From 5fc2fd763b07b735c2895f9375c2dfa6c43fe86a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 15 Jan 2023 10:46:17 +0100
Subject: Expand knowledge base, renamed "Boussinesq wave theory"

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 .../concept/korteweg-de-vries-equation/index.md    | 631 +++++++++++++++++++++
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----
-title: "Boussinesq wave equations"
-sort_title: "Boussinesq wave equations"
-date: 2023-01-07
-categories:
-- Physics
-- Mathematics
-- Fluid mechanics
-- Fluid dynamics
-layout: "concept"
----
-
-In fluid mechanics, **Boussinesq's wave theory**
-consists of several equations to describe surface waves of a liquid.
-It was the first attempt to explain the nonlinear phenomenon of solitons,
-which were not predicted by the linear theories existing at the time.
-
-
-
-## Boundary conditions
-
-Consider the [Euler equations](/know/concept/euler-equations/)
-for an incompressible fluid with negligible viscosity:
-
-$$\begin{aligned}
-    \va{g} - \frac{\nabla p}{\rho}
-    &= \pdv{\va{u}}{t} + (\va{u} \cdot \nabla) \va{u}
-    \qquad \qquad
-    \nabla \cdot \va{u}
-    = 0
-\end{aligned}$$
-
-We rewrite the former using the velocity potential $$\Psi$$ for $$\va{u}$$
-and the gravitational potential $$\Phi$$ for $$\va{g}$$,
-such that $$\va{u} = \nabla \Psi$$ and $$\va{g} = - \nabla \Phi$$.
-We also use a vector identity:
-
-$$\begin{aligned}
-    - \nabla \Phi - \frac{\nabla p}{\rho}
-    &= \nabla \pdv{\Psi}{t} + \frac{1}{2} \nabla |\va{u}|^2 + (\nabla \cross \nabla \Psi) \cross \va{u}
-\end{aligned}$$
-
-Recall that the curl of a gradient is always zero, so the last term disappears.
-Integrating in space yields integration constants $$C(t)$$ and $$p_0$$,
-the latter representing atmospheric pressure:
-
-$$\begin{aligned}
-    \pdv{\Psi}{t} + \frac{1}{2} |\va{u}|^2
-    &= -\Phi + \frac{p_0 - p}{\rho} + C
-\end{aligned}$$
-
-Consider a rectangular channel of depth $$h$$
-extending infinitely far along the $$x$$-axis,
-with a finite width along the $$y$$-axis.
-We choose our coordinate system so that
-$$z = 0$$ is the equilibrium water level,
-and the bottom is at $$z = -h$$.
-Let $$\eta(x, t)$$ be the deformation of the surface,
-for which we want to find a wave equation.
-All quantities are assumed to be constant in $$y$$,
-so we only consider a 2D flow $$\va{u} = \big(u^{(x)}, u^{(z)}\big)$$.
-At the surface $$z = \eta$$ we thus have:
-
-$$\begin{aligned}
-    \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
-    &= - g \eta + \frac{p_0 - p}{\rho} + C
-\end{aligned}$$
-
-Where $$g \eta = \Phi$$ with $$g \approx 9.81 \:\mathrm{m}/\mathrm{s}^2$$ on Earth.
-Later, we will differentiate this formula in $$x$$,
-so we can already set $$C = 0$$ now, since it will vanish then anyway.
-Furthermore, we assume that at the surface $$z = \eta$$
-the pressures are in equilibrium $$p_0 = p$$, leaving:
-
-$$\begin{aligned}
-    \boxed{
-        \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \eta
-        = 0
-    }
-\end{aligned}$$
-
-This is called the **free surface boundary condition**.
-Obviously, if $$z = \eta$$, then $$\eta \!-\! z = 0$$.
-Taking the [material derivative](/know/concept/material-derivative/)
-of this fact gives the following relation:
-
-$$\begin{aligned}
-    0
-    = \frac{\mathrm{D}}{\mathrm{D} t} (\eta - z)
-    = \pdv{\eta}{t} + \pdv{z}{t} + \va{u} \cdot \nabla \eta - \va{u} \cdot \nabla z
-\end{aligned}$$
-
-Since $$\eta$$ only depends on $$x$$ and $$t$$,
-this becomes the **kinematic boundary condition**:
-
-$$\begin{aligned}
-    \boxed{
-        \pdv{\eta}{t} + u^{(x)} \pdv{\eta}{x} - u^{(z)}
-        = 0
-    }
-\end{aligned}$$
-
-The equations will be derived from these two fundamental boundary conditions.
-
-
-
-## Boussinesq's approximation
-
-Let us take a Taylor expansion of the velocity potential $$\Psi(x, z, t)$$
-at the bottom $$z = -h$$:
-
-$$\begin{aligned}
-    \Psi(x, z)
-    = \Psi(x, -h) + (z + h) \: \Psi_z(x, -h) + \frac{(z + h)^2}{2} \: \Psi_{zz}(x, -h) + ...
-\end{aligned}$$
-
-Because the fluid is incompressible, this can be rewritten.
-Laplace's equation tells us:
-
-$$\begin{aligned}
-    \nabla \cdot \va{u}
-    = \nabla^2 \Psi
-    = 0
-    \qquad \implies \qquad
-    \:\:\:\Psi_{zz}
-    = - \Psi_{xx}
-\end{aligned}$$
-
-Which we use for the expansion's second-order term.
-Similarly, for the fourth-order term:
-
-$$\begin{aligned}
-    0
-    = \nabla^2 \nabla^2 \Psi
-    &= \Psi_{zzzz} + \pdvn{2}{}{z} \Psi_{xx} + \pdvn{2}{}{x} \Psi_{zz} + \Psi_{xxxx}
-    \\
-    &= \Psi_{zzzz} + \pdvn{2}{}{x} \Psi_{zz} - \pdvn{2}{}{x} \Psi_{xx} + \Psi_{xxxx}
-    \\
-    &= \Psi_{zzzz} - \Psi_{xxxx}
-\end{aligned}$$
-
-And so on for higher orders.
-In the Taylor expansion, all even derivatives can be rewritten in this way,
-and all odd derivatives can be split into a single $$\ipdv{}{z}$$ and an even $$x$$-derivative:
-
-$$\begin{aligned}
-    \Psi(x, z)
-    &= \Psi + (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^2}{2} \Psi_{xx}
-    - \frac{(z \!+\! h)^3}{6} \pdv{}{z} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} + ...
-    \\
-    &= \bigg(\! \Psi - \frac{(z \!+\! h)^2}{2} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} - ... \bigg)
-    \!+\! \bigg(\! (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^3}{6} \pdvn{2}{}{x} \Psi_{z} + ... \bigg)
-\end{aligned}$$
-
-By definition $$\Psi_z = u^{(z)}$$,
-but the bottom is solid, so at $$z = -h$$ we need $$\Psi_z = 0$$,
-leaving:
-
-$$\begin{aligned}
-    \boxed{
-        \Psi(x, z)
-        = \Psi(x, -h) - \frac{(z \!+\! h)^2}{2} \Psi_{xx}(x, - h) + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx}(x, - h) - ...
-    }
-\end{aligned}$$
-
-This result is exact for an inviscid incompressible fluid,
-but once the Taylor series is truncated at a finite number of terms,
-this is known as the **Boussinesq approximation**.
-In effect, this removes all $$z$$-derivatives from the problem,
-and will enable us to describe the surface dynamics
-based on $$\Psi$$'s behaviour near the channel's bottom.
-
-This expression for $$\Psi$$ gives the flow components,
-where we define $$f(x, t) \equiv \Psi_x(x, -h, t)$$
-as the value of $$u^{(x)}$$ at the bottom:
-
-$$\begin{aligned}
-    u^{(x)}(x, z)
-    = \pdv{\Psi}{x}
-    &= f(x) - \frac{(z \!+\! h)^2}{2} f_{xx}(x) + \frac{(z \!+\! h)^4}{24} f_{xxxx}(x) - ...
-    \\
-    u^{(z)}(x, z)
-    = \pdv{\Psi}{z}
-    &= - (z \!+\! h) f_x(x) + \frac{(z \!+\! h)^3}{6} f_{xxx}(x) - ...
-\end{aligned}$$
-
-The Boussinesq approximation is the basis of many other shallow-water wave theories,
-most notably the *Korteweg-de Vries equation*.
