How to calculate the averaged equations for the weakly nonlinear oscillator $ddot x+x+varepsilon (xdot...












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$begingroup$


This is Strogatz exercise $7.6.5:$



For the system $ddot x+x+varepsilon h(x,dot x)=0$, where $h(x,dot x)=xdot x^2$ with $0 < ε << 1$, calculate the averaged equations and if possible, solve the averaged equations explicitly for $x(t,ε)$, given the initial conditions $x(0)=a$, $dot x(0)=0$.






ordinary-differential-equations dynamical-systems nonlinear-system non-linear-dynamics







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    $begingroup$


    This is Strogatz exercise $7.6.5:$



    For the system $ddot x+x+varepsilon h(x,dot x)=0$, where $h(x,dot x)=xdot x^2$ with $0 < ε << 1$, calculate the averaged equations and if possible, solve the averaged equations explicitly for $x(t,ε)$, given the initial conditions $x(0)=a$, $dot x(0)=0$.






    ordinary-differential-equations dynamical-systems nonlinear-system non-linear-dynamics







    share|cite|improve this question









    $endgroup$















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      -1


      2



      $begingroup$


      This is Strogatz exercise $7.6.5:$



      For the system $ddot x+x+varepsilon h(x,dot x)=0$, where $h(x,dot x)=xdot x^2$ with $0 < ε << 1$, calculate the averaged equations and if possible, solve the averaged equations explicitly for $x(t,ε)$, given the initial conditions $x(0)=a$, $dot x(0)=0$.






      ordinary-differential-equations dynamical-systems nonlinear-system non-linear-dynamics







      share|cite|improve this question









      $endgroup$




      This is Strogatz exercise $7.6.5:$



      For the system $ddot x+x+varepsilon h(x,dot x)=0$, where $h(x,dot x)=xdot x^2$ with $0 < ε << 1$, calculate the averaged equations and if possible, solve the averaged equations explicitly for $x(t,ε)$, given the initial conditions $x(0)=a$, $dot x(0)=0$.






      ordinary-differential-equations dynamical-systems nonlinear-system non-linear-dynamics




      ordinary-differential-equations dynamical-systems nonlinear-system non-linear-dynamics






      share|cite|improve this question













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      asked Jan 9 at 21:47









      HeptapodHeptapod

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      593317






















          2 Answers
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          $begingroup$

          $0)$



          $$r=sqrt{x^2+y^2},$$



          $$phi=arctanbiggr(frac{y}{x}biggr)-t (theta=t+phirightarrowphi=theta-t).$$



          $1)$



          $$h(x,dot x)=xdot x^2=rcos(t+phi)r^2sin^2(t+phi)=r^3cos(t+phi)sin^2(t+phi).$$



          $2)$



          $$dot{bar r}=Bigr<varepsilon hsin(t+phi)Bigr>_t,$$



          $$dot{bar phi}=Bigr<frac{varepsilon h}{r}cos(t+phi)Bigr>_t.$$



          begin{align}
          rightarrow dot r & = Bigr<varepsilon r^3cos(t+phi)sin^2(t+phi)Bigr[sin(t+phi)Bigr]Bigr>_t \
          & = varepsilon r^3Bigr<cos(t+phi)sin^3(t+phi)Bigr>_t \
          & = varepsilon r^3frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos(t+phi)sin^3(t+phi)dt=0 \
          & rightarrow dot r=0rightarrow frac{dr}{dt}=0rightarrow dr=0rightarrow r=r_0.
          end{align}



          begin{align}
          rightarrow dot phi & = Bigr<frac{varepsilon}{r} r^3cos(t+phi)sin^2(t+phi)Bigr[cos(t+phi)Bigr]Bigr>_t \
          & = varepsilon r^2Bigr<cos^2(t+phi)sin^2(t+phi)Bigr>_t \
          & = varepsilon r^2frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos^2(t+phi)sin^2(t+phi)dt \
          & = varepsilon r^2frac{1}{2pi}frac{pi}{4}=frac{varepsilon r^2}{8} \
          & rightarrow dot phi= frac{varepsilon r^2}{8}rightarrow frac{dphi}{dt}=frac{varepsilon r^2}{8} \
          & rightarrow dphi=frac{varepsilon r^2}{8}dt rightarrow phi=frac{varepsilon r^2}{8}t+phi_0.
          end{align}



          The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.



