matrix dimension problem with integral

1 Ansicht (letzte 30 Tage)
Shan  Chu
Shan Chu am 16 Aug. 2017
Bearbeitet: Torsten am 17 Aug. 2017
Dear all,
I have devloped this code and don't think there is any problem with is. However, Matlab returned an error on the matrix dimensions but I didn't use an array.
Could you please help? I also attached the file for the function expression.
Thanks
Ra_i=10.5e-3;
Ra_o=11.5e-3;
l1=1e-3;
mu=4*pi*1e-7;
n1=1;
A=@(x) integral(@(y) y.*besselj(1,x.*y),Ra_i,Ra_o);
fun_M=@(alp,gam) (gam.*(A(gam)).^2.*(sin(alp.*l1/2)).^2)./(alp.^2.*(alp.^2+gam.^2))
L=8*mu.*n1.*integral2(fun_M,0,Inf,0,Inf,'AbsTol', 1e-12,'RelTol',1e-12);

Akzeptierte Antwort

Torsten
Torsten am 16 Aug. 2017
A=@(x) integral(@(y) y.*besselj(1,x.*y),Ra_i,Ra_o,'ArrayValued',true);
Best wishes
Torsten.
  2 Kommentare
Shan  Chu
Shan Chu am 16 Aug. 2017
Thank you so much. it works Could you please explain why we need to assign this as an array?
Torsten
Torsten am 17 Aug. 2017
Bearbeitet: Torsten am 17 Aug. 2017
Take a look at the example "Vector-Valued Function" under
https://de.mathworks.com/help/matlab/ref/integral.html
Note that "gam" from integral2 will be a vector. So in order to calculate A(gam), you will have to activate the "ArrayValued=true" option.
Best wishes
Torsten.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Debugging and Analysis finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by