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5 files changed, 621 insertions, 24 deletions

diff --git a/content/know/concept/blasius-boundary-layer/index.pdc b/content/know/concept/blasius-boundary-layer/index.pdc
new file mode 100644
index 0000000..d9563c2
--- /dev/null
+++ b/content/know/concept/blasius-boundary-layer/index.pdc
@@ -0,0 +1,119 @@
+---
+title: "Blasius boundary layer"
+firstLetter: "B"
+publishDate: 2021-05-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-05-10T18:41:28+02:00
+draft: false
+markup: pandoc
+---
+
+# Blasius boundary layer
+
+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:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    = \nu \pdv[2]{v_x}{y}
+\end{aligned}$$
+
+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.
diff --git a/content/know/concept/fredholm-alternative/index.pdc b/content/know/concept/fredholm-alternative/index.pdc
new file mode 100644
index 0000000..8f20e57
--- /dev/null
+++ b/content/know/concept/fredholm-alternative/index.pdc
@@ -0,0 +1,67 @@
+---
+title: "Fredholm alternative"
+firstLetter: "F"
+publishDate: 2021-05-29
+categories:
+- Mathematics
+
+date: 2021-03-09T20:36:01+01:00
+draft: false
+markup: pandoc
+---
+
+# Fredholm alternative
+
+The **Fredholm alternative** is a theorem regarding equations involving
+a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/),
+and is useful in the context of multiple-scale perturbation theory.
+It is an *alternative* because it gives two mutually exclusive options,
+given here in [Dirac notation](/know/concept/dirac-notation/):
+
+1.  $\hat{L} \ket{u} = \ket{f}$ has a unique solution $\ket{u}$ for every $\ket{f}$.
+2.  $\hat{L}^\dagger \ket{w} = 0$ has non-zero solutions.
+    Then regarding $\hat{L} \ket{u} = \ket{f}$:
+    1.  If $\braket{w}{f} = 0$ for all $\ket{w}$, then it has infinitely many solutions $\ket{u}$.
+    2.  If $\braket{w}{f} \neq 0$ for any $\ket{w}$, then it has no solutions $\ket{u}$.
+
+Where $\hat{L}^\dagger$ is the adjoint of $\hat{L}$.
+In other words, $\hat{L} \ket{u} = \ket{f}$ has non-trivial solutions if
+and only if for all $\ket{w}$ (including the trivial case $\ket{w} = 0$)
+it holds that $\braket{w}{f} = 0$.
+
+As a specific example,
+if $\hat{L}$ is a matrix and the kets are vectors,
+this theorem can alternatively be stated as follows using the determinant:
+
+1.  If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$
+    has a unique solution $\vec{u}$ for every $\vec{f}$.
+2.  If $\mathrm{det}(\hat{L}) = 0$,
+    then $\hat{L}^\dagger \vec{w} = \vec{0}$ has non-zero solutions.
+    Then regarding $\hat{L} \vec{u} = \vec{f}$:
+    1.  If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has
+        infinitely many solutions $\vec{u}$.
+    2.  If $\vec{w} \cdot \vec{f} \neq 0$ for any $\vec{w}$, then it has
+        no solutions $\vec{u}$.
+
+Consequently, the Fredholm alternative is also brought up
+in the context of eigenvalue problems.
+Define $\hat{M} = (\hat{L} - \lambda \hat{I})$,
+where $\lambda$ is an eigenvalue of $\hat{L}$
+if and only if $\mathrm{det}(\hat{M}) = 0$.
+Then for the equation $\hat{M} \ket{u} = \ket{f}$, we can say that:
+
+1.  If $\lambda$ is *not* an eigenvalue,
+    then there is a unique solution $\ket{u}$ for each $\ket{f}$.
+2.  If $\lambda$ is an eigenvalue, then $\hat{M}^\dagger \ket{w} = 0$
+    has non-zero solutions. Then:
+    1.  If $\braket{w}{f} = 0$ for all $\ket{w}$, then there are
+        infinitely many solutions $\ket{u}$.
+    2.  If $\braket{w}{f} \neq 0$ for any $\ket{w}$, then there are no
+        solutions $\ket{u}$.
+
+
+
+## References
+1.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.
diff --git a/content/know/concept/optical-wave-breaking/index.pdc b/content/know/concept/optical-wave-breaking/index.pdc
index 757a633..2ab3ff1 100644
--- a/content/know/concept/optical-wave-breaking/index.pdc
+++ b/content/know/concept/optical-wave-breaking/index.pdc
@@ -75,11 +75,10 @@ the instantaneous frequencies for these separate effects:
 $$\begin{aligned}
     \omega_i(z,t)
     &\approx \omega_\mathrm{GVD}(z,t) + \omega_\mathrm{SPM}(z,t)
-    \\
 %    &= \frac{\beta_2 z / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} \frac{t}{T_0^2}
 %    + \frac{2\gamma P_0 z}{T_0^2} t \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big)
 %    \\
-    &= \frac{tz}{T_0^2} \bigg( \frac{\beta_2 / T_0^2}{1 + \beta_2^2 z^2 / T_0^4}
+    = \frac{tz}{T_0^2} \bigg( \frac{\beta_2 / T_0^2}{1 + \beta_2^2 z^2 / T_0^4}
     + 2\gamma P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
 \end{aligned}$$
 
