Discontinuities when computing integration of error functions using integral function
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I am trying to integrate a function over a region in different time intervals. The integration looks something like this.
fun_uz = @(u)1./sqrt(u).*exp(-Z.^2./(2.*u));
fun_Y = @(u)(erf((Y+B)./sqrt(2.*u))-erf((Y-B)./sqrt(2.*u)));
fun_Z = @(u)(erf((X+L+u)./sqrt(2.*u))-erf((X-L+u)./sqrt(2.*u)));
fun = @(u)inc.*fun_uz(u).*fun_Y(u).*fun_Z(u);
fint = integral(fun,0,upperl);
The variable 'upperl' is the upper limit of the integral function. I have to perform this integration over different X,Y, and Z regions and different 'upperl' values. I am getting profiles which are discontinuous for different 'upperl' values. I have shown here profiles at few different 'upperl' values.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/278091/image.jpeg)
I am not able to understand why the discontinuity are occuring, any help is greatly appreciated. Thanks.
6 件のコメント
darova
2020 年 3 月 19 日
time = 1e-2; % The variable which is changed to generate different contour plots
xs = (-500:10:10)./1000;
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/278262/image.png)
採用された回答
darova
2020 年 3 月 19 日
The function you are trying to integrate looks like following
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/278274/image.png)
Put these lines inside for loops
ezplot(fun,[0 upperl])
pause(0.01)
When time > 0.1 upperl is big. When you call integral you don't know how many points it takes. Maybe it misses something
5 件のコメント
その他の回答 (1 件)
Walter Roberson
2020 年 3 月 20 日
Change the integral to
fint = integral(fun,0,upperl, 'waypoints', L-X);
You have two erf that only have an input near 0 (and so a measurable output) near-ish -(X+L) to -(X-L) . Some of your integral() calls just happened to evaluate near there, and some of them did not happen to evaluate near there and predicted that there was nothing interesting in that area. The above forces evaluation near that area.
1 件のコメント
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