Past exam of the optimal transport and partial differential equations course of the University of Oxford / 2025 (C4.9, A17515H1)

Original solutions to all three questions. Source: released paper, parent PDF pages 27–30. The paper's notation $d_p$ denotes the p-Wasserstein distance. Question 1(b) contains two printed errors; Questions 2(b–c) and 3(c) require the qualifications explained in their solutions.

Compact support of a measure gives finite second absolute moments for both probability measures. Since the source satisfies absolute continuity of measures with respect to Lebesgue measure, Brenier's theorem identifies the quadratic Kantorovich optimal transport problem with the Monge optimal transport problem:Here $S_{\#}\mu$ is the pushforward measure. The optimal transport plan is unique and equals $(\mathrm{Id},T{)}_{\#}\mu$. The optimal transport map is unique up to a $\mu$-null set, withThe gradient of this convex function is c-cyclically monotone for $c(x,y)=|x-y{|}^2$ on a full-measure set. In particular, summing the two supporting inequalities for the convex function gives $(T(x)-T(x^{\prime }))\cdot (x-x^{\prime })\ge 0$ wherever both gradients exist. No regularity of the target beyond the stated probability measure assumptions is needed.

Write $P=\omega {\omega }^T$. Since $P^T=P$ and $P^2=P$, the orthogonal reflection $R_{\omega }=I-2P$ satisfies $R_{\omega }^T{R}_{\omega }=I$ and $R_{\omega }^{-1}=R_{\omega }$. Thus, for every Borel set $E$, invariance under the orthogonal group gives

The spherical average must use normalized surface measure without an additional area factor. Let $d\sigma$ be the normalized surface measure on a sphere, and setThe correct radial disintegration of an orthogonally invariant measure isIndeed, take normalized Haar measure $H$ on the compact orthogonal group $O(d)$. Invariance and Fubini's theorem giveRight invariance of Haar measure and transitivity of the orthogonal group on each sphere show that the inner average depends only on $r=|x|$. Averaging it once more over $x=r\theta$, using invariance of surface measure on a sphere and Fubini's theorem, identifies its value as $A\eta (r)$. At $x=0$ the same identity reads $\eta (0)=\eta (0)$. The definition of pushforward measure now proves the boxed formula. This argument also works for invariant probability measures with an atom of a measure at the origin.

For the printed normalization, choose a smooth cutoff function $\eta$ equal to one on a ball containing the support of a measure $\mu$. Its left integral is one, whereas the printed right integral is ${\sigma }_d$. Thus the extra ${\sigma }_d$ cannot be retained. Equivalently, with unnormalized area measure $dS$, the average is ${\sigma }_d^{-1}\int \eta (r\theta )dS(\theta )$.

The reflection-averaging hint also fails in general dimension. For $x=e_1$,The first equality uses $\int {\omega }_i{\omega }_{j}d\sigma ={\delta }_{ij}/d$, which follows from coordinate reflections, coordinate permutations, and ${\sum }_i{\omega }_i^2=1$. Taking a test function equal to the first coordinate near the unit sphere disproves the hint when $d\ne 2$. Averaging over the full orthogonal group, as above, proves the required corrected result without that hint.

Let $\alpha ={\Phi }_{\#}{\mu }_0={\Phi }_{\#}{\mu }_1$. The corrected radial disintegration of an orthogonally invariant measure, whose proof did not require absolute continuity of measures, gives for every test function $\eta$Finite Borel measures on Euclidean space are determined by their integrals against test functions: the difference defines the zero distribution, or one can approximate continuous compactly supported functions uniformly by smooth functions and use regularity of the measures. Therefore

Write $d\mu =f(x)dx$, with $f\ge 0$. Applying polar coordinates and Fubini's theorem gives, for bounded nonnegative measurable $h$,Also $\tilde{\mu }(\{0\})=\mu (\{0\})=0$ by absolute continuity of measures. Extending $g$ by zero on the negative half-line proves

