800
Appendix: All Solutions to the Problems
J [u] =
¨
D
u r
∂r
∂ x
+ u θ
∂θ
∂ x
2
+
u r
∂r
∂ y
+ u θ
∂θ
∂ y
2
r dr dθ
=
¨
D
u r cos θ − u θ
sin θ
r
2
+
u r sin θ + u θ
cos θ
r
2
r dr dθ
=
¨
D
ru
2
r +
1
r
u
2
θ
dr dθ
The Euler–Ostrogradsky equation of the functional is written, the Laplace
equation expressed in polar coordinates can be obtained
ru rr +
1
r
u θθ + u r = 0
Of course, according to Eq. (1.4.12), the preceding Laplace equation also can be
directly written.
2.85 Find the Euler equation of the functional J [u] =
˜
D
1
r 3 (u
2
r + u
2
z − 4au)dr dz.
Solution Putting F =
1
r 3 (u
2
r + u
2
z − 4au), the various partial derivatives are F u =
−
4a
r 3 , F u r =
2u r
r 3 , F u z =
2u z
r 3 , the Euler equation of the functional is
2a
r 3 +
u rr
r 3 −
3u r
r 4 +
u zz
r 3 = 0
or
u rr + u zz −
3u r
r
+ 2a = 0
2.86 Write the Ostrogradsky equations of the potential energy functional of a circular
arch
J [u, w] =
1
2
θ 1
θ 0
E A
R
(u
− w)
2
+
E I
R 3 (u
+ w
)
2
− 2R( pu + qw)
− p(u
2
+ uw
) − q(uw
+ w
2
)
dθ
where, E is the modulus of elasticity for the material; A is the cross-sectional area;
R is the radius of the circular arch; I is the moment of inertia of the cross section
in a bending plane; p and q are the tangential load and normal load on the arch
respectively.
Solution The integrand of the functional is