Solve and plot a loop

2020 5 19
1 个回答
更新时间：2020 5 20
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Hi. I'm try to plot ODE 2 equations system for different parameter D values. My system:
That's why I tried, but I receiving error:
Qmax=100;
tet=10;
sig=3;
A=1.3;
vm=-2.1;
mv=-1.8;
tm=0.1;
tv=0.1;
sys=@(t,x, D)[tm*(-x(1)+mv*(Qmax/(1+exp(-(x(2)-10)./sig)))+A); tv*(-x(2)+vm*(Qmax/(1+exp(-(x(1)-10)./sig)))+D)];
tspan = linspace(0, 1, 25);
D = linspace(0, 0.1, 5);
for k1 = 1:length(D)
[t,x] = ode45(@(t,x) sys(t,x,D(k1)), tspan, [0 0]);
ty1(k1,:,:) = [t x];
end
Also, I want to plot graph plot(D, x(1)). Maybe somebody can help to fix and update this code?

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回答（1 个）

Walter Roberson
Walter Roberson 2020-5-19

1 个投票

Unrecognized variable c1, and you do not use c1 anywhere in the loop. It would seem to make more sense to iterate to length(D)
Also, I want to plot graph plot(x(1), D)
Ummm, okay...
plot(ty1(:,:,2),D)
This will give you 25 lines, one for each tspan entry. And since D is the independent variable and x(1) is the dependent variable, it seems strange to use the x(1) value as the independent (first) coordinate and D as the dependent (second) coordinate.

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Ewona CZ
Ewona CZ 2020-5-19
Hi Walter. Yes, it should be length(D), it was a typo and yes should be (D, x(1)).
ty1(k1,:,:) = [t x]; - Matlab gives error in this line.
Thanks for comments. Is it possible to plot such graph?
@Walter
Walter Roberson
Walter Roberson 2020-5-19
I ran your code changing only c1 to D. After that
plot(D, ty1(:,:,2))
You might have an old ty1 in memory that needs to be removed
Ewona CZ
Ewona CZ 2020-5-19
编辑：Ewona CZ 2020-5-19
Yes, it works, but seems I wrote wrong code. I actually wanted to plot solutions depending on paramater D.
This graph is in my book for this system. Is it possible to get? :D I try something like this for the first time.
Walter Roberson
Walter Roberson 2020-5-20
At the moment, I cannot tell which variable is Vm in terms of your x output.
To get that kind of graph, with multiple Vm values for a given D value, D would have to be the dependent variable, which is not the case in your code.
Ewona CZ
Ewona CZ 2020-5-20
编辑：Ewona CZ 2020-5-20
Not sure, but actually in my article D is given like a parameter, but I really saw such graphs in books where D - different values, not a function. I think Vm here is a solution for the first equation of the system. I added original system to my question, please have a look if you are able. Thanks in advance!
Walter Roberson
Walter Roberson 2020-5-20
I do not seem to understand how that diagram is constructed, sorry.
The only guess I have is that maybe there are different initial values for Vm that in combination with a D value lead to some particular outcome, but I have no guesses about what they might be looking for.

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