We study the linearization of 2D Navier-Stokes around a boundary layer , in which 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 , the boundary layer varies slowly in variables:

For this reason, we start the study of the linearized problem around the boundary layer of the form . The linearization reads

with on the boundary . Let be the vorticity. Then, solves

We write this equation in term of the stream function defined through and . We note that . The vorticity equation then reads

Following the classical linear stability theory (see, for instance, the nice book by Drazin and Reid), one searches for the solution ansatz: , in which is a real-valued positive wave number, denotes the eigenvalue of the linearized Navier-Stokes operator (and so, is a complex number), and is a complex-valued function. If is a solution to the vorticity equation, then its real part is also a solution. The pressure is then simply recovered by taking the divergence of (1), giving

Putting into the vorticity equation yields the well-known Orr-Sommerfeld fourth-order ODE equations:

Here, the boundary conditions are exactly from on the boundary. Note that the stream function is defined up to a constant, and so we simply take on the boundary. The interest is to determine whether there is a triple of solutions to (2) so that the complex number has a positive imaginary part, which corresponds to the existence of a growing solution of the linearized Navier-Stokes problem (in the original coordinates:, yielding instability in a vanishing time!). Clearly, when (that is, in the low Reynolds number regime), there is no such an unstable solution: is spectrally stable. We are interested in the singular perturbation problem: (or equivalently, large Reynolds number limit ). Two scenarios:

**Unstable profiles:**The boundary layer is unstable in the limit : there exists a triple , with the imaginary part , that solves the limiting Rayleigh problem:A perturbative analysis can be used to construct a solution to the singular perturbation problem (2) so that ; in particular, .

**Stable profiles:**The boundary layer is stable to the Rayleigh problem. For instance, this is the case when has no inflection point. For such a stable profile, all the spectrum of the Rayleigh equation is imbedded on the imaginary axis: , and thus it is not clear whether a perturbative argument to construct solutions to (2) would yield stability () or instability (). 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 so that whenever , there exist an unstable eigenvalue and an eigenfunction to the Orr-Sommerfeld problem. For large Reynolds number , the asymptotic behavior of and is estimated, depending on the background profile:-
- a. Plane Poiseuille flow:
- b. Boundary layer profiles:

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

for each fixed parameters and . We shall also construct the inverse of the Rayleigh operator .

**1.1. Inviscid stability criteria**

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

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**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 solving (4) so that . Multiplying by to the Rayleigh equation (4) and taking integration by parts yield

Thus, the instability condition must imply that changes its sign. This gives the Rayleigh criterium. Consequently, boundary layer profiles with no inflection point are spectrally stable to the linearized Euler equations. Additionally, there are stable flows with inflection point:

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**Fjortoft criterium (1950):** *A necessary condition for instability is that somewhere in the flow, where is a point at which .* Indeed, let be the inflection point and consider the real part of the identity (5):

Adding to this the identity (6):

we obtain

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

for each fixed , under the assumption that

for some positive constants , and some constant . Such an assumption holds for the unstable profile , since . We shall use the function spaces , for , to denote the spaces consisting of functions that are smooth and satisfy

for all and all . The best constant that satisfies the above inequality defines a norm, denoted by , on the Banach space . We denote in place of , respectively.

**1.2.1 Homogenous solutions.**

We obtain the following lemma:

Lemma 1There are two independent solutions to the homogenous equation such that

for all .

*Proof:*

Let us denote . The homogenous equation becomes

We denote by the limit of as . Eigenvalues of are . Let be the associated eigenvectors. We construct two independent solutions of (8) of the form

Let us first consider the case: . Plugging the above ansatz into (8), solves

Since all eigenvalues of have nonnegative real parts, we have for , for arbitrarily small . Thanks to the decay assumption on the boundary layer profile , there holds . Here, denotes the standard matrix norm. The Duhamel formula yields

Denote by the right hand side of the above identity. We shall show that is a contractive map from into itself, for sufficiently large . Indeed, for , we get

for all . Taking large enough so that , the map is contractive. Hence, there exists a solution such that , or equivalently solves (8) on . By the standard ode theory, the solution on can be extended to be a global solution on , since the right-hand side of (8) is uniformly bounded.

Similarly, with the ansatz solution: , there holds

in which the matrix has only nonpositive eigenvalues. It follows that for , for arbitrarily small . The solution is then constructed via the Duhamel formula:

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

**1.2.2 Exact Rayleigh solver.** The Green function of the operator can be defined by

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

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

We have the following simple convolution estimate.

Lemma 2For any positive constant and for each , there holds

*Proof:* By Lemma 1 and the assumption (7) on the profile , we have

for all , and . This yields

This proves the first estimate stated in the lemma. For the second estimate, using the equation , we write

Similarly, involves precisely and its first derivative, multiplied by derivatives of the profile . This proves at once the second estimate in the lemma, upon using the exponential localization of derivatives of .

