数理解析研究所講究録
第 1985 巻 2016 年 96-115

96

BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW IN
HALF SPACECHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

1. INTRODUCTION
In this paper, we consider the asymptotic behavior of inviscid compressible non-isentropic flow in half$d=2$ , 3, as the heat-conductivity tends to zero. As Prandtl found in his pioneering work [30],space ofaway from the boundary the flow is mainly driven by the inertia while the friction force is negligible, where
the flow can be described approximately as inviscid, and the friction force plays a key role in determining
the behavior of flow near physical boundaries, in which it generates large vortices. On the other hand, as
observed by many physicians and mechanicians, cf. [31, 32, 28, 2, 17, 37], the heat conduction also has large
gradient near physical boundaries, thus it is challenging and very important to analyse the flow behavior
near boundaries. In our case, the boundary layers can be describe by the following initial-boundary value
problem in $\{(t, x’, y) : t>0, x’\in \mathbb{R}^{d-1}, y\in \mathbb{R}_{+}\}$ :$\mathbb{R}^{d},$(1.1)$\{\begin{array}{l}\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}=0,\partial_{t}\theta+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})\theta=\frac{\kappa}{P}\theta\partial_{y}^{2}\theta+\kappa_{\vec{P}}P\theta,\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\frac{\kappa}{P}\partial_{y}^{2}\theta-\frac{(1-\kappa)P_{t}}{P},(u_{d},\theta)|_{y=0}(0,\theta^{0}(t, x , \lim_{yarrow+\infty}\theta(t, x, y)=\Theta(t, x(u_{h}, \theta)|_{t=0}=(u_{h0},\theta_{0})(x’,y) ,\end{array}$where $P=P(t)$ andare positive known functions, and $\kappa>0$ is a constant. One can find thederivation of (1.1) in Appendix. This paper is devote to show the local existence of smooth solution ofboundary layers problem (1.1) without monotonicity condition for initial data, and give a simple solution
which forms a singularity in finite time. The limit of smooth solutions is studied when the temperature tends
to a constant state. Then, we consider the linear stability of the problem (1.1) around a shear flow.
The mathematical theory and stability analysis of boundary layers in fluids are classical and chalenging
problems. The concept of boundary layers was introduced in Prandtl’s seminal work [30] in 1904, and its
importance in mechanics was first studied by Prandtl in [31, 32]. It was also studied throughly by manymechanicians in [2, 17, 28, 35, 37] and references therein. Oleinik and her collaborators in [29], obtained thewell-posedness of the two-dimensional Prandtl boundary layer equations in viscous incompressible flow in theclass of tangential velocity being strictly monotonic with respect to the normal variable by introducing the vonMiss or Crocco transformation. Recently, Oleinik’s well-posedness result was reproved by the energy method$\Theta(t_{\rangle}x’)$

without using the Crocco transformation by [1] and [27] independently. Under the Oleinik monotonicity
assumption and an additional condition of favorite pressure, Xin and Zhang ([43]) obtained a global weaksolution to the two-dimensional Prandtl equations. Without the monotonicity assumption of the velocity
field, the Prandtl equations are ill-posedness in general in the finite order Sobolev spaces of solutions, it isrelated with the separation of boundary layers, it is studied by many mathematicians, cf. [3, 4, 5, 6, 8, 10, 12].
The wel-posedness of the Prandtl equations in the frame of analytic solutions was studied in [16, 23, 33],
and the almost global existence of analytic solutions was obtained recently by [13, 44]. Recently, the authors[20, 21, 22] have studied the stability and instability of the Prandtl equations in three space variables. Amongthe mathematical problems of boundary layers, the convergence from the Navier-Stokes equations to thecomposition of the Euler equations for outflow and equations of boundary layers in the small viscosity li1nit
is almost completely un-known, except some special cases, such as under additional diﬀusion conditions ofenergy [14, 15, 38], for circularly symmetric flows [24] or anisotropic viscous flows [18], in the space of analytic

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CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

solutions [34], under the assumption of support of vortices being away from boundary [25], a steady flow overa moving plate [11], the problems of non-characteristic boundary layers [9, 26], the flows with Navier-slip
boundary conditions [41] and so on. All above theoretical works were focused on the incompressible flow,
for which there are only viscous boundary layers near the boundaries. As studied in [2, 17, 35, 37], thecompressible flow near boundary is much complicated, there are not only viscous layers but also thermal
layers, in which the heat transfers quickly, and there exists interaction between viscous layers and thermal
layers. There are a few works on compressible viscous flows, the linearized compressible Prandtl equations
were studied in [42], the well-posedness of the Prandtl equations in two-dimensional isentropic compressible
flows in the monotonic class of tangential velocity was studied in [39] and [7] independently, and the large
Mach number limit of the compressible Prandtl equations was also considered in [7], the small viscosity limitof the compressible isentropic viscous flow with the Navier-slip condition was obtained in [40], and the Kato
theory of the limit from the Navier-Stokes equations to the Euler equations was extended to the compressible
flow in [36]. These results are only limited to the isentropic flow which avoids the thermal layers, and so far
there are few mathematical theories on the thermal layers. In [19], we have studied the behavior of viscous
layers and thermal layers in nonisentropic compressible circularly symmetric flows with the viscosity and heat
conductivity having the same scale.
2. MAIN RESULTSFirstly, we show the local existence of solutions to problem (1.1). Let$I_{k}$

be the$k\cross k$

identity matrix forsome integer , and denote $det(A)$ by the determinant of the matrix . Then, we improve the method of [12]and have the following local existence result for the above problem (1.1), where no monotonicity condition isrequired on the initial data.
$A$

$k$

Proposition 2.1. Let $u_{h0}(x’, y)$ ,(2.1)$\theta_{0}(x’, y)$,$\theta^{0}(t, x’)$

, $P(t)$ and$\Theta(t, x’)$be smooth functions satisfying that$t^{*}:= \inf\{t:\inf_{(x,y)\in \mathbb{R}_{+}^{d}}det(I_{d-1}+\mathcal{S}\nabla_{h}u_{h0}(x’, y))>0, \forall s\in[0, t]\}>0$

and there exists a positive constant(2.2)$C_{0}$

such that for $t\in[O, t^{*}$ ) and$(x’, y)\in \mathbb{R}_{+}^{d-1},$$\Vert u_{h0}\Vert_{C^{2}}, \Vert\theta_{0}\Vert_{C^{1}}, |P_{t}|\leq C_{0}, C_{0}^{-1}\leq\theta_{0}(x’, y) , \theta^{0}(t, x \Theta(t, x P(t)\leq C_{0}.$Then, there exists a$t_{0}$: $0<t_{0}\leq t^{*}$ and a unique classical solution to (1.1) in$[0, t_{0}$)$\cross \mathbb{R}_{+}^{d}$$u_{h}(t, x’,y)=u_{h0}(\xi(t, x’, \eta(t,x’,y)), \eta(t,x’, y)) , \theta(t,x’, y)=\tilde{\theta}(t,x’, \eta(t, x’, y))$given by,(2.3)
$u_{d}(t, x’,y)= \int_{0}^{\eta(t,x’,y)}\partial_{t}(\frac{\tilde{\theta}}{a})(t,x’, z)dz+\int_{0}^{\eta(t,x’,y)}[b(t, x’, \eta(t,x’, y))\cdot\nabla_{h}(\frac{\tilde{\theta}}{a})(t, x’, z)]dz.$

Here, the vector function(2.4)$\xi(t, x’, z)\in \mathbb{R}^{d-1}$

is the solution of$x’=\xi+tu_{h0}(\xi, z)$ ;the scale function $a(t, x’, z)$ and vector function $b(t, x’, z)$ are given by the initial data and $\xi(t, x’, z)$ :(2.5)

the junction(2.6)$\{\begin{array}{l}a(t,x’, z):=\frac{P(t)}{P(0)}\theta_{0}(\xi(t, x’, z), z)\cdot det(I_{d-1}+t\nabla_{h}u_{h0})(\xi(t, x’, z), z) ,b(t, x’, z):=u_{h0}(\xi(t, x’, z), z) ,\end{array}$$\tilde{\theta}(t, x’, z)\dot{u}$a positive smooth solution to the following problem in $\{0<t\leq t_{0}, z>0, x’\in \mathbb{R}^{d-1}\},$$\{\begin{array}{l}\partial_{t}\tilde{\theta}+b\cdot\nabla_{h}\tilde{\theta}_{\vec{P}}^{\kappa P}-\tilde{\theta}-\frac{\kappa a}{P}\partial_{z}(\frac{a}{\overline{\theta}}\partial_{z}\tilde{\theta})=0,\tilde{\theta}|_{z=0}=\theta^{0}(t, x zarrow+\infty hm\tilde{\theta}=\Theta(t,x\tilde{\theta}|_{t=0}=\theta_{0}(x’, z) ;\end{array}$and then the scale junction $\eta(t, x’, y)$ is the solution of(2.7)$y= \int_{0}^{\eta}\frac{\tilde{\theta}(t,x’,z)}{a(t,x,z)}dz.$

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Proof. We use the method of characteristics to solve the problem (1.1). We assume thatsolution to (1.1), and introduce characteristic coordinates:$(u, v,\theta)$

is a smooth$t=\tau, x’=x’(\tau, \xi, \eta) , y=y(\tau,\xi, \eta)$with$\xi=(\xi_{1}, \cdots, \xi_{d-1})\in \mathbb{R}^{d-1}$(2.8), which satisfies that$\{\begin{array}{l}\partial_{\tau}(x’,y)(\tau, \xi, \eta)=(u_{h}, u_{d})(\tau, x’(\tau, \xi,\eta), y(\tau, \xi, \eta)) ,(x’,y)(0, \xi, \eta)=(\xi, \eta) .\end{array}$Then, let
(2.9)$(\overline{u}_{h},\overline{u}_{d},\overline{\theta})(\tau, \xi, \eta):=(u_{h},u_{d}, \theta)(\tau, x’(\tau, \xi, \eta), y(\tau,\xi, \eta))$,the equations of (1.1) are deduced as(2.10)$\{\begin{array}{l}\partial_{\tau}\overline{u}_{h}=0,\partial_{\tau}\overline{\theta}=\frac{\kappa}{P}\overline{\theta}\partial_{y}^{2}\overline{\theta}+\frac{\kappa P}{P}L\overline{\theta},\end{array}$with the initial data:(2.11)$(\overline{u}_{h},\overline{\theta})|_{\tau=0}=(u_{h0},\theta_{0})(\xi, \eta)$.

Combining (2.10) with (2.11), it is easy to obtain(2.12)$\overline{u}_{h}(\tau, \xi,\eta)\equiv u_{h0}(\xi, \eta)$,

Plugging (2.12) into (2.8) yields(2.13)$x’=\xi+\tau u_{h0}(\xi, \eta)$.

