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## Math 597F, Notes 6: Linear inviscid stability theory

We study the linearization of 2D Navier-Stokes around a boundary layer ${\vec v_\mathrm{app} =[\bar u, \sqrt \nu \bar v](t', x', y'/\sqrt \nu)}$, in which ${(t',x',y')}$ denotes the original coordinates for Navier-Stokes. Materials in this lecture are drawn from the joint paper(s) with Grenier and Guo (here and here).

In the scaled variables ${(t,x,z): = (t',x',y') / \sqrt \nu}$, the boundary layer varies slowly in ${(x,t)}$ variables:

$\displaystyle \vec v_\mathrm{app} = [\bar u, \sqrt \nu \bar v](\sqrt \nu t, \sqrt \nu x, z).$

For this reason, we start the study of the linearized problem around the boundary layer of the form ${\vec v_\mathrm{app} = [U(z),0]}$. The linearization reads

\displaystyle \begin{aligned} \vec v_t + \vec v_\mathrm{app} \cdot \nabla \vec v + \vec v\cdot \nabla \vec v_\mathrm{app} + \nabla q &= \frac 1R \Delta \vec v, \qquad R = \frac{1}{\sqrt \nu} \\ \nabla \cdot \vec v & =0, \end{aligned}\ \ \ \ \ (1)

with ${\vec v = [u,v] = 0}$ on the boundary ${z=0}$. Let ${\omega = \nabla \times \vec v = u_z - v_x}$ be the vorticity. Then, ${\omega}$ solves

$\displaystyle (\partial_t + U \partial_x)\omega + v U'' = \frac 1R \Delta \omega.$

We write this equation in term of the stream function defined through ${u = \psi_z}$ and ${v = -\psi_x}$. We note that ${\omega = u_z - v_x = \Delta \psi}$. The vorticity equation then reads

$\displaystyle (\partial_t + U \partial_x) \Delta \psi - U'' \psi_x = \frac 1R \Delta^2 \psi.$

Following the classical linear stability theory (see, for instance, the nice book by Drazin and Reid), one searches for the solution ansatz: ${\psi = e^{i\alpha (x-c t)} \phi(z)}$, in which ${\alpha}$ is a real-valued positive wave number, ${\lambda = -i\alpha c}$ denotes the eigenvalue of the linearized Navier-Stokes operator (and so, ${c}$ is a complex number), and ${\phi(z)}$ is a complex-valued function. If ${\psi}$ is a solution to the vorticity equation, then its real part is also a solution. The pressure ${q}$ is then simply recovered by taking the divergence of (1), giving

$\displaystyle -\Delta q = \nabla \cdot (U \nabla^\perp \psi_x - \psi_x [U',0]) , \qquad \nabla^\perp = [\partial_z, -\partial_x].$

Putting ${\psi = e^{i\alpha (x-c t)} \phi(z)}$ into the vorticity equation yields the well-known Orr-Sommerfeld fourth-order ODE equations:

\displaystyle \left \{ \begin{aligned} (U-c) (\partial_z^2 - \alpha^2) \phi - U'' \phi &= \epsilon (\partial_z^2 - \alpha^2)^2 \phi , \qquad \epsilon = \frac{1}{i\alpha R} \\ \phi_{\vert_{z=0}} = \phi'_{\vert_{z=0}} &= 0, \qquad \lim_{z\rightarrow \infty} \phi(z) =0. \end{aligned} \right. \ \ \ \ \ (2)

Here, the boundary conditions are exactly from ${[u,v] =0}$ on the boundary. Note that the stream function is defined up to a constant, and so we simply take ${\phi =0}$ on the boundary. The interest is to determine whether there is a triple of solutions ${(c,\alpha, \phi)}$ to (2) so that the complex number ${c}$ has a positive imaginary part, which corresponds to the existence of a growing solution ${\vec v \approx e^{\theta_0 t}}$ of the linearized Navier-Stokes problem (in the original coordinates:${\vec v \approx e^{\theta_0 t'/\sqrt{\nu}}}$, yielding instability in a vanishing time!). Clearly, when ${\epsilon \gg 1}$ (that is, in the low Reynolds number regime), there is no such an unstable solution: ${U(z)}$ is spectrally stable. We are interested in the singular perturbation problem: ${\epsilon \rightarrow 0}$ (or equivalently, large Reynolds number limit ${R = \frac{1}{\sqrt \nu}\rightarrow \infty}$). Two scenarios:

• Unstable profiles: The boundary layer ${U(z)}$ is unstable in the limit ${\epsilon=0}$: there exists a triple ${(\alpha_0, c_0, \phi_0)}$, with the imaginary part ${\Im~c_0>0}$, that solves the limiting Rayleigh problem:

\displaystyle \left \{ \begin{aligned} (U-c) (\partial_z^2 - \alpha^2) \phi - U'' \phi &= 0 \\ \phi_{\vert_{z=0}} = 0, \qquad \lim_{z\rightarrow \infty} \phi(z) &=0. \end{aligned} \right. \ \ \ \ \ (3)

A perturbative analysis can be used to construct a solution ${(\alpha_\epsilon, c_\epsilon, \phi_\epsilon)}$ to the singular perturbation problem (2) so that ${c_\epsilon = c_0 + \mathcal{O}(\epsilon)}$; in particular, ${\Im~c_\epsilon >0}$.

