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For fixed terminal time and fixed endpoint, use the maximizing form of the Pontryagin maximum principle. For dynamics $\dot{q}=f(q,u,t)$ and running cost $L$, there are multipliers $(p,p_0)$, not all zero, with $p_0\ge 0$, and Hamiltonian of an optimal-control problem $\mathcal{H}=p\cdot f-p_{0}L$. Along an optimum, $\dot{q}={\mathcal{H}}_p$, $\dot{p}=-{\mathcal{H}}_q$, and the chosen control maximizes $\mathcal{H}$ over admissible controls almost everywhere. Fixed terminal state imposes no prescribed terminal value of $p$; fixed time imposes no free-time condition $\mathcal{H}(T)=0$. The normal case scales $p_0=1$; abnormal extremals have $p_0=0$.

For this problem the normal Hamiltonian of an optimal-control problem isMaximization gives $u=p_x+ky$, $v=p_y-kx$. The costate equations give ${\dot{p}}_x=kv$, ${\dot{p}}_y=-ku$, hence $\dot{u}=2kv$, $\dot{v}=-2ku$. Nonzero velocities rotate at constant angular rate, and rotational symmetry permits $u(0)=A$, $v(0)=0$. For a closed path, $2k=2\pi n$, with integer $n\ne 0$. Then $z(1)=A^2/(2\pi n)$, so the required positive endpoint has $n>0$, $A^2=2\pi n$ and cost $\pi n$.

To prove a global minimum rather than only classify stationary extremals, write $r=(x,y)$ and subtract its time average; this changes neither its derivative nor $\int (y\dot{x}-x\dot{y})dt$. The resulting closed mean-zero path satisfies the periodic Wirtinger inequality $\int |r{|}^2\le (4{\pi }^2{)}^{-1}\int |\dot{r}{|}^2$. This follows immediately by its Fourier expansion: each nonzero mode of the derivative has squared multiplier at least $4{\pi }^2$. The Cauchy-Schwarz inequality now givesThe argument extends to square-integrable controls by the periodic Sobolev/Fourier interpretation. Thus the cost is at least $\pi$ for every feasible control. Equality is attained byThese satisfy every endpoint and have cost $\pi$. Therefore the minimum is $\pi$, an area-constrained quadratic control cost. The global inequality also prevents an overlooked abnormal candidate from lowering it; in fact an abnormal finite maximizing Hamiltonian of an optimal-control problem would force both linear control coefficients to vanish and then force zero velocity, incompatible with $z(1)=1$.