-
-
-
-## Two coupled equations
-
-Inserting this result (without truncating)
-into the kinematic boundary condition with $$z = \eta$$:
-
-$$\begin{aligned}
-    0
-    &= \pdv{\eta}{t} + \pdv{\eta}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
-    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
-\end{aligned}$$
-
-And into the free surface boundary condition after differentiating it with respect to $$x$$:
-
-$$\begin{aligned}
-    0
-    &= \pdv{u^{(x)}}{t} + \frac{1}{2} \pdv{}{x} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \pdv{\eta}{x}
-    \\
-    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
-    + \frac{1}{2} \pdv{}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)^2 \\
-    &\qquad + \frac{1}{2} \pdv{}{x} \bigg( \!-\! (\eta \!+\! h) f_x + \frac{(\eta \!+\! h)^3}{6} f_{xxx} - ... \bigg)^2 + g \pdv{\eta}{x}
-    \\
-    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
-    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \pdv{\eta}{x}
-\end{aligned}$$
-
-Switching to a shorter notation for derivatives,
-we now have the following set of equations:
-
-$$\begin{aligned}
-    0
-    &= \eta_t + \eta_x \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
-    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
-    \\
-    0
-    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
-    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \eta_x
-\end{aligned}$$
-
-Now we must decide which terms to keep, i.e. where to truncate the expansion.
-Let us therefore introduce the characteristic length scales $$\eta \sim a$$ and $$x \sim \lambda$$,
-and assume that the water is shallow compared to the waves' length ($$\lambda \gg h$$ by a lot),
-and that the waves' amplitude is small compared to the channel's depth ($$h \gg a$$).
-Specifically, we assume:
-
-$$\begin{aligned}
-    \frac{a}{h}
-    \gg \frac{h}{\lambda}
-    \gg \frac{a}{\lambda}
-\end{aligned}$$
-
-We also introduce characteristic horizontal velocity scales $$u_0$$ and $$f_0$$
-respectively at the surface and at the bottom.
-Finally, let there be a characteristic time scale $$\lambda / u_0$$ for the surface dynamics.
-Inserting all these scales into the two boundary conditions yields:
-
-$$\begin{aligned}
-    0
-    &\sim \frac{a u_0}{\lambda}
-    + \frac{a}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
-    + \bigg( \frac{(a \!+\! h)}{\lambda} f_0 - \frac{(a \!+\! h)^3}{6 \lambda^3} f_0 + ... \bigg)
-    \\
-    0
-    &\sim \frac{u_0}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
-    + \frac{1}{2 \lambda} \bigg( f_0^2 + \frac{(a \!+\! h)^2}{\lambda^2} f_0^2 + ... \bigg)
-    + \frac{g a}{\lambda}
-\end{aligned}$$
-
-Intuitively, we expect that $$u_0 \gg f_0$$, and this is indeed true:
-from a linearization of this problem (given in the next section),
-it turns out that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.
-Multiplying the former equation by $$1 / u_0$$
-and the latter by $$\lambda / u_0^2$$ this leads to:
-
-$$\begin{aligned}
-    0
-    &\sim \frac{a}{\lambda}
-    + \frac{a}{\lambda} \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
-    + \bigg( \frac{(a \!+\! h)}{\lambda} \frac{a}{h} - \frac{(a \!+\! h)^3}{6 \lambda^3} \frac{a}{h} + ... \bigg)
-    \\
-    0
-    &\sim \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
-    + \frac{1}{2} \bigg( \frac{a^2}{h^2} + \frac{(a \!+\! h)^2}{\lambda^2} \frac{a^2}{h^2} + ... \bigg)
-    + \frac{a}{h}
-\end{aligned}$$
-
-The smallest term we will include is $$a h^2 / \lambda^3$$;
-anything smaller (specifically containing $$a^2 / \lambda^2$$) will be discarded.
-Of course, this decision is arbitrary:
-higher-order approximations exist for deeper water and/or taller waves,
-but we stick with Boussinesq's original choice, leaving:
-
-$$\begin{aligned}
-    0
-    &= \eta_t + \eta_x f + (\eta \!+\! h) f_{x} - \frac{h^3}{6} f_{xxx}
-    \\
-    0
-    &= \pdv{}{t} \bigg( f - \frac{h^2}{2} f_{xx} \bigg)
-    + \frac{1}{2} \pdv{}{x} \big( f^2 \big) + g \eta_x
-\end{aligned}$$
-
-Rearranging this gives the **Boussinesq equations**
-for nonlinear waves on shallow water:
-
-$$\begin{aligned}
-    \boxed{
-        \begin{aligned}
-            \pdv{\eta}{t} + \pdv{}{x} \Big( (\eta \!+\! h) f \Big)
-            &= \frac{h^3}{6} \pdvn{3}{f}{x}
-            \\
-            \pdv{f}{t} + f \pdv{f}{x} + g \pdv{\eta}{x}
-            &= \frac{h^2}{2} \pdv{}{t} \pdvn{2}{f}{x}
-        \end{aligned}
-    }
-\end{aligned}$$
-
-If we instead took $$\lambda$$ to be even larger compared to $$h$$,
-the right-hand sides of these equations would have vanished,
-yielding a form of the so-called *shallow water equations*.
-
-
-
-## Single equation
-
-We would like to combine these two equations into a single one for $$\eta$$,
-but their nonlinear nature makes it hard to eliminate $$f$$ directly.
-To get around this, Boussinesq opted to make a lower-order version of these equations,
-to use as a guide for some additional approximations
-to help handle the higher-order version.
-
-In the above discussion of the terms' relative sizes,
-let us instead choose $$a / \lambda$$ as the highest order to include,
-thereby reducing the equations to:
-
-$$\begin{aligned}
-    \eta_t + h f_x
-    = 0
-    \qquad \qquad
-    f_t + g \eta_x
-    = 0
-\end{aligned}$$
-
-Respectively differentiating them with respect to $$t$$ and $$x$$,
-and then substituting $$f_{xt}$$ in the former using the latter,
-we get Lagrange's linear wave equation:
-
-$$\begin{aligned}
-    \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x}
-    = 0
-\end{aligned}$$
-
-It is well-known that such a problem has a general solution
-consisting of an arbitrary forward-moving part $$\eta_{+}$$
-and a backward-moving part $$\eta_{-}$$,
-both going at a constant velocity $$\sqrt{g h}$$,
-and neither of which change shape over time:
-
-$$\begin{aligned}
-    \eta(x, t)
-    = \eta_{+}\big(x - \sqrt{g h} t\big) + \eta_{-}\big(x + \sqrt{g h} t\big)
-\end{aligned}$$
-
-Let us consider only forward-moving waves $$\eta(x \!-\! \sqrt{g h} t)$$,
-such that we can rewrite $$t$$-derivatives as $$x$$-derivatives with a factor $$-\sqrt{g h}$$.
-Our linearized free-surface equation thus becomes:
-
-$$\begin{aligned}
-    0
-    = \pdv{\eta}{t} + h \pdv{f}{x}
-    = - \sqrt{g h} \pdv{\eta}{x} + h \pdv{f}{x}
-    \qquad \implies \qquad
-    f
-    = \sqrt{\frac{g}{h}} \eta + C
-\end{aligned}$$
-
-The integration constant $$C$$ can be removed by absorbing it into $$\eta$$.
-Effectively, we have seen that, at least as a first-order approximation,
-$$\eta$$ is proportional to $$f$$.
-Note that this analysis justifies our earlier assumption
-that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.
-
-Armed with this knowledge, we return
-to the higher-order equations after some rearranging:
-
-$$\begin{aligned}
-    \eta_{t} + h f_x + \pdv{}{x} \bigg( \eta f - \frac{h^3}{6} f_{xx} \bigg)
-    &= 0
-    \\
-    f_{t} + g \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( f^2 - h^2 f_{xt} \bigg)
-    &= 0
-\end{aligned}$$
-
-Inserting $$f = \sqrt{g / h} \eta$$ into both equations
-and multiplying the latter by $$h$$ yields:
-
-$$\begin{aligned}
-    \eta_{t} + \sqrt{g h} \eta_x + \sqrt{\frac{g}{h}} \pdv{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
-    &= 0
-    \\
-    \sqrt{g h} \eta_{t} + g h \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( g \eta^2 - h^2 \sqrt{g h} \eta_{xt} \bigg)
-    &= 0
-\end{aligned}$$
-
-Respectively differentiating by $$t$$ and $$x$$
-and assuming a travelling wave $$\eta(x \!-\! \sqrt{g h} t)$$
-such that we can rewrite $$\ipdv{}{t} = - \sqrt{g h} \ipdv{}{x}$$:
-
-$$\begin{aligned}
-    \eta_{tt} + \sqrt{g h} \eta_{xt} - g \pdvn{2}{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
-    &= 0
-    \\
-    \sqrt{g h} \eta_{tx} + g h \eta_{xx} + \frac{g}{2} \pdvn{2}{}{x} \bigg( \eta^2 + h^3 \eta_{xx} \bigg)
-    &= 0
-\end{aligned}$$
-
-Subtracting the latter from the former yields the following equation containing only $$\eta$$:
-
-$$\begin{aligned}
-    \eta_{tt} + \sqrt{g h} \eta_{xt} - \sqrt{g h} \eta_{tx} - g h \eta_{xx}
-    - \pdvn{2}{}{x} \bigg( g \Big( \eta^2 - \frac{h^3}{6} \eta_{xx} \Big) + \frac{g}{2} \Big( \eta^2 + h^3 \eta_{xx} \Big) \bigg)
-    &= 0
-\end{aligned}$$
-
-After cleaning up, this becomes the **Boussinesq equation** for the shape of travelling waves:
-
-$$\begin{aligned}
-    \boxed{
-        \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x} - g h \pdvn{2}{}{x} \bigg( \frac{3}{2} \frac{\eta^2}{h} + \frac{h^2}{3} \pdvn{2}{\eta}{x} \bigg)
-        = 0
-    }
-\end{aligned}$$
-
-Clearly, the assumption of non-deforming waves $$\eta(x \!-\! \sqrt{g h} t)$$
-was essential to get this equation.
