Double calculus of variations 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.
 A: 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
$$\,\,\,\left( \int_0^1 \partial_4\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_1 \right)  - \frac{\mathrm d}{\mathrm d t_2} \left( \int_0^1 \partial_6\mathcal{L} (t_1, t_2, x (t_1), x (t_2), \dot x (t_1), \dot x (t_2)) \, \mathrm d t_1 \right) = 0$$
