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







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












$begingroup$


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






share|cite|improve this question













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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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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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