Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 8

For a second-order linear partial differential equation with principal part $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}$, the discriminant is $B^2-AC$. Here $A=1$, $B=0$, $C=-x^2$, soThe equation is therefore a hyperbolic partial differential equation throughout its domain. The lower-order term $-u_x/x$ does not affect this classification.

The characteristic curves satisfyIntegrating gives the two families $y\mp x^2/2=\text{constant}$. Choose coordinates constant along these families:Their Jacobian determinant is $8x\ne 0$, so they define an invertible change of variables on $x>0$, with image $\xi +\eta >0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesBoth the pure second partial derivatives and the first partial derivatives cancel in the original equation, leavingThis is the full canonical form of a second-order partial differential equation, including the effect of the lower-order term. Integrating shows $U_{\xi }$ is a function of $\xi$ alone, hencewith arbitrary twice differentiable functions $f,g$. The representation is global on $x>0$: for each fixed $\xi$, the allowed $\eta >-\xi$ form an interval, so $U_{\xi }$ is independent of $\eta$ throughout that interval; after subtracting its antiderivative, the same reasoning in the other coordinate leaves only $g(\eta )$. These are quadratic characteristic coordinates for a variable-speed wave equation.

For the data on $x=1$, set $s=1-2y$, so the other argument is $2-s$. The two initial conditions becomeDifferentiating the first equation with respect to $s$ also gives $f^{\prime }(s)-g^{\prime }(2-s)=0$. Solving these two linear equations yields $f^{\prime }(s)=g^{\prime }(2-s)=1/2$. Since $s$ ranges over all real numbers,ConsequentlyIndeed $u_{xx}=2$, $u_{yy}=0$, $u_x=2x$, so the equation and both initial conditions hold directly.