By COHONGTRAN at 2:18 am, May 19, 2007

Numerical and Graphical Solutions of Duffing Equation by CO.H . TRAN & PHONG . T . NGO - University of Natural Sciences , HCMC Vietnam coth123@math.com & coth123@yahoo.com Copyright 2006 Jan 06 2006 -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------** Abstract : Duffing quation is solved by Runge-Kutta approximating method . ** Subjects: Vibration Mechanics , The Differential equation . ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------NOTE: This worksheet demonstrates Maple's capabilities in the design of a dynamic system and finding the numerical solution of the Duffing equation .

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**********LOI

GIAI SO VA DO THI PHUONG TRﾃ君H DUFFING********** TRAN HONG CO & NGO THANH PHONG - Dai hoc Khoa hoc tu nhien - tp HCM Vietnam coth123@math.com & coth123@yahoo.com

A . Xac dinh he thong . [ System Definition ] > restart: > with(plots): > interface(warnlevel=0):

B. Mo hinh dao dong phi tuyen . [ Non-linear Vibration Model ]

Khao sat mau vat the dang giai tich co hinh mo phong nhu tren [ Consider an analytical model which has the simulation figure above ] Phuong trinh vi phan chuyen dong : [ Differential equation of vibration ] > ptcd:= m*diff(y(t),t,t) + b*diff(y(t),t) + c1*y(t) + c3*y(t)^3 F(t);

m

=

0 dtd 0 dtd y ( t ) 1 1 C b 0 dtd y ( t ) 1 C c1 y ( t ) C c3 y ( t ) 3 = F (t )

Xac dinh cac dieu kien dau . [ Define initial conditions ] > dkdau:= y(0)=0.85,D(y)(0)=0 ;

y (0 ) = 0.85, (D(y ) ) (0 ) = 0 Thay luc F phi tuyen va cac gia tri cua tham so m , b , c1 , c3 . [ Substitute non-linearity force F and parameter values m , b , c1 , c3 ] . > ptcdcuthe:= 10*diff(y(t),`\$`(t,2))+1000*diff(y(t),t)+10000000*y(t)9000000*y(t)^3 = 100*cos(Pi*t);

10

0 dtd 0 dtd y ( t ) 1 1 C 1000 0 dtd y ( t ) 1 C 10000000 y (t ) K 9000000 y ( t ) 3 = 100 cos( p t )

Loi giai so cua phuong trinh vi phan phi tuyen . [ Numerical solution of the non-linear differential equation ] > loigiai:=dsolve({ptcdcuthe,dkdau}, y(t), type=numeric, method=rkf45);with(DEtools):with(plots):u:=t->rhs(loigiai(t)[2]);n:= 5;for i from 0 to n do loigiaiso[i]:=normal(loigiai(i/n)); od;plot(u,-1..5,axes=box);

proc(x_rkf45 ) ...end proc t/rhs( ( loigiai( t ) ) 2 ) 5

⎡ d ⎤ ⎢ t = 0., y (t ) = 0.85000000000000, dt y ( t ) = 0. ⎥ ⎣ ⎦ ⎡ ⎢ t = 0.20000000000000, y ( ⎣ t ) = K0.0000192743776969386379, ⎤ K0.0189311978181220356 ⎥ ⎦

d y(t ) = dt

⎡ ⎢ t = 0.40000000000000, y ( ⎣ t ) = 0.00000309430997202991374, ⎤ K0.0000314943550511259446 ⎥ ⎦

d y(t ) = dt

⎡ ⎢ t = 0.60000000000000, y ( ⎣ t ) = K0.00000308741697915723870, ⎤ K0.0000298814748639385770 ⎥ ⎦

d y(t ) = dt

⎡ ⎢ t = 0.80000000000000, y ( ⎣ t ) = K0.00000808842606692218408, ⎤ K0.0000186448661419193254 ⎥ ⎦

d y(t ) = dt

⎡ d ⎢ t = 1., y ( t ) = K0.0000100001048405655681, dt y ( t ) ⎣ ⎤ = 2.21276548367689596 10-7 ⎥ ⎦

Do thi dao dong voi t = 0 (giay) den t = 1 (giay) [ Graph of vibration from t = 0 (s) to t = 1 (s) > m:=5;for j from 1 to m do plot(u,j/(m)..0,axes=box); od;

