2.5.2 Non-integral constraints

We now suppose that instead of the integral constraint (2.45) we have an equality constraint which must hold pointwise:

$$M(x,y(x),y'(x))=0$$ (2.51)

for all $x\in [a,b]$ . A vector-valued function $M$ can be used to describe multiple constraints, but here we assume for simplicity that there is only one constraint.

Let $y$ be a test curve. It turns out that the first-order necessary condition for optimality in this case is similar to that for integral constraints, but the Lagrange multiplier is now a function of $x$ . In other words, the Euler-Lagrange equation must hold for the augmented Lagrangian

$$L+\lambda^*(x) M$$

where $\lambda^*:[a,b]\to\mathbb{R}$ is some function. As in Section 2.5.1, we need a technical assumption to rule out degenerate cases. Here we need to assume that there are at least two degrees of freedom and that everywhere along the curve we have ${M}_{y'}\ne 0$ or, if $y'$ does not appear in (2.52), ${M}_{y}\ne 0$ . (This permits us, via the Implicit Function Theorem, to locally solve for one dependent variable component in (2.52) in terms of the others and to guarantee the existence of other curves near $y$ satisfying the constraint.)

We will not give a proof of the above result, and instead limit ourselves to an intuitive explanation. The integral constraint (2.45) is global, in the sense that it applies to the entire curve. In contrast, the non-integral constraint (2.52) is local, i.e., applies to each point on the curve. Locally around each point, there is no difference between the two. This suggests that for each $x$ there should exist a Lagrange multiplier, and these can be pieced together to give the desired function $\lambda^*=\lambda^*(x)$ . Another way to see the correspondence between the two problems is to follow Lagrange's idea of considering an augmented cost functional which coincides with the original one for curves satisfying the constraint. For integral constraints this was accomplished by (2.51), while here we can consider the minimization of

$$\int_a^b Ldx+\int_a^b \lambda(x) Mdx$$ (2.52)

over $y$ and $\lambda$ . Note that $\lambda$ no longer needs to be constant, since $M$ is identically 0. (The role of integration in the second term is simply to obtain a cost for the whole curve.)

What if we want to solve the constraint (2.52) for $y'$ as a function of $x$ and $y$ ? In general, we have fewer constraints than the dimension of $y'$ , i.e., the system of equations is under-determined. This means that we will have some free parameters; denoting by $u$ the vector of these parameters, we can write the solution in the form

$$y'=f(x,y,u).$$

The reader hopefully recognizes this as a control system, in the disguise of a somewhat unfamiliar notation compared to (1.1). For example, if $n=2$ (two degrees of freedom) and $M(x,y,y')=y'_1-y_2$ , then we have $y'_1=y_2$ and $y'_2=u$ (free). Incidentally, the case of no constraints (our Basic Calculus of Variations Problem) corresponds to $y'=u$ . We will turn our attention to control problems later in the book, and the present lack of depth in our treatment of non-integral constraints will be amply compensated for. In particular, the ``distributed" Lagrange multiplier $\lambda^*(\cdot)$ will correspond to the adjoint vector, or costate, in the maximum principle.