@@ -97,14 +96,14 @@ and $N_\mathrm{sol}$ is the **soliton number**,
 which is defined as:
 
 $$\begin{aligned}
-    N_\mathrm{sol}^2 = \frac{L_D}{L_N} = \frac{\gamma P_0 T_0^2}{|\beta_2|}
+    N_\mathrm{sol}^2 \equiv \frac{L_D}{L_N} = \frac{\gamma P_0 T_0^2}{|\beta_2|}
 \end{aligned}$$
 
 This quantity is very important in anomalous dispersion,
 but even in normal dispesion, it is still a useful measure of the relative strengths of GVD and SPM.
 As was illustrated earlier, $\omega_i$ overtakes itself at the edges,
-so OWB only occurs when $\omega_i$ is not monotonic,
-which is when its $t$-derivative,
+so OWB occurs when $\omega_i$ oscillates there,
+which starts when its $t$-derivative,
 the **instantaneous chirpyness** $\xi_i$,
 has *two* real roots for $t^2$:
 
@@ -122,11 +121,17 @@ leading to the following exact minimum value $N_\mathrm{min}^2$ for $N_\mathrm{s
 such that OWB can only occur when $N_\mathrm{sol}^2 > N_\mathrm{min}^2$:
 
 $$\begin{aligned}
-    N_\mathrm{min}^2 = \frac{1}{4} \exp\!\Big(\frac{3}{2}\Big) \approx 1.12
+    \boxed{
+        N_\mathrm{min}^2 = \frac{1}{4} \exp\!\Big(\frac{3}{2}\Big) \approx 1.12
+    }
 \end{aligned}$$
 
