Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 14

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.

Differentiate the equation for $f$ with respect to $x$, commuting mixed partial derivatives for a sufficiently smooth solution. With $y=f_x$ this givesThis is the viscous Burgers equation when $k>0$.

For $k\ne 0$, the chain rule gives $g_t=-f_{t}g/(2k)$ and $g_{xx}=(-f_{xx}/(2k)+f_x^2/(4k^2))g$. ConsequentlyThusThis is the potential form of the Cole-Hopf transformation, with $y=-2k{g}_x/g$. For $k>0$ it is a forward diffusion equation; at $k=0$ the exponential transformation is undefined, while negative $k$ gives a backward heat equation.