Lesson plan
Singular control
In optimal control, problems of singular control are problems that are difficult to solve because a straightforward application of Pontryagin's minimum principle fails to yield a complete solution. Only a few such problems have been solved, such as Merton's portfolio problem in financial economics or trajectory optimization in aeronautics. A more technical explanation follows. The most common difficulty in applying Pontryagin's principle arises when the Hamiltonian depends linearly on the control u , i.e., is of the form: H(u)=\phi(x,\lambda,t)u+\cdots and the control is restricted to being between an upper and a lower bound: a\le u(t)\le b . To minimize H(u) , we need to make u as big or as small as possible, depending on the sign of \phi(x,\lambda,t) , specifically: : u(t) = \begin{cases} b, & \phi(x,\lambda,t) 0.\end{cases} If \phi is positive at some times, negative at others and is only zero instantaneously, then the solution is straightforward and is a bang-bang control that switches from b to a at times when \phi switches from negative to positive. The case when \phi remains at zero for a finite length of time t_1\le t\le t_2 is called the singular control case. Between t_1 and t_2 the maximization of the Hamiltonian with respect to u gives us no useful information and the solution in that time interval is going to have to be found from other considerations.…
Article summary
In optimal control, problems of singular control are problems that are difficult to solve because a straightforward application of Pontryagin's minimum principle fails to yield a complete solution. Only a few such problems have been solved, such as Merton's portfolio problem in financial economics or trajectory optimization in aeronautics. A more technical explanation follows. The most common difficulty in applying Pontryagin's principle arises when the Hamiltonian depends linearly on the control u , i.e., is of the form: H(u)=\phi(x,\lambda,t)u+\cdots and the control is restricted to being between an upper and a lower bound: a\le u(t)\le b . To minimize H(u) , we need to make u as big or as small as possible, depending on the sign of \phi(x,\lambda,t) , specifically: : u(t) = \begin{cases} b, & \phi(x,\lambda,t) 0.\end{cases} If \phi is positive at some times, negative at others and is only zero instantaneously, then the solution is straightforward and is a bang-bang control that switches from b to a at times when \phi switches from negative to positive. The case when \phi remains at zero for a finite length of time t_1\le t\le t_2 is called the singular control case. Between t_1 and t_2 the maximization of the Hamiltonian with respect to u gives us no useful information and the solution in that time interval is going to have to be found from other considerations.…
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Discussion questions
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