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Set $N={\int }_V{P}^{2}dV$, $D={\int }_{{\mathbb{R}}^3}|\nabla P{|}^{2}dV$, $M={\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$, and $q={\max }_V|Q|$. The confined flow satisfies the impermeability condition $\mathbf{u}\cdot \mathbf{n}=0$, so $Q=0$ at $r=a$. Multiplying the radial magnetic induction scalar equation by $P$ and integrating, the advection term vanishes by the divergence theorem:The same integration by parts, using $\nabla \cdot \mathbf{B}=0$, gives

For diffusion, Green's first identity in the sphere gives ${\int }_{V}P\Delta P=-{\int }_V|\nabla P{|}^2+{\int }_{\partial V}P{\partial }_{r}P$. In the exterior, $\Delta P=0$ and the normal on its inner boundary is $-\hat{\mathbf{r}}$, soThe infinity term vanishes for the isolated decaying field. The insulating boundary condition for the radial magnetic scalar permits matching the two traces, and therefore the exterior contribution is essential: it turns the diffusive term into $-\eta D$. Altogether,Applying the pointwise Cauchy-Schwarz inequality and then its integral version,Thus a nonzero field with stationary or growing $N$ must satisfy the radial-flow dynamo energy bound:

The temporal qualification matters. For general time-dependent dynamo action, nondecay does not make $N^{\prime }\ge 0$ at every instant; the boxed bound applies at nondecreasing instants, to a marginal or growing mode, or in an appropriate long-time sense. A uniform strict violation, $q\sqrt{M/D}\le \eta -\epsilon$ with $\epsilon >0$, gives $N^{\prime }\le -2\epsilon D$. The whole-space Sobolev inequality and Holder inequality give $N\le |V{|}^{2/3}\Vert P{\Vert }_6^2\le C_{a}D$, forcing exponential decay of $N$. Alternatively, for a bounded statistically steady field, time averaging the energy balance and applying Cauchy-Schwarz once more gives $q_{\ast }^2\ge {\eta }^2⟨D⟩/⟨M⟩$, where $q_{\ast }={\sup }_{t}q(t)$. Sustained dynamo action cannot have a uniform negative dissipation gap.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. The resistive induction equation, together with the solenoidal conditions on both fields, isSince $D_t\mathbf{x}=\mathbf{u}$, the product rule givesAlso $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}=\mathbf{x}\cdot \Delta \mathbf{B}$. Substitution proves the radial magnetic induction scalar equation:

In the insulating exterior the quasistatic magnetic field is current-free, so $\nabla \times \mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$. Thus $\Delta \mathbf{B}=0$ and $\Delta P=0$ for $r>a$, with decay at infinity for an isolated field. This is an instantaneous exterior Laplace equation, rather than a diffusion equation with the interior value of $\eta$.

Use the usual nonmagnetic-interface assumptions: equal magnetic permeability on both sides and no surface current. The electromagnetic boundary conditions then make $B_r$ and the tangential magnetic field continuous. Hence $P=r{B}_r$ is continuous. In spherical coordinates, the solenoidal magnetic-field constraint readswhere the last term is the angular divergence on the unit sphere. Its right-hand side is also continuous, giving the insulating boundary condition for the radial magnetic scalar:Equivalently, the decaying exterior degree-$l$ spherical harmonic is proportional to $r^{-(l+1)}$, so its boundary amplitude satisfies ${\partial }_r{P}_{lm}(a)=-(l+1)P_{lm}(a)/a$.