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For constant isotropic $\alpha ,\beta$ and zero mean flow, the mean induction equation isLet $\overline{\mathbf{B}}=\hat{\mathbf{B}}{e}^{\sigma t+i\mathbf{k}\cdot \mathbf{x}}$, with $\mathbf{k}\cdot \hat{\mathbf{B}}=0$. On this two-dimensional transverse space, $i\mathbf{k}\times$ has eigenvalues $\pm k$. For example, if $\mathbf{k}=k{\mathbf{e}}_z$, the vectors ${\mathbf{e}}_x\pm i{\mathbf{e}}_y$ have these curl eigenvalues. The homogeneous alpha-squared dynamo growth criterion follows fromChoose the circular polarization of the mode to match the sign of $\alpha$. This is an alpha-squared dynamo: the alpha effect supplies both parts of the regeneration loop without differential rotation.

If arbitrarily long wavelengths are permitted, every nonzero alpha has a growing band $0<k<|\alpha |/{\eta }_T$. Maximizing the quadratic givesprovided $k_{\ast }\ell \ll 1$ so the mean-field approximation is valid. A uniform field has $k=0$, zero curl and zero growth; a constant alpha cannot amplify that uniform mode. In a periodic domain of size $L$, the lowest nonzero wavenumber is $2\pi /L$, giving $|\alpha |L/{\eta }_T>2\pi$. Other boundaries and geometries give different critical constants, but the general scale criterion is $|\alpha |L/{\eta }_T$ sufficiently large. Nonzero kinetic helicity alone is therefore not a sufficient finite-domain dynamo criterion: its induced alpha must overcome diffusion on an available mean-field scale.

Because $v$ is a divergence-free vector field, the product rule givesThe integral of a divergence over the periodic domain vanishes. Hence integration by parts yieldsThis is the skew-symmetry of incompressible transport.

A periodic domain identifies opposite faces of a fundamental cell, producing a flat torus. On an interval $[0,L]$, smooth periodic boundary conditions require $u(0,t)=u(L,t)$ and matching spatial derivatives. These endpoints are one location, rather than two independently controlled walls. Periodic integration by parts has no boundary term, and zero-mean functions have no zero Fourier mode.

Use dimensionless $x$ and $t$, the periodic domain $[0,2\pi ]$, and the same initial conditionWith zero PDE source term, compare the heat equation $u_t=u_{xx}$ and the transport equation $u_t+u_x=0$. Their exact solutions areEach cosine satisfies ${\partial }_x^2\cos (nx)=-n^2\cos (nx)$, so the heat equation reduces its amplitude according to $A_n^{\prime }=-n^2{A}_n$. The superposition principle combines the constant and two Fourier modes. The transport equation shifts the argument of the entire profile. Both formulas meet the initial condition and matching periodic boundary conditions exactly.^[ref]

At $t=1$, the heat amplitudes are approximately $0.368$ and $0.0000494$: the small ripples disappear much sooner than the broad variation. The transport equation retains amplitudes $1$ and $0.4$, shifting the profile right by one unit with wraparound. Both preserve spatial average $1$, but only the heat equation approaches the uniform profile. This gives diffusion its usefulness for smoothing uneven temperatures, while the transport equation models a profile carried by a flow. The signed profile can represent a scaled temperature deviation.

To restore units, let physical position be $X=\ell x$ and time be $T=\tau t$. For $U_T=D{U}_{XX}$, the dimensionless coefficient is $D\tau /{\ell }^2$, with $D$ measured in length squared per time. For $U_T+c{U}_X=0$, it is $c\tau /\ell$, with $c$ measured in length per time. Both plotted coefficients equal one when $D={\ell }^2/\tau$ and $c=\ell /\tau$. Thus the example compares specified models on the same scales, rather than identifying diffusivity with velocity.