Problem with fsolve for non linear system

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Julio Caye
Julio Caye el 15 de Jul. de 2014
Comentada: Julio Caye el 15 de Jul. de 2014
Hi there, I'm trying to solve this system with fsolver:
F1 = P(1) - P(3) - (- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)))^2 * sign(- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)));
F2 = P(1) - P(2) - (dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)))^2 * sign(dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)));
F3 = P(1) + ((V1(i+1)*(P(1) - P1(i,1)) + V2(i+1)*(P(2) - P2(i,1)) + V3(i+1)*(P(3) - P3(i,1)))/Dtheta(i+1) + P(2)*dV2(i+1) + P(3)*dV3(i+1))/dV1(i+1);
F = [F1 ; F2 ; F3];
P0 = [P1(i,1); P2(i,1); P3(i,1)];
options = optimoptions('fsolve','Display','iter');
[P,fval] = fsolve(F,P0,options)
And it returns:
Error using lsqfcnchk (line 108)
FUN must be a function, a valid string expression, or an inline function object.
Error in fsolve (line 197)
funfcn = lsqfcnchk(FUN,'fsolve',length(varargin),funValCheck,gradflag);
Error in wankinc (line 81)
[P,fval] = fsolve(F,P0,options)
Could anyone tell me what the problem is? I don't see why the values of F1, F2 and F3 are not valid.
Thank you!
PS: all the variables within the functions are calculated in the same code, previously. The complete code is copied below, if anyone finds it necessary:
e = 3/1000; %geometry
R = 21/1000;
W = 5/1000;
G=3*sqrt(3)*2000*W*R^2*(3*R^2/8 + e^2)/4; %Rotational inertia
P1(1,1) = 1.01325e5 - 1e3; %starting fluid pressures, from equilibrium
P2(1,1) = 1.01325e5 + 1e3;
P3(1,1) = 1.01325e5;
P= [P1(1,1) P2(1,1) P3(1,1)];
dP1(1) = 0; %pressure change over one timestep
dP2(1) = 0;
dP3(1) = 0;
X(1) = 0; %pressure gradient between chambers. Properly defined ahead.
Y(1) = 0;
Z(1) = 0;
T_app(1) = 1; %T_app(1) is the torque to be ultimately imposed on the system
T_app(2) = T_app(1)/20; %T_app(2) is a variable torque that increases from 0 up to the imposed torque
T_res(1) = 0; %Torque caused by the pressure in the chambers
T = zeros(1010);
theta(1) = 0; %Displacement over the entire simultaion
Dtheta(1) = 0; %Displacement over one time step
vel(1) = 0; %Angular velocity
alpha(1) = 0; %Angular acceleration
A = pi*(e^2 + R^2/3) - sqrt(3)*R^2/4; %A, B, C and D are constants defined here for computational speed
B = 3*sqrt(3)*e*R/4;
C = 3*sqrt(3)*e*R/2;
D = 0.6 * pi*0.1^2/4 * sqrt(2/1000); %This is the flow constant for the orifices
E = 1e3; %Error of 1kPa is accepted before continuing to the next step
f = zeros(1500,4);
f(:,:) = 2*E;
V1(1) = A - B * (sqrt(3)*sin(2*(theta(1) - 2*pi/3) + cos(2*(theta(1) - 2*pi/3)))); %Chamber volume, a function of theta
dV1(1) = C*(sin(2*(theta(1) - 2*pi/3)) - sqrt(3)*cos(2*(theta(1) - 2*pi/3))); %dV/dtheta
V2(1) = A - B * (sqrt(3)*sin(2*theta(1)) + cos(2*theta(1)));
dV2(1) = C*(sin(2*theta(1)) - sqrt(3)*cos(2*theta(1)));
V3(1) = A - B * (sqrt(3)*sin(2*(theta(1) + 2*pi/3) + cos(2*(theta(1) + 2*pi/3))));
dV3(1) = C*(sin(2*(theta(1) + 2*pi/3)) - sqrt(3)*cos(2*(theta(1) + 2*pi/3)));
ts = 1e-3; %Time step of 0.1 ms
alpha(1) = T_app(2)/G;
for i = 1:1:1
if T_app(2)<T_app(1) %Ramps up the torque in each iteration
T_app(2) = T_app(2) + T_app(1)/20;
end
vel(i+1) = vel(i) + alpha(i)*ts;
theta(i+1) = theta(i) + vel(i)*ts + alpha(i)*ts^2/2;
Dtheta(i+1) = theta(i+1) - theta(i);
V1(i+1) = A - B * (sqrt(3)*sin(2*(theta(i+1)) - 2*pi/3) + cos(2*(theta(i+1) - 2*pi/3)));
V2(i+1) = A - B * (sqrt(3)*sin(2*theta(i+1)) + cos(2*theta(i+1)));
V3(i+1) = A - B * (sqrt(3)*sin(2*(theta(i+1)) + 2*pi/3) + cos(2*(theta(i+1) + 2*pi/3)));
dV1(i+1) = C*(sin(2*(theta(i+1) - 2*pi/3)) - sqrt(3)*cos(2*(theta(i+1) - 2*pi/3)));
dV2(i+1) = C*(sin(2*theta(i+1)) - sqrt(3)*cos(2*theta(i+1)));
dV3(i+1) = C*(sin(2*(theta(i+1) + 2*pi/3)) - sqrt(3)*cos(2*(theta(i+1) + 2*pi/3)));
F1 = P(1) - P(3) - (- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)))^2 * sign(- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)));
F2 = P(1) - P(2) - (dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)))^2 * sign(dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)));
F3 = P(1) + ((V1(i+1)*(P(1) - P1(i,1)) + V2(i+1)*(P(2) - P2(i,1)) + V3(i+1)*(P(3) - P3(i,1)))/Dtheta(i+1) + P(2)*dV2(i+1) + P(3)*dV3(i+1))/dV1(i+1);
F = [F1 ; F2 ; F3];
P0 = [P1(i,1); P2(i,1); P3(i,1)];
options = optimoptions('fsolve','Display','iter');
[P,fval] = fsolve(F,P0,options)
P1(i+1,1) = P(1);
P2(i+1,1) = P(2);
P3(i+1,1) = P(3);
T_res(i+1) = sqrt(3)*R*W*e * (-P1(i+1,1)*cos(2*theta(i+1) - pi/6) + P2(i+1,1)*cos(2*theta(i+1) + pi/6) + P3(i+1,1)*sin(2*theta(i+1)));
T(i+1) = T_app(2) - T_res(i+1);
alpha(i+1) = T(i+1)/G;
end

