1

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?

user2913869
  • 227

1 Answers1

0

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?

Juan Ospina
  • 2,257
  • I can't see how you got left = right on the first line. I tested it in Mathcad with randomly chosen functions: $x_1(t) = t^2$, $x2(t) = 0.5t^3$, $\delta x_1(t) = t^4$, and $\delta x_2(t) = (0.5t)^5$, and plotted from x = -2 to 2 and y = 0 to 200, and they did not look the same... – user2913869 Sep 16 '14 at 01:09
  • If you have $e^{x+\delta x}$ then at first order in $\delta$ we have $$e^xe^{\delta x}=e^x(1+\delta x)=e^x +e^x \delta x$$ – Juan Ospina Sep 16 '14 at 01:15
  • I'm just not seeing how you got that the left equals the middle. The theory of variations say that the perturbation delta-x is supposed to be arbitrary....and I arbitrarily chose $x_1(t) = t^2$ and $\delta x_1(t) = t^4$. Plotting your left equation and middle equation gives two different graphs. Maybe you thought that delta-x was a constant delta multiplied by function x(t)? – user2913869 Sep 16 '14 at 01:22
  • Given that $\delta x$ is a small value then $e^{\delta x} = 1+ \delta x$ at first order of $\delta x$. And $\delta x$ is not $\delta$ multiplied by $x$. Do you agree? – Juan Ospina Sep 16 '14 at 01:27
  • Okay, I see your point if delta-x is very small – user2913869 Sep 16 '14 at 01:33