5

>

C . Chuong trinh tinh toan mo hinh dao dong phi tuyen

. [ Calculation procedures for the non-linear vibration model ] Bang hoat trinh Maple nay gom 2 thu tuc . VĂ­ du ve cach su dung trong cac bai toan thuc te , hay xem phan tiep sau . * daodong(khoi luong M , hang so can nhot b, he so lo xo c1, he so lo xo c3, luc F , dieu kien dau , hoanh do) This Maple worksheet contains 2 procedures . For examples of applying them to real problems, see the following action . * daodong(mass M ,viscous damping constant b,spring constant c1, spring constant c3, force F , initial conditions ,absciss) > > daodong:=proc(Mf,bf,c1f,c3f,Ff,Nf,dkd,hd)

"--------------------------------------------------------------------------------------------------------------------------------"

> daodong(10,1000,10000000,9000000,100*cos(Pi*t),5,[0.5,0],5.8) ;

" Cac tham so : m = ,"10, "b = ", 1000, "c = ", c, "F = ", 100 cos( p t ) , "N = ", 5 " Phuong trinh vi phan co dang

10

0 dtd 0 dtd y ( t ) 1 1 C 1000 0 dtd y ( t ) 1 C 10000000 y ( t ) C 9000000 y ( t ) 3 = 100 cos( p t )

" Dieu kien dau : "

" y(0) = ", y0, "=", 0.5

" y'(0) = ", (D(y ) ) (0 ), "=", [2 ] " Loi giai so bang phuong phap RUNGE - KUTTA

⎡ d ⎤ ⎢ t = 0., y (t ) = 0.50000000000000, dt y ( t ) = 0. ⎥ ⎣ ⎦ ⎡ d ⎢ t = 1., y ( t ) = K0.0000100001922036141262, dt y ( t ) ⎣ ⎤ = 2.00259407474319750 10-7 ⎥ ⎦

⎡ d ⎢ t = 2., y ( t ) = 0.0000100005096748307881, dt y ( t ) ⎣ ⎤ = 1.15874619398825230 10-7 ⎥ ⎦

⎡ d ⎢ t = 3., y ( t ) = K0.00000999979035875673534, dt y ( t ) ⎣ ⎤ = 3.11729864059405209 10-7 ⎥ ⎦ ⎡ d ⎢ t = 4., y ( t ) = 0.00000999991013686392712, dt y ( t ) ⎣ ⎤ = K5.03447765508274115 10-8 ⎥ ⎦ ⎡ d ⎢ t = 5., y ( t ) = K0.0000100001052242845409, dt y ( t ) ⎣ ⎤ = K1.63032715969934755 10-7 ⎥ ⎦

> mohinh(0.8);

> mohinh(6);

> daodong(10,1000,10000000,9000000,100*cos(Pi*t),10,[0.15,0],1.8);

" Cac tham so : m = ,"10, "b = ", 1000, "c = ", c, "F = ", 100 cos ( p t ) , "N = ", 10 " Phuong trinh vi phan co dang

10

0 dtd 0 dtd y ( t ) 1 1 C 1000 0 dtd y ( t ) 1 C 10000000 y ( t ) C 9000000 y ( t ) 3 = 100 cos( p t )

" Dieu kien dau : "

" y(0) = ", y0, "=", 0.15

" y'(0) = ", (D(y ) ) (0 ), "=", [2 ] " Loi giai so bang phuong phap RUNGE - KUTTA

⎡ d ⎤ ⎢ t = 0., y (t ) = 0.15000000000000, dt y ( t ) = 0. ⎥ ⎣ ⎦ ⎡ d ⎢ t = 1., y ( t ) = K0.0000100002100965684427, dt y ( t ) ⎣ ⎤ = 1.74445319969807240 10-7 ⎥ ⎦

⎡ d ⎢ t = 2., y ( t ) = 0.0000100003219245372214, dt y ( t ) ⎣ ⎤ = K7.58163623338622300 10-8 ⎥ ⎦ ⎡ d ⎢ t = 3., y ( t ) = K0.0000100002374849772969, dt y ( t ) ⎣ ⎤ = K1.90236480746275206 10-7 ⎥ ⎦ ⎡ d ⎢ t = 4., y ( t ) = 0.0000100004144928036077, dt y ( t ) ⎣ ⎤ = 7.70508220812301378 10-8 ⎥ ⎦ ⎡ d ⎢ t = 5., y ( t ) = K0.0000100003957106223904, dt y ( t ) ⎣ ⎤ = K8.90413129754806662 10-8 ⎥ ⎦ ⎡ d ⎢ t = 6., y ( t ) = 0.0000100002617106955826, dt y ( t ) ⎣ ⎤ = K1.53364006735818149 10-7 ⎥ ⎦ ⎡ d ⎢ t = 7., y ( t ) = K0.0000100000618081845140, dt y ( t ) ⎣ ⎤ = 2.64609326752723164 10-7 ⎥ ⎦ ⎡ d ⎢ t = 8., y ( t ) = 0.0000099999993278359298, dt y ( t ) ⎣ ⎤ = K1.30065877267496021 10-7 ⎥ ⎦ ⎡ d ⎢ t = 9., y ( t ) = K0.0000099999913174129808, dt y ( t ) ⎣ ⎤ = 8.24227836181808212 10-8 ⎥ ⎦

