MaplePrimes - answers and comments on Question, Correlation coefficient for system of nonlinear differential equations and data

By Correlation command

Markiyan Hirnyk — Thu, 13 Dec 2012 01:20:21 Z

It can be done as follows.
> a := map(rhs, [seq(N(t)[2], t = 1 .. 28)]);
   [HFloat(76.97385347315416), HFloat(118.09583623704273),

     HFloat(180.2471996657171), HFloat(272.95692438189803),

     HFloat(408.54714237771907), HFloat(601.1382355077297),

     HFloat(863.3371155654564), HFloat(1199.5809147794562),

     HFloat(1597.0203346321264), HFloat(2018.9567293830953),

     HFloat(2408.8033314612817), HFloat(2707.60026184862),

     HFloat(2875.9336996469806), HFloat(2905.5254183027214),

     HFloat(2815.065049942941), HFloat(2637.238685371348),

     HFloat(2406.4381982423683), HFloat(2151.6059834671246),

     HFloat(1893.8840418742589), HFloat(1647.0630056512841),

     HFloat(1419.1196709578), HFloat(1213.8839177442349),

     HFloat(1032.4421159172612), HFloat(874.1877455459112),

     HFloat(737.5548184994766), HFloat(620.5043121460309),

     HFloat(520.8334118576129), HFloat(436.3626703088189)]
> b := convert(iV, list);
[HFloat(265.0), HFloat(385.0), HFloat(328.0), HFloat(434.0),

HFloat(593.0), HFloat(807.0), HFloat(1122.0), HFloat(1303.0),

HFloat(1536.0), HFloat(1948.0), HFloat(2401.0), HFloat(2859.0),

HFloat(3190.0), HFloat(2846.0), HFloat(2218.0), HFloat(2312.0),

HFloat(2120.0), HFloat(2121.0), HFloat(1828.0), HFloat(1717.0),

HFloat(1623.0), HFloat(1431.0), HFloat(1216.0), HFloat(1018.0),

HFloat(1112.0), HFloat(734.0), HFloat(794.0), HFloat(501.0)]
> Correlation(a, b);
                   HFloat(0.9796259029889479)
correlation.mw

PS. The related reference http://www.mapleprimes.com/questions/140898-Data-Fitting-To-SIR-Model#comment140929 for the MaplePrimes users' convenience.

Determining value of t at i`(t)=0 (Maxima)

kierstejn — Fri, 14 Dec 2012 04:56:55 Z

This is awesome!! just what i needed

I just have one more question regarding the SIR model, out of topic although.

I need to find the value of t were the curve of i(t) is peaking, and the value of i at this point.

Usually i would solve it like this: solve(i´(t)=0,t) , but i have no idea what to do when it is a system of differential equations.

The solution is probably very easy, but i hope you can help me with this??

Sir_model.mw

finding the max

PatrickT — Fri, 14 Dec 2012 10:23:29 Z

One approach is to get the solution of your system in the form of a list of points and then find the maximum from that list. The answer will be more precise the finer the grid you choose.

There may be other, more efficient methods. This is what I got:

#rewriting your input slightly. Note: I recommend you use 1D input in classic/standard worksheet.
restart;
beta := 0.660707762617268e-4:
alpha := .229305575688023:
ode1 := diff(s(t), t) = -beta*s(t)*i(t):
ode2 := diff(i(t), t) = beta*s(t)*i(t)-alpha*i(t):
ode3 := diff(r(t), t) = alpha*i(t):
sys := [ode1,ode2,ode3];
ini := [i(0) = 50, r(0) = 0, s(0) = 10000];

sol := dsolve([op(sys),op(ini)], 'numeric', 'output' = listprocedure);

ps := plots:-odeplot(sol, [t, s(t)], t = 0 .. 50, 'colour' = red):
pi := plots:-odeplot(sol, [t, i(t)], t = 0 .. 50, 'colour' = green):
pr := plots:-odeplot(sol, [t, r(t)], t = 0 .. 50, 'colour' = blue):
plots:-display([ps,pi,pr]);

# eval(i(t),sol); plot(%, 0 ..50, 'discont'=true);# alternative way to plot

L := [ seq( eval([t,i(t)],sol) (k), k = 0 .. 50) ]:
max(L);

2902.84847845621

L := [ seq( eval([t,i(t)],sol) (k), k = 0 .. 50, 0.1) ]: # finer grid
max(L);

2906.47960109601

It is easy

Markiyan Hirnyk — Fri, 14 Dec 2012 10:25:05 Z

> Optimization:-Maximize(proc (t) options operator, arrow; rhs(sol(t)[2]) end proc, 1 .. 30);

opti2.mw