Respuesta aceptada

Sara
Sara el 15 de Jul. de 2014
Do this:
- in your main function, after calculating P1,P2,P3, put:
P0 = [P1(i,1); P2(i,1); P3(i,1)];
options = optimoptions('fsolve','Display','iter');
[P,fval] = fsolve(@myfun,P0,options)
- create a second function
function F = myfun(P)
F1 = P(1) - P(3) - (- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)))^2 * sign(- dV1(i+1)*vel(i+1)/D - sqrt(abs(P(1) - P(2)))*sign(P(1) - P(2)));
F2 = P(1) - P(2) - (dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)))^2 * sign(dV2(i+1)*vel(i+1)/D + sqrt(abs(P(2) - P(3)))*sign(P(2) - P(3)));
F3 = P(1) + ((V1(i+1)*(P(1) - P1(i,1)) + V2(i+1)*(P(2) - P2(i,1)) + V3(i+1)*(P(3) - P3(i,1)))/Dtheta(i+1) + P(2)*dV2(i+1) + P(3)*dV3(i+1))/dV1(i+1);
F = [F1 ; F2 ; F3];
To pass the parameters, you can either use a global or do:
[P,fval] = fsolve(@(x)myfun(x,param1,param2),P0,options)
function F = myfun(P,param1,param2)
  6 comentarios
Julio Caye
Julio Caye el 15 de Jul. de 2014
Can I ask just one more thing, please? How do I make it not plot out the number of iterations and error and stuff? I have to run that system some 10 thousand steps and that information kinda overwhelms me haha
Julio Caye
Julio Caye el 15 de Jul. de 2014
nevermind, found the option to do it here (and btw, the eqations were right, just had to reduce the timestep a bit)
thanks again!

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