Используя (4.2), выразим $V^{\prime \prime \prime}(0)$ через производные поля $\boldsymbol{X}_{0}$ в точке $(0,0)$. Предположим, что координаты были выбраны так, что
\[
\begin{aligned}
d \widehat{X}_{0}(0,0)= & \left(\begin{array}{ll}
\frac{\partial \widehat{X}_{0}^{1}}{\partial x_{1}}(0,0,0) & \frac{\partial \widehat{X}_{0}^{1}}{\partial x_{2}}(0,0,0) \\
\frac{\partial \widehat{X}_{0}^{2}}{\partial x_{1}}(0,0,0) & \frac{\partial \widehat{X}_{0}^{2}}{\partial x_{2}}(0,0,0)
\end{array}\right)= \\
& =\left(\begin{array}{cc}
0 & |\lambda(0)| \\
-|\lambda(0)| & 0
\end{array}\right) .
\end{aligned}
\]

Такой выбор переменных не необходим, но он значительно упрощает вычисления, и хотя наш метод нахождения $V^{\prime \prime \prime}(0)$ и будет работать, если переменные специально не выбирать, однако в этом случае нашей формулой нельзя пользоваться.

Из (4.2) видно, что
\[
\begin{aligned}
V^{\prime}\left(x_{1}\right) & =\int_{0}^{T\left(x_{1}\right)} \frac{d}{d x_{1}}\left[\hat{X}^{1}\left(a_{t}\left(x_{1}, 0\right), b_{t}\left(x_{1}, 0\right)\right)\right] d t+ \\
& +T^{\prime}\left(x_{1}\right) \hat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right) . \\
V^{\prime \prime}\left(x_{1}\right) & =\int_{0}^{T\left(x_{1}\right)} \frac{d^{2}}{d x_{1}^{2}}\left[\hat{X}^{1}\left(a_{t}\left(x_{1}, 0\right), b_{t}\left(x_{1}, 0\right)\right)\right] d t+ \\
& +T^{\prime}\left(x_{1}\right) \frac{d}{d x_{1}}\left[\hat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)\right]+ \\
& +T^{\prime \prime}\left(x_{1}\right) \hat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)+ \\
& +T^{\prime}\left(x_{1}\right) \frac{d}{d x_{1}}\left[\hat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)\right] .
\end{aligned}
\]

Используя правило дифференцирования сложной функции, получаем
\[
\begin{aligned}
V^{\prime \prime}\left(x_{1}\right) & =\int_{0}^{T\left(x_{1}\right)} \frac{d^{2}}{d x_{1}^{2}}\left[\widehat{X}^{1}\left(a_{t}\left(x_{1}, 0\right), b_{t}\left(x_{1}, 0\right)\right)\right] d t+ \\
& +T^{\prime}\left(x_{1}\right)\left[\left.\frac{\partial \widehat{X}^{1}}{\partial a} \frac{\partial a_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\left.\frac{\partial \widehat{X}^{1}}{\partial b} \frac{\partial b_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}\right]+ \\
& +T^{\prime \prime}\left(x_{1}\right) \widehat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)+ \\
& +T^{\prime}\left(x_{1}\right)\left[\left.T^{\prime}\left(x_{1}\right) \frac{\partial \widehat{X}^{1}}{\partial a} \frac{\partial a_{t}}{\partial t}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\left.\frac{\partial \widehat{X}^{1}}{\partial a} \frac{\partial a_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\right. \\
& \left.+\left.T^{\prime}\left(x_{1}\right) \frac{\partial \widehat{X}^{1}}{\partial b} \frac{\partial b_{t}}{\partial t}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\left.\frac{\partial \widehat{X}^{1}}{\partial b} \frac{\partial b_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}\right]+ \\
& +T^{\prime}\left(x_{1}\right)^{2}\left[\left.\frac{\partial \widehat{X}^{1}}{\partial a} \frac{\partial a_{t}}{\partial t}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\left.\frac{\partial \widehat{X}^{1}}{\partial b} \frac{\partial b_{t}}{\partial t}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}\right]+ \\
& +T^{\prime \prime}\left(x_{1}\right) \widehat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right) .
\end{aligned}
\]

