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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+---
+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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