Let the stationary path of the functional $S[y]=\int_{0}^{v}(y'^2+y^2)dx, y(0)=1, y(v)=v, v>0$ be $y=\cosh(x)+B\sinh(x), 0\leq x\leq v$, where $B$ and $v$ are given by the solutions of the equations $v=\cosh(v)+B\sinh(v)$ and $B^2-1=2\sinh(v)+2B\cosh(v)$. Suppose that $v$ is given by the real solution(s) of $f(v)=0$, where $f(v)=v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v)$. Show that there is only one stationary path. Note that a graphical argument alone is insufficient. Please give rigorous analysis to back up your explanation.
My plan is to show that the equation of $f(v)=v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v)=0$ has a unique solution but I don't know how to. Should I find $f'(v)$ first? If that's the case, then I have $f'(v)=2v-2(1+\sinh(v))\cosh(v)-2v\sinh(v)+2\cosh(v)=0$. But what should I do from here?
Best Answer
The function $$ f(v)=v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v) \tag{1} $$ is continuous, $f(0)=1$ and $\lim_{v\to\infty}f(v)=-\infty$; therefore, $f$ must have at least one positive real root. On the other hand, since \begin{align} f'(v)=-2\sinh(v)(\cosh(v)+v+2v\sinh(v))<0\quad\text{if $v>0$}, \tag{2} \end{align} $f$ is strictly decreasing in the interval $(0,\infty)$, hence it can have at most one positive real root. We conclude, therefore, that $f$ has exactly one positive real root, which implies that $S$ has only one stationary path.