Chapter XI. Solution of Ordinary Differential Equations

Transcription

1 dy d 7 Capter XI Solution of Ordinary Differential Equations Several approaces are described for te solution of ODE (ordinary differential equations). Tese include: Te Taylor series Metod Te Euler Metod Te Improved Euler Metod Te Runge-Kutta Metods. We will begin our presentation on te numerical solution of a first-order ODE and later on etend te concepts to any order ODE.. Problem Statement Given = f (, y); y( ) = y 0 0 (.) were y( 0 ) = 0 is te initial condition needed to solve te problem. Tat is, y = y 0 at = 0. Determine y as a function of in eiter tabulated form or grap. Table 0. sows te tabulated solution required. Table. Tabulated results n- n- y y 0 y... y n- y n- Were i = i + and ) is given. 0. Te Taylor Series Metod Epand y() about 0 using Taylor series:

2 74 y ( 0) y ( ) = y ( 0) + y ( 0)( 0) + ( 0) +! ( 4) y ( 0) y ( ) +!! ( ) ! 4 Let 0 =, terefore: 4 y ( ) = y0 + f ( 0, y0 ) + f ( 0, y 0) + f ( 0, y 0) + f ( 0, y 0) +!! 4! Note tat f ( 0, y0) f ( 0, y0) df ( 0, y0) = d + dy y We will drop te ( 0,y 0 ) from ere on. Dividing by d we get! (.) Similarly one can derive: df d = f f + y f = F (.) d f d d f d " i d f i d " F F = + y f = F F F = + y f = F Fi Fi = + y f = F i (.4) Eample. Epress te solution of

3 dy d 75 = y wit y( 0) =. Solution. From Equations (.) and (.4) we can write: f = 0 y0 = F f = ( 0 y0) = y0 = F f = + ( 0 y0) = 0 y0 = F f = ( y ) = + + y = F " Substituting 0 = 0 and y 0 = - we get: Tat is: f =, f =, f =, f =,! 4 y ( ) = + +!!! 4! For small we can neglect terms iger tan 4 and ence given one can calculate y(). Setting 0 = and y 0 = y() we can repeat te above steps to calculate y() and so on.. Euler Metod In te Euler metod te first two terms of te Taylor series are used. y ( ) = y ( ) + f(, y) or y ( ) = y ( ) + f(, y) i+ i i i were i+ = i +, i = 0,,,! Euler metod can be grapically described as sown in Fig... Te slope at ( 0, y 0 ) is used to calculate y( ). Using and y( ) calculate f(, y ) and from tere calculate y( ) and so on.

4 76 Figure. Euler Metod..4 Improved Euler Metod In te improved Euler metod we use tree terms of te Taylor series: y ( i+ ) = y ( i) + f( i, yi) + f ( i, y i)! i = 0!,,,, Te derivative of f( i, y i ) can be calculated using forward difference as follows: were f (, y ) * y i+ is calculated using Euler metod as follows: i i * f ( i+, yi+ ) f ( i, yi) Hence: * yi+ = yi + f ( i, yi) y ( ) = y ( ) + f(, y) + i+ i i i ( ) f i+, yi + f ( i, yi) f ( i, yi)

5 77 wic can be placed in te form: [ ( )] y ( i+ ) = y ( i) + f ( i, y i) + f i+, y i + f ( i, y i) were = y ( i ) + ( k + k ) k = f (, y ) i (, ) k = f + y + k i i i Procedure: 5. Calculate k = f ( i, yi) 6. Calculate y * i = yi + k * 7. Calculate k = f ( i +, yi ) 8. Calculate yi+ = yi + k + k ( ) Te improved Euler metod can be described grapically as sown Figure.. Figure. Improved Euler Metod.

6 78 From Fig.. te average slope f dy d = * f (, y ) + f (, y ) 0 0 = y wit y( 0) =. is used in Euler metod to calculate te net value yi+ = yi + f, wic is te same approac derived in tis section for te Improved Euler metod. Eample. Calculate y(0.) and y(0.) using =0. for te following ODE Solution. Using f (, y) = y and te initial conditions 0 = 0, y0 = we follow te procedure described above.. Calculate k = f (, 0 ) = () 0 ( ) = *. Calculate y = y + k 0 = + 0.* = 09.. * Calculate k = f ( , y ) = f ( 0., 09. ) = Calculate y y 0 y 0. 0 k k 0. ( ) ( ) = Repeating te procedure to obtain y(0.).. Calculate k = f (., ) = * Calculate y = y + k = ( 0. 75) = * Calculate k = f ( + 0., y ) = f ( 0., ) = Calculate 0. y = y(.) 0 = y(.) 0 + ( k + k ) =. + ( ) = Hence:

