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The set of equation \begin{cases} (X^2+Y^2+Z^2)^{5/2}X''+(X^2+Y^2-2Z^2)Y'+3YZZ'&=0\\ (X^2+Y^2+Z^2)^{5/2}Y''-(X^2+Y^2-2Z^2)X'-3XZZ'&=0\\ (X^2+Y^2+Z^2)^{5/2}Z''+3Z(XY'-YX')&=0 \end{cases} with initial condition \begin{cases} X(0)=1;&X'(0)=0.005\\ Y(0)=1;&Y'(0)=0.005\\ Z(0)=1;&Z'(0)=0.01 \end{cases} is "analogy to charged particles from the solar wind trapped the magnetic field of Earth". Solved Problems in Classical Electromagnetism by J. Pierrus Question 4.23.

The "common sense" and numerical solution showed that both $X,Y,Z$ has stable global periodic solution(a $\sin,\cos$ of large $\omega$ and large amplitude composite with $\sin,\cos$ of small $\omega$ and small amplitude ), much like the closed orbit in 3d.

In terms of protentional, the kinetic energy $E_k=\frac{1}{2}m(x'^2+y'^2+z'^2)$ is conserved, namely $E_k'=0$.

However, how to solve it analytically, to show that it has global periodic solutions?

Notice in spherical or cylinder system $ (X^2+Y^2+Z^2)^{5/2}=r^5$

Lutz Lehmann
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ShoutOutAndCalculate
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  • Could you explain the "common sense" reasoning for a periodic solution? As it is, the solution has to lie inside an energy hypersurface in the 6-dimensional phase space. To conclude for a periodic solution by Poincaré-Bendixson you need to find at least 3 more conserved quantities. – Lutz Lehmann Jan 25 '19 at 15:08
  • You get some kind of invariance under rotation about the Z axis, which by the Noether theorems gives another invariant of motion. The graph in the solution of the problem https://books.google.de/books?id=wY5lDwAAQBAJ&pg=PA233 looks exactly like that, a path in the projection of some hypersurface. – Lutz Lehmann Jan 25 '19 at 15:27
  • @LutzL I used common sense for "magmatic field does not work" and there is bunch of symmetry lies in the model so it's expected to show some periodic motions, and the numerical solution confirmed so. It's 6 dimensional phase space, but the graphs in the book clearly indicated a global $w_x,w_y,w_z$(angular frequency) for $X,Y,Z$ separately. I'm thinking about using cylinder or spherical coordinate system but not sure where to start. – ShoutOutAndCalculate Jan 25 '19 at 20:42
  • You get a mostly decoupled behavior. Each component is a mix of all the frequencies, they are not pure sine waves. If you have it programmed, try slight and large variations of the initial conditions to check if the claimed behavior is true for all solutions. – Lutz Lehmann Jan 25 '19 at 20:50

1 Answers1

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The Hamiltonian for this Lorentz force is $$ H=\frac12(p+A)^2 $$ with here $A=e_3\times \nabla f(r)$ where $f(r)=\frac1r$, so that $A=\frac{(-y,x,0)}{r^3}$. That is $$ H=\frac12(p_1-\frac{y}{r^3})^2+\frac12(p_2+\frac{x}{r^3})^2+\frac12p_3^2 $$ With $\dot x=H_p$ one finds $\dot x_k=p_k+A_k$ so that indeed the square sum of the velocities is invariant, $$H=\frac12(\dot x^2+\dot y^2+\dot z^2).$$

The rotation about the $z$ axis leaves the Hamiltonian invariant. Thus the angular momentum in the Hamiltonian impulse coordinates is also invariant, $$I_3=p_2x-p_1y=(\dot y-A_2)x-(\dot x-A_1)y=\dot yx-\dot xy-\frac{x^2+y^2}{r^3}.$$

Thus if the initial velocities and thus $H$ are relatively small against the coordinates, we get $$ \left(I_3+\frac{x^2+y^2}{r^3}\right)^2=(x\dot y-\dot xy)^2\le 2H(x^2+y^2) $$ which means that the motion is restricted so some small neighborhood of the curve $$-I_3+\frac{a^2}{(a^2+z^2)^{3/2}}=0\implies \begin{cases} s\in[0,1]\\ r=\frac{s}{|I_3|}\\a = \sqrt{|I_3|r^3}=r\sqrt{s}\\z=\sqrt{r^2-a^2}=r\sqrt{1-s} \end{cases}$$

For your initial values ($(x_0,y_0,z_0)=(1,1,1)$ in contrast to the source which usese $(1,1,0)$), the actual curve $(\sqrt{x^2+y^2},z)$ overlays the theoretical curve in the prediceted fashion as depicted in the plot below enter image description here

The 3-dimensional plot gives enter image description here

As for the nearly decoupling of the frequencies, it does not really happen here. In the next plot, the first two graphs are the polar coordinates of the pair $(x,y)$, while the last is the $z$ graph. As is visible, there is more happening than just the linear superposition of 3 sinusoid functions. enter image description here

Lutz Lehmann
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