-But what if solving it yields a wave without that property?
-Can it be trusted?
-Fortunately yes: the first two terms ($$\eta_{tt}$$ and $$g h \eta_{xx}$$),
-were not affected by that assumption (this is easy to see),
-and the others are small according to the characteristic scales:
-
-$$\begin{aligned}
-    0
-    \sim \frac{a u_0^2}{\lambda^2} - g h \frac{a}{\lambda^2}
-    - g h \frac{1}{\lambda^2} \bigg( \frac{3}{2} \frac{a^2}{h} + \frac{h^2}{3} \frac{a}{\lambda^2} \bigg)
-\end{aligned}$$
-
-After dividing out $$a / \lambda$$,
-we see that the last two terms are roughly $$a / \lambda$$ and $$h^3 / \lambda^3$$,
-meaning they are much smaller than the first two,
-which are both on the order of $$h / \lambda$$.
-
-
-
-## References
-1.  J. Boussinesq,
-    [Théorie des ondes et des remous qui se propagent le long d'un canal rectangulaire horizontal, en communiquant au liquide contenu dans ce canal des vitesses sensiblement pareilles de la surface au fond](http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16416&Deb=63&Fin=116&E=PDF),
-    1872, Bibliothèque nationale de France.
-2.  E.M. de Jager,
-    [On the origin of the Korteweg-de Vries equation](https://arxiv.org/abs/math/0602661),
-    University of Amsterdam.
-3.  D. Dutykh, F. Dias,
-    [Dissipative Boussinesq equations](https://doi.org/10.1016/j.crme.2007.08.003),
-    2007, Elsevier.
diff --git a/source/know/concept/boussinesq-wave-theory/index.md b/source/know/concept/boussinesq-wave-theory/index.md
new file mode 100644
index 0000000..31228ba
--- /dev/null
+++ b/source/know/concept/boussinesq-wave-theory/index.md
@@ -0,0 +1,453 @@
+---
+title: "Boussinesq wave theory"
+sort_title: "Boussinesq wave theory"
+date: 2023-01-07
+categories:
+- Physics
+- Mathematics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+In fluid mechanics, **Boussinesq wave theory**
+consists of several equations to describe waves on a liquid's surface.
+It was the first attempt to explain the nonlinear phenomenon of solitons,
+which were not predicted by the linear theories existing at the time.
+
+
+
+## Boundary conditions
+
+Consider the [Euler equations](/know/concept/euler-equations/)
+for an incompressible fluid with negligible viscosity:
+
+$$\begin{aligned}
+    \va{g} - \frac{\nabla p}{\rho}
+    &= \pdv{\va{u}}{t} + (\va{u} \cdot \nabla) \va{u}
+    \qquad \qquad
+    \nabla \cdot \va{u}
+    = 0
+\end{aligned}$$
+
+We rewrite the former using the velocity potential $$\Psi$$ for $$\va{u}$$
+and the gravitational potential $$\Phi$$ for $$\va{g}$$,
+such that $$\va{u} = \nabla \Psi$$ and $$\va{g} = - \nabla \Phi$$.
+We also use a vector identity:
+
+$$\begin{aligned}
+    - \nabla \Phi - \frac{\nabla p}{\rho}
+    &= \nabla \pdv{\Psi}{t} + \frac{1}{2} \nabla |\va{u}|^2 + (\nabla \cross \nabla \Psi) \cross \va{u}
+\end{aligned}$$
+
+Recall that the curl of a gradient is always zero, so the last term disappears.
+Integrating in space yields integration constants $$C(t)$$ and $$p_0$$,
+the latter representing atmospheric pressure:
+
+$$\begin{aligned}
+    \pdv{\Psi}{t} + \frac{1}{2} |\va{u}|^2
+    &= -\Phi + \frac{p_0 - p}{\rho} + C
+\end{aligned}$$
+
+Consider a rectangular channel of depth $$h$$
+extending infinitely far along the $$x$$-axis,
+with a finite width along the $$y$$-axis.
+We choose our coordinate system so that
+$$z = 0$$ is the equilibrium water level,
+and the bottom is at $$z = -h$$.
+Let $$\eta(x, t)$$ be the deformation of the surface,
+for which we want to find a wave equation.
+All quantities are assumed to be constant in $$y$$,
+so we only consider a 2D flow $$\va{u} = \big(u^{(x)}, u^{(z)}\big)$$.
+At the surface $$z = \eta$$ we thus have:
+
+$$\begin{aligned}
+    \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
+    &= - g \eta + \frac{p_0 - p}{\rho} + C
+\end{aligned}$$
+
+Where $$g \eta = \Phi$$ with $$g \approx 9.81 \:\mathrm{m}/\mathrm{s}^2$$ on Earth.
+Later, we will differentiate this formula in $$x$$,
+so we can already set $$C = 0$$ now, since it will vanish then anyway.
+Furthermore, we assume that at the surface $$z = \eta$$
+the pressures are in equilibrium $$p_0 = p$$, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \eta
+        = 0
+    }
+\end{aligned}$$
+
+This is called the **free surface boundary condition**.
+Obviously, if $$z = \eta$$, then $$\eta \!-\! z = 0$$.
+Taking the [material derivative](/know/concept/material-derivative/)
+of this fact gives the following relation:
+
+$$\begin{aligned}
+    0
+    = \frac{\mathrm{D}}{\mathrm{D} t} (\eta - z)
+    = \pdv{\eta}{t} + \pdv{z}{t} + \va{u} \cdot \nabla \eta - \va{u} \cdot \nabla z
+\end{aligned}$$
+
+Since $$\eta$$ only depends on $$x$$ and $$t$$,
+this becomes the **kinematic boundary condition**:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\eta}{t} + u^{(x)} \pdv{\eta}{x} - u^{(z)}
+        = 0
+    }
+\end{aligned}$$
+
+The equations will be derived from these two fundamental boundary conditions.
+
+
+
+## Boussinesq approximation
+
+Let us take a Taylor expansion of the velocity potential $$\Psi(x, z, t)$$
+at the bottom $$z = -h$$:
+
+$$\begin{aligned}
+    \Psi(x, z)
+    = \Psi(x, -h) + (z + h) \: \Psi_z(x, -h) + \frac{(z + h)^2}{2} \: \Psi_{zz}(x, -h) + ...
+\end{aligned}$$
+
+Because the fluid is incompressible, this can be rewritten.
+Laplace's equation tells us:
+
+$$\begin{aligned}
+    \nabla \cdot \va{u}
+    = \nabla^2 \Psi
+    = 0
+    \qquad \implies \qquad
+    \:\:\:\Psi_{zz}
+    = - \Psi_{xx}
+\end{aligned}$$
+
+Which we use for the expansion's second-order term.
+Similarly, for the fourth-order term:
+
+$$\begin{aligned}
+    0
+    = \nabla^2 \nabla^2 \Psi
+    &= \Psi_{zzzz} + \pdvn{2}{}{z} \Psi_{xx} + \pdvn{2}{}{x} \Psi_{zz} + \Psi_{xxxx}
+    \\
+    &= \Psi_{zzzz} + \pdvn{2}{}{x} \Psi_{zz} - \pdvn{2}{}{x} \Psi_{xx} + \Psi_{xxxx}
+    \\
+    &= \Psi_{zzzz} - \Psi_{xxxx}
+\end{aligned}$$
+
+And so on for higher orders.
+In the Taylor expansion, all even derivatives can be rewritten in this way,
+and all odd derivatives can be split into a single $$\ipdv{}{z}$$ and an even $$x$$-derivative:
+
+$$\begin{aligned}
+    \Psi(x, z)
+    &= \Psi + (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^2}{2} \Psi_{xx}
+    - \frac{(z \!+\! h)^3}{6} \pdv{}{z} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} + ...
