Using polar coordinates to find the critical points












2












$begingroup$


I have written the following DE system
$$dot{x} = x+y-x(x^2+y^2)\
dot{y} = -x+3y-y(x^2+y^2) $$
as follows in Polar form:
$$dot{r} = -2rcos^2theta+r(3-r^2) $$
$$dot{theta} = 2sinthetacostheta-1$$
What I wonder is could I also find all the critical points of this system by setting $dot{theta}$ and $dot{r}$ equal to zero? Since these two systems are actually equivalent to each other this seemed like a logical thing to me and at the same time finding the critical points through setting $dot{x}$ and $dot{y}$ equal to zero seems like much more tedious in this state.



Solving for $dot{theta}=0$ leads to $2sinthetacostheta = 1$ which is the same as $sin2theta=1$ leading to $theta = frac{pi}{4}$ or $theta = -frac{3pi}{4}$. Substituting these within $dot{r}$ leads to:
$$-2rcos^2theta+r(3-r^2) = -2r(pmfrac{sqrt{2}}{2})^2 +r(3-r^2)=r(2-r^2)$$
So $dot{r}=0$ if $r=0$ or $r^2=2$. This leads us to the following critical points: $(0,0)$, $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.
However what bugs me is that entering these points into the non-polar system does not lead to $dot{x}=dot{y}=0$ for $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.



What am I missing here? Is the whole thought of trying to find critical points through putting $dot{theta}$ and $dot{r}$ equal to zero unwise? Any help or hint with this will be much appreciated.






ordinary-differential-equations systems-of-equations polar-coordinates







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$endgroup$












  • $begingroup$
    As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
    $endgroup$
    – John Omielan
    Jan 4 at 5:12
















2












$begingroup$


I have written the following DE system
$$dot{x} = x+y-x(x^2+y^2)\
dot{y} = -x+3y-y(x^2+y^2) $$
as follows in Polar form:
$$dot{r} = -2rcos^2theta+r(3-r^2) $$
$$dot{theta} = 2sinthetacostheta-1$$
What I wonder is could I also find all the critical points of this system by setting $dot{theta}$ and $dot{r}$ equal to zero? Since these two systems are actually equivalent to each other this seemed like a logical thing to me and at the same time finding the critical points through setting $dot{x}$ and $dot{y}$ equal to zero seems like much more tedious in this state.



Solving for $dot{theta}=0$ leads to $2sinthetacostheta = 1$ which is the same as $sin2theta=1$ leading to $theta = frac{pi}{4}$ or $theta = -frac{3pi}{4}$. Substituting these within $dot{r}$ leads to:
$$-2rcos^2theta+r(3-r^2) = -2r(pmfrac{sqrt{2}}{2})^2 +r(3-r^2)=r(2-r^2)$$
So $dot{r}=0$ if $r=0$ or $r^2=2$. This leads us to the following critical points: $(0,0)$, $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.
However what bugs me is that entering these points into the non-polar system does not lead to $dot{x}=dot{y}=0$ for $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.



What am I missing here? Is the whole thought of trying to find critical points through putting $dot{theta}$ and $dot{r}$ equal to zero unwise? Any help or hint with this will be much appreciated.






ordinary-differential-equations systems-of-equations polar-coordinates







share|cite|improve this question









$endgroup$












  • $begingroup$
    As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
    $endgroup$
    – John Omielan
    Jan 4 at 5:12














2












2








2





$begingroup$


I have written the following DE system
$$dot{x} = x+y-x(x^2+y^2)\
dot{y} = -x+3y-y(x^2+y^2) $$
as follows in Polar form:
$$dot{r} = -2rcos^2theta+r(3-r^2) $$
$$dot{theta} = 2sinthetacostheta-1$$
What I wonder is could I also find all the critical points of this system by setting $dot{theta}$ and $dot{r}$ equal to zero? Since these two systems are actually equivalent to each other this seemed like a logical thing to me and at the same time finding the critical points through setting $dot{x}$ and $dot{y}$ equal to zero seems like much more tedious in this state.



