Hi everyone,

I have the following functional

S[y]=

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

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

I am now trying to show thatv = cosh v + B sinh vand thatB^{2}− 1 = 2 sinh v + 2B cosh v

For the first part can I just put my condition ofy(v) = vinto my equation to obtainv = 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