Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 2 / 6B (Differential Equations)

The chain rule gives ${\partial }_x={\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y=\alpha {\partial }_{\xi }+{\partial }_{\eta }$. The coefficients of $u_{\xi \xi }$ and $u_{\eta \eta }$ are $2+3{\alpha }^2-7\alpha$ and $2{\beta }^2+3-7\beta$. Their integer roots are $\alpha =2$ and $\beta =3$. Thus choose characteristic coordinates $\xi =x+2y$, $\eta =3x+y$, whose Jacobian determinant is $-5\ne 0$. The coefficient of $u_{\xi \eta }$ is $4\beta +6\alpha -7(1+\alpha \beta )=-25$, so the equation becomes $-25u_{\xi \eta }=0$, equivalently $u_{\xi \eta }=0$.

Integrating the transformed hyperbolic partial differential equation gives $u=F(\xi )+G(\eta )$. Along $x=-2y$ the coordinates are $(0,-5y)$, so the zero boundary condition gives $G(\eta )=-F(0)$ for every real $\eta$. Along $x=0$ they are $(2y,y)$, so $F(2y)-F(0)=4y^2$, hence $F(\xi )={\xi }^2+F(0)$. The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and $2u_{xx}+3u_{yy}-7u_{xy}=4+24-28=0$.