, which iswith the derivative operator
Note that from (2.13),$\xi=\xi(\tau, x’, \eta)$$0\leq\tau<t^{*}$(2.13)
givetoequation
impliesisinvertible(2.1),
that theitpositive forby virtue ofwhen $0\leq\tau<t^{*}.$:
Next, denote by
the Jacobian of the transformation between $(x’, y)$ and$\nabla_{\xi}=(\partial_{\xi_{1}}, \cdots, \partial_{\xi_{d-1}})^{T}$$\nabla_{\xi}x’=I_{d-1}+\tau\nabla_{\xi}u_{h0}$$(\xi, \eta)$$J(\tau,\xi, \eta)$$J( \tau, \xi, \eta) :=\frac{\partial(x’,y)}{\partial(\xi,\eta)}=\det(\nabla_{\xi}x’)\cdot\partial_{\eta}y-\sum_{\iota=1}^{d-1}[det(\nabla_{\iota}x’)\cdot\partial_{\xi}.y],$where the derivative operators$\nabla_{i}=(\cdots, \partial_{\xi.-1}, \partial_{\eta}, \partial_{\xi_{l+1}}, \cdots)^{T}, 1\leq i\leq d-1.$

From (2.8), we have $J(O, \xi, \eta)=1$ , and by combining with the divergence condition in (1.1),$\partial_{\tau}J(\tau,\xi, \eta)=J(\tau,\xi, \eta)\cdot(\nabla_{h}\cdot u_{h}+\partial_{y}u_{d})(\tau, x’(\tau,\xi’, \eta),y(\tau, \xi’, \eta))$$=J( \tau,\xi, \eta) [\frac{\kappa}{P(\tau)}\partial_{y}^{2}\overline{\theta}(\tau,\xi_{\rangle}\eta)-(1-\kappa)\frac{P_{\tau}(\tau)}{P(\tau)}],$

which implies that by virtue of (2.10) and (2.11),(2.14)$J( \tau, \xi, \eta)=\frac{P(0)}{P(\tau)\theta_{0}(\xi,\eta)}\overline{\theta}(\tau,\xi, \eta)$.

From (2.13), we have$\det(\nabla_{\xi}x’)(\tau, \xi, \eta)=det(I_{d-1}+\tau\nabla_{\xi}u_{h0}(\tau,\xi, \eta))>0$,

for$\tau\leq t^{*},$

then combining with the above relation (2.14), it follows that
(2.15)$\partial_{\eta}y-\sum_{i=1}^{d-1}[\frac{det(\nabla_{i}x’)}{det(\nabla_{\xi}x)}\cdot\partial_{\xi}.y]=\frac{P(0)}{P(\tau)\theta_{0}(\xi,\eta)\cdot det(\nabla_{\xi}x)(\tau,\xi_{)}\eta)}\overline{\theta}(\tau, \xi, \eta)$.

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We obtain that by calculation, the characteristics of the equation (2.15) are $x’=$ constant or $\xi=\xi(\tau, x’, \eta)$
given in (2.13), then denote by(2.16)$\tilde{\theta}(\tau,x’, \eta) :=\overline{\theta}(\mathcal{T}, \xi(\tau,x’, \eta), \eta)$,

and it implies that from (2.15),$\frac{\partial}{\partial_{\eta}}y(\tau, \xi(\tau, x_{\rangle}’\eta), \eta)=\frac{P(0)}{\theta_{0}(\xi(\tau,x’,\eta),\eta)\cdot det(\nabla_{\xi}x’)(\tau,\xi(\tau,x’,\eta),\eta)}\overline{\theta}(\tau, \xi(\tau, x’, \eta), \eta)$(2.17)$= \frac{\tilde{\theta}(\tau,x’,\eta)}{a(\tau,x’,\eta)},$where the scale function
is given in (2.5). Moreover, as $u_{d}|_{y=0}=0$ and the characteristic equation(2.8), the boundary $y=0$ is a particle path, thus we may set $y=0$ when $\eta=0$ . Therefore, integrating (2.17)
along characteristics, we obtain$a(\mathcal{T}, x’, \eta)$(2.18)$y=y( \tau, \xi(\tau, x’, \eta), \eta)=\int_{0}^{\eta}\frac{\tilde{\theta}(\tau,x’,z)}{a(\tau,x’,z)}dz.$

Consequently, when $0\leq\tau<t^{*}$ andof (2.18),$\tilde{\theta}>0$, we have that $a>0$ from the definition (2.5), and then by virtue$\partial_{\eta}y=\frac{\tilde{\theta}(\tau,x’,\eta)}{a(\tau,x,\eta)}>0,$

the equation (2.18) is invertible to give $\eta=\eta(\tau, x’, y)$ with
(2.19)$\eta_{y}=\frac{a(\tau_{\rangle}x’,\eta)}{\tilde{\theta}(\tau,x,\eta)}>$O.

Moreover, the domain $\{y>0\}$ is changed as $\{\eta>0\}$ with the boundary $\{y=0\},$ $yarrow+\infty$ respectively,
being changed as $\{\eta=0\},$respectively.
Now, we will derive the formula (2.3) and the problem (2.16) for. Note that the inverse functionof $(x’, y)(\tau, \xi, \eta)$ is$\etaarrow+\infty$$\tilde{\theta}(\tau, x’, \eta)$$(\xi(\tau, x’, \eta(\tau, x’, y)), \eta(\tau, x’, y))$given by (2.13) and (2.18). Thus, combining (2.9), (2.12) and (2.16) yields that$u_{h}(\tau,x’, y)=u_{h0}(\rangle, \theta(\tau, x’, y)=\tilde{\theta}(\tau,x’, \eta(\tau,x’, y))$ ,which gives the formulas of $u_{h}(t, x’, y)$ and$\theta(t, x’, y)$in (2.3). Denote by$\tilde{y}(\tau, x’, \eta):=\int_{0}^{\eta}\frac{\tilde{\theta}(\tau,x’,z)}{a(\tau,x,z)}dz,$then from (2.18) and (2.13) we have
(2.20)$y(\tau, \xi, \eta)=\tilde{y}(\tau,\xi+\tau u_{h0}(\xi, \eta), \eta)$, which yields that$y_{\tau}(\tau,\xi, \eta)=\partial_{\tau}\tilde{y}(\tau, \xi+\tau u_{h0}(\xi, \eta), \eta)+u_{h0}(\xi, \eta)\cdot\nabla_{h}\tilde{y}(\tau, \xi+\tau u_{h0}(\xi, \eta), \eta)$.

Combining (2.8) with (2.20), we get that$u_{d}(\tau, x’(\tau, \xi, \eta), y(\tau,\xi, \eta))=y_{\tau}(\tau, \xi, \eta)$$= \int_{0}^{\eta}\partial_{\tau}(\frac{\tilde{\theta}}{a})(\tau, \xi+\tau u_{h0}(\xi, \eta), z)dz+\int_{0}^{\eta}u_{h0}(\xi, \eta)\cdot\nabla_{h}(\frac{\tilde{\theta}}{a})(\tau, \xi+\tau u_{h0}(\xi, \eta), z)dz,$

which implies the formula of $u_{d}(t, x, y)$ in (2.3) by using that (2.13) and (2.18). Next, from (2.16) and therelation (2.13) we have$\overline{\theta}(\tau, \xi, \eta)=\tilde{\theta}(\tau, \xi+\tau u_{h0}(\xi, \eta), \eta)$,

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and then,$\partial_{\tau}\overline{\theta}=\partial_{\tau}\tilde{\theta}+u_{h0}\cdot\nabla\tilde{\theta}, \nabla_{\xi}\overline{\theta}=(I_{d-1}+\tau\nabla_{\xi}u_{h0})\cdot\nabla_{h}\tilde{\theta},$$\partial_{\eta}\overline{\theta}=\partial_{\eta}\tilde{\theta}+\tau\partial_{\eta}u_{h0}\cdot\nabla_{h}\tilde{\theta}.$Moreover, from (2.13) it follows that$(I_{d-1}+\tau\nabla_{\xi}u_{h0})\cdot\xi_{\eta}+\tau\partial_{\eta}u_{h0}=0,$

thus we obtain that by virtue of (2.19),$\partial_{y}\overline{\theta}(\tau,\xi(\tau, x’, \eta), \eta)=[(\xi_{\eta}\cdot\nabla_{\xi}+\partial_{\eta})\overline{\theta}](\tau,\xi(\tau,x’, \eta),\eta)\cdot\eta_{y}=\partial_{\eta}\tilde{\theta}(\tau, x’, \eta) \frac{a(\tau,x’,\eta)}{\tilde{\theta}(\tau,x,\eta)}.$Therefore, the problem for$\overline{\theta}$

given in (2.10) and (2.11) can be reduced as follows,$\{\begin{array}{ll}\partial_{\tau}\tilde{\theta}+u_{h0}(\xi(\tau,x’, \eta), \eta)\cdot\nabla_{h}\tilde{\theta}=\frac{\kappa a(\tau,x’,\eta)}{P(\tau)}\partial_{\eta}(\frac{a(\tau,x’,\eta)}{\overline{\theta}}\partial_{\eta}\tilde{\theta})+\frac{\kappa P_{\tau}(\tau)}{P(\tau)}\tilde{\theta}, in [0, t^{*})x\mathbb{R}_{+}^{d},\tilde{\theta}|_{\tau=0}=\theta_{0}(x’, \eta) .

Then, from the boundary conditions of$\theta$\end{array}$in (1.1), we get$\tilde{\theta}|_{\eta=0}=\theta^{0}(\tau,x \lim_{\etaarrow+\infty}\tilde{\theta}(\tau,x’, \eta)=\Theta(\tau, x$. Then, by the standard Picard method, we may know that theso we obtain the problem (2.6) for
Finally, One can checkfor some
problem (2.6) admits a positive classical solution in , to)directly that $(2.3)-(2.7)$ defines a smooth solution to the problem (1.1).
$\tilde{\theta}(\tau, x’,\eta)$$[0$$t_{0}\leq t^{*}$$\cross \mathbb{R}_{+}^{d}$$\square$Remark 2.1. Rnom (2.3) and (2.7) with the definition of the function in (2.5), one can deduce that there
may be a loss of derivatives in the horizontal variables for the solution to (1.1).
$a$

$x’$

2.1. Singularity formation. In this subsection, we establish a singular solution of problem (1.1) based on
the inviscid Prandtl equations. For this, we consider a simple case of the problem (1.1) with the constant
outflow, i.e., functions $P(t)$ andare constants. More precisely, we consider the following problem$\Theta(t, x’)$$\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}=0,$$\partial_{t}\theta+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})\theta=\theta\partial_{y}^{2}\theta,$(2.21)$\{$$\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\partial_{y}^{2}\theta,$$(u_{d}, \theta)|_{y=0}=(0, \theta^{0}(t, x , \lim_{yarrow+\infty}\theta(t,x, y)=1,$$(u_{h}, \theta)|_{t=0}=(u_{h0}, \theta_{0})(x’, y)$.

Then, the following proposition shows the singularity formation of the solution to (2.21).

Proposition 2.2. Assume that the initial-boundary data of problem (2.21) is given by$(u_{h0},\theta_{0})(x’,y)=(U(y+f_{0}(x’)), 1) , \theta^{0}(t,x’)=1,$

where $U(y)=(U_{1}(y), U_{2}(y))$ andand$f_{0}(x’)$

are smooth functions. Then, there exists a solution to (2.21) with$\theta(t, x’, y)\equiv 1$$u_{h}(t, x’, y)=U(y+f(t,x , u_{d}(t, x’, y)=-f_{t}(t, x’)-U(y+f(t, x \cdot\nabla_{h}f(t,x$where the function $f(t, x’)$ is the solution of$\{\begin{array}{l}f_{t}+U(f)\cdot\nabla_{h}f=0,f(0, x’)=f_{0}(x’) .\end{array}$If$t^{*} :=-[ \inf_{x\in R^{2}}U(f_{0}(x’))\cdot\nabla_{h}f_{0}(x’)]^{-1}>0,$then$\nabla_{h}u_{h}$and$u_{d}$

blow up as$t\uparrow t^{*}$

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The proof of the above proposition is similar to the one of Proposition 3.1 in [12], so we omit it here.
2.2. Convergence to the inviscid Prandtl equations. In this subsection, we investigate the asymptotic
behavior of solution to (1.1), as tends to a positive constant. We take the simple case (2.21) for brevity,
and the general case can be studied by similar arguments. More precisely, we will consider the asymptotic
behavior of solution under the condition$\theta$(2.22)

with$\theta^{0}(t,x’)=1+\epsilon\tilde{\theta}^{0}(t, x \theta_{0}(x’, y)=1+\epsilon\tilde{\theta}_{0}(x’, y)$$\epsilon<<1$. Formally, (2.21) tends to the following inviscid Prandtl system for small$\epsilon,$$\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}=0,$(2.23)$\{$$\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=0,$$u_{h}|_{t=0}=u_{h0}(x’,y)$$u_{d}|_{y=0}=0,$.

