We seek a solution of systems (
1.1)(
1.5) in the form
$$ x(t,\mu)= \textstyle\textstyle\begin{cases} \sum_{i=0}^{\infty}\mu^{i}\{\bar{x}_{i}(t)+L_{i}x(\tau _{0})+Q_{i}^{()}x(\tau_{c})\},& 0\leq t\leq\sigma, \\ \sum_{i=0}^{\infty}\mu^{i}\{\bar{\bar {x}}_{i}(t)+Q_{i}^{(+)}x(\tau_{c})+R_{i}x(\tau_{1})\}, &\sigma\leq t\leq T ,\end{cases} $$
(2.1)
where
\(\tau_{0}=\frac{t}{\mu}\),
\(\tau_{c}=\frac{t\sigma}{\mu}\),
\(\tau _{1}=\frac{tT}{\mu}\). By the initial and boundary value conditions, we obtain
$$\begin{aligned}& \bar{u}_{0}(0)+L_{0} u(0)=\theta(0),\qquad \bar{u}_{k}(0)=L_{k} u(0), \\& \bar{z}_{0}(0)+L_{0} z(0)=\rho(0),\qquad \bar{z}_{k}(0)=L_{k} z(0), \\& \bar{\bar{z}}_{0}(T)+R_{0} z(0)=z^{T},\qquad \bar{\bar{z}}_{k}(T)=R_{k} z(0). \end{aligned}$$
In order to make
\(x(t,\mu)\) continuous at
\(t=\sigma\), the following equation must be satisfied:
$$ x^{()} (\sigma,\mu)=x^{(+)} (\sigma,\mu). $$
(2.2)
From the continuity condition (
2.2), we get
$$\begin{aligned}& \bar{z}_{0} (\sigma)+Q^{()}_{0} z(0)=\bar{ \bar{z}}_{0} (\sigma )+Q^{(+)}_{0} z(0)=p_{0}, \\& \bar{z}_{k} (\sigma)+Q^{()}_{k} z(0)=\bar{ \bar{z}}_{k} (\sigma )+Q^{(+)}_{k} z(0)=p_{k}, \end{aligned}$$
where
\(p_{k}\) (
\(k\geq0\)) are unknown parameters.
2.1 The construction of asymptotic solutions on the interval \([0,\sigma]\)
First, we consider the regular part of the solution. Substituting (
1.12) into the original problem and setting
\(\mu=0\), we obtain the socalled degenerate problems of
\(\bar{x}_{0}(t)\):
$$\begin{aligned}& A\bigl(\bar{u}_{0}(t),\theta(t\sigma),t\bigr) \bar{y}_{0}(t)=0, \\& f\bigl(\bar{z}_{0}(t),\rho(t\sigma),t\bigr)=0, \\& C\bigl(\bar{u}_{0}(t),\theta(t\sigma),t\bigr)\bar{y}_{0}(t)=0. \end{aligned}$$
(2.3)
The root of (
2.3) is
$$ \bar{z}_{0}(t)=\bar{\alpha}(t),\qquad \bar{y}_{0}(t)=0, \qquad \bar{u}_{0}(t)=\bar{\boldsymbol{\gamma}}(t), $$
(2.4)
where
\(\bar{{\boldsymbol{\gamma}}}(t)\) is an arbitrary
kdimensional vector function.
The equations of
\(L_{0}x(\tau_{0})\) are
$$\begin{aligned}& \frac{d}{d \tau_{0}}L_{0} z=\tilde{A}\bigl(\bar{{ \boldsymbol{\gamma}}}(0)+L_{0} u(\tau_{0}),\theta(\sigma),0 \bigr)L_{0} y(\tau_{0}), \\& \frac{d}{d \tau_{0}}L_{0} y=\tilde{f}\bigl(\bar{\alpha}(0)+L_{0} z(\tau _{0}),\rho(\sigma),0\bigr)\bar{f}\bigl(\bar{\alpha}(0),\rho( \sigma),0\bigr), \\& \frac{d}{d \tau_{0}}L_{0} u=\tilde{C}\bigl(\bar{{\boldsymbol{ \gamma}}}(0)+L_{0} u(\tau_{0}),\theta(\sigma),0 \bigr)L_{0} y(\tau_{0}), \end{aligned}$$
(2.5)
with the initial and boundary conditions
$$ L_{0} z(0)=\rho(0)\bar{\alpha}(0),\qquad L_{0} u(0)= \theta(0)\bar{\boldsymbol{\gamma}}(0),\qquad L_{0} x(+\infty )=0, $$
(2.6)
where
\(\bar{\boldsymbol{\gamma}}(0)\) is unknown, and the initial value of
\(L_{0} y(\tau_{0})\) is as yet arbitrary. We will use this arbitrariness to ensure that
\(L_{0}x(\tau_{0})\) satisfies condition
\(L_{0} x(+\infty)=0\).
From (
2.5), we have
$$ \frac{dL_{0} u}{dL_{0} z}=\frac{\tilde{C}(\bar{{\boldsymbol{\gamma}}}(0)+L_{0} u(\tau_{0}),\theta(0),0)}{\tilde{A}(\bar{{\boldsymbol{\gamma}}}(0)+L_{0} u(\tau_{0}),\theta(0),0)}. $$
(2.7)
Integrating the above formula after separating variable and using implicit function theorem, we find that (
2.7) has a solution. Let us denote by
$$ L_{0} u=\bar{U}_{0}\bigl(\bar{\boldsymbol{ \gamma}}(0),L_{0} z\bigr) $$
(2.8)
the solution of this system such that
\(L_{0} u=0\) for
\(L_{0} z=0\), that is,
\(\bar{U}_{0}(\bar{\boldsymbol{\gamma}}(0),0)=0\).
The condition (
\(\mathrm{H}_{1}'\)) shows that there exists a unique solution in a certain neighborhood of the point
\(L_{0} z=0\). Then substituting (
2.8) into (
2.5), we obtain the system of equations
$$ \begin{aligned} &\frac{d}{d \tau_{0}}L_{0} z=\tilde{A}\bigl(\bar{\boldsymbol{ \gamma}}(0)+\bar{U}_{0}\bigl(\bar{\gamma}(0),L_{0} z\bigr), \theta(0),0\bigr)L_{0} y(\tau _{0}), \\ &\frac{d}{d \tau_{0}}L_{0} y=\tilde{f}\bigl(\bar{\alpha}(0)+L_{0} z(\tau _{0}),\rho(\sigma),0\bigr)\bar{f}\bigl(\bar{\alpha}(0),\rho( \sigma ),0\bigr). \end{aligned} $$
(2.9)
The equilibrium point
\((0,0)\) of (
2.9) is a saddle, since the roots of the corresponding characteristic equation are clearly equal to
\(\pm\sqrt{\tilde{A}\tilde{f}_{L_{0} z}}\) and, by virtue of the condition (
\(\mathrm{H}_{1}'\)), are real and have opposite signs.
