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diff --git a/source/know/concept/boussinesq-wave-equations/index.md b/source/know/concept/boussinesq-wave-theory/index.md index 928d365..31228ba 100644 --- a/source/know/concept/boussinesq-wave-equations/index.md +++ b/source/know/concept/boussinesq-wave-theory/index.md @@ -1,6 +1,6 @@ --- -title: "Boussinesq wave equations" -sort_title: "Boussinesq wave equations" +title: "Boussinesq wave theory" +sort_title: "Boussinesq wave theory" date: 2023-01-07 categories: - Physics @@ -10,8 +10,8 @@ categories: layout: "concept" --- -In fluid mechanics, **Boussinesq's wave theory** -consists of several equations to describe surface waves of a liquid. +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. @@ -104,7 +104,7 @@ The equations will be derived from these two fundamental boundary conditions. -## Boussinesq's approximation +## Boussinesq approximation Let us take a Taylor expansion of the velocity potential $$\Psi(x, z, t)$$ at the bottom $$z = -h$$: @@ -185,7 +185,7 @@ $$\begin{aligned} \end{aligned}$$ The Boussinesq approximation is the basis of many other shallow-water wave theories, -most notably the *Korteweg-de Vries equation*. +most notably the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/). diff --git a/source/know/concept/korteweg-de-vries-equation/index.md b/source/know/concept/korteweg-de-vries-equation/index.md new file mode 100644 index 0000000..4a050fe --- /dev/null +++ b/source/know/concept/korteweg-de-vries-equation/index.md @@ -0,0 +1,631 @@ +--- +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. |