Past exam of the differential equations and applications course of the University of Oxford / 2008 / B

The square-root reduction of a quadratic eikonal equation explains why two initial normal derivatives are possible. Set $F=p^2+q^2-4u$. Charpit's equations are $\dot{x}=2p$, $\dot{y}=2q$, $\dot{u}=2p^2+2q^2=8u$, $\dot{p}=4p$, $\dot{q}=4q$. Parametrize the initial semicircle by $(x_0,y_0)=(\cos s,\sin s)$, $-\pi /2<s<\pi /2$. Tangential compatibility gives $-p_0\sin s+q_0\cos s=0$, and $F=0$ gives $p_0^2+q_0^2=4$. Thus there are two normal-gradient choices, $p_0=2\epsilon \cos s$, $q_0=2\epsilon \sin s$, $\epsilon =\pm 1$.

Integration gives the requested parametric solutionsFor $\epsilon =+1$, $r=\sqrt{x^2+y^2}=e^{4t}$, so $u_{+}=r^2$ on the whole right half-plane. For $\epsilon =-1$, before focusing at the origin, $r=2-e^{4t}$, so $u_{-}=(2-r{)}^2$ on $x>0$, $0<r<2$. The origin is a caustic for this inward family. The circle $r=2$ is approached only as characteristic time tends to negative infinity, where $u$ and its gradient vanish. It cannot be crossed by this initial characteristic chart. Thus the initial data alone permit two branches; each is locally unique only after its normal-gradient sign is chosen.

On the added half-line write $x_0=a\le 0$, $y_0=1$, $u_0=1$. Tangential differentiation gives $p_0=0$ and $q_0=2\delta$, $\delta =\pm 1$. Its characteristics giveThe plus branch is $u=y^2$ on $x\le 0,y>0$; the minus branch is $u=(2-y{)}^2$ on $x\le 0,y<2$. On the positive $y$-axis, the semicircle traces are $y^2$ and $(2-y{)}^2$, respectively. Therefore only matching sign choices are continuous throughout their common positive-axis coverage. Mixed choices agree merely at the initial junction $y=1$.

For the matching-plus branch the data-covered open domain isFor the matching-minus branch it isIts formula is $(2-\sqrt{x^2+y^2}{)}^2$ in the right portion and $(2-y{)}^2$ in the left portion. The origin is excluded. The negative $y$-axis must also be a cut: the two formulas have different traces there, so including it would destroy continuity. Both matching solutions are $C^1$ across the included positive-axis seam; generally their second transverse derivatives differ, which is compatible with this first-order problem. Values beyond these characteristic charts can sometimes be continued, but are not selected by the initial data through a nonsingular characteristic graph.

With $A$ invertible, write the system as ${\mathbf{u}}_x+M{\mathbf{u}}_y=\mathbf{d}$, $M=A^{-1}B$. Real distinct eigenvalues of $M$ give strict hyperbolicity; a real complete eigenbasis gives the usual diagonalizable hyperbolic form. The characteristic curves satisfy $dy/dx={\lambda }_i$. Left eigenvectors form characteristic dependent combinations $w_i={\ell }_i\mathbf{u}$; for constant coefficients they obey separate transport equations. Characteristic coordinates are functions constant on those curves, such as $y-{\lambda }_{i}x$ in the constant case. For variable coefficients extra derivatives of the eigenvectors can couple the combinations; local characteristic directions still follow the real matrix pencil. If $A$ is singular one first chooses a noncharacteristic coordinate direction with an invertible coefficient combination.

For the specified constants,Its speeds are $2$ and $-1$, with left eigenvectors $(2,1)$ and $(1,1)$. HenceThe transported Riemann invariants are $\boxed{R_{+}=2u+v-4x,R_{-}=u+v-x}$, constant on $y-2x=\text{constant}$ and $y+x=\text{constant}$, respectively. Initial data give $w_{+}(0,s)=s-\sin (s/2)$ and $w_{-}(0,s)=-e^s$, soInvert these two combinations:Both characteristic feet must lie on the supplied positive initial half-line. ThusOutside this sector at least one invariant needs unsupplied initial data. The closed expressions give one possible extension there but are not uniquely forced by the question's initial half-line.

