Change to polar coordinate in ODEs

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Consider a planar ODE (e.g., $dotx=f_1(x,y),doty=f_2(x,y)$). According to page $68$ of ODE with Applications by Chicone, a change to polar coordinates (i.e., $x=rcos(theta),y=rsin(theta)$) in this system introduces a singularity on the line $(r,theta): r=0 $. Furthermore, if the original ODE has a rest point at the origin, this singularity is removable. Why is this true? What is the intuition? How can we prove this result?
Thank you.

real-analysis differential-equations differential-geometry polar-coordinates control-theory

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asked Jul 21 at 20:05

Arthur

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Consider a planar ODE (e.g., $dotx=f_1(x,y),doty=f_2(x,y)$). According to page $68$ of ODE with Applications by Chicone, a change to polar coordinates (i.e., $x=rcos(theta),y=rsin(theta)$) in this system introduces a singularity on the line $(r,theta): r=0 $. Furthermore, if the original ODE has a rest point at the origin, this singularity is removable. Why is this true? What is the intuition? How can we prove this result?
Thank you.

real-analysis differential-equations differential-geometry polar-coordinates control-theory

share|cite|improve this question

asked Jul 21 at 20:05

Arthur

19812

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Consider a planar ODE (e.g., $dotx=f_1(x,y),doty=f_2(x,y)$). According to page $68$ of ODE with Applications by Chicone, a change to polar coordinates (i.e., $x=rcos(theta),y=rsin(theta)$) in this system introduces a singularity on the line $(r,theta): r=0 $. Furthermore, if the original ODE has a rest point at the origin, this singularity is removable. Why is this true? What is the intuition? How can we prove this result?
Thank you.

real-analysis differential-equations differential-geometry polar-coordinates control-theory

share|cite|improve this question

asked Jul 21 at 20:05

Arthur

19812

Consider a planar ODE (e.g., $dotx=f_1(x,y),doty=f_2(x,y)$). According to page $68$ of ODE with Applications by Chicone, a change to polar coordinates (i.e., $x=rcos(theta),y=rsin(theta)$) in this system introduces a singularity on the line $(r,theta): r=0 $. Furthermore, if the original ODE has a rest point at the origin, this singularity is removable. Why is this true? What is the intuition? How can we prove this result?
Thank you.

real-analysis differential-equations differential-geometry polar-coordinates control-theory

share|cite|improve this question

asked Jul 21 at 20:05

Arthur

19812

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 20:05

Arthur

19812

asked Jul 21 at 20:05

Arthur

19812

asked Jul 21 at 20:05

Arthur

19812

19812

add a comment | 

add a comment | 

1 Answer
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From the equations $x= rcostheta$ and $y = rsin theta$ we see that the derivatives of $r$ and $theta$ are related to the derivatives of $x$ and $y$ by the equations
begineqnarray*
dot x & = & dot r cos theta - dot theta rsin theta\
dot y & = & dot r sin theta + dot theta rcos theta.
endeqnarray*
Thus, using the notation $g_i(r, theta) = f_i(rcos theta, rsintheta)$ (i = 1, 2), the given system of ODEs reads
begineqnarray*
dot r cos theta - dot theta rsin theta & = & g_1(r, theta) \
dot r sin theta + dot theta rcostheta & = & g_2(r, theta) .
endeqnarray*
Solving this for $dot r $ and $dot theta$ gives
begineqnarray*
dot r & = & g_1(r, theta)cos theta + g_2(r, theta) sin theta\
dot theta & = & fracg_2(r, theta)rcostheta - fracg_1(r, theta)r sin theta.
endeqnarray*
Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,theta) = (0,0)$. We have
begineqnarray*
g_i(r, theta)
& = & g_i(0,0) + partial_rg_i(0,0)r + partial_theta g_i(0,0)theta + rm HOT’s\
& = & f_i(0,0) + left(partial_x f_i(0,0) costheta + partial_y f_i(0,0) sinthetaright)r + left(- partial_x f_i(0,0) rsin theta + partial_y f_i(0,0) rcosthetaright)theta + rm HOT’s\
& = &
f_i (0,0) + rbig[partial_x f_i(0,0) costheta + partial_y f_i(0,0) sintheta + left(- partial_x f_i(0,0)sin theta + partial_y f_i(0,0)costhetaright)thetabig] + rm HOT’s
endeqnarray*
From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form
beginequation*
g_i(r, theta) = rh_i(r, theta)
endequation*
for some smooth function $h_i$. In particular, $fracg_i(r, theta)r$ causes no trouble.

