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Elliptic Monge–Ampère Equation in Dimension Two 53
V 0h = V h ∩ H 1 0 (Ω) (={v | v ∈ V h , v =0onΓ }). (35)
The functions Dhij 2 (Ω) areuniquely defined by the relations (33) and (34).
However, in order to simplify the computation of the above discrete second order
partial derivatives we will use the trapezoidal rule to evaluate the integrals
in the left hand sides of (33) and (34). Owing to their practical importance,
let us detail these calculations:
1. First we introduce the set Σ h of the vertices of T h and then Σ 0h =
{P | P ∈ Σ h , P /∈ Γ }. Next, we define the integers N h and N 0h by
N h = Card(Σ h )andN 0h = Card(Σ 0h ). We have then dim V h = N h
and dim V 0h = N 0h . We suppose that Σ 0h = {P k } N 0h
k=1 and Σ h =
Σ 0h ∪{P k } N h
k=N 0h +1 .
2. To P k ∈ Σ h we associate the function w k uniquely defined by
w k ∈ V h , w k (P k )=1, w k (P l )=0, if l =1, ···N h ,l≠ k. (36)
It is well known (see, e.g., [Glo84, Appendix 1]) that the sets B h =
{w k } N h
k=1 and B 0h = {w k } N 0h
k=1 are vector bases of V h and V 0h , respectively.
3. Let us denote by A k the area of the polygonal which is the union of
those triangles of T h which have P k as a common vertex. Applying the
trapezoidal rule to the integrals in the left hand side of the relations (33)
and (34), we obtain:
⎧
⎨ ∀i =1, 2, Dhii(ϕ) 2 ∈ V 0h ,
⎩ Dhii(ϕ)(P 2 k )=− 3 ∫
∂ϕ ∂w k
(37)
dx, ∀k =1, 2, ··· ,N 0h ,
A k Ω ∂x i ∂x i
⎧
Dh12(ϕ)(= 2 Dh21(ϕ)) 2 ∈ V 0h ,
⎪⎨
Dh12(ϕ)(P 2 k )=− 3 [ ∂ϕ ∂w k
+
2A k
∫Ω
∂ϕ ]
∂w k
dx, (38)
∂x 1 ∂x 2 ∂x 2 ∂x 1
⎪⎩
∀k =1, 2, ··· ,N 0h .
Computing the integrals in the right hand sides of (37) and (38) is quite
simple since the first order derivatives of ϕ and w k are piecewise constant.
Taking the above relations into account, approximating (E-MA-D) is now a
fairly simple issue. Assuming that the boundary function g is continuous over
Γ , we approximate the affine space V g by
V gh = {ϕ | ϕ ∈ V h , ϕ(P )=g(P ), ∀P ∈ Σ h ∩ Γ }, (39)
and then (E-MA-D) by
{
Find ψ h ∈ V gh such that for all k =1, 2,...,N 0h ,
D 2 h11 (ψ h)(P k )D 2 h22 (ψ h)(P k ) −|D 2 h12 (ψ h)(P k )| 2 = f h (P k ).
(E-MA-D) h
The iterative solution of the problem (E-MA-D) h will be discussed in the
following paragraph.