          $3)$



          $$x(0)=a,$$



          $$dot x(0)=0.$$



          $$rightarrow r_0=sqrt{x^2+y^2}=sqrt{0+a^2}=arightarrow r=a.$$



          $$rightarrow phi_0=arctanBigr(frac{0}{a}Bigr)-0=0rightarrow phi=frac{varepsilon r^2}{8}t.$$



          $4)$



          $$x(t)=r(t)cos(t+phi)rightarrow x(t)=acosBigr(t+frac{varepsilon r^2}{8}tBigr).$$






          share|cite|improve this answer











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            1












            $begingroup$

            This equation, that is the perturbation term, has a curious structure that allows to find a first integral via
            begin{align}
            ddot x + x(1+εdot x^2)&=0\[1em]
            implies
            frac{2dot xddot x}{1+εdot x^2}+2xdot x=0\[1em]
            implies
            frac1εln|1+εdot x^2|+x^2 = C
            end{align}
            For $(x(0),dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.




            To compute the period, first isolate the derivative
            $$
            dot x=pmsqrt{frac{exp(ε(a^2-x^2)-1}{ε}}=pmsqrt{a^2-x^2}sqrt{1+fracε2(a^2-x^2)+frac{ε^2}6(a^2-x^2)^2+...}
            $$
            A quarter period then computes as
            begin{align}
            frac{T}4&=int_0^afrac{sqrtε,dx}{sqrt{exp(ε(a^2-x^2)-1}}\
            &=int_0^{fracpi2}frac{ds}{sqrt{1+fracε2a^2cos^2s+frac{ε^2}6a^4cos^4s+...}}\
            &=fracpi2-fracε4a^2int_0^{fracpi2}cos^2s,ds+ frac{ε^2}{96}a^4int_0^{fracpi2}cos^4s,ds + frac{ε^3}{384}a^6int_0^{fracpi2}cos^6s,ds - frac{ε^3}{10240}a^8int_0^{fracpi2}cos^8s,ds mpdots\
            &=fracpi2left(1-fracε8a^2+ frac{ε^2}{256}a^4 + frac{5ε^3}{6144}a^6 - frac{7ε^4}{262144}a^8mpdotsright)
            end{align}
            The perturbed frequency is then
            $$
            frac{2pi}{T}=frac{pi/2}{T/4}=1 + frac{ε}8a^2 + frac{3ε^2}{256}a^4 + frac{ε^3}{6144}a^6 - frac{79ε^4}{786432}a^8+dots
            $$




            Series expansion using the CAS Magma (online calculator), with $z=εa^2$,



            PS<z>:=PowerSeriesRing(Rationals());
            
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
            
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
            
"frequency factor:",1/iq;





            share|cite|improve this answer











            $endgroup$













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              2 Answers
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              $begingroup$

              $0)$



              $$r=sqrt{x^2+y^2},$$



              $$phi=arctanbiggr(frac{y}{x}biggr)-t (theta=t+phirightarrowphi=theta-t).$$



              $1)$



              $$h(x,dot x)=xdot x^2=rcos(t+phi)r^2sin^2(t+phi)=r^3cos(t+phi)sin^2(t+phi).$$



              $2)$



              $$dot{bar r}=Bigr<varepsilon hsin(t+phi)Bigr>_t,$$



              $$dot{bar phi}=Bigr<frac{varepsilon h}{r}cos(t+phi)Bigr>_t.$$



              begin{align}
              rightarrow dot r & = Bigr<varepsilon r^3cos(t+phi)sin^2(t+phi)Bigr[sin(t+phi)Bigr]Bigr>_t \
              & = varepsilon r^3Bigr<cos(t+phi)sin^3(t+phi)Bigr>_t \
              & = varepsilon r^3frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos(t+phi)sin^3(t+phi)dt=0 \
              & rightarrow dot r=0rightarrow frac{dr}{dt}=0rightarrow dr=0rightarrow r=r_0.
              end{align}



              begin{align}
              rightarrow dot phi & = Bigr<frac{varepsilon}{r} r^3cos(t+phi)sin^2(t+phi)Bigr[cos(t+phi)Bigr]Bigr>_t \
              & = varepsilon r^2Bigr<cos^2(t+phi)sin^2(t+phi)Bigr>_t \
              & = varepsilon r^2frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos^2(t+phi)sin^2(t+phi)dt \
              & = varepsilon r^2frac{1}{2pi}frac{pi}{4}=frac{varepsilon r^2}{8} \
              & rightarrow dot phi= frac{varepsilon r^2}{8}rightarrow frac{dphi}{dt}=frac{varepsilon r^2}{8} \
              & rightarrow dphi=frac{varepsilon r^2}{8}dt rightarrow phi=frac{varepsilon r^2}{8}t+phi_0.
              end{align}



              The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.



