How to plot space of bistable solutions?
$begingroup$
I have the following ODE system
dx1 = -0.7 x1 y2
dy1 = 0.7 x1 y2 + 1.7 (1-x1-y1) y2 - y1
dx2 = -0.7 x2 y1
dy2 = 0.7 x2 y1 + 1.7 (1-x2-y2) y1 - y2
I have found that, for initial conditions of the type $x_1=1-y_1, x_2=1-y_2$, solutions either:
- reach a steady-state where $y_1=y_2=0$
- or reach a steady-state where $y_1,y_2>0$.
Now, I want to plot the plane of initial conditions $(y_1(0),y_2(0))$ and color each point in the plane blue or red depending on whether the steady-state falls in the first or second bucket.
I know how to use streamplot to explore the phase-space of solutions but I am not sure how to go about plotting the regions depending on the steady-state value. Any ideas would be much appreciated.
plotting differential-equations regions
asked 6 hours ago
rparpa
506
$endgroup$
add a comment |
$begingroup$
I have the following ODE system
dx1 = -0.7 x1 y2
dy1 = 0.7 x1 y2 + 1.7 (1-x1-y1) y2 - y1
dx2 = -0.7 x2 y1
dy2 = 0.7 x2 y1 + 1.7 (1-x2-y2) y1 - y2
I have found that, for initial conditions of the type $x_1=1-y_1, x_2=1-y_2$, solutions either:
- reach a steady-state where $y_1=y_2=0$
- or reach a steady-state where $y_1,y_2>0$.
Now, I want to plot the plane of initial conditions $(y_1(0),y_2(0))$ and color each point in the plane blue or red depending on whether the steady-state falls in the first or second bucket.
I know how to use streamplot to explore the phase-space of solutions but I am not sure how to go about plotting the regions depending on the steady-state value. Any ideas would be much appreciated.
plotting differential-equations regions
asked 6 hours ago
rparpa
506
$endgroup$
$begingroup$
By the way, what's the system a model of?
$endgroup$
– Chris K
4 hours ago
1
$begingroup$
It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
I have the following ODE system
dx1 = -0.7 x1 y2
dy1 = 0.7 x1 y2 + 1.7 (1-x1-y1) y2 - y1
dx2 = -0.7 x2 y1
dy2 = 0.7 x2 y1 + 1.7 (1-x2-y2) y1 - y2
I have found that, for initial conditions of the type $x_1=1-y_1, x_2=1-y_2$, solutions either:
- reach a steady-state where $y_1=y_2=0$
- or reach a steady-state where $y_1,y_2>0$.
Now, I want to plot the plane of initial conditions $(y_1(0),y_2(0))$ and color each point in the plane blue or red depending on whether the steady-state falls in the first or second bucket.
I know how to use streamplot to explore the phase-space of solutions but I am not sure how to go about plotting the regions depending on the steady-state value. Any ideas would be much appreciated.
plotting differential-equations regions
asked 6 hours ago
rparpa
506
$endgroup$
I have the following ODE system
dx1 = -0.7 x1 y2
dy1 = 0.7 x1 y2 + 1.7 (1-x1-y1) y2 - y1
dx2 = -0.7 x2 y1
dy2 = 0.7 x2 y1 + 1.7 (1-x2-y2) y1 - y2
I have found that, for initial conditions of the type $x_1=1-y_1, x_2=1-y_2$, solutions either:
- reach a steady-state where $y_1=y_2=0$
- or reach a steady-state where $y_1,y_2>0$.
Now, I want to plot the plane of initial conditions $(y_1(0),y_2(0))$ and color each point in the plane blue or red depending on whether the steady-state falls in the first or second bucket.
I know how to use streamplot to explore the phase-space of solutions but I am not sure how to go about plotting the regions depending on the steady-state value. Any ideas would be much appreciated.
plotting differential-equations regions
plotting differential-equations regions
asked 6 hours ago
rparpa
506
asked 6 hours ago
rparpa
506
asked 6 hours ago
rparpa
506
asked 6 hours ago
rparpa
506
asked 6 hours ago
rparpa
506
506
$begingroup$
By the way, what's the system a model of?
$endgroup$
– Chris K
4 hours ago
1
$begingroup$
It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
By the way, what's the system a model of?
$endgroup$
– Chris K
4 hours ago
1
$begingroup$
It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
$endgroup$
– rpa
4 hours ago
$begingroup$
By the way, what's the system a model of?
$endgroup$
– Chris K
4 hours ago
$begingroup$
By the way, what's the system a model of?
$endgroup$
– Chris K
4 hours ago
1
1
$begingroup$
It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
$endgroup$
– rpa
4 hours ago
$begingroup$
It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
$endgroup$
– rpa
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's an approach based on this answer by @ssch.
