I want to find a curve $y=f(x)$ satisfy these conditions:
- $f(0)=0$ and $f(1)=0$
- $f(x)\ge 0$ for $0\le x\le 1$
- The area between the curve of $f$ and X axis is equal to 1.
- The length L of the curve $f$ ($0\le x\le 1$) is smallest.
So in these conditions, I know that I need to find the function $f$ that satisfy:
- $L=\int_{0}^{1}\sqrt{1+y'^2}\text dx $ is smallest.
- $f(0)=0$ and $f(1)=0$
- $\int_{0}^{1}y\text dx=1$
In the past, the extreme problems that I learned are to find a $x$ so that $f$ has a minimum or a maximum.It's about derivation. But this is to find a $f$ so that $L$ has a minimum. I was confused at first. But after some research, I already know this is about the calculus of variations and functional analysis.(And some interesting stories about Newton and Bernoulli). And I know that maybe I could use the Euler-Lagrange equation to solve the problem. In that case, it has condition like $y(x_0)=y_0$ and $y(x_1)=y_1$. But I have an extra condition $\int_{0}^{1}y\text dx=1$.
So I want to know am I doing this in a right way. Does E-L equation work? Or it has other deformation.
The problem is not well-posed.
If we abandon the requirement that the curve $C$ from $(0,0)$ to $(0,1)$ should be a graph of a function, then the optimal curve is an arc of a circle with radius $R\approx 0.6131 > 1/2$. This arc will not fit inside the strip $S=\{0\leq x\leq 1\}$.
Therefore if we insists that the curve $C\subseteq S$ is inside the strip, the optimal curve $C$ will consists of parts of the two vertical line $x=0$ and $x=1$, which in particular are not graphs of a function.
This effectively invalidates the boundary conditions $f(0)=0$ and $f(1)=0$ if one insists that $C$ is a graph for $f$.