How to plot experimental data at special points? - maple

Question

I have a set of experimental data (P), and I want to obtain plot "experimental vs predicted". In order to do so, I use another set of data which depend on P (Q), plot ScatterPlot, use appropriate fit, then obtain regression line, and use its coefficients in appropriate differential equation. Plot of P looks good, but I need to add there experimental data. For simplicity, I've used interval t=0..150.
How can I plot experimental data so that P(0) = Pvals[1], P(10)=Pvals[2], etc.? Besides, how can I distribute data (say, I have t=0..800 and want to plot Pvals so that P(0) = Pvals[1] and P(800) = Pvals[16])?
Pvals := [3.929, 5.308, 7.24, 9.638, 12.866, 17.069, 23.192, 31.433, 38.558, 50.156, 62.948,
75.996, 91.972, 105.711, 122.775, 131.669]:
for i to 15 do Qval[i] := .1*(Pvals[i+1]/Pvals[i]-1); end do:
Qvals := [seq(Qval[i], i = 1 .. 15), 0.144513895e-1]:
with(Statistics);
ScatterPlot(Pvals, Qvals, fit = [a*v^2+b*v+c, v], thickness = 3,
legend = [points = "Point data", fit = typeset("fit to a", 2^nd, "degree polynomial")]);
with(CurveFitting);
LeastSquares(Pvals, Qvals, v, curve = a*v^2+b*v+c);
de := diff(P(t), t) = (0.370152282598477e-1-0.272504103112702e-3*P(t))*P(t);
sol := dsolve({de, P(0) = 3.929}, P(t));
P := plot(rhs(sol), t = 0 .. 160);

Answer 1

I'm not sure that I entirely follow your methodology. But is this something like what you are trying to accomplish?
restart;
with(Statistics):
Pvals := [3.929, 5.308, 7.24, 9.638, 12.866, 17.069, 23.192, 31.433,
38.558, 50.156, 62.948, 75.996, 91.972, 105.711, 122.775, 131.669]:
for i to 15 do Qval[i] := .1*(Pvals[i+1]/Pvals[i]-1); end do:
Qvals := [seq(Qval[i], i = 1 .. 15), 0.144513895e-1]:
form := a*v^2+b*v+c:
CF := CurveFitting:-LeastSquares(Pvals, Qvals, v, curve = form);
CF := 0.0370152282598477 - 0.000272504103112702 v
-7 2
+ 5.60958249026713 10 v
Now I use CF in the DE (since I don't understand why you dropped the v^2 term),
#de := diff(P(t), t) = (0.370152282598477e-1-0.272504103112702e-3*P(t))*P(t);
de := diff(P(t), t) = eval(CF, v=P(t))*P(t);
d /
de := --- P(t) = \0.0370152282598477 - 0.000272504103112702 P(t)
dt
-7 2\
+ 5.60958249026713 10 P(t) / P(t)
I'll use the numeric option of the dsolve command, and obtain a procedure that computes P(t) for numeric t values.
sol := dsolve({de, P(0) = 3.929}, P(t), numeric, output=listprocedure ):
Pfunc := eval(P(t), sol);
Pfunc := proc(t) ... end;
Pfunc(0.0), Pvals[1];
3.92900000000000, 3.929
Now some rescaling (which, again, is my guess as to your goal),
endpt := fsolve(Pfunc(t)-Pvals[16]);
endpt := 135.2246055
Pfunc(endpt), Pvals[16];
131.669000003321, 131.669
plot(Pfunc(t), t=0 .. endpt, size=[500,200]);
a,b,N := 0.0, 800.0, nops(Pvals);
a, b, N := 0., 800.0, 16
Pfuncscaled := proc(t)
if not t::numeric then
return 'procname'(args);
end if;
Pfunc(t*endpt/b);
end proc:
Pfuncscaled(0), Pvals[1];
3.92900000000000, 3.929
Pfuncscaled(800), Pvals[N];
131.669000003321, 131.669
PLscaled := plot( Pfuncscaled(t), t=a .. b,
color=red, size=[500,200] );
Now to display the Pdata against 0 .. 800 as well,
V := Vector(N, (i)->a+(i-1)*(b-a)/(N-1)):
V[1], V[-1];
0., 800.0000000
Pdatascaled := plot( < V | Vector(Pvals) >,
color=blue, size=[500,200],
style=pointline, symbol=solidcircle );
And, displaying the rescaled data together with the rescaled procedure from dsolve,
plots:-display( PLscaled, Pdatascaled, size=[500,500] );