• Stable profiles: The boundary layer ${U(z)}$ is stable to the Rayleigh problem. For instance, this is the case when ${U(z)}$ has no inflection point. For such a stable profile, all the spectrum of the Rayleigh equation is imbedded on the imaginary axis: ${\Re~(-i\alpha_0 c_0) = \alpha_0 \Im ~c_0 = 0}$, and thus it is not clear whether a perturbative argument to construct solutions ${(c_\epsilon,\phi_\epsilon)}$ to (2) would yield stability (${\Im~c_\epsilon <0}$) or instability (${\Im ~c_\epsilon >0}$). It were Heisenberg (1924) and then C. C. Lin (1940s) and Tollmien (1940s) who are the first to use asymptotic expansions to study the singular perturbation problem. There in the book of Drazin and Reid, it is documented that there are lower and upper marginal stability branches ${\alpha_\mathrm{low}(R), \alpha_\mathrm{up}(R)}$ so that whenever ${\alpha\in [\alpha_\mathrm{low}(R),\alpha_\mathrm{up}(R)]}$, there exist an unstable eigenvalue ${c_R}$ and an eigenfunction ${\phi_R(z)}$ to the Orr-Sommerfeld problem. For large Reynolds number ${R}$, the asymptotic behavior of ${\alpha_\mathrm{low}}$ and ${\alpha_\mathrm{up}}$ is estimated, depending on the background profile:
• a. Plane Poiseuille flow:
• $\displaystyle \alpha_\mathrm{low}(R) \approx R^{-1 / 7}, \quad \alpha_\mathrm{up}(R) \approx R^{-1/11} .$

• b. Boundary layer profiles:

$\displaystyle \alpha_\mathrm{low}(R) \approx R^{-1 / 4}, \quad \alpha_\mathrm{up}(R) \approx R^{-1/6} .$

~

Before we study the singular perturbation problem: the Orr-Sommerfeld ODEs. Let us focus on the Rayleigh problem (inviscid stability theory):

\displaystyle \left \{ \begin{aligned} Ray_\alpha(\phi): = (U-c) (\partial_z^2 - \alpha^2) \phi - U'' \phi &= 0 \\ \phi_{\vert_{z=0}} = 0, \qquad \lim_{z\rightarrow \infty} \phi(z) &=0, \end{aligned} \right. \ \ \ \ \ (4)

for each fixed parameters ${\alpha}$ and ${c}$. We shall also construct the inverse of the Rayleigh operator ${Ray_\alpha(\cdot)}$.

1.1. Inviscid stability criteria

We recall two well-known instability criteria for the basic velocity profiles ${U(z)}$ of the Rayleigh problem (4).

~
Rayleigh’s inflexion-point criterium (Rayleigh, 1880): A necessary condition for instability is that the basic velocity profile must have an inflection point. Indeed, assume that there is a triple ${(c, \alpha,\phi)}$ solving (4) so that ${\Im ~c>0}$. Multiplying by ${\bar \phi/(U-c)}$ to the Rayleigh equation (4) and taking integration by parts yield

$\displaystyle \int_0^\infty (|\partial_z \phi|^2 + \alpha^2 |\phi|^2) \;dz+ \int_0^\infty \frac{U''}{U-c}|\phi|^2 \;dz=0,\ \ \ \ \ (5)$

$\displaystyle \Im ~c \int_0^\infty \frac{U''}{|U-c|^2}|\phi|^2 \;dz =0.\ \ \ \ \ (6)$

Thus, the instability condition ${\Im ~c>0}$ must imply that ${U''}$ changes its sign. This gives the Rayleigh criterium. Consequently, boundary layer profiles ${U(z)}$ with no inflection point are spectrally stable to the linearized Euler equations. Additionally, there are stable flows with inflection point:

~
Fjortoft criterium (1950): A necessary condition for instability is that ${U'' (U - U(z_c))<0}$ somewhere in the flow, where ${z_c}$ is a point at which ${U''(z_c) =0}$. Indeed, let ${z_c}$ be the inflection point and consider the real part of the identity (5):

$\displaystyle \int_0^\infty (|\partial_z \phi|^2 + \alpha^2 |\phi|^2) \;dz+ \int_0^\infty \frac{U'' (U-\Re ~c)}{|U-c|^2}|\phi|^2 \;dz=0.$

Adding to this the identity (6):

$\displaystyle (\Re ~c - U(z_c))\int_0^\infty \frac{U''}{|U-c|^2}|\phi|^2 \;dz =0,$

we obtain

$\displaystyle \int_0^\infty \frac{U'' (U-U(z_c))}{|U-c|^2}|\phi|^2 \;dz=-\int_0^\infty (|\partial_z\phi|^2 + \alpha^2 |\phi|^2) \;dz <0,$

from which the Fjortoft criterium follows.

1.2. Rayleigh problem for unstable profiles

To construct Orr-Sommerfeld solutions, we shall need to introduce appropriate iteration. We start with the Rayleigh problem for unstable profiles. Precisely, we shall construct an exact inverse for the Rayleigh equation and so find the complete solution to

$\displaystyle Ray_\alpha(\phi) := (U-c)(\partial_z^2 - \alpha^2)\phi - U'' \phi = f, \qquad z\ge 0, \ \ \ \ \ (7)$

for each fixed ${\alpha, c}$, under the assumption that

$\displaystyle |U(z) - c|\ge \theta , \qquad |\partial_z^k[U(z) - U_+] |\le C e^{-\eta_0 z}, \qquad z\ge 0,\quad k\ge 0\ \ \ \ \ (8)$

for some positive constants ${\theta, \eta_0, C}$, and some constant ${U_+}$. Such an assumption holds for the unstable profile ${U(z)}$, since ${\Im~c \not =0}$. We shall use the function spaces ${X_{\eta,p}}$, for ${p\ge 0, \eta > 0}$, to denote the spaces consisting of functions ${f = f(z)}$ that are ${C^p}$ smooth and satisfy

$\displaystyle |\partial_z^k f(z)| \le C e^{-\eta z}$

for all ${z\ge 0}$ and all ${k = 0,\cdots,p}$. The best constant that satisfies the above inequality defines a norm, denoted by ${\|\cdot\|_{\eta,p}}$, on the Banach space ${X_{\eta,p}}$. We denote ${X_\eta, \|\cdot \|_\eta}$ in place of ${X_{\eta,0}, \|\cdot \|_{\eta,0}}$, respectively.

