SKMM 3023 Applied Numerical Methods

Transcription

1 UNIVERSITI TEKNOLOGI MALAYSIA SKMM 3023 Applied Numerical Methods Ordinary Differential Equations ibn Abdullah Faculty of Mechanical Engineering Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 1 / 67

2 Outline 1 Introduction 2 Sample Engineering Problems 3 Order of ODE 4 Linearization of ODE 5 Non-computer Methods for Solving ODE 6 ODE - Initial Value Problem Euler s Method Heun s Method Runge-Kutta (RK) Methods Problems of Order n 7 ODE - Boundary Value Problem Shooting Method Finite Difference Method 8 References Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 2 / 67

3 Introduction Newton s second law to compute velocity v of a falling mass m as a function of time t can be written as dv dt = g c m v where g is gravitational constant, and c is drag coefficient. Eq. (1) is composed of unknown function, v, and its derivatives, dv/dt Such equation is called differential equation or rate equation. Also, in Eq. (1) v is the dependent variable t is the independent variable When function involves ONE independent variable, the equation is called the ordinary differential equation, e.g. dv dt = g c m v When function involves TWO or MORE independent variables, the equation is called the partial differential equation, e.g. ρ Du Dt = p x + τxx x + τyx y + τzx z + ρfx (1) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 3 / 67

4 Sample Engineering Problems Newton s second law of motion dv dt = F m where v is velocity, F is force and m is mass Fourier s heat law q = k dt dx where q is heat flux, k is thermal conductivity and T is temperature Fick s law of diffusion J = D dc dx where J is mass flux, D is diffusion coefficient and c is concentration Faraday s law (voltage drop across an inductor) V t = L di dt where V t is voltage drop, L is inductance and i is current (2) (3) (4) (5) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 4 / 67

5 Order of ODE Differential equations are classified as to the order: First order equation highest derivative is first derivative dv dt = g c m v Second order equation highest derivative is second derivative, for example equation describing position x of mass-spring system with damping m d2 x dt 2 + c dx + kx = 0 (6) dt where c is gravitational constant and k is mass and c is drag coefficient Similarly, n th order differential equation would include n th derivative. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 5 / 67

6 Order of ODE Higher order differential equation can be reduced to a system of first order equations. For Eq. (6) above, m d2 x dt 2 + cdx dt + kx = 0 we do it by defining y = dx dt which is differentiated to yield (7) dy dt = d2 x dt 2 (8) Eqs. (7) and (8) are then substituted into Eq. (6) to yield m dy + cy + kx = 0 (9) dt Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 6 / 67

7 Order of ODE Eq. (9) can be re-arranged into dy + kx = cy dt m = 0 (10) Thus Eqs. (7) and (10) are two first order equations that are equivalent to the original second order equation, i.e. Eq. (6). Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 7 / 67

8 Linearization of ODE A linear ODE is one that fit the general form a n(x)y (n) a 1(x)y + a 0(x)y = f(x) (11) where y (n) is n th derivative of y w.r.t. x, a is specified function of x and c is specified function of x. Eq. (11) is linear because there are NO products or nonlinear function of dependent variable y and its derivatives. Motion of a swinging pendulum, Figure 1, is governed by d 2 θ dt + g sin θ = 0 (12) 2 l Eq. (12) is nonlinear because of the term sin θ. Figure 1: Motion of swinging pendulum. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 8 / 67

9 Linearization of ODE To obtain solution to Eq. (12) we linearize it by assuming, for small values of θ, sin θ θ (13) which, on substitution into Eq. (12), yields a linear form d 2 θ dt 2 + g l θ = 0 (14) Why linear ODE are important? Because they can be solved analytically. In contrast, most nonlinear equations cannot be solved exactly!! Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 9 / 67

10 Non-computer Methods for Solving ODE Without computers, ODE are solved with analytical integration techniques in which Eq. (1) could be multiplied by dt and then integrated to yield Z v = g cm «v dt (15) RHS of Eq. (15) is an indefinite integral because limits of integration are not specified Eq. (15) can be solved analytically, assuming v = 0 at t = 0, to yield v(t) = gm «1 e (c/m)t c Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 10 / 67

11 Mathematical Background A solution of an ODE is a specific function of independent variable and parameters that satisfy the original differential equation. Let say we were given a forth order polynomial y = 0.5x 4 + 4x 3 10x x + 1 (16) Differentiating Eq. (16) gives us an ODE dy dx = 2x3 + 12x 2 20x (17) Eq. (17) also describes the behaviour of a polynomial but in a different manner than Eq. (16) i.e. it gives rate of change of y w.r.t. x, which is a slope. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 11 / 67

