Assignment 3
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The document describes assignments for implementing and comparing numerical integration methods. It provides formulations for the explicit midpoint and trapezoidal methods and instructs students to write modules for these methods. Students are asked to compare the accuracy of Forward Euler, Backward Euler, midpoint, and trapezoidal methods by adjusting the timestep and plotting angular displacement against the exact solution. For homework, students are tasked with generating a module for the implicit trapezoidal method and comparing the performance of explicit and implicit trapezoidal methods in terms of absolute error over time and computational time.
![ME 372 Computer-Aided Mechanical Engineering
Instructor: Emre Alpman
Assignment #3
Due Date: Mar 16th, 2011
SECOND ORDER ACCURATE SCHEMES
Summary of the Previous Assignment
In Assignment #2, the main goal was to analyze the effect of local error on the solution accuracy with
the aim of controlling it. A desired tolerance on the local error was set by dtol which was then used to
adjust the stepsize that produces it.
For this purpose the module accuracy.m was developed according to the algorithm given in
Assignment #2. This module is given in Table 1.
Table 1. Module accuracy.m
% This module adjusts the step size according to the given
% error tolerance dtol
% Inputs: Step sizei error tolerance, initial time and condition
% Outputs: Adjusted time step
function [hnew] = accuracy(h,dtol,t0,y0)
hnew = h ;
yh = y0 + hnew*deriv(t0,y0) ; % one step solution
ymid = y0 + hnew/2*deriv(t0,y0); % take half an euler step
tmid = t0+hnew/2;
yh2 = ymid + hnew/2*deriv(tmid,ymid) ; % two step solution
local_err = norm((yh2yh),Inf)/norm(yh2,Inf) % Local error
% Repeat the above steps by dividing the step size by 2 until
% relative error becomes small enough
while local_err > dtol
hnew = hnew/2 ;
yh = ymid ;
ymid = y0 + hnew/2*deriv(t0,y0);
tmid = t0+hnew/2;
yh2 = ymid + hnew/2*deriv(tmid,ymid) ;
local_err = norm((yh2yh),Inf)/norm(yh2,Inf)
end](https://image.slidesharecdn.com/assignment3-141223124337-conversion-gate01/85/Assignment-3-1-320.jpg)
![The above routine is called by hw2_main.m before the solution loop starts. The modified
hw2_main.m is given in Table 2.
Table 2. Module hw2_main.m
% main program for hw2
clear all;
% assignments
t0 = 0.0 ; % starting time
tf = 20. ; % final time
h = 0.2 ; % time step size
ctol = 0.001; % Error tolerance for BE
% Open a file to write the data in
filename = 'data.txt' ;
fid = fopen (filename, 'w+');
fprintf (fid,'time approximate exactn');
% initial conditions
y0 = [10.*pi/180; 0.0];
% initialization
i = 0 ; % loop step counter
y = y0 ; % current state = initial state
t = t0 ; % current time = initial time
fprintf (fid,'%3i %6.4f %6.4fn',t,y(1),y0(1));
% Exact Solution
yexact = exact(t0, y0);
figure(1) ;
plot(t,y(1), 'ro',t,yexact,'b*');
grid;
axis([0 20 0.2 0.2])
hold;
% Adjust time step by controlling the local error
h = accuracy(h,0.01,t0,y0)
tic
% time loop
while ( t < tf )
% Obtain the solution
[t,y] = beuler(h,t,y, ctol) ;
% Exact Solution
yexact = exact(t,y0);
% Write data to the file](https://image.slidesharecdn.com/assignment3-141223124337-conversion-gate01/85/Assignment-3-2-320.jpg)
Assignment 3
