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 $y(1)=0$. 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 $s(t)=\int_{0}^{t}\sqrt{\dot{x}^2+2\gamma\dot{x}\dot{y}+\dot{y}^2}dt.$ Note that $(\frac{dx}{ds})^2+2\gamma\frac{dx}{ds}\frac{dy}{ds}+(\frac{dy}{ds})^2=1.$
Question: Show that the equations for $x$ and $y$ can be written in the form $(1-\gamma^2)x'=D-X$ where $D=2(c-\gamma d), X=2\lambda(\gamma x+y), (1-\gamma^2)y'=C+Y$ where $C=2(d-\gamma c), Y=2\lambda(x+\gamma y)$, where $x'=\frac{dx}{ds}, y'=\frac{dy}{ds}.$ Hence show that $X$ and $Y$ satisfy $X^2+Y^2-2\gamma XY-4c(1-\gamma^2)X+4d(1-\gamma^2)Y+C^2+D^2+2\gamma CD=(1-\gamma^2)^2.$ Then use the boundary conditions to show that $C^2+2\gamma CD+D^2=(1-\gamma^2)^2$ and $C=-\lambda R.$ Show that if $\gamma^2>1$, then there are always two real solutions of these equations, and if $\gamma^2<1$, then there are two real solutions if $\lambda R<\sqrt{1-\gamma^2}$ and none if $\lambda R>\sqrt{1-\gamma^2}.$
Here's my work:
Consider the equations $\frac{dx}{ds}+\gamma\frac{dy}{ds}=2(c-\lambda y)$ and $\gamma\frac{dx}{ds}+\frac{dy}{ds}=2(d+\lambda x)$.
Let $x'=\frac{dx}{ds}, y'=\frac{dy}{ds}$.
Then we have $x'+\gamma y'=2(c-\lambda y)$ and $\gamma x'+y'=2(d+\lambda x)$.
By solving the system of equations $x'+\gamma y'=2(c-\lambda y)$ and $-\gamma(\gamma x'+y')=-2\gamma(d+\lambda x)$ produce $(1-\gamma^2)x'=D-X$ where $D=2(c-\gamma d), X=2\lambda(\gamma x+y).$
Similarly, we obtain $(1-\gamma^2)y'=C+Y$ where $C=2(d-\gamma c), Y=2\lambda(x+\gamma y)$.
I just don't know how to show that $C=-\lambda R$. Also, how should I prove that if $\gamma^2>1$, then there are always two real solutions of these equations and if $\gamma^2<1$, then there are two real solutions if $\lambda R<\sqrt{1-\gamma^2}$ and none if $\lambda R>\sqrt{1-\gamma^2}?$
Best Answer
This is getting almost unbearably tedious. Let's go step by step.
Do you know how to show the algebraic quadratic equation relating $X$ an $Y$? Just get $x'$ and $y'$ from the two equations you showed in the post using $X$ and $Y$, and plug them into the condition $$ x'^2 + 2\gamma x' y' + y'^2=1. $$
Now the initial conditions on $x(0), y(0)$ and $x(1), y(1)$ imply conditions on $X(0), Y(0)$ and $X(1), Y(1)$. Plug them into you quadratic equation to get the two conditions on $C$ and $D$.
Now consider the 3 equations \begin{gather} C=-\lambda R,\\ C^2+2\gamma CD + D^2 = (1-\gamma^2)^2,\\ X^2 + Y^2 -2\gamma XY - ... = (1-\gamma^2)^2. \end{gather} If $\gamma^2>1$, the quadratic forms have determinant $1-\gamma^2<0$, so there are real solutions. If $\gamma^2<1$, then the quadratic form is positive definite. To have solutions, we need the minimal values to be < RHS. Apply this to the second equation, the quadratic function is minimal when $D=-\gamma C$, and the minimal value is $(1-\gamma^2)C^2$. For this to be < RHS, we need $\lambda R<\sqrt{1-\gamma^2}$. If $>$ here, then no solution.
This requires a lot of calculation, I hope you can carry this out as much as possible.
In general, this question is related to the isoperimetric problem. To enclose an area, what is the shortest curve needed? But here we are using a more general metric $$ \sqrt{\dot x^2 + 2\gamma \dot x \dot y + \dot y^2}. $$ The standard case is when $\gamma = 0$. The solutions is a circle. The quadratic equations relating $X$ and $Y$ is supposed to represent a quadric curve solution in general. But the details are very hard to totally nail down.