Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 8C (Differential Equations)

For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.