**1.3. Rayleigh problem for stable profiles**

In this part, we shall construct an exact inverse for the Rayleigh operator for small in the case that is stable. Precisely, we construct the inverse of the problem:

in which and are sufficiently small. To do so, we first invert the Rayleigh operator when by exhibiting an explicit Green function. We then use this inverse to construct an approximate inverse to operator through the construction of an approximate Green function. Finally, the construction of the exact inverse of follows by an iterative procedure. We introduce the function spaces. Let be some complex number so that , for each fixed complex number . We will use the function spaces , for , to denote the spaces consisting of measurable functions such that the norm

is bounded. In case , we simply write in places of , respectively. We also introduce the function spaces , , such that for any , the function additionally satisfies

for all and for . The best constant in the previous bounds defines the norm . We will prove in this section the following proposition.

Proposition 3Let be in and . Assume that and is sufficiently small. Then, there exists an operator from to (defined by (30)) so that

for all .

**1.3.1 Case **

As mentioned, we begin with the Rayleigh operator when . We will find the inverse of . More precisely, we will construct the Green function of and solve

We first prove the following lemma.

Lemma 4Assume that . There are two independent solutions of with the Wronskian determinant

Furthermore, there are analytic functions with and so that the asymptotic descriptions

as , for some positive constants and for . Here when is on the negative real axis, we take the value of to be .

*Proof:* First, we observe that

is an exact solution of . In addition, the claimed asymptotic expansion for clearly holds for near since . We then construct a second particular solution , imposing the Wronskian determinant to be one:

From this, the variation-of-constant method then yields

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

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 is not properly defined for when , it is coherent to choose the determination of the logarithm which is defined on .

With such a choice of the logarithm, is holomorphic in . In particular if , is holomorphic in excepted on the half line . For , is holomorphic as a function of excepted if is real and negative, namely excepted if . For a fixed , is an holomorphic function of provided does not cross , and provided does not cross . The Lemma then follows from the explicit expression (1) of .

Let be constructed as in Lemma 4. Then the Green function of the operator can be defined by

Here we note that is complex with and so the Green function is a well-defined function in , continuous across , and its first derivative has a jump across . Let us now introduce the inverse of as

The following lemma asserts that the operator is in fact well-defined from to , which in particular shows that gains two derivatives, but losses the fast decay at infinity.

Lemma 5Assume that . For any , the function is a solution to the Rayleigh problem(15). In addition, , and there holds

for some universal constant .

*Proof:* As long as it is well-defined, the function solves the equation (15) at once by a direct calculation, upon noting that

for each fixed .

Next, by scaling, we assume that . By Lemma 4, it is clear that and are uniformly bounded. Thus, by direct computations, we have

That is, grows linearly in for large and has a singularity of order when is near , for arbitrary . Since , the integral (18) is well-defined and satisfies

in which we used the fact that . Finally, as for derivatives, we need to check the order of singularities for near . We note that , and hence

Thus, behaves as near the critical layer. In addition, from the equation, we have

This proves that by definition of the function space .

*Proof:* This is Lemma 5 when . When or , the lemma follows directly from the identity (20).

**1.3.2 Case : an approximate Green function**

Let and be the two solutions of that are constructed above in Lemma 4. We note that solutions of tend to a constant value as since . We now construct normal mode solutions to the Rayleigh equation with

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 at infinity. In order to study the mode which behaves like we introduce

A direct calculation shows that the Wronskian determinant

is non zero. In addition, we can check that

We are then led to introduce an approximate Green function defined by

Again, like , the Green function is “singular” near with two sources of singularities: one arising from for near and the other coming from the singularity of . By a view of (23), it is clear that

for each fixed . Here the error term is defined by

We then introduce an approximate inverse of the operator defined by

Lemma 7Assume that , and let be or . For any , with , the function is well-defined in , satisfying

*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 and assume . Similarly to the estimate (19), Lemma 4 and the definition of yield

Hence, by definition,

which is clearly bounded by . This proves the right exponential decay of at infinity, for all .

Next, by definition, we have

Since decay exponentially at infinity, the exponential decay of follows directly from the above integral representation. It remains to check the order of singularity near the critical layer. Clearly, for bounded , we have

The lemma then follows at once, using the extra factor of in the front of the integral (26) to bound the factor. The estimates for derivatives follow similarly.

**1.3.3 Case : the exact solver for Rayleigh**

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

It then follows that . Inductively, we define

for . It is then clear that for all ,

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

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

which implies that in as as long as is sufficiently small. In addition, by a view of (27),

This shows that the series

converges in , assuming that is small.

Next, by taking the limit of in (29), the equation (14) holds by definition at least in the distributional sense. The estimates when is near follow directly from the similar estimates on ; see Lemma 7. The proof of Proposition 3 is thus complete.

**1.3.4 Exact Rayleigh solutions**

We shall construct two independent exact Rayleigh solutions by iteration, starting from the approximate Rayleigh solutions defined as in (22).

Lemma 8For small enough so that , there exist two independent functions such that

Furthermore, we have the following expansions in :

as . At , there hold

with .

*Proof:* Let us start with the decaying solution , which is now constructed by induction. Let us introduce

and inductively for ,

We also introduce

By definition, it follows that

We observe that and . Inductively for , by the estimate (28), we have

and by Lemma 7,

Thus, for sufficiently small , the series converges in and the error term in . This proves the existence of the exact decaying Rayleigh solution in , or in .

As for the growing solution, we simply define

By definition, solves the Rayleigh equation identically. Next, since tends to , is of order as .

Finally, at , we have

From the definition, we have . This proves the lemma, upon using that .

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).

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