Korteweg-De Vries Equation

Following the procedure applied in the preceding section, we final that for the KdV equation

$$u_t+uu_x+\sigma u_{xxx}=0\ ,\eqno (4.1)$$

$$u=\phi^{-2}\sum_{j=0}^{\infty} u_j\phi_j\ .\eqno (4.2)$$

The recursion relations for the $u_j$ are presented elsewhere [12]. It is found that the ``resonances" occur at

$$j=-1,\ 4,\ 6\ .\eqno (4.3)$$

The compatability conditions at $j=4,6$ are satisfied identically and the KdV equation possesses the Painlevé property. For instance, wen

truecm $j=0\ ,\qquad u_0=-12\sigma\phi_x^2\ ,$ (4.4)

truecm $j=1\ ,\qquad u_1=12\sigma\phi_{xx}\ ,$ (4.5)

truecm $j=2\ ,\qquad\phi_x\phi_t+\phi_x^2 u_2+4 \sigma\phi_x\phi_{xxx}-3\sigma\phi_{xx}^2=0\ ,$(4.6)

truecm $j=3\ ,\qquad\phi_{xt}+\phi_{xx} u_2- \phi_x^2 u_3+\sigma\phi_{xxxx}=0\ ,$(4.7)

truecm $j=4\ ,\qquad\hbox{compatibility\ condition:}$

$${\partial\over\partial x}(\phi_{xt}+\sigma\phi_{xxxx}+\phi_{xx} u_2- \phi_x^2 u_3)=0\ .\eqno (4.8)$$

By (4.7) the compatibility condition (4.8) at $j=4$ is satisfied identically. The compatibility condition at $j=6$ involves extensive calculation and is, therefore, presented elsewhere [12].

We now specialize (4.2) by setting the ``resonance" functions

$$u_4=u_6=0\ .\eqno (4.9)$$

Furthermore, by requiring

$$u_3=0\ ,\eqno (4.10)$$

it is easily demonstrated that

$$u_j=0\ ,\qquad j\ge 3\ ,\eqno (4.11)$$

if

$$u_{2t}+u_2u_{2x}+\sigma u_{2xxx}=0\ .\eqno (4.12)$$

Collecting (4.4)-(4.12), we find that

(i) $u_n=0\ ,\qquad j\ge 3\ ,$

(ii) $u_0=-12\sigma\phi_x^2\ ,\qquad u_2=12\sigma\phi_{xx}\ ,$

(iii)(a)     $\phi_x\phi_t+\phi_x^2 u_2+ 4\sigma\phi_x\phi_{xxx}-3\sigma\phi_{xx}^2=0\ ,$

(b)     $\phi_{xt}+\phi_{xx} u_2+\sigma \phi_{xxxx}=0\ ,$

(iv) $u_{2t}+u_2u_{2x}+\sigma u_{2xxx}=0\ ,$

(v) $u=-12\sigma\phi_x^2/\phi^2+12\sigma \phi_{xx}/\phi+u_2\ ,$

or

$$u=12\sigma(\partial^2/\partial x^2)\ln\phi+u_2\ ,$$

where

$$u_t+uu_x+\sigma u_{xxx}=0\ .\eqno (4.13)$$

(4.13) will define a Bäcklund transform for the KdV equation if equations (4.13.iii) and (4.13.iv) are consistent. By solving Eq. (4.13.iiia) for $\phi$ then differentiating w.r.t. $x$, using Eq. (4.13.iiib), and making the change of variable

$$\phi_x=V^2\eqno (4.14)$$

it is found that

(i) $6\sigma V_{xx}+u_2V=\lambda V\ ,$

(ii) $2V_t+u_2V_x+\lambda V_x+2\sigma V_{xxx}=0 \ .$(4.15)

Thus, (4.13.iii) reduces to (4.15), which is just the Lax pair [13] for (4.13.iv), as may be verified by direct calculation [12]. In summary, our result reads

$$u=12\sigma {\partial^2\over\partial x^2}\ln\phi+u_2\ ,$$

where

$$\begin{array}{l} \phi_x=V^2\ ,\\ \noalign{\vskip 5pt} 6\s... ...5pt} 2V_t+u_2V_x+\lambda V_x+2\sigma V_{xxx}=0\ , \end{array}$$

and

$$\null\,\vcenter{\openup\jot \ialign{\strut\hfil$\displaysty... ...=0\ ,\cr &u_t+uu_x+\sigma u_{xxx}=0\ .\cr\crcr}}\,\eqno (4.16)$$

This Bäcklund transform connects two solutions of the KdV equation via the squared eigenfunction associated with one solution by the IST transform. Although a Bäcklund transform for the KdV is hardly novel, the construction of the transform from a finite Painlevé expansion about the singularity manifold is obtained in a remarkably straightforward way.