For , the 3D curl-operator is defined as
Let I ⊆ {1, 2, 3} =: I0, define scaling functions
with
The corresponding wavelets are
Here and after, denotes the non-zero apexes of the unite cube and
Let , which is the tensor product of corresponding interpolatory scaling functions on the interval. The corresponding duals are given similarly. Furthermore, define
and the projection operators:
Lemma 2.1[8]. For smooth functions .
Proposition 2.1. For , there has .
Proof. Note that Lemma 2.1, then
which is by definition.
Proposition 2.1 is important, because it tells us that keeps curl-free property. In general, vector-valued wavelets and wavelet spaces are given, respectively, by
For and m ∈ {1, 2}3, we define
Clearly, .
To give a decomposition for , take
for i ∈ I0\{i
e
}. Here, we choose i
e
such that .
Proposition 2.2. The vector-valued function system is complete in .
Proof. It is sufficient to show the statement for j = 0. Let satisfy
for all and i ≠ i
e
. Here, the inner product is in L2([0,1]3). Without loss of generality, one assumes i
e
= 1. Then, leads to
By the definition of ψIand differential relations (1.1), one knows
Moreover, reduces to . Now, it follows that from i
e
= 1. Finally, and follows from the definition of .
To give the bi-orthogonal decomposition, we define
(2.1)
Assume I = {i, i
e
, i'}, then
with |ε1| = |ε2| = 1 and ε1ε2 = -1. Now, define
(2.2)
Here, the derivatives are meant in the sense of distributions. Now, we state the main result:
Proposition 2.3. The set is a bi-orthogonal wavelet basis of L2([0,1]3)3 with duals defined in (2.1) and (2.2).
Proof. According to Proposition 2.2, one only need show
(i) ;
(ii) ;
(iii) ;
(iv) .
The identity (i) holds obviously for i ≠ i
e
. For i = i
e
, since i ≠ ie', then i
e
≠ ie', which means e ≠ e'. Finally, the result (i) follows from the bi-orthogonality of and .
Note that . Then (ii) follows from curl⋅grad = 0. Furthermore,
Then by the fact and the bi-orthogonality of , one obtains
Now, it remains to prove (iv), which is equivalent to
(2.3)
It is easily proved, when e1 = e2: In fact, since , one can assume i1 = i2 and , because leads to (2.3) obviously. In that case, the left-hand side of (2.3) reduces to , which is the desired. To the end, it is sufficient to prove that for e1 ≠ e2, that is
(2.4)
Note that . Then the conclusion is obvious when . When , then and the left-hand side of (2.4) reduces to . Hence one only need to show (2.4), when . However, (2.4) becomes
(2.5)
in that case. Since , two cases should be considered: and . Using , the left-hand side of (2.5) is
in the first case; In the second one, the left-hand side of (2.5) becomes . According to the differential relation (1.1), is a linear combination of . By the bi-orthogonality of and , one receives the desired conclusion.