Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 20 / b

For $t>0$, differentiate the printed ansatz:Matching the independent coefficients in $h_t=(h{h}_x{)}_x-1$ givesThere are two parameter pairs, including the degenerate spatially constant case:Their corresponding solutions areThe request for “the general form” cannot mean the complete solution set of this nonlinear partial differential equation: substitution of one ansatz cannot provide arbitrary initial data, and the superposition principle does not apply.

A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, $h=A(t)+B(t)x^2$. Coefficient comparison gives $B^{\prime }=6B^2$ and $A^{\prime }=2AB-1$. For $B\ne 0$, write $\tau =t-t_0>0$ to obtain $B=-1/(6\tau )$ and $A=C{\tau }^{-1/3}-3\tau /4$. Spatial translation, also a symmetry of the equation, therefore givesHere $C,x_0,t_0$ are arbitrary constants; the displayed nontrivial ansatz corresponds to $C=1/6$ and $x_0=t_0=0$. There are also affine solutions $h=px+(p^2-1)t+q$, including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.