Why there are two plots here when I ran the following code?

Hi every one,

when I run the following code, I get an extra line (circled in red). This line is not expected to occur but I think that this part of the code (215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2))) is responsible of this extra line becuase if it is deleted , the extra line will be removed. This extra line means that there are two values for a single x value on the x axis, which is wrong. How can I remove this extra line without deleting that part?

V1 = @(r,w) -acosh(10*(w/(1600*r + 21))^(1/2))/20000000000
V2 = @(r,w) acosh(10*(w/(1600*r + 21))^(1/2))/20000000000
% Define function to be integrated
fun = @(x,r,w)2.3*r.*0.0018./((w./((cosh(10^10.*x./0.5)).^2)-(r.*16+0.21)).^(1/2));
www = @(w,r)5.124+6.4*10^-6.*215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)).*exp(-215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)))./(integral(@(x)fun(x,r,w),V1(r,w),V2(r,w)))-(exp(-37.45.*r).*(70.31));
fimplicit(www,[0 5 0.001 0.075],'MeshDensity',500, 'LineWidth',1.5),grid

[Torsten]

Find the equation of the line as r = f(w), evaluate "www" at (w,f(w)) and you will find out the reason.

[Abdallah Qaswal]

May you please help me in this issue? I want just to remove this extra line! or to exclude it!

[Torsten]

Did you notice that your integral in the definition of "www" returns complex numbers ?

Who divides an expression by 20000000000 ?

Who multiplies an expression by 10^10 ?

With such a function, everything can happen if you try to graph it.

[Abdallah Qaswal]

No, the integral should not returns complex numbers!

Dividing by 2000000000 and multiplying by 10^10 is part of the code! what do you mean excatly? is there any error here?

[Torsten]

V1 = @(r,w) -acosh(10*(w/(1600*r + 21))^(1/2))/20000000000;
V2 = @(r,w) acosh(10*(w/(1600*r + 21))^(1/2))/20000000000;
% Define function to be integrated
fun = @(x,r,w)2.3*r.*0.0018./((w./((cosh(10^10.*x./0.5)).^2)-(r.*16+0.21)).^(1/2));
www = @(w,r)5.124+6.4*10^-6.*215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)).*exp(-215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)))./(integral(@(x)fun(x,r,w),V1(r,w),V2(r,w)))-(exp(-37.45.*r).*(70.31));
r = 0.001:0.001:0.07;
wguess = 0.5;
w = zeros(size(r));
for i = 1:numel(r)
    w(i) = fsolve(@(w)www(w,r(i)),wguess);
    wguess = w(i);
end

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

plot(w,r)

[Abdallah Qaswal]

thank you, but something wired happens when the range is extended to 0.08 , see here :

why?

V1 = @(r,w) -acosh(10*(w/(1600*r + 21))^(1/2))/20000000000;
V2 = @(r,w) acosh(10*(w/(1600*r + 21))^(1/2))/20000000000;
% Define function to be integrated
fun = @(x,r,w)2.3*r.*0.0018./((w./((cosh(10^10.*x./0.5)).^2)-(r.*16+0.21)).^(1/2));
www = @(w,r)5.124+6.4*10^-6.*215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)).*exp(-215.6*0.5*(w.^(1/2)-(r.*16+0.21).^(1/2)))./(integral(@(x)fun(x,r,w),V1(r,w),V2(r,w)))-(exp(-37.45.*r).*(70.31));
r = 0.001:0.001:0.08;
wguess = 0.5;
w = zeros(size(r));
for i = 1:numel(r)
    w(i) = fsolve(@(w)www(w,r(i)),wguess);
    wguess = w(i);
end
plot(w,r)

[Torsten]

From the implicit plot, you should know that no zeros exists above r = 0.07.

"fsolve" also indicates this by the message that no solution was found.

[Abdallah Qaswal]

I mean the graph is wired when w values are larger than 2, the line should be straight horizontal line not step shape graph!

[Torsten]

Did you check the return message from "fsolve" ? I bet that it is "No solution found" in the region 0.07 < r <= 0.08.

[Abdallah Qaswal]

I am sorry, but this code also did not solve my issue unfortunately!

[Sam Chak]

Hi @Abdallah Qaswal, take it easy. You need be mathematically competent in order to solve this complex problem, because we don't know what your equations do as we can only tell you what's wrong with the code, or the limitations of the code. You are the expert in your own field.

Can you identify the specfic mathematical components in your equations that cause the unwanted issue in the plot?

Once identified, I think the MATLAB MVPs can probably help or advise on the coding part, or check if there is a workaround or trade-off that you can make without taking compromise on the performance or the accuracy of the result.