Past exam of the differential equations and applications course of the University of Oxford / 2009 / A

Define the auxiliary variable $y=ϵ\dot{x}-x+x^3/3$. Differentiating gives $\dot{y}=ϵ\ddot{x}+(x^2-1)\dot{x}=-x$, while rearrangement gives $ϵ\dot{x}=x-x^3/3+y$. Conversely eliminating $y$ recovers the second-order equation, so this phase plane system is equivalent. At an equilibrium $\dot{y}=0$ forces $x=0$, and the first equation then forces $y=0$. Its Jacobian matrix isThus $ϵ{\lambda }^2-\lambda +1=0$. Expanding $\sqrt{1-4ϵ}=1-2ϵ-2{ϵ}^2-4{ϵ}^3+O({ϵ}^4)$ givesFor sufficiently small positive $ϵ$ both are positive: the equilibrium is an unstable node. For negative $ϵ$ the fast root is negative and the slow root positive, so it becomes a saddle equilibrium.

The nullcline is $\boxed{y=h(x)=x^3/3-x}$. Write $F=x-x^3/3+y$. Away from it at an order-one distance, $F$ is order one on the indicated finite phase-space scales. Thus $\dot{x}=O({ϵ}^{-1})$ while $\dot{y}=-x=O(1)$: the trajectory changes $x$ on the fast time in a singular perturbation $t/ϵ$, with only an $O(ϵ)$ change in $y$. Equivalently $dy/dx=-ϵx/F=O(ϵ)$, which explains the nearly horizontal motion.

For positive $ϵ$, $F_x=1-x^2$ makes the branches $|x|>1$ fast attracting and the middle branch $|x|<1$ fast repelling. The arrows point right where $y>h(x)$ and left where $y<h(x)$, carrying a generic off-nullcline point towards an attracting branch. The reduced description changes near a fold, where $F_x=0$; the rapid/slow distinction there needs a separate transition layer.

The left panel shows positive $ϵ$; the right panel shows negative $ϵ$, including the three specified starting points. Horizontal arrows indicate the leading fast motion.

On the attracting right branch, $x>1$ and $\dot{y}=-x<0$, so the motion goes from $(2,2/3)$ towards the fold $(1,-2/3)$. At essentially fixed $y=-2/3$ it jumps to the other attracting root $x=-2$. The left branch then carries it to $(-1,2/3)$, where it jumps right to $x=2$. This gives the clockwise limiting relaxation oscillation shown in the preceding figure.

Along a slow branch $y=h(x)$, differentiation gives $(x^2-1)\dot{x}=-x$. The right-branch time isReflection symmetry gives the same time on the left. The fast jumps take a time tending to zero as $ϵ\to 0^{+}$, as do the fold-transition corrections at leading order. HenceThis is a leading asymptotic period, not an exact finite-$ϵ$ equality.

For negative $ϵ$ the fast arrows reverse: the outer branches become repelling and the middle branch attracting. On the middle branch the slow equation is $\dot{x}=x/(1-x^2)$, so the slow motion goes away from the saddle equilibrium at zero towards a fold.

At $(3,1)$, $F=3-9+1=-5$, so $\dot{x}>0$: the trajectory immediately escapes to the right. At $(1,-2)$, $F=1-1/3-2=-4/3$, so it also moves rapidly right and escapes. In each case the large-$x$ balance is $\dot{x}\sim x^3/(3|ϵ|)$, with $y$ decreasing; it is not a bounded relaxation cycle.

At $(-1,-1/3)$, $F=-1$, so the initial fast motion is rightward. At nearly fixed $y=-1/3$ it approaches the attracting middle root of $x-x^3/3=1/3$, approximately $x=0.347$. It then moves slowly down/right along the middle branch towards $(1,-2/3)$, leaves that fold to the right, and eventually escapes. These are the small-negative-$ϵ$ limiting descriptions; the fold locations and fast landing point receive small corrections. The right panel of the figure distinguishes this slow excursion from the two immediate escapes.

Put $y=e^{x}w$. Then $y^{\prime \prime }-2y^{\prime }+y=e^x{w}^{\prime \prime }$ and the endpoint conditions become $w(0)=w(\pi )=0$. The transformed eigenvalue equation is $-w^{\prime \prime }=\lambda w$, soZero and negative $\lambda$ yield no nonzero solution with both endpoint values zero.

For the usual unweighted $L^2(0,\pi )$ inner product, the operator $D^2-2D+1$ is not self-adjoint: its formal adjoint is $D^2+2D+1$. The mismatch is already visible on compactly supported test functions, so endpoint conditions cannot remove it. With the different weight $e^{-2x}$ one does obtain a symmetric Sturm-Liouville form, since $e^{-2x}(y^{\prime \prime }-2y^{\prime }+y)=(e^{-2x}{y}^{\prime }{)}^{\prime }+e^{-2x}y$. Thus the inner-product convention matters; the real spectrum alone does not prove self-adjointness in the unweighted space.

Use the complex $L^2$ inner product $⟨f,g⟩={\int }_0^{\pi }\overline{f}g$, linear in its second argument. Exchanging the order of integration givesSince the kernel is real, the adjoint operator is thereforeor, renaming variables, $L^{\ast }f(x)={\int }_0^{\pi }k(t,x)f(t)dt$. Continuity of the kernel on the compact integration square justifies these integrals and the exchange.

Symmetry $k(x,t)=k(t,x)$ makes the adjoint kernel equal to the original kernel, so $L^{\ast }=L$. For a nonzero eigenfunction $Ly=\lambda y$,Thus $\boxed{\lambda \in \mathbb{R}}$. The argument includes the zero eigenvalue and does not require normalizing the eigenfunction.