By integrating (
2.9), we get the separate equation of the saddle point
$$ L_{1}\mbox{: } L_{0}y(\tau_{0})= \biggl(2\int _{0}^{L_{0}z}\frac{\tilde {f}(\bar{\alpha}(0)+\xi, \rho(0),0)\bar{f}(\bar{\alpha}(0),\rho(0),0)}{\tilde{A}(\bar {{\boldsymbol{\gamma}}}(0)+\bar{U}_{0}(\bar{{\boldsymbol{\gamma}}}(0),\xi), \rho(0),0)}\, d\xi \biggr)^{\frac{1}{2}}. $$
(2.10)
Equations (
2.8) and (
2.10) give an analytic expression of the onedimensional manifold
\(\bar{\Omega}_{0}\) having the property that if
\(L_{0} x(0)\in\bar{\Omega}_{0}\), then
\(L_{0} x(\tau_{0})\in\bar {\Omega}_{0}\) for
\(\tau_{0}>0\), and the inequality
$$ \bigl\Vert L_{0} x(\tau_{0})\bigr\Vert \leq ce^{\sigma_{0} \tau_{0}}\quad (\tau_{0}\geq0) $$
(2.11)
is satisfied. In order to make the solution of system (
2.5) satisfy an exponential decay estimation, we need the following assumption.
 (\(\mathrm{H}_{2}\)):

Suppose that
\(L_{0} z(0)\cap\bar {\Omega}_{0}\neq\emptyset\).
Substituting (
2.6) into (
2.8), we have
$$ \theta(0)\bar{\boldsymbol{\gamma}}(0)=\bar{U}_{0} \bigl(\bar{\boldsymbol{\gamma}}(0),\rho(0)\bar{\alpha}(0)\bigr), $$
(2.12)
which represents a system of
k scalar equations in the
k unknown components of the vector
\(\bar{\boldsymbol{\gamma}}(0)\). The following condition is concerned with the solvability of
\(\bar {\boldsymbol{\gamma}}(0)\).
 (\(\mathrm{H}_{3}\)):

Suppose that (2.12) has a root
\(\bar{\boldsymbol{\gamma}}(0)=\bar{\boldsymbol{\gamma}}^{0}\).
By means of (\(\mathrm{H}_{3}\)) and by taking (2.10) into account, we can determine the initial value \(L_{0} y(0)\), and consequently, \(L_{0} x(\tau_{0})\) satisfies an exponential decay. For determining \(L_{0} x(\tau_{0})\), it is necessary to substitute (2.10) into (2.9) and to solve the resulting scalar equation for \(L_{0} z(\tau_{0})\) with the initial value condition \(L_{0} z(0)=\rho(0)\bar{\alpha}(0)\). By virtue of (2.8) and (2.10) we obtain \(L_{0} u(\tau_{0})\) and \(L_{0} y(\tau_{0})\). Thus \(L_{0} x(\tau_{0})\) are completely determined, while for the as yet unknown function \(\bar{\boldsymbol{\gamma}}(t)\) we only know its initial value \(\bar{\boldsymbol{\gamma}}^{0}\). The function \(\bar{\boldsymbol{\gamma}}(t)\) need to be determined in the first approximation.
For
\(\bar{x}_{1}(t)\), we get
$$\begin{aligned}& \bar{A}\bigl(\bar{\boldsymbol{\gamma}}(t),\theta(t\sigma),t\bigr) \bar{y}_{1} (t)+\bar{B}\bigl(\bar{\boldsymbol{\gamma}}(t),\theta(t \sigma),t\bigr)=\bar {\alpha}'(t), \\& f_{z} \bigl(\bar{\alpha}(t),\rho(t\sigma),t\bigr) \bar{z}_{1}(t)=0, \\& \bar{C}\bigl(\bar{\boldsymbol{\gamma}}(t),\theta(t\sigma),t\bigr) \bar{y}_{1} (t)+\bar{D}\bigl(\bar{{\boldsymbol{\gamma}}}(t),\theta(t \sigma),t\bigr)=\bar {\boldsymbol{\gamma}}'(t), \end{aligned}$$
which has a solution
$$ \bar{z}_{1}(t)=0,\qquad w\bar{y}_{1}(t)= \frac{\bar{\alpha}'(t)\bar{B}}{\bar{A}},\qquad \frac{d\bar{{\boldsymbol{\gamma}}}}{dt}=\frac{\bar{C}(\bar{\alpha }'(t)\bar{B})}{\bar{A}}+ \bar{D}, $$
(2.13)
where
\(\bar{A}\),
\(\bar{B}\),
\(\bar{C}\), and
\(\bar{D}\) are all taken value at the point
\((\bar{\boldsymbol{\gamma}}(t),\theta(t\sigma),t)\). From the existence of a solution for the initial value problem, (
2.13) together with the initial condition
\(\bar{\boldsymbol{\gamma}}(0)=\bar{\boldsymbol{\gamma}}^{0}\) has the solution
\(\bar{\boldsymbol{\gamma}}=\bar{\boldsymbol{\gamma}}(t)\) for
\(0\leq t \leq\sigma\). Therefore
\(\bar{x}_{0}(t)\) can be completely determined. Equation (
2.13) has only determined
\(\bar{z}_{1}(t)\) and
\(\bar {y}_{1}(t)\), while
\(\bar{u}_{1}(t)\) is yet unknown. We will take advantage of the first approximation to determine
\(\bar{u}_{1}(t)\).
The equations for
\(L_{1}x(\tau_{0})\) have the form
$$\begin{aligned}& \frac{d}{d\tau_{0}}L_{1} z=\tilde{A}\bigl(\bar{{ \boldsymbol{\gamma}}}(0)+L_{0} u(\tau_{0}),\theta(\sigma),0 \bigr)L_{1} y(\tau_{0}) \\& \hphantom{\frac{d}{d\tau_{0}}L_{1} z=}{}+\tilde{A_{u}}\bigl(\bar{u}_{1}(0)+L_{1} u(\tau_{0})\bigr)L_{0} y(\tau _{0})+ \varphi_{1}(\tau_{0}), \\& \frac{d}{d\tau_{0}}L_{1} y=\tilde{f}_{z} L_{1} z(\tau_{0})+\varphi_{2}(\tau _{0}), \\& \frac{d}{d\tau_{0}}L_{1} u=\tilde{C}\bigl(\bar{{\boldsymbol{ \gamma}}}(0)+L_{0} u(\tau_{0}),\theta(\sigma),0 \bigr)L_{1} y(\tau_{0}) \\& \hphantom{\frac{d}{d\tau_{0}}L_{1} u=}{}+\tilde{C_{u}}\bigl(\bar{u}_{1}(0)+L_{1} u(\tau_{0})\bigr)L_{0} y(\tau _{0})+ \varphi_{3}(\tau_{0}) , \end{aligned}$$
(2.14)
where
$$\begin{aligned}& \varphi_{1}(\tau_{0}) = \tilde{A}\bar{y}_{1}( \mu\tau_{0})+\bigl[\tilde{A}_{u} {\boldsymbol{ \gamma}}'(0)\tau_{0}+\tilde{A}_{\theta} \theta'(0)\tau _{0}+\tilde{A}_{t} \tau_{0}\bigr]L_{0} y(\tau_{0}) \\& \hphantom{\varphi_{1}(\tau_{0}) =}{} \bar{A}\bar{y}_{1}(\mu\tau_{0})+( \tilde{B}\bar{B}), \\& \varphi_{2}(\tau_{0}) = \tilde{f}_{z} \bar{ \alpha}'(0)\tau_{0}+\tilde {f}_{\rho} \rho'(0)\tau_{0} +\tilde{f}_{t} \tau_{0} , \\& \varphi_{3}(\tau_{0}) = \tilde{C}\bar{y}_{1}( \mu\tau_{0})+\bigl[\tilde{C}_{u} {\boldsymbol{ \gamma}}'(0)\tau_{0}+\tilde{C}_{\theta} \theta'(0)\tau _{0}+\tilde{C}_{t} \tau_{0}\bigr]L_{0} y(\tau_{0}) \\& \hphantom{\varphi_{3}(\tau_{0}) =}{} \bar{C}\bar{y}_{1}(\mu\tau_{0})+( \tilde{D}\bar{D}), \end{aligned}$$
here
\(\bar{A}\),
\(\bar{B}\),
\(\bar{C}\),
\(\bar{D}\) take values at the point
\((\bar{\boldsymbol{\gamma}}(0),\theta(\sigma),0)\),
\(\tilde {A}\),
\(\tilde{A}_{u}\),
\(\tilde{A}_{\theta}\),
\(\tilde{A}_{t}\),
\(\tilde{B}\),
\(\tilde {C}\),
\(\tilde{C}_{u}\),
\(\tilde{C}_{\theta}\),
\(\tilde{C}_{t}\),
\(\tilde{D}\) take values at the point
\((\bar{\boldsymbol{\gamma}}(0)+L_{0} u(\tau _{0}),\theta(\sigma), 0)\), and
\(\tilde{f}_{z}\),
\(\tilde{f}_{\rho}\),
\(\tilde{f}_{t}\) take values at the point
\((\bar{\alpha}(0)+L_{0} z(\tau _{0}),\rho(\sigma),0)\).
