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

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

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