Calculus of variations
The boundary conditions are0 = ^1
6+c 1 +c 2 1 = ^8
3+ 2 c 1 +c 2that is
c 1 +c 2 =1
6 ,^2 c^1 +c^2 =2
6
(^)
1 1
2 1
(^)
(^) = 1 ,
(^)
1 /6 1
2 /6 1
(^)
(^) = 1 / 6 ,
(^)
1 1/^6
2 2/ 6
(^)
(^) = 0
hencec 1 = 1 / 6 ,c 2 = 0 .The kernel is convex in
x,xso it is a minimum.