Euler Equation in Optimal Control Problem

Question

Let $c$ a continuous convex function.

And we consider the following problem

$$\int_0^1[x(t)E(t)-c(E(t))]dt$$ $$x(0)=1000 \quad x(1)=500$$ $$x'(t)=-x(t)E(t)$$

the exercise asks for Euler-Lagrange equation ?

Is it in this cas : if we consider $f(t,x,E)=xE-c(E)$

then the Euler Equation is : $\frac{d}{dt}(\frac{\partial f}{\partial E})=\frac{\partial f}{\partial x}$

which is equivalent to : $-E(t).(x(t)+1)=E'(t)c''(E(t))$ ?

is it true.

Because in optimal control, I think that we don't use Euler Equation, we just use the hamiltonian method.

Answer 1

Due to the dynamic restriction you should consider

$$ f(x,\dot x,e,\lambda) = x(t)e(t)-c(e(t))+\lambda(t)(\dot x(t)+x(t)e(t)) $$

giving

$$ e(t)(1+\lambda(t))-\dot\lambda(t) = 0\\ (1+\lambda(t))x(t)-\dot c(e(t)) = 0\\ \dot x(t)+x(t)e(t) = 0 $$

Answer 2

I would replace the restriction inside the integrand. From the restriction: $$ E(t) = -\frac{x^\prime(t)}{x(t)} $$

Replacing in the integrand $$ \int_0^1 \Biggl[ -x^\prime(t) - c \Bigl( \frac{-x^\prime(t)}{x(t)} \Bigr) \Biggr] dt $$

Now $$ f(t,x,x^\prime) = -x^\prime(t) - c \Bigl( \frac{-x^\prime(t)}{x(t)} \Bigr) $$

Euler equation is $$ \frac{\partial}{\partial t} \Biggl[ \frac{\partial f(\cdot)}{\partial x^\prime} \Biggr] = \frac{\partial f(\cdot)}{\partial x} $$

Now computing those partial derivatives is hard work, but if I did it correctly (thanks Gonzalo) I get $$ x^{\prime\prime}(t) x(t) = (x^\prime(t))^2 $$

That's the Euler equation, which is nonlinear. Good thing is that the c function disappears.