+    \\
+    &= \bigg(\! \Psi - \frac{(z \!+\! h)^2}{2} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} - ... \bigg)
+    \!+\! \bigg(\! (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^3}{6} \pdvn{2}{}{x} \Psi_{z} + ... \bigg)
+\end{aligned}$$
+
+By definition $$\Psi_z = u^{(z)}$$,
+but the bottom is solid, so at $$z = -h$$ we need $$\Psi_z = 0$$,
+leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \Psi(x, z)
+        = \Psi(x, -h) - \frac{(z \!+\! h)^2}{2} \Psi_{xx}(x, - h) + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx}(x, - h) - ...
+    }
+\end{aligned}$$
+
+This result is exact for an inviscid incompressible fluid,
+but once the Taylor series is truncated at a finite number of terms,
+this is known as the **Boussinesq approximation**.
+In effect, this removes all $$z$$-derivatives from the problem,
+and will enable us to describe the surface dynamics
+based on $$\Psi$$'s behaviour near the channel's bottom.
+
+This expression for $$\Psi$$ gives the flow components,
+where we define $$f(x, t) \equiv \Psi_x(x, -h, t)$$
+as the value of $$u^{(x)}$$ at the bottom:
+
+$$\begin{aligned}
+    u^{(x)}(x, z)
+    = \pdv{\Psi}{x}
+    &= f(x) - \frac{(z \!+\! h)^2}{2} f_{xx}(x) + \frac{(z \!+\! h)^4}{24} f_{xxxx}(x) - ...
+    \\
+    u^{(z)}(x, z)
+    = \pdv{\Psi}{z}
+    &= - (z \!+\! h) f_x(x) + \frac{(z \!+\! h)^3}{6} f_{xxx}(x) - ...
+\end{aligned}$$
+
+The Boussinesq approximation is the basis of many other shallow-water wave theories,
+most notably the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/).
+
+
+
+## Two coupled equations
+
+Inserting this result (without truncating)
+into the kinematic boundary condition with $$z = \eta$$:
+
+$$\begin{aligned}
+    0
+    &= \pdv{\eta}{t} + \pdv{\eta}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
+\end{aligned}$$
+
+And into the free surface boundary condition after differentiating it with respect to $$x$$:
+
+$$\begin{aligned}
+    0
+    &= \pdv{u^{(x)}}{t} + \frac{1}{2} \pdv{}{x} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \pdv{\eta}{x}
+    \\
+    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \frac{1}{2} \pdv{}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)^2 \\
+    &\qquad + \frac{1}{2} \pdv{}{x} \bigg( \!-\! (\eta \!+\! h) f_x + \frac{(\eta \!+\! h)^3}{6} f_{xxx} - ... \bigg)^2 + g \pdv{\eta}{x}
+    \\
+    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \pdv{\eta}{x}
+\end{aligned}$$
+
+Switching to a shorter notation for derivatives,
+we now have the following set of equations:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + \eta_x \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
+    \\
+    0
+    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \eta_x
+\end{aligned}$$
+
+Now we must decide which terms to keep, i.e. where to truncate the expansion.
+Let us therefore introduce the characteristic length scales $$\eta \sim a$$ and $$x \sim \lambda$$,
+and assume that the water is shallow compared to the waves' length ($$\lambda \gg h$$ by a lot),
+and that the waves' amplitude is small compared to the channel's depth ($$h \gg a$$).
+Specifically, we assume:
+
+$$\begin{aligned}
+    \frac{a}{h}
+    \gg \frac{h}{\lambda}
+    \gg \frac{a}{\lambda}
+\end{aligned}$$
+
+We also introduce characteristic horizontal velocity scales $$u_0$$ and $$f_0$$
+respectively at the surface and at the bottom.
+Finally, let there be a characteristic time scale $$\lambda / u_0$$ for the surface dynamics.
+Inserting all these scales into the two boundary conditions yields:
+
+$$\begin{aligned}
+    0
+    &\sim \frac{a u_0}{\lambda}
+    + \frac{a}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
+    + \bigg( \frac{(a \!+\! h)}{\lambda} f_0 - \frac{(a \!+\! h)^3}{6 \lambda^3} f_0 + ... \bigg)
+    \\
+    0
+    &\sim \frac{u_0}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
+    + \frac{1}{2 \lambda} \bigg( f_0^2 + \frac{(a \!+\! h)^2}{\lambda^2} f_0^2 + ... \bigg)
+    + \frac{g a}{\lambda}
+\end{aligned}$$
+
+Intuitively, we expect that $$u_0 \gg f_0$$, and this is indeed true:
+from a linearization of this problem (given in the next section),
+it turns out that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.
+Multiplying the former equation by $$1 / u_0$$
+and the latter by $$\lambda / u_0^2$$ this leads to:
+
+$$\begin{aligned}
+    0
+    &\sim \frac{a}{\lambda}
+    + \frac{a}{\lambda} \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
+    + \bigg( \frac{(a \!+\! h)}{\lambda} \frac{a}{h} - \frac{(a \!+\! h)^3}{6 \lambda^3} \frac{a}{h} + ... \bigg)
+    \\
+    0
+    &\sim \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
+    + \frac{1}{2} \bigg( \frac{a^2}{h^2} + \frac{(a \!+\! h)^2}{\lambda^2} \frac{a^2}{h^2} + ... \bigg)
+    + \frac{a}{h}
+\end{aligned}$$
+
+The smallest term we will include is $$a h^2 / \lambda^3$$;
+anything smaller (specifically containing $$a^2 / \lambda^2$$) will be discarded.
+Of course, this decision is arbitrary:
+higher-order approximations exist for deeper water and/or taller waves,
+but we stick with Boussinesq's original choice, leaving:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + \eta_x f + (\eta \!+\! h) f_{x} - \frac{h^3}{6} f_{xxx}
+    \\
+    0
+    &= \pdv{}{t} \bigg( f - \frac{h^2}{2} f_{xx} \bigg)
+    + \frac{1}{2} \pdv{}{x} \big( f^2 \big) + g \eta_x
+\end{aligned}$$
+
+Rearranging this gives the **Boussinesq equations**
+for nonlinear waves on shallow water:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \pdv{\eta}{t} + \pdv{}{x} \Big( (\eta \!+\! h) f \Big)
+            &= \frac{h^3}{6} \pdvn{3}{f}{x}
+            \\
+            \pdv{f}{t} + f \pdv{f}{x} + g \pdv{\eta}{x}
+            &= \frac{h^2}{2} \pdv{}{t} \pdvn{2}{f}{x}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+If we instead took $$\lambda$$ to be even larger compared to $$h$$,
+the right-hand sides of these equations would have vanished,
+yielding a form of the so-called *shallow water equations*.
+
+
+
+## Single equation
+
+We would like to combine these two equations into a single one for $$\eta$$,
+but their nonlinear nature makes it hard to eliminate $$f$$ directly.
+To get around this, Boussinesq opted to make a lower-order version of these equations,
+to use as a guide for some additional approximations
+to help handle the higher-order version.
+
+In the above discussion of the terms' relative sizes,
+let us instead choose $$a / \lambda$$ as the highest order to include,
+thereby reducing the equations to:
+
+$$\begin{aligned}
+    \eta_t + h f_x
+    = 0
+    \qquad \qquad
+    f_t + g \eta_x
+    = 0
+\end{aligned}$$
+
+Respectively differentiating them with respect to $$t$$ and $$x$$,
+and then substituting $$f_{xt}$$ in the former using the latter,
+we get Lagrange's linear wave equation:
+
+$$\begin{aligned}
+    \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x}
+    = 0
+\end{aligned}$$
+
+It is well-known that such a problem has a general solution
+consisting of an arbitrary forward-moving part $$\eta_{+}$$
+and a backward-moving part $$\eta_{-}$$,
+both going at a constant velocity $$\sqrt{g h}$$,
+and neither of which change shape over time:
+
+$$\begin{aligned}
+    \eta(x, t)
+    = \eta_{+}\big(x - \sqrt{g h} t\big) + \eta_{-}\big(x + \sqrt{g h} t\big)
+\end{aligned}$$
+
+Let us consider only forward-moving waves $$\eta(x \!-\! \sqrt{g h} t)$$,
+such that we can rewrite $$t$$-derivatives as $$x$$-derivatives with a factor $$-\sqrt{g h}$$.
+Our linearized free-surface equation thus becomes:
+
+$$\begin{aligned}
+    0
+    = \pdv{\eta}{t} + h \pdv{f}{x}
+    = - \sqrt{g h} \pdv{\eta}{x} + h \pdv{f}{x}
+    \qquad \implies \qquad
+    f
+    = \sqrt{\frac{g}{h}} \eta + C
+\end{aligned}$$
+
+The integration constant $$C$$ can be removed by absorbing it into $$\eta$$.
+Effectively, we have seen that, at least as a first-order approximation,
+$$\eta$$ is proportional to $$f$$.
+Note that this analysis justifies our earlier assumption
+that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.
