Summing results of ode45

Hello, using ode45 I solved 3 differential equations. I now want to create a function x that adds all the 3 components of the diff equations. Could you help me set it up?

Fx=@(t,xqxc)[ (-xqxc(1)+G_q .*(q-q_b)./q_b)./tau_q ;               
                                (-xqxc(2) +0.3+3.*tanh(Pa_co2./Pa_co2_b -1.1))./tau_co2 ]

[t, xqxc]=ode45(F,[0 100], [0 0 0]);
xq= xqxc(:,1);
xc= xqxc(:,2);

syms xm(t)
[V] = odeToVectorField(diff(xm, 2) == (-tau2*diff(xm) -xm)/tau1^2)
M = matlabFunction(V,'vars', {'t','Y'})
sol = ode45(M,[0 20],[2 0]);

x= sol+ xc -xq;

the error I get is Undefined operator '+' for input arguments of type 'struct'.

Error in CBF_v2 (line 56) x= sol+ xc -xq;

[Walter Roberson]

gorilla3: our policy is not to delete questions that have a meaningful response unless the questions were abusing the system.

You can edit your question to clarify it.

[gorilla3]

I asked this in multiple ways but I really don't know how to solve the problem. I need a system of two 1st order diff eq and one 2nd order diff eq. I tried numerous approaches, please help me. This is how far I got:

Fx=@(t,xqxc)[ (-xqxc(1)+G_q .*(q-q_b)./q_b)./tau_q ;               
                                (-xqxc(2) +0.3+3.*tanh(Pa_co2./Pa_co2_b -1.1))./tau_co2;
      %2nd order diff:    xm'' = (-tau2* xm' -xm)/tau1^2 
                         %xm' ??

                          %xm'' 
                              (-tau2.*xqxcxmxm1(4) - xqxcxmxm1(3))./tau1.^2]

[t, xqxc]=ode45(F,[0 100], [0 0 0]);
xq= xqxc(:,1);
xc= xqxc(:,2);
xm= xqxc(:,3);
xm1= xqxc(:,4);

x=xm+ xc -xq;

[Torsten]

Fx=@(t,xqxc)[ (-xqxc(1)+G_q .*(q-q_b)./q_b)./tau_q ; (-xqxc(2) +0.3+3.*tanh(Pa_co2./Pa_co2_b -1.1))./tau_co2; xqxc(4) ; (-tau2*xqxc(4)-xqxc(3))/tau1^2];
[t, xqxc]=ode45(Fx,[0 100], [0 0 0 0]);
xq= xqxc(:,1);
xc= xqxc(:,2);
xm= xqxc(:,3);
xm1= xqxc(:,4);

x=xm+ xc -xq;

Best wishes

Torsten.

[gorilla3]

Thank you for your reply but I still get the error:

Error using odearguments (line 92)
@(T,XQXC)[(-XQXC(1)+G_Q.*(Q-Q_B)./Q_B)./TAU_Q;(-XQXC(2)+0.3+3.*TANH(PA_CO2./PA_CO2_B-1.1))./TAU_CO2;XQXC(4);(-TAU2*XQXC(4)-XQXC(3))/TAU1^2]
returns a vector of length 288, but the length of initial conditions vector is 4. The vector returned
by
@(T,XQXC)[(-XQXC(1)+G_Q.*(Q-Q_B)./Q_B)./TAU_Q;(-XQXC(2)+0.3+3.*TANH(PA_CO2./PA_CO2_B-1.1))./TAU_CO2;XQXC(4);(-TAU2*XQXC(4)-XQXC(3))/TAU1^2]
and the initial conditions vector must have the same number of elements.

Error in ode45 (line 115)
    odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in CBF_v4 (line 48)
[t, xqxc]=ode45(Fx,[0 100], [0 0 0 0]);

[Torsten]

Error message should be clear. We can't tell why one of your inputs (G_Q, Q, Q_B, ...) is a vector and not a scalar and why you generate 288 and not only 4 equations in the function Fx. Does the vector contain a profile of the variable over time ?

[gorilla3]

I believe it's because before that part of the code I have another system of diff eq, which I use to calculate a variable q:

F = @(t, V_saP1P2) [ Ca.*(V_saP1P2(2)-P_ic);
                            ((P_a_b-V_saP1P2(2))./(R_la + 0.5 .*R_sa_b) - (V_saP1P2(2)-V_saP1P2(3))./(0.5 .*R_sa_b+R_sv))./Ca;
                                     ((V_saP1P2(2)-V_saP1P2(3))./(0.5 .*R_sa +R_sv)-(V_saP1P2(3)-P_v)./R_lv)./ (1./(k_ven.*(V_saP1P2(3)-P_ic-P_v1))) ];
 [t, V_saP1P2] = ode45(F, [0 100], [0 0 0]);
 V_sa = V_saP1P2(:,1);
 P1 = V_saP1P2(:,2);
 P2 = V_saP1P2(:,3);

q= (P1-P2)/0.5 *R_sa + R_sv ;

I had to separate the ode45 in 2 systems instead of one because I needed the variable q for the second system. I believe it is this that generates the error but I don't know how else to organize the code.

[Torsten]

Solve all equations simultaneously (7 as far as I can see).

This will resolve all your problems.

Best wishes

Torsten.

[gorilla3]

Oh, I made such a silly mistake! Thank you, Torsten!!

Answer 1

[Jan]

sol = ode45(...) replies a struct, as the error message tells clearly. Either use:

x = sol.y + xc - xq;

Or use two outputs

[t2, x2] = ode45(...)

You obtain xc from t=0 to 100, but the 2nd integration goes from 0 to 20. Are you sure that you can add them directly? Is this meaningful?

You provide an initial with 3 elements in:

[t, xqxc] = ode45(F,[0 100], [0 0 0]);

But F replies a [2 x 1] vector only.

[gorilla3]

Hi,

thank you for your reply. The initial issue was that I separated xm from xc xq because xm is a 2nd order differential and I wasn't sure on how to input it in the rest of the system. Could you help with that?

And subsequently I'd like to perform x=xm+xc-xq

[Torsten]

Print M to a file and you'll see how to set up the 2nd order ODE using ODE45.

Best wishes

Torsten.

[gorilla3]

I asked this in multiple ways but I really don't know how to solve the problem. I need a system of two 1st order diff eq and one 2nd order diff eq. I tried numerous approaches, please help me. This is how far I got:

Fx=@(t,xqxc)[ (-xqxc(1)+G_q .*(q-q_b)./q_b)./tau_q ;               
                                (-xqxc(2) +0.3+3.*tanh(Pa_co2./Pa_co2_b -1.1))./tau_co2;
      %2nd order diff:    xm'' = (-tau2* xm' -xm)/tau1^2 
                         %xm' ??
                          %xm'' 
                              (-tau2.*xqxcxmxm1(4) - xqxcxmxm1(3))./tau1.^2]
[t, xqxc]=ode45(F,[0 100], [0 0 0]);
xq= xqxc(:,1);
xc= xqxc(:,2);
xm= xqxc(:,3);
xm1= xqxc(:,4);
x=xm+ xc -xq;

[Jan]

You can convert the 2nd order system to a 1st order system easily. Then you can integrate both trajectories together. Or you define tspan as a vector, such the output of the integrator is calculated for the same time points. Then you can add the trajectories directly.

But currently you integrate one trajectory from 0 to 100, and the other from 0 to 20. Then it does not look logical to sum the trajectories.

Summary: I still do not understand exactly, what you are asking for.