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

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!

Answer 1

[Star Strider]

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]);

.

[Jani]

Well in the first part I try to calculate chi_b, and chi_e, and after that I want to solve the differential equations, which contains these constants. I could solve it separetaly, but one of the constants in chi_b (K_be) will contain the chi_p parameter, which is the Y(1). So this is pretty much a mess for me, I kind of a beginner in Matlab, but need to modell a fluid bed dryer.

I wrote into the code, what you suggested, but still doesn't wanna work :(

[Star Strider]

I calculated ‘chi_b’ and ‘chi_e’ on my computer as:

chi_b=double(sol.chi_b)

chi_e=double(sol.chi_e)

and pasted the results here. I then commented-out all the symbollic code in ‘diffeq_fv’ (since it re-calculates those same values that are now passed as extra parameters), and ran it to get this result:

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_e;
Fv(2,1)=-4*Y(1)+3*Y(2);
end

I am not certain what result you want, however the code now runs without error.

.

[Jani]

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]

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.

.