-Now, consider two times $t_1$ and $t_2$ in the pulse, separated by
-a small initial interval $(t_2 - t_1)$.
+If this condition $N_\mathrm{sol}^2 > N_\mathrm{min}^2$ is not satisfied,
+$\xi_i$ cannot have two roots for $t^2$, meaning $\omega_i$ cannot overtake itself.
+GVD is unable to keep up with SPM, so OWB will not occur.
+
+Next, consider two points at $t_1$ and $t_2$ in the pulse,
+separated by a small initial interval $(t_2 - t_1)$.
 The frequency difference between these points due to $\omega_i$
 will cause them to displace relative to each other
 after a short distance $z$ by some amount $\Delta t$,
@@ -136,21 +141,21 @@ $$\begin{aligned}
     \Delta t
     &\approx z \Delta\beta_1
     \qquad
-    &&\Delta\beta_1 = \beta_1(\omega_i(z,t_2)) - \beta_1(\omega_i(z,t_1))
+    &&\Delta\beta_1 \equiv \beta_1(\omega_i(z,t_2)) - \beta_1(\omega_i(z,t_1))
     \\
     &\approx z \beta_2 \Delta\omega_i
     \qquad
-    &&\Delta\omega_i = \omega_i(z,t_2) - \omega_i(z,t_1)
+    &&\Delta\omega_i \equiv \omega_i(z,t_2) - \omega_i(z,t_1)
     \\
     &\approx z \beta_2 \Delta\xi_i \,(t_2 - t_1)
     \qquad \quad
-    &&\Delta\xi_i = \xi_i(z,t_2) - \xi_i(z,t_1)
+    &&\Delta\xi_i \equiv \xi_i(z,t_2) - \xi_i(z,t_1)
 \end{aligned}$$
 
 Where $\beta_1(\omega)$ is the inverse of the group velocity.
 OWB takes place when $t_2$ and $t_1$ catch up to each other,
 which is when $-\Delta t = (t_2 - t_1)$.
-The distance where this happens, $z = L_\mathrm{WB}$,
+The distance where this happens first, $z = L_\mathrm{WB}$,
 must therefore satisfy the following condition
 for a particular value of $t$:
 
@@ -161,7 +166,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The time $t$ of OWB must be where $\omega_i(t)$ has its steepest slope,
-which is at the minimum value of $\xi_i(t)$, and, by extension $f(x)$.
+which is at the minimum value of $\xi_i(t)$, and by extension $f(x)$.
 This turns out to be $f(3/2)$:
 
 $$\begin{aligned}
@@ -170,16 +175,17 @@ $$\begin{aligned}
     = 1 - N_\mathrm{sol}^2 / N_\mathrm{min}^2
 \end{aligned}$$
 
-Clearly, $f_\mathrm{min} \ge 0$ when
-$N_\mathrm{sol}^2 \le N_\mathrm{min}^2$, which, when inserted in the
-condition above, confirms that OWB cannot occur in that case. Otherwise,
-if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then:
+Clearly, $f_\mathrm{min} \ge 0$ when $N_\mathrm{sol}^2 \le N_\mathrm{min}^2$,
+which, when inserted above, leads to an imaginary $L_\mathrm{WB}$,
+confirming that OWB cannot occur in that case.
+Otherwise, if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then:
 
 $$\begin{aligned}
-    L_\mathrm{WB}
-    = - \frac{T_0^2}{\beta_2 \, \sqrt{f_\mathrm{min}}}
-    = \frac{L_D}{\sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}}
-    = \frac{L_D}{\sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}}
+    \boxed{
+        L_\mathrm{WB}
+        = \frac{T_0^2}{\beta_2 \, \sqrt{- f_\mathrm{min}}}
+        = \frac{L_D}{\sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}}
+    }
 \end{aligned}$$
 
 This prediction for $L_\mathrm{WB}$ appears to agree well
@@ -196,7 +202,7 @@ Filling $L_\mathrm{WB}$ in into $\omega_\mathrm{SPM}$ gives:
 
 $$\begin{aligned}
     \omega_{\mathrm{SPM}}(L_\mathrm{WB},t)
-    = \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big)
+    = \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
 \end{aligned}$$
 
 Assuming that $N_\mathrm{sol}^2$ is large in the denominator, this can
@@ -205,8 +211,8 @@ be approximately reduced to:
 $$\begin{aligned}
     \omega_\mathrm{SPM}(L_\mathrm{WB}, t)
 %   = \frac{2 \gamma P_0 t \exp(-t^2 / T_0^2)}{\beta_2 \sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}}
-    \approx \frac{2 \gamma P_0 t}{\beta_2 N_\mathrm{sol}} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big)
-    = 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\frac{t^2}{T_0^2}\Big)
+    \approx \frac{2 \gamma P_0 t}{\beta_2 N_\mathrm{sol}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
+    = 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
 \end{aligned}$$
 
