A quatity is being solved by a self consistent integration
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
How to find Z from the below equation
here \mathcal{P} means the principal value integration. I was tried in the following way, but couldn't figure out how to solve this,
A = 2000; a = 500; tolerance = 10^-4; Z = 0;
for i = 1 : 10
result = integral(@(x) (x.^2.*((A^2+Z^2)./(A^2+((x.^2+a^2)))) .* (sqrt(x.^2+a^2).*(x.^2+a^2-Z^2)).^(-1)), 0,A, 'PrincipalValue', true);
new_Z = sqrt(result);
if abs(new_Z - Z) < tolerance
Z = new_Z;
break;
end
Z = new_Z;
end
disp(new_Z);
Thank you in advance!
2 Commenti
Risposta accettata
Torsten
il 9 Mag 2024
Modificato: Torsten
il 9 Mag 2024
format long
syms x
A = 2000;
a = 500;
b = 1000;
Z = 0;
for i=1:20
f = x^4*((A^2+Z^2)/(A^2+4*(x^2+a^2)))^4 / (sqrt(x^2+a^2)*(x^2+a^2-Z^2));
I = double(int(f,x,0,A,'PrincipalValue',true));
Zpi = sqrt(b^2-I)
Z = real(Zpi)
end
f = x^4*((A^2+Z^2)/(A^2+4*(x^2+a^2)))^4 / (sqrt(x^2+a^2)*(x^2+a^2-Z^2));
double(Z^2 - b^2 + real(int(f,x,0,A,'PrincipalValue',true)))
2 Commenti
Torsten
il 12 Mag 2024
Modificato: Torsten
il 12 Mag 2024
If the values for A don't change much, you should use the result for Z of the call for A(i) as initial guess for the call with A(i+1).
Further, you could try to solve your equation directly without fixed-point iteration using the "vpasolve" function:
syms Z x
A = 2000;
a = 500;
b = 1000;
f = x^4*((A^2+Z^2)/(A^2+4*(x^2+a^2)))^4 / (sqrt(x^2+a^2)*(x^2+a^2-Z^2));
eqn = Z^2 - b^2 + real(int(f,x,0,A,'PrincipalValue',true)) == 0;
vpasolve(eqn,Z)
Più risposte (0)
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!