t∈T x (t)e−j2piut∆t. This article examines an important component of the Fourier transform on the time scales. 2. Cosine and sine integral transform on the time scales Definition 2.1 ([9]). Time scale Th is determined by Th :=  0 if h = ∞ hZ if h > 0 R if h = 0 . Received November 2, 2018. Revised: November 19, 2018. Accepted November 26, 2018 Contact Nguyen Xuan Thao, e-mail address: thao.nguyenxuan@hust.edu.vn 3 Nguyen Xuan Thao and Tran Kim Huong Definition 2.2. Suppose that f : Th → R. The cosine and sine transforms of the function f on the time scales Th are defined as follow: f˜(y) = (Tcf) (y) = h ∞∑ n=−∞ f (nh) cos (ynh) , ∀y ∈ [ − 1 2h , 1 2h ] , h > 0, (2.1) and (Tsf) (y) = h ∞∑ n=−∞ f (nh) sin (ynh) ,∀y ∈ [ − 1 2h , 1 2h ] , h > 0, . (2.2) Definition 2.3. The space L1 (Th) are the set of all functions f : Th → R such that h ∞∑ n=−∞ |f (nh)| <∞ with the corresponding norm ‖f‖ 1 = h ∞∑ n=−∞ |f (nh)|, h > 0. Characteristic. Suppose that f, g ∈ L1(Th). The following properties hold. (i) Linearity (Tc (αf + βg)) (y) = α (Tcf) (y) + β (Tcg) (y) , ∀α, β ∈ C,∀y ∈ [ − 1 2h , 1 2h ] , (Ts (αf + βg)) (y) = α (Tsf) (y) + β (Tsg) (y) . (ii) Displacement (Tcf) (x+ y) = ((Tcf) (x) cos yt)− ((Tsf) (x) sin yt) ,∀t ∈ Th, ∀x, y ∈ [ − 1 2h , 1 2h ] ; x+ y ∈ [ − 1 2h , 1 2h ] , (Tcf) (x− y) = ((Tcf) (x) cos yt) + ((Tsf) (x) sinyt) , x− y ∈ [ − 1 2h , 1 2h ] , (Tsf) (x+ y) = ((Tsf) (x) cos yt) + ((Tcf) (x) sinyt) , (Tsf) (x− y) = ((Tsf) (x) cos yt)− ((Tcf) (x) sinyt) . (iii) If f is an even function, we have (Tcf (−t)) (y) = (Tcf (t)) (y) ,∀t ∈ Th,∀y ∈ [ − 1 2h , 1 2h ] , (Tsf (−t)) (y) = 0. (iv) If f is an odd function, we have (Tcf (−t)) (y) = 0,∀t ∈ Th,∀y ∈ [ − 1 2h , 1 2h ] , (Tsf (−t)) (y) = (Tsf (t)) (y) . 4 Cosine transform on time scales Comment: When T0 = R, we have (i) (Tcf) (y) + (Tsf) (y) = (Hf) (y), in which: (Hf) (y) = 1 2pi ∞∫ −∞ f (x)cas (xy) dx, is Hartley transform [10], with cas (xy) = cos (xy) + sin (xy) . (ii) (Tcf) (y) + i (Tsf) (y) = (Ff) (y) , where (Ff) (y) denotes the Fourier transform of f [11]. 3. Convolution for the cosine integral transform on the time scales Th Definition 3.1. The convolution of two functions f, g for integral transform Tc is defined as follows (f ∗ g) (t) = h ∞∑ n=−∞ f (nh) [g (t− nh) + g (t+ nh)] , t ∈ Th. (3.1) Theorem 3.1. Suppose that f, g ∈ L1 (Th) , then the convolution (f ∗ g) is well-defined on Th. Moreover, (f ∗ g) ∈ L1 (Th) and the following inequality is true ‖f ∗ g‖ 1 ≤ 2‖f‖ 1 ‖g‖ 1 (3.2) Beside, this convolution satisfies the factorization equality Tc (f ∗ g) (y) = 2 (Tcf) (y) · (Tcg) (y) , ∀y ∈ [ − 1 2h , 1 2h ] . (3.3) Proof. First of all, we prove: f ∗ g ∈ L1(Th). Using definition 3 for equation (3.1), we have ‖f ∗ g‖ 1 = h ∞∑ n=−∞ ∣∣∣∣h ∞∑ m=−∞ f (mh) [g ((n−m) h) + g ((n+m)h)] ∣∣∣∣ ≤ h2 ∞∑ n=−∞ ∣∣∣∣ ∞∑ m=−∞ f (mh)g ((n−m) h) ∣∣∣∣+ h2 ∞∑ n=−∞ ∣∣∣∣ ∞∑ m=−∞ f (mh)g ((n+m)h) ∣∣∣∣ = I1 + I2. (3.4) Put k = n−m, we get I1 = h 2 ∞∑ k=−∞ ∣∣∣∣∣ ∞∑ m=−∞ f (mh)g (kh) ∣∣∣∣∣. (3.5) Similar, put j = n−m, we have I2 = h 2 ∞∑ j=−∞ ∣∣∣∣∣ ∞∑ m=−∞ f (mh)g (jh) ∣∣∣∣∣. (3.6) 5 Nguyen Xuan Thao and Tran Kim Huong Combining equalities (3.5), (3.6), and (3.4), we have I1 + I2 = 2h 2 ∞∑ k=−∞ |g (kh)| ∣∣∣∣∣ ∞∑ m=−∞ f (mh) ∣∣∣∣∣. (3.7) Since (3.4), (3.7) and definition 3, we have ‖f ∗ g‖ 1 ≤ 2 ( h ∞∑ k=−∞ |g (kh)| )( h ∞∑ m=−∞ |f (mh)| ) = 2‖f‖ 1 ‖g‖ 1 . It shows that f ∗ g ∈ L1(Th). From (2.1) and (3.1), we have Tc (f ∗ g) (y) = h ∞∑ n=−∞ (f ∗ g) (nh) cos (ynh) = h2 ∞∑ n=−∞ ∞∑ m=−∞ f (mh) [g ((n−m) h) + g ((n+m) h)] cos (ynh) = I1 + I2. (3.8) Changing variables, put k = −n−m and j = n+m, we get I1 = h 2 ∞∑ k=−∞ ∞∑ m=−∞ f (mh)g (kh) cos ((m+ k) yh) , (3.9) I2 = h 2 ∞∑ j=−∞ ∞∑ m=−∞ f (mh)g (jh) cos ((j −m) yh) . (3.10) Substituting (3.9), (3.10) to (3.8), we get I1 + I2 = h 2 ∞∑ k=−∞ ∞∑ m=−∞ f (mh)g (kh) [cos ((m+ k) yh) + cos ((m− k) yh)] = 2h2 ∞∑ k=−∞ ∞∑ m=−∞ f (mh)g (kh) cos (myh) cos (kyh) = 2 ( h ∞∑ k=−∞ g (kh) cos (kyh) )( h ∞∑ m=−∞ f (mh) cos (myh) ) = 2 (Tcf) (y) (Tcg) (y) . The proof is completed. Remark. Time scales Th equived by the convolution product operator is a commutative ring. 6 Cosine transform on time scales Definition 3.2. The space L1 ( Th, e nh ) is defined as follows L1 ( Th, e nh ) : = { f : Th → R |h ∞∑ n=−∞ enh |f (nh)| <∞ } . Theorem 3.2 (Titchmarsh’s type theorem). . Suppose that f, g ∈ L1 ( Th, e nh ) . If (f ∗ g) (t) ≡ 0,∀t ∈ Th, then either f(t) ≡ 0 or g(t) ≡ 0, t ∈ Th. Proof. Suppose that (f ∗ g) (t) = 0, ∀t ∈ Th, combining (2.1) and (3.1) we have: Tc (f ∗ g) (y) = 0, ∀y ∈ [ − 1 2h , 1 2h ] . (3.11) Using (3.3) for equation (3.1) we get (Tcf) (y) (Tgf) (y) ≡ 0, ∀y ∈ [ − 1 2h , 1 2h ] . Consider is integral transform Tc (Tcf) (y) = h ∞∑ n=−∞ f (nh) cos (ynh) , ∀y ∈ [ − 1 2h , 1 2h ] . We have, ∣∣∣∣ dkdyk cos (ynh) f (nh) ∣∣∣∣ = ∣∣∣f (nh) (nh)k cos(ynh+ kpi2)∣∣∣ ≤ ∣∣∣(nh)kf (nh)∣∣∣ = hk ∣∣∣e−nh(nh)kenhf (nh)∣∣∣ ≤ ∣∣∣e−nh(nh)k∣∣∣ ∣∣∣enhf (nh)∣∣∣ , k = 1, 2, 3 . . . , (3.12) and 0 ≤ e−nh(nh)k = e−nh (nh)k k! k! < e−nhenhk! = k!. (3.13) Combining (3.12), (3.13) and definition 5, we have∣∣∣∣dk (Tcf) (y)dyk ∣∣∣∣ ≤ k! ( h ∞∑ n=−∞ enh |f (nh)| ) ≤ Ck! (3.14) Consider Taylor’s expansion (Tcf) (y) = ∞∑ n=0 1 k! dk (Tcf) (y0) dyk (y − y0) k. (3.15) 7 Nguyen Xuan Thao and Tran Kim Huong From (3.12)-(3.15) we obtain∣∣∣∣ 1k! dk (Tcf) (y)dyk (y − y0)k ∣∣∣∣ ≤ C 1k!k!