TITLE: How to fit the parameters of differential equations with known data?
QUESTION [6 upvotes]: I have the following data from chemical kinetics research to fit the parameters of ordinary differential equations:
$$
\left[
\begin{array}{ccccccc}
\text{No.}& t & y_1(t)&y_2(t) & y_3(t) & y_4(t) & y_5(t)\\
 1&30.0000 & 9.1300 & 0.0931 & 0.0899 & 0.1000 & 0.0000 \\
 2&60.0000 & 8.9300 & 0.1270 & 0.1230 & 0.2270 & 0.0049 \\
 3&90.0000 & 8.6000 & 0.1510 & 0.1390 & 0.4920 & 0.0153 \\
 4&120.0000 & 8.2800 & 0.1540 & 0.1490 & 0.7780 & 0.0249 \\
 5&150.0000 & 7.9700 & 0.1540 & 0.1570 & 1.0700 & 0.0329 \\
 6&180.0000 & 7.8600 & 0.1540 & 0.1600 & 1.1700 & 0.0348 \\
 7&210.0000 & 7.8100 & 0.1530 & 0.1530 & 1.2100 & 0.0404 \\
 8&240.0000 & 7.7700 & 0.1400 & 0.1420 & 1.2800 & 0.0432 \\
\end{array}
\right]
$$
The ordinary differential equations to fit have $k_1,k_2,k_3,k_4,k_5,k_6$ to be determined.
$$
\left\{
\begin{array}{l}
{y_1}'(t)=-{k_1} {y_1}(t)-{k_2} {y_1}(t),\\
{y_2}'(t)={k_2} {y_1}(t)-{k_3} {y_2}(t),\\
{y_3}'(t)={k_1}   {y_1}(t)+{k_3} {y_2}(t)-{k_4} {y_3}(t),\\
{y_4}'(t)={k_4} {y_3}(t)-{k_5} {y_2}(t) {y_4}(t)+{k_6} {y_5}(t),\\
{y_5}'(t)={k_5} {y_2}(t) {y_4}(t)-{k_6} {y_5}(t)\\
\end{array}
\right.$$
In order to solve it from conventional numerical optimization methods, my original thoughts are: first convert it into least square problems, then apply numerical optimization to it, but this requires symbolically solve a nonlinear system of ordinary differential equations into explicit solutions first, which seems difficult.
My questions are:
(1)Is it possible to determine the global (least square or similarly converted) solution to all the parameters $k_i(i=1,\cdots, 6)$ ? 
(2)Is the solution unique?
(3) are there general approaches to solve such problems (globally if possible)?
Update: Further data from replicates:
$$
\begin{array}{ccccccc}
\text{No}&t& y_1(t) &y_2(t) &y_3(t) & y_4(t) &y_5(t)\\
 9&30.0000 & 9.0400 & 0.1190 & 0.1040 & 0.1390 & 0.0044 \\
10& 60.0000 & 8.8000 & 0.1640 & 0.1120 & 0.3210 & 0.0097 \\
 11&90.0000 & 8.5300 & 0.1640 & 0.1140 & 0.5630 & 0.0219 \\
 12&120.0000 & 8.1800 & 0.1600 & 0.1250 & 0.8730 & 0.0369 \\
13&150.0000 & 7.9700 & 0.1550 & 0.1380 & 1.0600 & 0.0459 \\
 14&180.0000 & 7.7900 & 0.1580 & 0.1510 & 1.2000 & 0.0545 \\
 15&210.0000 & 7.4900 & 0.1480 & 0.1430 & 1.5000 & 0.0636 \\
 16&240.0000 & 7.0800 & 0.1390 & 0.1380 & 1.9100 & 0.0756 \\
\end{array}
$$
Update 2:
I am still seeking for a working approach to obtaining solutions to the problem.

REPLY [2 votes]: If you have access to Maple and the Global Optimization Toolbox, you could try something like this.
data:=[[0.0000,9.1300,0.0931,0.0899,0.1000,0.0000],
[30.0000,8.9300,0.1270,0.1230,0.2270,0.0049],
[60.0000,8.6000,0.1510,0.1390,0.4920,0.0153],
[90.0000,8.2800,0.1540,0.1490,0.7780,0.0249],
[120.0000,7.9700,0.1540,0.1570,1.0700,0.0329],
[150.0000,7.8600,0.1540,0.1600,1.1700,0.0348],
[180.0000,7.8100,0.1530,0.1530,1.2100,0.0404],
[210.0000,7.7700,0.1400,0.1420,1.2800,0.0432]]:
des:=[diff(y1(t),t) = -k1*y1(t) - k2*y1(t),
diff(y2(t),t) = k2*y1(t) - k3*y2(t),
diff(y3(t),t) = k1*y1(t) +k3*y2(t) - k4*y3(t),
diff(y4(t),t) = k4*y3(t) - k5*y2(t)*y4(t) + k6*y5(t),
diff(y5(t),t) = k5*y2(t)*y4(t) - k6*y5(t)]:
ics := seq((y || i)(0) = data[1, i+1], i = 1 .. 5):
res := dsolve({ics, des[]}, numeric, parameters = [k1, k2, k3, k4, k5, k6]):
timeList := [0, 30, 60, 90, 120, 150, 180, 210];
sse := proc (k1, k2, k3, k4, k5, k6) 

   res(parameters = [k1, k2, k3, k4, k5, k6]); 

   add(
   (rhs(select(has, res(timeList[i]), y1)[])-data[i, 2])^2+
   (rhs(select(has, res(timeList[i]), y2)[])-data[i, 3])^2+
   (rhs(select(has, res(timeList[i]), y3)[])-data[i, 4])^2+
   (rhs(select(has, res(timeList[i]), y4)[])-data[i, 5])^2+
   (rhs(select(has, res(timeList[i]), y5)[])-data[i, 6])^2, 
   i = 2 .. 8);

end proc;
c := GlobalOptimization:-GlobalSolve('sse'(k1, k2, k3, k4, k5, k6), k1 = 0 .. 1, k2 = 0 .. 1, k3 = 0 .. 1, k4 = 0 .. 1, k5 = 0 .. 1, k6 = 0 .. 1, timelimit = 10);
   [0.219132447080011505, [k1 = 8.52482740113834e-4, 

     k2 = 5.2683998680924474e-5, k3 = 0.0, 

     k4 = 0.05113239298267808, 

     k5 = 0.004363021255887466, k6 = 0.0]]
res(parameters = c[2]):
p:=Array(1..5):
for n from 1 to 5 do
   p[n]:=plots:-display(plots:-odeplot(res, [t, (y || n)(t)], t = 0 .. 210),plots:-pointplot([seq([data[i, 1], data[i, n+1]], i = 1 .. 8)]));
end do;
plots:-display(p)

The code is a 
parameterized numeric solution of the differential equations (I shifted all the data down by 30s, so I could get the ICs at t=0)
followed by a global optimization of the least-squares problem.
I haven't tested the code for correctness, and if you let the optimizer run for longer or change the optimization bounds, you may get better results.