 The expression $x \exp(-x^2)$ has its global extrema
diff --git a/content/know/concept/prandtl-equations/index.pdc b/content/know/concept/prandtl-equations/index.pdc
new file mode 100644
index 0000000..00f7773
--- /dev/null
+++ b/content/know/concept/prandtl-equations/index.pdc
@@ -0,0 +1,211 @@
+---
+title: "Prandtl equations"
+firstLetter: "P"
+publishDate: 2021-05-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-05-10T18:41:20+02:00
+draft: false
+markup: pandoc
+---
+
+# Prandtl equations
+
+In fluid dynamics, the **Prandtl equations** or **boundary layer equations**
+describe the movement of a [viscous](/know/concept/viscosity/) fluid
+with a large [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \gg 1$
+close to a solid surface.
+
+Fluids with a large Reynolds number
+are often approximated as having zero viscosity,
+since the simpler [Euler equations](/know/concept/euler-equations)
+can then be used instead of the [Navier-Stokes equations](/know/concept/navier-stokes-equations/).
+
+However, in reality, a viscous fluid obeys the *no-slip* boundary condition:
+at every solid surface the local velocity must be zero.
+This implies the existence of a **boundary layer**:
+a thin layer of fluid "stuck" to solid objects in the flow,
+where viscosity plays an important role.
+This is in contrast to the ideal flow far away from the surface.
+
+We consider a simple theoretical case in 2D:
+a large flat surface located at $y = 0$ for all $x \in \mathbb{R}$,
+with a fluid *trying* to flow parallel to it at $U$.
+The 2D treatment can be justified by assuming that everything is constant in the $z$-direction.
+We will not solve this case,
+but instead derive general equations
+to describe the flow close to a flat surface.
+
+At the wall, there is a very thin boundary layer of thickness $\delta$,
+where the fluid is assumed to be completely stationary $\va{v} = 0$.
+We are mainly interested in the region $\delta < y \ll L$,
+where $L$ is the distance at which the fluid becomes practically ideal.
+This the so-called **slip-flow** region,
+in which the fluid is not stationary,
+but still viscosity-dominated.
+
+In 2D, the steady Navier-Stokes equations are as follows,
+where the flow $\va{v} = (v_x, v_y)$:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    &= - \frac{1}{\rho} \pdv{p}{x} + \nu \Big( \pdv[2]{v_x}{x} + \pdv[2]{v_x}{y} \Big)
+    \\
+    v_x \pdv{v_y}{x} + v_y \pdv{v_y}{y}
+    &= - \frac{1}{\rho} \pdv{p}{y} + \nu \Big( \pdv[2]{v_y}{x} + \pdv[2]{v_y}{y} \Big)
+    \\
+    \pdv{v_x}{x} + \pdv{v_y}{y}
+    &= 0
+\end{aligned}$$
+
+The latter represents the fluid's incompressibility.
+We non-dimensionalize these equations,
+and assume that changes along the $y$-axis
+happen on a short scale (say, $\delta$),
+and along the $x$-axis on a longer scale (say, $L$).
+Let $\tilde{x}$ and $\tilde{y}$ be dimenionless variables of order $1$:
+
+$$\begin{aligned}
+    x
+    = L \tilde{x}
+    \qquad \quad
+    y
+    = \delta \tilde{x}
+    \qquad \quad
+    \pdv{x}
+    = \frac{1}{L} \pdv{\tilde{x}}
+    \qquad \quad
+    \pdv{y}
+    = \frac{1}{\delta} \pdv{\tilde{y}}
+\end{aligned}$$
+
+Furthermore, we choose velocity scales
+to be consistent with the incompressibility condition,
+and a pressure scale inspired
+by [Bernoulli's theorem](/know/concept/bernoullis-theorem/):
+
+$$\begin{aligned}
+    v_x
+    = U \tilde{v}_x
+    \qquad \quad
+    v_y
+    = \frac{U \delta}{L} \tilde{v}_y
+    \qquad \quad
+    p
+    = \rho U^2 \tilde{p}
+\end{aligned}$$
+
+We insert these scalings into the Navier-Stokes equations, yielding:
+
+$$\begin{aligned}