Solving for $dot{theta}=0$ leads to $2sinthetacostheta = 1$ which is the same as $sin2theta=1$ leading to $theta = frac{pi}{4}$ or $theta = -frac{3pi}{4}$. Substituting these within $dot{r}$ leads to:
$$-2rcos^2theta+r(3-r^2) = -2r(pmfrac{sqrt{2}}{2})^2 +r(3-r^2)=r(2-r^2)$$
So $dot{r}=0$ if $r=0$ or $r^2=2$. This leads us to the following critical points: $(0,0)$, $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.
However what bugs me is that entering these points into the non-polar system does not lead to $dot{x}=dot{y}=0$ for $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.



What am I missing here? Is the whole thought of trying to find critical points through putting $dot{theta}$ and $dot{r}$ equal to zero unwise? Any help or hint with this will be much appreciated.






ordinary-differential-equations systems-of-equations polar-coordinates







share|cite|improve this question









$endgroup$




I have written the following DE system
$$dot{x} = x+y-x(x^2+y^2)\
dot{y} = -x+3y-y(x^2+y^2) $$
as follows in Polar form:
$$dot{r} = -2rcos^2theta+r(3-r^2) $$
$$dot{theta} = 2sinthetacostheta-1$$
What I wonder is could I also find all the critical points of this system by setting $dot{theta}$ and $dot{r}$ equal to zero? Since these two systems are actually equivalent to each other this seemed like a logical thing to me and at the same time finding the critical points through setting $dot{x}$ and $dot{y}$ equal to zero seems like much more tedious in this state.



Solving for $dot{theta}=0$ leads to $2sinthetacostheta = 1$ which is the same as $sin2theta=1$ leading to $theta = frac{pi}{4}$ or $theta = -frac{3pi}{4}$. Substituting these within $dot{r}$ leads to:
$$-2rcos^2theta+r(3-r^2) = -2r(pmfrac{sqrt{2}}{2})^2 +r(3-r^2)=r(2-r^2)$$
So $dot{r}=0$ if $r=0$ or $r^2=2$. This leads us to the following critical points: $(0,0)$, $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.
However what bugs me is that entering these points into the non-polar system does not lead to $dot{x}=dot{y}=0$ for $(sqrt{2},sqrt{2})$, $(-sqrt{2},-sqrt{2})$.



What am I missing here? Is the whole thought of trying to find critical points through putting $dot{theta}$ and $dot{r}$ equal to zero unwise? Any help or hint with this will be much appreciated.






ordinary-differential-equations systems-of-equations polar-coordinates




ordinary-differential-equations systems-of-equations polar-coordinates






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asked Jan 3 at 21:04









BluedogBluedog

257




257












  • $begingroup$
    As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
    $endgroup$
    – John Omielan
    Jan 4 at 5:12


















  • $begingroup$
    As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
    $endgroup$
    – John Omielan
    Jan 4 at 5:12
















$begingroup$
As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
$endgroup$
– John Omielan
Jan 4 at 5:12




$begingroup$
As I state in my answer, I'm sorry I didn't answer what you were asking. Thus, I suggest you change your accepted answer to a different one so I may then delete my answer. Thanks.
$endgroup$
– John Omielan
Jan 4 at 5:12










2 Answers
2






active

oldest

votes


















4












$begingroup$

If $r=sqrt 2$ and $theta=pi/4$ you get $x=y=sqrt 2frac 1{sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    It is not that difficult to proceed from $dot x=0=dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then
    $$
    0=xy+y^2+x^2-3xy=(x-y)^2
    $$
    has solutions only for $x=y$. Insert that in either equation to get
    $$
    0=2x -x(x^2+x^2)=2x(1-x^2)
    $$
    to conclude that the values taken are $x=yin{-1,0,1}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
      $endgroup$
      – Bluedog
      Jan 4 at 6:04










    • $begingroup$
      @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
      $endgroup$
      – LutzL
      Jan 4 at 9:03













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    2 Answers
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    2 Answers
    2






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    4












    $begingroup$

    If $r=sqrt 2$ and $theta=pi/4$ you get $x=y=sqrt 2frac 1{sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      If $r=sqrt 2$ and $theta=pi/4$ you get $x=y=sqrt 2frac 1{sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        If $r=sqrt 2$ and $theta=pi/4$ you get $x=y=sqrt 2frac 1{sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.






