On Thermal Boundary Layer in a Viscous Non-Newtonian Medium

Abstract

An existence and uniqueness theorem for a classical solution to the system of equations describing thermal boundary layers in viscous media with the Ladyzhenskaya rheological law is generalized.

INTRODUCTION

If the surface of a plate and the incoming flow have different temperatures, then, near the plate surface, a thermal boundary layer is formed in which the temperature varies from its value at the wall to the free-stream temperature at the edge of the thermal boundary layer. Moreover, heat transfer occurs between the plate surface and the fluid flow (see [1]).

Thus, near the surface of a body placed in fluid flow, dynamic and thermal boundary layers are formed, which represent the boundaries of the corresponding perturbation fronts separating the perturbed and unperturbed flows.

A system of equations governing a thermal boundary layer developing in a plane-parallel steady forced convective flow of a Newtonian fluid was considered in [2].

In this paper, we study a system of equations describing thermal boundary layers in plane-parallel fluid flows with Ladyzhenskaya rheology (see Fig. 1¹). A thermal boundary layer is assumed to develop at the interface between two fluids or a fluid and a gas (Marangoni boundary layer).

Fig. 1.

Plane-parallel thermal boundary layer with convection.

1 FORMULATION OF THE PROBLEM

Consider the system of equations governing the thermal boundary layer developing in a plane-parallel fluid flow with Ladyzhenskaya rheological law:

$$\begin{gathered} \nu \left( {1 + 3k{{{\left( {\frac{{\partial u}}{{\partial y}}} \right)}}^{2}}} \right)\frac{{{{\partial }^{2}}u}}{{\partial {{y}^{2}}}} - u\frac{{\partial u}}{{\partial x}} - v\frac{{\partial u}}{{\partial y}} = - U{{U}_{x}}, \\ \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0, \\ \end{gathered} $$

(1)

$$a\frac{{{{\partial }^{2}}T}}{{\partial {{y}^{2}}}} - u\frac{{\partial T}}{{\partial x}} - v\frac{{\partial T}}{{\partial y}} = - \frac{\nu }{c}{{\left( {\frac{{\partial u}}{{\partial y}}} \right)}^{2}}.$$

(2)

Assume that system (1), (2) is defined in the domain

$$Q = \{ (x,y) \in {{\mathbb{R}}^{2}}{\text{:}}\,0 < x < X,0 < y < \infty \} ,\quad X > 0.$$

System (1), (2) is supplemented with boundary conditions of the form

$$\begin{gathered} {{\left. {\frac{{\partial u}}{{\partial y}}} \right|}_{{y = 0}}} = \widehat A(x),\quad u{{{\text{|}}}_{{x = 0}}} = {{u}_{0}}(y),\quad v{{{\text{|}}}_{{y = 0}}} = {{v}_{0}}(x), \\ u(x,y) \to U(x) \\ {\text{uniformly with respect to}}\;\;x \in [0,X]\;\;{\text{as}}\;\;y \to + \infty , \\ \end{gathered} $$

(3)

$$\begin{gathered} T{{{\text{|}}}_{{y = 0}}} = {{T}_{w}}(x),\quad T(x,y) \to {{T}_{\infty }} \\ {\text{uniformly with respect to}}\;\;x \in [0,X]\;\;{\text{as}}\;\;y \to + \infty . \\ \end{gathered} $$

(4)

The unknowns in problem (1)–(4) are the streamwise and spanwise velocities $u(x,y)$, $v(x,y)$ of the flow at the point (x, y) and the temperature $T(x,y)$ of the medium at this point. The constants $\nu$, a, and c are physical parameters of the considered fluid, which are assumed to be prescribed. The constant ${{T}_{\infty }}$ is the free-stream temperature. The functions $U(x)$, $\widehat A(x)$, ${{v}_{0}}(x)$, and ${{T}_{w}}(x)$ are also assumed to be prescribed; they denote the free-stream velocity, surface tension at the boundary $\{ y = 0\}$, the blowing (suction) rate at the point x on the lower wall of the domain, and the wall temperature at the point x, respectively.

Since the functions $u(x,y)$ and $v(x,y)$ in problem (1), (3) do not depend on the temperature $T(x,y)$, problem (1), (3) can be solved separately.

