The exchange model (Video)

We consider a fully isotropic ferromagnet in two space dimensions, $\bm{m}=\bm{m}(x,y,t)$, thus we have the model that contains only the exchange interaction

$$ \dot{\bm{m}} = -\bm{m}\times\Delta\bm{m}. $$

Remark. Any global rotation of $\bm{m}$ around any axis in space leaves the equation invariant.

For static (time-independent) solutions, we have

$$ \begin{aligned} \bm{m}\times\Delta\bm{m} = 0 & \Rightarrow \bm{m}\times(\partial_1^2\bm{m} + \partial_2^2\bm{m}) = 0 \\ & \Rightarrow \partial_1(\bm{m}\times \partial_1\bm{m}) + \partial_2(\bm{m}\times \partial_2\bm{m}) = 0. \end{aligned} $$

We note that the equation is satisfied if the following simpler equations are satisfied

$$ \bm{m}\times \partial_1\bm{m} = \partial_2\bm{m} \qquad\text{and}\qquad \bm{m}\times \partial_2\bm{m} = -\partial_1\bm{m}. \tag{Bgl} $$

We may indeed verify that

$$ \partial_1\underbrace{(\bm{m}\times \partial_1\bm{m})}{\partial_2\bm{m}} + \partial_2\underbrace{(\bm{m}\times \partial_2\bm{m})}{-\partial_1\bm{m}} = 0. $$

Note that Eqs. (Bgl) are, in fact, a single equation because the one implies the other. For example, take the cross product of the first with $\bm{m}$,

$$ \bm{m}\times(\bm{m}\times \partial_1\bm{m}) = \bm{m}\times\partial_2\bm{m} \Rightarrow -\partial_1\bm{m} = \bm{m}\times\partial_2\bm{m}. $$

In conclusion, the forms satisfying the first order equation

$$ \bm{m}\times \partial_1\bm{m} = \partial_2\bm{m} $$