How should I show that $B$ is given by the solution of the equation $B^2-1=2\sinh(v)+2B\cosh(v)$

calculuscalculus-of-variations

Show that the stationary path of the functional $S[y]=\int_{0}^{v}(y'^2+y^2)dx, y(0)=1, y(v)=v, v>0$ is given by $y=\cosh(x)+B\sinh(x), 0\leq x\leq v$, where $B$ and $v$ are given by the solutions of the equations $v=\cosh(v)+B\sinh(v)$ and $B^2-1=2\sinh(v)+2B\cosh(v)$.

Here's my work:

From the definition, Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$ for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$.

Let $F(x, y, y')=y'^2+y^2$.

Then $\frac{\partial F}{\partial y'}=2y'$ and $\frac{\partial F}{\partial y}=2y$.

This gives $\frac{d}{dx}(\frac{\partial F}{\partial y'})=2y''$.

Thus, the Euler-Lagrange equation is $2y''-2y=0\implies y''-y=0$.

Note that the general solution is $y(x)=A\cosh(x)+B\sinh(x)$, where $A$ and $B$ are constants.

Applying the boundary-value conditions $y(0)=1$ and $y(v)=v, v>0$ produce:

$y(0)=1\implies 1=A\cosh(0)+B\sinh(0)\implies 1=A+0\implies A=1$ and

$y(v)=v\implies v=\cosh(v)+B\sinh(v)$.

Above is my work, although I've shown the given stationary path with the constant v and the equation associated with it, I still don't know how to show that the constant $B$ is given by the solution of the equation $B^2-1=2\sinh(v)+2B\cosh(v)$. How should I do this?

Best Answer

To obtain the other equation, one has to view $S[y]$ not only as functional of $y(x)$, but also as a function of the endpoint $v$. Thus, given the stationary path $y_v(x)=\cosh(x)+B(v)\sinh(x)$ of $$ S[y]=\int_0^v(y'^2+y^2)dx, \quad y(0)=1, \quad y(v)=v, \tag{1} $$ let's define$^{(*)}$ \begin{align} s(v):=S[y_v]&=\int_0^v[(\sinh(x)+B\cosh(x))^2+(\cosh(x)+B\sinh(x))^2]\,dx \\ &=\int_0^v[(B^2+1)\cosh(2x)+2B\sinh(2x)]\,dx \\ &= \frac{1}{2}(B^2+1)\sinh(2v)+B(\cosh(2v)-1). \tag{2} \end{align} The critical points of $s(v)$ are given by the roots of $s'(v)=0$, i.e. $$ (B^2+1)\cosh(2v)+2B\sinh(2v)+BB'\sinh(2v)+B'(\cosh(2v)-1)=0. \tag{3} $$ To find $B'$, we differentiate the relation $v=\cosh(v)+B\sinh(v)$ with respect to $v$: $$ 1=\sinh(v)+B\cosh(v)+B'\sinh(v) \implies B'\sinh(v)=1-\sinh(v)-B\cosh(v). \tag{4} $$ Plugging $(4)$ into $(3)$, we obtain, after some tedious algebra, the relation $$ B^2-1=2B\cosh(v)+2\sinh(v). \tag{5} $$

$^{(*)}$ The following identities are used in the calculations of this answer:

  • $\cosh^2(x)-\sinh^2(x)=1$;
  • $\cosh(2x)=\cosh^2(x)+\sinh^2(x)$;
  • $\sinh(2x)=2\sinh(x)\cosh(x)$.
Related Question