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### Derivation of the time-independent Landau-Lifshitz equation (Video)

Static configurations of the magnetisation $\mathbf{m}(\mathbf{x})$ minimize the magnetic energy $E(\mathbf{m})$.

We will consider only the exchange energy. We have the Lagrangian density

$$\mathcal{L} = \frac{1}{2}\, \partial_x\mathbf{m}\cdot\partial_x\mathbf{m}.$$

We have to consider the constraint $\mathbf{m}^2 = 1$. The constraint is imposed via a Lagrange multiplier $\lambda(x)$ [Raj, Sec. 3.3][Yan, Sec. 2.1.1][FG, Sec. 12.2].

We have to extremize the functional

$$L[\mathbf{m}] = \int dx \underbrace{\left[ \frac{1}{2}\,\partial_x\mathbf{m}\cdot\partial_x\mathbf{m} + \frac{\lambda(x)}{2} (1-\mathbf{m}^2) \right]}_{\mathcal{L}}.$$

The Euler-Lagrange equation

The functional $L$ is minimized for $\mathbf{m}(x)$ that satisfies the Euler-Lagrange equation

$$-\frac{\delta L}{\delta \mathbf{m}} = 0 \Rightarrow \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial_x\mathbf{m}}\right) - \frac{\partial\mathcal{L}}{\partial\mathbf{m}} = 0.$$

We calculate

$$-\frac{\delta L}{\delta \mathbf{m}} = \frac{d}{dx}\left( \partial_x\mathbf{m} \right) + \lambda \mathbf{m} = 0$$

or

$$\partial_x^2 \mathbf{m} + \lambda \mathbf{m} = 0.$$