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diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md
index de80f96..bca933f 100644
--- a/source/know/concept/blasius-boundary-layer/index.md
+++ b/source/know/concept/blasius-boundary-layer/index.md
@@ -12,88 +12,76 @@ 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 $$\mathrm{Re} \gg 1$$
-close to a surface.
+at large [Reynolds number](/know/concept/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 $$x \ge 0$$.
-To describe this, we make an ansatz
-for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$:
+Let the half-plane lie in the $$(x,z)$$-plane (i.e. at $$y = 0$$)
+and exist for all $$x \ge 0$$, such that its edge lies on the $$z$$-axis.
+A fluid with velocity field $$\va{v} = U \vu{e}_x$$
+approaches the half-plane's edge head-on.
+To describe the fluid's movements around the plane,
+we make an ansatz for the so-called **slip-flow region**'s $$x$$-velocity $$v_x(x, y)$$:
 
 $$\begin{aligned}
     v_x
     = U f'(s)
-    \qquad \quad
+    \qquad \qquad
     s
     \equiv \frac{y}{\delta(x)}
 \end{aligned}$$
 
+Where $$\delta(x) \equiv \sqrt{\nu x / U}$$ is the boundary layer thickness
+estimate that was used to derive the Prandtl equations.
 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,
-which we will stick with.
-For later use, it is worth writing the derivatives of $$s$$:
+and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$,
+i.e. the fluid is stationary at the half-plane's surface $$s = 0$$,
+and has velocity $$U$$ far away at $$s \to \infty$$.
 
-$$\begin{aligned}
-    \pdv{s}{x}
-    = - y \frac{\delta'}{\delta^2}
-    = - s \frac{\delta'}{\delta}
-    \qquad \quad
-    \pdv{s}{y}
-    = \frac{1}{\delta}
-\end{aligned}$$
-
-Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:
+Inserting the ansatz into the incompressibility condition $$\nabla \cdot \va{v} = 0$$ yields:
 
 $$\begin{aligned}
     \pdv{v_y}{y}
     = - \pdv{v_x}{x}
-    = U s f'' \frac{\delta'}{\delta}
+    = - \pdv{v_x}{s} \pdv{s}{x}
+    = U \frac{\delta'}{\delta} s f''
 \end{aligned}$$
 
-Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:
+Which we integrate by parts to get an expression for the $$y$$-velocity $$v_y$$, namely:
 
 $$\begin{aligned}
     v_y
     = U \frac{\delta'}{\delta} \int s f'' \dd{y}
+    = U \delta' \bigg( s f' - \int f' \dd{s} \bigg)
     = U \delta' \: (s f' - f)
 \end{aligned}$$
 
 Now, consider the main Prandtl equation,
-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:
-
-$$\begin{aligned}
-    - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
-    = \nu U \frac{1}{\delta^2} f'''
-\end{aligned}$$
-
-After  multiplying it by $$\delta^2 / U$$ and cancelling out some terms,
-it reduces to:
+assuming the attack velocity $$U$$ is constant.
+Inserting our expressions for $$v_x$$ and $$v_y$$ into it gives:
 
 $$\begin{aligned}
-    \nu f''' + U \delta' \delta f'' f
-    = 0
+    \nu \pdvn{2}{v_x}{y}
+    &= v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    \\
+    \nu \pdvn{2}{v_x}{s} \pdvn{2}{s}{y}
+    &= v_x \pdv{v_x}{s} \pdv{s}{x} + v_y \pdv{v_x}{s} \pdv{s}{y}
+    \\
+    \nu U \frac{1}{\delta^2} f'''
+    &= - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
 \end{aligned}$$
 
-Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:
+We multiply by $$\delta^2 / U$$, cancel out some terms,
+and substitute $$\delta(x) \equiv \sqrt{\nu x / U}$$, leaving:
 
 $$\begin{aligned}
-    \nu f''' + U \frac{\nu}{2 U} f'' f
-    = 0
+    \nu f'''
+    &= - U \delta' \delta f'' f
+    = - U \frac{\nu}{2 U} f'' f
 \end{aligned}$$
 
-Simplifying this leads us to the **Blasius equation**,
+This leads us to the **Blasius equation**,
 which is a nonlinear ODE for $$f(s)$$:
 
 $$\begin{aligned}