--- 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.