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.



Double calculus of variations
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](https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation) 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


$$\,\,\,\left( \int\_0^1 \partial\_4\mathcal{L} (t\_1, t\_2, x (t\_1), x (t\_2), \dot x (t\_1), \dot x (t\_2)) \, \mathrm d t\_1 \right) - \frac{\mathrm d}{\mathrm d t\_2} \left( \int\_0^1 \partial\_6\mathcal{L} (t\_1, t\_2, x (t\_1), x (t\_2), \dot x (t\_1), \dot x (t\_2)) \, \mathrm d t\_1 \right) = 0$$


