Error while integrating bessel functions

1 view (last 30 days)
Hi,
I am trying to solve the integral equation with Bessel functions as shown in the attached image.
I get a error "Operator '+' is not supported for operands of type 'function_handle'".
Can anyone specify how to rectify the error.
The code is as follows,
gammalv = 72e-3;
gammasv = 0.06;
h = 50e-6;
nu = 0.5;
E = 3000;
R = 0.2e-3;
r = 0.0002;
a = @(s) (5 - 12*nu + 8*nu.^2 + 2*s.^2*h.^2 + (3 - 4*nu)*cosh(2*s.*h))./((3 - 4*nu)*sinh(2*s.*h) - 2*s.*h);
b = @(s) (2*(1 - nu.^2).*s.*gammasv)./E;
Q = @(s) 2*(1 - (nu.^2))./(s.*E).*(1./(a + b));
f = @(s) s.*(R.*besselj(0,s.*R) - ((2*besselj(1,s.*R))./s)).*Q(s).*besselj(0,s.*r);
ans = integral(f,0,inf);
Thanks,
Alwar

Accepted Answer

Star Strider
Star Strider on 14 Sep 2021
The problem is:
Q = @(s) 2*(1 - (nu.^2))./(s.*E).*(1./(a(s) + b(s)));
↑ ← HERE
These both need to be evaluated with their arguments (that I supplied here).
gammalv = 72e-3;
gammasv = 0.06;
h = 50e-6;
nu = 0.5;
E = 3000;
R = 0.2e-3;
r = 0.0002;
a = @(s) (5 - 12*nu + 8*nu.^2 + 2*s.^2*h.^2 + (3 - 4*nu)*cosh(2*s.*h))./((3 - 4*nu)*sinh(2*s.*h) - 2*s.*h);
b = @(s) (2*(1 - nu.^2).*s.*gammasv)./E;
Q = @(s) 2*(1 - (nu.^2))./(s.*E).*(1./(a(s) + b(s)));
f = @(s) s.*(R.*besselj(0,s.*R) - ((2*besselj(1,s.*R))./s)).*Q(s).*besselj(0,s.*r);
ans = integral(f,0,inf);
Warning: Inf or NaN value encountered.
There are still problems with infinite results. The second term of ‘f’ (with ‘s’ in the numerator and denominator) is likely the cause. I leave that for you to resolve.
.
  2 Comments
Alwar Samy Ramasamy
Alwar Samy Ramasamy on 14 Sep 2021
Edited: Alwar Samy Ramasamy on 14 Sep 2021
Thanks Star Strider. I will try to figure out the infinite result part.
Star Strider
Star Strider on 14 Sep 2021
My pleasure!
The result returned is NaN not Inf since the result will be of the 0/0 calculation, while an Inf result would be the result of a x/0 calculation with ‘x’ being any non-zero value.
If my Answer helped you solve your problem, please Accept it!
.

Sign in to comment.

More Answers (0)

Products


Release

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!