path: root/source/know/concept/boussinesq-wave-theory/index.md

---
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](/know/concept/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 $$h / \lambda$$.



## Dimensionless form

Let us non-dimensionalize the equation by introducing
dimensionless quantities $$\tilde{\eta}$$, $$\tilde{t}$$ and $$\tilde{x}$$:

$$\begin{aligned}
    \tilde{\eta}(\tilde{x}, \tilde{t})
    = \frac{\eta(x, t)}{\eta_c}
    \qquad \qquad
    \tilde{t}
    = \frac{t}{t_c}
    \qquad \qquad
    \tilde{x}
    = \frac{x}{x_c}
\end{aligned}$$

Where $$\eta_c$$, $$t_c$$ and $$x_c$$ are unspecified scale parameters.
We rewrite the Boussinesq equation with these quantities
by using the chain rule of differentiation,
and divide by $$\eta_c / t_c^2$$:

$$\begin{aligned}
    0
    &= \tilde{\eta}_{\tilde{t} \tilde{t}}
    - \frac{g h t_c^2}{x_c^2} \tilde{\eta}_{\tilde{x} \tilde{x}}
    - \pdvn{2}{}{x} \bigg( \frac{3 g \eta_c t_c^2}{2 x_c^2} \tilde{\eta}^2 + \frac{g h^3 t_c^2}{3 x_c^4} \tilde{\eta}_{\tilde{x} \tilde{x}} \bigg)
\end{aligned}$$

Now we must choose values for $$\eta_c$$, $$t_c$$ and $$x_c$$
such that the prefactors become simple constants.
Conventionally it is demanded that:

$$\begin{aligned}
    \frac{g h t_c^2}{x_c^2}
    = 1
    \qquad \qquad
    \frac{3 g \eta_c t_c^2}{2 x_c^2}
    = 3
    \qquad \qquad
    \frac{g h^3 t_c^2}{3 x_c^4}
    = 1
\end{aligned}$$

Solving this system of equations yields the following values for the scale parameters:

$$\begin{aligned}
    \eta_c
    = 2 h
    \qquad \qquad
    t_c
    = \sqrt{\frac{h}{3 g}}
    \qquad \qquad
    x_c
    = \frac{h}{\sqrt{3}}
\end{aligned}$$

And the Boussinesq equation is reduced to its standard dimensionless form:

$$\begin{aligned}
    \boxed{
        \pdvn{2}{\tilde{\eta}}{\tilde{t}} - \pdvn{2}{\tilde{\eta}}{\tilde{x}}
        - \pdvn{2}{}{x} \bigg( 3 \tilde{\eta}^2 + \pdvn{2}{\tilde{\eta}}{\tilde{x}} \bigg)
        = 0
    }
\end{aligned}$$

Many authors flip the sign of $$\tilde{\eta}_{\tilde{x}\tilde{x}\tilde{x}\tilde{x}}$$
to get the so-called "good" Boussinesq equation
(as opposed to the "bad" one above).
For fluid surface waves, this is unphysical,
but it makes the problem more well-behaved mathematically;
the details are beyond the scope of this article.



## Soliton solution

Let us make an ansatz for $$\tilde{\eta}$$ that describes a wave
with a fixed shape propagating in the positive $$\tilde{x}$$-direction
at dimensionless velocity $$v$$:

$$\begin{aligned}
    \tilde{\eta}(\tilde{x}, \tilde{t})
    = \phi(\xi)
    \qquad
    \xi
    \equiv \tilde{x} - v \tilde{t}
    \qquad \implies \qquad
    \pdv{}{\tilde{t}}
    = -v \pdv{}{\xi}
    \qquad
    \pdv{}{\tilde{x}}
    = \pdv{}{\xi}
\end{aligned}$$

With this, the Boussinesq equation becomes a nonlinear ordinary differential equation:

$$\begin{aligned}
    0
    &= (v^2 - 1) \phi_{\xi \xi} - \pdvn{2}{}{\xi} (3 \phi^2 + \phi_{\xi \xi})
\end{aligned}$$

We abbreviate $$w \equiv v^2 \!-\! 1$$ and integrate twice,
introducing integration constants $$A$$ and $$B$$:

$$\begin{aligned}
    w \phi - 3 \phi^2 - \phi_{\xi \xi}
    = A \xi + B
\end{aligned}$$

We restrict ourselves to localized solutions
by demanding that $$\phi \to 0$$ for $$\xi \to \pm \infty$$.
This implies that also $$\phi_\xi \to 0$$ and $$\phi_{\xi \xi} \to 0$$,
meaning that we must set $$A = B = 0$$ to satisfy the equation at infinity.
The remaining terms are multiplied by $$\phi_\xi$$ to give:

$$\begin{aligned}
    0
    &= w \phi \phi_\xi - 3 \phi^2 \phi_\xi - \phi_{\xi \xi} \phi_\xi
    = \frac{1}{2} \pdv{}{\xi} \Big( w \phi^2 - 2 \phi^3 - (\phi_\xi)^2 \Big)
\end{aligned}$$

Integrating (and dropping the integration constant due to localization) yields:

$$\begin{aligned}
    (\phi_\xi)^2
    = \phi^2 (w - 2 \phi)
\end{aligned}$$

Because $$\phi_\xi$$ is real, we need the right-hand side to be positive,
so $$w > 2 \phi$$; for $$\phi \to 0$$, this means that $$w > 0$$.
This equation is similar to the one encountered when solving
the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/)
and is integrated in the same way; look there for details.
The result is:

$$\begin{aligned}
    \boxed{
        \tilde{\eta}(\tilde{x}, \tilde{t})
        = \frac{w}{2} \sech^2\!\bigg( \frac{\sqrt{w}}{2} \big( \tilde{x} - \sqrt{1 \!+\! w} \: \tilde{t} - \tilde{x}_0 \big) \bigg)
    }
\end{aligned}$$

This is known as a **soliton**.
Reintroducing units by replacing $$\tilde{\eta} = \eta / \eta_c$$ etc.
leads to:

$$\begin{aligned}
    \boxed{
        \eta(x, t)
        = w h \sech^2\!\bigg( \frac{\sqrt{3 w}}{2 h} \Big( x - \sqrt{(1 \!+\! w) g h} \: t - x_0 \Big) \bigg)
    }
\end{aligned}$$

Note that Boussinesq's original calculation had $$(1 \!+\! w/2)$$ instead of $$\sqrt{1 \!+\! w}$$;
the former is simply a first-order approximation of the latter.
Recall that $$\sqrt{g h}$$ is the phase velocity of Lagrange's linear theory:
this shows that nonlinear waves are faster,
and speed up with amplitude.



## References
1.  J. Boussinesq,
    [Théorie des ondes et des remous qui se propagent le long