No error is there, but not getting a finite value after doing triple integration. Output is just the function with integral sign. I need an exact value of the integral.
Not getting finite result
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Athira T Das
am 9 Jun. 2022
Kommentiert: Athira T Das
am 15 Jun. 2022
clear all
clc
syms theta
syms z K
L=40
lambda=0.532*10^(-6)
a=10*(lambda*L)^0.5
A=1
n=0
m=0
k=2*pi/(lambda)
e =(k*a)/((k*a^2)-(1i*L))
ee =(k*a)/((k*a^2)+(1i*L))
G=A*ee*a*hermiteH(n,0)*hermiteH(m,0)
f=norm(G)
B=(1i*ee*(L-z)/(2*k^2*a))*((k*a^2)+(1i*z))*K^2
Hn=hermiteH(n,(z-L)/k*e*K*cos(theta))
Hm=hermiteH(m,(z-L)/k*e*K*sin(theta))
N=A*1i*k*a*ee*exp(B)*Hn*Hm
NNN=subs(N, 1i, -1i)
NN=subs(N, k, -k)
W1=N*NN/G^2
W2=N*NNN/f^2
etta=1*10^-3
delta=8.284*(K*etta)^(4/3)+12.978*(K*etta)^2
chiT=10^-10
w=-2
epsilon=10^-1
ff=chiT*(exp(-1.863*10^-2*delta)+w.^(-2)*exp(-1.9*10^-4*delta)-2*w.^(-1)*exp(-9.41*10^-3*delta))
Phi=0.388*10^(-8)*epsilon^(-1/3)*K^(-11/3)*(1+2.35*(K*etta)^(2/3))*ff
F=K*(W1+W2)*Phi
fun = matlabFunction(F)
q1 = int(F, theta, 0, 2*pi)
q2 = int(K*q1, K, 0, 1)
q3 = int(q2, z, 0, L)
m=4*pi*real(q3)
5 Kommentare
Akzeptierte Antwort
Torsten
am 13 Jun. 2022
Bearbeitet: Torsten
am 13 Jun. 2022
syms theta
syms z K
L=40;
lambda=0.532*10^(-6);
a=10*(lambda*L)^0.5;
A=1;
n=0;
m=0;
k=2*pi/(lambda);
e =(k*a)/((k*a^2)-(1i*L));
ee =(k*a)/((k*a^2)+(1i*L));
G=A*ee*a*hermiteH(n,0)*hermiteH(m,0);
f=norm(G);
B=(1i*ee*(L-z)/(2*k^2*a))*((k*a^2)+(1i*z))*K^2;
Hn=hermiteH(n,(z-L)/k*e*K*cos(theta));
Hm=hermiteH(m,(z-L)/k*e*K*sin(theta));
N=A*1i*k*a*ee*exp(B)*Hn*Hm;
NNN=subs(N, 1i, -1i);
NN=subs(N, k, -k);
W1=N*NN/G^2;
W2=N*NNN/f^2;
etta=1*10^-3;
delta=8.284*(K*etta)^(4/3)+12.978*(K*etta)^2;
chiT=10^-10;
w=-2;
epsilon=10^-1;
ff=chiT*(exp(-1.863*10^-2*delta)+w.^(-2)*exp(-1.9*10^-4*delta)-2*w.^(-1)*exp(-9.41*10^-3*delta));
Phi=0.388*10^(-8)*epsilon^(-1/3)*K^(-11/3)*(1+2.35*(K*etta)^(2/3))*ff;
F=K*(W1+W2)*Phi;
F = K*F; % Maybe superfluous and already done in the line before (F=K*(W1+W2)*Phi;) ; I refer here to the line q2 = int(K*q1, K, 0, 1) in your old code
fun = matlabFunction(F,'Vars',[theta,K,z]);
r=integral3(fun,0,2*pi,0,1,0,L)
m=4*pi*real(r)
5 Kommentare
Torsten
am 13 Jun. 2022
Removed the line
F = K*F;
But still the integration is unsuccessful.
I didn't want you to remove this line. I only wanted you to check whether the multiplication with K is correct.
Since you integrate over theta, maybe you have to take care of the functional determinant of a coordinate transformation.
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