Hi everyone,

I have the following functional

S[y] = \int_{0}^{v}\! \left( {\frac {\rm d}{{\rm d}x}}y \left( x \right)  \right) ^{2}+ \left( y \left( x \right)  \right) ^{2}\,{\rm d}x

Where y(0) = 1, y(v) = v, v > 0,

I then showed that the stationary path of the functional is y = cosh x + B sinh x, 0 ≤ x ≤ v, by solving the differential equation to obtain y = A cosh x + B sinh x, and then using my condition of y(0) = 1 to prove that A = 1.

I am now trying to show that v = cosh v + B sinh v and that B2 − 1 = 2 sinh v + 2B cosh v

For the first part can I just put my condition of y(v) = v into my equation to obtain v = cosh v + B sinh v?

For the second part I really dont know where to start.... from looking at a similar question i suspect it might be something to do with a Transversality condition, but i'm really not sure how to do this.

Any ideas?

Thanks