In this section we consider the following nonlinear dynamical system of ordinary differential equations:
$$ \frac{du}{dt}+f(\mu,u)=\left [\textstyle\begin{array}{@{}c@{}} \dot{u}_{1} \\ \dot{u}_{2} \\ \dot{u}_{3} \\ \dot{u}_{4} \end{array}\displaystyle \right ]+ \left [\textstyle\begin{array}{@{}c@{}} -u_{2}+\mu u_{1}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ u_{1}+\mu u_{2}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ u_{4}+\mu u_{3}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ -u_{3}+\mu u_{4}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3} \end{array}\displaystyle \right ]= \left [\textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ], $$
(23)
where
$$ u=\left [\textstyle\begin{array}{@{}c@{}} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{array}\displaystyle \right ],\qquad f(\mu,u)= \left [\textstyle\begin{array}{@{}c@{}} f_{1} \\ f_{2} \\ f_{3} \\ f_{4} \end{array}\displaystyle \right ]= \left [\textstyle\begin{array}{@{}c@{}} -u_{2}+\mu u_{1}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ u_{1}+\mu u_{2}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ u_{4}+\mu u_{3}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ -u_{3}+\mu u_{4}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3} \end{array}\displaystyle \right ], $$
and \(u(0)=u(2\pi)\), \(\mu\in\mathbb{R}\).
An equivalent representation to the above system is
$$ \left \{ \textstyle\begin{array}{l} \dot{u}_{1}-u_{2}+\mu u_{1}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}=0, \\ \dot{u}_{2}+u_{1}+\mu u_{2}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}=0, \\ \dot{u}_{3}+u_{4}+\mu u_{3}+(1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}=0, \\ \dot{u}_{4}-u_{3}+\mu u_{4}+ (1+\mu)(u_{1}+u_{2}+u_{3}+u_{4})^{3}=0, \end{array}\displaystyle \right . $$
(24)
where \(\dot{u}_{l}=\frac{du_{l}}{dt}\), \(u_{l}(0)=u_{l}(2\pi)\), \(l=1,2,3,4\), \(\mu\in\mathbb{R}\) is the parameter. The problem is now to find periodic solutions of the above system with period 2π.
Note that this example was studied by Tan in [6] by an algebraic geometry method. It is a particular case of the equation
$$ \frac{du}{dt}+T(u)+\lambda L(u)+H(\lambda,u)+K(\lambda,u)=0, $$
(25)
where \((\lambda,u)\in R \times \mathcal{C}_{2\pi}^{1}(\mathbb{R},\mathbb{R}^{4})\) and
$$\begin{aligned}& u=\left [\textstyle\begin{array}{@{}c@{}} u_{1}(t) \\ u_{2}(t) \\ u_{3}(t) \\ u_{4}(t) \end{array}\displaystyle \right ],\qquad T= \left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} -1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & -1 & -1 \end{array}\displaystyle \right ],\qquad L= \left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\displaystyle \right ], \\& H(\lambda,u)=\lambda \left [\textstyle\begin{array}{@{}c@{}} (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3} \end{array}\displaystyle \right ], \qquad K(\lambda,u)=0. \end{aligned}$$
Taking \(\lambda=\overline{\mu}+\mu=1+\mu\), we turn to the equation
$$ \frac{du}{dt}+T(u)+(1+\mu)L(u)+H(1+\mu,u)=0, $$
or, considering the linearity of the mappings T and L, to the equation
$$ \frac{du}{dt}+(T+L) (u)+\mu L(u)+H(1+\mu,u)=0, $$
i.e.,
$$ \left [\textstyle\begin{array}{@{}c@{}} \dot{u}_{1} \\ \dot{u}_{2} \\ \dot{u}_{3} \\ \dot{u}_{4} \end{array}\displaystyle \right ]+ \left [\textstyle\begin{array}{@{}c@{}} -{u}_{2} \\ {u}_{1} \\ {u}_{4} \\ -{u}_{3} \end{array}\displaystyle \right ]+ \left [\textstyle\begin{array}{@{}c@{}} \mu{u}_{1} \\ \mu{u}_{2} \\ \mu{u}_{3} \\ \mu{u}_{4} \end{array}\displaystyle \right ]+ (1+\mu) \left [\textstyle\begin{array}{@{}c@{}} (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3}\\ (u_{1}+u_{2}+u_{3}+u_{4})^{3} \end{array}\displaystyle \right ]=0. $$
Note also that for the parameter \(\overline{\mu}=1\) the matrix
$$ T+L=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array}\displaystyle \right ] $$
has twice eigenvalues ±i. This is also an equivalent of the matrix \(L_{0}=\frac{\partial f(0,0)}{\partial u}\) in Buchner et al. [5] and in the present work, because further we will also denote the matrix \(T+L\) by \(L_{0}\). Therefore,
$$ \operatorname{Ker}(L_{0}-iI)=\operatorname{span}(w_{1},w_{2}), $$
(26)
where
$$w_{1}=\left [\textstyle\begin{array}{@{}c@{}} 1 \\ -i \\ 1 \\ i \end{array}\displaystyle \right ] \quad \mbox{and}\quad w_{2}=\left [\textstyle\begin{array}{@{}c@{}} i \\ 1 \\ -i \\ 1 \end{array}\displaystyle \right ] $$
are linearly independent eigenvectors corresponding to the eigenvalue i [6].
