The isoperimetric problem is the following:
Among all curves of length L in the upper half-plane passing through the points $( - a, 0)$ and $(a, 0)$, find the one which together with the interval $[ - a, a]$ encloses the largest area.
Solution from Formin and Gelfand's book of calculus of variations (page 49):
We are looking for the function $y = y(x)$ for which the integral
$$J[y]= \int_{-a}^{a}ydx $$
takes the the lagest value subject to the conditions
$$y( - a) = y(a) = 0$$ $$ K[y] = \int_{-a}^{a} \sqrt{1 + y'^2}dx = L$$
Thus, we are dealing with an isoperimetric problem. Then, we form the functional:
$$ J[y] + \lambda K[y] = \int_{-a}^{a}(y + \lambda \sqrt{ 1+ y'^2}) dx $$
and write the corresponding Euler equation
$$1+ \lambda \frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=0$$
which implies
$$ x+ \lambda \frac{y'}{\sqrt{1+y'^2}}= C_1 $$
integratin, we obtain the equation
$$(x - C_1)^2 + (y - C_2)^2 = \lambda^2$$
The values of $C_l$, $C_2$ and $\lambda$ are then determined from the conditions
$$y( - a) = y(a) = 0 $$ $$ K[y] = L$$
My doubt is if there are other solutions in which the curves obtained can't be represented by a function $y=y(x)$ or $y'$ doesn't exists in some points. For example, if the answer is really a circumference and $L$ is large enough, thus we have the following figure (in the figure $a=2$):
There're points on the curve which $y'$ doesn't exist. Besides, $y$ isn't a function of $x$. Could someone help me? If doesn't exist other solutions, how you prove that?
I thought of considering $y$ and $x$ as a function of another variable $t$, but I don't know how calculate the area as a integral in this case.
Remark: There're other questions in which I came across the same problem. If you can generalize your argument for other problems I would appreciate.