12 Mathematical Background But our objective here is to determine the original function (Eq. (16)) given the ODE (Eq. (17)). So we go back to Eq. (17) and solve analytically Z y = ( 2x x 2 20x + 8.5)dx which leads to y = 0.5x 4 + 4x 3 10x x + C (18) Eq. (18) is identical to Eq. (16) with one notable exception 1 in Eq. (16) being replaced by C in Eq. (18) and C is the constant of integration which is arbitrary. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 12 / 67

13 Mathematical Background Therefore, to specify the solution completely, an ODE is usually accompanied by auxiliary condition(s) For first order ODE, this auxiliary condition, called initial value, is required to determine the constant of integration and obtain a unique solution. In order to get back Eq. (16), Eq. (18) needs the initial value that x = 0, y = 1. Substituing the initial value into Eq.(18) yields or Hence, 1 = 0.5(0) 4 + 4(0) 3 10(0) x + C (19) C = 1 (20) y = 0.5x 4 + 4x 3 10x x + 1 Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 13 / 67

14 Basis of Numerical Solution Basis of numerical method devoted to solving ODE of the form is dy = f(x, y) dx New Value = Old Value which is mathematically + (Slope Step size) y i+1 = y i + φh (21) where φ is the slope estimate. This method is known as one-step method, Figure 2. Figure 2: One-step method. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 14 / 67

15 Euler s Method The first derivative provides a direct estimate of the slope at x i φ = f(x i, y i ) where f(x i, y i ) is differential equation evaluated at x i and y i. Substituting this into Eq. (21) yields y i+1 = y i + f(x i, y i )h (22) Eq. (22) represents the Euler s (or Cauchy-Euler) method. Figure 3: Euler s method. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 15 / 67

16 Example 1: Euler s Method Problem Statement: Use Euler s method to solve dy dx = 2x3 + 12x 2 20x from x = 0 to x = 4 with step size of 0.5. Initial condition at x = 0 is y = 1. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 16 / 67

17 Example 1: Euler s Method Solution: At x = 0.5, Eq. (22), with initial condition y(0) = 1, yield y(0.5) = y(0) + f(0, 1) 0.5 = [ 2(0) (0) 2 20(0) + 8.5] 0.5 = 5.25 True solution at x = 0.5 is y = 0.5(0.5) 4 + 4(0.5) 3 10(0.5) (0.5) + 1 = E x = = or R x = 63.1% Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 17 / 67

18 Example 1: Euler s Method Solution: continued... At x = 1.0, y(1.0) = y(0.5) + f(0.5, 5.25) (0.5) = [ 2(0.5) (0.5) 2 20(0.5) + 8.5] 0.5 = True solution at x = 1.0 is y = 0.5(1.0) 4 + 4(1.0) 3 10(1.0) (1.0) + 1 = 3.0 E x = = or R x = 95.8% Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 18 / 67

19 Example 1: Euler s Method Solution: continued... Computation is repeated and results tabulated Show the effect of reduced step size. x y true y Euler Error (%) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 19 / 67

20 Example 1: Euler s Method Solution: continued... Using Matlab to solve the same involves the following steps: 1 Write the ODE into Ý ÜºÑ 2 Optionally, write the known solution into ÝØÖÙ ºÑ 3 Matlab script to solve the ODE is written in ÑÝ ÙÐ ÖºÑ Ý ÜºÑ ÙÒØ ÓÒ Ý Ü Üµ ¹¾ ܺ ½¾ ܺ ¾ ¹ ¾¼ Ü º ÝØÖÙ ºÑ ÙÒØ ÓÒ Ý ÝØÖ٠ܵ Ý ¹¼º ܺ ܺ ¹ ½¼ ܺ ¾ º Ü ½ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 20 / 67

21 Example 1: Euler s Method Solution: continued... ÑÝ ÙÐ ÖºÑ Ü¼ ¼º¼ ÜÒ º¼ Ü Ø Ô ¼º¼½ Ü Ü¼ Ü Ø Ô ÜÒ Ò Ð Ò Ø Üµ ÜØ Ð Ò Ô Ü ½µ Ü Òµ ½¼¼µ ÝØ ÝØÖ٠ܵ ÔÐÓØ Ü ÝØ ³Ö³µ ÓÐ ÓÒ Ý ½µ ½º¼ ± Ì ÁÆÁÌÁ Ä Î ÄÍ ÓÖ ¾ Ò Ý µ Ý ¹½µ Ý Ü Ü ¹½µµ Ü Ø Ô Ò ÖÖ Ýعݵº»Ýص ½¼¼ ÔÐÓØ Ü Ý ³ ³µ ܳ ݳ Ýس ÖÖ³ P.S. First order technique such as Euler s method demands great computational effort to obtain acceptable error level Simplicity of Euler s method makes it attractive, easy to program, suitable for quick analyses. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 21 / 67

