Error in TransformedField












6












$begingroup$


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











$endgroup$

















    6












    $begingroup$


    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











    $endgroup$















      6












      6








      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











      $endgroup$




      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




      coordinate-transformation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 4 hours ago









      MarcoB

      36.9k556113




      36.9k556113










      asked 4 hours ago









      rparpa

      856




      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











          $endgroup$









          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            12 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
            9 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
            6 mins ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            17 secs ago











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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











          $endgroup$









          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            12 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
            9 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
            6 mins ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            17 secs ago
















          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











          $endgroup$









          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            12 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
            9 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
            6 mins ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            17 secs ago














          5












          5








          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











          $endgroup$



          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














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          MooMoo

          7511515




          7511515








          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            12 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
            9 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
            6 mins ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            17 secs ago














          • 1




            $begingroup$
            Have you reported it to the Wolfram tech support?
            $endgroup$
            – Alexey Popkov
            12 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
            9 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
            6 mins ago












          • $begingroup$
            @AlexeyPopkov: I sent them an email per your suggestion.
            $endgroup$
            – Moo
            17 secs ago








          1




          1




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




          $begingroup$
          Have you reported it to the Wolfram tech support?
          $endgroup$
          – Alexey Popkov
          12 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
          9 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
          9 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
          6 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
          6 mins ago














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




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


















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