              $3)$



              $$x(0)=a,$$



              $$dot x(0)=0.$$



              $$rightarrow r_0=sqrt{x^2+y^2}=sqrt{0+a^2}=arightarrow r=a.$$



              $$rightarrow phi_0=arctanBigr(frac{0}{a}Bigr)-0=0rightarrow phi=frac{varepsilon r^2}{8}t.$$



              $4)$



              $$x(t)=r(t)cos(t+phi)rightarrow x(t)=acosBigr(t+frac{varepsilon r^2}{8}tBigr).$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $0)$



                $$r=sqrt{x^2+y^2},$$



                $$phi=arctanbiggr(frac{y}{x}biggr)-t (theta=t+phirightarrowphi=theta-t).$$



                $1)$



                $$h(x,dot x)=xdot x^2=rcos(t+phi)r^2sin^2(t+phi)=r^3cos(t+phi)sin^2(t+phi).$$



                $2)$



                $$dot{bar r}=Bigr<varepsilon hsin(t+phi)Bigr>_t,$$



                $$dot{bar phi}=Bigr<frac{varepsilon h}{r}cos(t+phi)Bigr>_t.$$



                begin{align}
                rightarrow dot r & = Bigr<varepsilon r^3cos(t+phi)sin^2(t+phi)Bigr[sin(t+phi)Bigr]Bigr>_t \
                & = varepsilon r^3Bigr<cos(t+phi)sin^3(t+phi)Bigr>_t \
                & = varepsilon r^3frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos(t+phi)sin^3(t+phi)dt=0 \
                & rightarrow dot r=0rightarrow frac{dr}{dt}=0rightarrow dr=0rightarrow r=r_0.
                end{align}



                begin{align}
                rightarrow dot phi & = Bigr<frac{varepsilon}{r} r^3cos(t+phi)sin^2(t+phi)Bigr[cos(t+phi)Bigr]Bigr>_t \
                & = varepsilon r^2Bigr<cos^2(t+phi)sin^2(t+phi)Bigr>_t \
                & = varepsilon r^2frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos^2(t+phi)sin^2(t+phi)dt \
                & = varepsilon r^2frac{1}{2pi}frac{pi}{4}=frac{varepsilon r^2}{8} \
                & rightarrow dot phi= frac{varepsilon r^2}{8}rightarrow frac{dphi}{dt}=frac{varepsilon r^2}{8} \
                & rightarrow dphi=frac{varepsilon r^2}{8}dt rightarrow phi=frac{varepsilon r^2}{8}t+phi_0.
                end{align}



                The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.



                $3)$



                $$x(0)=a,$$



                $$dot x(0)=0.$$



                $$rightarrow r_0=sqrt{x^2+y^2}=sqrt{0+a^2}=arightarrow r=a.$$



                $$rightarrow phi_0=arctanBigr(frac{0}{a}Bigr)-0=0rightarrow phi=frac{varepsilon r^2}{8}t.$$



                $4)$



                $$x(t)=r(t)cos(t+phi)rightarrow x(t)=acosBigr(t+frac{varepsilon r^2}{8}tBigr).$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $0)$



                  $$r=sqrt{x^2+y^2},$$



                  $$phi=arctanbiggr(frac{y}{x}biggr)-t (theta=t+phirightarrowphi=theta-t).$$



                  $1)$



                  $$h(x,dot x)=xdot x^2=rcos(t+phi)r^2sin^2(t+phi)=r^3cos(t+phi)sin^2(t+phi).$$



                  $2)$



                  $$dot{bar r}=Bigr<varepsilon hsin(t+phi)Bigr>_t,$$



                  $$dot{bar phi}=Bigr<frac{varepsilon h}{r}cos(t+phi)Bigr>_t.$$



                  begin{align}
                  rightarrow dot r & = Bigr<varepsilon r^3cos(t+phi)sin^2(t+phi)Bigr[sin(t+phi)Bigr]Bigr>_t \
                  & = varepsilon r^3Bigr<cos(t+phi)sin^3(t+phi)Bigr>_t \
                  & = varepsilon r^3frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos(t+phi)sin^3(t+phi)dt=0 \
                  & rightarrow dot r=0rightarrow frac{dr}{dt}=0rightarrow dr=0rightarrow r=r_0.
                  end{align}



