To plot the absolute percent error

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Naveen Krish 2022-3-18

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https://ww2.mathworks.cn/matlabcentral/answers/1674959-to-plot-the-absolute-percent-error

评论： Naveen Krish 2022-3-18

Hi, I'm doing a project on formation of disinfection byproducts during water treatment to solve this problem i have to plot the absolute percent error in the concentration A at each timestep of 3600 secs from ti = 0 to tf=12 hours. I need to compare the results between Euler's method and 4th order RK. I have wrriten the code but need to determine the absolute percent error. I have tried searching in MATLAB Answers pertaining to my topics of study, but to no avail. The MATLAB code and the error message are shown below:

%%Eulers method

nsteps = 12;

t = zeros (nsteps,1);

A = zeros (nsteps,1);

B = zeros(nsteps, 1);

P = zeros(nsteps,1);

A(1) = 1;

B(1) = 3;

C(1) = 0;

K = 5*10^-5

for k = 2:13

t(k) = t(k-1)+3600

A(k) = A(k-1)+(-K*A(k-1)*B(k-1))*3600;

B(k) = B(k-1)+(-Yb*(K*A(k-1)*B(k-1)))*3600;

P(k)= P(k-1)+ Yp*(K*A(k-1)*B(k-1))*3600;

end

%%RK Method

h=3600;

A0=1;

B0=3;

P0=0;

K=5*10^-5;

Yb=1;

Yp=0.15;

t = 0:h:43200;

fyt = @(t,y) [(-K*y(1)*y(2));

(-Yb*(K*y(1)*y(2)));

(Yp*(K*y(1)*y(2)))];

Y = zeros(3,numel(t))

Y(1,1) = 1.0;

Y(2,1) = 3.0;

Y(3,1) = 0;

for i=1 : numel(t)-1

k1 = fyt(t(i),Y(:,i));

k2 = fyt(t(i)+0.5*h,Y(:,i)+0.5*h*k1);

k3 = fyt(t(i)+0.5*h,Y(:,i)+0.5*h*k2);

k4 = fyt(t(i)+h,Y(:,i)+h*k3);

Y(:,i+1) = Y(:,i) + (h/6)*(k1+2*k2+2*k3+k4);

end

For the following parameter values: 𝑌𝐵 = 1, 𝑌𝑃 = 0.15, 𝐴0 = 1 mg/L, 𝐵0 = 3 mg/L, 𝑃0 = 0, 𝐾 = 5E − 5

Use the analytical solution to plot the absolute percent error in the concentration of A at each timestep for both Euler’s method and 4th order RK. (Notice how the error grows with each time step).

I have also attached the code/data for your convenience. I appreciate your help with troubleshooting the problem. Thanks for considering my request.