A uniqueness theorem for the solution of problem (1), (3) was proved in [3]. Assume that the functions $u$ and ${v}$ satisfy Eqs. (1) in the domain

$$Q = \{ 0 < x < X,0 < y < \infty \} ,$$

are continuous in $\overline Q ,$ and obey conditions (3); additionally, suppose that

$$0 < u < {{C}_{1}},\quad \psi > 0,$$

$${{C}_{2}}y \leqslant u \leqslant {{C}_{3}}y,\quad 0 < y < {{y}_{0}},$$

$$\frac{{{{\partial }^{2}}u}}{{\partial {{y}^{2}}}} \leqslant {{C}_{4}},\quad (x,y) \in Q,$$

where ${{C}_{1}}$, ${{C}_{2}}$, ${{C}_{3}}$, ${{C}_{4}}$, and y₀ are positive constants. Then $u$, ${v}$ are a unique solution of problem (1), (3) with the indicated properties.

2 EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A THERMAL BOUNDARY LAYER

In this section, taking into account the conditions imposed on the functions $u(x,y)$ and $v(x,y)$, we prove the existence and uniqueness of a classical solution $T(x,y) \in {{C}^{2}}(Q) \cup C(\overline Q )$ to problem (2), (4). Without loss of generality, the coefficient a is assumed to be equal to 1. For convenience, the right-hand side of Eq. (2) is denoted by $f(x,y)$.

Theorem 1. Suppose that functions $u(x,y),\,\,v(x,y) \in C(\overline Q )$ satisfy system (1), conditions (3), and the solution uniqueness conditions. Assume that $f(x,y) \in C({{Q}_{X}})$ is a locally Hölder continuous function in $Q$ with a Hölder exponent $0 < \gamma \leqslant 1$, while ${{T}_{w}}$ is a differentiable function on $[0,X]$ and its derivative is bounded on $[0,X]$. Additionally, suppose that

$$\int\limits_0^{ + \infty } \exp \left( {\int\limits_0^t \beta (\tau )d\tau } \right)dt < \infty ,$$

where $\beta (y) = \mathop {\max }\limits_{0 \leqslant x \leqslant X} v(x,y)$, and the function $g(y) = \mathop {\max }\limits_{0 \leqslant x \leqslant X} {\text{|}}\,f(x,y){\text{|}}$ satisfies

$$\int\limits_0^{ + \infty } g(\tau )\exp \left( { - \int\limits_0^\tau \beta (s)ds} \right)d\tau < \infty .$$

Then problem (2), (4) has a unique bounded classical solution $T(x,y)$ in Q.

Proof. By using the change of variables

$$x = x,\quad \eta = 1 - \frac{1}{{1 + y}},$$

(5)

the domain $Q = \{ (x,y) \in {{\mathbb{R}}^{2}}{\text{:}}\,\,0 < x < X,\,\,0 < y < \infty \}$ is transformed into the open rectangle $D = \{ (x,\eta ) \in {{\mathbb{R}}^{2}}{\text{:}}\,\,0 < x < X,\,\,0 < \eta < 1\}$. To rewrite problem (2), (4) in the new variables (see (5)), we consider the partial derivatives involved in Eq. (2):

$$\frac{{\partial \eta }}{{\partial y}} = \frac{1}{{{{{(1 + y)}}^{2}}}} = (1 - \eta {{)}^{2}},$$

$$\frac{{{{\partial }^{2}}\eta }}{{\partial {{y}^{2}}}} = - \frac{2}{{{{{(1 + y)}}^{3}}}} = - 2(1 - \eta {{)}^{3}},$$

$$\frac{{\partial T}}{{\partial y}} = \frac{{\partial T}}{{\partial \eta }}\frac{{\partial \eta }}{{\partial y}} = (1 - \eta {{)}^{2}}\frac{{\partial T}}{{\partial \eta }},$$

$$\begin{gathered} \frac{{{{\partial }^{2}}T}}{{\partial {{y}^{2}}}} = {{\left( {\frac{{\partial \eta }}{{\partial y}}} \right)}^{2}}\frac{{{{\partial }^{2}}T}}{{\partial {{\eta }^{2}}}} + \frac{{{{\partial }^{2}}\eta }}{{\partial {{y}^{2}}}}\frac{{\partial T}}{{\partial \eta }} \\ \, = {{(1 - \eta )}^{4}}\frac{{{{\partial }^{2}}T}}{{\partial {{\eta }^{2}}}} - 2(1 - \eta {{)}^{3}}\frac{{\partial T}}{{\partial \eta }}. \\ \end{gathered} $$

