Past exam of the differential equations 1 course of the University of Oxford / 2023 / 2

An equilibrium point $p=(x_0,y_0)$ of an autonomous planar system is a point where its vector field vanishes:For a locally well-posed ordinary differential equation, Lyapunov stability means that every prescribed neighbourhood can be protected by choosing a sufficiently small initial neighbourhood. More precisely, for every $\epsilon >0$ there exists $\delta >0$ such that every trajectory starting with $\Vert p(0)-p\Vert <\delta$ exists for all $t\ge 0$ and satisfiesThe number $\delta$ may depend on $\epsilon$ but not on the particular starting point. Asymptotic stability additionally requires convergence to the equilibrium point for all sufficiently close starting points.

At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.

A characteristic curve in $(x,y,z)$ follows the vector field defining the first-order partial differential equation. Along such a curve the chain rule turns the equation intoIts characteristic projection is the curve $(x(s),y(s))$ in the independent-variable plane.

Writing $x(0)=\xi$, $y(0)=\eta$, $z(0)=\zeta$, separation of variables givesThese formulas hold on the maximal interval through $s=0$ where the denominators remain nonzero, using the real cube root. They include $\eta =0$ (a stationary characteristic projection) and $\zeta =0$ (the zero height). For $\zeta \ne 0$, the last formula follows from $d(z^{-3})/ds=1$. The alternative zero solution must be retained when deriving this by division by $z^4$.

Parametrize the data by $(\xi ,\eta ,\zeta )=(\cos \theta ,\sin \theta ,-1)$. The determinant testing Noncharacteristic Cauchy data isThus the open upper and lower semicircles are noncharacteristic, but the entire circle is not a valid noncharacteristic data curve: the determinant vanishes at $(1,0)$ and $(-1,0)$.

There is a stronger obstruction at those two points. On $y=0$, the differential equation forces $0=-z^4/3$, and hence $z=0$. This contradicts the prescribed value $z=-1$ at both endpoints. ThereforeThe obstruction is already present if the equation is imposed next to the circle and the solution and its first partial derivatives extend continuously to it.

Away from the endpoints, the characteristic projections are radial half-lines:For $\sin \theta >0$, $r$ increases with $s$; for $\sin \theta <0$, $r$ decreases with $s$. Each projection meets its own open semicircle exactly once and never crosses $y=0$. The endpoint projections themselves are single stationary points, although their lifted characteristic curves have varying height.

Use the compatible open semicircles and set $r=\sqrt{x^2+y^2}$. The characteristic curve through $(\cos \theta ,\sin \theta ,-1)$ hasThe condition $s<1$ is essential: at $s=1$ there is finite-time blowup, so the opposite branch cannot be reached from the data. From $r=(1-s\sin \theta {)}^{-1}$ and $y=r\sin \theta$,Eliminating $s$ yieldsThe ratio inside the cube root is positive precisely on the two data-determined regionsOn $D_{+}$ each point traces to the upper semicircle, and on $D_{-}$ to the lower semicircle, with $s<1$. To check the parameter ranges explicitly, for $\eta =\sin \theta >0$ they are $-\infty <s<1$, whereas for $\eta <0$ they are $1/\eta <s<1$. At the latter lower endpoint the planar projection escapes to infinity. The inverse function theorem and uniqueness for the characteristic ordinary differential equations give a unique smooth solution on each region. Equivalently, each point has the unique foot $(x/r,y/r)$ and parameter $(r-1)/y$.

A direct check is particularly short. For $s=(r-1)/y$, let $\mathcal{D}=xy{\partial }_x+y^2{\partial }_y$. Then $\mathcal{D}r=yr$, $\mathcal{D}y=y^2$, andSince $dz/ds=-z^4/3$, the displayed formula solves the partial differential equation, and $r=1$ gives $z=-1$ on both open semicircles. The finite boundary where $r=1+y$, away from the axis, is the parabola $y=(x^2-1)/2$ and corresponds to blowup. Points on $y=0$ cannot be added to a classical continuation of these regions: where they are approached from a data-determined region, the height behaves like a nonzero multiple of $|y{|}^{1/3}$, or has incompatible limits at the endpoints. The formula is therefore the complete classical answer for the compatible arcs, while the full-circle problem as printed has no classical solution.