On the Solutions of First and Second Order Nonlinear Initial Value Problems

Transcription

1 Proceedings of he World Congress on Engineering 13 Vol I, WCE 13, July 3-5, 13, London, U.K. On he Soluions of Firs and Second Order Nonlinear Iniial Value Problems Sia Charkri Absrac In his paper, we show a simple new mehod of solving firs and second order nonlinear differenial equaions in he form and y () + a y() = f(, y),, (1) y() = α, () ay () + by () + cy() = g(, y, y ),, (3) y() = β, y () = β 1, (4) a,, a, b, c, α, β, β 1 are given consans, and f(, y) and g(, y, y ) are given nonlinear funcions. By subsiuing he given consans and funcions, hen using an ieraion mehod, soluions are eily obained. Moreover, some examples are shown exac soluions. Keywords: iniial value problem, successive approximaion, differenial equaion, volerra inegral equaion, laplace ransform 1 Inroducion Finding exac soluions of nonlinear iniial value problems (IVPs) is a goal for mahemaicians, engineers, and scieniss, and i plays an imporan role in real world applicaions. In recen years, firs and second order nonlinear IVPs were considered by many auhors. For insance, [1-] used he Adomian decomposiion mehod (ADM) o solve nonlinear differenial equaions such Duffing- Vanderpole equaion, [3-5] solved nonlinear IVPs by he Laplace Adomian decomposiion mehod (LADM), [6-7] obained approximae soluions by he mehod of differenial ransforms (DTM), and he variaional ieraion mehods (VIM) were used by many auhors [8-9]. Alhough ADM, LADM and DTM are effecive and famous mehods for solving nonlinear equaions, here are limiaions for using. For example, ADM, LADM and DTM require infinie series o ge soluions which someimes i is difficul o invesigae closed form soluion from infinie series. And we have o use some analyical mehods o complee hose schemes by inverse ransformaions This work h been suppored by Maejo universiy research fund (MJU-55-43). Sia Charkri is wih he Deparmen of Mahemaics, Faculy of Sciences, Maejo Universiy, Chiangmai, 59 Thailand (Tel: ; sia@mju.ac.h). ISBN: ISSN: (Prin); ISSN: (Online) of infinie series in order o obain soluions. Furhermore, VIM needs Lagrange muliplier before using ieraion formula. Recenly, Sia [1] inroduced an alernaive mehod for finding soluions of nonlinear higher order IVPs by convering IVP ino Volerra inegral equaion. Then by he use of he successive approximaion, a high accuracy soluion will be obained. However, in he work of [1], i is inroduced in a general form and requires he inverse Laplace ransforms of some funcions in main resuls of [1]. Moivaed by he abovemenioned work, in his work, a new mehod needs no ransformaion or linearizaion or Lagrange muliplier, and i shows some formul for solving he firs order IVP (1)-() and he second order IVP (3)-(4). A soluion is eily obained by jus having some bic knowledge of inegraions and subsiuing given consans ino he formula. Then an ieraive mehod is needed o seek an approximae or exac soluion. Finally, some examples esablish ha his alernaive mehod is very simple and high performance. Bic idea of Laplace ransforms In his secion, we are going o review some bic idea of he Laplace ransforms o use imporan ools of our main resuls. Definiion.1 The Laplace ransform of a funcion f(), defined for all real numbers, is he funcion F (s), defined by F (s) = Lf()} = e s f()d. Definiion. Le he Laplace ransform of f() is Lf()} = F (s), hen we say ha he inverse Laplace ransform of F (s) is f(). Or i is defined by L 1 F (s)} = f(). Propery.1 Le ω, c 1 and c be given consans. (P1) Inverse Laplace ransforms of some funcions L 1 1 sin ω s } = + ω ω, s L 1 s } = cos ω, + ω L 1 1 sinh ω s } =, L 1 s ω ω s } = cosh ω. ω WCE 13

