I've been given the calculus of variations problem of extremizing $$I=\int_a^bf(x,y,y')dx=\int_a^by\sqrt{1+y'^2}dx$$ where of course $y$ is a function of $x$. I came up with what I thought was a pretty sharp solution method that got around nonlinearities in the application of the Euler-Lagrange equation, but it leads to an apparent contradiction:
Since there is no direct $x$ dependence in the integrand, the Hamiltonian/first-integral must be constant for the extremizing $y$, i.e.
$$f-y'\frac{\partial f}{\partial y'}=c_1\Leftrightarrow y\sqrt{1+y'^2}-\frac{yy'^2}{\sqrt{1+y'^2}}=c_1\Leftrightarrow y+yy'^2-yy'^2=c_1\sqrt{1+y'^2}$$
$$\Leftrightarrow y^2=c_1^2(1+y'^2)\Leftrightarrow 1+y'^2= \frac{y^2}{c_1^2}.\tag{1}$$
Extremal $y$ must also satisfy the normal Euler-Lagrange equations
$$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}\Leftrightarrow \sqrt{1+y'^2}=\frac{d}{dx}\left[ \frac{yy'}{\sqrt{1+y'^2}} \right]$$
$$\Leftrightarrow \sqrt{1+y'^2}=\frac{y'^2}{\sqrt{1+y'^2}}+yy''\left(\frac{1}{\sqrt{1+y'^2}}- \frac{y'^2}{(1+y'^2)^{3/2}} \right)$$
$$\Leftrightarrow (1+y'^2)^2=y'^2+y'^4+yy''+yy'^2y''-yy'^2y''$$
$$\Leftrightarrow y'^4+2y'^2+1=y'^2+y'^4+yy''$$ $$\Leftrightarrow y'^2-yy''+1=0.\tag{2}$$
Substituting the first-integral expression for $1+y'^2$,
$$\frac{y^2}{c_1^2}=yy''\Leftrightarrow y''-\frac{y}{c_1^2}=0.\tag{3}$$
However, the solution to this linear ODE is not a solution to the Euler-Lagrange ODE. The only way I can see that the argument is unjustified is if $c_1=0$ and therefore $y=0$, but boundary conditions were given in the problem that this solution cannot satisfy. I've even written out proofs of the E-L equations and Hamiltonian conservation that don't appear to make any unjustified assumptions.
Any ideas?
Best Answer
You manipulate two pretty much equivalent (if we neglect the constant solutions) differential equations into the third equation. The manipulation is not an equivalence, i.e. the new equation may have more solutions that the original one(s). The only thing you know is that the extremals are going to satisfy the new equation (indeed, they do as they are $\cosh$), but you won't be able to purify the solutions of the third equation to become (spurious) extremals as well.
Example: neglecting constants $y'=y$ and $y''=y'$ have the same solution $Ce^x$, however, $y''=y$ have more solutions as $e^{-x}$ is not a solution to the first two equations.