I have to find the function $q(t)$ that optimizes the expected value of the integral $\int{V(q(t),t)dq}$ where t is a randomly distributed value with probability density $f(t)$. Thus, the functional to optimize would be $\int{f\int{Vq'dt}dt}$ My textbook just integrates by parts with the constant term for $F$ being $-1$ without any further explanation, though it makes sense to me why it should be either $0$ or $-1$. I've tried using the chain rule to get $\int{fδ\int{Vq'dt}dt}$ but that would just allow me to cancel out the $f$ after setting the first variation equal to zero, which doesn't make sense to me, as the problem would be independent from the probability distribution.
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In the formula $\int{f\int{Vq'dt}dt}$ you have $dt$ twice. That's not allowed. Can you clarify? – md2perpe Sep 29 '21 at 20:21
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Simply put, I have to optimize the expected value of the integral of V over q, but given that q is a function of t, I just replaced dq with q'dt – Unix Sep 29 '21 at 21:35
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The variable that is integrated over in the inner integral is not available in the outer integral. Is the outer variable found in the limits of the inner integral or should the inner integral somehow be over $q$ not $q',dt$? – md2perpe Sep 29 '21 at 21:45
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By the way, please use MathJax (~ LaTeX). For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – md2perpe Sep 29 '21 at 21:52
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The outer variable $t$ is both an argument of $V$ and $q$. However, $q$ is the control function, so it's replaced by $q'dt$. The purpose of the inner integral is that the outer one does not integrate along $q$ due to it being a function of $t$. Instead, I'd like to optimize the expected value antiderivative of $V$ on $q$, which integrates the antiderivative of V only on its second argument. – Unix Sep 29 '21 at 21:58
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Antiderivative, you say... Should the outer integral variable be the upper limit of the inner integral? – md2perpe Sep 29 '21 at 22:14
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It's the expected value of the antiderivative of $V$ on $q$ for a probability density function $f(t)$, where $t$ is also an argument on $V$ and $q$ – Unix Sep 30 '21 at 01:26
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The variable $t$ on the outer integral should be the upper limit, yes. In actuality, the antiderivative is already given, but given that I have to avoid integrating $q(t)$ on the outer integral, I have to add the inner integral, thus I can't just use the total antiderivative of $V(q(t),t)$ over $dt$, even though it's the original function I'm given. – Unix Sep 30 '21 at 06:51
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So the integral is something like $$\int_{s_0}^{s_1} f(s) \left( \int_{t_0}^{s} V(q(t), t) , q'(t) , dt \right) ds$$ – md2perpe Sep 30 '21 at 13:15
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Yes, I guess that should be it. If I'm not mistaken, the Euler Lagrange equation should be $f(s) δ\left( \int_{t_0}^{s} V(q(t), t) , q'(t) , dt \right)=0$, but it seems wrong to me that I should just ignore the probability distribution. – Unix Sep 30 '21 at 17:12
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Taking the functional to be $$ S[q] = \int_{s_0}^{s_1} f(s) \left( \int_{t_0}^{s} V(q(t), t) \, q'(t) \, dt \right) ds $$ and calculating the functional derivative w.r.t. $q$ I get $$\begin{align} \langle S'[q], \delta q \rangle &= \int_{s_0}^{s_1} f(s) \left( \int_{t_0}^{s} \left( \frac{\partial V(q(t), t)}{\partial q(t)} \delta q(t) \, q'(t) + V(q(t), t) \, \delta q'(t) \right) \, dt\right) ds \\ &= \int_{s_0}^{s_1} f(s) \left( \int_{t_0}^{s} \left( \frac{\partial V(q(t), t)}{\partial q(t)} q'(t) - \frac{dV(q(t), t)}{dt} \right) \delta q(t) \, dt\right) ds \end{align}$$ which suggests the Euler-Lagrange equations be $$ \frac{\partial V(q(t), t)}{\partial q(t)} q'(t) - \frac{dV(q(t), t)}{dt} = 0. $$ But $dV/dt$ can be expanded which results in $$ \frac{\partial V(q(t), t)}{\partial q(t)} q'(t) - \frac{\partial V(q(t), t)}{\partial q(t)} q'(t) - \frac{\partial V(q(t), t)}{\partial t} = 0. $$ Here the first two terms cancel and I'm left with $$ \frac{\partial V(q(t), t)}{\partial t} = 0. $$
md2perpe
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You see, that's my problem. I get the same result as if I only optimized the inner integral. Why is it legal to just ignore the probability distribution? – Unix Sep 30 '21 at 20:40
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@Unix. $f>0,$ isn't it? Then, for $\langle S'[q], \delta q\rangle$ to vanish, the inner integral must vanish. And then, if the innermost parenthesis is continuous and doesn't vanish everywhere, we can choose $\delta q$ to have the same sign as that at every $t.$ But that would give a positive inner integral, which contradicts that it must vanish. So the innermost parenthesis must vanish. – md2perpe Sep 30 '21 at 21:21
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