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A standard formulation of the one-dimensional variational problem is to find necessary and sufficient conditions for the functional $x:\mathbb R\rightarrow \mathbb R$ that minimizes

$ \int_0^1 L[t, x_t, \dot x_t] dt $

For a given $L: \mathbb R^3\rightarrow \mathbb R$ that is sufficiently well-behaved.

I am encountering applications that result in a slightly different problem:

$ \int_{[0,1]^2} L[s, t, x_s, x_t, \dot x_s, \dot x_t] dsdt $

I believe this must have either been solved, or it has been covered in some paper; but I couldn't find anything. Perhaps control theorists are very familiar. Any recommendation/pointer/way to attack the problem is welcome.

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closed as off-topic by Michael Renardy, Jan-Christoph Schlage-Puchta, Sebastian Goette, Ben McKay, José Figueroa-O'Farrill Mar 21 '17 at 12:34

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  • $\begingroup$ There is an enormous literature on calculus of variations for multiple integrals. Even some of the Hilbert problems came from this field! $\endgroup$ – Michael Renardy Mar 20 '17 at 18:06
  • $\begingroup$ Necessary and sufficient conditions are not easy in either case, but the same procedure which produces necessary conditions (the Euler equations) in the one variable case also works in the two variable case - you just have to replace the integration by parts step with the divergence theorem. There are similar sufficient conditions involving the Legendre test etc. but they are generally a bit more complicated because boundary value problems are hard. $\endgroup$ – Paul Siegel Mar 20 '17 at 19:37
  • $\begingroup$ I think you replace integration by part with divergence when you deal with problems in many variables. In this case the problem is in a single variable, $x$. $\endgroup$ – gappy3000 Mar 20 '17 at 21:50
  • $\begingroup$ If you have never studied calculus of variations with multiple independent variables, it's hard to say where to start. You can find quite a thorough treatment in Giaquinta & Hildebrandt's Calculus of Variations (vol. I)[dx.doi.org/10.1007/978-3-662-03278-7] and (vol. II)[dx.doi.org/10.1007/978-3-662-06201-2]. However, depending on your background, this may or may not be easy reading. $\endgroup$ – Igor Khavkine Mar 20 '17 at 23:17
  • $\begingroup$ Thanks. Very helpful. The only book I had is Ewing's Calculus of Variations. $\endgroup$ – gappy3000 Mar 21 '17 at 14:41
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Let

$$S (x) := \iint_{[0,1]^2} \mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_1 \mathrm d t_2$$

Hence,

$$\begin{array}{rl} \delta S := S (x + \delta x) - S (x) &= \displaystyle\iint_{[0,1]^2} \partial_3\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \delta x (t_1) \, \mathrm d t_1 \mathrm d t_2\\\\ &+ \displaystyle\iint_{[0,1]^2} \partial_4\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \delta x (t_2) \, \mathrm d t_1 \mathrm d t_2\\\\ &+ \displaystyle\iint_{[0,1]^2} \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \delta \dot x (t_1) \, \mathrm d t_1 \mathrm d t_2\\\\ &+ \displaystyle\iint_{[0,1]^2} \partial_6\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \delta \dot x (t_2) \, \mathrm d t_1 \mathrm d t_2\end{array}$$

Integrating by parts,

$$\begin{array}{rl} &\quad \displaystyle\iint_{[0,1]^2} \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \delta \dot x (t_1) \, \mathrm d t_1 \mathrm d t_2\\\\ &= \displaystyle\int_0^1 \left(\int_0^1 \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_2 \right) \delta \dot x (t_1) \, \mathrm d t_1\\\\ &= \left(\displaystyle\int_0^1 \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_2 \right) \delta x (t_1) \, \bigg|_0^1 \\\\ &- \displaystyle\int_0^1 \frac{\mathrm d}{\mathrm d t_1}\left(\int_0^1 \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_2 \right) \delta x (t_1) \, \mathrm d t_1\end{array}$$

Thus, the Euler-Lagrange equations are

$$\,\,\,\left( \int_0^1 \partial_3\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_2 \right) - \frac{\mathrm d}{\mathrm d t_1} \left( \int_0^1 \partial_5\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_2 \right) = 0$$

and

$$\,\,\,\left( \int_0^1 \partial_4\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_1 \right) - \frac{\mathrm d}{\mathrm d t_2} \left( \int_0^1 \partial_6\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_1 \right) = 0$$

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  • 1
    $\begingroup$ Thanks. I had worked out the same equations after asking the question. $\endgroup$ – gappy3000 Mar 21 '17 at 14:42

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