Second order system with different time scales

lchaves75 · August 27, 2015, 2:26am

Hi, folks,

I’m trying to simulate this system below. It´s a system where each equation has a different time scale. “Is” is in function of “t”, and “Iz” is in function of “tau”, where tau=t*beta). And beta=5.

Clear[t];
a = 0.00001381594775713100;
c = 0.00007261477318255780;
j = 0.00004126704565565830;
d = 0.00006126704565565830;
b = 0.498812483;
r = 0.498812483;
Population = 40000;
mi = 0.4;
Na = Population*mi;
Ns = Population*(1 - mi);
beta = 5;
tau = beta*t;
ts = 400;
system = {
   Iz'[tau] == ((a*(Na - Iz[tau])*Iz[tau]) +
                               (j*(Ns - Is[tal])*Iz[tau]) -
                               ((b)*Iz[tau] )),
   Is'[t] == ((c*(Ns - Is[t])*Is[t]) +
                     (d*(Na - Iz[t])*Is[t]) -
                     ((r)*Is[t]))
   };
initialvalues = {Is[0] == 160, Iz[0] == 40};
solution = 
  NDSolve[Join[system, initialvalues], {Iz, Is }, {t, 0, ts}, {tau, 
    beta, (ts/beta)}];

The result of simulation is “5 t cannot be used as variable.”

Some tricks?

Thanks.

gianlucagorni · August 27, 2015, 9:53am

First I replace the function Iz[tau] with its equivalent Izt[t] such that Iz[tau]=Izt[tau/beta]

Clear[tau, beta, a, c, j, d, b, r, mi, Na, Ns, Population, ts];
system = {Iz'[
     tau] == ((a (Na - Iz[tau]) Iz[tau]) + (j (Ns - Is[tau]) Iz[
         tau]) - ((b)*Iz[tau])), 
   Is'[t] == ((c (Ns - Is[t]) Is[t]) + (d (Na - Iz[t]) Is[t]) - ((r)*
        Is[t]))};
newSystem = 
 system /. {Iz -> Function[\[Tau], Izt[\[Tau]/\[Beta]]]} /. 
  tau -> \[Beta]*t

What I get is not a classical differential system, because the unknown functions are evaluated at different times:

{Derivative[1][Izt][t]/\[Beta] == -b Izt[t] + 
   j (Ns - Is[t \[Beta]]) Izt[t] + a (Na - Izt[t]) Izt[t], 
 Derivative[1][Is][t] == -r Is[t] + c (Ns - Is[t]) Is[t] + 
   d Is[t] (Na - Izt[t/\[Beta]])}

samcarrettie · August 27, 2015, 11:16am

You have a mess of parenthesis. I think your system should look like:

Clear[t];
a = 0.00001381594775713100;
c = 0.00007261477318255780;
j = 0.00004126704565565830;
d = 0.00006126704565565830;
b = 0.498812483;
r = 0.498812483;
Population = 40000;
mi = 0.4;
Na = Population*mi;
Ns = Population*(1 - mi);
ts = 400;

system = {
   		Iz'[5t]/5 == a Iz[5t](Na - Iz[5t])  + j Iz[5t](Ns - Is[5t]) - b Iz[5t],
   		Is'[t] == c Is[t](Ns - Is[t]) + d Is[t](Na - Iz[t]) - r Is[t]
       	};

initialvalues = {Is[0] == 160, Iz[0] == 40};

solution = NDSolve[Join[system, initialvalues], {Iz, Is}, {t, 0, ts}];

But then it is a sort of Delay Differential Equation. So if you run it you get an error:

Initial history needs to be specified for all variables for delay-differential equations. >>

And if you look at the help DDLs have usually this sort of initial conditions:

ndssol = First[
   NDSolve[{x'[t] == y[t], y'[t] == -x[t - 1], x[t /; t <= 0] == t^2, 
     y[t /; t <= 0] == 2 t}, {x, y}, {t, 0, 5}]];
Plot[Evaluate[{x[t], y[t]} /. ndssol], {t, 0, 5}]

But then if here for example you change time shift into time rescaling like:

ndssol = First[
   NDSolve[{x'[t] == y[t], y'[t] == -x[2 t], x[t /; t <= 0] == t^2, 
     y[t /; t <= 0] == 2 t}, {x, y}, {t, 0, 5}]];
Plot[Evaluate[{x[t], y[t]} /. ndssol], {t, 0, 5}]

