MP 3

ES 204

Submitted by:

Timothy John S. Acosta

Submitted on:

May 26, 2015

Executive Summary:

A. Define the Problem

Solve the following equation:

= 50

50 (0) = 2

B. Problems Encountered

The trapezoidal method was used to solve this differential equation. Since the method solves for

the values implicitly and the differential equation is nonlinear, the Newton Rhapson Method

was used to solve for the roots.

C. References

Burden R., Faires J. (2011). Numerical Analysis 9th Edition. Brooks/Cole, 20 Channel Center

Street. Boston, MA 02210, USA.

D. Results

Figure 1. Solutions for different values of h

In the graph above, the blue plot shows the solution for a h value of 0.005. The purple

plot shows the solution for a h value of 0.02. And the orange plot shows the solution for a value

of 0.5. As you can see, all the choses values where within the region of absolute stability since no

overflow can be observed. Therefore any value between 0.5 and below can be used but the

problem would be the accuracy of the solution. For a low relative error, we can use a value of 0.05

for the h values. This would give us relative errors of around 0.001% to 0.820921% which may be

acceptable.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.02 0.04 0.06 0.08 0.1 0.12

time value

0 1.414214

0.005 1.266092

0.01 1.167563

0.015 1.103876

0.02 1.063665

0.025 1.038717

0.03 1.023425

0.035 1.014127

0.04 1.008503

0.045 1.005111

0.05 1.00307

time value

0.055 1.001843

0.06 1.001106

0.065 1.000664

0.07 1.000398

0.075 1.000239

0.08 1.000143

0.085 1.000086

0.09 1.000052

0.095 1.000031

0.1 1.000019

The tables above show the values for each time step from 0 to 0.1.

Executive Summary:

A. Define the Problem

Solve the differential equation:

= 5 6 (0) = 1.0 0 < < 5

B. Problems Encountered

Like the first problem. The trapezoidal method was also used with the newton rhapson

method to solve for the stiff differential equation.

C. Reference

Burden R., Faires J. (2011). Numerical Analysis 9th Edition. Brooks/Cole, 20 Channel Center

Street. Boston, MA 02210, USA.

D. Results

Figure 2. Solution to the differential equation with different values of h

As observed in the figure above, solutions with a time steps above 1e-6 do not converge to a single

value. It can be said the region of stability for this problem is for time steps 1e-6 and below. It can also be

observed the differences in the values are of 1e-3 for the later values of x. But for the earlier values of x,

there is almost 0% relative error.

-12

-10

-8

-6

-4

-2

0

2

0 1 2 3 4 5 6

h=1e-8

h=1e-7

h=1e-6

h=5e-6

h=1e-4

H Values

time 1.00E-04 1.00E-06 1.00E-07 1.00E-08 5.00E-09

0 1 1 1 1 1

0.2 0.818731 0.818731 0.818731 0.818731 0.818731

0.4 0.67032 0.67032 0.67032 0.67032 0.67032

0.6 0.548812 0.548812 0.548812 0.548812 0.548812

0.8 0.449329 0.449329 0.449329 0.449329 0.449329

1 0.367879 0.367879 0.367879 0.367879 0.367879

1.2 0.301194 0.301194 0.301194 0.301194 0.301194

1.4 0.246597 0.246597 0.246597 0.246597 0.246597

1.6 0.201896 0.201897 0.201897 0.201897 0.201897

1.8 0.165298 0.165299 0.165299 0.165299 0.165299

2 0.135332 0.135335 0.135335 0.135335 0.135335

2.2 0.110795 0.110803 0.110803 0.110803 0.110803

2.4 0.090695 0.090718 0.090718 0.090718 0.090718

2.6 0.074212 0.074274 0.074274 0.074274 0.074274

2.8 0.060643 0.06081 0.06081 0.06081 0.06081

3 0.049333 0.049787 0.049787 0.049787 0.049787

3.2 0.039528 0.040762 0.040762 0.040762 0.040761

3.4 0.030018 0.033372 0.033372 0.033371 0.03337

3.6 0.018204 0.027319 0.027321 0.027319 0.027316

3.8 -0.00242 0.022358 0.022363 0.022358 0.02235

4 -0.04907 0.018281 0.018295 0.01828 0.018259

4.2 -0.16817 0.014901 0.014941 0.014898 0.014842

4.4 -0.48562 0.01202 0.012128 0.012011 0.011859

4.6 -1.34337 0.009354 0.009646 0.009329 0.008915

4.8 -3.67077 0.006332 0.007126 0.006265 0.00514

5 -9.99381 0.001579 0.003737 0.001396 -0.00166

Table 2. Computed values for different H sizes

Appendix: CC++ Code #include #include #include "header.h" #include #include using namespace std; double evalfunc(double t,double y){ return 5*y-6*exp(-t); } double dfunc(double t,double y){ return 5; } int main(){ double h=5e-9; double t=0; double w=1; double n=(5)/h; double k1,wo,FLAG,TOL=1e-8; int j,i,writz=0.04*n; ofstream writer; writer.open("data.txt"); cout