How to use integral2 when the integrand is a array?

(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 Kommentar

>> 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));

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Walter Roberson
Walter Roberson am 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.

Weitere Antworten (0)

Kategorien

Mehr zu MATLAB finden Sie in Hilfe-Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by