CODEWITHSUNDEEP
matlabvectorizationintegral
fpe
fpe

Reputation: 2750

Bi-dimensional integral

I suppose I have a straightforward question to answer.

I have this piece of code:

clc, clear all, close all

Nx = 8192;
Ny = 16;
Nz = 16;
delta = 6;
UHub = 11.4;
IECturbC = 'A';
HubHt = 90;

%%INITIALISATION

% definition of constants
twopi=2*pi;
%constants and derived parameters from IEC
gamma = 3.9; %IEC, (B.12)
alpha = 0.2; %IEC, sect. 6.3.1.2

%set delta1 according to guidelines (chap.6)
if HubHt<=60,
    delta1=0.7*HubHt;
else
    delta1=42;
end;

%IEC, Table 1, p.22
if IECturbC == 'A',
    Iref=0.16;
elseif IECturbC == 'B',
    Iref=0.14; 
elseif IECturbC == 'C',
    Iref=0.12;
else
    error('IECturbC can be equal to A,B or C;adjust the input value')
end;

%IEC, sect. 6.3.1.3
b=5.6;
sigma1=Iref*(0.75*UHub+b);
sigma2=0.7*sigma1;
sigma3=0.5*sigma1;
%derived constants
l=0.8*delta1; %IEC, (B.12)
sigmaiso=0.55*sigma1; %IEC, (B.12)

Cij2=zeros(3,3,Nx,Ny,Nz);
ikx=cat(2,-Nx/2:-1,1:Nx/2-1);
[x y z]=ndgrid(ikx,1:Ny,1:Nz);
k1=(x)*l/(Nx*delta)*twopi;
k2=(y-Ny/2)*l/(Ny*delta)*twopi;
k3=(z-Nz/2)*l/(Nz*delta)*twopi;
kabs=sqrt(k1.^2+k2.^2+k3.^2);
beta= gamma./((kabs.^(2/3).*(hypergeometric2f1(1/3, 17/6,4/3,-kabs.^(-2),11))).^(0.5));
k03=k3+beta.*k1;
k0abs=sqrt(k1.^2+k2.^2+k03.^2);
Ek0=1.453*k0abs.^4./(1+k0abs.^2).^(17/6);
C1=beta.*k1.^2.*( k0abs.^2 - 2*k03.^2 + beta.*k1.*k03 )./( kabs.^2.*( k1.^2 + k2.^2 ));
C2=k2.*k0abs.^2./ (exp( (3/2).*log( k1.^2 + k2.^2 ) )) .* atan2( beta.*k1.* sqrt( k1.^2 + k2.^2 ) ,( k0abs.^2 - k03.*k1.*beta));
xhsi1=C1 - k2.*C2./k1;
xhsi2=k2.*C1./k1 + C2;
CC=sigmaiso*sqrt(twopi*pi*l^3.*Ek0./(Nx*Ny*Nz*delta^3.*k0abs.^4));

Then, I define

phi_11 = (Ek0./4*pi.*kabs.^4).*(k0abs.^2 - k1.^2 - 2*k1.*k03.*xhsi1 + (k1.^2 + k2.^2).*xhsi1.^2);

Now I would like to perform this integral

enter image description here

with phi_11 instead of phi_ii.

How can I perform it in Matlab, since I have all vectorized variables?

I thank you in advance for the support.

WKR, Francesco

Upvotes: 1

Views: 103

Answers (1)

Rody Oldenhuis
Rody Oldenhuis

Reputation: 38032

I don't have hypergeometric2f1 (function? variable?), so I can't test this.

Anyway, the normal way to go about this is to define phi_11 like this: 

phi_11 = @(k1,k2) (Ek0./4*pi.*kabs.^4).*(k0abs.^2 - k1.^2 - 2*k1.*k03.*xhsi1 + (k1.^2 +  k2.^2).*xhsi1.^2);

And then use a double-quadrature method:

F_11 = dblquad(phi_11, -inf,inf, -inf,inf);

You might need to adjust a whole bunch of the variables, since k1 and k2 will be dimensioned rather arbitrarily (dblquad is an adaptive method, and it will evaluate the function at points that depend on the function's local behaviour).

Upvotes: 2

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