1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
|
---
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
that was originally derived to describe water waves.
It is usually given in its dimensionless form:
$$\begin{aligned}
\boxed{
\pdv{u}{t} - 6 u \pdv{u}{x} + \pdvn{3}{u}{x}
= 0
}
\end{aligned}$$
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. $$\eta \gg \alpha \gg \gamma$$.
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:
$$\begin{aligned}
\sigma
\equiv \frac{h^3}{3} - \frac{h T}{g \rho}
\end{aligned}$$
But what about $$\alpha$$?
Looking at the ansatz for $$f$$, we see that
the body of water is assumed to be moving at $$q_0 - g \alpha / q_0$$,
and $$q_0$$ is set to $$\pm \sqrt{g h}$$ by almost all authors,
so $$\alpha$$ controls the velocity of our reference frame.
Nonlinear waves do not travel at the same speed as linear waves,
so we can choose $$\alpha$$ to make the wave stationary
without breaking the $$q_0$$ "tradition".
That term in the KdV equation simply corrects for our chosen value of $$\alpha$$.
## 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:
$$\begin{aligned}
\boxed{
0
= \tilde{\eta}_{\tilde{t}} - 6 \tilde{\eta} \tilde{\eta}_{\tilde{x}} + \tilde{\eta}_{\tilde{x} \tilde{x} \tilde{x}}
}
\end{aligned}$$
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 demonstrates the purpose 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
|