+    \frac{U^2}{L} \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \frac{U^2}{L} \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \frac{U^2}{L} \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{U}{L^2} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{U}{\delta^2} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{U^2 \delta}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{U^2 \delta}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \frac{U^2}{\delta} \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{U \delta}{L^3} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{U}{L \delta} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+For future convenience,
+we multiply the former equation by $L / U^2$, and the latter by $\delta / U^2$:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{1}{U L} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{L}{U \delta^2} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{\delta^2}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\delta^2}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{\delta^2}{U L^3} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+We would like to estimate $\delta$.
+Intuitively, we expect that higher viscosities $\nu$ give thicker layers,
+and that faster velocities $U$ give thinner layers.
+Furthermore, we expect *downstream thickening*:
+with distance $x$, viscous stresses slow down the slip-flow,
+leading to a gradual increase of $\delta(x)$.
+Some dimensional analysis thus yields the following estimate:
+
+$$\begin{aligned}
+    \delta
+    \approx \sqrt{\frac{\nu x}{U}}
+    \sim \sqrt{\frac{\nu L}{U}}
+\end{aligned}$$
+
+We thus insert $\delta = \sqrt{\nu L / U}$ into the Navier-Stokes equations, giving us:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{1}{U L} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \frac{1}{\nu} \pdv[2]{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{\nu}{U L} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\nu}{U L} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{\nu}{U^2 L^2} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdv[2]{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+Here, we recognize the definition of the Reynolds number $\mathrm{Re} = U L / \nu$:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \frac{1}{\mathrm{Re}} \pdv[2]{\tilde{v}_x}{\tilde{x}} + \pdv[2]{\tilde{v}_x}{\tilde{y}}
+    \\
+    \frac{1}{\mathrm{Re}} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \frac{1}{\mathrm{Re}^2} \pdv[2]{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \pdv[2]{\tilde{v}_y}{\tilde{y}}
+\end{aligned}$$
+
+Recall that we are only considering large Reynolds numbers $\mathrm{Re} \gg 1$,
+in which case $\mathrm{Re}^{-1} \ll 1$,
+so we can drop many terms, leaving us with these redimensionalized equations:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    = - \frac{1}{\rho} \pdv{p}{x} + \nu \pdv[2]{v_x}{y}
+    \qquad \quad
+    \pdv{p}{y}
+    = 0
+\end{aligned}$$
+
+The second one tells us that for a given $x$-value,
+the pressure is the same at the surface
+as in the main flow $y > L$, where the fluid is ideal.
+In the latter regime, we apply Bernoulli's theorem to rewrite $p$,
+using the *Bernoulli head* $H$ and the mainstream velocity $U(x)$:
+
+$$\begin{aligned}
+    p
+    = \rho H - \frac{1}{2} \rho U^2
+    = p_0 - \frac{1}{2} \rho U^2
+\end{aligned}$$
+
+Inserting this into the reduced Navier-Stokes equations,
+we arrive at the Prandtl equations:
+
+$$\begin{aligned}
+    \boxed{
+        v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+        = U \dv{U}{x} + \nu \pdv[2]{v_x}{y}
+        \qquad \quad
+        \pdv{v_x}{x} + \pdv{v_y}{y}
+        = 0
+    }
