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Adding a limit in NDSolve to avoid division by zero

Return partial result when MemoryConstrained aborts NDSolveInfinite Expression Error from NDSolveusing Module in NDSolveNDSolve solution does not satisfy boundary conditionsGetting error NDSolve:ndnum when solving a nonlinear coupled differential equationErrors in Solving Heat EquationSolving 2nd order ODE with NDSolveNDSolve unable to handle discontinuitiesDynamically updating parameter each iteration of NDSolveUsing WhenEvent to limit the derivative

0

$begingroup$

I'm trying to solve a set of differential equations in which one of the functions that describe the time derivative gets values which make it divide by zero

x'[t] = (Exp[x] - 1)/(Exp[x] - 1 + x) 

So what happens is that when NDSolve gets values of x=0 you get that an Infinite expression of 1/0 encountered.

However, when I have x=0 I would actually like to replace it with the limit of x->0

 Limit[(Exp[x] - 1)/(Exp[x] - 1 + x), x -> 0]

which is 1/2.

Any suggestions of how to implement the idea in NDSolve?

Addition

Look for simplicity at the following case

NDSolve[x'[t] == (Exp[x[t]] - 1)/(Exp[x[t]] - 1 + x[t]), 
 x[0] == 0, x, t, 0, 1]

Here x'[t] encounters 1/0 in the initial condition, but I would like it to get the limit of x->0 which is 1/2.
Note that in my problem which is far more complicated, x'[t] encounters this limit many times and the value of the limit is varied with respect to other state variables, therefore I would like the limit to be calculated in each iteration.

differential-equations singularity

share|improve this question

edited May 2 at 14:43

jarhead

asked May 2 at 14:08

jarheadjarhead

674615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does your real problem involve 1/0 in the initial conditions or elsewhere? I find that your simple example runs fine as long as you don't use x[0] == 0.
  $endgroup$
  – Chris K
  May 2 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisK, elsewhere, actually all along the solution periodically. I gave the initial condition as an example.
  $endgroup$
  – jarhead
  May 2 at 15:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give an example where it doesn't work x[0] != 0?
  $endgroup$
  – Chris K
  May 2 at 16:14
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to solve a set of differential equations in which one of the functions that describe the time derivative gets values which make it divide by zero

x'[t] = (Exp[x] - 1)/(Exp[x] - 1 + x) 

So what happens is that when NDSolve gets values of x=0 you get that an Infinite expression of 1/0 encountered.

However, when I have x=0 I would actually like to replace it with the limit of x->0

 Limit[(Exp[x] - 1)/(Exp[x] - 1 + x), x -> 0]

which is 1/2.

Any suggestions of how to implement the idea in NDSolve?

Addition

Look for simplicity at the following case

NDSolve[x'[t] == (Exp[x[t]] - 1)/(Exp[x[t]] - 1 + x[t]), 
 x[0] == 0, x, t, 0, 1]

Here x'[t] encounters 1/0 in the initial condition, but I would like it to get the limit of x->0 which is 1/2.
Note that in my problem which is far more complicated, x'[t] encounters this limit many times and the value of the limit is varied with respect to other state variables, therefore I would like the limit to be calculated in each iteration.

differential-equations singularity

share|improve this question

edited May 2 at 14:43

jarhead

asked May 2 at 14:08

jarheadjarhead

674615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does your real problem involve 1/0 in the initial conditions or elsewhere? I find that your simple example runs fine as long as you don't use x[0] == 0.
  $endgroup$
  – Chris K
  May 2 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisK, elsewhere, actually all along the solution periodically. I gave the initial condition as an example.
  $endgroup$
  – jarhead
  May 2 at 15:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give an example where it doesn't work x[0] != 0?
  $endgroup$
  – Chris K
  May 2 at 16:14
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to solve a set of differential equations in which one of the functions that describe the time derivative gets values which make it divide by zero

x'[t] = (Exp[x] - 1)/(Exp[x] - 1 + x) 

So what happens is that when NDSolve gets values of x=0 you get that an Infinite expression of 1/0 encountered.

However, when I have x=0 I would actually like to replace it with the limit of x->0

 Limit[(Exp[x] - 1)/(Exp[x] - 1 + x), x -> 0]

which is 1/2.

Any suggestions of how to implement the idea in NDSolve?

Addition

Look for simplicity at the following case

NDSolve[x'[t] == (Exp[x[t]] - 1)/(Exp[x[t]] - 1 + x[t]), 
 x[0] == 0, x, t, 0, 1]

Here x'[t] encounters 1/0 in the initial condition, but I would like it to get the limit of x->0 which is 1/2.
Note that in my problem which is far more complicated, x'[t] encounters this limit many times and the value of the limit is varied with respect to other state variables, therefore I would like the limit to be calculated in each iteration.

differential-equations singularity

share|improve this question

edited May 2 at 14:43

jarhead

asked May 2 at 14:08

jarheadjarhead

674615

$endgroup$

x'[t] = (Exp[x] - 1)/(Exp[x] - 1 + x) 

 Limit[(Exp[x] - 1)/(Exp[x] - 1 + x), x -> 0]