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edited Jul 22 at 16:12

Arthur

19812

answered Jul 22 at 13:26

BindersFull

498110

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  Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
  – Arthur
  Jul 22 at 16:15
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

From the equations $x= rcostheta$ and $y = rsin theta$ we see that the derivatives of $r$ and $theta$ are related to the derivatives of $x$ and $y$ by the equations
begineqnarray*
dot x & = & dot r cos theta - dot theta rsin theta\
dot y & = & dot r sin theta + dot theta rcos theta.
endeqnarray*
Thus, using the notation $g_i(r, theta) = f_i(rcos theta, rsintheta)$ (i = 1, 2), the given system of ODEs reads
begineqnarray*
dot r cos theta - dot theta rsin theta & = & g_1(r, theta) \
dot r sin theta + dot theta rcostheta & = & g_2(r, theta) .
endeqnarray*
Solving this for $dot r $ and $dot theta$ gives
begineqnarray*
dot r & = & g_1(r, theta)cos theta + g_2(r, theta) sin theta\
dot theta & = & fracg_2(r, theta)rcostheta - fracg_1(r, theta)r sin theta.
endeqnarray*
Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,theta) = (0,0)$. We have
begineqnarray*
g_i(r, theta)
& = & g_i(0,0) + partial_rg_i(0,0)r + partial_theta g_i(0,0)theta + rm HOT’s\
& = & f_i(0,0) + left(partial_x f_i(0,0) costheta + partial_y f_i(0,0) sinthetaright)r + left(- partial_x f_i(0,0) rsin theta + partial_y f_i(0,0) rcosthetaright)theta + rm HOT’s\
& = &
f_i (0,0) + rbig[partial_x f_i(0,0) costheta + partial_y f_i(0,0) sintheta + left(- partial_x f_i(0,0)sin theta + partial_y f_i(0,0)costhetaright)thetabig] + rm HOT’s
endeqnarray*
From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form
beginequation*
g_i(r, theta) = rh_i(r, theta)
endequation*
for some smooth function $h_i$. In particular, $fracg_i(r, theta)r$ causes no trouble.

share|cite|improve this answer

edited Jul 22 at 16:12

Arthur

19812

answered Jul 22 at 13:26

BindersFull

498110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
  – Arthur
  Jul 22 at 16:15
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

From the equations $x= rcostheta$ and $y = rsin theta$ we see that the derivatives of $r$ and $theta$ are related to the derivatives of $x$ and $y$ by the equations
begineqnarray*
dot x & = & dot r cos theta - dot theta rsin theta\
dot y & = & dot r sin theta + dot theta rcos theta.
endeqnarray*
Thus, using the notation $g_i(r, theta) = f_i(rcos theta, rsintheta)$ (i = 1, 2), the given system of ODEs reads
begineqnarray*
dot r cos theta - dot theta rsin theta & = & g_1(r, theta) \
dot r sin theta + dot theta rcostheta & = & g_2(r, theta) .
endeqnarray*
Solving this for $dot r $ and $dot theta$ gives
begineqnarray*
dot r & = & g_1(r, theta)cos theta + g_2(r, theta) sin theta\
dot theta & = & fracg_2(r, theta)rcostheta - fracg_1(r, theta)r sin theta.
endeqnarray*
Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,theta) = (0,0)$. We have
begineqnarray*
g_i(r, theta)
& = & g_i(0,0) + partial_rg_i(0,0)r + partial_theta g_i(0,0)theta + rm HOT’s\
& = & f_i(0,0) + left(partial_x f_i(0,0) costheta + partial_y f_i(0,0) sinthetaright)r + left(- partial_x f_i(0,0) rsin theta + partial_y f_i(0,0) rcosthetaright)theta + rm HOT’s\
& = &
f_i (0,0) + rbig[partial_x f_i(0,0) costheta + partial_y f_i(0,0) sintheta + left(- partial_x f_i(0,0)sin theta + partial_y f_i(0,0)costhetaright)thetabig] + rm HOT’s
endeqnarray*
From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form
beginequation*
g_i(r, theta) = rh_i(r, theta)
endequation*
for some smooth function $h_i$. In particular, $fracg_i(r, theta)r$ causes no trouble.

share|cite|improve this answer

edited Jul 22 at 16:12

Arthur

19812

answered Jul 22 at 13:26

BindersFull

498110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
  – Arthur
  Jul 22 at 16:15
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