Дифференцируя еще раз, получим
\[
\begin{array}{l}
V^{\prime \prime \prime}\left(x_{1}\right)=\int_{0}^{T\left(x_{1}\right)} \frac{d^{3}}{d x_{1}^{3}}\left[\widehat{X}^{1}\left(a_{t}\left(x_{1}, 0\right), b_{t}\left(x_{1}, 0\right)\right)\right] d t+ \\
+T^{\prime}\left(x_{1}\right) \frac{d^{2}}{d x_{1}^{2}}\left[\widehat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)\right]+ \\
+2 T^{\prime \prime}\left(x_{1}\right)\left[\left.\frac{\partial \widehat{X}^{1}}{\partial a} \cdot \frac{\partial a_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+\left.\frac{\partial \widehat{X}}{\partial b} \frac{\partial b_{t}}{\partial x_{1}}\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}\right]+
\end{array}
\]

\[
\begin{array}{l}
+2 T^{\prime}\left(x_{1}\right)\left[\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}} \cdot \frac{\partial a_{t}}{\partial x_{1}}\left[\frac{\partial a_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial a_{t}}{\partial x_{1}}\right]+\right. \\
+\frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b} \frac{\partial a_{t}}{\partial x_{1}}\left(\frac{\partial b_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial b_{t}}{\partial x_{1}}\right)+\frac{\partial \widehat{X}^{1}}{\partial a}\left(\frac{\partial^{2} a_{t}}{\partial t \partial x_{1}} T^{\prime}\left(x_{1}\right)+\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}\right)+ \\
+\frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}} \frac{\partial b_{t}}{\partial x_{1}}\left(\frac{\partial b_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial b_{t}}{\partial x_{1}}\right)+ \\
\left.+\frac{\partial \widehat{X}^{1}}{\partial b}\left(\frac{\partial^{2} b_{t}}{\partial t \partial x_{1}} T^{\prime}\left(x_{1}\right)+\frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}\right)\right]\left.\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+ \\
+\left.2 T^{\prime}\left(x_{1}\right) T^{\prime \prime}\left(x_{1}\right)\left(\frac{\partial \widehat{X}^{1}}{\partial a} \cdot \frac{\partial a_{t}}{\partial t}+\frac{\partial \widehat{X}^{1}}{\partial b} \cdot \frac{\partial b_{t}}{\partial t}\right)\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+ \\
+T^{\prime}\left(x_{1}\right)^{2}\left[\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}} \cdot \frac{\partial a_{t}}{\partial t}\left(\frac{\partial a_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial a_{t}}{\partial x_{1}}\right)+\right. \\
+\frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b} \frac{\partial a_{t}}{\partial t}\left(\frac{\partial b_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial b_{t}}{\partial x_{1}}\right)+\frac{\partial \widehat{X}^{1}}{\partial a}\left(\frac{\partial^{2} a_{t}}{\partial t^{2}} T^{\prime}\left(x_{1}\right)+\frac{\partial^{2} a_{t}}{\partial x_{1} \partial t}\right)+ \\
+\frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}} \frac{\partial b_{t}}{\partial t}\left(\frac{\partial b_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\frac{\partial b_{t}}{\partial x_{1}}\right)+ \\
\left.+\frac{\partial \widehat{X}^{1}}{\partial b}\left(\frac{\partial^{2} b_{t}}{\partial t^{2}} T^{\prime}\left(x_{1}\right)+\frac{\partial^{2} b_{t}}{\partial x_{1} \partial t}\right)\right]\left.\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)}+ \\
+T^{\prime \prime \prime}\left(x_{1}\right) \widehat{X}^{1}\left(a_{T\left(x_{1}\right)}\left(x_{1}, 0\right), b_{T\left(x_{1}\right)}\left(x_{1}, 0\right)\right)+ \\
+T^{\prime \prime}\left(x_{1}\right)\left[\frac{\partial \widehat{X}^{1}}{\partial a} \frac{\partial a_{t}}{\partial t} \cdot T^{\prime}\left(x_{1}\right)+\frac{\partial \widehat{X}^{1}}{\partial a} \cdot \frac{\partial a_{t}}{\partial x_{1}}+\frac{\partial \widehat{X}^{1}}{\partial b} \cdot \frac{\partial b_{t}}{\partial t} T^{\prime}\left(x_{1}\right)+\right. \\
\left.+\frac{\partial \widehat{X}^{1}}{\partial b} \cdot \frac{\partial b_{t}}{\partial x_{1}}\right]\left.\right|_{\left(T\left(x_{1}\right), x_{1}, 0\right)} \cdot
\end{array}
\]

При $x_{1}=0$ это уравнение можно значительно упростить. Относительно поля в точке $(0,0)$ известно следующее:
\[
a_{t}(0,0)=b_{t}(0,0)=0 \text { для всех } t \text {, }
\]
\[
\frac{\partial a_{t}}{\partial t}(0,0)=\frac{\partial b_{t}}{\partial t}(0,0)=0,
\]
\[
d \hat{X}\left(a_{t}(0,0), b_{t}(0,0)\right)=d \hat{X}(0,0)=\left(\begin{array}{cc}
0 & |\lambda(0)| \\
-|\lambda(0)| & 0
\end{array}\right),
\]
\[
d \hat{\varphi}_{t}(0,0)=e^{t d \widehat{X}(0,0)}=\left(\begin{array}{rr}
\cos |\lambda(0)| t & \sin |\lambda(0)| t \\
-\sin |\lambda(0)| t & \cos |\lambda(0)| t
\end{array}\right),
\]
\[
T(0)=\frac{2 \pi}{|\lambda(0)|}
\]

