Question 3.5.7: Determine a steady profile of the non-linear dispersive soli......

Consider the solution of the KdV equation in the form of an impulse of steady shape that propagates with a constant speed: V = f(x − ut). Its substitution into equation (3.104) yields −uf^{′} + ff^{′} + βf^{′′′} = 0, which after one integration results in

\frac{∂V}{∂t} + V \frac{∂V}{∂x} + β \frac{∂^{3}V}{∂^{3}x} = 0,        (3.104)

f^{′′} = \frac{u}{β} f −  \frac{f^{2}}{2β} + C        (3.105)

Here C is a constant of integration, which can be made equal to zero by an appropriate choice of the reference frame. This equation can be analyzed with the help of the mechanical analogy by writing (3.105) (with C = 0) in the form

\frac{∂^{2}f}{∂^{2}x} = \frac{u}{β} f − \frac{f^{2}}{2β} ≡ F(f)        (3.106)

By interpreting f as the coordinate and x as time, it becomes the equation of motion for a point of unit mass under a force F(f). Hence, the respective potential energy, W(f), which is defined as F = −dW/df, is equal to

W(f) = − \frac{u}{2β} f^{2} + \frac{1}{6β} f^{3}        (3.107)

It is plotted in Figure 3.17, where u > 0 is assumed. The interest here is in the “bounded” trajectories, which correspond to the energy interval −2u^{3}/3β < E ≤ 0. For a low energy the “particle” oscillates at the bottom f = 2u of the potential well, where the shape of W(f) is close to parabolic. This yields

V (x, t) ≈ 2u + V_{0} \exp \left[i \left(\frac{u}{β} \right)^{1/2}  (x − ut) \right] ,        (3.108)

which represents a linear wave with the dispersion law that follows from the “linearized” KdV equation

\frac{∂V}{∂t} + 2u \frac{∂V}{∂x} + β \frac{∂^{3}V}{∂^{3}x} = 0

(note that in equation (3.108) k = \sqrt{u/β}). A role of the non-linearity increases with the increasing energy E, when a non-symmetric shape of the potential energy W(f) (steeper at f > 2u, and flatter at f < 2u) becomes important.

Therefore, in the non-linear periodic wave the maxima of V become relatively narrow (a “particle” passes through respective interval of f “quickly”), while the minima are wider (a “particle” moves there “slowly”). In the limit of E → 0 the left-hand side “reflection” point f = 0 is reached asymptotically at t → ∞, as the period of oscillation diverges at this energy. In terms of f(x) it means that while x varies from −∞ to +∞, the solution f(x) varies from f(−∞) = 0 to the maximum f(0) = 3u, and then back to f = 0 at x → +∞.

This particular solution is called the “soliton” (from a solitary wave). Thus, on the surface of the fluid, which is at rest at infinity, such a soliton propagates with the “supersonic” speed equal to u_{0} + u, where u = f_{max}/3.

Consider now the shape of the soliton. The first integral of equation (3.106) reads

\frac{1}{2} \left(\frac{df}{dx} \right)^{2} +W(f) = const = E = 0        (3.109)

It follows then from equations (3.109) and (3.107) that

\frac{df}{dx} = ±\left(\frac{u}{β} \right)^{1/2} f \left( 1 − \frac{f}{3u}\right)^{1/2}        (3.110)

If the soliton apex, where f = f_{max} = 3u, is located at x = 0, in the region x > 0 the derivative df/dx < 0, and one gets from equation (3.110) that

\int_{f/f_{max}}^{1} \frac{dz}{z \sqrt{1 − z}} = \left(\frac{u}{β} \right)^{1/2} x,

which after a standard integration yields

\frac{1 + \sqrt{1 − f/f_{max}}}{\sqrt{f/f_{max}}} = \exp \left[\left(\frac{u}{4β} \right)^{1/2} x \right]

Thus, the shape of the soliton is discribed by the following expression

f(x) = \frac{3u}{cosh^{2}(x/l)}, l = \sqrt{4β/u}        (3.111)

As seen from expression (3.111), the width of the soliton, l, decreases with the increasing amplitude, u, as l ∝ 1/ \sqrt{u}. Such a scaling can be explained by the following consideration. A steady shape of the soliton is established by competition of two processes: the non-linearity, that leads to the steepening of the impulse, and the dispersion, that causes its spreading. Thus, for the soliton, where these two effects balance each other, the non-linear and the dispersive terms in the KdV equation (3.104) should be of the same order of magnitude. Hence,

V \frac{∂V}{∂x} ∼ \frac{u^{2}}{l} ≈ β \frac{∂^{3}V}{∂^{3}x} ∼ β \frac{u}{l^{3}},

which yields l ∼ \sqrt{β/u}.