22 Heun s Method An attempt to improve Euler s method Determines two derivatives for the interval or steps at beginning of interval at end of interval These two derivatives are then averaged to get an improved estimate of the slope. Figure 4: Heun s method. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 22 / 67

23 Heun s Method Slope at beginning of interval y i = f(x i, y i ) (23) is used to extrapolate linearly to y o i+1 y o i+1 = y i + f(x i, y i ) h (24) y o i+1 is an intermediate prediction called predictor Slope at end of interval is estimated using the predictor in Eq. (24) y i+1 = f(x i+1, y o i+1) (25) The two slopes Eqs. (23) and (25) are combined to obtain average slope for the interval ȳ = y i + y i+1 2 = f(x i, y i ) + f(x i+1, y o i+1) 2 (26) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 23 / 67

24 Heun s Method The average slope, Eq. (26), is then used to extrapolate linearly from y i to y i+1 using Euler s method» f(xi, y i ) + f(x i+1, y o y i+1 = y i + i+1) h (27) 2 Eq. (27) is known as the corrector equation Heun s method is the basis of many other predictor-corrector methods Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 24 / 67

25 Example 2: Heun s Method Problem Statement: Use Heun s method to solve dy dx = 4e0.8x 0.5y from x = 0 to x = 4 with step size of 1.0. Initial condition at x = 0 is y = 2. Note: Analytical solution is y = 4 «e 0.8x e 0.5x + 2e 0.5x 1.3 Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 25 / 67

26 Example 2: Heun s Method Solution: Slope at initial point (x 0, y 0) y 0 = 4e 0 0.5(2) = 3 Use predictor equation (Eq. (24)) to obtain estimate of y at 1.0 y o 1 = 2 + 3(1) = 5 Use value y o i to predict slope at end of interval y 1 = f(x 1, y o 1) = 4e 0.8(1) 0.5(5) = Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 26 / 67

27 Example 2: Heun s Method Solution: continued... Average slope over the interval 0 x 1 x = 0 and x = 1 is y = = This average slope is then fed into the corrector equation (Eq. (27)) to give an estimate at x = 1 y 1 = (1) = Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 27 / 67

28 Example 2: Heun s Method Solution: continued... Computation is repeated and results tabulated x y true y Heun Error (%) Show the effect of reduced step size. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 28 / 67

29 Example 2: Heun s Method Solution: continued... Using Matlab to solve the same involves the following steps: 1 Write the ODE into Ý Ü½ºÑ 2 Optionally, write the known solution into ÝØÖÙ ½ºÑ 3 Matlab script to solve the ODE is written in ÑÝ ÙÒºÑ Ý Ü½ºÑ ÙÒØ ÓÒ Ý Ü½ Ü Ýµ ÜÔ ¼º ܵ ¹ ¼º Ý ÝØÖÙ ½ºÑ ÙÒØ ÓÒ Ý ÝØÖÙ ½ ܵ Ý»½º µ ÜÔ ¼º ܵ ¹ ÜÔ ¹¼º ܵµ ¾ ÜÔ ¹¼º ܵ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 29 / 67

30 Example 2: Heun s Method Solution: continued... ÑÝ ÙÒºÑ Ü¼ ¼º¼ ÜÒ º¼ Ü Ø Ô ¼º¾ Ü Ü¼ Ü Ø Ô ÜÒ Ò Ð Ò Ø Üµ ÝØ ÝØÖÙ ½ ܵ ÔÐÓØ Ü ÝØ ³Ö³µ ÓÐ ÓÒ Ü ½µ ¼º¼ Ý ½µ ¾º¼ ± ÁÆÁÌÁ Ä ÇÆ ÁÌÁÇÆ ÓÖ ¾ Ò ÐÓÔ Ý Ü½ Ü ¹½µ Ý ¹½µµ ÔÖ ØÓÖ Ý ¹½µ ÐÓÔ Ü Ø Ô ÐÓÔ Ý Ü½ Ü µ ÔÖ ØÓÖµ Ú ÐÓÔ ÐÓÔ ÐÓÔ µ»¾ Ý µ Ý ¹½µ Ú ÐÓÔ Ü Ø Ô Ò ÖÖ Ýعݵº»Ýص ½¼¼ ÔÐÓØ Ü Ý ³ ³µ ܳ ݳ Ýس ÖÖ³ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 30 / 67

