With the spin-orbit force, we can modify our Woods-Saxon potential to $$ \hat{u}_{\mathrm{ext}}(r)=-\frac{V_0}{1+\exp{(r-R)/a}}+V_{so}(r)\boldsymbol{ls}, $$ with $$ V_{so}(r) = V_{so}\frac{1}{r}\frac{d f_{so}(r)}{dr}, $$ where we have $$ f_{so}(r) = \frac{1}{1+\exp{(r-R_{so})/a_{so}}}. $$