1effect said:
I know how to use LaTex, I just preferred to use an editable file that people could write into.
Alright, although I still don't see why you don't attach it to your post instead of linking to an external site.
I made some progress, see http://www.savefile.com/files/1458817 .
What you did was transform a system of 2 first order equations into a single second order equation. This is always possible with any system of n differential-algebraic equations ofp^{th}-order (for n > 1): by differentiating you can transform it into a system of n-1 differential-algebraic equations of order p+1.
Since it is possible to do this in general, I am not surprised that you did it in a particular case. Although I am not sure from your document that you realize that the final equation will be second order. When you write:
\frac{d v_y}{dt} = f(v_x,\frac{d v_x}{dt})
I am sure you mean to differentiate the expression immediately above it, although you do not say this. If I am not mistaken and you did in fact take a time derivative, then wouldn't it be more appropriate to write:
\frac{d v_y}{dt} = g(v_x,\frac{d v_x}{dt},\frac{d^2 v_x}{dt^2})
Since differentiating will in fact cause the appearance of various \frac{d^2 v_x}{dt^2}) terms by the chain rule. Here is your expression:
<br />
v_y = \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}<br />
This is time derivative of that which you do not show:
<br />
\frac{\frac{-\frac{8 c^2 \text{vx}'(t) \text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2<br />
\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)}{2<br />
\sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}}<br />
So what we have now is the second-order equation involving only one unknown function v_x (where I have no substituted your expression for \gamma, for the sake of brevity):
<br />
\text{}\leftB \text{vx}(t) q-\frac{\text{E} q}{\gamma ^2}+\frac{m \left(\frac{-\frac{8 c^2 \text{vx}'(t)<br />
\text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2<br />
\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)\right)<br />
\gamma }{2 \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2<br />
q^2}}}}=0\right<br />
Personally, I consider this to be a regress rather then a progress. It can always be done, but rarely would you want to. We are certainly no closer to finding an exact solution in terms of known functions.
It seems that a symbolic (not analytic) solution can be found. Analytic means something else.
A symbolic solution? In that case how about y = f(t). Or did you want to be even more pedantic and say "a solution that can be represented in terms of arithmetic operations, root extractions, trigonometric/exponential functions, or other cataloged special functions." The term analytic is often used to describe solutions that can be represented in this way, and in this context the meaning was clear. Did you really think I was talking about functions defined on a complex domain?
In any case, I stand by my original statement that this equation cannot be solved "in the way that you are looking for." Did you consider my advice to apply perturbation theory? I am sure the results would be interesting.