Through analogous arguments in Proposition 2.1, it’s not diﬃcult to show the local exsitence of solution to(2.23), where we don’t need monotonicity condition on the initial data, either.
Proposition 2.3. Let $v_{O}(x’, y)$ be smooth initial data of problem (2.23) such that
(2.24)$t_{1}^{*}:= \inf\{t :\inf_{(x,y)\in R_{+}^{d}}det(I_{d-1}+s\nabla_{h}u_{h0}(x’, y))>0, \forall s\in[0, t]\}>0$

Then, (2.23) has a unique classical solution in$[0_{\rangle}t^{*}$) given by$u_{h}(t, x’, y)=u_{h0}(\xi_{1}(t, x’, \eta_{1}(t, x’, y)), \eta_{1}(t, x’, y))$,(2.25)$u_{d}(t, x’, y)= \int_{0}^{\eta_{1}(t,x’,y)}\partial_{t}(\frac{1}{a_{1}})(t, x’, z)dz+\int_{0}^{\eta_{1}(t,x’,y)}[b_{1}(t, x’, \eta_{1}(t, x’, y))\cdot\nabla_{h}(\frac{1}{a_{1}})(t, x’, z)]dz.$

Here, the vector function$\xi_{1}(t, x’, z)\in \mathbb{R}^{d-1}\dot{u}$(2.26)

the solution of$x’=\xi_{1}+tu_{h0}(\xi_{1}, z)$;

the scale function $a_{1}(t, x’, z)$ and vector function $b_{1}(t, x’, z)$ are given by the initial data and $\xi_{1}(t,x’, z)$ :$a_{1}(t,x’, z):=det(I_{d-1}+t\nabla_{h}u_{h0})(\xi_{1}(t,x’, z), z)$(2.27),$\{$$b_{1}(t, x’, z)$$:=u_{h0}(\xi_{1}(t,x’, z), z)$;

and the scale function $\eta_{1}(t,x’, y)$ is the solution of(2.28)$y= \int_{0}^{\eta_{1}}\frac{1}{a_{1}(t,x,z)}dz.$

Now, we show that the solution of (2.21) given in Proposition 2.1 converges tois a solution of (2.23) given in Proposition 2.3.

$(u_{h}, u_{d}, 1)$, where$(u_{h}, u_{d})$Proposition 2.4. For the problem (2.21) with smooth initial-boundary dataand satisfying (2.22)
and the assumptions of Proposition 3.12, let $(u_{h}, u_{d}, \theta)(t, x’, y)$ be the solution of (2.21). Also, we assumethat$(u_{h0}, \theta_{0})$(2.29)

for some constant$\theta^{0}$$(1+y)^{k}\tilde{\theta}_{0}(x’, y)\in H_{x}^{2}, (R^{d-1}, H_{y}^{1}(\mathbb{R}_{+}))$$k> \frac{1}{2}$. Let$(u_{h1}, u_{d1})(t, x’, y)$be the solution of problem (2.23) with smooth initial data$u_{h0}.$Then, for suﬃciently small there is a constant $C>0$ independent
with being given in Proposition 2.1,$\epsilon$$of\epsilon$, such that for $(t, x’, y)\in[0, t_{0}$ )$t_{0}$(2.30)$|(u_{h}, u_{d}, \theta)(t, x’, y)-(u_{h1}, u_{d1},1)(t, x’, y)|\leq C\epsilon.$$\cross \mathbb{R}_{+}^{d}$

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Proof. Firstly, by (2.22) the representation of in (2.3) can be rewritten as$\theta$(2.31)

where$\theta(t, x’,y)=1+\epsilon\tilde{\theta}(t, x’, \eta(t,x’, y))$$\tilde{\theta}(t, x’, z)$(2.32),

satisfies$\{\begin{array}{l}\partial_{t}\tilde{\theta}+b\cdot\nabla_{h}\tilde{\theta}-a\partial_{z}(\frac{a}{1+\epsilon\tilde{\theta}}\partial_{z}\tilde{\theta})=0, in\{0\leq t<t^{*}, z>0,x’\in \mathbb{R}^{d-1}\},\tilde{\theta}|_{z=0}=\tilde{\theta}^{0}(t, x \tilde{\theta}|_{t=0}=\tilde{\theta}_{0}(x’, z) .\end{array}$Then, through classical Picard method we may have the local existence of solution to (2.32). Moreover, under
the assumption (2.29), by the standard energy method it’s not diﬃcult to obtain that there is a constant$C_{1}>0$ independent of , such that$\epsilon$(2.33)$\Vert(1+z)^{k}\tilde{\theta}|_{L_{t}^{\infty}(H_{x}^{2},H_{z}^{1})}\leq C_{1},$which also shows the mponent of in (2.30) by the Sobolev embedding inequality.
Secondly, compare Propositions 2.1 with 2.4, we know that the auxilary function $\xi(t, x’, z)$ given by (2.4)
coincides with $\xi_{1}(t,x’, z)$ in (2.26), which implies that by combining (2.5) with (2.27),$\theta$$\infty$(2.34)$a(t,x’, z)=a_{1}(t, x’, z)[1+\epsilon\tilde{\theta}_{0}(\xi(t,x’, z), z)], b(t,x’, z)=b_{1}(t, x’, z)$.

Then, from (2.7) and (2.28) we have$\int_{0}^{\eta(t,x’,z)}\frac{1+\epsilon\tilde{\theta}(t,x’,z)}{a(t,x,z)}dz=\int_{0}^{\eta_{1}(t,x’,z)}\frac{1}{a_{1}(t,x’,z)}dz,$

which implies that by (2.34),(2.35)$\epsilon\int_{0}^{\eta(t,x’,z)}\frac{\overline{\theta}(t,x’,z)-\tilde{\theta}_{0}(\xi(t,x’,z),z)}{a(t_{\rangle}x,z)}dz=\int_{\eta(t,x,z)}^{\eta_{1}(t,x’,z)}\frac{1}{a_{1}(t,x,z)}dz,$

From (2.27) we know that the functions andindependent of such thatis a constant$a$$C_{2}$$a_{1}$

are bounded and have positive lower bounds, that is, there$\epsilon$$C_{2}^{-1}\leq a(t, x’, z) , a_{1}(t, x’, z)\leq C_{2}, (t,x’, z)\in[0, t_{0})\cross \mathbb{R}_{+}^{d},$then the right-hand side of the above equality gives that
(2.36)$| \int_{\eta(t,x,z)}^{\eta_{1}(t,x’,z)}\frac{1}{a_{1}(t,x,z)}dz|\geq\frac{|\eta-\eta_{1}|(t,x’,z)}{C_{2}}.$On the other hand, we have that for the left-hand side term of (2.38),$| \int_{0}^{\eta(t,x’,z)}\frac{\tilde{\theta}(t,x’,z)-\tilde{\theta}_{0}(\xi(t,x’,z),z)}{a(t,x,z)}dz|\leq C_{2}(\Vert\tilde{\theta}(t, x’, z)\Vert_{L_{z}^{1}}+\Vert\tilde{\theta}_{0}(\xi(t,x’, z), z)\Vert_{L_{z}^{1}})$$\leq C_{3}(\Vert(1+z)^{k}\tilde{\theta}(t, x’, z)\Vert_{L_{z}^{2}}+\Vert\Vert(1+z)^{k}\tilde{\theta}_{0}(\xi(t, x’, z), z)\Vert_{L_{z}^{2}})$

Note that for $t\in[0, t_{0}$ ) and.

$x’\in \mathbb{R}^{d-1},$$|\tilde{\theta}(t, x’, z)|\leq\Vert\tilde{\theta}(t,x’, z)\Vert_{H_{x}^{2}}, \rangle$

and$|\tilde{\theta}_{0}(\xi(t, x’, z), z)|\leq\Vert\tilde{\theta}_{0}(\xi(t,x’, z),z)\Vert_{H_{x}^{2}}, \leq C_{4}\Vert\tilde{\theta}_{0}(x’, z)\Vert_{H_{x}^{2}},$,has bounded derivatives up to order two, provided that the smooth initial data

where we use that
. From the above three expressions we obtain that$\xi(t, x’, z)$$u_{h0}$(2.37)$| \int_{0}^{\eta(t,x’,z)}\frac{\tilde{\theta}(t,x’,z)-\tilde{\theta}_{0}(\xi(t,x’,z),z)}{a(t,x,z)}dz|\leq C_{5}\Vert(1+z)^{k}\tilde{\theta}\Vert_{L_{t}^{\infty}(H_{x}^{2},L_{z}^{2})}.$

for some constant $C_{5}>0$ independent of . Plugging (2.36) and (2.37) into (2.35) implies that$\epsilon$(2.38)$|\eta-\eta_{1}|(t_{\}}x’, z)\leq C_{2}C_{5}\Vert(1+z)^{k}\tilde{\theta}\Vert_{L_{t}^{\infty}(H_{x}^{2},L_{z}^{2})}\epsilon, \forall(t, x’, z)\in[0, t_{0})\cross \mathbb{R}_{+}^{d}.$103
CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANGof (2.30). Since $\xi(t, x’, z)=\xi_{1}(t, x’, z)$ , it follows that fromandNow we will prove the components ofgiven by (2.3) and (2.25) respectively,
the formulas of and$u_{h}$$u$$u_{d}$$u_{1}$$|u_{h}(t, x’, y)-u_{h1}(t, x’, y)|\leq\Vert\xi_{z}(t, x’, z)\cdot\nabla_{h}u_{h0}(\xi(t, x’, z), z)+\partial_{y}u_{h0}(\xi(t, x’, z), z)\Vert_{L\infty}$$|\eta-\eta_{1}|(t, x’, y)$Combining (2.38) with the above inequality yields that there is a constant $C_{6}>0$ independent ofthat(2.39)$\epsilon_{\rangle}$

.

such

$|u_{h}(t, x’, y)-u_{h1}(t,x’, y)|\leq C_{6}\epsilon,$provided the smooth initial data
the proof of this proposition.

$u_{h0}$. Similarly, we can show the component of$u_{d}$in (2.30), and complete$\square$2.3. The linearization around a shear flow. In this subsection, we study the well-posedness and stabilityof the linearization of problem (2.21) around a shear flow. It is easy to know that under the proper initialboundary values, (2.21) has a shear flow solution:(2.40)$(u_{h}, u_{d}, \theta)(t, x’, y)=(U_{h}(y), 0,1)$.

