# How do I solve this?

1. Mar 23, 2008

### 1effect

I have a system of differential equatiuons (see http://www.savefile.com/files/1458343 [Broken]). How can I solve it? Is it even solvable? Thank you

Last edited by a moderator: May 3, 2017
2. Mar 23, 2008

### Crosson

Did you notice that you can attach .doc files to your posts at physicsforums?

Even better, you could learn LaTeX which is the main way we communicate formulas on these forums.

In this case, you even could have stated the problem in words:

"Is it possible to solve the relativistic equations of motion for a charged particle in a constant electric and magnetic field?"

In this case no one has found analytical solutions to the full nonlinear system. Since we know the exact solution for the case $\gamma = 1$ the best hope for analytical results in your case is to use perturbation theory together with the first or second term of the power series expansion for gamma. Maybe someone has done this before.

Last edited: Mar 23, 2008
3. Mar 23, 2008

### 1effect

I know how to use LaTex, I just preferred to use an editable file that people could write into.

I made some progress, see http://www.savefile.com/files/1458817 [Broken]. It seems that a symbolic (not analytic) solution can be found. Analytic means something else.

Last edited by a moderator: May 3, 2017
4. Mar 23, 2008

### nicksauce

Most people on these boards are not going to download a file in .doc format like that. It's just a virus waiting to happen.

Last edited: Mar 23, 2008
5. Mar 23, 2008

### tiny-tim

Yup. I took one look at your first post several hours ago, and opted out.

6. Mar 23, 2008

### 1effect

Sorry, the file is getting very long and complicated, there is no virus , you can scan it with any of the virus protection programs. I am not here to play games, I am trying to get help in solving a tough problem.

7. Mar 23, 2008

### Crosson

Alright, although I still don't see why you don't attach it to your post instead of linking to an external site.

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 of$p^{th}$-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:

$$v_y = \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}$$

This is time derivative of that which you do not show:

$$\frac{\frac{-\frac{8 c^2 \text{vx}'(t) \text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2 \sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)}{2 \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}}$$

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):

$$\text{}\leftB \text{vx}(t) q-\frac{\text{E} q}{\gamma ^2}+\frac{m \left(\frac{-\frac{8 c^2 \text{vx}'(t) \text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2 \sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)\right) \gamma }{2 \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}}=0\right$$

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.

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.

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8. Mar 23, 2008

### 1effect

Thank you, there is a small error in your derivative but I get the gist. It doesn't look promising.

This one has a few small errors as well but it doesn't matter. It doesn't appear solvable in terms of finding a symbolic solution.

This is what symbolic means. Analytic means that the function is continous with continous derivatives. I didn't mean to be pedantic, I meant to convey more precisely what I was looking for.

I am not familiar with the term, do you mean using Taylor expansions for $$v_x(t)$$ and $$v_y(t)$$? Can you take me thru the first steps?

9. Mar 23, 2008

### Crosson

This is a good introduction to the method:

http://www.sm.luth.se/~johanb/applmath/chap2en/index.html [Broken]

I might try to work it out another time.

I neglected to mention that I took only the positive branch of your multi-valued solution, being concerned about murky existence and uniqueness issues.

If I say the solution is an analytic function then I mean it's $C^\infty$ but if I say that the solution is analytic than I only means that it has the potential to be analyzed (natural language meaning, since there is no formal mathematical definition for the concept we are talking about).

Which one is a better natural language word to describe "by-hand" solutions, in contrast with the solutions which are only guaranteed by existence theorems?

Analytic: Dividing into elemental parts or basic principles, i.e. has the potential to be analyzed

Symbolic: Of, relating to, or expressed by means of symbols or a symbol.

It seems to me that any time we know that there is a solution we can express it symbolically as $y = f(t)$ but what we are looking for in this setting is an analytic solution i.e. one that is amenable to further analysis.

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10. Mar 23, 2008

### 1effect

Thank you.
The immediate question that arises is how to extend the method to systems of equations. The link shows only the application to single equations. Even more worrisome, by the author's own admission, the approach doesn't always work. This seems to evolve into a very tough problem.

Last edited by a moderator: May 3, 2017