SME 3023 Applied Numerical Methods

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1 UNIVERSITI TEKNOLOGI MALAYSIA SME 3023 Applied Numerical Methods Solution of Nonlinear Equations Abu Hasan Abdullah Faculty of Mechanical Engineering Sept 2012 Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

2 Outline 1 Introduction Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

3 Outline 1 Introduction 2 Engineering Applications Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

4 Outline 1 Introduction 2 Engineering Applications 3 Methods Available Incremental Search Method Bisection Method Newton-Raphson Method Secant Method Fixed Point Iteration Method Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

5 Outline 1 Introduction 2 Engineering Applications 3 Methods Available Incremental Search Method Bisection Method Newton-Raphson Method Secant Method Fixed Point Iteration Method 4 Root Finding with Matlab Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

6 Outline 1 Introduction 2 Engineering Applications 3 Methods Available Incremental Search Method Bisection Method Newton-Raphson Method Secant Method Fixed Point Iteration Method 4 Root Finding with Matlab 5 Roots of Nonlinear Polynomials Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

7 Outline 1 Introduction 2 Engineering Applications 3 Methods Available Incremental Search Method Bisection Method Newton-Raphson Method Secant Method Fixed Point Iteration Method 4 Root Finding with Matlab 5 Roots of Nonlinear Polynomials 6 Bibliography Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

8 Introduction Many engineering analyses require determination of value(s) of variable x that satisfy a nonlinear equation f(x) = 0 (1) where x is known as roots of Eq. (1) or zeros of function f(x). Number of roots maybe finite or infinite depending on nature of problem and physical problem Examples of f(x) are x 4 80x = 0 tan x tanh x = 0 polynomial transcendental equation Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

9 Engineering Applications Example Problem 1 Figure 1 : Discharge of water from reservoir. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

10 Engineering Applications Example Problem 1 Problem Statement: Water is discharge from a reservoir through a long pipe as shown in Figure 1.By neglecting the change in the level of the reservoir, the transient velocity of the water flowing from pipe, v(t), can be expressed as: v(t) t p p = tanh 2gh 2gh 2L where h is the height of the fluid in the reservoir, L is the length of the pipe, g is the acceleration due to gravity, and t is the time elapsed from the beginning of the flow. Find the value of h necessary for achieving a velocity of v = 5 m/s at time t = 3 s when L = 5 m and g = 9.81 m/s 2. Solution: Work through the example see Rao (2002). Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

11 Engineering Applications Example Problem 2 Figure 2 : Open belt drive. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

12 Engineering Applications Example Problem 2 Problem Statement: The length of a belt in an open-belt drive, L, is given by L = p 4c 2 (D d) `DθD + dθ d (E1) where θ D = π + 2 sin 1 D d 2c ««θ d = π 2 sin 1 D d 2c (E2,E3) c is the centre distance, D is the diameter of the larger pulley, d is the diameter of the smaller pulley, θ D is the angle of contact of the belt with the larger pulley, and θ d is the angle of contact of the belt with the smaller pulley, Figure 2. If a belt having a length 11 m is used to connect the two pulleys with diameters 0.4 m and 0.2 m, determine the centre distance between the pulleys. Solution: Work through the example see Rao (2002). Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

13 Engineering Applications Example Problem 3 Figure 3 : Contact stress between spheres. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

14 Engineering Applications Example Problem 3 Problem Statement: The shear stress induced along the z-axis when two spheres are in contact with each other, while carrying a load F, is given by where h(λ) = λ tan λ2 1 λ «0.65 h(λ) = τzx p max and λ = z a in which τ zx is the shear force, (E1) p max = 3F 2πa 2 is the maximum pressure deveoped at the centre of the contact area, and (E2) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

15 Engineering Applications Example Problem 3 the radius of the contact area, Figure 3, is 8 1 >< + 1 «91/3 E 1 E >= 2 a = F 1 >: + 1 «>; d 1 d 2 (E3) where E 1 and E 2 are Young s moduli of the two spheres, and d 1 and d 2 are diameters of the two spheres. Poisson s ratios of the two spheres was assume to be 0.3 in deriving Eqs. (E1) and (E3). Determine the value of λ at which the shear stress, given by Eq. (E1), attains its maximum value. Solution: Work through the example see Rao (2002). Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

16 Methods Available Incremental Search Method Bisection Method Newton-Raphson Method Secant Method Regula Falsi Method Fixed Point Iteration Method Bairstow Method Muller s Method Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

