The Bousinesq Equation

The Bousinesq equation

$$u_{tt}+{\partial^2\over\partial x^2}\left({u_{xx}\over 3}+u^2\right) =0\eqno (4.1)$$

possesses the Painlevé property [1].

In general, it is found that

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

where the resonances occur at

$$j=-1,4,5,6 .\eqno (4.3)$$

The compatability conditions at $j=4,5$, and 6 are satisfied identically. We will require that

$$u_3=0\eqno (4.4)$$

and set

$$u_4=u_5=u_6=0\eqno (4.5)$$

(implying $u_j=0\ge j\ge 3$). We find the associated Bäcklund transform

$$u=2{\partial^2\over\partial x^2}\ln\phi+u_2 ,\eqno (4.6)$$

where $(u,u_2)$ satisfy (4.1), and

$$\phi_t^2-\phi_{xx}^2+\textstyle{4\over 3}\phi_x\phi_{xxx}+2u_2\phi_x^2=0 , \eqno (4.7)$$

$$\phi_{tt}+\textstyle{1\over 3}\phi_{xxxx}+2\phi_{xx} u_2=0 .\eqno (4.8)$$

Now, by eliminating $u_2$ in (4.7) and (4.8), it is found that

$${\partial\over\partial t}\left({\phi_t\over\phi_x}\right)+{1\... ...{1\over 3} {\partial\over\partial x}\{\phi;x\}=0 ,\eqno (4.9)$$

where $\{\phi;x\}$ is the Schwarzian derivative.

As before, this equation is invariant under the Moebius group. We let

$$\phi=v_1/v_2\eqno (4.10)$$

and find

$${\partial\over\partial t}\left({\psi_t\over\theta_x}\right)+{... ...eta_{xx}\over\theta_x}{v_{2x}\over v_2}\right)=0 ,\eqno (4.11)$$

where

$$\theta_x=v_2v_{1x}-v_1v_{2x} ,\eqno (4.12)$$

$$\psi_t=v_2v_{1t}-v_1v_{2t} ,\eqno (4.13)$$

and

$$\{\theta;x\}={\partial\over\partial x}\left({\theta_{xx}\over... ...er 2}\left({\theta_{xx}\over\theta_x}\right)^2 . \eqno (4.14)$$

To solve (4.11), first we let $(v_1,v_2)$ satisfy the following:

Second-order scattering problem:

$$v_{xx}=av ,\eqno (4.15)$$

$$v_t=bv_x+cv .\eqno (4.16)$$

Then,

$$\theta_{xx}=0 ,$$

$$\psi_t=b\theta_x ,$$

and from (4.11)

$$b_t+bb_x=\textstyle{2\over 3}ax ,\eqno (4.17)$$

while by compatability of (4.15) and (4.16)

$$\null \vcenter{\openup\jot \ialign{\strut\hfil$\displaysty... ...{xx}=0 ,\cr &a_t=c_{xx}+2ab_x+ba_x ,\cr\crcr}} \eqno (4.18)$$

or, collecting equations,

$$\null \vcenter{\openup\jot \ialign{\strut\hfil$\displaysty... ... b_t&=-bb_x+\textstyle{2\over 3}a_x .\cr\crcr}} \eqno (4.19)$$

Unfortunately, it is not possible to introduce an arbitrary,spectral parameter

$$a=u+\lambda\eqno (4.20)$$

into Eq. (4.15) without obtaining from (4.19) equations that will depend explicitly on that parameter. Therefore, we shall consider the following:

Third-order scattering problem:

$$v_{xxx}=av_x+bv ,\eqno (4.21)$$

$$v_t=cv_{xx}+dv_x+ev .\eqno (4.22)$$

Then,

$$\psi_t=c\theta_{xx}+d\theta_x ,\eqno (4.23)$$

and substitution into (4.11) determines

$$c^2=1 ,\qquad d=0\eqno (4.24)$$

$$ce_x+\textstyle{2\over 3}a_x=0 .\eqno (4.25)$$

By the consistency condition for (4.21) and (4.22) we find

$$\null \vcenter{\openup\jot \ialign{\strut\hfil$\displaysty... ...xx} ,\cr &b_t+ae_x=e_{xxx}+cb_{xx} .\cr\crcr}} \eqno (4.26)$$

Letting

$$a=-\textstyle{3\over 2}ce ,\eqno (4.27)$$

we find

$$ce_{tt}+{\partial^2\over\partial x^2}\left(e^2+c{e^{xx}\over 3} \right)=0 ,\eqno (4.28)$$

where

$$c^2=1 .$$

Let

$$c=1 ;$$

then

truecm $e$ is a solution of the Bousinesq equation

truecm $a=-\textstyle{3\over 2}e ,$

truecm $b=\lambda-\textstyle{3\over 4}e_x^2-\textstyle{3\over 4} \displaystyle{\int^x} e_t ,$

truecm $c=1 ,\qquad d=0 .$(4.29)

Equations (4.29) determine the Lax pair for the Bousinesq equation [7]. We note the appearance of the spectral parameter $\lambda$.