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## Math 597F, Notes 7: unstable Orr-Sommerfeld solutions for unstable profiles

(this post was also posted here on my new blog address: http://blog.toannguyen.org/ )

We now return to the Orr-Sommerfeld equations (the linearized Navier-Stokes equations around a boundary layer ${U(z)}$; see this lecture):

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

For convenience, we denote ${Ray_\alpha(\phi) = (U-c) (\partial_z^2 - \alpha^2) \phi - U'' \phi}$ and ${\Delta_\alpha = \partial_z^2 - \alpha^2}$. The Orr-Sommerfeld equation simply reads

$\displaystyle Ray_\alpha(\phi) = \epsilon \Delta_\alpha ^2 \phi.$

We shall construct solutions for each fixed pair ${(\alpha,c)}$ and for sufficiently small ${\epsilon}$.

As mentioned, there are unstable and stable profiles ${U}$. We warm up ourselves by considering the unstable case, namely: we assume that there is a pair ${(\alpha_0, c_0)}$, with ${\Im c_0>0}$, so that the Rayleigh problem has a nonzero solution:

\displaystyle \left \{ \begin{aligned} (U-c_0) (\partial_z^2 - \alpha_0^2) \phi_0 - U'' \phi_0 &= 0 \\ {\phi_0}_{\vert_{z=0}} = 0, \qquad \lim_{z\rightarrow \infty} \phi_0(z) &=0. \end{aligned} \right. \ \ \ \ \ (2)

In addition, we assume that

$\displaystyle \partial_c {\phi_0}_{\vert_{z=0}} \not =0.$

This latter assumption means precisely that the unstable eigenvalue of the linearized Euler operator is simple. This is equivalent to the condition that the Wronskian determinant, or the Evans function, ${E(\alpha,c)}$ satisfies ${\partial_c E(\alpha_0,c_0) \not =0}$. We introduce a perturbative analysis to construct a solution ${(\alpha, c, \phi)}$ to the Orr-Sommerfeld problem, when ${\epsilon}$ is sufficiently small, so that ${c \rightarrow c_0}$ as ${\epsilon \rightarrow 0}$. This yields a growing mode to the linearized Navier-Stokes problem.

7.1. A toy model

Consider the problem

$\displaystyle L\phi: = (1 - \partial_z - \epsilon \partial_z^2) \phi = f, \qquad \phi_{\vert_{z=0}} = 0.$

Let ${\phi_0}$ be the (bounded) inviscid solution: ${\phi_0 = (1-\partial_z)^{-1} f}$. Due to the outward characteristic to the boundary ${\{z=0\}}$, bounded inviscid solutions need not to satisfy the zero boundary condition and the viscous equation. Boundary layers are needed, solving

$\displaystyle -(\partial_z + \epsilon \partial_z^2) \phi_{bl} = g,$

for some source ${g}$, which is due to the error introduced by the inviscid solution. Precisely,

$\displaystyle L\phi_0 = f - \epsilon \partial_z^2 \phi_0 = f - \epsilon \partial_z^2 (1-\partial_z)^{-1} f ,$

leaving the error term: ${\epsilon \partial_z^2 (1-\partial_z)^{-1} f }$, which might not be well-defined (and so not contractive for the iteration to work!). We then correct this by introducing the iteration operator:

$\displaystyle Iter: = \underbrace{(\epsilon \partial_z^2 + \partial_z )^{-1}}_{\mbox{boundary layer}} \circ \quad \underbrace{\epsilon \partial_z^2}_{\mbox{error}} \quad \circ \quad \underbrace{(1-\partial_z)^{-1}}_{\mbox{inviscid}}.$

The Boundary Layer operator added is to regularize the possible singularity coming from the loss of derivatives in the error: ${\partial_z^2 (1-\partial_z)^{-1}f}$. We can check that the new solution ${\phi_1: = \phi_0 + Iter(f)}$ solves

