# How to solve a system of coupled first order differential equations using ode45?

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Md. Golam Zakaria on 1 Feb 2023
Commented: dpb on 25 Oct 2023
Hello everyone,
I have a system of four coupled first oder differential equation, which needs to be solved. While the equations are long, its pretty straightforward. I have written the following code and getting some error.
function dydt=ccp_ode(t,y)
A_E=.00707;
A_G=0.1650;
e=1.60217663e-19;
n=3.15e16;
l_B=0.1;
epsilon_0=8.85418782e-12;
v_e = 594.0413;
m_e=9.10938e-31;
C_B=2e-9;
m_i=40;
T_e=1.78;
u_B=9.8e3*sqrt(T_e/m_i);
T_ar=300;
p=1;
k_B=1.380649e-23;
n_g=(p/(k_B*T_ar));
I_iP=e*n*u_B*A_E;
I_iG=e*n*u_B*A_G;
L_pl=(l_B*m_e)/(e^2*n*A_E);
K_iz= 2.34e-14*T_e^0.59*exp(-17.44/T_e);
K_ex=2.48e-14*T_e^0.33*exp(-12.78/T_e);
K_el=2.336e-14*T_e^1.609*exp(0.0618*(log(T_e))^2-0.1171*(log(T_e))^3);
Km=K_iz+K_ex+K_el;
v_m=Km*n_g;
v_eff=v_m+(v_e/l_B);
A=2*e*epsilon_0*n*A_E^2;
B=2*e*epsilon_0*n*A_G^2;
C=e*n*v_e*A_E;
D=e*n*v_e*A_G;
f=13.56e6;
omega_1=2*pi*f;
V0=300;
dydt(1)=((-(sqrt(A/y(1)))^-1)*(y(3)+I_iP-(C*exp((-e*y(1))/T_e))));
dydt(2)=((-(sqrt(B/y(2)))^-1)*(-y(3)+I_iG-(D*exp((-e*y(2))/T_e))));
dydt(3)=((L_pl^-1)*((V0*cos(omega_1*t))+y(1)-y(2)+y(4))-(v_eff^y(3)));
dydt(4)=((-1/C_B)*y(3));
end
To call this function I used [t,y]=ode45('ccp_ode', [0 .1], [0 0 2 0]). However I am getting the following error:
Error using odearguments (line 93)
CCP_ODE must return a column vector.
Can anyone please , point me the problem with my code and what is the solution.
Thank You.
MOSLI KARIM on 25 Oct 2023
function d
[t,y]=ode45(@ccp_ode, [0 .1], [0 0 2 0])
plot(t,y)
function DYDT=ccp_ode(t,y)
A_E=.00707;
A_G=0.1650;
e=1.60217663e-19;
n=3.15e16;
l_B=0.1;
epsilon_0=8.85418782e-12;
v_e = 594.0413;
m_e=9.10938e-31;
C_B=2e-9;
m_i=40;
T_e=1.78;
u_B=9.8e3*sqrt(T_e/m_i);
T_ar=300;
p=1;
k_B=1.380649e-23;
n_g=(p/(k_B*T_ar));
I_iP=e*n*u_B*A_E;
I_iG=e*n*u_B*A_G;
L_pl=(l_B*m_e)/(e^2*n*A_E);
K_iz= 2.34e-14*T_e^0.59*exp(-17.44/T_e);
K_ex=2.48e-14*T_e^0.33*exp(-12.78/T_e);
K_el=2.336e-14*T_e^1.609*exp(0.0618*(log(T_e))^2-0.1171*(log(T_e))^3);
Km=K_iz+K_ex+K_el;
v_m=Km*n_g;
v_eff=v_m+(v_e/l_B);
A=2*e*epsilon_0*n*A_E^2;
B=2*e*epsilon_0*n*A_G^2;
C=e*n*v_e*A_E;
D=e*n*v_e*A_G;
f=13.56e6;
omega_1=2*pi*f;
V0=300;
dydt(1)=((-(sqrt(A/y(1)))^-1)*(y(3)+I_iP-(C*exp((-e*y(1))/T_e))));
dydt(2)=((-(sqrt(B/y(2)))^-1)*(-y(3)+I_iG-(D*exp((-e*y(2))/T_e))));
dydt(3)=((L_pl^-1)*((V0*cos(omega_1*t))+y(1)-y(2)+y(4))-(v_eff^y(3)));
dydt(4)=((-1/C_B)*y(3));
DYDT=dydt'
end
end

