Skip to main content

Estimating parameters for pair of differential equations efficiently


I'm trying to estimate the parameters for the following pair of differential equations


meq = gamma*v[t]*m[t] - mu*m[t];
veq = v[t]*(k - epsilon - mu) - v[t]^2*k + m[t]*l - m[t]^2*l -
m[t]*v[t]*(k + l + gamma);
equations = {m'[t] == meq, v'[t] == veq};

where gamma, k, l and epsilon are parameters to be estimated, and mu is a known parameter with value 4.58*10^(-5). Also I have initial conditions given by


initials = {m[0] == 2.03, v[0] == 3.09};
mu = 4.58*10^(-5)


and the data for the function m is the following


data = {{60, 2.13597}, {300, 2.27247}, {390, 2.29472}, {420, 2.40096}, {630, 
2.59312}, {660, 2.84918}, {780, 2.93677}, {1020, 3.02945}, {1110,
3.04794}, {1140, 3.05796}, {1140, 3.08739}, {1380, 3.21218}, {1380,
3.2873}, {1500, 3.3347}, {1680, 3.44467}, {1710, 3.47574}, {1710,
3.48421}, {1830, 3.50433}}

I defined a parametric solution for the pair of equations with


pfun = ParametricNDSolveValue[Join[equations, initials], m, {t, 0, 2000}, 

{epsilon, k, l, gamma}]

And then I tried to use FindFit as


fit = FindFit[data, pfun[epsilon, k, l, gamma][t], {epsilon, k, l, gamma}, 
t]

However this method seems to be super inefficient as it seems to take unreasonable amount of time to complete the fit. Is there a way maybe to form the parametric function in a way that the FindFit works faster or use the FindFit more efficiently? Of course if you have another faster method in mind it's very welcome.


I also tried to solve this with some different programs than Mathematica and they seems to be working much faster but in those I have no idea what algorithms they use and also in them I have serious data issues.



Answer



Here you need to use a different model. In this model, too, not everything is smooth, but the result is clear and fast



meq = gamma*v[t]*m[t] - mu*m[t];
veq = v[t]*(k - epsilon - mu) - v[t]^2*k + m[t]*l - m[t]^2*l -
m[t]*v[t]*(k + l + gamma);
equations = {m'[t] == meq, v'[t] == veq};
initials = {m[0] == 2.03, v[0] == 3.09};
mu = 4.58*10^(-5);
data = {{60, 2.13597}, {300, 2.27247}, {390, 2.29472}, {420,
2.40096}, {630, 2.59312}, {660, 2.84918}, {780, 2.93677}, {1020,
3.02945}, {1110, 3.04794}, {1140, 3.05796}, {1140,
3.08739}, {1380, 3.21218}, {1380, 3.2873}, {1500, 3.3347}, {1680,

3.44467}, {1710, 3.47574}, {1710, 3.48421}, {1830, 3.50433}};


model[epsilon_?NumberQ, k_?NumberQ, l_?NumberQ,
gamma_?NumberQ] := (model[epsilon, k, l, gamma] =
First[m /.
NDSolve[{v'[t] ==
v[t]*(k - epsilon - mu) - v[t]^2*k + m[t]*l - m[t]^2*l -
m[t]*v[t]*(k + l + gamma),
m'[t] == gamma*v[t]*m[t] - mu*m[t], m[0] == 2.03,

v[0] == 3.09}, {m, v}, {t, 2000}]]);

fit =
FindFit[data,
model[epsilon, k, l, gamma][
t], {{epsilon, -4.8}, {k, 4.9}, {l, 0.}, {gamma, .1}}, t,
PrecisionGoal -> 4, AccuracyGoal -> 4]

(*Out[]= {epsilon -> -4.78195, k -> 4.89863, l -> -0.00336569,
gamma -> 0.0769293}*)


Show[
Plot[model[epsilon, k, l, gamma][t] /. fit, {t, 0, 2000}],
ListPlot[data, PlotStyle -> Orange]]

fig1


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...