2 Proceedings of he World Congress on Engineering 13 Vol I, WCE 13, July 3-5, 13, London, U.K. (P) Lineariy propery Lc 1 f() + c g()} = c 1 Lf()} + c Lg()}, L 1 c 1 F (s) + c G(s)} = c 1 L 1 F (s)} + c L 1 G(s)}. (P3) Shifing propery Le ω f()} = F (s ω), L 1 F (s ω)} = e ω f(). (P4) Laplace ransform of derivaives Lf ()} = slf()} f(), Lf ()} = s Lf()} sf() f (). Definiion.3 The convoluion of f() and g() is defined f() g() = f(x)g( x)dx. Propery. Properies of he convoluion (P5) f g = g f, (P6) L 1 F (s).g(s)} = f g. For more deails abou he mehod of Laplace ransform, see [11]. 3 Main Resuls 3.1 Lemma Lemma 3.1 Suppose ha aεr +,and b, c, m, pεr. Le = b c, hen he inverse Laplace ransforms saisfy he followings: (H1) for >, + bs + c } = k 1e α1 + k e α, k 1 = ( b)m + ap a (H) for <, α 1 = b a, k = ( + b)m ap a,, α = b, a + bs + c } = eα (c 1 cos ω + c sin ω), ap bm a, ω = ISBN: ISSN: (Prin); ISSN: (Online) α = b a, c 1 = m a, c = a. Proof Since ms + p + bs + c = b bm m(s + a ) + (p a ) a a ) + (c b ) = m a [ a ) a ) + ( c b ) ] ap bm +( a )( 1 a ) + ( c b ) ). By aking inverse Laplace ransforms and using (P), we obain + bs + c } = L 1 m a [ a ) a ) + ( c b ) ]} By using (P3), we have +L 1 ap bm 1 ( a ) a ) + ( c b ) }. b m s } = e a [ + bs + c a L 1 s + ( c b ) }] +e b ap bm a [( a )L 1 1 s + ( c b ) }]. Le = b c > and by using (P1), hen b m } = e a [ + bs + c a cosh a bm + (ap a ) sinh Since cosh x = ex +e x and sinh x = ex e x, we have ha b)m + ap + bs + c } = (( a b )e a + ( ( + b)m ap a b )e a. On he oher hand, by seing <, we have ha + bs + c The proof is compleed. 3. Theorems b } = e a [ m a cos a ap bm + ( a ) sin a ]. Theorem 3.1 Suppose ha εr +,and a, α εr, and f : [o, ) R R. A nonlinear iniial value problem y () + a y() = f(, y),, y() = α, is equal o he Volerra inegral equaion y() = α e a a e a e a x f(x, y(x))dx. a ]. WCE 13

3 Proceedings of he World Congress on Engineering 13 Vol I, WCE 13, July 3-5, 13, London, U.K. Theorem 3. A nonlinear iniial value problem (1)- () h a closed form soluion if y() = lim y i (), y i+1 () = α e a a e a for i =, 1,,..., and y () = α e a. e a x f(x, y i (x))dx, Theorem 3.3 Suppose ha aεr +,and b, c, β, β 1 εr, and g : [o, ) R R. Given = b c, a nonlinear iniial value problem ay () + by () + cy() = g(, y, y ),, y() = β, y () = β 1, is equal o he Volerra inegral equaion followings: Le m = aβ and p = aβ 1 + bβ, (A1) for >, y() = k 1 e α1 + k e α + 1 (e α1( x) e α( x) )g(x, y(x), y (x))dx, k 1 = ( b)m + ap a b α 1 = a (A) for <, y() = e α (c 1 cos ω + c sin ω), k = ( + b)m ap a,, α = b, a + e α( x) sin ω( x)g(x, y(x), y (x))dx, α = b a, c 1 = m a, c = ap bm a and ω = a. 3.3 Proofs of Theorems A proof of Theorem 3.1 Consider he differenial equaion (1), and by aking Laplace ransforms and using (P) and (P4), we have Ly ()} + a Ly()} = Lf(, y)}, (sly} y()) + a Ly} = Lf(, y)}. From he iniial condiion (), we ge ha i.e., ( s + a )Ly} = α + Lf(, y)}, Ly} = α + Lf}. s + a s + a By aking inverse Laplace ransforms and using (P) and (P6), we have y() = L 1 α } + L 1 Lf} } s + a s + a = α L 1 1 } + f L 1 1 } s + a s + a = α e a + f 1 e a. This complees he proof by he definiion of convoluion and (P5). A proof of Theorem 3.3 Consider he differenial equaion (3), and by aking Laplace ransforms and using (P) and (P4), we have ha aly ()} + bly ()} + cly()} = Lg(, y, y )}, a(s Ly} sy() y ()) + b(sly} y()) + cly} From he iniial condiion (4), we ge ha = Lg(, y, y )}. Theorem 3.4 A nonlinear iniial value problem (3)- (4) h a closed form soluion if y() = lim y i (), i =, 1,,..., (B1) for >, and y () = k 1 e α1 + k e α, y i+1 () = e α (c 1 cos ω + c sin ω) + e α( x) sin ω( x)g(x, y i (x), y i(x))dx. Ly} = ms + p + bs + c + Lg(, y, y )} + bs + c, m = aβ and p = aβ 1 + bβ. By aking he inverse Laplace ransforms and using (P), we have y i+1 () = k 1 e α1 + k e α y() = bs + c } + Lg} L 1 }. (5) + bs + c (e α1( x) e α( x) )g(x, y i (x), y i(x))dx, Consider he second erm of Eq.(5), and using (P5), (P6) and (H1), and by seing m =, p = 1 and >, we (B) for <, and y () = e α (c 1 cos ω + c sin ω), have ISBN: ISSN: (Prin); ISSN: (Online) L 1 Lg} + bs + c } = g 1 L 1 + bs + c } = g 1 (e α1 e α ). WCE 13