First, a function to solve the system and report the final values of y1[tmax] and y2[tmax]:
res[y10_?NumericQ, y20_?NumericQ] := Module[{sol},
sol = NDSolve[{
x1'[t] == -0.7 x1[t] y2[t],
y1'[t] == 0.7 x1[t] y2[t] + 1.7 (1 - x1[t] - y1[t]) y2[t] - y1[t],
x2'[t] == -0.7 x2[t] y1[t],
y2'[t] == 0.7 x2[t] y1[t] + 1.7 (1 - x2[t] - y2[t]) y1[t] - y2[t],
x1[0] == 1 - y10, y1[0] == y10, x2[0] == 1 - y20, y2[0] == y20},
{x1, y1, x2, y2}, {t, 0, tmax}][[1]];
Return[{y1[tmax], y2[tmax]} /. sol]
];
Then RegionPlot where the final values are close enough to {0, 0}:
tmax = 100;
tol = 10^-3;
RegionPlot[{Norm[res[y10, y20]] < tol, Norm[res[y10, y20]] > tol},
{y10, 0, 0.2}, {y20, 0, 0.2},
PlotPoints -> 20, MaxRecursion -> 2, PlotStyle -> {Blue, Red}, FrameLabel -> {y10, y20}]
See also this question.
answered 4 hours ago
Chris KChris K
6,72421841
$endgroup$
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
Here's what I ended up doing in case anyone else has a similar question:
I used NDSolve to numerically solve the system. I then extracted the values of $y_1$ and $y_2$ at time $t=100$, which I know is sufficiently long for the system to reach steady-state. I then threshold the average of $y_1$ and $y_2$ and set the threshold close to zero. Then, I did this for multiple values of the initial conditions using Table and plotted the resulting array using arrayplot.
ArrayPlot[
Table[(i1[100] + i2[100])/2 /.
NDSolve[{s1'[t] == -0.7 s1[t] i2[t],
i1'[t] == 0.7 s1[t] i2[t] + 1.7 (1 - s1[t] - i1[t]) i2[t] - i1[t],
s2'[t] == -0.7 s2[t] i1[t],
i2'[t] == 0.7 s2[t] i1[t] + 1.7 (1 - s2[t] - i2[t]) i1[t] - i2[t],
i1[0] == p1, i2[0] == p2, s1[0] == 1 - p1, s2[0] == 1 - p2},
{i1, i2}, {t, 0, 100}][[1]], {p1, 0, 1, 0.005}, {p2, 0, 1, 0.005}],
ColorFunction -> (If[# > 0.02, Red, Blue] &), DataReversed -> True]
answered 4 hours ago
rparpa
506
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an approach based on this answer by @ssch.
First, a function to solve the system and report the final values of y1[tmax] and y2[tmax]:
res[y10_?NumericQ, y20_?NumericQ] := Module[{sol},
sol = NDSolve[{
x1'[t] == -0.7 x1[t] y2[t],
y1'[t] == 0.7 x1[t] y2[t] + 1.7 (1 - x1[t] - y1[t]) y2[t] - y1[t],
x2'[t] == -0.7 x2[t] y1[t],
y2'[t] == 0.7 x2[t] y1[t] + 1.7 (1 - x2[t] - y2[t]) y1[t] - y2[t],
x1[0] == 1 - y10, y1[0] == y10, x2[0] == 1 - y20, y2[0] == y20},
{x1, y1, x2, y2}, {t, 0, tmax}][[1]];
Return[{y1[tmax], y2[tmax]} /. sol]
];
Then RegionPlot where the final values are close enough to {0, 0}:
tmax = 100;
tol = 10^-3;
RegionPlot[{Norm[res[y10, y20]] < tol, Norm[res[y10, y20]] > tol},
{y10, 0, 0.2}, {y20, 0, 0.2},
PlotPoints -> 20, MaxRecursion -> 2, PlotStyle -> {Blue, Red}, FrameLabel -> {y10, y20}]
See also this question.
answered 4 hours ago
Chris KChris K
6,72421841
$endgroup$
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
Here's an approach based on this answer by @ssch.
