# Stability of the nonlinear Milne Problem for radiative heat transfer system

Mohamed Ghattassi  Department of Mathematics, New York University in Abu
Dhabi, Saadiyat Island, P.O. Box 129188, Abu Dhabi, United Arab Emirates,
<EMAIL_ADDRESS>,  Xiaokai Huo  Institute for Analysis and Scientific
Computing, Vienna University of Technology, Wiedner Hauptstraße 8–10, 1040
Wien, Austria<EMAIL_ADDRESS>and  Nader Masmoudi  Department of
Mathematics, New York University in Abu Dhabi, Saadiyat Island, P.O. Box
129188, Abu Dhabi, United Arab Emirates- Courant Institute of Mathematical
Sciences, New York University, 251 Mercer Street, New York, NY 10012, USA,
<EMAIL_ADDRESS>
###### Abstract.

This paper focuses on the nonlinear Milne problem of the radiative heat
transfer system on the half-space. The nonlinear model is described by a
second order ODE for temperature coupled to transport equation for radiative
intensity. The nonlinearity of the fourth power Stefan-Boltzmann law of black
body radiation, bring additional difficulty in mathematical analysis, compared
to the well-developed theory for Milne problem of linear transport equation.
With the help of the monotonicity property of the second order ODE, we prove
the existence of the nonlinear Milne problem on a finite interval using
monotonic convergence theorems. Then the solution is extended to the half-
space using a uniform weighted estimate and the compactness method. Moreover,
the solutions are proved to converge to constants as $x\to\infty$. Therefore,
the linear stability analysis is used to study the uniqueness of the nonlinear
Milne problem. The existence and uniqueness for the linearized system is
established under a spectral assumption on the solution of the nonlinear
problem. The spectral assumption is shown to be satisfied when the boundary
data is close to the well-prepared case by using a generalized Hardy’s
inequality. The uniqueness of the solution to the half-line nonlinear Milne
problem is established in a neighborhood of solutions satisfying a spectral
assumption and the energy estimate. The current work extends the study of
Milne problem for linear transport equations and provides a comprehensive
study on the nonlinear Milne problem of radiative heat transfer systems.

###### Key words and phrases:

Radiative transfer, Milne problem, Maximum principle, Fixed point theorem,
Spectral stability

###### Contents

  1. 1 Introduction
    1. 1.1 Main results
    2. 1.2 Organizations
  2. 2 Existence on the bounded interval
    1. 2.1 Monotonicity of the second order ODE
    2. 2.2 Existence and uniqueness
    3. 2.3 Weighted estimate
  3. 3 Existence and decay results for the nonlinear Milne problem
    1. 3.1 Existence
    2. 3.2 Weighted estimate
    3. 3.3 Exponential decay
  4. 4 Linearized problem
    1. 4.1 The spectral assumption
    2. 4.2 Existence on the bounded interval
    3. 4.3 Existence on the half-space
    4. 4.4 Weighted estimate and exponentially decay
    5. 4.5 Uniqueness
  5. 5 Uniqueness of the nonlinear Milne problem
    1. 5.1 Sufficient condition for the spectral assumption
    2. 5.2 Uniqueness for almost well-prepared boundary data
    3. 5.3 Uniqueness under small perturbations
  6. A Proof of Theorem 2 for the general case
  7. B Monotonicity of linear transport equation
  8. C Formula for the transport equation
  9. D Generalized Hardy’s inequality
  10. E Leray-Schauder Theorem

##  1\. Introduction

We consider the following system

$\displaystyle\partial_{x}^{2}T+\langle\psi-T^{4}\rangle=~{}$ $\displaystyle
0,$ (1.1) $\displaystyle\mu\partial_{x}\psi+(\psi-T^{4})=~{}$ $\displaystyle
0,$ (1.2)

in the space $(x,\mu)\in\mathbb{R}_{+}\times[-1,1]$, where $T=T(x)$ and
$\psi=\psi(x,\mu)$ are the temperature and radiative intensity, respectively.
The bracket $\langle\cdot\rangle$ denotes the integral over $\mu\in[-1,1]$,
i.e. $\langle f\rangle:=\int_{-1}^{1}f(\mu)d\mu$. The boundary conditions are
taken to be the non-homogeneous Dirichlet conditions

$\displaystyle T(0)=~{}$ $\displaystyle T_{b},$ (1.3)
$\displaystyle\psi(0,\mu)=~{}$ $\displaystyle\psi_{b}(\mu),\text{ for any
}\mu\in(0,1],$ (1.4)

where $T_{b}\geq 0$ is a given constant and $\psi_{b}=\psi_{b}(\mu)\geq 0$ is
a given function of $\mu\in(0,1].$ Here $\psi_{b}$ describes the radiative
intensity transmitted into the medium from outside. In this paper, the above
problem is called the _nonlinear Milne problem of radiative heat transfer_.
The main objective of our work is to study the existence, uniqueness, long
time behavior and stability of this nonlinear Milne problem.

System (1.1)-(1.2) was derived in the boundary layer problem for the diffusive
limit of the following radiative heat transfer system

$\displaystyle\partial_{t}T^{\varepsilon}=\Delta
T^{\varepsilon}+\frac{1}{\varepsilon^{2}}\langle\psi^{\varepsilon}-(T^{\varepsilon})^{4}\rangle,$
$\displaystyle\partial_{t}\psi^{\varepsilon}+\frac{1}{\varepsilon}\omega\cdot\nabla\psi^{\varepsilon}=-\frac{1}{\varepsilon^{2}}(\psi^{\varepsilon}-(T^{\varepsilon})^{4}),$

where $T^{\varepsilon}=T^{\varepsilon}(t,X)$,
$\psi^{\varepsilon}=\psi(t,X,\omega)$ with $X\in\mathbb{R}^{3}$,
$\omega\in\mathbb{S}^{2}$. System (1.1)-(1.2) is the boundary layer problem of
the corresponding steady state problem of the above radiative heat transfer
system, for more details we refer the reader to [19] and references therein.
When the boundary data is well-prepared such that $\psi_{b}=T_{b}^{4}$, system
(1.1)-(1.2) has the constant solution $(T_{b},\psi_{b}1_{\mu\in[0,1]})$, where
$1_{\mu\in[0,1]}$ denotes the characteristic function on $[0,1]$, and no
boundary layer exists. In this case the diffusive limit for the radiative
transfer system was rigorously justified in our previous work [13] based on
the weak convergence method and the relative entropy method. In general, when
$\psi_{b}\neq T_{b}^{4}$, it is crucial to study system (1.1)-(1.2) because
the boundary layer effects make the interior solution obtained by asymptotic
analysis invalid near the boundary. The solution to (1.1)-(1.2) provides a
better description of the radiative transfer system near the boundary.
Moreover, the boundary condition for the interior solutions needs to be
determined by the limit of the solutions to the nonlinear Milne problem
(1.1)-(1.2) as $x\to\infty$. Based on this work, we are able to provide a
rigorous justification of the diffusive limit [11, 12]. Hence it is important
to study the nonlinear Milne problem and show the behavior of its solutions at
infinity.

There has been a well-developed theory of the Milne problem for the linear
transport equations. When the equation for temperature is not considered,
system (1.1)-(1.2) becomes

$\displaystyle\mu\partial_{x}\psi+\psi-\frac{1}{2}\langle\psi\rangle=0.$ (1.5)

This problem, referred to as the Milne problem for linear transport equation,
dates back to the work [5, 4]. It was derived in [5] as the boundary layer
problem of the linear transport equation. The equation was shown to decay
exponentially to a constant therein by using the theory of stochastic
processes. Later, this was also proved by using compactness method and
weighted estimates in [4]. Since $0$ belongs to the spectral of the linear
operator
$\mathcal{L}\psi:=-\mu\partial_{x}\psi-\psi+\tfrac{1}{2}\langle\psi\rangle$,
one cannot use semigroup theory to show the existence of solutions for the
above problem. Instead, the maximum principle is used to show the existence of
solutions in $L^{\infty}$. Moreover, using a weighted estimate, the
exponential convergence and the uniqueness of solutions is shown in [4].
Recently, a geometric correction to the above Milne problem is proposed in
[25] to better describe the behavior of solutions near the boundary when the
boundary is not flat. A weighted estimate accounting for the geometric
corrections was provided therein to show the exponential convergence of
solutions.

Compared to the Milne problem for linear transport equations, there are only
few works on the nonlinear Milne problem. For example, in [19], the existence
of weak solutions in $L^{\infty}_{\rm loc}(\mathbb{R}_{+})\times
L^{\infty}_{\rm loc}(\mathbb{R}_{+}\times[-1,1])$ was shown under the
assumption that $\gamma_{1}\leq T_{b}\leq\gamma_{2}$ for some constants
$\gamma_{1}>0$, $\gamma_{2}\geq 0$ by using the Leray-Schauder fixed point
theorem. However, whether the solutions converges to some finite constants is
not known. Moreover, the uniqueness of weak solutions are not known. In [18],
existence and uniqueness are shown for system (1.1)-(1.2) with Dirichlet
boundary data in a bounded interval by using the maximum principle. For the
half-space nonlinear Milne problem, the boundary condition at infinity is not
known and such method cannot be adapted to our problem.

The major difficulty of the study of the nonlinear Milne problem (1.1)-(1.4)
lies in two aspects. Firstly, the second order ODE (1.1) and the transport
equation (1.2) are coupled and need to be studied together. Maximum principle
holds for equation (1.1) and (1.2) separately, but is not known to hold for
the system. Secondly, the $T^{4}$ term is nonlinear, which is due to the
Stefan-Boltzmann law of black body radiation. This physical term, makes the
Milne problem nonlinear and the theory of linear operators not applicable. The
lack of maximum principle and the nonlinearity nature of the problem, make the
study of our problem challenging.

To overcome the above difficulty, several techniques are used. Firstly, the
maximum principle of the second order ODE (1.1) and the transport equation
(1.2) also implies the monotonicity of solutions with respect to source term
and boundary data. Such monotonicity property could help to construct an
approximate sequence converging to the solutions to (1.1)-(1.4). Such a
technique was used in [9] to study the Milne problem for a similar system but
without the second order derivative $\partial_{x}^{2}T$. Here we need to use
the monotonicity property of the second order ODE to construct functions that
converge to solutions of the nonlinear Milne problem. Secondly, to prove the
exponential decay property of the solution, we cannot directly follow [4]
since the property that $\partial_{x}\psi$ satisfies the same equation as
$\psi$ is used to show the exponential decay of solution for the Milne problem
(1.5). However, in our case, due to nonlinearity,
$(\partial_{x}T,\partial_{x}\psi)$ satisfies a different system. Fortunately,
we are able to derive a weighted estimate in a similar way as [4] that leads
to the exponential decay of $T$ as well as $\partial_{x}T$, which also implies
the decay property of $\psi$ by directly using the formulas for linear
transport equation. Thirdly, unlike the linear Milne problem (1.5), here the
uniqueness of nonlinear Milne problem (1.1)-(1.4) does not follow from the
weighted estimates and decay property of solutions due to the nonlinearity. We
solve this issue by using linear stability analysis and propose a spectral
assumption that leads to weighted estimates and the uniqueness of the
linearized system. Lastly, a sufficient condition is proposed for the spectral
assumption by using the generalized Hardy’s inequality. This allows us to
prove the uniqueness of the nonlinear Milne problem in the vicinity of
solutions satisfying the sufficient condition or near well-prepared boundary
data where $\psi_{b}-T_{b}^{4}$ is small.

System (1.1)-(1.2) could be extended by considering the dependence on
frequency of the radiative density. The corresponding frequency dependent
model reads as

$\displaystyle\partial_{x}^{2}T+\int_{0}^{\infty}\int_{-1}^{1}(\psi-
B_{\nu}(T))d\mu d\nu=0,$ $\displaystyle\mu\partial_{x}\psi+\psi-B_{\nu}(T)=0,$

where $T=T(x),\psi=\psi(x,\mu,\nu)$ and $\nu\in(0,\infty)$ is the frequency.
The nonlinear function $B_{\nu}(T)$ is usually taken to be the Planck function

$B_{\nu}(T)=\frac{2h\nu^{3}}{c^{2}}\frac{1}{e^{\frac{h\nu}{kT}}-1},$

where $c$ is the speed of light, $k$ is the Boltzmann constant and $h$ is the
Planck constant. When the diffusive operator $\partial_{x}^{2}T$ is neglected,
the Milne problem is studied in [20, 23, 9]. The existence of the above
frequency dependent half-space problem was proved in [23], where more later a
simplified demonstration was provided in [9]. However, the existence and
uniqueness for the nonlinear Milne problem (1.1)-(1.2) and the above frequency
dependent problem has not been proved in the literature. Our result of this
paper could be extended to the frequency dependent case without any
difficulty.

Our work could be treated as a development of the method of weighted estimates
first developed in [4] to the nonlinear coupled Milne problem. The work [4]
paves the way to use rigorous mathematical analysis to study the boundary
layer problem of kinetic PDEs. The same technique was used in [25] to study
the Milne problem of linear transport equation with geometric corrections.
Such techniques have been also used and developed for the other kinetic
equations, in particular Boltzmann equations. For example, existence,
uniqueness and asymptotic behavior of the linearized Boltzmann equation for a
gas of hard spheres is studied by using more complicated weighted estimates in
[2]. The analysis is extended to gas mixtures in [1] and to the Vlasov-
Boltzmann-Poisson equation in [7]. The nonlinear Milne problem for Boltzmann
equation was considered in [15], where existence and exponential decay
property of solutions are proved for the Milne problem for the nonlinear
Boltzmann equation in one-dimension with a slightly perturbed reflection
boundary condition using weighted estimates and the Banach fixed-point
theorem. For a boundary condition that the gas is in contact with its
condensed phase, the existence and uniqueness of a uniformly decaying solution
is shown in a neighborhood of Maxwellian equilibrium in [6], where a penalized
problem is studied in order to consider the phase transition at the interface.
Besides these works, there are also some other ways to study the Milne
problems of Boltzmann equations, for example [1, 7, 16, 21, 24]. Among them,
In [21], the authors use the theory of invariant manifolds and use Green
functions to construct linear invariant manifolds for the half-space Milne
problem of Boltzmann equation with Dirichlet boundary conditions. Based on the
macro-micro or hydrodynamics-kinetic decomposition of solutions, existence of
solutions to the same nonlinear problem is proved in [24] and it is found that
the Mach number of Maxwellian affects the existence. For the case of
reflective boundary conditions, the existence is shown recently in [16], where
uniqueness is also established under some constraint conditions. For a review
of the boundary layer problem for Boltzmann equation, we refer the reader to
[3]. Since techniques and methods for the Milne problem of linear transport
equations have been extended to study the Milne problem for the Boltzmann
equation, for example from [4] to [2] and [6], the study of the nonlinear
Milne problem (1.1)-(1.4) could provide tools and insights for other linear
and nonlinear Milne problem for system of kinetic equations.

###  1.1. Main results

The existence and exponential decay properties of solutions to the nonlinear
Milne problem (1.1)-(1.4) is stated in the following theorem.

######  Theorem 1 (Existence of the nonlinear Milne problem).

Given $T_{b}\geq 0$, $\psi_{b}=\psi_{b}(\mu)\geq 0$ for any $\mu\in(0,1]$,
there exists a bounded non-negative weak solution $\left(T,\psi\right)\in
L_{\rm loc}^{2}(\mathbb{R}_{+})\times L_{\rm
loc}^{2}(\mathbb{R}_{+}\times[-1,1])$ to system (1.1)-(1.2), and the solution
satisfies the following estimate:

$\displaystyle\int_{0}^{\infty}e^{2\alpha
x}4T^{3}|\partial_{x}T|^{2}dx+(1-\alpha)\int_{0}^{\infty}\int_{-1}^{1}e^{2\alpha
x}(\psi-T^{4})^{2}d\mu dx$
$\displaystyle\quad+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi(0,\mu)-T_{b}^{4})^{2}d\mu\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$
(1.6)

for any $0\leq\alpha<1$. Moreover, there exists a constant $T_{\infty}\geq 0$
such that

$\displaystyle|T(x)-T_{\infty}|\leq M_{\alpha}e^{-\alpha x},$ (1.7)
$\displaystyle|\psi(x,\mu)-T_{\infty}^{4}|\leq|\psi_{b}-T_{b}^{4}|e^{-\frac{x}{\mu}}+4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{1-\mu\alpha}e^{-\alpha
x},~{}\text{for }\mu>0,$ $\displaystyle|\psi(x,\mu)-T_{\infty}^{4}|\leq
4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{1-\mu\alpha}e^{-\alpha
x},\quad\text{for }\mu<0,$

for any $x\in\mathbb{R}_{+}$ and $0\leq\alpha<1$, where

$M_{\alpha}=\frac{1}{\sqrt{6\alpha(1-\alpha)}}\left(\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{\frac{1}{2}}.$

As a consequence, we can get the estimate

$\displaystyle T_{\infty}\leq T_{b}+M_{\alpha},\quad|T(x)|\leq
T_{b}+2M_{\alpha}.$ (1.8)

The above theorem gives the existence of bounded solutions for system
(1.1)-(1.2) and shows that $T$ converges exponentially to some nonnegative
constant $T_{\infty}$ as $x\to\infty$, and provides an estimate of the limit
of solutions at infinity. However, the uniqueness of solutions is not
guaranteed by this theorem.

The uniqueness of solutions to system (1.1)-(1.2) is related to the linear
stability of the system. The linear stability relies on the following spectral
assumption.

Spectral Assumption :

  1. (A)

We say that the function $T\in C^{1}(\mathbb{R}_{+})$ satisfies the spectral
assumption if there exists constants $\beta_{0}>0$ and $0<M<1$, such that

$\displaystyle
M\int_{0}^{\infty}e^{2\beta_{0}x}(2T^{\frac{3}{2}})^{2}|\partial_{x}f|^{2}dx\geq
4\int_{0}^{\infty}e^{2\beta_{0}x}|\partial_{x}(2T^{\frac{3}{2}})|^{2}f^{2}dx$
(1.9)

for all measurable function $f\in C^{1}(\mathbb{R}_{+})$ with $f(0)=0$.

To show how this spectral assumption implies linear stability, we study the
following system, which is the linearization of system (1.1)-(1.2).

$\displaystyle\partial_{x}^{2}g+\langle\phi-4T^{3}g\rangle=\langle
S_{1}\rangle,$ (1.10) $\displaystyle\mu\partial_{x}\phi+(\phi-4T^{3}g)=S_{1},$
(1.11)

supplemented with the following boundary conditions

$\displaystyle g(0)=0,$ (1.12) $\displaystyle\phi(0,\mu)=\phi_{b}(\mu),\text{
for any }\mu\in(0,1],$ (1.13)

where $T$ is the solution to system (1.1)-(1.2) and the source term
$S_{1}=S_{1}(x)$ is a given function that vanishes as $x\to\infty$. Define the
linear operator $\mathcal{F}:\,\,\,\,L^{p}(\mathbb{R}_{+})\times
L^{p}(\mathbb{R}_{+}\times[-1,1])\longrightarrow L^{p}(\mathbb{R}_{+})\times
L^{p}(\mathbb{R}_{+}\times[-1,1])$ (with $1\leq p\leq\infty$), given by

$\displaystyle\mathcal{F}(g,\phi):=\left(\begin{array}[]{cc}\partial_{x}^{2}g+\langle\phi-4T^{3}g\rangle\\\
-\mu\partial_{x}\phi-(\phi-4T^{3}g)\end{array}\right).$ (1.16)

Then it will be shown later that the spectral assumption implies that the
spectrum of $\mathcal{F}$ all lie in the space $\Re z\leq 0$. However, similar
as the linear transport operator studied in [4], we can derive a weighted
estimate on the solutions of (1.10)-(1.11) that leads to the exponentially
decay. The weighted estimate relies on the following inequality

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx+\int_{0}^{\infty}\partial_{x}e^{2\beta
x}(4T^{3})f\partial_{x}fdx\geq 0,$ (1.17)

for any $f\in C^{1}_{\rm loc}(\mathbb{R}_{+})$ satisfying $f(0)=0$, where
$\beta$ is the same as in the spectral assumption (A). It will be proved that
the above inequality is a direct consequence of the spectral assumption (A)
(see Lemma 4 for the details). The weighted estimates allows us to prove the
existence, uniqueness and exponential decay for solutions to the linearized
system (1.10)-(1.11).