Then, the linearization of (2.21) at the shear flow (2.40) is(2.41)$\{\begin{array}{l}\partial_{t}u_{h}+U_{h}(y)\cdot\nabla_{h}u_{h}+U_{h}’(y)u_{d}=0,\partial_{t}\theta+U_{h}(y)\cdot\nabla_{h}\theta=\partial_{y}^{2}\theta,\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\partial_{y}^{2}\theta,(u_{d}, \theta)|_{\fbox{Error::0x0000}-0}=0, (u_{h}, \theta)|_{t=0}=(u_{h0}, \theta_{0})(x’, y) .\end{array}$

We may solve the problem (2.41) by two steps. Firstly, we determine
initial-boundary value problem:(2.42)$\theta(t, x’, y)$through the following linear$\{\begin{array}{l}\partial_{t}\theta+U_{h}(y)\cdot\nabla_{h}\theta=\partial_{y}^{2}\theta,\theta|_{\fbox{Error::0x0000}-0}=0, \theta|_{t=0}=\theta_{0}(x’, y) .\end{array}$It’s easy to know that the problem (2.42) with smooth initial data has a global classical solution and thegiven by (2.42), we solve $(u_{h}, u_{d})(t, x’, y)$ by
solution is unique. Secondly, with the known function
the problem:$\theta(t, x’, y)$$\partial_{t}u_{h}+U_{h}(y)\cdot\nabla_{h}u_{h}+U_{h}’(y)u_{d}=0,$(2.43)$\{$$\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\partial_{y}^{2}\theta,$$u_{d}|_{\fbox{Error::0x0000}-0}=0,$$u_{h}|_{t=0}=u_{h0}(x’, y)$.

Therefore, we have the following result for the problem (2.41).

Proposition 2.5. Let $U_{h}(y)$ , $u_{h0}(x’, y)$ and$(u_{h}, u_{d}, \theta)(t, x’, y)$ to the problem (2.41), whereis given by(2.44)$\theta_{0}(x’,y)$be smooth functions, then there exists a classical solution
is solved by the problem (2.42), and $(u_{h}, u_{d})(t, x’, y)$$\theta(t, x’, y)$$u_{h}(t,x’, y)=u_{h0}(x’-tU_{h}(y), y)+tU_{h}(y) \int_{0}^{y}(\nabla_{h}\cdot u_{h0})(x’-tU_{h}(z), z)dz$$+U_{h}(y) \int_{0}^{y}\theta_{0}(x’-tU_{h}(z), z)dz-U_{h}’(y)\int_{0}^{y}\theta(t, x’, z)dz_{\rangle}$$u_{d}(t_{\rangle}x’, y)=- \int_{0}^{y}\{(\nabla_{h}\cdot u_{h0})(x’-tU_{h}(z), z)+t[U_{h}(y)-U_{h}(z)]\cdot\nabla_{h}(\nabla_{h}\cdot u_{h0})(x’-tU_{h}(z), z)\}dz$$- \int_{0}^{y}\{[U_{h}(y)-U_{h}(z)]\cdot\nabla_{h}\theta_{0}(x’-tU_{h}(z), z)\}dz+\theta_{y}(t, x’, y)-\theta_{y}(t, x’, 0)$$+ \int_{0}^{y}\{[U_{h}(y)-U_{h}(z)]\cdot\nabla_{h}\theta(t, x’, z)\}dz.$

104
BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW

. From theProof. According to the above arguments, we only need to show the expressions (2.44) of
satisfy a transport equation, and from the equations of (2.41), it follows$(u_{h},u_{d})$problem (2.41), we know that
that
satisfies the equation$u_{h}$$\partial_{y}u_{d}$(2.45)$(\partial_{t}+U_{h}(y)\cdot\nabla_{h})(\partial_{y}u_{d}-\partial_{y}^{2}\theta)-(U_{h}’(y)\cdot\nabla_{h})u_{d}=0,$

and the initial data
(2.46)$\partial_{y}u_{d}(0,x’,y)=\partial_{y}^{2}\theta_{0}(x’,y)-(\nabla_{h}\cdot u_{h0})(x’, y)$.

Thereby, we introduce the following coordinate transformation:(2.47)

with$\tau=t, \xi=x’-tU_{h}(y) , \eta=y$$\xi=(\xi_{1}, \cdots, \xi_{d-1})\in \mathbb{R}^{d-1}$(2.48), and obtain the corresponding partial derivatives as follows:$\partial_{\tau}=\partial_{t}+U_{h}(y)\cdot\nabla_{h}, \nabla_{\xi}=\nabla_{h}, \partial_{\eta}=\partial_{y}+tU_{h}’(y)\cdot\nabla_{h}.$Then the first two equations of (2.41) and (2.45) are reduced as(2.49)

where$\{\begin{array}{l}\partial_{\tau}u_{h}+U_{h}’(\eta)u_{d}=0, \partial_{\tau}\theta=\partial_{y}^{2}\theta,\partial_{\tau}(\partial_{y}u_{d}-\partial_{y}^{2}\theta)-U_{h}’(\eta)\cdot\nabla_{\xi}u_{d}=0,\end{array}$$\nabla_{\xi}=(\partial_{\xi_{1}}, \cdots, \partial_{\xi_{d-1}})^{T}$(2.50)

Moreover, we have the initial data:$(u_{h}, \theta, \partial_{y}u_{d})(0, \xi,\eta)=(u_{h0}, \theta_{0},\partial_{y}^{2}\theta_{0}-\nabla_{h}\cdot u_{h0})(\xi, \eta)$

Set a Lagrangian streamfunction$\Psi$(2.51).

satisfying that$\Psi_{\eta}=\partial_{y}u_{d}, \Psi|_{\eta=0}=0,$

then, from (2.50) we have the following initial data:(2.52)$\Psi_{\eta}|_{r=0}=(\partial_{y}^{2}\theta_{0}-\nabla_{h}\cdot u_{h0})(\xi, \eta)$.

Combining (2.51) with the third equation of (2.49), and using the second equation of (2.49) we have
(2.53)$\partial_{\tau}\Psi_{\eta}-U_{h}’(\eta)\cdot\nabla_{\xi}u_{d}=\partial_{\tau}^{2}\theta,$

which implies that by virtue of (2.47)(2.54), $or$
and using the boundary values of$\partial_{\tau}(\tau\Psi_{\eta})=\partial_{\eta}u_{d}+\tau\partial_{\tau}^{2}\theta$

Integrating (2.54) in$\eta$$\partial_{\eta}u_{d}=\partial_{\eta}(\tau\Psi)_{\tau}-\tau\partial_{\tau}^{2}\theta.$$u_{d}$

and$\Psi$, it yields that$u_{d}=( \tau\Psi)_{\tau}-\tau\partial_{\tau}^{2}(\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta),\zeta)d\zeta)$(2.55)$= \partial_{\tau}(\tau\Psi-\tau\partial_{\tau}\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta), \zeta)d\zeta+\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)$

Then, substituting the expression (2.55) ofdata (2.50), we have$u_{d}$

into the equation of$u_{h}$.

in (2.49) and combining with the initial(2.56)$u_{h}=u_{h0}-U_{h}’(\eta)[\tau\Psi-\tau\partial_{\tau}\int_{0}^{\eta_{\theta(\mathcal{T},\xi+\tau U_{h}(\zeta))}}\zeta)d\zeta+\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta), \zeta)d\zeta-\int_{0}^{\eta}\theta_{0}(\xi, \zeta)d\zeta)$.

Also, plugging the expression (2.55) of $ud$ into (2.53), we get$\partial_{\tau}(\Psi_{\eta}-\tau U_{h}’(\eta)\cdot\nabla_{\xi}\Psi)$$= \partial_{\tau}^{2}\theta-\partial_{\tau}(\tau U_{h}’(\eta)\cdot\nabla_{\xi}(\partial_{\tau}\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta))+U_{h}’(\eta)\cdot\nabla_{\xi}(\partial_{\tau}\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta),\zeta)d\zeta)$$= \partial_{\tau}[\partial_{\eta}(\partial_{\tau}\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta),\zeta)d\zeta)-\tau U_{h}’(\eta)\cdot\nabla_{\xi}(\partial_{\tau}\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)]$$+ \partial_{\tau}[U_{h}’(\eta)\cdot\nabla_{\xi}(\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)],$

105
CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

which implies that by virtue of(2.48),(2.57)$\partial_{\tau}(\Psi-\partial_{\tau}\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)_{y}=\partial_{\tau}[U_{h}’(\eta)\cdot\nabla_{\xi}(\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)].$Integrating this equation with respect to$\tau$

and using the initial data (2.52), we have$( \Psi-\partial_{\tau}\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta), \zeta)d\zeta)_{y}$$=-( \nabla_{h}\cdot u_{h0})(\xi, \eta)-U_{h}’(\eta)\cdot\nabla_{\xi}(\int_{0}^{\eta}\theta_{0}(\xi, \zeta)d\zeta)+U_{h}’(\eta)\cdot\nabla_{\xi}(\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{h}(\zeta), \zeta)d\zeta)$$=-( \nabla_{h}\cdot u_{h0})(x’-tU_{h}(y), y)-U_{h}’(y)\cdot\nabla_{h}(\int_{0}^{y}\theta_{0}(x’-tU_{h}(y), z)dz)$$+ U_{h}’(y)\cdot\nabla_{h}(\int_{0}^{y}\theta(t, x’-tU_{h}(y)+tU_{h}(z), z)dz)$

Then, integrating the above quality in$y$.

we obtain that by using the boundary condition $\Psi|_{y=0}=0,$$\Psi-\partial_{\tau}\int_{0}^{\eta}\theta(\tau,\xi+\tau U_{\hslash}(\zeta),\zeta)d\zeta$$=- \int_{0}^{y}(\nabla_{h}\cdot u_{h0})(x’-tU_{h}(z), z)dz-\int_{0}^{y}\int_{0}^{\tilde{y}}(U_{h}’(\tilde{y})\cdot\nabla_{h}\theta_{0}(x’-tU_{h}(\tilde{y}), z))dzd\tilde{y}$$+ \int_{0}^{y}\int_{0}^{\tilde{y}}(U_{h}’(\tilde{y})\cdot\nabla_{h}\theta(t, x’-tU_{h}(\tilde{y})+tU_{h}(z), z))dzd\tilde{y},$which implies that
(2.58)$\tau(\Psi-\partial_{\tau}\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta), \zeta)d\zeta)+\int_{0}^{\eta}\theta(\tau, \xi+\tau U_{h}(\zeta), \zeta)d\zeta$$=-t \int_{0}^{y}(\nabla_{h}\cdot u_{h0})(x’-tU_{h}(z), z)dz+\int_{0}^{y}\theta_{0}(x’-tU_{h}(y), z)dz-\int_{0}^{y}\theta_{0}(x’-tU_{h}(z), z)dz+\int_{0}^{y}\theta(t, x’, z)dz.$

Therefore, we substitute (2.58) into (2.55) and obtain the expression of$u_{d}$:$u_{d}(t, x’, y)=- \int_{0}^{y}[\nabla_{\hslash}\cdot u_{h0}+(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}(t\nabla_{h}\cdot u_{h0}+\theta_{0})](x’-tU_{h}(z), z)dz$$+ \int_{0}^{y}(\theta_{t}+U_{h}(y)\cdot\nabla_{h}\theta)(t, x’, z)dz$(2.59)$=- \int_{0}^{y}[\nabla_{h}\cdot u_{h0}+(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}(t\nabla_{h}\cdot u_{h0}+\theta_{0})](x’-tU_{h}(z), z)dz$$+ \int_{0}^{y}[(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}\theta](t_{\rangle}x’, z)dz+\theta_{y}(t, x’, y)-\theta_{y}(t, x’, 0)$

where we use the equation of$\theta$,in (2.41) Meanwhile, plugging (2.58) into (2.56) yields that$u_{h}(t, x’, y)=u_{h0}(x’-tU_{h}(y), y)+U_{h}’(y)\int_{0}^{y}[t\nabla_{h}\cdotu_{h0}+\theta_{0}](x’-tU_{h}(z), z)dz-U_{h}’(y)\int_{0}^{y}\theta(t, x’, z)dz.$

Consequently, we obtain the expressions of$(u_{h}, u_{d})$and complete this proof.