31 Runge-Kutta (RK) Methods Advantage achieve accuracy of a Taylor series approach without requiring calculation of higher derivatives. Generalized form of Runger-Kutta methods is y i+1 = y i + φ(x i, y i, h) h (28) where φ(x i, y i, h) is an increment function which is a representative slope over the interval and given by φ = a 1k 1 + a 2k a nk n (29) k s in Eq. (29) are recurrence relationships i.e. k 1 appears in k 2 which appears in k 3 and so forth k 1 = f(x i, y i ) (30) k 2 = f(x i + p 1h, y i + q 11k 1h) (31) k 3 = f(x i + p 2h, y i + q 21k 1h + q 22k 2h) (32)... k n = f(x i + p n 1h, y i + q n 1,1k 1h + q n 2,2k 2h q n 1,n 1k n 1h) (33) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 31 / 67

32 Runge-Kutta (RK) Methods Various types of RK methods can be devised by employing different numbers of terms in the increment function as specified by n Once n is chosen, values for a s, p s and q s are evaluated by setting Eq. (28) equal to terms in a Taylor series expansion. First order RK method, when n = 1, is actually Euler s method. Second order RK method, when n = 2, is exact if solution to ODE is quadratic. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 32 / 67

33 Runge-Kutta Order 1 (RK1) Method For n = 1, we can write, using Eq. (28), the first order RK (RK1) method as y i+1 = y i + (a 1k 1)h (34) where k 1 = f(x i, y i ) With a 1 = 1, the first order RK method y i+1 = y i + f(x i, y i )h (35) can be seen to be the same as Euler s method. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 33 / 67

34 Runge-Kutta Order 2 (RK2) Methods For n = 2, we can write, using Eq. (28), the second order RK (RK2) method as y i+1 = y i + (a 1k 1 + a 2k 2)h (36) where k 1 = f(x i, y i ) (37) k 2 = f(x i + p 1h, y i + q 11k 1h) (38) To use Eq. (36), values for a 1, a 2, p 1 and q 11 must first be determined. Recall second order Taylor series for y i+1 in terms of y i and f(x i, y i ) y i+1 = y i + f(x i, y i )h + f (x i, y i ) h2 2! (39) where f (x i, y i ), by chain-rule differentiation, yields f (x i, y i ) = f(x, y) x + f(x, y) dy y dx (40) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 34 / 67

35 Runge-Kutta Order 2 (RK2) Methods Substituting Eq. (40) into Eq. (39) yields f y i+1 = y i + f(x i, y i )h + x + f «dy h 2 y dx 2! (41) Taylor series for two-variable function is g(x + r, y + s) = g(x, y) + r g x + s g y +... and apply this to expand Eq. (38) f(x i + p 1h, y i + q 11k 1h) = f(x i, y i ) + p f f 1h + q11k1h x y + O(h2 ) (42) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 35 / 67

36 Runge-Kutta Order 2 (RK2) Methods Eqs. (42) and (37) are then substituted into Eq. (36) to yield y i+1 = y i + a 1hf(x i, y i ) + a 2hf(x i, y i ) + a 2p 1h 2 f x + a 2q 11h 2 f(x i, y i ) f y + O(h3 ) (43) and collecting terms we get» y i+1 = y i + a 1f(x i, y i ) + a 2f(x i, y i ) h+» f a 2p 1 x + a2q11f(x i, y i ) f h 2 + O(h 3 ) (44) y Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 36 / 67

37 Runge-Kutta Order 2 (RK2) Methods Comparing like terms in Eqs. (41) and (44), the following must hold a 1 + a 2 = 1 a 2p 1 = 1 2 a 2q 11 = 1 2 We now have 3 simultaneous equations containing 4 unknown constants hence, there s no unique solution. However, assuming a value for one of the constants, we can determine the other 3! Consequently, there is a family of second order RK methods rather than a single version! Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 37 / 67

38 Runge-Kutta Order 2 (RK2) Methods The second order RK method version of Eq. (28) is y i+1 = y i + (a 1k 1 + a 2k 2)h (45) where k 1 = f(x i, y i ) (46) k 1 = f(x i + p 1h, y i + q 11k 1h) (47) a 1 + a 2 = 1 (48) a 2p 1 = 1 2 (49) a 2q 11 = 1 2 (50) Suppose we specify a value for a 2, then Eqs. (48) (50) can be solved simultaneously for a 1 = 1 a 2 (51) p 1 = q 11 = 1 2a 2 (52) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 38 / 67

