# Thread: Euler's equation and transversality condition

1. ## Euler's equation and transversality condition

Hi, this is my first post and I have two past paper questions I'm going over and finding difficulty in the last part.

The 1st question is,

i) If F = y^2 -2xy - (y')^2 , [a,b]=[0,(pi/2)]
y(0)=1
y(pi/2)=0, show that the extremal is

Yo(x)= x -(pi/2)sinx + cosx

now, i done that bit, however, the next part asks for me to find the extremal(s) in which y(pi/2) is not specified.

So, my working has yielded this,

Yo(x) = Acosx +Bsinx + x y(0)=1
y(0)=1 T.F. A=1

so, Yo(x)= x+ cosx +Bsinx

T.C Fy'| x=(pi/2) Fy' = -2y'

t.f. -2y'(pi/2) = 0 t.f. y'(pi/2) = 0

sub this into y'(x)= 1 -sinx + Bcosx

i get 1 -1 + Bcos(pi/2) = 0

the "ones" cancel out and since cos(pi/2)=0, I'm left with 0 = 0 and no value for B.

this is the other question I have had a problems with.

ii) F = (y')^2 + 2y - y^2 [a,b] = [0, pi/2]
y(0)=2
y(pi/2)=0

and this will show that the extremal is
Yo(x) = cosx - sinx + 1

however, the tricky part for me is that I have to write down the transversaility conditions if BOTH y(0) and y(pi/2) are not specifed, and find the extremal in this case.

So, what I have is Yo(x)= Acosx + Bsinx + 1 and
for Fy'| x=0, I get y'(0)=0
Fy'| x= pi/2 I get y'(pi/2) = 0

T.f. y'(0)=y'(pi/2) and t.f.

-Asinx + Bcosx| x=0 = -Asinx + Bcosx| x=pi/2

t.f. B = -A

Or is it that A and B are equal to 0 when plugging these in separetly?

I will appreciate all the help I can get.

Thank you

Jas

2. Can I ask? Is Euler's equation not on the syllabus in courses in US or other parts of the world, since I've posted the same question on two other maths forums, without any replies.

It is part of calculus, but I guess not many people have encountered this problem.

Anyway, If someone is aware of how to approach this problem, I would appreciate your help.

Thanks,

Jas