= Solution
Work with an incompressible <Newtonian fluid> of constant <kinematic viscosity>. The <vorticity equation> is $D\boldsymbol\omega/Dt=(\boldsymbol\omega\cdot\nabla)\mathbf u+\nu\nabla^2\boldsymbol\omega$. In the axisymmetric <Burgers vortex> the swirl does not advect the axial <vorticity>, and the stretching rate is $\partial_z u_z=\alpha$. Put $s(t)=\delta^2(t)$ and $W=\omega_z$. The Gaussian profile has
$$
\frac{\partial_tW}{W}=\dot s\left(-\frac1s+\frac{r^2}{s^2}\right),\quad
\frac{u_r\partial_rW}{W}=\frac{\alpha r^2}{s},\quad
\frac{\nabla^2W}{W}=-\frac4s+\frac{4r^2}{s^2}.
$$
Substitution into $\partial_tW+u_r\partial_rW=\alpha W+\nu\nabla^2W$ matches both the constant and $r^2$ terms precisely when
$$
\boxed{\dot s+\alpha s=4\nu}.
$$
For every finite positive initial $s_0$,
$$
\boxed{\delta^2(t)=\frac{4\nu}{\alpha}+\left(s_0-\frac{4\nu}{\alpha}\right)e^{-\alpha t}},\qquad
\delta(t)\longrightarrow\sqrt{4\nu/\alpha}.
$$
The strain compresses the cross-section and stretches its <vorticity>, while <kinematic viscosity> spreads the core. The attracting radius is the balance of these effects; its relaxation time is $\alpha^{-1}$. Total <circulation> stays fixed because stretching increases the <vorticity> as the cross-sectional area decreases.
At the steady core radius, the swirl speed is of order $\Gamma_0/\delta$, and its <velocity gradient tensor> is of order $\Gamma_0/\delta^2$. Its <viscous dissipation> integrated over a cross-section therefore scales as
$$
\mathcal D_{\rm tube}\sim\nu\left(\frac{\Gamma_0}{\delta^2}\right)^2\delta^2
\sim\boxed{\alpha\Gamma_0^2}.
$$
This is per unit length and per unit density. More precisely, for the swirl contribution, <integration by parts> relates the integrated strain <viscous dissipation> to $\nu\int W^2\,dA$; the boundary term vanishes for the $1/r$ swirl velocity. Thus
$$
\boxed{\mathcal D_{\rm tube}=\frac{\nu\Gamma_0^2}{2\pi\delta^2}=\frac{\alpha\Gamma_0^2}{8\pi}}.
$$
Multiply by density for power per physical length. Here the azimuthal velocity's <circulation> $\Gamma$ is $\Gamma_0$, as follows by integrating $W$ over the cross-section. The result is finite as $\nu\to0$ with strain and <circulation> held fixed. Local core <viscous dissipation> grows like $\nu^{-1}$ while its area shrinks like $\nu$. For $\Gamma_0/\nu\gg1$, the swirl gradients dominate the background strain within the core.
This finite quantity is the excess <viscous dissipation> due to the vortex. The uniform straining flow alone dissipates $3\nu\alpha^2$ per unit volume, and integrating that over an infinite cross-section diverges. The total <viscous dissipation> of the complete idealized flow is therefore not a finite tube power unless the background is subtracted or spatially bounded.
A <Burgers vortex sheet> instead smooths a fixed velocity jump $U$ across a layer of thickness $\delta_s\sim\sqrt{\nu/\alpha}$. Its <viscous dissipation> per sheet area is of order
$$
\mathcal D_{\rm sheet}\sim\nu(U/\delta_s)^2\delta_s\sim U^2\sqrt{\nu\alpha}\longrightarrow0.
$$
\b[A tube with fixed <circulation> supports finite <viscous dissipation> in a vanishingly small core; a sheet with fixed velocity jump does not.] This makes strained <vortex tubes> plausible intense dissipative structures at high <Reynolds number>. It does not assert that sheets are absent from <turbulence>: sheets can carry <inertial range> organization and become unstable, producing finer vortical structures.
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