 Solution

Choose a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$. The resistive induction equation permits the evolution equationA gauge transformation changes $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ by a surface integral of the normal magnetic field, which is zero here. Thus the integrated magnetic helicity has gauge invariance under the stated boundary conditions.

Differentiate both factors in $H$. Use the stated vanishing-boundary assumption to remove the helicity surface flux; this can be implemented in the usual closed-boundary gauge $\mathbf{n}\times \mathbf{A}=0$, or by extending the localized field and integrating over all space with sufficient decay. The curl integration by parts identity then givesThe motional term is zero pointwise because $(\mathbf{u}\times \mathbf{B})\cdot \mathbf{B}=0$. The scalar-potential term integrates to zero by $\nabla \cdot \mathbf{B}=0$ and the boundary condition. The resistive magnetic-helicity balance is thereforeUnlike the resistive magnetic energy contribution, this signed integral need not have a fixed sign.