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Numerical integrals and MATLAB precision

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carlos g
carlos g 2017 年 7 月 10 日
コメント済み: Karan Gill 2017 年 7 月 11 日
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 件のコメント
carlos g
carlos g 2017 年 7 月 10 日
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 2017 年 7 月 11 日
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.

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