1.2.1 Homogenous solutions.

We obtain the following lemma:

Lemma 1 There are two independent solutions ${\phi_{\alpha,\pm} \in C^\infty ([0,\infty))}$ to the homogenous equation ${Ray_\alpha(\phi) = 0}$ such that

$\phi_{\alpha,\pm}(z) =e^{\pm \alpha z} \Big[ 1 + \mathcal{O}(e^{-\eta_0 z})\Big]$

for all ${z\ge 0}$.

Proof:

Let us denote ${W = (\phi, \partial_z\phi)^{tr}}$. The homogenous equation ${Ray_\alpha(\phi) = 0}$ becomes

$\displaystyle W' = A W, \qquad A = \begin{pmatrix} 0&1 \\ \alpha^2 + \frac{U''}{U-c} &0\end{pmatrix} . \ \ \ \ \ (8)$

We denote by ${A_+}$ the limit of ${A(z)}$ as ${z\rightarrow \infty}$. Eigenvalues of ${A_+}$ are ${\mu_\pm = \pm \alpha}$. Let ${V_\pm= (1,\pm \alpha)^{tr}}$ be the associated eigenvectors. We construct two independent solutions of (8) of the form

$\displaystyle W_\pm = e^{\mu_\pm z} V(z) , \qquad \lim_{z\rightarrow \infty}V(z) = V_\pm.$

Let us first consider the ${(-)}$ case: ${\mu_-=-\alpha}$. Plugging the above ansatz into (8), ${V(z)}$ solves

$\displaystyle V' = (A_+ - \mu_-) V + (A - A_+) V.$

Since all eigenvalues of ${A_+ - \mu_-}$ have nonnegative real parts, we have ${\|e^{(A_+ - \mu_-)z} \|\le C_\delta e^{\delta|z|}}$ for ${z\le 0}$, for arbitrarily small ${\delta>0}$. Thanks to the decay assumption on the boundary layer profile ${U(z)}$, there holds ${\|A (z)- A_+\| \le C e^{-\eta_0 |z|}}$. Here, ${\|\cdot \|}$ denotes the standard matrix norm. The Duhamel formula yields

$\displaystyle V(z) = V_- - \int_z^\infty e^{(A_+ - \mu_-) (z-y)} (A(y) - A_+) V(y) \; dy .$

Denote by ${TV(z)}$ the right hand side of the above identity. We shall show that ${T}$ is a contractive map from ${L^\infty([M,\infty))}$ into itself, for sufficiently large ${M}$. Indeed, for ${V_1,V_2 \in L^\infty([M,\infty))}$, we get

\displaystyle \begin{aligned} |TV_1(z) - TV_2(z)| &\le \int_z^\infty \|e^{(A_+ - \mu_-) (z-y)} \| \|(A(y) - A_+)\| \|V_1- V_2\|_{L^\infty} \; dy \\ &\le C\|V_1- V_2\|_{L^\infty}\int_z^\infty e^{\delta (y-z)}e^{-\eta_0 y} \; dy \\ &\le C e^{-\eta_0 z}|V_1- V_2\|_{L^\infty} \le C e^{-\eta_0 M}\|V_1- V_2\|_{L^\infty}, \end{aligned}

for all ${z\ge M}$. Taking ${M}$ large enough so that ${C e^{-\eta_0 M} <1/2}$, the map ${T}$ is contractive. Hence, there exists a solution ${V\in L^\infty([M,\infty))}$ such that ${V = TV}$, or equivalently ${e^{-\alpha z}V(z)}$ solves (8) on ${[M,\infty)}$. By the standard ode theory, the solution ${V}$ on ${[M,\infty)}$ can be extended to be a global solution on ${[0,\infty)}$, since the right-hand side of (8) is uniformly bounded.

Similarly, with the ansatz solution: ${W_+ = e^{\mu_+ z} V(z)}$, there holds

$\displaystyle V' = (A_+ - \mu_+) V + (A - A_+) V,$

in which the matrix ${A_+ - \mu_+}$ has only nonpositive eigenvalues. It follows that ${\|e^{(A_+ - \mu_+)z} \|\le C_\delta e^{\delta|z|}}$ for ${z\ge 0}$, for arbitrarily small ${\delta>0}$. The solution is then constructed via the Duhamel formula:

$\displaystyle V(z) = V_+ + \int_M^z e^{(A_+ - \mu_+) (z-y)} (A(y) - A_+) V(y) \; dy .$

It is easy to show that such a solution exists and satisfies the claimed bound. The lemma is proved. $\Box$

1.2.2 Exact Rayleigh solver. The Green function ${G_{\alpha}(x,z)}$ of the ${Ray_\alpha}$ operator can be defined by

$\displaystyle G_{\alpha}(x,z) = \frac{1}{ E(\alpha,c) (U(x)-c) }\left\{ \begin{array}{rrr} \phi_{\alpha,-}(z) \phi_{\alpha,+}(x), \quad \mbox{if}\quad z>x,\\ \phi_{\alpha,-}(x) \phi_{\alpha,+}(z), \quad \mbox{if}\quad z

in which the so-called Evans function is defined by the Wronskian determinant

$\displaystyle E(\alpha, c): = W[\phi_{\alpha,-},\phi_{\alpha,+}](0) ,$

for each ${\alpha, c}$. Here, we note that the Evans function does not depend on ${z}$ (since the derivative of the Wronskian with respect to ${z}$ is zero). The exact Rayleigh solver can be defined by

\displaystyle \begin{aligned} RaySolver_\alpha(f) (z) &: = \int_0^\infty G_{\alpha}(x,z) f(x) dx. \end{aligned} \ \ \ \ \ (9)

We have the following simple convolution estimate.

