Date: Jan 7, 2013 4:06 AM
Author: Francesco Perrone
Subject: Handle function implicitly accounting for independent variables
I have that

clc, clear all, close all

tic

k1 = 1E-02:0.1:1E+02;

k2 = 1E-02:0.1:1E+02;

k3 = 1E-02:0.1:1E+02;

k = sqrt(k1.^2+k2.^2+k3.^2);

c = 1.476;

gamma = 3.9;

colors = {'b'};

Ek = (1.453*k.^4)./((1 + k.^2).^(17/6));

E = @(k) (1.453*k.^4)./((1 + k.^2).^(17/6));

E_int = zeros(1,numel(k));

E_int(1) = 1.5;

for i = 2:numel(k)

if k(i) < 400

E_int(i) = E_int(i-1) - integral(E,k(i-1),k(i));

elseif k(i) > 400

E_int(i) = 2.180/(k(i)^(2/3));

end %end if

end %end i

beta = (c*gamma)./(k.*sqrt(E_int));

figure

plot(k,beta,colors{1})

count = 0;

%F_11 = zeros(1,numel(k1));

F_33 = zeros(1,numel(k1));

Afterwards, I compute F_33 as

for i = 1:numel(k1)

count = count + 1;

phi_33 = @(k2,k3) (1.453./(4.*pi)).*(((k1(i)^2+k2.^2+(k3 + beta(i).*k1(i)).^2).^2)./((k1(i)^2+k2.^2+k3.^2).^2)).*((k1(i)^2+k2.^2)./((1+k1(i)^2+k2.^2+(k3+beta(i).*k1(i)).^2).^(17/6)));

F_33(count) = 4*integral2(phi_33,0,1000,0,1000);

end

Now let's come to my question. I know from a paper that:

k = sqrt(k1.^2+k2.^2+k3.^2);

k30 = k3 + beta.*k1;

k0 = sqrt(k1.^2+k2.^2+k30.^2);

E_k0 = 1.453.*(k0.^4./((1+k0.^2).^(17/6)));

Therefore the expression for phi_33 would result in

phi_33 = (E_k0./(4*pi.*(k.^4))).*(k1.^2+k2.^2);

The question is: how can I make use of this final expression insted of the long one I'm using at the moment (within the for loop)?

The last expression for phi_33 is easier to handle (especially because of reckless mistakes in writing the former) and it would "pass by reference" (k2,k3), which are the independent variables.

Any hint is more than welcome.

Best regards, fpe