In what follows, we assume that $u = u(x,y(\eta ))$, $v = v(x,y(\eta ))$, and $f = f(x,y(\eta ))$. Taking into account the above expressions for the derivatives and the change of variables (5), Eq. (2) becomes

$$\begin{gathered} L(T) \equiv \left( {{{{(1 - \eta )}}^{4}}\frac{{{{\partial }^{2}}T}}{{\partial {{\eta }^{2}}}} - 2(1 - \eta {{)}^{3}}\frac{{\partial T}}{{\partial \eta }}} \right) \\ \, - u\frac{{\partial T}}{{\partial x}} - v{{(1 - \eta )}^{2}}\frac{{\partial T}}{{\partial \eta }} = f. \\ \end{gathered} $$

Grouping terms in this equation, we finally represent the problem in terms of variables (5):

$$\begin{array}{*{20}{l}} {L(T) \equiv c(\eta )\frac{{{{\partial }^{2}}T}}{{\partial {{\eta }^{2}}}} - u\frac{{\partial T}}{{\partial x}} + b(x,\eta )\frac{{\partial T}}{{\partial \eta }} = f,} \end{array}$$

(6)

$$T{{{\text{|}}}_{{\eta = 0}}} = {{T}_{w}},\quad T{{{\text{|}}}_{{\eta = 1}}} = {{T}_{\infty }}.$$

(7)

Here,

$$c(\eta ) = (1 - \eta {{)}^{4}},\quad b(x,\eta ) = - {{(1 - \eta )}^{2}}(2(1 - \eta ) + v).$$

Let us prove the uniqueness of a solution to problem (6), (7). Suppose that ${{T}_{1}}(x,\eta )$ and ${{T}_{2}}(x,\eta )$ are two solutions of problem (6), (7). Their difference is denoted by $T(x,\eta ) = {{T}_{1}}(x,\eta ) - {{T}_{2}}(x,\eta )$.

For an arbitrary $\varepsilon > 0$, we choose $\delta = \delta (\varepsilon ) > 0$ such that ${\text{|}}T(x,\eta ){\text{|}} \leqslant \varepsilon$ for $x \in [0,X]$ and $\eta = 1 - \delta (\varepsilon )$. Then the function $V(x,\eta ) = T(x,\eta ) - \varepsilon$ satisfies the equation $L(V) = 0$ in D and the inequality $V(x,\eta ) \leqslant 0$ for $x \in [0,X]$, $\eta = 0$, and $\eta = 1 - \delta (\varepsilon )$. Let us show that $V \leqslant 0$ in the rectangle

$${{D}_{\varepsilon }} = \{ (x,\eta ) \in {{\mathbb{R}}^{2}}{\text{:}}\,\,0 < x < X,\,\,0 < \eta < 1 - \delta (\varepsilon )\} .$$

Let

$$V(x,\eta ) = H(\eta )R(x,\eta ),$$

where $H(\eta )$ is assumed to be a positive function from ${{C}^{1}}([0,1 - \delta (\varepsilon )])$. Assume that $H = H(\eta )$ and $R = R(x,\eta )$. Consider the operator L(V). The partial derivatives involved in the equation $L(v) = 0$ can be written as

$$\frac{{\partial V}}{{\partial \eta }} = H{\kern 1pt} 'R + H\frac{{\partial R}}{{\partial \eta }},$$

$$\frac{{{{\partial }^{2}}V}}{{\partial {{\eta }^{2}}}} = H{\kern 1pt} '{\kern 1pt} 'R + 2H{\kern 1pt} '\frac{{\partial R}}{{\partial \eta }} + H\frac{{{{\partial }^{2}}R}}{{\partial {{\eta }^{2}}}},$$

