Constrained optimization
In pure optimization problems, one defines the Lagrangian as
$$\mathcal{L}(x,\lambda) = f(x) + \lambda^\top g(x) \,,$$
where $f(x)$ is the cost function we whish to minimize, $g(x)\leq0$ are inequality constraints, $x$ are the decision variables and $\lambda$ are the Lagrange multipliers.
This formulation implies
$$\lambda\geq0 \tag{1}\label{1}$$
with
$$
\lambda
\begin{cases}
= 0, \quad g(x) \lt 0\\
\geq 0, \quad g(x) = 0
\end{cases}
$$
being part of the optimality conditions.
Optimal control
In optimal control, one defines the Hamiltonian as
$$\mathcal{H}(t,x(t),u(t),\lambda(t)) = f(t,x(t),u(t)) + \lambda^\top(t) \, g(t,x(t),u(t))$$
where $f(\cdot)$ is a distributed cost, $g(\cdot) – \dot{x}(t)=0$ are the equations of motion and $\lambda(t)$ are the costates.
In this case, the optimality condition on the costate is
$$ \dfrac{\partial \mathcal{H}}{ \partial x(t) }+ \dot{\lambda}^\top(t) = 0 \tag{2} \label{2}$$
Question
Given that the costates in optimal control inherit the principles of Lagrange multipliers in optimization, does the condition $\lambda(t)\geq0$ from Eq. \eqref{1} still apply to the costates in optimal control? Why so or why not?
Best Answer
The condition that the Lagrange multipliers must be non-negative only applies if the associated constraint is an inequality. If the constraint is an equality there is no such condition. This is true for constrained parametric optimization as well as for optimal control.
In answer to the question, the condition $\lambda(t) \geq 0$ does not apply for the constraint $g(\cdot) - \dot{x} = 0$, because this is an equality constraint. It is not because some property/principle is lost when one switches from contrained optimization to optimal control.