и
\[
T^{\prime}(0)=0 \text {. }
\]

Доказательство (4.10). Пусть $S\left(x_{1}\right)=T\left(x_{1}, \mu\left(x_{1}\right)\right)$. Тогда $S^{\prime}(0)=0$, потому что для данного малого $x>0$ существует малое $y<0$, такое, что $S(x)=S(y)$. Таким образом, $\frac{S(x)-S(0)}{x}$ и $\frac{S(y)-S(0)}{y}$ имеют противоположные знаки. Выбирая $x_{n} \downarrow 0$, получаем требуемый результат. Далее, $S^{\prime}(0)=\frac{\partial T}{\partial x_{1}}(0,0)+\mu^{\prime}(0) \frac{\partial T}{\partial \mu}(0,0)$ Но $\mu^{\prime}(0)=0$; как было показано в доказательстве теоремы 3.1 (см. стр. 59). Поэтому
\[
V^{\prime \prime \prime}(0)=\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{3} \widehat{X}^{1}}{\partial x_{1}^{3}}\left(a_{t}\left(x_{1}, 0\right), b_{t}\left(x_{1}, 0\right)\right) d t .
\]

Вычисляя $\left.\frac{\partial^{3} \widehat{X}^{1}}{\partial x_{1}^{3}}\right|_{a_{t}(0,0), b_{t}(0,0)}$, будем иметь
\[
\begin{array}{l}
V^{\prime \prime \prime}(0)=\int_{0}^{2 \pi /|\lambda(0)|}\left[\frac{\partial^{3} \widehat{X}}{\partial a^{3}}(0,0) \cos ^{3}|\lambda(0)| t-\frac{\partial^{3} \widehat{X}}{\partial b^{3}}(0,0) \sin ^{3}|\lambda(0)| t+\right. \\
+3 \frac{\partial^{3} \widehat{X}^{1}}{\partial a \partial b^{2}}(0,0) \cos |\lambda(0)| t \sin ^{2}|\lambda(0)| t- \\
-3 \frac{\partial^{3} \hat{X}^{1}}{\partial a^{2} \partial b}(0,0) \cos ^{2}|\lambda(0)| t \sin |\lambda(0)| t+ \\
+3 \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t- \\
-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0) \cdot \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t+ \\
+3 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}(0,0)\left(\frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t-\right. \\
\left.\left.-\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t\right)+|\lambda(0)| \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0)\right] d t= \\
=\int_{0}^{2 \pi / 1 \hat{(0) \mid}}\left[3 \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t-\right. \\
-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0) \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t+ \\
+3 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}(0,0)\left(\frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t-\right. \\
\left.\left.-\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t\right)+|\lambda(0)| \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0 ; 0)\right] d t \text {. } \\
\end{array}
\]

Чтобы получить формулу для $V^{\prime \prime \prime}(0)$, зависящую лишь от производных $\mathcal{X}$ в начале координат, мы должны выразить производные потока (например, $\frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0)$ ) через производные поля $\mathcal{X}$ в точке $(0,0)$. Это сделать можно потому, что начало координат является неподвижной точкой потока, порожденного векторным полем $\mathbb{X}$. В силу важности этой идеи установим этот факт в более общем случае.
(4.1) Теорема. Пусть $X$-гладкое векторное поле класса C $^{k}$ на $R^{n}$, такое, что $X(0)=0$ (или $X(p)=0$ ). Пусть 甲 $_{t}-$ поток поля $X$. Тогда первые три (или первые $j$ ) производные $\varphi_{t}$ в 0 могут быть вычислены через первые три (или первые j) производные векторного поля $X$ в точке 0 .
Доказательство. Рассмотрим $\frac{\partial \varphi_{i}^{i}}{\partial x_{i}}(0):$
\[
\begin{array}{r}
\frac{\partial}{\partial t} \frac{\partial \varphi_{t}^{i}}{\partial x_{j}}(0)=\frac{\partial}{\partial x_{j}} \frac{\partial \varphi_{t}^{i}}{\partial t}(0)=\frac{\partial}{\partial x_{j}} X^{i} \circ \varphi_{t}(0)=\frac{\partial X^{i}}{\partial x_{k}} \circ \varphi_{t}(0) \frac{\partial \varphi_{t}^{k}}{\partial x_{j}}(0)= \\
=\frac{\partial X^{i}}{\partial x_{k}}(0) \frac{\partial \varphi_{t}^{k}}{\partial x_{j}}(0),
\end{array}
\]

