# Apply Euler-Lagrange equation with a integral condition

I want to find a curve $y=f(x)$ satisfy these conditions:

1. $f(0)=0$ and $f(1)=0$
2. $f(x)\ge 0$ for $0\le x\le 1$
3. The area between the curve of $f$ and X axis is equal to 1.
4. 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.

• From what you described, you do not seem to know anything about calculus of variations. It is hard to write an answer for you. – user99914 Jan 31 '18 at 19:47
• @JohnMa Thank you for notice and comment.I am not urgent to get the answer.I already know the application to solve the brachistochrone.But integral constraints confuse me.So any conference or chapter from textbooks would help. Maybe a direct answer is better, so I can know where should I go. – Jaqen Chou Jan 31 '18 at 21:31

1. 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 $S=\{0\leq x\leq 1\}$.
2. 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.
3. This effectively invalidates the boundary conditions $f(0)=0$ and $f(1)=0$ if one insists that $C$ is a graph for $f$.
• Notes for later: The half-chord is $R\sin\theta=\frac{1}{2}$ and the half-area is $\frac{R^2}{2}(\pi-\theta)+\frac{R^2\sin2\theta}{4}=\frac{1}{2}$. – Qmechanic Feb 4 '18 at 14:38