+
+Armed with this knowledge, we return
+to the higher-order equations after some rearranging:
+
+$$\begin{aligned}
+    \eta_{t} + h f_x + \pdv{}{x} \bigg( \eta f - \frac{h^3}{6} f_{xx} \bigg)
+    &= 0
+    \\
+    f_{t} + g \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( f^2 - h^2 f_{xt} \bigg)
+    &= 0
+\end{aligned}$$
+
+Inserting $$f = \sqrt{g / h} \eta$$ into both equations
+and multiplying the latter by $$h$$ yields:
+
+$$\begin{aligned}
+    \eta_{t} + \sqrt{g h} \eta_x + \sqrt{\frac{g}{h}} \pdv{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
+    &= 0
+    \\
+    \sqrt{g h} \eta_{t} + g h \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( g \eta^2 - h^2 \sqrt{g h} \eta_{xt} \bigg)
+    &= 0
+\end{aligned}$$
+
+Respectively differentiating by $$t$$ and $$x$$
+and assuming a travelling wave $$\eta(x \!-\! \sqrt{g h} t)$$
+such that we can rewrite $$\ipdv{}{t} = - \sqrt{g h} \ipdv{}{x}$$:
+
+$$\begin{aligned}
+    \eta_{tt} + \sqrt{g h} \eta_{xt} - g \pdvn{2}{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
+    &= 0
+    \\
+    \sqrt{g h} \eta_{tx} + g h \eta_{xx} + \frac{g}{2} \pdvn{2}{}{x} \bigg( \eta^2 + h^3 \eta_{xx} \bigg)
+    &= 0
+\end{aligned}$$
+
+Subtracting the latter from the former yields the following equation containing only $$\eta$$:
+
+$$\begin{aligned}
+    \eta_{tt} + \sqrt{g h} \eta_{xt} - \sqrt{g h} \eta_{tx} - g h \eta_{xx}
+    - \pdvn{2}{}{x} \bigg( g \Big( \eta^2 - \frac{h^3}{6} \eta_{xx} \Big) + \frac{g}{2} \Big( \eta^2 + h^3 \eta_{xx} \Big) \bigg)
+    &= 0
+\end{aligned}$$
+
+After cleaning up, this becomes the **Boussinesq equation** for the shape of travelling waves:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x} - g h \pdvn{2}{}{x} \bigg( \frac{3}{2} \frac{\eta^2}{h} + \frac{h^2}{3} \pdvn{2}{\eta}{x} \bigg)
+        = 0
+    }
+\end{aligned}$$
+
+Clearly, the assumption of non-deforming waves $$\eta(x \!-\! \sqrt{g h} t)$$
+was essential to get this equation.
+But what if solving it yields a wave without that property?
+Can it be trusted?
+Fortunately yes: the first two terms ($$\eta_{tt}$$ and $$g h \eta_{xx}$$),
+were not affected by that assumption (this is easy to see),
+and the others are small according to the characteristic scales:
+
+$$\begin{aligned}
+    0
+    \sim \frac{a u_0^2}{\lambda^2} - g h \frac{a}{\lambda^2}
+    - g h \frac{1}{\lambda^2} \bigg( \frac{3}{2} \frac{a^2}{h} + \frac{h^2}{3} \frac{a}{\lambda^2} \bigg)
+\end{aligned}$$
+
+After dividing out $$a / \lambda$$,
+we see that the last two terms are roughly $$a / \lambda$$ and $$h^3 / \lambda^3$$,
+meaning they are much smaller than the first two,
+which are both on the order of $$h / \lambda$$.
+
+
+
+## References
+1.  J. Boussinesq,
+    [Théorie des ondes et des remous qui se propagent le long d'un canal rectangulaire horizontal, en communiquant au liquide contenu dans ce canal des vitesses sensiblement pareilles de la surface au fond](http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16416&Deb=63&Fin=116&E=PDF),
+    1872, Bibliothèque nationale de France.
+2.  E.M. de Jager,
+    [On the origin of the Korteweg-de Vries equation](https://arxiv.org/abs/math/0602661),
+    University of Amsterdam.
+3.  D. Dutykh, F. Dias,
+    [Dissipative Boussinesq equations](https://doi.org/10.1016/j.crme.2007.08.003),
+    2007, Elsevier.
diff --git a/source/know/concept/korteweg-de-vries-equation/index.md b/source/know/concept/korteweg-de-vries-equation/index.md
new file mode 100644
index 0000000..4a050fe
--- /dev/null
+++ b/source/know/concept/korteweg-de-vries-equation/index.md
@@ -0,0 +1,631 @@
+---
+title: "Korteweg-de Vries equation"
+sort_title: "Korteweg-de Vries equation"
+date: 2023-01-14
+categories:
+- Physics
+- Mathematics
+layout: "concept"
+---
+
+The **Korteweg-de Vries (KdV) equation** is
+a nonlinear 1+1D partial differential equation
+derived to describe water waves.
+It is usually given in its dimensionless form, namely:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{u}{t} - 6 u \pdv{u}{x} + \pdvn{3}{u}{x}
+        = 0
+    }
+\end{aligned}$$
+
+Where $$u(x, t)$$ is the wave's profile,
+with $$x$$ being the transverse coordinate.
+The KdV equation notably has **soliton** solutions,
+which can travel long distances without changing shape.
+
+
+
+## Derivation
+
+The derivation of the KdV equation starts in the same way as for
+the [Boussinesq wave equations](/know/concept/boussinesq-wave-theory/);
+the common parts will be discussed only briefly here.
+Recall that Boussinesq set up two boundary conditions
+at the liquid's surface $$z = \eta(x, t)$$.
+Firstly, the *kinematic boundary condition*:
+
+$$\begin{aligned}
+    \eta_t + u^{(x)} \eta_x - u^{(z)}
+    = 0
+\end{aligned}$$
+
+And secondly, the *free surface boundary condition*
+from integrating the main [Euler equation](/know/concept/euler-equations/):
+
+$$\begin{aligned}
+    \Psi_t + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
+    &= -g \eta + \frac{p_0 - p}{\rho}
+\end{aligned}$$
+
+Where $$\Psi$$ is the velocity potential $$\va{u} = \nabla \Psi$$,
+with $$\va{u} = \big( u^{(x)}, u^{(z)} \big)$$ being 2D
+due to the assumed symmetry along the $$y$$-axis.
+Unlike Boussinesq, who assumed that $$p_0 = p$$ at the surface,
+de Vries decided to include surface tension using
+the [Young-Laplace law](/know/concept/young-laplace-law/):
+
+$$\begin{aligned}
+    p_0 - p
+    = T \kappa
+    = \frac{T \eta_{xx}}{\big( 1 + \eta_x^2\big)^{3/2}}
+    \approx T \eta_{xx}
+\end{aligned}$$
+
+Where $$T$$ is the energy cost per unit area,
+and $$\eta$$ is assumed to be slowly-varying such that $$\eta_{x}^2$$
+can be neglected in the [curvature](/know/concept/curvature/) formula.
+His free surface condition was thus:
+
+$$\begin{aligned}
+    \Psi_t + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
+    &= -g \eta + \frac{T}{\rho} \eta_{xx}
+\end{aligned}$$
+
+Then, like Boussinesq, de Vries differentiated this with respect to $$x$$, yielding:
+
+$$\begin{aligned}
+    \pdv{u^{(x)}}{t} + \frac{1}{2} \pdv{}{x} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
+    &= -g \eta_x + \frac{T}{\rho} \eta_{xxx}
+\end{aligned}$$
+
+And he made the *Boussinesq approximation*
+to eliminate all $$z$$-derivatives from the problem:
+
+$$\begin{aligned}
+    u^{(x)}(x, z)
+    = \pdv{\Psi}{x}
+    &= f(x) - \frac{(z \!+\! h)^2}{2} f_{xx}(x) + \frac{(z \!+\! h)^4}{24} f_{xxxx}(x) - ...
+    \\
+    u^{(z)}(x, z)
+    = \pdv{\Psi}{z}
+    &= - (z \!+\! h) f_x(x) + \frac{(z \!+\! h)^3}{6} f_{xxx}(x) - ...
+\end{aligned}$$
+
+Where $$f(x, t) \equiv \Psi_{x}(x, -h, t)$$ is the $$x$$-velocity
+at the channel's bottom $$z = -h$$.
+Inserting this expansion into the two boundary conditions
+yields these coupled equations:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + \eta_x \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
+    \\
+    0
+    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
+    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 \!-\! f f_{xx}) + ... \bigg) + g \eta_x - \frac{T}{\rho} \eta_{xxx}
+\end{aligned}$$
+
+These are simply the *Boussinesq equations* before truncation
+and with surface tension.
