A quatity is being solved by a self consistent integration

Question

pritha 2024-5-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2116926-a-quatity-is-being-solved-by-a-self-consistent-integration

编辑： Torsten 2024-5-12

采纳的回答： Torsten

在 MATLAB Online 中打开

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!

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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 2024-5-9

Hi Torsten,

Z is completely unknown and there was no specified situation for that.

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Answer 1

Torsten 2024-5-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2116926-a-quatity-is-being-solved-by-a-self-consistent-integration#answer_1455091

编辑：Torsten 2024-5-9

在 MATLAB Online 中打开

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

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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 2024-5-12

编辑：Torsten 2024-5-12

在 MATLAB Online 中打开

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)