Both radial pushforward measures have compact support of a measure, so Brenier's theorem on the line supplies a unique optimal transport plan and a transport map $\tau$, unique $\tilde{\mu }$-almost everywhere, withIn one dimension, the gradient of a convex function is a nondecreasing function. Nonnegativity holds $\tilde{\mu }$-almost everywhere because ${\tau }_{\#}\tilde{\mu }=\tilde{\nu }$ is carried by $[0,\infty )$. To choose a nonnegative, nondecreasing representative everywhere, let $F,G$ be the radial cumulative distribution functions and use the monotone rearrangementAt $F(r)=0$ set $\tau (r)=0$, and at $F(r)=1$ use the largest point of the support of a measure $\tilde{\nu }$. These choices preserve monotonicity and boundedness. Since $F$ is continuous, $F(R)$ has the uniform distribution on $(0,1)$ when $R$ has law $\tilde{\mu }$, and the quantile function $G^{-1}$ sends this uniform distribution to $\tilde{\nu }$. Thus this representative has the required pushforward measure, even if the target has atoms.

Choose the bounded, nonnegative nondecreasing function $\tau$ from part (d), and defineThe function $h$ is convex and nondecreasing. For $0\le a\le 1$, the triangle inequality, monotonicity, and convexity giveHence $U$ is a convex function. By the gradient of a radial function formula and the fundamental theorem of calculus,for $x\ne 0$ whenever $h^{\prime }(|x|)=\tau (|x|)$. A monotone function has at most countably many discontinuities, so part (d) makes the exceptional radii $\tilde{\mu }$-null. The origin is also $\mu$-null. Define $\mathcal{T}(0)=0$; thus $\mathcal{T}=\nabla U$ $\mu$-almost everywhere.

For every $Q$ in the orthogonal group, $\mathcal{T}(Qx)=Q\mathcal{T}(x)$, including at zero. Consequently ${\mathcal{T}}_{\#}\mu$ is invariant under the orthogonal group. Moreover $|\mathcal{T}(x)|=\tau (|x|)$ away from the null origin, and composition of pushforward measures givesPart (c) therefore gives ${\mathcal{T}}_{\#}\mu =\nu$. The Knott–Smith optimality criterion makes a transport map given by a convex function's gradient optimal for the quadratic transport cost. Uniqueness from Brenier's theorem yieldsSince directions agree, this also identifies the optimal value:

Put $D=d_2(\mu ,\nu )$ and $z_t(x,y)=(1-t)x+ty$. The marginal distributions of the optimal transport plan $\Pi$ give ${\mu }_0=\mu$ and ${\mu }_1=\nu$. Alsoso ${\mu }_t$ has a finite second absolute moment. For $0\le s<t\le 1$, the pushforward measure $(z_s,z_t{)}_{\#}\Pi$ is a coupling of ${\mu }_s,{\mu }_t$. ThereforeApply the triangle inequality for the second Wasserstein distance along ${\mu }_0,{\mu }_s,{\mu }_t,{\mu }_1$:All inequalities are equalities, including when $D=0$. Thus this displacement interpolation is a constant speed metric geodesic:

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

Use an independent copy $(x^{\prime },y^{\prime })$ of $(x,y)$ under the product measure $\Pi \otimes \Pi$. Write $q=y-x$, $q^{\prime }=y^{\prime }-x^{\prime }$, andSince the pushforward measure of $\Pi \otimes \Pi$ under $(z_t,z_t^{\prime })$ is ${\mu }_t\otimes {\mu }_t$, the interaction energy isTwo applications of the chain rule giveThe factor $1/2$ is retained. This classical identity requires differentiation under the integral sign to be justified, just as in part (b). Bounded Hessian matrix of $W$ is sufficient, since $|q-q^{\prime }{|}^2\le 2|q{|}^2+2|q^{\prime }{|}^2$ is integrable; compactly supported endpoints are another sufficient case. The printed quadratic-growth condition alone does not provide this domination.