$\square$

The representation (2.44) in Proposition 2.5 shows that there is a loss of derivatives with respect to thehorizontal variables for the solution of problem (2.41). Denote by$x’$(2.60)$\Vert u_{h}\Vert(t, y) :=(\int_{\mathbb{R}^{d-1}}|u_{h}(t, x’, y)|^{2}dx’)^{2}l,$106
BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW

and we use the following anisotropic space:$L^{p,q}:=\{f=f(x’, y)$$H^{m,k}$measurable:$:=\{f=f(x’,y)$ measurable :$\Vert f\Vert_{L^{p,q}}:=\Vert\Vert f\Vert_{L^{p}(dx’)}\Vert_{L^{q}(dy)}<\infty\},$$\Vert f\Vert_{H^{n,k}}$$1\leq p,$ $q\leq\infty,$$:=( \sum_{|\alpha|\leq m,0\leq\iota\leq k}\Vert\partial_{x}^{\alpha},\partial_{y}^{t}f\Vert_{L^{2}(dx’dy)}^{2})^{2}\perp<\infty\}$with$\partial_{x}^{\alpha}, =\partial_{x_{1}}^{\alpha_{1}}\cdots\partial_{x_{d-1}}^{\alpha_{d-1}}, \alpha=(\alpha_{1}, \cdots, \alpha_{d-1}) , |\alpha|=\alpha_{1}+\cdots+\alpha_{d-1}.$

Then, from Proposition 2.5 we have the following result.

Proposition 2.6. Assume thatthat

$U_{h}(y)\in W^{2,\infty}(\mathbb{R}_{+})$

is smooth and the initial data of problem (2.41) satisfies$\Vert u_{h0}\Vert(y) , \Vert\nabla_{h}\cdot u_{h0}\Vert(y) , \Vert\nabla_{h}(\nabla_{h}\cdot u_{h0})\Vert(y) , \Vert\theta_{0}\Vert(y) , \Vert\nabla_{h}\theta_{0}\Vert(y)<\infty.$Let$(u_{h}, u_{d}, \theta)(t,x’, y)$$M_{0}$$U_{h}(y)\Vert_{L\infty(R_{+})})$

be the solution
and $C_{2}=C_{2}$(2.61)

Then, there exist positive constants) independent of , such thatof (2.41).

$U_{h}(y)\Vert_{W^{2\infty}(R_{+})}$$M_{0}$$=$$t$$\Vert\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}\leq\Vert\theta_{0}\Vert_{L^{2}(R_{+}^{d})}, \Vert\nabla_{h}\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}\leq\Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(R_{+}^{d})},$

and$\Vert\theta\Vert(t, y) , \Vert\theta_{y}\Vert(t,y)\leq M_{0}(\Vert\theta_{0}\Vert_{H^{1,0}}+\Vert\theta_{0}\Vert_{H^{0,2}}) , \Vert\nabla_{h}\theta\Vert(t, y)\leq M_{0}(\Vert\theta_{0}\Vert_{H^{2,0}}+\Vert\theta_{0}\Vert_{H^{1,2}})$(2.62)$\Vert\partial_{y}^{2}\theta\Vert(t, y)\leq M_{1}(\Vert\theta_{0}\Vert_{H^{2,0}}+\Vert\theta_{0}\Vert_{H^{1,2}}+\Vert\theta_{0}\Vert_{H^{0,4}})$,,moreover,$\Vert u_{h}\Vert(t, y)\leq\Vert u_{h0}\Vert(y)+t|lf_{h}(y)|\cdot\int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)dz+2\Vert\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})} |\sqrt{y}U_{h}(y)|,$(2.63)$\Vert u_{d}\Vert(t, y)\leq\int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)dz+t\int_{0}^{y}[|U_{h}(y)-U_{h}(z)| \Vert\nabla_{h}(\nabla_{h}\cdot u_{h0})\Vert(t, z)]dz$$+2 \Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(R_{+}^{d})}(\int_{0}^{y}|U_{h}(y)-U_{h}(z)|^{2}dz)^{2}\iota+M_{0}(\Vert\theta_{0}\Vert_{H^{1,0+}}\Vert\theta_{0}\Vert_{H^{0,2}})$

Proof. Firstly, from Proposition 2.5 we know thateasy to obtain that by energy estimate,$\theta(t, \xi’, y)$satisfies the linear problem (2.42). Then, it is$\frac{d}{2dt}\Vert\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}^{2}+\Vert\partial_{y}\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}^{2}=0,$

which implies that
(2.64)$\Vert\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}\leq\Vert\theta(0, \cdot)\Vert_{L^{2}(R_{+}^{d})}=\Vert\theta_{0}\Vert_{L^{2}(R_{+}^{d})}.$

Denote by the operator$\partial_{\mathcal{T}}^{\alpha}:=\partial_{t}^{\alpha_{1}}\partial_{x_{1}}^{\alpha_{2}}\cdot\partial_{x_{d-1}}^{\alpha_{d}}, \alpha=(\alpha_{1}, \cdots, \alpha_{d}) , |\alpha|=\alpha_{1}+\cdots+\alpha_{d}.$Applying the operator(2.65)$\partial_{\tau}^{\alpha},$$|\alpha|=1$.

to the equation of (2.42), and similarly we have that,$\Vert\partial_{\mathcal{T}}^{\alpha}\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}\leq\Vert\partial_{T}^{\alpha}\theta(0, \cdot)\Vert_{L^{2}(R_{+}^{d})}.$

Combining the equation of (2.42) with the estimates in (2.65), and using that$\theta_{t}(0, x’,y)=\partial_{y}^{2}\theta_{0}(x’,y)-U_{h}(y)\cdot\nabla_{h}\theta_{0}(x’, y) , \theta_{x}.(0,x’,y)=\theta_{0x}.(x’, y) , 1\leq i\leq d-1,$

it follows that$\Vert\partial_{y}^{2}\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}\leq\Vert\theta_{t}(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}+\Vert U_{h}(y)\Vert_{L\infty(R_{+})} \Vert\nabla_{h}\theta(t, \cdot)\Vert_{L^{2}(R_{+}^{d})}$(2.66)$\leq 2\Vert U_{h}(y)\Vert_{L\infty(R_{+})} \Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(R_{+}^{d})}+\Vert\partial_{y}^{2}\theta_{0}\Vert_{L^{2}(R_{+}^{d})}.$

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CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

By the classical interpolation inequality we obtain that from (2.64) and (2.66),$\Vert\theta_{y}(t, \cdot)\Vert_{L^{2}(\mathbb{R}_{+}^{d})}\leq C(\Vert\theta(t,\cdot)\Vert_{L^{2}(\mathbb{R}_{+}^{d})}+\Vert\partial_{y}^{2}\theta(t, \cdot)\Vert_{L^{2}(\mathbb{R}_{+}^{d})})$(2.67)$\leq C(\Vert\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})}+\Vert U_{h}(y)\Vert_{L\infty(\mathbb{R}_{+})} \Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})}+\Vert\partial_{y}^{2}\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})})$,

where is a positive constant independent of . Then, from the estimates (2.64), (2.66) and (2.67) it implies
independent ofby the imbedding inequality that there is a positive constantsuch that,(2.68)
$C$

$t$

$t,$

$C_{0}=C_{0}(\Vert U_{h}(y)\Vert_{L^{\infty}(\mathbb{R}_{+})})$$\Vert\theta(t_{\}}\cdot)\Vert_{L^{2},\infty}\leq\Vert\theta(t, \cdot)\Vert_{H^{0,1}}\leq C_{0}(\Vert\theta_{0}\Vert_{H^{1,0}}+\Vert\theta_{0}\Vert_{H^{0,2}})$$\Vert\theta_{y}(t, \cdot)\Vert_{L^{2},\infty}\leq\Vert\theta_{y}(t, \cdot)\Vert_{H^{0,1}}\leq C_{0}(\Vert\theta_{0}\Vert_{H^{1,0}}+\Vert\theta_{0}\Vert_{H^{0,2}})$

Next, we apply$\partial_{\mathcal{T}}^{\alpha},$$|\alpha|=1$.

to the equation in (2.42) and get$\partial_{t}\partial_{\mathcal{T}}^{\alpha}\theta+U_{h}(y)\cdot\nabla_{h}\partial_{\mathcal{T}}^{\alpha}\theta-\partial_{y}^{2}\partial_{\mathcal{T}}^{\alpha}\theta=0,$

moreover, we have the following boundary value of$\partial_{\mathcal{T}}^{\alpha}\theta(t, x’, y)$$\partial_{\mathcal{T}}^{\alpha}\theta_{x}.(t, x’, 0)=$:O.

Thus, through the above analogous arguments we can obtain that there exist positive constants $C_{1}=$independent of , such that
and

$C_{1}(\Vert U_{h}(y)\Vert_{L\infty(\mathbb{R}_{+})})$$t$$C_{2}=C_{2}(\Vert U_{h}(y)\Vert_{W^{2\infty}(\mathbb{R}_{+})})$$\Vert\nabla_{h}\theta(t, \cdot)\Vert_{L^{2,\infty}}\leq C_{1}(\Vert\theta_{0}\Vert_{H^{2.0}}+\Vert\theta_{0}\Vert_{H^{1,2}})$(2.69),$\Vert\theta_{t}(t, \cdot)\Vert_{L^{2,\infty}}\leq C_{2}(\Vert\theta_{0}\Vert_{H^{2,0}}+\Vert\theta_{0}\Vert_{H^{1,2}}+\Vert\theta_{0}\Vert_{H^{0,4}})$Thus, from the equation in (2.42) we obtain that there is a positive constant
independent of , such that.
$C_{3}=C_{3}(\Vert U_{h}(y)\Vert_{W^{2\infty}(\mathbb{R}_{+})})$$t$$\Vert\partial_{y}^{2}\theta(t, \cdot)\Vert_{L^{2},\infty}\leq\Vert\theta_{t}(t,\cdot)\Vert_{L^{2,\infty}}+\Vert U_{h}(y)\Vert_{L^{\infty}(\mathbb{R}_{+})} \Vert\nabla_{h}\theta\Vert_{L^{2,\infty}}$(2.70)$\leq C_{3}(\Vert\theta_{0}\Vert_{H^{2,0}}+\Vert\theta_{0}\Vert_{H^{1,2}}+\Vert\theta_{0}\Vert_{H^{0,4}})$.