It is worth to emphasize that the authors of [5, 6] search periodic solutions with period near 2π. Our purpose is to find solutions of the system (24) with a fixed period which is equal to 2π.
Below we prove the following theorem.
Theorem 23
For sufficiently small
\(\alpha\in(-\varepsilon,\varepsilon)\)
the system (23) has nontrivial solutions of the form
$$\begin{aligned} x(\alpha,t) =&\bigl(-48\alpha^{2}+r_{1}(\alpha),2\alpha\cos t+r_{2}(\alpha,t),-2\alpha\sin t+r_{2}(\alpha,t), \\ &-2\alpha\sin t+r_{3}(\alpha,t), 2\alpha\cos t+r_{4}( \alpha,t)\bigr), \end{aligned}$$
where
\(\|r_{1}(\alpha)\|=o(\alpha^{2})\)
and
\(\|r_{l}(\alpha,t)\|=o(\alpha^{2})\), \(l=2,3,4,5\).
Proof
The system (23) can be considered as
$$\begin{aligned} F(x) =&F(\mu,u) \\ =&F(\mu, u_{1},u_{2},u_{3},u_{4}) \\ =&\bigl(\dot{u}_{1}-u_{2}+\mu u_{1}+(1+\mu) (u_{1}+u_{2}+u_{3}+u_{4})^{3}, \\ &\dot{u}_{2}+u_{1}+\mu u_{2}+ (1+\mu ) (u_{1}+u_{2}+u_{3}+u_{4})^{3}, \\ &\dot{u}_{3}+u_{4}+\mu u_{3}+(1+\mu ) (u_{1}+u_{2}+u_{3}+u_{4})^{3}, \\ &\dot{u}_{4}-u_{3}+\mu u_{4}+ (1+\mu) (u_{1}+u_{2}+u_{3}+u_{4})^{3} \bigr)=0, \end{aligned}$$
(27)
where \(F\in\mathcal{C}^{3}(\mathbb{R}\times \mathcal{C}^{1}(\mathbb{R},\mathbb{R}^{4}), \mathcal{C}(\mathbb{R},\mathbb{R}^{4}))\) and \(u_{l}(0)=u_{l}(2\pi)\), \(l=1,2,3,4\). Note that \(x^{\ast}=(0,0)=(0,0,0,0,0)\) is a trivial solution of this system. Let us evaluate the first derivative of the mapping F at the point \((0,0)\),
$$ F'(0,0)=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 & \frac{d}{d\tau} & -1 & 0 & 0 \\ 0 & 1 & \frac{d}{d\tau} & 0 & 0 \\ 0 & 0 & 0 & \frac{d}{d\tau} & 1 \\ 0 & 0 & 0 & -1 & \frac{d}{d\tau} \end{array}\displaystyle \right ]:= \biggl( 0, \frac{d}{d \tau}+ L_{0} \biggr), $$
(28)
where
$$L_{0}:=\frac{\partial f(0,0)}{\partial u}=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array}\displaystyle \right ] \quad \mbox{and}\quad \biggl(\frac{d}{d \tau}+ L_{0}\biggr)u:=\frac{du}{d \tau}+ L_{0}u. $$
From this we can describe the kernel of the first derivative:
$$ \operatorname{Ker} F'(0, 0)=R\oplus\operatorname{Ker} \biggl( \frac{d}{d \tau}+ L_{0}\biggr). $$
The problem is now to find the space \(\operatorname{Ker} (\frac{d}{d \tau}+ L_{0})\). To this end, one has to solve the following system of equations:
$$ \frac{du}{d \tau}+ L_{0}u=0, $$
with condition \(u(0)=w_{k}\), \(k=1\mbox{ or }2\) (see (26)). The solution of the above system is the following:
$$ u(\tau)=e^{-L_{0}\tau}u(0)=e^{-i\tau}u(0). $$
Thus we obtain
$$ \operatorname{Ker}\biggl(\frac{d}{d \tau}+ L_{0}\biggr)= \operatorname{span} (\phi_{1},\phi_{2},\phi_{3}, \phi_{4}), $$
where \(\phi_{2k-1}(\tau)=\operatorname{Re}(e^{-i\tau}w_{k})\), \(\phi_{2k}(\tau)=\operatorname{Im}(e^{-i\tau}w_{k})\), \(k=1,2\) (see [6]).
Hence for the vector \(h=[h_{\mu},h_{u_{1}},h_{u_{2}},h_{u_{3}},h_{u_{4}}]\in\mathbb{R}\times \mathcal{C}^{1}(\mathbb{R},\mathbb{R}^{4}) \), taking into account that \(0\cdot h_{\mu}=0\) and solving the following system of differential equations:
$$ F'(0,0)[h]= \biggl(\frac{dh_{u_{1}}}{d\tau}-h_{u_{2}}, \frac{dh_{u_{2}}}{d\tau }+h_{u_{1}},\frac{dh_{u_{3}}}{d\tau}+h_{u_{4}}, \frac{dh_{u_{4}}}{d\tau }-h_{u_{3}}\biggr)=0, $$
(29)
subject to the conditions \(h_{u_{l}}(0)= h_{u_{l}}(2\pi)\), \(l=1,2,3,4\), we obtain
$$ \operatorname{Ker} F'(0,0)=\mathbb{R}\times \operatorname{span}( \phi_{1},\phi_{2},\phi_{3},\phi_{4})= \mathbb{R}\times \operatorname{Ker} F'_{u}(0,0), $$
(30)
where
$$ \phi_{1}=\left [\textstyle\begin{array}{@{}c@{}} \cos\tau\\ -\sin\tau\\ \cos\tau\\ \sin\tau \end{array}\displaystyle \right ],\qquad \phi_{2}=\left [\textstyle\begin{array}{@{}c@{}} -\sin\tau\\ -\cos\tau\\ -\sin\tau\\ \cos\tau \end{array}\displaystyle \right ],\qquad \phi_{3}=\left [\textstyle\begin{array}{@{}c@{}} \sin\tau\\ \cos\tau\\ -\sin\tau\\ \cos\tau \end{array}\displaystyle \right ],\qquad \phi_{4}=\left [\textstyle\begin{array}{@{}c@{}} \cos\tau\\ -\sin\tau\\ -\cos\tau\\ -\sin\tau \end{array}\displaystyle \right ]. $$
(31)
Define now the space \(Y_{1}= \operatorname{Im} (\frac{d}{d\tau}+ L_{0})=(\operatorname{Ker} (\frac{d}{d\tau}+ L_{0})^{\ast})^{\bot}\). Note that the adjoint operator has the form \((\frac{d}{d\tau}+ L_{0})^{\ast}=-\frac{d}{d\tau}+ L_{0}^{\ast}=-\frac{d}{d\tau}+ L_{0}^{\top}\). A basis \(\{\psi_{1},\psi_{2},\psi_{3},\psi_{4}\}\) of the space \(\operatorname{Ker} (\frac{d}{d\tau}+ L_{0})^{\ast}\) can be found with \(\langle\psi_{l}, \phi_{j}\rangle=\delta_{lj}\), where \(\langle g,h\rangle=\int_{0}^{2\pi}(g(\tau),h(\tau))\, d\tau\) and \((g(\tau),h(\tau))\) is the standard scalar product in \(\mathbb{R}^{4}\). Then
$$\begin{aligned} Y_{1} =&\bigl\{ g\in \mathcal{C}_{2\pi}\bigl( \mathbb{R},\mathbb{R}^{4}\bigr):\langle g,\psi_{l}\rangle=0, l=1,2,3,4\bigr\} \\ =& \biggl\{ g\in \mathcal{C}_{2\pi}\bigl(\mathbb{R}, \mathbb{R}^{4}\bigr):\int_{0}^{2\pi} \bigl(g(\tau ),\psi_{l}\bigr)\, d\tau=0, l=1,2,3,4\biggr\} . \end{aligned}$$
