更改

</math>

</math>

|{{EquationRef|1}}}}

|{{EquationRef|1}}}}

其中<math>f\left(x^{*}, r_{c}\right)=0</math>，且因为<math>x^*</math>是不动点，且由相切情况可得<math>\left \frac {\partial f}{\partial x}\right|_{\left(x^{*} ; r_{c}\right)}=0</math>，再将<math>\left \frac {\partial f}{\partial r}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>a</math>，将<math>\left \frac 1/2{\partial^{2} f}{\partial x^{2}}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>b</math>。忽略后面的高阶项，即可得到<math>\dot{x} = a(r-r_c)+b(x-x^*)^2</math>

其中<math>f\left(x^{*}, r_{c}\right)=0</math>，且因为<math>x^*</math>是不动点，且由相切情况可得<math>\left. \frac {\partial f}{\partial x}\right|_{\left(x^{*} ; r_{c}\right)}=0</math>，再将<math>\left. \frac {\partial f}{\partial r}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>a</math>，将<math>\left. \frac 1/2{\partial^{2} f}{\partial x^{2}}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>b</math>。忽略后面的高阶项，即可得到<math>\dot{x} = a(r-r_c)+b(x-x^*)^2</math>

=案例=

=案例=