798
Appendix: All Solutions to the Problems
= −2
˚
V
(w xx + w yy + w zz )δϕdxdydz = 0
Since δu is arbitrary, there is
w =
∂
2 w
∂ x 2 +
∂
2 w
∂ y 2 +
∂
2 w
∂z 2 = 0
Using the vector method to solve below. The functional can be written as
J [w] =
˚
V
(∇w · ∇w)dV
Taking the first variation
δ J = 2
˚
V
(∇δw · ∇w)dV = 2
˚
V
[∇ · (δw∇w) − δww]dV
= 2
S
∂w
∂n
δwdS − 2
˚
V
wδwdV = −2
˚
V
wδwdV = 0
Since δw is arbitrary, there is
w =
∂
2 w
∂ x 2 +
∂
2 w
∂ y 2 +
∂
2 w
∂z 2 = 0
It is the same as the above method, but it is convenient much more.
2.82 Write the Euler–Ostrogradsky equation of the functional
J [u(x 1 , . . . , x n )] =
¨
. . .
⎡
⎣
n
j=1
a j (x 1 , . . . , x n )u
2
x j
− b(x 1 , . . . , x n )u
2
+2u f (x 1 , . . . , x n )]dx 1 . . . dx n
Solution The Euler–Ostrogradsky equation of the functional is
n
j=1
a j (x 1 , . . . , x n )
∂
2 u
∂ x
2
j
+ b(x 1 , . . . , x n )u = f (x 1 , . . . , x n )
2.83 In polar coordinates find the extremal curve of the functional J [y] =
x 1
x 0
x 2 + y 2
1 + y 2 dx.
Solution Let the polar coordinate equation be r = r (θ ), then
x = r cos θ, y = r sin θ, r
2
= x
2
+ y
2