Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 14 / a

To distinguish the old and new coordinate descriptions, write $u(x,y)=U(\xi (x,y),\eta (x,y))$. Since ${\xi }_x={\eta }_x=1$, the multivariable chain rule givesEquivalently, the differential operator is ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$.

The coordinate partial derivatives are ${\xi }_y=1/\sqrt{y}$ and ${\eta }_y=-1/\sqrt{y}$. The multivariable chain rule therefore givesSince $\xi -\eta =4\sqrt{y}>0$, the expression entirely in the new coordinates isThe inverse coordinate formulas are $x=(\xi +\eta )/2$ and $y=(\xi -\eta {)}^2/16$, with domain $\xi >\eta$.

Apply ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ once more. The chain rule and symmetry of second derivatives giveHere and in the next part the second partial derivatives are understood with the usual smoothness required for this partial differential equation.

Differentiate $u_y=y^{-1/2}(U_{\xi }-U_{\eta })$, retaining the derivative of the prefactor. The product rule and multivariable chain rule giveUsing $y=(\xi -\eta {)}^2/16$, this becomesThe last term is essential because the coordinate change is nonlinear.