The initial and boundary conditions of
\(L_{1} x(\tau_{0})\) are
$$ L_{1} z(0)=0,\qquad L_{1} u(0)= \bar{u}_{1}(0),\qquad L_{1} x(+\infty)=0. $$
(2.15)
Like the case of
\(L_{0} x(\tau_{0})\), we will take advantage of the arbitrariness of
\(L_{1} y(0)\) and choose the appropriate value to guarantee that
\(L_{1} y(0)\) satisfies condition (
2.15). For this purpose, we introduce the diagonalization transform
$$ L_{1} z=\delta_{1}, \qquad L_{1} y= \delta_{2}, \qquad L_{1} u=\delta_{3}+ \frac{\delta _{1} \tilde{C}}{\tilde{A}}. $$
It is easy to verify that we obtain the system
$$\begin{aligned}& \frac{d\delta_{1}}{d\tau_{0}}=\frac{\tilde{A}_{u} \tilde{C}}{\tilde {A}}L_{0} y(\tau_{0}) \delta_{1} +\tilde{A}\delta_{2}+\tilde{A}_{u} L_{0} y(\tau_{0}) \bigl(\delta_{3}+\bar {u}_{1}(0)\bigr)+\varphi_{1}(\tau_{0}), \\& \frac{d\delta_{2}}{d\tau_{0}}=\tilde{f}_{z} \delta_{1}+ \varphi_{2}(\tau _{0}), \\& \frac{d\delta_{3}}{d\tau_{0}}=\biggl(\tilde{C}_{u}\frac{\tilde{C}\tilde {A}_{u}}{\tilde{A}} \biggr)L_{0} y(\tau_{0}) \bigl(\delta_{3}+ \bar{u}_{1}(0)\bigr)+\biggl(\varphi _{3}( \tau_{0})\frac{\tilde{C}}{\tilde{A}}\varphi_{1}(\tau_{0}) \biggr). \end{aligned}$$
(2.16)
Obviously, (
2.16) has already separated
\(\delta_{3}\) from
\(\delta _{1}\) and
\(\delta_{2}\) completely. The initial and boundary conditions of
\(\delta_{1}\),
\(\delta_{2}\), and
\(\delta_{3}\) are
\(\delta_{1} (0)=0\),
\(\delta_{3} (0)=\bar{u}_{1} (0)\), and
\(\delta_{i} (+\infty)=0\) (
\(i=1,2,3\)). Together with initial condition, the solution of
\(\delta_{3}\) in (
2.16) is
$$ \delta_{3} (\tau_{0})=\bar{u}_{1} (0)+\int _{0}^{\tau_{0}}\Phi(\tau _{0}) \Phi^{1}(s)\varphi_{4}(s)\, ds, $$
where
\(\varphi_{4}(\tau_{0})=\varphi_{3}(\tau_{0})\frac{\varphi_{1}(\tau _{0}) \tilde{C}}{\tilde{A}}\),
\(\Phi(\tau_{0})\) is the fundamental matrix of the corresponding homogeneous equation with
\(\Phi(0)=E_{k}\).
The condition
\(\delta_{3} (+\infty)=0\) uniquely determines
\(\bar{u}_{1}(0)\):
$$ \bar{u}_{1} (0)=\Phi(\tau_{0})\int_{0}^{+\infty} \Phi^{1}(s)\varphi_{4}(s)\, ds. $$
Since
\(\Phi(\tau_{0})\) exponentially converges to
\(\Phi(+\infty)\) as
\(\tau_{0}\rightarrow+\infty\), the exponential estimate of
\(\delta_{3}(\tau_{0})\) can be obtained. After determining
\(\delta_{3}(\tau_{0})\), the first two questions in (
2.16) can be rewritten as
$$ \begin{aligned} &\frac{d\delta_{1}}{d\tau_{0}}=\frac{\tilde{A}_{u} \tilde{C}}{\tilde {A}}L_{0} y(\tau_{0})\delta_{1} +\tilde{A}\delta_{2}+ \varphi_{5}(\tau_{0}), \\ &\frac{d\delta_{2}}{d\tau_{0}}=\tilde{f}_{z} \delta_{1}+ \varphi_{2}(\tau _{0}), \end{aligned} $$
(2.17)
where
\(\varphi_{5}(\tau_{0})=\tilde{A}_{u} L_{0} y(\tau_{0})(\delta_{3}+\bar {u}_{1}(0))+\varphi_{1}(\tau_{0})\), and
\(\varphi_{5}(\tau_{0})\) is an exponentially decreasing function. The homogeneous equation of (
2.17) is the variational equation of (
2.9). Therefore, by Lemma 4.5 in [
9], (
2.17) has a unique solution which is satisfied with conditions
\(\delta_{1} (0)=0\),
\(\delta_{i} (+\infty)=0\) (
\(i=1,2,3\)), and which is exponentially decreasing. Thus
\(L_{1} x(\tau_{0})\) is determined.
As for the unknown function \(\bar{u}_{1}(t)\), the determination of \(\bar{u}_{1}(t)\) is entirely similar to the case of \(\bar{\boldsymbol{\gamma}}(t)\). We have already solved its initial value \(\bar{u}_{1}(0)\). The fundamental difference between the system of \(\bar{u}_{1}(t)\) and the system of \(\bar{\boldsymbol{\gamma}}(t)\) is that the equation of \(\bar{u}_{1}(t)\) is linear, while the equation of \(\bar{\boldsymbol{\gamma}}(t)\) is nonlinear. Hence we can construct an asymptotic solution up to an arbitrary order.
