Consider if there is some position vector $\mathbf{r}=(x,y,0)$, with its velocity given by the pointwise derivative. For some external frequency vector $\boldsymbol{\Omega}=(0,0,\Omega)$, suppose the Lagrangian is given by $$\mathscr{L}=\frac{1}{2}|\dot{\mathbf{r}}|^2+\boldsymbol{\Omega}\cdot(\mathbf{r}\times\dot{\mathbf{r}})+\frac{1}{2}\omega_0^2|\mathbf{r}|^2.\tag{1}$$ So the Lagrangian can then be expressed as $$\mathscr{L}=\frac{1}{2}(\dot{x}^2+\dot{y}^2)+(x\dot{y}-y\dot{x})\Omega+\frac{1}{2}\omega_0^2(x^2+y^2).\tag{2} $$ And by the First Euler-Lagrange Equation, that is, $$\partial_{q^i}\mathscr{L}-\frac{d \partial_{\dot{q}^i}\mathscr{L}}{dt}=0\tag{3} $$ it turns out the equations of motion are $$x''(t)=2y'(t)\Omega +\omega_0^2 x(t)$$ $$y''(t) = -2x'(t)\Omega+\omega_0^2 y(t).\tag{4}$$ I am lost at how to decouple these equations to solve for the equations of motion in $x(t)$ and $y(t)$. One approach I tried was a Laplace transform, but this involved a lot of constants to determine when I was inverse transforming. I also tried letting $z=x+iy$, noting that this is a solution to the equation $$z''(t)=\Omega z(t)-i\omega_0^2 z'(t).$$ This resulted in what I suspect is too simple an answer; namely, that $$x''+iy''=\Omega x+\Omega iy-i\omega_0^2x'+\omega_0^2y',$$ then by taking the real part (this is a physical system), $x''(t)=\Omega x$ and $y''(t)=\omega_0^2 y'(t).$ The problem at hand also mentions that there is a stability condition $$\mathbf{r}=\mathbf{r'}=0\iff \Omega^2>\omega^2.$$ I am not even sure where the $\Omega^2$ comes in. If anyone can give some tips, I'd appreciate it!

  • $\begingroup$ Similar question. math.stackexchange.com/questions/1801940/… $\endgroup$ Apr 16 at 6:50
  • $\begingroup$ Let $x' = X, y' = Y$ then try to diagonalise the corresponding matrix on the RHS. $\endgroup$ Apr 16 at 6:51
  • $\begingroup$ @MatthewCassell the issue with this is, I get a 2x2 matrix of the functions $X, Y, x$. I thought this is usually with the constants. $\endgroup$ Apr 16 at 7:25
  • $\begingroup$ ^ and $y$ as well $\endgroup$ Apr 16 at 7:31
  • $\begingroup$ Your complexified equation should be $z''=i\Omega z'+\omega_0^2z$. Everything below that is wrong. Especially if you separate real and imaginary parts, you should get back the original equations. $\endgroup$ Apr 16 at 8:08

1 Answer 1


Adding both equations as real and imaginary part gives $$ z''=x''+iy''=\Omega(y'-ix')+\omega_0^2(x+iy) \\=-i\Omega z'+\omega_0^2z $$ The term coupling the real and imaginary parts is the first-order one. There are standard methods to eliminate this. Consider $w=e^{i\Omega t/2}z$. Then $$ w'=e^{i\Omega t/2}(z'+i(\Omega/2)z)\\ w''=e^{i\Omega t/2}(z''+i\Omega z'-(\Omega^2/4)z)\\ $$ Inserting the differential equation gives $$ w''=(\omega_0^2-\Omega^2/4)w $$ As the factor on the right side is real, this decouples the equations for the real and imaginary part of $w$.


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