= Solution
Interpret unit-intensity Gaussian white noise as $dW_t/dt$. Differentiating $z=x+(T-t)y$ cancels the two $y\,dt$ terms and gives
$$
\boxed{dz=(T-t)u\,dt+(T-t)dW_t,\qquad z(T)=x(T).}
$$
The cost is terminal $z(T)^2$ plus the same running $u^2$. Write $h=T-t$ and guess the value $V(t,z)=P(t)z^2+C(t)$. The <Hamilton-Jacobi-Bellman equation> is
$$
V_t+\tfrac12h^2V_{zz}+\min_u(u^2+huV_z)=0,\qquad V(T,z)=z^2.
$$
Completing the square gives $u_*=-hPz$. Equating powers yields $P'=h^2P^2$, $C'=-h^2P$, with $P(T)=1,C(T)=0$. Integration gives
$$
\boxed{P(t)=\frac1{1+(T-t)^3/3},\quad C(t)=\log[1+(T-t)^3/3],\quad
u_*=-\frac{(T-t)z}{1+(T-t)^3/3}.}
$$
Indeed $V_t+\mathcal L^uV+u^2=(u+hPz)^2\geq0$, and the Itô verification identity attains equality at this control. This proves optimality for admissible controls of finite expected quadratic cost. The additive noise changes only $C$, not the feedback gain, so \b[this is certainty-equivalence control]. A different constant noise intensity multiplies $C$ but leaves the control unchanged, as in the <terminal-position reduction of double-integrator control>.
Back to article page