Next, we will construct the right boundary layer terms. The determination of
\(Q^{()}_{i} x(\tau_{c})\) is similar to that of
\(L_{i} x(\tau_{0})\). For
\(Q^{()}_{0} x(\tau_{c})\), we have
$$\begin{aligned}& \frac{d}{d \tau_{c}}Q^{()}_{0} z=\hat{A}\bigl(\bar{{ \boldsymbol{\gamma}}}(\sigma)+Q^{()}_{0} u( \tau_{c}),\theta(0),\sigma\bigr)Q^{()}_{0} y( \tau _{c}), \\& \frac{d}{d \tau_{c}}Q^{()}_{0} y=\hat{f}\bigl(\bar{\alpha}( \sigma )+Q^{()}_{0} z(\tau_{c}),\rho(0),\sigma \bigr)\check{f}\bigl(\bar{\alpha }(\sigma),\rho(0),\sigma\bigr), \\& \frac{d}{d \tau_{c}}Q^{()}_{0} u=\hat{C}\bigl(\bar{{ \boldsymbol{\gamma}}}(\sigma)+Q^{()}_{0} u( \tau_{c}),\theta(0),\sigma\bigr)Q^{()}_{0} y(\tau _{c}) , \end{aligned}$$
(2.18)
and the initial and boundary value conditions
$$ Q^{()}_{0} z(0)=p_{0}\bar{\alpha}(\sigma),\qquad Q^{()}_{0} x(\infty)=0, $$
(2.19)
where
\(p_{0}\) is an undetermined parameter. The difference between (
2.18) and (
2.5) is that
\(\bar {\boldsymbol{\gamma}}(\sigma)\) in (
2.18) is a known quantity, while
\(\bar{\boldsymbol{\gamma}}(0)\) in (
2.5) is unknown.
From (
2.18), we obtain
$$ \frac{dQ^{()}_{0} u}{dQ^{()}_{0} z}=\frac{\hat{C}(\bar{{\boldsymbol{\gamma}}}(\sigma)+L_{0} u(\tau_{c}),\theta(\sigma),\sigma)}{\hat {A}(\bar{{\boldsymbol{\gamma}}}(\sigma)+L_{0} u(\tau_{c}),\theta(\sigma ),\sigma)}. $$
(2.20)
Integrating the above formula after separating the variable and using the implicit function theorem, one can claim that the system (
2.20) has a solution. Let us denote by
$$ Q^{()}_{0} u=\hat{U}_{0}\bigl(\bar{{\boldsymbol{ \gamma}}}(\sigma),Q^{()}_{0} z\bigr) $$
(2.21)
the solution of (
2.20) satisfying the condition
\(Q^{()}_{0} u=0\) for
\(Q^{()}_{0} z=0\), namely
\(\hat{U}_{0}(\bar{\boldsymbol{\gamma}}(\sigma), 0)=0\).
Then substituting (
2.21) into the first two equations of (
2.18), we get
$$ \begin{aligned} &\frac{d}{d \tau_{c}}Q^{()}_{0} z= \hat{A}\bigl(\bar{{\boldsymbol{\gamma}}}(\sigma)+\hat{U}_{0}\bigl( \bar{{\boldsymbol{\gamma}}}(\sigma),Q^{()}_{0} z\bigr), \theta(0),\sigma\bigr)Q^{()}_{0} y(\tau_{c}), \\ &\frac{d}{d \tau_{c}}Q^{()}_{0} y=\hat{f}\bigl(\bar{\alpha}( \sigma )+Q^{()}_{0} z(\tau_{c}),\rho(0),\sigma \bigr)\check{f}\bigl(\bar{\alpha }(\sigma),\rho(0),\sigma\bigr). \end{aligned} $$
(2.22)
From this we obtain the separate equation of the saddle point
$$ L_{2}\mbox{: } Q^{()}_{0}y(\tau_{c})= \biggl(2\int_{0}^{Q^{()}_{0}z}\frac {\hat{f}(\bar{\alpha}(\sigma)+\xi, \rho(0),\sigma)\check{f}(\bar{\alpha}(\sigma),\rho(0),\sigma )}{\hat{A}(\bar{\boldsymbol{\gamma}}(\sigma) +\hat{U}_{0}(\bar{\boldsymbol{\gamma}}(\sigma),\xi),\theta(0),\sigma )}\,d\xi \biggr)^{\frac{1}{2}}. $$
(2.23)
Equations (
2.21) and (
2.23) give an analytic expression of the onedimensional manifold
\(\bar{\Omega}_{1}\), which is similar to that for the manifold
\(\bar{\Omega}_{0}\). Therefore we require that the following condition is established.
 (\(\mathrm{H}_{4}\)):

Suppose that
\(Q^{()}_{0} z(0)\cap \bar{\Omega}_{1}\neq\emptyset\).
Substituting \(Q^{()}_{0} z(0)\) into (2.21) and (2.23), we get the initial values \(Q^{()}_{0}u(0)\) and \(Q^{()}_{0}y(0)\), and these initial values are related to the unknown parameter \(p_{0}\).