потому что $\varphi_{t}(0)=0$. Далее, $\frac{\partial \varphi_{0}^{i}}{\partial x_{j}}(0)=\delta_{i j}$, так как $\varphi_{0}(x)=x$ для всех $x$. Поэтому $d \varphi_{t}(0)$ удовлетворяет дифференциальному уравнению $\frac{\partial}{\partial t}\left(d \varphi_{t}(0)\right)=d X(0) \cdot d \varphi_{t}(0)$, и $\quad d \varphi_{0}(0)=I$. Поэтому $d \varphi_{t}(0)=e^{t d X(0)}$.
Рассмотрим $\frac{\partial^{2} \varphi_{i}^{i}}{\partial x_{j} \partial x_{k}}(0)$ :
\[
\begin{array}{c}
\frac{\partial}{\partial t} \frac{\partial^{2} \varphi_{t}^{i}}{\partial x_{j} \partial x_{k}}(0)=\frac{\partial^{2}}{\partial x_{j} \partial x_{k}} \frac{\partial \varphi_{t}^{l}}{\partial t}(0)=\frac{\partial^{2}}{\partial x_{i} \partial x_{k}} X^{i} \circ \varphi_{t}(0)= \\
=\frac{\partial}{\partial x_{j}}\left(\frac{\partial X^{i}}{\partial x_{l}} \circ \varphi_{t} \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}\right)(0)= \\
=\frac{\partial^{2} X^{i}}{\partial x_{p} \partial x_{l}} \circ \varphi_{t} \frac{\partial \varphi_{t}^{p}}{\partial x_{j}} \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}+\frac{\partial X^{i}}{\partial x_{l}} \circ \varphi_{t} \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k}}(0)= \\
=\frac{\partial^{2} X^{i}}{\partial x_{p} \partial x_{l}}(0) \frac{\partial \varphi_{t}^{p}}{\partial x_{j}}(0) \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}(0)+\frac{\partial X^{i}}{\partial x_{l}}(0) \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{j} x_{t}}(0) .
\end{array}
\]

Кроме того, $\frac{\partial^{2} \varphi_{0}^{i}}{\partial x_{i} \partial x_{k}}(0)=0$. Получили дифференциальное уравнение:
\[
\frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k}}(0)=d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{j}}(0), \frac{\partial \varphi_{t}}{\partial x_{k}}(0)\right)+d X(0) \cdot \frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{k}} .
\]

Его решением является
\[
\begin{array}{c}
\frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{k}}(0)=e^{t d X(0)} \int_{0}^{t} e^{-s d X(0)} d^{2} X(0)\left(\frac{\partial \varphi_{s}}{\partial x_{j}}(0), \frac{\partial \varphi_{s}}{\partial x_{k}}(0)\right) d s+ \\
+e^{t d X(0)} \frac{\partial^{2} \varphi_{0}}{\partial x_{j} \partial x_{k}}(0), \\
\frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{k}}(0)=e^{t d X(0)} \int_{0}^{t} e^{-s d X(0)} d^{2} X(0)\left(\frac{\partial \varphi_{s}}{\partial x_{j}}(0), \frac{\partial \varphi_{s}}{\partial x_{k}}(0)\right) d s .
\end{array}
\]

