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

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 ##?

Math100 said:

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

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 [itex]B[/itex] from the above two equations will yield this.

pasmith said:

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