How i can maximizing parameter in ODE

I solve one ODE dx/dt=Na0/((-rA*V)) using code:

function dtimedxa = ode_sys (xa,time)
    Nao=5;
    k=0.023;
    Vo=10;
    epsilon=1;
    Cao=Nao/Vo;
    Ca=Cao*(1-xa)/(1+epsilon*xa);
    p=time*Ca;   
    ra=-k*Ca^2;
    V=Vo*(1+epsilon*xa);
    dtimedxa=Nao/((-ra)*V);
end

and script:

range=[0,0.75];
ICs=[1];

[xa,time]=ode23(@ode_sys,range,ICs);

figure
plot (xa,time(:,1))
title('dependence')
xlabel('conversion')
ylabel ('time')

Is it posible to solve equation above dt/dx=Na0/(-rA*V) maximizing t or p?

Thanks in advance.

[Walter Roberson]

What is t in this situation? I see p but not t ? Other than in the sense of dx/dt in which case the maximum t would be range(end) which is not something to optimize.

You have dx/dt . The independent variable, t, is always passed as the first parameter in an ode*() call, and the boundary conditions are passed as the second parameter. But your ode_sys function has xa then time, so inside your ode_sys, xa is the independent variable, t, and time is the dependent variable x . This is very confusing for readers. I think you have made a mistake.

[Bosnian Kingdom]

Sorry, you are right.

xa is conversion, and t si time, and i need to find xa (dependence variable) in dependence of t (time_independence variable).

dxa/dt. How i can maximize xa or p. Is it posible?.

[darova]

It depends on your initial conditions?

[Bosnian King]

Hello Darova,

what do you mean?

[darova]

What do you mean by 'maximize'?

After ode45 solve you have xa(t) (curve). You want all values of this curve to be bigger?

[Bosnian King]

For example, i want to maximize p (p=time*Ca). Is it posible..

[Walter Roberson]

What parameter or constant would be altered in order to optimize?

Are you just looking for the time at which p is maximum ?

Taking into account the earlier discussion, could you clarify whether you really do intend

function dtimedxa = ode_sys (xa,time)

or if it instead should be

function dtimedxa = ode_sys (time,xa)

[Bosnian King]

The function is:

function dtimedxa = ode_sys (xa,time)

I want to find, for example value vor Nao is 5, can i find value for Nao (or k, or Vo) for which the xa will be maximal?

[Walter Roberson]

Calling the independent variable xa and the dependent variable time when you are trying to work with dx/dt is too confusing for me. I am not going to be able to assist with this.

[Bosnian King]

Sorry! Sorry! You are right! I want to find, for example valuer for Nao (or k, or Vo) for which the t (time) will be maximal? xa is independent variable and i know range of xa (0-0.75).

Forget my previous answe

[Walter Roberson]

xa being an independent variable does not work with the system dx/dt=Na0/((-rA*V)) : the independent variable for dx/dt=Na0/((-rA*V)) is t.

[Bosnian King]

xa is actualy x. when you start program you'll get t (time) in function of x ( xa). So my system is dt/dx=Na0/((*rA*V)). Did you try to start code from the begining? Is it posible to optimize values for Nao (or k, or Vo) for which the t (time) will be maximal?

[Torsten]

Sure. Use "fminsearch".