Past exam of the differential equations and applications course of the University of Oxford / 2008 (B5, DMAB 2644, DMST 2644)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

Set $y=\dot{x}$. The phase-plane system is $\dot{x}=y$, $\dot{y}=-9x+ϵ(1-4x^2)y$. The origin is its only equilibrium, and its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -9& ϵ\end{array}\right)$. ThusFor the stated small positive $ϵ$, the eigenvalues have positive real part and nonzero imaginary part: the origin is an unstable focus.

Use the Poincaré-Lindstedt method. Introduce $\theta =\Omega t$ and $x=x_0(\theta )+ϵx_1(\theta )+\cdots$. The leading equation gives ${\omega }_0=3$ and $x_0=A_0\cos \theta$, after fixing the arbitrary phase. At order $ϵ$,Here $4{\cos }^2\theta \sin \theta =\sin \theta +\sin 3\theta$. Periodicity requires the two resonant coefficients to vanish. With $A_0>0$, this givesTherefore the leading limit cycle is $x(t)=\cos (3t)+O(ϵ)$, with frequency $3+O({ϵ}^2)$. For example the nonresonant first correction contains $-ϵ\sin (9t)/24$; its homogeneous sine/cosine pieces depend on phase and amplitude conventions. The absence of a first-order frequency shift is a solvability condition, not an assumption.

Use the method of multiple scales with independent fast and slow variables $t$ and $\tau =ϵt$, and write $x_0=R(\tau )\cos \Theta$, $\Theta =3t+\Phi (\tau )$. At the next order, $x_{1,tt}+9x_1=-2x_{0,t\tau }-(4x_0^2-1)x_{0,t}$. Its resonant terms areSuppressing secular terms yieldsThe leading initial data are $\boxed{R_0(0)=4,{\Phi }_0(0)=0}$, up to adding an integer multiple of $2\pi$ to the phase. Thus $x(t)=R_0(ϵt)\cos [3t+{\Phi }_0(ϵt)]+O(ϵ)$. The amplitude decreases from four towards the stable value one, consistent with the selected limit cycle; the zero-amplitude state is unstable. Order-$ϵ$ corrections enforce the exact initial derivative after differentiating the slow amplitude term.

Let $L=D^2+1$ on the mixed endpoint domain $y(0)=0$, $y^{\prime }(1)=0$ in $L^2(0,1)$. Integration by parts gives the boundary formAt zero, $y^{\prime }(0)$ is free, forcing $z(0)=0$; at one, $y(1)$ is free, forcing $z^{\prime }(1)=0$. Therefore the formal adjoint expression and full adjoint endpoint conditions coincide with the original ones:With the usual second-order operator domain this is a self-adjoint problem. The consequences for its eigenvalues and eigenfunctions follow below.

For $Ly=-\lambda y$, self-adjointness gives $-\lambda \Vert y{\Vert }^2=⟨y,Ly⟩=\overline{⟨y,Ly⟩}$. Since an eigenfunction is nonzero, $\boxed{\lambda \in \mathbb{R}}$. One may also see reality directly from $\lambda \int |y{|}^2=\int |y^{\prime }{|}^2-\int |y{|}^2$, whose boundary term vanishes under the mixed conditions.

If $L{y}_j=-{\lambda }_j{y}_j$, then $⟨y_1,L{y}_2⟩=⟨L{y}_1,y_2⟩$ gives $({\lambda }_1-{\lambda }_2)⟨y_1,y_2⟩=0$. Thus eigenfunctions belonging to distinct eigenvalues are orthogonal. Normalization can make them orthonormal. Equality of eigenvalues alone would not force arbitrary eigenvectors to be orthogonal, although in this particular scalar separated problem each eigenspace is one-dimensional.

The equation is $y^{\prime \prime }+(1+\lambda )y=0$. If $1+\lambda \le 0$, the linear or hyperbolic branches cannot meet both homogeneous endpoint conditions nontrivially. Otherwise $y=C\sin (kx)$, $k=\sqrt{1+\lambda }$, and $y^{\prime }(1)=Ck\cos k=0$ gives $k_n=(n+1/2)\pi$. ThereforeThe factor $\sqrt{2}$ makes the eigenfunctions unit normalized. All the eigenvalues are positive, in particular none is zero.

The difference of two solutions would satisfy $Ly=0$ with $y(0)=0$, $y^{\prime }(1)=0$. This would be a zero eigenvalue of the homogeneous problem, but part (b) shows there is none. Hence at most one solution exists. For continuous forcing the standard second-order existence construction also works: the free initial slope is fixed by an equation with coefficient $\cos 1\ne 0$. Thus for the stated regular boundary-value problem the solution is unique even with nonzero endpoint data.

Expand a forcing in the complete orthonormal basis ${\varphi }_n$. Since $L{\varphi }_n=-{\lambda }_n{\varphi }_n$, its inverse multiplies the $n$th coefficient by $-1/{\lambda }_n$. ConsequentlyThe negative sign follows from the eigenvalue convention $Ly=-\lambda y$. This kernel satisfies the mixed endpoint conditions in its first argument and $L_{x}g=\delta (x-t)$.