$$\frac{{\partial V}}{{\partial x}} = H\frac{{\partial R}}{{\partial x}}.$$

Taking into account these expressions for the derivatives, the operator L(V) is given by

$$\begin{array}{*{20}{l}} \begin{gathered} L(V) \equiv c\left( {H{\kern 1pt} '{\kern 1pt} 'R + 2H{\kern 1pt} '\frac{{\partial R}}{{\partial \eta }} + H\frac{{{{\partial }^{2}}R}}{{\partial {{\eta }^{2}}}}} \right) \\ \, - uH\frac{{\partial R}}{{\partial x}} + b\left( {H{\kern 1pt} 'R + H\frac{{\partial R}}{{\partial \eta }}} \right) = 0. \\ \end{gathered} \end{array}$$

Transforming terms in this equation yields

$$\begin{array}{*{20}{l}} \begin{gathered} L(V) \equiv cH\frac{{{{\partial }^{2}}R}}{{\partial {{\eta }^{2}}}} + (2cH{\kern 1pt} '\; + bH)\frac{{\partial R}}{{\partial \eta }} \\ \, - uH\frac{{\partial R}}{{\partial x}} + (bH{\kern 1pt} '\; + cH{\kern 1pt} '{\kern 1pt} ')R = 0. \\ \end{gathered} \end{array}$$

Dividing both sides of the equation $L(V) = 0$ by the function $H(\eta )$, we derive the equation

$$\begin{array}{*{20}{l}} {L(R) \equiv c\frac{{{{\partial }^{2}}R}}{{\partial {{\eta }^{2}}}} - u\frac{{\partial R}}{{\partial x}} + B\frac{{\partial R}}{{\partial \eta }} + FR = 0,} \end{array}$$

where

$$B(x,\eta ) = \frac{{2cH{\kern 1pt} '}}{H} + b,\quad F(x,\eta ) = \frac{{bH{\kern 1pt} '\; + cH{\kern 1pt} '{\kern 1pt} '}}{H}.$$

The function $H(\eta ) > 0$ is chosen so that, for some constant ${{F}_{0}} < 0$, the inequality $F(x,\eta ) \leqslant {{F}_{0}}$ holds for all $\eta \in [0,1 - \delta (\varepsilon )]$.

Next, we introduce the function

$$\Phi (x) = \int\limits_x^X \frac{{dt}}{{U(t)}},\quad 0 < x < X.$$

Then, taking into account the form of $\Phi (x)$ and the conditions imposed on $u(x,y(\eta ))$, we have in ${{D}_{\varepsilon }}$ the inequality

$$L(\Phi ) = F(x,\eta )\Phi (x) + \frac{{u(x,y(\eta ))}}{{U(x)}} \leqslant {{F}_{0}}\Phi (x) + 1.$$

Since ${{F}_{0}} < 0$ and $\Phi (x) \to + \infty$ as $x \to 0 + 0$, it is possible to find a number $0 < {{x}_{0}} < X$ such that $L(\Phi ) \leqslant 0$ for all $0 < x \leqslant {{x}_{0}}$ and $0 < \eta < 1 - \delta (\varepsilon )$. Then, for any $\lambda > 0,$ the linear operator L satisfies the inequality

$$\begin{array}{*{20}{l}} \begin{gathered} L(\lambda \Phi (x) - R(x,\eta )) = L(\lambda \Phi (x)) - L(R(x,\eta )) \leqslant 0, \\ 0 < x \leqslant {{x}_{0}},\quad 0 < \eta < 1 - \delta (\varepsilon ). \\ \end{gathered} \end{array}$$

