= Solution
Let $K$ be the <kinetic energy>, $M$ the <magnetic energy> and $E=K+M$:
$$
K=\frac\rho2\int_D|\mathbf v|^2\,dV,\qquad M=\frac1{2\mu_0}\int_D|\mathbf B|^2\,dV.
$$
Dot the <magnetohydrodynamic momentum equation> with $\mathbf v$ and integrate. <Incompressibility> and the zero boundary <velocity> remove the advective and <pressure> work. The viscous term integrates to $-\rho\nu\int_D|\nabla\mathbf v|^2$. Thus
$$
\dot K=\int_D\mathbf v\cdot(\mathbf j\times\mathbf B)\,dV-\rho\nu\int_D|\nabla\mathbf v|^2\,dV.
$$
The <ideal magnetohydrodynamic induction equation> gives, with no boundary contribution,
$$
\dot M=\frac1{\mu_0}\int_D\mathbf B\cdot\nabla\times(\mathbf v\times\mathbf B)\,dV=-\int_D\mathbf v\cdot(\mathbf j\times\mathbf B)\,dV.
$$
The <Lorentz force density> therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla\cdot\mathbf v=0$ and $\mathbf v=0$ on the boundary, another <integration by parts> gives $\int|\nabla\mathbf v|^2=\int|\boldsymbol\omega_v|^2$, where $\boldsymbol\omega_v=\nabla\times\mathbf v$ is <vorticity>. The <viscous magnetic-relaxation energy identity> is
$$
\boxed{\frac{dE}{dt}=-\rho\nu\int_D|\boldsymbol\omega_v|^2\,dV,\qquad E(0)=\frac1{2\mu_0}\int_D|\mathbf B_0|^2\,dV.}
$$
The conserved nonzero <magnetic helicity> prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the <magnetic vector potential>, so that $\|\mathbf A\|_2\leq C_D\|\mathbf B\|_2$. The <Cauchy-Schwarz inequality> gives the <magnetic-energy lower bound from helicity>,
$$
M\geq\frac{|H_M|}{2\mu_0C_D}>0.
$$
Consequently $E$ decreases to a finite positive limit and
$$
\rho\nu\int_0^\infty\|\boldsymbol\omega_v(t)\|_2^2,dt=E(0)-E_\infty<\infty.
$$
Under the intended regular long-time interpretation of smooth <vorticity>, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\|\boldsymbol\omega_v(t)\|_2^2$ is <uniformly continuous>, <uniformly continuous integrable dissipation tends to zero> proves $d(t)\to0$. The <Poincaré inequality> and the identity above then imply $\|\mathbf v(t)\|_2\to0$.
To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces $\boldsymbol\omega_v=0$, and the zero boundary <velocity> forces $\mathbf v=0$. The induction equation is then stationary, and the momentum equation becomes
$$
\boxed{\mathbf j_\infty\times\mathbf B_\infty=\nabla p_\infty,\qquad \nabla\cdot\mathbf B_\infty=0,\qquad \mathbf B_\infty\cdot\mathbf n=0.}
$$
This is <magnetostatic equilibrium>, with nonzero field allowed and required here by conserved <magnetic helicity>. It is not necessarily a <force-free magnetic field>: a <pressure gradient> can balance the magnetic force. If the field develops a <current sheet>, equilibrium is understood with the corresponding interface force balance.
There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth <vorticity> does not require smooth <electric current density> in the limiting state.
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