A closed model force field on an annular domain without a global potential energy: force-generating potentials exist only locally

The model. Take $N=1$, a mass $m:=1\mathrm{k}\mathrm{g}$, time window $\mathbb{R}$, write $x=(x_1,x_2,x_3)$ in metres, with $x_i=x(i-1)$, ${\mathrm{\Phi }}_i=\mathrm{\Phi }(i-1)$ and ${\mathrm{\partial }}_i$ denoting differentiation in native coordinate $i-1$ for $i=1,2,3$, and let the open annular configuration domain be $Q:=\{x\in {\mathbb{R}}^3:\frac{1}{2}\mathrm{m}<\sqrt{x_1^2+x_2^2}<2\mathrm{m}\}\mathrm{.}$ Set $\beta :=1\mathrm{J}$ and define the time- and velocity-independent force map $F(t,x,v)=\beta \mathrm{\Phi }(x),\mathrm{\Phi }(x):=\frac{(-x_2,x_1,0)}{x_1^2+x_2^2}\mathrm{.}$ Since $\mathrm{\Phi }$ has dimension ${\mathrm{\ell }}^{-1}$, $F$ has force dimension $\mu \mathrm{\ell }{\tau }^{-2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). This one-label field is the external single-label force of the potential definition's clause (d), and it is smooth on $Q$, which does not meet the axis.

Standing regularity. Put $r(x):=\sqrt{x_1^2+x_2^2}$ and let $A(x_1,x_2,x_3)=(-x_2,x_1,0)$; then $\mathrm{\parallel }A\mathrm{\parallel }=1$, $\mathrm{\parallel }Ax{\mathrm{\parallel }}_2=r(x)$ and $\mathrm{\Phi }(x)=Ax\mathrm{/}r(x{)}^2$, so $\mathrm{\parallel }\mathrm{\Phi }(x){\mathrm{\parallel }}_2=1\mathrm{/}r(x)$. Differentiating this formula gives $\mathrm{\parallel }D\mathrm{\Phi }(x)\mathrm{\parallel }\le 3\mathrm{/}r(x{)}^2$. For $x,y\in Q$ with $\mathrm{\parallel }x-y{\mathrm{\parallel }}_2\le \frac{1}{4}\mathrm{m}$, every point of the segment from $x$ to $y$ has radial coordinate at least $r(x)-\mathrm{\parallel }x-y{\mathrm{\parallel }}_2>\frac{1}{4}\mathrm{m}$; hence along $g(u)=\mathrm{\Phi }(x+u(y-x))$ one has $\mathrm{\parallel }{g}^{\mathrm{\prime }}(u){\mathrm{\parallel }}_2\le 48{\mathrm{m}}^{-2}\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$, and [L3] gives $\mathrm{\parallel }\mathrm{\Phi }(x)-\mathrm{\Phi }(y){\mathrm{\parallel }}_2\le 48{\mathrm{m}}^{-2}\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$. For $\mathrm{\parallel }x-y{\mathrm{\parallel }}_2>\frac{1}{4}\mathrm{m}$, $\mathrm{\parallel }\mathrm{\Phi }(x)-\mathrm{\Phi }(y){\mathrm{\parallel }}_2\le 1\mathrm{/}r(x)+1\mathrm{/}r(y)<4{\mathrm{m}}^{-1}\le 16{\mathrm{m}}^{-2}\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$. Thus $F$ is globally Lipschitz in its position argument with constant $48\mathrm{N}{\mathrm{m}}^{-1}$; because it is independent of $v$, it is locally Lipschitz in the state variable on compact state cylinders in the numerical SI coordinates, as required by the force branch (Local Lipschitz continuity in the state variable, locally uniform in time and parameters).

Closedness. Write $s:=x_1^2+x_2^2$, so that $\mathrm{\Phi }=(-x_2{s}^{-1},x_1{s}^{-1},0)$. The quotient rule [L3] gives ${\mathrm{\partial }}_1{\mathrm{\Phi }}_2={\mathrm{\partial }}_1\left(x_1{s}^{-1}\right)=\frac{s-x_1\cdot 2x_1}{s^2}=\frac{x_2^2-x_1^2}{s^2},{\mathrm{\partial }}_2{\mathrm{\Phi }}_1={\mathrm{\partial }}_2(-x_2{s}^{-1})=-\frac{s-x_2\cdot 2x_2}{s^2}=\frac{x_2^2-x_1^2}{s^2},$ so ${\mathrm{\partial }}_1{\mathrm{\Phi }}_2={\mathrm{\partial }}_2{\mathrm{\Phi }}_1$; the remaining pairs vanish because ${\mathrm{\Phi }}_3=0$ and ${\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2$ do not depend on $x_3$ (the partial derivatives are those of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case). Multiplying all components by the constant $\beta$ preserves these identities, so $F$ is closed on $Q$.