Furthermore, for each $\lambda > 0,$ there exists a number $0 < {{x}_{1}}(\lambda ) < {{x}_{0}}$ such that $\lambda \Phi (x) - R(x,\eta ) \geqslant 0$ for $0 \leqslant \eta \leqslant 1 - \delta (\varepsilon )$, $0 < x \leqslant {{x}_{1}}$. Additionally, $\lambda \Phi (x) - R(x,\eta ) \geqslant 0$ for $\eta = 0$, $0 < x \leqslant X$ and for $\eta = 1 - \delta (\varepsilon )$, $0 < x \leqslant X$. By the maximum principle, $\lambda \Phi (x) - R(x,\eta ) \geqslant 0$ for $0 < x \leqslant {{x}_{0}}$, $0 \leqslant \eta \leqslant 1 - \delta (\varepsilon )$. Since $\lambda > 0$ is arbitrary, we obtain $R(x,\eta ) \leqslant 0$ for $0 \leqslant x \leqslant {{x}_{0}}$, $0 \leqslant \eta \leqslant 1 - \delta (\varepsilon )$. Once again applying the maximum principle yields $R(x,\eta ) \leqslant 0$ in $\overline {{{D}_{\varepsilon }}}$. Therefore, $T(x,\eta ) \leqslant \varepsilon$ in $\overline {{{D}_{\varepsilon }}}$. In view of the symmetry between ${{T}_{1}}$ and ${{T}_{2}}$, we obtain ${\text{|}}T(x,\eta ){\text{|}} \leqslant \varepsilon$ in $\overline {{{D}_{\varepsilon }}}$. Since $\varepsilon > 0$ is arbitrary, it follows that ${{T}_{1}} = {{T}_{2}}$ in $\overline D$. The uniqueness of a solution is established.

Now we prove the existence of a solution to problem (6), (7). For each $0 < \delta < \frac{1}{4}$, Eq. (6) is considered in the rectangle

$${{D}_{\delta }} = \{ (x,\eta ) \in {{\mathbb{R}}^{2}}{\kern 1pt} :\;\delta < x \leqslant X,\,\,\delta < \eta < 1 - \delta \} $$

with boundary conditions

$$T{{{\text{|}}}_{{\eta = \delta }}} = {{T}_{w}}(x),\quad T{{{\text{|}}}_{{\eta = 1 - \delta }}} = {{T}_{\infty }},\quad T{{{\text{|}}}_{{x = \delta }}} = T_{1}^{\delta }(\eta ),$$

(8)

where $T_{1}^{\delta }(\eta )$ is a continuous function on the interval $[\delta ,1 - \delta ]$ having the properties

$$T_{1}^{\delta }(\eta ) = {{T}_{w}}(\delta )\;\;{\text{if}}\;\;\eta \leqslant \frac{1}{3},$$

$$T_{1}^{\delta }(\eta ) = {{T}_{\infty }}\;\;{\text{if}}\;\;{\kern 1pt} \eta \geqslant \frac{2}{3},$$

$$\begin{gathered} {\text{|}}T_{1}^{\delta }(\eta ){\text{|}} \leqslant M = \max \{ {\text{|}}{{T}_{\infty }}{\text{|}},\mathop {\max }\limits_{0 \leqslant x \leqslant X} {\text{|}}{{T}_{w}}(x){\text{|}}\} \\ {\text{for}}\;\;\delta \leqslant \eta \leqslant 1 - \delta . \\ \end{gathered} $$

The solution ${{T}^{\delta }}(x,\eta )$ of problem (6), (8) exists and, according to the maximum principle for nondegenerate parabolic equations (see, e.g., [4]),

$${\text{|}}{{T}^{\delta }}(x,\eta ){\text{|}} \leqslant M,$$

(9)

where M does not depend on $\delta$. By virtue of Schauder-type estimates [4], in any fixed rectangle ${{D}_{\delta }}$ for each $0 < \delta < \frac{1}{4}$, the Hölder norms of the solution T ^δ and its derivatives $\frac{{\partial {{T}^{\delta }}}}{{\partial \eta }}$, $\frac{{{{\partial }^{2}}{{T}^{\delta }}}}{{\partial {{\eta }^{2}}}}$, and $\frac{{\partial {{T}^{\delta }}}}{{\partial x}}$ satisfy the estimate

$${{\left\| {{{T}^{\delta }}} \right\|}_{{2 + \alpha }}} \leqslant \overline K ({{\left\| \psi \right\|}_{{2 + \alpha }}} + {{\left\| f \right\|}_{\alpha }}),\quad \overline K = {\text{const}},$$

