From a Linear Hamiltonian to an Endpoint Control
When a scalar control is constrained to an interval, optimizing a Hamiltonian that is affine in that control is an elementary but consequential step in solving an HJB equation. The optimizer is usually an endpoint of the interval. Which endpoint is selected depends on the sign of the control’s coefficient; the states and times where that coefficient vanishes form the switching surface. At those points the Hamiltonian is indifferent among all admissible controls.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a controlled scalar diffusion on a finite horizon \([0,T]\): \(dX_t=b(t,X_t,u_t)\,dt+\sigma(t,X_t,u_t)\,dB_t\). Here \(u_t\) is an adapted, progressively measurable control taking values in the fixed interval \([\underline u,\overline u]\), where \(\underline u<\overline u\). The stochastic integral is the Ito integral, with progressively measurable integrand \(\sigma(t,X_t,u_t)\) satisfying \(\mathbb{E}\int_0^T \sigma(t,X_t,u_t)^2\,dt<\infty\). We consider a minimization problem with running cost \(\ell(t,x,u)\) and terminal cost \(g(x)\).
For a sufficiently smooth candidate value function \(V(t,x)\), the minimization HJB equation is \(V_t+\min_{u\in[\underline u,\overline u]}\{b(t,x,u)V_x+\tfrac12\sigma(t,x,u)^2V_{xx}+\ell(t,x,u)\}=0\), with terminal condition \(V(T,x)=g(x)\). The expression in braces is the Hamiltonian being minimized. The drift \(b\) multiplies the first spatial derivative, while the diffusion coefficient \(\sigma\) contributes through the second derivative and its square. The bang-bang rule below applies when the entire control-dependent part of this Hamiltonian is affine in \(u\).
The Endpoint-Selection Result
Suppose the control-dependent part of the Hamiltonian has the form \(A(t,x)u\), with all other terms independent of \(u\). For example, this form holds when the drift is \(b_0(t,x)+b_1(t,x)u\), the diffusion coefficient is independent of \(u\), and the running cost is \(\ell_0(t,x)+c(t,x)u\). In that case \(A=b_1V_x+c\). The same rule applies if other control-dependent terms combine to give an affine function of \(u\).
Proof. If \(A>0\), then for every \(u\in[\underline u,\overline u]\) we have \(Au-A\underline u=A(u-\underline u)\ge0\). Equality holds exactly when \(u=\underline u\), so this is the unique minimizer. If \(A<0\), then \(Au-A\overline u=A(u-\overline u)\ge0\), because \(u-\overline u\le0\); equality holds exactly when \(u=\overline u\). If \(A=0\), then \(Au=0\) for every admissible \(u\), so all controls minimize. For maximization, the same comparisons reverse: when \(A>0\), \(Au\le A\overline u\), and when \(A<0\), \(Au\le A\underline u\); equality holds only at the stated endpoint. When \(A=0\), every value still ties. \(\square\)
For a minimization HJB equation, this gives the feedback rule \(u^*(t,x)=\underline u\) where \(A(t,x)>0\), and \(u^*(t,x)=\overline u\) where \(A(t,x)<0\). Where \(A(t,x)=0\), any admissible value minimizes the pointwise Hamiltonian. Reversing the optimization objective reverses the endpoint choices. The theorem is a pointwise optimization result: it identifies a Hamiltonian minimizer, but by itself does not establish that a candidate \(V\) solves the HJB equation or that the resulting control is admissible for the original problem.
Proof. The zero set is closed because it is the inverse image of the closed set \(\{0\}\) under the continuous function \(A\). Now take a point \((t_0,x_0)\) where \(A(t_0,x_0)>0\). By continuity, there is a neighborhood on which \(A(t,x)>0\), so the minimizing endpoint there is \(\underline u\). If \(A(t_0,x_0)<0\), continuity gives a neighborhood on which \(A(t,x)<0\), so the minimizing endpoint there is \(\overline u\). These are all points outside the zero set. \(\square\)
Worked Examples of Switching Feedback
Worked Example: A State-Dependent Switching Threshold
Suppose the control interval is \([-1,2]\), the problem is a minimization, and the Hamiltonian is \(H(t,x,u)=x^2+u(2x-5)\). Determine the feedback and switching point.
The coefficient of \(u\) is \(A(t,x)=2x-5\). Its zero occurs at \(2x-5=0\), hence at \(x=\tfrac52\). If \(x<\tfrac52\), then \(2x-5<0\), and the theorem selects the upper endpoint \(u^*=2\). If \(x>\tfrac52\), then \(2x-5>0\), and it selects the lower endpoint \(u^*=-1\). At \(x=\tfrac52\), the coefficient is zero, so every control in \([-1,2]\) minimizes.
Check: at \(x=0\), the control-dependent term is \(-5u\), which is minimized at \(u=2\), giving \(-10\). At \(x=3\), it is \(u\), which is minimized at \(u=-1\), giving \(-1\). At \(x=\tfrac52\), it is \(0u=0\) for all admissible \(u\). These evaluations agree with the rule and the tie at the switching point.
