# How should I show that the stationary paths of this parametric functional are given by the solutions of the following equations?

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?

• Would a solution using Mathematica be admissible? Commented Jan 30 at 3:16

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.
• Yes, but your first term $\frac{\partial \Phi}{\partial \dot x}$ is longer, it has an extra $\lambda y$, that is why you call the constant $2c$. Commented Feb 1 at 1:02
• So the Euler-Langrange eqn is $\frac{d}{dt}(\frac{\dot x+\gamma \dot y}{\sqrt{\dot x^2+2\gamma\dot x\dot y+\dot y^2}} + \lambda y + \lambda y)=0$, so the whole function is $2c$ for some constant $c$. This would give one of the answer equations. The other one is similar. Commented Feb 1 at 2:09
• $\frac{dx}{ds}$ is not $\dot x$, by chain rule, it is $\frac{dx}{ds}=\frac{1}{\sqrt{\dot x^2 +2\gamma\dot x\dot y+\dot y^2}}\frac{dx}{dt}$, so there is no denominator to multiply through. The $\frac{dx}{ds}+\gamma \frac{d\gamma}{ds} = \frac{\dot x + \gamma \dot y}{\sqrt{\dot x^2+2\gamma \dot x\dot y + \dot y^2}}$. Nothing more to be done. Commented Feb 1 at 3:03