(10)

where, for an arbitrary Holder continuous function $\varphi (x,\eta ),$ the norms ${\text{||}}\varphi {\text{|}}{{{\text{|}}}_{\alpha }}$ and ${\text{||}}\varphi {\text{|}}{{{\text{|}}}_{{2 + \alpha }}}$ in (10) are defi-ned as

$$\begin{array}{*{20}{l}} \begin{gathered} {{\left\| \varphi \right\|}_{\alpha }} = \mathop {\sup }\limits_{(x,\eta ) \in {{D}_{\delta }}} \left| \varphi \right| + \mathop {\sup }\limits_{{{P}_{1}},{{P}_{2}} \in {{D}_{\delta }},{{P}_{1}} \ne {{P}_{2}}} \frac{{\left| {\varphi ({{P}_{1}}) - \varphi ({{P}_{2}})} \right|}}{{{{{\left| {{{P}_{1}} - {{P}_{2}}} \right|}}^{\alpha }}}}, \\ {{P}_{1}} = (x,\eta ),{{P}_{2}} = (\bar {x},\bar {\eta }), \\ \end{gathered} \\ {{{{\left\| \varphi \right\|}}_{{2 + \alpha }}} = {{{\left\| \varphi \right\|}}_{\alpha }} + {{{\left\| {\frac{{\partial \varphi }}{{\partial x}}} \right\|}}_{\alpha }} + {{{\left\| {\frac{{\partial \varphi }}{{\partial \eta }}} \right\|}}_{\alpha }} + {{{\left\| {\frac{{{{\partial }^{2}}\varphi }}{{\partial {{\eta }^{2}}}}} \right\|}}_{\alpha }}} \end{array}$$

(see, e.g., [4]). The function $\psi$ in (10) denotes boundary conditions (8).

Relying on estimates (10), we extract a subsequence ${{T}^{{{{\delta }_{m}}}}}$, $m = 1,2,...$, that converges uniformly as $m \to \infty$, together with the derivatives involved in Eq. (6), in each closed domain lying strictly inside D. Passing to the limit as $m \to \infty$ in the equation for ${{T}^{{{{\delta }_{m}}}}}$, we conclude that the limit function $T(x,\eta )$ satisfies Eq. (6) in the rectangle D.

To prove the fulfillment of the condition $T(x,0) = {{T}_{w}}(x)$, we estimate the difference

$${{S}^{\delta }}(x,\eta ) = {{T}^{\delta }}(x,\eta ) - {{T}_{w}}(x)$$

in the case of small $\eta$. In the domain

$$D_{\delta }^{'} = \left\{ {\delta \leqslant x \leqslant X,\,\,\delta < \eta < \frac{1}{3}} \right\},$$

the function ${{S}^{\delta }}(x,\eta )$ satisfies the equation

$$L({{S}^{\delta }}) = f(x,\eta ) + u(x,\eta ){{T}_{{w'}}}(x).$$

Suppose that

$${\text{|}}\,f(x,\eta )\,{\text{|}} \leqslant {{M}_{1}},\quad u(x,\eta )\,{\text{|}}{{T}_{{w{\kern 1pt} '}}}(x){\text{|}} \leqslant {{M}_{2}}$$

in $D_{\delta }^{'}$.

Next, we introduce the auxiliary function Y(η) = $K(1 - {{e}^{{ - N\eta }}})$. Here, the constant N > 0 is chosen using the condition $c(\eta )N \geqslant {\text{|}}b(x,\eta ){\text{|}} + 1$. This is possible, since the coefficient $v(x,y)$ is bounded for bounded y or, according to (5), for $\eta \leqslant {{\eta }_{0}} < 1$. The constant K > 0 is chosen so that

$$K \geqslant \max \left\{ {\frac{{2M}}{{1 - {{e}^{{\frac{{ - N}}{3}}}}}},\,\,\frac{{{{M}_{1}} + {{M}_{2}}}}{{N{{e}^{{ - N}}}}}} \right\},$$

(11)

where M is the same as in inequality (9). Given $N > 0$, we calculate $L(Y)$, namely,

$$L(Y) = - KN(c(\eta )N\, - \,b(x,\eta )){{e}^{{ - N\eta }}} \leqslant - KN{{e}^{{ - N}}}\, < \,\,0.$$

