Ordinary Differential Equations (ODE)

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1 Ordinary Differential Equations (ODE)

2 Why study Differential Equations? Many physical phenomena are best formulated mathematically in terms of their rate of change. Motion of a swinging pendulum Bungee-jumper Equation d dt g l sin 0 dv dt g c d m v CE 06: Engg. Computation Sessional

3 Euler s Method The first derivative provides a direct estimate of the slope at t i : dy dt ti f t i, y i Euler method uses that estimate as the increment function: f t i, y i y i1 y i f t i, y i h CE 06: Engg. Computation Sessional

4 MATLAB Code: Euler s Method function [t,y] = eulode(tspan,y0,h) % input: dydt = name of the M-file that evaluates the ODE % tspan = [ti, tf], ti and tf limits of independent variable % y0 = initial value of dependent variable h = step size % output:y = solution vector if nargin<3,error( less than 3 input arguments'),end ti = tspan(1); tf = tspan(); t = (ti:h:tf)'; n = length(t); % if necessary, add an additional value of t % so that range goes from t = ti to tf if t(n)<tf t(n+1) = tf; n = n+1; end y = y0*ones(n,1); %preallocate y to improve efficiency for i = 1:n-1 %implement Euler's method y(i+1) = y(i) + dydt(t(i),y(i))*(t(i+1)-t(i)); end yi 1 yi f ti, y i h CE 06: Engg. Computation Sessional

5 v (m/s) Exercise: Euler s method Solve the bungee jumper problem using Euler s method (use a step size of s). Compare the results with the analytical solution dv dt g cd m v Given, g = 9.81 m/s, c d = 0.5 kg/m, m = 68.1 kg Analytical solution: v( t) gm c d tanh gc m d t CE 06: Engg. Computation Sessional t (sec)

6 Heun s Method h y t f y t f y y h y t f y y i i i i i i i i i i ), ( ), ( ), ( Predictor: Corrector: CE 06: Engg. Computation Sessional Exercise: Modify eulode.m to implement Heun s method. Solve the same bungee-jumper problem.

7 4 th order Runge-Kutta method where k 1 f t i, y i k f t i 1 h, y i 1 k 1h k 3 f t i 1 h, y 1 i k h k 4 f t i h, y i k 3 h y i1 y i 1 6 k k k k h CE 06: Engg. Computation Sessional

8 MATLAB Code: Runge-Kutta Method function [tp,yp] = rk4(tspan,y0,h) % input: dydt = name of the M-file that evaluates the ODEs % tspan = [ti, tf]; initial and final times % y0 = initial values of dependent variables % h = step size % output: tp = vector of independent variable % yp = vector of solution for dependent variables if nargin<3,error( 3 input arguments required'), end ti = tspan(1); tf = tspan(); tp = (ti:h:tf)'; n = length(tp); % if necessary, add an additional value of t % so that range goes from tp = ti to tf if tp(n)<tf tp(n+1) = tf; n = n+1; end CE 06: Engg. Computation Sessional

9 MATLAB Code: Runge-Kutta Method y = y0*ones(n,1); %preallocate y to improve efficiency for i = 1:n-1 %implement RK method end hh = tp(i+1)-tp(i); tt = tp(i); yy = yp(i); k1 = dydt(tt,yy); k = dydt(tt + hh/, yy + 0.5*k1*hh); k3 = dydt(tt + hh/, yy + 0.5*k*hh); k4 = dydt(tt + hh, yy + k3*hh); yp(i+1) = yy + hh*(k1 + *k + *k3 + k4)/6; CE 06: Engg. Computation Sessional

10 Practice Problem Solve the following initial value problem over the interval from t = 0 to where y(0) = 1. Display all your results in the same graph dy dt yt 3 1.5y (a) Analytically (b) Euler s method with h = 0.5 and h = 0.5 (c) Heun s method with h = 0.5 (d) 4 th order RK method with h = 0.5 CE 06: Engg. Computation Sessional

11 MATLAB s common ODE solvers MATLAB s ode3 function uses second- and third-order RK functions to solve the ODE and adjust step sizes. MATLAB s ode45 function uses fourth- and fifth-order RK functions to solve the ODE and adjust step sizes. This is recommended as the first function to use to solve a problem. MATLAB s ode113 function is a multistep solver useful for computationally intensive ODE functions. CE 06: Engg. Computation Sessional

12 Example of ODE solver: ode45 The functions are generally called in the same way [t, y] = ode45(odefun, tspan, y0) y: solution array, where each column represents one of the variables and each row corresponds to a time in the t vector odefun: function returning a column vector of the righthand-sides of the ODEs tspan: time over which to solve the system If tspan has two entries, the results are reported for those times as well as several intermediate times based on the steps taken by the algorithm If tspan has more than two entries, the results are reported only for those specific times y0: vector of initial values CE 06: Engg. Computation Sessional

