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How can I calculate the functional derivative of this functional?

$$F[x](t) = \int_{0}^{t}x(t_1)a(t_1)\left \{ \int_{0}^{t_1}x(t_2)b(t_2) \,dt_2\right \} dt_1 .$$

Where $a(t)$ and $b(t)$ are real fixed functions. Following the usual method, I'm getting

$$\frac{\delta F\left [ x \right ]}{\delta x(s) } = \int_{0}^{s}\left [ a(t')b(s)+a(s)b(t') \right ]x(t')dt'.$$

which seems odd since that is also a functional, and I think that I should be getting a function.

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    $\begingroup$ I cannot edit because it's just a 3 character-change, but in the first equation it should be $F[x]$ instead of $F[x(t)]$. $\endgroup$ – anderstood Sep 4 '14 at 2:55
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Use the identity \begin{align} \frac{\delta x(t)}{\delta x(s)} = \delta(t-s) \end{align} to obtain \begin{align} \frac{\delta}{\delta x(s)}\Big(x(t_1)x(t_2)a(t_1)b(t_2)\Big) &= \delta(t_1-s)x(t_2)a(t_1)b(t_2) + x(t_1)\delta(t_2-s)a(t_1)b(t_2) \end{align} from which it follows that \begin{align} \frac{\delta F}{\delta x(s)}[x] = a(s)\int_0^s x(t_2)b(t_2)\,dt_2 + b(s)\int_s^tx(t_1)a(t_1)\, dt_1 \end{align} where we have assumed $0<s<t$ and $0<s<t_1$.

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  • $\begingroup$ I think that the lower limit of the second integral is $s$, because $t_1>t_2=s$ $\endgroup$ – Trimok Sep 4 '14 at 10:41
  • $\begingroup$ @Trimok Thanks; edited. $\endgroup$ – joshphysics Sep 4 '14 at 18:18
  • $\begingroup$ Thanks for the answer! One more thing, is it possible now to take the derivative of this new expression? Which would be the second order derivative of the functional? Would that be $$\frac{\delta ^2 F\left [ x \right ]}{\delta x(y)\delta x(s) } = b(s)a(y)+a(s)b(y)$$ ??? $\endgroup$ – Martin Drech Sep 4 '14 at 18:39
  • $\begingroup$ Any chance for assistance here: math.stackexchange.com/questions/1066495/… $\endgroup$ – Royi Dec 14 '14 at 6:48

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