(12)

Consider the function $Y \pm {{S}^{\delta }}(x,\eta )$. Taking into account (11) and (12), $L(Y \pm {{S}^{\delta }})$ in $D_{\delta }^{'}$ is estimated as

$$\begin{array}{*{20}{l}} \begin{gathered} L(Y \pm {{S}^{\delta }}) \leqslant - KN{{e}^{{ - N}}} \pm f(x,\eta ) \pm u(x,\eta ){{T}_{{w{\kern 1pt} '}}}(x) \\ \, \leqslant - KN{{e}^{{ - N}}} + {{M}_{1}} + {{M}_{2}} \leqslant 0. \\ \end{gathered} \end{array}$$

(13)

Combining boundary conditions (8) with inequalities (9) and (13), we conclude that $Y \pm {{S}^{\delta }} \geqslant 0$ on the boundary of $D_{\delta }^{'}$ lying on the straight lines $x = \delta$, $\eta = \delta$, and $\eta = \frac{1}{3}$. According to the maximum principle, it follows that $Y \pm {{S}^{\delta }} \geqslant 0$ everywhere in $D_{\delta }^{'}$. Therefore,

$${\text{|}}{{S}^{\delta }}(x,\eta ){\text{|}} \leqslant Y(\eta ),$$

which is uniform with respect to $\delta$ and x. Passing to the limit as $\delta \to 0$ and $\eta \to 0$ yields the condition $T(x,0) = {{T}_{w}}(x)$.

To prove the fulfillment of the second boundary condition in (7), we estimate the difference

$${{J}^{\delta }}(x,\eta ) = {{T}^{\delta }}(x,\eta ) - {{T}_{\infty }}$$

for small $1 - \eta$. The equation $L({{J}^{\delta }}) = f(x,\eta )$ satisfied by the function ${{J}^{\delta }}(x,\eta )$ is considered in the domain

$$D_{\delta }^{{''}} = \left\{ {\delta < x \leqslant X,\,\,\frac{2}{3} \leqslant \eta < 1 - \delta } \right\}.$$

Next, we introduce the new function

$$Z(\eta ) = {{K}_{1}}\int\limits_\eta ^1 G(t){{G}_{1}}(t)dt,$$

(14)

where

$$\begin{gathered} G(t) = \frac{1}{{{{{(1 - t)}}^{2}}}}\exp \left\{ {\int\limits_0^t \frac{{\beta (\tau )d\tau }}{{{{{(1 - \tau )}}^{2}}}}} \right\}, \\ {{G}_{1}}(t) = 1 + \int\limits_0^t \frac{{g(\tau )d\tau }}{{{{{(1 - \tau )}}^{4}}G(\tau )}}, \\ \end{gathered} $$

and the functions $\beta (\tau )$, $g(\tau )$ are the same as in Theorem 1. Integral (14) exists by virtue of the assumptions of the theorem and defines a positive function for $\eta < 1$. The constant ${{K}_{1}} \geqslant 1$ is determined by the condition $Z\left( {\frac{2}{3}} \right) \geqslant 2M$, where M is the constant from inequality (9). The function $L(Z)$ is calculated as

$$L(Z)\, = \, - {{K}_{1}}{{G}_{1}}(\eta )[c(\eta )G{\kern 1pt} '(\eta )\, + \,b(x,\eta )G(\eta )]\, - \,{{K}_{1}}g(\eta ).$$

(15)

Since

$$\begin{gathered} c(\eta )G{\kern 1pt} '(\eta ) + b(x,\eta )G(\eta ) \\ \, = \exp \left\{ {\int\limits_0^\eta \frac{{\beta (\tau )d\tau }}{{{{{(1 - \tau )}}^{2}}}}} \right\}[\beta (\eta ) - v(x,\eta )] \geqslant 0 \\ \end{gathered} $$

by the definition of the function $\beta (\eta )$, it follows from (15) that $L(Z) \leqslant - {{K}_{1}}g(\eta ).$ On the boundary of $D_{\delta }^{{''}}$ lying on the straight lines $x = \delta$, $\eta = \frac{2}{3}$, and $\eta = 1 - \delta$, it is true that $Z(\eta ) \pm {{J}^{\delta }}(x,\eta ) \geqslant 0.$ Inside $D_{\delta }^{{''}}$, by virtue of the choice of ${{K}_{1}} \geqslant 1$, we have

$$L(Z \pm {{J}^{\delta }}) \leqslant - {{K}_{1}}g(\eta ) \pm f(x,\eta ) \leqslant - {{K}_{1}}g(\eta ) + g(\eta ) \leqslant 0.$$