7 79 y( 0) = y(.) 0 = y(.) 0 = Runge Kutta Metods Te main reason for te development of alternative metods to te Taylor series approac is to avoid te differentiations required by te series. Neiter Euler nor Improved metods require tat we differentiate any function. Te Runge Kutta Metods are basically generalizations of te Improved Euler Metod. For second order Runge-Kutta metod te following is assumed: y ( + ) y ( ) + ak + ak were k = f (, y) k = f ( + m, y + mk ) (.5) Te problem is now stated as follows: obtain te unknowns a, a, m suc tat y( + ) y ( + ) = y ( ) + y ( ) + y ( )! } obtained using te above equation matces te Taylor Series epansion of y( + ) truncated after te first derivative. Te Taylor Series epansion wit te terms truncated after te second derivative is: f y y f y f (, y ) (, ) (, ) y f (, y ) + + = + + y f (, y) f (, y) f (, y) + ( ) + y + ( y)! y y +! (.6) Since te derivatives in te Taylor series epansion are in terms of f(,y) we need to epand f ( + m, y + mk) as a series in terms of f(,y). Tis calls for te use of Taylor series epansion in two variables wic is given by: Hence we can write:

8 80 k = [ f (, y) + mf (, y) + mk f (, y) +!] (.7) y f (, y) Were f(, y) = and fy(, y) = [ y] [ y] y( + ) = y( ) + a f (, y) + a f (, y) + mf (, y) + mk f = y + a f + a f + mf + mff = y + ( a + a ) f + a m ( f + ff ) y ( ) = f df (, y) f f y ( ) = = + = + d y f f ff y f (, y) y Replacing k wit f(,y) and k wit te rigt-and side of Eq.(.7) in Equation (.5) we get: Notice tat we dropped te (,y) just for compactness sake. In te Taylor series epansion (Eq..6) replace: Hence y (.8) ( ff y) y ( + ) = y ( ) + f + f + (.9) Comparing (.8) and (.9) we get: a + a = am= Tese are two equations in tree unknowns and ence we ave te coice to select one of te variables. Coosing m=, terefore Hence a second order Runge-Kutta metod is: a = a =

9 8 y ( + ) = y ( ) + ( k + k) wic turns out to be te Improved Euler Metod. Te Improved Euler metod is actually a special case of te Runge-Kutta metods. We will now tackle te fourt order Runge-Kutta metod. In tis case we assume te Runge-Kutta formulas are: Wat is required is to find te coefficients y ( + ) y ( ) + ak + bk + ck + dk k = f (, y) k = f ( + m, y+ mk) k = f ( + n, y+ mk) k = f ( + p, y+ pk ) 4 4 abcdmnp,,,,,, } duplicate te Taylor series troug te term in 4. Tat is: Since dy d suc tat te above formulas 4 y ( ) y ( ) y ( ) y ( ) y ( ) y ( iv) + = ( )!! 4! = y ( ) = f (, y) ten y ( ) = f f f ( ) = + = + y f f f f y Let F = f + fyf y ( ) = F Similarly and y ( ) = f + ff + f f + f ( f + ff ) y yy y y Let F = f + ff + f f y ( ) = F + f F y yy y

10 8 (iv) y ( ) = f + ff + f f + f f + f ( f + ff + f f ) Let y yy yyy y y yy + ( f + ff )( f + ff ) + f ( f + ff ) y y yy y y F = f + ff + f f + f f y yy yyy (iv) y ( ) = F + f F + F( f + ff ) + f F y y yy y wic allows te Taylor series to be written as: y ( + ) = y ( ) + F + ( F + ff y ) [ y ( y yy) y ] 4 F + f F + f + ff F + f F +! Turning now to te various k values, similar computations produce k = f k = f + mf + m F + m F +! 6 k = f + nf + n F + mnf yf + ( ) n F + m nfyf+ 6mn ( fy + ffyy) F } +! 6 k4 = f + pf + p F + npf yf + ( ) p F + n pf yf + 6np ( f y + ff yy) F + 6mnpf yf } +! 6 Combining tese equations as suggested by te Runge-Kutta formula,

