How to use integral2 when the integrand is a array?
2 views (last 30 days)
Show older comments
(x,y) is a generated integrand as the following codes. d is a parameter in Tuu. I want to get a set of value of the integration "integral2(@Tuu,0,pi/2,0,pi/4)" with different values of d. And thus I set d=1e-9:1e-10:3e-9 as in the codes. However the codes "integral2(@Tuu,0,pi/2,0,pi/4)" gives the error "insufficient number of inputs". Why? How to solve this problem? Many thanks!
The codes of Tuu(x,y) are as following:
function U=Tuu(x,y)
d=1e-9:1e-10:3e-9;
mu=8;
delta=10;
vh=16;
HBAR=1.05457266e-34;
ME=9.1093897e-31;
ELEC=1.60217733e-19;
Kh=2.116e10;
kc=sqrt(2.*ME.*ELEC./HBAR.^2);
ku=kc.*sqrt(mu+delta);
kd=kc.*sqrt(mu-delta);
puu1=sqrt(ku.^2-ku.^2.*sin(x).^2+kc.^2.*vh);
puu2=sqrt(ku.^2-ku.^2.*sin(x).^2-kc.^2.*vh);
quu1=sqrt(ku.^2-(Kh-ku.*sin(x).*cos(y)).^2-ku.^2.*sin(x).^2.*sin(y).^2+kc.^2.*vh);
quu2=sqrt(ku.^2-(Kh-ku.*sin(x).*cos(y)).^2-ku.^2.*sin(x).^2.*sin(y).^2-kc.^2.*vh);
U=1/4.*ku.*sin(x).*(real(puu1).*exp(-2.*imag(puu1).*d)+real(puu2).*exp(-2.*imag(puu2).*d)+real(quu1).*exp(-2.*imag(quu1).*d)+real(quu2).*exp(-2.*imag(quu2).*d)+((real(puu1)+real(puu2)).*cos((real(puu1)-real(puu2)).*d)-(imag(puu1)-imag(puu2)).*sin((real(puu1)-real(puu2)).*d)).*exp(-(imag(puu1)+imag(puu2)).*d)-((real(quu1)+real(quu2)).*cos((real(quu1)-real(quu2)).*d)-(real(quu1)-imag(quu2)).*sin((real(quu1)-real(quu2)).*d)).*exp(-(imag(quu1)+imag(quu2)).*d));
end
1 Comment
Walter Roberson
on 2 Nov 2018
>> integral2(@Tuu,0,pi/2,0,pi/4)
Matrix dimensions must agree.
Error in Tuu (line 17)
U=1/4.*ku.*sin(x).*(real(puu1).*exp(-2.*imag(puu1).*d)+real(puu2).*exp(-2.*imag(puu2).*d)+real(quu1).*exp(-2.*imag(quu1).*d)+real(quu2).*exp(-2.*imag(quu2).*d)+((real(puu1)+real(puu2)).*cos((real(puu1)-real(puu2)).*d)-(imag(puu1)-imag(puu2)).*sin((real(puu1)-real(puu2)).*d)).*exp(-(imag(puu1)+imag(puu2)).*d)-((real(quu1)+real(quu2)).*cos((real(quu1)-real(quu2)).*d)-(real(quu1)-imag(quu2)).*sin((real(quu1)-real(quu2)).*d)).*exp(-(imag(quu1)+imag(quu2)).*d));
Accepted Answer
Walter Roberson
on 2 Nov 2018
d=1e-9:1e-10:3e-9;
output = arrayfun(@(D) integral2(@(x,y) Tuu(x, y, D), 0,pi/2,0,pi/4,'reltol', 2e-4), d);
function U=Tuu(x, y, d)
mu=8;
delta=10;
vh=16;
HBAR=1.05457266e-34;
ME=9.1093897e-31;
ELEC=1.60217733e-19;
Kh=2.116e10;
kc=sqrt(2.*ME.*ELEC./HBAR.^2);
ku=kc.*sqrt(mu+delta);
kd=kc.*sqrt(mu-delta);
puu1=sqrt(ku.^2-ku.^2.*sin(x).^2+kc.^2.*vh);
puu2=sqrt(ku.^2-ku.^2.*sin(x).^2-kc.^2.*vh);
quu1=sqrt(ku.^2-(Kh-ku.*sin(x).*cos(y)).^2-ku.^2.*sin(x).^2.*sin(y).^2+kc.^2.*vh);
quu2=sqrt(ku.^2-(Kh-ku.*sin(x).*cos(y)).^2-ku.^2.*sin(x).^2.*sin(y).^2-kc.^2.*vh);
U=1/4.*ku.*sin(x).*(real(puu1).*exp(-2.*imag(puu1).*d)+real(puu2).*exp(-2.*imag(puu2).*d)+real(quu1).*exp(-2.*imag(quu1).*d)+real(quu2).*exp(-2.*imag(quu2).*d)+((real(puu1)+real(puu2)).*cos((real(puu1)-real(puu2)).*d)-(imag(puu1)-imag(puu2)).*sin((real(puu1)-real(puu2)).*d)).*exp(-(imag(puu1)+imag(puu2)).*d)-((real(quu1)+real(quu2)).*cos((real(quu1)-real(quu2)).*d)-(real(quu1)-imag(quu2)).*sin((real(quu1)-real(quu2)).*d)).*exp(-(imag(quu1)+imag(quu2)).*d));
end
If you try to use a smaller relative tolerance then you will get warning messages about using too many iterations. Your integrals are in the range of 2E20 so they do not converge well.
More Answers (0)
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!