and $m_{s,0}^{\ast }$ can lead to a theoretical uncertainty of about $\pm 16$ MeV for the extraction of $K_{0}$. We further find the $E_{\mathrm{ISGMR}}$ difference between $^{100}$Sn and $^{132}$Sn displays a strong correlation with $L$ and thus provides a potential probe of the symmetry energy.

Methods {#Theory}
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Skyrme-Hartree-Fock approach and macroscopic properties of asymmetric nuclear matter
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The EOS of isospin asymmetric nuclear matter, given by its binding energy per nucleon, can be expanded to $2$nd-order in isospin asymmetry $\delta $ as $$E(\rho ,\delta )=E_{0}(\rho )+E_{\mathrm{sym}}(\rho )\delta ^{2}+O(\delta
^{4}),  \label{EOSANM}$$where $\rho =\rho _{n}+\rho _{p}$ is the baryon density with $\rho _{n}$ and $\rho _{p}$ denoting the neutron and proton densities, respectively; $\delta
=(\rho _{n}-\rho _{p})/(\rho _{p}+\rho _{n})$ is the isospin asymmetry; $E_{0}(\rho )=E(\rho ,\delta =0)$ is the binding energy per nucleon in symmetric nuclear matter, and the nuclear symmetry energy is expressed as $$E_{\mathrm{sym}}(\rho )=\frac{1}{2!}\frac{\partial ^{2}E(\rho ,\delta )}{\partial \delta ^{2}}|_{\delta =0}.  \label{Esym}$$Around $\rho _{0}$, the symmetry energy can be characterized by using the value of $E_{\text{\textrm{sym}}}({\rho _{0}})$ and the density slope parameter $L=3{\rho _{0}}\frac{\partial E_{\mathrm{sym}}(\rho )}{\partial
\rho }|_{\rho ={\rho _{0}}}$, i.e.,

$$E_{\mathrm{sym}}(\rho )=E_{\text{\textrm{sym}}}({\rho _{0}})+\frac{L}{3}(\frac{\rho -{\rho _{0}}}{{\rho _{0}}})+O((\frac{\rho -{\rho _{0}}}{{\rho _{0}}})^{2}).$$

In the standard Skyrme Hartree-Fock approach, the nuclear effective 