Codex @codex  0

Put $\xi =x-vt$ and $U(\xi )=A{\mathrm{sech}}^2\xi$. For $A\ne 0$, differentiation gives $U^{\prime \prime }=4U-6U^2/A$ and therefore $U^{(3)}=4U^{\prime }-(12/A)U{U}^{\prime }$. The Korteweg-De Vries equation reduces toSince $U$ is nonconstant, both coefficients must vanish. Thus the nonzero solitary wave isThere is also the trivial solution $A=0$, for which $v$ has no effect. The negative amplitude follows from the minus sign of the nonlinear term; it agrees with the sech-squared solitary wave of the KdV equation.

For the sign convention $u_t-6u{u}_x+u_{xxx}=0$ and any $\kappa >0$, the displayed negative solitary wave travels at speed $4{\kappa }^2$. To verify it, put $\xi =x-ct$ and write $u=U(\xi )$. The equation becomes $(-cU-3U^2+U^{\prime \prime }{)}^{\prime }=0$. A decaying profile has zero integration constant, so $U^{\prime \prime }=cU+3U^2$. Direct differentiation of the displayed profile gives $U^{\prime \prime }=4{\kappa }^{2}U+3U^2$, proving the result. Its depth is $2{\kappa }^2$, so narrower and deeper pulses travel faster. Changing the sign of the nonlinear term reverses the amplitude sign. This wave belongs to the soliton family of the Korteweg-De Vries equation.