+\end{aligned}$$
+
+A notable application of these equations is
+the [Blasius boundary layer](/know/concept/blasius-boundary-layer/),
+where the surface in question
+is a semi-infinite plane.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/wicks-theorem/index.pdc b/content/know/concept/wicks-theorem/index.pdc
new file mode 100644
index 0000000..824885a
--- /dev/null
+++ b/content/know/concept/wicks-theorem/index.pdc
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+---
+title: "Wick's theorem"
+firstLetter: "W"
+publishDate: 2021-05-29
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-05-29T14:41:55+02:00
+draft: false
+markup: pandoc
+---
+
+# Wick's theorem
+
+In the [second quantization](/know/concept/second-quantization/) formalism,
+**Wick's theorem** helps to evaluate products
+of creation and annihilation operators by
+breaking them down into smaller products.
+
+Firstly, let us define the **normal product** or **normal order** as
+a product of second quantization operators
+reordered such that
+all creation operators are on the left of
+all annihilation operators.
+For two operators this is written as follows,
+at least in the case of bosons:
+
+$$\begin{aligned}
+    \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger}
+    \equiv \hat{b}_\beta^\dagger \hat{b}_\alpha
+\end{aligned}$$
+
+For fermions, the result must be negated for each swapping of adjacent operators
+(and every reordering of operators can be treated as a sequence of such swaps):
+
+$$\begin{aligned}
+    \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger}
+    \equiv - \hat{f}_\beta^\dagger \hat{f}_\alpha
+\end{aligned}$$
+
+The normal product of three or more operators works in the same way,
+but might not be unique depending,
+on how many of each type there are.
+
+Next, the **contraction** of the operators $A$ and $B$
+is defined as the vacuum matrix element,
+i.e. the expectation value of $\ket{0}$:
+
+$$\begin{aligned}
+    \expval{A B}_0
+    \equiv \matrixel{0}{A B}{0}
+\end{aligned}$$
+
+Unsurprisingly, a contraction can only be nonzero if
+$A = \hat{c}_\alpha$ is an annihilation and $B = \hat{c}_\alpha^\dagger$
+a creation for the same state $\alpha$.
+
+Wick's theorem states:
+**any product of second quantization operators can be
+rewritten as a sum of normal products,
+from which 0, 1, 2, etc. contractions have been removed
+in every possible way.**
+For fermions, the sign of a term must also be swapped
+every time two adjacent operators are swapped.
+As an example, for four operators:
+
+$$\begin{aligned}
+    A B C D
+    = \underline{A B C D}
+    &+ \underline{A B} \expval{C D}_0 \pm \underline{A C} \expval{B D}_0 + \underline{A D} \expval{B C}_0
+    \\
+    &+ \underline{B C} \expval{A D}_0 \pm \underline{B D} \expval{A C}_0 + \underline{C D} \expval{A B}_0
+    \\
+    &+ \expval{A B}_0 \expval{C D}_0 \pm \expval{A C}_0 \expval{B D}_0 + \expval{A D}_0 \expval{B C}_0
+\end{aligned}$$
+
+Where the negative signs apply to fermions only.
+We take the normal product with 0 contractions removed ($\underline{ABCD}$),
+then with 1 contraction removed in every possible way (first two lines),
+then with 2 contractions removed in every possible way (last line), and so on.
+
+
+## Proof
+
+We will prove this by induction, with the base case being two operators,
+where Wick's theorem becomes as follows:
+
+$$\begin{aligned}
+    A B
+    = \underline{AB} + \expval{A B}_0
+\end{aligned}$$
+
+This must be proven separately for fermions and bosons.
+For fermions, a general consequence of the definition of the anticommutator is:
+
+$$\begin{aligned}
+    \hat{f}_\alpha \hat{f}_\beta^\dagger