Наконец, рассмотрим $\frac{\partial^{3} \varphi_{t}^{l}}{\partial x_{f} \partial x_{k} \partial x_{h}}(0)$ :
\[
\begin{array}{l}
\frac{\partial}{\partial t}\left(\frac{\partial^{3} \varphi_{t}^{l}}{\partial x_{i} \partial x_{k} \partial x_{h}}\right)(0)=\frac{\partial^{3}}{\partial x_{j} \partial x_{k} \partial x_{h}}\left(\frac{\partial \varphi_{t}^{l}}{\partial t}\right)(0)= \\
\quad=\frac{\partial^{3}}{\partial x_{j} \partial x_{k} \partial x_{h}} X^{i} \circ \varphi_{t}(0)=\frac{\partial}{\partial x_{h}}\left(\frac{\partial^{2} X^{l}}{\partial x_{p} \partial x_{t}} \circ \varphi_{t} \frac{\partial \varphi_{t}^{p}}{\partial x_{j}} \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}+\right. \\
\left.\quad+\frac{\partial X^{l}}{\partial x_{t}} \circ \varphi_{t} \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{t} \partial x_{k}}\right)(0)=\left[\frac{\partial^{3} X^{i}}{\partial x_{p} \partial x_{l} \partial x_{q}} \circ \varphi_{t} \frac{\partial \varphi_{t}^{q}}{\partial x_{h}} \frac{\partial \varphi_{t}^{p}}{\partial x_{t}} \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}+\right. \\
\quad+\frac{\partial^{2} X^{i}}{\partial x_{p} \partial x_{l}} \circ \varphi_{t}\left(\frac{\partial^{2} \varphi_{t}^{p}}{\partial x_{j} \partial x_{h}} \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}+\frac{\partial \varphi_{t}^{p}}{\partial x_{l}} \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{k} \partial x_{h}}\right)+ \\
\left.\quad+\frac{\partial^{2} X^{j}}{\partial x_{p} \partial x_{l}} \circ \varphi_{t} \frac{\partial \varphi_{t}^{p}}{\partial x_{h}} \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k}}+\frac{\partial X^{i}}{\partial x_{l}} \circ \varphi_{t} \frac{\partial^{3} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k} \partial x_{h}}\right](0)= \\
\quad=\frac{\partial^{3} X^{l}}{\partial x_{p} \partial x_{l} \partial x_{q}}(0) \frac{\partial \varphi_{t}^{q}}{\partial x_{h}}(0) \frac{\partial \varphi_{t}^{p}}{\partial x_{j}}(0) \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}(0)+ \\
\quad+\frac{\partial^{2} X^{l}}{\partial x_{p} \partial x_{l}}(0)\left(\frac{\partial^{2} \varphi_{t}^{p}}{\partial x_{j} \partial x_{h}}(0) \frac{\partial \varphi_{t}^{l}}{\partial x_{k}}(0)+\frac{\partial \varphi_{t}^{p}}{\partial x_{j}}(0) \frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{k} \partial x_{h}}(0)+\right. \\
\left.\quad+\frac{\partial^{2} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k}}(0) \frac{\partial \varphi_{t}^{p}}{\partial x_{h}}(0)\right)+\frac{\partial X^{l}}{\partial x_{l}}(0) \frac{\partial^{3} \varphi_{t}^{l}}{\partial x_{j} \partial x_{k} \partial x_{k}}(0) .
\end{array}
\]

При этом $\frac{\partial^{3} \varphi_{0}^{i}}{\partial x_{j} \partial x_{k} \partial x_{h}}(0)=0$. Следовательно, мы получили дифференциальное уравнение
\[
\begin{array}{r}
\frac{\partial}{\partial t}\left(\frac{\partial^{3} \varphi_{t}^{i}}{\partial x_{j} \partial x_{k} \partial x_{h}}(0)\right)=d^{3} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{l}}(0), \frac{\partial \varphi_{t}}{\partial x_{k}}(0), \frac{\partial \varphi_{t}}{\partial x_{h}}(0)\right)+ \\
+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{k}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{i} \partial x_{h}}(0)\right)+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{j}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{k} \partial x_{h}}(0)\right)+ \\
\quad+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{h}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{k}}(0)\right)+d X(0)\left(\frac{\partial^{3} \varphi_{t}}{\partial x_{j} \partial x_{k} \partial x_{h}}(0)\right)
\end{array}
\]

и $\frac{\partial^{3} \varphi_{0}}{\partial x_{j} \partial x_{k} \partial \dot{x}_{h}}(0)=0$. Его решением является следующее выражение:
\[
\begin{array}{l}
\frac{\partial^{3} \varphi_{t}}{\partial x_{l} \partial x_{k} \partial x_{h}}(0)= \\
=e^{t d X(0)} \int_{0}^{t} e^{-s d X(0)}\left\{d^{3} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{j}}(0), \frac{\partial \varphi_{t}}{\partial x_{k}}(0), \frac{\partial \varphi_{t}}{\partial x_{h}}(0)\right)+\right. \\
+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{k}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{h}}(0)\right)+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{j}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{k} \partial x_{h}}(0)\right)+ \\
\left.+d^{2} X(0)\left(\frac{\partial \varphi_{t}}{\partial x_{h}}(0), \frac{\partial^{2} \varphi_{t}}{\partial x_{j} \partial x_{k}}\right)\right\} d s .
\end{array}
\]

В рассматриваемом нами случае
\[
d \hat{X}(0,0)=\left(\begin{array}{cc}
0 & |\lambda(0)| \\
-|\lambda(0)| & 0
\end{array}\right),
\]

