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Error in TransformedField

Using spherical derivatives with NDSolveIs there an easy way to transform unit vectors from spherical to Cartesian coordinates?Why does a transform from polar coordinates using `TransformedField` not agree with by-hand calculation?ListDensityPlot in Polar coordinates with a higher efficiencyChanging from Cartesian to polar coordinates in partial differential equationA little confused about coordinate conversionInverse substitution polar-cartesianIntegration using polar coordinatesPlotting a function $psi(rho,theta,phi)$ in spherical coordinatesTransform a field with derivatives

6

1

$begingroup$

I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:

TransformedField[

  "Cartesian" -> "Polar",

  {μ x1 - x2 -  σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)}, 

  {x1, x2} -> {r, θ}

] // Simplify

and I get the result

{r μ - r^3 σ, r}

but I am pretty sure that the right answer should be

{r μ - r^3 σ, 1}

Where is the error?

coordinate-transformation

share|improve this question

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

$endgroup$

add a comment | 

6

1

$begingroup$

I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:

TransformedField[

  "Cartesian" -> "Polar",

  {μ x1 - x2 -  σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)}, 

  {x1, x2} -> {r, θ}

] // Simplify

and I get the result

{r μ - r^3 σ, r}

but I am pretty sure that the right answer should be

{r μ - r^3 σ, 1}

Where is the error?

coordinate-transformation

share|improve this question

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

$endgroup$

add a comment | 

$begingroup$

I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:

TransformedField[

  "Cartesian" -> "Polar",

  {μ x1 - x2 -  σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)}, 

  {x1, x2} -> {r, θ}

] // Simplify

and I get the result

{r μ - r^3 σ, r}

but I am pretty sure that the right answer should be

{r μ - r^3 σ, 1}

Where is the error?

coordinate-transformation

share|improve this question

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

$endgroup$

TransformedField[

  "Cartesian" -> "Polar",

  {μ x1 - x2 -  σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)}, 

  {x1, x2} -> {r, θ}

] // Simplify

{r μ - r^3 σ, r}

{r μ - r^3 σ, 1}

share|improve this question

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

share|improve this question

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

edited 4 hours ago

MarcoB

36.9k556113

edited 4 hours ago

MarcoB

36.9k556113

asked 4 hours ago

rparpa

856

asked 4 hours ago

rparpa

856

1 Answer
1

active

oldest

votes

5

$begingroup$

We can test the built-in TransformedField by defining our own functions.

From $x',y'$ to $r',theta'$, we derive:
$$
r' = left(sqrt{x^2 +y^2} right)'
= frac{(x^2 +y^2)'}{2
sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
$$
and
$$
theta' = left(arctan frac{y}{x} right)'
= frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
$$

First, we define

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

We now make the substitution and simplify

rdot[r Cos[t], r Sin[t]] // FullSimplify

This yields (matches Mathematica)

$$r' = mu r-r^3 sigma$$

We now do the same for the other

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

We now make the substitution and simplify

thetadot[r Cos[t], r Sin[t]] // FullSimplify

This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)

$$theta'= 1$$

I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!

share|improve this answer

edited 1 hour ago

answered 2 hours ago

MooMoo

7511515

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you reported it to the Wolfram tech support?
  $endgroup$
  – Alexey Popkov
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
  $endgroup$
  – Moo
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
  $endgroup$
  – Alexey Popkov
  7 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I sent them an email per your suggestion.
  $endgroup$
  – Moo
  1 min ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

We can test the built-in TransformedField by defining our own functions.

From $x',y'$ to $r',theta'$, we derive:
$$
r' = left(sqrt{x^2 +y^2} right)'
= frac{(x^2 +y^2)'}{2
sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
$$
and
$$
theta' = left(arctan frac{y}{x} right)'
= frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
$$

First, we define

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

We now make the substitution and simplify

rdot[r Cos[t], r Sin[t]] // FullSimplify

This yields (matches Mathematica)

$$r' = mu r-r^3 sigma$$

We now do the same for the other

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

We now make the substitution and simplify

thetadot[r Cos[t], r Sin[t]] // FullSimplify

This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)

$$theta'= 1$$

I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!

share|improve this answer

edited 1 hour ago

answered 2 hours ago

MooMoo

7511515

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you reported it to the Wolfram tech support?
  $endgroup$
  – Alexey Popkov
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
  $endgroup$
  – Moo
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
  $endgroup$
  – Alexey Popkov
  7 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I sent them an email per your suggestion.
  $endgroup$
  – Moo
  1 min ago
  

  
  
  
  

add a comment | 

5

$begingroup$

We can test the built-in TransformedField by defining our own functions.

From $x',y'$ to $r',theta'$, we derive:
$$
r' = left(sqrt{x^2 +y^2} right)'
= frac{(x^2 +y^2)'}{2
sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
$$
and
$$
theta' = left(arctan frac{y}{x} right)'
= frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
$$

First, we define

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

We now make the substitution and simplify

rdot[r Cos[t], r Sin[t]] // FullSimplify

This yields (matches Mathematica)

$$r' = mu r-r^3 sigma$$

We now do the same for the other

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

We now make the substitution and simplify

thetadot[r Cos[t], r Sin[t]] // FullSimplify

This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)

$$theta'= 1$$

I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!

share|improve this answer

edited 1 hour ago

answered 2 hours ago

MooMoo

7511515

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you reported it to the Wolfram tech support?
  $endgroup$
  – Alexey Popkov
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I have not. I have had many issues with it when transforming between different methods. These days, I don't trust it and just create my own transformation rules to do it.
  $endgroup$
  – Moo
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is worth to write them about it in order to get it finally fixed. You can even write a short letter to support@wolfram.com with a link to this post.
  $endgroup$
  – Alexey Popkov
  7 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexeyPopkov: I sent them an email per your suggestion.
  $endgroup$
  – Moo
  1 min ago
  

  
  
  
  

add a comment | 

$begingroup$

We can test the built-in TransformedField by defining our own functions.

From $x',y'$ to $r',theta'$, we derive:
$$
r' = left(sqrt{x^2 +y^2} right)'
= frac{(x^2 +y^2)'}{2
sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
$$
and
$$
theta' = left(arctan frac{y}{x} right)'
= frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
$$

First, we define

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

We now make the substitution and simplify

rdot[r Cos[t], r Sin[t]] // FullSimplify

This yields (matches Mathematica)

$$r' = mu r-r^3 sigma$$

We now do the same for the other

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

We now make the substitution and simplify

thetadot[r Cos[t], r Sin[t]] // FullSimplify

This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)

$$theta'= 1$$

I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!

share|improve this answer

edited 1 hour ago

answered 2 hours ago

MooMoo

7511515

$endgroup$

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

rdot[r Cos[t], r Sin[t]] // FullSimplify

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

thetadot[r Cos[t], r Sin[t]] // FullSimplify

share|improve this answer

edited 1 hour ago

answered 2 hours ago

MooMoo

7511515

answered 2 hours ago

MooMoo

7511515

answered 2 hours ago

MooMoo

7511515