$\displaystyle L\phi_1 = f + Iter(f) .$

Inductively, let ${E_n = Iter(f)^n}$ be the error at the ${n^{th}}$ step and define

$\displaystyle \phi_{n+1} = \phi_{n} + (1-\partial_z)^{-1} E_{n} + Iter(E_n)$

to be the next iterative solution. There holds

$\displaystyle L\phi_{n+1} = f + Iter(f)^{n+1}.$

It suffices to show that ${Iter(\cdot)}$ is well-defined and contractive in some function space, and so ${E_n \rightarrow 0}$ and ${\phi_n \rightarrow \phi_\infty}$ as ${n\rightarrow \infty}$, giving an exact solution ${\phi_\infty}$ to the problem ${L \phi =f}$. Let us work with the function space ${X = L^{\infty}}$. The iteration solves ${(\epsilon \partial_z^2 + \partial_z ) Iter(f) = \epsilon \partial^2_z g}$, with ${g = (1-\partial_z)^{-1}f}$, which gives

$\displaystyle Iter(f)(z) = e^{-z/\epsilon} Iter(f)(0) + \int_0^z e^{-(z-y) /\epsilon} \partial_y g \; dy.$

The boundary value ${Iter(f)(0)}$ can be taken to be the minus of that of the inviscid solution. For simplicity, we take ${Iter(f)(0)=0}$. The iteration is contractive in ${X}$, since

$\displaystyle |Iter(f)(z)| = \Big| \int_0^z e^{-(z-y) /\epsilon} \partial_y g \; dy \Big| \le \| \partial_z g \|_\infty \int_0^\infty e^{-|z-y|/\epsilon} \; dy = \epsilon \| \partial_z g \|_\infty$

and

$\displaystyle \partial_z g = \partial_z (1-\partial_z)^{-1} f = f + (1-\partial_z)^{-1} f ,$

which is clearly bounded by ${C \|f\|_{L^\infty}}$. This proves that the Iter operator is contractive for sufficiently small ${\epsilon}$, precisely: there holds ${\|Iter(f)\|_{L^\infty} \le C \epsilon \|f\|_{L^\infty}}$ for all bounded functions ${f}$. It is worth noticing that the contraction is due to the localizedness of the boundary layers of size ${\epsilon}$ (and hence, ${L^1}$ norm is of order ${\epsilon}$ as seen above).

7.2. Asymptotic behavior as ${z \rightarrow + \infty}$

In order to construct the independent solutions of Orr-Sommerfeld, let us study their possible behavior at infinity. One observes that as ${z \rightarrow +\infty}$, solutions must behave like solutions of constant-coefficient limiting equation:

$\displaystyle \epsilon \partial_z^4 \phi = (U_+ - c + 2 \epsilon \alpha^2) \partial_z^2 \phi - \alpha^2 (\epsilon \alpha^2 + U_+ - c)\phi,$

with ${U_+ = U(+\infty)}$. Solutions are thus of the form ${ e^{\mu z}}$ with ${\mu = \mu_{s,\pm}}$ or ${\mu = \mu_{f,\pm}}$, where

$\displaystyle \mu_{s,\pm} = \pm \alpha + \mathcal{O}(\alpha^2 \sqrt \epsilon ), \qquad \mu_{f,\pm} = \pm \frac {1}{\sqrt \epsilon} (U_+-c)^{1/2} + \mathcal{O}(\alpha).$