и поэтому $d \hat{\varphi}_{t}(0,0)=\left(\begin{array}{cc}\cos |\lambda(0)| t & \sin |\lambda(0)| t \\ -\sin |\lambda(0)| t & \cos |\lambda(0)| t\end{array}\right)$.
Мы хотим вычислить
\[
\frac{\partial^{2} \hat{\varphi}_{t}^{\mathrm{A}}}{\partial x_{1}^{2}}(0,0)=\left(\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0), \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0)\right) \text { и } \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0) .
\]

$=\left(\begin{array}{rr}\cos |\lambda(0)| s & -\sin |\lambda(0)| s \\ \sin |\lambda(0)| s & \cos |\lambda(0)| s\end{array}\right)$. Далее,
\[
\begin{array}{l}
d^{2} \widehat{X}(0)\left(\frac{\partial \Phi_{s}}{\partial x_{1}}(0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0)\right)=\left(\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}(0,0) \cos ^{2}|\lambda(0)| s-\right. \\
\quad-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}(0,0) \cos |\lambda(0)| s \cdot \sin |\lambda(0)| s+ \\
\quad+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}(0,0) \sin ^{2}|\lambda(0)| s, \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}(0,0) \cos ^{2}|\lambda(0)| s- \\
\quad-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}(0,0) \cos |\lambda(0)| s \cdot \sin |\lambda(0)| s+ \\
\left.\quad+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}(0,0) \sin ^{2}|\lambda(0)| s\right)
\end{array}
\]

Следовательно,
\[
\begin{array}{l}
e^{-s d \hat{X}(0)} d^{2} \widehat{X}(0)\left(\frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0)\right)= \\
=\left[\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}(0) \cos ^{3}|\lambda(0)| s-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}(0) \sin ^{3}|\lambda(0)| s+\right. \\
+\left(-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} d x_{2}}(0)-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}(0)\right) \sin |\lambda(0)| s \cdot \cos ^{2}|\lambda(0)| s+ \\
+\left(2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2} \partial x_{2}^{\prime}}(0)+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}(0)\right) \cos |\lambda(0)| s \cdot \sin ^{2}|\lambda(0)| s, \\
+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}(0) \cos ^{3}|\lambda(0)| s+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}(0) \sin ^{3}|\lambda(0)| s+ \\
+\left(-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}(0)+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}(0)\right) \sin |\lambda(0)| s \cdot \cos ^{2}|\lambda(0)| s+ \\
\left.+\left(-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}(0)+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}(0)\right) \cos |\lambda(0)| s \cdot \sin ^{2}|\lambda(0)| s\right] \\
\end{array}
\]

и, таким образом,
\[
\begin{array}{l}
\int_{0}^{t} e^{-s d \widehat{X}(0)} d^{2} \widehat{X}(0)\left(\frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0)\right) d s= \\
=\frac{1}{3|\lambda(0)|}\left[\left(-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right)+3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \sin |\lambda(0)| t+\right. \\
+3 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}} \cos |\lambda(0)| t+\left(-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}+2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}\right) \sin ^{3}|\lambda(0)| t+ \\
+\left(-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right) \cos ^{3}|\lambda(0)| t, \\
\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right)+3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \sin |\lambda(0)| t- \\
-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}} \cos |\lambda(0)| t+\left(-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}\right) \sin ^{3}|\lambda(0)| t+ \\
\left.+\left(\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right) \cos ^{3}|\lambda(0)| t\right]
\end{array}
\]
(где все производные вычислены в начале координат).
Собирая все вместе, получим
\[
\begin{array}{l}
e^{t d \widehat{X}(0)} \int_{0}^{t} e^{-s d \widehat{X}(0)} d^{2} \widehat{X}(0)\left(\frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0)\right) d s= \\
=\left(\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0), \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0)\right)= \\
=\frac{1}{3|\lambda(0)|}\left[\left(-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right) \cos |\lambda(0)| t+\right. \\
+\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right) \sin |\lambda(0)| t+ \\
+\left(3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}\right) \sin |\lambda(0)| t \cdot \cos |\lambda(0)| t+ \\
+3 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}} \cos ^{2}|\lambda(0)| t+3 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}} \sin ^{2}|\lambda(0)| t+ \\
+\left(-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}+2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}\right) \cos |\lambda(0)| t \cdot \sin ^{3}|\lambda(0)| t+ \\
\end{array}
\]