######  Theorem 2.

Suppose $T$ satisfies the spectral assumption (A) for some constants
$0<\beta<1$ and $M>0$. Assume $S_{1}$ decays to zero exponentially such that
$\int_{0}^{\infty}e^{2\beta x}|S_{1}|^{2}dx<\infty$. Then, there exists a
unique bounded solution $\left(g,\phi\right)\in L^{2}_{\rm
loc}(\mathbb{R}_{+})\times L^{2}_{\rm loc}(\mathbb{R}_{+}\times[-1,1])$ to
system (1.10)-(1.11) with boundary conditions (1.12)-(1.13), and the solution
satisfies

$\displaystyle\int_{0}^{\infty}\int_{-1}^{1}e^{2\beta x}(\phi-4T^{3}g)^{2}d\mu
dx+\frac{1}{1-\beta}\int_{-1}^{0}|\mu|(\phi(0,\cdot))^{2}d\mu$
$\displaystyle\qquad\leq\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{2}{(1-\beta)^{2}}\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+})}^{2},$ (1.18)

for any $x\in\mathbb{R}_{+}$. Moreover, there exists a constant
$g_{\infty}\in\mathbb{R}$ such that

$\displaystyle|g(x)-g_{\infty}|\leq N_{\beta}e^{-\beta x},$ (1.19)
$\displaystyle|\phi(x,\mu)-4T_{\infty}^{3}g_{\infty}|\leq|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+4(T_{b}+2M_{\alpha})^{2}(4M_{\alpha}+T_{b})N_{\beta}e^{-\beta
x}$ $\displaystyle\qquad\qquad\qquad\qquad+\frac{1}{1-2\mu\beta}e^{-\beta
x}\|e^{\beta x}S_{1}\|_{L^{2}(\mathbb{R}_{+})},~{}\text{for }\mu>0,$
$\displaystyle|\phi(x,\mu)-4T_{\infty}^{3}g_{\infty}|\leq
4(T_{b}+2M_{\alpha})^{2}(4M_{\alpha}+T_{b})N_{\beta}e^{-\beta x},$
$\displaystyle\qquad\qquad\qquad\qquad+\frac{1}{1-2\mu\beta}e^{-\beta
x}\|e^{\beta x}S_{1}\|_{L^{2}(\mathbb{R}_{+})},~{}\text{for }\mu<0.$

where

$N_{\beta}=\frac{1}{\sqrt{2\beta}}\left(\frac{2}{3(1-\beta)}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{3(1-\beta)^{2}}\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+})}^{2}\right)^{\frac{1}{2}},$

where $\alpha$ is the constant in Theorem 1. As a consequence,

$\displaystyle g_{\infty}\leq N_{\beta},\quad|g(x)|\leq 2N_{\beta}.$ (1.20)

Although the spectral assumption (A) implies the uniqueness for the linearized
system (1.10)-(1.11), we are not able to establish such a uniqueness result
for the nonlinear Milne problem (1.1)-(1.4). This is partly because the
spectral assumption (A) is not additive, which means if $T_{1}$ and $T_{2}$
satisfy the spectral assumption, $T_{1}+T_{2}$ is not known to satisfy the
spectral assumption. In order to show the uniqueness of the nonlinear Milne
problem, we need a more restrictive condition. Inspired by the generalized
Hardy’s inequality, we propose the following condition:

(A0) $\displaystyle
A_{0}^{2}:=\sup_{r\in(0,\infty)}\left(\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\cdot\int_{0}^{r}\frac{e^{-2\beta
x}}{4T^{3}}dx\right)<\frac{1}{2}.$ (1.21)

By using the generalized Hardy’s inequality, this condition is shown to be a
sufficient condition for the spectral assumption (A) (see Lemma 6). Moreover,
when $A_{0}^{2}\geq 2$, the spectral assumption fails to hold.

The benefits of the condition (A0) is that if $T_{1}$ and $T_{2}$ satisfies
(A0), then
$w:=(T_{1}^{4}-T_{2}^{4})/(T_{1}-T_{2})=T_{1}^{3}+T_{2}^{3}+T_{1}^{2}T_{2}+T_{2}^{2}T_{1}$
satisfies the spectral assumption with $T^{\frac{3}{2}}$ replaced by
$\tfrac{1}{2}w^{\frac{1}{2}}$. This property, together with the uniqueness of
(1.10)-(1.11), implies the uniqueness for the nonlinear Milne problem
(1.1)-(1.2).

######  Theorem 3.

Let $(T,\psi)$ and $(T_{1},\psi_{1})$ be two solutions to the nonlinear Milne
problem (1.1)-(1.2) satisfying the weighted estimate (1). Suppose the function
$T$ satisfy the spectral assumption (A). Then if
$(T_{1},\psi_{1})\in\mathcal{V}$ with

$\displaystyle\mathcal{V}=\\{(G,\phi)$ $\displaystyle\in L^{2}_{\rm
loc}(\mathbb{R}_{+})\times L^{2}_{\rm loc}(\mathbb{R}_{+}\times[-1,1]):$
$\displaystyle\|G-T\|_{L^{2}(\mathbb{R}_{+})}\leq\varepsilon\\},$ (1.22)

for $\varepsilon>0$ sufficiently small. Then, $T_{1}(x)=T(x)$ and
$\psi_{1}(x,\mu)=\psi(x,\mu)$ for $(x,\mu)\in\mathbb{R}_{+}\times[-1,1]$
almost everywhere.

Furthermore, if the condition $(T_{1},\psi_{1})\in\mathcal{V}$ is replaced by
the condition

$\displaystyle\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\leq
C_{b},$ (1.23)

for some constant $C_{b}$ (given in (5.15)). Then, $T_{1}(x)=T(x)$ and
$\psi_{1}(x,\mu)=\psi(x,\mu)$ for $(x,\mu)\in\mathbb{R}_{+}\times[-1,1]$
almost everywhere, and the solution of the nonlinear Milne problem (1.1)-(1.2)
is unique.

Here we are only able to establish the uniqueness of the solution of the
nonlinear Milne problem for solutions satisfying the weighted estimate (1) or
under the smallness assumption (1.23). Indeed, the assumption (1.23) is
satisfied for almost well-prepared boundary data and the uniqueness of the
nonlinear Milne problem (1.1)-(1.2) is proved for this case. However, the
uniqueness of the solution of the nonlinear Milne problem for general boundary
data remains an open problem.

###  1.2. Organizations

The paper is organized as follows. In section 2, the existence of solutions to
system (1.1)-(1.2) is proved in a bounded interval, in Theorem 4. To prove
this theorem, the monotonicity for the second order ODE is shown in (1.1) in
section 2.1 and Theorem 2 is proved in section 2.2 for existence and
uniqueness, and section 2.3, for the weighted estimate. Section 3 is devoted
to the proof of Theorem 1. The existence of solutions and weighted estimates
are shown in Section 3.1 and section 3.2, respectively, both by passing to the
limit $B\to\infty$ in the solutions on the bounded interval $[0,B]$. The
exponential decay property (LABEL:eq:thm.ex) of solutions is shown in section
3.3. Section 4 is devoted to the study of the linearized Milne problem.
Followed by a discussion on the spectral assumption (A) in section 4.1, the
existence of weak solutions to the linearized Milne problem on a bounded
interval $[0,B]$ is shown in section 4.2. The solution is then extended to the
half-space, with existence proved in section 4.3, weighted estimate and decay
proved in section LMB4, uniqueness in 4.5, thus proving Theorem 2. Finally, we
prove Theorem 3 in section 5, where the sufficient condition 1.21 is shown to
imply the spectral assumption (A) in section 5.1 and the uniqueness of
solutions for system (1.1)-(1.2) is established for the case (1.23) in section
5.2 and the case $T_{2}\in\mathcal{V}$ in section 5.3 .

##  2\. Existence on the bounded interval

In this section, we consider system (1.1)-(1.2) on the bounded interval
$x\in[0,B]$:

$\displaystyle\partial_{x}^{2}T^{B}+\langle\psi^{B}-(T^{B})^{4}\rangle=0,$
(2.1) $\displaystyle\mu\partial_{x}\psi^{B}+\psi^{B}-(T^{B})^{4}=0,$ (2.2)

associated to the following boundary conditions

$\displaystyle T^{B}(0)=T_{b},\quad\partial_{x}T^{B}(B)=0,$ (2.3)
$\displaystyle\psi^{B}(0,\mu)=\psi_{b}(\mu),\quad\psi^{B}(B,\mu)=\psi^{B}(B,-\mu),\quad\text{
for any }\mu\in(0,1].$ (2.4)

The boundary conditions at $x=B$ are motivated by the fact that solutions to
the nonlinear Milne problem (1.1)-(1.2) are expected to converge to some
constants at infinity. If the right boundary conditions at $x=B$ are replaced
by Dirichlet boundary conditions, the existence for the above system was
proved in [18] by using the fixed point theorem together with the maximum
principle. However, here we do not know the value of $T^{B},\psi^{B}$ on the
boundary $x=B$. The method of [18] cannot be used to show the existence for
the above system. A different approach based on the theory of subsolutions and
fixed point theorems is used here to prove the existence of the solution for
system (2.1)-(2.2). The theory of subsolutions is used in [9] to show the
existence of solutions to system (1.1)-(1.2) without the diffusion operator
$\partial_{x}^{2}T$. Here we will study the property of solutions for (2.1)
and show the existence for system (2.1)-(2.2). The existence theorem reads as
follows.

######  Theorem 4.

Assume $0\leq T_{b}\leq\gamma$, $0\leq\psi_{b}=\psi_{b}(\mu)\leq\gamma^{4}$
for any $\mu\in(0,1]$ and for some constant $\gamma>0$. Then there exists a
unique solution $\left(T^{B},\psi^{B}\right)\in C^{2}([0,B])\times
C^{1}([0,B]\times[-1,1])$ to system (2.1)-(2.2) with boundary conditions
(2.3)-(2.4), and the solution satisfies

$0\leq T^{B}(x)\leq\gamma,\quad
0\leq\psi^{B}(x,\mu)\leq\gamma^{4},\quad\text{for any
}x\in[0,B],\,\mu\in[-1,1].$

Furthermore, the following estimate

$\displaystyle\int_{0}^{B}$ $\displaystyle e^{2\alpha
x}4(T^{B})^{3}|\partial_{x}T^{B}|^{2}dx+(1-\alpha)\int_{0}^{B}\int_{-1}^{1}e^{2\alpha
x}(\psi^{B}-(T^{B})^{4})^{2}d\mu dx$
$\displaystyle+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$
(2.5)

holds for any $\alpha\in[0,1),$ where
$\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu\geq 0$ for any
$x\in[0,B]$.

The above existence theorem is proved by constructing a monotonic sequence of
functions that converges to the solution of system (2.1)-(2.2). Before we
construct the sequence, we first show the monotonicity property of equation
(2.1).

###  2.1. Monotonicity of the second order ODE

The monotonicity property of the second order ODE (2.1) is given in the
following lemma.

######  Lemma 1.

Given $0\leq T_{b}\leq\gamma$ for some constant $\gamma>0$. Let $\phi$ be a
strictly increasing bounded function on $\mathbb{R}$ and $g\in C([0,B])$ be a
continuous bounded function satisfying $0\leq g\leq\phi(\gamma)$. Then there
exists a unique bounded solution $T\in C^{2}([0,B])$ to the equation

$\displaystyle-\partial_{x}^{2}T(x)+\phi(T(x))=g(x),\quad\text{for any
}x\in[0,B],$ (2.6)

supplemented to the following boundary conditions

$\displaystyle T(0)=T_{b},\quad\partial_{x}T(B)=0,$

and the solution satisfies $0\leq T(x)\leq\gamma$ for any $x\in[0,B]$.
Moreover, let $T_{1}$, $T_{2}$ be two solutions to the above equation with
source terms $g_{1}$, $g_{2}$ and boundary data $T_{b1}$, $T_{b2}$,
respectively. If $0\leq g_{1}(x)\leq g_{2}(x)\leq\phi(\gamma)$ for all
$x\in[0,B]$ and $0\leq T_{b1}\leq T_{b2}\leq\gamma$, then $0\leq T_{1}(x)\leq
T_{2}(x)\leq\gamma$ for all $x\in[0,B]$.

###### Proof.

_Existence_. Let $h\in C[0,B]$, then the solution to the problem

$\displaystyle\left\\{\begin{array}[]{ll}-\partial_{x}^{2}f=h,\quad\text{on
}[0,B],\\\ f(0)=0,\quad\partial_{x}f(B)=0\end{array}\right.$

can be explicitly written as

$\displaystyle f(x)=\int_{0}^{x}\int_{t}^{B}h(s)dsdt.$ (2.7)

Denote the above mapping by $f=\mathcal{G}h$. Then $\mathcal{G}$ is a linear
operator from $C([0,B])$ to $C^{2}([0,B])$. We next show the mapping
$\mathcal{N}$ defined by

$\mathcal{N}T:=\mathcal{G}(g-\phi(T))+T_{b}$

has a fixed point. To show this, we recall the statement of the Leray-Schauder
Theorem in appendix E, [14, Theorem 11.3].

  * •

We consider the Banach space $D=C([0,B])$ equipped with the supremum norm

$\|f\|_{D}:=\sup_{x\in[0,B]}|f(x)|.$

Since $\mathcal{N}:C([0,B])\to C^{2}([0,B])$ and $C^{2}([0,B])$ is compactly
embedded in $C([0,B])$, where $C^{2}([0,B])$ is equipped with the norm

$\|f\|_{C^{2}([0,B])}:=\sup_{x\neq
y}\dfrac{|f(x)-f(y)|}{|x-y|^{2}}+\|f\|_{C([0,B])},$

$\mathcal{N}$ is a compact map from $C([0,B])$ to itself.

  * •

We now need to show that the set

$\mathcal{A}:=\Big{\\{}T\in C([0,B]):T=\sigma\mathcal{N}T,\,\,\,\text{for some
}\sigma\in[0,1]\Big{\\}},$

is bounded. To show this, suppose $T=\sigma\mathcal{N}T$, then

$\displaystyle T=\sigma(\mathcal{G}(g-\phi(T))+T_{b}),$ (2.8)

which satisfies

$\displaystyle-\partial_{x}^{2}T+\sigma\phi(T(x))=\sigma
g(x),\,x\in[0,B]\quad\text{ and }T(0)=\sigma T_{b},\,\partial_{x}T(B)=0.$
(2.9)

We first show $T\geq 0$. We use a contradiction argument. Suppose the minimum
of $T$ is less than zero. Then let $x_{m}\in[0,B]$ be the minimum point,
$T(x_{m})<0$. By the monotonicity of $\phi$, $\phi(T(x_{m}))<\phi(0)<0$. Hence

$\displaystyle-\partial_{x}^{2}T(x_{m})=\sigma(g(x_{m})-\phi(x_{m}))>0,$

i.e. around $x_{m}$, $T(x)$ is concave, which implies $x_{m}$ cannot be a
local minimum point. Hence $x_{m}\not\in(0,B).$ If $x_{m}=B$, due to
$-\partial_{x}^{2}T(B)>0$ and $\partial_{x}T(B)=0$, $x_{m}=B$ should be a
maximum point rather than a minimum point. Therefore, $x_{m}\neq B$. Otherwise
if $x_{m}=0$, $T(0)=\sigma T_{b}\geq 0$, which contradicts to the assumption
$T(x_{m})<0$, so $x_{m}\neq 0$. Therefore, $x_{m}\not\in[0,B]$ and thus the
set of minimum points with negative value is empty. Therefore, we conclude
that $T(x)\geq 0$ for all $x\in[0,B].$

We can follow a similar contradiction argument to show $T\leq\gamma$. Suppose
$T$ attains its maximum at $x=x_{M}$ with $T(x_{M})>\gamma$. Due to
monotonicity of $\phi$, $\phi(T(x_{M}))>\phi(\gamma)$ and thus

$-\partial_{x}^{2}T(x_{M})=\sigma(g(x_{M})-\phi(T(x_{M})))<\sigma(\phi(\gamma)-\phi(\gamma))<0,$

which implies $x_{M}$ is a loal minimum. Hence $x_{M}\not\in(0,B)$. If
$x_{M}=0$, then $T(x_{M})=\sigma T_{b}\leq\gamma$, so $x_{M}\neq 0$. If
$x_{M}=B$, due to $\partial_{x}T(B)=0$, $\partial_{x}^{2}T(B)\geq 0$ also
implies that $B$ is a minimum point, which contradicts the assumption that
$x_{M}=B$ is a maximum point. Therefore, we conclude that the set of maximum
point with values bigger than $\gamma$ is empty. Therefore,
$T(x_{M})\leq\gamma$ and so $T(x)\leq\gamma$ for all$x\in[0,B]$. Finally, from
$0\leq T\leq\gamma$, we conclude that

$0\leq\|T\|_{C([0,B])}\leq\gamma\quad\text{for any}\,\,\,\sigma\in[0,1].$

We can thus conclude that the set $\mathcal{A}$ is bounded for any
$\sigma\in[0,1]$.

Consequently, the hypotheses of Leray-Schauder fixed point theorem are
verified, which implies that $\mathcal{N}$ has a fixed point in $C([0,B])$.
Therefore, there exists a solution in $C([0,B])$ for equation (2.6). Moreover,
from (2.8) and the formula (2.7), $T\in C^{2}([0,B])$.

_Uniqueness._ Suppose $T_{1}$ and $T_{2}$ are two distinct solutions to (2.6),
then $T_{1}-T_{2}$ satisfies

$\displaystyle\partial_{x}^{2}(T_{1}-T_{2})-(\phi(T_{1})-\phi(T_{2}))=0,$

with

$\displaystyle(T_{1}-T_{2})(0)=0,\,\partial_{x}(T_{1}-T_{2})(B)=0.$

We first show any maximum point $x_{M}\in[0,B]$ of $T_{1}(x)-T_{2}(x)$ cannot
be positive. Otherwise, $T_{1}(x_{M})-T_{2}(x_{M})>0$ and then by the
monotonicity of $\phi$, $\phi(T_{1}(x_{M}))>\phi(T_{2}(x_{M}))>0$, hence
$\partial_{x}^{2}(T_{1}-T_{2})(x_{M})=\phi(T_{1}(x_{M}))-\phi(T_{2}(x_{M}))>0$,
which implies $T_{1}-T_{2}$ is convex near $x_{M}$. Thus $x_{M}$ can only be a
local minimum or on the boundary. By the assumption, $x_{M}$ is a maximum
point, $x_{M}\not\in(0,B)$. If $x_{M}=0$, $T_{1}(0)-T_{2}(0)=0$ contradicts
the assumption $(T_{1}-T_{2})(x_{M})>0$, so $x_{M}\neq 0$. If $x_{M}=B$, then
$\partial_{x}(T_{1}-T_{2})(B)=0$ and $\partial_{x}^{2}(T_{1}-T_{2})(B)=0$
implies $B$ is a local minimum and thus cannot be a maximum point. Hence the
set of $\\{x:(T_{1}-T_{2})(x)>0\\}$ is empty.

Similarly, we can show any minimum point $x_{m}\in[0,B]$ of
$T_{1}(x)-T_{2}(x)$ cannot have values $T_{1}(x_{m})-T_{2}(x_{m})<0$.
Otherwise, $T_{1}(x_{m})-T_{2}(x_{m})<0$ and
$\partial_{x}^{2}(T_{1}-T_{2})(x_{m})=\phi(T_{1}(x_{m}))-\phi(T_{2}(x_{m}))<0$
and $T_{1}-T_{2}$ is locally concave at $x_{m}$. So $x_{m}$ cannot be a local
minimum, $x_{m}\not\in(0,B)$. The fact that $(T_{1}-T_{2})(0)=0$ implies
$x_{m}\neq 0$. The condition $\partial_{x}(T_{1}-T_{2})(B)=0$ and
$\partial_{x}^{2}(T_{1}-T_{2})(B)<0$ implies that $B$ is a local maxium point
and thus cannot be a minimum point, i.e. $x_{m}\neq B$. Therefore, we conclude
that the set of $\\{x:(T_{1}-T_{2})(x)<0\\}$ is empty.

Combining the above arguments, the set $\\{x:(T_{1}-T_{2})(x)\neq 0\\}$ is
empty. That is, $T_{1}=T_{2}$ on $[0,B]$ and the solution is unique.

_Monotonicity._ Assume $g_{1}(x)\leq g_{2}(x)$ on $x\in[0,B]$ and $T_{b1}\leq
T_{b2}$, and $T_{1}$, $T_{2}$ are two solutions with source terms
$g_{1},\,g_{2}$ and boundary data $T_{b1},T_{b2}$, respectively. Suppose the
maximum of $T_{1}(x)-T_{2}(x)$ occurs at $x=x_{M}\in(0,B)$ with
$T_{1}(x)-T_{2}(x)>0$. Then
$\partial_{x}^{2}(T_{1}-T_{2})(x_{M})=\phi(T_{1}(x_{M}))-\phi(T_{2}(x_{M}))>0$
and so $x=x_{M}$ can not be a local maximum, $x_{M}\not\in(0,B)$. If the
maximum occurs at $x=B$ and $(T_{1}-T_{2})(B)>0$, then
$\partial_{x}^{2}(T_{1}-T_{2})(B)>0$ and so $x=B$ is a minimum point due to
$\partial_{x}T_{1}(B)=\partial_{x}T_{2}(B)$, which contradicts the assumption
that $x=B$ is a maximum point. If the maximum occurs at $0$, we have
$T_{1}(0)-T_{2}(0)=T_{b1}-T_{b2}\leq 0$, which contradicts the assumption
$T_{1}(x_{M})-T_{2}(x_{M})>0$. Therefore, we have $T_{1}\leq T_{2}$ on
$[0,B]$. ∎

###  2.2. Existence and uniqueness

With the monotonicity of solutions to the second order ODE in Lemma 1 and the
monotonicity of solutions for the linear transport equation in Lemma 9 in
Appendix A, we are prepared to prove the existence for system (2.1)-(2.2).
Before we give the proof, we first introduce the concept of subsolutions.

######  Definition 1.

We call $(T,\psi)\in C^{2}([0,B])\times C^{1}([0,B]\times[-1,1])$ a
subsolution of (2.1)-(2.2) if the following inequalities hold

$\displaystyle-\partial_{x}^{2}T-\langle\psi-T^{4}\rangle\leq~{}$
$\displaystyle 0,$ (2.10)
$\displaystyle\mu\partial_{x}\psi+(\psi-T^{4})\leq~{}$ $\displaystyle 0,$
(2.11)

and on the boundary they satisfy $0\leq T(0)\leq T_{b}$,
$0\leq\psi(0,\mu)\leq\psi_{b}(\mu)$ for $\mu\in(0,1]$.

