Appendix: All Solutions to the Problems
793
2.73 Write the Ostrogradsky equation of the functional J [u] =
˜
D (u
2
x − u
2
y )dxdy.
Solution Let the integrand F = u
2
x − u
2
y , there are F u = 0,
∂
∂ x
F u x = 2
∂
2 u
∂ x 2 ,
∂
∂ y
F u y =
−2
∂
2 u
∂ y 2 , substituting them into the Ostrogradsky equation, we get
∂
2 u
∂ x 2 −
∂
2 u
∂ y 2 = 0
2.74 Find the Ostrogradsky equation of the functional J [u]
=
˜
D [ p(x)u
2
x − q(y)u
2
y ]dxdy.
Solution The Ostrogradsky equation of the functional is
F u −
∂
∂ x
F u x −
∂
∂ y
F u y = −
∂
∂ x
[2 p(x)u x ] +
∂
∂ y
[2q(y)u y ] = 0
or
∂
∂ x
[ p(x)u x ] −
∂
∂ y
[q(y)u y ] = 0
2.75 Write the Ostrogradsky equation of the functional J [u]
=
˜
D
d
2
(u xx + u yy )
2
− f (x, y)u
dxdy.
Solution The integrand is F =
d
2
(u xx + u yy )
2
− f (x, y)u, calculating the various
partial derivatives
F u = − f (x, y),
∂ 2
∂ x 2 F u xx = D
∂ 4 u
∂ x 4 +
∂ 4 u
∂ x 2 ∂ y 2
,
∂ 2
∂ y 2 F u yy = D
∂ 4 u
∂ x 2 ∂ y 2 +
∂ 4 u
∂ y 4
Substituting them into the Ostrogradsky equation, we get
D
2 u = D
∂
4 u
∂ x 4 + 2
∂
4 u
∂ x 2 ∂ y 2 +
∂
4 u
∂ y 4
= f (x, y)
When f (x, y) = 0, it can be turned into biharmonic equation.
2.76 Find the Ostrogradsky equation of the functional J [u]
=
˜
D
1 + u 2
x + u 2
y dxdy.
Solution The integrand F =
1 + u 2
x + u 2
y does not depend on u, F u x =
u x
√
1+u 2
x +u 2
y
,
F u y =
u y
√
1+u 2
x +u 2
y
, the Ostrogradsky equation of the functional is