Solving for Root of a Non-linear Integral Equation

 TR = 170+520;
 Pc = 692; 
 Tc = 374; 
 g = 0.71; 
 qsc = 4; 
 kh = 29.8; 
 phih = 1.9;
 rw = 0.5; 
 Pinit = 3546
 pb = 14.65;

 %Calculations
 syms p 

 ci = 1/Pinit; 

n=0.038-0.026*sqrt(TR/Tc);
m=0.51*(TR/Tc)^-4.133;
zi=1-m*(Pinit/Pc)+n*(Pinit/Pc)^2+0.0003*(Pinit/Pc)^3;

Ki=(9.4+0.02*29*g)*TR^1.5/(209+19*29*g+TR);
Xi=3.5+986/TR+0.29*g;
Yi=2.4-0.2*Xi;
ugi=1e-4*Ki*exp(Xi*(29*g*Pinit/(zi*10.732*TR*62.4))^Yi);

 psif=@(p) p./((1-m*(p/Pc)+n*(p/Pc).^2+0.0003*(p/Pc).^3).*(1e-4*Ki*exp(Xi*(29*g*p./((1-m*(p/Pc)+n*(p/Pc).^2+0.0003*(p/Pc).^3)*10.732*TR*62.4)).^Yi)));
  psifun=2*int(psif,pb,p)-20000;
  p0=2000;
  dp=diff(psifun,p);
  ddp=diff(dp,p);
  dp0=double(subs(dp,p,p0));
  ddp0=double(subs(ddp,p,p0));
  psifun0=double(subs(psifun,p,p0));
  while(((psifun0*ddp0)/(psifun0^2))>1)
  p0=2000;
  psifun0=double(subs(psifun,p,p0));
  dp0=double(subs(dyp,p,p0));
  ddp0=double(subs(ddp,p,p0));
  end
  p1=p0-(psifun0/dp0);
  while (abs(p1-p0)>0.0001)
  p0=p1;
  psifun0=double(subs(psifun,p,p0));
  dp0=double(subs(dp,p,p0));
  p1=p0-(psifun0/dp0);
  end

[Isaac]

Please, I would kindly need assistance to solve for p...with the current code, the derivative of psifun (a nonlinear integral equation) is zero, which ought not. Thanks

[Isaac]

Alternatively, I used MATLAB solver; still getting no solution:

psifun=2*int(psif,p); psoln=vpasolve(psifun==2000,p)

[Walter Roberson]

my calculation is that psoln is between 15.5618609866312 and 15.5618609866313

Answer 1

[Walter Roberson]

int() is for symbolic integration, and should be applied to a symbolic expression. You are applying it to anonymous function instead.

[Isaac]

Using a simpler expression like exp(x), it works. I don't think the prob is the use of handle function but the complexity of the integrand. MATLAB can't solve the integration symbolically