###### Proof of Theorem 4, existence and uniqueness.

The proof is divided into four steps. An approximate sequence is constructed
in the first step. Then, we prove the existence of solutions for system
(2.1)-(2.2). After that, the uniqueness of solutions is shown using a
contradiction argument. Finally, a weighted energy estimate is derived.

_Step 1: Construction of an approximate sequence._ Let $\phi(T)$ be a strictly
increasing function defined by

$\displaystyle\phi(T)=\left\\{\begin{array}[]{cl}\frac{T}{1-T}&\text{if
}T<0,\\\ T^{4},&\text{if }0\leq T\leq\gamma,\\\
\gamma^{4}+\frac{T-\gamma}{1+T-\gamma},&\text{if }T>\gamma.\end{array}\right.$

The above function satisfies $\phi(T)<0$ if $T<0$ and $\phi(T)>\gamma^{4}$ if
$T>\gamma$.

We construct a sequence of functions
$\\{T^{k}=T^{k}(x),\,\psi^{k}=\psi^{k}(x,\mu)\\}_{k=0}^{\infty}$ starting with
$T^{0}$, $\psi^{0}$ being a subsolution given by Definition 1111 We may take
$T^{0}=0$ and $\psi^{0}=0$, which is a subsolution. , i.e.

$\displaystyle-\partial_{x}^{2}T^{0}+2\phi(T^{0})\leq\langle\psi^{0}\rangle,$
(2.12) $\displaystyle\mu\partial_{x}\psi^{0}+\psi^{0}\leq(T^{0})^{4}.$ (2.13)

with

$\displaystyle 0\leq T^{0}(0)\leq T_{b},\quad\partial_{x}T^{0}(B)=0,$
$\displaystyle
0\leq\psi^{0}(0,\mu)\leq\psi_{b}(\mu),\quad\psi(B,\mu)=\psi(B,-\mu),\quad\text{for
any }\mu\in(0,1].$

Given $(T^{k},\psi^{k})$, $(T^{k+1},\psi^{k+1})$ is obtained by solving

$\displaystyle-\partial_{x}^{2}T^{k+1}+2\phi(T^{k+1})=\langle\psi^{k}\rangle,$
(2.14) $\displaystyle\mu\partial_{x}\psi^{k+1}+\psi^{k+1}=(T^{k})^{4},$ (2.15)

with boundary conditions

$\displaystyle T^{k+1}(0)=T_{b},\quad\partial_{x}T^{k+1}(B)=0,$
$\displaystyle\psi^{k+1}(0,\mu)=\psi_{b},\quad\psi^{k+1}(B,\mu)=\psi^{k+1}(B,-\mu),\quad\text{for
any }\mu>0.$

_Step 2: Existence._ Assume $(T^{k},\psi^{k})\in C^{2}([0,B])\times
C^{1}([0,B]\times[-1,1])$ satisfy $0\leq T^{k}\leq\gamma$ and
$0\leq\psi^{k}\leq\gamma^{4}$, then Lemma 1 and Lemma 9 imply that there
exists a unique solution $(T^{k+1},\psi^{k+1})\in C^{2}([0,B])\times
C^{1}([0,B]\times[-1,1])$ to system (2.14)-(2.15) and the solution satisfies
$0\leq T^{k+1}\leq\gamma$ and $0\leq\psi^{k+1}\leq\gamma^{4}$. By induction,
we conclude that for any $k\geq 1$ and $k\in\mathbb{N}$,

$\displaystyle 0\leq T^{k}(x)\leq\gamma,\quad
0\leq\psi^{k}(x,\mu)\leq\gamma^{4},\text{ for any }x\in[0,B],\,\mu\in[-1,1].$
(2.16)

Next we prove by induction that $T^{k+1}(x)\geq
T^{k}(x),\psi^{k+1}(x,\mu)\geq\psi^{k}(x,\mu)$ for any $x\in[0,B]$ and
$\mu\in[-1,1]$. First $T^{1},\psi^{1}$ solves

$\displaystyle-\partial_{x}^{2}T^{1}+2\phi(T^{1})=\langle\psi^{0}\rangle,$
$\displaystyle\mu\partial_{x}\psi^{1}+\psi^{1}=(T^{0})^{4}.$

Comparing this with equations (2.12)-(2.13) for $(T^{0},\psi^{0})$, we get
from Lemma 1 and Lemma 9 that $T^{1}(x)\geq T^{0}(x)$ and
$\psi^{1}(x,\mu)\geq\psi^{0}(x,\mu)$ for $x\in[0,B]$ and $\mu\in[-1,1]$.

Suppose $T^{k}(x)\geq T^{k-1}(x)$ and $\psi^{k}(x,\mu)\geq\psi^{k-1}(x,\mu)$,
then

$-\partial_{x}^{2}T^{k+1}+2\phi(T^{k+1})=\langle\psi^{k}\rangle,$
$-\partial_{x}^{2}T^{k}+2\phi(T^{k})=\langle\psi^{k-1}\rangle,$

and by Lemma 1, $T^{k+1}(x)\geq T^{k}(x)$ for any $x\in[0,B]$. Similarly, we
have

$\displaystyle\mu\partial_{x}\psi^{k+1}+\psi^{k+1}=(T^{k})^{4},$
$\displaystyle\mu\partial_{x}\psi^{k}+\psi^{k}=(T^{k-1})^{4},$

and Lemma 9 implies that $\psi^{k+1}(x,\mu)\geq\psi^{k}(x,\mu)$ for any
$x\in[0,B]$ and $\mu\in[-1,1]$. Therefore, we conclude that for any $k\geq 1$
and $k\in\mathbb{N}$,

$\displaystyle T^{k+1}(x)\geq
T^{k}(x),\quad\psi^{k+1}\geq\psi^{k}(x),\quad\text{for any
}x\in[0,B],\,\mu\in[-1,1].$ (2.17)

Combining this with (2.16), we conclude that
$\\{T^{k}\\}_{k=0}^{\infty},\\{\psi^{k}\\}_{k=0}^{\infty}$ are bounded
increasing sequences. Hence there exists a converging subsequence such that

$\displaystyle\lim_{k\to\infty}T^{k}(x)=T^{B}(x),\quad\lim_{k\to\infty}\psi^{k}(x,\mu)=\psi^{B}(x,\mu),$

for any $x\in[0,B],\mu\in[-1,1]$. Moreover, by the Beppo Levi’s lemma, it
holds that

$\displaystyle\lim_{k\to\infty}\int_{0}^{x}T_{k}(y)dy=\int_{0}^{x}T^{B}(y)dy,\quad\lim_{k\to\infty}\int_{0}^{x}\psi_{k}(y,\mu)dy=\int_{0}^{x}\psi^{B}(y,\mu)dy.$

Hence, we can pass to the limit $k\to\infty$ in

$\displaystyle
T^{k+1}=\mathcal{G}\left(\langle\psi^{k}\rangle-\phi(T^{k+1})\right)+T_{b},$
(2.18)

and get

$\displaystyle
T^{B}=\lim_{k\to\infty}\mathcal{G}\left(\langle\psi^{k}\rangle-\phi(T^{k})\right)+T_{b}=\mathcal{G}\left(\langle\psi\rangle-\phi(T)\right)+T_{b}$
(2.19)

and by the continuity of the operator $\mathcal{G}$, $T^{B}\in C^{2}([0,B]$.
In a similar way we can also get $\psi\in C^{1}([0,B]\times[-1,1])$. Hence, we
can apply the Dini’s theorem and conclude that the convergence of
$(T_{k},\psi_{k})$ to $(T^{B},\psi^{B})$ is uniform in $C^{2}([0,B])\times
C^{1}([0,B]\times[-1,1])$.

Moreover, since $0\leq T_{k},\psi_{k}\leq\gamma$, $(T^{B},\psi^{B})$ satisfies
$0\leq T\leq\gamma$, $0\leq\psi\leq\gamma^{4}$. We can pass to the limit in
(2.14)-(2.15) and get that $(T^{B},\psi^{B})$ satisfies the system

$\displaystyle-\partial_{x}^{2}T^{B}+2\phi(T^{B})=\langle\psi^{B}\rangle,$
$\displaystyle\mu\partial_{x}\psi^{B}+\psi^{B}=(T^{B})^{4},$

with

$\displaystyle T^{B}(0)=T_{b},\quad\partial_{x}T^{B}(B)=0,$
$\displaystyle\psi^{B}(0,\mu)=\psi_{b},\quad\psi^{B}(B,\mu)=\psi(B,-\mu),\quad\text{for
any }\mu>0.$

Since $0\leq T^{B}\leq\gamma$, $\phi^{B}(T)=T^{4}$ and the above system is
equivalent to (2.1)-(2.2). Therefore, there exists a solution
$(T^{B},\psi^{B})\in C^{2}([0,B])\times C^{1}([0,B]\times[-1,1])$ to system
(2.1)-(2.2) with boundary conditions (2.3)-(2.4).

_Step 3: Uniqueness._ Suppose
$(T_{1}^{B},\psi_{1}^{B}),(T_{2}^{B},\psi_{2}^{B})$ are two solutions to
(2.1)-(2.2) with boundary conditions (2.3)-(2.4). Define

$\displaystyle
T^{0}=\max(T_{1}^{B},T_{2}^{B}),\quad\psi^{0}=\max(\psi_{1}^{B},\psi_{2}^{B}),$

which satisfies (2.12)-(2.13) and is thus a subsolution. We construct a
sequence of functions $\\{T^{k}(x),\psi^{k}(x,\mu)\\}_{k=0}^{\infty}$ by
solving iteratively

$\displaystyle-\partial_{x}^{2}T^{k+1}+2\phi(T^{k+1})=\langle\psi^{k}\rangle,$
$\displaystyle\mu\partial_{x}\psi^{k+1}+\psi^{k+1}=(T^{k})^{4}.$

Then Lemma 9 implies that there exists a subsequence such that
$(T^{k},\psi^{k})$ converges to a solution $(T^{B},\psi^{B})$ of system
(1.1)-(1.2) as $k\to\infty$ and $T^{B}\geq T^{0},\psi^{B}\geq\psi^{0}$. Hence
$g^{B}=T^{B}-T_{1}^{B},\varphi^{B}=\psi^{B}-\psi_{1}^{B}$ is a nonnegative
solution to the following system

$\displaystyle-\partial_{x}^{2}g^{B}+2((T^{B})^{4}-(T_{1}^{B})^{4})=\langle\varphi\rangle,$
(2.20)
$\displaystyle\mu\partial_{x}\varphi^{B}+\varphi^{B}=((T^{B})^{4}-(T_{1}^{B})^{4}),$
(2.21)

with

$\displaystyle g^{B}(0)=0,\quad\partial_{x}g^{B}(B)=0,$
$\displaystyle\varphi^{B}(0,\mu)=0,\quad\varphi^{B}(B,\mu)=\varphi^{B}(B,-\mu),\text{
for any }\mu>0.$

Comparing (2.20) to (2.21) gives

$\displaystyle\partial_{x}(\partial_{x}g^{B}-\langle\mu\varphi^{B}\rangle)=0.$

Due to the boundary conditions $\partial_{x}g^{B}(B)=0$ and
$\langle\mu\varphi^{B}(B,\cdot)\rangle=0$,

$\displaystyle\partial_{x}g^{B}(x)=\langle\mu\varphi^{B}(x,\cdot)\rangle,\text{
for any }x\in[0,B].$

Using (B.4), we have

$\displaystyle\varphi^{B}(0,\mu)=$
$\displaystyle-\int_{0}^{B}\frac{1}{\mu}e^{\frac{2B-s}{\mu}}((T^{B})^{4}-(T_{1}^{B})^{4})(s)ds$
$\displaystyle-\int_{x}^{B}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}((T^{B})^{4}-(T_{1}^{B})^{4})(s)ds,\text{
for }\mu<0.$

If $T^{B}(x)>T_{1}^{B}(x)$ on some interval $(a,b)\subset[0,B]$, then the
above equation implies that $\varphi^{b}(0,\mu)>0$ for $\mu<0$. This, together
with the boundary condition $\varphi^{B}(0,\mu)=0$ for $\mu>0$, implies

$\displaystyle\partial_{x}g^{B}(0)=\langle\mu\varphi^{B}(0,\cdot)\rangle=\int_{-1}^{0}\mu\varphi^{B}(0,\mu)d\mu<0.$
(2.22)

This contradicts to the fact that $g^{B}(x)\geq 0$ for any $x\in[0,B]$ and
$g^{B}(0)=0$. Hence $T^{B}(x)=T_{1}^{B}(x)$ for $x\in[0,B]$ almost everywhere.

Due to $T^{B}=T_{1}^{B}$, equation (2.21) becomes
$\mu\partial_{x}\varphi^{B}+\varphi^{B}=0$, with boundary conditions
$\varphi^{B}(0,\mu)=0,\varphi(B,\mu)=\varphi(B,-\mu)$ for $\mu>0$. Therefore,
$\psi^{B}(x,\mu)=\psi_{1}^{B}(x,\mu)$ for $x\in[0,B],\mu\in[-1,1]$. By the
same argument we also obtain $T^{B}=T_{2}^{B},\psi^{B}=\psi_{2}^{B}$. Hence
$T_{1}^{B}=T_{2}^{B}$ and $\psi_{1}^{B}=\psi_{2}^{B}$, i.e. the solution to
system (2.1)-(2.4) is unique.

∎

###  2.3. Weighted estimate

Next we derive the weighted estimate (4).

###### Proof of Theorem 4, weighted estimate.

We multiply (2.1) by $(T^{B})^{4}$, integrate over $[0,x]$ and multiply (2.2)
by $\psi^{B}$, integrate over $(z,\mu)\in[0,x]\times[-1,1]$, and combine the
results to get

$\displaystyle-\int_{0}^{x}(T^{B})^{4}\partial_{z}^{2}T^{B}dz+\frac{1}{2}\int_{0}^{x}\int_{-1}^{1}\mu\partial_{z}(\psi^{B})^{2}d\mu
dz+\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu dz=0.$ (2.23)

Using integration-by-parts, we obtain

$\displaystyle-\int_{0}^{x}(T^{B})^{4}\partial_{z}^{2}T^{B}dz$
$\displaystyle=\int_{0}^{x}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dz-(T^{B})^{4}(z)\partial_{z}T^{B}(z)\bigg{|}_{z=0}^{x}.$
(2.24)

and

$\displaystyle\int_{0}^{x}\int_{-1}^{1}\mu\partial_{z}(\psi^{B})^{2}d\mu
dz=\int_{-1}^{1}\mu(\psi^{B})^{2}(z,\cdot)d\mu\bigg{|}_{z=0}^{x}.$ (2.25)

Comparing equation (2.1) to (2.2) gives

$\displaystyle\partial_{x}^{2}T^{B}=\partial_{x}\langle\mu\psi^{B}\rangle,$
(2.26)

this together with $\partial_{x}T^{B}(B)=\langle\mu\psi^{B}(B,\cdot)\rangle=0$
implies

$\displaystyle\partial_{x}T^{B}(x)=\langle\mu\psi^{B}(x,\cdot)\rangle,\quad\text{for
any }x\in[0,B].$ (2.27)

Using this and taking (2.24) and (2.25) into (2.23), we get

$\displaystyle\int_{0}^{x}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dz+\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu
dz$
$\displaystyle\quad+\frac{1}{2}\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu-\frac{1}{2}\int_{-1}^{1}\mu(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu=0.$

The boundary condition (2.4) implies

$\displaystyle\int_{-1}^{1}\mu(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu=\int_{-1}^{0}\mu(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu+\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$
(2.28)

Taking it into the previous equation leads to the estimate (4) with
$\alpha=0$:

$\displaystyle\int_{0}^{x}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dx+\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu
dz$
$\displaystyle\quad+\frac{1}{2}\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu$
$\displaystyle\qquad=\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$
(2.29)

Note here we have

$\displaystyle\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu\geq
0$

for any $x\in[0,B]$, which is due to the fact that
$\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu$ is a non-
increasing function of $x$ and

$\displaystyle\int_{-1}^{1}\mu(\psi^{B}(B,\cdot)-(T^{B}(B))^{4})^{2}d\mu=0,$
(2.30)

which is due to the boundary condition $\psi^{B}(B,\mu)=\psi^{B}(B,-\mu)$. The
non-increasing fact of
$\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu$ can be seen if we
integrate over $[x_{1},x_{2}]$ with $0\leq x_{1}<x_{2}\leq B$ in the
derivation above, the inequality

$\displaystyle\frac{1}{2}\int_{-1}^{1}\mu(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu\bigg{|}_{x_{1}}^{x_{2}}$
$\displaystyle\quad=-\int_{x_{1}}^{x_{2}}4(T^{B})^{3}|\partial_{x}T^{B}|^{2}dx-\int_{x_{1}}^{x_{2}}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu
dx\leq 0$

is satisfied.

Next, let $\alpha\in(0,1)$ be a constant. We multiply (2.1) by $e^{2\alpha
x}(T^{B})^{4}$, and (2.2) by $e^{2\alpha x}\psi^{B}$, and integrate to get

$\displaystyle-\int_{0}^{x}e^{2\alpha
z}(T^{B})^{4}\partial_{z}^{2}T^{B}dx+\frac{1}{2}\int_{0}^{x}\int_{-1}^{1}\mu
e^{2\alpha z}\partial_{z}(\psi^{B})^{2}d\mu dx$
$\displaystyle\quad+\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi^{B}-(T^{B})^{4})^{2}d\mu dx=0.$ (2.31)

Using integration-by-parts, we get

$\displaystyle-\int_{0}^{x}e^{2\alpha z}(T^{B})^{4}\partial_{z}^{2}T^{B}dz$
$\displaystyle=\int_{0}^{x}e^{2\alpha
z}4(T^{B})^{3}|\partial_{x}T^{B}|^{2}dx+2\alpha\int_{0}^{x}e^{2\alpha
z}(T^{B})^{4}\partial_{z}T^{B}dx$ $\displaystyle\quad-e^{2\alpha
z}(T^{B})^{4}(z)\partial_{z}T^{B}(z)\bigg{|}_{z=0}^{x},$

and

$\displaystyle\int_{0}^{x}\int_{-1}^{1}\mu e^{2\alpha
z}\partial_{z}(\psi^{B})^{2}d\mu dz$
$\displaystyle=-2\alpha\int_{0}^{x}\int_{-1}^{2}\mu e^{2\alpha
z}(\psi^{B})^{2}d\mu dz+\int_{-1}^{1}\mu e^{2\alpha
z}(\psi^{B})^{2}(z,\cdot)d\mu\bigg{|}_{0}^{x}.$

Taking the above equations into (2.3) and using the relation (2.27) gives

$\displaystyle\int_{0}^{x}e^{2\alpha
z}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dx+\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi^{B}-(T^{B})^{4})^{2}d\mu dz$
$\displaystyle\quad-\alpha\int_{0}^{x}\int_{-1}^{1}\mu e^{2\alpha
z}(\psi^{B}-(T^{B})^{4})^{2}d\mu dz=-\frac{1}{2}\int_{-1}^{1}\mu e^{2\alpha
z}(\psi^{B}(z,\cdot)-(T^{B}(z))^{4})^{2}d\mu\bigg{|}_{0}^{x}.$

Using the relation (2.28) and $1-\alpha\mu\geq 1-\alpha$ for $0\leq\alpha<1$
in the above equation leads to

$\displaystyle\int_{0}^{x}e^{2\alpha
z}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dx+(1-\alpha)\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi^{B}-(T^{B})^{4})^{2}d\mu dz$
$\displaystyle\quad+\frac{1}{2}\mu\int_{-1}^{1}e^{2\alpha
x}(\psi^{B}(x,\cdot)-(T^{B}(x))^{4})^{2}d\mu+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi^{B}(0,\cdot)-T_{b})^{4})^{2}d\mu$
$\displaystyle\qquad\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$
(2.32)

and finishes the proof of Theorem 4. ∎

##  3\. Existence and decay results for the nonlinear Milne problem

In this section, we consider the existence of the nonlinear Milne problem and
prove Theorem 1. We first pass to the limit $B\to\infty$ in system (2.1)-(2.2)
and show the existence of weak solutions to system (1.1)-(1.2) as well as the
weighted estimate (1). Then the exponentially decay of solutions is shown.

###  3.1. Existence

We first pass to the limit $B\to\infty$ on system (2.1)-(2.2) and show the
existence of weak solutions for system (1.1)-(1.2).

###### Proof of Theorem 1, existence.

First we show the convergence of $T^{B}(x)$ as $B\to\infty$. To show this, we
first prove that there exists a subsequence $\\{T^{B_{k}}(B_{k})\\}$
converging to some constant $T_{\infty}$. From the weighted estimate (4)
(taking $x=B$ in (4)), we can get

$\displaystyle(T^{B})^{5/2}(B)-T_{b}^{5/2}=\int_{0}^{B}\partial_{x}(T^{B})^{5/2}dx=\int_{0}^{B}e^{-\alpha
x}e^{\alpha x}\partial_{x}(T^{B})^{5/2}dx$
$\displaystyle\quad\leq\left(\int_{0}^{B}e^{2\alpha
x}|\partial_{x}(T^{B})^{5/2}|^{2}dx\right)^{1/2}\left(\int_{0}^{B}e^{-2\alpha
x}dx\right)^{1/2}$
$\displaystyle\quad\leq\frac{1}{(2\alpha)^{1/2}}\sqrt{(1-e^{-2\alpha
B})}\left(\frac{25}{16}\int_{0}^{B}e^{2\alpha
x}4(T^{B})^{3}|\partial_{x}(T^{B})|^{2}dx\right)^{1/2}$
$\displaystyle\quad\leq\frac{5}{4(2\alpha)^{1/2}}\sqrt{(1-e^{-2\alpha
B})}\left(\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{1/2}$
$\displaystyle\quad\leq\frac{5}{4(2\alpha)^{1/2}}\left(\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{1/2}.$

Hence $(T^{B})^{5/2}(B)$ is bounded for any $B>0$. By the Bolzano–Weierstrass
theorem, there exists a subsequence $\\{B_{k}\\}$ such that

$\displaystyle T^{B_{k}}(B_{k})\to T_{\infty},\quad\text{as }k\to\infty,$

where $T_{\infty}$ is some constant. Since $0\leq
T^{k}(B_{k})(B_{k})\leq\gamma$ according to Theorem 4, $0\leq
T_{\infty}\leq\gamma$.