17 Methods Available Incremental Search Method Value of x is incremented, by x, from an initial value, x 1, successively until a change in the sign of the function f(x) is observed. f(x) changes sign between x i and x i+1, if it has root in the interval [x i, x i+1 ] which implies f(x i ) f(x i+1 ) < 0 wherever a root is crossed. Plot of the function is usually very useful in guiding the task of finding the interval. A potential problem is the choice of increment length, x: too small, the search can be very time-consuming, too great, closely spaced roots might be missed. Algorithm 1 Sets an initial guess for x i=1, and a stepsize x. 2 Call the function f(x) to calculate its value at x i=1. 3 Increment x i+1 = x i + x. 4 Call the function f(x) to calculate its value at x i+1. 5 Compares the sign of the returned function value f(x i+1 ) to the previous f(x i ). * If the sign of f(x i+1 ) does NOT change, repeat from Step 3 again. * If the sign of f(x i+1 ) does change, the root lies between x i and x i+1. Reduce stepsize x, set x i = x i 1 and repeat from Step 3. Iterate until f(x i+1 ) f(x i ) ε where ε is a specified very small number called tolerance. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

18 Methods Available Incremental Search Method Example 1 Problem Statement: Find the root of the equation f(x) = 1.5x 0.65 (1 + x 2 tan 1 ) 2 « x x 1 + x = 0 (E1) 2 using the incremental search method with x 1 = 0.0 and x (1) = 0.1. Solution: Work through the example. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

19 Methods Available Bisection Method If f(x) is real and continuous in the interval prescribed by a lower bound, x L and upper bound, x U and f(x L) and f(x U) have opposite signs, such that f(x L) f(x U) < 0 (2) then there is at least one real root in the interval [x L, x U] Algorithm 1 Choose lower x L and upper x U guesses for the root such that the function changes sign over the interval. This can be checked by ensuring that f(x L) f(x U) < 0 2 An estimate of the root x R is determined by xl + xu x R = 2 3 Make the following evaluations to determine in which subinterval the root lies: * if f(x L) f(x R) = 0, the root equals x R. Terminate computation. * if f(x L) f(x R) < 0, the root lies in the lower subinterval. Therefore set x U = x R and return to step 2 * if f(x L) f(x R) > 0, the root lies in the upper subinterval. Therefore set x L = x R and return to step 2 Iterate until f(x R) ε where ε is a specified very small number called tolerance. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

20 Methods Available Bisection Method Example 1 Problem Statement: Find the root of the equation f(x) = 1.5x 0.65 (1 + x 2 tan 1 ) 2 « x x 1 + x = 0 (E1) 2 using the bisection method with x (1) l = 0.0, x (1) u = 2.0 and ε = Solution: Modify the Fortran code, Ü½ ½º, and Matlab code, ØºÑ, to solve Eq. (E1) above. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

21 Methods Available Newton-Raphson Method Function f(x) is expressed using Taylor s series about an arbitrary point, x 1 f(x) = f(x 1) + (x x 1)f (x 1) + 1 2! (x x1)2 f (x 1) +... (3) where f, f, f,... are evaluated at x 1 Consider only the first two terms in the expansion f(x) = f(x 1) + (x x 1)f (x 1) = 0 (4) and set f(x) = 0 to give f(x 1) + (x x 1)f (x 1) = 0 (5) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

22 Methods Available Newton-Raphson Method Since higher order derivative terms were neglected in the approximation of f(x) in Eq. (4), the solution of Eq. (5) yields the next approximation to the root (instead of the exact root) as x = x 2 = x 1 f(x1) f (x 1) (6) x 2 denotes improved approximation to the root. For the next improvement we use x 2 in place of x 1 on the RHS of Eq. (6) to obtain x 3 The iterative procedure of Newton-Raphson method is generalized as x i+1 = x i f(x i) : i = 1, 2,... (7) f (x i ) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

23 Methods Available Newton-Raphson Method Example 1 Problem Statement: Find the root of the equation f(x) = 1.5x 0.65 (1 + x 2 tan 1 ) 2 « x x 1 + x = 0 (E1) 2 using the Newton-Raphson method with the starting point x 1 = 0.0 and the convergence criteria ε = Solution: Modify the Fortran code, Ü½ ¾º, and Matlab code, Ò ÛØÓÒºÑ, to solve Eq. (E1) above. Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

24 Methods Available Secant Method Similar to Newton s method but is different in that the derivative f is approximated using two consecutive iterative values of f f (x i ) f(x i) f(x i 1 ) x i x i 1 (8) and Eq. (7) can be re-written as x i+1 = x i f(x i) f (x i ) = x i f(x i )» x i x i 1 f(x i ) f(x i 1 ) (9) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

