magnetohydrodynamics - How is the MHD magnetic field time evolution equation transformed to the vector potential time evolution equation?

Starting from the time evolution equation of the magnetic field for incompressible MHD (magnetohydrodynamics)

$$\frac{\partial \vec{B}}{\partial t} = \nabla \times (\vec{v} \times \vec{B}) + \frac{\eta}{\mu_{0}} \nabla^{2} \vec{B}$$

and the definition of the vector potential $\vec{A}$

$$\nabla \times \vec{A} = \vec{B}$$

How is it that one can arrive at the time evolution equation of the vector potential? Which is

$$\frac{\partial \vec{A}}{\partial t} + (\vec{v} \cdot \nabla) \vec{A} = \frac{\eta}{\mu_{0}} \nabla^{2} \vec{A}$$

according to these lecture notes (NB: PDF) from Rony Keppens.

I have derived that

$$\begin{align} \nabla \times (\vec{v} \times \vec{B}) &= -(\nabla \cdot \vec{v})\vec{B} - (\vec{v} \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{v} + (\nabla \cdot \vec{B})\vec{v}\\ &= -(\vec{v} \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{v} \end{align}$$ where I have used the Maxwell equation that $\nabla \cdot \vec{B} = 0$ and the continuity equation for an incompressible fluid $\nabla \cdot \vec{v} = 0$. However, I don't think this helps me at all.

Chiefly, I think my difficulty is understanding how to recover $\frac{\partial \vec{A}}{\partial t}$ from setting $\frac{\partial \vec{B}}{\partial t} = \frac{\partial (\nabla \times \vec{A})}{\partial t}$

But in general my question is: how does one derive the time evolution equation for the vector potential in the form written above?

Answer

You are making the problem too difficult for yourself. You should be looking for vector calculus identities and space-time orthogonality. Specifically,

$$\frac{\partial}{\partial t}\nabla\times\mathbf A=\nabla\times\frac{\partial\mathbf A}{\partial t}\\ \nabla^2\left(\nabla\times\mathbf A\right)=\nabla\times\left(\nabla^2\mathbf A\right)$$ You'll then have three terms with $\nabla\times$ in front (ignoring constants): $$\nabla\times\frac{\partial\mathbf A}{\partial t}=\nabla\times\mathbf u\times\nabla\times\mathbf A+\nabla\times\nabla^2\mathbf A\tag{1}$$ This is a vector relation of the form, $$\nabla\times\mathbf f=\nabla\times\mathbf g$$ which implies $$\mathbf f=\mathbf g+\nabla h$$ where $h$ is some scalar.

Thus, Equation (1) can be 'uncurled' by considering the equivalent relation

$$\frac{\partial\mathbf A}{\partial t}=\mathbf u\times\nabla\times\mathbf A+\nabla^2\mathbf A+\nabla\phi$$ (taking the curl of this returns (1)) where $\phi$ is your gauge. Then using the BAC-CAB rule, you get $$\frac{\partial\mathbf A}{\partial t}=\nabla\left(\mathbf u\cdot\mathbf A\right)-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A+\nabla\phi\tag{2}$$ You can then fix the gauge, choosing $\phi=-\mathbf u\cdot\mathbf A$ such that (2) becomes $$\begin{align} \frac{\partial\mathbf A}{\partial t}&=\nabla\left(\mathbf u\cdot\mathbf A\right)-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A-\nabla\left(\mathbf u\cdot\mathbf A\right)\\ &=-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A\tag{$\star$} \end{align}$$ where ($\star$) is the result you are after.