Next we show $T^{B}(x)-T^{B}(B)$ is uniformly bounded in $L^{2}([0,B])$. From
estimate (4), it holds the inequality

$\displaystyle(1-\alpha)\int_{0}^{B}\int_{-1}^{1}e^{2\alpha
x}(\psi^{B}-(T^{B})^{4})^{2}d\mu
dx\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$

Using this estimate and the relation

$\displaystyle\left(\int_{-1}^{1}\mu\psi^{B}d\mu\right)^{2}$
$\displaystyle=\left(\int_{-1}^{1}\mu(\psi^{B}-(T^{B})^{4})d\mu\right)^{2}\leq\int_{-1}^{1}\mu^{2}d\mu\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu$
$\displaystyle\leq\frac{2}{3}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu,$

we can get

$\displaystyle\int_{0}^{B}e^{2\alpha x}|\langle\mu\psi^{B}\rangle|^{2}dx$
$\displaystyle\leq\frac{2}{3}\int_{0}^{B}\int_{-1}^{1}e^{2\alpha
x}(\psi^{B}-(T^{B})^{4})^{2}d\mu dx$
$\displaystyle\leq\frac{1}{3}\frac{1}{1-\alpha}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$

Due to (2.27), the above inequality implies

$\displaystyle\int_{0}^{B}e^{2\alpha
x}|\partial_{x}T^{B}|^{2}dx\leq\frac{1}{3}\frac{1}{1-\alpha}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$
(3.1)

Therefore,

$\displaystyle\int_{0}^{B}(T^{B}(x)-T^{B}(B))^{2}dx$
$\displaystyle=\int_{0}^{B}\left(\int_{x}^{B}\partial_{z}T^{B}(z)dz\right)^{2}dx$
$\displaystyle=\int_{0}^{B}\left(\int_{x}^{B}e^{-\alpha z}e^{\alpha
z}\partial_{z}T^{B}(z)dz\right)^{2}dx$
$\displaystyle\leq\int_{0}^{B}\left(\int_{x}^{B}e^{2\alpha
z}|\partial_{z}T^{B}|^{2}(z)dz\cdot\int_{x}^{B}e^{-2\alpha z}dz\right)dx$
$\displaystyle\leq\frac{1}{2\alpha}\int_{0}^{B}e^{2\alpha
x}|\partial_{x}T^{B}|^{2}dx\cdot\int_{0}^{B}(e^{-2\alpha x}-e^{-2\alpha B})dx$
$\displaystyle\leq\frac{1}{6}\frac{1}{1-\alpha}\frac{e^{2\alpha B}-2\alpha
B-1}{2\alpha^{2}}e^{-2\alpha B}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu$
$\displaystyle\leq\frac{1}{6}\frac{1}{2\alpha^{2}(1-\alpha)}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$

where $(e^{2\alpha B}-2\alpha B-1)e^{-2\alpha B}=1-2\alpha Be^{-2\alpha
B}-e^{-2\alpha B}\leq 1$. Hence $T^{B}(x)-T^{B}(B)\in L^{2}_{\rm
loc}(\mathbb{R}_{+})$ is uniformly bounded. Moreover, due to the boundary
condition (2.3), $\partial_{x}T^{B}(B)=0$ and so the estimate (3.1) implies

$\displaystyle\int_{0}^{B}|\partial_{x}T^{B}(x)-\partial_{x}T^{B}(B)|^{2}dx\leq\frac{1}{3(1-\alpha)}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$
(3.2)

thus $T^{B}(x)-T^{B}(B)$ is uniformly bounded in $H^{1}_{\rm
loc}(\mathbb{R}_{+})$. Moreover,
$\|\partial_{x}^{2}T^{B}\|_{L^{2}([0,B])}=\|\langle\psi^{B}-(T^{B})^{4}\rangle\|_{L^{2}([0,B])}\leq\sqrt{2}\|\psi^{B}-(T^{B})^{4}\|_{L^{2}([0,B]\times[-1,1])}$
is uniformly bounded. Therefore, there exists a subsequence $B_{k}$ such that

$\displaystyle T^{B_{k}}(x)-T^{B_{k}}(B_{k})\rightharpoonup
T(x)-T_{\infty},\quad\text{weakly in }H^{2}_{\rm loc}(\mathbb{R}_{+}).$ (3.3)

By the continuous embedding $C_{\rm loc}^{1}\subset H^{2}_{\rm
loc}(\mathbb{R}_{+})$, we get from the above convergence

$\displaystyle T^{B_{k}}(x)\to T(x),\quad\text{strongly in }C^{1}_{\rm
loc}(\mathbb{R}_{+}).$ (3.4)

Next we show the convergence of $\psi^{B}(x)$. Due to the uniform estimate
(2.3), $\|\psi^{B}-(T^{B})^{4}\|_{L^{2}([0,B]\times[-1,1])}$ is uniformly
bounded in $B$. This, together with (3.4) implies that there exists a
subsequence $\\{\psi^{B_{k}}\\}$ such that as $k\to\infty$,

$\displaystyle\psi^{B_{k}}(x,\mu)-(T^{B_{k}})^{4}(x)\rightharpoonup\psi(x,\mu)-T^{4},\quad\text{
weakly in }L^{2}_{\rm loc}(\mathbb{R}_{+}\times[-1,1]).$ (3.5)

Next we show the limit $T,\psi$ solves system (1.1)-(1.2) with boundary
conditions (1.3)-(1.4). First we show the limit $T(x)$, $\psi(x)$ satisfies
the boundary condition (1.3)-(1.4). Due to the trace operator $u\mapsto u(0)$
is continous (see [8, Ex 8.18]) for any $u\in H^{1}_{\rm
loc}(\mathbb{R}_{+})$, we can get from (3.3) that
$T(0)=\lim_{k\to\infty}T^{B_{k}}(0)=T_{b}$. One can also check that
$\varphi\to\varphi(0,\cdot)$ is also continous for any $\varphi\in
L^{2}_{\operatorname{loc}}(\mathbb{R}_{+}\times[-1,1])\cap\\{\varphi:\mu\partial_{x}\varphi+\varphi\in
C_{\rm loc}(\mathbb{R}_{+})\\}$ (one can use the formulas (B.3)-(B.4) to show
this). Hence, $\psi^{B_{k}}(0,\mu)\rightharpoonup\psi^{k}(0,\mu)$ as
$B_{k}\to\infty$. To show $(T,\psi)$ satisfies (1.1)-(1.2), we apply a test
function $h=h(x)\in C^{1}([0,x])$, $\varphi\in C^{1}([0,x]\times[-1,1])$ to
(2.1) and (2.2), integrate by parts, and obtain

$\displaystyle\int_{0}^{x}\partial_{z}T^{B}\partial_{z}hdz-\partial_{z}T^{B}(z)h(z)\bigg{|}_{0}^{x}+\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})hd\mu
dz=0,$ (3.6)
$\displaystyle\int_{0}^{x}\int_{-1}^{1}\mu\psi^{B}\partial_{z}\varphi d\mu
dz-\int_{-1}^{1}\mu\varphi(z,\cdot)\psi^{B}(z,\cdot)d\mu\bigg{|}_{0}^{x}+\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})\varphi
d\mu dz=0,$ (3.7)

Due to (3.4), we can pass to the limit

$\displaystyle\int_{0}^{x}\partial_{z}T^{B}\partial_{z}hdz\to\int_{0}^{x}\partial_{z}T\partial_{z}hdz,\quad\text{
as }B\to\infty.$ (3.8)

Due to (3.4)-(3.5), we can pass to the limit

$\displaystyle\int_{0}^{x}\int_{-1}^{1}\mu\psi^{B}\partial_{z}\varphi d\mu
dz\to\int_{0}^{x}\int_{-1}^{1}\mu\psi\partial_{z}\varphi d\mu dz,\quad\text{as
}B\to\infty,$ (3.9)

Using (3.5), we can pass to the limit as $B\to\infty$,

$\displaystyle\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})hd\mu
dz\to\int_{0}^{x}\int_{-1}^{1}(\psi-T^{4})hd\mu dz,$ (3.10)
$\displaystyle\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-T^{4})\varphi d\mu
dz\to\int_{0}^{x}\int_{-1}^{1}(\psi-T^{4})\varphi d\mu dz.$ (3.11)

By the continuity of the trace operator, we get from the above facts,

$\displaystyle\partial_{x}T^{B}(x)\to\partial_{x}T(x),\quad\text{almost
everywhere.}$ (3.12)

Due to $\mu\partial_{x}\psi+\psi\in C_{\rm loc}(\mathbb{R}_{+})$,
$\psi^{B}(x,\mu)\rightharpoonup\psi(x,\mu)$ in $L_{\mu}^{2}([-1,1])$ almost
everywhere,

$\displaystyle\int_{-1}^{1}\mu\varphi\psi^{B}d\mu\to\int_{-1}^{1}\mu\varphi\psi
d\mu.$ (3.13)

Combining the above limits, we can pass to the limit $B\to\infty$ in
(3.6)-(3.7) and

$\displaystyle\int_{0}^{x}\partial_{z}T\partial_{z}h-\partial_{z}T(z)h(z)\bigg{|}_{0}^{x}+\int_{0}^{x}\int_{-1}^{1}(\psi-(T)^{4})hd\mu
dz=0,$ (3.14)
$\displaystyle\int_{0}^{x}\int_{-1}^{1}\mu\psi\partial_{z}\varphi d\mu
dz-\int_{-1}^{1}\mu\varphi(z,\cdot)\psi(z,\cdot)d\mu\bigg{|}_{0}^{x}+\int_{0}^{x}\int_{-1}^{1}(\psi-T^{4})\varphi
d\mu dz=0$ (3.15)

holds. Therefore, we have proved that there exists a weak solution
$(T,\psi)\in C_{\rm loc}(\mathbb{R}_{+})\times L^{2}_{\rm
loc}(\mathbb{R}_{+}\times[-1,1])$ to system (1.1)-(1.2). ∎

###  3.2. Weighted estimate

Next we show the weighted estimate (1).

###### Proof of Theorem 1, weighted estimate.

Due to the lower semi-continuous of the norm $\|\cdot\|_{L^{2}([0,x])}$, the
estimate (2.3) and the strong convergence (3.4) of $T$ and the weak
convergence (3.3) of $\partial_{x}T^{B}$ imply

$\displaystyle\int_{0}^{x}4T^{3}|\partial_{z}T|^{2}dz\leq\liminf_{B\to\infty}\int_{0}^{x}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dz.$
(3.16)

Due to the weak convergence (3.5) of $\psi^{B}-(T^{B})^{4}$,

$\displaystyle\int_{0}^{x}\int_{-1}^{1}(\psi-T^{4})^{2}d\mu
dz\leq\liminf_{B\to\infty}\int_{0}^{x}\int_{-1}^{1}(\psi^{B}-(T^{B})^{4})^{2}d\mu
dz.$ (3.17)

From (2.3), $\langle\mu(\psi^{B}-(T^{B})^{2})^{2}\rangle$ is non-negative and
uniformly bounded in $L^{\infty}_{\rm loc}(\mathbb{R}_{+})$. Hence

$\displaystyle\langle\mu(\psi^{B}-(T^{B})^{4})^{2}\rangle\rightharpoonup^{*}\langle\mu(\psi-T^{4})^{2}\rangle,\quad\text{weakly-*
in }L^{\infty}_{\rm loc}(\mathbb{R}_{+}).$

Therefore,

$\displaystyle\limsup_{B\to\infty}\langle\mu(\psi^{B}-(T^{B})^{4})^{2}\rangle\geq\langle\mu(\psi-T^{4})^{2}\rangle\geq
0.$ (3.18)

Combining the above inequality with (3.16)-(3.17), we can take $\limsup$ on
(2.3) and using
$\limsup_{n\to\infty}(a_{n}+b_{n})\geq\limsup_{n\to\infty}a_{n}+\liminf_{n\to\infty}b_{n}$
to get

$\displaystyle\int_{0}^{x}4T^{3}|\partial_{z}T|^{2}dz+\int_{0}^{x}\int_{-1}^{1}(\psi-T^{4})^{2}d\mu
dz+\frac{1}{2}\langle\mu(\psi(x,\cdot)-T^{4}(x))^{2}\rangle$
$\displaystyle\quad+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi(0,\cdot)-T_{b}^{4})^{2}d\mu\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$
(3.19)

holds for any $x\in\mathbb{R}_{+}$. Note in the above
$\langle\mu(\psi(x,\cdot)-T^{4}(x))^{2}\rangle\geq 0$ for any
$x\in\mathbb{R}_{+}$.

Similarly, we can pass to the limit $B\to\infty$ in (4) and use the weak
convergence up to a subsequence implied by (4):

$\displaystyle e^{\alpha x}|\partial_{x}(T^{B})^{\frac{5}{2}}|\rightharpoonup
e^{\alpha x}|\partial_{x}T^{\frac{5}{2}}|,\quad\text{weakly in }L^{2}_{\rm
loc}(\mathbb{R}_{+}),$ $\displaystyle e^{\alpha
x}(\psi^{B}-(T^{B})^{4})\rightharpoonup e^{\alpha
x}(\psi-T^{4}),\quad\text{weakly in }L^{2}_{\rm
loc}(\mathbb{R}_{+}\times[-1,1]),$ $\displaystyle e^{2\alpha
x}\langle\mu(\psi^{B}-(T^{B})^{4})^{2}\rangle\rightharpoonup^{*}e^{2\alpha
x}\langle\mu(\psi-T^{4})^{2}\rangle,\quad\text{weakly-* in }L^{\infty}_{\rm
loc}(\mathbb{R}_{+}),$
$\displaystyle\int_{-1}^{0}|\mu|(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu\to\int_{-1}^{0}|\mu|(\psi(0,\cdot)-T_{b}^{4})^{2}d\mu,\quad\text{weakly-*
in }L^{\infty}_{\rm loc}(\mathbb{R}_{+}),$

Consequently, we obtain

$\displaystyle\int_{0}^{x}e^{2\alpha
z}4T^{3}|\partial_{z}T|^{2}dz\leq\liminf_{B\to\infty}\int_{0}^{x}e^{2\alpha}4(T^{B})^{3}|\partial_{z}T^{B}|^{2}dz,$
$\displaystyle\int_{0}^{x}\int_{-1}^{1}e^{2\alpha z}(\psi-T^{4})^{2}d\mu
dz\leq\liminf_{B\to\infty}\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi^{B}-(T^{B})^{4})^{2}d\mu dz,$
$\displaystyle\limsup_{B\to\infty}\langle\mu(\psi^{B}-(T^{B})^{4})^{2}\rangle\geq\langle\mu(\psi-T^{4})^{2}\rangle\geq
0,$
$\displaystyle\lim_{B\to\infty}\int_{-1}^{0}|\mu|(\psi^{B}(0,\cdot)-T_{b}^{4})^{2}d\mu=\int_{-1}^{0}|\mu|(\psi(0,\cdot)-T_{b}^{4})^{2}d\mu.$

Therefore, we take $\limsup_{B\to\infty}$ in (4) and use the above
inequalities to get

$\displaystyle\int_{0}^{x}e^{2\alpha
z}4T^{3}|\partial_{z}T|^{2}dx+(1-\alpha)\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi-T^{4})^{2}d\mu dz$ $\displaystyle\quad+\frac{1}{2}e^{2\alpha
x}\langle\mu(\psi(x,\cdot)-T^{4}(x))^{2}\rangle+\frac{1}{2}\int_{-1}^{0}|\mu|(\psi(0,\cdot)-T_{b}^{4})^{2}d\mu$
$\displaystyle\quad\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,\quad\text{for
any }x\in\mathbb{R}_{+}.$ (3.20)

From the above estimate, $e^{2\alpha x}\langle\mu(\psi-T^{4})^{2}\rangle$ is
bounded, hence $\langle\mu(\psi-T^{4})^{2}\rangle$ vanishes as $x\to\infty$.
Since $\alpha<1$ is arbitrary, we also have $e^{2\alpha
x}\langle\mu(\psi-T^{4})^{2}\rangle\to 0$ as $x\to\infty$. We thus let
$x\to\infty$ in the above inequality and get the estimate (1).

∎

###  3.3. Exponential decay

Now, we will establish the exponential decay property of solutions to system
(1.1)-(1.2) using the estimate (1).

###### Proof of Theorem 1, exponential decay.

Let $x_{2}>x_{1}\geq 0$, we have from (1),

$\displaystyle|T^{\frac{5}{2}}(x_{1})-T^{\frac{5}{2}}(x_{2})|$
$\displaystyle=\left|\int_{x_{1}}^{x_{2}}\partial_{z}T^{\frac{5}{2}}dz\right|=\left|\int_{x_{1}}^{x_{2}}e^{-\alpha
x}e^{\alpha z}\partial_{z}T^{\frac{5}{2}}dz\right|$
$\displaystyle\leq\left(\int_{x_{1}}^{x_{2}}e^{-2\alpha
z}dz\right)^{\frac{1}{2}}\left(\frac{25}{16}\int_{x_{1}}^{x_{2}}e^{2\alpha
z}4T^{3}|\partial_{z}T|^{2}dz\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{5}{4}\frac{1}{\sqrt{2\alpha}}\left(e^{-2\alpha
x_{1}}-e^{-2\alpha
x_{2}}\right)^{\frac{1}{2}}\left(\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{\frac{1}{2}}.$

Taking $x_{1},x_{2}\geq-(2\log\varepsilon)/2\alpha$ with $M>0$ large enough,
we can get

$\displaystyle(e^{-\alpha x_{1}}-e^{-\alpha x_{2}})\leq e^{-\alpha
x_{1}}+e^{-\alpha x_{2}}\leq 2\varepsilon^{2},$

and so

$\displaystyle|T^{\frac{5}{2}}(x_{1})-T^{\frac{5}{2}}(x_{2})|\leq\varepsilon\frac{5}{4}\frac{1}{\sqrt{2\alpha}}\left(\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{\frac{1}{2}}$

hence

$\displaystyle\lim_{x\to\infty}T(x)=T_{\infty}\text{ exists and $0\leq
T_{\infty}\leq\gamma$ is finite.}$

Next we show $|T(x)-T_{\infty}|$ decays to zero exponentially as $x$ goes to
infinity. We first show that $\partial_{x}T(x)=\langle\mu\psi(x,\cdot)\rangle$
holds for any $x\in\mathbb{R}_{+}$. To see this, we follow the work [19] and
compare (1.1) with (1.2) to get

$\displaystyle\partial_{x}(\partial_{x}T-\langle\mu\psi\rangle)=0.$ (3.21)

Let $j:=\partial_{x}T-\langle\mu\psi\rangle$. Multiplying (1.2) by $\mu$ and
integrating over $[-1,1]$ gives

$\displaystyle\partial_{x}\langle\mu^{2}\psi\rangle=\langle\mu\psi\rangle.$
(3.22)

We thus get $j=\partial_{x}(T+\langle\mu^{2}\psi\rangle).$ By the boundness of
$T,\psi$, $j=0$ and $T(x)+\langle\mu^{2}\psi(x,\cdot)\rangle$ is a constant,
i.e.

$\displaystyle\partial_{x}T(x)=\langle\mu\psi(x,\cdot)\rangle,\quad\text{ for
all}\,\,\,\forall x\in\mathbb{R}_{+}.$ (3.23)

From the estimate (1), we obtain

$\displaystyle(1-\alpha)\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi-T^{4})^{2}d\mu
dz\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu,$

Therefore,

$\displaystyle\int_{0}^{x}e^{2\alpha z}|\langle\mu\psi\rangle|^{2}dz$
$\displaystyle=\int_{0}^{x}e^{2\alpha
z}\left(\int_{-1}^{1}\mu(\psi-T^{4})d\mu\right)^{2}dz$
$\displaystyle\leq\int_{0}^{x}e^{2\alpha
z}\left(\int_{-1}^{1}\mu^{2}d\mu\right)\left(\int_{-1}^{1}(\psi-T^{4})^{2}d\mu\right)dz$
$\displaystyle\leq\frac{2}{3}\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi-T^{4})^{2}d\mu dz$
$\displaystyle\leq\frac{1}{3(1-\alpha)}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$
(3.24)

Due to the relation (3.23), the above inequality is equivalent to

$\displaystyle\int_{0}^{x}e^{2\alpha
z}|\partial_{z}T|^{2}dz\leq\frac{1}{3(1-\alpha)}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$
(3.25)

We get from the above inequality that

$\displaystyle|T(x)-T_{\infty}|$
$\displaystyle=\left|-\int_{x}^{\infty}\partial_{z}Tdz\right|=\left|\int_{x}^{\infty}e^{-\alpha
z}e^{\alpha z}\partial_{z}Tdz\right|$
$\displaystyle\leq\left(\int_{x}^{\infty}e^{-2\alpha
z}dx\right)^{\frac{1}{2}}\left(\int_{x}^{\infty}e^{2\alpha
z}|\partial_{z}T|^{2}dx\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{1}{\sqrt{2\alpha}}e^{-\alpha
x}\left(\frac{1}{3(1-\alpha)}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{\frac{1}{2}},$
(3.26)

and thus $T$ decays exponentially to some constant $T_{\infty}$ and the first
inequality of (LABEL:eq:thm.ex) holds. A direct consequence of the above
estimate is (1.8).

