Isoperimetric inequality via Lagrange multipliers in infinite dimensions. End of a proof.

Question

Let $\gamma:[0,2\pi]\rightarrow\mathbb{R}^2$ be an injective $C^2$ curve with $\gamma(0)=\gamma(2\pi)$ and velocity $1$ (parametrized by the arc length). Write it as $\gamma(t)=(x(t),y(t))$. We know that its length is given by $$ l(\gamma)=\int_0^{2\pi} \| \dot{\gamma}(t)\|dt=2\pi$$ and the area inside it is given, by Green's formula, by $$ A(\gamma)=\int_0^{2\pi} x(t)\dot{y}(t)dt.$$ Thus, the isoperimetric problem can be seen as finding $$ \max_{l(\gamma)=2\pi} A(\gamma). $$ Using Lagrange multpliers, one gets that there exists $\lambda\in\mathbb{R}$ such that $$ \dot{y}=\lambda\frac{d}{dt}\left(\frac{\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}}\right),\quad -\dot{x}=\lambda\frac{d}{dt}\left(\frac{\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}\right).$$ Using $\sqrt{\dot{x}^2+\dot{y}^2}=1$ and integrating in time, $$ y=\lambda \dot{x}+C,\quad -x=\lambda\dot{y}+D.$$ Solving this, one derives $$ x(t)=-D+\alpha\cos(t/\lambda)+\beta\sin(t/\lambda),\quad y(t)=C+\delta\cos(t/\lambda)+\eta\sin(t/\lambda),\quad t\in [0,2\pi].$$ I would like to obtain that $\gamma$ is a circle. I tried by setting some assumptions: $y(0)=y(2\pi)=0$ (by translating $\gamma$) and $y(\pi)=0$ (by rotating $\gamma$).

Answer 1

$(x,y)$ traces a circle centered at $(-D,C)$ with radius $\lambda$:

From $$y = \lambda \dot{x} + c\text{ and}\\-x=\lambda \dot{y} + D $$ follows $$(D + x)^2 + (y - C)^2 = \lambda^2\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right) = \lambda^2$$