 Solution

Assume $\kappa ,c_s,{\Sigma }_0>0$, and use the positive Toomre parameter. The dimensionless dispersion relation isAt fixed dimensionless gravitational softening $\delta$, maximum growth corresponds to maximizing $F(s)=2s{e}^{-\delta s}-s^2$. Differentiating gives the most unstable softened disk wavenumberFor $\delta >0$, the right side is strictly decreasing on $0<s<1/\delta$, while the left side increases; there is exactly one positive solution. The equation implies $\delta s_{\ast }<1$ and $s_{\ast }<1-\delta s_{\ast }$, soInstability occurs precisely when $Q^2<F(s_{\ast })$. Thus the critical Toomre parameter with exponential softening isThe printed description of a minimum $Q$ for instability reverses the threshold: $Q_c$ is the upper boundary of unstable values and the lower boundary of stable ones. Because $s_{\ast }<1/(1+\delta )\le 1$ and $2s-s^2$ increases on $[0,1]$,Both strict comparisons become equality in the unsoftened limit $\delta =0$: then $s_{\ast }=1$ and $Q_c=1$. Gravitational softening suppresses short-wave gravity, shifts the most dangerous mode to a longer wavelength and requires stronger self-gravity, or smaller $Q$, for instability.

The plot holds $Q$ fixed and changes gravitational softening. If $Q$ is varied physically at fixed $\kappa ϵ/c_s$, remember $\delta =(\kappa ϵ/c_s)/Q$: the marginal curve must be evaluated at its corresponding $\delta$, rather than treating these two dimensionless parameters as independently fixed along that physical variation.