Past exam of the differential equations 1 course of the University of Oxford / 2021 / 1 / b

Parametrize the initial curve by $(x_0(r),y_0(r))=(1,r)$, $-1\le r\le 1$. Its tangent is $(0,1)$, whereas the projected characteristic vector field there is $(1,2r+\beta )$. The determinantproves that these are Noncharacteristic Cauchy data: the initial curve is everywhere transverse to the characteristic curves.

There is a regularity qualification. The prescribed function $g(r)=\min \{r,0\}$ is continuous and Lipschitz continuous, but is not differentiable at $r=0$. Thus the data are Cauchy in this geometric sense, with data given by a smooth function on each of $r<0$ and $r>0$; the usual theorem giving a global classical solution from continuously differentiable data does not apply across $r=0$.

Use the initial conditions $x(0,r)=1$, $y(0,r)=r$, $u(0,r)=g(r)$ in the characteristic equations for a transport equation. The resulting parametrized solution surface isThe Jacobian determinant of $(s,r)\mapsto (x,y)$ is $e^{3s}>0$. In fact the inverse is explicit:Consequently the explicit solution of a differential equation on the region reached from the data isTo verify it away from $q=0$, set $D=x{\partial }_x+(2y+\beta x){\partial }_y$. Direct differentiation gives $Dq=0$ and $D(x-1+2\log x)=x+2$. At $x=1$ we have $q=y$, giving precisely the prescribed boundary condition.

Across $q=0$ the formula remains continuous and locally Lipschitz, is a smooth function on each side, and solves the partial differential equation along every characteristic curve. It is also a weak solution, since its first distributional derivatives agree almost everywhere with its ordinary derivatives. It is not a classical solution throughout the region because of the inherited corner.

Every point with $x>0$ has exactly one backward characteristic curve meeting $x=1$, at the point $(1,q(x,y))$. It meets the specified data precisely when $-1\le q\le 1$. Thus the whole region determined by these data, including the two bounding characteristic curves, isIts interior is the corresponding open domain for the partial differential equation. Uniqueness here means uniqueness in the class of continuous functions satisfying the characteristic equations for a transport equation, including the piecewise smooth function defining the solution of a differential equation above. The endpoints of the initial segment determine the two boundary values by continuity.

This region is maximal for determination by the given data: on $x>0$, arbitrary extensions of $g$ outside $[-1,1]$ give different values on the characteristic curves with $q\notin [-1,1]$. No characteristic curve from $x=1$ reaches $x\le 0$.

For $\beta =1$,The lower boundary is a straight line, the upper boundary is a parabola, and the characteristic carrying the corner is $y=x^2-x$. All three approach the excluded origin as $x\downarrow 0$.

For $\beta =1$, $q=(y+x-x^2)/x^2$. On the two sides of the corner-carrying characteristic curve, the partial derivative isAt every point $(x,x^2-x)$ with $x>0$, the two one-sided partial derivatives are $x^{-2}$ and $0$. Therefore $u_y$ does not exist on this curve; its piecewise values have a jump discontinuity of magnitude $x^{-2}$. There are no other interior discontinuities. The values on the outer boundaries are interpreted from within ${\mathcal{D}}_1$.

This is exactly propagation of a derivative jump along a characteristic. Already on $x=1$, the data have slopes $1$ and $0$ on the two sides of $y=0$. The transport equation transports this corner along the characteristic curve from $(1,0)$, which is $y=x^2-x$; it provides no smoothing. This is consistent with Noncharacteristic Cauchy data and does not contradict the smooth-data theorem, whose differentiability hypothesis fails here.