How do we find the equation for the gyrating motion of a particle in a uniform magnetic field and a non-uniform Electric field? [closed]

Considering the gyrating motion is not negligible and also retaining the guiding center drift, how do we get the trajectories x(t),y(t),z(t) of the particle?

In this case is the variation in the Larmor frequency and Larmor radius significant enough to consider them as variables.

How would the particle behave if they were considered as constants ?

$$E(x)=E_0 x$$, for some $$E_0$$ and $$B = B_0$$ for some constant Bo.

I came across the following differential equation while solving for the trajectory of a particle in a non-uniform electric and uniform magnetic field.

$$x'''+x''+xx' = 0, \quad x' = \frac{\mathrm{d}x}{\mathrm{d}t}.$$

I chose the constants such as to get a simpler differential equation.

All the standard textbooks on Plasma Physics seem to skip over this part and only consider the guiding center drift.

• Comments are not for extended discussion; this conversation has been moved to chat. – Chris Jul 1 '19 at 18:20
• These edits as far as I can tell essentially invalidate the posted answer. It's generally not a good idea to edit the question in a way that makes already posted answers invalid. – JMac Jul 2 '19 at 11:46
• Please do not edit questions in a way that invalidates existing answers. – ACuriousMind Jul 2 '19 at 17:25
• @ACuriousMind ok . Noted – Manoj Jul 2 '19 at 18:00

The equation can be integrated once, then we have second order equation $$x''+x'+\frac {x^2}{2}=C_1$$ Make a substitution $$x'=y(x)$$, then $$x''=y\frac {dy}{dx}$$ as a result, we obtain the Abel equation , for which an analytical solution is known, see https://www.hindawi.com/journals/ijmms/2011/387429/#sec2 $$y\frac {dy}{dx}+y=C_1-x^2/2$$ The numerical solution of the original equation depends on the boundary conditions. We can put the initial data as follows $$x(0)=x_0,x'(0)=v_0,x''(0)=0$$. Figure 1 shows the solution of the equation when changing $$x_0$$ (left) and $$v_0$$ (right)
• what do you mean by $y[x]$? – AccidentalFourierTransform Jun 30 '19 at 19:38
• I don't understand how you get your second equation from your first equation. I suspect the second equation is wrong. I believe $x'$ is a derivative with respect to some variable $t$, not $x$. – akhmeteli Jun 30 '19 at 20:13
• @AccidentalFourierTransform This is a standard trick, you are trying to find $x'$ as a function of $x$. Alex is denoting this function by $y[x]$. Once $y$ is found, you solve the equation $x'=y[x]$ by standard methods. – Peter Kravchuk Jun 30 '19 at 20:16
• @akhmeteli, it is not wrong. Take $t$ derivative of the substitution, use the substitution again. – Peter Kravchuk Jun 30 '19 at 20:18
• @PeterKravchuk Thanks. Alex is a user of Mathematica, where brackets denote function evaluation, so I wasnt sure he meant $y[x]=y(x)$, or perhaps $y[x]=xy(x)$. Anyway, it might be useful to clarify the notation in the OP. – AccidentalFourierTransform Jun 30 '19 at 20:22