For
\(Q^{()}_{1} z(\tau_{c})\), we have
$$\begin{aligned}& \frac{d}{d\tau_{c}}Q^{()}_{1} z= \hat{A}Q^{()}_{1} y(\tau_{C})+\hat {A_{u}}\bigl(\bar{u}_{1}(\sigma)+Q^{()}_{1} u(\tau_{c})\bigr)Q^{()}_{0} y(\tau _{c})+\psi_{1}(\tau_{c}), \\& \frac{d}{d\tau_{c}}Q^{()}_{1} y=\hat{f}_{z} Q^{()}_{1} z(\tau_{c})+\psi _{1}( \tau_{c}), \\& \frac{d}{d\tau_{c}}Q^{()}_{1} u=\hat{C}Q^{()}_{1} y(\tau_{c})+\hat {C_{u}}\bigl(\bar{u}_{1}( \sigma)+Q^{()}_{1} u(\tau_{c})\bigr)Q^{()}_{0} y(\tau _{c})+\psi_{3}(\tau_{c}) , \end{aligned}$$
(2.24)
where
$$\begin{aligned}& \psi_{1}(\tau_{c}) = \hat{A}\bar{y}_{1}(\sigma+ \mu\tau_{c})+\bigl[\hat{A}_{u} \bar{{\boldsymbol{ \gamma}}}'(\sigma)\tau_{c}+\hat{A}_{\theta}\theta '(\sigma)\tau_{c}+\hat{A}_{t} \tau_{c}\bigr]Q^{()}_{0} y(\tau_{c}) \\& \hphantom{\psi_{1}(\tau_{c}) =}{} \check{A}\bar{y}_{1}(\sigma+\mu \tau_{c})+(\hat{B}\check{B}), \\& \psi_{2}(\tau_{c}) = \hat{f}_{z} \bar{ \alpha}'(\sigma)\tau_{c}+\hat {f}_{\rho} \rho'(\sigma)\tau_{c} +\hat{f}_{t} \tau_{c} , \\& \psi_{3}(\tau_{c}) = \hat{C}\bar{y}_{1}(\sigma+ \mu\tau_{c})+\bigl[\hat{C}_{u} \bar{{\boldsymbol{ \gamma}}}'(\sigma)\tau_{c}+\hat{C}_{\theta}\theta '(\sigma)\tau_{c}+\hat{C}_{t} \tau_{c}\bigr]Q^{()}_{0} y(\tau_{c}) \\& \hphantom{\psi_{3}(\tau_{c}) =}{} \check{C}\bar{y}_{1}(\sigma+\mu \tau_{c})+(\hat{D}\check{D}), \end{aligned}$$
here
\(\check{A}\),
\(\check{B}\),
\(\check{C}\),
\(\check{D}\) take values at the point
\((\bar{{\boldsymbol{\gamma}}}(\sigma),\theta(0),\sigma)\),
\(\hat {A}\),
\(\hat{A}_{u}\),
\(\hat{A}_{\theta}\),
\(\hat{A}_{t}\),
\(\hat{B}\),
\(\hat{C}\),
\(\hat {C}_{u}\),
\(\hat{C}_{\theta}\),
\(\hat{C}_{t}\),
\(\hat{D}\) take values at the point
\((\bar{\boldsymbol{\gamma}}(\sigma )+Q^{()}_{0} u(\tau_{c}),\theta(0), \sigma)\), and
\(\hat{f}_{z}\),
\(\hat {f}_{\rho}\),
\(\hat{f}_{t}\) take values at the point
\((\bar{\alpha }(\sigma)+Q^{()}_{0} z(\tau_{c}),\rho(0),\sigma)\). The initial and boundary conditions of
\(Q^{()}_{1} z(\tau_{c})\) are
$$ Q^{()}_{1} z(0)=p_{1} \bar{z}_{1} (\sigma), \qquad Q^{()}_{1} x( \infty)=0. $$
(2.25)
Analogously to the determination of
\(L_{1} x(\tau_{0})\), performing the diagonalization transform,
\(Q^{()}_{1} z=\delta_{4}\),
\(Q^{()}_{1} y=\delta_{5}\),
\(Q^{()}_{1} u=\delta _{6}+\frac{\delta_{4} \hat{C}}{\hat{A}}\). We obtain from (
2.24) the system
$$\begin{aligned}& \frac{d\delta_{4}}{d\tau_{c}}=\frac{\hat{A}_{u} \hat{C}}{\hat {A}}Q^{()}_{0} y( \tau_{c})\delta_{4} +\hat{A}\delta_{5}+ \hat{A}_{u} Q^{()}_{0} y(\tau_{c}) \bigl(\delta_{6}+\bar {u}_{1}(\sigma)\bigr)+ \psi_{1}(\tau_{c}), \\& \frac{d\delta_{5}}{d\tau_{c}}=\hat{f}_{z} \delta_{4}+ \psi_{2}(\tau_{c}), \\& \frac{d\delta_{6}}{d\tau_{c}}=\biggl(\hat{C}_{u}\frac{\hat{C}\hat {A}_{u}}{\hat{A}} \biggr)Q^{()}_{0} y(\tau_{c}) \bigl( \delta_{6}+\bar{u}_{1}(\sigma )\bigr)+\biggl( \varphi_{3}(\tau_{c})\frac{\hat{C}}{\hat{A}} \varphi_{1}(\tau_{c})\biggr), \end{aligned}$$
(2.26)
with the initial and boundary value conditions
$$ \delta_{4} (0)=p_{1}, \qquad \delta_{6} (0)=Q_{1}^{()}u(0)\frac{p_{1} \hat {C}}{\hat{A}}, \qquad \delta_{i} (\infty)=0 \quad (i=4,5,6). $$
Under the initial and boundary value conditions, the solution of the third equation of (
2.26) is
\(\delta_{6} (\tau_{c})=Q_{1}^{()}u(0)\frac{p_{1} \hat{C}}{\hat{A}}+\int_{0}^{\tau_{c}}\hat{\Phi}(\tau_{c})\hat{\Phi}^{1}(s)\psi_{4}(s)\, ds\), where
\(\psi_{4}(\tau_{c})=\psi_{3}(\tau_{c})\frac{\psi_{1}(\tau_{c}) \hat {C}}{\hat{A}}\),
\(\hat{\Phi}(\tau_{c})\) is the fundamental matrix of the corresponding homogeneous equation with
\(\Phi(0)=E_{k}\).
The condition
\(\delta_{6} (+\infty)=0\) uniquely determines
\(Q_{1}^{()}u(0)\):
$$ Q_{1}^{()}u(0)=\frac{p_{1} \hat{C}}{\hat{A}}\hat{\Phi}(+\infty )\int _{0}^{+\infty}\hat{\Phi}^{1}(s) \psi_{4}(s)\, ds. $$
After computing
\(\delta_{6}(\tau_{c})\), we transform the expression for the first two equations of (
2.26)
$$ \begin{aligned} &\frac{d\delta_{4}}{d\tau_{c}}=\frac{\hat{A}_{u} \hat{C}}{\hat {A}}Q^{()}_{0} y(\tau_{c})\delta_{4} +\hat{A}\delta_{5}+ \psi_{5}(\tau_{c}), \\ &\frac{d\delta_{5}}{d\tau_{c}}=\hat{f}_{z} \delta_{4}+ \psi_{2}(\tau_{c}), \end{aligned} $$
(2.27)
where
\(\varphi_{5}(\tau_{c})=\hat{A}_{u} Q^{()}_{0} y(\tau_{c})(\delta _{6}+\bar{u}_{1}(\sigma))+\psi_{1}(\tau_{c})\). The homogeneous equation of (
2.27) is the variational equation of (
2.22). Therefore, by Lemma 4.5 in [
9], (
2.17) has a unique solution which is satisfied with the conditions
\(\delta_{4} (0)=p_{1}\) and
\(\delta _{i}(\infty)=0\) (
\(i=4,5,6\)). It should be noted that
\(Q^{()}_{0} x(\tau _{c})\) is related to
\(p_{1}\), and we will use the continuous condition to determine
\(p_{k}\) (
\(k\geq0\)).
2.2 The construction of asymptotic solutions on the interval \([\sigma,T]\)
By analogy with the regular part on the interval
\([0,\sigma]\), one can get the degenerate problem
$$\begin{aligned}& A\bigl(\bar{\bar{u}}_{0}(t),\bar{{\boldsymbol{ \gamma}}}(t),t\bigr)\bar{\bar {y}}_{0}(t)=0, \\& f\bigl(\bar{\bar{z}}_{0}(t),\bar{\alpha}(t),t\bigr)=0, \\& C\bigl(\bar{\bar{u}}_{0}(t),\bar{{\boldsymbol{\gamma}}}(t),t\bigr) \bar{\bar {y}}_{0}(t)=0. \end{aligned}$$
(2.28)
Clearly, the solution of (
2.28) is
$$ \bar{\bar{z}}_{0}(t)=\bar{\bar{\alpha}}(t),\qquad \bar{ \bar{y}}_{0}(t)=0,\qquad \bar{\bar{u}}_{0}(t)=\bar{\bar{ \boldsymbol{\gamma}}}(t), $$
(2.29)
where
\(\bar{\bar{\boldsymbol{\gamma}}}(t)\) is an arbitrary
kdimensional vector function.
