Maximising $\int_{0}^{T} v(t)\, dt$, subject to constraints $|v(t)| \leq a; v(0)=0; v(T)=0$

Question

Besides those constraints, we know nothing else about $v(t)$. Interpreting the integral as the distance travelled by a particle, a little geometry tells us that the answer should be $aT^{2}/4$

However, I'm interested in knowing whether there's an analytic approach, which hopefully allows for generalisation. I don't have a background in variational calculus, but I do know one variable calculus. Would I be able to maximise the above integral without appeal to variational calculus? If the answer is no, I'd like to know how one would go about doing it with the full machinery.

Answer 1

By the mean value theorem, $v(t) \le v'_\mathrm{max} t = at$ for all $t$. It follows that $I(v) = \int_0^T v(t) \le aT^2/2$. So this is an upper bound.

This bound cannot be reached. Indeed suppose that you have some $v$ with $v(0) = 0$, $v'(t) \le a$ for all $t$, and $I(v) = a T^2 / 2$. Then $\int_0^T (at - v(t)) dt = 0$; but $at - v(t) \ge 0$, which implies that $v(t) = at$ for all $t$. In particular $v(T) = aT \neq 0$, which means that $v$ does not fulfill your conditions.

The best we can hope for is some sequence of functions $v_n$ such that all the $v_n$ satisfy the conditions and $\int_0^T v_n(t) dt \to aT^2/2$ when $n \to +\infty$. Let $x_n = 1 - 1/n$, then you can for example take the piecewise linear function: $$v_n(t) = \begin{cases} at & 0 \le t \le x_n \\ \frac{t-T}{x_n-T} x_n & x_n \le t \le 1 \end{cases}$$ and smooth it out to get a differentiable function. Since $\int_0^T v_n(t) dt$, this shows that $aT^2/2$ really is the sup of possible integrals of functions satisfying your conditions.

Answer 2

If you have $v(0)=v(T)=0$, and $\lvert v'(t)\rvert \leqslant a$ for all $t\in[0,T]$, then the fundamental theorem of calculus ensures, for all $t\in[0,T/2]$, that $$ v(t)=v(t)-v(0) = \int_0^t v'(t)dt \leqslant at, $$ and for all $t\in[T/2,T]$, that $$ v(t) =-(v(T)-v(t))=-\int_t^T v'(t)dt \leqslant a(T-t). $$

Integrating then yields $$ \int _0^T v(t) dt\leqslant \frac{1}{4}aT^2. $$

On the other hand, the choice $$ v(t)=\begin{cases} at,&t\in[0,T/2],\\ a(T-t),&t\in[T/2,T], \end{cases} $$ satisfies $v(0)=v(T)=0$, $$ v'(t)=\begin{cases} a,&t\in[0,T/2),\\ -a,&t\in(T/2,T], \end{cases} $$ and $$ \int_0^T v(t)dt=\frac{1}{4}aT^2. $$

Note that this choice of $v$ is only piecewise differentiable, so whether or not you agree that $v$ maximizes your functional depends on the class in which you are looking for a maximizer.