39 Runge-Kutta Order 2 (RK2) Methods Heun s Method with a Single Corrector: a 2 = 1 2 If a 2 = 1, Eqs. (51) and (52) can be solved for 2 a1 = 1 and p1 = q11 = 1. These 2 parameters, when substituted into Eq. (45) yield y i+1 = y i + ( 1 2 k k2)h (53) where k 1 = f(x i, y i ) : slope at begining of interval (54) k 2 = f(x i + h, y i + k 1h) : slope at end of interval (55) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 39 / 67

40 Runge-Kutta Order 2 (RK2) Methods The Midpoint Method: a 2 = 1 If a 2 = 1, then a 1 = 0 and p 1 = q 11 = 1 and Eq. (45) becomes 2 y i+1 = y i + k 2h (56) where k 1 = f(x i, y i ) k 2 = f(x i h, y i k1h) (57) (58) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 40 / 67

41 Runge-Kutta Order 2 (RK2) Methods Ralston s Method: a 2 = 2 3 Choosing a 2 = 2 provides a minimum bound on the truncation error for second 3 order RK algorithms. It leads to a 1 = 1 and 3 p1 = q11 = 3, and yields 4 1 y i+1 = y i + «h 3 k k2 (59) where k 1 = f(x i, y i ) k 2 = f(x i h, y i k1h) (60) (61) Your homework! For details see Section on pp of STEVEN C. CHAPRA, RAYMOND P. CANALE (2006): Numerical Methods for Engineers, 5ed, ISBN , McGraw-Hill Also see Section on page 658 of SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN X, Prentice Hall Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 41 / 67

42 Runge-Kutta Order 3 (RK3) Methods For n = 3, we can write, using Eq. (28), the third order RK (RK3) method as where y i+1 = y i + (a 1k 1 + a 2k 2 + a 3k 3)h (62) k 1 = f(x i, y i ) (63) k 2 = f(x i + p 1h, y i + q 11k 1h) (64) k 3 = f(x i + p 2h, y i + q 21k 1h + q 22k 2h) (65) To use Eq. (62), values for a 1, a 2, a 3, p 1, p 2, q 11, q 21 and q 22 must first be determined. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 42 / 67

43 Runge-Kutta Order 3 (RK3) Methods One of the more popular version of Eq. (62) is where y i+1 = y i + 1 (k1 + 4k2 + k3)h (66) 6 k 1 = f(x i, y i ) k 2 = f(x i h, y i hk1) k 3 = f(x i + h, y i hk 1 + 2hk 2) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 43 / 67

44 Runge-Kutta Order 4 (RK4) Methods RK4 methods are by far the most popular, with the classical fourth order RK method being the most commonly used: where y i+1 = y i + 1 (k1 + 2k2 + 2k3 + k4)h (67) 6 k 1 = f(x i, y i ) k 2 = f(x i h, y i hk1) k 3 = f(x i h, y i hk2) k 4 = f(x i + h, y i + hk 3) Your homework! For details see Section on page 707 of STEVEN C. CHAPRA, RAYMOND P. CANALE (2006): Numerical Methods for Engineers, 5ed, ISBN , McGraw-Hill Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 44 / 67

45 Example 3: Runge-Kutta Order 4 (RK4) Methods Problem Statement: Use the classical RK4 method to solve f(x) = dy dx = 2x3 + 12x 2 20x using step size of h = 0.5 and initial condition of y = 1 at x = 0. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 45 / 67

46 Example 3: Runge-Kutta Order 4 (RK4) Methods Solution: At x = 0.0 k 1 = 2(0.0) (0.0) 2 20(0.0) = k 2 = 2( ) ( ) 2 20( ) = k 3 = 2( ) ( ) 2 20( ) = k 4 = 2( ) ( ) 2 20( ) = Hence, estimate at x = 0.5 is y i+1 = y i (k 1 + 2k 2 + 2k 3 + k 4 )h y(0.5) = ˆ ( ) + 2( ) «0.5 = Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 46 / 67

47 Example 3: Runge-Kutta Order 4 (RK4) Methods Solution: continued... At x = 0.5 k 1 = 2(0.5) (0.5) 2 20(0.5) = k 2 = 2( ) ( ) 2 20( ) = k 3 = 2( ) ( ) 2 20( ) = k 4 = 2( ) ( ) 2 20( ) = Hence, estimate at x = 1.0 is y i+1 = y i (k 1 + 2k 2 + 2k 3 + k 4 )h 1 y(1.0) = 1 + ˆ ( ) + 2( ) «0.5 = Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 47 / 67