        share|cite|improve this answer











        $endgroup$



        If $r=sqrt 2$ and $theta=pi/4$ you get $x=y=sqrt 2frac 1{sqrt2}=1$. Then $(1,1)$ is a critical point. Similarly, you can check the other solution $(-1,-1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 21:46

























        answered Jan 3 at 21:39









        AndreiAndrei

        11.8k21026




        11.8k21026























            3












            $begingroup$

            It is not that difficult to proceed from $dot x=0=dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then
            $$
            0=xy+y^2+x^2-3xy=(x-y)^2
            $$
            has solutions only for $x=y$. Insert that in either equation to get
            $$
            0=2x -x(x^2+x^2)=2x(1-x^2)
            $$
            to conclude that the values taken are $x=yin{-1,0,1}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
              $endgroup$
              – Bluedog
              Jan 4 at 6:04










            • $begingroup$
              @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
              $endgroup$
              – LutzL
              Jan 4 at 9:03


















            3












            $begingroup$

            It is not that difficult to proceed from $dot x=0=dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then
            $$
            0=xy+y^2+x^2-3xy=(x-y)^2
            $$
            has solutions only for $x=y$. Insert that in either equation to get
            $$
            0=2x -x(x^2+x^2)=2x(1-x^2)
            $$
            to conclude that the values taken are $x=yin{-1,0,1}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
              $endgroup$
              – Bluedog
              Jan 4 at 6:04










            • $begingroup$
              @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
              $endgroup$
              – LutzL
              Jan 4 at 9:03
















            3












            3








            3





            $begingroup$

            It is not that difficult to proceed from $dot x=0=dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then
            $$
            0=xy+y^2+x^2-3xy=(x-y)^2
            $$
            has solutions only for $x=y$. Insert that in either equation to get
            $$
            0=2x -x(x^2+x^2)=2x(1-x^2)
            $$
            to conclude that the values taken are $x=yin{-1,0,1}$.






            share|cite|improve this answer









            $endgroup$



            It is not that difficult to proceed from $dot x=0=dot y$. Multiply the first with $y$, the second equation with $x$ and subtract. Then
            $$
            0=xy+y^2+x^2-3xy=(x-y)^2
            $$
            has solutions only for $x=y$. Insert that in either equation to get
            $$
            0=2x -x(x^2+x^2)=2x(1-x^2)
            $$
            to conclude that the values taken are $x=yin{-1,0,1}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 21:48









            LutzLLutzL

            57.7k42054




            57.7k42054












            • $begingroup$
              I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
              $endgroup$
              – Bluedog
              Jan 4 at 6:04










            • $begingroup$
              @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
              $endgroup$
              – LutzL
              Jan 4 at 9:03




















            • $begingroup$
              I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
              $endgroup$
              – Bluedog
              Jan 4 at 6:04










            • $begingroup$
              @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
              $endgroup$
              – LutzL
              Jan 4 at 9:03


















            $begingroup$
            I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
            $endgroup$
            – Bluedog
            Jan 4 at 6:04




            $begingroup$
            I proceded to find the same points in your answer by inserting $frac{x+y}{x}=x^2+y^2$ (rewritten from $dot{x}=0$) into $dot{y}$. I thought that for two equations, one cannot simply combine them into one and say the original system is equivalent to the result and that one must keep at least one of the original two equations. So I tried going the "find a solution in one equation, insert in the other one and so on" and it was a gruesome process. But from your answer I am to understand that combining the equations in this situation is fine?
            $endgroup$
            – Bluedog
            Jan 4 at 6:04












            $begingroup$
            @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
            $endgroup$
            – LutzL
            Jan 4 at 9:03






            $begingroup$
            @Bluedog : This is effectively the same as my first step. Yes, any combination of equations is again an equation that must be satisfied, in the algebra this is the idea of ideals generated by some polynomials whose roots one is seeking. You must only be careful in declaring a previous equation no longer needed, that is, a new reduced set of equations equivalent to the original one. But the set of all obtained and original equations is obviously equivalent to the original set.
            $endgroup$
            – LutzL
            Jan 4 at 9:03




















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