I have the following function:
$$H(x,y,\dot{x},\dot{y})=\frac{1}{2}(y\dot{x}-x\dot{y})+\lambda\sqrt{\dot{x}^2+\dot{y}^2}$$
Where $\gamma:(a,b)\to\mathbb{R}^2,t\mapsto(x(t),y(t))^T$ is a regular, closed, simple curve with period K, $\lambda\in\mathbb{R}$.
That apparently if I solve its Euler-Lagrange equations I will find curves of fixed length that maximise area (circles). However, I have tried solving these equations and have not been able to find the equation of a circle. The two Euler Lagrange equations I found were:
$$-\frac{\dot{y}}{2}=\frac{\dot{y}}{2}+\frac{\lambda\ddot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}}-\frac{\lambda\dot{x}(\dot{x}\ddot{x}+\dot{y}\ddot{y})}{\left(\sqrt{\dot{x}^2+\dot{y}^2}\right)^3}
$$
$$\frac{\dot{x}}{2}=-\frac{\dot{x}}{2}+\frac{\lambda\ddot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}-\frac{\lambda\dot{y}(\dot{x}\ddot{x}+\dot{y}\ddot{y})}{\left(\sqrt{\dot{x}^2+\dot{y}^2}\right)^3}$$
Any help would be appreciated!
Best Answer
You can avoid second derivatives here.
$$H(x,y,\dot{x},\dot{y})=\frac{1}{2}(y\dot{x}-x\dot{y})+\lambda\sqrt{\dot{x}^2+\dot{y}^2}$$
$$\frac{d}{dt}\frac{\partial H}{\partial \dot{x}}-\frac{\partial H}{\partial x}=0$$
$$\implies (d/dt)(y/2+\frac{\lambda \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}})+\dot{y}/2=0$$
$$y+\frac{\lambda \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}}=C_1$$
$$(d/dt)(-x/2+\frac{\lambda \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}})-\dot{x}/2=0$$
$$-x+\frac{\lambda \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}=C_2$$
$$\frac{C_2+x}{C_1-y}=\frac{dy}{dx}$$
$$2C_2x +x^2=2C_1y-y^2+C_3$$
$(x+C_2)^2+(y-C_1)^2=C_1^2+C_2^2+C_3$