Finding approximation to stable manifold of saddle point












2












$begingroup$


I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).




Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
has a single fixed point, $(-1, 0)$. This is a saddle point. The
unstable manifold is $y=0$, but the stable manifold is some non-linear
curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
expressions for $dy/du$ and equate them.




I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.



I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
enter image description here



Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.



Basically, I'm a bit lost, could someone perhaps give me some hints?






ordinary-differential-equations taylor-expansion dynamical-systems







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).




    Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
    has a single fixed point, $(-1, 0)$. This is a saddle point. The
    unstable manifold is $y=0$, but the stable manifold is some non-linear
    curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
    > 0)$     and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
    expressions for $dy/du$ and equate them.




    I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.



    I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
    enter image description here



    Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.



    Basically, I'm a bit lost, could someone perhaps give me some hints?






    ordinary-differential-equations taylor-expansion dynamical-systems







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).




      Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
      has a single fixed point, $(-1, 0)$. This is a saddle point. The
      unstable manifold is $y=0$, but the stable manifold is some non-linear
      curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
      > 0)$       and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
      expressions for $dy/du$ and equate them.




      I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.



      I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
      enter image description here



      Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.



      Basically, I'm a bit lost, could someone perhaps give me some hints?






      ordinary-differential-equations taylor-expansion dynamical-systems







      share|cite|improve this question









      $endgroup$




      I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).




      Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
      has a single fixed point, $(-1, 0)$. This is a saddle point. The
      unstable manifold is $y=0$, but the stable manifold is some non-linear
      curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
      > 0)$       and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
      expressions for $dy/du$ and equate them.




      I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.



      I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
      enter image description here



      Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.



      Basically, I'm a bit lost, could someone perhaps give me some hints?






      ordinary-differential-equations taylor-expansion dynamical-systems




      ordinary-differential-equations taylor-expansion dynamical-systems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 21:44









      DashermanDasherman

      1,152818




      1,152818






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Remember that



          $$
          frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
          $$



          If you replace



          $$
          frac{{rm d}y}{{rm d}t} = -y tag{2}
          $$



          and



          $$
          frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
          $$



          In (1) you will get



          $$
          frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
          $$



          Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is



          begin{eqnarray}
          frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
          &=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
          end{eqnarray}



          You solve for coefficients, and you should get



          $$
          a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
          $$



          The black line below is the result



          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:10








          • 1




            $begingroup$
            @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:17










          • $begingroup$
            Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:27






          • 1




            $begingroup$
            @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:35






          • 1




            $begingroup$
            @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
            $endgroup$
            – caverac
            Dec 20 '18 at 13:26











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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









          2












          $begingroup$

          Remember that



          $$
          frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
          $$



          If you replace



          $$
          frac{{rm d}y}{{rm d}t} = -y tag{2}
          $$



          and



          $$
          frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
          $$



          In (1) you will get



          $$
          frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
          $$



          Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is



          begin{eqnarray}
          frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
          &=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
          end{eqnarray}



          You solve for coefficients, and you should get



          $$
          a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
          $$



          The black line below is the result



          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:10








          • 1




            $begingroup$
            @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:17










          • $begingroup$
            Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:27






          • 1




            $begingroup$
            @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:35






          • 1




            $begingroup$
            @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
            $endgroup$
            – caverac
            Dec 20 '18 at 13:26
















          2












          $begingroup$

          Remember that



          $$
          frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
          $$



          If you replace



          $$
          frac{{rm d}y}{{rm d}t} = -y tag{2}
          $$



          and



          $$
          frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
          $$



          In (1) you will get



          $$
          frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
          $$



          Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is



          begin{eqnarray}
          frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
          &=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
          end{eqnarray}



          You solve for coefficients, and you should get



          $$
          a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
          $$



          The black line below is the result



          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:10








          • 1




            $begingroup$
            @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:17










          • $begingroup$
            Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:27






          • 1




            $begingroup$
            @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:35






          • 1




            $begingroup$
            @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
            $endgroup$
            – caverac
            Dec 20 '18 at 13:26














          2












          2








          2





          $begingroup$

          Remember that



          $$
          frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
          $$



          If you replace



          $$
          frac{{rm d}y}{{rm d}t} = -y tag{2}
          $$



          and



          $$
          frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
          $$



          In (1) you will get



          $$
          frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
          $$



          Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is



          begin{eqnarray}
          frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
          &=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
          end{eqnarray}



          You solve for coefficients, and you should get



          $$
          a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
          $$



          The black line below is the result



          enter image description here






          share|cite|improve this answer









          $endgroup$



          Remember that



          $$
          frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
          $$



          If you replace



          $$
          frac{{rm d}y}{{rm d}t} = -y tag{2}
          $$



          and



          $$
          frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
          $$



          In (1) you will get



          $$
          frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
          $$



          Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is



          begin{eqnarray}
          frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
          &=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
          end{eqnarray}



          You solve for coefficients, and you should get



          $$
          a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
          $$



          The black line below is the result



          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 22:29









          caveraccaverac

          14.8k31130




          14.8k31130












          • $begingroup$
            Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:10








          • 1




            $begingroup$
            @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:17










          • $begingroup$
            Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:27






          • 1




            $begingroup$
            @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:35






          • 1




            $begingroup$
            @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
            $endgroup$
            – caverac
            Dec 20 '18 at 13:26


















          • $begingroup$
            Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:10








          • 1




            $begingroup$
            @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:17










          • $begingroup$
            Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
            $endgroup$
            – Dasherman
            Dec 16 '18 at 14:27






          • 1




            $begingroup$
            @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
            $endgroup$
            – caverac
            Dec 16 '18 at 14:35






          • 1




            $begingroup$
            @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
            $endgroup$
            – caverac
            Dec 20 '18 at 13:26
















          $begingroup$
          Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
          $endgroup$
          – Dasherman
          Dec 16 '18 at 14:10






          $begingroup$
          Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
          $endgroup$
          – Dasherman
          Dec 16 '18 at 14:10






          1




          1




          $begingroup$
          @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
          $endgroup$
          – caverac
          Dec 16 '18 at 14:17




          $begingroup$
          @Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
          $endgroup$
          – caverac
          Dec 16 '18 at 14:17












          $begingroup$
          Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
          $endgroup$
          – Dasherman
          Dec 16 '18 at 14:27




          $begingroup$
          Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
          $endgroup$
          – Dasherman
          Dec 16 '18 at 14:27




          1




          1




          $begingroup$
          @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
          $endgroup$
          – caverac
          Dec 16 '18 at 14:35




          $begingroup$
          @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
          $endgroup$
          – caverac
          Dec 16 '18 at 14:35




          1




          1




          $begingroup$
          @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
          $endgroup$
          – caverac
          Dec 20 '18 at 13:26




          $begingroup$
          @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
          $endgroup$
          – caverac
          Dec 20 '18 at 13:26


















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