(32)
Easy computations show that \(\psi_{l}=\frac{1}{4\pi}\phi_{l}\), for \(l=1,2,3,4\), and we have the following identity of subspaces:
$$ \operatorname{span}(\psi_{1},\psi_{2},\psi_{3}, \psi_{4})=\operatorname{span}(\phi _{1},\phi_{2}, \phi_{3},\phi_{4}). $$
Therefore, we obtain the relation
$$\begin{aligned}& Y_{1}=\operatorname{Im} F'(0,0)=\bigl( \operatorname{Ker} F'_{u}(0,0)^{\ast} \bigr)^{\bot}=\bigl(\operatorname{Ker} F'_{u}(0,0) \bigr)^{\bot} \quad \mbox{and} \\& Y_{1}= \bigl\{ g\in \mathcal{C}_{2\pi}\bigl(\mathbb{R}, \mathbb{R}^{4}\bigr):\langle g,\phi_{l} \rangle=0, l=1,2,3,4 \bigr\} \\& \hphantom{Y_{1}}= \biggl\{ g\in \mathcal{C}_{2\pi}\bigl(\mathbb{R}, \mathbb{R}^{4}\bigr):\int_{0}^{2\pi} \bigl(g(\tau ),\phi_{l}\bigr)\, d\tau=0, l=1,2,3,4\biggr\} . \end{aligned}$$
(33)
Now choose the vectors
$$\begin{aligned}& h = [0,\phi_{1}+\phi_{2}+\phi_{3}+ \phi_{4}]=2[0,\cos\tau,-\sin\tau, -\sin\tau, \cos\tau], \\& \overline{h} = [\varepsilon,0,0,0,0]. \end{aligned}$$
We proceed to show that all assumptions of Theorem 21 hold, i.e.,
$$\begin{aligned}& \Pi_{Y_{1}}F\bigl(x^{\ast}+\alpha h+\alpha^{2} \overline{h}\bigr) = o\bigl(\alpha ^{2}\bigr), \\& \Pi_{Y_{2}}F\bigl(x^{\ast}+\alpha h+\alpha^{2} \overline{h}\bigr) = o\bigl(\alpha ^{3}\bigr), \\& P_{Y_{2}^{\bot}} F\bigl(x^{\ast}+\alpha h+\alpha^{2} \overline{h}\bigr) = o\bigl(\alpha^{4}\bigr), \end{aligned}$$
where
$$\begin{aligned}& Y_{1} = \operatorname{Im} F'\bigl(x^{\ast}\bigr),\qquad Y_{2} = \operatorname{Im} \Pi_{Y_{1}^{\bot}} F'' \bigl(x^{\ast}\bigr)[h], \\& Y_{2}^{\bot} = \operatorname{Im} P_{(Y_{1}\oplus Y_{2})^{\bot}} \biggl(F''\bigl(x^{\ast}\bigr)[\overline{h}]+ \frac {1}{2}F'''\bigl(x^{\ast} \bigr)[h]^{2}\biggr),\qquad Y = Y_{1}\oplus Y_{2}\oplus Y_{2}^{\bot}. \end{aligned}$$
Note that
$$\begin{aligned}& F\bigl(\alpha h+\alpha^{2}\overline{h}\bigr) \\& \quad =F\bigl(\alpha^{2}\varepsilon,2\alpha\cos\tau,-2\alpha\sin \tau,-2\alpha\sin \tau,2\alpha\cos\tau\bigr) \\& \quad =\bigl(2\alpha^{3}\varepsilon\cos\tau+A,-2\alpha^{3} \varepsilon\sin\tau +A,-2\alpha^{3}\varepsilon\sin \tau+A,2 \alpha^{3}\varepsilon\cos\tau+A\bigr), \end{aligned}$$
where \(A=64\alpha^{3}(1+\alpha^{2}\varepsilon)(\cos\tau-\sin\tau)^{3}\). Therefore, \(F(\alpha h+\alpha^{2}\overline{h})=o(\alpha^{2})\) and from this, we have \(\Pi_{Y_{1}}F(\alpha h+\alpha ^{2}\overline{h})=o(\alpha^{2})\).
