How should I show that ## B ## is given by the solution of this?

[Math100]

Homework Statement

a) 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 $B^2-1=2\sinh(v)+2B\cosh(v)$.

b) Use these equations to show that $v$ is given by the real solution(s) of $f(v)=0$, where $f(v)=v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v)$.

Relevant Equations

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

a)
Consider the functional $S[y]=\int_{0}^{v}(y'^2+y^2)dx, y(0)=1, y(v)=v, v>0$.
By definition, the 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
$y(v)=v\implies v=A\cosh(v)+B\sinh(v)\implies v=\cosh(v)+B\sinh(v)$.

Based on my work above, how can I show that $B$ is given by the solution of the equation $B^2-1=2\sinh(v)+2B\cosh(v)$? I've tried to solve for $B$ first by getting $B=\frac{v-\cosh(v)}{\sinh(v)}$, but this expression didn't help me at all in showing this. Also, for part b) of this problem, I've got $f(\cosh(v)+B\sinh(v))=(\cosh(v)+B\sinh(v))^2-2(\cosh(v)+B\sinh(v))(1+\sinh(v))\cosh(v)+1+2\sinh(v)=\cosh^2(v)+2\cosh(v)\cdot B\sinh(v)+B^2\sinh^2(v)-2\cosh^2(v)-2B\sinh(v)\cdot \cosh(v)-2\sinh(v)\cdot \cosh^2(v)-2B\sinh^2(v)\cdot \cosh(v)+1+2\sinh(v)=0$ after I substituted the equation of $v=\cosh(v)+B\sinh(v)$ but how should I solve for $v$ in here by solving the equation $f(v)=0$?

 

[pasmith]

Homework Statement: a) 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 $B^2-1=2\sinh(v)+2B\cosh(v)$.

This seems incomplete. If $v$ is an arbitrary constant, then the only unknown is $B$, and we can find it from the first equation. There must be something else, some other condition on $y$, which elevates $v$ into an unknown; that condition would then give us the second equation linking $v$ and $B$.

b) Use these equations to show that $v$ is given by the real solution(s) of $f(v)=0$, where $f(v)=v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v)$.

I imagine that eliminating $B$ from the above two equations will yield this.

 

[Math100]

This seems incomplete. If $v$ is an arbitrary constant, then the only unknown is $B$, and we can find it from the first equation. There must be something else, some other condition on $y$, which elevates $v$ into an unknown; that condition would then give us the second equation linking $v$ and $B$.
I imagine that eliminating $B$ from the above two equations will yield this.

Okay, so for part b) of this problem, I've got that $v=\cosh(v)+B\sinh(v)\implies B\sinh(v)=v-\cosh(v)\implies B=\frac{v-\cosh(v)}{\sinh(v)}$. After I substitute $B=\frac{v-\cosh(v)}{\sinh(v)}$ into $B^2-1=2\sinh(v)+2B\cosh(v)$, I have $(\frac{v-\cosh(v)}{\sinh(v)})^2-1=2\sinh(v)+2(\frac{v-\cosh(v)}{\sinh(v)})\cdot \cosh(v)\implies \frac{(v-\cosh(v))^2-\sinh^2(v)}{\sinh^2(v)}=2\sinh(v)+\frac{2(v-\cosh(v))\cdot \cosh(v)}{\sinh(v)}\implies (v-\cosh(v))^2-\sinh^2(v)=2\sinh^3(v)+2v\sinh(v)\cosh(v)-2\sinh(v)\cosh^2(v)\implies v^2-2v\cosh(v)+\cosh^2(v)=2\sinh^3(v)+2v\sinh(v)\cosh(v)-2\sinh(v)\cosh^2(v)+\sinh^2(v)$. And after simplifying this, I got $v^2-2v(1+\sinh(v))\cosh(v)+1+2\sinh(v)=0$. Does this mean that $v$ is given by the real solution(s) of $f(v)=0$? Also, for part a) of the problem, you mentioned that there must be something else, some other condition on $y$, but after I took another closer look at this problem again, I've noticed that there's no more additional condition on $y$. That was the whole problem statement, so what should we do in here, in order to show that $B^2-1=2\sinh(v)+2B\cosh(v)$?