Lemma 2 For any positive constant ${\eta\le \eta_0}$ and for each ${f \in X_\eta}$, there holds

$\displaystyle |\partial_z^k RaySolver_\alpha(f) (z) |\le C E(\alpha,c)^{-1} \|f\|_\eta e^{-\alpha z}, \ \ \ \ \ (10)$

for some positive constant ${C}$, and ${k = 0,1}$. In addition, we have

$\displaystyle |(\partial_z^2 - \alpha^2 )^k RaySolver_\alpha(f) (z) | \le CE(\alpha,c)^{-1} \|f\|_\eta e^{-\alpha z} e^{-\eta z}, \ \ \ \ \ (11)$

for ${k = 1,2}$.

Proof: By Lemma 1 and the assumption (7) on the profile ${U}$, we have

$\displaystyle |\partial_z^k G_{\alpha}(x,z)| \le C E(\alpha,c)^{-1} e^{-\alpha|x-z|},$

for all ${x,z\ge 0}$, and ${k\ge 0}$. This yields

\displaystyle \begin{aligned} \Big| \partial_z^k \int_0^\infty G_{\alpha}(x,z) f(x) dx \Big| \le C E(\alpha,c)^{-1} \|f\|_\eta \int_0^{\infty} e^{-\alpha |x-z|}e^{-\eta x}\; dx \le C E(\alpha,c)^{-1} \|f\|_\eta e^{-\alpha z}. \end{aligned}

This proves the first estimate stated in the lemma. For the second estimate, using the equation ${Ray_\alpha (\phi) = 0}$, we write

$\displaystyle (\partial_z^2 - \alpha^2) RaySolver_\alpha(f) (z) = \frac{U''}{U-c} RaySolver_\alpha(f) (z) .$

Similarly, ${(\partial_z^2 - \alpha^2)^2 RaySolver_\alpha(f)}$ involves precisely ${RaySolver_\alpha(f)}$ and its first derivative, multiplied by derivatives of the profile ${U}$. This proves at once the second estimate in the lemma, upon using the exponential localization of derivatives of ${U}$. $\Box$

1.3. Rayleigh problem for stable profiles

In this part, we shall construct an exact inverse for the Rayleigh operator ${Ray_\alpha}$ for small ${\alpha}$ in the case that ${U(z)}$ is stable. Precisely, we construct the inverse of the problem:

$\displaystyle Ray_\alpha(\phi) = (U-c)(\partial_z^2 - \alpha^2)\phi - U'' \phi = f \ \ \ \ \ (13)$

in which ${\alpha}$ and ${\Im~c}$ are sufficiently small. To do so, we first invert the Rayleigh operator ${Ray_0}$ when ${\alpha =0}$ by exhibiting an explicit Green function. We then use this inverse to construct an approximate inverse to ${Ray_\alpha}$ operator through the construction of an approximate Green function. Finally, the construction of the exact inverse of ${Ray_\alpha}$ follows by an iterative procedure. We introduce the function spaces. Let ${z_c}$ be some complex number so that ${U(z_c) = c}$, for each fixed complex number ${c}$. We will use the function spaces ${X_{p}^\eta}$, for ${p\ge 0}$, to denote the spaces consisting of measurable functions ${f = f(z)}$ such that the norm

$\displaystyle \| f\|_{X_p^\eta} : = \sup_{|z-z_c|\le 1 } \sum_{k=0}^p| (z-z_c)^k \partial_z^k f(z) | + \sup_{|z-z_c|\ge 1} \sum_{k=0}^p| e^{\eta z} \partial_z^k f(z) |$

is bounded. In case ${p=0}$, we simply write ${X_\eta, \|\cdot \|_\eta}$ in places of ${X^{\eta}_0, \|\cdot \|_{X_0^\eta}}$, respectively. We also introduce the function spaces ${Y_p^\eta \subset X_p^\eta}$, ${p\ge 0}$, such that for any ${f\in Y_p^\eta}$, the function ${f}$ additionally satisfies

$\displaystyle |f(z)| \le C, \qquad | \partial_z f(z) | \le C (1 + | \log (z - z_c) | ) , \qquad | \partial_z^k f(z) | \le C (1 + | z - z_c |^{1 - k} )$

for all ${|z-z_c|\le 1}$ and for ${2\le k \le p}$. The best constant ${C}$ in the previous bounds defines the norm ${\| f \|_{Y_p^\eta}}$. We will prove in this section the following proposition.

Proposition 3 Let ${p}$ be in ${\{0,1,2\}}$ and ${\eta>0}$. Assume that ${\Im c \not =0}$ and ${\alpha |\log \Im c|}$ is sufficiently small. Then, there exists an operator ${RaySolver_{\alpha,\infty} (\cdot) }$ from ${X_p^\eta}$ to ${Y^\alpha_{p+2}}$ (defined by (30)) so that

\displaystyle \begin{aligned} Ray_\alpha (RaySolver_{\alpha,\infty} (f)) &= f. \end{aligned} \ \ \ \ \ (14)

$\displaystyle \| RaySolver_{\alpha,\infty}(f)\|_{Y^\alpha_{p+2}} \le C \|f\|_{X_p^\eta}(1+|\log (\Im c)|) ,$

for all ${f \in X_p^\eta}$.

1.3.1 Case ${\alpha = 0}$

As mentioned, we begin with the Rayleigh operator ${Ray_0}$ when ${\alpha = 0}$. We will find the inverse of ${Ray_0}$. More precisely, we will construct the Green function of ${Ray_0}$ and solve

$\displaystyle Ray_0 (\phi) = (U-c) \partial_z^2 \phi - U'' \phi = f.\ \ \ \ \ (15)$

We first prove the following lemma.

