I was attempting to use the calculus of variations to find the unique function $f$ such that
- f is smooth
- $f(1) = 1, f(2) = 3$
- $f(x+1) - f(x) = (x+1)$
- $f$ minimizes its arclength from $1$ to $2$.
Now I know that in general $$f(x) = \frac{1}{2}x(x+1) + \sum_{n=0}^{\infty} \left[ a_n \sin(2\pi nx) \right] $$ (for $a_n \in \mathbb{R} $) can be made to satisfy conditions 1 through 3. But I imagine that condition (4) would cut down our solution space strictly to $\frac{1}{2}x(x+1)$.
So I begin setting up our problem. We have a constraint $g(x,f)=0$ given by:
$$ g(x,f) = f(x+1) - f(x) - (x+1)$$
So we define our lagrangian
$$ \mathcal{L}(x,f) = \sqrt{1+(f')^2} - \lambda(x)(f(x+1) - f(x) - (x+1)) $$ So we setup a calculus of variations problems with lagrange multiplier $\lambda(x)$ that is we want to find a smooth choice of $f(x), \lambda(x)$ which minimizes
$$ \int_{1}^{2} \sqrt{1 + (f'(x))^2} - \lambda(x)(f(x+1) - f(x) - (x+1)) dx$$
The stationary solutions to this will be those satisfying
$$ \frac{\delta \mathcal{L}}{\delta f} = 0 \\ \frac{\delta \mathcal{L}}{\delta \lambda} = 0 $$
Recall that:
$$\frac{\delta \mathcal{L}}{\delta f} = \frac{\partial \mathcal{L} }{\partial f} - \frac{d}{dx} \left[ \frac{\partial \mathcal{L} }{\partial f'}\right] + \frac{d^2}{dx^2} \left[ \frac{\partial \mathcal{L} }{\partial f''}\right] - .... $$
And Recall that: $$f(x+1)-f(x)= f'(x) + \frac{1}{2!}f''(x) + \frac{1}{3!}f'''(x) .... $$
So putting this together and substituting into our integral we have
$$\frac{\delta \mathcal{L}}{\delta f} = \frac{d}{dx} \left[ \frac{f'}{\sqrt{1+(f')^2}} \right] +\lambda'(x) - \frac{1}{2!} \lambda''(x) + \frac{1}{3!} \lambda'''(x) ... = 0$$
$$ \frac{\delta \mathcal{L}}{\delta \lambda} = f(x+1)-f(x)-(x+1)=0$$
And now here is the problem. For ANY choice of $f$, it will always be possible to find a choice of $\lambda$ so that the first equation is true. And the second equation, after adding the boundary conditions on integer $f$ still has the same large solution set we mentioned at the start of this problem.
Where did I go wrong here? I know that these solutions CANNOT ALL be locally optimal since I can perturb any one solution of our original functional equation by a small SPECIFIC test function: $\epsilon \sin(2\pi nx) $ and get another solution, and these solutions will have obviously different arclengths.
You are looking for a thing which is simply not there.
Obviously, the non-smooth solution is the piecewise linear function ($2x-1$ for $1<x\leqslant2$, then $3x-3$ for $2<x\leqslant3$, and so on). Also obviously, we can approximate it with smooth functions on $[1,2]$ as good as we want, both in terms of $f$ and the arc length. Together, these facts mean that the desired minimum is never reached in the class of smooth functions. This is not uncommon for a variational problem.
So it goes.