The failure can occur for the interaction energy too. Take the even interaction potential $W(x)=1-\cos (2\pi x^3)$ and replace the endpoint probability measures in part (b) by $\frac{1}{2}{\delta }_0+\frac{1}{2}\mu$ and $\frac{1}{2}{\delta }_0+\frac{1}{2}\nu$. Keep the origin fixed and translate every other atom by one. This coupling is increasing on the line, hence is optimal: its assigned values extend to a nondecreasing transport map, which is the gradient of a convex function. Differences between translated atoms remain integers, so their mutual interaction vanishes. Only interactions with the origin contribute, giving $\mathcal{W}[{\mu }_t]=F(t)/4$, with $F$ from part (b). It therefore also lacks a finite classical second derivative at $t=1/2$.

Define the barycenter of a measure by $m_{\mu }=\int xd\mu (x)$. Under $\Pi$, the mean displacement is $\bar{q}=m_{\nu }-m_{\mu }$, and optimality gives $\int |q{|}^{2}d\Pi =D^2$. Expanding the square under the product measure yields the variance identity for independent copies:In the classical regime of part (c), the lower bound on the Hessian matrix of the interaction potential therefore givesFor equal barycenters, this becomes

For the full stated class of probability measures, prove the displacement convexity inequality directly. Pointwise strong convexity of $W$ givesIntegrating, including the $1/2$ in the interaction energy, and using the displayed variance identity for independent copies givesAll terms are finite by quadratic growth and finite second absolute moments. Equal barycenters give the claimed $\lambda$-displacement convexity. Also $m_{{\mu }_t}=(1-t)m_{\mu }+t{m}_{\nu }$, so this metric geodesic remains in the prescribed fixed-mean class. Applying the same argument to subintervals proves the second derivative bound in the sense of distributions even when a classical derivative has not been justified. The mean restriction is natural: translating all particles together leaves their pairwise differences and hence the interaction energy unchanged.

Choose a uniform lower bound $\kappa \in \mathbb{R}$ for the Hessian matrix of the confinement potential, so $D^{2}V(x)\ge \kappa I$ for every $x$. A sufficient condition isIndeed, convexity of $V(x)-\kappa |x{|}^2/2$ gives the confinement energy inequality from part (b), now with $\kappa$ possibly negative. Adding it to the equal-mean interaction energy inequality from part (d) givesThis proves positive displacement convexity on every class with fixed barycenter. In particular, if $D^{2}V\ge -KI$ with $K\ge 0$, the sufficient condition is $\lambda >K$ and one may take $\tilde{\lambda }=\lambda -K$. Wherever classical second derivatives exist, the equivalent differential calculation is ${\mathcal{E}}^{\prime \prime }[{\mu }_t]\ge (\kappa +\lambda )D^2$; the integrated proof also covers the printed growth assumptions.

SetSymmetry here means $W(-z)=W(z)$, so $\nabla W(0)=0$ and $K(0)=0$. The bounded Hessian matrix and the fact that $J$ is an orthogonal matrix giveThus the particle system is ${\dot{X}}^i=N^{-1}{\sum }_{j}K(X^i-X^j)$. On the configuration space with norm $\Vert X{\Vert }_{\max }={\max }_i|X^i|$, its vector field has Lipschitz constant at most $2L$:The Picard-Lindelöf theorem gives local existence and uniqueness. Linear growth and the Gronwall inequality give $\Vert X(t){\Vert }_{\max }\le e^{2Lt}\Vert X(0){\Vert }_{\max }$, which rules out escape to infinity in finite time and extends the solution to every $t\ge 0$.

The associated empirical measure isHere each ${\delta }_{X^i(t)}$ is a Dirac measure. The finite first absolute moment follows from $\int |x|d{\mu }_t^N=N^{-1}{\sum }_i|X^i(t)|<\infty$. Pairing equal particle labels defines a coupling, soConsequently

For a candidate curve $\rho$, define the velocity field by the convolutionIt is well defined with a finite first absolute moment, because $|K(x-z)|\le L(|x|+|z|)$. Equality of mixed partial derivatives givesThe Hessian matrix bound justifies moving a spatial derivative through this convolution. For smooth densities, the transport equation consequently becomes the continuity equationThis form defines the equation for singular probability measures as well. With initial probability measure ${\rho }_0$, a weak solution is a curve in $C([0,\infty ),{\mathcal{P}}_1)$ such that every $\phi \in C_c^1([0,\infty )\times {\mathbb{R}}^2)$ satisfiesUsing smooth test functions gives the same definition by approximation on compact sets. Continuity in the Wasserstein distance bounds first absolute moments on compact time intervals, so the displayed integrals exist.

