Past exam of the differential equations and applications course of the University of Oxford / 2010 / B

Use the initial-curve parameter $s$ and a characteristic parameter $t$. Charpit's equations areAt $t=0$ prescribe $x=x_0(s)$, $y=y_0(s)$ and $u=u_0(s)$. The two initial derivatives $p_0,q_0$ must satisfyThe second equation is the chain rule along the initial curve. For the usual local Cauchy construction also require the non-characteristic condition $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It ensures both local determination of an allowed branch of $p_0,q_0$ and local inversion of the projected characteristic coordinates.

Along a characteristic, the chain rule and the five equations giveSince $F=0$ initially, $\boxed{F\equiv 0}$ along each constructed curve. With the contact compatibility $du=pdx+qdy$ propagated by this system, an invertible characteristic chart reconstructs $p=u_x,q=u_y$ and hence a local classical solution.

Set $F=xp+q^2/4+u-1$. Parametrize the data by $x_0=s$, $y_0=0$, $u_0=1-s^2/3$. Tangential differentiation gives $p_0=-2s/3$; substitution into $F=0$ then gives $q_0^2=4s^2$. Thus the data allow two normal-derivative branches, indexed by $\epsilon =\pm 1$, with $q_0=2\epsilon s$. For $s>0$ each branch is non-characteristic; the endpoint $s=0$ is degenerate.

The Charpit's equations reduce to $\dot{x}=x$, $\dot{y}=q/2$, $\dot{p}=-2p$, $\dot{q}=-q$ and $\dot{u}=xp+q^2/2$. Integrating with these initial values givesDirect substitution checks $xp+q^2/4+u-1=0$, and $t=0$ recovers the initial data. The projection onto the $(x,y)$ plane has Jacobian determinantIt first vanishes at $t=\log 2$. These points form the characteristic caustic $y=\epsilon x/4$, $0<x\le 2$; a single classical solution cannot in general be continued through the fold by treating all projected characteristic curves as a graph.

For a precise description put $r=\epsilon y$. On the chart containing the initial curve and lying before the fold, $t<\log 2$, elimination givesThe plus square-root is forced by $s/x>1/2$, rather than a free second choice. The conditions $0<s\le 1$ give the complete unfolded chartInterior points of the domain have strict inequality at the boundary generated by $s=1$ as well. The other boundary is the characteristic caustic; the original endpoint $s=0$ contributes no open chart. On the portion obtained by following characteristics forward from the data, $0\le t<\log 2$, the region is more specificallyThe $s=1$ boundary is $y=\epsilon (1-1/x)$ and meets the characteristic caustic at $(2,\epsilon /2)$. Negative characteristic times give the continuation across the initial segment into the opposite half-plane. Each sign branch is a distinct local solution; the initial values alone do not select one of them.

The solid shading follows $t\ge 0$; the lighter shading shows the same smooth chart at $t<0$. Reflected panels correspond to the two allowed signs of the initial normal derivative. Beyond the dashed characteristic caustic the characteristic parametrization no longer gives this smooth graph.

Write the system as $U_x+A(U)U_y=0$ with $U=(u,v{)}^T$ andIts two characteristic speeds are ${\lambda }_{\pm }=v\pm \sqrt{u}$, so it is strictly hyperbolic precisely when $u>0$. At $u=0$ they coincide and the matrix is not diagonalizable; for $u<0$ the speeds are nonreal.

For $u>0$ define $R_{\pm }=v\pm 2\sqrt{u}$ and $D_{\pm }={\partial }_x+(v\pm \sqrt{u}){\partial }_y$. Substitution from the two differential equations givesThereforeThese are the two Riemann invariants; their constancy applies within smooth regions, not across a shock.

For a smooth solution, the first equation is $u_x+(uv{)}_y=0$. Multiplying it by $v$ and adding $u$ times the second equation givesThus the appropriate shallow water equations conservation variables and flux areUse this conservative extension to define a weak solution across $y=F(x)$. A jump contributes $-F^{\prime }[W]\delta (y-F(x))$ to $W_x$ and $[H]\delta (y-F(x))$ to $H_y$. Cancellation of the singular parts gives the Rankine-Hugoniot conditionsEquivalently this follows by integrating each conservation law over a thin space-time control region crossing the discontinuity.

