I consider the functional $J(y)= \int_{0}^{a}y(x)dx$ where $a$ is fixed and $v\in H^{1}(0,a)$. We have a set of constraints
$$ K = \{ y\in H^{1}(0,a) : F(y) = \int_{0}^{a}\sqrt{1+y’(x)^{2}}dx - l=0\} $$
Where $l\in\mathbb{R}_{+}$ is fixed. Now I would like to use Lagrange multiplier, I need to find the derivative of $J$ which is $J’(v)(h)= \int_{0}^{a} h(x)dx$ and the derivative of $F$, and here is my main difficulty :
$$ F(y+h) = \int_{0}^{a}\sqrt{1 + y’(x)^{2} +2y’(x)h’(x) + h’(x)^{2}}dx - l $$
From there, I don’t know how to purse with this square root, I would like to have your help through some hints please.
Thank you a lot
Thanks to Ted Shriffin, I propose the following update
First we put $g(x)=\sqrt{1+x}$ and $f(x) = (x’)^{2}$ where $x$ is a function from $\mathbb{R}$ to $\mathbb{R}$.
$$ L(y+h)=\int_{0}^{a}g(f(y+h))dx - l = \int_{0}^{a}g(f(y) + f’(y)(h) + o_1(h))dx - l = \int_{0}^{a}g(f(y)) + g’(f(y))(f’(y)(h) + o_1(h)) + o_2(f’(y)(h) + o_1(h))dx - l = L(y) + \int_{0}^{a}g’(f(y))(f’(y)(h))dx + \int_{0}^{a}g’(f(y))(o_1(h)) + o_2(f’(y)(h) + o_1(h))dx $$
It can be show that the last term is a small o (though I did not check yet). Then we have
$$ L’(y)(h) = \int_{0}^{a}\frac{1}{2\sqrt{1+(y’(x))^{2}}}2y’(x)h’(x)dx $$
We notice that $J’(y)(h)=\int_{0}^{a}h(x)dx$. Now this maximization problem has a solution so the Lagrange theorem provides the existence of a $\lambda$ such that at the minimizer $y$ we have
$$ J’(y)(h) + \lambda L’(y)(h) = \int_{0}^{a} \frac{1}{\sqrt{1+(y’(x))^{2}}}y’(x)h’(x) +\lambda h(x)dx = 0 $$
From there, I think I should do an integration by part but when I tried this either with $h$ fixed or $y$ I find nothing that permits me to extract the form of $y$. Is this the good path please ?
Second update
Doing an integration by part for $L’(y)(h)$ we get
$$ L’(y)(h) = \int_{0}^{a} \left(\frac{y’’(x)}{\sqrt{1+y’(x)^{2}}} - \frac{y’’(x)y’(x)}{(1+y’(x)^{2})^{3/2}}\right)h(x)dx $$
(Since $h(x)$ vanishes at the boundary)
Thus we get
$$ J’(y)(h) + \lambda L’(y)(h) = \int_{0}^{a} \left(1+ \lambda\frac{y’’(x)}{\sqrt{1+y’(x)^{2}}} - \lambda\frac{y’’(x)y’(x)}{(1+y’(x)^{2})^{3/2}}\right)h(x)dx = 0 $$
From there, the fundamental lemma of calculus of variations allows me to get
$$ 1 +\lambda\left(\frac{y’’(x)}{\sqrt{1+y’(x)^{2}}} - \ \frac{y’’(x)y’(x)}{(1+y’(x)^{2})^{3/2}}\right) = 0 $$
The standard integration by parts procedure leads you to the Euler-Lagrange equation for the (constrained) extremum: If we're considering the functional $F(y) = \int_a^b \Phi(x,y,y')\,dx$, then at a local extremum we have $$\frac{\partial\Phi}{\partial y} = \frac d{dx}\left(\frac{\partial\Phi}{\partial y'}\right).$$ In this case, including the Lagrange multiplier, we should get the equation $$\frac d{dx}\left(\frac{y'}{\sqrt{1+y'{}^2}}\right) = -\frac1\lambda.$$