Therefore, the maximum principle implies that $Z \pm {{J}^{\delta }} \geqslant 0$ everywhere in $D_{\delta }^{{''}}$, or ${\text{|}}{{J}^{\delta }}(x,\eta ){\text{|}} \leqslant Z(\eta )$ in this domain. Passing to the limit as $\delta \to 0$ in the last inequality, we conclude that $T(x,\eta ) \to {{T}_{\infty }}$ uniformly as $\eta \to 1$, as required.

Remark 1. In problem (2), (4), the free-stream temperature ${{T}_{\infty }}$ is assumed to be constant. It turns out that, for a solution $T(x,y)$ of Eq. (2) that is bounded in the strip Q, a value other than a constant cannot be specified at infinity.

Proof. Assume that the solution $T(x,y)$ of problem (2), (4) is such that $\mathop {\lim }\limits_{y \to \infty } T(x,y) = {{T}_{\infty }}(x)$. In view of substitution (5), this is equivalent to the condition $T(x,\eta ){{{\text{|}}}_{{\eta = 1}}}$ = T_∞(x). For any $\varepsilon > 0$, there exists $\delta (\varepsilon )$ such that

$${\text{|}}T(x,1 - \delta ) - {{T}_{\infty }}(x){\text{|}} < \varepsilon .$$

Let ${{J}^{\delta }}(x,\eta ) = {{T}^{\delta }}(x,\eta ) - {{T}_{\infty }}(\delta )$. Equation (6) for the function ${{J}^{\delta }}(x,\eta )$ is considered in the domain

$$D_{\delta }^{{''}} = \left\{ {\delta < x \leqslant X,\,\,\frac{2}{3} \leqslant \eta < 1 - \delta (\varepsilon )} \right\}.$$

Consider the new function $Z(\eta ) + \varepsilon \pm {{J}^{\delta }}(x,\eta )$, where $Z(\eta )$ is defined by equality (14). Repeating word for word the entire argument used to prove the fulfillment of the second condition in (4), we obtain an estimate in $\overline D _{\delta }^{{''}}$ that is uniform with respect to $\delta$ and $x$, namely,

$${\text{|}}{{J}^{\delta }}(x,\eta ){\text{|}} \leqslant \varepsilon + Z(\eta ).$$

Sending $\delta$ to zero and noting that $Z(\eta ) \to 0$ as $\eta \to 1$, we have

$${\text{|}}T(x,1) - {{T}_{\infty }}(0){\text{|}} \leqslant \varepsilon .$$

Since $\varepsilon$ is arbitrary, it follows that $T(x,1) \equiv {{T}_{\infty }}(0)$, i.e., ${{T}_{\infty }}(x) \equiv {{T}_{\infty }}(0) = {\text{const}}$.

Remark 2. If only the first condition is set in (4), while the second condition is dropped, then a bounded solution satisfying Eq. (2) and the boundary condition $T{{{\text{|}}}_{{y = 0}}} = {{T}_{0}}(x)$, where ${{T}_{0}}(x)$ is a continuous function defined in $0 \leqslant x \leqslant X$, may not be unique.

Proof. Equation (2) with $a = 1$, $u = {{x}^{k}}$, $v = \lambda y$, and $f \equiv 0$ $(k \geqslant 1,{\text{ }}\lambda = {\text{const}} < 0)$ and the boundary condition $T{{{\text{|}}}_{{y = 0}}} = 0$ are satisfied by the function

$$T(y) = \int\limits_0^y {{e}^{{\frac{{\lambda {{t}^{2}}}}{2}}}}dt,$$

which is bounded and nonzero in the strip Q. It follows that, under the considered conditions, the solution of problem (2), (4) is not unique.