Lemma 4 Assume that ${\Im c \not =0}$. There are two independent solutions ${\phi_{1,0},\phi_{2,0}}$ of ${Ray_0(\phi) =0}$ with the Wronskian determinant

$\displaystyle W(\phi_{1,0}, \phi_{2,0}) := \partial_z \phi_{2,0} \phi_{1,0} - \phi_{2,0} \partial_z \phi_{1,0} = 1.$

Furthermore, there are analytic functions ${P_1(z), P_2(z), Q(z)}$ with ${P_1(z_c) = P_2(z_c) = 1}$ and ${Q(z_c)\not=0}$ so that the asymptotic descriptions

$\displaystyle \phi_{1,0}(z) = (z-z_c) P_1(z) ,\qquad \phi_{2,0}(z) = P_2(z) + Q(z) (z-z_c) \log (z-z_c) \ \ \ \ \ (16)$

hold for ${z}$ near ${z_c}$, and

$\displaystyle | \phi_{1,0}(z) - V_+| \le Ce^{-\eta_0 |z|} , \qquad |\partial_z \phi_{2,0}(z) - \frac{1}{V_+} |\le Cze^{-\eta_0|z|}, \ \ \ \ \ (17)$

as ${|z|\rightarrow \infty}$, for some positive constants ${C,\eta_0}$ and for ${V_+ = U_+ - c}$. Here when ${z-z_c}$ is on the negative real axis, we take the value of ${\log (z-z_c)}$ to be ${ \log |z-z_c| - i \pi}$.

Proof: First, we observe that

$\displaystyle \phi_{1,0}(z) = U(z)-c$

is an exact solution of ${Ray_0(\phi) =0}$. In addition, the claimed asymptotic expansion for ${\phi_{1,0}}$ clearly holds for ${z}$ near ${z_c}$ since ${U(z_c) = c}$. We then construct a second particular solution ${\phi_{2,0}}$, imposing the Wronskian determinant to be one:

$\displaystyle W[\phi_{1,0},\phi_{2,0}] = \partial_z \phi_{2,0} \phi_{1,0} - \phi_{2,0} \partial_z \phi_{1,0} = 1.$

From this, the variation-of-constant method ${\phi_{2,0} (z)= C(z) \phi_{1,0}(z)}$ then yields

$\displaystyle \phi_{1,0} C \partial_z \phi_{1,0} + \ \phi_{1,0}^2 \partial_z C -\partial_z \phi_{1,0} C \phi_{1,0} = 1.$

This gives ${ \partial_z C(z) = 1/ \phi^2_{1,0}(z)}$ and therefore \beq \phi_{2,0}(z) = (U(z) – c) \int_{1/2}^z {1 \over (U(y) – c)^2} dy. \eeq Note that ${\phi_{2,0}}$ is well defined if the denominator does not vanishes, hence if ${\Im c \not = 0}$ or if ${\Im c = 0}$ and ${z> z_c}$. More precisely,

$\displaystyle {1 \over (U(c) - U(z_c))^2} = {1 \over ( U'(z_c) (z - z_c) + U''(z_c) (z - z_c)^2 / 2 + ...)^2}$

$\displaystyle = {1 \over U'(z_c)^2 (z-z_c)^2} - {U''(z_c) \over U'(z_c)^3} {1 \over z - z_c} + holomorphic .$

Hence \beq \phi_{2,0} = – {U(z) – c \over U'(z_c)^2 (z – z_c)} – {U”(z_c) \over U'(z_c)^3} (U(z) – c)\log (z – z_c) + holomorphic . \eeq As ${\phi_{2,0}}$ is not properly defined for ${z < z_c}$ when ${z_c \in \mathbb{R}^+}$, it is coherent to choose the determination of the logarithm which is defined on ${\mathbb{C} - \mathbb{R}^-}$.

With such a choice of the logarithm, ${\phi_{2,0}}$ is holomorphic in ${\mathbb{C} - \{ z_c + \mathbb{R}^- \}}$. In particular if ${\Im z_c = 0}$, ${\phi_{2,0}}$ is holomorphic in ${z}$ excepted on the half line ${z_c + \mathbb{R}^- }$. For ${z \in \mathbb{R}}$, ${\phi_{2,0}}$ is holomorphic as a function of ${c}$ excepted if ${z - z_c}$ is real and negative, namely excepted if ${z < z_c}$. For a fixed ${z}$, ${\phi_{2,0}}$ is an holomorphic function of ${c}$ provided ${z_c}$ does not cross ${\mathbb{R}^+}$, and provided ${z - z_c}$ does not cross ${\mathbb{R}^-}$. The Lemma then follows from the explicit expression (1) of ${\phi_{2,0}}$. $\Box$

Let ${\phi_{1,0},\phi_{2,0}}$ be constructed as in Lemma 4. Then the Green function ${G_{R,0}(x,z)}$ of the ${Ray_0}$ operator can be defined by

$\displaystyle G_{R,0}(x,z) = \left\{ \begin{array}{rrr} (U(x)-c)^{-1} \phi_{1,0}(z) \phi_{2,0}(x), \quad \mbox{if}\quad z>x,\\ (U(x)-c)^{-1} \phi_{1,0}(x) \phi_{2,0}(z), \quad \mbox{if}\quad z

Here we note that ${c}$ is complex with ${\Im c \not=0}$ and so the Green function ${G_{R,0}(x,z)}$ is a well-defined function in ${(x,z)}$, continuous across ${x=z}$, and its first derivative has a jump across ${x=z}$. Let us now introduce the inverse of ${Ray_0}$ as

\displaystyle \begin{aligned} RaySolver_0(f) (z) &: = \int_0^{+\infty} G_{R,0}(x,z) f(x) dx. \end{aligned} \ \ \ \ \ (18)

The following lemma asserts that the operator ${RaySolver_0(\cdot)}$ is in fact well-defined from ${X_0^\eta}$ to ${Y^0_2}$, which in particular shows that ${RaySolver_0(\cdot)}$ gains two derivatives, but losses the fast decay at infinity.

Lemma 5 Assume that ${\Im c \not =0}$. For any ${f\in {X_0^\eta}}$, the function ${RaySolver_0(f)}$ is a solution to the Rayleigh problem(15). In addition, ${RaySolver_0(f) \in Y^0_2}$, and there holds

$\displaystyle \| RaySolver_0(f)\|_{Y^0_2} \le C (1+|\log \Im c|) \|f\|_{{X_0^\eta}},$

for some universal constant ${C}$.