Moreover, using (1.1) and the estimate (1), we have

$\displaystyle\int_{0}^{x}e^{2\alpha z}|\partial_{z}^{2}T|^{2}dz$
$\displaystyle=\int_{0}^{x}e^{2\alpha
z}|\langle\psi-T^{4}\rangle|^{2}dz\leq\int_{0}^{x}\int_{-1}^{1}e^{2\alpha
z}(\psi-T^{4})^{2}d\mu dz$
$\displaystyle\leq\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu.$

By (3.25) and the above inequality, the Barbalat’s lemma ([10]) implies

$\partial_{x}T(x)\to 0,\quad\text{ as }x\to\infty.$

We can thus get

$\displaystyle|\partial_{x}T(x)|$
$\displaystyle=\left|\int_{x}^{\infty}\partial_{z}^{2}Tdz\right|=\left|\int_{x}^{\infty}e^{-\alpha
z}e^{\alpha z}\partial_{z}^{2}Tdz\right|$
$\displaystyle\leq\left(\int_{x}^{\infty}e^{-2\alpha
z}dz\right)^{\frac{1}{2}}\left(\int_{x}^{\infty}e^{2\alpha
z}|\partial_{z}^{2}T|^{2}dz\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{1}{\sqrt{2\alpha}}e^{-\alpha
z}\left(\frac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu\right)^{\frac{1}{2}}.$
(3.27)

Finally, we show the decay of $\psi$. We can pass to the limit $B\to\infty$ in
the formula (B.3)-(B.4) and show $\psi$ satisfies the formula (C.3)-(C.4).
Therefore, for $\mu>0$,

$\displaystyle|\psi(x,\mu)-T_{\infty}^{4}|$ (3.28)
$\displaystyle=\left|\psi_{b}e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}T^{4}(s)ds-
T_{\infty}^{4}\right|$
$\displaystyle=\left|\psi_{b}e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}(T^{4}(s)-T_{\infty}^{4})ds-T^{4}_{\infty}e^{-\frac{x}{\mu}}\right|$
$\displaystyle\leq|\psi_{b}-T_{\infty}^{4}|e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|T^{4}(s)-T_{\infty}^{4}|ds$
$\displaystyle\leq|\psi_{b}-T_{b}^{4}|e^{-\frac{x}{\mu}}+|T_{b}^{4}-T_{\infty}^{4}|e^{-\frac{x}{\mu}}$
$\displaystyle\quad+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|T(s)-T_{\infty}|\cdot(T^{2}(s)+T_{\infty}^{2})(T(s)+T_{\infty})ds$
$\displaystyle\leq|\psi_{b}-T_{b}^{4}|e^{-\frac{x}{\mu}}+4(T_{b}+M_{\alpha})^{3}M_{\alpha}e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}M_{\alpha}e^{-\alpha
x}4(T_{b}+2M_{\alpha})^{3}ds$
$\displaystyle=|\psi_{b}-T_{b}^{4}|e^{-\frac{x}{\mu}}+4(T_{b}+M_{\alpha})^{3}M_{\alpha}e^{-\frac{x}{\mu}}+4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{1-\mu\alpha}(e^{-\alpha
x}-e^{-\frac{x}{\mu}})$
$\displaystyle\leq|\psi_{b}-T_{b}^{4}|e^{-\frac{x}{\mu}}+4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{1-\mu\alpha}e^{-\alpha
x},$ (3.29)

which gives the second inequality of (LABEL:eq:thm.ex), and for $\mu<0$,

$\displaystyle|\psi(x,\mu)-T_{\infty}^{4}|$
$\displaystyle=\left|-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}T^{4}(s)ds-
T_{\infty}^{4}\right|$
$\displaystyle=\left|-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}T^{4}(s)ds+\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}T^{4}_{\infty}ds\right|$
$\displaystyle\leq-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|T^{4}-T_{\infty}^{4}|ds$
$\displaystyle\leq-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|T(s)-T_{\infty}|\cdot(T^{2}(s)+T_{\infty}^{2})(T(s)+T_{\infty})ds$
$\displaystyle\leq-\int_{x}^{\infty}e^{-\frac{x-s}{\mu}}M_{\alpha}e^{-\alpha
x}4(T_{b}+2M_{\alpha})^{3}ds$
$\displaystyle=-4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{-1+\mu\alpha}e^{-\alpha
x}$ $\displaystyle\leq
4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{1-\mu\alpha}e^{-\alpha x},$ (3.30)

which gives the last inequality of (LABEL:eq:thm.ex) and finishes the proof of
Theorem 1. ∎

######  Remark 1.

For the case $T_{b}=0,\psi_{b}=0$, it is obvious that zero is the solution to
system (1.1)-(1.2). For the case $T_{b}>0$, $\psi_{b}=0$, we assume $T(x)=0$
on $(a,b)\in\mathbb{R}_{+}$. Then the formula (see (B.3)-(B.4))

$\displaystyle\langle\psi(x,\cdot)\rangle=\int_{0}^{1}\psi_{b}(\mu)e^{-\frac{x}{\mu}}d\mu+\int_{0}^{1}\int_{0}^{\infty}\frac{1}{\mu}e^{-\frac{|x-s|}{\mu}}T^{4}(s)dsd\mu$

implies $\langle\psi(x,\cdot)\rangle>0$ for any $x\in(a,b)$. From (1.1),
$\partial_{x}^{2}T(x)=-\langle\psi(x,\cdot)-T^{4}(x)\rangle=-\langle\psi(x,\cdot)\rangle<0$
and thus $T(x)$ is concave on the interval $(a,b)$, which contradicts to the
assumption $T(x)=0$ on this interval. Therefore, $T(x)>0$ for any
$x\in\mathbb{R}_{+}$. For the case $T_{b}=0$ and $\psi_{b}(0,\mu)\neq 0$ on
the interval $\mu\in(c,d)\subset(0,1).$ Then $\int_{0}^{1}\psi_{b}(\mu)d\mu>0$
and by the above formula, $\langle\psi(x,\cdot)\rangle>0$. Hence by the same
contradiction argument, we have $T(x)>0$ for $x\in\mathbb{R}_{+}$. Finally for
the case $T_{b},\psi_{b}$ both are not zero, $\int_{0}^{1}\psi_{b}d\mu>0$ and
so $\langle\psi(x,\cdot)\rangle>0$. Therefore, following the same
contradiction argument, $T(x)>0$ for all $x\in\mathbb{R}_{+}$. We conclude
that in all cases $T(x)>0$ for all $x\in\mathbb{R}_{+}$ if $T_{b},\psi_{b}$
are not both zero. Here whether $\lim_{x\to\infty}T(x)$ is positive or not is
not known.

##  4\. Linearized problem

This section focus on the study of the linearized system (1.10)-(1.11) of
(1.1)-(1.2). We will present the spectral assumption and prove Theorem 2. Let
$T$ be a non-trivial solution to system (1.1)-(1.2). We consider system
(1.10)-(1.11):

$\displaystyle\partial_{x}^{2}g+\langle\phi-4T^{3}g\rangle=\langle
S_{1}\rangle,$ (4.1) $\displaystyle\mu\partial_{x}\phi+(\phi-4T^{3}g)=S_{1},$
(4.2)

with boundary conditions

$\displaystyle g(0)=0,$ (4.3) $\displaystyle\phi(0,\mu)=\phi_{b}(\mu),\text{
for any }\mu\in(0,1],$ (4.4)

where the source terms $S_{1}$ is a function that decays to zero at infinity.
Since $S_{1}$ does not have a definite sign, the solutions to the above system
are not necessarily positive. Moreover, $T$ may not be monotonic. Therefore,
the monotonicity technique used in section 2 to show the existence of
solutions for the nonlinear system (1.1)-(1.2) is not applicable here for the
linearized model. Instead, we use the Banach fixed point theorem to show the
existence of the linearized model on a bounded interval $[0,B]$. Then, a
uniform weighted estimate is derived, allowing us to pass to the limit
$B\to\infty$ and show the existence for system (4.1)-(4.2). Finally, a
contradiction argument is used for showing the uniqueness of solutions.

To establish the stability estimates and show the existence and uniqueness for
system (4.1)-(4.2), a spectral assumption is proposed. With the help of the
spectral assumption, we derive the weighted estimate on solutions of system
(4.1)-(4.2). An explanation of this assumption is given in section 4.1 and the
assumption will be shown later in the last section to hold when boundary
conditions are close to the well-prepared case.

###  4.1. The spectral assumption

We consider the spectrum of the nonlinear operator $\mathcal{F}$ defined in
(1.16) with absorbing boundary conditions. First, according to the theory of
linear transport operator, there exists $(g,\phi)\neq(0,0)$ such that

$\displaystyle\mathcal{F}(g,\phi)=0,$ (4.5)

hence $0$ is an eigenvalue (see [4, Appendix 1]). Next we show that all
eigenvalues lie in $\Re z\leq 0$ under the spectral assumption.

######  Lemma 2.

Under the spectral assumption (A), the spectrum of $\mathcal{F}$ in
$L^{p}(\mathbb{R}_{+})\times L^{p}(\mathbb{R}_{+}\times[-1,1])$ (with $1\leq
p\leq\infty$), is contained in the half-plane. i.e., $\Re z\leq 0$ where $0$
belongs to the spectrum.

Before giving the proof of this lemma, we will show some inequalities related
to the spectral assumption (A).

######  Lemma 3.

Assume $T\in C^{1}_{\rm loc}(\mathbb{R}_{+})$ satisfies the spectral
assumption (A). Then the following inequality

$\displaystyle
M\int_{0}^{\infty}(2T^{\frac{3}{2}})^{2}|\partial_{x}g|^{2}dx\geq
4\int_{0}^{\infty}|\partial_{x}(2T^{\frac{3}{2}})|^{2}g^{2}dx,$ (4.6)

holds for the same constant $M<1$ given in (A) and any function $g\in C_{\rm
loc}^{1}(\mathbb{R}_{+})$ with $g(0)=0$.

###### Proof.

Let $F\in C_{\rm loc}(\mathbb{R}_{+})$ be any function defined on the half-
line and $F(x)\geq 0$ for any $x\in\mathbb{R}_{+}$. Taking
$f=\int_{0}^{x}F(t)dt$ in (1.9) leads to

$\displaystyle M\int_{0}^{\infty}e^{2\beta
x}(2T^{\frac{3}{2}})^{2}F^{2}(x)dx\geq 4\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}(2T^{\frac{3}{2}})|^{2}\left(\int_{0}^{x}F(t)dt\right)^{2}dx,$
(4.7)

for the same constant $M<1$ as in the spectral assumption (A). Let
$G(x)=e^{\beta x}F(x)\geq 0$, the above inequality implies

$\displaystyle M\int_{0}^{\infty}(2T^{\frac{3}{2}})^{2}G^{2}(x)dx$
$\displaystyle\geq 4\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}(2T^{\frac{3}{2}})|^{2}\left(\int_{0}^{x}e^{-\beta
t}G(t)dt\right)^{2}dx$
$\displaystyle=4\int_{0}^{\infty}|\partial_{x}(2T^{\frac{3}{2}})|^{2}\left(\int_{0}^{x}e^{\beta(x-t)}G(t)dt\right)^{2}dx$
$\displaystyle\geq
4\int_{0}^{\infty}|\partial_{x}(2T^{\frac{3}{2}})|^{2}\left(\int_{0}^{x}G(t)dt\right)^{2}dx,$
(4.8)

since $e^{\beta(x-t)}\geq 1$ for any $t\leq x$. Due to
$\int_{0}^{x}|G(t)|dt\geq\int_{0}^{x}G(t)dt$, the above inequality also holds
for all $G(x)\in C_{\rm loc}(\mathbb{R}_{+})$. Let $g(x)=\int_{0}^{x}G(t)dt$,
then (4.6) holds for any $g\in C^{1}_{\rm loc}(\mathbb{R}_{+})$ satisfying
$g(0)=0$. ∎

Finally we give some inequalities implied by the spectral assumption that are
used in the proof of existence for system 4.1-(4.2).

######  Lemma 4.

Assume the spectral assumption (A) is fulfilled, then

$\displaystyle\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx+\int_{0}^{\infty}\partial_{x}(4T^{3})f\partial_{x}fdx\geq
0$ (4.9)

and

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx+\int_{0}^{\infty}e^{2\beta
x}\partial_{x}(4T^{3})f\partial_{x}fdx\geq 0$ (4.10)

hold for any function $f\in C^{1}$ such that $f(0)=0$ and for
$\beta\in[0,\beta_{0}]$ where $\beta_{0}$ is given in Lemma 6.

###### Proof.

The spectral assumption (A) implies

$\displaystyle\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx\geq\int_{0}^{\infty}36T|\partial_{x}T|^{2}f^{2}dx.$
(4.11)

By Young’s inequality, we deduce that

$\displaystyle\left|\int_{0}^{\infty}\partial_{x}(4T^{3})f\partial_{x}fdx\right|$
$\displaystyle\leq\frac{1}{2}\int_{0}^{\infty}\frac{1}{4T^{3}}|\partial_{x}(4T^{3})|^{2}f^{2}dx+\frac{1}{2}\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx$
$\displaystyle=\frac{1}{2}\int_{0}^{\infty}36T|\partial_{x}T|^{2}f^{2}dx+\frac{1}{2}\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx.$

Combining this with the previous inequality, we obtain

$\displaystyle\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx+\int_{0}^{\infty}\partial_{x}(4T^{3})f\partial_{x}fdx$
$\displaystyle\quad\geq\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx-\frac{1}{2}\int_{0}^{\infty}36T|\partial_{x}T|^{2}f^{2}dx-\frac{1}{2}\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx$
$\displaystyle\quad=\frac{1}{2}\int_{0}^{\infty}4T^{3}|\partial_{x}f|^{2}dx-\frac{1}{2}\int_{0}^{\infty}36T|\partial_{x}T|^{2}f^{2}dx\geq
0,$

and thus (4.9) holds.

Similarly, the spectral assumption (A) also implies for
$\beta\in[0,\beta_{0}]$,

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx\geq\int_{0}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}f^{2}dx.$ (4.12)

This implies

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx+\int_{0}^{\infty}e^{2\beta
x}\partial_{x}(4T^{3})f\partial_{x}fdx$
$\displaystyle\quad\geq\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx-\frac{1}{2}\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx-\frac{1}{2}\int_{0}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}f^{2}dx$
$\displaystyle\quad=\frac{1}{2}\int_{0}^{\infty}e^{2\beta
x}4T^{3}|\partial_{x}f|^{2}dx-\frac{1}{2}\int_{0}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}f^{2}dx\geq 0,$ (4.13)

which is (4.10) and the proof is finished. ∎

We next give the proof of Lemma 2.

###### Proof of Lemma 2.

Suppose $\lambda$ is an eigenvalue of $\mathcal{F}$ associated to the
eigenvector $(g,\phi)$. Then

$\displaystyle\lambda\left(\begin{array}[]{cc}g\\\
\phi\end{array}\right)=\mathcal{F}(g,\phi)=\left(\begin{array}[]{cc}\partial_{x}^{2}g+\langle\phi-4T^{3}g\rangle\\\
-\mu\partial_{x}\phi-(\phi-4T^{3}g)\end{array}\right).$ (4.18)

We multiply the above equation by $(4T^{3}g,\phi)$ and integrate over
$\mathbb{R}_{+}\times[-1,1]$ to get

$\displaystyle\lambda\int_{0}^{\infty}4T^{3}g^{2}dx-+\lambda\int_{0}^{\infty}\phi^{2}dx$
$\displaystyle\quad=\int_{0}^{\infty}4T^{3}g\partial_{x}^{2}gdx-\int_{0}^{\infty}\int_{-1}\mu\partial_{x}\frac{\phi^{2}}{2}d\mu
dx-\int_{0}^{\infty}\int_{-1}^{1}(\phi-4T^{3}g)^{2}d\mu dx.$ (4.19)

The spectral assumption implies

$\displaystyle\int_{0}^{\infty}4T^{3}g\partial_{x}^{2}gdx=-\int_{0}^{\infty}4T^{3}|\partial_{x}g|^{2}dx-\int_{0}^{\infty}\partial_{x}(4T^{3})g\partial_{x}gdx\leq
0.$ (4.20)

The absorbing boundary condition implies

$\displaystyle\int_{0}^{\infty}\int_{-1}\mu\partial_{x}\frac{\phi^{2}}{2}d\mu
dx=0.$ (4.21)

Taking the above two inequalities into (4.1), we obtain

$\displaystyle\lambda\int_{0}^{\infty}4T^{3}g^{2}dx+\lambda\int_{0}^{\infty}\phi^{2}dx\leq\int_{0}^{\infty}\int_{-1}^{1}(\phi-4T^{3}g)^{2}d\mu
dx\leq 0.$ (4.22)

Hence if $(g,\phi)\neq(0,0)$, then $\lambda\leq 0$ and finishes the proof of
Lemma 2. ∎

Assuming $S_{1}$, $S_{2}$ are two given functions decaying exponentially to
zero, by Lemma 2, there exists a unique solution to the following problem

$\displaystyle\varepsilon(g_{\varepsilon}^{1},\phi_{\varepsilon}^{1})^{T}+\mathcal{F}(g_{\varepsilon}^{1},\phi_{\varepsilon}^{1})=(-S_{1},S_{2})^{T},$
(4.23) $\displaystyle
g_{\varepsilon}^{1}(0)=0,\quad\phi_{\varepsilon}^{1}(0,\mu)=0,\quad\text{for
any }\mu>0.$ (4.24)

By a superposition argument, there also exists a unique solution to

$\displaystyle\varepsilon(g_{\varepsilon}^{2},\phi_{\varepsilon}^{2})^{T}+\mathcal{F}(g_{\varepsilon}^{2},\phi_{\varepsilon}^{2})=0,$
(4.25) $\displaystyle
g_{\varepsilon}^{2}(0)=g_{b},\quad\phi_{\varepsilon}^{2}(0,\mu)=\phi_{b},\quad\text{for
any }\mu>0.$ (4.26)

Then $g_{\varepsilon}=g_{\varepsilon}^{1}+g_{\varepsilon}^{2}$,
$\phi_{\varepsilon}:=\phi_{\varepsilon}^{1}+\phi_{\varepsilon}^{2}$ solves

$\displaystyle-\varepsilon
g_{\varepsilon}+\partial_{x}^{2}g_{\varepsilon}+\langle\phi_{\varepsilon}-4T^{3}g_{\varepsilon}\rangle=S_{1},$
(4.27)
$\displaystyle\varepsilon\phi_{\varepsilon}+\mu\partial_{x}\phi_{\varepsilon}+\phi_{\varepsilon}-4T^{3}g_{\varepsilon}=S_{2},$
(4.28) $\displaystyle
g_{\varepsilon}(0)=0,\quad\phi_{\varepsilon}(0,\mu)=0,\quad\text{for any
}\mu>0.$ (4.29)

Formally, the above system converges to (1.1)-(1.4). However, we need to
derive a uniform estimate on the solutions to the above problem. This can done
for the Milne problem of linear transport equation using a Maximum principle
([4]). However, due to $T$ not being a constant function, we cannot use the
Maximum principle to show the uniform boundness of
$(g_{\varepsilon},\phi_{\varepsilon})$. Here we overcome this problem by first
showing existence of (1.1)-(1.4) in a bounded interval and extend the
solutions to half-space using uniform estimates.

###  4.2. Existence on the bounded interval

We now consider system (4.1)-(4.2) on the bounded interval $[0,B]$:

$\displaystyle\partial_{x}^{2}g^{B}+\langle\phi^{B}-4T^{3}g^{B}\rangle=\langle
S_{1}\rangle,$ (4.30)
$\displaystyle\mu\partial_{x}\phi^{B}+(\phi^{B}-4T^{3}g^{B})=S_{1},$ (4.31)

with boundary conditions

$\displaystyle g^{B}(0)=0,\quad\partial_{x}g^{B}(B)=0,$ (4.32)
$\displaystyle\phi^{B}(0,\mu)=\phi_{b}(\mu),\quad\phi^{B}(B,\mu)=\phi^{B}(B,-\mu),\text{
for any }\mu>0,$ (4.33)

where $T^{B}$ is the solution to the system (2.1)-(2.2).

Define the weighted space

$\displaystyle
L^{2}_{m(x)}([0,B]):=\left\\{f:\int_{0}^{B}m(x)f^{2}dx<\infty\right\\}$

where $m(x)>0$ is a given function. We prove the following existence lemma.

######  Lemma 5.

Let $S_{1}=S_{1}(x)$ be a continuous function on $[0,B]$. Then there exists a
unique solution $(g^{B},\phi^{B})\in C^{1}([0,B])\times C([0,B]\times[-1,1])$
to system (4.30)-(4.31) with boundary conditions (4.32)-(4.33), and the
solution satisfies

$\displaystyle\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4(T^{B})^{3}g^{B})^{2}d\mu
dx+\frac{1}{1-\beta}\int_{-1}^{0}|\mu|(\phi^{B})^{2}(0,\cdot)d\mu$
$\displaystyle\qquad\leq\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{2}{(1-\beta)^{2}}\int_{0}^{B}e^{2\beta
x}|S_{1}|^{2}dx.$ (4.34)

###### Proof.

The proof is divided into two steps. In the first step, we prove existence. We
then derive the uniform estimate in the second step.