- 1. ME 372 Computer-Aided Mechanical Engineering Instructor: Emre Alpman Assignment #3 Due Date: Mar 16th, 2011 SECOND ORDER ACCURATE SCHEMES Summary of the Previous Assignment In Assignment #2, the main goal was to analyze the effect of local error on the solution accuracy with the aim of controlling it. A desired tolerance on the local error was set by dtol which was then used to adjust the stepsize that produces it. For this purpose the module accuracy.m was developed according to the algorithm given in Assignment #2. This module is given in Table 1. Table 1. Module accuracy.m % This module adjusts the step size according to the given % error tolerance dtol % Inputs: Step sizei error tolerance, initial time and condition % Outputs: Adjusted time step function [hnew] = accuracy(h,dtol,t0,y0) hnew = h ; yh = y0 + hnew*deriv(t0,y0) ; % one step solution ymid = y0 + hnew/2*deriv(t0,y0); % take half an euler step tmid = t0+hnew/2; yh2 = ymid + hnew/2*deriv(tmid,ymid) ; % two step solution local_err = norm((yh2yh),Inf)/norm(yh2,Inf) % Local error % Repeat the above steps by dividing the step size by 2 until % relative error becomes small enough while local_err > dtol hnew = hnew/2 ; yh = ymid ; ymid = y0 + hnew/2*deriv(t0,y0); tmid = t0+hnew/2; yh2 = ymid + hnew/2*deriv(tmid,ymid) ; local_err = norm((yh2yh),Inf)/norm(yh2,Inf) end
- 2. The above routine is called by hw2_main.m before the solution loop starts. The modified hw2_main.m is given in Table 2. Table 2. Module hw2_main.m % main program for hw2 clear all; % assignments t0 = 0.0 ; % starting time tf = 20. ; % final time h = 0.2 ; % time step size ctol = 0.001; % Error tolerance for BE % Open a file to write the data in filename = 'data.txt' ; fid = fopen (filename, 'w+'); fprintf (fid,'time approximate exactn'); % initial conditions y0 = [10.*pi/180; 0.0]; % initialization i = 0 ; % loop step counter y = y0 ; % current state = initial state t = t0 ; % current time = initial time fprintf (fid,'%3i %6.4f %6.4fn',t,y(1),y0(1)); % Exact Solution yexact = exact(t0, y0); figure(1) ; plot(t,y(1), 'ro',t,yexact,'b*'); grid; axis([0 20 0.2 0.2]) hold; % Adjust time step by controlling the local error h = accuracy(h,0.01,t0,y0) tic % time loop while ( t < tf ) % Obtain the solution [t,y] = beuler(h,t,y, ctol) ; % Exact Solution yexact = exact(t,y0); % Write data to the file
- 3. fprintf (fid,' %4.2f %6.4f %6.4fn',t,y(1),yexact(1)); % Plot data plot(t,y(1),'ro', t,yexact(1),'b*') ; end toc fclose(fid) ; Here h started from 0.2 and dropped to 0.05. The drop of the local error is given in Table 3. Table 3. Local error local_err = 0.074895 local_err = 0.025799 local_err = 0.0063613 h = 0.050000 The Backward Euler solution with the adjusted step-size is given in Figure 1. Figure 1. Comparison of numerical solution ( red circles) with analytical solution.
- 4. Lab Assignment In this assignment, you are expected to implement the explicit midpoint and explicit trapezoidal methods. The formulations for the above methods are given in equations 1 and 2, respectively. yk1/2= yk h 2 f tk , yk yk1=ykh f tk1/2, yk1/2 (1) yk1 0 = ykh f tk , yk yk1=yk h 2 f tk , yk f tk1, yk1 0 (2) You will write two separate modules named midp.m and trap.m that use Eq. 2 and set of Eqs. 3, respectively. These modules should be able to replace previous beuler.m module, just by changing its name in the main program. In this lab we will compare the methods we have studied for their accuracy. Start with h = 0.02 and adjust the stepsize with accuracy.m using dtol = 0.01. Then plot and compare angular displacement with the exact solution for Forward Euler, Backward Euler, Midpoint and Trapezoidal methods. Homework Assignment Generate a module for the implicit Trapezoidal method: yk1=yk h 2 f tk , yk f tk1 , yk1 (3) Since the above method is implicit you would need to generate an iterative solution procedure similar to the one we have done for the Backward Euler method. Compare the performance of explicit and implicit trapezoidal methods by plotting and tabulating the exact absolute error with respect to time. We can define the exact absolute error as: Et =∣exact t−approximate t∣ (4) Also compare them in terms of the computational time.