Switched linear systems

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up vote
1
down vote

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I'm curious whether it is possible to solve switched linear systems within the framework of NDSolve. For example a system of linear ode's like
$$x'(t) = left{beginarrayll A_1 x(t),& textif ,, x_1x_2leq 0 \ A_2 x(t), & textif ,, x_1x_2>0 endarrayright.$$

where $A_1$ and $A_2$ are two constant matrices with appropriate size (namely $A_1,A_2 in mathbbR^2times 2$) and $x(t) = left(x_1(t),x_2(t)right)^top$.

I tried WhenEvent but received an error message saying

"Warning: the rule !(*SuperscriptBox["x", "[Prime]",
MultilineFunction->None][t] -> A1 . x[t]) will not directly set the
state because the left-hand side is not a list of state variables."

Here is the code

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;
x[t_] = x1[t], x2[t]

NDSolve[x'[t] == A2.x[t], x1[0] == 6, x2[0] == 3, 
WhenEvent[x1[t] x2[t] <= 0, x'[t] -> A1.x[t]], x1, x2, t, 0, 
100, Method -> "EquationSimplification" -> "Residual"]

differential-equations

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

asked 3 hours ago

freddy90

34317

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I wonder if it might work with the rhs expressed using Piecewise?
  – Daniel Lichtblau
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
  – J. M. is somewhat okay.♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
  – freddy90
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
  – J. M. is somewhat okay.♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
  – Chris K
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

favorite

I'm curious whether it is possible to solve switched linear systems within the framework of NDSolve. For example a system of linear ode's like
$$x'(t) = left{beginarrayll A_1 x(t),& textif ,, x_1x_2leq 0 \ A_2 x(t), & textif ,, x_1x_2>0 endarrayright.$$

where $A_1$ and $A_2$ are two constant matrices with appropriate size (namely $A_1,A_2 in mathbbR^2times 2$) and $x(t) = left(x_1(t),x_2(t)right)^top$.

I tried WhenEvent but received an error message saying

"Warning: the rule !(*SuperscriptBox["x", "[Prime]",
MultilineFunction->None][t] -> A1 . x[t]) will not directly set the
state because the left-hand side is not a list of state variables."

Here is the code

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;
x[t_] = x1[t], x2[t]

NDSolve[x'[t] == A2.x[t], x1[0] == 6, x2[0] == 3, 
WhenEvent[x1[t] x2[t] <= 0, x'[t] -> A1.x[t]], x1, x2, t, 0, 
100, Method -> "EquationSimplification" -> "Residual"]

differential-equations

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

asked 3 hours ago

freddy90

34317

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I wonder if it might work with the rhs expressed using Piecewise?
  – Daniel Lichtblau
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
  – J. M. is somewhat okay.♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
  – freddy90
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
  – J. M. is somewhat okay.♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
  – Chris K
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm curious whether it is possible to solve switched linear systems within the framework of NDSolve. For example a system of linear ode's like
$$x'(t) = left{beginarrayll A_1 x(t),& textif ,, x_1x_2leq 0 \ A_2 x(t), & textif ,, x_1x_2>0 endarrayright.$$

where $A_1$ and $A_2$ are two constant matrices with appropriate size (namely $A_1,A_2 in mathbbR^2times 2$) and $x(t) = left(x_1(t),x_2(t)right)^top$.

I tried WhenEvent but received an error message saying

"Warning: the rule !(*SuperscriptBox["x", "[Prime]",
MultilineFunction->None][t] -> A1 . x[t]) will not directly set the
state because the left-hand side is not a list of state variables."

Here is the code

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;
x[t_] = x1[t], x2[t]

NDSolve[x'[t] == A2.x[t], x1[0] == 6, x2[0] == 3, 
WhenEvent[x1[t] x2[t] <= 0, x'[t] -> A1.x[t]], x1, x2, t, 0, 
100, Method -> "EquationSimplification" -> "Residual"]

differential-equations

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

asked 3 hours ago

freddy90

34317

I'm curious whether it is possible to solve switched linear systems within the framework of NDSolve. For example a system of linear ode's like
$$x'(t) = left{beginarrayll A_1 x(t),& textif ,, x_1x_2leq 0 \ A_2 x(t), & textif ,, x_1x_2>0 endarrayright.$$

where $A_1$ and $A_2$ are two constant matrices with appropriate size (namely $A_1,A_2 in mathbbR^2times 2$) and $x(t) = left(x_1(t),x_2(t)right)^top$.

