A bounded-actuator LQR control and its exact objective

Put $m={\int }_0^{1}u(t)dt$. Since ${\int }_0^1(u-m{)}^{2}dt={\int }_0^1{u}^{2}dt-m^2$, direct expansion gives $J(u)=\frac{1}{2}\mathrm{\parallel }u-m{\mathrm{\parallel }}_2^2+\frac{1}{2}[m^2+(m-1{)}^2]=\frac{1}{2}\mathrm{\parallel }u-m{\mathrm{\parallel }}_2^2+(m-\frac{1}{2}{)}^2+\frac{1}{4}$.

Both squared terms are nonnegative, so $J(u)\ge \frac{1}{4}$. Equality holds exactly when $u=m$ almost everywhere and $m=\frac{1}{2}$, hence exactly for $u^{\ast }(t)=\frac{1}{2}$ almost everywhere. This control satisfies $[-1,1]$ and direct substitution gives $J(u^{\ast })=\frac{1}{4}$. Therefore it is the unique minimizer.

The state is $x(t)=t\mathrm{/}2$. By [F2], the adjoint equation is $-\dot{p}=0$ with terminal value $p(1)=x(1)-1=-1\mathrm{/}2$. The constant curve $p(t)=-1\mathrm{/}2$ solves it, and uniqueness in [F2] makes this the adjoint. Under [A1], [F3] gives the reduced gradient $g=p+u^{\ast }=0$. Thus $g(t)(c-u^{\ast }(t))=0$ for every $c\in [-1,1]$, so the control-only pointwise variational inequality holds with equality for this example. The optimizer and objective calculation in steps 1.1–2.1 used no choice axiom.