Therefore, we can find two solutions ${\phi_{s,\pm}}$ with a “slow behavior” ${\mu \approx \pm \alpha}$ (one decaying and the other growing) and two solutions ${\phi_{f,\pm}}$ with a “fast behavior” where ${\mu}$ is of order ${\pm 1 / \sqrt{\epsilon}}$ (one decaying and the other growing). The first two slow-behavior solutions ${\phi_{s,\pm}}$ will be perturbations of eigenfunctions of the Rayleigh equation. The other two, ${\phi_{f,\pm}}$, are specific to the Orr Sommerfeld equation and will be linked to the solutions of ${\epsilon \partial_z^4 \phi = (U-c + 2 \epsilon \alpha^2) \partial_z^2 \phi}$. The standard conjugation method for ODEs does not apply directly due to the ${\epsilon}$-dependence in the equation, which makes the coefficients of the ODE decay exponentially only at a vanishing rate (in the fast variable) when tracking the fast modes. In fact, since the fast eigenvalues of the corresponding ODE system are well-separated from the slow ones, we will track these fast modes by diagonalization, keeping their variable-coefficients in our setting.

7.3. Slow modes

In this section, we iteratively construct “slow” modes ${\phi_{s,\pm}\approx e^{\pm \alpha z}}$ of the Orr-Sommerfeld equations, starting from the Rayleigh solutions: ${\phi_{\alpha,\pm} \approx e^{\pm \alpha z}}$ constructed in the previous lecture. Our iteration follows similarly to that for the toy model problem. We start from a solution solving the Rayleigh equation: ${Ray_\alpha (\phi_0) = 0}$. Let ${Orr(\cdot)}$ denote the Orr-Sommerfeld operator:

$\displaystyle Orr(\phi): =Ray_\alpha(\phi) - \epsilon \Delta_\alpha^2\phi .$

Then, together with the definition of ${Ray_\alpha}$ operator to solve for ${\Delta_\alpha \phi_0}$, there holds

$\displaystyle Orr(\phi_0) = - \epsilon \Delta_\alpha^2 \phi_0 = -\epsilon \Delta_\alpha \Big[ \frac{U'' \phi_0}{U-c} \Big] =: E_0(z)$

in which the error term on the right is bounded by ${C \epsilon e^{-\eta z} \| \phi_0\|_{W^{1,\infty}}}$. Next, by induction, let us assume that we have constructed ${\phi_n}$ so that

$\displaystyle Orr(\phi_n) = E_n(z),$

with a sufficiently small error term ${E_n}$ in ${X_\eta}$. We then improve the error term by constructing a new approximate solution ${\phi_{n+1}}$ so that it solves the Orr-Sommerfeld equations with a better error in ${X_{\eta }}$. Precisely, we introduce

$\displaystyle \phi_{n+1}: = \phi_{n} - Ray_{\alpha}^{-1} ( E_n)$

which clearly solves the Orr-Sommerfeld with a new error: ${Orr (\phi_{n+1} ) = E_{n+1}: = - \epsilon \Delta^2_\alpha Ray_\alpha^{-1}(E_n)}$. Again, by the definition of ${Ray}$ operator, we compute

$\displaystyle E_{n+1} = - \epsilon \Delta^2_\alpha Ray_\alpha^{-1}(E_n) = - \epsilon \Delta_\alpha \Big[ \frac{U'' Ray_\alpha^{-1} E_n}{U-c} \Big]$

which is bounded by ${C \epsilon e^{-\eta z} \| Ray_\alpha^{-1} E_n\|_{W^{1,\infty}}}$. By Lemma 2 in the last lecture, we have

$\displaystyle \| E_{n+1} \|_{X_\eta} \le C \epsilon E(\alpha,c)^{-1}\| E_{n}\|_{X_\eta} \le \Big[ C \epsilon E(\alpha,c)^{-1}\Big]^{n+1} \| \phi_0\|_{W^{1,\infty}},$

in which ${E(\alpha,c)}$ denotes the Evans function (or precisely, the Wronskian determinant) for the inviscid equations. Hence, whenever ${\epsilon E(\alpha,c)^{-1} \ll 1}$, ${E_n}$ converges to zero in ${L^\infty}$ and thus the series ${\phi_n}$ also converges to an exact solution of the Orr-Sommerfeld equations. We summarize the above into the following lemma, providing two independent slow modes of the Orr-Sommerfeld equations:

Lemma 1 Let ${\phi_{\alpha,\pm}}$ be the two independent Rayleigh solutions constructed in Lemma 2 in the previous lecture, and let ${E(\alpha,c)}$ be the Wronskian determinant of the two solutions (which is independent of ${z}$). For sufficiently small ${\epsilon}$ such that ${\epsilon E(\alpha,c)^{-1} \ll 1}$, there exist two independent solutions ${\phi_{s,\pm}(z)}$ which solve the Orr-Sommerfeld equations

$\displaystyle Orr(\phi_{s,\pm}) = 0,$

so that ${\phi_{s,\pm}}$ is close to the Rayleigh solutions ${\phi_{\alpha,\pm}}$ in ${X_\alpha}$. Precisely, we have

$\displaystyle \| \phi_{s,\pm} - \phi_{\alpha,\pm}\|_{X_\alpha} \le C\epsilon E(\alpha,c)^{-1} ,$

for some positive constant ${C}$ independent of ${\alpha,c, \epsilon}$.

7.4. Fast modes

In this section, we shall construct two independent solutions, which asymptotically behave as ${e^{\pm z/\sqrt{\epsilon}}}$, of the Orr-Sommerfeld equation: ${Orr(\phi) = 0}$.

Lemma 2 For sufficiently small ${\epsilon}$, there exist two independent solutions ${\phi_{f,\pm}(z)}$ which solve the Orr-Sommerfeld equations

$\displaystyle Orr(\phi_{f,\pm}) = 0,$

so that ${\phi_{f,\pm}}$ satisfy

$\displaystyle \phi_{f,\pm} (z) = e^{\pm \int_0^z\mu_{f}(y) \; dy} \Big[ 1 + \mathcal{O}(\epsilon e^{-\eta_0 z})\Big] ,$

in which ${\mu_f(z)^2=\frac1\epsilon ( U - c + \mathcal{O}(\alpha^2 \epsilon))}$. In addition, the Wronskian determinant satisfies the estimate

$\displaystyle W[\phi_{f,-},\phi_{f,+}](z) \approx \epsilon^{-1/2}.$

As we are in the case of unstable profiles, we can assume that ${c}$ is away from the real axis and so ${U-c}$ is bounded below away from zero. Hence, ${\mu_f \approx 1/\sqrt \epsilon}$. The Lemma 2 thus yields two independent solutions ${\phi_{f,\pm}}$ that behave as ${e^{\pm z/\sqrt\epsilon}}$ near the infinity.

Proof: We rewrite the Orr-Sommerfeld equation as

$\displaystyle Orr(\phi): = - \epsilon \partial_z^4 \phi + b(z) \partial_z^2 \phi - a(z)\phi= 0,$

in which for convenience we have denoted ${a(z): = \alpha^2 (\epsilon \alpha^2 + U - c) + U''}$ and ${b(z): = (U - c + 2 \epsilon \alpha^2)}$. We make a change of variables:

$\displaystyle \tilde z = \frac{z}{\sqrt \epsilon}, \qquad \phi(z) = \tilde \phi(\frac{z}{\sqrt \epsilon}).$

The equation for ${\tilde \phi}$ reads

$\displaystyle - \partial_{\tilde z}^4 \tilde \phi + b(\epsilon \tilde z)\partial_{\tilde z}^2 \tilde \phi - \epsilon a(\epsilon \tilde z)\tilde \phi = 0.$

It is convenient to write the above equation as the first order ode system. We introduce ${W = (\tilde \phi, \tilde \phi', \tilde \phi'', \tilde \phi''')}$, with prime denoting the derivative with respect to $\tilde z$. The system for ${W}$ reads

$\displaystyle \partial_{\tilde z} W = \mathcal{A}(\tilde z) W, \qquad \mathcal{A}(\tilde z) : = \begin{pmatrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1\\ -\epsilon a(\epsilon \tilde z) &0& b(\epsilon \tilde z) & 0\end{pmatrix} .\ \ \ \ \ (3)$