I tried WhenEvent but received an error message saying

"Warning: the rule !(*SuperscriptBox["x", "[Prime]",
MultilineFunction->None][t] -> A1 . x[t]) will not directly set the
state because the left-hand side is not a list of state variables."

Here is the code

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;
x[t_] = x1[t], x2[t]

NDSolve[x'[t] == A2.x[t], x1[0] == 6, x2[0] == 3, 
WhenEvent[x1[t] x2[t] <= 0, x'[t] -> A1.x[t]], x1, x2, t, 0, 
100, Method -> "EquationSimplification" -> "Residual"]

differential-equations

differential-equations

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

asked 3 hours ago

freddy90

34317

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

asked 3 hours ago

freddy90

34317

share|improve this question

share|improve this question

edited 26 mins ago

Carl Woll

59.9k279154

edited 26 mins ago

Carl Woll

59.9k279154

edited 26 mins ago

Carl Woll

59.9k279154

59.9k279154

asked 3 hours ago

freddy90

34317

asked 3 hours ago

freddy90

34317

asked 3 hours ago

freddy90

34317

34317

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I wonder if it might work with the rhs expressed using Piecewise?
  – Daniel Lichtblau
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
  – J. M. is somewhat okay.♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
  – freddy90
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
  – J. M. is somewhat okay.♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
  – Chris K
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I wonder if it might work with the rhs expressed using Piecewise?
  – Daniel Lichtblau
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
  – J. M. is somewhat okay.♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
  – freddy90
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
  – J. M. is somewhat okay.♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
  – Chris K
  2 hours ago
  

  
  
  
  

1

1

I wonder if it might work with the rhs expressed using Piecewise?
– Daniel Lichtblau
3 hours ago

I wonder if it might work with the rhs expressed using Piecewise?
– Daniel Lichtblau
3 hours ago

Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
– J. M. is somewhat okay.♦
2 hours ago

Indeed, if memory serves, NDSolve will set up the WhenEvent objects on your behalf if you use Piecewise. Still, it is useful to know how to adapt WhenEvent in case the automatic method fails.
– J. M. is somewhat okay.♦
2 hours ago

1

1

Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
– freddy90
2 hours ago

Tried this: system = x'[t] == Piecewise[A1.x[t], x1[t] x2[t] < 0, A2.x[t], x1[t] x2[t] > 0]; NDSolve[system, x1[t] == 3, x2[0] == 2, x1, x2, t, 0, 10]. I got the error: "Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions"
– freddy90
2 hours ago

1

1

Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
– J. M. is somewhat okay.♦
2 hours ago

Why are the initial conditions for both x1 and x2 scalars and not vectors? Try using Indexed if you want to refer to a vector-valued function componentwise, just like in your inequality conditions.
– J. M. is somewhat okay.♦
2 hours ago

1

1

It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
– Chris K
2 hours ago

It doesn't seem that defining x[t_] = x1[t], x2[t] is enough for Mathematica to know the relationship between x[t] and x1[t],x2[t].
– Chris K
2 hours ago

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

up vote
4
down vote

Like the error message says, I believe WhenEvent needs to change a state variable, not its (highest) derivative. Here's an approach that sets A as a DiscreteVariable that can be changed when needed.

listProduct[x_List] := Times @@ x;

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;

sol = NDSolve[x'[t] == A[t].x[t], x[0] == 6, 3, A[0] == A2,
 WhenEvent[listProduct[x[t]] <= 0, A[t] -> A1],
 WhenEvent[listProduct[x[t]] > 0, A[t] -> A2], x, A, t, 0, 100,
 DiscreteVariables -> A][[1]];

Plot[Sign[listProduct[x[t] /. sol]], t, 0, 100]

listProduct is by rm -rf from this answer.