+Of course we want to reduce the number of terms,
+so we discard everything above $$(h \!+\! \eta)^3$$:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + \eta_x f + (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^2}{2} \eta_x f_{xx} - \frac{(\eta \!+\! h)^3}{6} f_{xxx}
+    \\
+    0
+    &= f_t + f f_x - (\eta \!+\! h) \Big( \eta_t f_{xx} - \eta_x f_x^2 + \eta_x f f_{xx} \Big)
+    \\
+    &\qquad - \frac{(\eta \!+\! h)^2}{2} \Big( f_{xxt} - f_x f_{xx} + f f_{xxx} \Big)
+    + g \eta_x - \frac{T}{\rho} \eta_{xxx}
+\end{aligned}$$
+
+The goal is to reduce the number of terms even further,
+and then to combine these equations into one.
+To do this, the method of successive approximations is used:
+first, a linearized version of the problem is solved,
+which is easily shown to give Lagrange's result:
+
+$$\begin{aligned}
+    \eta_{tt} - g h \eta_{xx}
+    = 0
+    \qquad \implies \qquad
+    \eta
+    = \eta^{+}(x - \sqrt{g h} t) + \eta^{-}(x + \sqrt{g h} t)
+\end{aligned}$$
+
+Where $$\eta^{+}$$ and $$\eta^{-}$$ are arbitrary functions
+that respectively represent forward- and backward-propagating waves.
+Then this result is used to derive a higher-order equation.
+
+At this point, the calculations of Boussinesq and de Vries diverge.
+Boussinesq kept using static Cartesian coordinates
+and assumed a forward-moving wave $$\eta(x \!-\! \sqrt{g h} t)$$,
+whereas de Vries chose a reference frame moving at a speed $$q_0$$.
+
+However, the way de Vries did this is somewhat unusual:
+rather than transform the coordinate system,
+the velocity is incorporated into his ansatz for $$f$$;
+in other words, he assumed that the entire liquid is moving at $$q_0$$.
+For a wave going in the positive $$x$$-direction,
+the linearized problem then predicts a profile $$\eta(x \!-\! (\sqrt{g h} \!+\! q_0))$$,
+so de Vries chose $$q_0 = -\sqrt{g h}$$ to make it stationary.
+Analogously, $$q_0 = \sqrt{g h}$$ for a backward-moving wave.
+With this in mind, the ansatz is:
+
+$$\begin{aligned}
+    f(x, t)
+    = q_0 - \frac{g}{q_0} \Big( \eta(x, t) + \alpha + \gamma(x, t) \Big)
+\end{aligned}$$
+
+Where $$\alpha$$ is a constant parameter
+(which we will use to handle velocity discrepancies
+between the linear and nonlinear theories).
+The correction represented by $$\gamma$$ is much smaller,
+i.e. $$\eta \sim \alpha \gg \gamma$$.
+We insert this ansatz into the above equations, yielding:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + \eta_x \Big( q_0 - \frac{g}{q_0} (\eta + \alpha + \gamma) \Big)
+    - \frac{g}{q_0} (\eta \!+\! h) (\eta_x + \gamma_x)
+    \\
+    &\qquad + \frac{g}{q_0} \frac{(\eta \!+\! h)^2}{2} \eta_x (\eta_{xx} + \gamma_{xx})
+    + \frac{g}{q_0} \frac{(\eta \!+\! h)^3}{6} (\eta_{xxx} + \gamma_{xxx})
+    \\
+    0
+    &= g \eta_x - \frac{T}{\rho} \eta_{xxx}
+    - \frac{g}{q_0} (\eta_t + \gamma_t)
+    - \frac{g}{q_0} \Big( q_0 - \frac{g}{q_0} (\eta + \alpha + \gamma) \Big) (\eta_x + \gamma_x)
+    \\
+    &\qquad + \frac{g}{q_0} (\eta \!+\! h)
+    \bigg( \eta_t (\eta_{xx} + \gamma_{xx}) + \frac{g}{q_0} \eta_x (\eta_x + \gamma_x)^2
+    \\
+    &\qquad\qquad + \Big( q_0 - \frac{g}{q_0} (\eta + \alpha + \gamma) \Big) \eta_x (\eta_{xx} + \gamma_{xx}) \bigg)
+    \\
+    &\qquad + \frac{g}{q_0} \frac{(\eta \!+\! h)^2}{2}
+    \bigg( \eta_{xxt} + \gamma_{xxt} + \frac{g}{q_0} (\eta_x + \gamma_x) (\eta_{xx} + \gamma_{xx})
+    \\
+    &\qquad\qquad + \Big( q_0 - \frac{g}{q_0} (\eta + \alpha + \gamma) \Big) (\eta_{xxx} + \gamma_{xxx}) \bigg)
+\end{aligned}$$
+
+We keep terms on the order of $$\alpha \eta$$,
+but neglect anything smaller ($$\eta \gamma$$ etc.),
+because by assumption we have $$h \gg \eta \gg \alpha \gg \gamma$$.
+Furthermore, each $$x$$-derivative is roughly equivalent to dividing by $$\lambda$$,
+and since the water is shallow ($$\lambda \gg h$$)
+successive differentiations reduce terms' magnitudes,
+so terms like $$\alpha \eta$$ and $$\eta^2$$ are kept
+only if they contain at most one $$x$$-derivative:
+e.g. $$\eta \eta_x$$ stays, but $$\eta_x^2$$ does not.
+This reduces the equations to the following:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + q_0 \eta_x - \frac{g}{q_0} (\eta + \alpha) \eta_x - \frac{g h}{q_0} (\eta_x + \gamma_x)
+    - \frac{g}{q_0} \eta \eta_x + \frac{g h^3}{6 q_0} (\eta_{xxx} \!+\! \gamma_{xxx})
+    \\
+    0
+    &= g \eta_x - \frac{T}{\rho} \eta_{xxx} - \frac{g}{q_0} (\eta_t + \gamma_t) - g (\eta_x + \gamma_x)
+    + \frac{g^2}{q_0^2} (\eta + \alpha) \eta_x + \frac{g h^2}{2 q_0} (\eta_{xxt} \!+\! \gamma_{xxt} \!+\! q_0 \eta_{xxx})
+\end{aligned}$$
+
+Our reference frame moves with the wave at velocity $$q_0$$,
+so all $$t$$-derivatives describe deformation rather than transport,
+and are hence quite small.
+Therefore we discard all except for $$\eta_t$$:
+
+$$\begin{aligned}
+    0
+    &= \eta_t + q_0 \eta_x - \frac{g h}{q_0} (\eta_x + \gamma_x) - \frac{g}{q_0} (\eta + \alpha) \eta_x
+    - \frac{g}{q_0} \eta \eta_x + \frac{g h^3}{6 q_0} \eta_{xxx}
+    \\
+    0
+    &= - \frac{g}{q_0} \eta_t - g \gamma_x + \frac{g^2}{q_0^2} (\eta + \alpha) \eta_x + \frac{g h^2}{2} \eta_{xxx} - \frac{T}{\rho} \eta_{xxx}
+\end{aligned}$$
+
+Multiplying the first equation by $$-g / q_0$$, and inserting $$q_0 = \pm\sqrt{g h}$$ into both:
+
+$$\begin{aligned}
+    \frac{g}{q_0} \eta_{t}
+    &= g \gamma_x + \frac{g}{h} (2 \eta + \alpha) \eta_x - \frac{g h^2}{6} \eta_{xxx}
+    \\
+    \frac{g}{q_0} \eta_t
+    &= - g \gamma_x + \frac{g}{h} (\eta + \alpha) \eta_x + \Big( \frac{g h^2}{2} - \frac{T}{\rho} \Big) \eta_{xxx}
+\end{aligned}$$
+
+Note that some authors set $$q_0$$ to $$\sqrt{g h}$$, others to $$-\sqrt{g h}$$;
+we preserve $$q_0$$ on the left-hand side to cover both cases.
+Adding up these two equations:
+
+$$\begin{aligned}
+    2 \frac{g}{q_0} \eta_{t}
+    &= \frac{g}{h} (3 \eta + 2 \alpha) \eta_x + \Big( \frac{g h^2}{3} - \frac{T}{\rho} \Big) \eta_{xxx}
+    \\
+    &= \frac{g}{h} \pdv{}{x} \bigg( \frac{3}{2} \eta^2 + 2 \alpha \eta + \Big( \frac{h^3}{3} - \frac{h T}{g \rho} \Big) \eta_{xx} \bigg)
+\end{aligned}$$
+
+This leads to the original **Korteweg-de Vries equation** for waves on shallow water:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\eta}{t}
+        = \frac{3}{2} \frac{q_0}{h} \pdv{}{x} \bigg( \frac{1}{2} \eta^2 + \frac{2}{3} \alpha \eta + \frac{1}{3} \sigma \pdvn{2}{\eta}{x} \bigg)
+    }
+\end{aligned}$$
+
+Where we have defined the dispersion parameter $$\sigma$$ as follows:
+
+$$\begin{aligned}
+    \sigma
+    \equiv \frac{h^3}{3} - \frac{h T}{g \rho}
+\end{aligned}$$
+
+What about $$\alpha$$?
