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Along either coordinate axis, the restriction is $h(t,0)=h(0,t)=t^6$, which is positive for $t\ne 0$. Along either diagonal it iswhich is negative for sufficiently small nonzero $t$. Therefore every neighbourhood of the origin contains both positive and negative values, whereas $h(0,0)=0$. ConsequentlyThe Hessian matrix vanishes there, so the ordinary second-derivative test is inconclusive; the restrictions settle the question.

Near a nonzero stationary point $\mathbf{p}$, Taylor expansion of the gradient gives $\nabla h(\mathbf{p}+\mathbf{v})=H(\mathbf{p})\mathbf{v}+O(|\mathbf{v}{|}^2)$. Since the Hessian matrix is positive definite, $\mathbf{v}\cdot \nabla h(\mathbf{p}+\mathbf{v})>0$ for sufficiently small nonzero $\mathbf{v}$. Thus the gradient arrows point away from each of the four local minima, with different strengths in the two eigenvector directions.

At the origin, on the $x$-axis the gradient is $(6x^5,0)$ and on the $y$-axis it is $(0,6y^5)$, so arrows point outward. On $y=x$ the gradient is $6x^3(x^2-8)(1,1)$, and on $y=-x$ it is $6x^3(x^2-8)(1,-1)$, so arrows point toward the origin for small $|x|$. Away from the axes the leading terms are $(-48x{y}^2,-48x^{2}y)$, pointing inward in both coordinate signs. The narrow outward sectors close to the axes are visible in the exact vector field.

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

The gradient isIt is perpendicular to each nonzero level set and points towards increasing $f$. Near the two positive local maxima, the Hessian matrix is $-2e^{-1}I$, so the gradient arrows point inward. Near the two negative local minima, the Hessian matrix is $2e^{-1}I$, so they point outward. These arrows are shown on the figure in part (c).

At the saddle point, $\nabla f\approx (v,u{)}^T$. Along $v=u$, the arrows point away from the centre on both sides; along $v=-u$, they point towards the centre on both sides. On the positive $u$ axis they point up, on the negative $u$ axis down, on the positive $v$ axis right, and on the negative $v$ axis left. This specifies the four intervening sectors of the gradient pattern.

For a harmonic function, put $a=u_{xx}$ and $b=u_{xy}=u_{yx}$. Since $u_{yy}=-u_{xx}$, its Hessian matrix isA nonzero determinant is therefore negative, and the two real Hessian eigenvalues are $\pm \sqrt{a^2+b^2}$. At a stationary point the quadratic term of the Taylor expansion has both signs. Consequently every such nondegenerate stationary point is a saddle point. This is the nondegenerate stationary points of a planar harmonic function criterion. The same argument applies to $v$, whose stationary points coincide with those of $u$. A zero Hessian determinant is outside this second-order classification.

Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

The quadratic model has gradient $g(x_k)+B_{k}p$ and positive-definite Hessian matrix $B_k$. Its unique minimizer is thereforeIf $g(x_k)\ne 0$, then $p_k\ne 0$ and $g(x_k{)}^T{p}_k=-p_k^T{B}_k{p}_k<0$, proving that it is a descent direction.

Backtracking line search starts with a trial step and repeatedly reduces it until the Armijo condition holds. Part (ii) guarantees finite termination because sufficiently small steps are acceptable. The test guarantees a specified positive fraction of the decrease predicted by the linear model, namely $f(x_k+\alpha p_k)\le f(x_k)+\alpha \beta g(x_k{)}^T{p}_k<f(x_k)$ for an accepted positive step. It does not require solving the line-minimization problem exactly.

At an unconstrained local minimizer of a twice continuously differentiable objective function, the first-order optimality condition is $\boxed{\nabla f(x_{\ast })=0}$. The second-order necessary condition is $d^T{\nabla }^{2}f(x_{\ast })d\ge 0$ for every direction $d$, so the Hessian matrix is positive semidefinite.

For the Euclidean norm, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ when $g=\nabla f(x)\ne 0$; the algorithm usually uses the unnormalized direction $p=-g$. It is a descent direction because $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$.

Two limitations of unpreconditioned steepest descent method in high-accuracy smooth optimization are its slow zigzagging on ill-conditioned valleys and its sensitivity to scaling of the coordinates. It uses no curvature information to correct those directions, so it can require many function and gradient evaluations compared with a suitable Newton or quasi-Newton method. These are reasons to prefer curvature-aware methods in that setting, rather than a claim that gradient-based methods are never useful.

Use the optimization Lagrangian $\mathcal{L}(x,y)=f(x)-y^{T}c(x)$, with $A(x)=Dc(x)$ and $H(x,y)={\nabla }_{xx}^2\mathcal{L}(x,y)$. Under the usual linear independence constraint qualification, a local minimizer has multipliers $y_{\ast }$ satisfying the equality-constrained Karush-Kuhn-Tucker conditionsThe second-order necessary condition isOnly feasible tangent directions enter this last test; the Hessian matrix need not be positive semidefinite on the entire ambient space.

The linearized constraints may be inconsistent, for example when $-c(x_k)$ is not in the image of a linear map defined by $A(x_k)$. Even if feasible, the quadratic objective may have negative curvature along its feasible tangent space, making it unbounded below and leaving no local minimizer.

Fletcher's Sℓ1QP replaces the hard linearized constraints by an L1 exact penalty and bounds the step inside a trust region. One usual form isThe zero step is always admissible. The trust region is compact, so a minimum exists even when the model Hessian matrix is indefinite; the constraint penalty measures infeasibility rather than making the subproblem impossible to solve. Adjusting its weight promotes feasibility, while the radius keeps the local model within a controlled neighbourhood. Introducing nonnegative positive- and negative-part slack variables converts the absolute values to an equivalent smooth quadratic program with box constraints.^[ref]

The Jacobian matrix of $F_{\mu }$ in $(x,y)$ isThe top-left block is the Hessian matrix of $f-y^{T}c$ with $y$ held fixed. In the second block, differentiating $C(x)y$ in $x$ gives $YA$ and in $y$ gives $C$. The explicit derivative with respect to $\mu$ is $(0,-e{)}^T$. Differentiate $F_{\mu }(v(\mu ))=0$ by the chain rule:All blocks are evaluated on the current central path point.

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.