11 8 y( + ) = y( ) + ( a + b+ c+ d) f + ( bm+ cn+ dp) F 4 + ( bm + cn + dp ) F + ( bm + cn + dp ) F ( cmn + dnp) f yf + ( cm n + dn p) f yf ( cmn + dnp ) ( f + ff ) F + dmnp f F +! y yy y Comparison wit te Taylor series results in te eigt conditions a + b+ c+ d = cmn+ dnp = bm + cn + dp = cmn + dnp = bm + cn + dp = cm n + dn p = bm + cn + dp = dmnp = Since we ave eigt equations in seven unknowns we can select one of te unknowns and solve for te rest. A classical solution set is m = n = p = a = d = b = c = 6 leading to te Runge-Kutta formulas y ( + ) = y ( ) + ( k + k + k + k4) 6 k = f (, y) k = f ( +, y + k ) k = f ( +, y + k ) k = f ( +, y + k ) 4

12 84.7 Solving Higer Order ODE An n t order ODE can be broken into n simultaneous first order ODE using a state variable approac wic will be eplained troug an eample. Eample. Using =0. determine y at t=0. and 0. for te following ODE: d y dy + + y = 0. dt dt given y() 0 = y () 0 = 00. Solve using (a) Euler. (b) Improved Euler. (c) Fourt order Runge-Kutta. Solution. Let = y = y = y = = y = 0. y y = 0. or in matri form, d 0. dt = 0. dx or = G (, t X ) dt X( 0) = = t = 0 y() 0 0 () 0 = y 0 Tis approac applies to any order ODE as follows. For an n t order ODE n d y a d n y a d n y n + n + n +! + ay n = f(, ty) dt dt dt ( n ) given y() 0 = α, y () 0 = α,! y () 0 = α n

13 85 To place in te state variable form let n = y = y = y " = y ( n ) (a) Euler We can terefore write: = = = 4 " = f (, t ) a a! a n n n n Te coefficients of te ODE can be eiter constants or functions of t and/or y. Procedure At t= t i. Calculate K = G( t i, X i ). Calculate Xi+ = Xi + G( ti, Xi) Hence at t=0. K = 0. = t = = At t= (.) 0. X = X 0 + K =. 0 + = = y 0. = y (.) 0 y(.) 0 = 00.

14 86. K = = t = = X = X + K = 0. + y(.) 0 = (.) = = y 09. = y (.) 0 (b) Improved Euler Metod Procedure At t= t i. Calculate K = G( t i, X i ). Calculate XEST = Xi + K. Calculate K G(, XEST) = t i + 4. Calculate Xi+ = Xi + [ K + K] Hence at t=0. K = G( t0, X 0 ) = = t = = XEST = X 0 + =. 0 + = Κ. K = G( t = t0 +, XEST) = #$% = 09. At t= Xi = X0 + ( K + K) = 0 + y(.) 0 = =

15 K = G( t, X) = G(., ) = = XEST = X + K = = 085. & K = G( t +, XEST) = G(., 085. = = X = X + ( K + K) = = y(.) 0 = (c) Runge-Kutta 4 t order Metod Procedure At t= t i. Calculate K = G( t i, X i ). Calculate XEST = Xi + K. Calculate K = G( ti +, XEST) 4. Calculate XEST = Xi + K 5. Calculate K = G( ti +, XEST)

16 88 6. Calculate XEST = Xi + K 7. Calculate K4 = G( ti +, XEST) 6 8. Calculate Xi+ = Xi + [ K + K + K + K4] Te calculations follow a similar procedure to te Improved Euler metod and are left to te reader as an eercise. Eercise. Use te Runge-Kutta metod to solve te following ODE for te points y(0.), y(0.) and y(0.). Use =0.. d y ty dy t + + e y = 4y e dt dt y( 0) = 05. y ( 0) = 0. t Now we return back to program development. Te following is a C++ program for solving ODE using eiter Euler or Fourt order Runge-Kutta metod. To verify tat te program works it is tested against an ODE wit known analytical solution. Te ODE used is: Its analytical solution is given by: d y dy + + y = 00., y( 0) = y ( 0) = 0. dt dt 05. yt () = e cos( t) + sin( t) #include <iostream.> #include <iomanip.> //input/output manipulation for formatting using setw, etc. #include <conio.> #include <mat.> class ODE

