434
7 Variational Principles
= 2
D
[ p(∇δu)
2
+ q(δu)
2
]d D + 2
Γ 2
σ (δu)
2 dΓ > 0
Therefore the solution of the equation is the minimum when u = u 0 . Quod erat
demonstrandum.
Example 7.8.1 Try to establish the variational problem equivalent to first boundary
value problem for the Poisson equation
−u = −
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2 +
∂
2 u
∂z 2
= f (x, y, z) (x, y, z ∈ V )
(7.8.24)
u| S = 0 (S ∈ the boundary of V )
(7.8.25)
where, f (x, y, z) ∈ C[V + S].
Below using respectively Theorem 7.6.1 and Theorem 7.8.1 to solve.
Solution 1 Using Theorem 7.6.1 to solve. First to prove that operator T = − is a
symmetric operator and a positive definite operator. According to the Green’s first
formula (1.3.63), there is
(−u, v) =
˚
V
(−u)vdV =
˚
V
∇u · ∇vdV −
S
v
∂u
∂n
dS
Because the value of u on the boundary is zero, there is
(−u, v) =
˚
V
∇u · ∇vdV
Transposing u and v, there is also
(−v, u) =
˚
V
∇u · ∇vdV
From the above two expressions it can be obtained
(−u, v) = (−v, u)
Hence the operator T = − is a symmetric operator.
Putting v = u, when u = 0 时, then there is
(−u, u) =
V
∇u · ∇udV =
V
(∇u)
2 dV
=
V
∂u
∂ x
2
+
∂u
∂ y
2
+
∂u
∂z
2
dV > 0