                  begin{align}
                  rightarrow dot phi & = Bigr<frac{varepsilon}{r} r^3cos(t+phi)sin^2(t+phi)Bigr[cos(t+phi)Bigr]Bigr>_t \
                  & = varepsilon r^2Bigr<cos^2(t+phi)sin^2(t+phi)Bigr>_t \
                  & = varepsilon r^2frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos^2(t+phi)sin^2(t+phi)dt \
                  & = varepsilon r^2frac{1}{2pi}frac{pi}{4}=frac{varepsilon r^2}{8} \
                  & rightarrow dot phi= frac{varepsilon r^2}{8}rightarrow frac{dphi}{dt}=frac{varepsilon r^2}{8} \
                  & rightarrow dphi=frac{varepsilon r^2}{8}dt rightarrow phi=frac{varepsilon r^2}{8}t+phi_0.
                  end{align}



                  The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.



                  $3)$



                  $$x(0)=a,$$



                  $$dot x(0)=0.$$



                  $$rightarrow r_0=sqrt{x^2+y^2}=sqrt{0+a^2}=arightarrow r=a.$$



                  $$rightarrow phi_0=arctanBigr(frac{0}{a}Bigr)-0=0rightarrow phi=frac{varepsilon r^2}{8}t.$$



                  $4)$



                  $$x(t)=r(t)cos(t+phi)rightarrow x(t)=acosBigr(t+frac{varepsilon r^2}{8}tBigr).$$






                  share|cite|improve this answer











                  $endgroup$



                  $0)$



                  $$r=sqrt{x^2+y^2},$$



                  $$phi=arctanbiggr(frac{y}{x}biggr)-t (theta=t+phirightarrowphi=theta-t).$$



                  $1)$



                  $$h(x,dot x)=xdot x^2=rcos(t+phi)r^2sin^2(t+phi)=r^3cos(t+phi)sin^2(t+phi).$$



                  $2)$



                  $$dot{bar r}=Bigr<varepsilon hsin(t+phi)Bigr>_t,$$



                  $$dot{bar phi}=Bigr<frac{varepsilon h}{r}cos(t+phi)Bigr>_t.$$



                  begin{align}
                  rightarrow dot r & = Bigr<varepsilon r^3cos(t+phi)sin^2(t+phi)Bigr[sin(t+phi)Bigr]Bigr>_t \
                  & = varepsilon r^3Bigr<cos(t+phi)sin^3(t+phi)Bigr>_t \
                  & = varepsilon r^3frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos(t+phi)sin^3(t+phi)dt=0 \
                  & rightarrow dot r=0rightarrow frac{dr}{dt}=0rightarrow dr=0rightarrow r=r_0.
                  end{align}



                  begin{align}
                  rightarrow dot phi & = Bigr<frac{varepsilon}{r} r^3cos(t+phi)sin^2(t+phi)Bigr[cos(t+phi)Bigr]Bigr>_t \
                  & = varepsilon r^2Bigr<cos^2(t+phi)sin^2(t+phi)Bigr>_t \
                  & = varepsilon r^2frac{1}{2pi}displaystyle int_{t-pi}^{t+pi}cos^2(t+phi)sin^2(t+phi)dt \
                  & = varepsilon r^2frac{1}{2pi}frac{pi}{4}=frac{varepsilon r^2}{8} \
                  & rightarrow dot phi= frac{varepsilon r^2}{8}rightarrow frac{dphi}{dt}=frac{varepsilon r^2}{8} \
                  & rightarrow dphi=frac{varepsilon r^2}{8}dt rightarrow phi=frac{varepsilon r^2}{8}t+phi_0.
                  end{align}



                  The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.



                  $3)$



                  $$x(0)=a,$$



                  $$dot x(0)=0.$$



                  $$rightarrow r_0=sqrt{x^2+y^2}=sqrt{0+a^2}=arightarrow r=a.$$



                  $$rightarrow phi_0=arctanBigr(frac{0}{a}Bigr)-0=0rightarrow phi=frac{varepsilon r^2}{8}t.$$



                  $4)$



                  $$x(t)=r(t)cos(t+phi)rightarrow x(t)=acosBigr(t+frac{varepsilon r^2}{8}tBigr).$$







                  share|cite|improve this answer














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                  edited Jan 10 at 5:39









                  Dylan

                  12.7k31026




                  12.7k31026










                  answered Jan 9 at 21:47









                  HeptapodHeptapod

                  593317




                  593317























                      1












                      $begingroup$

                      This equation, that is the perturbation term, has a curious structure that allows to find a first integral via
                      begin{align}
                      ddot x + x(1+εdot x^2)&=0\[1em]
                      implies
                      frac{2dot xddot x}{1+εdot x^2}+2xdot x=0\[1em]
                      implies
                      frac1εln|1+εdot x^2|+x^2 = C
                      end{align}
                      For $(x(0),dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.