The conservative extension is essential here. Piecewise smoothness by itself does not imply any jump law. Moreover the smooth second equation can also be written $v_x+(u+v^2/2{)}_y=0$; imposing that different conservation law distributionally would give a different shock condition. Weak conservation laws depend on conserved variables. The requested jump law specifically selects mass $u$ and momentum $uv$, and this is the extension used below.

Test a single discontinuity connecting the two constant initial states. With state 1 below the discontinuity and state 2 above it, their jumps areThe first Rankine-Hugoniot condition gives $F^{\prime }=\sqrt{3}$, and the second independently gives $(-\sqrt{3})F^{\prime }=-3$, the same value. Since the initial jump is at the origin,Values on the shock itself do not affect the weak solution. No intermediate state or rarefaction is needed: the prescribed states already lie on the same compressive shock curve.

For admissibility, the upper-family characteristic speed is $\sqrt{3}/2+\sqrt{2}$ behind the shock and $1$ ahead of it. They satisfyBoth lower-family speeds, $-1$ and $\sqrt{3}/2-\sqrt{2}$, are less than the shock speed. These are the Lax shock admissibility conditions for a family-plus shock. This is the entropy-admissible solution for the mass/momentum conservative extension specified by the preceding jump law; imposing an additional velocity conservation law across the same jump would describe a different problem.

For $\eta =x/y^{\beta }$ the chain rule gives $u_y=-\beta \eta f^{\prime }/y$, $u_x=y^{-\beta }{f}^{\prime }$ and $(u{u}_x{)}_x=y^{-2\beta }[f{f}^{\prime \prime }+(f^{\prime }{)}^2]$. Multiplication of the equation by $y$ givesA nontrivial scale-invariant balance therefore selects $\boxed{\beta =1/2}$ and yieldsIts left side is the derivative of $f{f}^{\prime }+\eta f/2$. The boundary conditions give $f(0)=1$ and $f^{\prime }(0)=0$, so the integration constant is zero:On the positive branch before the first zero, divide by $f$ and integrate, obtaining $f=1-{\eta }^2/4$. Its first positive zero is ${\eta }_F=2$. HenceA direct check gives $u_y=x^2/(4y^2)$ and $(u{u}_x{)}_x+u/(2y)=x^2/(4y^2)$. The zero extension past this front is also a natural weak compact-support continuation: $u$ is continuous and the degenerate diffusive flux $u{u}_x$ tends to zero at the front. This is a source-driven porous-medium similarity front.

Assume a classical solution $u\in C^2(D)\cap C(\overline{D})$, as needed to interpret the boundary values. At an interior local minimum its Hessian is positive semidefinite, hence $\Delta u\ge 0$. But the equation gives $\Delta u=-e^{-x^2}<0$, a contradiction. Since $D$ is bounded and $u=0$ on its boundary, a nonpositive interior value would either furnish an interior negative minimum or itself be a zero minimum. Thus $\boxed{u>0\text{ in }D}$.

For the upper bound use the barrier $w(x,y)=(1-x^2-y^2)/4$, which satisfies $-\Delta w=1$. Set $z=w-u$. On the boundary $z=w\ge 0$, because the domain lies inside the unit disk, whileThe quadratic perturbation proof of the minimum principle gives $z\ge 0$. The strong minimum principle for elliptic operators makes this strict in the connected interior unless $z$ is constant. Constancy is impossible since $1-e^{-x^2}$ is not identically zero on any open domain. HenceThis last strong-principle step handles points with $x=0$, where the comparison forcing itself vanishes. Simple connectedness is not needed for the comparison argument.

With source point $\xi \in D$ and field point $X=(x,y)$, the Dirichlet Green function for the operator $-\Delta$ satisfiesClassically it is harmonic off $\xi$, with the two-dimensional normalization $G(X,\xi )=-(2\pi {)}^{-1}\log |X-\xi |+H(X,\xi )$, where $H$ is regular near the source. Equivalently $-{\lim }_{r\downarrow 0}{\int }_{|X-\xi |=r}{\partial }_{r}Gds=1$. This sign convention gives the representation $u(X)={\int }_{D}G(X,\xi )e^{-{\xi }_1^2}d\xi$.