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Freezing coefficients locally in the Keplerian viscous diffusion equation gives ${\Sigma }_{1,t}\simeq 3{\partial }_{\Sigma }(\bar{\nu }\Sigma ){\Sigma }_{1,rr}$. A radial Fourier mode grows at rate $-3{\partial }_{\Sigma }(\bar{\nu }\Sigma )k_r^2$. Negative response reverses ordinary smoothing and gives a backward heat equation, within the wavelength range where the thin disc transport closure applies.

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 ${\Sigma }_0\propto r^{-p}$, soConsequently ${\bar{\nu }}_0\propto r^{p-1/2}$. On this steady background, introduce the square-root-radius diffusion transformSince $q$ is constant and ${\partial }_r=(2x{)}^{-1}{\partial }_x$, the linear equation becomesThus $A(x)\propto x^{2p-3}$, and the required choice isFor a Fourier mode $g\propto e^{\lambda t+i{k}_{x}x}$, $\lambda =-A{k}_x^2$. 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.

The negative-diffusion criterion for viscous disk instability is ${\partial }_{\Sigma }(\bar{\nu }\Sigma )<0$. A surface density enhancement then transports angular momentum less effectively and grows. Freezing the Keplerian viscous diffusion equation locally gives growth proportional to the square of radial wavenumber, within the thin-disc transport range; wavelengths must still be sufficiently long compared with the disk scale height for that closure. The formal short-wave limit is a backward heat equation.