Applied Mthemticl Sciences, Vol. 8, 214, no. 11, 525-53 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/1.12988/ms.214.312715 The Solution of Volterr Integrl Eqution of the Second Kind by Using the Elzki Trnsform Yechn Song Inh Technicl College Incheon 42-752, Kore Hwjoon Kim Kyungdong University Goseong 219-75, Gngwon, Kore Corresponding uthor Copyright c 214 Yechn Song nd Hwjoon Kim. This is n open ccess rticle distributed under the Cretive Commons Attribution License, which permits unrestricted use, distribution, nd reproduction in ny medium, provided the originl work is properly cited. Abstrct We hve checked the Volterr integrl equtions of the second kind with n integrl of the form of convolution by using the Elzki trnsform. Mthemtics Subject Clssifiction: 45B5, 44A5 Keywords: Volterr integrl eqution, Elzki trnsform 1 Introduction The Volterr integrl equtions re specil type of integrl equtions, nd they re divided into the first kind nd the second[6]. A liner Volterr eqution of the second kind hs the form of x(t) =y(t)+ k(t, s)x(s)ds,
526 Yechn Song nd Hwjoon Kim where k is the kernel. The second(first) kind mens tht the unknown function y occurs(does not occur) outside of the integrl[7]. In this rticle, we would like to del with the form of x(t) =y(t)+ t k(t s)x(s)ds, where k is the kernel. On the other hnd, the Elzki trnsform[2, 4], modified Lplce[3]/Sumudu[1, 9, 11] trnsform, ws introduced by Elzki in 211 to solve initil vlue problems in controlling engineering problems, nd it is defined by T (u) =u e t/u f(t)dt, for E[f(t)] = T (u)[1]. Elzki insists tht the Elzki trnsform should be esily pplied to the initil vlue problems with less computtionl work, nd solve vrious exmples which is not solved by the Lplce or the Sumudu trnsform[5]. By the reson, we would like to check some Volterr integrl equtions of the second kind by using the Elzki trnsform. 2 The solution of Volterr integrl eqution of the second kind by using the Elzki trnsform We proved the convolution of Elzki trnsform in [1] by the different method with Elzki. Tht is E(f g) = 1 u E(f)E(g) for E(f) is the Elzki trnsform of f. In generl, we cn find the solution of differentil equtions with vrible coefficients by using the Elzki trnsform s following; Theorem 2.1 The solution of Volterr integrl eqution of the second kind y(t) =x(t)+ K(t τ)y(τ)dτ ( ) is expressed by y(t) =E 1 (T (u)) = E 1 ( ux u K ), where K is the kernel nd E[y(t)] = T (u).
Volterr integrl eqution by using Elzki trnsform 527 Proof. Let us consider the generl cse. Let E[y(t)] = T (u), E(r) = R nd E(q) =Q. If f(y(t)) = r(t) is given, tking the Elzki trnsform on both sides, we hve T (u) = 1 u E(r)E(q) = 1 u RQ for Q is the trnsfer function. Let us tke the inverse Elzki trnsform on both sides, we get y = T 1 (u) =r q = T 1 ( 1 u RQ) for * is the stndrd nottion of convolution. With the bove ide, let us tke the Elzki trnsform on the eqution (*). Then we hve T (u) =X + E(y k) =X + 1 u T (u)k for X = E(x) nd for K = E(k). Orgnizing the equlity, we hve T (u) = for the kernel k. Therefore we hve ux u K y(t) =E 1 (T (u)) = E 1 ( ux u K ). Let us check the following exmples. Exmple 2.2 Let us consider y + y =with the initil condition y() =, y () = 1. Solution. We cn esily find tht the solution is y = sin t. Next, we would like to check this by the integrl eqution. Since the eqution is equivlent to y(t) =t + (τ t)y(τ)dτ, we cn rewrite the eqution with y = t + {y ( I)} for * is the stndrd nottion of convolution nd for I is the identity function. Tking the Elzki trnsform on both sides, we hve T (u) =u 3 1 u T (u)u3 = u 3 u 2 T (u) for E[y(t)] = T (u), becuse of E(t) =u 3. Thus we hve T (u) = u3 1+u 2.
528 Yechn Song nd Hwjoon Kim As we scn tble of the Elzki trnsform[5], we hve the solution y = sin t becuse of E[sin(t)] = u3 1+ 2 u. 2 Finlly, let us directly pproch by the Elzki trnsform. Note tht [y (t)] = T (u)/u 2 y() uy () holds, by the definition. Tking the Elzki trnsform on both sides, we hve T (u)/u 2 y() uy () + T (u) = for E[y(t)] = T (u). Collecting the T (u)-terms, we hve T (u)( 1 u +1)=u 2 nd so, we hve the solution y = sin t. Now tht let us consider the Volterr integrl eqution of the second kind. Exmple 2.3 Solve the Volterr integrl eqution of the second kind y(t) y(τ)sin(t τ)dτ = t [7]. Solution. The given eqution cn be rewritten by y y sint = t. Let us write E[y(t)] = T (u) nd pply the convolution theorem. Then we hve T (u) 1 u T (u) u 3 1+u = 2 u3. Orgnizing the equlity, we hve T (u) = u 3 (1 + u 2 )=u 3 + u 5. As we scn tble of Elzki trnsforms[5], we hve the nswer This is the sme result s tht of [7]. y(t) =t + t3 6.
Volterr integrl eqution by using Elzki trnsform 529 Exmple 2.4 Solve the Volterr integrl eqution y(t) for h is hyperbolic function. τy(t τ)dτ =1 Solution. Writing y t y =1, we hve T (u) 1 E(t)T (u) =E(1) u for E[y(t)] = T (u). From the tble of Elzki trnsform[5], we hve Arrnging the equlity, we hve T (u) 1 u u3 T (u) =u 2. T (u) = u2 1 u. 2 Tking the inverse Elzki trnsform, we hve the solution for h is hyperbolic function. y(t) = cos ht It is well known fct tht the first order ODE dy = f(x, y) dx with the condition y() =y is rewritten to φ(x) =y + x f(t, φ(t))dt, where f is continuous nd contins the point (, y ). Similrly, n initil vlue problem y + A(t)y + B(t)y = with the condition y() =y, y () =y 1 is rewritten to the Volterr integrl eqution of the second kind y(t) =f(t)+ k(t, τ)y(τ)dτ, ( ) where K(t, τ) = A(τ)+(τ t)(b(τ) A (τ)). And the bove f(t) is continuous on [, b] nd the kernel k is continuous on the tringulr region R in the tτ-plne given by τ t, t b. Then we know tht (**) hs unique solution[8] y on [, b].
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