Why my graph not same as research paper?

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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 Comments
Torsten
Torsten on 8 Jun 2022
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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Accepted Answer

Torsten
Torsten on 8 Jun 2022
Edited: Torsten on 8 Jun 2022
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 Comments
Asif Solanki
Asif Solanki on 25 Aug 2022
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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More Answers (2)

Sam Chak
Sam Chak on 8 Jun 2022
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 Comments
Sam Chak
Sam Chak on 8 Jun 2022
Edited: Sam Chak on 8 Jun 2022
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:
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 on 8 Jun 2022
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 Comments
SALAH ALRABEEI
SALAH ALRABEEI on 8 Jun 2022
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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