= Solution
For $\bar\nu=C r^a\Sigma^b$, the response is $q=b+1$. A nonaccreting background has $\bar\nu_0\Sigma_0\propto r^{-1/2}$, while $\Sigma_0\propto r^{-p}$, so
$$
\boxed{a-(b+1)p=-\tfrac12,\qquad a+\tfrac12=qp.}
$$
Consequently $\bar\nu_0\propto r^{p-1/2}$. On this steady background, introduce the <square-root-radius diffusion transform>
$$
x=\sqrt r,\qquad g=\sqrt r\,\bar\nu_0\Sigma_1.
$$
Since $q$ is constant and $\partial_r=(2x)^{-1}\partial_x$, the <linear equation> becomes
$$
\partial_tg=3q\bar\nu_0 r^{-1/2}\partial_r(r^{1/2}\partial_rg)
=\frac{3q\bar\nu_0(r)}{4r}\partial_x^2g.
$$
Thus $A(x)\propto x^{2p-3}$, and the required choice is
$$
\boxed{p=\tfrac32,\qquad a=1+\tfrac32b,\qquad A=\frac{3q}{4}\frac{\bar\nu_0}r=\text{constant}.}
$$
For a <Fourier mode> $g\propto e^{\lambda t+ik_xx}$, $\lambda=-Ak_x^2$. Positive <kinematic viscosity> makes $A$ have the sign of $q$, so $q<0$ gives growing modes whose rate increases with $k_x^2$. The formal equation is a <backward heat equation> and predicts arbitrarily rapid small-scale amplification. Physically the <thin disc> diffusion closure applies only to <wavelengths> sufficiently larger than the thickness and stress-relaxation scales; this formal limit identifies the need for a cutoff, rather than a finite fastest <wavelength> absent from the model. The boundary case $q=0$ has vanishing linear transport response.
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