_Step 1: Existence._ For a given $g^{B}$, the solution to (4.31), according to
(B.3)-(B.4), can be expressed as

$\displaystyle\phi^{B}=\left\\{\begin{array}[]{lc}\phi_{b}(\mu)e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4(T^{B})^{3}(s)g^{B}(s)ds,&\text{for
}\mu>0,\\\
\phi_{b}(-\mu)e^{\frac{2B-x}{\mu}}-\int_{0}^{B}\frac{1}{\mu}e^{\frac{2B-x-s}{\mu}}4(T^{B})^{3}(s)g^{B}(s)dx\\\
\qquad-\int_{x}^{B}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4(T^{B})^{3}(s)g^{B}(s)ds,&\text{for
}\mu<0.\end{array}\right.$ (4.38)

We denote the above relation by the mapping
$\phi^{B}={\Phi}_{\phi_{b}}(g^{B}).$ Then we have

$\displaystyle\langle\phi^{B}(x,\cdot)\rangle$
$\displaystyle=\int_{0}^{1}\phi^{B}(x,\mu)d\mu+\int_{-1}^{0}\phi^{B}(x,\mu)d\mu$
$\displaystyle=\int_{0}^{1}\phi_{b}(\mu)e^{-\frac{x}{\mu}}d\mu+\int_{0}^{1}\phi_{b}(\mu)e^{-\frac{2B-x}{\mu}}d\mu$
$\displaystyle\quad+\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{|x-s|}{\mu}}4(T^{B})^{3}(s)g^{B}(s)d\mu
ds$
$\displaystyle\quad+\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}4(T^{B})^{3}(s)g^{B}(s)d\mu
ds.$ (4.39)

System (4.30)-(4.31) is equivalent to the following equation

$\displaystyle-\partial_{x}^{2}g^{B}+8T^{3}g^{B}-\langle{\Phi}_{\phi_{b}}(g^{B})\rangle=2S_{1},$
(4.40)

with

$\displaystyle g^{B}(0)=0,\quad\partial_{x}g^{B}(B)=0.$

To prove the existence for the above equation, we construct a sequence
$\\{g_{k}\\}_{k=0}^{\infty}$ with $g^{0}(x)=g_{b}$ and $g^{k}$ is solved via

$\displaystyle-\partial_{x}^{2}g_{k}+8T^{3}g_{k}-\langle{\Phi}_{\phi_{b}}(g_{k-1})\rangle=2S_{1},$
(4.41)

with

$\displaystyle g_{k}(0)=0,\quad\partial_{x}g_{k}(B)=0.$

Denote the above mapping by $\mathcal{G}$ with $g_{k}=\mathcal{G}g_{k-1}$.
Then according to [17, Chpater 2.2.3], $\mathcal{G}$ is invertible on the
space $C([0,B])$ and thus the above equation has a solution in $g_{k}\in
C([0,B])$. Moreover, by (4.41), $g_{k}\in C^{2}([0,B])$. Next we show
$\mathcal{G}$ is a contraction mapping in the weighted space
$L^{2}_{(16T^{6})}([0,B])$.

We first show $\mathcal{G}$ maps the space $L^{2}_{(16(T^{B})^{6})}([0,B])$
onto itself. We multiply (4.41) by $4(T^{B})^{3}g_{k}$ and integrate over
$[0,B]$ to get

$\displaystyle-\int_{0}^{B}4T^{3}g_{k}\partial_{x}^{2}g_{k}dx+\int_{0}^{B}32T^{6}g_{k}^{2}dx=\int_{0}^{B}4T^{3}g_{k}\langle\Phi_{\phi_{b}}(g_{k-1})\rangle
dx.$ (4.42)

Since $T^{B}$ satisfies the spectral assumption (A), by Lemma 4, inequality
(4.9) holds, which implies

$\displaystyle-\int_{0}^{B}4T^{3}g_{k}\partial_{x}^{2}g_{k}dx=\int_{0}^{B}4T^{3}|\partial_{x}g_{k}|^{2}dx+\int_{0}^{B}\partial_{x}(4T^{3})g_{k}\partial_{x}g_{k}dx\geq
0.$ (4.43)

Then, the Young’s inequality implies

$\displaystyle\int_{0}^{B}4T^{3}g_{k}\langle\Phi_{\phi_{b}}(g_{k-1})\rangle
dx\leq\int_{0}^{B}16T^{6}g_{k}^{2}dx+\frac{1}{4}\int_{0}^{B}|\langle\Phi_{\phi_{b}}(g_{k-1})\rangle|^{2}dx.$
(4.44)

Using the formula (4.39) and the Young’s convolution inequality we obtain

$\displaystyle\left\|\int_{0}^{x}f(s)\int_{0}^{1}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}d\mu
ds\right\|_{L^{2}([0,B])}^{2}$
$\displaystyle\leq\left\|\int_{0}^{1}\frac{1}{\mu}e^{-\frac{x}{\mu}}d\mu\right\|_{L^{1}([0,B])}^{2}\|f\|_{L^{2}([0,B])}^{2}$
$\displaystyle\leq C(B)\|f\|_{L^{2}([0,B])}^{2},$

and

$\displaystyle\left\|\int_{x}^{B}f(s)\int_{0}^{1}\frac{1}{\mu}e^{\frac{x-s}{\mu}}d\mu
ds\right\|_{L^{2}([0,B])}^{2}$
$\displaystyle\leq\left\|\int_{0}^{1}\frac{1}{\mu}e^{\frac{x}{\mu}}d\mu\right\|_{L^{1}([0,B])}^{2}\|f\|_{L^{2}([0,B])}$
$\displaystyle\leq C(B)\|f\|_{L^{2}([0,B])},$

where $C(B)>0$ is a constant depending only on $B$. Similarly,

$\displaystyle\left\|\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}f(s,\mu)d\mu\right\|_{L^{2}([0,B])}\leq
C(B)\|f\|_{L^{2}([0,B]\times[-1,1])}$ (4.45)

By the above inequalities and (4.39), we deduce

$\displaystyle\|\langle|\Phi_{\phi_{b}}(g_{k-1})\rangle\|_{L^{2}([0,B])}^{2}\leq
C(B,\phi_{b})+C(B)(\|4T^{3}g_{k-1}\|_{L^{2}([0,B])}^{2}.$

Taking this into (4.44) leads to

$\displaystyle\int_{0}^{B}4T^{3}g_{k}\langle\Phi_{\phi_{b}}(g_{k-1})\rangle
dx\leq\int_{0}^{B}16(T^{B})^{6}g_{k}^{2}dx+C(B)\int_{0}^{B}16T^{6}g_{k-1}^{2}dx.$
(4.46)

Combining the above inequality with (4.43), (4.42) implies

$\displaystyle\int_{0}^{B}16T^{6}g_{k}^{2}dx\leq
C(B)\int_{0}^{B}16T^{6}g_{k-1}^{2}dx.$ (4.47)

Hence $\mathcal{G}$ maps from $L^{2}_{16T^{6}}([0,B])$ onto itself.

We next show that $\mathcal{G}$ is a contraction mapping. Let
$h_{k}=g_{k}-g_{k-1}$, we have

$\displaystyle\partial_{x}^{2}h_{k}=8T^{3}h_{k}-\langle\Phi_{0}(h_{k-1})\rangle,$
(4.48)

with

$\displaystyle h_{k}=0,\quad\partial_{x}h_{k}(B)=0.$

We multiply (4.48) by $4(T^{B})^{3}h_{k}$ and integrate over $[0,B]$ to get

$\displaystyle-\int_{0}^{B}4T^{3}h_{k}\partial_{x}^{2}h_{k}dz+\int_{0}^{B}32T^{6}h_{k}^{2}dx-\int_{0}^{B}\langle\Phi_{0}(h_{k-1})\rangle
4T^{3}h_{k}dx=0.$ (4.49)

The inequality (4.9) implied by the spectral assumption on $T^{B}$ leads to

$\displaystyle-\int_{0}^{B}4T^{3}h_{k}\partial_{x}^{2}h_{k}dx=\int_{0}^{B}4T^{3}|\partial_{x}h_{k}|^{2}dz+\int_{0}^{B}\partial_{x}(4T^{3})h_{k}\partial_{x}h_{k}dz\geq
0.$ (4.50)

Recalling (4.39), we have

$\displaystyle\int_{0}^{B}4$ $\displaystyle
T^{3}h_{k}\langle\Phi_{\phi_{b}=0}(h_{k-1})\rangle dx$ (4.51)
$\displaystyle=~{}$
$\displaystyle\int_{0}^{B}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{|x-s|}{\mu}}16T^{3}(s)h_{k-1}(s)T^{3}(x)h_{k}(x)d\mu
dsdx$
$\displaystyle+\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}16T^{3}(s)h_{k-1}(s)T^{3}(x)h_{k}(x)d\mu
dsdx$ $\displaystyle=~{}$
$\displaystyle-\frac{1}{2}\int_{0}^{B}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{|x-s|}{\mu}}(4T^{3}(s)h_{k-1}(s)-4T^{3}(x)h_{k}(x))^{2}d\mu
dsdx$
$\displaystyle-\frac{1}{2}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}(4T^{3}(s)h_{k-1}(s)-4T^{3}(x)h_{k}(x))^{2}d\mu
dsdx$
$\displaystyle+\frac{1}{2}\int_{0}^{B}\int_{0}^{B}\int_{0}^{1}e^{-\frac{|x-s|}{\mu}}(16T^{6}(s)h_{k-1}^{2}(s)+16T^{3}(x)h_{k}^{2}(x))d\mu
dsdx$
$\displaystyle+\frac{1}{2}\int_{0}^{B}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}(16T^{6}(s)h_{k-1}^{2}(s)+16T^{3}(x)h_{k}^{2}(x))d\mu
dsdx$
$\displaystyle=-\frac{1}{2}\int_{0}^{B}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{|x-s|}{\mu}}(4T^{3}(s)h_{k-1}(s)-4T^{3}(x)h_{k}(x))^{2}d\mu
dsdx$
$\displaystyle-\frac{1}{2}\int_{0}^{B}\int_{0}^{1}\frac{1}{\mu}e^{-\frac{2B-x-s}{\mu}}(4T^{3}(s)h_{k-1}(s)-4T^{3}(x)h_{k}(x))^{2}d\mu
dsdx$
$\displaystyle+\frac{1}{2}\int_{0}^{B}\int_{0}^{1}(1-e^{-\frac{x}{s}}+1-e^{-\frac{B-x}{\mu}}+e^{-\frac{B-x}{\mu}}-e^{-\frac{2B-x}{\mu}})$
$\displaystyle\cdot 16T^{6}(x)(h_{k}^{2}(x)+h_{k-1}^{2}(x))d\mu dx$
$\displaystyle\leq~{}$
$\displaystyle\frac{1}{2}\int_{0}^{B}\int_{0}^{1}16(2-e^{-\frac{x}{\mu}}-e^{-\frac{2B-x}{\mu}})T^{6}(x)(h_{k}^{2}(x)+h_{k-1}^{2}(x))d\mu
dx.$

Taking (4.50) and (4.51) into (4.49) gives

$\displaystyle\frac{1}{2}\int_{0}^{B}\int_{0}^{1}16(2+e^{-\frac{x}{\mu}}+e^{-\frac{2B-x}{\mu}})T^{6}(x)h_{k}^{2}(x)d\mu
dx$
$\displaystyle\quad-\frac{1}{2}\int_{0}^{B}\int_{0}^{1}16(2-e^{-\frac{x}{\mu}}-e^{-\frac{2B-x}{\mu}})T^{6}(x)h_{k-1}^{2}(x)d\mu
dx\leq 0.$

Due to

$\displaystyle\min_{x\in[0,B]}(e^{-\frac{x}{\mu}}+e^{-\frac{2B-x}{\mu}})=e^{-\frac{B}{\mu}}+e^{-\frac{B}{\mu}}=2e^{-\frac{B}{\mu}},$

The previous inequality implies

$\displaystyle\int_{0}^{B}16T^{6}(x)h_{k}^{2}(x)dz\leq\delta\int_{0}^{B}16T^{6}(x)h_{k-1}^{2}(x)dz,$
(4.53)

with

$\displaystyle\delta=\frac{\int_{0}^{1}(1-e^{-\frac{B}{\mu}})d\mu}{\int_{0}^{1}(1+e^{-\frac{B}{\mu}})d\mu}<1.$

Then (4.53) can be written as

$\displaystyle\|h_{k}\|^{2}_{L^{2}_{16T^{6}}([0,B])}\leq\delta\|h_{k-1}\|^{2}_{L^{2}_{16T^{6}}([0,B])},$

which is

$\displaystyle\|g_{k}-g_{k-1}\|_{L^{2}_{16T^{6}}([0,B])}^{2}\leq\delta\|g_{k-1}-g_{k-2}\|_{L^{2}_{16T^{6}}([0,B])}^{2},$

with $\delta<1$. Thus the map $\mathcal{G}:L^{2}_{16T^{6}}([0,B])\mapsto
L^{2}_{16T^{6}}([0,B])$ is a contraction mapping. By the Banach fixed point
theorem, there exists a unique solution to the equation (4.40) in
$L^{2}_{16T^{6}}([0,B])$, i.e. there exists a unique solution $g^{B}\in
L^{2}_{16T^{6}}([0,B])$ to the equation (4.40). By equation (4.40),
$\partial_{x}^{2}g^{B}\in L^{2}([0,B])$ and thus $g^{B}\in C^{1}([0,B])$.
Since $g^{B}\in L^{2}_{16T^{6}}([0,B])$ is equivalent to
$4(T^{B})^{3}g^{B}\in{L^{2}([0,B])}$. By the formula (4.38), $\phi^{B}\in
C([0,B]\times[-1,1])$.

_Step 2. The weighted estimates._ We first multiply (4.30) by $4T^{3}g^{B}$
and integrate over $[0,B]$, we also multiply (4.31) by $\phi^{B}$ and
integrate over $[0,B]$ and $[-1,1]$ to get

$\displaystyle-\int_{0}^{B}4T^{3}g^{B}\partial_{x}^{2}g^{B}dx+\int_{0}^{B}\int_{-1}^{1}\mu\partial_{x}\frac{(\phi^{B})^{2}}{2}d\mu
dx$
$\displaystyle\quad+\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx=\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})S_{1}d\mu dx.$ (4.54)

By the boundary condition (4.32) and the inequality (4.9) implied by the
spectral assumption (4.10), we can get

$\displaystyle-\int_{0}^{B}4T^{3}g^{B}\partial_{x}^{2}g^{B}dx=~{}$
$\displaystyle\int_{0}^{B}4T^{3}|\partial_{x}g^{B}|^{2}dx+\int_{0}^{B}\partial_{x}(4T^{3})g^{B}\partial_{x}(g^{B})\geq
0.$ (4.55)

Using integration by parts and the boundary conditions (4.33), we have

$\displaystyle\int_{0}^{B}\int_{-1}^{1}\mu\partial_{x}\frac{(\phi^{B})^{2}}{2}d\mu
dx$
$\displaystyle=\frac{1}{2}\int_{-1}^{1}\mu(\phi^{B})^{2}d\mu\bigg{|}_{0}^{B}=-\frac{1}{2}\int_{-1}^{1}\mu(\phi^{B})^{2}d\mu$
$\displaystyle=-\frac{1}{2}\int_{-1}^{0}\mu(\phi^{B})^{2}(0,\cdot)d\mu-\frac{1}{2}\int_{0}^{1}\mu\phi_{b}^{2}d\mu.$
(4.56)

Using Young’s inequality on the last term of (4.2) gives

$\displaystyle\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})S_{1}d\mu dx$
$\displaystyle\quad\leq\xi\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx+\frac{1}{4\xi}\int_{0}^{B}|S_{1}|^{2}dx\int_{-1}^{1}d\mu$
$\displaystyle\quad=\xi\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx+\frac{1}{2\xi}\int_{0}^{B}|S_{1}|^{2}dx,$ (4.57)

where $\xi>0$ is a positive constant and will be chosen later. Taking the
above inequality and (4.55), (4.2) into (4.2) leads to the estimate

$\displaystyle(1-\xi)\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx+\frac{1}{2}\int_{-1}^{0}|\mu|(\phi^{B})^{2}(0,\cdot)d\mu$
$\displaystyle\quad\leq\frac{1}{2}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{1}{2\xi}\int_{0}^{B}|S_{1}|^{2}dx.$
(4.58)

Taking $\xi=\frac{1}{2}$, the above inequality becomes the weighted estimate
(5) with $\beta=0$.

We next derive the weighted estimate on system (4.30)-(4.31) for $\beta>0$.
Multiplying (4.30) by $e^{2\beta x}4T^{3}g^{B}$ and (4.31) by $e^{2\beta
x}\phi^{B}$ and integrating over $[0,B]$ and $[-1,1]$, we obtain

$\displaystyle-\int_{0}^{B}e^{2\beta
x}4T^{3}g^{B}\partial_{x}^{2}g^{B}dx+\int_{0}^{B}\int_{-1}^{1}\mu e^{2\beta
x}\partial_{x}\frac{(\phi^{B})^{2}}{2}d\mu dx$
$\displaystyle\quad+\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})^{2}d\mu dx=\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})S_{1}dx.$ (4.59)

Using integration by parts, the inequality (4.10) implied by the spectral
assumption (A), and the boundary condition (4.32), we get

$\displaystyle-\int_{0}^{B}e^{2\beta x}4T^{3}g^{B}\partial_{x}^{2}g^{B}dx$
$\displaystyle\quad=\int_{0}^{B}e^{2\beta
x}4T^{3}|\partial_{x}g^{B}|^{2}dx+\int_{0}^{x}e^{2\beta
x}\partial_{x}(2T^{3})\partial_{x}(g^{B})^{2}dx$
$\displaystyle\qquad+2\beta\int_{0}^{x}e^{2\beta
x}4T^{3}g^{B}\partial_{x}g^{B}dx-e^{2\beta
x}4T^{3}g^{B}\partial_{x}g^{B}\bigg{|}_{0}^{B}$ $\displaystyle\qquad\geq
2\beta\int_{0}^{x}e^{2\beta x}4T^{3}g^{B}\partial_{x}g^{B}dx.$ (4.60)

Moreover, using integration by parts, we have

$\displaystyle\int_{0}^{B}\int_{-1}^{1}\mu e^{2\beta
x}\partial_{x}\frac{(\phi^{B})^{2}}{2}d\mu dx$
$\displaystyle\quad=-\beta\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}\mu(\phi^{B})^{2}d\mu dx+\frac{1}{2}\int_{-1}^{1}e^{2\beta
x}\mu(\phi^{B})^{2}d\mu\bigg{|}_{0}^{B}$
$\displaystyle\quad=-\beta\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}\mu(\phi^{B})^{2}d\mu
dx-\frac{1}{2}\int_{-1}^{1}\mu(\phi^{B})^{2}(0,\cdot)d\mu.$ (4.61)

Comparing (4.30) with (4.31) and using the assumption $\langle
S_{2}\rangle=S_{1}$, we obtain

$\displaystyle\partial_{x}(\partial_{x}g^{B}-\langle\mu\phi^{B}\rangle)=0.$
(4.62)

Due to the boundary conditions (4.32)-(4.33),

$\partial_{x}g^{B}(B)=0,\quad\langle\mu\phi^{B}(B,\cdot)\rangle=0.$

Therefore, by (4.62), we deduce that

$\displaystyle\partial_{x}g^{B}(x)=\langle\mu\phi^{B}(x,\cdot)\rangle,\quad\text{
for any }x\in[0,B].$ (4.63)

Using the above relation, we have

$\displaystyle\langle\mu(\phi^{B})^{2}\rangle-4T^{3}g^{B}=\langle\mu(\phi^{B})^{2}\rangle-4T^{3}\langle\mu\phi^{B}\rangle=\langle\mu(\phi^{B}-4T^{3}g^{B})^{2}\rangle.$
(4.64)

Consequently, it follows that

$\displaystyle 2\beta\int_{0}^{B}e^{2\beta
x}4T^{3}g^{B}\partial_{x}g^{B}dx-\beta\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}\mu(\phi^{B})^{2}d\mu dx$
$\displaystyle\quad=-\beta\int_{0}^{B}\int_{-1}^{1}\mu(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx.$ (4.65)

For the right terms of (4.2), we can use Young’s inequality to get

$\displaystyle\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})S_{1}d\mu dx$
$\displaystyle\quad\leq\xi\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})^{2}d\mu dx+\frac{1}{2\xi}\int_{0}^{B}e^{2\beta
x}|S_{1}|^{2}dx.$

Combing the above inequality with (4.2), (4.2) and (4.2), the equality (4.2)
becomes

$\displaystyle\int_{0}^{B}\int_{-1}^{1}(1-\xi-\mu\beta)e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})^{2}d\mu dx$
$\displaystyle\quad+\frac{1}{2}\int_{-1}^{0}|\mu|(\phi^{B})^{2}(0,\cdot)d\mu\leq\frac{1}{2}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{1}{2\xi}\int_{0}^{B}e^{2\beta
x}S_{1}^{2}dx.$

Taking $\beta<1$, then $1-\xi-\mu\beta\geq 1-\xi-\beta$. Taking
$\xi=(1-\beta)/2$, the above inequality implies (5) and finishes the proof of
Lemma 5. ∎

###  4.3. Existence on the half-space

Next we pass to the limit $B\to\infty$ in system (4.30)-(4.31) and show the
existence for system (4.1)-(4.2).

