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

(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

Answer 1

[Walter Roberson]

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)

[Henan Fang]

@Walter Roberson If I define a simple function I=abc(x,y), and I=@(x,y) x+y, I have tested that integral2(abc,0,1,0,2) will work. When I tried integral2(@Tuu,0.01,pi/2,0,pi/4), more errors appear. Why? Thank you very much!

[Torsten]

U=subs(Tuux,d,dd);

instead of

U=@(x,y) subs(Tuux,d,dd);

in the last line.

[Henan Fang]

@Torsten I have done as your suggestion, however, the error that "wrong use of symengine Array size must match" appears, why, how to solve this problem? Thank you very much!

[Walter Roberson]

U = double(subs(Tuux,d,dd));

and

integral2(@(X,Y) arrayfun(@(x,y) Tuu(x,y),X,Y),0.01,pi/2,0,pi/4)

[Henan Fang]

@Walter Roberson Thank you very much! Can you tell me why use arrayfun? why x and y are array, not double number?

[Walter Roberson]

integral2() passes in two arrays the same size for its arguments, and the function is responsible for calculating the value at each corresponding location. However, your Tuu is only able to take in one pair of values, so you need to invoke Tuu over and over again.

This code is too slow to be usable. You need to move away from the symbolic toolbox as much as is feasible.

[Henan Fang]

@Walter Roberson Thank you very much! I have accepted this answer and voted it.

[Walter Roberson]

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]

@Walter Roberson Thank you very much! I will revised the codes as your suggestions. If there is any problem, may I have your guidance?