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Does nudging an exact differential equation nudge or destroy the identity integrating factor?

2

$begingroup$

This question will be related to this one, if for no other reason because a positive answer to the latter would likely help to solve the former.

Consider the differential equation $(y) dx + (x) dy = 0$. It is already exact, so we can think of the multiplicative identity, $u(x) = 1$, as an integrating factor. We can also observe by separating variables that $u(x) = frac{1}{xy}$ is an integrating factor, yielding $(frac{1}{x}) dx + (frac{1}{y}) dy = 0$. From here, $u(x) = xy$ allows us to return to the original form, so we can toggle freely between the two, and both produce the same solution, $y = frac{C}{x}$.

Now instead consider the inexact equation $(y) dx + (x + epsilon x) dy = 0$, for some small value of $epsilon$. By separating variables, we find the integrating factor $u(x) = frac{1}{(x + epsilon x)y}$, and the solution $y = frac{C}{x^{frac{1}{1 + epsilon}}}$, which respectively are very close to $u(x) = frac{1}{xy}$ and $ = frac{C}{x}$ from the previous problem. However, the original inexact form has $frac{partial N}{partial x} - frac{partial M}{partial y} = epsilon$, which is very close to $0$. Does this imply (not only for this problem, but in general) that an integrating factor very close to $u(x) = 1$ exists, or is the nearby exact form destroyed entirely when we nudge $frac{partial N}{partial x} - frac{partial M}{partial y}$ even by a tiny amount?

ordinary-differential-equations partial-derivative integrating-factor

share|cite|improve this question

edited Mar 24 at 20:36

user10478

asked Mar 23 at 18:35

user10478user10478

497413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Consider slightly changing the title of the question. If I were to answer the question in title, I would say that exactness is a fragile property (non structurally stable, using terminology from dynamical systems): you can always perturb the exact equation with non-degenerate extremum of potential and get an equation which can't be exact at all. However, the question in your post is a bit different.
  $endgroup$
  – Evgeny
  Mar 24 at 18:20
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, I think you're right. Is this any better?
  $endgroup$
  – user10478
  Mar 24 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems good to me :) Also, don't forget that user gets a notification about a comment is when you mention them with @ .
  $endgroup$
  – Evgeny
  Mar 26 at 18:20
  

  
  
  
  

add a comment | 

2

$begingroup$

This question will be related to this one, if for no other reason because a positive answer to the latter would likely help to solve the former.

Consider the differential equation $(y) dx + (x) dy = 0$. It is already exact, so we can think of the multiplicative identity, $u(x) = 1$, as an integrating factor. We can also observe by separating variables that $u(x) = frac{1}{xy}$ is an integrating factor, yielding $(frac{1}{x}) dx + (frac{1}{y}) dy = 0$. From here, $u(x) = xy$ allows us to return to the original form, so we can toggle freely between the two, and both produce the same solution, $y = frac{C}{x}$.

Now instead consider the inexact equation $(y) dx + (x + epsilon x) dy = 0$, for some small value of $epsilon$. By separating variables, we find the integrating factor $u(x) = frac{1}{(x + epsilon x)y}$, and the solution $y = frac{C}{x^{frac{1}{1 + epsilon}}}$, which respectively are very close to $u(x) = frac{1}{xy}$ and $ = frac{C}{x}$ from the previous problem. However, the original inexact form has $frac{partial N}{partial x} - frac{partial M}{partial y} = epsilon$, which is very close to $0$. Does this imply (not only for this problem, but in general) that an integrating factor very close to $u(x) = 1$ exists, or is the nearby exact form destroyed entirely when we nudge $frac{partial N}{partial x} - frac{partial M}{partial y}$ even by a tiny amount?

ordinary-differential-equations partial-derivative integrating-factor

share|cite|improve this question

edited Mar 24 at 20:36

user10478

asked Mar 23 at 18:35

user10478user10478

497413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Consider slightly changing the title of the question. If I were to answer the question in title, I would say that exactness is a fragile property (non structurally stable, using terminology from dynamical systems): you can always perturb the exact equation with non-degenerate extremum of potential and get an equation which can't be exact at all. However, the question in your post is a bit different.
  $endgroup$
  – Evgeny
  Mar 24 at 18:20
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, I think you're right. Is this any better?
  $endgroup$
  – user10478
  Mar 24 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems good to me :) Also, don't forget that user gets a notification about a comment is when you mention them with @ .
  $endgroup$
  – Evgeny
  Mar 26 at 18:20
  

  
  
  
  

add a comment | 

$begingroup$

This question will be related to this one, if for no other reason because a positive answer to the latter would likely help to solve the former.

Consider the differential equation $(y) dx + (x) dy = 0$. It is already exact, so we can think of the multiplicative identity, $u(x) = 1$, as an integrating factor. We can also observe by separating variables that $u(x) = frac{1}{xy}$ is an integrating factor, yielding $(frac{1}{x}) dx + (frac{1}{y}) dy = 0$. From here, $u(x) = xy$ allows us to return to the original form, so we can toggle freely between the two, and both produce the same solution, $y = frac{C}{x}$.

Now instead consider the inexact equation $(y) dx + (x + epsilon x) dy = 0$, for some small value of $epsilon$. By separating variables, we find the integrating factor $u(x) = frac{1}{(x + epsilon x)y}$, and the solution $y = frac{C}{x^{frac{1}{1 + epsilon}}}$, which respectively are very close to $u(x) = frac{1}{xy}$ and $ = frac{C}{x}$ from the previous problem. However, the original inexact form has $frac{partial N}{partial x} - frac{partial M}{partial y} = epsilon$, which is very close to $0$. Does this imply (not only for this problem, but in general) that an integrating factor very close to $u(x) = 1$ exists, or is the nearby exact form destroyed entirely when we nudge $frac{partial N}{partial x} - frac{partial M}{partial y}$ even by a tiny amount?

ordinary-differential-equations partial-derivative integrating-factor

share|cite|improve this question

edited Mar 24 at 20:36

user10478

asked Mar 23 at 18:35

user10478user10478

497413

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:36

user10478

asked Mar 23 at 18:35

user10478user10478

497413

share|cite|improve this question

edited Mar 24 at 20:36

user10478

asked Mar 23 at 18:35

user10478user10478

497413

asked Mar 23 at 18:35

user10478user10478

497413

asked Mar 23 at 18:35

user10478user10478

497413