25 Methods Available Fixed Point Iteration Method Also known as Successive Substitution Method, in which function f(x) = 0 is re-written in the form of x = g(x) and iterative procedure is used on x i+1 = g(x i ) : i = 1, 2, 3,... (10) where new approximation to root, x i+1, is found using the previous one, x i. Iterative process is stopped with a convergence criterion 10 3 < ε < 10 6 x i+1 g(x i+1 ) ε (11) This method is simple but may NOT always converge with an arbitrarily chosen form of the function g(x) but condition to be satisfied for convergence to the correct root is given by g (x i+1 ) < 1 (12) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

26 Methods Available Fixed Point Iteration Method Example 1 Problem Statement: Find the root of the equation f(x) = 1.5x (1 + x tan 1 ) 2 « x x 1 + x 2 = 0 (E1) using the fixed point iteration with the starting point x 1 = 0.0 and the convergence criteria ε = Solution: Re-arrange Eq. (E1) 1.5x (1 + x 2 = 0.65 tan 1 ) 2 «1 0.65x x 1 + x 2 = 0 1 «13 30 x(1 + x2 ) x = (1 + x2 ) 2 tan 1 x {z } g(x) so the RHS is g(x). Thus, from Eq. (E2), the iterative process can be expressed as x i+1 = 13 «1 30 (1 + x2 i )2 tan 1 13 x i 30 x i(1 + xi 2 ) (E2) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

27 Root Finding with Matlab Example 1 Problem Statement: Find the root of the equation f(x) = tan 1 x = 0 using Matlab. Solution: Matlab Code Ü ¹¾ Ô ¼º¼½ ¾ Ô Ü ³ Ø Ò Üµ³ ÔÐÓØ Ü ¹¾ Ô ¾ Ô µ ÜÐ Ð ³Ü³µ ÝÐ Ð ³Ý Üµ³µ Ø ØÐ ³ÊÓÓØ Ò Ò ³µ ÖÓÓØ Þ ÖÓ Ü ½º µ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

28 Root Finding with Matlab Example 2 Problem Statement: Find the root of the equation f(x) = sin 2x = 0 using Matlab. Solution: Matlab Code Ü ¹¾ Ô ¼º¼½ ¾ Ô Ü ³ Ò ¾ Üµ³ ÔÐÓØ Ü ¹¾ Ô ¾ Ô µ ÜÐ Ð ³Ü³µ ÝÐ Ð ³Ý Üµ³µ Ø ØÐ ³ÊÓÓØ Ò Ò ³µ ÖÓÓØ Þ ÖÓ Ü ¼º µ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

29 Root Finding with Matlab Example 3 Problem Statement: Find the root of the equation f(x) = 1.5x 0.65 (1 + x 2 tan 1 ) 2 using Matlab. Solution: Matlab Code « x x 1 + x = 0 2 Ü ¹¾º¼ ¼º¼½ ¾ Ü ³½º Üº» ½ Üº ¾µº ¾¹¼º Ø Ò ½º»Üµµ ¼º Üº» ½ Üº ¾µ³ ÔÐÓØ Ü ¹¾ ¾ µ ÜÐ Ð ³Ü³µ ÝÐ Ð ³Ý Üµ³µ Ø ØÐ ³ÊÓÓØ Ò Ò ³µ Ü¾ Þ ÖÓ Ü ¹¼º½µ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

30 Roots of Nonlinear Polynomials There is theorem which states that there is no such formula for general polynomials of degree higher than four. In practice we use numerical method to solve polynomial equations of degree higher than two. As a special case of f(x) = 0, consider a polynomial equation f(x) = a nx n + a n 1x n a 2x 2 + a 1x + a 0 = 0 (13) where n denote degree of polynomial, a 0, a 1, a 2... a n are coefficients of polynomial. Eq. (13) in general will have n roots Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

31 Roots of Nonlinear Polynomials If x 1, x 2... x n are roots of Eq. (13), they are related to the coefficients of polynomial as nx i=1 nx x i nx i=1 j=1,j 1 nx x i nx nx i=1 j=1,j i k=1,k j x i = a n 1 a n x i x j = a n 2 a n x i x j x k = a n 3 a n... x 1 x 2... x n 1 x n = ( 1) n a 0 a n (14) Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

32 Roots of Nonlinear Polynomials Muller s Method Your homework! Read Section 2.11, pp of SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN X, Prentice Hall Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