Circulation. Let $L_0:=1\mathrm{m}$ and $\gamma (\theta ):=L_0(\cos \theta ,\sin \theta ,0)$ for the dimensionless angle $\theta \in [0,2\pi ]$. This is a $C^1$ path in $Q$ with ${\gamma }^{\mathrm{\prime }}(\theta )=L_0(-\sin \theta ,\cos \theta ,0)$. Along it, $F(\gamma (\theta ))=(\beta \mathrm{/}{L}_0)(-\sin \theta ,\cos \theta ,0)$, so $⟨F(\gamma (\theta )),{\gamma }^{\mathrm{\prime }}(\theta )⟩=\beta ({\sin }^2\theta +{\cos }^2\theta )=\beta$ by bilinearity of the inner product and the Pythagorean identity [L3]. By the definition of the vector line integral [L2], the single piece $[0,2\pi ]$ with derivative extension ${\gamma }^{\mathrm{\prime }}$ gives ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{2\pi }⟨F(\gamma (\theta )),{\gamma }^{\mathrm{\prime }}(\theta )⟩d\theta ={\int }_0^{2\pi }\beta d\theta =2\pi \beta =2\pi \mathrm{J}\mathrm{\ne }0,$ the middle integral by [L3].

No global potential. Suppose there were a $C^2$ function $U$ on $Q$ with $F=-\mathrm{\nabla }U$. The line integral of [L2] is linear in the field by its definition as a sum of parameter integrals, so ${\int }_{\gamma }F\cdot d\mathbf{r}=-{\int }_{\gamma }\mathrm{\nabla }U\cdot d\mathbf{r}$; the gradient theorem of [L2] applied to the $C^1$ function $U$ gives ${\int }_{\gamma }\mathrm{\nabla }U\cdot d\mathbf{r}=U(\gamma (2\pi ))-U(\gamma (0))=U(L_0,0,0)-U(L_0,0,0)=0$ because $\gamma$ is closed. Hence ${\int }_{\gamma }F\cdot d\mathbf{r}=0$, contradicting the value $2\pi \beta \mathrm{\ne }0$ of step 2.3. Therefore no such global potential exists, the model is not a global potential model in the sense of clause (a) of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets, and no global potential presentation with zero residual force is available. A global presentation with $U=0$ and $d=F$ is nevertheless admitted by the same definition and has mechanical energy $E=K$; it asserts no conservation.

Local potentials and local energy. Let $p\in Q$ and let $\epsilon :=\frac{1}{2}\min \{r(p)-\frac{1}{2}\mathrm{m},2\mathrm{m}-r(p)\}>0,B:=\{z:\mid z-p\mid <\epsilon \}$; here $\epsilon$ has units of length. For $z\in B$ one has $r(z)<r(p)+\epsilon <2\mathrm{m}$ and $r(z)>r(p)-\epsilon >\frac{1}{2}\mathrm{m}$, so $B\subseteq Q$. $B$ is open and star-shaped with respect to its centre (Star-shaped open subsets of Euclidean space). Since $-F{\mathrm{\mid }}_B$ is also a closed $C^1$ field, the Poincaré lemma of [L1] applied to $-F{\mathrm{\mid }}_B$ gives a $C^2$ function $U_B:B\to \mathbb{R}$ with $\mathrm{\nabla }{U}_B=-F{\mathrm{\mid }}_B$, that is, $F{\mathrm{\mid }}_B=-\mathrm{\nabla }{U}_B$. The restriction of the model's force map to the configuration domain $B$ is then a global potential model with a time-independent potential and no residual forces, so clause 2 of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form gives that $K+U_B$ is constant along every motion of the restricted model, hence along every subinterval of a motion of the annular model on which the position remains in $B$.

Conclusion. The force field of the model is closed and not exact on $Q$; the refuted statement is false. The time-independent zero-residual presentations constructed in step 4.1 provide conserved mechanical energy on those star-shaped subdomains. Globally no such presentation exists by step 3.1, so that conservation clause cannot be invoked for it; the global presentation $U=0,d=F$ remains available with $E=K$.