+Looking at the ansatz for $$f$$, we see that
+the body of water is already assumed to be moving at $$q_0$$,
+minus $$g \alpha / q_0$$, so by varying $$\alpha$$
+we are modifying the water's velocity.
+The term in the KdV equation simply corrects for our chosen value of $$\alpha$$.
+It has no deeper meaning than that: for any value of $$\alpha$$,
+the full range of KdV solutions can still be obtained.
+
+
+
+## Dimensionless form
+
+Let us derive the standard non-dimensionalized form
+of the KdV equation seen in most literature.
+To do so, we make the following coordinate transformation,
+where $$\tilde{\eta}$$, $$\tilde{x}$$ and $$\tilde{t}$$ are dimensionless,
+and $$\eta_c$$, $$x_c$$, $$t_c$$ and $$v_c$$ are free dimensioned scale parameters:
+
+$$\begin{aligned}
+    \tilde{\eta}(\tilde{x}, \tilde{t})
+    = \frac{\eta(x, t)}{\eta_c}
+    \qquad \qquad
+    \tilde{t}
+    = \frac{t}{t_c}
+    \qquad \qquad
+    \tilde{x}
+    = \frac{x - v_c t}{x_c}
+\end{aligned}$$
+
+The original derivatives with respect to $$x$$ and $$t$$ are then rewritten like so:
+
+$$\begin{aligned}
+    \pdv{}{t}
+    &= \pdv{\tilde{t}}{t} \pdv{}{\tilde{t}} + \pdv{\tilde{x}}{t} \pdv{}{\tilde{x}}
+    = \frac{1}{t_c} \pdv{}{\tilde{t}} - \frac{v_c}{x_c} \pdv{}{\tilde{x}}
+    \\
+    \pdv{}{x}
+    &= \pdv{\tilde{t}}{x} \pdv{}{\tilde{t}} + \pdv{\tilde{x}}{x} \pdv{}{\tilde{x}}
+    = \frac{1}{x_c} \pdv{}{\tilde{x}}
+\end{aligned}$$
+
+Writing out the KdV equation and inserting our transformation, we arrive at:
+
+$$\begin{aligned}
+    0
+    &= \eta_t - \frac{3 q_0}{2 h} \eta \eta_x - \frac{q_0 \alpha}{h} \eta_x - \frac{q_0}{2 h} \sigma \eta_{xxx}
+    \\
+    &= \frac{\eta_c}{t_c} \tilde{\eta}_{\tilde{t}}
+    - \frac{v_c \eta_c}{x_c} \tilde{\eta}_{\tilde{x}}
+    - \frac{3 q_0 \eta_c^2}{2 h x_c} \tilde{\eta} \tilde{\eta}_{\tilde{x}}
+    - \frac{q_0 \alpha \eta_c}{h x_c} \tilde{\eta}_{\tilde{x}}
+    - \frac{q_0 \sigma \eta_c}{2 h x_c^3} \tilde{\eta}_{\tilde{x} \tilde{x} \tilde{x}}
+\end{aligned}$$
+
+Multiplying by $$t_c / \eta_c$$ to make all terms unitless
+and bring the first to the desired form:
+
+$$\begin{aligned}
+    0
+    &= \tilde{\eta}_{\tilde{t}}
+    - \frac{t_c}{x_c} \bigg( v_c + \frac{q_0 \alpha}{h} \bigg) \tilde{\eta}_{\tilde{x}}
+    - \frac{3 q_0 \eta_c t_c}{2 h x_c} \tilde{\eta} \tilde{\eta}_{\tilde{x}}
+    - \frac{q_0 \sigma t_c}{2 h x_c^3} \tilde{\eta}_{\tilde{x} \tilde{x} \tilde{x}}
+\end{aligned}$$
+
+Now we must choose the scale parameters' values.
+By convention, the second term is removed,
+the third has a factor $$6$$, and the last has a factor $$-1$$,
+yielding equations:
+
+$$\begin{aligned}
+    v_c + \frac{q_0 \alpha}{h}
+    = 0
+    \qquad \qquad
+    \frac{3 q_0 \eta_c t_c}{2 h x_c}
+    = 6
+    \qquad \qquad
+    \frac{q_0 \sigma t_c}{2 h x_c^3}
+    = -1
+\end{aligned}$$
+
+This is pure convention; other choices are valid too.
+Reducing these equations:
+
+$$\begin{aligned}
+    v_c
+    = - \frac{q_0 \alpha}{h}
+    \qquad \qquad
+    t_c
+    = \frac{4 h x_c}{q_0 \eta_c}
+    \qquad \qquad
+    x_c^2
+    = -\frac{2 \sigma}{\eta_c}
+\end{aligned}$$
+
+To proceed, we need to take the square root of $$x_c^2$$,
+but we must make sure that $$x_c^2 > 0$$, because all quantities are real.
+We enforce this in our choice of $$\eta_c$$, where $$s \equiv \sgn(\sigma)$$:
+
+$$\begin{aligned}
+    \eta_c
+    = - s h
+    \qquad \qquad
+    v_c
+    = - \frac{q_0}{h} \alpha
+    \qquad \qquad
+    x_c
+    = \sqrt{\frac{2 \sigma}{s h}}
+    \qquad \qquad
+    t_c
+    = - \frac{1}{s q_0} \sqrt{\frac{32 \sigma}{s h}}
+\end{aligned}$$
+
+These are the final scale parameter values,
+leading to the desired dimensionless form:
+
+$$\begin{aligned}
+    0
+    &= \tilde{\eta}_{\tilde{t}} - 6 \tilde{\eta} \tilde{\eta}_{\tilde{x}} + \tilde{\eta}_{\tilde{x} \tilde{x} \tilde{x}}
+\end{aligned}$$
+
+Recall that $$\alpha$$ sets the background fluid velocity,
+and $$v_c$$ controls the coordinate system's motion:
+our choice of $$v_c$$ simply cancels out the effect of $$\alpha$$.
+This reveals the point of $$\alpha$$:
+the KdV equation has solutions moving at various speeds,
+so, for a given $$\eta$$, we can always choose $$\alpha$$ (and hence $$v_c$$)
+such that the wave appears stationary.
+
+
+
+## Soliton solution
+
+Let us make the following ansatz for the dimensionless wave profile $$\tilde{\eta}$$,
+assuming there exists a solution that maintains its shape
+while propagating at a constant "velocity" $$v$$:
+
+$$\begin{aligned}
+    \tilde{\eta}(\tilde{x}, \tilde{t})
+    = \phi(\xi)
+    \qquad
+    \xi
+    \equiv \tilde{x} - v \tilde{t}
+    \qquad \implies \qquad
+    \pdv{}{\tilde{t}}
+    = - v \pdv{}{\xi}
+    \qquad
+    \pdv{}{\tilde{x}}
+    = \pdv{}{\xi}
+\end{aligned}$$
+
+Inserting this into the dimensionless KdV equation
+tells us that $$\phi$$ must satisfy:
+
+$$\begin{aligned}
+    0
+    &= - v \phi_{\xi} - 6 \phi \phi_{\xi} + \phi_{\xi\xi\xi}
+    = \pdv{}{\xi} (- v \phi - 3 \phi^2 + \phi_{\xi\xi})
+\end{aligned}$$
+
+Integrating this equation and introducing an integration constant $$A/2$$ gives:
+
+$$\begin{aligned}
+    0
+    = - 3 \phi^2 - v \phi + \phi_{\xi\xi} + \frac{1}{2} A
+\end{aligned}$$
+
+Let us restrict our search further by demanding
+that $$\phi \to 0$$ and $$\phi_{\xi} \to 0$$ for $$\xi \to \pm \infty$$.
+Clearly, that implies $$\phi_{\xi\xi} \to 0$$, so we must set $$A = 0$$.
+We will do so shortly; first multiply by $$\phi_{\xi}$$:
+
+$$\begin{aligned}
+    0
+    = - 3 \phi^2 \phi_{\xi} - v \phi \phi_{\xi} + \phi_{\xi\xi} \phi_{\xi} + \frac{1}{2} A \phi_{\xi}
+    = \pdv{}{\xi} \bigg(\!-\! \phi^3 - \frac{v}{2} \phi^2 + \frac{1}{2} (\phi_{\xi})^2 + \frac{1}{2} A \phi \bigg)
+\end{aligned}$$
+
+By integrating this again and introducing $$B/2$$,
+we arrive at an equivalent of the KdV equation
+for all solutions of the form $$\phi(\tilde{x} \!-\! v \tilde{t})$$:
+
+$$\begin{aligned}
+    \boxed{
+        (\phi_{\xi})^2
+        = 2 \phi^3 + v \phi^2 - A \phi - B
+        \equiv P(\phi)
+    }
+\end{aligned}$$
+
+Informally, this can be said to describe a pseudoparticle
+with kinetic energy $$(\phi_{\xi})^2$$ and potential energy $$-P(\phi)$$.
