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How to solve a set of 4 first-order non-linear coupled ODEs?

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Shashank
Shashank on 3 Jul 2013
I am trying to solve a 4th order differential equation using shooting method by disintegrating the ODE into four coupled first order ODEs. I only have initial conditions. So I am trying to perturb y(1) and y(4) and use them as a parameter to achieve a target value for y(3). But all I get is NaNs on my output matrices for y(1), y(2), y(3) and y(4). Any inputs will be appreciated.
Here is my code:
%%Constants
A = 1e-20; % Hamaker Constant
R_FC = 24.597; % Gas Constant of FC-72 (J/kg-K)
c = 1; % Accomodation Coefficient
sigma = 0.01; % Surface Tension Coefficient (N/m)
rho_l = 1593.84; % Liquid Density (kg/m3)
k_l = 0.0544; % Liquid Conductivity (W/m-K)
mu_l = 0.0004683; % Liquid Viscosity (kg/m2-s)
rho_v = 11.61; % Vapor Density (kg/m3)
h_lv = 88000; % Enthalpy of Vaporization (J/kg)
Twall = 334; % Wall Temperature (K), 10 degrees superheat
Tsat = 329; % Saturation Temperature of FC-72 at 1 atm (K)
R_int = ((2 - c)/(2*c))*(Tsat^(1.5))*((2*pi*R_FC)^(0.5))/(rho_v*(h_lv^2));
delta_ad = (A/(rho_l*h_lv*((Twall/Tsat) - 1)))^(1/3);
e1 = delta_ad/1000;
deltaP_ad = A/((delta_ad)^3);
deltaP_ad_corr = A/((delta_ad+e1)^3);
eQ = 1;
%%Function
tpcl = @(xi,y) [y(2); (1/sigma)*((1+(y(2)*y(2)))^(1.5))*(y(3)-(A/(y(1)^3))); (3*mu_l/(rho_l*h_lv))*(-y(4)/(y(1)^3)); (Twall-Tsat-(Tsat*y(3)/(rho_l*h_lv)))/((y(1)/k_l)+R_int)];
xi_end = 5e-7;
N = 1000;
[xi,y1] = rk4(tpcl,[0 xi_end],[delta_ad 0 deltaP_ad 0],N);
[xi,y2] = rk4(tpcl,[0 xi_end],[delta_ad+e1 0 deltaP_ad_corr eQ],N);
delta_a = delta_ad;
delta_b = delta_ad + e1;
Q_a = 0;
Q_b = eQ;
slope_a = y1(2,end);
slope_b = y2(2,end);
%%Shooting Method
tol = 1e-5;
iter = 10;
slope_target = tan(pi/18);
for i = 1:iter
Q_new = Q_a + (Q_b-Q_a)/(slope_b-slope_a)*(slope_target-slope_a);
delta_new = delta_a + (delta_b-delta_a)/(slope_b-slope_a)*(slope_target-slope_a);
deltaP_new = A/((delta_new)^3);
[xi,y] = rk4(tpcl,[0 xi_end],[delta_new 0 deltaP_new 0],N);
fprintf('iter:%2d, delta(0)=%17.15f, slope(xi_end)=%17.15f\n',i,delta_new,y(2,end));
if (abs(y(2,end)-slope_target) <= tol)
break;
end
Q_a = Q_b;
Q_b = Q_new;
delta_a = delta_b;
delta_b = delta_new;
slope_a = slope_b;
slope_b = y(2,end);
end

Accepted Answer

Jan
Jan on 4 Jul 2013
Edited: Jan on 4 Jul 2013
Anonymous functions are hard to debug. Better define the ODE as a function and use http://www.mathworks.de/matlabcentral/answers/1971 to define the constants on demand.
Then you can use the debugger to step through the program line by line until you find the source of the NaN's by your own. This is more efficient to let all forum users debug the code for you, especially when the function rk4() is not known.
  5 Comments
Jan
Jan on 7 Jul 2013
This could have different reasons: A bug in the code, a division by zero or another mathematical effect. It is your turn to find the exact cause for the creation of the NaNs. E.g. this might be useful:
dbstop if naninf

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More Answers (1)

Gentian Zavalani
Gentian Zavalani on 7 Jul 2013
When you construct an anonymous function, the part directly after the @ must be pure variable names and not expressions or indexed variables. Your code is tpcl = @(xi,y) [y(2); (1/sigma)*((1+(y(2)*y(2)))^(1.5))*(y(3)-(A/(y(1)^3))); (3*mu_l/(rho_l*h_lv))*(-y(4)/(y(1)^3)); (Twall-Tsat-(Tsat*y(3)/(rho_l*h_lv)))/((y(1)/k_l)+R_int)];

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