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  v11.3 -- did you try a fresh kernel?
  – Chris K
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  jep, tried that but unfortunately doesn't change anything…
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, my colleague's v10.3 has the same problem. Bug?
  – Chris K
  14 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

You can use Piecewise in the vector form of the ODE, the only tricky part is how to create the condition. Here are two possibilities:

pm1 = 0, 1, 1, 0;

sol1 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], x[t].pm1.x[t]>0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

sol2 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], Indexed[x[t], 1] Indexed[x[t], 2] > 0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

Visualizations:

Plot[
 Indexed[sol1[t],1], Indexed[sol1[t], 2],
 t,0,100,
 PlotRange->All
]

Plot[
 Indexed[sol2[t],1], Indexed[sol2[t], 2],
 t,0,100,
 PlotRange->All
]

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

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

up vote
4
down vote

Like the error message says, I believe WhenEvent needs to change a state variable, not its (highest) derivative. Here's an approach that sets A as a DiscreteVariable that can be changed when needed.

listProduct[x_List] := Times @@ x;

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;

sol = NDSolve[x'[t] == A[t].x[t], x[0] == 6, 3, A[0] == A2,
 WhenEvent[listProduct[x[t]] <= 0, A[t] -> A1],
 WhenEvent[listProduct[x[t]] > 0, A[t] -> A2], x, A, t, 0, 100,
 DiscreteVariables -> A][[1]];

Plot[Sign[listProduct[x[t] /. sol]], t, 0, 100]

listProduct is by rm -rf from this answer.

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  v11.3 -- did you try a fresh kernel?
  – Chris K
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  jep, tried that but unfortunately doesn't change anything…
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, my colleague's v10.3 has the same problem. Bug?
  – Chris K
  14 mins ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

Like the error message says, I believe WhenEvent needs to change a state variable, not its (highest) derivative. Here's an approach that sets A as a DiscreteVariable that can be changed when needed.

listProduct[x_List] := Times @@ x;

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;

sol = NDSolve[x'[t] == A[t].x[t], x[0] == 6, 3, A[0] == A2,
 WhenEvent[listProduct[x[t]] <= 0, A[t] -> A1],
 WhenEvent[listProduct[x[t]] > 0, A[t] -> A2], x, A, t, 0, 100,
 DiscreteVariables -> A][[1]];

Plot[Sign[listProduct[x[t] /. sol]], t, 0, 100]

listProduct is by rm -rf from this answer.

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  v11.3 -- did you try a fresh kernel?
  – Chris K
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  jep, tried that but unfortunately doesn't change anything…
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, my colleague's v10.3 has the same problem. Bug?
  – Chris K
  14 mins ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Like the error message says, I believe WhenEvent needs to change a state variable, not its (highest) derivative. Here's an approach that sets A as a DiscreteVariable that can be changed when needed.

listProduct[x_List] := Times @@ x;

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;

sol = NDSolve[x'[t] == A[t].x[t], x[0] == 6, 3, A[0] == A2,
 WhenEvent[listProduct[x[t]] <= 0, A[t] -> A1],
 WhenEvent[listProduct[x[t]] > 0, A[t] -> A2], x, A, t, 0, 100,
 DiscreteVariables -> A][[1]];

Plot[Sign[listProduct[x[t] /. sol]], t, 0, 100]

listProduct is by rm -rf from this answer.

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

Like the error message says, I believe WhenEvent needs to change a state variable, not its (highest) derivative. Here's an approach that sets A as a DiscreteVariable that can be changed when needed.

listProduct[x_List] := Times @@ x;

A1 = 0, -1, 2, 0;
A2 = 0, -2, 1, 0;

sol = NDSolve[x'[t] == A[t].x[t], x[0] == 6, 3, A[0] == A2,
 WhenEvent[listProduct[x[t]] <= 0, A[t] -> A1],
 WhenEvent[listProduct[x[t]] > 0, A[t] -> A2], x, A, t, 0, 100,
 DiscreteVariables -> A][[1]];

Plot[Sign[listProduct[x[t] /. sol]], t, 0, 100]

listProduct is by rm -rf from this answer.

