Plotting bacterial growth using odes
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Billy Bob
2019 年 10 月 16 日
コメント済み: Emmanuel Ayodeji-Ogunsanya
2022 年 4 月 16 日
Hi,
I would like to plot the following functions using Matlab:
dx/dt v. time and ds/dt v. time (with dx/dt on the y axis and time on the x axis)
The expression for dx/dt is given as the following:
---- where S and X are unknown
there is an expression for S in this case:
--- where S and X are unknown.
I know how to plot the above system by converting the differentials to first order and then solving them using 'ode45'. However, this gives me the plots for X v. t and S v. t.
What I would like instead are plot of the differential equations themselves against time. Any help in this regard would be much appreciated!
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採用された回答
Star Strider
2019 年 10 月 16 日
編集済み: Star Strider
2019 年 10 月 16 日
I believe Monod kinetics and curve fitting can help. You are not fitting data, so just use the ODE and ode45 call syntax.
EDIT — (16 Oct 2019 at 17:30)
If you want to plot the differential equations themselves, substitute ‘t’ and ‘y’ back into ‘M’ to get the derivatives:
% Change to First Order%
[V,Sbs] = odeToVectorField([ode1,ode2,ode3])
M = matlabFunction(V,'vars',{'t','Y'})
% Setting Intervals%
interval = [0 1000];
% Initial Conditions%
y0 = [X0 S0 P0];
ysol = ode45(M,interval,y0)
% Graph Plots%
fplot(@(x)deval(ysol,x,1), [0, 35]);
hold on
fplot(@(x)deval(ysol,x,2), [0, 35]);
fplot(@(x)deval(ysol,x,3), [0, 35]);
hold off
legend(string(Sbs))
t = ysol.x;
ys = ysol.y;
for k = 1:numel(t)
dy(:,k) = M(t(k),ys(:,k));
end
figure
plot(t, dy)
grid
legend(string(Sbs))
xlim([0 35])
title('Derivatives')
Alternatively, you can use the gradient function on the solved values of ‘y’, although you would need to specify ‘interval’ as a vector of more than two elements using the linspace function for best results, since gradient prefers regularly-sampled vectors. It all depends on what you want to do.
EDIT — (16 Oct 2019 at 17:59)
The plot image is not appearing when I paste it in, so I attached it as well.
2 件のコメント
Star Strider
2019 年 10 月 21 日
As always, my pleasure.
You can take the absolute value of if you like (or the absolute values of all of the derivatives), however I strongly advise against that. The integrated value of decreases for the entire time, then evenutally levels off at 0. The derivative reflects that.
I would not change anything.
その他の回答 (2 件)
darova
2019 年 10 月 16 日
How do you know that it is the correct order (why not S,X,P ?)
y0 = [X0 S0 P0];
Try this to plot X vs S
[t,ysol] = ode45(M,interval,y0)
% Graph Plots%
plot(ysol(:,1),ysol(:,2))
0 件のコメント
Shivya Shrivastava
2020 年 10 月 29 日
An investigator has reported the data tabulated below for an experiment to determine the growth rate of bacteria k (per d), as a function of oxygen concentration c (mg/L). Find which degree of polynomial is the best fit for given data using MATLAB.
c (mg/L)
0.5
0.8
1.5
2.5
4
k (per d)
1.1
2.4
5.3
7.6
8.9
Plot the best fit curve by continuous line along with the given data points by ‘o’ on the same graph. Print the equation on command prompt after getting the coefficient.
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