NDSolve[x'[t] == (Exp[x[t]] - 1)/(Exp[x[t]] - 1 + x[t]), 
 x[0] == 0, x, t, 0, 1]

share|improve this question

edited May 2 at 14:43

jarhead

asked May 2 at 14:08

jarheadjarhead

674615

share|improve this question

edited May 2 at 14:43

jarhead

asked May 2 at 14:08

jarheadjarhead

674615

asked May 2 at 14:08

jarheadjarhead

674615

asked May 2 at 14:08

jarheadjarhead

674615

1 Answer
1

active

oldest

votes

4

$begingroup$

If:

eq = With[x = x[t], D[x, t] == If[x == 0, 1/2, (Exp[x] - 1)/(Exp[x] - 1 + x)]]

sol = NDSolveValue[eq, x[0] == -1, x, t, 0, 6]

Plot[sol[t], t, 0, 6]

---

Update

If the limit needs to be calculated each time it encounters zero:

eq = With[x = x[t], 
 With[expr = (Exp[x] - 1)/(Exp[x] - 1 + x), 
 D[x, t] == If[x == 0, Limit[expr, x -> 0], expr]]]

share|improve this answer

edited May 2 at 14:57

answered May 2 at 14:38

xzczdxzczd

28.2k577261

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer, please look at my addition to the question. It is important that the limit will be calculated each time it encounters zero.
  $endgroup$
  – jarhead
  May 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please also keep the answer restricted to 'NDSolve' if possible.
  $endgroup$
  – jarhead
  May 2 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jarhead Check my update.
  $endgroup$
  – xzczd
  May 2 at 14:57
  

  
  
  
  

add a comment | 

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4

$begingroup$

If:

eq = With[x = x[t], D[x, t] == If[x == 0, 1/2, (Exp[x] - 1)/(Exp[x] - 1 + x)]]

sol = NDSolveValue[eq, x[0] == -1, x, t, 0, 6]

Plot[sol[t], t, 0, 6]

---

Update

If the limit needs to be calculated each time it encounters zero:

eq = With[x = x[t], 
 With[expr = (Exp[x] - 1)/(Exp[x] - 1 + x), 
 D[x, t] == If[x == 0, Limit[expr, x -> 0], expr]]]

share|improve this answer

edited May 2 at 14:57

answered May 2 at 14:38

xzczdxzczd

28.2k577261

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer, please look at my addition to the question. It is important that the limit will be calculated each time it encounters zero.
  $endgroup$
  – jarhead
  May 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please also keep the answer restricted to 'NDSolve' if possible.
  $endgroup$
  – jarhead
  May 2 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jarhead Check my update.
  $endgroup$
  – xzczd
  May 2 at 14:57
  

  
  
  
  

add a comment | 

4

$begingroup$

If:

eq = With[x = x[t], D[x, t] == If[x == 0, 1/2, (Exp[x] - 1)/(Exp[x] - 1 + x)]]

sol = NDSolveValue[eq, x[0] == -1, x, t, 0, 6]

Plot[sol[t], t, 0, 6]

---

Update

If the limit needs to be calculated each time it encounters zero:

eq = With[x = x[t], 
 With[expr = (Exp[x] - 1)/(Exp[x] - 1 + x), 
 D[x, t] == If[x == 0, Limit[expr, x -> 0], expr]]]

share|improve this answer

edited May 2 at 14:57

answered May 2 at 14:38

xzczdxzczd

28.2k577261

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer, please look at my addition to the question. It is important that the limit will be calculated each time it encounters zero.
  $endgroup$
  – jarhead
  May 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please also keep the answer restricted to 'NDSolve' if possible.
  $endgroup$
  – jarhead
  May 2 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jarhead Check my update.
  $endgroup$
  – xzczd
  May 2 at 14:57
  

  
  
  
  

add a comment | 

$begingroup$

If:

eq = With[x = x[t], D[x, t] == If[x == 0, 1/2, (Exp[x] - 1)/(Exp[x] - 1 + x)]]

sol = NDSolveValue[eq, x[0] == -1, x, t, 0, 6]

Plot[sol[t], t, 0, 6]

---

Update

If the limit needs to be calculated each time it encounters zero:

eq = With[x = x[t], 
 With[expr = (Exp[x] - 1)/(Exp[x] - 1 + x), 
 D[x, t] == If[x == 0, Limit[expr, x -> 0], expr]]]

share|improve this answer

edited May 2 at 14:57

answered May 2 at 14:38

xzczdxzczd

28.2k577261

$endgroup$

eq = With[x = x[t], D[x, t] == If[x == 0, 1/2, (Exp[x] - 1)/(Exp[x] - 1 + x)]]

sol = NDSolveValue[eq, x[0] == -1, x, t, 0, 6]

Plot[sol[t], t, 0, 6]

eq = With[x = x[t], 
 With[expr = (Exp[x] - 1)/(Exp[x] - 1 + x), 
 D[x, t] == If[x == 0, Limit[expr, x -> 0], expr]]]

share|improve this answer

edited May 2 at 14:57

answered May 2 at 14:38

xzczdxzczd

28.2k577261

answered May 2 at 14:38

xzczdxzczd

28.2k577261

answered May 2 at 14:38

xzczdxzczd

28.2k577261