You get an error:

The method currently implemented for delay differential equations does not support delays that depend directly on the time variable or dependent variables. >>

So I think we might be in trouble here. An expert opinion is needed.

kayh · August 27, 2015, 2:08pm

Hi,
Why not transform your first equation to have the same scale as your second equation tau/beta->t f[tau]-> f[t] and f’[tau]-> beta f’[t]

Clear[t];
a = 0.00001381594775713100;
c = 0.00007261477318255780;
j = 0.00004126704565565830;
d = 0.00006126704565565830;
b = 0.498812483;
r = 0.498812483;
Population = 40000;
mi = 0.4;
Na = Population*mi;
Ns = Population*(1 - mi);
beta = 5;
tau = beta*t;
ts = 400;

system = {Iz'[
     t] == ((a*(Na - Iz[t])*Iz[t]) + (j*(Ns - Is[t])*Iz[t]) - ((b)*
         Iz[t]))/beta, 
   Is'[t] == ((c*(Ns - Is[t])*Is[t]) + (d*(Na - Iz[t])*Is[t]) - ((r)*
        Is[t]))};

initialvalues = {Is[0] == 160, Iz[0] == 40};

solution = NDSolve[Join[system, initialvalues], {Iz, Is}, {t, 0, ts}]

Plot[Iz[t] /. solution, {t, 0, 10}]

Plot[Is[t] /. solution, {t, 0, 10}]

that gives me those results:

lchaves75 · August 27, 2015, 3:14pm

Yes, I agree it´s not a classical differential system.
So, probabily It´s not possible to integrate those with NDSolve, is it?
Thanks.

anon21378091 · August 27, 2015, 4:48pm

Yes, I agree it´s not a classical differential system.
So, probabily It´s not possible to integrate those with NDSolve, is it?
Thanks.

lchaves75 · August 27, 2015, 4:50pm

Sam,

I´m not sure if this system has a kind of delay, despite the error message from Mathematica.

Actually, when we think about a system with delay, all equations have the same time scale for the derivatives. In others words, the left side of equations are in function of “t”, and in the rigth side we have something in “t - x” or “t + x”.

Thanks.

lchaves75 · August 31, 2015, 5:07am

Dear Kay,

In a first place, thanks about your reply. I was thinking about it. I’m not sure about your suggested transformation.

I was carrying simulations with a cellular automata model, and trying to do an equivalent model with differential equations.

My cellular automata simulation brings me some oscillation and I hope the differential equation model either.

What do you think about Gianluca reply (the first one reply)?

Thanks.

gianlucagorni · September 1, 2015, 9:43am

As I said, I introduced a new unknown function Izt[t] that makes the diagram in the picture commute Iz[tau]=Izt[tau/beta], that is, Izt[t]=Iz[beta*t]. The replacement is made as Iz -> Function[\[Tau], Izt[\[Tau]/\[Beta]] so that the derivatives are done properly. I get system of two differential equation, where the two unknown functions Izt and Is are evaluated sometimes at t and sometimes at beta t. I can solve esplicitly one of the two equations and replace into the other, obtaining an integro-differential equation:

Clear[tau, beta, a, c, j, d, b, r, mi, Na, Ns, Population, ts];
system = {Iz'[
     tau] == ((a (Na - Iz[tau]) Iz[tau]) + (j (Ns - Is[tau]) Iz[
         tau]) - ((b)*Iz[tau])), 
   Is'[t] == ((c (Ns - Is[t]) Is[t]) + (d (Na - Iz[t]) Is[t]) - ((r)*
        Is[t]))};
newSystem = 
  system /. {Iz -> Function[\[Tau], Izt[\[Tau]/\[Beta]]]} /. 
   tau -> \[Beta]*t;
sol2 = DSolve[newSystem[[2]], Is, t]
newSystem[[1]] /. sol2

I am not sure what to do next.

gianlucagorni · September 1, 2015, 9:52am

You get an integro-differential system directly, using the original two functions Is and Iz:

system[[1]] /. DSolve[system[[2]], Is, t]

kayh · September 1, 2015, 6:28pm

Note this is not a delay equation. This is just s stretching of the time variable, therefore going from Tau to t you have Iz’[tau] → beta Iz’[t]
So that’s why the first equation has a factor of 1/beta on the right hand side in my solution.
A simple transformation of the equation from tau to t with tau=beta t

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Second order system with different time scales

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