Now we consider the left boundary system
$$\begin{aligned}& \frac{d}{d \tau_{c}}Q^{(+)}_{0} z=\tilde{ \tilde{A}}\bigl(\bar{\bar {\boldsymbol{\gamma}}}(\sigma)+Q^{(+)}_{0} u(\tau_{c}),\bar{{\boldsymbol{\gamma}}}(\sigma)+L_{0} u, \sigma\bigr)Q^{(+)}_{0} y(\tau_{c}), \\& \frac{d}{d \tau_{c}}Q^{(+)}_{0} y=\tilde{\tilde{f}}\bigl(\bar{ \bar{\alpha }}(\sigma)+Q^{(+)}_{0} z(\tau_{c}),\bar{ \alpha}(\sigma)+L_{0} z,\sigma \bigr)\bar{\bar{f}}\bigl(\bar{\bar{ \alpha}}(\sigma),\bar{\alpha}(\sigma ),\sigma\bigr), \\& \frac{d}{d \tau_{c}}Q^{(+)}_{0} u=\tilde{\tilde{C}}\bigl(\bar{ \bar {\boldsymbol{\gamma}}}(\sigma)+Q^{(+)}_{0} u( \tau_{c}),\bar{{\boldsymbol{\gamma}}}(\sigma)+L_{0} u,\sigma \bigr)Q^{(+)}_{0} y(\tau_{c}), \end{aligned}$$
(2.30)
and the initial and boundary conditions
$$ Q^{(+)}_{0} z(0)=p_{0}\bar{\bar{\alpha}}(\sigma), \qquad Q^{(+)}_{0} x(\infty)=0. $$
(2.31)
Note that
\(p_{0}\) and
\(\bar{\bar{\boldsymbol{\gamma}}}(0)\) in (
2.30)(
2.31) are both unknown and the initial value of
\(Q^{(+)}_{0} y(\tau_{c})\) is arbitrary.
We first consider the stability manifold
\(\bar{\bar{\Omega }}_{0}\). From (
2.30), we have
$$ \frac{dQ^{(+)}_{0} u}{dQ^{(+)}_{0} z}=\frac{\tilde{\tilde{C}}(\bar {\bar{{\boldsymbol{\gamma}}}}(\sigma)+Q^{(+)}_{0} u(\tau_{c}),\bar {\boldsymbol{\gamma}}(\sigma)+L_{0} u,\sigma)}{\tilde{\tilde{A}}(\bar {\bar{\boldsymbol{\gamma}}}(\sigma)+Q^{(+)}_{0} u(\tau_{c}),\bar{\boldsymbol{\gamma} }(\sigma)+L_{0} u,\sigma)}. $$
(2.32)
Similarly, we denote the solution of (
2.32) by
$$ Q^{(+)}_{0} u=\bar{\bar{U}}_{0}\bigl(\bar{\bar{{ \boldsymbol{\gamma}}}}(\sigma ),Q^{(+)}_{0} z \bigr), $$
(2.33)
satisfying the condition
\(Q^{(+)}_{0} u=0\) for
\(Q^{(+)}_{0} z=0\), namely
\(\bar{\bar{U}}_{0}(\bar {\bar{\boldsymbol{\gamma}}}(\sigma),0)=0\).
By assumption (
\(\mathrm{H}'_{1}\)), one can claim that there exists a unique solution in a certain neighborhood of the point
\(Q^{(+)}_{0} z=0\). Then substituting (
2.33) into (
2.30), we get
$$ \begin{aligned} &\frac{dQ^{(+)}_{0} z}{d \tau_{c}}=\tilde{\tilde{A}}\bigl(\bar{\bar {{ \boldsymbol{\gamma}}}}(\sigma)+\bar{\bar{U}}_{0}\bigl(\bar{\bar{{ \boldsymbol{\gamma}}}}(\sigma),Q^{(+)}_{0} z\bigr),\bar{{ \boldsymbol{\gamma}}}(\sigma )+L_{0} u,\sigma\bigr)Q^{(+)}_{0} y(\tau_{c}), \\ &\frac{dQ^{(+)}_{0} y}{d \tau_{c}}=\tilde{\tilde{f}}\bigl(\bar{\bar{\alpha }}( \sigma)+Q^{(+)}_{0} z(\tau_{c}),\bar{\alpha}( \sigma)+L_{0} z,\sigma \bigr)\bar{\bar{f}}\bigl(\bar{\bar{\alpha}}( \sigma),\bar{\alpha}(\sigma ),\sigma\bigr). \end{aligned} $$
(2.34)
The equilibrium point
\((0,0)\) on the phase plane
\((Q^{(+)}_{0} z,Q^{(+)}_{0} y)\) is also a saddle point. Integrating (
2.34), we can get a separate equation for the saddle point,
$$ L_{3}\mbox{: }Q^{(+)}_{0}y(\tau_{c})= \biggl(2\int_{0}^{Q^{(+)}_{0}z}\frac {\tilde{\tilde{f}}(\bar{\bar{\alpha}}(\sigma)+\xi,\bar{\alpha }(\sigma)+L_{0} z,\sigma)\bar{\bar{f}}(\bar{\bar{\alpha}}(\sigma ),\bar{\alpha}(\sigma),\sigma)}{ \tilde{\tilde{A}}(\bar{\bar{\boldsymbol{\gamma}}}(\sigma)+\bar {\bar{U}}_{0}(\bar{\bar{\boldsymbol{\gamma}}}(\sigma),\xi),\bar {\boldsymbol{\gamma}}(\sigma)+L_{0} u,\sigma)}\,d\xi \biggr)^{\frac{1}{2}}. $$
(2.35)
Equations (
2.35) and (
2.33) give an analytic representation of the onedimensional manifold
\(\tilde{\Omega}_{0}\). In order to make sure (
2.32) has a solution, we need the condition that follows:
 (\(\mathrm{H}_{5}\)):

Suppose that
\(Q^{(+)}_{0} z(0)\cap \tilde{\Omega}_{0}\neq0\).
Putting (
2.31) into (
2.33), we have
$$ Q_{0}^{(+)}u(0)=\bar{\bar{U}}_{0}\bigl(\bar{\bar{{ \boldsymbol{\gamma}}}}(\sigma),p_{0}\bar{\bar{\alpha}}(\sigma) \bigr), $$
(2.36)
which represents a system of
k scalar equations in the
k unknown components of the vector
\(\bar{\bar{\boldsymbol{\gamma}}}(\sigma)\).
 (\(\mathrm{H}_{6}\)):

Suppose that (2.36) has a solution
\(\bar{\bar{\boldsymbol{\gamma}}}(\sigma)=\bar{\bar {\boldsymbol{\gamma}}}^{0}\).
