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:

ut6uux+3ux3=0\begin{aligned} \boxed{ \pdv{u}{t} - 6 u \pdv{u}{x} + \pdvn{3}{u}{x} = 0 } \end{aligned}

Where u(x,t)u(x, t) is the wave’s profile, with xx 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=η(x,t)z = \eta(x, t). Firstly, the kinematic boundary condition:

ηt+u(x)ηxu(z)=0\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:

Ψt+12((u(x))2+(u(z))2)=gη+p0pρ\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 Ψ\Psi is the velocity potential u=Ψ\va{u} = \nabla \Psi, with u=(u(x),u(z))\va{u} = \big( u^{(x)}, u^{(z)} \big) being 2D due to the assumed symmetry along the yy-axis. Unlike Boussinesq, who assumed that p0=pp_0 = p at the surface, de Vries decided to include surface tension using the Young-Laplace law:

p0p=Tκ=Tηxx(1+ηx2)3/2Tηxx\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 TT is the energy cost per unit area, and η\eta is assumed to be slowly-varying such that ηx2\eta_{x}^2 can be neglected in the curvature formula. His free surface condition was thus:

Ψt+12((u(x))2+(u(z))2)=gη+Tρηxx\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 xx, yielding:

u(x)t+12x((u(x))2+(u(z))2)=gηx+Tρηxxx\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 zz-derivatives from the problem:

u(x)(x,z)=Ψx=f(x)(z ⁣+ ⁣h)22fxx(x)+(z ⁣+ ⁣h)424fxxxx(x)...u(z)(x,z)=Ψz=(z ⁣+ ⁣h)fx(x)+(z ⁣+ ⁣h)36fxxx(x)...\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)Ψx(x,h,t)f(x, t) \equiv \Psi_{x}(x, -h, t) is the xx-velocity at the channel’s bottom z=hz = -h. Inserting this expansion into the two boundary conditions yields these coupled equations:

0=ηt+ηx(f(η ⁣+ ⁣h)22fxx+...)+((η ⁣+ ⁣h)fx(η ⁣+ ⁣h)36fxxx+...)0=t(f(η ⁣+ ⁣h)22fxx+...)+12x(f2+(η ⁣+ ⁣h)2(fx2 ⁣ ⁣ffxx)+...)+gηxTρηxxx\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 ⁣+ ⁣η)3(h \!+\! \eta)^3:

0=ηt+ηxf+(η ⁣+ ⁣h)fx(η ⁣+ ⁣h)22ηxfxx(η ⁣+ ⁣h)36fxxx0=ft+ffx(η ⁣+ ⁣h)(ηtfxxηxfx2+ηxffxx)(η ⁣+ ⁣h)22(fxxtfxfxx+ffxxx)+gηxTρηxxx\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:

ηttghηxx=0    η=η+(xght)+η(x+ght)\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 η+\eta^{+} and η\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 η(x ⁣ ⁣ght)\eta(x \!-\! \sqrt{g h} t), whereas de Vries chose a reference frame moving at a speed q0q_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 ff; in other words, he assumed that the entire liquid is moving at q0q_0. For a wave going in the positive xx-direction, the linearized problem then predicts a profile η(x ⁣ ⁣(gh ⁣+ ⁣q0))\eta(x \!-\! (\sqrt{g h} \!+\! q_0)), so de Vries chose q0=ghq_0 = -\sqrt{g h} to make it stationary. Analogously, q0=ghq_0 = \sqrt{g h} for a backward-moving wave. With this in mind, the ansatz is:

f(x,t)=q0gq0(η(x,t)+α+γ(x,t))\begin{aligned} f(x, t) = q_0 - \frac{g}{q_0} \Big( \eta(x, t) + \alpha + \gamma(x, t) \Big) \end{aligned}

Where α\alpha is a constant parameter (which we will use to handle velocity discrepancies between the linear and nonlinear theories). The correction represented by γ\gamma is much smaller, i.e. ηαγ\eta \sim \alpha \gg \gamma. We insert this ansatz into the above equations, yielding:

0=ηt+ηx(q0gq0(η+α+γ))gq0(η ⁣+ ⁣h)(ηx+γx)+gq0(η ⁣+ ⁣h)22ηx(ηxx+γxx)+gq0(η ⁣+ ⁣h)36(ηxxx+γxxx)0=gηxTρηxxxgq0(ηt+γt)gq0(q0gq0(η+α+γ))(ηx+γx)+gq0(η ⁣+ ⁣h)(ηt(ηxx+γxx)+gq0ηx(ηx+γx)2+(q0gq0(η+α+γ))ηx(ηxx+γxx))+gq0(η ⁣+ ⁣h)22(ηxxt+γxxt+gq0(ηx+γx)(ηxx+γxx)+(q0gq0(η+α+γ))(ηxxx+γxxx))\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 αη\alpha \eta, but neglect anything smaller (ηγ\eta \gamma etc.), because by assumption we have hηαγh \gg \eta \gg \alpha \gg \gamma. Furthermore, each xx-derivative is roughly equivalent to dividing by λ\lambda, and since the water is shallow (λh\lambda \gg h) successive differentiations reduce terms’ magnitudes, so terms like αη\alpha \eta and η2\eta^2 are kept only if they contain at most one xx-derivative: e.g. ηηx\eta \eta_x stays, but ηx2\eta_x^2 does not. This reduces the equations to the following:

0=ηt+q0ηxgq0(η+α)ηxghq0(ηx+γx)gq0ηηx+gh36q0(ηxxx ⁣+ ⁣γxxx)0=gηxTρηxxxgq0(ηt+γt)g(ηx+γx)+g2q02(η+α)ηx+gh22q0(ηxxt ⁣+ ⁣γxxt ⁣+ ⁣q0ηxxx)\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 q0q_0, so all tt-derivatives describe deformation rather than transport, and are hence quite small. Therefore we discard all except for ηt\eta_t:

0=ηt+q0ηxghq0(ηx+γx)gq0(η+α)ηxgq0ηηx+gh36q0ηxxx0=gq0ηtgγx+g2q02(η+α)ηx+gh22ηxxxTρηxxx\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/q0-g / q_0, and inserting q0=±ghq_0 = \pm\sqrt{g h} into both:

gq0ηt=gγx+gh(2η+α)ηxgh26ηxxxgq0ηt=gγx+gh(η+α)ηx+(gh22Tρ)ηxxx\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 q0q_0 to gh\sqrt{g h}, others to gh-\sqrt{g h}; we preserve q0q_0 on the left-hand side to cover both cases. Adding up these two equations:

2gq0ηt=gh(3η+2α)ηx+(gh23Tρ)ηxxx=ghx(32η2+2αη+(h33hTgρ)ηxx)\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:

ηt=32q0hx(12η2+23αη+13σ2ηx2)\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 σ\sigma as follows:

σh33hTgρ\begin{aligned} \sigma \equiv \frac{h^3}{3} - \frac{h T}{g \rho} \end{aligned}

What about α\alpha? Looking at the ansatz for ff, we see that the body of water is already assumed to be moving at q0q_0, minus gα/q0g \alpha / q_0, so by varying α\alpha we are modifying the water’s velocity. The term in the KdV equation simply corrects for our chosen value of α\alpha. It has no deeper meaning than that: for any value of α\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 η~\tilde{\eta}, x~\tilde{x} and t~\tilde{t} are dimensionless, and ηc\eta_c, xcx_c, tct_c and vcv_c are free dimensioned scale parameters:

η~(x~,t~)=η(x,t)ηct~=ttcx~=xvctxc\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 xx and tt are then rewritten like so:

t=t~tt~+x~tx~=1tct~vcxcx~x=t~xt~+x~xx~=1xcx~\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:

0=ηt3q02hηηxq0αhηxq02hσηxxx=ηctcη~t~vcηcxcη~x~3q0ηc22hxcη~η~x~q0αηchxcη~x~q0σηc2hxc3η~x~x~x~\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 tc/ηct_c / \eta_c to make all terms unitless and bring the first to the desired form:

0=η~t~tcxc(vc+q0αh)η~x~3q0ηctc2hxcη~η~x~q0σtc2hxc3η~x~x~x~\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 66, and the last has a factor 1-1, yielding equations:

vc+q0αh=03q0ηctc2hxc=6q0σtc2hxc3=1\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:

vc=q0αhtc=4hxcq0ηcxc2=2σηc\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 xc2x_c^2, but we must make sure that xc2>0x_c^2 > 0, because all quantities are real. We enforce this in our choice of ηc\eta_c, where ssgn(σ)s \equiv \sgn(\sigma):

ηc=shvc=q0hαxc=2σshtc=1sq032σsh\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:

0=η~t~6η~η~x~+η~x~x~x~\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 α\alpha sets the background fluid velocity, and vcv_c controls the coordinate system’s motion: our choice of vcv_c simply cancels out the effect of α\alpha. This reveals the point of α\alpha: the KdV equation has solutions moving at various speeds, so, for a given η\eta, we can always choose α\alpha (and hence vcv_c) such that the wave appears stationary.

Soliton solution

Let us make the following ansatz for the dimensionless wave profile η~\tilde{\eta}, assuming there exists a solution that maintains its shape while propagating at a constant “velocity” vv:

η~(x~,t~)=ϕ(ξ)ξx~vt~    t~=vξx~=ξ\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 ϕ\phi must satisfy:

0=vϕξ6ϕϕξ+ϕξξξ=ξ(vϕ3ϕ2+ϕξξ)\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/2A/2 gives:

0=3ϕ2vϕ+ϕξξ+12A\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 ϕ0\phi \to 0 and ϕξ0\phi_{\xi} \to 0 for ξ±\xi \to \pm \infty. Clearly, that implies ϕξξ0\phi_{\xi\xi} \to 0, so we must set A=0A = 0. We will do so shortly; first multiply by ϕξ\phi_{\xi}:

0=3ϕ2ϕξvϕϕξ+ϕξξϕξ+12Aϕξ=ξ( ⁣ ⁣ϕ3v2ϕ2+12(ϕξ)2+12Aϕ)\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/2B/2, we arrive at an equivalent of the KdV equation for all solutions of the form ϕ(x~ ⁣ ⁣vt~)\phi(\tilde{x} \!-\! v \tilde{t}):

(ϕξ)2=2ϕ3+vϕ2AϕBP(ϕ)\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 (ϕξ)2(\phi_{\xi})^2 and potential energy P(ϕ)-P(\phi). In any case, it is a powerful result.

