Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 8 / Solution

First consider the usual local reduction with sufficiently smooth coefficients. The hyperbolic partial differential equation has two distinct characteristic directions because $S^2-RT>0$. Choose independent local characteristic coordinates $\xi ,\eta$ whose gradients satisfyLet $J={\xi }_x{\eta }_y-{\xi }_y{\eta }_x\ne 0$ andThe determinant of the transformed principal coefficient matrix gives $B^2=J^2(S^2-RT)>0$. By the chain rule, the left side of the original partial differential equation becomeswith the analogous expression for $A_{\eta }$. The original first derivatives are $z_x=u_{\xi }{\xi }_x+u_{\eta }{\eta }_x$ and $z_y=u_{\xi }{\xi }_y+u_{\eta }{\eta }_y$. Substitution and division by $2B$ yield the requested mixed-derivative canonical form. The construction is local, in a neighborhood where the two characteristic foliations give a diffeomorphism.

There is a regularity qualification in the printed hypotheses: merely continuous coefficients need not give twice continuously differentiable characteristic coordinates. Such coordinates are needed for the second-derivative chain rule above. For example, take $R=0$, $S=1/2$ and $T=|x|$ near the origin. The characteristic equation factors as ${\xi }_y({\xi }_x+|x|{\xi }_y)=0$. One family is $x=\text{constant}$ and the other has first integrals $\eta =h(y-x|x|/2)$. For an independent coordinate $h^{\prime }\ne 0$, but ${\eta }_{xx}$ has opposite one-sided limits at $x=0$. Hence no twice continuously differentiable characteristic coordinate system exists there. This illustrates that continuous hyperbolic coefficients need not admit smooth characteristic coordinates. The formal general argument requires additional smoothness, for example twice continuously differentiable coefficients. The concrete coefficients below are smooth, so this issue does not affect their solution.

For the concrete partial differential equation, putThe Jacobian determinant is $1/y^2\ne 0$. Direct use of the chain rule givesThereforeTo impose the data efficiently, let $v=z_x$. Then $2x{v}_x+y{v}_y=0$, whose characteristic curves preserve $x/y^2$, so $v=H(x/y^2)$. On the initial line $x=1$, the value $v(1,y)=y^2$ gives $H(\xi )=1/\xi$ for every $\xi >0$. Thus $z_x=y^2/x$, and integration in $x$ gives $z=y^2\log x+C(y)$. The remaining initial condition gives $C(y)=0$:Every point in this region is reached from the data by a characteristic with positive $x/y^2$. The line $x=0$ separates these characteristics from those with $x<0$, and the logarithmic divergence prevents a smooth extension of this solution through $x=0$. The original condition restricts us to $y>0$.