# Optimizing a functional with initial value problem constraint

Let $$y$$ be a smooth function $$[0,L] \to \mathbb{R}$$. I am interested in finding the values of $$\alpha\in[0,2\pi)$$ such that the solution of the initial value problem $$\tag{1}\label{eq1} \begin{cases} g(t,y,y')=0,\\ y(0)=\alpha \end{cases}$$ is a local extremum of the functional $$J(y) = \int_{0}^{L}F(t,y)\,dt.$$

I thought I could solve this problem using calculus of variations. As far as I understood (and also looking at this question), if $$y$$ is both a solution of \eqref{eq1} and an extremum of $$J$$, then there exists a function $$\lambda = \lambda(t)$$ such that the following equation holds: $$\frac{\partial F}{\partial y} = \lambda \left(\frac{\partial g}{\partial y} - \frac{d}{dt} \frac{\partial g}{\partial y'}\right).$$ Am I correct?

In particular, in my case the second term inside the parenthesis vanishes, so that I only have $$\tag{2}\label{eq2} \frac{\partial F}{\partial y} = \lambda \frac{\partial g}{\partial y}.$$

Now, I can solve \eqref{eq2} for $$y= y(\lambda)$$. However, being $$\lambda$$ a function, substitution of $$y(\lambda)$$ into \eqref{eq1} gives a new initial value problem (this time in $$\lambda$$), and so I have not reduced the problem at all.

What am I missing?

• Don't forget the condition $g(t,y,y')=0$. With this and $F_y = \lambda g_y$ you have two equations and two unknowns $\lambda$ and $y$. – Cesareo Feb 8 at 12:38

The calculus of variation generally solves problems of the form

What arbitrary function $$y$$ extremizes $$I = \int {\cal F}[y](\mathbf{x}) \ {\rm d} \mathbf{x}$$, possibly given some constraints on $$y$$ and fixed boundary values?

You have a problem of the form

What arbitrary boundary value $$\alpha$$ extremizes $$I = \int {\cal F}[y_\alpha](\mathbf{x}) \ {\rm d} \mathbf{x}$$, given that $$y_\alpha(\mathbf{x})$$ is a uniquely fixed function given by $$\alpha$$ (by a procedure involving $$g$$)?

Now you could rephrase your problem to involve an arbitrary function $$y$$ and impose the constraint $$g(t,y,y')=0$$ using a Lagrange multiplier -- but the constrained calculus of variations is only ever going to tell you that the only possible choice of $$y$$ is $$y_\alpha$$, the thing satisfying the $$g$$ constraint. And then you're still left wondering what $$\alpha$$ you should choose.

You need to actually have a varying function of $$\alpha$$ which you extremize at some point. Clearly, $$\alpha$$ determines $$y_\alpha$$ (via $$g$$ constraint) and hence $$\alpha$$ determines $$j(\alpha) := J(y_\alpha)$$. You need to vary $$\alpha$$ in this expression and optimize.

(Note that you may be able to rewrite $$J(y_\alpha)$$ in a useful way using the equation $$y_\alpha$$ obeys, in order to extract its $$\alpha$$ dependence, maybe without ever needing to actually work out $$y_\alpha$$.)