dpb on 1 Feb 2023
...
dydt(1)=((-(sqrt(A/y(1)))^-1)*(y(3)+I_iP-(C*exp((-e*y(1))/T_e))));
dydt(2)=((-(sqrt(B/y(2)))^-1)*(-y(3)+I_iG-(D*exp((-e*y(2))/T_e))));
dydt(3)=((L_pl^-1)*((V0*cos(omega_1*t))+y(1)-y(2)+y(4))-(v_eff^y(3)));
dydt(4)=((-1/C_B)*y(3));
The error message tells you what the problem is -- the above will return a row vector because you didn't preallocate the variable nor expressly define the terms as column vector members nor reshape() the result to ensure the function returns a column vector.
You could have found this easily by just using the debugger to inpsect the result in your function before it returns or looking at what it returned if called from the command line standalone.
To solve it is simple; several ways...
1. Preallocate the vector before populating it...
dydt=zeros(4,1);
dydt(1)= ....
...
2. Explicitly store into first column...
dydt(1,1)=...
dydt(2,1)=...
...
3. Ensure result is column vector...
...
dydt(1)=...
dydt(2)=...
...
dydt=dydt(:);
return
end
Of these, the cleanest is the first as it avoids the dynamic reallocation that occurs with any of the others without having done so. For only four elements the performance difference will be negligible except the function will be called every solution pass and multiple times for evaluation for every time step so performance here may well be significant. Hence, get into the habit of preallocating arrays where can.
##### 3 CommentsShow 1 older commentHide 1 older comment
MOSLI KARIM on 25 Oct 2023
Edited: dpb on 25 Oct 2023
my proposed solution
function d
[t,y]=ode45(@ccp_ode, [0 .1], [0 0 2 0])
plot(t,y)
function DYDT=ccp_ode(t,y)
A_E=.00707;
A_G=0.1650;
e=1.60217663e-19;
n=3.15e16;
l_B=0.1;
epsilon_0=8.85418782e-12;
v_e = 594.0413;
m_e=9.10938e-31;
C_B=2e-9;
m_i=40;
T_e=1.78;
u_B=9.8e3*sqrt(T_e/m_i);
T_ar=300;
p=1;
k_B=1.380649e-23;
n_g=(p/(k_B*T_ar));
I_iP=e*n*u_B*A_E;
I_iG=e*n*u_B*A_G;
L_pl=(l_B*m_e)/(e^2*n*A_E);
K_iz= 2.34e-14*T_e^0.59*exp(-17.44/T_e);
K_ex=2.48e-14*T_e^0.33*exp(-12.78/T_e);
K_el=2.336e-14*T_e^1.609*exp(0.0618*(log(T_e))^2-0.1171*(log(T_e))^3);
Km=K_iz+K_ex+K_el;
v_m=Km*n_g;
v_eff=v_m+(v_e/l_B);
A=2*e*epsilon_0*n*A_E^2;
B=2*e*epsilon_0*n*A_G^2;
C=e*n*v_e*A_E;
D=e*n*v_e*A_G;
f=13.56e6;
omega_1=2*pi*f;
V0=300;
dydt(1)=((-(sqrt(A/y(1)))^-1)*(y(3)+I_iP-(C*exp((-e*y(1))/T_e))));
dydt(2)=((-(sqrt(B/y(2)))^-1)*(-y(3)+I_iG-(D*exp((-e*y(2))/T_e))));
dydt(3)=((L_pl^-1)*((V0*cos(omega_1*t))+y(1)-y(2)+y(4))-(v_eff^y(3)));
dydt(4)=((-1/C_B)*y(3));
DYDT=dydt'
end
end
dpb on 25 Oct 2023
function DYDT=ccp_ode(t,y)
...
dydt(1)=((-(sqrt(A/y(1)))^-1)*(y(3)+I_iP-(C*exp((-e*y(1))/T_e))));
dydt(2)=((-(sqrt(B/y(2)))^-1)*(-y(3)+I_iG-(D*exp((-e*y(2))/T_e))));
dydt(3)=((L_pl^-1)*((V0*cos(omega_1*t))+y(1)-y(2)+y(4))-(v_eff^y(3)));
dydt(4)=((-1/C_B)*y(3));
DYDT=dydt';
end
NOTA BENE: -- The above doesn't preallocate the array and also creates a new temporary variable just to reshape the existing vector. The ' operator also is conjugate transpose; while it should not generate complex values here, it is the .' operator for nonconjugate transpose that is wanted here; it would change the meaning of DYDT from that of dydt if were complex.

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