48 Example 4: Runge-Kutta Order 4 (RK4) Methods Problem Statement: Use the classical RK4 method to solve f(x, y) = dy dx = 4e0.8x 0.5y from x = 0 to x = 0.5 with step size of 0.5 and initial condition of y(0) = 2. Solution: Using Matlab to solve the above involves the following steps: 1 Write the ODE into Ý Ü½ºÑ 2 Optionally, write the known solution into ÝØÖÙ ½ºÑ 3 Matlab script to solve the ODE is written in ÑÝÖ ÜÝºÑ Ý Ü½ºÑ ÙÒØ ÓÒ Ý Ü½ Ü Ýµ ÜÔ ¼º ܵ ¹ ¼º Ý Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 48 / 67

49 Example 4: Runge-Kutta Order 4 (RK4) Methods Solution: continued... ÝØÖÙ ½ºÑ ÙÒØ ÓÒ Ý ÝØÖÙ ½ ܵ Ý»½º µ ÜÔ ¼º ܵ ¹ ÜÔ ¹¼º ܵµ ¾ ÜÔ ¹¼º ܵ ÑÝÖ ÜÝºÑ Ü¼ ¼º¼ ÜÒ ¼º ¼º¼ Ü Ü¼ ÜÒ Ò Ð Ò Ø Üµ ÝØ ÝØÖÙ ½ ܵ ÔÐÓØ Ü ÝØ ³Ö³µ ÓÐ ÓÒ Ü ½µ ¼º¼ Ý ½µ ¼º ± ÁÆÁÌÁ Ä ÇÆ ÁÌÁÇÆ ÓÖ ¾ Ò ½ Ý Ü½ Ü ¹½µ Ý ¹½µµ ¾ Ý Ü½ Ü ¹½µ ¼º Ý ¹½µ ¼º ½ µ Ý Ü½ Ü ¹½µ ¼º Ý ¹½µ ¼º ¾ µ Ý Ü½ Ü ¹½µ Ý ¹½µ µ Ô ½ ¾ ¾ ¾ µ» Ý µ Ý ¹½µ Ô Ò ÖÖ Ýعݵº»Ýص ½¼¼ ÔÐÓØ Ü Ý ³Ó³µ ÔÐ Ý ³Ü Ý ÝØ ÖÖ³µ ܳ ݳ Ýس ÖÖ³ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 49 / 67

50 Example 5: Using Matlab s Ó ¾ Function Problem Statement: Use built-in Matlab function Ó ¾ to numerically solve first order ODE f(x, y) = dy dx = xy2 + y Initial condition of at x = 0 is y = 1. Solution: Using Matlab to solve the same involves the following steps: 1 Write the ODE into ÑÝÓ ½ºÑ 2 Matlab script to solve ÑÝÓ ½ºÑ is written in ÓÐÚ ÑÝÓ ½ºÑ ÑÝÓ ½ºÑ ÙÒØ ÓÒ ÑÝÓ ½ Ü Ýµ ÝÔÖ Ñ Ü Ýº ¾ Ý ÓÐÚ ÑÝÓ ½ºÑ ÜÖ ¼ ¼º ± Ø Ü¹Ö Ò Ý¼ ½ ± Ò Ø Ð ÓÒ Ø ÓÒ Ü Ý Ó ¾ ³ÑÝÓ ½³ ÜÖ Ý¼µ ÔÐÓØ Ü Ýµ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 50 / 67

51 Example 6: Using Matlab s Ó Function Problem Statement: Use built-in Matlab function Ó to solve the first order ODE f(t, y) = dy dt = y + t + 1 Initial condition of at x = 0 is y = 1. Solution: Using Matlab to solve the above involves the following steps: 1 Write the ODE into Ú½ºÑ 2 Matlab script to solve Ú½ºÑ is written in ÓÐÚ Ú½ºÑ Ú½ºÑ ÙÒØ ÓÒ Ý ÓØ Ú½ Ø Ýµ Ý ÓØ ¹Ý Ø ½ ÓÐÚ Ú½ºÑ Ø Ô Ò ¼ ½º¼ ± Ø Ø Ñ Ô Ò Ý¼ ½ ± Ò Ø Ð ÓÒ Ø ÓÒ Ø Ý Ó ³ Ú½³ Ø Ô Ò Ý¼µ ÔÐÓØ Ø Ýµ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 51 / 67

52 Problems of Order n We will show this kind of problem through an example: d 2 x dt dx dt + x = 5 This is n th order problem where, in this case n = 2. It can be re-written as a system of n first order ODE. This can be done in the following steps: 1 Let x 1 = x, then d 2 x 1 dt 2 or, re-written as + 3 dx1 dt + x1 = 5 d 2 x 1 dt 2 = 3 dx1 dt x1 + 5 (68) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 52 / 67