33 Roots of Nonlinear Polynomials Example 1 Problem Statement: Use Matlab to determine the roots of the following polynomial f(x) = x 4 12x 3 + 0x x = 0 Solution: Matlab Code Ô ½ ¹½¾ ¼ ¾ ½½ Ö ÖÓÓØ Ôµ ÔÔ ÔÓÐÝ Öµ ± Ò Ó ÒØ ØÓ Ñ ØÖ Ü Ô ± Ò ÖÓÓØ Ò Ò Ø Ñ ØÓ Ö ± Ë ÓÙÐ Ö ØÙÖÒ ÔÔ Ôº Ì ± ÓÙ Ñ Ø Û ÒØ ØÓ ÔÐÓØ Ø ÔÓÐÝÒÓÑ Ð Ö Ø Ü Ð Ò Ô ¼ ½¾µ ± Ë Ø Ö Ò ÓÖ Ü Ý ÔÓÐÝÚ Ð Ô Üµ ± Ú ÐÙ Ø ÔÓÐÝÒÓÑ Ð Ø ÐÐ Ú ÐÙ Ó Ü Ü Ü Ü ¼ ± ¹ Ü ÔÐÓØ Ü Ü Ü Ü Ýµ ± ÈÐÓØ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

34 Roots of Nonlinear Polynomials Example 2 Problem Statement: Use Matlab to determine the roots of the following polynomial f(x) = x 5 3.5x x x x which has three real roots: 0.5, 1.0 and 2.0, and a pair of complex roots: 1 ± 0.5i. Solution: Matlab Code ±± ÒØ Ö Ó ÒØ Ó ÔÓÐÝÒÓÑ Ð ÒØÓ Ñ ØÖ Ü Ô ½ ¹ º ¾º ¾º½¾ ¹ º ½º¾ ±± ÓÙ Ñ Ý Ú ÐÙ Ø ÔÓÐÝÒÓÑ Ð Üµ Ý Ø Ü ½ ÓÖ Ö ÒØ Ø Øººº ÔÓÐÝÚ Ð Ô ½µ ÔÓÐÝ Ö Ôµ ±± Ö Ø ÕÙ Ö Ø ÕÒ ÖÓÑ ØÛÓ ÒÓÛÒ ÖÓÓØ ±± Ü¹¼º µ Ü ½µ Ü ¾ ¼º Ü ¹ ¼º Ö ¼º ¹½º¼ Õ ÔÓÐÝ Öµ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

35 Roots of Nonlinear Polynomials Example 2 Matlab Code (continued) ± Ú Ø ÓÖ Ò Ð ÔÓÐÝÒÓÑ Ð Üµ Ý Ø ÕÙ Ö Ø ± ÕÒ Ò Ò Ö ÙÐØ ÒØÓ ÕÙÓØ ÒØ Ò Ö Ñ Ò Ö ÓÒÚ Ô Õµ ÖÓÓØ µ ± ÊÓÓØ Ó ÕÙÓØ ÒØ ÔÓÐÝÒÓÑ Ð ÓÒÚ Õµ ± ÅÙÐØ ÔÐÝ Ý Õ ØÓ ÓÑ ÙÔ Û Ø Ø ÓÖ Ò Ð Üµ ± ÇÖ ÄÄ ÖÓÓØ Ñ Ý Ø ÖÑ Ò Ý Ö ÖÓÓØ Ôµ ± Ù ØÓ Û Ö ÓÑ Ó Ø ÖÓÓØ Ñ Ø ÝÓÙ Ñ Ý ± Û ÒØ ØÓ ÔÐÓØ ÔÓÐÝÒÓÑ Ð Üµ ÓÚ Ö Ö Ò Ý ¹½¼ Ü ½¼ Ü ¹½¼ ¼º½ ½¼ ± Ì Ò Ø Ý Üµ Ò Ü¹ Ü ººº Ò ÔÐÓØ Ý Üº ¹ º Üº ¾º Üº ¾¹ º Ü ½º¾ Ý ¼ Ü ¼ ÔÐÓØ Ü Ý ¼ Ü Ýµ Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

36 Bibliography 1 STEVEN C. CHAPRA, RAYMOND P. CANALE (2006): Applied Numerical Methods for Engineers, 5ed, ISBN , McGraw-Hill 2 SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN X, Prentice Hall 3 DAVID KINCAID AND WARD CHENEY (1991): Numerical Analysis: Mathematics of Scientific Computing, ISBN , Brooks/Cole Publishing Co. 4 STEVEN C. CHAPRA (2005): Applied Numerical Methods with MATLAB for Engineers and Scientists, ISBN , McGraw-Hill 5 WILLIAM J. PALM III (2005): Introduction to Matlab 7 for Engineers, ISBN , McGraw-Hill 6 JOHN D. ANDERSON, JR. (1995): Computational Fluid Dynamics The Basics with Applications, ISBN , McGraw-Hill Abu Hasan Abdullah (FME) SME 3023 Applied Numerical Methods Sept / 31

SKMM 3023 Applied Numerical Methods

SKMM 3023 Applied Numerical Methods Solution of Nonlinear Equations ibn Abdullah Faculty of Mechanical Engineering Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Solution of Nonlinear Equations