Antisymmetry $k(x,t)=-k(t,x)$ instead gives $L^{\ast }=-L$, so $L$ is skew-adjoint. Its quadratic form satisfies $⟨y,Ly⟩=-\overline{⟨y,Ly⟩}$. Hence an eigenpair satisfiesNonzero eigenvalues are purely imaginary in the complexified function space; zero is allowed as well. On a strictly real vector space this means there are no nonzero real eigenvalues.

Write $c(x)=\cos 2x$. The separated kernel givesThus its image is $E=\mathrm{span}\{1,c\}$. Since $L1=\pi c$ and $Lc=\pi /2$, its matrix on this basis is $\left(\begin{array}{cc}0& \pi /2\\ \pi & 0\end{array}\right)$. ThereforeBoth eigenspaces have dimension one, and there are no further nonzero eigenvalues because every such eigenfunction belongs to the image. The zero eigenspace is $E^{\perp }$, which is infinite-dimensional. One explicit nonzero example is $\boxed{y(x)=\sin 2x}$, orthogonal to both $1$ and $\cos 2x$.

The image of $L$ consists only of $a+b\cos 2x$. If $\sin x$ were such a function, its endpoint values would give $a+b=0$ and its midpoint value $a-b=1$, forcing $a=1/2$, $b=-1/2$. That combination is ${\sin }^{2}x$, which differs from $\sin x$, for example at $x=\pi /6$. Continuity makes the same argument valid for equality almost everywhere. Thus the forcing is outside the image and no solution exists.

Here ${\sin }^{2}x=(1-\cos 2x)/2$ belongs to the image. Matching the two coefficients using $L(A+B\cos 2x)=B\pi /2+A\pi \cos 2x$ gives $A=-1/(2\pi )$, $B=1/\pi$. ConsequentlyThere are infinitely many solutions, since any element of the infinite-dimensional kernel may be added.

The eigenvalues of $L$ are $\pi /\sqrt{2}$, $-\pi /\sqrt{2}$ and zero, none equal to one. On $E=\mathrm{span}\{1,\cos 2x\}$ the map $L-I$ is an invertible two-dimensional operator; on $E^{\perp }$ it is $-I$. Hence it is invertible on all of $L^2(0,\pi )$. Thus $(L-I)y=\sin x$ has exactly one solution. This is the finite-rank Fredholm alternative; no range obstruction survives the nonresonant identity term.

The quantity inside a fixed interval changes by influx at its left endpoint minus outflux at its right endpoint:In a smooth region the fundamental theorem of calculus turns this into ${\int }_a^b[{\rho }_t+{\partial }_{x}q(\rho )]dx=0$. Since the interval is arbitrary, the local conservation law is $\boxed{{\rho }_t+{\partial }_{x}q(\rho )=0}$.

For a jump at $x=S(t)$, write the two smooth pieces using a step function. The singular part of ${\rho }_t$ is $-S^{\prime }[\rho ]\delta (x-S)$ and that of ${\partial }_{x}q$ is $[q]\delta (x-S)$. Cancellation gives the Rankine-Hugoniot conditionThe same identity follows by applying conservation to a shrinking moving control interval across the shock.

In smooth regions the characteristic equations are $\dot{x}=q^{\prime }(\rho )={\rho }^3$ and $\dot{\rho }=0$. If $\xi$ labels the initial point, their parametrization isFor the two constant states, characteristics from the left have speed one and those from the right speed zero; they would overlap. The entropy solution replaces the overlap by a compressive shock. Its speed isThusThe left parametrization $x=\xi +t$ is retained only for $x<t/4$, and the right parametrization $x=\xi$ only for $x>t/4$. The characteristic speeds obey $0<c<1$, so characteristics enter the shock from both sides.

Before intersection, the characteristic map is $x=\xi +{\rho }_0(\xi {)}^{3}t$, with derivative $1+3{\rho }_0^2{\rho }_0^{\prime }t$. Because the initial profile decreases, higher-density characteristics behind move faster than lower-density characteristics ahead. Compression therefore forms shocks, which absorb more characteristics. The far states remain one and zero, so a shock joining their limiting traces has speed tending to $[q]/[\rho ]=1/4$.

The precise large-scale interpretation uses $\zeta =x/t$. The rescaled initial transition tends to a step, and entropy comparison for the convex flux givesOne way to see the comparison is to choose distant initial points beyond which the profile is within $\delta$ of each far state. Step profiles with those nearby states bracket it; their shock speeds tend to $1/4$ as $\delta \downarrow 0$, while their fixed initial locations disappear after division by $t$.

A fixed translation of the shock can persist. If the deviation from a reference step is integrable, its signed mass fixes a phase shift. Without such a tail assumption one should assert convergence on the $x/t$ scale and $S(t)/t\to 1/4$, rather than claim that the exact line $x=t/4$ is the unshifted asymptote for every smooth initial transition.

Mass preservation of a nonnegative finite-mass profile requires $\alpha +\beta =0$. Balancing ${\rho }_t$ and ${\partial }_x({\rho }^4/4)$ requires $\alpha -1=4\alpha -\beta$. ThereforeThe similarity ODE isFor a source-type profile adjoining vacuum, the integration constant is zero. The positive branch is $f={\eta }^{1/3}$ for positive $\eta$, terminated by a shock into zero. Normalize the mass:Thus a unit-mass similarity solution isAt the moving right edge, $S^{\prime }={\eta }_F{t}^{-3/4}/4$ equals ${\rho }_L^3/4$, so the Rankine-Hugoniot condition is satisfied. The characteristic speeds satisfy $0<S^{\prime }<{\rho }_L^3$, making it an entropy shock. The left edge meets vacuum continuously. As $t\downarrow 0$ the mass concentrates at the origin; this is a source solution, not a claim that every finite-mass initial profile equals it at finite time.