Calculus of variations
Example
Find the extremal of the functional
Z 3
1
( 3 t x)xdt, x( 1 )= 1 ,x( 3 )= 4
1
2.
The kernel function is
F
t,x,x
=( 3 t x)x.
Fx^0 = 3 t 2 x,Fx^0 = 0.
The EulerñLagrange equation is 3t 2 x= 0 ,sox= 3 / 2 t.But att= 1
x( 1 )= 3 /2 the Örst boundary condition is not valid. Hence there is no
extremal solution.