+In any case, it is a powerful result.
+
+We already argued that $$A = 0$$ based on our localization requirement;
+likewise, because we want $$\phi_{\xi} \to 0$$ when $$\phi \to 0$$,
+we must set $$B = 0$$ too.
+This just leaves:
+
+$$\begin{aligned}
+    (\phi_{\xi})^2
+    = P(\phi)
+    = \phi^2 (2 \phi + v)
+\end{aligned}$$
+
+Because $$\phi_{\xi}$$ is real, the right-hand side
+must always be positive, meaning $$v > - 2 \phi$$.
+Taking the limit $$\phi \to 0$$, this tells us that $$v > 0$$
+is needed for the solution we want.
+
+We now have the necessary knowledge to find $$\phi$$.
+Taking the equation's square root:
+
+$$\begin{aligned}
+    \phi_{\xi}
+    = \pdv{\phi}{\xi}
+    = \pm \sqrt{\phi^2 (2 \phi + v)}
+\end{aligned}$$
+
+We rearrange this such that $$\dd{\xi}$$ is on one side,
+and then integrate from arbitrary constants $$\xi_0$$ and $$\phi_0$$
+up to the coordinates $$\xi$$ and $$\phi$$:
+
+$$\begin{aligned}
+    \dd{\xi}
+    = \pm \frac{1}{\phi \sqrt{2 \phi + v}} \dd{\phi}
+    \qquad \implies \qquad
+    \int_{\xi_0}^{\xi} \dd{\zeta}
+    = \pm \int_{\phi_0}^{\phi} \frac{1}{\psi \sqrt{2 \psi + v}} \dd{\psi}
+\end{aligned}$$
+
+We proceed with integration by substitution:
+define a new variable $$f$$ such that $$\psi = - \frac{1}{2} v f^2$$,
+and update the integration limits to $$\chi \equiv \sqrt{-2 \phi / v}$$
+and $$\chi_0 \equiv \sqrt{-2 \phi_0 / v}$$:
+
+$$\begin{aligned}
+    \xi - \xi_0
+    &= \pm \int_{\chi_0}^{\chi} \frac{-2}{v f^2 \sqrt{- v f^2 + v}} \dv{\psi}{f} \dd{f}
+    \\
+    &= \pm \frac{2}{\sqrt{v}} \int_{\chi_0}^{\chi} \frac{1}{f \sqrt{1 - f^2}} \dd{f}
+\end{aligned}$$
+
+The integrand can be looked up: it is the derivative of the inverse hyperbolic secant:
+
+$$\begin{aligned}
+    \xi - \xi_0
+    &= \pm \frac{2}{\sqrt{v}} \int_{\chi_0}^{\chi} \dv{}{f} \Big( \sech^{-1}(f) \Big) \dd{f}
+    \\
+    &= \pm \frac{2}{\sqrt{v}} \Big[ \sech^{-1}(f) \Big]_{\chi_0}^{\chi}
+\end{aligned}$$
+
+Evaluating this further,
+and combining the integration constants $$\xi_0$$ and $$\chi_0$$ into $$\tilde{x}_0$$:
+
+$$\begin{aligned}
+    \sech^{-1}(\chi)
+    &= \pm \frac{\sqrt{v}}{2} \Big( \xi - \xi_0 + \sech^{-1}(\chi_0) \Big)
+    = \pm \frac{\sqrt{v}}{2} \big( \xi - \tilde{x}_0 \big)
+\end{aligned}$$
+
+We rearrange, write out $$\chi$$, and discard $$\pm$$
+(since $$\sech$$ is symmetric and $$x_0$$ is arbitrary):
+
+$$\begin{aligned}
+    \sqrt{-\frac{2 \phi}{v}}
+    = \sech\!\bigg( \frac{\sqrt{v}}{2} \Big( \xi - \tilde{x}_0 \Big) \bigg)
+\end{aligned}$$
+
+Isolating this for $$\phi$$ yields a dimensionless soliton solution,
+whose speed, amplitude and width are all determined by a single parameter $$v > 0$$:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{\eta}(\tilde{x}, \tilde{t})
+        = -\frac{v}{2} \sech^2\!\bigg( \frac{\sqrt{v}}{2} \big( \tilde{x} - v \tilde{t} - \tilde{x}_0 \big) \bigg)
+    }
+\end{aligned}$$
+
+What does this look like in units?
+Let us replace $$\tilde{\eta}$$, $$\tilde{x}$$ and $$\tilde{t}$$
+with their dimensioned counterparts $$\eta$$, $$x$$ and $$t$$,
+and appropriate scale parameters:
+
+$$\begin{aligned}
+    \frac{\eta}{\eta_c}
+    = -\frac{v}{2} \sech^2\!\bigg( \frac{\sqrt{v}}{2} \Big( \frac{x - v_c t}{x_c} - v \frac{t}{t_c} - \frac{x_0}{x_c} \Big) \bigg)
+\end{aligned}$$
+
+Inserting the expressions for $$\eta_c$$, $$x_c$$ and $$t_c$$
+we found during non-dimensionalization:
+
+$$\begin{aligned}
+    -\frac{\eta}{s h}
+    = -\frac{v}{2} \sech^2\!\Bigg( \frac{\sqrt{v}}{2} \bigg( \sqrt{\frac{s h}{2 \sigma}} x
+    + \frac{q_0 \alpha}{h} \sqrt{\frac{s h}{2 \sigma}} t + s v q_0 \sqrt{\frac{s h}{32 \sigma}} t - \sqrt{\frac{s h}{2 \sigma}} x_0 \bigg) \Bigg)
+\end{aligned}$$
+
+Cleaning up and isolating for $$\eta$$ gives the form below.
+Remember that $$v$$ is dimensionless:
+
+$$\begin{aligned}
+    \eta
+    &= \frac{s v h}{2} \sech^2\!\Bigg( \sqrt{\frac{s v h}{8 \sigma}}
+    \bigg( x + q_0 \Big( \frac{\alpha}{h} + \frac{s v}{4} \Big) t - x_0 \bigg) \Bigg)
+\end{aligned}$$
+
+We are almost finished, and could leave the solution in this form if we wanted to.
+However, this function contains two free parameters, $$v$$ and $$\alpha$$,
+and it would be nice to combine them into one
+(which is indeed possible without losing information).
+
+From looking at the expression, it is clear that both $$v$$ and $$\alpha$$
+control how fast the soliton moves in our reference frame.
+As discussed earlier, $$\alpha$$ simply modifies the bulk fluid velocity,
+so could we relate $$v$$ and $$\alpha$$ such that the soliton appears stationary?
+Yes, by demanding:
+
+$$\begin{aligned}
+    \frac{\alpha}{h} + \frac{s v}{4}
+    = 0
+    \qquad \implies \qquad
+    \boxed{
+        v
+        = - \frac{4 \alpha}{h \sgn(\sigma)}
+    }
+\end{aligned}$$
+
+Recall that $$v > 0$$ to get a stable dimensionless solution:
+this result therefore tells us that $$\alpha$$ and $$\sigma$$ should have opposite signs.
+That requirement is actually equivalent to $$v > 0$$, and can be found directly
+by deriving the $$\phi_{\xi}^2 = P(\phi)$$ equation without non-dimensionalization.
+At last, this brings us to the general stationary soliton, given by:
+
+$$\begin{aligned}
+    \boxed{
+        \eta(x)
+        = -2 \alpha \sech^2\!\bigg( \sqrt{\frac{-\alpha}{2 \sigma}} (x - x_0) \bigg)
+    }
+\end{aligned}$$
+
+For $$\sigma > 0$$ and $$\alpha < 0$$ the amplitude is positive;
+the wave is a "bump" on the water, as you would expect.
+However, for $$\sigma < 0$$ and $$\alpha > 0$$ the amplitude is negative,
+so then the wave is actually a "dip", which may be surprising.
+For water, the condition $$\sigma < 0$$ equates to $$h \lesssim 0.5\:\mathrm{cm}$$,
+so such waves are indeed hard to observe.
+
+
+
+## References
+1.  D.J. Korteweg, G. de Vries,
+    [On the change of form of long waves advancing in a rectangular canal, and on a new type of long stationary waves](https://doi.org/10.1080/14786449508620739),
+    1895, Philosophical Magazine 39 (240).
+2.  G. de Vries,
+    [Bijdrage tot de kennis der lange golven](https://books.google.nl/books?id=x7sI8lbzxwUC),
+    1894, University of Amsterdam.
+3.  E.M. de Jager,
+    [On the origin of the Korteweg-de Vries equation](https://arxiv.org/abs/math/0602661),
+    University of Amsterdam.
+4.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.
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