Why my graph not same as research paper?

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nur
nur 2022-6-8
评论： Asif Solanki 2022-8-25
Ran in:
Hi, i want to ask why i did not get same graph as above?
This is my code that i have try:
%for t1=5%
x=0:0.2:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t1=5;
a=((((u_0)^2)/(4*D_0))+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t1/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t1/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C1=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C1)
hold on
%for t2=10%
x=0:0.2:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t2=10;
a=(((u_0)^2)/(4*D_0)+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t2/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t2/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C2=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C2)
hold off
hold on
%for t3=15%
x=0:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t3=15;
a=(((u_0)^2)/(4*D_0)+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t3/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t3/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C3=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C3)
hold on
legend({'t=5','t=10','t=15'},'Location','southwest')
xlabel('Distance,x (meter)')
ylabel('Concentration,C(x,t)')
%ylim([0, 1])
xlim([0, 10])
  6 个评论
显示 4更早的评论
Torsten
Torsten 2022-6-8
The only thing that helps:
Compare the formulas from the article and yours.
Torsten
Torsten 2022-6-8
@nur
I corrected the three errors in your code detected by @SALAH ALRABEEI
Looks better now, but concentrations still become negative. So your search must go on.

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采纳的回答

Torsten
Torsten 2022-6-8
编辑：Torsten 2022-6-8
Ran in:
Sometimes it's better to start anew:
x=(0:0.1:10).';
C0 = 1.0;
tt = [5.0,10.0,15,0];
u0 = 0.25;
D0 = 0.45;
m = 0.1;
gamma0 = 0.02;
mu0 = 0.001;
C = zeros(numel(x),numel(tt));
for i=1:numel(tt)
t = tt(i);
T = 1/(sqrt(2)*m)*(log(2+sqrt(2)+(4+3*sqrt(2))*tanh(m*t/2))-...
log(2+sqrt(2)-sqrt(2)*tanh(m*t/2)));
F = 0.5*exp((u0-sqrt(u0^2+4*gamma0*D0))*x/(2*D0)).*...
erfc((x-sqrt(u0^2+4*gamma0*D0)*T)./(2*sqrt(D0*T)))+...
0.5*exp((u0+sqrt(u0^2+4*gamma0*D0))*x/(2*D0)).*...
erfc((x+sqrt(u0^2+4*gamma0*D0)*T)./(2*sqrt(D0*T)));
G = exp(-gamma0*T)*(1-0.5*erfc((x-u0*T)./(2*sqrt(D0*T)))-...
0.5*exp(u0*x/D0).*erfc((x+u0*T)./(2*sqrt(D0*T))));
C(:,i) = mu0/gamma0 + (C0-mu0/gamma0)*F - mu0/gamma0*G;
end
plot(x,C)
  2 个评论
nur
nur 2022-6-9
thank you so much!!! this graph exactly same as in the research paper. :)
Asif Solanki
Asif Solanki 2022-8-25
Dear Sir,
My name is Asif Solanki I am a student persuing master's program in Italy. I am currently working on a small project where I do need to find the concentration of the pollutants using the Bear analytical method.
Therefore, would you like to assist me to find the solution?
Thank you very much
Best regards
Asif Solanki

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更多回答（2 个）

Sam Chak
Sam Chak 2022-6-8
Hi @nur
One of the ways to find out is to determine the equilibrium points from the advection-dispersion equation.
I haven't checked the long equations. Can you verify if you plotted C or ? Sometimes, the transformations can be a little tricky.
  2 个评论
nur
nur 2022-6-8
I plotted C. I hope you can help me ;(
Sam Chak
Sam Chak 2022-6-8
编辑：Sam Chak 2022-6-8
Hi @nur
Try to make a little bit of value-added effort to the problem.
Another way is to plot the concentration C from the numerical solution of the advection-dispersion differential equation.
This way you can compare with the analytical solution obtained from the paper. Bear in mind that sometimes misprints can occur due to authors' mistake, or the production crew's mistake. So, what was shown on the paper might not be truly the analytical solution. Therefore, yes you have to check.
But I'd suggest you to take numerical solution approach because that is directly from the governing advection-dispersion law. The analytical solution, which is "human-processed equation" from the law.
Edit: Hey, check this out:
https://nnesmo2019.webspace.durham.ac.uk/practicals/session42/
One of the objectives is to Solve the one-dimensional advection dispersion equation using ode45 or ode15s.

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SALAH ALRABEEI
SALAH ALRABEEI 2022-6-8
You have two mistakes here
s=(exp(p_0)*T);
The three s in the code should be this
s=(exp(-p_0*T));
  3 个评论
显示 1更早的评论
nur
nur 2022-6-8
thank youuu it much better but can i know why the end line of graph not end at 0 on y axis?
SALAH ALRABEEI
SALAH ALRABEEI 2022-6-8
The reason is that the curves have data less than zero. Unlike those in the paper.
Anyway, you still can do this by adding this
axis([0 10 0 1])

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