53 Problems of Order n 2 Next, let dx 1 dt = x 2 (69) This is the first equation of the system. 3 Differentiate x 2 and substitute it into Eq. (69) dx 2 dt = d2 x 1 dt 2 = 3 dx1 dt x1 + 5 = 3x2 x1 + 5 (70) This is the second equation of the system. 4 System of ODE represented by the first order Eqs. (69) and (70) can be then be written into a user-defined function in Matlab see Matlab codes Ú¼ºÑ. Ú¼ºÑ ÙÒØ ÓÒ Ý ÓØ Ú¼ Ø Ýµ ݽ ÓØ Ý ¾µ ݾ ÓØ ¹ Ý ¾µ ¹ Ý ½µ Ý ÓØ Ý½ ÓØ Ý¾ ÓØ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 53 / 67

54 Example 7: Problems of Order n = 2 Problem Statement: Use built-in Matlab function(s) to solve the classic van de Pol equation ẍ µ(1 x 2 )ẋ + x = 0 where µ is a parameter greater than zero. Solution: Choose y 1 = x = y 1 = dx dt = y2 y 2 = dx dt = y 2 = d2 x dt = µ(1 2 y2 1)y 2 y 1 Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 54 / 67

55 Example 7: Problems of Order n = 2 Solution: (continued... ) Using Matlab to solve the above involves the following steps: 1 Write the system of first order ODE into Ú ÔÓÐºÑ 2 Matlab script to solve Ú ÔÓÐºÑ is written in ÓÐÚ Ú ÔÓÐºÑ Ú ÔÓÐºÑ ÙÒØ ÓÒ Ý ÓØ Ú ÔÓÐ Ø Ýµ ÑÙ ¾ ± Ø ØÓ Ú ÐÙ Ö Ø Ö Ø Ò Þ ÖÓ Ý½ ÓØ Ý ¾µ ݾ ÓØ ÑÙ ½¹Ý ½µ ¾µ Ý ¾µ¹Ý ½µ Ý ÓØ Ý½ ÓØ Ý¾ ÓØ ÓÐÚ Ú ÔÓÐºÑ Ø Ô Ò ¼ ¾¼ ± Ø Ñ Ô Ò ÝÓ ¾ ¼ ± Ò Ø Ð ÓÒ Ø ÓÒ Ø Ý Ó ³Ú ÔÓг Ø Ô Ò ÝÓµ ± ÓÐÚ Ø Ç ¹ÁÎÈ ÔÐÓØ Ø Ý ½µ Ø Ý ¾µ ³¹¹³µ ÜÐ Ð ³Ø Ñ ³µ ÝÐ Ð ³Ý½ ÓØ Ò Ý¾ Óسµ Ø ØÐ ³Ú Ò Ö ÈÓÐ ËÓÐÙØ ÓÒ³µ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 55 / 67

56 Example 8: Problems of Order n = 3 Problem Statement: Use built-in Matlab function(s) to solve the following third order ODE d 3 y dt d2 y dt dy dt + 7y = 6 This is n th order problem with n = 3. It can be re-written as a system of 3 first order ODE. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 56 / 67

57 Example 8: Problems of Order n = 3 Solution: 1 Let x 1 = y, then d 3 x 1 dt + 3 d2 x 1 3 dt + 5 dx1 2 dt + 7x1 = 6 or, re-written as d 3 x 1 = 3 d2 x 1 dt 3 dt 5 dx1 2 7x1 + 6 (71) dt Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 57 / 67

58 Example 8: Problems of Order n = 3 Solution: continued... 2 Let dx 1 dt = x 2 (72) This is the first equation of the system. Differentiate x 2, twice dx 2 dt = d2 x 1 dt 2 (73a) d 2 x 2 = d3 x 1 dt 2 dt 3 (73b) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 58 / 67

59 Example 8: Problems of Order n = 3 Solution: continued... 3 Let dx 2 dt = x 3 This is the second equation of the system. But from Eq. (73a) Thus dx 2 dt = d2 x 1 dt 2 x 3 = d2 x 1 dt 2 (74) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 59 / 67

60 Example 8: Problems of Order n = 3 Solution: continued... 4 Differentiate x 3 dx 3 dt = d3 x 1 dt 3 (75) Subsituting Eqs. (75), (74) and (72) into Eq. (71) and making substitution dx 3/dt = d 3 x 1/dt 3, etc. yields dx 3 dt = 3x 3 5x 2 7x (76) This is the third equation of the system. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 60 / 67

