Korteweg - de Vries equation

The Korteweg - de Vries equation is a partial differentialequation defined as

$$u_t+6u{u}_x+u_{xxx}=0$$(1)

where $u=u(x,t)$ and the subscripts indicate derivatives. Thisequation arises in hydrodynamics and was originally proposed tomodel waves in a canal. In addition to its practical applications,this equation is quite interesting as an object of mathematicalstudy. It exhibits interesting soliton solutions, has a largealgebra of conserved quantities, and can be solved using methodsof inverse scattering.

1 Travelling Wave Solution

It is easy to exhibit a solution which describes a traveling wave.To find this solution, one substitutes the ansatz $u(x,t)=f(x-ct)$into the equation to obtain the following equation for $f$:

$$-c{f}^{\prime }+6f{f}^{\prime }+f^{\prime \prime \prime }=0$$

This equation can be written as

$${(-cf+3f^2+f^{\prime \prime })}^{\prime }=0$$

and, hence,

$$-cf+3f^2+f^{\prime \prime }=k$$

for some constant $k$. Multiplying by $f^{\prime }$, we can repeat the sametrick:

$$-cf{f}^{\prime }+3f^2{f}^{\prime }+f^{\prime }{f}^{\prime \prime }-k{f}^{\prime }={(-c{f}^2/2+f^3+{(f^{\prime })}^2-kf)}^{\prime }=0$$

hence

$$-c{f}^2/2+f^3+{(f^{\prime })}^2-kf=h$$

or

$$f^{\prime }=\sqrt{kf+c{f}^2/2-f^3}$$

This can be solved implicitly by an integral:

$$y={\int }^{f(y)}\sqrt{kf+c{f}^2/2-f^3}𝑑y+C$$

Since this is an elliptic integral, the result is an elliptic function.

The solution obtained above is know as a solitary wave, or soliton.This term “solitary wave” refers to the fact that this solution describesa single wave pulse traveling with velocity $c$. Note that the amplitude ofthe pulse is determined by its velocity, unlike in the case of linear waveequations where the velocity of propagation does not depend upon the amplitude.There are also solutions which describe more than one solitary wave. Inparticular, there are solutions in which two of these waves collide and thenre-emerge from the collision.

2 Conserved Currents

It is possible to exhibit integrals whose value is conserved under theevolution. To construct these quantities, we begin with vector fields.Suppose that $u$ is a thrice differentiable function of $x$ and $t$ andconsider the following vector fields:

$v^{(1)}$$=(u,3u^2+u_{xx})$

$v^{(2)}$$=(u^2,4u^3-u_x^2+2u{u}_{xx})$

Computing their divergences and rearranging the result,

${\displaystyle \frac{\partial v_t^{(1)}}{\partial t}}+{\displaystyle \frac{\partial v_t^{(1)}}{\partial x}}$$=u_t+6u{u}_x+u_{xxx}$

${\displaystyle \frac{\partial v_t^{(2)}}{\partial t}}+{\displaystyle \frac{\partial v_t^{(2)}}{\partial x}}$$=2u(u_t+6u{u}_x+u_{xxx}).$

If $u$ happens to satisfy our differential equation, then thedivergence of these vector fields will go zero.

To obtain the conserved quantities, we will integrate them over a rectangle inthe $t,x$ plane with sides parallel to the $x$ and $t$ axes use Green’s theorem .

${\displaystyle {\int }_{t_1}^{t_2}}𝑑t{\displaystyle {\int }_{x_1}^{x_2}}𝑑x$$(u_t(t,x)+6u(t,x)u_x(t,x)+u_{xxx}(t,x))$

$=$${\displaystyle {\int }_{x_1}^{x_2}}𝑑x(3u^2(x,t_2)+u_{xx}(x,t_2))+{\displaystyle {\int }_{t_2}^{t_1}}𝑑tu(t,x_2)+$

${\displaystyle {\int }_{x_1}^{x_2}}𝑑x(3u^2(x,t_1)+u_{xx}(x,t_1))+{\displaystyle {\int }_{t_1}^{t_2}}𝑑tu(t,x_1)$

$=$$3{\displaystyle {\int }_{x_1}^{x_2}}𝑑x{u}^2(x,t_2)-3{\displaystyle {\int }_{x_1}^{x_2}}𝑑x{u}^2(x,t_1)$

$+$${\displaystyle {\int }_{t_1}^{t_2}}𝑑tu(t,x_1)-{\displaystyle {\int }_{t_1}^{t_2}}𝑑tu(t,x_2)+u_x(x_2,t_2)-u_x(x_1,t_2)-u_x(x_2,t_1)+u_x(x_1,t_1)$

${\displaystyle {\int }_{t_1}^{t_2}}𝑑t{\displaystyle {\int }_{x_1}^{x_2}}𝑑x$$2u(u_t(t,x)+6u(t,x)u_x(t,x)+u_{xxx}(t,x))$

$=$${\displaystyle {\int }_{x_1}^{x_2}}𝑑x(4u^3(x,t_2)-u_x^2(x,t_2)+2u(x,t_2)u_{xx}(x,t_2))+{\displaystyle {\int }_{t_2}^{t_1}}𝑑t{u}^2(t,x_2)+$

${\displaystyle {\int }_{x_2}^{x_2}}𝑑x(4u^3(x,t_1)-u_x^2(x,t_1)+2u(x,t_1)u_{xx}(x,t_1))+{\displaystyle {\int }_{t_1}^{t_2}}𝑑t{u}^2(t,x_1)$

$=$${\displaystyle {\int }_{x_1}^{x_2}}𝑑x(4u^3(x,t_2)-3u_x^2(x,t_2))-{\displaystyle {\int }_{x_1}^{x_2}}𝑑x(4u^3(x,t_1)-3u_x^2(x,t_1))$

$+$${\displaystyle {\int }_{t_1}^{t_2}}𝑑t{u}^2(t,x_1)-{\displaystyle {\int }_{t_1}^{t_2}}𝑑t{u}^2(t,x_2)$

$+$$u(x_2,t_2)u_x(x_2,t_2)-u(x_1,t_2)u_x(x_1,t_2)-u(x_2,t_1)u_x(x_2,t_1)+u(x_1,t_1)u_x(x_1,t_1)$

We consider now several popular boundary conditions for our equation:

• Periodic Suppose that $u$ is periodic in $x$ with period $p$ andsatisfies the Kortwieg - de Vries equation. Then we integrals over $t$cancel against each other as do the endpoint terms, leaving us with

  $${\int }_{x_1}^{x_2}𝑑x{u}^2(x,t_1)={\int }_{x_1}^{x_2}𝑑x{u}^2(x,t_2).$$

• Whole line Suppose that $u$ satisfies the Kortwieg - de Vries equationand that $u$ and its derivatives tend towards zero as $x\to \pm \mathrm{\infty }$. Then,taking the limit as $x_1\to -\mathrm{\infty }$ and $x_2\to +\mathrm{\infty }$, we conclude that

  $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}𝑑x{u}^2(x,t_1)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}𝑑x{u}^2(x,t_2).$$

• Finite interval We may also consider our differential equation ona finite interval and impose suitable boundary conditions at the endpoints.In this case, the integrals with respect to $t$ do not cancel or automaticallyvanish as a consequence of the boundary conditions, so they will not, in general,give conserved quantities. Nevertheless, it is at least still possible to computetheir value solely from the boundary data without having to know how the solutionbehaves in the interior.

Starting with the two vector fields given above, it is possible to generatemore such vector fields and more conserved quantities. Since the Lie bracketof two vector fields with zero divergence also has zero divergence, we may