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

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 $t>0$.