61 Example 8: Problems of Order n = 3 Solution: continued... Thus, the third order ODE d 3 y dt + 3 d2 y 3 dt + 5 dy 2 dt + 7y = 6 can be re-written as a system of 3 first order ODE: dx 1 dt = x 2; dx 2 dt = x 3; dx 3 dt = 3x 3 5x 2 7x Using Matlab to solve the above involves the following steps: 1 Write a Matlab user-defined function called, say Ú ºÑ 2 Matlab script to solve Ú ºÑ is written in ÓÐÚ Ú ºÑ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 61 / 67

62 Example 8: Problems of Order n = 3 Solution: continued... Ú ºÑ ÙÒØ ÓÒ Ý ÓØ Ú Ø Üµ ܽ ÓØ Ü ¾µ ܾ ÓØ Ü µ Ü ÓØ ¹ Ü µ ¹ Ü ¾µ ¹ Ü ½µ Ü ÓØ Ü½ ÓØ Ü¾ ÓØ Ü ÓØ ÓÐÚ Ú ºÑ Ø Ô Ò ¼ ¾¼ ± Ø Ñ Ô Ò ÜÓ ¾ ¼ ± Ò Ø Ð ÓÒ Ø ÓÒ Ø Ü Ó ³ Ú ³ Ø Ô Ò ÜÓµ ± ÓÐÚ Ø Ç ¹ÁÎÈ ÔÐÓØ Ø Ü ½µ Ø Ü ¾µ Ø Ü µ ³¹¹³µ ÜÐ Ð ³Ø Ñ ³µ ÝÐ Ð ³Ü½ ÓØ Ò Ü¾ ÓØ Ü Óسµ Ø ØÐ ³Ç Ó ÇÖ Ö ³µ Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 62 / 67

63 ODE - Boundary Value Problem Differences between IVP and BVP In ODE - Initial Value Problem (ODE-IVP) conditions are specified at the same value of the independent variable for n th order ODE, n conditions are required. In ODE - Boundary Value Problem (ODE-BVP) conditions are specified at the different values of the independent variable because these conditions are often specified at the extreme points or boundaries of a system, they are referred to as ODE-BVP. Methods of Numerical Solution ODE-BVP may be classified into linear or nonlinear, separated or mixed, specified at two points or more. Two general approaches to numerical solution of ODE-BVP are shooting method, finite-difference method Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 63 / 67

64 ODE - Boundary Value Problem Numerical Solution Take conservation of energy to develop a heat balance for a long, thin rod. Mathematical model is represented by where d 2 T + h(ta T) = 0 (77) dx2 h : heat transfer coefficient T a : temperature of surrounding air To solve Eq. (77), appropriate boundary conditions (BC) must be specified: T(0) = T 1; T(L) = T 2 With these two BC, Eq. (77) can be solved analytically using calculus. If L = 10 m, T a = 20 C, T 1 = 40 C, T 2 = 200 C and h = 0.01, the analytical solution is T = e 0.1x e 0.1x + 20 (78) Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 64 / 67

65 ODE - Boundary Value Problem Numerical Solution Problem Statement: To show both solution approaches for ODE-BVP, we will solve Eq. (77) d 2 T + h(ta T) = 0 dx2 for a 10 m rod with h = 0.001, T a = 20 and BC: T(0) = 40 and T(10) = 200 Solution: Shooting methods is 1 based on converting BVP into equivalent IVP. Here, the second order ODE of Eq. (77) is transformed into two first order ODE dt dx = z; dz = h(t Ta) dx We need initial value for z to solve this system of ODE, using, for example, RK4. 2 A trial-and-error approach is then implemented to solve IVP. Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 65 / 67

66 ODE - Boundary Value Problem Numerical Solution: Finite Difference Method Solution: continued... Finite difference methods is... Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 66 / 67

67 Bibliography 1 STEVEN C. CHAPRA & RAYMOND P. CANALE (2009): Numerical Methods for Engineers, 6ed, ISBN , McGraw-Hill 2 SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN X, Prentice Hall 3 DAVID KINCAID & WARD CHENEY (1991): Numerical Analysis: Mathematics of Scientific Computing, ISBN , Brooks/Cole Publishing Co. 4 STEVEN C. CHAPRA (2012): Applied Numerical Methods with MATLAB for Engineers and Scientists, 3ed, ISBN , McGraw-Hill 5 JOHN H. MATHEWS & KURTIS D. FINK (2004): Numerical Methods Using Matlab, 4ed, ISBN , Prentice Hall 6 WILLIAM J. PALM III (2011): Introduction to MATLAB for Engineers, 3ed, ISBN , McGraw-Hill Òº ÙÐÐ ÚºÒÙÐÐ ¾¼½ SKMM 3023 Applied Numerical Methods Ordinary Differential Equations 67 / 67