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

share|improve this answer

share|improve this answer

answered 2 hours ago

Chris K

5,76221738

answered 2 hours ago

Chris K

5,76221738

answered 2 hours ago

Chris K

5,76221738

5,76221738

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  v11.3 -- did you try a fresh kernel?
  – Chris K
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  jep, tried that but unfortunately doesn't change anything…
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, my colleague's v10.3 has the same problem. Bug?
  – Chris K
  14 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  v11.3 -- did you try a fresh kernel?
  – Chris K
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  jep, tried that but unfortunately doesn't change anything…
  – freddy90
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, my colleague's v10.3 has the same problem. Bug?
  – Chris K
  14 mins ago
  

  
  
  
  

What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
– freddy90
1 hour ago

What version of mathematica do you use? I tried your code with version 10.2 and I can only integrate till t = 4 without waiting forever
– freddy90
1 hour ago

2

2

v11.3 -- did you try a fresh kernel?
– Chris K
1 hour ago

v11.3 -- did you try a fresh kernel?
– Chris K
1 hour ago

jep, tried that but unfortunately doesn't change anything…
– freddy90
1 hour ago

jep, tried that but unfortunately doesn't change anything…
– freddy90
1 hour ago

Yeah, my colleague's v10.3 has the same problem. Bug?
– Chris K
14 mins ago

Yeah, my colleague's v10.3 has the same problem. Bug?
– Chris K
14 mins ago

add a comment | 

up vote
1
down vote

You can use Piecewise in the vector form of the ODE, the only tricky part is how to create the condition. Here are two possibilities:

pm1 = 0, 1, 1, 0;

sol1 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], x[t].pm1.x[t]>0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

sol2 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], Indexed[x[t], 1] Indexed[x[t], 2] > 0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

Visualizations:

Plot[
 Indexed[sol1[t],1], Indexed[sol1[t], 2],
 t,0,100,
 PlotRange->All
]

Plot[
 Indexed[sol2[t],1], Indexed[sol2[t], 2],
 t,0,100,
 PlotRange->All
]

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

add a comment | 

up vote
1
down vote

You can use Piecewise in the vector form of the ODE, the only tricky part is how to create the condition. Here are two possibilities:

pm1 = 0, 1, 1, 0;

sol1 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], x[t].pm1.x[t]>0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

sol2 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], Indexed[x[t], 1] Indexed[x[t], 2] > 0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

Visualizations:

Plot[
 Indexed[sol1[t],1], Indexed[sol1[t], 2],
 t,0,100,
 PlotRange->All
]

Plot[
 Indexed[sol2[t],1], Indexed[sol2[t], 2],
 t,0,100,
 PlotRange->All
]

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

add a comment | 

up vote
1
down vote

up vote
1
down vote

You can use Piecewise in the vector form of the ODE, the only tricky part is how to create the condition. Here are two possibilities:

pm1 = 0, 1, 1, 0;

sol1 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], x[t].pm1.x[t]>0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

sol2 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], Indexed[x[t], 1] Indexed[x[t], 2] > 0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

Visualizations:

Plot[
 Indexed[sol1[t],1], Indexed[sol1[t], 2],
 t,0,100,
 PlotRange->All
]

Plot[
 Indexed[sol2[t],1], Indexed[sol2[t], 2],
 t,0,100,
 PlotRange->All
]

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

You can use Piecewise in the vector form of the ODE, the only tricky part is how to create the condition. Here are two possibilities:

pm1 = 0, 1, 1, 0;

sol1 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], x[t].pm1.x[t]>0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

sol2 = NDSolveValue[
 
 x'[t] == Piecewise[A2.x[t], Indexed[x[t], 1] Indexed[x[t], 2] > 0, A1.x[t]],
 x[0] == 6, 3
 ,
 x,
 t, 0, 100
];

Visualizations:

Plot[
 Indexed[sol1[t],1], Indexed[sol1[t], 2],
 t,0,100,
 PlotRange->All
]

Plot[
 Indexed[sol2[t],1], Indexed[sol2[t], 2],
 t,0,100,
 PlotRange->All
]

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

share|improve this answer

share|improve this answer

answered 28 mins ago

Carl Woll

59.9k279154

answered 28 mins ago

Carl Woll

59.9k279154

answered 28 mins ago

Carl Woll

59.9k279154

59.9k279154

add a comment | 

add a comment | 

 

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