# Strange differential equation

1. May 4, 2008

### BeauGeste

I'm not sure where to put this but I though DEQ would be a good start.
I have been dealing with the following differential equations:

$$\frac{d \mathbf{m_1}}{dt} = \gamma_1 \mathbf{m_1} \times \mathbf{B} - \frac{n_{2}}{\gamma_{cr}} \mathbf{m_1} + \frac{n_1}{\gamma_{cr}} \mathbf{m_2}$$

$$\frac{d \mathbf{m_2}}{dt} = \gamma_2 \mathbf{m_2} \times \mathbf{B} - \frac{n_{1}}{\gamma_{cr}} \mathbf{m_2} + \frac{n_2}{\gamma_{cr}} \mathbf{m_1}$$

The weird things is that $$\frac{d (\mathbf{m_1} + \mathbf{m_2})}{dt}$$ is not independent of $$\gamma_{cr}$$ even though when you add bottom and top deq's the $$\gamma_{cr}$$ terms cancel out. All I know is that it has something to do with $$\gamma_{1} \neq \gamma_{2}$$ .

Also when you convert to polar coordinates (i assume the m-vectors are only in the x-y direction and B is in the z-direction), the 2 diff. eq.'s take on a form where it's obvious that the $$\gamma_{cr}$$'s stick around.

I'm not too fluent in differential equations so I'm not sure if there's an obvious answer here.
Thanks.

2. May 4, 2008

### Parlyne

$$\frac{d(\boldsymbol{m_1}+\boldsymbol{m_2})}{dt}$$ certainly has no explicit dependence on $$\gamma_{cr}$$. However, the differential equation you get for it is not sufficient to uniquely specify $$\boldsymbol{m_1}+\boldsymbol{m_2}$$, even beyond the usual integration constant. We can see this by a change a variable.

So, let $$\boldsymbol{m_+} = \boldsymbol{m_1}+\boldsymbol{m_2}$$ and $$\boldsymbol{m_-} = \boldsymbol{m_1}-\boldsymbol{m_2}$$.

Then, it's not too hard to see that

$$\frac{d\boldsymbol{m_+}}{dt} = (\gamma_+ \boldsymbol{m_+} + \gamma_- \boldsymbol{m_-})\times \boldsymbol{B}$$

where $$\gamma_+ = \frac{\gamma_1 + \gamma_2}{2}$$ and $$\gamma_- = \frac{\gamma_1 - \gamma_2}{2}$$. This immediately means that it is necessary to know the dynamics of $$\boldsymbol{m_-}$$, as well. But, since that will be found by subtracting the two equations, it will automatically depend, explicitly, on $$\gamma_{cr}$$. Thus, $$\boldsymbol{m_+}$$ will depend on $$\gamma_{cr}$$ implicitly through its dependence on $$\boldsymbol{m_-}$$.

3. May 4, 2008

### BeauGeste

Wow. Thanks a lot for that. I've been banging my head for a while about that

A follow up question, if I may:

Will your reasoning be true whenever

$$\frac{d(\boldsymbol{m_1}+\boldsymbol{m_2})}{dt} \neq 0$$?

4. May 4, 2008

### Parlyne

No, it won't. If the equation for $$\frac{d\boldsymbol{m_+}}{dt}$$ is totally independent of $$\boldsymbol{m_-}$$, then $$\boldsymbol{m_+}$$ will not depend on $$\gamma_{cr}$$. As your system is defined, this will happen whenever $$\gamma_- = 0$$. And, this will happen in exactly those cases where $$\gamma_1 = \gamma_2$$.

5. May 5, 2008

### BeauGeste

OK, I see. I guess the better question to ask is
When in general will you know that a solution will be independent of some parameter? For this case you made a change in variable and realized that the parameter in question will be an important part of the solutions. Is that something you would do for any system of equations? Or is there a more general way to tell?

Is there a general name for these type of systems? Would I find this info in a Diff. Eq. textbook?

Thanks again.