Worked Example: Maximization Reverses the Endpoints
Maximize \(H(t,x,u)=3x+u(4-2x)\) over \(u\in[0,3]\). Find the switching surface and evaluate the maximizing control at \(x=1\), \(x=2\), and \(x=4\).
Here \(A(t,x)=4-2x\), which vanishes at \(x=2\). For \(x<2\), \(A>0\), so maximization selects \(u=3\). For \(x>2\), \(A<0\), so it selects \(u=0\). At \(x=2\), all controls in \([0,3]\) maximize.
At \(x=1\), the coefficient is \(4-2=2\), and \(H=3+2u\), whose maximum on \([0,3]\) is \(9\) at \(u=3\). At \(x=2\), \(H=6+0u=6\) for every \(u\). At \(x=4\), the coefficient is \(4-8=-4\), so \(H=12-4u\), whose maximum is \(12\) at \(u=0\).
Check: substituting the selected controls gives \(H(1,3)=3+2(3)=9\) and \(H(4,0)=12-4(0)=12\); the switching-state value is \(H(2,u)=6\) regardless of \(u\). The coefficient signs and direct evaluations agree.
Worked Example: A Time-Dependent Switching Surface
For a minimization Hamiltonian \(H(t,x,u)=x^2+u(x+t-1)\) with \(u\in[0,2]\), find the feedback and test it at \(t=\tfrac14\) for \(x=0\), \(x=\tfrac34\), and \(x=1\).
The coefficient is \(A(t,x)=x+t-1\). The switching set is the line \(x=1-t\). When \(x<1-t\), \(A<0\), so the minimizing control is \(u=2\). When \(x>1-t\), \(A>0\), so it is \(u=0\). On the line, every value in \([0,2]\) minimizes.
At \(t=\tfrac14\), the threshold is \(x=1-\tfrac14=\tfrac34\). For \(x=0\), \(A=0+\tfrac14-1=-\tfrac34\), so the minimizing control is \(2\), and \(H=\left(0\right)^2+2(-\tfrac34)=-\tfrac32\). For \(x=\tfrac34\), \(A=\tfrac34+\tfrac14-1=0\), so every control ties and \(H=(\tfrac34)^2=\tfrac9{16}\). For \(x=1\), \(A=1+\tfrac14-1=\tfrac14\), so the minimizing control is \(0\), and \(H=1^2+0(\tfrac14)=1\).
Check: at the switching state, direct substitution gives \(H(\tfrac14,\tfrac34,u)=\tfrac9{16}+u(0)=\tfrac9{16}\) for every admissible control. On either side, the selected endpoint is the one that makes the linear term smaller.
When the Bang-Bang Rule Does Not Apply
The essential hypothesis is affine dependence on the control, not merely a bounded control interval. If the diffusion coefficient depends on \(u\), the term \(\tfrac12\sigma(t,x,u)^2V_{xx}\) can be nonlinear in \(u\). A nonlinear running cost can have the same effect. In either case, minimizing the Hamiltonian need not amount to selecting an endpoint.
For instance, minimizing \(H(u)=(u-1)^2\) over \([0,2]\) gives \(u=1\): the value is \(0\) there, while both endpoint values are \(1\). This is not bang-bang. The affine theorem cannot be applied because the Hamiltonian is quadratic, not linear, in \(u\). More generally, endpoint comparison alone is insufficient for a nonlinear Hamiltonian; one must optimize its actual dependence on the control.
There are also two separate issues in applying the feedback. First, a feedback value must remain in the admissible interval; the endpoint rule guarantees this pointwise. Second, the chosen feedback must meet the problem’s admissibility requirements along the resulting state process. The pointwise Hamiltonian calculation does not establish those process-level properties. When a sufficiently smooth HJB solution and the required admissibility hypotheses are available, the Verification Theorem for classical HJB solutions supplies the connection between such a minimizing feedback and the value function.
A Common Sign Error
A frequent mistake is to choose the lower endpoint whenever the coefficient is negative in a minimization problem. For \(A<0\), that reverses the comparison: since \(u-\overline u\le0\), we have \(A(u-\overline u)\ge0\), so \(Au\ge A\overline u\). The upper endpoint, not the lower endpoint, minimizes.
For a concrete check, minimize \(H(u)=-2u\) over \([-1,2]\). The incorrect choice \(u=-1\) gives \(H=2\), while the correct choice \(u=2\) gives \(H=-4\). Since \(-4<2\), direct substitution confirms the theorem. At a zero coefficient, do not assign a unique endpoint: every admissible control has the same Hamiltonian value there.
Check Your Understanding
Use the sign of the control coefficient to justify each endpoint choice, and treat zero coefficients separately.
- For minimization of \(H=5+u(3-x)\) over \([-2,1]\), determine the feedback regions and switching state.
- For maximization of \(H=u(t-2x)\) over \([0,4]\), identify the switching set and describe the tie there.
- Give a condition on the drift, diffusion, and running cost that makes the control dependence of the HJB Hamiltonian affine.
- Why does the endpoint-selection theorem not justify a bang-bang policy for \(H(u)=(u-1)^2\) on \([0,2]\)?
- At a point where the affine coefficient vanishes, what additional information would be needed to select a unique feedback value?