Homework Help: Find the general solution of a coupled differential equation:

1. Sep 2, 2010

vorcil

1. The problem statement, all variables and given/known data

I want to find the general solution of these two equations,

$$\ddot{y}=\omega\dot{z}$$

$$\ddot{z}=\omega\left(\frac{\mathbf{E}}{\mathbf{B}} - \dot{y}\right)$$

2. Relevant equations
These two equations are the result of quantitatively solving to find the trajectory of a charged particle in a uniform electric field, perpendicular to a magnetic one,

Suppose that the B field points in the x direction, and the E field in the z direction, and the particle is released from rest,

the lorentz force law,
$$\mathbf{F}_{mag} = Q(\mathbf{v} \times \mathbf{B})$$
3. The attempt at a solution

(i'm just going to show you how I got to the coupled differential equations

there is no force in the x direction due to the magnetic field being in that direction,

the position of the particle at t can be described by the vector, (0, y(t), z(t))

so $$\mathbf{v}=(0,\dot{y},\dot{z})$$
(the dots are the derivatives with respect to time)
calculating the cross product from the Lorentz force law,

$$\mathbf{v} \times \mathbf{B} = \left| \begin{array}{ccc} \mathbf{\hat{x}} & \mathbf{\hat{y}} & \mathbf{\hat{z}} \\ 0 & \dot{y}& \dot{z} \\ \mathbf{B}& 0& 0 \\ \end{array} \right| =B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}}$$

and applying newtons second law, F=ma=QE=Q(v\timesB)=Q(E+(v\timesB))
$$\mathbf{F} = Q(\mathbf{E}+(\mathbf{v}\times\mathbf{B})) = Q(E\mathbf{\hat{z}} + B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}} = m\mathbf{a} = m(\ddot{y}\mathbf{\hat{y}}+\ddot{z}\mathbf{\hat{z}}$$

separating the $$\hat{y} & \hat{z}$$ components

$$\mathbf{QB}\dot{z} = m\ddot{y}$$ & $$\mathbf{QE - QB}\dot{y} = m\ddot{z}$$

substituting in the cyclotron frequency formula, $$\omega = \frac{\mathbf{QB}}{m}$$

I get the two coupled differential equations that I showed earlier,

1: $$\ddot{y}=\omega\dot{z}$$

2: $$\ddot{z}=\omega\left(\frac{\mathbf{E}}{\mathbf{B}} - \dot{y}\right)$$

I want to figure out the general solution, but I can never get it,
I'm supposed to Differentiate the first function and then use the second to eliminate $$\ddot{z}$$

-

well if I differentiate the first equation,

I get a triple dot y, $$\dot{\ddot{y}} = w\ddot{z}$$
Is that how I differentiate the first equation?

Last edited: Sep 3, 2010
2. Sep 3, 2010

Sourabh N

Yes. And you substitute the value of $$\ddot{z}$$ from second equation there.

3. Sep 3, 2010

vorcil

so,

$$\dot{\ddot{y}} = \omega \omega(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y})=\omega^2 (\frac{\mathbf{E}}{\mathbf{B}}-\dot{y})$$

$$\omega^2 = (\frac{QB}{m})^2$$

$$\mathbf{QE - QB}\dot{y} = mw(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}) = QB*(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}$$

I don't understand what to do from here

4. Sep 3, 2010

Sourabh N

Differentiate 2nd equation. Plug in y$$^{..}$$ from first equation. It looks like a sinusoidal equation, with z$$^{.}$$ as your variable. Find z$$^{.}$$ from here, and z. Plug it back in the equations and similarly find y.

The approach you are using, certainly works. Its just slightly messier.

5. Sep 3, 2010

vorcil

ok if equation2= $$\ddot{z} = \omega (\frac{E}{B} - \dot{y})$$
then expanding

$$\ddot{z} = \omega (\frac{E}{B}) - \omega(\dot{y})$$

and differentiating

$$\dot{\ddot{z}} =\frac{E}{B} - \omega(\ddot{y})$$

then plugging in $$\ddot{y}$$ from the first equation

$$\dot{\ddot{z}} = \frac{E}{B} - \omega(\omega\dot{z}) = \frac{E}{B}-\omega^2\dot{z}$$

I don't see how this looks like a sinusoidal equation?

so to get rid of those dots, I integrate?

6. Sep 3, 2010

HallsofIvy

Well, you can't just integrate. Do you know anything at all about solving "linear differential equations with constant coefficients"?

7. Sep 3, 2010

vorcil

Well I know how to integrate, and i've dealed with hard differential/coupled equations in physics before, but haven't started proper, differential equations in mathematics yet, I don't think?
is it just the same as integrating normal linear equations?

Last edited: Sep 3, 2010
8. Sep 3, 2010

klondike

So it's essentially an ODE with constant coeff
$$y'''+\omega ^2y'=K$$
Perhaps the easiest way to solve it is using LapLace transform. To make life further easier, assuming you have the system completely at rest for t=0, i.e. y(0)=y'(0)=y''(0). Take LT of both sides and rearrange one would get:
$$Y(s)=\frac{K}{s^2(s^2+\omega^2)}=\frac{K}{\omega^2}(\frac{1}{s^2}-\frac{1}{s^2+\omega^2})$$

So the solution is a ramp minus a sinusoid.

9. Sep 4, 2010

HallsofIvy

I really dislike the use of the Laplace transform for problems like this!

If you were to try a solution of the form $y= e^{rx}$, then you would have $y'= re^{rx}$, y"=$r^2e^{rx}$, $y'''= r^3e^{rx}$ and the "associated homogeneous equation" (with the "K" dropped) becomes $r^3e^{rx}+ \omega^2 r e^{rx}= (r^3+ \omega r)e^{rx}= r(r^2+ \omega)e^{rx}= 0$. Since $e^{rx}$ can't be 0, we must have $r(r^2+ \omega)= 0$ which gives r= 0 and $r= \pm i\sqrt{\omega}$. Of course, $e^{ix}= cos(x)+ i sin(x)$ so we can write $e^{it\sqrt{\omega}}$ and $it\sqrt{\omega}}$ in terms of $cos(t\sqrt{\omega})$ and $sin(t\sqrt{\omega})$. Of course, $e^{0t}= 1$: the general solution to the associated homogeneous equation is
$$C+ Dcos(t\sqrt{\omega})+ Esin(t\sqrt{\omega})$$.

To find the solution to the entire equation, try a constant solution, y= A, and add to that previous solution.

10. Sep 4, 2010

vorcil

What should I do here Ivy? I think I'll just go down to the physics department on tuesday and ask one of the tutors to help me, I don't know how they expect me to solve this, sad thing is, I think everyone else in my class can do this really quick, but I can't, bugger it.

11. Sep 4, 2010

klondike

Ivy's way is more "mathematician". The main idea is assuming the solution in form of complex exponential. Since God makes $$(e^{st})'=se^{st}$$, the DE will be converted to a algebra equations. I think this hinted Oliver Heaviside to invent using LT to solve this kind of DE, by letting s=a+jb.

As engineer, we were trained to use LT for linear ODE with constant coeffs which is the guts of LTI systems. We did learn Ivy's way before LT was taught. So I think you should know both.

Last edited: Sep 4, 2010