这是一份uwa西澳大学PHYS3012的成功案例

which is equivalent to
$$
\frac{d^{3} \hat{\gamma}}{d t^{3}} \mp \sqrt{-1} \kappa_{2} \frac{d^{2} \hat{\gamma}}{d t^{2}}+\left(\kappa_{1}^{2}+1\right) \frac{d \hat{\gamma}}{d t} \mp \sqrt{-1} \kappa_{2} \hat{\gamma}=0
$$
where the double signs take the same signature. Since its characteristic equation
$$
\lambda^{3} \mp \sqrt{-1} \kappa_{2} \lambda^{2}+\left(\kappa_{1}^{2}+1\right) \lambda \mp \sqrt{-1} \kappa_{2}=0
$$
should have pure imaginary solutions, by setting $\lambda=\sqrt{-1} \Lambda$ we have
$$
\Lambda^{3} \mp \kappa_{2} \Lambda^{2}-\left(\kappa_{1}^{2}+1\right) \Lambda \pm \kappa_{2}=0,
$$
which turns to
$$
\left(\Lambda \mp \frac{\kappa_{2}}{3}\right)^{3}-\frac{1}{3}\left(3 \kappa_{1}^{2}+\kappa_{2}^{2}+3\right)\left(\Lambda \mp \frac{\kappa_{2}}{3}\right) \mp \frac{\kappa_{2}}{27}\left(9 \kappa_{1}^{2}+2 \kappa_{2}^{2}-18\right)=0 .
$$

PHYS3012 COURSE NOTES ：

Finally, in (c) is supposed that non-autonomus vector field $\xi_{h}^{t}$ on $M$ is integrable up to time one, i.e. there exists a family of curves $h \rightarrow f_{h}^{t}$ in the diffeomorphism group of $M$ with $f_{0}^{t}=i d$ and
$$
\frac{d}{d h} f_{h}^{t}=\xi_{h}^{t} \circ f_{h^{}}^{t} $$ Then since Lie derivative $\mathcal{L}{\xi{h}^{t}}$ satisfies $\mathcal{L}{\xi{h}^{t}}=d i_{\xi_{h}^{t}}+i_{\xi_{h}^{t}} d$ and the forms $\omega_{t}^{h}$ are closed, one concludes that
$$
\frac{d}{d h}\left(\left(f_{h}^{t}\right)^{} \omega_{t}^{h}\right)=\left(f_{h}^{t}\right)^{}\left(\mathcal{L}{\xi{h}^{t}} \omega_{t}^{h}+\frac{d}{d h} \omega_{t}^{h}\right)=\left(f_{h}^{t}\right)^{}\left(d i_{\xi_{h}} \omega_{t}^{h}+\sigma_{t}\right)=0
$$
for all $h \in[0,1]$. This shows that $f^{t}:=f_{h}^{t}$ fulfills the requirement $\left(f^{t}\right)^{} \tilde{\omega}{t}=$ $\tilde{\omega}{t}^{a}$ since
$$
\left(f^{t}\right)^{} \tilde{\omega}{t}-\tilde{\omega}{t}^{a}=\left(f_{1}^{t}\right)^{} \tilde{\omega}{t}^{1}-\left(f{0}^{t}\right)^{} \tilde{\omega}{t}^{0}=\int{0}^{1} \frac{d}{d h}\left(\left(f_{h}^{t}\right)^{*} \omega_{t}^{h}\right) d h=0 .
$$
Putting the parameter $t=1$ and $\varepsilon \rightarrow 0$ we obtain the existing part of the theorem.