MATLAB: Matrix dimension problem with integral

integral

Dear all,

I have devloped this code and don't think there is any problem with is. However, Matlab returned an error on the matrix dimensions but I didn't use an array.

Could you please help? I also attached the file for the function expression.

Thanks

Ra_i=10.5e-3;
Ra_o=11.5e-3;
l1=1e-3;
mu=4*pi*1e-7;
n1=1;
A=@(x) integral(@(y) y.*besselj(1,x.*y),Ra_i,Ra_o);
fun_M=@(alp,gam) (gam.*(A(gam)).^2.*(sin(alp.*l1/2)).^2)./(alp.^2.*(alp.^2+gam.^2))
L=8*mu.*n1.*integral2(fun_M,0,Inf,0,Inf,'AbsTol', 1e-12,'RelTol',1e-12);

Best Answer

A=@(x) integral(@(y) y.*besselj(1,x.*y),Ra_i,Ra_o,'ArrayValued',true);

Best wishes

Torsten.

Related Solutions

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As soon as you write the formula a function instead of an anonymous function, you can debug it:

% Create anonymous function to provide parameters:
fun4 = @(y) fun4fcn(y, d1i, d2i, bx, s, tau);
...  Call the integrator
% The actual function:
function dy = fun4fcn(y, d1i, d2i, bx, s, tau)
dy = (y .^ 2 - d1i' * y + d2i) * f1 * exp(-(y-bx) .^ 2 ./ (2 * s * tau));
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Now use the debugger:

dbstop if error

When Matlab stops at the error, you can check the dimensions of the variables. Because there is only one "+" operator, I guess that d2i has an unexpected shape.

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When you use integral(), the argument passed will be a vector of arbitrary length. You are then using that vector of arbitrary length in the nested integral() call, which itself gets passed a vector of arbitrary length that might well be different.

int_fun = @(x) arrayfun( @(X) integral(@(y) X + y, 0, 1), x);

Better, of course, would be to use integral2()

integral2( @(x,y) x+y, 1, 100, 0, 1)

Best Answer

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