How to define transfer variables when solving equations in 1stOpt?

Question

Yuting el 15 de Ag. de 2023

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https://es.mathworks.com/matlabcentral/answers/2008892-how-to-define-transfer-variables-when-solving-equations-in-1stopt

Comentada: Torsten el 21 de Ag. de 2023

I am trying to solve a non-linear integration equations in 1stOpt. The parameters to be solved are x1, y1, Uratio, and 5 transfer variables are used to build the equations: Left, Right, C, pl, pr.

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Problems

1 The code doesn't run at all

2 I don't know how to difine the transfer variables

3 I'm not sure the use of int() is correct

Original code

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Constant  alph=pi*15/180, L=0.116, Tice=18, addweight=0.212, T0=15, dL=-0.02, tilt=0.1, heightofweight=12.5, forcefromcable=0, muL=0.001, rhoS=920*0.95, rhoL=1000, Cps=2049.41, hm=334000+Cp_s*T_ice, KL=0.57, Pe=102770, g1=-addweight*9.8, g2=-0.38*9.8, heightofmasscenter=0.07, heightofcable=0.13, d = pi*L*sin(alph)/2, M = g1*(dL*cos(tilt)+heightofweight*sin(tilt))-g2*heightofmasscenter*sin(tilt)-forcefromcable*heightofcable*sin(tilt), G = g1+g2;
Parameter  x1[0,], y1[,0], Uratio;
//transfer variables
Left = x1^2+y1^2-2*x*(x1*sin(alph)-y1*cos(alph)+x^2);
Right = x1^2+y1^2-2*x*(x1*sin(alph)+y1*cos(alph)+x^2);
C = (int(x*Right^2,x=0:L)-int(x*Left^2,x=0:-L))/(int(Left^1.5,x=0:-L)-int(Right^1.5,x=0:L));
pl = 12*muL*rhoS^4*hm^3*Uratio^4*(int(x*Left^2,x=y:-L)+C*int(Left^1.5,x=y:-L))/(rhoL*T0^3*KL^3)+Pe;
pr = 12*muL*rhoS^4*hm^3*Uratio^4*(int(x*Right^2,x=y:L)+C*int(Right^1.5,x=y:L))/(rhoL*T0^3*KL^3)+Pe;
Function  int(pl*y,y=-L:0)+int(pr*y,y=0:L)+M/d=0;
          int(pl,y=-L:0)*sin(alph+tilt)+int(pr,y=0:L)*sin(alph-tilt)+G/d-2*L*Pe*sin(alph)*cos(tilt)=0;
          int(pl,y=-L:0)*cos(alph+tilt)-int(pr,y=0:L)*cos(alph-tilt)+2*L*Pe*sin(alph)*sin(tilt)=0;