Next we show that
$$ P_{Y_{1}^{\bot}}F\bigl(\alpha h+\alpha^{2}\overline{h} \bigr)=o\bigl(\alpha^{4}\bigr) $$
(34)
for \(\overline{h}=[-48,0,0,0,0]\), where \(Y_{1}^{\bot}=\operatorname{Ker}F'_{u}(0,0)\).
The projection \(P_{Y_{1}^{\bot}}F(\alpha h+\alpha^{2}\overline{h})\) has the following form:
$$\begin{aligned}& P_{Y_{1}^{\bot}}F\bigl(\alpha^{2}\varepsilon,2\alpha\cos\tau,-2\alpha \sin\tau ,-2\alpha\sin\tau,2\alpha\cos\tau\bigr) \\& \quad = \sum_{i=1}^{4}\bigl\langle F\bigl( \alpha^{2}\varepsilon,2\alpha\cos\tau,-2\alpha\sin\tau,-2\alpha\sin \tau,2\alpha\cos\tau\bigr), \phi_{i} \bigr\rangle \phi_{i} \\& \quad = \phi_{1}\int_{0}^{2\pi}\bigl[ \bigl(2\alpha^{3}\varepsilon\cos\tau+A\bigr)\cos\tau +\bigl(-2 \alpha^{3}\varepsilon\sin\tau+A\bigr) (-\sin\tau) \\& \qquad {} + \bigl(-2\alpha^{3}\varepsilon\sin\tau+A\bigr)\cos\tau+ \bigl(2\alpha^{3}\varepsilon \cos\tau+A\bigr)\sin\tau\bigr]\, d\tau \\& \qquad {} + \phi_{2}\int_{0}^{2\pi}\bigl[ \bigl(2\alpha^{3}\varepsilon\cos\tau+A\bigr) (-\sin\tau )+\bigl(-2 \alpha^{3}\varepsilon\sin\tau+A\bigr) (-\cos\tau) \\& \qquad {} + \bigl(-2\alpha^{3}\varepsilon\sin\tau+A\bigr) (-\sin\tau)+ \bigl(2\alpha ^{3}\varepsilon\cos\tau+A\bigr)\cos\tau\bigr]\, d\tau \\& \qquad {} + \phi_{3}\int_{0}^{2\pi}\bigl[ \bigl(2\alpha^{3}\varepsilon\cos\tau+A\bigr)\sin\tau +\bigl(-2 \alpha^{3}\varepsilon\sin\tau+A\bigr)\cos\tau \\& \qquad {} + \bigl(-2\alpha^{3}\varepsilon\sin\tau+A\bigr) (-\sin\tau)+ \bigl(2\alpha ^{3}\varepsilon\cos\tau+A\bigr)\cos\tau\bigr]\, d\tau \\& \qquad {} + \phi_{4}\int_{0}^{2\pi} \bigl[\bigl(2\alpha^{3}\varepsilon\cos\tau+A\bigr)\cos\tau +\bigl(-2 \alpha^{3}\varepsilon\sin\tau+A\bigr) (-\sin\tau) \\& \qquad {} + \bigl(-2\alpha^{3}\varepsilon\sin\tau+A\bigr) (-\cos\tau)+ \bigl(2\alpha ^{3}\varepsilon\cos\tau+A\bigr) (-\sin\tau)\bigr]\, d\tau. \end{aligned}$$
The last expression can be represented as
$$\begin{aligned}& P_{Y_{1}^{\bot}}F\bigl(\alpha^{2}\varepsilon,2\alpha\cos\tau,-2\alpha \sin \tau,-2\alpha\sin\tau,2\alpha\cos\tau\bigr) \\& \quad = \phi_{1}\int_{0}^{2\pi}\bigl[2 \alpha^{3}\varepsilon+64\alpha^{3}\bigl(1+\alpha ^{2}\varepsilon\bigr) \bigl(2\cos^{4}\tau+6 \cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \qquad {} + \phi_{2}\int_{0}^{2\pi} \bigl[2\alpha^{3}\varepsilon+64\alpha^{3}\bigl(1+\alpha ^{2}\varepsilon\bigr) \bigl(2\sin^{4}\tau+6 \cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \qquad {} + \phi_{3}\int_{0}^{2\pi} \bigl[2\alpha^{3}\varepsilon+64\alpha^{3}\bigl(1+\alpha ^{2}\varepsilon\bigr) \bigl(2\cos^{4}\tau+6 \cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \qquad {} + \phi_{4}\int_{0}^{2\pi} \bigl[2\alpha^{3}\varepsilon+64\alpha^{3}\bigl(1+\alpha ^{2}\varepsilon\bigr) \bigl(2\sin^{4}\tau+6 \cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \end{aligned}$$