###### Proof of Theorem 2, existence..

By the relation (4.63), we have

$\displaystyle\int_{0}^{B}e^{2\beta
x}|\partial_{x}g^{B}|^{2}dx=\int_{0}^{B}e^{2\beta
x}|\langle\mu\phi^{B}\rangle|^{2}dx.$ (4.66)

Using the fact $\langle\mu 4T^{3}g^{B}\rangle=0$ and Hölder’s inequality, the
above equation equals

$\displaystyle\int_{0}^{B}e^{2\beta x}|\partial_{x}g^{B}|^{2}dx$
$\displaystyle=\int_{0}^{B}e^{2\beta
x}\left(\int_{-1}^{1}\mu(\phi^{B}-4T^{3}g^{B})d\mu\right)^{2}dx$
$\displaystyle\leq\int_{0}^{B}\left(\int_{-1}^{1}\mu
d\mu\right)\left(\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu\right)dx$
$\displaystyle\leq\frac{2}{3}\int_{0}^{B}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4T^{3}g^{B})^{2}d\mu dx$
$\displaystyle\leq\frac{2}{3}\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{3(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx,$ (4.67)

which is uniformly bounded. Hence

$\displaystyle|g^{B}(x)|$
$\displaystyle=|g^{B}(x)-g^{B}(0)|=\left|\int_{0}^{x}\partial_{z}g^{B}(z)dz\right|$
$\displaystyle=\left|\int_{0}^{x}e^{-\beta z}e^{\beta
z}\partial_{z}g^{B}(z)dz\right|$ $\displaystyle\leq\left(\int_{0}^{x}e^{2\beta
z}|\partial_{z}g|^{2}dz\right)^{\frac{1}{2}}\left(\int_{0}^{x}e^{-2\beta
z}dz\right)^{\frac{1}{2}}$ $\displaystyle\leq\frac{1}{\sqrt{2\beta}}e^{-\beta
x}\left(\frac{2}{3}\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{3(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx\right)^{\frac{1}{2}}$ $\displaystyle=N_{\beta}e^{-\beta x}.$
(4.68)

Hence $g^{B}\in L_{\rm loc}^{2}(\mathbb{R}_{+})$. By the equation (4.30),

$\displaystyle\int_{0}^{B}e^{2\beta
x}|\partial_{x}^{2}g^{B}|^{2}dx\leq\frac{2}{3}\int_{0}^{B}\int_{-1}^{2}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx+2\int_{0}^{B}e^{2\beta x}|S_{1}|^{2}dx$ (4.69)

and is also uniformly bounded. Hence $g^{B}\in H^{2}_{\rm
loc}(\mathbb{R}_{+})$ is uniformly bounded and we can find a subsequence that

$\displaystyle g^{B}\rightharpoonup g,\quad\text{weakly in }H^{2}_{\rm
loc}(\mathbb{R}_{+}),$ (4.70) $\displaystyle g^{B}\to g,\quad\text{strongly in
}C^{1}_{\rm loc}(\mathbb{R}_{+}).$

Moreover, by the continuity of the trace operator, we can pass to the limit
$B\to\infty$ in (4.32) and get that $g(0)=0$.

By the above convergence result and the uniform boundness of
$\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu dx$, there exists
subsequence such that

$\displaystyle\phi^{B}-4T^{3}g^{B}\rightharpoonup\phi-4T^{3}g,\quad\text{weakly
in }L^{2}_{\rm loc}(\mathbb{R}_{+}\times[-1,1]).$ (4.71)

Hence

$\displaystyle\phi^{B}\rightharpoonup\phi,\quad\text{weakly in }L^{2}_{\rm
loc}(\mathbb{R}_{+}\times[-1,1]).$ (4.72)

Moreover, by equation (4.31),

$\displaystyle\int_{0}^{B}\int_{-1}^{1}|\partial_{x}(\mu\psi)|^{2}d\mu
dx\leq\int_{0}^{B}\int_{-1}^{1}(\phi^{B}-4T^{3}g^{B})^{2}d\mu
dx+2\int_{0}^{B}|S_{1}|^{2}dx,$ (4.73)

is uniformly bounded, hence $\partial_{x}(\mu\psi)\in L^{2}_{\rm
loc}(\mathbb{R}_{+}\times[-1,1])$. Thus we can use the trace theorem and pass
to the limit in (4.33) to get that $\phi(0,\mu)=\phi_{b}(\mu)$ for any
$\mu>0$. Moreover, we can pass the weak limit in (4.30)-(4.31) and use (4.70)
and (4.72) and obtain that $(g,\phi)$ satisfies the system (4.1)-(4.2) with
boundary conditions (4.3)-(4.4).

The proof is divided in to four steps. In the first step, we pass to the limit
$B\to\infty$ in (4.32)-(4.33) and prove the limit satisfies system (4.1)-(4.2)
in the weak sense. Then we show the uniform estimate and weighted decay
property (1.19).

###  4.4. Weighted estimate and exponentially decay

Next we derive the uniform estimate for system (4.1)-(4.2) on the half-space
and show the exponentially decay properties.

###### Proof of Theorem 2, weighted estimate and decay.

The weak convergence of $\phi^{B}-4(T^{B})^{3}g^{B}$ in $L^{2}_{\rm
loc}(\mathbb{R}_{+})$ and the weak lower semi-continuity of the norm
$\|\cdot\|_{L^{2}}$ implies that

$\displaystyle\|e^{\beta
x}(\phi-4T^{3}g)\|_{L^{2}(\mathbb{R}_{+}\times[-1,1])}^{2}\leq\liminf_{B\to\infty}\|e^{\beta
x}(\phi^{B}-4(T^{B})^{3}g^{B})\|_{L^{2}([0,B]\times[-1,1])}^{2}.$ (4.74)

Moreover, by the trace theorem we have

$\displaystyle\lim_{B\to\infty}\int_{-1}^{0}|\mu|(\phi^{B})^{2}(0,\cdot)d\mu=\int_{-1}^{0}|\mu|\phi^{2}(0,\cdot)d\mu.$
(4.75)

Therefore, we can take the $\liminf_{B\to\infty}$ in (5) and obtain

$\displaystyle\int_{0}^{\infty}\int_{-1}^{1}e^{2\beta
x}(\phi^{B}-4(T^{B})^{3}g^{B})^{2}d\mu
dx+\frac{1}{1-\beta}\int_{-1}^{0}|\mu|\phi^{2}(0,\cdot)d\mu$
$\displaystyle\qquad\leq\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{2}{(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx,$ (4.76)

which is the estimate (2) in Theorem 2.

To show the exponentially decay property, we first note that the condition
(4.63) still holds for $g,\phi$ on the half-line. We can pass to the limit in
(4.63) due to (4.70) and (4.72), and

$\displaystyle\partial_{x}g(x)=\langle\mu\phi(x,\cdot)\rangle,\quad\text{ for
}x\in\mathbb{R}_{+}\quad\text{almost everywhere}.$ (4.77)

In order to show the above relation still holds at infinity, we need to show
this quantity are bounded. To show this, first comparing the equation (4.1) to
$\langle\eqref{eq:l-2}\rangle$ gives

$\displaystyle\partial_{x}(\partial_{x}g-\langle\mu\phi\rangle)=0.$ (4.78)

By equation (4.2) and the estimate (4.4), similarly as (4.3),

$\displaystyle\int_{0}^{\infty}e^{2\beta x}|\langle\mu\phi\rangle|^{2}dx$
$\displaystyle\leq\frac{2}{3}\int_{0}^{\infty}\int_{-1}^{1}(\phi-4T^{3}g)^{2}d\mu
dx$
$\displaystyle\leq\frac{2}{3}\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{3(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx.$ (4.79)

Moreover, we have

$\displaystyle\partial_{x}\langle\mu\phi\rangle=-\langle\phi-4T^{3}g\rangle+\langle
S_{1}\rangle,$ (4.80)

and so

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}\langle\mu\phi\rangle|^{2}dx$
$\displaystyle\quad\leq\int_{0}^{\infty}e^{2\beta
x}|\langle\phi-4T^{3}g\rangle|^{2}dx+\int_{0}^{\infty}e^{2\beta x}|\langle
S_{1}\rangle|^{2}dx$ $\displaystyle\quad\leq
2\int_{0}^{\infty}\int_{-1}^{1}e^{2\beta
x}(\phi-4T^{3}g)^{2}dx+2\int_{0}^{\infty}e^{2\beta x}|S_{1}|^{2}dx$
$\displaystyle\quad\leq 2\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx+\frac{2}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx$ (4.81)

From (4.4) and (4.4), we can apply the Barbalat’s lemma [10] and deduce

$\langle\mu\phi(x,\cdot)\rangle\to 0,\quad\text{as}\quad x\to\infty$

Hence

$\displaystyle|\langle\mu\phi(x,\cdot)\rangle|=\left|\int_{x}^{\infty}e^{-\beta
x}e^{\beta x}\partial_{x}\langle\mu\phi\rangle
dx\right|\leq\frac{1}{\sqrt{2\beta}}e^{-\beta
x}\left(\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}\langle\mu\phi\rangle|^{2}dx\right)^{\frac{1}{2}},$ (4.82)

and is uniformly bounded. Thus (4.77) holds for all $x\in[0,\infty]$. Using
the relation (4.77) and (4.4), we obtain

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}g|^{2}dx\leq\frac{2}{3}\frac{1}{1-\beta}\int_{0}^{1}\mu\phi_{b}^{2}d\mu+\frac{4}{3(1-\beta)^{2}}\int_{0}^{\infty}e^{2\beta
x}|S_{1}|^{2}dx=2\beta N_{\beta}^{2}.$ (4.83)

Hence

$\displaystyle|g(x)|=\left|\int_{0}^{x}\partial_{z}g(z)dz\right|$
$\displaystyle=\left|\int_{0}^{x}e^{-\beta x}e^{\beta
x}\partial_{x}gdx\right|$
$\displaystyle\leq\frac{1}{\sqrt{2\beta}}\left(\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}g|^{2}dx\right)^{\frac{1}{2}}=N_{\beta}.$ (4.84)

We can let $x\to\infty$ and obtain $g_{\infty}:=\lim_{x\to\infty}g(x)$ exists
and

$\displaystyle|g_{\infty}|\leq N_{\beta}.$ (4.85)

We also have

$\displaystyle|g(x)-g_{\infty}|$
$\displaystyle=\left|\int_{x}^{\infty}e^{-\beta x}e^{\beta
x}\partial_{x}gdx\right|$ $\displaystyle\leq\frac{1}{\sqrt{2\beta}}e^{-\beta
x}\left(\int_{0}^{\infty}e^{2\beta
x}|\partial_{x}g|^{2}dx\right)^{\frac{1}{2}}=N_{\beta}e^{-\beta x},$ (4.86)

which gives the first inequality in (1.19). The other two inequalities in
(1.19) can be derived by using the formula (C.3)-(C.4) in a similar manner as
(3.28)-(3.3). For $\mu>0$,

$\displaystyle\left|\psi(x,\mu)-4T_{\infty}^{3}g_{\infty}-\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}S_{1}(s)ds\right|$
(4.87)
$\displaystyle=\left|\psi_{b}e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}(4T^{3}(s)g(s)+S_{1}(s))ds-4T_{\infty}^{3}g_{\infty}\right|$
$\displaystyle=\left|\psi_{b}e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}(4T^{3}(s)g(s)-T_{\infty}^{4})ds-4T^{3}_{\infty}g_{\infty}e^{-\frac{x}{\mu}}ds\right|$
$\displaystyle\leq|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|4T^{3}(s)g(s)-4T_{\infty}^{3}g_{\infty}|ds$
$\displaystyle\leq|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4|T^{3}(s)-T_{\infty}^{3}|g(s)ds$
$\displaystyle\quad+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4T_{\infty}^{3}|g(s)-g_{\infty}|ds$
$\displaystyle\leq|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4\cdot
3(T_{b}+2M_{\alpha})^{2}M_{\alpha}e^{-\alpha x}N_{\beta}ds$
$\displaystyle\quad+\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4(T_{b}+M_{\alpha})^{3}N_{\beta}e^{-\beta
x}dx$
$\displaystyle=|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+12(T_{b}+2M_{\alpha})^{2}M_{\alpha}N_{\beta}\frac{1}{1-\mu\alpha}(e^{-\alpha
x}-e^{-\frac{x}{\mu}})$
$\displaystyle\quad+4(T_{b}+M_{\alpha})^{3}N_{\beta}\frac{1}{1-\mu\beta}(e^{-\beta
x}-e^{-\frac{x}{\mu}})$
$\displaystyle\leq|\psi_{b}-4T_{\infty}^{3}g_{\infty}|e^{-\frac{x}{\mu}}+4(T_{b}+2M_{\alpha})^{2}(4M_{\alpha}+T_{b})N_{\beta}e^{-\beta
x},$ (4.88)

where in the last inequality we take $\alpha=\beta$. For $\mu<0$,

$\displaystyle\left|\psi(x,\mu)-4T_{\infty}^{3}g_{\infty}+\int_{x}^{\infty}\frac{1}{\mu}e^{\frac{x-s}{\mu}}S_{1}(s)ds\right|$
$\displaystyle=\left|-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4T^{3}(s)g(s)ds-4T_{\infty}^{3}g_{\infty}\right|$
$\displaystyle=\left|-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4T^{3}(s)g(s)ds+\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4T^{3}_{\infty}g_{\infty}ds\right|$
$\displaystyle\leq-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}|4T^{3}(s)g(s)-4T_{\infty}^{3}g_{\infty}|ds$
$\displaystyle\leq-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4|T^{3}(s)-T_{\infty}^{3}|g(s)ds-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4T_{\infty}^{3}|g(s)-g_{\infty}|ds$
$\displaystyle\leq-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4\cdot
3(T_{b}+2M_{\alpha})^{2}M_{\alpha}e^{-\alpha x}N_{\beta}ds$
$\displaystyle\quad-\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}4(T_{b}+M_{\alpha})^{3}N_{\beta}e^{-\beta
x}dx$
$\displaystyle=12(T_{b}+2M_{\alpha})^{2}M_{\alpha}N_{\beta}\frac{1}{1-\mu\alpha}e^{-\alpha
x}+4(T_{b}+M_{\alpha})^{3}N_{\beta}\frac{1}{1-\mu\beta}e^{-\beta x}$
$\displaystyle=-4(T_{b}+2M_{\alpha})^{3}M_{\alpha}\frac{1}{-1+\mu\alpha}e^{-\alpha
x}$ $\displaystyle\leq
4(T_{b}+2M_{\alpha})^{2}(4M_{\alpha}+T_{b})N_{\beta}e^{-\beta x},$ (4.89)

where in the last inequality we also take $\alpha=\beta$. Since for $\mu>0$

$\displaystyle\left|\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}S_{1}(s)ds\right|$
$\displaystyle=\left|\int_{0}^{x}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}e^{-\beta
s}e^{\beta s}S_{1}(s)ds\right|$
$\displaystyle\leq\left(\int_{0}^{x}\frac{1}{\mu^{2}}e^{-\frac{2(x-s)}{\mu}}e^{-2\beta
s}ds\right)^{\frac{1}{2}}\left(\int_{0}^{x}e^{2\beta
s}|S_{1}(s)|^{2}ds\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{1}{1-2\mu\beta}(e^{-2\beta
x}-e^{-2\frac{x}{\mu}})^{\frac{1}{2}}\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+})}$
$\displaystyle\leq\frac{1}{1-2\mu\beta}\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+})}e^{-\beta x},$ (4.90)

and similarly for $\mu<0$,

$\displaystyle\left|\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}S_{1}(s)ds\right|$
$\displaystyle=\left|\int_{x}^{\infty}\frac{1}{\mu}e^{-\frac{x-s}{\mu}}e^{-\beta
s}e^{\beta s}S_{1}(s)ds\right|$
$\displaystyle\leq\left(\int_{x}^{\infty}\frac{1}{\mu^{2}}e^{-\frac{2(x-s)}{\mu}}e^{-2\beta
s}ds\right)^{\frac{1}{2}}\left(\int_{x}^{\infty}e^{2\beta
s}|S_{1}(s)|^{2}ds\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{1}{1-2\mu\beta}e^{-\beta x}\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+})}.$ (4.91)

Combing the above inequalities with (4.87) and (4.4), we obtain the last two
inequalities in (1.19). ∎

###  4.5. Uniqueness

Suppose $(g_{1},\phi_{1})$ and $(g_{2},\phi_{2})$ are two solutions of the
system (4.1)-(4.2) with boundary conditions (4.3)-(4.4). Then $h=g_{1}-g_{2}$
and $\varphi=\phi_{1}-\phi_{2}$ satisfy the system

$\displaystyle\partial_{x}^{2}h+\langle\varphi-4T^{3}h\rangle$
$\displaystyle=0,$ (4.92)
$\displaystyle\mu\partial_{x}\varphi+(\varphi-4T^{3}h)$ $\displaystyle=0,$
(4.93)

and the boundary conditions

$\displaystyle h(0)=0,\quad\varphi(0,\mu)=0,\text{ for }\mu>0.$ (4.94)

From the previous section, we have that $h,\varphi$ also verify

$\displaystyle\lim_{x\to\infty}\partial_{x}h=0,\lim_{x\to\infty}\langle\mu\varphi(x,\cdot)\rangle=0,$
(4.95)

and

$\displaystyle\partial_{x}h(x)=\langle\mu\varphi(x,\cdot)\rangle,\quad\text{for
any}\quad x\in\mathbb{R}_{+}.$ (4.96)

We now multiply (4.92) by $4T^{3}h$ and (4.93) by $\varphi$ and integrate to
get

$\displaystyle-\int_{0}^{\infty}4T^{3}h\partial_{x}^{2}hdx+\int_{0}^{\infty}\mu\partial_{x}\frac{\varphi^{2}}{2}d\mu
dx+\int_{0}^{\infty}\int_{-1}^{1}(\varphi-4T^{3}h)^{2}d\mu dx=0.$

We integrate by parts and by using (4.94)-(4.95) and the inequality (4.10) due
to the spectral assumption (A), we obtain

$\displaystyle-\int_{0}^{\infty}4$ $\displaystyle
T^{3}h\partial_{x}^{2}hdx+\int_{0}^{\infty}\mu\partial_{x}\frac{\varphi^{2}}{2}d\mu
dx$ $\displaystyle=~{}$
$\displaystyle-4T^{3}h\partial_{x}h\bigg{|}_{0}^{\infty}+\left(\int_{0}^{\infty}4T^{3}|\partial_{x}h|^{2}dx+\frac{1}{2}\int_{0}^{\infty}\partial_{x}(4T^{3})\partial_{x}g^{2}dx\right)+\frac{1}{2}\langle\mu\varphi^{2}\rangle\bigg{|}_{0}^{\infty}$
$\displaystyle=~{}$
$\displaystyle-\frac{1}{2}\langle\mu(\varphi(0,\cdot)-4T^{3}h(0))^{2}dx+\left(\int_{0}^{\infty}4T^{3}|\partial_{x}h|^{2}+\frac{1}{2}\int_{0}^{\infty}\partial_{x}(4T^{3})\partial_{x}g^{2}dx\right)$
$\displaystyle\geq~{}$
$\displaystyle\frac{1}{2}\int_{-1}^{0}|\mu|(\varphi(0,\cdot)|^{2}d\mu.$

The relation (4.96) is used in the second equality. Taking the above
inequality into the previous equation leads to

$\displaystyle\frac{1}{2}\int_{-1}^{0}|\mu|(\varphi(0,\cdot)|^{2}d\mu+\int_{0}^{\infty}\int_{-1}^{1}(\varphi-4T^{3}h)^{2}d\mu
dx\leq 0.$

This implies that

$\displaystyle\varphi-4T^{3}h=0,\varphi(0,\mu)=0,\text{ for any
}\mu\in[-1,1],$

and so

$\displaystyle\partial_{x}^{2}h=0,\partial_{x}\langle\mu\varphi\rangle=0,\langle\mu\varphi(0,\cdot)\rangle=0.$

Therefore, we obtain

$\displaystyle\langle\mu\varphi(x,\cdot)\rangle=0,\text{ for any }x\geq 0$

and

$\displaystyle\partial_{x}h(x)=\langle\mu\varphi(x,\cdot)\rangle=0,\quad\text{for
any}\quad x\geq 0.$

Consequently, we have $h\equiv 0$ and $\varphi\equiv 4T^{3}h\equiv 0$.
Finally, system (4.92)-(4.93) has only zero solutions. Therefore, the solution
to system (4.1)-(4.2) is unique.

∎

##  5\. Uniqueness of the nonlinear Milne problem

In this section, we prove Theorem 3. We first show the uniqueness for small
boundary data and then for small perturbations in a neighborhood of functions
satisfying the spectral assumptions.

###  5.1. Sufficient condition for the spectral assumption

We next give some sufficient and necessary conditions for the spectral
assumption (A) using the Hardy’s inequality.

