Why there is the error that the number of input parameters is insufficient?

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Henan Fang
Henan Fang 2018-9-18
评论： Henan Fang 2018-9-19
(x,y) is a function generated as the following codes. If I input,e.g., vpa(Tuu(0.2,0.3)), it will give a correct result. However, when I calculate the integration that "integral2(Tuu,0.01,pi/2,0,pi/4)", the error that the number of input parameters is insufficient appears, and the first error comes from "kuu =[-ku.*sin(x).*cos(y), -ku.*sin(x).*sin(y), kz]". Why there is the error, how to solve this problem? Many thanks!
function U=Tuu(x,y)
syms kz d
m = 2;
dd=2.106*(m+1);
vh = 4;
mu = 11;
delta = 8;
HBAR = 1.05457266e-34;
ME = 9.1093897e-31;
ELEC = 1.60217733e-19;
Kh = 2.106;
vKh = [0,0,0;Kh,0,0;-Kh,0,0;0,Kh,0;0,-Kh,0];
kc = sqrt(2.*ME.*ELEC/HBAR^2).*1e-10;
ku = kc.*sqrt(mu+delta);
kd = kc.*sqrt(mu-delta);
a3 = [pi/Kh,pi/Kh,sqrt(2).*pi/Kh];
kuu =[-ku.*sin(x).*cos(y), -ku.*sin(x).*sin(y), kz];
n=0:m;
for p=1:5;
for q=1:5;
tuu(p,q)= (sum((kuu + vKh(p,:)).^2)-ku^2).*(p==q)+ kc^2*vh*sum(exp(i.*n.*sum((vKh(q,:)-vKh(p,:)).*a3)))/(m+1).*(p~=q);
end
end
dtuu=det(tuu);
kz0=vpasolve(dtuu,kz);
kzz=kz0(real(vpa(kz0))>=0&imag(vpa(kz0))>=0);
tuu1=subs(tuu,kz,kzz(1));
tuu2=subs(tuu,kz,kzz(2));
tuu3=subs(tuu,kz,kzz(3));
tuu4=subs(tuu,kz,kzz(4));
tuu5=subs(tuu,kz,kzz(5));
tuu11=double(tuu1);
tuu22=double(tuu2);
tuu33=double(tuu3);
tuu44=double(tuu4);
tuu55=double(tuu5);
nuu1=null(tuu11);
nuu2=null(tuu22);
nuu3=null(tuu33);
nuu4=null(tuu44);
nuu5=null(tuu55);
piuu=[nuu1,nuu2,nuu3,nuu4,nuu5];
pei=[1;0;0;0;0];
A=piuu\pei;
psiuu1=A(1).*nuu1(1)*exp(i*kzz(1)*d)+A(2)*nuu2(1)*exp(i*kzz(2)*d)+A(3)*nuu3(1)*exp(i*kzz(3)*d)+A(4)*nuu4(1)*exp(i*kzz(4)*d)+A(5)*nuu5(1)*exp(i*kzz(5)*d);
psiuu2=A(1).*nuu1(2)*exp(i*kzz(1)*d)+A(2)*nuu2(2)*exp(i*kzz(2)*d)+A(3)*nuu3(2)*exp(i*kzz(3)*d)+A(4)*nuu4(2)*exp(i*kzz(4)*d)+A(5)*nuu5(2)*exp(i*kzz(5)*d);
psiuu3=A(1).*nuu1(3)*exp(i*kzz(1)*d)+A(2)*nuu2(3)*exp(i*kzz(2)*d)+A(3)*nuu3(3)*exp(i*kzz(3)*d)+A(4)*nuu4(3)*exp(i*kzz(4)*d)+A(5)*nuu5(3)*exp(i*kzz(5)*d);
psiuu4=A(1).*nuu1(4)*exp(i*kzz(1)*d)+A(2)*nuu2(4)*exp(i*kzz(2)*d)+A(3)*nuu3(4)*exp(i*kzz(3)*d)+A(4)*nuu4(4)*exp(i*kzz(4)*d)+A(5)*nuu5(4)*exp(i*kzz(5)*d);
psiuu5=A(1).*nuu1(5)*exp(i*kzz(1)*d)+A(2)*nuu2(5)*exp(i*kzz(2)*d)+A(3)*nuu3(5)*exp(i*kzz(3)*d)+A(4)*nuu4(5)*exp(i*kzz(4)*d)+A(5)*nuu5(5)*exp(i*kzz(5)*d);
Tuux=ku.*sin(x).*(imag(conj(psiuu1)*diff(psiuu1,d)+conj(psiuu2)*diff(psiuu2,d)+conj(psiuu3)*diff(psiuu3,d)+conj(psiuu4)*diff(psiuu4,d)+conj(psiuu5)*diff(psiuu5,d)));
U=@(x,y) subs(Tuux,d,dd);
end

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采纳的回答

Walter Roberson
Walter Roberson 2018-9-18
integral2(Tuu,0.01,pi/2,0,pi/4)
is equivalent for this purpose to
integral2(Tuu(),0.01,pi/2,0,pi/4)
which is to say that Tuu is called with no parameters, and is expected to output a function handle that will then be integrated with integral2.
You need
integral2(@Tuu,0.01,pi/2,0,pi/4)
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Walter Roberson
Walter Roberson 2018-9-19
Note what I said about the code being too slow to be usable. I ran the integral2 for hours without producing a result.
At the very least you need to take out of the routine everything that can be pre-calculated, right down to the formula that you vpasolve on: solve() of that formula gives a symbolic form that you can pre-calculate. Then you would substitute the actual x and y values into that symbolic form.
Henan Fang
Henan Fang 2018-9-19
@Walter Roberson Thank you very much! I will revised the codes as your suggestions. If there is any problem, may I have your guidance?

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