and condition (34) holds, if
$$\begin{aligned}& \phi_{1}\int_{0}^{2\pi}\bigl[2 \alpha^{3}\varepsilon+64\alpha^{3}\bigl(2\cos ^{4} \tau+6\cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \quad {} + \phi_{2}\int_{0}^{2\pi}\bigl[2 \alpha^{3}\varepsilon+64\alpha^{3}\bigl(2\sin ^{4} \tau+6\cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \quad {} + \phi_{3}\int_{0}^{2\pi}\bigl[2 \alpha^{3}\varepsilon+64\alpha^{3}\bigl(2\cos ^{4} \tau+6\cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d\tau \\& \quad {} + \phi_{4}\int_{0}^{2\pi} \bigl[2\alpha^{3}\varepsilon+64\alpha^{3}\bigl(2\sin ^{4}\tau+6\cos^{2}\tau\sin^{2}\tau\bigr)\bigr]\, d \tau =o\bigl(\alpha^{4}\bigr). \end{aligned}$$
Therefore, we have to solve the following equation:
$$ 4\pi\varepsilon+64 \biggl(2\frac{3\pi}{4}+6\frac{\pi}{4} \biggr)=0, $$
and we obtain \(\varepsilon=-48\). From this we have \(\overline{h}=[-48,0,0,0,0]\) and condition (34) is satisfied.
By \(Y_{2}\subset Y_{1}^{\bot}\) (while \(Y_{1}^{\bot}=\operatorname{Ker}F'_{u}(0,0)\)), we have
$$ \Pi_{Y_{2}}F\bigl(\alpha h+\alpha^{2}\overline{h}\bigr)=o\bigl( \alpha^{3}\bigr) $$
and
$$ P_{Y_{2}^{\bot}} F\bigl(\alpha h+\alpha^{2}\overline{h}\bigr)=o\bigl( \alpha^{4}\bigr). $$
In the next step we verify that the operator
$$\begin{aligned} \Psi_{2}(h,\overline{h}) =&\Pi_{Y_{1}}F'(0,0)+\Pi _{Y_{2}}F''(0,0)[h] \\ &{}+P_{Y_{2}^{\bot}} \biggl(F''(0,0)[\overline{h}]+\frac {1}{2}F'''(0,0)[h]^{2} \biggr) \end{aligned}$$
is a surjection. Of course, this operator is
$$\begin{aligned} \Psi_{2}(h,\overline{h}) =&F'(0,0)+ \Pi_{Y_{2}}F''(0,0)[h] \\ &{}+P_{Y_{2}^{\bot }} \biggl(F''(0,0)[\overline{h}]+\frac{1}{2}F'''(0,0)[h]^{2} \biggr). \end{aligned}$$
(35)
Let us note that a consequence of Lemma 15 is the following remark. If, for any \(s\in(0,\delta)\), where \(\delta>0 \) is sufficiently small, the operator \(\Pi_{Y_{2}}F''(0,0)[h]+sP_{Y_{1}^{\bot}}F''(0,0)[\overline{h}]\) is a surjection, where \(\Pi_{Y_{2}}:Y\rightarrow Y_{2}\) is the projection operator from the space Y onto \(Y_{2}\) along \(Y_{1}\) and \(P_{Y_{1}^{\bot}}\) is the projection operator from the space Y onto \(Y_{1}^{\bot}\) along \(Y_{1}\), then the operator \(\Pi_{Y_{2}}F''(0,0)[h]+P_{Y_{2}^{\bot}}F''(0,0)[\overline{h}]\) is a surjection too.