For fixed source $(\xi ,\eta )\in D$, define the Dirichlet Green function by $-{\Delta }_{x,y}G={\delta }_{(\xi ,\eta )}$ in $D$, $G=0$ on $\partial D$. It is harmonic away from the source and has singularity $-(2\pi {)}^{-1}\log \sqrt{(x-\xi {)}^2+(y-\eta {)}^2}$. The outward normal vector derivative is ${\partial }_{n}G=\mathbf{n}\cdot {\nabla }_{x,y}G$ at a boundary point of arc-length parameter $s$.

Green's second identity gives ${\int }_D(u\Delta G-G\Delta u)={\int }_{\partial D}(u{\partial }_{n}G-G{\partial }_{n}u)ds$. Substituting the two PDEs and the boundary data yieldsThe singular version follows by excising a small source disk and taking its radius to zero, or directly by the distributional identity.

For the upper half-plane use the opposite-sign image source at $(\xi ,-\eta )$:The image lies outside the domain, so it is harmonic inside; the true source has the required unit singularity, and the two distances agree at $y=0$, giving zero boundary value. On that boundary $\mathbf{n}=-{\mathbf{e}}_y$, soThe assumed far-field decay makes the large semicircle contribution vanish. With $f=0$, the representation becomesFinally set $s=\xi +\eta \tan \theta$, $ds=\eta {\sec }^2\theta d\theta$. The Jacobian matrix cancels the denominator, givingIf $g$ is even, the substitution $s\mapsto -s$ in the Poisson integral gives $u(-\xi ,\eta )=u(\xi ,\eta )$. At $\eta =0$ this agrees with the continuous even boundary trace.

At a positive interior maximum, $u_x=0$, $u_{xx}\le 0$, and $u_t\ge 0$ if the point is compared with its past. The PDE instead gives $u_t\le -u^3<0$, a contradiction. Thus a positive global/past-strip maximum must come from the spatial boundary or initial-time limit. The far value is positive, so any attained global maximum is positive; the argument does not prohibit a negative local maximum in isolation.

For uniqueness put $w=u_1-u_2$. It satisfiesThe parabolic comparison principle with zero difference data gives $w\le 0$ and then $-w\le 0$, hence $w=0$. As usual on this noncompact singular-time domain, this uses an admissible comparison class in which the initial and far-field limits justify truncation; for example bounded/subquadratic differences on finite past strips with the corresponding uniform zero traces suffice. A barrier $\delta (1+x^2+2t)$ controls spatial infinity, since ${\partial }_t-{\partial }_x^2+[x/(2t)]{\partial }_x+a$ applied to it is nonnegative. The printed fixed-point limits alone should not silently be substituted for all uniform/growth hypotheses of this comparison argument.

Use drift cancellation in a parabolic similarity reduction. For the similarity ansatz, derivative scaling givesThus $\boxed{\beta =1/2}$, the drift terms cancel, and $f^{\prime \prime }=f^3-f/2$ with $f(\infty )=1/\sqrt{2}$. Use the positive-equilibrium separatrix for a cubic profile equation. Multiply by $f^{\prime }$ and use the limiting equilibrium to determine the integration constant:The branch approaching the positive far equilibrium satisfies $f^{\prime }=(1/2-f^2)/\sqrt{2}$; the constant equilibrium is included as well. Separation givesFor a finite boundary value at $x=0$ and no interior pole, take $A>-1$. Values $-1<A<0$ describe the corresponding coth-type approach from above; $A=0$ is constant, and positive $A$ gives the tanh-type branch. The other first-integral sign cannot approach the positive equilibrium nontrivially as $\eta \to \infty$.

The requested deficit integral is, after $x=\sqrt{t}\eta$,Therefore $\boxed{A=1}$, and the selected solution is $\boxed{u(x,t)=(2t{)}^{-1/2}\tanh [x/(2\sqrt{t})]}$.