Let $\gamma$ be a real constant with $\gamma^2\neq 1$ for the parametric functional $S[x, y]=\int_{0}^{1}[\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}-\lambda(x\dot{y}-\dot{x}y)]dt, \lambda>0,$ with the boundary conditions $x(0)=y(0)=0, x(1)=R>0$ and $y(1)=0$.
a) Show that the stationary paths of this parametric functional are given by the solutions of the equations $\frac{dx}{ds}+\gamma\frac{dy}{ds}=2(c-\lambda y)$ and $\gamma\frac{dx}{ds}+\frac{dy}{ds}=2(d+\lambda x)$, where $c$ and $d$ are constants and $s(t)=\int_{0}^{t}\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}dt$.
b) Show that $(\frac{dx}{ds})^2+2\gamma\frac{dx}{ds}\frac{dy}{ds}+(\frac{dy}{ds})^2=1.$
Here's my work for part a):
Note that the associated Euler-Lagrange equations are $\frac{d}{dt}(\frac{\partial\Phi}{\partial\dot{x}})-\frac{\partial\Phi}{\partial x}=0$ and $\frac{d}{dt}(\frac{\partial\Phi}{\partial\dot{y}})-\frac{\partial\Phi}{\partial y}=0$ where $\Phi=\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}-\lambda(x\dot{y}-\dot{x}y)$ such that $\lambda>0.$
Then $\frac{\partial\Phi}{\partial\dot{x}}=\frac{2\dot{x}+2\gamma\dot{y}}{2\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}}+\lambda y=\frac{\dot{x}+\gamma\dot{y}}{\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}}+\lambda y, \frac{\partial\Phi}{\partial x}=-\lambda\dot{y}.$
Similarly, I found $\frac{\partial\Phi}{\partial\dot{y}}=\frac{2\gamma\dot{x}+2\dot{y}}{2\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}}-\lambda x=\frac{\gamma\dot{x}+\dot{y}}{\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}}-\lambda x, \frac{\partial\Phi}{\partial y}=\lambda\dot{x}.$
From here, how should I take the derivatives and find $\frac{d}{dt}(\frac{\partial\Phi}{\partial\dot{x}}), \frac{d}{dt}(\frac{\partial\Phi}{\partial\dot{y}})$ so I can find those Euler-Lagrange equations in order to solve for the stationary paths?
Best Answer
You again are on the right tack. It is about organizing the information. You don't take a further derivative. As you say, \begin{align} \frac{d}{dt}\frac{\partial \Phi}{\partial \dot x} &= \frac{\partial \Phi}{\partial x}=-\lambda \dot y = \frac{d}{dt}(-\lambda y),\\ \frac{d}{dt}\frac{\partial \Phi}{\partial \dot y} &= \frac{\partial \Phi}{\partial y}=\lambda \dot x = \frac{d}{dt}(\lambda x). \end{align} You integrate this already, since you know $\frac d{dt}$ of things are equal iff they differ by constants, so there exist $2c$ and $2d$ such that \begin{align} \frac{\partial \Phi}{\partial \dot x}=2c - \lambda y,\\ \frac{\partial \Phi}{\partial \dot y}=2d - \lambda y. \end{align} Now plugging in your calculations for the left, you would get part a) as long as you notice that $$ \frac{d}{ds}=\frac{dt}{ds}\frac{d}{dt}=\frac{1}{\frac{ds}{dt}}\frac{d}{dt}=\frac{1}{\sqrt{\dot x^2 +2\gamma\dot x\dot y+\dot y^2}}\frac{d}{dt}, $$ by the chain rule and the definition of $s$.
Now part b) has nothing more. It is just chain rule. Note that $$ (\frac{dx}{ds})^2 + 2\gamma (\frac{dx}{ds})(\frac{dy}{ds}) + (\frac{dy}{ds})^2 = (\dot x^2 + 2\gamma\dot x\dot y+\dot y^2)(\frac{dt}{ds})^{2}=\frac{\dot x^2 + 2\gamma\dot x\dot y+\dot y^2}{(\frac{ds}{dt})^{2}}=1, $$ by the above.
(This question is so concrete, I wonder if there is more geometric meaning to it. It seems to be related to geodesics.)