Categories: Mathematics, Physics.

# Korteweg-de Vries equation

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; 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:

\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:

\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 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.

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, 1895, Philosophical Magazine 39 (240).
2. G. de Vries, Bijdrage tot de kennis der lange golven, 1894, University of Amsterdam.
3. E.M. de Jager, On the origin of the Korteweg-de Vries equation, University of Amsterdam.
4. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.

© 2023 Marcus R.A. Newman, CC BY-NC-SA 4.0.