To begin, note that for the vector \(H=[h_{\mu},h_{u_{1}},h_{u_{2}},h_{u_{3}},h_{u_{4}}]\) we obtain (see Lemma 14)
$$ F''(0,0)[H]^{2}=2!(h_{\mu}h_{u_{1}},h_{\mu}h_{u_{2}},h_{\mu }h_{u_{3}},h_{\mu}h_{u_{4}}), $$
and, in matrix form, we get the following representation of the operator \(F''(0,0)[H]\):
$$ F''(0,0)[H]=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} h_{u_{1}} & h_{\mu} & 0 & 0 & 0 \\ h_{u_{2}} & 0 & h_{\mu} & 0 & 0 \\ h_{u_{3}} & 0 & 0 & h_{\mu} & 0 \\ h_{u_{4}} & 0 & 0 & 0 & h_{\mu} \end{array}\displaystyle \right ]. $$
It follows that
$$ F''(0,0)[h]=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 2\cos\tau& 0 & 0 & 0 & 0 \\ -2\sin\tau& 0 & 0 & 0 & 0 \\ -2\sin\tau& 0 & 0 & 0 & 0 \\ 2\cos\tau& 0 & 0 & 0 & 0 \end{array}\displaystyle \right ] $$
and
$$ F''(0,0)[\overline{h}]=\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 & -48 & 0 & 0 & 0 \\ 0 & 0 & -48 & 0 & 0 \\ 0 & 0 & 0 & -48 & 0 \\ 0 & 0 & 0 & 0 & -48 \end{array}\displaystyle \right ]. $$
We apply Lemma 10 to examine the surjectivity of operator (35) on the kernel of the first derivative at the point \((0,0)\).
Take an arbitrary vector \([\lambda, v_{1},v_{2},v_{3},v_{4}]\in \operatorname{Ker} F'(0,0)\). We have \(\lambda\in\mathbb{R}\) and
$$ \left [\textstyle\begin{array}{@{}c@{}} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]= \left [\textstyle\begin{array}{@{}c@{}} a\cos\tau-b\sin\tau+c\sin\tau+d\cos\tau\\ -a\sin\tau-b\cos\tau+c\cos\tau-d\sin\tau\\ a\cos\tau-b\sin\tau-c\sin\tau-d\cos\tau\\ a\sin\tau+b\cos\tau+c\cos\tau-d\sin\tau \end{array}\displaystyle \right ], $$
where \(a,b,c,d\in\mathbb{R}\). From this, we obtain
$$ F'(0,0)\left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]=0,\qquad F''(0,0)[h]\left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]=2\lambda \left [\textstyle\begin{array}{@{}c@{}} \cos\tau\\ -\sin\tau\\ -\sin\tau\\ \cos\tau \end{array}\displaystyle \right ]. $$
Next we have
$$ \Pi_{Y_{2}}F''(0,0)[h](\lambda,v_{1},v_{2},v_{3},v_{4})=4 \pi\lambda(\phi _{1}+\phi_{2}+\phi_{3}+ \phi_{4})= 4\pi\lambda \left [\textstyle\begin{array}{@{}c@{}} 2\cos\tau\\ -2\sin\tau\\ -2\sin\tau\\ 2\cos\tau \end{array}\displaystyle \right ] $$
(the calculations are analogous to (34)).
Note that the element
$$4\pi\lambda \left [\textstyle\begin{array}{@{}c@{}} 2\cos\tau\\ -2\sin\tau\\ -2\sin\tau\\ 2\cos\tau \end{array}\displaystyle \right ] $$
belongs to \(\operatorname{Ker}F'_{u}(0,0)=Y_{1}^{\bot}\).