Combining (2.68), (2.69) and (2.70), we obtain the estimates (2.62).
Finally, it is easy to establish the following estimates for $(u_{h}, u_{d})(t, x’, y)$ by the expression (2.44) in
Proposition 2.5,$\Vert u_{h}\Vert(t, y)\leq\Vert u_{h0}\Vert(y)+|U_{h}’(y)|\int_{0}^{y}[\Vert\theta_{0}\Vert(z)+\Vert\theta\Vert(t, z)]dz+t|U_{h}’(y)|\int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)dz$(2.71)$\leq\Vert u_{h0}\Vert(y)+t|U_{h}’(y)|\int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)d_{Z}+|\sqrt{y}U_{h}’(y)|(\Vert\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})}+\Vert\theta(t, \cdot)\Vert_{L^{2}(\mathbb{R}_{+}^{d})})$,

and$\Vert u_{d}\Vert(t, y)\leq 2\Vert\theta_{y}(t, \cdot)\Vert_{L^{2},\infty+}\int_{0}^{y}[|U_{h}(y)-U_{h}(z)| (\Vert\nabla_{h}\theta_{0}\Vert(z)+\Vert\nabla_{h}\theta\Vert(t, z))]dz$$+ \int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)dz+t\int_{0}^{y}[|U_{h}(y)-U_{h}(z)| \Vert\nabla_{h}(\nabla_{h}\cdot u_{h0})\Vert(t, z)]dz$(2.72)$\leq 2\Vert\theta_{y}(t, \cdot)\Vert_{L^{2},\infty}+(\int_{0}^{y}|U_{h}(y)-U_{h}(z)|^{2}dz)^{2}\perp (\Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(\mathbb{R}_{+}^{d})}+\Vert\nabla_{h}\theta(t, \cdot)\Vert_{L^{2}(\mathbb{R}_{+}^{d})})$$+ \int_{0}^{y}\Vert\nabla_{h}\cdot u_{h0}\Vert(z)dz+t\int_{0}^{y}[|U_{h}(y)-U_{h}(z)| \Vert\nabla_{h}(\nabla_{h}\cdot u_{h0})\Vert(t, z)]dz.$

Combining (2.64) with (2.71), we obtain the estimate for(2.72), the estimate forin (2.63) follows immediately.

$u_{h}$in (2.63). Substituting (2.62) and (2.65) into$\square$$u_{d}$

Now, we show that under some certain conditions of $U_{h}(y)$ , the shear flow solution $(U_{h}(y), 0,1)$ to problem(2.21) is linearly unstable, that is, solutions to the linearized problem (2.41) of (2.21) grow algebraically intime. More precisely, we have the following result:

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BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW

and the initial data are smooth, and $u_{h}(x’,y)$
Proposition 2.7. Suppose that for the problem (2.41),
be the solution of (2.41).
decays rapidly in . Let(1) For $d=2$ , if $U(y)$ has no critical points, thenis bounded uniformly in :$U_{h}(y)$$x’$$(u_{h}, u_{d_{\rangle}}\theta)(t, x, y)$$\Vert u_{h}\Vert(t,y)$$t$$\Vert u_{h}\Vert(t, y)\leq|\frac{U_{h}(y)}{U_{h}(0)}| \Vert u_{h0}\Vert(0)+\int_{0}^{y}\{|\frac{U_{h}(y)}{U_{h}(z)}|\Vert\partial_{y}u_{h0}\Vert(z)+|\frac{U_{h}’(z)}{(U_{h}(z))^{2}}|\Vert u_{h0}\Vert(z)\}dz$(2.73)$+2\Vert\theta_{0}\Vert_{L^{2}(R_{+}^{d})}\sqrt{y}|U_{h}(y)|,$and as $tarrow+\infty,$(2.74)$\Vert u_{h}+U_{h}’(y)\int_{0}^{y}\theta(t,x’, z)dz\Vert(t, y)arrow\Vert u_{h0}\Vert(0) |\frac{U_{h}(y)}{U_{h}(0)}|.$

If $U(y)$ has a single, non-degenerate critical point at(2.75)$y=y_{0}$, then when $y>y_{0}$ , we have that when $tarrow+\infty,$$\Vert u_{h}+U_{h}(y)\int_{0}^{y}\theta(t, x’, z)dz\Vert(t,y)\sim\sqrt{2\pi t}\Vert u_{h0}\Vert_{1}2(y_{0})\frac{|U_{h}(y)|}{\sqrt{|U_{h}(y_{0})|}},$

where$\Vert u_{h0}\Vert_{z}\iota(y):=(\int_{R^{2}}|\xi| |\hat{u}_{h0}|^{2}(\xi_{\rangle}y)d\xi)^{2}\iota$

Also, for suﬃciently large$t$

we have(2.76)$\Vert u_{h}\Vert(t, y)\geq\sqrt{\frac{\pi t}{2}}\Vert u_{h0}\Vert_{1}2(y_{0})\frac{|U_{h}(y)|}{\sqrt{|U_{1}’(y_{0})|}}.$

Furthermore, we have similar results as above forand(2) For $d=3$ , suppose that $U_{h}(y)$ satisfies that $U_{i}(y)(i=1$ or 2 has no critical points (we assume that$i=1)$ , and $U_{2}(y)=kU_{1}(y)$ for some constant . Thenis bounded uniformly in :$\partial_{y}u_{h},$$u_{d}$$\partial_{y}u_{d}.$

$)$

$k$

$\Vert u_{d}\Vert(t, y)$$t$$\Vert u_{d}\Vert(t, y)\leq|\frac{U_{1}(y)-U_{1}(0)}{U_{1}’(0)}| \Vert\nabla_{h}\cdot u_{h0}\Vert(0)+\int_{0}^{y}\{|\frac{U_{1}(y)-U_{1}(z)}{U_{1}’(z)}| \Vert\nabla_{h}\cdot\partial_{y}u_{h0}\Vert(z)$(2.77)$+| \frac{U_{1}"(z)(U_{1}(y)-U_{1}(z))}{(U_{1}’(z))^{2}}|\Vert\nabla_{h}\cdot u_{h0}\Vert(z)\}dz$$+2 \Vert\nabla_{h}\theta_{0}\Vert_{L^{2}(R_{+}^{d})}(\int_{0}^{y}|U_{h}(y)-U_{h}(z)|^{2}dz)^{2}\iota+M_{0}(\Vert\theta_{0}\Vert_{H^{1,0}}+\Vert\theta_{0}\Vert_{H^{0,2}})$$Moreover_{Z}$as.

$tarrow+\infty,$(2.78)$\Vert u_{d}-\theta_{y}(t, x’, y)+\theta_{y}(t, x’, 0)-\int_{0}^{y}[(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}\theta(t, x’, z)]dz\Vert(t, y)arrow\Vert\nabla_{h}\cdot u_{h0}\Vert(0)|\frac{U_{1}(y)-U_{1}(0)}{U_{1}(0)}|.$

Assume that $U_{i}(y)(i=1$ or 2 (we assume that $i=1$ ) has a single, non-degenerate critical point at $y=y_{0z}$and $U_{2}(y)=kU_{1}(y)$ for some constant . Then when $y>y_{0}$ , we have that when $tarrow+\infty$ :
$)$

$k$

$\Vert u_{d}-\theta_{y}(t,x’,y)+\theta_{y}(t, x’,0)-\int_{0}^{y}[(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}\theta(t,x’, z)]dz\Vert(t, y)$(2.79)$\sim\sqrt{2\pi t}\Vert\nabla_{h}\cdot u_{h0}\Vert_{1_{k}},(y_{0})\frac{|U_{1}(y)-U_{1}(y_{0})|}{\sqrt{|U_{1}"(y_{0})|}}2,$

where$\Vert\nabla_{h}\cdot u_{h0}\Vert_{1_{k}},(y)2:=(\int_{\mathbb{R}^{2}}|\xi_{1}+k\xi_{2}||\xi\cdot\hat{u}_{h0}|^{2}(\xi,y)d\xi)^{2}\iota$

Also, for suﬃciently large(2.80)$t$

we have$\Vert u_{d}\Vert(t,y)\geq\sqrt{\frac{\pi t}{2}}\Vert\nabla_{h}\cdot u_{h0}\Vert_{1_{k}},(y_{0})\frac{|U_{1}(y)-U_{1}(y_{0})|}{\sqrt{|U_{1}’(y_{0})|}}2,\cdot$

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CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

If

$U_{2}(y)\neq kU_{1}(y)$large

for any$k\in \mathbb{R}$, then there is a pointsuch that, when $y>y_{0}$ we have that for uﬃciently$\mathcal{S}$$y_{0}$$t,$(2.81)$\Vert u_{d}\Vert(t, y)\geq c\sqrt{t}$with the constant $C=C(y, y_{0_{\rangle}}U_{h}, u_{h0})>0$ independent of . Moreover, we have similar results as above forand$t$$\nabla_{h}\cdot u_{h}.$$\partial_{y}u_{d}$Proof. We will prove the second part of this proposition, the first part of two-dimensional case is similar
to Proposition 6.1 of [12]. From Proposition 2.5, we have the expressions (2.44) of the solution to problem(2.41), then we take the Fourier transform with respect to for the expression of $u_{d}(t, x’, y)$ to obtain that$x’$$\hat{u}_{d}(t, \xi, y)=-\int_{0}^{y}\{i\xi\cdot\hat{u}_{h0}(\xi, y)+[i\xi\cdot(U_{h}(y)-U_{h}(z))] [it\xi\cdot\^{u}_{h0}(\xi, z)+\hat{\theta}_{0}(\xi, z)]\}e^{-\iota t\xi\cdot U_{h}(z)}dz$(2.82)$+ \int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))]\hat{\theta}(t, \xi, z)dz+\hat{\theta}_{y}(t, \xi, y)-\hat{\theta}_{y}(t, \xi, 0)$.

When $U_{2}(y)=kU_{1}(y)$ for some constant , then (2.82) is reduced as$k$$\hat{u}_{d}(t, \xi, y)=-\int_{0}^{y}[1+it(\xi_{1}+k\xi_{2})(U_{1}(y)-U_{1}(z))][i\xi\cdot\^{u}_{h0}(\xi, z)]e^{-it(\xi_{1}+k\xi_{2})U_{1}(z)}dz$(2.83)$- \int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))] [\hat{\theta}_{0}(\xi, z)e^{-\iotat\xi\cdot U_{h}(z)}+\hat{\theta}(t,\xi, z)]dz+\hat{\theta}_{y}(t, \xi, y)-\hat{\theta}_{y}(t,\xi, 0)$(we may assume that $i=1$ ) has no critical points and for
If
that by integration by parts,$U_{i}(y)$$\xi_{1}\neq-k\xi_{2}$.

in the above equality, we obtain$\hat{u}_{d}(t, \xi,y)=-\frac{U_{1}(y)-U_{1}(0)}{U_{1}’(0)}[i\xi\cdot\^{u}_{h0}(\xi, 0)]e^{-\iota t(\xi_{1}+k\xi_{2})U_{1}(0)}-\int_{0}^{y}\{\frac{U_{1}(y)-U_{1}(z)}{U_{1}’(z)}$$[i \xi\cdot\overline{\partial_{y}u}_{h0}(\xi, z)]-\frac{U_{1}"(z)(U_{1}(y)-U_{1}(z))}{(U_{1}’(z))^{2}}[i\xi\cdot\hat{u}_{h0}(\xi, z)]\}e^{-\iota t(\xi_{1}+k\xi_{2})U_{1}(z)}dz$(2.84)$- \int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))] [\hat{\theta}_{0}(\xi, z)e^{-it\xi\cdot U_{h}(z)}-\hat{\theta}(t, \xi, z)]dz$$+\hat{\theta}_{y}(t, \xi,y)-\hat{\theta}_{y}(t, \xi,0)$.