For the empirical measure, the chain rule and the particle equations giveIntegrating in time, with zero terminal contribution because $\phi$ has compact support, gives precisely the weak identity and the correct initial empirical measure. Thus ${\mu }^N$ is a global weak solution. The self-interaction term vanishes since $K(0)=0$.

A finite first moment suffices for the velocity, but may not suffice for the printed potential. For example, $W(z)=|z{|}^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\ge 1$, then $\mathbb{E}S=2$ but $\mathbb{E}{S}^2=\infty$, and $(W\ast \mu )(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second absolute moment, or another hypothesis ensuring integrability of $W$.

For the specified class ${\mathcal{P}}_1$, define the velocity fields directly by $b_{\mu }=K\ast \mu$ and $b_{\hat{\mu }}=K\ast \hat{\mu }$. Equivalently, use the renormalized interaction potentialThe bounded Hessian matrix and $\nabla W(0)=0$ implyThus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W\ast \mu$, by dominated differentiation under the integral sign. Its rotated negative gradient is exactly $b_{\mu }$. Whenever $W\ast \mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same transport equation.

We check the hypotheses for characteristic representation of a measure-valued continuity equation. For every $x,y$,For any coupling $\gamma \in \Gamma ({\mu }_t,{\hat{\mu }}_t)$, the Lipschitz condition on $K$ also givesTaking the infimum over couplings proves the uniform boundApplying the same estimate at two times of one curve proves time continuity of $b_{\mu }$, uniformly in $x$. Also $|m_1({\mu }_t)-m_1({\mu }_s)|\le d_1({\mu }_t,{\mu }_s)$, so $m_1({\mu }_t)$ is bounded on compact time intervals. Spatial partial derivatives of $b_{\mu }$ are continuous and bounded by $L$, using dominated convergence in space and weak convergence of probability measures in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\hat{\mu }$.

The characteristic representation of a measure-valued continuity equation says that a continuous velocity field, continuously differentiable and globally Lipschitz in space and with locally uniform linear growth, has a global characteristic flow map $X_{\mu }(t;s,a)$ solvingFor each $f\in {\mathcal{P}}_1$, its pushforward measure is the unique weak solution in $C([0,\infty ),{\mathcal{P}}_1)$ of the associated continuity equation. The hypotheses have all been verified above, and the fields have zero divergence, so these are also the two required transport equations. Explicitly,For completeness, the Gronwall inequality givesso these pushforward measures have finite first absolute moments. Integrating $|X_{\mu }(t;0,a)-X_{\mu }(s;0,a)|$ against $f$ proves continuity in the Wasserstein distance. The chain rule proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward test function $\phi (s,x)=\psi (X_{\mu }(t;s,x))$, where $\psi$ is continuously differentiable with compact support. It satisfies ${\partial }_s\phi +b_{\mu }\cdot \nabla \phi =0$ and has compact spatial support uniformly for $0\le s\le t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int \psi d{\rho }_t=\int \psi (X_{\mu }(t;0,a))df(a)$, identifying every weak solution with the displayed pushforward measure.

Finally compare the two characteristic curves with the same initial point $a$. Put ${\Delta }_t=X_{\mu }(t;0,a)-X_{\hat{\mu }}(t;0,a)$. The spatial Lipschitz condition and the boxed velocity estimate givefor almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The Gronwall inequality impliesThe pushforward measure $(X_{\mu }(t;0,\cdot ),X_{\hat{\mu }}(t;0,\cdot ){)}_{\#}f$ is a coupling of ${\rho }_t,{\hat{\rho }}_t$. Integrating the last bound against the probability measure $f$ provesThe same constant $L$ depends only on the uniform bound for the Hessian matrix of $W$, and is independent of the prescribed curves and their common initial data.