Why my graph not same as research paper?

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])

[SALAH ALRABEEI]

what is equation in the research paper!!

[nur]

This is the research paper that i did. Please help me ;(

[Torsten]

@nur

Is C given by an infinite series in x and t ?

[nur]

I think no. i attached my paper. I hope you can help me ;(

[Torsten]

The only thing that helps:

Compare the formulas from the article and yours.

[Torsten]

@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.

Answer 1

[Torsten]

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)  

[nur]

thank you so much!!! this graph exactly same as in the research paper. :)

[Asif Solanki]

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

Answer 2

[Sam Chak]

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.

[nur]

I plotted C. I hope you can help me ;(

[Sam Chak]

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.

Answer 3

[SALAH ALRABEEI]

You have two mistakes here

s=(exp(p_0)*T);

The three s in the code should be this

s=(exp(-p_0*T));

[nur]

thank youuu it much better but can i know why the end ine of graph not end at 0 on y axis?

[nur]

thank youuu it much better but can i know why the end line of graph not end at 0 on y axis?

[SALAH ALRABEEI]

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])