Here \(\bar{\bar{\boldsymbol{\gamma}}}^{0}\) is related to \(p_{0}\), namely \(\bar{\bar{\boldsymbol{\gamma}}}^{0}=\bar{\bar{\boldsymbol{\gamma}}}^{0}(p_{0},\bar{\bar{\alpha}}(\sigma))\). \(Q^{(+)}_{0} x(\tau_{c})\) can be determined in the same manner as \(L_{0} x(\tau_{0})\). \(Q^{(+)}_{0} x(\tau_{c})\) is also related to \(p_{0}\). For \(\bar{\bar{\boldsymbol{\gamma}}}(t)\), we only know its initial value \(\bar{\boldsymbol{\gamma}}^{0}\). The determination of \(\bar{\bar{\boldsymbol{\gamma}}}(t)\) should be in the next step.
For
\(\bar{\bar{x}}_{1} (t)\), we have
$$\begin{aligned}& \bar{\bar{A}}\bigl(\bar{\bar{\boldsymbol{\gamma}}}(t),\bar{\boldsymbol{ \gamma}}(t),t\bigr)\bar{\bar{y}}_{1} (t)+\bar{\bar{B}}\bigl(\bar{\bar { \boldsymbol{\gamma}}}(t),\bar{\boldsymbol{\gamma}}(t),t\bigr)=\bar{\bar { \alpha}}'(t), \\& f_{z} \bigl(\bar{\bar{\alpha}}(t),\bar{\alpha}(t),t\bigr)\bar{\bar {z}}_{1}(t)=0, \\& \bar{\bar{C}}\bigl(\bar{\bar{{\boldsymbol{\gamma}}}}(t),\bar{\boldsymbol{ \gamma}}(t),t\bigr)\bar{\bar{y}}_{1} (t)+\bar{\bar{D}}\bigl(\bar{\bar { \boldsymbol{\gamma}}}(t),\bar{{\boldsymbol{\gamma}}}(t),t\bigr)=\bar {\boldsymbol{ \gamma}}'(t), \end{aligned}$$
whose solution is
$$ \bar{\bar{z}}_{1}(t)=0,\qquad \bar{\bar{y}}_{1}(t)= \frac{\bar{\bar{\alpha}}'(t)\bar{\bar {B}}}{\bar{\bar{A}}},\qquad \frac{d\bar{\bar{{\boldsymbol{\gamma}}}}}{dt}=\frac{\bar{\bar {C}}(\bar{\bar{\alpha}}'(t)\bar{\bar{B}})}{\bar{\bar{A}}}+\bar { \bar{D}}. $$
(2.37)
By the existence of solution for the initial value problem, (
2.37) together with the initial condition
\(\bar{\bar{ \boldsymbol{\gamma}}}(\sigma)=\bar{\bar{\boldsymbol{\gamma}}}^{0}\) has a solution
\(\bar{\bar{\boldsymbol{\gamma}}}=\bar {\bar{\boldsymbol{\gamma}}}(t)\) for
\(\sigma\leq t\leq T\). Therefore
\(\bar{\bar{x}}_{0}(t)\) can be completely determined. For
\(\bar{\bar{x}}_{1}(t)\), the determination of it is similar to that of
\(\bar{x}_{1}(t)\). Equation (
2.37) has only determined
\(\bar {\bar{z}}_{1}(t)\) and
\(\bar{\bar{y}}_{1}(t)\), while
\(\bar{\bar {u}}_{1}(t)\) is yet unknown. We need use the first approximation of the left boundary term to determine
\(\bar{\bar{u}}_{1}(t)\).
For
\(Q^{(+)}_{1}(\tau_{c})\), we obtain
$$\begin{aligned}& \frac{d}{d\tau_{c}}Q^{(+)}_{1} z=\tilde{ \tilde{A}}\bigl(\bar{\bar {{\boldsymbol{\gamma}}}}(\sigma)+Q^{(+)}_{0} u(\tau_{c}),\bar{{\boldsymbol{\gamma}}}(\sigma)+L_{0} u( \tau_{c}),\sigma\bigr)Q^{(+)}_{1} y( \tau_{c}) \\& \hphantom{\frac{d}{d\tau_{c}}Q^{(+)}_{1} z=}{}+\tilde{\tilde{A_{u}}}\bigl(\bar{ \bar{u}}_{1}(\sigma )+Q^{(+)}_{1} u( \tau_{c})\bigr)Q^{(+)}_{0} y(\tau_{c})+ \tilde{\varphi}_{1}(\tau _{c}), \\& \frac{d}{d\tau_{c}}Q^{(+)}_{1} y=\tilde{\tilde{f}}_{z} Q^{(+)}_{1} z(\tau _{c})+\tilde{ \varphi}_{2}(\tau_{c}), \\& \frac{d}{d\tau_{c}}Q^{(+)}_{1} u=\tilde{\tilde{C}}\bigl(\bar{ \bar {{\boldsymbol{\gamma}}}}(0)+Q^{(+)}_{0} u( \tau_{c}),\bar{{\boldsymbol{\gamma}}}(\sigma),\sigma \bigr)Q^{(+)}_{1} y(\tau_{c}) \\& \hphantom{\frac{d}{d\tau_{c}}Q^{(+)}_{1} u=}{}+\tilde{\tilde{C_{u}}}\bigl(\bar{ \bar{u}}_{1}(0)+Q^{(+)}_{1} u(\tau_{c}) \bigr)Q^{(+)}_{0} y(\tau_{c})+\tilde{ \varphi}_{3}(\tau_{c}) , \end{aligned}$$
(2.38)
where the expression of
\(\tilde{\varphi}_{i}(\tau_{c})\) (
\(i=1,2,3\)) is similar to that of
\(\varphi_{i}(\tau_{c})\) (
\(i=1,2,3\)). The initial and boundary conditions of
\(Q^{(+)}_{1} x(\tau_{c})\) are
$$ Q^{(+)}_{1} z(0)=p_{1}\bar{ \bar{z}}_{1}(\sigma),\qquad Q^{(+)}_{1} x(+ \infty)=0. $$
(2.39)
Similar to
\(L_{1} x(\tau_{0})\), we introduce the diagonalization transform
\(Q^{(+)}_{1} z=\tilde{\delta}_{1}\),
\(Q^{(+)}_{1} y=\tilde{\delta }_{2}\),
\(Q^{(+)}_{1} u=\tilde{\delta}_{3}+\frac{\tilde{\delta}_{1} \tilde {\tilde{C}}}{\tilde{\tilde{A}}}\). The system (
2.38) takes the form
$$\begin{aligned}& \frac{d\tilde{\delta}_{1}}{d\tau_{c}}=\frac{\tilde{\tilde{A}}_{u} \tilde{\tilde{C}}}{\tilde{\tilde{A}}}Q^{(+)}_{0} y( \tau_{c})\tilde {\delta}_{1} +\tilde{\tilde{A}}\tilde{ \delta}_{2}+\tilde{\tilde{A}}_{u} Q^{(+)}_{0} y(\tau_{c}) \bigl(\tilde{\delta}_{3}+\bar{ \bar{u}}_{1}(\sigma)\bigr)+\tilde {\varphi}_{1}( \tau_{c}), \\& \frac{d\tilde{\delta}_{2}}{d\tau_{c}}=\tilde{\tilde{f}}_{z} \tilde { \delta}_{1}+\tilde{\varphi}_{2}(\tau_{c}), \\& \frac{d\tilde{\delta}_{3}}{d\tau_{c}}=\biggl(\tilde{\tilde{C}}_{u}\frac {\tilde{\tilde{C}}\tilde{\tilde{A}}_{u}}{\tilde{\tilde {A}}} \biggr)Q^{(+)}_{0} y(\tau_{c}) \bigl(\tilde{ \delta}_{3}+\bar{\bar{u}}_{1}(\sigma )\bigr)+\biggl(\tilde{ \varphi}_{3}(\tau_{c}) \frac{\tilde{\tilde{C}}}{\tilde{\tilde{A}}}\tilde{\varphi }_{1}(\tau_{c})\biggr), \end{aligned}$$
(2.40)
with the initial and boundary value conditions
$$ \tilde{\delta}_{1} (0)=p_{1},\qquad \tilde{ \delta}_{3} (0)=Q_{1}^{(+)}u(0)\frac {p_{1}\tilde{\tilde{C}}}{\tilde{\tilde{A}}}, \qquad \tilde{\delta}_{i} (+\infty)=0\quad (i=1,2,3). $$
Then the solution of the third equation of (
2.40) is
$$ \tilde{\delta}_{3}(\tau_{c})= Q_{1}^{(+)}u(0) \frac{p_{1}\tilde{\tilde {C}}}{\tilde{\tilde{A}}}+\int_{0}^{\tau_{c}}\tilde{\Phi}(\tau _{c})\tilde{\Phi}^{1}(s)\tilde{\varphi}_{4}(s)\, ds, $$
where
\(\tilde{\varphi}_{4}(\tau_{c})=\tilde{\varphi}_{3}(\tau_{c})\frac {\tilde{\varphi}_{1}(\tau_{c}) \tilde{\tilde{C}}}{\tilde{\tilde {A}}}\),
\(\tilde{\Phi}(\tau_{c})\) is the fundamental solution matrix of the corresponding homogeneous equations with
\(\tilde{\Phi}(0)=E_{k}\).