\[
\begin{array}{l}
+\left(\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right) \sin |\lambda(0)| t \cdot \cos ^{3}|\lambda(0)| t+ \\
+\left(-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right) \cos ^{4}|\lambda(0)| t+ \\
+\left(-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}\right) \sin ^{4}|\lambda(0)| t \\
\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{4} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right) \cos |\lambda(0)| t+ \\
+\left(2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right) \sin |\lambda(0)| t+ \\
+\left(3 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}-3 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}\right) \sin |\lambda(0)| t \cdot \cos |\lambda(0)| t- \\
-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \sin ^{2}|\lambda(0)| t-3 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}} \cos { }^{2}|\lambda(0)| t+ \\
+\left(\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}\right) \sin ^{2}|\lambda(0)| t \cdot \cos ^{3}|\lambda(0)| t+ \\
+\left(-\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}\right) \cos |\lambda(0)| t \cdot \sin ^{3}|\lambda(0)| t+ \\
+\left(\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}\right) \sin ^{4}|\lambda(0)| t+ \\
+\left(\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}+2 \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}}-\frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}}\right) \cos ^{4}|\lambda(0)| t \cdot
\end{array}
\]

Перед вычислением $\frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0)$ (являющимся очень громоздким) используем полученную выше информацию, чтобы упростить выражение для $V^{\prime \prime \prime}(0)$. Чтобы это сделать, мы должны сосчитать:
\[
\begin{array}{c}
\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t \cdot d t, \\
\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t \cdot d t, \\
\int_{0}^{2 \pi /|(0)|} \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cdot \cos |\lambda(0)| t \cdot d t
\end{array}
\]

и
\[
\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cdot \sin |\lambda(0)| t d t .
\]

Результаты этого вычисления следующие:
\[
\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) d t=\frac{\pi}{|\lambda(0)|^{2}}\left[\frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}+\frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}\right] .
\]
\[
\begin{aligned}
\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t d t & \\
& =\frac{\pi}{3|\lambda(0)|^{2}}\left(2 \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}-\frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}\right) .
\end{aligned}
\]
$\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t d t=$
\[
=\frac{\pi}{\left.3 \backslash \lambda(0)\right|^{2}}\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial a \partial b}+\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}\right) .
\]
$\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t d t=$
\[
=\frac{\pi}{3|\lambda(0)|^{2}}\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial a \partial b}+\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}\right) .
\]
$\int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t d t=$
\[
=\frac{\pi}{3|\lambda(0)|^{2}}\left(2 \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}+\frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}\right) .
\]

Поэтому
\[
\begin{aligned}
V^{\prime \prime \prime}(0) & =|\lambda(0)| \int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0) d t+ \\
& +3 \int_{0}^{2 \pi /|\lambda(0)|}\left\{\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t-\right. \\
& -\frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0) \frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t+ \\
& +\frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}(0,0)\left[\frac{\partial^{2} b_{t}}{\partial x_{1}^{2}}(0,0) \cos |\lambda(0)| t-\right. \\
& \left.\left.-\frac{\partial^{2} a_{t}}{\partial x_{1}^{2}}(0,0) \sin |\lambda(0)| t\right]\right\} d t,
\end{aligned}
\]

T. e.
\[
\begin{aligned}
V^{\prime \prime \prime}(0) & =|\lambda(0)| \int_{0}^{2 \pi /|\lambda(0)|} \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0) d t+ \\
& +\left\lvert\, \lambda \frac{\pi}{\left.(0)\right|^{2}} \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}\left(-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}-\frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}\right)+\right. \\
& +\frac{\pi}{|\lambda(0)|^{2}} \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}\left(-2 \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}-2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}-\frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}\right)= \\
& =|\lambda(0)| \int_{0}^{2 \pi /|(0)|} \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0) d t- \\
& -\frac{\pi}{|\lambda(0)|^{2}}\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}} \cdot \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}+\frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}} \cdot \frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}} \cdot \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}+\right. \\
& \left.+2 \frac{\partial^{2} \hat{X}^{1}}{\partial b^{2}} \cdot \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}+\frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}} \cdot \frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}+2 \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}} \cdot \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}\right),
\end{aligned}
\]

где все производные взяты в начале координат. Чтобы вычислить $\frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0)$, нам нужно пользоваться уравнением
\[
\begin{array}{l}
\left(\frac{\partial^{3} a_{t}}{\partial x_{1}^{3}}(0,0), \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0)\right) \\
=e^{t d \hat{X}(0)} \int_{0}^{t} e^{-s d \widehat{X}(0)}\left\{d^{3} \hat{X}(0)\left(\frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0,0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0,0), \frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0,0)\right)+\right. \\
\left.+3 d^{2} \hat{X}(0)\left(\frac{\partial \hat{\varphi}_{s}}{\partial x_{1}}(0,0), \frac{\partial^{2} \hat{\varphi}_{s}}{\partial x_{1}^{2}}(0,0)\right)\right\} d s
\end{array}
\]

Вычисления являются чрезвычайно длинными ${ }^{1}$ ), но простыми, поэтому мы укажем, как их проводить, и приведем окончательные результаты. После очень длинных вычислений по-
1) Мы не шутим! Чтобы пыполнить предыдущие вычисления, нужно быть готовым к потере нескольких дней. Детали могут быть высланы, но лишь по серьезному поводу.

