Clear Filters
Clear Filters

Numerical integrals and MATLAB precision

1 view (last 30 days)
carlos g
carlos g on 10 Jul 2017
Commented: Karan Gill on 11 Jul 2017
The following code
clc
clear
M=4.5
lambda_0=0.332;
betats=0.05;
betan=betats/(lambda_0^(5/4));
f=@(x,y) y^(1/3)*(y^2+x^2)-1.001*sqrt((1-M^2)*y^2 + x^2);
alphan=real(fsolve(@ (y) y^(1/3)*(y^2+betan^2)-1.001*sqrt((1-M^2)*y^2 + betan^2),3));
alphats=alphan*(lambda_0^(5/4));
omegan=2.299*(alphan^(2/3));
omegats=(lambda_0^(3/2))*omegan;
gammats=sqrt((M^2-1)*alphats^2-betats^2);
gammats=-complex(0,1)*gammats;
beta1=0;
alpha1=1;
omega1=6;
gamma1=sqrt((M^2-1)*alpha1^2-beta1^2);
alpha2=alpha1-alphats;
beta2=beta1-betats;
omega2=omega1-omegats;
gamma2=sqrt((M^2-1)*alpha2^2-beta2^2);
N=25;
YMAX=150;
eta02=-i*omega2/((i*alpha2*lambda_0)^(2/3));
eta_inf2=((i*alpha2*lambda_0)^(1/3))*YMAX+eta02;
syms lu
aiprime(lu) = airy(1,lu);
aisecond(lu)= diff(airy(1,lu));
airy_D2=airy(1,eta02);
airy_DD2=vpa(aisecond(eta02));
airy_INT2=integral(@(n) airy(n),eta02,eta_inf2);
aa2=2*pi*complex(0,1)*gamma2*beta2*lambda_0*airy_D2/(alpha2*(alpha2^2+beta2^2)*airy_INT2-gamma2*lambda_0*airy_D2*(i*alpha2*lambda_0)^(2/3));
bb2=-aa2*airy(2,eta02)*airy_INT2/airy(eta02);
ai2=@(x) x*airy(x);
bi2=@(x) x*airy(2,x);
eta2=@(y) ((i*alpha2*lambda_0)^(1/3))*y+eta02;
Gi2=@(x) -(airy(2,x)*integral(@(n) airy(n),eta_inf2,x)-airy(x)*integral(@(n) airy(2,n),eta02,x));
W2=@(eta) aa2*Gi2(eta)+bb2*airy(eta);
W2VEC=arrayfun(W2,arrayfun(eta2,0:0.01:N));
plot(abs(W2VEC),0:0.01:N)
produces a noisy `W2VEC`. I think there is something very wrong either in the numerical integration or in the Airy functions.
The constants `aa2` and `bb2` have been chosen so that the two terms cancel each other at `0`, so a good result will verify `W2VEC(1)=0`.
What is going on here? Is it the accuracy of the integrals?
  2 Comments
carlos g
carlos g on 10 Jul 2017
Any idea? It seems to me that it has to do with the precision for the integral calculation being lower than the magnitude of the numbers involved (10^15). Is this right? If so, how could I overcome such problem?
Karan Gill
Karan Gill on 11 Jul 2017
The code is hard to read. If there was some explanation, that would help.
  • Try checking the value of intermediate expressions
  • High-precision integration is available using the vpasolve function in Symbolic Math Toolbox.

Sign in to comment.

Answers (0)

Community Treasure Hunt

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

Start Hunting!