A little confused about finding the variation of the functional
J = $\int_{t0}^{tf}(e^{x_1(t)+x_2(t)})dt$
When I perturb and find the increment, I get:
$\Delta J = \int_{t0}^{tf} (e^{x_1(t) + \delta x_1(t) + x_2(t) + \delta x_2(t)} - e^{x_1(t) + x_2(t)}$)dt
To find the variation, I must eliminate any terms that are non-linear in $\delta x$, which pretty much eliminates the left term due to the exponential:
$\delta J = \int_{t0}^{tf} (-e^{x_1(t) + x_2(t)})dt$
I'm not sure where to go from here. I tried integration by parts, but I got stuck in an infinite loop. Am I missing something?
At first order:
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) +\delta\, x_{{1}} \left( t \right) +\delta\,x_{{2}} \left( t \right) }}={{\rm e} ^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}+{{\rm e}^{x_{{ 1}} \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1 }} \left( t \right) +\delta\,x_{{2}} \left( t \right) \right) $$
Then
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) +\delta\, x_{{1}} \left( t \right) +\delta\,x_{{2}} \left( t \right) }}-{{\rm e} ^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}={{\rm e}^{x_{{ 1}} \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1 }} \left( t \right) +\delta\,x_{{2}} \left( t \right) \right) $$
and the variation is
$$\delta J =\int _{t_{{0}}}^{t_{{f}}}\!{{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1}} \left( t \right) +\delta\,x_{{2}} \left( t \right) \right) {dt} $$
Then the corresponding Euler-Lagrange equation is
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}=0$$
Do you agree?