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+---
+title: "Blasius boundary layer"
+date: 2021-05-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+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.
+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 $v_x(x, y)$:
+
+$$\begin{aligned}
+    v_x
+    = U f'(s)
+    \qquad \quad
+    s
+    \equiv \frac{y}{\delta(x)}
+\end{aligned}$$
+
+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.
+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 $s$:
+
+$$\begin{aligned}
+    \pdv{s}{x}
+    = - y \frac{\delta'}{\delta^2}
+    = - s \frac{\delta'}{\delta}
+    \qquad \quad
+    \pdv{s}{y}
+    = \frac{1}{\delta}
+\end{aligned}$$
+
+Inserting the ansatz for $v_x$ into the incompressibility condition then yields:
+
+$$\begin{aligned}
+    \pdv{v_y}{y}
+    = - \pdv{v_x}{x}
+    = U s f'' \frac{\delta'}{\delta}
+\end{aligned}$$
+
+Which we integrate to get an expression for the $y$-velocity $v_y$, namely:
+
+$$\begin{aligned}
+    v_y
+    = U \frac{\delta'}{\delta} \int s f'' \dd{y}
+    = U \delta' \: (s f' - f)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \nu f''' + U \delta' \delta f'' f
+    = 0
+\end{aligned}$$
+
+Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields:
+
+$$\begin{aligned}
+    \nu f''' + U \frac{\nu}{2 U} f'' f
+    = 0
+\end{aligned}$$
+
+Simplifying this leads us to the **Blasius equation**,
+which is a nonlinear ODE for $f(s)$:
+
+$$\begin{aligned}
+    \boxed{
+        2 f''' + f'' f = 0
+    }
+\end{aligned}$$
+
+Unfortunately, this cannot be solved analytically, only numerically.
+Nevertheless, the result shows a boundary layer $\delta(x)$
+exhibiting the expected downstream thickening.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.