######  Lemma 6.

The spectral assumption (A) holds for function $T\in C^{1}(\mathbb{R}_{+})$ if

$\displaystyle A_{0}:=\sup_{r\in(0,\infty)}\left(\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\right)^{\frac{1}{2}}\left(\int_{0}^{r}\frac{1}{e^{2\beta
x}4T^{3}}dx\right)^{\frac{1}{2}}<\infty$ (5.1)

satisfies $A_{0}<\frac{1}{2}$ for some constant $\beta>0$. Moreover, if the
spectral assumption (A) holds, then $A_{0}<1$. Furthermore, if $T$ satisfies
$T_{m}\leq|T|\leq C$, $|\partial_{x}T|\leq Ce^{-\alpha x}$ for some constants
$T_{m}>0$, $C>0$ and $\alpha>0$. Then the spectral assumption (A) holds if

$\displaystyle
A_{1}:=\sup_{r\in(0,\infty)}\left(\int_{r}^{\infty}36T|\partial_{x}T|^{2}dx\right)^{\frac{1}{2}}\left(\int_{0}^{r}\frac{1}{4T^{3}}dx\right)^{\frac{1}{2}}<\frac{1}{2}.$
(5.2)

###### Proof.

This lemma is a direct consequence of the the generalized Hardy’s inequality
(Lemma 10 in Appendix D). We take $u(x)=e^{2\beta x}36T|\partial_{x}T|^{2}$,
$v(x)=e^{2\beta x}4T^{3}$, $F(x)=\partial_{x}f$ and $b=\infty$,
$p=p^{\prime}=2$ in (D.1) and get that

$\displaystyle\int_{0}^{\infty}f^{2}e^{2\beta x}36T|\partial_{x}T|^{2}dx\leq
M_{0}\int_{0}^{\infty}|\partial_{x}f|^{2}e^{2\beta x}4T^{3}dx,$ (5.3)

holds for any $f\in C^{1}(\mathbb{R}_{+})$ with $f(0)=0$, with the best
constant $M_{0}$ satisfying $A_{0}^{2}\leq M_{0}\leq 4A_{0}^{2}$, where
$A_{0}$ is given by (5.1).

On the other hand, if the spectral assumption (A) holds, then the best
constant $M_{0}$ in (5.3) satisfies $M_{0}<1$. Since $A_{0}^{2}\leq M_{0}$ due
to Lemma D, $A_{0}<1$.

Finally, if $|T|\leq C$, $|\partial_{x}T|\leq Ce^{-\alpha x}$, then

$\displaystyle T(x)|\partial_{x}T(x)|^{2}\leq Ce^{-\alpha x}.$ (5.4)

This implies

$\displaystyle\int_{r}^{\infty}(e^{2\beta x}-1)T|\partial_{x}T|^{2}dx$
$\displaystyle\leq C\int_{r}^{\infty}(e^{2\beta x}-1)e^{-\alpha
x}dx=Ce^{-\alpha r}\frac{2\beta-\alpha(1-e^{2\beta
r})}{\alpha(\alpha-2\beta)}$ $\displaystyle\leq
C\frac{1}{(\alpha-2\beta)}e^{-(\alpha-2\beta)r}-C\frac{1}{\alpha}e^{-\alpha
r}$ (5.5)

for $\beta<\alpha/2$.

Assume $T\geq T_{m}>0$ for some constant $T_{m}>0$. Then

$\displaystyle\int_{0}^{r}\frac{1}{4T^{3}}dx\leq\frac{1}{4T_{m}^{3}}r$ (5.6)

Therefore,

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\int_{0}^{r}\frac{1}{e^{2\beta x}4T^{3}}dx$ (5.7)
$\displaystyle\quad\leq\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\int_{0}^{r}\frac{1}{4T^{3}}dx$
$\displaystyle\quad=\int_{r}^{\infty}(e^{2\beta
x}-1)36T|\partial_{x}T|^{2}dx\int_{0}^{r}\frac{1}{4T^{3}}dx+\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\int_{0}^{r}\frac{1}{4T^{3}}dx$
$\displaystyle\quad\leq\frac{1}{4T_{m}^{3}}r(C\frac{1}{(\alpha-2\beta)}e^{-(\alpha-2\beta)r}-C\frac{1}{\alpha}e^{-\alpha
r})+A_{1}^{2}.$ (5.8)

for $\beta$ sufficiently small, the first term on the last line of the above
inequality can be sufficiently small and so

$\displaystyle A_{0}<A_{1}+\varepsilon<\frac{1}{2}.$ (5.9)

Hence by the previous argument, the spectral assumption (A) holds.

∎

###  5.2. Uniqueness for almost well-prepared boundary data

We first show that the spectral assumption is not empty and its holds when
$\int_{0}^{1}(\psi_{b}-T_{b}^{4})^{2}d\mu$ is small using the Hardy’s
inequality, [Appendix D]. We then give the uniqueness of the solution for
system (1.1)-(1.2).

######  Lemma 7.

Assume the boundary data $(T_{b},\psi_{b})$ satisfies (1.23) for some constant
$C_{b}$ satisfying (5.15). Let $(T,\psi)$ be the corresponding solution
obtained in Theorem 1. Then $T$ satisfies the inequality (1.21). As a
consequence, $T$ also satisfies the spectral assumption (A).

###### Proof.

By Lemma 6, the spectral assumption (A) holds if the condition (5.1) is
fulfilled. We next show (5.1) holds when
$\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu$ is sufficiently small. Define
$C_{b}:=\tfrac{1}{2}\int_{0}^{1}\mu(\psi_{b}-T_{b}^{4})^{2}d\mu$,

$\displaystyle|\partial_{x}T|\leq\frac{1}{\sqrt{2\alpha}}e^{-\alpha
x}C_{b}^{\frac{1}{2}}.$ (5.10)

Theorem 1 implies that $(T,\psi)$ are uniformly bounded such that

$\displaystyle 0\leq T(x)\leq\gamma,\quad 0\leq\psi(x)\leq\gamma^{4},$

for any $x\geq 0$. Hence for $\beta<\alpha$,

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\leq\frac{36}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma
C_{b}e^{-2(\alpha-\beta)r}.$ (5.11)

On the other hand, due to (3.25), we get

$\displaystyle|T(x)-T_{b}|$
$\displaystyle=\left|\int_{0}^{x}\partial_{x}Tdx\right|=\left|\int_{0}^{x}e^{-\alpha
z}e^{\alpha z}\partial_{z}Tdx\right|$
$\displaystyle\leq\left(\int_{0}^{x}e^{-2\alpha
z}dx\right)^{\frac{1}{2}}\left(\int_{0}^{x}e^{2\alpha
z}|\partial_{z}T|^{2}dx\right)^{\frac{1}{2}}$
$\displaystyle\leq\frac{1}{\sqrt{2\alpha}}\sqrt{1-e^{-2\alpha
x}}\sqrt{\frac{2}{3(1-\alpha)}}C_{b}$
$\displaystyle\leq\sqrt{\frac{1-e^{-2\alpha x}}{3\alpha(1-\alpha)}}C_{b},$

and so

$\displaystyle T(x)\geq T_{b}-\sqrt{\frac{1-e^{-2\alpha
r}}{3\alpha(1-\alpha)}}C_{b},\quad\text{ for any }0\leq x\leq r,$

if $C_{b}$ is sufficiently small. For example, we can take
$C_{b}=\tfrac{1}{2}T_{b}\sqrt{3\alpha(1-\alpha)}$ and the above inequality
implies

$\displaystyle T(x)\geq\frac{1}{2}T_{b}.$ (5.12)

Hence

$\displaystyle\int_{0}^{r}\frac{1}{{e^{2\beta x}4T^{3}}}dx\leq{\frac{1}{\beta
T_{b}^{3}}}.$ (5.13)

Combining this inequality with (5.11) gives

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\cdot\int_{0}^{r}\frac{1}{{e^{2\beta
x}4T^{3}}}dx\leq\frac{36}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma
C_{b}{\frac{1}{\beta T_{b}^{3}}}e^{-2(\alpha-\beta)r}.$

Since $e^{-2(\alpha-\beta)r}\leq 1$, inequality (5.1) is realized if

$\displaystyle\frac{36}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma
C_{b}{\frac{1}{\beta T_{b}^{3}}}<\frac{1}{4}.$ (5.14)

Therefore, inequality (1.21) is fulfilled if we take

$\displaystyle
C_{b}=\min\left\\{\frac{1}{2}T_{b}\sqrt{3\alpha(1-\alpha)},\frac{1}{72}\gamma^{-1}T_{b}^{3}\beta(\alpha-\beta)\sqrt{\alpha}\right\\}$
(5.15)

in (1.23). By Lemma 6, the spectral assumption (A) also holds. ∎

With this lemma, we can prove Theorem 3 for the case when boundary data is
almost well-prepared.

###### Proof of Theorem 3, near well-prepared boundary data.

Let $\left(T,\psi\right)$ and $\left(T_{1},\psi_{1}\right)$ be two solutions
to system (1.1)-(1.2) with boundary conditions (1.3)-(1.4) satisfying the
assumption of Theorem 3. Then $h:=T_{1}-T,\phi=\psi_{1}-\psi$ satisfies

$\displaystyle\partial_{x}^{2}h+\langle\phi-wh\rangle=0,$ (5.16)
$\displaystyle\mu\partial_{x}\phi+\phi-wh=0,$ (5.17)

with boundary conditions

$\displaystyle h(0)=0,\phi(0,\mu)=0,\quad\text{for }\mu>0,$

where $w=(T_{1}^{4}-T^{4})/(T_{1}-T).$ By Lemma 7, $T$ and $T_{1}$ both
satisfy the spectral assumption (A) and thus inequality (4.10) is satisfied
for both $T$ and $T_{1}$. Moreover, one can show in the same way that
$T_{1}^{2}T$(or $T_{1}^{2}T$) also satisfies the inequality

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}4T^{2}T_{1}|\partial_{x}f|^{2}dx\geq\int_{0}^{\infty}e^{2\beta
x}\frac{1}{4T^{2}T_{1}}|\partial_{x}(4T^{2}T_{1})|^{2}f^{2}dx$ (5.18)

which holds if

$\displaystyle\int_{0}^{\infty}e^{2\beta
x}\left(32T_{2}|\partial_{x}T|^{2}+8\frac{T^{2}}{T_{1}}|\partial_{x}T_{1}|^{2}\right)f^{2}dx\leq\int_{0}^{\infty}e^{2\beta
x}4T^{2}T_{1}|\partial_{x}g|^{2}dx.$

According to the Hardy’s inequality (D.1), a sufficient condition for the
above inequality to hold is (see Lemma 6)

$\displaystyle\sup_{r\in(0,\infty)}\left(\int_{r}^{\infty}e^{2\beta
x}\left(32T_{1}|\partial_{x}T|^{2}+8\frac{T^{2}}{T_{1}}|\partial_{x}T_{1}|^{2}\right)dx\cdot\int_{0}^{r}e^{2\beta
x}\frac{1}{{4T^{2}T_{1}}}dx\right)\leq\frac{1}{4}.$ (5.19)

Taking $C_{b}=\tfrac{1}{2}T_{b}\sqrt{3\alpha(1-\alpha)}$, (5.12) implies
$T(x),T_{1}(x)\geq\tfrac{1}{2}T_{b}$. Due to the fact that
$T,T_{1}\leq\gamma$, we get

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}\left(32T_{1}|\partial_{x}T|^{2}+8\frac{T^{2}}{T_{1}}|\partial_{x}T_{1}|^{2}\right)dx$
$\displaystyle\quad\leq\frac{32}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma
C_{b}e^{-2(\alpha-\beta)r}+\frac{8}{(\alpha-\beta)\sqrt{2\alpha}T_{b}}\gamma^{2}C_{b}e^{-2(\alpha-\beta)r}.$

On the other hand, $T(x),\,T_{1}(x)\geq\frac{1}{2}T_{b}$ implies

$\displaystyle\int_{0}^{r}\frac{1}{{4T^{2}T_{1}}e^{2\beta
x}}dx\leq{\frac{1}{\beta T_{b}^{3}}}.$

Hence (5.13) is also realized. Then

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}\left(32T_{1}|\partial_{x}T|^{2}+8\frac{T^{2}}{T_{1}}|\partial_{x}T_{1}|^{2}\right)dx\cdot\int_{0}^{r}\frac{1}{{4T^{2}T_{1}e^{2\beta
x}}}dx$
$\displaystyle\quad\leq\left(\frac{32}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma+\frac{8}{(\alpha-\beta)\sqrt{2\alpha}T_{b}}\gamma^{2}\right)\frac{C_{b}}{\beta
T_{b}^{3}}e^{-2(\alpha-\beta)r}$
$\displaystyle\quad\leq\left(\frac{32}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma+\frac{8}{(\alpha-\beta)\sqrt{2\alpha}T_{b}}\gamma^{2}\right)\frac{C_{b}}{\beta
T_{b}^{3}}.$ (5.20)

Hence (5.19) is fulfilled if

$\displaystyle C_{b}\leq\min\bigg{\\{}$
$\displaystyle\frac{1}{2}T_{b}\sqrt{3\alpha(1-\alpha)},\frac{1}{72}\gamma^{-1}T_{b}^{3}\beta(\alpha-\beta)\sqrt{\alpha},$
$\displaystyle\frac{1}{4}\sqrt{\alpha}(\alpha-\beta)T_{b}^{3}\gamma^{-1}(32\gamma+16\gamma
T_{b}^{-1})^{-1}\bigg{\\}}.$

Then (5.18) holds and $w=(T_{1}^{4}-T^{4})/(T_{1}-T)$ satisfies the spectral
assumption (A). Moreover, by the proof of Lemma 6, $M\leq
4\left(\frac{32}{2(\alpha-\beta)\sqrt{2\alpha}}\gamma+\frac{8}{(\alpha-\beta)\sqrt{2\alpha}T_{b}}\gamma^{2}\right)\frac{C_{b}}{\beta
T_{b}^{3}}$. With a smaller $C_{b}$, the constant $M$ in (1.9) of the spectral
assumption (A) is expected to the smaller.

Therefore, by Theorem 2, the solution to the linear problem (5.16)-(5.17) is
unique. Consequently, we have $h=0,\psi=0$, which is $T_{1}=T,\psi_{1}=\psi$,
i.e. the solution to system (1.1)-(1.2) is unique.

∎

###  5.3. Uniqueness under small perturbations

First we show that the spectral assumption holds under small perturbations.

######  Lemma 8.

Assume $T$ satisfies the inequality (1.21). Let $\mathcal{V}$ be the function
space defined in (3). Then there exists a small constant $\varepsilon>0$ such
that the inequality (1.21) holds for any $T_{1}\in\mathcal{V}$. As a
consequence, the spectral assumption (A) is realized for any
$T_{1}\in\mathcal{V}$.

###### Proof.

Let $A_{0}$ be given by (5.1). Let $h=(T_{1}-T)/\varepsilon$. We then
calculate

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}36T_{1}|\partial_{x}T_{1}|^{2}dx\cdot\int_{0}^{r}\frac{1}{4T_{1}^{3}e^{2\beta
x}}dx$ (5.21) $\displaystyle\quad=\int_{r}^{\infty}e^{2\beta
x}36(T+\varepsilon
h)|\partial_{x}T+\varepsilon\partial_{x}h|^{2}dx\cdot\int_{0}^{r}\frac{1}{4(T+\varepsilon
h)^{3}e^{2\beta x}}dx$ $\displaystyle\quad=\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\cdot\int_{0}^{r}\frac{1}{4T^{3}e^{2\beta x}}dx$
$\displaystyle\qquad+\varepsilon\int_{r}^{\infty}e^{2\beta
x}36h|\partial_{x}T+\varepsilon\partial_{x}h|^{2}dx\cdot\int_{0}^{r}\frac{1}{4(T+\varepsilon
h)^{3}e^{2\beta x}}dx$
$\displaystyle\qquad+\varepsilon\int_{r}^{\infty}e^{2\beta
x}36T|\partial_{x}T|^{2}dx\cdot\int_{0}^{r}\frac{-3T^{2}h-3\varepsilon
Th^{2}-\varepsilon^{2}h^{3}}{4(T+\varepsilon h)^{3}T^{3}e^{2\beta x}}dx$
$\displaystyle\qquad+\varepsilon\int_{r}^{\infty}e^{2\beta
x}36(2\partial_{x}T\partial_{x}h+\varepsilon|\partial_{x}h|^{2})dx\cdot\int_{0}^{r}\frac{1}{4(T+\varepsilon
h)^{3}e^{2\beta x}}dx.$ (5.22)

Consequently,

$\displaystyle(A^{\prime})^{2}$
$\displaystyle:=\sup_{r\in(0,\infty)}\left(\int_{r}^{\infty}e^{2\beta
x}36(T+\varepsilon
h)|\partial_{x}T+\varepsilon\partial_{x}h|^{2}dx\cdot\int_{0}^{r}\frac{1}{4(T+\varepsilon
h)^{3}e^{2\beta x}}dx\right)$ $\displaystyle\leq A_{0}^{2}+C_{1}\varepsilon.$
(5.23)

for some constant $C_{1}>0$. By the Hardy’s inequality, we have

$\displaystyle C_{hd}^{\prime}\int_{0}^{\infty}e^{2\beta x}4(T+\varepsilon
h)^{3}|\partial_{x}f|^{2}dx\geq\int_{0}^{\infty}e^{2\beta x}36(T+\varepsilon
h)|\partial_{x}(T+\varepsilon h)|^{2}f^{2}dx$ (5.24)

where $C_{hd}^{\prime}\leq 4(A^{\prime})^{2}\leq
4A_{0}^{2}+4C_{1}\varepsilon$. Therefore if $A_{0}^{2}<1/4$, then
$(A^{\prime})^{2}<1/4$ for $\varepsilon$ sufficiently small, and the
inequality 1.21 holds for $T+\varepsilon h$ if $T$ satisfies (5.1) with
$A_{0}<1/2$. Therefore, we conclude that the inequality (1.21) holds for any
$T_{1}\in C^{1}(\mathbb{R}_{+})$ such that
$\|T_{1}-T\|_{H^{1}(\mathbb{R}_{+})}\leq C_{2}\varepsilon$ for some constant
$C_{2}>0$, where $T$ satisfies (5.1) with $A_{0}<1/2$.

Moreover, for a smaller $A_{0}$, $\varepsilon$ could be larger in order for
$4A_{0}^{2}+4C_{1}\varepsilon<1$ to hold. A smaller $A_{0}$ implies a smaller
$M$ in (1.9). Therefore, it is expected that a smaller $M$ in the spectral
assumption is more stable with respect to perturbations. We can also show when
$\varepsilon$ is large, the spectral assumption does not hold. Assume
$\varepsilon$ is large, then the dominant term is

$\displaystyle\int_{r}^{\infty}e^{2\beta
x}36h|\partial_{x}h|^{2}dx\cdot\int_{0}^{r}\frac{1}{4h^{3}e^{2\beta x}}dx,$
(5.25)

which can become large for $h\in L^{2}_{\rm loc}(\mathbb{R}_{+})$. For
example, we may take $\beta=0$, $h(x)=xe^{-x}$, the integral
$\int_{r}^{\infty}36h|\partial_{x}h|^{2}dx$ is finite positive while
$\int_{0}^{r}\frac{1}{4h^{3}}dx=\int_{0}^{r}\frac{e^{3x}}{4x^{3}}dx$ diverges.
Therefore, when $\varepsilon\to\infty$, $(A^{\prime})^{2}\to\infty$ and
$C_{hd}^{\prime}\geq(A^{\prime})^{2}$ also goes to $\infty$. Hence Lemma 8
only holds for $\varepsilon$ sufficiently small. ∎

We proceed to prove Theorem 3.

###### Proof of Theorem 3: The case of small perturbations..

Suppose $T,T_{1}$ are two functions satisfying the assumptions of Theorem 3
and $T_{1}\in\mathcal{V}$ with $\mathcal{V}$ given in (3). Setting
$h=(T_{1}-T)/\varepsilon$, then $w:=(T_{1}-T^{4})/(T_{1}-T)$ is

$\displaystyle w$ $\displaystyle=\frac{(T+\varepsilon
h)^{4}-T^{4}}{\varepsilon h}=4T^{3}+6\varepsilon
T^{2}h+4\varepsilon^{2}h^{2}T+\varepsilon^{3}h^{3}$

is a small perturbation of $4T^{3}$, thus by the proof of Lemma 8 and Lemma 4,
(4.10) still holds with $4T^{3}$ replaced by $w$. Thus by Theorem 2, $T_{1}=T$
and the solution is unique in $\mathcal{V}$. This finishes the proof of
Theorem 3.

∎

##  Appendix A Proof of Theorem 2 for the general case

Consider

$\displaystyle\partial_{x}^{2}g+\langle\phi-4T^{3}g\rangle=\langle
S_{1}\rangle,$ (A.1) $\displaystyle\mu\partial_{x}\phi+(\phi-4T^{3}g)=S_{2},$
(A.2)

supplemented with the following boundary conditions

$\displaystyle g(0)=g_{b},$ (A.3)
$\displaystyle\phi(0,\mu)=\phi_{b}(\mu),\text{ for any }\mu\in(0,1],$ (A.4)

Then the following theorem holds.

######  Theorem 5.

Suppose $T$ satisfies the spectral assumption (A) for some constants
$0<\beta<1$ and $M>0$. Assume $S_{1}=S_{1}(x,\beta),S_{2}=S_{2}(x,\beta)$
decays to zero exponentially such that $\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+}\times[-1,1])},\|e^{\beta
x}S_{1}\|_{L^{2}(\mathbb{R}_{+}\times[-1,1])}<\infty$. Then there exists a
unique bounded solution $\left(g,\phi\right)\in L^{2}_{\rm
loc}(\mathbb{R}_{+})\times L^{2}_{\rm loc}(\mathbb{R}_{+}\times[-1,1])$ to
system (A.1)-(A.2) with boundary conditions (A.3)-(A.4), and the solution
satisfies

$\displaystyle|g(x)-g_{\infty}|\leq Ce^{-\beta
x},\quad|\phi(x,\mu)-4T_{\infty}^{3}g_{\infty}|\leq Ce^{-\beta x},$ (A.5)

for some constant $C>0$ and $g_{\infty}\in\mathbb{R}$, where
$T_{\infty}=\lim_{x\to\infty}T(x)$.

###### Proof.

First we can take $h=g-g_{b}e^{-x}$ and system (A.1)-(A.2) becomes

$\displaystyle\partial_{x}^{2}h+\langle\phi-4T^{3}g\rangle=\langle