Proof: As long as it is well-defined, the function ${RaySolver_0(f)(z)}$ solves the equation (15) at once by a direct calculation, upon noting that

$\displaystyle Ray_0 (G_{R,0}(x,z) ) = \delta_x(z),$

for each fixed ${x}$.

Next, by scaling, we assume that ${ \| f\|_{X_0^\eta} = 1}$. By Lemma 4, it is clear that ${\phi_{1,0}(z)}$ and ${\phi_{2,0}(z)/(1+z)}$ are uniformly bounded. Thus, by direct computations, we have

$\displaystyle |G_{R,0}(x,z)| \le C \max\{ (1+x), |x-z_c|^{-1} \}.\ \ \ \ \ (19)$

That is, ${G_{R,0}(x,z)}$ grows linearly in ${x}$ for large ${x}$ and has a singularity of order ${|x-z_c|^{-1}}$ when ${x}$ is near ${z_c}$, for arbitrary ${z \ge 0}$. Since ${|f(z)|\le e^{-\eta z}}$, the integral (18) is well-defined and satisfies

$\displaystyle |RaySolver_0(f) (z)| \le C \int_0^\infty e^{-\eta x} \max\{ (1+x), |x-z_c|^{-1} \} \; dx\le C (1+|\log \Im c|),$

in which we used the fact that ${\Im z_c \approx \Im c}$. Finally, as for derivatives, we need to check the order of singularities for ${z}$ near ${z_c}$. We note that ${|\partial_z \phi_{2,0}| \le C (1+|\log(z-z_c)|)}$, and hence

$\displaystyle |\partial_zG_{R,0}(x,z)| \le C \max\{ (1+x), |x-z_c|^{-1} \} (1+|\log(z-z_c)|).$

Thus, ${\partial_z RaySolver_0(f)(z)}$ behaves as ${1+|\log(z-z_c)|}$ near the critical layer. In addition, from the ${Ray_0}$ equation, we have

$\displaystyle \partial_z^2 (RaySolver_0(f)) = \frac{U''}{U-c} RaySolver_0(f) + \frac{f}{U-c}.\ \ \ \ \ (20)$

This proves that ${RaySolver_0(f) \in Y_2^0}$ by definition of the function space ${Y_2^0}$. $\Box$

Lemma 6 Assume that ${\Im c \not =0}$. Let ${p}$ be in ${\{0,1,2\}}$. For any ${f \in X_p^\eta}$, we have

\displaystyle \begin{aligned} \| RaySolver_0(f) \|_{Y_{p+2}^0} \le C \|f\|_{X_p^\eta}(1+|\log (\Im c)| ) \end{aligned}

Proof: This is Lemma 5 when ${p=0}$. When ${p=1}$ or ${2}$, the lemma follows directly from the identity (20). $\Box$

1.3.2 Case ${\alpha \ne 0}$: an approximate Green function

Let ${\phi_{1,0}}$ and ${\phi_{2,0}}$ be the two solutions of ${Ray_0(\phi) = 0}$ that are constructed above in Lemma 4. We note that solutions of ${Ray_0(\phi) = f}$ tend to a constant value as ${z \rightarrow + \infty}$ since ${\phi_{1,0} \rightarrow U_+-c}$. We now construct normal mode solutions to the Rayleigh equation with ${\alpha \not=0}$

$\displaystyle Ray_\alpha(\phi) = (U-c)(\partial_z^2 - \alpha^2)\phi - U'' \phi = f \ \ \ \ \ (21)$

By looking at the spatially asymptotic limit of the Rayleigh equation, we observe that there are two normal mode solutions of (21) whose behaviors are as ${e^{\pm\alpha z}}$ at infinity. In order to study the mode which behaves like ${e^{-\alpha z}}$ we introduce

$\displaystyle \phi_{1,\alpha } = \phi_{1,0} e^{-\alpha z} ,\qquad \phi_{2,\alpha} = \phi_{2,0} e^{-\alpha z}. \ \ \ \ \ (22)$

A direct calculation shows that the Wronskian determinant

$\displaystyle W[\phi_{1,\alpha},\phi_{2,\alpha}] = \partial_z \phi_{2,\alpha} \phi_{1,\alpha} - \phi_{2,\alpha} \partial_z \phi_{1,\alpha} = e^{-2\alpha z}$

is non zero. In addition, we can check that

$\displaystyle Ray_\alpha(\phi_{j,\alpha}) = - 2 \alpha (U-c) \partial_z \phi_{j,0} e^{-\alpha z} \ \ \ \ \ (23)$

We are then led to introduce an approximate Green function ${G_{R,\alpha}(x,z)}$ defined by

$\displaystyle G_{R,\alpha}(x,z) = \left\{ \begin{array}{rrr} (U(x)-c)^{-1} e^{-\alpha (z-x)} \phi_{1,0}(z) \phi_{2,0}(x), \quad \mbox{if}\quad z>x\\ (U(x)-c)^{-1} e^{-\alpha (z-x)} \phi_{1,0}(x) \phi_{2,0}(z), \quad \mbox{if}\quad z< x.\end{array}\right.$

Again, like ${G_{R,0}(x,z)}$, the Green function ${G_{R,\alpha}(x,z)}$ is “singular” near ${z = z_c}$ with two sources of singularities: one arising from ${1/ (U(x) - c)}$ for ${x}$ near ${ z_c}$ and the other coming from the ${(z - z_c) \log (z- z_c)}$ singularity of ${\phi_{2,0}(z)}$. By a view of (23), it is clear that

