Past exam of the differential equations 1 course of the University of Oxford / 2023 (A1, A15294W1)

Source: Oxford Part A paper A1, A15294W1, 14 June 2023; parent PDF pages 276–279, paper pages 1–4. These original solutions cover all three questions and every subpart without reproducing the question statements. In Question 2(b), the prescribed data are incompatible at two points; the solutions on the compatible open arcs are given explicitly.

Fix any $t_0\ge 0$ and compare $y=R$ with $y=0$. The quotient required in a global Lipschitz condition in the dependent variable isConsequently,It is nevertheless locally Lipschitz in $y$: for $|y|,|v|\le R$,Thus the Picard-Lindelöf theorem gives local existence and uniqueness for the initial value problem; the remaining work establishes existence on the entire half-line for the specified small data.

Write $A(t)={\int }_0^{t}a(s)ds$. Multiplying the ordinary differential equation by the integrating factor $e^{A(t)}$ and using the product rule givesIntegration from $0$ to $t$ therefore yields the Volterra integral equationConversely, a continuous function satisfying this integral equation becomes continuously differentiable after multiplication by $e^{A(t)}$, by the fundamental theorem of calculus, and satisfies the original initial value problem.

For $s>t$, the coefficient bound gives $A(s)-A(t)\ge C_1(s-t)$. Thus for $0\le s\le t$ the positive kernel satisfiesThe starting Picard iterate is continuous and has supremum norm $M_0=|b|$. If the $n$th Picard iterate is bounded and continuous, its integral defines another continuous function, andTaking the supremum norm and applying mathematical induction proves boundedness of every iterate and

Put $B=|b|>0$ and $\gamma =4B/C_1<1$. Initially $M_0=B<2B$. If $M_n<2B$, the preceding supremum norm estimate givesHence mathematical induction gives

The difference of two successive Picard iterates factors asThe preceding bound and the kernel estimate therefore giveMoreover $D_1:=\Vert y_1-y_0{\Vert }_{\infty }\le B^2/C_1$. Summing the resulting geometric series, for $m>n$,Thus there is a uniform limit $y$ on $[0,\infty )$, continuous by the uniform limit theorem, with $\Vert y{\Vert }_{\infty }\le 2B$. The integral passes to this limit becauseTherefore $y$ satisfies the Volterra integral equation, and hence the initial value problem. This proves global existence by Global Picard iteration with a decaying kernel. The same estimate gives the useful error boundIf $b=0$, every iterate is zero and the convergence assertion is immediate.

Let $y$ be any continuous function solving the integral equation on $[0,\infty )$, and retain $B=|b|>0$. A first-exit argument avoids assuming in advance that its supremum norm is finite. If $|y|$ ever reached $2B$, continuity would give a first such time $T>0$. On $[0,T]$ we would have $|y|\le 2B$, so the kernel estimate would implya contradiction. Therefore $|y(t)|<2B$ for every finite $t$.

That pointwise strict inequality alone would not give a strict bound on the supremum norm. Apply the integral equation once more, now using $|y|\le 2B$ globally:

Existence was proved by Picard iteration. If $y$ and $v$ are any two continuous solutions with the same $b\ne 0$, the previous part makes both bounded by $2B$. Subtracting their Volterra integral equations givesSince $\gamma <1$, this forces $y=v$.

For $b=0$, zero is a solution. On any finite interval $[0,T]$, another continuous solution has a finite bound $K_T={\max }_{[0,T]}|y|$. Since $K(t,s)\le 1$, its integral equation impliesThe Gronwall inequality gives $y=0$ on every such interval. Hence

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.

Let $\mathcal{P}u=u_t-\Delta u+a{u}_x+b{u}_y$ on the unit disk over $0<t\le 1$. Smoothness is understood to include continuity on the closed cylinder. If $u$ exceeded its nonnegative parabolic boundary maximum $A$, it would attain a positive global maximum $M>A$ at some spatial interior point at time $t_0>0$. There,At $t_0=1$ the last inequality uses the derivative from the past; at an interior time it is equality. These signs imply $\mathcal{P}u+u^3\ge M^3>0$, contradicting the given differential inequality. ThusIf $A>0$, equality at any point of $Q$ gives the same contradiction with $M=A$, so $u<A$ throughout $Q$. If $A=0$, equality is possible: $u\equiv 0$ is an example. More precisely, write the inequality as $\mathcal{P}u+cu\le 0$ with $c=u^2\ge 0$. The strong parabolic maximum principle shows that an interior zero at time $t_0>0$ forces $u\equiv 0$ on $B\times [0,t_0]$.

That conclusion need not extend into the future for a subsolution. For $t_0<1$, take a spatially constant function which is zero for $t\le t_0$ and equals $-\exp (-1/(t-t_0))$ for $t>t_0$. It is smooth, nonincreasing and nonpositive, so $\mathcal{P}u+u^3\le 0$, but it becomes negative after $t_0$.

With zero drift, choose the smooth functionIt equals $-1$ on the entire parabolic boundary, so $A=-1$. On the closed cylinder, $-1\le u\le -7/8$, whileConsequentlyNevertheless the maximum at $(0,0,1)$ isThe negative cubic term permits an interior increase while the boundary stays fixed, explaining why the sign condition in the preceding part matters.

The parabolic boundary values are zero. An exact solution satisfies the preceding differential inequality, so its maximum estimate with $A=0$ gives $u\le 0$. The function $v=-u$ satisfies the same equation because both the linear differential operator and the cubic term change sign. Applying the same maximum principle to $v$ gives $-u\le 0$. Hence the complete solution set isSubstitution verifies existence, and the two inequalities prove uniqueness among the assumed smooth solutions for every continuous choice of the drift coefficients.

Define the squared L2 norm $E(t)={\int }_0^{1}u(x,t{)}^{2}dx$. Multiplication of the semilinear partial differential equation by $u$, followed by integration by parts, givesAt $x=1$ the Dirichlet boundary condition makes the boundary term zero. At $x=0$ the Robin boundary condition is $u_x(0,t)=\alpha u(0,t)$, so subtraction of the lower endpoint gives $-\alpha u(0,t{)}^2$. In particular the sign is dissipative. Thus the exact energy identity iswhich implies

For every real $r$ and integer $n\ge 1$, $|\sin r|\le \min (1,|r|)$ givesThis estimate covers both even and odd $n$, without any assumption on the sign of $u$. Since $a(t)\ge 0$, the energy estimate from part (i) becomesEquivalently, an integrating factor givesThe bracket is nonnegative and starts at zero, so it vanishes identically. This is also the zero-data case of the Gronwall inequality. The vanishing L2 norm and continuity of $u$ imply