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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.

Fix $x$ and put $g={\nabla }_{\mathcal{B}}f(x)$, the block coordinate gradient. For $t\ge 0$, define the one-variable function $q(t)=f(x-tg)$. By the chain rule,The last equality holds because the coordinates of $g$ outside $\mathcal{B}$ are zero. The fundamental theorem of calculus and the Cauchy-Schwarz inequality therefore giveFor $\alpha \ge 0$, the supplied condition (6), applied at the fixed starting point $x$ with parameter $t$, bounds the integrand by $L_{\mathcal{B}}t\Vert g{\Vert }^2$. Integration yieldsThis argument needs only the given directional Lipschitz continuity and continuous differentiability; convexity is not needed for this part. If the block coordinate gradient is zero, the inequality is equality.

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

Applying the chain rule to both inverse identities givesFor the unnumbered calculation, the requested point is $f(c,c)$. The Jacobian there isTo apply its inverse to $(1/2,1/2)$, solve $e^c(v_1+v_2)=1/2$, $e^c{v}_1-e^{-c}{v}_2=1/2$. Subtraction gives $v_2=0$, and henceFor precision, the displayed example maps bijectively onto $V=\{(s,t):s>0,t>0,st>1\}$, rather than the entire positive quadrant. Indeed writing $q=e^x$, the equations are $(s-q)(t-q)=1$ with $0<q<\min (s,t)$. This left side decreases continuously from $st$ to zero, giving exactly one $q$ if and only if $st>1$, after which $e^y=s-q$ is unique. Its Jacobian never vanishes, so the inverse is differentiable on $V$. The requested point lies in $V$, making the calculation well-defined.

For a twice continuously differentiable stream function, equality of mixed derivatives givesThis is the incompressible flow condition in two dimensions. A streamline is a curve whose tangent at each regular point is parallel to the instantaneous velocity field. Parameterize it by ${\gamma }^{\prime }(s)=c(s)(u,v)$. The chain rule yieldsTherefore the stream function is constant on each streamline. This proof includes arbitrary regular parametrizations; at a stagnation point the velocity vanishes and the usual tangent direction is not determined.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

For differentiable maps with compatible domains, the multivariable chain rule isIn Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.

A $C^1$ diffeomorphism $F:U\to V$ is a bijection for which both $F$ and $F^{-1}$ are continuously differentiable. If a higher regularity class is specified, both maps must belong to that class. In particular, a diffeomorphism is a homeomorphism, and the chain rule gives $D(F^{-1})(F(x))=(DF(x){)}^{-1}$.

A $C^1$ diffeomorphism $F:U\to V$ is a bijection for which both $F$ and $F^{-1}$ are continuously differentiable. If a higher regularity class is specified, both maps must belong to that class. In particular, a diffeomorphism is a homeomorphism, and the chain rule gives $D(F^{-1})(F(x))=(DF(x){)}^{-1}$.

The Fréchet derivative at $a$ is the linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ satisfyingIt is unique. When both derivatives exist, the chain rule is $D(g\circ f{)}_a=D{g}_{f(a)}\circ D{f}_a$.

For the square map,The remainder has norm at most $\Vert \Delta {\Vert }^2$ in a submultiplicative matrix norm, hence is $o(\Vert \Delta \Vert )$. ThereforeMatrix multiplication is not commutative, so the two terms cannot generally be combined as $2A\Delta$.

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.

The chain rule states that $f\circ g$ is differentiable at $x$ andIn terms of Jacobian matrices,The factors have dimensions $k\times m$ and $m\times n$, respectively, giving the required $k\times n$ Jacobian matrix.

The total derivative of $g$ sends a scalar increment $s$ toSince ${\sin }^{2}t+{\cos }^{2}t=1$, the function composition $f\circ g$ is constantly $1/2$, so $D(f\circ g)(t)=0$. Directly, the Jacobian matrix product isThis verifies the chain rule at every $t$.

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and

If $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$, and $G$ is defined on a neighborhood of $F(x)$ and differentiable there, the chain rule statesIn standard coordinates the same statement isThe evaluation point in the first factor is $F(x)$, and the matrix order follows the order in which the linear maps act on increments.

For an increment $h$, direct expansion and the chain rule giveFor example, $\Vert x+h{\Vert }^2-\Vert x{\Vert }^2=2x^{T}h+\Vert h{\Vert }^2$, so the quadratic remainder divided by $\Vert h\Vert$ tends to zero. The corresponding gradients are $\nabla g(x)=2x$ and $\nabla k(x)=2A^{T}Ax$; the derivative of $F$ is the constant matrix $A$.

If $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$, and $G$ is defined on a neighborhood of $F(x)$ and differentiable there, the chain rule statesIn standard coordinates the same statement isThe evaluation point in the first factor is $F(x)$, and the matrix order follows the order in which the linear maps act on increments.

For an increment $h$, direct expansion and the chain rule giveFor example, $\Vert x+h{\Vert }^2-\Vert x{\Vert }^2=2x^{T}h+\Vert h{\Vert }^2$, so the quadratic remainder divided by $\Vert h\Vert$ tends to zero. The corresponding gradients are $\nabla g(x)=2x$ and $\nabla k(x)=2A^{T}Ax$; the derivative of $F$ is the constant matrix $A$.