We note that ${ b(\epsilon \tilde z) = U(\epsilon \tilde z) - c + 2 \epsilon \alpha^2}$ is bounded away from zero, for ${\tilde z\ge 0}$. Hence, four eigenvalues of ${\mathcal{A}}$, two slow and two fast, satisfy

$\displaystyle \mu_{s}^2(\tilde z) = \epsilon a(\epsilon \tilde z) /b(\epsilon \tilde z)+ \mathcal{O}(\epsilon^2), \qquad \mu_{f}^2(\tilde z) = b(\epsilon \tilde z)+ \mathcal{O}( \epsilon).$

Let ${\mu_{f,\pm} = \pm \mu_f}$ be the two fast eigenvalues with positive/negative real part, respectively. Notice that ${\mu_{s} = \mathcal{O}(\sqrt \epsilon)}$, ${\mu_{f,\pm} \approx \pm 1}$ and ${\partial_{\tilde z} \mu_{f,\pm} = \mathcal{O}(\epsilon e^{-\eta_0 \epsilon \tilde z})}$. We let

$\displaystyle \mathcal{T}_\pm = \begin{pmatrix} 1&0&0&0\\ \mu_{f,\pm} &1 &0&0\\ \mu^2_{f,\pm} &0&1&0\\ \mu^3_{f,\pm} &0&0&1 \end{pmatrix} \qquad \mathcal{T}_\pm^{-1} = \begin{pmatrix} 1&0&0&0\\ -\mu_{f,\pm} &1 &0&0\\ -\mu^2_{f,\pm} &0&1&0\\ -\mu^3_{f,\pm} &0&0&1 \end{pmatrix}$

and for each ${+/-}$ case, we introduce

$\displaystyle W = \mathcal{T}_\pm V.$

The function ${V}$ then solves

$\displaystyle \partial_{\tilde z} V = \begin{pmatrix} \mu_{f,\pm} &0 \\ 0& \mathcal{A}_1\end{pmatrix} V - \mathcal{T}_\pm^{-1} \partial_{\tilde z} \mathcal{T}_\pm V,$

in which ${\mathcal{A}_1}$ is defined as the lower ${3\times 3}$ block of the matrix ${\mathcal{T}_\pm^{-1} \mathcal{A} \mathcal{T}_\pm}$, and the first row of the matrix ${\mathcal{T}_\pm^{-1} \partial_{\tilde z} \mathcal{T}_\pm}$ is zero and ${ \mathcal{T}_\pm^{-1} \partial_{\tilde z} \mathcal{T}_\pm = \mathcal{O}(\epsilon e^{-\eta_0 \epsilon \tilde z})}$ by direct computation.

We further introduce

$\displaystyle V = e^{\int_0^{\tilde z} \mu_{f,\pm}(y)\; dy} Z .$

Then, ${Z}$ solves

$\displaystyle \partial_{\tilde z} Z = \begin{pmatrix} 0 &0 \\ 0& \mathcal{A}_1 - \mu_{f,\pm} I\end{pmatrix} Z - \mathcal{T}_\pm^{-1} \partial_{\tilde z} \mathcal{T}_\pm Z.\ \ \ \ \ (4)$

We write ${Z = (Z_1,\tilde Z)}$. Then, the above system yields

$\displaystyle Z_1(\tilde z) = 1, \qquad \tilde Z(\tilde z) = \tilde Z(M_\pm) + \int_{M_\pm}^{\tilde z} e^{ (\mathcal{A}_1 - \mu_{f,\pm}) (\tilde z - y)} \mathcal{O}(\epsilon e^{-\eta_0 \epsilon y}) Z(y) \; dy,\ \ \ \ \ (5)$

for ${M_\pm}$ depending on each ${+/-}$ case.

