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How can a differential equation be solved, if its constants are in a system of equation relation?

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So the whole equations are here, but for the sake of simplicity I reduced many of the constants, so I can focus only on the relevant things.
So as far as I can see, chi_b and chi_e are two unkwons in a system of equation. In the equation chi_b(z) there is chi_e and in the equation chi_e there is the avarage of chi_b(z). I can solve this with this script (simplified version):
syms chi_b chi_e
K_be=7
delta_b=4
nu_b=33
eqn1=2*chi_b+3*chi_e+exp(K_be * delta_b / nu_b)==chi_b;
eqn2=3*chi_b+2*chi_e==chi_e;
sol=solve([eqn1,eqn2],[chi_b,chi_e]);
chi_b=sol.chi_b
chi_e=sol.chi_e
And I can solve differential equation with this script: Y(1)=chi_p, Y(2)=T_p
[t,Y]=ode23('diffeq_fv',[0 5],[0;1]);
plot(t,Y(:,1),'+',t,Y(:,2),'o')
function Fv=diffeq_fv(t,Y);
Fv(1,1)=3*Y(1);
Fv(2,1)=-4*Y(1)+3*Y(2);
But the d/dt chi_p diff.eq. inculdes chi_e, and when I write chi_e into the diff equations it doesn't work.
[t,Y]=ode23('diffeq_fv',[0 5],[0;1]);
plot(t,Y(:,1),'+',t,Y(:,2),'o')
function Fv=diffeq_fv(t,Y);
syms chi_b chi_e
K_be=7
delta_b=4
nu_b=33
eqn1=2*chi_b+3*chi_e+exp(K_be * delta_b / nu_b)==chi_b;
eqn2=3*chi_b+2*chi_e==chi_e;
sol=solve([eqn1,eqn2],[chi_b,chi_e]);
chi_b=sol.chi_b
chi_e=sol.chi_e
Fv(1,1)=3*Y(1)*chi_e;
Fv(2,1)=-4*Y(1)+3*Y(2);
and gives me this error message: "Inputs must be floats, namely single or double."
So my quesion is how can a diff equation which includes a system of equation be solved?
Many thanks in advance!

Respuestas (1)

Star Strider
Star Strider el 24 de Ag. de 2021
It is difficult to follow the posted code, so I am not certain what you are doing.
If you want to pass those variables to ‘diffeq_fv’ it will be necessary to pas them as extra parameters:
function Fv=diffeq_fv(t,Y,chi_b,chi_e);
K_be=7
delta_b=4
nu_b=33
eqn1=2*chi_b+3*chi_e+exp(K_be * delta_b / nu_b)==13;
eqn2=3*chi_b+2*chi_e==12;
sol=solve([eqn1,eqn2],[chi_b,chi_e]);
chi_b=sol.chi_b
chi_e=sol.chi_e
Fv(1,1)=3*Y(1)*chi_e;
Fv(2,1)=-4*Y(1)+3*Y(2);
end
See Passing Extra Parameters for details.
The ode23 call then becomes:
[t,Y]=ode23(@(t,Y) diffeq_fv(t,Y,chi_b,chi_e),[0 5],[0;1]);
.
  4 comentarios
Jani
Jani el 24 de Ag. de 2021
Thanks for your answer!
Yes I thought of that, solving the chi_b-chi_e system first, and then put them into the diffeq.
However now I see that one of the constant in chi_b, contains the Y(1) (chi_p)
so i dont know if it can be solved at all, but for example here is a simple system of equation:
eqn1=2*chi_b+3*chi_e+3+Y(1)==chi_b;
eqn2=-3*chi_b+2*chi_e+6==chi_e;
and here is the diffeq system:
Fv(1,1)=3*Y(1)*chi_b;
Fv(2,1)=-4*Y(1)+3*Y(2)
Is there a solution for this problem?
Star Strider
Star Strider el 24 de Ag. de 2021
It would seem that it would only be necessary to switch the constants in the differential equation:
chi_b = 0.292013077080326;
chi_e = -0.876039231240978;
[t,Y]=ode23(@(t,Y) diffeq_fv(t,Y,chi_b,chi_e),[0 5],[0;1]);
figure
plot(t,Y)
grid
function Fv=diffeq_fv(t,Y,chi_b,chi_e);
% K_be=7
% delta_b=4
% nu_b=33
% eqn1=2*chi_b+3*chi_e+exp(K_be * delta_b / nu_b)==13;
% eqn2=3*chi_b+2*chi_e==12;
% sol=solve([eqn1,eqn2],[chi_b,chi_e]);
% chi_b=sol.chi_b
% chi_e=sol.chi_e
Fv(1,1)=3*Y(1)*chi_b;
Fv(2,1)=-4*Y(1)+3*Y(2);
end
Experiment to get different results.
.

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