From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 .../know/concept/blasius-boundary-layer/index.md   | 34 +++++++++++-----------
 1 file changed, 17 insertions(+), 17 deletions(-)

(limited to 'source/know/concept/blasius-boundary-layer')

diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md
index fd4af57..de80f96 100644
--- a/source/know/concept/blasius-boundary-layer/index.md
+++ b/source/know/concept/blasius-boundary-layer/index.md
@@ -12,15 +12,15 @@ layout: "concept"
 In fluid dynamics, the **Blasius boundary layer** is an application of
 the [Prandtl equations](/know/concept/prandtl-equations/),
 which govern the flow of a fluid
-at large Reynolds number $\mathrm{Re} \gg 1$
+at large Reynolds number $$\mathrm{Re} \gg 1$$
 close to a surface.
 Specifically, the Blasius layer is the solution
 for a half-plane approached from the edge by a fluid.
 
-A fluid with velocity field $\va{v} = U \vu{e}_x$ flows to the plane,
-which starts at $y = 0$ and exists for $x \ge 0$.
+A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane,
+which starts at $$y = 0$$ and exists for $$x \ge 0$$.
 To describe this, we make an ansatz
-for the *slip-flow* region's $x$-velocity $v_x(x, y)$:
+for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$:
 
 $$\begin{aligned}
     v_x
@@ -30,13 +30,13 @@ $$\begin{aligned}
     \equiv \frac{y}{\delta(x)}
 \end{aligned}$$
 
-Note that $f'(s)$ is the derivative of an unknown $f(s)$,
-and that it obeys the boundary conditions $f'(0) = 0$ and $f'(\infty) = 1$.
-Furthermore, $\delta(x)$ is the thickness of the stationary boundary layer at the surface.
+Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$,
+and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$.
+Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface.
 To derive the Prandtl equations,
-the estimate $\delta(x) = \sqrt{\nu x / U}$ was used,
+the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used,
 which we will stick with.
-For later use, it is worth writing the derivatives of $s$:
+For later use, it is worth writing the derivatives of $$s$$:
 
 $$\begin{aligned}
     \pdv{s}{x}
@@ -47,7 +47,7 @@ $$\begin{aligned}
     = \frac{1}{\delta}
 \end{aligned}$$
 
-Inserting the ansatz for $v_x$ into the incompressibility condition then yields:
+Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:
 
 $$\begin{aligned}
     \pdv{v_y}{y}
@@ -55,7 +55,7 @@ $$\begin{aligned}
     = U s f'' \frac{\delta'}{\delta}
 \end{aligned}$$
 
-Which we integrate to get an expression for the $y$-velocity $v_y$, namely:
+Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:
 
 $$\begin{aligned}
     v_y
@@ -64,21 +64,21 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Now, consider the main Prandtl equation,
-assuming that the attack velocity $U$ is constant:
+assuming that the attack velocity $$U$$ is constant:
 
 $$\begin{aligned}
     v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
     = \nu \pdvn{2}{v_x}{y}
 \end{aligned}$$
 
-Inserting our expressions for $v_x$ and $v_y$ into this leads us to:
+Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to:
 
 $$\begin{aligned}
     - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
     = \nu U \frac{1}{\delta^2} f'''
 \end{aligned}$$
 
-After  multiplying it by $\delta^2 / U$ and cancelling out some terms,
+After  multiplying it by $$\delta^2 / U$$ and cancelling out some terms,
 it reduces to:
 
 $$\begin{aligned}
@@ -86,7 +86,7 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields:
+Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:
 
 $$\begin{aligned}
     \nu f''' + U \frac{\nu}{2 U} f'' f
@@ -94,7 +94,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Simplifying this leads us to the **Blasius equation**,
-which is a nonlinear ODE for $f(s)$:
+which is a nonlinear ODE for $$f(s)$$:
 
 $$\begin{aligned}
     \boxed{
@@ -103,7 +103,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Unfortunately, this cannot be solved analytically, only numerically.
-Nevertheless, the result shows a boundary layer $\delta(x)$
+Nevertheless, the result shows a boundary layer $$\delta(x)$$
 exhibiting the expected downstream thickening.
 
 
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