The integral equation says precisely $h=L^{-1}z$ with the homogeneous mixed endpoint conditions. Since the supplied $h$ satisfies those conditions, apply $L$ directly:No integration is required. The absence of a zero eigenvalue established in (b)–(c) also proves uniqueness of this recovered forcing.

Write the integral part as $Ky=12b⟨a,y⟩-30a⟨b,y⟩$, where $a(x)=e^{-x}$ and $b(x)=e^{-2x}$. The kernel is real, so its adjoint transposes the two arguments. HenceThis follows by exchanging the two integrations in the $L^2(0,\infty )$ inner product; the exponential factors make the finite-rank operators bounded.

Every eigenvector with eigenvalue different from one lies in $E=\mathrm{span}\{a,b\}$, because $(\lambda -1)y=Ky\in E$. The elementary overlaps are $⟨a,a⟩=1/2$, $⟨a,b⟩=1/3$, $⟨b,b⟩=1/4$. ThusIts characteristic polynomial is $\lambda (\lambda +4)$, so the finite-multiplicity eigenpairs areEach eigenspace has dimension one. On $E^{\perp }$, both integrals defining $K$ vanish, so $Ly=y$. Thus $\boxed{\lambda =1}$ is also an eigenvalue, of infinite multiplicity, and there are no others.

The adjoint also leaves $E$ invariant, with $L^{\ast }a=5a-15b$ and $L^{\ast }b=3a-9b$. By adjoint in weighted coordinates, its matrix in the same basis is $\left(\begin{array}{cc}5& 3\\ -15& -9\end{array}\right)$. Solving the two eigenvector equations givesThe basis $a,b$ is not orthonormal, so one must not simply transpose the previous numerical matrix without accounting for its Gram matrix. Directly transposing the integral kernel avoids that error.

The nonzero kernel of $L$ rules out a unique solution for every value of $c$. By the Fredholm alternative, solvability requires orthogonality of the forcing to $\ker L^{\ast }=\mathrm{span}\{3a-5b\}$. Its pairing with $6a+cb$ isThus only $c=-4$ is compatible. In that case $6a-4b=2y_{-4}$ and a particular solution is $-y_{-4}/2$. All solutions areTherefore infinitely many solutions exist for $c=-4$, no solution exists for $c\ne -4$, and no parameter value gives a unique solution.

Integrate the conservation law over $[-R,R]$: $d{\int }_{-R}^R\rho dx/dt=-t[{\rho }^2{]}_{-R}^R$. The Gaussian data and their finite-time characteristic/entropy evolution have vanishing far-field flux. Taking $R\to \infty$ yields $\boxed{M^{\prime }(t)=0}$ and hence $\boxed{M(t)=\sqrt{2\pi }}$. After a shock forms, its Rankine-Hugoniot condition preserves the same integral mass; the formal multivalued characteristic surface itself must be replaced by the conservative entropy solution.

Expanding the flux gives ${\rho }_t+2t\rho {\rho }_x=0$. Along characteristics, $\dot{\rho }=0$ and $\dot{x}=2t\rho$. Labeling the initial point by $\xi$ givesEquivalently, time reparametrization for a separable conservation-law flux uses $T=t^2/2$. The projection derivative is $x_{\xi }=1-t^2\xi e^{-{\xi }^2/2}$. For large enough $t$ it becomes negative on an interval of positive $\xi$, although it tends to one at both far ends. The projection then folds and assigns several density values to one position. This establishes multivaluedness of the formal characteristic continuation, not of the admissible shock solution.

The first fold occurs when $t^2{\max }_{\xi >0}\xi e^{-{\xi }^2/2}=1$. Differentiation gives $(\xi e^{-{\xi }^2/2}{)}^{\prime }=e^{-{\xi }^2/2}(1-{\xi }^2)$, so the maximum is $e^{-1/2}$ at $\xi =1$. ThereforeAt that point the first singular projection is at $x_c=1+t_c^2{e}^{-1/2}=2$, with density $e^{-1/2}$.

The two characteristic families are $x=\xi +t^2$, $\rho =1$ for left-state labels, and $x=\xi$, $\rho =0$ for right-state labels. The explicit time-dependent flux is $q=t{\rho }^2$, so the shock speed isStarting at zero givesThe characteristic speeds are $2t$ behind and zero ahead, with shock speed $t$ between them. The left parabolas and right vertical lines terminate at the compressive shock.

Try $\rho =t^{a}f(\eta )$, $\eta =x/t^b$. The boundary data force $a=3$ and $f(0)=1$. The time derivative scales as $t^{a-1}$, advection as $t^{2a-b}$ and the source as $t^{a+c}$. Balancing all three gives $c=-1$ and $b=a+1=4$. Substitution then yieldsFor the pointwise initial trace at any fixed $x>0$, the far-profile requirement is $f(\eta )=o({\eta }^{3/4})$ as $\eta \to \infty$ on any smooth branch extending that far; a compact-support continuation also meets it. Solving or selecting a shock continuation was not required. The time-dependent source is singular at $t=0$, so the similarity reduction is stated on $t>0$.

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 $\boxed{u(x,t)=(2t{)}^{-1/2}\tanh [x/(2\sqrt{t})]}$.