                      To compute the period, first isolate the derivative
                      $$
                      dot x=pmsqrt{frac{exp(ε(a^2-x^2)-1}{ε}}=pmsqrt{a^2-x^2}sqrt{1+fracε2(a^2-x^2)+frac{ε^2}6(a^2-x^2)^2+...}
                      $$
                      A quarter period then computes as
                      begin{align}
                      frac{T}4&=int_0^afrac{sqrtε,dx}{sqrt{exp(ε(a^2-x^2)-1}}\
                      &=int_0^{fracpi2}frac{ds}{sqrt{1+fracε2a^2cos^2s+frac{ε^2}6a^4cos^4s+...}}\
                      &=fracpi2-fracε4a^2int_0^{fracpi2}cos^2s,ds+ frac{ε^2}{96}a^4int_0^{fracpi2}cos^4s,ds + frac{ε^3}{384}a^6int_0^{fracpi2}cos^6s,ds - frac{ε^3}{10240}a^8int_0^{fracpi2}cos^8s,ds mpdots\
                      &=fracpi2left(1-fracε8a^2+ frac{ε^2}{256}a^4 + frac{5ε^3}{6144}a^6 - frac{7ε^4}{262144}a^8mpdotsright)
                      end{align}
                      The perturbed frequency is then
                      $$
                      frac{2pi}{T}=frac{pi/2}{T/4}=1 + frac{ε}8a^2 + frac{3ε^2}{256}a^4 + frac{ε^3}{6144}a^6 - frac{79ε^4}{786432}a^8+dots
                      $$




                      Series expansion using the CAS Magma (online calculator), with $z=εa^2$,



                      PS<z>:=PowerSeriesRing(Rationals());
                      
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
                      
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
                      
"frequency factor:",1/iq;





                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        This equation, that is the perturbation term, has a curious structure that allows to find a first integral via
                        begin{align}
                        ddot x + x(1+εdot x^2)&=0\[1em]
                        implies
                        frac{2dot xddot x}{1+εdot x^2}+2xdot x=0\[1em]
                        implies
                        frac1εln|1+εdot x^2|+x^2 = C
                        end{align}
                        For $(x(0),dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.




                        To compute the period, first isolate the derivative
                        $$
                        dot x=pmsqrt{frac{exp(ε(a^2-x^2)-1}{ε}}=pmsqrt{a^2-x^2}sqrt{1+fracε2(a^2-x^2)+frac{ε^2}6(a^2-x^2)^2+...}
                        $$
                        A quarter period then computes as
                        begin{align}
                        frac{T}4&=int_0^afrac{sqrtε,dx}{sqrt{exp(ε(a^2-x^2)-1}}\
                        &=int_0^{fracpi2}frac{ds}{sqrt{1+fracε2a^2cos^2s+frac{ε^2}6a^4cos^4s+...}}\
                        &=fracpi2-fracε4a^2int_0^{fracpi2}cos^2s,ds+ frac{ε^2}{96}a^4int_0^{fracpi2}cos^4s,ds + frac{ε^3}{384}a^6int_0^{fracpi2}cos^6s,ds - frac{ε^3}{10240}a^8int_0^{fracpi2}cos^8s,ds mpdots\
                        &=fracpi2left(1-fracε8a^2+ frac{ε^2}{256}a^4 + frac{5ε^3}{6144}a^6 - frac{7ε^4}{262144}a^8mpdotsright)
                        end{align}
                        The perturbed frequency is then
                        $$
                        frac{2pi}{T}=frac{pi/2}{T/4}=1 + frac{ε}8a^2 + frac{3ε^2}{256}a^4 + frac{ε^3}{6144}a^6 - frac{79ε^4}{786432}a^8+dots
                        $$




                        Series expansion using the CAS Magma (online calculator), with $z=εa^2$,



                        PS<z>:=PowerSeriesRing(Rationals());
                        
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
                        
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
                        
"frequency factor:",1/iq;





                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          This equation, that is the perturbation term, has a curious structure that allows to find a first integral via
                          begin{align}
                          ddot x + x(1+εdot x^2)&=0\[1em]
                          implies
                          frac{2dot xddot x}{1+εdot x^2}+2xdot x=0\[1em]
                          implies
                          frac1εln|1+εdot x^2|+x^2 = C
                          end{align}
                          For $(x(0),dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.