|y − y0|k = C|y − y0|k. (3.16) Estimation (3.16) shows that series is convergent for |y−y0| < 1. Hence, (Tcf) (y) is analytic for all y ∈ [ − 1 2h , 1 2h ] , so that |y − y0| < 1. Similarly, (Tcg) (y) is analytic ∀y, y0 ∈ [ − 1 2h , 1 2h ] , when |y − y0| < 1. Since cos (ynh) = 0 at the discrete points which are not converged, where (Tcf) (y),(Tcg) (y) are analytic, we have (Tcf) (y) ≡ 0 or (Tcg) (y) ≡ 0, ∀y ∈ [ − 1 2h , 1 2h ] . That leads to f (nh) = 0, ∀n or g (mh) = 0,∀m. The proof is completed. 4. Application on solving linear difference equation In this section we consider the difference enquations of the first kind as well as the second kind. First, based on Wiener-Levy theorem [10], we have the proposition. Proposition 4.1 (Wiener-Levy’s type theorem). If f˜ is a transform Tc of function f belonging to L1(Th), φ(u) is an analytic function on the domain value f˜(x) and φ(0) = 0. As a result, φ(f˜(x)) is a transform Tc of function belonging to L1(Th). From the theorem on the inverse transform of the Fourier transform [11] on Th and the transform (Ff) (t) = (Tcf) (t) + i (Tsf) (t) , ∀t ∈ Th. We get we the proposition. Proposition 4.2. If f is an even function, then we have the inverse transform f (t) = ( T−1c f˜ ) (t) = B∫ −B (Tcf) (u) cos (2piut) du, ∀t ∈ Th. In which B = 1 2h , h > 0, u ∈ R, f˜(t) = (Tcf)(t). 4.1. Difference equation of the second kind Conside difference equation f (t) + h ∞∑ n=−∞ f (nh) [g (t− nh) + g (t+ nh)] = h (t) , (4.1) in which, g, h are known functions in L1(Th), f is unknown function in L1(Th). Theorem 4.1. With condition: 1 + 2 (Tcg) (y) 6= 0, ∀y ∈ [ − 1 2h , 1 2h ] , the equation (4.1) has the unique solution in L1(Th) is f = h − (h ∗ ϕ) , in which the function ϕ ∈ L1(Th) defined as follows: (Tcϕ) (y) = (Tcg) (y) 1 + 2 (Tcg) (y) . 8 Cosine transform on time scales Proof. Equation (4.1) can be rewritten in the form f + (f ∗ g) = h. (4.2) Using theorem 1 and theorem 3 for equation (4.2), we get (Tcf) (y) = (Tch) (y) 1 + 2 (Tcg) (y) (4.3) From (4.3), we have (Tcf) (y) = (Tch) (y) [ 1− 2 (Tcg) (y) 1 + 2 (Tcg) (y) ] . (4.4) Function φ (s) = s 1 + s has analytics based on s, φ(0) = 0. Therefore, according to proposition 1, there exist ϕ ∈ L1(Th) such that (Tcϕ) (y) = (Tcg) (y) 1 + 2 (Tcg) (y) (4.5) Substituting (4.5) to (4.4) we have (Tcf) (y) = (Tch) (y) [1− 2 (Tcϕ) (y)] . (4.6) According to (4.6), theorem 1 and combined with proposition 2, we get the unique solution f = h− (h ∗ ϕ) ∈ L1 (Th) . The proof is completed. 4.2. Difference equation of the first kind Conside difference equation h ∞∑ n=−∞ f (nh) [g (t− nh) + g (t+ nh)] = h (t) , (4.7) in which: g, h are known functions in L1(Th), f is hidden function in L1(Th). Theorem 4.2. Supposing that functions g, k ∈ L1(Th), satisfy the following conditions: (Tcg) (y) 6= 0 and 1 2 (Tcg) (y) = (Tck) (y) , ∀y ∈ [ −1 2h , 1 2h ] . 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