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## Math 597F, Notes 4: Prandtl boundary layer theory

With simple integration by parts, we were able to see in the last two lectures essentially the current “state of art” of the ${L^2}$ convergence of Navier-Stokes to Euler. Embarrassingly, the inviscid limit problem is widely open as discussed. It is noted that the ${L^2}$ energy norm is quite weak, and does not see in the inviscid limit the appearance of thin layers that might (and indeed will) occur near the boundary (for instance, the ${L^2}$ norm of Kato’s layer is of order ${\sqrt \nu}\to 0$). We will have to work with a different, stronger norm. Regarding the significance of viscosity despite being arbitrarily small (e.g., viscosity of air at zero temperature is about ${10^{-4}}$, which seems to be neglectable), d’Alembert in the 18th century has already argued out that ideal flows can’t explain well many of the physics, and the viscosity plays a crucial role near the boundary; for instance, one of his conclusions, known as d’Alembert’s paradox, asserts that solid body emerged in stationary ideal flows feels no drag acting on it (in the layman words, birds can’t fly!). Not until the beginning of the 20th century, Prandtl then postulated a solution Ansatz that revolutionized the previous understanding of slightly viscous flows near a boundary, later known as Prandtl boundary layer theory. The theory gave birth to the field of aerodynamics, and is regarded as one of the greatest achievements in fluid dynamics in the last century. Below, I’ll derive the Prandtl boundary layers.

1. Regular versus singular perturbations

Regular perturbations are perturbations that have the same quantitative nature in the equations. For instance, consider the problem

$\displaystyle A v - \epsilon B v = f,\qquad \epsilon \rightarrow 0.$

Assume that ${A,B}$ are two bounded linear operators from the same ${X}$ to ${Y}$. For simplicity, assume that ${A}$ is invertible. The quantitative nature of the problem is unchanged when ${\epsilon \rightarrow 0}$. The solution for the perturbed problem can be constructed in term of the series in ${\epsilon}$:

$\displaystyle v= (A-\epsilon B)^{-1}f = A^{-1}(1-\epsilon BA^{-1})^{-1}f = A^{-1} f + A^{-1} \sum_{k = 1}^\infty (\epsilon B A^{-1})^{k} f .\ \ \ \ \ (1)$

When ${\epsilon \rightarrow 0}$, the solution of the perturbed problem simply converges to the solution of the unperturbed problem. No new behavior arises in the limit.

Perturbation is singular when ${A,B}$ are quatitatively not the same type. For instance, as simple as the algebraic problem

$\displaystyle a x - \epsilon x^2 = f,$

with ${a}$ being a nonzero constant, beside the regular solution of the form ${x = \frac fa + \mathcal{O}(\epsilon)}$, there is a new solution due to the singular perturbation, which is of the form:

$\displaystyle x = \frac{a}{\epsilon} - \frac fa + \mathcal{O}(\epsilon) .$

(notice the first term in the expansion comes from balancing ${ax - \epsilon x^2 = 0}$). Clearly, this solution does not converge to the solution of the unperturbed problem as ${\epsilon \rightarrow 0}$.

The following singular perturbation example is more relevant to us. Consider the following ODE problem on the half-line ${x\ge 0}$:

\displaystyle \begin{aligned} v + \epsilon \partial_x v &= 1, \\ v_{\vert_{x=0}} &= v_0. \end{aligned}

A few observations. When ${\epsilon =0}$, the constant function ${v^E = 1}$ is the only solution to the limiting equation, but it does not solve the boundary (or rather, initial) condition at ${x=0}$. When ${\epsilon>0}$, the ODE problem has a unique smooth solution, which is equal to the prescribed boundary condition ${v_0}$ and is well approximated by ${v^E =1}$ away from the boundary. As ${v_0}$ might be different from ${v^E=1}$, a new layer appears near the boundary to correct the discrepancy. To find the equation for this new layer, we postulate a solution Ansatz: ${v = v^E(x) + v^\mathrm{bl} (\frac x\delta)}$ for some ${\delta \rightarrow 0}$ as ${\epsilon \rightarrow 0}$. Plugging this ansatz into the equation, one gets the “boundary layer equation”:

$\displaystyle v^\mathrm{bl} + \frac{\epsilon}{\delta} \partial_z v^\mathrm{bl} (z) = 0, \qquad v^\mathrm{bl}_{\vert_{z=0}} = v_0 - v^E_{\vert_{x=0}}, \qquad \lim_{z\rightarrow \infty} v^\mathrm{bl} (z) =0,$

with ${z}$ denoting the fast variable ${z = x/\delta}$. To balance the above equation, it’s clear to take the “boundary layer thickness” ${\delta = \epsilon}$. We get the solution ${v^\mathrm{bl} = (v_0 - 1) e^{-x/\epsilon}}$, which is called a boundary layer. By construction, this solution ansatz ${v = v^E(x) + v^\mathrm{bl} (\frac x\epsilon)}$ satisfies the boundary condition and approximately solves the ODE equation (in this example, since ${\partial_x v^E =0}$, the ansatz in fact solves the equation exactly). We remark that we would miss out this boundary layer, if we were doing the ${L^2}$ theory:

$\displaystyle \int_0^\infty | v - v^E|^2\; dx = - \epsilon \int_0^\infty \partial_x (v-v^E) (v - v^E)\; dx = \frac \epsilon 2 |v_0 - v^E|^2$

which converges to zero as ${\epsilon \rightarrow 0}$, for arbitrary initial data ${v_0}$.

Let me give another example of a singular perturbation, which will later come up in our business (for instance, when studying the resolvent equation of some linearized operators around a given profile solution). Consider the ODE problem on the half-line ${x\ge 0}$:

\displaystyle \begin{aligned} v + a \partial_x v - \epsilon \partial^2_x v &= f(x) \\ v_{\vert_{x=0}} &=0. \end{aligned}

Assume that the given source ${f(x)}$ is sufficiently smooth and integrable. Here, ${a}$ is some nonzero constant, and ${\epsilon}$ is positive and sufficiently small. When ${\epsilon =0}$, the unique bounded solution of the ODE equation is defined by

\displaystyle v^E(x) = \left \{ \begin{aligned} \frac 1a \int_0^x e^{-(x-y)/a} f(y)\; dy , &\qquad a>0 \\ - \frac 1a \int_x^\infty e^{-(x-y)/a} f(y)\; dy , &\qquad a<0, \end{aligned}\right.

Consider the case ${a>0}$. In this case, the inviscid solution ${v^E}$ satisfies the zero boundary condition and there is no need to add a boundary layer to correct the boundary condition. The solution of the perturbed ODE can be constructed via the formal series (1), yielding the estimate ${v = v^E(x) + \mathcal{O}(\epsilon)}$ in the ${L^\infty}$ norm.

Next, consider the case ${a<0}$. The inviscid solution ${v^E}$ defined as above does not in general satisfy the zero boundary condition, and so a boundary layer is needed. Again, let us denote the boundary layer solution by ${v^\mathrm{bl}(\frac x\delta)}$. Then, the boundary layer thickness is ${\delta = \epsilon}$ and ${v^\mathrm{bl}(z)}$ solves the boundary layer problem

$\displaystyle a \partial_z v^{\mathrm{bl}} - \partial_z^2 v^\mathrm{bl} = 0, \qquad v^\mathrm{bl}_{\vert_{z=0}} = - v^E_{\vert_{x=0}}, \qquad \lim_{z\rightarrow \infty} v^\mathrm{bl} (z) =0 ,$

which simply gives ${v^\mathrm{bl} = -v^E(0) e^{a x / \epsilon}}$. It can be proved that the exact solution of the perturbed ODE problem satisfies ${v = v^E (x)+ v^\mathrm{bl}(\frac x \epsilon) + \mathcal{O}(\epsilon)}$ in the ${L^\infty}$ norm.

