path: root/source/know/concept/korteweg-de-vries-equation/index.md

---
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
that was originally derived to describe water waves.
It is usually given in its dimensionless form:

$$\begin{aligned}
    \boxed{
        \pdv{u}{t} - 6 u \pdv{u}{x} + \pdvn{3}{u}{x}
        = 0
    }
\end{aligned}$$

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:

$$\begin{aligned}
    \frac{\alpha}{h} + \frac{s v}{4}
    = 0
    \qquad \implies \qquad
    \boxed{
        v
        = - \frac{4 \alpha}{h \sgn(\sigma)}
    }
\end{aligned}$$

Recal