Finding First Variation Given the functional $ I(U) = \int_0^1 F(u'(x),u(x), x) dx$ 
and: $F(p,u,x)= \sqrt{p^2+u^2}$
how would you find the first variation using say the definition on Wikipedia (Gateaux Derivative) and what's the purpose of it? I know the Euler-Lagrange tell you the functions which make the given functional stationary, but I'm not quite certain how to relate these two ideas. Thanks
 A: Let
$$\begin{align}I[r]&=\int_{\theta_0}^{\theta_1}L(\theta,r,r')\mathrm{d}\theta\\
L(\theta,r,r')&=\sqrt{(r')^2+r^2}\text{.}\end{align}$$
Write
$$L(h)=L(\theta,r+h\delta r,r'+h\delta r')$$
where $h$ is a scalar. If $L$ is differentiable at $h=0$ for constant $\theta$, $r$, $r'$ $\delta r$, and $\delta r'$, then
$$L(h)=L(0)+h L'(0)+o(h)$$
where
$$L'(0)=L_{,r}(\theta,r,r')\delta r+L_{,r'}(\theta,r,r')\delta r'\text{.}$$
Then 
$$\begin{split}I[r+h\delta r]&=\int_{\theta_0}^{\theta_1}L(h)\,\mathrm{d}\theta\\
&=\int_{\theta_0}^{\theta_1}\left(L(0)+hL'(0)+o(h)\right)\,\mathrm{d}\theta\\
&=I[r]+h\int_{\theta_0}^{\theta_1}L'(0)\mathrm{d}\theta +o(h)\text{.}
\end{split}$$
and
$$\begin{split}\int_{\theta_0}^{\theta_1}L'(0)\mathrm{d}\theta&=
\int_{\theta_0}^{\theta_1}\left(L_{,r}(\theta,r,r')\delta r+L_{,r'}(\theta,r,r')\delta r'\right)\mathrm{d}\theta \\
&=\int_{\theta_0}^{\theta_1}\left(L_{,r}(\theta,r,r')-\frac{\mathrm{d}}{\mathrm{d}\theta}L_{,r'}(\theta,r,r')\right)\delta r\mathrm{d}\theta+\left.L_{,r'}(\theta,r,r')\delta r\right\rvert_{\theta_0}^{\theta_1}\text{.}
\end{split}$$
For each function $r$, the right side of this last equality is a linear functional in the function $\delta r$: call it $\nabla I[r]$:
$$\langle \nabla I[r],\delta r\rangle\stackrel{\text{def}}{=}\int_{\theta_0}^{\theta_1}\left(L_{,r}-\frac{\mathrm{d}}{\mathrm{d}\theta}L_{,r'}\right)\delta r\mathrm{d}\theta+\left.L_{,r'}\delta r\right\rvert_{\theta_0}^{\theta_1}\text{.}$$
Then
$$I[r+h\delta r]=I[r]+h\langle\nabla I[r],\delta r\rangle + o(h)\text{;}$$
$\nabla I[r]$ is precisely the Gâteaux derivative (or first variation) of $I$ with respect to $r$. Note that $\nabla I[r]$ is composed of two terms:


*

*a "bulk" term $\int_{\theta_0}^{\theta_1}\left(L_{,r}-\frac{\mathrm{d}}{\mathrm{d}\theta}L_{,r'}\right)\delta r\mathrm{d}\theta$—this vanishes for all $\delta r$ if and only if the Euler—Lagrange equations hold; and

*a "boundary" term $\left.L_{,r'}\delta r\right\rvert_{\theta_0}^{\theta_1}$—setting this to zero gives us boundary conditions for the Euler—Lagrange equation if the endpoints of the variation are not fixed.


Frequently, one encounters minimization problems in mechanics in which boundary conditions are not fixed—using all of the first variation instead of just the Euler—Lagrange equations tells us what those boundary conditions should be.