2. Prandtl’s boundary layer theory

Let us come back to the singular perturbation problem of the inviscid limit ${\nu \rightarrow 0}$. For sake of presentation, consider the NS equations in the half-plane ${\Omega = \mathbb{R}^{n-1} \times \mathbb{R}_+}$ with the flat boundary ${\{ y=0\}}$. We then write ${\vec x = (x, y)\in \Omega}$ and the velocity vector field ${\vec v = (u,v)\in \mathbb{R}^{n-1}\times \mathbb{R}}$, with horizontal component ${u}$ and vertical component ${v}$ of the velocity. The NS equations written in the component-wise read

\displaystyle \left \{ \begin{aligned} u_t + (u\cdot \nabla_{x} + v \partial_y) u + \nabla_x p & = \nu \Delta u \\ v_t + ( u \cdot \nabla_x + v \partial_y) v + \partial_y p &= \nu \Delta v \\ \nabla_x \cdot u + \partial_y v &= 0 \\ (u,v)_{\vert_{y=0}} &= 0. \end{aligned}\right. \ \ \ \ \ (2)

Let ${\vec v^E = (u^E, v^E)}$ be the solution to the corresponding Euler equation (that is, the equation (2) above, with ${\nu =0}$ and with zero boundary condition on ${v^E}$). There are many solutions to the Euler equation. For instance, ${\vec v^E = (u^E(y),0)}$ is an exact solution to Euler equation, for arbitrary vector function ${u^E}$.

Clearly, ${u^E}$ needs not to vanish on the boundary ${y=0}$, and hence a boundary layer is needed to correct this discrepancy of the zero boundary conditions for Navier-Stokes. In 1904, Prandtl postulated the following solution Ansatz:

$\displaystyle \begin{pmatrix} u\\v \end{pmatrix} = \begin{pmatrix} u^E\\v^E \end{pmatrix} (t,x,y)+ \begin{pmatrix} u^P \\ v^P \end{pmatrix}(t,x,\frac y \delta).$

Let us denote the fast variable by ${z=\frac y \delta}$. The boundary conditions for ${(u^P, v^P)}$ are

$\displaystyle u^P_{\vert_{z=0}} = - u^E_{\vert_{y=0}}, \qquad v^P_{\vert_{z=0}} = 0, \qquad \lim_{z\rightarrow \infty} u^P(t,x,z) = 0.$

Observe that ${\partial_y = \frac 1\delta \partial_z}$. From the divergence-free condition: ${\nabla_x\cdot u^P + \frac 1\delta \partial_z v^P = 0.}$ This balance shows that the vertical velocity of ${V^P}$ is of order ${\delta}$, as compared to order one of the horizontal component ${u^P}$. Thus, we rescale ${v^P}$ by writing ${v^P = \delta \tilde v^P}$, for some new ${\tilde v^P}$. The divergence-free condition becomes

$\displaystyle \nabla_x \cdot u^P + \partial_z \tilde v^P = 0.$

Since ${v^P = \mathcal{O}(\delta)}$, to leading order in small ${\delta}$ and ${\nu}$, the NS equation for the vertical component of velocity simply reads

$\displaystyle \frac 1 \delta \partial_z p^P = 0,$

if we denote by ${p^P}$ the pressure for the Prandtl layer. This yields at once that ${p^P = p^P(t,x)}$, the pressure does not depend on the vertical direction within the boundary layer. Next, to leading order, the NS equation for the horizontal component of velocity becomes

$\displaystyle u^P_t + (u^E + u^P)\cdot \nabla_{x} u^P + u^P \cdot \nabla_x u^E + (\frac{v^E}{\delta} + \tilde v^P) \partial_z u^P + \nabla_x p^P =\frac \nu{\delta^2} \partial_z^2 u^P$