17 89 // Te default member classification in classes is private. //Private variables double *; //state space vector [0],[],... //Te k's in te Runge-Kutta metod. // eac k is a vector of dimension = order of ODE double *k; double *k; double *k; double *k4; double *est; void(*g)(double tp,double *p,double *dot); //function for calculating te k's double t; //independent variable int neqns; //number of equations=order of ODE double ; //step lengt public: ODE(void (*GP)(double, double *,double *), int n) G=GP; neqns=n; =new double[neqns]; k=new double[neqns]; k=new double[neqns]; k=new double[neqns]; k4=new double[neqns]; est=new double[neqns]; } ~ODE() delete []; delete []k; delete []k; delete []k; delete []k4; delete []est; } void Init(double *p, double tp, double dt); double get_t()

18 90 return t; } double* get() return ; } void Euler(); void RK4(); // 4t order Runge-Kutta }; void ODE::Init(double *p, double tp, double dt) for(int i=0; i<neqns; i++) [i]=p[i]; t=tp; =dt; } void ODE::Euler() (*G)(t,,k); for(int i=0; i<neqns; i++) [i]=[i]+*k[i]; t=t+; } //Fourt order Runge-Kutta void ODE::RK4() int i; (*G)(t,,k); for(i=0; i<neqns; i++) est[i]=[i]+0.5**k[i]; (*G)(t+0.5*,est,k); for(i=0; i<neqns; i++) est[i]=[i]+ 0.5**k[i];

19 9 (*G)(t+0.5*,est,k); for(i=0; i<neqns; i++) est[i]=[i]+*k[i]; (*G)(t+,est,k4); for(i=0; i<neqns; i++) [i]=[i]+ *(k[i]+.0*k[i]+.0*k[i]+k4[i])/6.0; t=t+; } //User supplied routine // ODE y''(t)+y'(t)+y = 0, intitial conditions y'(0)=y(0)= // Analytical solution y=ep(-0.5t)*(cos(sqrt()*t/)+sqrt()*sin(sqrt()*t/)) void Vector( double t, double *, double *dot) dot[0]=[]; dot[]=0.0-[0]-[]; } // int main() ODE eqn(vector, ); //Replace (*G) wit Vector and set neqns= double *, t,, yp[0], tp[0]; =new double[]; //initial conditions [0]=.0; []=.0; t=0.0; =0.; eqn.init(,t,); //set initial conditions tp[0]=0.0; yp[0]=[0]; int i; for(i=; i<0; i++)

20 9 eqn.rk4(); =eqn.get(); yp[i]=[0]; tp[i]=eqn.get_t(); } // print solution cout << "From Runge Kutta" << " " << "t" << " " << "From Analytical solution" << endl; for(i=0; i<0; i++) cout << setprecision(7) << setw(0) << yp[i] << setprecision() << setw(0) << tp[i] << setprecision(7) << setw(0) << (ep(-0.5*tp[i])*(cos(0.866*tp[i])+.7*sin(0.866*tp[i])))<< endl; } getc(); return ; } Print-out From Runge Kutta t From Analytical solution Problems. Solve te following ODE s in 0 # t # 5 using te improved Euler metod wit =0. (write your own program) d y a. + 8y = 0, y( 0) =, y'( 0) = 0 dt

21 b. 9 d y dy y = sin( t), y( 0) = 0, y'( 0) = dt dt c. ( e y) d y t + = t, y( 0) =, y'( 0) = 0 dt. Te number of bacterial cells (P) in a given reactor is related to time in days (t) as described by te following matematical model dp dt = 0. P P If at time t=0, P=0 6 determine te number of cells wen t=0 days. Use te fourt-order Runge-Kutta metod and a time increment of day.. Wen a veicle travels over a roug road, a vertical displacement will result. Consider te following igly idealized model: Figure. Problem. Suppose tat a manufacturer is interested in determining te largest vertical displacement for a veicle traveling at a velocity of 55 miles per our over a sinusoidal surface, given

22 94 m=0 lb-s /in. k=0 lb/in. c=88 lb-s/in. L=50 ft. )= in. Te matematical model describing tis beavior is given by: m d dt c d + + k = mϖ sinϖt dt were ϖ = πvlwere v is te velocity of te veicle. Determine te maimum displacement using te Runge-Kutta 4 t order metod and )t=0.0 s. Assume tat te veicle was initially at rest ( () 0 = '( 0) = 0). 4. Use Taylor series to solve te following second-order ordinary differential equation at t=0.: d d = 0 dt dt wit te initial conditions ( 0) =, '( 0) =.