Solving Nonlinear ODE: magnetism with varying particle charge

[Eloise]

Hello.

I have a set of ODE where
1) $$\frac{dv_x}{dt}=\frac{q(t)B}{m}v_y$$
2) $$\frac{dv_y}{dt}=\frac{q(t)B}{m}v_x$$
3) $$\frac{dv_z}{dt}=0$$

Following the strategy to solve a simple harmonic oscillator,
I differentiate (1) to get
4) $$\frac{d^2v_x}{dt^2}=\frac{q(t)B}{m}\frac{dv_y}{dt}+q'(t)v_y$$

and substitute (1) and (2) into it to get
5) $$\frac{d^2v_x}{dt^2}=(\frac{q(t)B}{m})^2v_x+\frac{q'(t)}{q(t)}\frac{dv_x}{dt}$$

which involve only functions of $t$ and $x$.
The initial position of the particle is at the origin.
I do not know if this is the correct way. I should solve it by inspired guessing?

Can you please help me by giving me some tips on solving this?
Thank you very much.

 

[hilbert2]

I differentiate (1) to get
4) $$\frac{d^2v_x}{dt^2}=\frac{q(t)B}{m}\frac{dv_y}{dt}+q'(t)v_y$$

¨
Shouldn't that be $\frac{d^2v_x}{dt^2}=\frac{q(t)B}{m}\frac{dv_y}{dt}+\frac{q'(t)B}{m}v_y$? Otherwise your method is correct, but you have to know the function $q(t)$ to solve the equation. The equation is not nonlinear, by the way.

 

[pasmith]

Hello.

I have a set of ODE where
1) $$\frac{dv_x}{dt}=\frac{q(t)B}{m}v_y$$
2) $$\frac{dv_y}{dt}=\frac{q(t)B}{m}v_x$$
3) $$\frac{dv_z}{dt}=0$$

I think, starting from
$$m\frac{d\vec v}{dt} = q(t)(\vec v \times \vec B)$$
with $\vec B = (0, 0, B)$, that you should have a minus sign on the right hand side of (2) above.

Following the strategy to solve a simple harmonic oscillator,
I differentiate (1) to get
4) $$\frac{d^2v_x}{dt^2}=\frac{q(t)B}{m}\frac{dv_y}{dt}+q'(t)v_y$$

and substitute (1) and (2) into it to get
5) $$\frac{d^2v_x}{dt^2}=(\frac{q(t)B}{m})^2v_x+\frac{q'(t)}{q(t)}\frac{dv_x}{dt}$$

I don't think that helps you here.

I do not know if this is the correct way. I should solve it by inspired guessing?

Can you please help me by giving me some tips on solving this?
Thank you very much.

It's generally best to keep a first order linear system (which yours is) as a first order system rather than turn it into a second order system.

Your system is essentially
$$\frac{dv_x}{dt} = f(t)v_y \\ \frac{dv_y}{dt} = f(t)v_x$$
Now if you set $u_1 = v_x + v_y$ and $u_2 = v_x- v_y$ then you have
$$\frac{du_1}{dt} = f(t)u_1 \\ \frac{du_2}{dt} = -f(t)u_2$$
and those first-order equations are separable and can be solved.

Alternatively, you can make the inspired guess that
$$v_x(t) = v_x(0) \cosh(F(t)) + v_y(0) \sinh(F(t)), \\ v_y(t) = v_y(0) \cosh(F(t)) + v_x(0) \sinh(F(t)).$$
for a suitable $F(t)$ with $F(0) = 0$, which is prompted by the observation that
$$\frac{d}{dt} \cosh(t) = \sinh (t), \\ \frac{d}{dt} \sinh(t) = \cosh (t)$$
so that if
$$P(t) = C \cosh(t) + D \sinh(t) \\ Q(t) = D \cosh(t) + C \sinh(t)$$
then
$$\frac{dP}{dt} = Q \\ \frac{dQ}{dt} = P$$
which is almost your system, but to get the factor of $f(t)$ on the right of each equation we have to exploit the chain rule and instead set
$$P(t) = C \cosh(F(t)) + D \sinh(F(t)) \\ Q(t) =D \cosh(F(t)) + C \sinh(F(t))$$
so that now
$$\frac{dP}{dt} = F'(t)Q \\ \frac{dQ}{dt} = F'(t)P$$
which is your system. Only $F'$ is forced on us, so we can choose $F(0)$ to be whatever we want, and the choice $F(0) = 0$ is convenient, since then $P(0) = C$ and $Q(0) = D$.

The initial position of the particle is at the origin.

I assume $v_x$ and $v_y$ are components of velocity, so you need to specify an initial velocity as well.

 

[hilbert2]

It's generally best to keep a first order linear system (which yours is) as a first order system rather than turn it into a second order system.

Actually, in my classical mechanics textbook, the trajectory of a charged particle in constant magnetic field is solved in that fashion (differentiating to increase the order of the system). That's the "brute force" method to solve the problem.

 

[pasmith]

Actually, in my classical mechanics textbook, the trajectory of a charged particle in constant magnetic field is solved in that fashion (differentiating to increase the order of the system). That's the "brute force" method to solve the problem.

And it makes things easier only in the constant-coefficient case, which is not this case.

 

[Eloise]

Thank you very much. I didn't thought of using hyperbolic functions. Thanks again for helping.