In which we note that ${\frac{v^E}{\delta}}$ can be approximated by ${z \partial_y v^E(t,x,0)}$, since ${v^E(t,x,0) = 0}$. As the above equation is defined near the boundary, one can replace ${u^E}$ by its value on the boundary ${u^E(t,x,0)}$ (to the leading order). The above set of equations is called the Prandtl boundary layer equation, a single equation for the horizontal component of velocity ${u^P}$, with its vertical velocity ${\tilde v^P}$ being defined through the divergence-free condition. Note that since ${u^P}$ vanishes away from the boundary, the above equation yields that ${p^P}$ is simply a constant. That is, there is no pressure (gradient) for the Prandtl boundary layer. It turns out to be convenient to denote the following Prandtl’s variables:

$\displaystyle \bar u = u^E (t,x,0)+ u^P(t,x,z), \qquad \bar v = z \partial_y v^E(t,x,0) + \tilde v^P(t,x,z) .\ \ \ \ \ (3)$

The Prandtl boundary layer problem simply becomes

\displaystyle \left \{ \begin{aligned} \bar u_t +\bar u \cdot \nabla_x\bar u + \bar v \partial_z\bar u &= \tilde \nu \partial_z^2\bar u - \nabla_x\bar p \\ \nabla_x \cdot \bar u + \partial_z \bar v &=0 \\ (\bar u,\bar v)_{\vert_{z=0}} & =0 \\ \lim_{z\rightarrow \infty} \bar u(t,x,z) & = u^E(t,x,0) \end{aligned} \right. \ \ \ \ \ (4)

in which the pressure is known: ${ \bar p_x= -u^E_t(t,x,0)- \frac{|u^E(t,x,0)|^2}{2} }$ and the Prandtl’s viscosity ${\tilde \nu = \frac{\nu}{\delta^2}}$. Classically, one takes the boundary layer thickness ${\delta}$ to be ${\sqrt \nu}$. One can also eliminate the vertical velocity ${\bar v}$ by writing ${\bar v = - \int_0^z \nabla_x \cdot \bar u \; dz}$, thanks to the divergence-free condition. The above set of equations and boundary conditions is complete.

3. Mathematical justification

Having introduced the Prandtl boundary layer ${(u^P, \delta \tilde v^P)}$ through (3) and (4), it is natural to ask the following question:
A mathematical problem: Under what circumstances, there holds the convergence of Navier-Stokes to Euler plus Prandtl layer in the inviscid limit? More precisely, when do we have the asymptotic expansion in the inviscid limit:

$\displaystyle \begin{pmatrix} u\\v \end{pmatrix} = \begin{pmatrix} u^E\\v^E \end{pmatrix} (t,x,y)+ \begin{pmatrix} u^P \\ \delta \tilde v^P \end{pmatrix}(t,x,\frac y \delta) + E_\nu(t,x,y),$

for which the remainder ${E_\nu (t,x,y)\rightarrow 0}$ in some norm as ${\nu\rightarrow 0}$ ?

The problem is solved, with the classical boundary layer thickness ${\delta = \sqrt \nu}$, in the analytic framework by Sanmartino and Caflisch in 1998. Here, analytic framework means that they work with initial data in the analytic space. As mentioned in the previous lecture, we’re interested in allowing initial data to have less regularity (for instance, in Sobolev spaces or even ${C^\infty}$ smooth data). The first issue to overcome is the existence of solutions for the Prandtl equation. This turns out to be very difficult, since no natural energy estimates are available for the Prandtl equation and there is a loss of ${x}$-derivative, precisely coming from the nonlocal ${v}$ term. Fortunately, back in the 60s, Oleinik was able to construct smooth solutions of the Prandtl equation, provided that the initial horizontal velocity ${u^P(0,x,z)}$ is monotonic in ${z}$. Monotonicity plays a crucial role in her proof, as it allows her to make certain change of variables (indeed, without the monotonicity, Gérard-Varet and Dormy in fact showed that the Prandtl equation is linearly ill-posed in Sobolev spaces). Given the known monotonic boundary layers constructed by Oleinik, the asymptotic expansion remains open, no matter how small, but nonzero, the initial perturbations ${E_\nu(0,x,y)}$ are. In the next few lectures, I’ll plan to touch on this issue of controlling the remainders, or more precisely to study the linearized operator around a boundary layer.