Consider the ${-}$ case. We observe that the eigenvalues of ${\mathcal{A}_1}$ are the same as those of ${\mathcal{A}}$ except for the eigenvalue ${\mu = \mu_{f,-}}$. Thus, the eigenvalues of ${\mathcal{A}_1 - \mu_{f,-}}$ have real parts that are all positive and bounded away from zero. This proves that there is a positive constant ${\theta_0}$ so that

$\displaystyle \| e^{(\mathcal{A}_1 - \mu_{f,-}) y } \|\le C e^{\theta_0 y}, \qquad y\le 0.$

Taking ${M_- = \infty}$ in (5), we get

$\displaystyle \tilde Z(\tilde z) = \tilde Z_- - \int_{\tilde z}^\infty e^{ (\mathcal{A}_1 - \mu_{f,-}) (\tilde z - y)} \mathcal{O}(\epsilon e^{-\eta_0 \epsilon y}) Z(y) \; dy.$

We denote by ${TZ}$ the right-hand side of the above integral form. We shall show that ${T}$ is contractive from ${L^\infty([0,\infty))}$ into itself. Indeed, for any ${Z,Y \in L^\infty([0,\infty))}$ with ${Z_1 = Y_1 = 1}$, one calculates

\displaystyle \begin{aligned} |(TZ - TY)(\tilde z)| &\le \int_{\tilde z}^\infty \|e^{ (\mathcal{A}_1 - \mu_{f,-}) (\tilde z - y)} \|\mathcal{O}(\epsilon e^{-\eta_0 \epsilon y}) |\tilde Z - \tilde Y|(y) \; dy \\ &\le C \epsilon e^{-\eta_0 \epsilon \tilde z} \int_{\tilde z}^\infty e^{-\frac{\theta_0}{2} (y-\tilde z)} |\tilde Z - \tilde Y|(y) \; dy \\ & \le C \epsilon e^{-\eta_0 \epsilon \tilde z} \| Z - Y \|_{L^\infty}. \end{aligned}

By taking ${\epsilon}$ small enough, the above proves that ${T}$ is a contractive map, and thus there exists a (unique) solution ${Z \in L^\infty([0,\infty))}$ solving (4) in the ${-}$ case. Furthermore, there holds

$\displaystyle |Z(\tilde z) - Z_-| \le C \epsilon e^{-\eta_0 \epsilon \tilde z} , \qquad \tilde z\ge 0.$

Next, consider the ${+}$ case in (5). In this case, the eigenvalues of ${\mathcal{A}_1 - \mu_{f,+}}$ have real parts that are all negative and bounded away from zero. This yields

$\displaystyle \| e^{(\mathcal{A}_1 - \mu_{f,+}) y } \|\le C e^{ -\theta_0 y}, \qquad y\ge 0,$

for some ${\theta_0>0}$. Taking ${M_+ = 0}$ in (5), we get

$\displaystyle \tilde Z(\tilde z) = \tilde Z_+ + \int_0^{\tilde z} e^{ (\mathcal{A}_1 - \mu_{f,+}) (\tilde z - y)} \mathcal{O}(\epsilon e^{-\eta_0 \epsilon y}) Z(y) \; dy.$

Again, if we denote by ${TZ}$ the right-hand side of the above integral form, we have for any ${Z,Y \in L^\infty([0,\infty))}$ with ${Z_1 = Y_1 = 1}$

\displaystyle \begin{aligned} |(TZ - TY)(\tilde z)| &\le \int_0^{\tilde z} \|e^{ (\mathcal{A}_1 - \mu_{f,+}) (\tilde z - y)} \|\mathcal{O}(\epsilon e^{-\eta_0 \epsilon y}) |\tilde Z - \tilde Y|(y) \; dy \\ &\le C \epsilon e^{-\eta_0 \epsilon \tilde z} \int_0^{\tilde z} e^{-\theta_0 (\tilde z-y)} |\tilde Z - \tilde Y|(y) \; dy \\ & \le C \epsilon e^{-\eta_0 \epsilon \tilde z} \| Z - Y \|_{L^\infty}. \end{aligned}

