Euler equation first integral of $f^* = (y-\lambda)\sqrt{1+y'^2}$ is $\frac{y-\lambda}{\sqrt{1+y'^2}}=A$?

Question

I'm looking at this isoperimetric problem with $$I = \int_{0}^a{y\sqrt{1+y'^2}}dx $$ subject to the constraint $$J = \int_{0}^a{y\sqrt{1+y'^2}}dx = l $$ Using Lagrange multipliers so $$I^* = I-\lambda J = \int_{0}^a{(y-\lambda)\sqrt{1+y'^2}}dx = min$$ $$y(0)=y(a)=0$$ And the book is saying that since $f^*=(y-\lambda)\sqrt{1+y'^2}$ doesn't contain any explicit x dependence, the Euler equation has the first integral $$\frac{y-\lambda}{\sqrt{1+y'^2}}=A$$

I'm not able to produce this result. The Euler equation is $\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y'})=0$, and integrating it should give $$\int{\left[\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)\right]dx}=x\frac{\partial f}{\partial y}-\left(\frac{\partial f}{\partial y'}\right)=x\sqrt{1+y'^2}-\frac{(y-\lambda)(y')}{\sqrt{1+y'^2}}=A$$

Can somebody show me what I'm misunderstanding?

Answer 1

The problem with your last integration is that $\partial f/\partial y$ is still a function of $x$, so you can't just integrate it as if it's constant.

The book is probably alluding to the Beltrami identity, which says that if $\partial f/\partial x=0$ (no explicit $x$-dependence), $$ f- y' \frac{\partial f}{\partial y'} $$ is constant (that is, independent of $x$) for a $y$ satisfying the Euler–Lagrange equation. (This is the obscure first integral: other special cases have more obvious ones, like $\partial f /\partial y=0$ means that $\partial f/\partial y'$ is constant.) One can verify this easily enough: $$ \frac{d}{dx} \left( f- y' \frac{\partial f}{\partial y'} \right) = \frac{\partial f}{\partial x} + y'\frac{\partial f}{\partial y} + y'' \frac{\partial f}{\partial y'} - y'' \frac{\partial f}{\partial y'} - y' \frac{d}{dx}\frac{\partial f}{\partial y'} \\ = 0 + \left( \frac{\partial f}{\partial y} - y' \frac{d}{dx}\frac{\partial f}{\partial y'} \right)y' = 0. $$

Using this then gives $$ A = (y-\lambda)\sqrt{1+y'^2} - y'(y-\lambda)\frac{y'}{1+y'^2} = \frac{y-\lambda}{\sqrt{1+y'^2}}. $$