Then, from the above equality (2.84) it implies that by Parseval’s identity,$\Vert u_{d}\Vert(t_{\rangle}y)\leq|\frac{U_{1}(y)-U_{1}(0)}{U_{1}’(0)}| \Vert\nabla_{h}\cdot u_{h0}\Vert(0)+\int_{0}^{y}[|\frac{U_{1}(y)-U_{1}(z)}{U_{1}’(z)}| \Vert\nabla_{h}\cdot\partial_{y}u_{h0}\Vert(z)$(2.85)$+| \frac{U_{1}"(z)(U_{1}(y)-U_{1}(z))}{(U_{1}(z))^{2}}| \Vert\nabla_{h}\cdot u_{h0}\Vert(z)]dz$$+ \int_{0}^{y}|U_{h}(y)-U_{h}(z)| [\Vert\nabla_{h}\theta_{0}\Vert(z)+\Vert\nabla_{h}\theta\Vert(t, z)]dz+2\Vert\theta_{y}(t, \cdot)\Vert_{L^{2},\infty},$thus, by using Proposition 2.6 in the above estimate (2.85) we obtain (2.77). Moreover, from (2.84) we have$\hat{u}_{d}(t, \xi, y)-\int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))]\cdot\hat{\theta}(t, \xi, z)dz-\hat{\theta}_{y}(t,\xi, y)+\hat{\theta}_{y}(t, \xi, 0)$(2.86)$arrow-\frac{U_{1}(y)-U_{1}(0)}{U_{1}’(0)}[i\xi\cdot\^{u}_{h0}(\xi, 0)]e^{-\iota t(\xi_{1}+k\xi_{2})U_{1}(0)}$,as$tarrow+\infty.$

Thereby, (2.78) follows from (2.86) and Parseval’s identity.
(we may assume that $i=1$ ) has a non-degenerate critical
If $U_{2}(y)=kU_{1}(y)$ for some constant and, by an application of the method of stationary phase topoint at $y=y_{0}$ , then for $y>y_{0}$ and$k$$U_{i}(y)$$\xi_{1}\neq-k\xi_{2}$110
BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW

(2.83) yields that when $tarrow+\infty,$$\hat{u}_{d}(t,\xi, y)-\int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))] \hat{\theta}(t,\xi, z)dz-\hat{\theta}_{y}(t_{\rangle}\xi, y)+\hat{\theta}_{y}(t, \xi, 0)$(2.87)$\sim sgn(\xi_{1}+k\xi_{2})\sqrt{2\pi t|\xi_{1}+k\xi_{2}|}\cdot[\xi\cdot\hat{u}_{h0}(\xi, y_{0})]\frac{U_{1}(y)-U_{1}(y_{0})}{\sqrt{|U_{1}’(y_{0})|}}$. $\exp\{-it(\xi_{1}+k\xi_{2})U_{1}(y_{0})-isgn((\xi_{1}+k\xi_{2})U_{1}"(y_{0}))\pi/4\}.$
By using Parseval’s identity in the above equality, we obtain (2.79). Then, from the uniform boundedness ofandwith respect to in Proposition 2.6, we obtain (2.80) for suﬃcently large
If $U_{2}(y)\neq kU_{1}(y)$ for any, then there is a pointsuch that
$t.$

$t$

$\Vert\nabla_{h}\theta\Vert(t, y)$$\Vert\theta_{y}\Vert(t, y)$$k\in \mathbb{R}$(2.88)$y_{0}$$U_{1}’(y_{0})U_{2}"(y_{0})\neq U_{2}’(y_{0})U_{1}"(y_{0})$.

We may assume that $U_{1}’(y_{0})>0$ and $U_{1}’(y_{0})U_{2}"(y_{0})-U_{2}’(y_{0})U_{1}"(y_{0})>0$ . Then, we aﬃrm that for any $\delta>0,$
there is a intervalsuch that$S_{\delta}\subseteq(y0-\delta, y_{0}+\delta)$(2.89)$U_{1}’(y)>0, U_{2}’(y)\neq 0, U_{1}’(y)U_{2}"(y)-U_{2}’(y)U_{1}"(y)>0, \forall y\in S_{\delta}.$Indeed, from (2.88) and the smoothness of $U_{1}(y)$ , $U_{2}(y)$ we know that there is a$y\in(y0-\delta_{0_{\rangle}}y_{0}+\delta_{0})$$\delta_{0}>0$such that for any,$U_{1}’(y)>0, U_{1}’(y)U_{2}"(y)-U_{2}’(y)U_{1}"(y)>0.$

Then, by virtue of (2.88) again we have $U_{2}(y_{1})\neq 0$ for some, thus we take some smallneighborhood of as the required set . Moreover, (2.89) implies that the functions
is monotonicaly
Next,
increasing in .
denote by$y_{1}\in(y_{0}-\delta_{0}, y_{0}+\delta_{0})$$\frac{U}{U}\omega’1(y)$$S_{\delta}$$y_{1}$$S_{\delta}$(2.90)$I_{\delta}^{R}$$:=\{\xi=(\xi_{1}, \xi_{2})\in \mathbb{R}^{2}\backslash \{0\}$;$|\xi|\leq R$, and$\exists y\in S_{\delta},$$s.t.$ $\xi\cdot U_{h}’(y)=0\},$in , we know that the point
nicityand from thesatisfying $\xi\cdot U_{h}’(y)=0$ for, it is easy to know that the measure ofis unique. Moreover, by virtue of the continuity ofispositive, i.e.,. For $y>y_{0}$ and any, there exists a uniquewithsuch that, and then, from (2.89) we have
. For suchin (2.82) it yields that by anapplication of the method of stationary phase,$mon\circ t\circ$$\circ f_{U_{1}}^{U’}\simeq_{(y)}\omega$$\xi\in I_{\delta}^{R}$$y\in S_{\delta}$$S_{\delta}$$U_{h}’(y)$$m(I_{\delta}^{R})>0$$\xi\in I_{\delta}^{R}$$\xi\cdot U_{h}’(y_{\xi})=0$$I_{\delta}^{R}$$\delta\leq y-y_{0}$$\xi\cdot U_{h}"(y_{\xi})\neq 0$$y_{\xi}\in S_{\delta}$$(\xi, y)$$\hat{u}_{d}(t,\xi, y)-\int_{0}^{y}[i\xi\cdot(U_{h}(y)-U_{h}(z))]\cdot\hat{\theta}(t,\xi, z)dz-\hat{\theta}_{y}(t, \xi, y)+\hat{\theta}_{y}(t, \xi, 0)$(2.91)$\sim\sqrt{}\frac{2\pi t}{|\xi\cdot U_{h}’(y_{\xi})|} [\xi\cdot(U_{h}(y)-U_{h}(y\epsilon))][\xi\cdot\hat{u}_{h0}(\xi,y_{\xi})]$

Note that when$\delta$

is small enough, we have that for any$\xi\in I_{\delta}^{R},$$| \frac{\xi\cdot(U_{h}(y)-U_{h}(y_{\xi}))}{\sqrt{|\xi U_{h}"(y_{\xi})|}}[\xi\cdot\hat{u}_{h0}(\xi,y_{\xi})]|\geq|\frac{\xi\cdot(U_{h}(y)-U_{h}(y_{0}))}{2\sqrt{|\xi U_{h}"(y_{0})|}}[\xi\cdot\hat{u}_{h0}(\xi, y_{0})]|.$

Thus, for suﬃciently large$t$

we obtain that by using Parseval’s identity in (2.91),$\Vert u_{d}-\theta_{y}(t, x’, y)+\theta_{y}(t, x’, 0)-\int_{0}^{y}[(U_{h}(y)-U_{h}(z))\cdot\nabla_{h}\theta(t,x’, z)]dz\Vert(t,y)$$\geq\frac{\sqrt{2\pi t}}{4}\Vert\frac{\xi\cdot(U_{h}(y)-U_{h}(y_{0}))}{\sqrt{|\xi U_{h}"(y_{0})|}}[\xi\cdot\^{u}_{h0}(\xi, y_{0})]\Vert_{L_{\xi}^{2}(I_{\delta}^{R})},$

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CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

and then, combining with the uniform boundedness of
sition 2.6, it implies that$\Vert\theta_{y}\Vert(t, y)$and$\Vert\nabla_{h}\theta\Vert(t, y)$with respect to in Propo$t$$\Vert u_{d}\Vert(t, y)\geq\frac{\sqrt{2\pi t}}{8}\Vert\frac{\xi\cdot(U_{h}(y)-U_{h}(y_{0}))}{\sqrt{|\xi U_{h}"(y_{0})|}}[\xi\cdot\^{u}_{h0}(\xi, y_{0})]\Vert_{L_{\xi}^{2}(I_{\delta}^{R})}.$

Consequently, we get the estimate (2.81). Through analogous arguments as above, we can obtain similar
results forandas the ones of$u_{d}$$\nabla_{h}\cdot u_{h}$$u_{d}.$

3. APPENDIX
Now, we will give a formal derivative of boundary layer problem (1.1) of compressible flow. Consider thewith $d=2$ , 3,following problem in the domain$\mathbb{R}_{+}\cross \mathbb{R}_{+}^{d}$(3.1)$\{\begin{array}{l}\partial_{t}\rho+\nabla\cdot(\rho u)=0,\rho\{\partial_{t}u+(u\cdot\nabla)u\}+\nabla p(\rho, \theta)=0,c_{V}\rho\{\partial_{t}\theta+(u\cdot\nabla)\theta\}+p(\rho, \theta)\nabla\cdot u=\epsilon\triangle\theta,\end{array}$with $x’=(x_{1}, \cdots, x_{d-1})\in \mathbb{R}^{d-1}$ and the scale variable
where $t>0$ , the spatial variables$x_{d}>0,$
is theis the density, $u=(u_{1}, \cdots, u_{d})^{T}$ is the velocity, is the absolute temperature,is the coeﬃcient of heat conduction.
pressure, the constant $c_{V}>0$ is the specific heat capacity,
For the equations (3.1), we endow they with the following boundary values:$x=(x’, x_{d})\in \mathbb{R}_{+}^{d}$$p(\rho, \theta)$$\theta$$\rho$$\epsilon=\epsilon(\rho, \theta)$(3.2)$u_{d}|_{x_{d}=0}=0, [\alpha\partial_{x_{d}}\theta+\beta\theta]|_{x_{d}=0}=\gamma,$

where $\alpha=\alpha(t, x \beta=\beta(t, x’)$ andmodel for the problem $(3.1)-(3.2)$ ,are given functions. In the paper, we consider the ideal gas$\gamma=\gamma(t, x’)$

(3.3)$p(\rho, \theta)=R\rho\theta$

to thewith a positive constant $R$ . We are concerned with the asymptotic behavior of solutions
problem $(3.1)-(3.2)$ when the heat conduction coeﬃcient tends zero, i.e.,, solutions
to problem $(3.1)-(3.2)$ tend to
Formally, we can obtain that when:
, which satisfy the following compressible non-isentropic Euler equations in$(\rho, u, \theta)(t, x)$$\epsilonarrow 0.$$\epsilonarrow 0$$(\rho, u, \theta)(t, x)$$(\rho^{e}, u^{e}, \theta^{e})(t, x)$$\mathbb{R}+\cross \mathbb{R}_{+}^{d}$(3.4)$\{\begin{array}{l}\partial_{t}\rho^{e}+\nabla\cdot(\rho^{e}u^{e})=0,\rho^{e}\{\partial_{t}u^{e}+(u^{e}\cdot\nabla)u^{e}\}+R\nabla(\rho^{e}\theta^{e})=0,c_{V}\rho^{e}\{\partial_{t}\theta^{e}+(u^{e}\cdot\nabla)\theta^{e}\}+R\rho^{e}\theta^{e}(\nabla\cdot u^{e})=0\end{array}$with the boundary condition(3.5)$u_{d}^{e}|_{x_{d}=0}=0.$

