7.8 Variational Problems of Equivalence with Boundary Value Problem …
435
therefore the operator T = − is a positive definite operator.
According to theorem 7.6.1, the quadratic functional can be established
J [u] = (−u, u) − 2(u, f ) =
˚
V
[(∇u)
2
− 2u f ]dV
(7.8.26)
Solution 2 Directly using Theorem 7.8.1 to solve. Equation (7.8.24) is equivalent to
the cases of p = 1, q = 0 in Eq. (7.8.6), The boundary (7.8.25) is equivalent to the
case of g = 0 in the boundary condition (7.8.10), thus according to the functional
(7.8.12) the functional can be established
J [u] =
˚
V
[(∇u)
2
− 2u f ]dV
(7.8.27)
It follows that the functional (7.8.26) is the same as the functional (7.8.27).
This example is solved on the three dimensional Dirichlet problem, For the twodimensional problem, it need only to change the space domain V into the plane
domain D, and to change the triple integral into the double integral, of course,
the closed surface S on the boundary should also be correspondingly changed to the
closed curve Γ , and closed surface integral is correspondingly change into the closed
curve integral.
Example 7.8.2 Let p(x, y, z) ∈ C
1
(V ), q(x, y, z), f (x, y, z), σ (x, y, z) be all
belong to C(V ), and p > 0, q 0, σ 0. Where V is a bounded domain in space,
the boundary is S. Find the minimum of the functional equivalent to the following
boundary value problem.
−
∂
∂ x
( pu x ) −
∂
∂ y
( pu y ) −
∂
∂z
( pu z ) + qu = f (x, y, z) (x, y, z ∈ V )
p
∂u
∂n
+ σ u
S
= 0
where, u ∈ C
2
(V ).
Solution The differential equation can be expressed as
−∇ · ( p∇u) + qu = f
According to the meaning of problem, it belongs to the case of the third kind of
boundary condition k = 0, directly using the functional (7.8.12) indiscriminately,
the corresponding functional is
˚
V
[ p(∇u)
2
+ qu
2
− 2u f ]dV +
¨
S
σ u
2 dS = 0
or
˚
V
[ p(u
2
x + u
2
y + u
2
z ) + qu
2
− 2u f ]dV +
¨
S
σ u
2 dS = 0