4 Proceedings of he World Congress on Engineering 13 Vol I, WCE 13, July 3-5, 13, London, U.K. Moreover, by using (H) wih m = and p = 1, for <, we have L 1 Lg} + bs + c } = g 1 L 1 + bs + c } = g e α sin ω. Finally, we complee he proof by Definiion.3 and Lemma 3.1. Proofs of Theorems 3. and 3.4 The proofs are compleed by he successive approximae heorem in [1]. 4 Examples Example 1 exac soluion : y() = e. y y = y; y() = 1, (6) By Theorem 3.1, we conver Eq.(6) o he inegral equaion y = e e e x xy(x)dx. To find a soluion, we use Theorem 3.. Thus y i+1 = e e e x xy i (x)dx, i =, 1,,... and y () = e. So we ge y 1 () = e e e x xe x dx = e (1 ), y () = e e e x xe x (1 x )dx = e (1 + 4 ), y 3 () = e (1 + 4! 6 3! ), y 4 () = e (1 + 4! 6 3! + 8 4! ),. y i () = e (1 + 4 i ( 1)i! i! ), lim y i() = y() = e e = e. This is an exac soluion. Example y y = y e 4 ; y() = 1, y () =, (7) ISBN: ISSN: (Prin); ISSN: (Online) exac soluion : y() = e. We use Theorem 3.3 o conver Eq.(7) o he inegral equaion y = e + 1 (e ( x) 1)(y (x) e 4x )dx. And we use Theorem 3.4 o propose an ieraion formula. So we ge y i+1 () = e + 1 i =, 1,,... and y () = e. Hence, we see ha y 1 () = e + 1 y () = e + 1 (e ( x) 1)(y i (x) e 4x )dx, (e ( x) 1)(e 4x e 4x )dx = e, (e ( x) 1)(e 4x e 4x )dx = e,. y i () = e, lim y i() = y() = e. Example 3 Vanderpole Oscillaor Problem y +y +y+y y = cos cos 3 ; y() =, y () = 1, (8) exac soluion : y() = sin. From Eq.(8), we have y + y = cos cos 3 (1 + y )y. (9) By Theorem 3.3, we conver Eq.(9) ino he inegral equaion and use Theorem 3.4. Then we have y i+1 = sin + sin( x)( cos x cos 3 x (1 + y i (x))y i(x))dx, i =, 1,,... and y () = sin. Thus we have y i () = sin, and 5 Conclusions lim y i() = y() = sin. A new simple way for solving a nonlinear IVP w proposed. I provided a formula of a soluion by jus using a bic knowledge of inegraion. Some examples were given o show he effeciveness of a new mehod. WCE 13

5 Proceedings of he World Congress on Engineering 13 Vol I, WCE 13, July 3-5, 13, London, U.K. References [1] G. Adomian, The Decomposiion Mehod, Kluwer, Boson, [] Gh. Asadi Cordshooli, A. R. Vahidi, Soluions of Duffing - van der Pol Equaion Using Decomposiion Mehod, Adv. Sudies Theor. Phys, V5, pp , 11. [3] Suheil A. Khuri, A Laplace decomposiion algorihm applied o a cls of nonlinear differenial equaion, J. Applied Mahemaics, V14, pp , 1. [4] Onur Kiymaz, An algorihm for solving iniial value problems using Laplace Adomian decomposiion mehod, Applied Mahemaical Sciences, V3, pp , 9. [5] Elcin Yusufoglu, Numerical soluion of Duffing equaion by he Laplace decomposiion algorihm, Applied Mahemaics and Compuaion, V177, pp , 6. [6] Veda Sua Erurk, Differenial Transformaion Mehod for Solving Differenial Equaions, Mahemaical and Compuaional Applicaions, V1, pp , 7. [7] J. Biaza, M. Eslami, Differenial Transform Mehod for Quadraic Riccai Differenial Equaion, Inernaional Journal of Nonlinear Science, V9, pp , 1. [8] B. Baiha, M. S. M. Noorani, I. Hhim, Applicaion of Variaional Ieraion Mehod o a General Riccai Equaion, Inernaional Mahemaical Forum, V, N56, pp , 7. [9] Ji-Huan He, Xu-Hong Wu, Variaional ieraion mehod: New developmen and applicaions, Compuers & mahemaics wih applicaions, V54, pp , 7. [1] Sia Charkri, An alernaive mehod for solving nonlinear differenial equaions via inegral equaions, o be submied. [11] Erwin Kreyszig, Advanced Engineering Mahemaics, Wiley, ISBN: ISSN: (Prin); ISSN: (Online) WCE 13