Oscillatory solution for a given Lagrangian

[Siberion]

Homework Statement

Consider the following Lagrangian:$$L = \frac{m}{2}(x'^2+y'^2+z'^2) + \frac{q}{2}(xy'-yx')$$

Where q denotes a charged particle.

a) Find the equations of motion
b) Find the solution for z
c) Find the solution in the x-y plane, and prove that it corresponds to an oscillatory motion along both axes.

The Attempt at a Solution

Considering the Euler-Lagrange equations, the solution for coordinate z is given by:

d/dt (mz') = 0
Thus, a solution for z corresponds to a uniform linear motion along z axis.

For x and y, I get the following systems of dif. equations:

mx'' - qy' = 0
my'' + qx' = 0

I tried solving this system by integrating and substituting,

i.e. mx'' = qy' /integrate
mx' = qy + C
x' = qy/m + C

Substituting, we get
y'' + y(q^2)/(m^2) = C

The inverse process could be done to get an equation for x.

Which I do not think is the right answer, and I don't know how many laws of mathematics I violated while doing that. Could anyone please show me what is the right procedure to solve this system of equations?

Also, I have the impression this Lagrangian corresponds to a charge q interacting with an electromagnetic field. It would be great if anyone could explain me a little bit further about that.Any help would be greatly appreciated.
Thanks a lot.

 

[TSny]

mx'' - qy' = 0
my'' - qx' = 0

Should both equations have a negative sign in the left side?

 

[Siberion]

Should both equations have a negative sign in the left side?

Sorry, my bad, I missed a minus sign in the second equation, then the system becomes:

mx'' - qy' = 0
my'' + qx' = 0

Following the same process I mentioned before, equation for y becomes:

y'' + y(q^2)/(m^2) = C

 

[TSny]

I'm not sure of your math background. If you've had a course in differential equations, then you know that the general solution to this type of equation is y = y_h + y_p where y_h is the general solution of the homogeneous equation y'' + (q/m)²y = 0 and y_p is any particular solution of y'' + (q/m)²y = C.

 

[Siberion]

I'm not sure of your math background. If you've had a course in differential equations, then you know that the general solution to this type of equation is y = y_h + y_p where y_h is the general solution of the homogeneous equation y'' + (q/m)²y = 0 and y_p is any particular solution of y'' + (q/m)²y = C.

I'm sorry if this becomes too obvious. I've taken Dif. Equations, but my math background is a bit blurry at the moment, because I've just re-taken classes. Once I recall the methods things go easier.

I can see that the solution for the homogeneus equation is indeed an oscillatory motion, both for x and y. I'm having trouble about what to do with the particular solution since I don't have any initial condition.

Would you say that the method I applied to solve the problem is correct?

Thanks for your help.

 

[TSny]

Yes, your method is a correct way to do it. You've found the general solution to y'' + (q/m)²y = 0. Now you just need any particular solution to y'' + (q/m)²y = C. You should be able to this by inspection. You do not need initial conditions here.