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Why there are two plots here when I ran the following code?
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Abdallah Qaswal
on 21 Jun 2022
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. T
his 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?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1040275/image.png)
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
11 Comments
Torsten
on 21 Jun 2022
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
on 21 Jun 2022
May you please help me in this issue? I want just to remove this extra line! or to exclude it!
Abdallah Qaswal
on 21 Jun 2022
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
on 21 Jun 2022
Edited: Torsten
on 21 Jun 2022
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)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1040545/image.png)
Abdallah Qaswal
on 21 Jun 2022
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
on 21 Jun 2022
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
on 21 Jun 2022
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1040585/image.png)
I mean the graph is wired when w values are larger than 2, the line should be straight horizontal line not step shape graph!
Sam Chak
on 21 Jun 2022
Edited: Sam Chak
on 21 Jun 2022
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.
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