($m_{s,0}^{\ast }$) from $30$ MeV ($0.6m$) to $90$ MeV ($0.9m$) shifts the single collective peak from about $14.6$ ($15.0$) MeV to $13.9$ ($13.9$) MeV. These results are consistent with the results shown in Fig. [XEaPb208]{}. In addition, the calculated width with MSL0 by varying individually $K_{0}$, $L$, and $m_{s,0}^{\ast }$ shows roughly the same value as that of experimental data [@You04; @Sag07b]. This agreement implies that the width of ISGMR is essentially determined by the Landau damping and the coupling to the continuum, which are properly taken into account in the present calculations.

The ISGMR energy $E_{\mathrm{ISGMR}}$ is conventionally related to a finite nucleus incompressibility $K_{A}(N,Z)$ for a nucleus with $N$ neutrons and $Z
$ protons ($A=N+Z$) by the following definition$$E_{\mathrm{ISGMR}}=\sqrt{\frac{\hbar ^{2}K_{A}(N,Z)}{m\left\langle
r^{2}\right\rangle }},  \label{EGMRKa}$$where $m$ is the nucleon mass and $\left\langle r^{2}\right\rangle $ is the mean square mass radius of the nucleus in the ground state. Similarly to the semi-empirical mass formula, the finite nucleus incompressibility $K_{A}(N,Z)
$ is usually expanded as [@Bla80] $$\begin{aligned}
K_{A}(N,Z) &=&K_{0}+K_{\mathrm{surf}}A^{-1/3}+K_{\mathrm{curv}}A^{-2/3}
\notag \\
&&+(K_{\tau }+K_{\mathrm{ss}}A^{-1/3})\left( \frac{N-Z}{A}\right) ^{2}
\notag \\
&&+K_{\mathrm{Coul}}\frac{Z^{2}}{A^{4/3}}+\cdot \cdot \cdot .  \label{KA1}\end{aligned}$$Neglecting the curvature term $K_{\mathrm{curv}}$, the surface symmetry term $K_{\mathrm{ss}}$ and the other higher-order terms of the finite nucleus incompressibility $K_{A}(N,Z)$ 