+    = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}
+\end{aligned}$$
+
+This anticommutator is known to be $\delta_{\alpha\beta}$,
+so we can inconsequentially take
+its inner product with the vacuum state $\ket{0}$:
+
+$$\begin{aligned}
+    \hat{f}_\alpha \hat{f}_\beta^\dagger
+    &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}}{0}
+    = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger + \hat{f}_\beta^\dagger \hat{f}_\alpha}{0}
+    \\
+    &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger}{0}
+    = \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} + \expval*{\hat{f}_\alpha \hat{f}_\beta^\dagger}_0
+\end{aligned}$$
+
+Which agrees with Wick's theorem. For bosons, we use the commutator:
+
+$$\begin{aligned}
+    \hat{b}_\alpha \hat{b}_\beta^\dagger
+    = \hat{b}_\beta^\dagger \hat{b}_\alpha + [\hat{b}_\alpha, \hat{b}_\beta^\dagger]
+\end{aligned}$$
+
+This commutator is known to be $\delta_{\alpha\beta}$,
+so we take the inner product with $\ket{0}$, like before:
+
+$$\begin{aligned}
+    \hat{b}_\alpha \hat{b}_\beta^\dagger
+    &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{[\hat{b}_\alpha, \hat{b}_\beta^\dagger]}{0}
+    = \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger - \hat{b}_\beta^\dagger \hat{b}_\alpha}{0}
+    \\
+    &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger}{0}
+    = \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger} + \expval*{\hat{b}_\alpha \hat{b}_\beta^\dagger}_0
+\end{aligned}$$
+
+Which again agrees with Wick's theorem.
+Next, we prove that if it holds for $N$ operators, then it also holds for $N + 1$.
+To begin with, consider the following statement about right-multiplying
+by an extra $A_{N+1}$, with $s = 1$ for bosons and $s = -1$ for fermions:
+
+$$\begin{aligned}
+    \underline{A_1 ... A_N} A_{N+1}
+    = \underline{A_1 ... A_N A_{N+1}}
+    + \sum_{n = 1}^N s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+\end{aligned}$$
+
+If $A_{N + 1}$ is an annihilation operator, then this is trivial:
+appending it does not break the existing normal order,
+and $\expval{A_n A_{N+1}}_0 = 0$ for all $A_n$.
+
+However, if $A_{N + 1}$ is a creation operator,
+then to restore the normal order,
+we move it to the front by swapping,
+which introduces a bunch of (anti)commutators:
+
+$$\begin{aligned}
+    \underline{A_1 ... A_N} A_{N+1}
+    &= s^N A_{N+1} \underline{A_1 ... A_N}
+    + \sum_{n} s^{n + N} \{[A_n, A_{N+1}]\} \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+    \\
+    &= \underline{A_1 ... A_N A_{N+1}}
+    + \sum_{n} s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+\end{aligned}$$
+
+Where $\{[]\}$ is the anticommutator or commutator,
+respectively for fermions or bosons.
+
+If we take Wick's theorem for $N$ operators $A_1 ... A_N$,
+and right-multiply it by $A_{N + 1}$,
+then each term will contain a product of the form $\underline{A_{v} ... A_{w}} A_{N+1}$.
+Using the relation that we just proved,
+each such product can be rewritten as follows:
+
+$$\begin{aligned}
+    \underline{A_v ... A_w} A_{N+1}
+    &= \underline{A_v ... A_w A_{N+1}}
+    + \sum_{n} s^{n + N} \expval{A_n A_{N+1}}_0 \underline{A_v ... A_{n-1} A_{n+1} ... A_w}
+\end{aligned}$$
+
+Inserting this back into Wick's theorem,
+we get new terms with contractions of $A_{N+1}$.
+After a lot of rearranging,
+the result turns out to just be Wick's theorem for $N\!+\!1$ operators.
+Therefore,
+if Wick's theorem holds for $N$ operators,
+it also holds for $N\!+\!1$.
+
+We showed that Wick's theorem holds for $N = 2$,
+so, by induction, it holds for all $N \ge 2$.
+
+
+
+## References
+1.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.