First, a function to solve the system and report the final values of y1[tmax] and y2[tmax]:
res[y10_?NumericQ, y20_?NumericQ] := Module[{sol},
sol = NDSolve[{
x1'[t] == -0.7 x1[t] y2[t],
y1'[t] == 0.7 x1[t] y2[t] + 1.7 (1 - x1[t] - y1[t]) y2[t] - y1[t],
x2'[t] == -0.7 x2[t] y1[t],
y2'[t] == 0.7 x2[t] y1[t] + 1.7 (1 - x2[t] - y2[t]) y1[t] - y2[t],
x1[0] == 1 - y10, y1[0] == y10, x2[0] == 1 - y20, y2[0] == y20},
{x1, y1, x2, y2}, {t, 0, tmax}][[1]];
Return[{y1[tmax], y2[tmax]} /. sol]
];
Then RegionPlot where the final values are close enough to {0, 0}:
tmax = 100;
tol = 10^-3;
RegionPlot[{Norm[res[y10, y20]] < tol, Norm[res[y10, y20]] > tol},
{y10, 0, 0.2}, {y20, 0, 0.2},
PlotPoints -> 20, MaxRecursion -> 2, PlotStyle -> {Blue, Red}, FrameLabel -> {y10, y20}]
See also this question.
answered 4 hours ago
Chris KChris K
6,72421841
$endgroup$
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
Here's an approach based on this answer by @ssch.
First, a function to solve the system and report the final values of y1[tmax] and y2[tmax]:
res[y10_?NumericQ, y20_?NumericQ] := Module[{sol},
sol = NDSolve[{
x1'[t] == -0.7 x1[t] y2[t],
y1'[t] == 0.7 x1[t] y2[t] + 1.7 (1 - x1[t] - y1[t]) y2[t] - y1[t],
x2'[t] == -0.7 x2[t] y1[t],
y2'[t] == 0.7 x2[t] y1[t] + 1.7 (1 - x2[t] - y2[t]) y1[t] - y2[t],
x1[0] == 1 - y10, y1[0] == y10, x2[0] == 1 - y20, y2[0] == y20},
{x1, y1, x2, y2}, {t, 0, tmax}][[1]];
Return[{y1[tmax], y2[tmax]} /. sol]
];
Then RegionPlot where the final values are close enough to {0, 0}:
tmax = 100;
tol = 10^-3;
RegionPlot[{Norm[res[y10, y20]] < tol, Norm[res[y10, y20]] > tol},
{y10, 0, 0.2}, {y20, 0, 0.2},
PlotPoints -> 20, MaxRecursion -> 2, PlotStyle -> {Blue, Red}, FrameLabel -> {y10, y20}]
See also this question.
answered 4 hours ago
Chris KChris K
6,72421841
$endgroup$
Here's an approach based on this answer by @ssch.
First, a function to solve the system and report the final values of y1[tmax] and y2[tmax]:
res[y10_?NumericQ, y20_?NumericQ] := Module[{sol},
sol = NDSolve[{
x1'[t] == -0.7 x1[t] y2[t],
y1'[t] == 0.7 x1[t] y2[t] + 1.7 (1 - x1[t] - y1[t]) y2[t] - y1[t],
x2'[t] == -0.7 x2[t] y1[t],
y2'[t] == 0.7 x2[t] y1[t] + 1.7 (1 - x2[t] - y2[t]) y1[t] - y2[t],
x1[0] == 1 - y10, y1[0] == y10, x2[0] == 1 - y20, y2[0] == y20},
{x1, y1, x2, y2}, {t, 0, tmax}][[1]];
Return[{y1[tmax], y2[tmax]} /. sol]
];
Then RegionPlot where the final values are close enough to {0, 0}:
tmax = 100;
tol = 10^-3;
RegionPlot[{Norm[res[y10, y20]] < tol, Norm[res[y10, y20]] > tol},
{y10, 0, 0.2}, {y20, 0, 0.2},
PlotPoints -> 20, MaxRecursion -> 2, PlotStyle -> {Blue, Red}, FrameLabel -> {y10, y20}]
See also this question.
answered 4 hours ago
Chris KChris K
6,72421841
answered 4 hours ago
Chris KChris K
6,72421841
answered 4 hours ago
Chris KChris K
6,72421841
answered 4 hours ago
Chris KChris K
6,72421841
6,72421841
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
$begingroup$
Thanks!! I ended up figuring out another way and I posted my solution as the same time as yours. However, I'm accepting your answer as I think it's better!
$endgroup$
– rpa
4 hours ago
add a comment |
$begingroup$
Here's what I ended up doing in case anyone else has a similar question:
I used NDSolve to numerically solve the system. I then extracted the values of $y_1$ and $y_2$ at time $t=100$, which I know is sufficiently long for the system to reach steady-state. I then threshold the average of $y_1$ and $y_2$ and set the threshold close to zero. Then, I did this for multiple values of the initial conditions using Table and plotted the resulting array using arrayplot.
ArrayPlot[
Table[(i1[100] + i2[100])/2 /.