                          To compute the period, first isolate the derivative
                          $$
                          dot x=pmsqrt{frac{exp(ε(a^2-x^2)-1}{ε}}=pmsqrt{a^2-x^2}sqrt{1+fracε2(a^2-x^2)+frac{ε^2}6(a^2-x^2)^2+...}
                          $$
                          A quarter period then computes as
                          begin{align}
                          frac{T}4&=int_0^afrac{sqrtε,dx}{sqrt{exp(ε(a^2-x^2)-1}}\
                          &=int_0^{fracpi2}frac{ds}{sqrt{1+fracε2a^2cos^2s+frac{ε^2}6a^4cos^4s+...}}\
                          &=fracpi2-fracε4a^2int_0^{fracpi2}cos^2s,ds+ frac{ε^2}{96}a^4int_0^{fracpi2}cos^4s,ds + frac{ε^3}{384}a^6int_0^{fracpi2}cos^6s,ds - frac{ε^3}{10240}a^8int_0^{fracpi2}cos^8s,ds mpdots\
                          &=fracpi2left(1-fracε8a^2+ frac{ε^2}{256}a^4 + frac{5ε^3}{6144}a^6 - frac{7ε^4}{262144}a^8mpdotsright)
                          end{align}
                          The perturbed frequency is then
                          $$
                          frac{2pi}{T}=frac{pi/2}{T/4}=1 + frac{ε}8a^2 + frac{3ε^2}{256}a^4 + frac{ε^3}{6144}a^6 - frac{79ε^4}{786432}a^8+dots
                          $$




                          Series expansion using the CAS Magma (online calculator), with $z=εa^2$,



                          PS<z>:=PowerSeriesRing(Rationals());
                          
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
                          
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
                          
"frequency factor:",1/iq;





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                          $endgroup$



                          This equation, that is the perturbation term, has a curious structure that allows to find a first integral via
                          begin{align}
                          ddot x + x(1+εdot x^2)&=0\[1em]
                          implies
                          frac{2dot xddot x}{1+εdot x^2}+2xdot x=0\[1em]
                          implies
                          frac1εln|1+εdot x^2|+x^2 = C
                          end{align}
                          For $(x(0),dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.




                          To compute the period, first isolate the derivative
                          $$
                          dot x=pmsqrt{frac{exp(ε(a^2-x^2)-1}{ε}}=pmsqrt{a^2-x^2}sqrt{1+fracε2(a^2-x^2)+frac{ε^2}6(a^2-x^2)^2+...}
                          $$
                          A quarter period then computes as
                          begin{align}
                          frac{T}4&=int_0^afrac{sqrtε,dx}{sqrt{exp(ε(a^2-x^2)-1}}\
                          &=int_0^{fracpi2}frac{ds}{sqrt{1+fracε2a^2cos^2s+frac{ε^2}6a^4cos^4s+...}}\
                          &=fracpi2-fracε4a^2int_0^{fracpi2}cos^2s,ds+ frac{ε^2}{96}a^4int_0^{fracpi2}cos^4s,ds + frac{ε^3}{384}a^6int_0^{fracpi2}cos^6s,ds - frac{ε^3}{10240}a^8int_0^{fracpi2}cos^8s,ds mpdots\
                          &=fracpi2left(1-fracε8a^2+ frac{ε^2}{256}a^4 + frac{5ε^3}{6144}a^6 - frac{7ε^4}{262144}a^8mpdotsright)
                          end{align}
                          The perturbed frequency is then
                          $$
                          frac{2pi}{T}=frac{pi/2}{T/4}=1 + frac{ε}8a^2 + frac{3ε^2}{256}a^4 + frac{ε^3}{6144}a^6 - frac{79ε^4}{786432}a^8+dots
                          $$




                          Series expansion using the CAS Magma (online calculator), with $z=εa^2$,



                          PS<z>:=PowerSeriesRing(Rationals());
                          
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
                          
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
                          
"frequency factor:",1/iq;






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                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 10 at 11:37

























                          answered Jan 10 at 11:20









                          LutzLLutzL

                          57.8k42054




                          57.8k42054






























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