How to calculate a function of multiple variables which also has an integral in its definition?

Dea All,

I have the following function whose definition needs an integral to be evaluated. The integral itself is dependent on the function input variables.

r0 = 0.5;
z0 = 0.5;
G(r,z,z-z0) = 1/2*r*r0^2 * integral(cos(lambda)/sqrt((r^2+r0^2-2*r*r0*cos(lambda)+(z-z0)^2)) dlambda, -pi, pi);

Could someone please help me how I can get for example G(0.75, 0.75, 0.25)? My final goal is to find G over a rectangular meshgrid.

Thanks,

Ahmad

Answer 1

[Matt J]

Create an anonymous function for the integrand as a function of lambda

G=@(r,z,z-z0)  1/2*r*r0^2 * integral(@(lambda) cos(lambda)/sqrt((r^2+r0^2-2*r*r0*cos(lambda)+(z-z0)^2)) , -pi, pi);

[AP]

when I run it in Matlab I get the following error:

   G=@(r,z,z-z0)  1/2*r*r0^2 * integral(@(lambda) cos(lambda)/sqrt((r^2+r0^2-2*r*r0*cos(lambda)+(z-z0)^2)) , -pi, pi);
           |
  Error: Unexpected MATLAB operator.

Then, I defined another variable:

z_minus_z0=z-z0

and redefined the function handle and it worked. The problem is now evaluating the function in for example G(0.75, 0.75, 0.25). I set:

r0=0.5; z0=0;

but I when I run the following

feval(G, 0.75, 0.75, 0.25):

I get the error:

Error using integralCalc/finalInputChecks (line 515)
Output of the function must be the same size as the input. If FUN is an array-valued integrand, set the 'ArrayValued'
option to true.

[Matt J]

It needs to be something like

 G=@(r,z,z0) ...

It's not clear what functional form you meant when you wrote G(r,z,z-z0). Where is the z argument being substituted in the integrand on the RHS and where is z-z0 being substituted? To understand clearly, you really would need to show us the form of G(x,y,z) as a function of 3 independent variables.

[AP]

This is the new definition:

r0=0.5;
z0=0.5;  
G=@(r,z,z_minus_z0)  1/2*r*r0^2 * integral(@(lambda) cos(lambda)/sqrt((r^2+r0^2-2*r*r0*cos(lambda)+z_minus_z0^2)) , -pi, pi);
feval(G, 0.75, 0.75, 0.25)

[Matt J]

I don't see z anywhere on the RHS. Why not just have

G=@(r,z_minus_z0)

[Matt J]

Replace all the * and / by elementwise operations .* and ./

G=@(r,z,z_minus_z0) 1/2.*r.*r0^2 .* ... integral(@(lambda) cos(lambda)./sqrt((r.^2+r0.^2-2.*r.*r0.*cos(lambda)+z_minus_z0.^2)) , -pi, pi);

Answer 2

[Star Strider]

You need to ‘vectorize’ it:

r0 = 0.5;
z0 = 0.5;
r = 1;
z = 1;
G = @(r,z,z0) 1/2.*r.*r0.^2 .* integral(@(lambda) cos(lambda)./sqrt((r.^2+r0.^2-2.*r.*r0.*cos(lambda)+(z-z0).^2)) , -pi, pi);

G(r,z,z0)

[AP]

Thank @Star Strider. I decided to put all the variables into the functioin definition and did very similar to what you did. I don't know why it does not evaluate G.

r0 = 0.5;
z0 = 0.5;
r = 1;
z = 1;
G=@(r, r0, z, z0)  1/2*r*r0^2 * integral(@(lambda) cos(lambda)/sqrt((r^2+r0^2-2*r*r0*cos(lambda)+(z-z0)^2)) , -pi, pi);
G(r,r0,z,z0)

I get this error:

Error using integralCalc/finalInputChecks (line 515)
Output of the function must be the same size as the input. If FUN is an array-valued integrand, set the 'ArrayValued'
option to true.

[AP]

Thanks all. Runs perfectly with elementwise operations.

r0 = 0.5;
z0 = 0.5;
r = 1;
z = 1;
G=@(r, r0, z, z0)  1/2.*r.*r0^2 .* integral(@(lambda) cos(lambda)./sqrt((r.^2+r0.^2-2*r.*r0.*cos(lambda)+(z-z0).^2)) , -pi, pi);
G(r,r0,z,z0)
ans =

      0.1390

[Star Strider]

You have to vectorize it using the ‘dot’ operators:

G = @(r, r0, z, z0) 1/2.*r.*r0.^2 .* integral(@(lambda) cos(lambda)./sqrt((r.^2+r0.^2-2.*r.*r0.*cos(lambda)+(z-z0).^2)) , -pi, pi);

See if that works as you want it to.