We already argued that A=0A = 0 based on our localization requirement; likewise, because we want ϕξ0\phi_{\xi} \to 0 when ϕ0\phi \to 0, we must set B=0B = 0 too. This just leaves:

(ϕξ)2=P(ϕ)=ϕ2(2ϕ+v)\begin{aligned} (\phi_{\xi})^2 = P(\phi) = \phi^2 (2 \phi + v) \end{aligned}

Because ϕξ\phi_{\xi} is real, the right-hand side must always be positive, meaning v>2ϕv > - 2 \phi. Taking the limit ϕ0\phi \to 0, this tells us that v>0v > 0 is needed for the solution we want.

We now have the necessary knowledge to find ϕ\phi. Taking the equation’s square root:

ϕξ=ϕξ=±ϕ2(2ϕ+v)\begin{aligned} \phi_{\xi} = \pdv{\phi}{\xi} = \pm \sqrt{\phi^2 (2 \phi + v)} \end{aligned}

We rearrange this such that dξ\dd{\xi} is on one side, and then integrate from arbitrary constants ξ0\xi_0 and ϕ0\phi_0 up to the coordinates ξ\xi and ϕ\phi:

dξ=±1ϕ2ϕ+vdϕ    ξ0ξdζ=±ϕ0ϕ1ψ2ψ+vdψ\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 ff such that ψ=12vf2\psi = - \frac{1}{2} v f^2, and update the integration limits to χ2ϕ/v\chi \equiv \sqrt{-2 \phi / v} and χ02ϕ0/v\chi_0 \equiv \sqrt{-2 \phi_0 / v}:

ξξ0=±χ0χ2vf2vf2+vdψdfdf=±2vχ0χ1f1f2df\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:

ξξ0=±2vχ0χddf(sech1(f))df=±2v[sech1(f)]χ0χ\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 ξ0\xi_0 and χ0\chi_0 into x~0\tilde{x}_0:

sech1(χ)=±v2(ξξ0+sech1(χ0))=±v2(ξx~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 χ\chi, and discard ±\pm (since sech\sech is symmetric and x0x_0 is arbitrary):

2ϕv=sech ⁣(v2(ξx~0))\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 ϕ\phi yields a dimensionless soliton solution, whose speed, amplitude and width are all determined by a single parameter v>0v > 0:

η~(x~,t~)=v2sech2 ⁣(v2(x~vt~x~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 η~\tilde{\eta}, x~\tilde{x} and t~\tilde{t} with their dimensioned counterparts η\eta, xx and tt, and appropriate scale parameters:

ηηc=v2sech2 ⁣(v2(xvctxcvttcx0xc))\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 ηc\eta_c, xcx_c and tct_c we found during non-dimensionalization:

ηsh=v2sech2 ⁣(v2(sh2σx+q0αhsh2σt+svq0sh32σtsh2σx0))\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 η\eta gives the form below. Remember that vv is dimensionless:

η=svh2sech2 ⁣(svh8σ(x+q0(αh+sv4)tx0))\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, vv and α\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 vv and α\alpha control how fast the soliton moves in our reference frame. As discussed earlier, α\alpha simply modifies the bulk fluid velocity, so could we relate vv and α\alpha such that the soliton appears stationary? Yes, by demanding:

αh+sv4=0    v=4αhsgn(σ)\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>0v > 0 to get a stable dimensionless solution: this result therefore tells us that α\alpha and σ\sigma should have opposite signs. That requirement is actually equivalent to v>0v > 0, and can be found directly by deriving the ϕξ2=P(ϕ)\phi_{\xi}^2 = P(\phi) equation without non-dimensionalization. At last, this brings us to the general stationary soliton, given by:

η(x)=2αsech2 ⁣(α2σ(xx0))\begin{aligned} \boxed{ \eta(x) = -2 \alpha \sech^2\!\bigg( \sqrt{\frac{-\alpha}{2 \sigma}} (x - x_0) \bigg) } \end{aligned}

For σ>0\sigma > 0 and α<0\alpha < 0 the amplitude is positive; the wave is a “bump” on the water, as you would expect. However, for σ<0\sigma < 0 and α>0\alpha > 0 the amplitude is negative, so then the wave is actually a “dip”, which may be surprising. For water, the condition σ<0\sigma < 0 equates to h0.5cmh \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, 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.
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