So, the inconsistent of boundary conditions between (3.2) and (3.5) leads to the appearance of boundary layer.
Since the diﬀusion terms are important in the boundary layer and should be balanced by the convective terms,
and note that the vertical component of velocity field vanishes at the boundary in the problem $(3.1)-(3.2)$ , wemay just consider the characteristic boundary layers, that is, the sizes of boundary layers are . Therefore,as
we express solutions to $(3.1)-(3.2)$ via$\sqrt{\epsilon}$$(\rho^{\epsilon}, u^{\epsilon}, \theta^{\epsilon})$$( \rho, u, \theta)(t, x)=(\rho(t, x’, \frac{x_{d}}{\sqrt{\epsilon}}), u_{h}(t, x’, \frac{x_{d}}{\sqrt{\epsilon}}), \sqrt{\epsilon}\{\frac{u_{d}(t,x^{\prime\underline{x}_{A}}\sqrt{\epsilon})}{\sqrt{\epsilon’}}\}, \theta(t,x’, \frac{x_{d}}{\sqrt{\epsilon}}))$$=(\rho^{\epsilon}, u_{h}^{\epsilon},u_{d}^{\epsilon}, \theta^{\epsilon})(t,x’, y)$,, the tangential component $u_{h}=(u_{1}, \cdots, u_{d-1})$ of the velocity
where we introduce the scale variable
field and note that the scale normal velocity
isof the original velocity . In these new variables$y=-x\sqrt{\epsilon}A$$u$$u_{d}^{\epsilon}$$\frac{1}{\sqrt{\epsilon}}$$u_{d}$

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BOUNDARY LAYERS OF INVISCID COMPRESSIBLE NON-ISENTROPIC FLOW

above, the problem (3.1) (3.2) reads that in the domain $\{(t,x’,y):t>0, x’\in \mathbb{R}^{d-1}, y>0\}$ :(3.6)

where the derivatives
andSimilar as the hypothesis of Prandtl boundary layers for the incompressible flow, we assume that solutionsof (3.6) can be approximated as follows:$\nabla_{h}=(\partial_{x_{1}}, \cdots, \partial_{x_{d-1}})^{T}$(3.7)$\Delta_{h}=\partial_{x_{1}}^{2}+\cdots+\partial_{x_{d-1}}^{2}.$$( \rho^{\epsilon}, u_{h}^{\epsilon}, u_{d}^{\epsilon}, \theta^{\epsilon})(t,x’,y)\approx(\rho^{e}, u_{h}^{e}, \frac{u_{d}^{e}}{\sqrt{\epsilon}}, \theta^{e})(t,x’, \sqrt{\epsilon}y)+(\rho^{b}, u_{h}^{b},u_{d}^{b},\theta^{b})(t, x’, y)$,, and the boundary
where
denotes the Euler flow given by $(3.4)-(3.5)$ withlayer profile
decrease rapidly as $yarrow+\infty$ . We plug the ansatz (3.7) into the problem(3.6) and take the leading terms with respect to . By virtue of (3.5) and then the asymptotic expansion for
the Euler flow$(\rho^{e}, u^{e}, \theta^{e})(t, x)$$u^{e}=(u_{h}^{e},u_{d}^{e})^{T}$$(\rho^{b}, u_{h}^{b}, u_{d}^{b}, \theta^{b})(t, x’, y)$$\epsilon$$( \rho^{e}, u_{h)}^{e}\frac{u_{d}^{e}}{\sqrt{\epsilon}}, \theta^{e})(t, x’, \sqrt{\epsilon}y)=(\rho^{e}, u_{h}^{e}, y\partial_{x_{d}}u_{d}^{e}, \theta^{e})(t, x_{\rangle}’0)+O(\sqrt{\epsilon})$,we obtain that the new boundary layer profile$(\rho, u_{h},u_{d}, \theta)(t_{\rangle}x’, y):=(\rho^{e}, u_{h}^{e}, y\partial_{x_{d}}u_{d}^{e}, \theta^{e})(t,x_{\rangle}’0)+(\rho^{b}, u_{h}^{b}, u_{d}^{b}, \theta^{b})(t,x’, y)$satisfies the following problem in

(3.8)

:

$\{\begin{array}{l}\partial_{t}\rho+\nabla_{h}\cdot(\rho u_{h})+\partial_{y}(\rho u_{d})=0,\rho\{\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}\}+R\nabla_{h}(\rho\theta)=0,\partial_{y}(\rho\theta)=0,cv\rho\{\partial_{t}\theta+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})\theta\}+R\rho\theta(\nabla_{h}\cdot u_{h}+\partial_{y}u_{d})=\partial_{y}^{2}\theta,u_{d}|_{\fbox{Error::0x0000}-0}=0, \lim_{yarrow+\infty}( \rho, u_{h}, \theta)=(\rho^{e}, u_{h}^{e}, \theta^{e})(t, x’, 0) ,\end{array}$with the boundary values for$\theta$

:

(3.9)

where

$\mathbb{R}+\cross \mathbb{R}_{+}^{d}$$\{\begin{array}{ll}\partial_{y}\theta|_{y=0}=0, when \alpha\neq 0,\theta|_{\ulcorner-0}=\theta^{0}(t,x’) , when \alpha=0,\end{array}$, provided
Then, from the third equation and boundary conditions in (3.8), it implies that$\theta^{0}(t, x’)$$\beta\neq 0.$$:= \frac{\gamma(t,x’)}{\beta(t,x}$(3.10)$( \rho\theta)(t,x’,y)\equiv(\rho^{e}\theta^{e})(t,x’,O)=\frac{p^{e}(t,x’,0)}{R},$where is the pressure of Euler flow and $p^{e}>0$ . The relation (3.10) shows that there isn’t boundary layerof size offor the pressure. Note that for the problem (3.11) endowed with the Neumann boundary, it is easy to check thatcondition for in (3.9), i.e.,$p^{e}$$\mathcal{O}(\sqrt{\epsilon})$$\theta$$\partial_{y}\theta|_{y=0}=0$$(\rho,u_{h}, \theta)(t, x’, y)=(\rho^{e}, u_{h}^{e},\theta^{e})(t,x’, 0) , u_{d}(t,x’, y)=y\partial_{x_{d}}u_{d}^{e}(t, x’, 0)$satisfies the problem (3.8). Indeed, we can investigate this by restricting the equations (3.4) on the boundary$\{x_{d}=0\}$ and using the boundary condition (3.5). In this case, it means that the state
doaen’t exist. Therefore, we focus on the problem (3.8) with the Dirichlet boundary
boundary layers of size ofcondition for in (3.9).
$(\rho, u, \theta)$$\mathcal{O}(\sqrt{\epsilon})$$\theta$113
CHENG-JIE LIU, YA-GUANG WANG, AND TONG YANG

Plugging (3.10) into the problem $(3.8)-(3.9)$ , which can be reduced as the following problem infor the profile $(u_{h}, u_{d}, \theta)(t, x’,y)$ :(3.11)$\mathbb{R}+\cross \mathbb{R}_{+}^{d}$$\{\begin{array}{l}\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}+\frac{R\theta}{P}\nabla_{h}P=0,\partial_{t}\theta+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})\theta=\frac{R}{(R+c_{V})P}\theta(\partial_{y}^{2}\theta+u_{h}\cdot\nabla_{h}P+P_{t}) ,\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\frac{R}{(R+c_{V})P}\partial_{y}^{2}\theta-\frac{c_{V}}{(R+c_{V})P}(u_{h}\cdot\nabla_{h}P+P_{t}) ,(u_{d}, \theta)|_{\fbox{Error::0x0000}-0}=(0, \theta^{0}(t, x , \lim_{yarrow+\infty}(u_{h}, \theta)=(U_{h}, \Theta)(t, x)\end{array}$where the known functions$(P, U_{h}, \Theta)(t, x’) :=(p^{e}, u_{h}^{e},\theta^{e})(t, x’,0)$

given by the Euler flow, and satisfy by using $(3.4)-(3.5)$ ,(3.12)$\{\begin{array}{l}\partial_{t}U_{h}+U_{h}\cdot\nabla_{h}U_{h}+\frac{R\Theta}{P}\nabla_{h}P=0,\partial,\Theta+U_{h}\cdot\nabla_{h}\Theta-\frac{R\Theta}{(R+c_{V})P}\cdot(P_{t}+U_{h}\cdot\nabla_{h}P)=0.\end{array}$We endow the problem (3.11) with the initial data

(3.13)$(u_{h}, \theta)(0, x’, y)=(u_{h0}, \theta_{0})(x’, y)$and under the compatibility condition of(3.14)$u_{h0}$,:$\lim_{yarrow+\infty}u_{h0}=U_{h}(0,x$we can remove the infinity condition forholds automatically from:problem in$(3.12)_{1}$$u_{h}$

as $yarrow+\infty$ in (3.11), since the conditionand (3.14).

$yarrow+\infty hmu_{h}=U_{h}(t, x’)$

Therefore, we obtain the following initial-boundary value$\mathbb{R}+\cross \mathbb{R}_{+}^{d}$(3.15)$\{\begin{array}{l}\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}+\frac{R\theta}{P}\nabla_{h}P=0,\partial_{t}\theta+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})\theta=\frac{\kappa\theta}{P}(\partial_{y}^{2}\theta+u_{h}\cdot\nabla_{h}P+P_{t}) ,\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\frac{\kappa}{P}\partial_{y}^{2}\theta-\frac{1-\kappa}{P}(u_{h}\cdot\nabla_{h}P+P_{t}) ,(u_{d}, \partial_{y}\theta)|_{y=0}=0, \lim_{yarrow+\infty}\theta(t, x, y)=\Theta(t, x(u_{h}, \theta)|_{t=0}=(u_{h0}, \theta_{0})(x’, y) ,\end{array}$with the constant. In this paper, we focus on a simple case of the problem (3.15), i.e., the pressureof the outflow is a positive function depending only on the variable$\kappa:=\frac{R}{R+c_{V}}$$P(t_{\rangle}x’)$$t,$$P(t, x’)\equiv P(t)>0.$

Consequently, the problem (3.15) is reduced as follows:(3.16)$\{\begin{array}{l}\partial_{t}u_{h}+(u_{h}\cdot\nabla_{h}+u_{d}\partial_{y})u_{h}=0,\partial_{t}\theta+(u_{h}\cdot\nabla_{\hslash}+u_{d}\partial_{y})\theta=\frac{\kappa}{P}\theta\partial_{y}^{2}\theta+\kappa_{\vec{P}}P\theta,\nabla_{h}\cdot u_{h}+\partial_{y}u_{d}=\frac{\kappa}{P}\partial_{y}^{2}\theta-\frac{(1-\kappa)P_{t}}{P},(u_{d}, \theta)|_{y=0}(0, \theta^{0}(t, x , \lim_{yarrow+\infty}\theta(t, x, y)=\Theta(t, x(u_{h}, \theta)|_{t=0}=(u_{h0}, \theta_{0})(x’, y) .\end{array}$

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$Z_{\rangle}$

CHENGJIE LIU

DEPARTMENT OF MATHEMATICS, CITY UNIVERSITY OF HONG KONG, HONG KONG, P. R. CHINA
address: cjliusjtuQgmail. com$E-ma\iota l$

YA-GUANG WANG
DEPARTMENT OF MATHEMATICS, MOE-LSC AND SHL-MAC, SHANGHAI JIAO TONG UNIVERSITY, SHANGHAI, 200240, P. R.
CHINA-mail address: ygwangQs tu. edu. cn
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TONG YANG
DEPARTMENT OF MATHEMATICS, SHANGHAI JIAO TONG UNIVERSITY, SHANGHAI, P. R. CHINA
DEPARTMENT OF MATHEMAT1CS, CITY UNIVERSITY OF HONG KONG, HONG KONG, P. R. CHINA: matyangQcityu. edu. hk
-mail
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