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Appendix: All Solutions to the Problems
=
¨
D
∂
2 u
∂ x 2
2
+ 2
∂
2 u
∂ x∂ y
2
+
∂
2 u
∂ x 2
2
− 2u f
dxdy
7.12 Let the first boundary value problem for the Poisson equation be
−u = −
∂
2 u
∂ x 2 +
∂
2 u
∂ y 2
= f (x, y) (x, y ∈ D, f (x, y) ∈ C(D))
u| Γ = 0
where, Γ is the boundary of D. Prove that the two-dimensional Laplace negative
operator − is a symmetric positive definite operator, and find out the equivalent
variational problem.
Solution First prove that the two-dimensional Laplace operator T = − =
−
∂
2
∂ x 2 +
∂
2
∂ y 2
is the symmetric operator and positive definite operator. According to
the Green’s first formula, there is
(−u, v) =
¨
D
(−u)vdD =
¨
D
∇u · ∇vdD −
Γ
v
∂u
∂n
dΓ
Since the value of u on the boundary is zero, there is
(−u, v) =
¨
D
∇u · ∇vdD
Transposing u and v, the above result remains unchanged, there is
(−u, v) = (−v, u)
Thus the operator T = − is a symmetric operator. Then putting v = u, when
∇u = 0, there is
(−u, u) =
¨
D
∇u · ∇udD =
¨
D
(∇u)
2 dD =
¨
D
∂u
∂ x
2
+
∂u
∂ y
2
dD > 0
Hence the operator 子T = − is also a positive definite operator. According to
Theorem 7.6.1, the quadratic functional can be established
J [u] = (−u, u) − 2(u, f ) =
¨
D
[(∇u)
2
− 2u f ]dD
7.13 Let the boundary of the closed domain D be Γ = Γ 1 + Γ 2 , where Γ 1 is the line
segment a ≤ x ≤ b on the Ox axis, Γ 2 is the curve arc in the upper half-plane on
the Ox y plane, and the two points of the arc are connected with the points of Γ 1 on