$\displaystyle Ray_\alpha (G_{R,\alpha}(x,z)) = \delta_{x} -2\alpha (U- c) E_{R,\alpha}(x,z), \ \ \ \ \ (24)$

for each fixed ${x}$. Here the error term ${E_{R,\alpha}(x,z)}$ is defined by

$\displaystyle E_{R,\alpha}(x,z) = \left\{ \begin{array}{rrr} (U(x)-c)^{-1} e^{-\alpha (z-x)} \partial_z \phi_{1,0}(z) \phi_{2,0}(x), \quad \mbox{if}\quad z>x\\ (U(x)-c)^{-1}e^{-\alpha (z-x)}\ \phi_{1,0}(x) \partial_z \phi_{2,0}(z), \quad \mbox{if}\quad z< x.\end{array}\right.$

We then introduce an approximate inverse of the operator ${Ray_\alpha}$ defined by

$\displaystyle RaySolver_\alpha(f)(z) := \int_0^{+\infty} G_{R,\alpha}(x,z) f(x) dx \ \ \ \ \ (25)$

and the error remainder

$\displaystyle Err_{R,\alpha}(f)(z) := 2\alpha (U(z) - c) \int_0^{+\infty} E_{R,\alpha}(x,z) f(x) dx \ \ \ \ \ (26)$

Lemma 7 Assume that ${\Im c \not =0}$, and let ${p}$ be ${0,1,}$ or ${2}$. For any ${f\in {X_p^\eta}}$, with ${\eta > \alpha}$, the function ${RaySolver_\alpha(f)}$ is well-defined in ${Y^\alpha_{p+2}}$, satisfying

$\displaystyle Ray_\alpha(RaySolver_\alpha(f)) = f + Err_{R,\alpha}(f).$

Furthermore, there hold

$\displaystyle \| RaySolver_\alpha(f)\|_{Y^\alpha_{p+2}} \le C (1+|\log \Im c|) \|f\|_{{X_p^\eta}}, \ \ \ \ \ (27)$

and

$\displaystyle \|Err_{R,\alpha}(f)\|_{Y_{p}^\eta} \le C\alpha (1+|\log (\Im c)|) \|f\|_{X_p^\eta} , \ \ \ \ \ (28)$

for some universal constant ${C}$.

Proof: The proof follows similarly to that of Lemmas 5 and 6. In fact, the proof of the right order of singularities near the critical layer follows identically from that of Lemmas 5 and 6.

Let us check the right behavior at infinity. Consider the case ${p=0}$ and assume ${\|f \|_{X_0^\eta} =1}$. Similarly to the estimate (19), Lemma 4 and the definition of ${G_{R,\alpha}}$ yield

$\displaystyle |G_{R,\alpha}(x,z)| \le C e^{-\alpha (z-x)} \max\{ (1+x), |x-z_c|^{-1} \}.$

Hence, by definition,

$\displaystyle |RaySolver_\alpha (f)(z) |\le C e^{-\alpha z} \int_0^\infty e^{\alpha x} e^{-\eta x}\max\{ (1+x), |x-z_c|^{-1} \}\; dx$

which is clearly bounded by ${C(1+|\log \Im c|) e^{-\alpha z}}$. This proves the right exponential decay of ${RaySolver_\alpha (f)(z)}$ at infinity, for all ${f \in X_0^\eta}$.

Next, by definition, we have

\displaystyle \begin{aligned} Err_{R,\alpha}(f)(z) &= -2\alpha (U(z) - c) \partial_z \phi_{2,0}(z) \int_z^\infty e^{-\alpha (z-x)} \phi_{1,0}(x)\frac{f(x)}{U(x)-c}\; dx \\ & \quad - 2\alpha (U(z) - c) \partial_z \phi_{1,0}(z) \int_0^ze^{-\alpha (z-x)} \phi_{2,0}(x) {f(x) \over U(x) - c} dx . \end{aligned}

Since ${f(z), \partial_z \phi_{1,0}(z)}$ decay exponentially at infinity, the exponential decay of ${Err_{R,\alpha}(f)(z)}$ follows directly from the above integral representation. It remains to check the order of singularity near the critical layer. Clearly, for bounded ${z}$, we have

$\displaystyle |E_{R,\alpha}(x,z) | \le C (1+ |\log (z-z_c)| ) e^{\alpha x} \max \{ 1, |x-z|^{-1}\} .$

The lemma then follows at once, using the extra factor of ${U-c}$ in the front of the integral (26) to bound the ${\log (z-z_c)}$ factor. The estimates for derivatives follow similarly. $\Box$

1.3.3 Case ${\alpha \ne 0}$: the exact solver for Rayleigh

We now construct the exact solver for the Rayleigh operator by iteration. Let us denote

$\displaystyle S_0(z) : = RaySolver_\alpha(f)(z),\qquad E_0(z): = Err_{R,\alpha}(f)(z).$

It then follows that ${Ray_\alpha (S_0) (z) = f(z) + E_0(z)}$. Inductively, we define

$\displaystyle S_n(z): = - RaySolver_\alpha(E_{n-1})(z), \qquad E_n(z): = - Err_{R,\alpha}(E_{n-1})(z) ,$

for ${n\ge 1}$. It is then clear that for all ${n\ge 1}$,

$\displaystyle Ray_\alpha \Big( \sum_{k=0}^n S_k(z)\Big) = f(z) + E_n(z).\ \ \ \ \ (29)$

This leads us to introduce the exact solver for Rayleigh defined by

$\displaystyle RaySolver_{\alpha,\infty} (f) := RaySolver_\alpha(f)(z) - \sum_{n\ge 0} (-1)^n RaySolver_{\alpha} (E_n)(z).\ \ \ \ \ (30)$

Proof: } By a view of (28), we have

$\displaystyle \| E_n \|_\eta = \| (Err_{R,\alpha})^n (f)\|_\eta \le C^n\alpha^n (1+|\log (\Im c)|)^n \|f\|_\eta,$

which implies that ${E_n\rightarrow 0}$ in ${X_\eta}$ as ${n \rightarrow \infty}$ as long as ${\alpha \log \Im c}$ is sufficiently small. In addition, by a view of (27),

$\displaystyle \|RaySolver_{\alpha} (E_n)\|_{Y^\alpha_2} \le C C^n\alpha^n (1+|\log (\Im c)|)^n \|f\|_\eta .$

This shows that the series

$\displaystyle \sum_{n\ge 0} (-1)^n RaySolver_{\alpha} (E_n)(z)$

converges in ${Y_2^\alpha}$, assuming that ${\alpha \log \Im c}$ is small.