This proves that ${T}$ is a contractive map for small ${\epsilon}$, and thus there exists a (unique) solution ${Z \in L^\infty([0,\infty))}$ solving (4) in the ${+}$ case. Furthermore, there holds

$\displaystyle |Z(\tilde z) - Z_+| \le C \epsilon e^{-\eta_0 \epsilon \tilde z} , \qquad \tilde z\ge 0.$

In summary, we have obtained two independent solutions ${W_\pm}$ to (3) such that

$\displaystyle W_\pm (\tilde z) = e^{\int_0^{\tilde z} \mu_{f,\pm}(y)\; dy} \mathcal{T}_\pm \Big [ Z_\pm + \mathcal{O}(\epsilon e^{-\theta_0 \epsilon \tilde z}) \Big ],$

for arbitrary constants ${Z_\pm}$. By rescaling to the original variables, the lemma follows at once. $\Box$

7.5. Dispersion relation

We are ready to construct a solution to the Orr-Sommerfeld problem, with ${\Im c>0}$. Indeed, we let ${\phi_s}$ and ${\phi_f}$ be two slow and fast modes that decay at infinity. A bounded solution to the Orr-Sommerfeld problem is constructed as a linear combination of these two decaying modes:

$\displaystyle \phi = A \phi_s + B \phi_f.$

The zero boundary conditions on ${\phi}$ yield the linear dispersion relation:

$\displaystyle \frac{\phi_s}{\phi'_s}{\Big\vert_{z=0}} = \frac{\phi_f}{\phi'_f}{\Big\vert_{z=0}} . \ \ \ \ \ (6)$

We need to show that there are choices of ${\alpha, c}$ so that the above relation holds for ${\epsilon}$ sufficiently small. From the construction of fast decaying modes, we have

$\displaystyle \frac{\phi_f }{ \phi_f'}{\Big\vert_{z=0}}\approx \frac{1}{\mu_f} \approx \sqrt \epsilon.$

Let us calculate the ratio for the slow modes. We recall that there holds the asymptotic expansion:

$\displaystyle \phi_s = \phi_0 + \mathcal{O}(\epsilon E(\alpha,c)^{-1})$

as long as ${\epsilon E(\alpha,c)^{-1} \ll 1}$. Here, we take the Rayleigh eigenfunction ${\phi_0}$ so that such that ${\phi_0 (0) = 0}$ and ${\phi'_0(0) \not =0}$ on the boundary ${z=0}$ (we note that if ${\phi'_0}$ vanishes on the boundary, then ${\phi_0 \equiv 0}$ identically, since by the Rayleigh equation, all derivatives of ${\phi_0}$ vanish on the boundary). The dispersion relation then yields

$\displaystyle E(\alpha,c) = \mathcal{O}(\sqrt \epsilon) \Big[ 1 +\mathcal{O}(\epsilon E(\alpha,c)^{-1}) \Big].$

Now, thanks to the assumption that the boundary layer is unstable to the inviscid problem, ${E(\alpha_0,c_0) = 0}$, with ${\Im c_0 >0}$. The condition ${\partial_c E(\alpha_0,c_0)}$ guarantees that there is a continuous dependence ${c(\epsilon)}$, with ${c(0) = c_0}$, so that the dispersion relation holds at ${(\alpha_0, c(\epsilon))}$ for all ${\epsilon}$ sufficiently small. Note in addition that ${E(\alpha_0,c(\epsilon)) \approx \sqrt \epsilon}$ and so the smallness assumption: ${\epsilon E(\alpha,c)^{-1}\ll 1}$ is valid. A growing mode of the linearized Navier-Stokes follows (for the unstable profiles ${U}$).

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