From the equations $x= rcostheta$ and $y = rsin theta$ we see that the derivatives of $r$ and $theta$ are related to the derivatives of $x$ and $y$ by the equations
begineqnarray*
dot x & = & dot r cos theta - dot theta rsin theta\
dot y & = & dot r sin theta + dot theta rcos theta.
endeqnarray*
Thus, using the notation $g_i(r, theta) = f_i(rcos theta, rsintheta)$ (i = 1, 2), the given system of ODEs reads
begineqnarray*
dot r cos theta - dot theta rsin theta & = & g_1(r, theta) \
dot r sin theta + dot theta rcostheta & = & g_2(r, theta) .
endeqnarray*
Solving this for $dot r $ and $dot theta$ gives
begineqnarray*
dot r & = & g_1(r, theta)cos theta + g_2(r, theta) sin theta\
dot theta & = & fracg_2(r, theta)rcostheta - fracg_1(r, theta)r sin theta.
endeqnarray*
Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,theta) = (0,0)$. We have
begineqnarray*
g_i(r, theta)
& = & g_i(0,0) + partial_rg_i(0,0)r + partial_theta g_i(0,0)theta + rm HOT’s\
& = & f_i(0,0) + left(partial_x f_i(0,0) costheta + partial_y f_i(0,0) sinthetaright)r + left(- partial_x f_i(0,0) rsin theta + partial_y f_i(0,0) rcosthetaright)theta + rm HOT’s\
& = &
f_i (0,0) + rbig[partial_x f_i(0,0) costheta + partial_y f_i(0,0) sintheta + left(- partial_x f_i(0,0)sin theta + partial_y f_i(0,0)costhetaright)thetabig] + rm HOT’s
endeqnarray*
From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form
beginequation*
g_i(r, theta) = rh_i(r, theta)
endequation*
for some smooth function $h_i$. In particular, $fracg_i(r, theta)r$ causes no trouble.

share|cite|improve this answer

edited Jul 22 at 16:12

Arthur

19812

answered Jul 22 at 13:26

BindersFull

498110

From the equations $x= rcostheta$ and $y = rsin theta$ we see that the derivatives of $r$ and $theta$ are related to the derivatives of $x$ and $y$ by the equations
begineqnarray*
dot x & = & dot r cos theta - dot theta rsin theta\
dot y & = & dot r sin theta + dot theta rcos theta.
endeqnarray*
Thus, using the notation $g_i(r, theta) = f_i(rcos theta, rsintheta)$ (i = 1, 2), the given system of ODEs reads
begineqnarray*
dot r cos theta - dot theta rsin theta & = & g_1(r, theta) \
dot r sin theta + dot theta rcostheta & = & g_2(r, theta) .
endeqnarray*
Solving this for $dot r $ and $dot theta$ gives
begineqnarray*
dot r & = & g_1(r, theta)cos theta + g_2(r, theta) sin theta\
dot theta & = & fracg_2(r, theta)rcostheta - fracg_1(r, theta)r sin theta.
endeqnarray*
Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,theta) = (0,0)$. We have
begineqnarray*
g_i(r, theta)
& = & g_i(0,0) + partial_rg_i(0,0)r + partial_theta g_i(0,0)theta + rm HOT’s\
& = & f_i(0,0) + left(partial_x f_i(0,0) costheta + partial_y f_i(0,0) sinthetaright)r + left(- partial_x f_i(0,0) rsin theta + partial_y f_i(0,0) rcosthetaright)theta + rm HOT’s\
& = &
f_i (0,0) + rbig[partial_x f_i(0,0) costheta + partial_y f_i(0,0) sintheta + left(- partial_x f_i(0,0)sin theta + partial_y f_i(0,0)costhetaright)thetabig] + rm HOT’s
endeqnarray*
From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form
beginequation*
g_i(r, theta) = rh_i(r, theta)
endequation*
for some smooth function $h_i$. In particular, $fracg_i(r, theta)r$ causes no trouble.

share|cite|improve this answer

edited Jul 22 at 16:12

Arthur

19812

answered Jul 22 at 13:26

BindersFull

498110

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 16:12

Arthur

19812

edited Jul 22 at 16:12

Arthur

19812

edited Jul 22 at 16:12

Arthur

19812

19812

answered Jul 22 at 13:26

BindersFull

498110

answered Jul 22 at 13:26

BindersFull

498110

answered Jul 22 at 13:26

BindersFull

498110

498110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
  – Arthur
  Jul 22 at 16:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
  – Arthur
  Jul 22 at 16:15
  

  
  
  
  

Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
– Arthur
Jul 22 at 16:15

Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^r-1$.
– Arthur
Jul 22 at 16:15

add a comment | 

 

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