лучим
\[
\begin{array}{l}
\int_{0}^{2 \pi /|\lambda(0)|} e^{t d \widehat{X}(0)} \int_{0}^{t} e^{-s d \widehat{X}(0)} d^{2} \widehat{X}(0)\left(\frac{\partial \varphi_{s}}{\partial x_{1}}\left(0 \frac{\partial^{2} \varphi_{s}}{\partial x_{1}^{2}}(0,0)\right) d s d t=\right. \\
=\left(\text { нечто, } \frac{2 \pi}{|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}+\frac{5 \pi}{4|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}}+\right. \\
+\frac{7 \pi}{4|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1}^{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}+\frac{5 \pi}{4|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}}+ \\
+\frac{3 \pi}{4|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}+\frac{5 \pi}{4|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}+ \\
\left.+\frac{\pi}{|\lambda(0)|^{3}} \frac{\partial^{2} \widehat{X}^{1}}{\partial x_{2}^{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1}^{2}}+\frac{3 \pi}{4 \mid \lambda(0){ }_{1}^{3}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{1} \partial x_{2}} \frac{\partial^{2} \widehat{X}^{2}}{\partial x_{2}^{2}}\right) \text { (в точке (0,0)). }
\end{array}
\]

и легко видеть, что
\[
\begin{array}{c}
\int_{0}^{2 \pi /|\lambda(0)|} e^{t d \dot{X}(0)} \int_{0}^{t} e^{-s d \hat{X}(0)} d^{3} \hat{X}(0) \times \\
\quad \times\left(\frac{\partial \varphi_{s}}{\partial x_{1}}(0,0), \frac{\partial \varphi_{s}}{\partial x_{1}}(0,0), \frac{\partial \varphi_{s}}{\partial x_{1}}(0,0)\right) d s d t= \\
=\frac{3 \pi}{4|\lambda(0)|^{2}}\left(\frac{\partial^{3} \widehat{X}^{1}}{\partial x_{1}^{3}}+\frac{\partial^{3} \widehat{X}^{1}}{\partial x_{1} \partial x_{2}^{2}}+\frac{d^{3} \widehat{X}^{1}}{d x_{1}^{2} d x_{2}}+\frac{\partial^{3} \widehat{X}^{2}}{\partial x_{2}^{3}}\right) .
\end{array}
\]

Поэтому конечный результат следующий:
\[
\begin{array}{l}
\int_{0}^{2 \pi / \mid \lambda_{(0) \mid}} \frac{\partial^{3} b_{t}}{\partial x_{1}^{3}}(0,0) d t= \\
=\frac{3 \pi}{4|\lambda(0)|^{2}}\left(\frac{\partial^{3} \widehat{X}^{1}}{\partial a^{3}}(0,0)+\frac{\partial^{3} \widehat{X}^{1}}{\partial a \partial b^{2}}(0,0)+\frac{\partial^{3} \widehat{X}^{2}}{\partial a^{2} \partial b}(0,0)+\frac{\partial^{3} \widehat{X}^{2}}{\partial b^{3}}(0,0)\right)+ \\
+\frac{\pi}{|\lambda(0)|^{3}}\left(2 \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}(0,0)+\frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}(0,0)+\right. \\
+\frac{5}{4} \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}(0,0)+\frac{3}{4} \frac{\partial^{2} \widehat{X}^{2}}{\partial b^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{2}}{\partial a \partial b}(0,0)+ \\
+\frac{7}{4} \frac{\partial^{2} \widehat{X}^{1}}{\partial a^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}(0,0)+\frac{\partial^{2} \widehat{X}^{2}}{4 b^{2}}(0,0) \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0)+ \\
\left.+\frac{5}{4} \frac{\partial^{2} \widehat{X}^{1}}{\partial a \partial b}(0,0) \frac{\partial^{2} \widehat{X}^{1}}{\partial b^{2}}(0,0)+\frac{3}{4} \frac{\partial^{2} \widehat{X}^{2}}{\partial a \partial b}(0,0) \frac{\partial^{2} \widehat{X}^{2}}{\partial a^{2}}(0,0)\right) .
\end{array}
\]

Таким образом, мы получили искомую формулу.

В двумерном случае, когда $X=X$, она может быть непосредственно использована для проверки условий устойчивости, правда, если $d X_{0}(0,0)$ имеет вид, как в $(4.4)$.