Question 7.2: LOCAL AND GLOBAL TRUNCATION ERRORS In Example 7.1, calculate...

Numerical Methods for Engineers and Scientists Using MATLAB®

LOCAL AND GLOBAL TRUNCATION ERRORS

In Example 7.1, calculate the local and global truncation errors at each point and tabulate the results.

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Starting with the initial condition $y_0=\frac{1}{2}$ , the Euler’s computed value at $x_1 = 0.1$ is $y^{c}_{1} = 0.5250$ while the actual value is $y^{a}_{1} = 0.522007$ . At this stage, the global and local truncation errors are the same because Euler’s method used the initial condition, which is exact, to find the estimate. At $x_2 = 0.2$ , the computed value is $y^{c}_{2}= 0.543992$ , which was calculated by Euler’s method using the estimated value $y^{c}_{2} = 0.5250$ from the previous step. If instead of $y^{c}_{1}$ we use the actual value $y^{a}_{1} = 0.522007$ , the computed value at $x_2$ is

\overset{\sim }{y_{2} }=y^{a}_{1}+hf(x_1,y^{a}_{1} )=0.522007+0.1f(0.1,0.522007)=0.541148

Therefore, local truncation error at $x_2$ is

y^{a}_{2}-\overset{\sim }{y_2}=0.538525-0.541148=-0.002623

The global truncation error at $x_2$ is simply calculated as

y^{a}_{2}-y^{c}_{2}=0.538525-0.543992=-0.005467

It is common to express these errors in the form of percent relative errors, hence at each point, we evaluate

\frac{(local   or   global)   truncation   error}{actual   value}\times 100

With this, the (local) percentage relative error at $x_2$ is

$\frac{y^{a}_{2}-\overset{\sim }{y_2} }{y^{a}_{2} }\times 100= \frac{-0.002623}{0.538525}\times 100\cong -0.49$ %

The (global) percentage relative error at $x_2$ is

$\frac{y^{a}_{2}-y^{c}_{2} }{y^{a}_{2} }\times 100= \frac{-0.005467}{0.538525}\times 100\cong -1.02$ %

The following MATLAB script uses this approach to find the percentage relative errors at all xi, and completes the table presented earlier in Example 7.1.

disp(' x yEuler yExact e_local e_global')
h = 0.1; x = 0:h:1; y0 = 1/2;
f = @(x,y)((exp(-x)-y)/2);
yEuler = EulerODE(f,x,y0);
y _ exact = matlabFunction(dsolve('2*Dy + y = exp(-x)','y(0)=1/2','x'));
ytilda = 0*x; ytilda(1) = y0;
for n = 1:length(x)−1,
ytilda(n+1) = y_exact(x(n)) + h*f(x(n),y_exact(x(n)));
end
for k = 1:length(x),
x_coord = x(k);
yE = yEuler(k);
yEx = y_exact(x(k));
e_local = (yEx-ytilda(k))/yEx*100;
e_global = (yEx-yE)/yEx*100;
fprintf('%6.2f %11.6f %11.6f %6.2f %6.2f\n',x _ coord,yE,yEx,e _ local,e _ global)
end
x yEuler yExact e_local e_global
0.00 0.500000 0.500000 0.00 0.00
0.10 0.525000 0.522007 −0.57 −0.57
0.20 0.543992 0.538525 −0.49 −1.02 Hand calculations (see above)
0.30 0.557729 0.550244 −0.42 −1.36
0.40 0.566883 0.557776 −0.36 −1.63
0.50 0.572055 0.561671 −0.31 −1.85
0.60 0.573779 0.562416 −0.27 −2.02
0.70 0.572531 0.560447 −0.23 −2.16
0.80 0.568733 0.556151 −0.20 −2.26
0.90 0.562763 0.549873 −0.17 −2.34
1.00 0.554953 0.541917 −0.15 −2.41 Max. % rel. err. reported earlier