13 Solving a system of ODE using ode45 Predator-prey problem Solve dy 1 dt 1.y 1 0.6y 1 y dy dt 0.8y 0.3y 1 y with y 1 (0)= and y (0)=1 for 0 seconds function yp = predprey(t, y) yp = [1.*y(1)-0.6*y(1)*y(); -0.8*y()+0.3*y(1)*y()]; predprey.m file tspan = [0 0]; y0 = [, 1]; [t, y] = ode45(@predprey, tspan, y0); figure(1); plot(t,y); figure(); plot(y(:,1),y(:,)); CE 06: Engg. Computation Sessional

14 Example using ode45 Predator-prey problem Solve dy 1 dt 1.y 1 0.6y 1 y dy dt 0.8y 0.3y 1 y with y 1 (0)= and y (0)=1 for 0 seconds CE 06: Engg. Computation Sessional

15 Practice Problem: motion of a pendulum d dt g l sin Analytical solution (assuming θ = sinθ): 0 ( t) cos 0 g t l If l = ft, g = 3. ft/s and θ 0 = π/4, Solve for θ from t = 0 to 1.6 s using (a) Euler s method with h = 0.05 (b) Using ode45 function with h = 0.05 (c) Plot your results and compare with the analytical solution d dt g l sin 0 d v dt dv dt g l = 0, θ 0 = = 0, v = 0 CE 06: Engg. Computation Sessional

16 Boundary value problem: the shooting method the boundary-value problem is converted into an equivalent initial-value problem. d T dx h T T dt z 0 dx dz ht dx T Generally, the equivalent system will not have sufficient initial conditions -A guess is made for any undefined values. - The guesses are changed until the final solution satisfies all the B.C. For linear ODEs, only two shots are required - the proper initial condition can be obtained as a linear interpolation of the two guesses. CE 06: Engg. Comp. Sessional

17 Problem: shooting method Solve d T dx h T T T 4 T 4 0 with σ =.7x10-9 K -3 m -, L=10 m, h =0.05 m-, T =00 K, T(0) = 300 K, and T(10) = 400 K. First - break into two equations: d T dx h 4 4 T T T T 0 dt dx dz dx z T T CE 06: Engg. Computation Sessional

18 Problem: shooting method Code for derivatives: function dy=dydxn(x,y) dy=[y(); -0.05*(00-y(1))-.7e-9*(1.6e9-y(1)^4)]; Code for residual: function r=res(za) [0 10], [300 za]); r=y(length(x),1)-400; Code for finding root of residual: -50) Code for solving system: [0 10], [ ) ]); CE 06: Engg. Computation Sessional

19 Practice Problem: shooting method The basic differential equation of the elastic curve for a uniformly loaded beam is given as EI d y dx wlx wx If E = ksi, I = 800 in 4, w = 1 kip/ft, L = 10 ft, solve for the deflection of the beam using the shooting method and compare the numerical results with the analytical solution wlx y 1EI 3 4 wx 4EI 3 wl x 4EI CE 06: Engg. Computation Sessional

20 Numerical solution: finite difference finite differences are substituted for the derivatives in the original equation d T dx h T T 0 d T dx T i1 T i T i1 x T i1 T i T i1 x T i1 h x h T T i 0 T i T i1 h x T CE 06: Engg. Comp. Sessional

21 Numerical solution: finite difference The linear differential equation is transformed into a set of simultaneous algebraic equations. Since T 0 and T n are known (Drichlet boundary conditions), they will be on the right-hand-side of the linear algebra system h x 1 1 T h x 1 1 T 1 h x T n1 h x T T 0 h x T h x T T n CE 06: Engg. Comp. Sessional

22 Numerical solution: finite difference If there is a derivative boundary condition (Neumann B.C.), the centered difference equation is solved at the point and rewriting the system equation accordingly. For a Neumann condition at T 0 point: CE 06: Engg. Comp. Sessional dt T 1 T 1 dx 0 x T 1 h x T 1 x dt h x dx 0 T 1 T 1 x dt dx 0 T 0 T 1 h x T T 0 T 1 h x T h x T 0 T 1 h x T x dt dx 0

23 Practice Problem: Finite difference Solve the nondimensionalized ODE using finite difference methods that describe the temperature distribution in a circular rod with internal heat source S d T dr 1 r dt dr S Over the range 0 r 1, with boundary conditions dt T( r 1) 1 0 dr For S = 1, 10 and 0 K/m plot the temperature versus radius r0 0 CE 06: Engg. Computation Sessional

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