Next, by taking the limit of ${n\rightarrow \infty}$ in (29), the equation (14) holds by definition at least in the distributional sense. The estimates when ${z}$ is near ${z_c}$ follow directly from the similar estimates on ${RaySolver_\alpha(\cdot)}$; see Lemma 7. The proof of Proposition 3 is thus complete. $\Box$

1.3.4 Exact Rayleigh solutions

We shall construct two independent exact Rayleigh solutions by iteration, starting from the approximate Rayleigh solutions ${\phi_{j,\alpha}}$ defined as in (22).

Lemma 8 For ${\alpha}$ small enough so that ${\alpha |\log \Im c| \ll 1}$, there exist two independent functions ${\phi_{Ray,\pm} \in e^{\pm \alpha z}L^\infty}$ such that

$\displaystyle Ray_\alpha ( \phi_{Ray,\pm} ) = 0, \qquad W[\phi_{Ray,+},\phi_{Ray,-}](z) = \alpha.$

Furthermore, we have the following expansions in ${L^\infty}$:

\displaystyle \begin{aligned} \phi_{Ray,-} (z)&= e^{-\alpha z} \Big (U-c + O(\alpha )\Big). \\ \phi_{Ray,+} (z)&= e^{\alpha z} \mathcal{O}(1), \end{aligned}

as ${z\rightarrow \infty}$. At ${z = 0}$, there hold

\displaystyle \begin{aligned} \phi_{Ray,-}(0) &= U_0 - c + \alpha (U_+-U_0) ^2 \phi_{2,0}(0) + \mathcal{O}(\alpha(\alpha + |z_c|)) \\ \phi_{Ray,+}(0) &= \alpha \phi_{2,0}(0) + \mathcal{O}(\alpha^2) \end{aligned}

with ${\phi_{2,0}(0) = {1 \over {U'_c}} - {2U''_c\over {U'_c}^2 } z_c \log z_c + \mathcal{O}(z_c)}$.

Proof: Let us start with the decaying solution ${\phi_{Ray,-}}$, which is now constructed by induction. Let us introduce

$\displaystyle \psi_{0} = e^{-\alpha z} (U-c), \qquad e_{0} = - 2\alpha (U-c) U' e^{-\alpha z},$

and inductively for ${k \ge 1}$,

$\displaystyle \psi_{k} = - RaySolver_\alpha (e_{k-1}), \qquad e_{k} = - Err_{R,\alpha} (e_{k-1}).$

We also introduce

$\displaystyle \phi_{N} = \sum_{k=0}^N \psi_{k} .$

By definition, it follows that

$\displaystyle Ray_\alpha (\phi_{N}) = e_{N}, \qquad \forall ~N\ge 1.$

We observe that ${\|e_{0}\|_{\eta+\alpha} \le C \alpha}$ and ${\|\psi_{0}\|_\alpha \le C}$. Inductively for ${k\ge 1}$, by the estimate (28), we have

$\displaystyle \| e_{k}\|_{\eta + \alpha} \le C\alpha (1+|\log (\Im c)|) \|e_{k-1}\|_{\eta+\alpha} \le C \alpha(C\alpha (1+|\log (\Im c)|))^{k-1} ,$

and by Lemma 7,

$\displaystyle \| \psi_{k} \|_\alpha \le C(1+|\log (\Im c)|) \| e_{k-1}\|_{\eta + \alpha} \le (C\alpha (1+|\log (\Im c)|))^{k} .$

Thus, for sufficiently small ${\alpha}$, the series ${\phi_{N}}$ converges in ${X_\alpha}$ and the error term ${e_{N}\rightarrow 0}$ in ${X_{\eta+\alpha}}$. This proves the existence of the exact decaying Rayleigh solution ${\phi_{Ray,-}}$ in ${X_\alpha}$, or in ${e^{-\alpha z}L^\infty}$.

As for the growing solution, we simply define

$\displaystyle \phi_{Ray,+} =\alpha \phi_{Ray,-}(z) \int_{1/2} ^ z \frac{1}{\phi^2_{Ray,-} (y) }\; dy.$

By definition, ${\phi_{Ray,+} }$ solves the Rayleigh equation identically. Next, since ${\phi_{Ray,-}(z)}$ tends to ${e^{-\alpha z} (U_+ - c + \mathcal{O}(\alpha))}$, ${\phi_{Ray,+} }$ is of order ${e^{\alpha z}}$ as ${z \rightarrow \infty}$.

Finally, at ${z=0}$, we have

\displaystyle \begin{aligned} \psi_1(0) &= - RaySolver_\alpha(e_0) (0) = - \phi_{2,\alpha}(0) \int_0^{+\infty} e^{2\alpha x}\phi_{1,\alpha}(x) {e_0(x) \over U(x) - c} dx \\ &= 2 \alpha \phi_{2,0}(0) \int_0^{+\infty} U' (U-c)dz = \alpha (U_+-U_0) (U_+ + U_0 - 2c) \phi_{2,0}(0) \\ & = \alpha (U_+-U_0)^2 (U_+ + U_0 - 2c) \phi_{2,0}(0) + 2\alpha (U_+-U_0) (U_0 - c) \phi_{2,0}(0). \end{aligned}

From the definition, we have ${\phi_{Ray,-}(0) = U_0 -c + \psi_1(0) + \mathcal{O}(\alpha^2)}$. This proves the lemma, upon using that ${U_0 - c = \mathcal{O}(z_c)}$. $\Box$

This completes the study on the Rayleigh problem (linearized Euler equations). We shall next study the singular perturbation problem of Orr-Sommerfeld equations (linearized Navier-Stokes equations).