The condition
\(\tilde{\delta}_{3} (\infty)=0\) uniquely determines
\(Q_{1}^{(+)}u(0)\):
$$ Q_{1}^{(+)}u(0)=\frac{p_{1} \tilde{\tilde{C}}}{\tilde{\tilde {A}}}\tilde{\Phi}(\infty)\int _{0}^{\infty}\tilde{\Phi }^{1}(s)\tilde{ \varphi}_{4}(s)\, ds. $$
After determining
\(\tilde{\delta}_{3}(\tau_{0})\), the first two equations of (
2.40) can be rewritten as
$$ \begin{aligned} &\frac{d\tilde{\delta}_{1}}{d\tau_{c}}=\frac{\tilde{\tilde{A}}_{u} \tilde{\tilde{C}}}{\tilde{\tilde{A}}}Q^{(+)}_{0} y(\tau_{c})\tilde {\delta}_{1} +\tilde{\tilde{A}}\tilde{ \delta}_{2}+\tilde{\varphi}_{5}(\tau_{c}), \\ &\frac{d\tilde{\delta}_{2}}{d\tau_{c}}=\tilde{\tilde{f}}_{z} \tilde { \delta}_{1}+\tilde{\varphi}_{2}(\tau_{c}). \end{aligned} $$
(2.41)
The homogeneous equations of (
2.41) are the variational equations of (
2.34). Thus (
2.41) has a unique solution, which satisfies the conditions
\(\tilde{\delta}_{1} (0)=p_{1}\),
\(\tilde{\delta}_{i} (+\infty )=0\), and which meets the exponential decay estimation. Hence
\(Q^{(+)}_{1} x(\tau_{c})\) is completely determined. As for
\(\bar{\bar {u}}_{1}(t)\), we have obtained its initial value
\(\bar{\bar {u}}_{1}(\sigma)\). The determination of
\(\bar{\bar{u}}_{1}(t)\) is in the next step, which is completely similar to that of
\(\bar{\bar{\boldsymbol{\gamma}}}(t)\). Then we will apply the continuity condition to solve
\(p_{k}\) (
\(k\geq0\)).
By means of the continuous property of solution, we obtain, in view of (
2.2),
$$\begin{aligned}& \mu^{0}\mbox{: } \bar{y}_{0}(\sigma)+Q_{0}^{()}y(0)= \bar{\bar{y}}_{0}(\sigma )+Q_{0}^{(+)}y(0), \\& \mu^{1}\mbox{: } \bar{y}_{1}(\sigma)+Q_{1}^{()}y(0)= \bar{\bar{y}}_{1}(\sigma )+Q_{1}^{(+)}y(0), \\& \ldots, \\& \mu^{k}\mbox{: }\bar{y}_{k}(\sigma)+Q_{k}^{()}y(0)= \bar{\bar{y}}_{k}(\sigma )+Q_{k}^{(+)}y(0), \\& \ldots. \end{aligned}$$
(2.42)
From the first relation of (
2.42), we get an equation involving
\(p_{0}\),
$$\begin{aligned} M(p_{0}): =&  \biggl(2\int_{0}^{p_{0}\bar{\alpha}(\sigma)} \frac {\hat{f}(\bar{\alpha}(\sigma)+\xi, \rho(0),\sigma)\check{f}(\bar{\alpha}(\sigma),\rho(0),\sigma )}{\hat{A}(\bar{{\boldsymbol{\gamma}}}(\sigma) +\hat{U}_{0}(\bar{{\boldsymbol{\gamma}}}(\sigma),\xi),\theta(0),\sigma )}\,d\xi \biggr)^{\frac{1}{2}} \\ &{}  \biggl(2\int_{0}^{p_{0}\bar{\bar{\alpha}}(\sigma)}\frac {\tilde{\tilde{f}}(\bar{\bar{\alpha}}(\sigma)+\xi,\bar{\alpha }(\sigma)+L_{0} z,\sigma)\bar{\bar{f}}(\bar{\bar{\alpha}}(\sigma ),\bar{\alpha}(\sigma),\sigma)}{ \tilde{\tilde{A}}(\bar{\bar{{\boldsymbol{\gamma}}}}^{0}(p_{0},\bar {\bar{\alpha}}(\sigma)) +\bar{\bar{U}}_{0}(\bar{\bar{{\boldsymbol{\gamma}}}}^{0}(p_{0},\bar{\bar {\alpha}}(\sigma)),\xi),\bar{\boldsymbol{\gamma}}(\sigma)+L_{0} u,\sigma)}\,d \xi \biggr)^{\frac{1}{2}} \\ =&0. \end{aligned}$$
(2.43)
 (\(\mathrm{H}_{7}\)):

Suppose that the equation
\(M(p_{0})=0\)
has a solution and
\(\frac{d M}{d p} _{p=p_{0}}\neq0\).
Similarly, by the kth continuous condition, \(p_{k}\) (\(k\geq1\)) can be solved in turn. Thus, \(Q^{(+)}x(\tau_{c})\) is determined. The solvable method of \(Rx(\tau_{1})\) is analogous to that of \(Q^{()}x(\tau_{c})\) and it is omitted here.