Hence
$$ P_{Y_{1}^{\bot}}F''(0,0)[h](\lambda ,v_{1},v_{2},v_{3},v_{4})=F''(0,0)[h]( \lambda,v_{1},v_{2},v_{3},v_{4}) $$
and \(\Pi_{Y_{2}}F''(0,0)[h](\lambda ,v_{1},v_{2},v_{3},v_{4})=F''(0,0)[h](\lambda,v_{1},v_{2},v_{3},v_{4})\), because \(Y_{2}\subset Y_{1}^{\bot}\).
Consider now
$$F''(0,0)[\overline{h}]\left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]= \left [\textstyle\begin{array}{@{}c@{}} -48v_{1} \\ -48v_{2} \\ -48v_{3} \\ -48v_{4} \end{array}\displaystyle \right ]. $$
Obviously
$$F''(0,0)[\overline{h}]\left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]\in\operatorname{Ker} F'_{u}(0,0)= Y_{1}^{\bot}. $$
Hence
$$P_{Y_{1}^{\bot}}F''(0,0)[\overline{h}] \left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]=F''(0,0)[ \overline{h}]\left [\textstyle\begin{array}{@{}c@{}} \lambda\\ v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\displaystyle \right ]. $$
Therefore the operator
$$\begin{aligned}& \Pi_{Y_{2}}F''(0,0)[h]+sP_{Y_{1}^{\bot}}F''(0,0)[ \overline{h}] \\& \quad =F''(0,0)[h]+sF''(0,0)[ \overline{h}] =\left [\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} \cos\tau& 48s & 0 & 0 & 0 \\ -\sin\tau& 0 & 48s & 0 & 0 \\ -\sin\tau& 0 & 0 & 48s & 0 \\ \cos\tau& 0 & 0 & 0 & 48s \end{array}\displaystyle \right ] \end{aligned}$$
is a surjection onto \(Y_{1}^{\bot}\) and the operator \(\Pi_{Y_{2}}F''(0,0)[h]+P_{Y_{2}^{\bot}}F''(0,0)[\overline{h}]\) is also a surjection. From this and by Lemma 10 the operator (35) is a surjection. Note that the examination of the operator \(P_{Y_{2}^{\bot}}(\frac{1}{2}F'''(0,0)[h]^{2})\) is not necessary.
We verified all assumptions of Theorem 21. Hence there exist nontrivial solutions of system (23). We can write them in the form \(x(\alpha,t)=x^{\ast}+\alpha h+\alpha^{2}\overline{h}+r(\alpha)\), with \(\|r(\alpha)\|=o(\alpha^{2})\), i.e.,
$$\begin{aligned} x(\alpha,t) =&\bigl(-48\alpha^{2}+r_{1}(\alpha),2\alpha\cos t+r_{2}(\alpha,t),-2\alpha\sin t+r_{2}(\alpha,t), \\ &-2\alpha\sin t+r_{3}(\alpha,t), 2\alpha\cos t+r_{4}( \alpha,t)\bigr), \end{aligned}$$
where \(\|r_{1}(\alpha)\|=o(\alpha^{2})\) and \(\|r_{l}(\alpha,t)\|=o(\alpha^{2})\), \(l=2,3,4,5\), which proves Theorem 23. □
Remark 24
For \(\mu=-48\alpha^{2}+r_{1}(\alpha)\) such that \(\mu<0\) and \(\|r_{1}(\alpha)\|=o(\alpha^{2})\) the solutions can be written as follows:
$$\begin{aligned} u(\mu, t) =&\biggl(\frac{1}{2}\sqrt{-\frac{\mu}{3}}\cos t+ \overline{r}_{1}(\mu,t),-\frac{1}{2}\sqrt{-\frac{\mu}{3}}\sin t+\overline{r}_{2}(\mu,t), \\ &-\frac{1}{2}\sqrt{-\frac{\mu}{3}}\sin t+\overline{r}_{3}( \mu,t),\frac{1}{2}\sqrt{-\frac{\mu}{3}}\cos t+\overline{r}_{4}( \mu,t)\biggr), \end{aligned}$$
where \(\|r_{l}(\mu,t)\|=o(\mu)\), \(l=1,2,3,4\).
From Theorem 23 it follows that there are no solutions in the case \(\mu> 0\).