⎡ d ⎢ t = 10., y ( t ) = 0.00000999991202646871412, dt y ( t ) ⎣ ⎤ = K1.71616608259521394 10-7 ⎥ ⎦

> mohinh(5);

Gia su ta co dieu kien dau khong tuong thich , dieu gi se xay ra ? [ Suppose we have the incompatible initial conditions , what then ? ] Ta co the thay ket qua bat thuong tu do thi va mo phong dao dong . [ We can see the extraordinary result from the graph and simulation of vibration model ] > daodong(10,1000,10000000,9000000,100*cos(Pi*t),10,[0,0.1],4.8);

" Cac tham so : m = ,"10, "b = ", 1000, "c = ", c, "F = ", 100 cos ( p t ) , "N = ", 10 " Phuong trinh vi phan co dang

10

0 dtd 0 dtd y ( t ) 1 1 C 1000 0 dtd y ( t ) 1 C 10000000 y ( t ) C 9000000 y ( t ) 3 = 100 cos( p t )

" Dieu kien dau : "

" y(0) = ", y0, "=", 0

" y'(0) = ", (D(y ) ) (0 ), "=", [2 ] " Loi giai so bang phuong phap RUNGE - KUTTA

⎡ d ⎤ ⎢ t = 0., y (t ) = 0., dt y ( t ) = 0.10000000000000 ⎥ ⎣ ⎦ ⎡ d ⎢ t = 1., y ( t ) = K0.0000100003031557671699, dt y ( t ) ⎣ ⎤ = 8.56054697571733248 10-8 ⎥ ⎦ ⎡ d ⎢ t = 2., y ( t ) = 0.0000100001186730557259, dt y ( t ) ⎣ ⎤ = 1.38567705758366402 10-7 ⎥ ⎦ ⎡ d ⎢ t = 3., y ( t ) = K0.0000100001881829770924, dt y ( t ) ⎣ ⎤ = 1.43173631414504646 10-7 ⎥ ⎦ ⎡ d ⎢ t = 4., y ( t ) = 0.00000999992373954906582, dt y ( t ) ⎣ ⎤ = 6.60995478059635895 10-8 ⎥ ⎦ ⎡ d ⎢ t = 5., y ( t ) = K0.0000100003314492074806, dt y ( t ) ⎣ ⎤ = 3.14077500376234288 10-8 ⎥ ⎦ ⎡ d ⎢ t = 6., y ( t ) = 0.00000999992793437681750, dt y ( t ) ⎣ ⎤ = 1.53023326407566206 10-7 ⎥ ⎦ ⎡ d ⎢ t = 7., y ( t ) = K0.0000100001940930439043, dt y ( t ) ⎣ ⎤ = K1.73254913410583658 10-7 ⎥ ⎦ ⎡ d ⎢ t = 8., y ( t ) = 0.0000100005562067931500, dt y ( t ) ⎣ ⎤ = K2.19553498185131744 10-7 ⎥ ⎦

⎡ d ⎢ t = 9., y ( t ) = K0.0000100009141784075014, dt y ( t ) ⎣ ⎤ = K3.27149509524763626 10-7 ⎥ ⎦ ⎡ d ⎢ t = 10., y ( t ) = 0.0000100003112333562182, dt y ( t ) ⎣ ⎤ = K6.82392457566916905 10-8 ⎥ ⎦

> mohinh(5);

>

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Disclaimer: While every effort has been made to validate the solutions in this worksheet, the authors are not responsible for any errors contained and are not liable for any damages resulting from the use of this material.

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. It has been made available for product evaluation purposes only and may not be used in any other context without the express permission of Maplesoft. Digitally signed by COHONGTRAN DN: cn=COHONGTRAN, c=VN, o=MMI, ou=NCU HUI, email=cohtran@math.com Reason: I am the author of this document Location: HCMC Date: 2009.06.21 10:30:44 +07'00'

TRẦN

Digitally signed by TRẦN HỒNG CƠ DN: cn=TRẦN HỒNG

HỒNG CƠ

CƠ, c=VN, o=MMI VN, ou=MATH-INF, email=coth123@math .com Reason: I am the author of this document Date: 2007.05.19 02:21:13 +07'00'

MATHEMATICS_NUMERICAL-GRAPHICAL SOLUTIONS OF DUFFING EQUATION

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