A quatity is being solved by a self consistent integration

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pritha
pritha 2024-5-9
编辑： Torsten 2024-5-12
How to find Z from the below equation
How to find Z with the known parameters A(=2000) and a(=500).
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 个评论
Torsten
Torsten 2024-5-9
编辑：Torsten 2024-5-9
Why is the Principal Value necessary to be taken ? In case a^2 - Z^2 <= 0 ?
pritha
pritha 2024-5-9
Hi Torsten,
Z is completely unknown and there was no specified situation for that.

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采纳的回答

Torsten
Torsten 2024-5-9
编辑：Torsten 2024-5-9
Ran in:
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
Zpi =
9.822515593894991e+02
Z =
9.822515593894991e+02
Zpi =
9.926239912427430e+02 -1.331193739501012e-147i
Z =
9.926239912427430e+02
Zpi =
9.942989166123056e+02 -4.915924275823428e-153i
Z =
9.942989166123056e+02
Zpi =
9.945686435588058e+02 +9.829182155008009e-152i
Z =
9.945686435588058e+02
Zpi =
9.946120599497613e+02 -1.207757180393064e-148i
Z =
9.946120599497613e+02
Zpi =
9.946190479156575e+02 -2.457171010268977e-153i
Z =
9.946190479156575e+02
Zpi =
9.946201726310163e+02 +1.228584115851398e-152i
Z =
9.946201726310163e+02
Zpi =
9.946203536539656e+02 +9.828671137972504e-153i
Z =
9.946203536539656e+02
Zpi =
9.946203827896022e+02 -2.061221673054303e-146i
Z =
9.946203827896022e+02
Zpi =
9.946203874789816e+02 +1.179440496446327e-151i
Z =
9.946203874789816e+02
Zpi =
9.946203882337369e+02 -1.106609953389711e-137i
Z =
9.946203882337369e+02
Zpi =
9.946203883552148e+02 -3.542724730738004e-147i
Z =
9.946203883552148e+02
Zpi =
9.946203883747665e+02 +1.006455889394421e-149i
Z =
9.946203883747665e+02
Zpi =
9.946203883779135e+02
Z =
9.946203883779135e+02
Zpi =
9.946203883784200e+02 -1.610329423025159e-147i
Z =
9.946203883784200e+02
Zpi =
9.946203883785015e+02 +1.228583849353811e-152i
Z =
9.946203883785015e+02
Zpi =
9.946203883785146e+02 -1.030610830736004e-145i
Z =
9.946203883785146e+02
Zpi =
9.946203883785167e+02 -2.061221661472003e-146i
Z =
9.946203883785167e+02
Zpi =
9.946203883785171e+02 +4.221381962694661e-143i
Z =
9.946203883785171e+02
Zpi =
9.946203883785171e+02 +1.376013911276247e-151i
Z =
9.946203883785171e+02
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)))
ans =
-6.035923459074367e-11
  2 个评论
pritha
pritha 2024-5-12
Hi Torsten,
Thank you. This works very well. However, when I run the code for 1500 values of 'a', it takes too much time, almost like 1hr. Could you please suggest me some wayout or any otherr process with which such kind of problem can be solved?
Torsten
Torsten 2024-5-12
编辑：Torsten 2024-5-12
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)

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