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The problem is to maximize $\int_0^1 y(t) + u(t)^2 dt$ where $y$ is state and $u$ is control.

Further we have $y' = u, y(0) = 5$.

I set up the Maximum Principle equations, but, in particular, I need to maximize the Hamiltonian in the $u$-variable.

My nstructor's solutions does this by differentiating and letting it equal zero, i.e. he gets $$2u(t) + \lambda = 0.$$

Then he goes on with the other equations, and sovles for $\lambda$, and then $u$.

However, if we differentiate the Hamiltonian again with respect to $u$, then we get $$2 > 0$$ which is not negative, hence surely the 1st equation is not warranted, as we have found a minimum, not maximum?

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  • $\begingroup$ Is your control unconstrained? $\endgroup$ – Boaten Oct 24 '17 at 17:01
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Hamiltonian \begin{align} H(y(t),u(t),\lambda(t)) = y(t)+u(t)^2 + \lambda(t) u(t) \end{align}

First order conditions read \begin{align} &H_u = 0 \quad \Longleftrightarrow \quad u(t) = -\frac{\lambda(t)}{2}\\[2mm] &\frac{d\lambda(t)}{dt} = -H_y = -1\\[2mm] &\lambda(1) = 0 \end{align} Integrate costate \begin{align} &\int \frac{d\lambda(t)}{dt} = \int -1\\[2mm] &\int d\lambda(t) = \int -dt\\[2mm] &\lambda(t) = -t + C \end{align} with \begin{align} &\lambda(1) = -1 + C = 0 \quad \Longleftrightarrow \quad C = 1\\[2mm] \Longrightarrow \quad & \lambda(t) = 1-t\\[2mm] \Longrightarrow \quad & u(t) = \frac{t-1}{2} \end{align} Now integrate state equation \begin{align} &\int \frac{dy(t)}{dt} = \int \frac{t-1}{2}\\[2mm] &\int dy(t) = \int \frac{t-1}{2}dt\\[2mm] &y(t) = \frac{t}{2}\left(\frac{t}{2}-1\right) + C \end{align} with \begin{align} &y(0) = C = 5\\ \Longrightarrow \quad & y(t) = \frac{t}{2}\left(\frac{t}{2}-1\right) + 5 \end{align}

If I'm not mistaken necessary conditions are sufficient if maximized Hamiltonian is concave in state (Arrow sufficiency if I recall correctly (see e.g. Caputo, 2005, Ch. 3)) \begin{align} H^*(y,\lambda) = \max_u H(y,u,\lambda) = y - \frac{\lambda^2}{2}. \end{align} Now we have \begin{align} H_y > 0 = H_{yy} \end{align} such that concavity is given.

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