NDSolve[{s1'[t] == -0.7 s1[t] i2[t],
i1'[t] == 0.7 s1[t] i2[t] + 1.7 (1 - s1[t] - i1[t]) i2[t] - i1[t],
s2'[t] == -0.7 s2[t] i1[t],
i2'[t] == 0.7 s2[t] i1[t] + 1.7 (1 - s2[t] - i2[t]) i1[t] - i2[t],
i1[0] == p1, i2[0] == p2, s1[0] == 1 - p1, s2[0] == 1 - p2},
{i1, i2}, {t, 0, 100}][[1]], {p1, 0, 1, 0.005}, {p2, 0, 1, 0.005}],
ColorFunction -> (If[# > 0.02, Red, Blue] &), DataReversed -> True]
answered 4 hours ago
rparpa
506
$endgroup$
add a comment |
$begingroup$
Here's what I ended up doing in case anyone else has a similar question:
I used NDSolve to numerically solve the system. I then extracted the values of $y_1$ and $y_2$ at time $t=100$, which I know is sufficiently long for the system to reach steady-state. I then threshold the average of $y_1$ and $y_2$ and set the threshold close to zero. Then, I did this for multiple values of the initial conditions using Table and plotted the resulting array using arrayplot.
ArrayPlot[
Table[(i1[100] + i2[100])/2 /.
NDSolve[{s1'[t] == -0.7 s1[t] i2[t],
i1'[t] == 0.7 s1[t] i2[t] + 1.7 (1 - s1[t] - i1[t]) i2[t] - i1[t],
s2'[t] == -0.7 s2[t] i1[t],
i2'[t] == 0.7 s2[t] i1[t] + 1.7 (1 - s2[t] - i2[t]) i1[t] - i2[t],
i1[0] == p1, i2[0] == p2, s1[0] == 1 - p1, s2[0] == 1 - p2},
{i1, i2}, {t, 0, 100}][[1]], {p1, 0, 1, 0.005}, {p2, 0, 1, 0.005}],
ColorFunction -> (If[# > 0.02, Red, Blue] &), DataReversed -> True]
answered 4 hours ago
rparpa
506
$endgroup$
add a comment |
$begingroup$
Here's what I ended up doing in case anyone else has a similar question:
I used NDSolve to numerically solve the system. I then extracted the values of $y_1$ and $y_2$ at time $t=100$, which I know is sufficiently long for the system to reach steady-state. I then threshold the average of $y_1$ and $y_2$ and set the threshold close to zero. Then, I did this for multiple values of the initial conditions using Table and plotted the resulting array using arrayplot.
ArrayPlot[
Table[(i1[100] + i2[100])/2 /.
NDSolve[{s1'[t] == -0.7 s1[t] i2[t],
i1'[t] == 0.7 s1[t] i2[t] + 1.7 (1 - s1[t] - i1[t]) i2[t] - i1[t],
s2'[t] == -0.7 s2[t] i1[t],
i2'[t] == 0.7 s2[t] i1[t] + 1.7 (1 - s2[t] - i2[t]) i1[t] - i2[t],
i1[0] == p1, i2[0] == p2, s1[0] == 1 - p1, s2[0] == 1 - p2},
{i1, i2}, {t, 0, 100}][[1]], {p1, 0, 1, 0.005}, {p2, 0, 1, 0.005}],
ColorFunction -> (If[# > 0.02, Red, Blue] &), DataReversed -> True]
answered 4 hours ago
rparpa
506
$endgroup$
Here's what I ended up doing in case anyone else has a similar question:
I used NDSolve to numerically solve the system. I then extracted the values of $y_1$ and $y_2$ at time $t=100$, which I know is sufficiently long for the system to reach steady-state. I then threshold the average of $y_1$ and $y_2$ and set the threshold close to zero. Then, I did this for multiple values of the initial conditions using Table and plotted the resulting array using arrayplot.
ArrayPlot[
Table[(i1[100] + i2[100])/2 /.
NDSolve[{s1'[t] == -0.7 s1[t] i2[t],
i1'[t] == 0.7 s1[t] i2[t] + 1.7 (1 - s1[t] - i1[t]) i2[t] - i1[t],
s2'[t] == -0.7 s2[t] i1[t],
i2'[t] == 0.7 s2[t] i1[t] + 1.7 (1 - s2[t] - i2[t]) i1[t] - i2[t],
i1[0] == p1, i2[0] == p2, s1[0] == 1 - p1, s2[0] == 1 - p2},
{i1, i2}, {t, 0, 100}][[1]], {p1, 0, 1, 0.005}, {p2, 0, 1, 0.005}],
ColorFunction -> (If[# > 0.02, Red, Blue] &), DataReversed -> True]
answered 4 hours ago
rparpa
506
answered 4 hours ago
rparpa
506
answered 4 hours ago
rparpa
506
answered 4 hours ago
rparpa
506
506
add a comment |
add a comment |
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By the way, what's the system a model of?
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– Chris K
4 hours ago
1
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It's an extension of an epidemic model that exhibits bistability in the initial conditions. See this paper
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– rpa
4 hours ago