Homework #2. (Due Sept. 17)
Math. 610:600
(Last modified on 2015-09-08 08:46)
For the following problem, it is simplest to expand the cross product as
a×b=
3
X
Pijk ai bj ek .
i,j,k=1
Here a, b ∈ R3 and ek , for k = 1, 2, 3, are the unit vectors in the xk directions,
respectively. Finally, Pijk takes on the values 1, 0, −1 and satisfies: Pijk = 0
if i, j, k are not distinct, P123 = 1 and Pijk = −Pi′ j ′ k′ if {i′ , j ′ , k ′ } results
from permuting two of the indices of {i, j, k}. Thus, P213 = −1, P231 = 1,
etc. The analogous expansion for the curl operator, ∇ × u, is
(0.1)
∇×u=
3
X
Pijk
i,j,k=1
∂uj
ek .
∂xi
Problem 1. Let u be a vector field in C 1 (Ω̄)3 where Ω ⊂ R3 .
(a) Verify the Green’s Theorem type formula
Z
Z
n × u ds
∇ × u dx =
∂Ω
Ω
by putting (0.1) into the left hand side above and integrating by parts.
(b) Prove the integration by parts formula u, v vector fields in C 1 (Ω̄)3 ,
Z
Z
Z
(∇ × u) · v dx =
(n × u) · v ds +
u · (∇ × v) dx
∂Ω
Ω
Ω
by putting (0.1) into the left hand side above and integrating by parts.
You will have to use the permutation identity on the right hand side.
These integration by parts formulas arise in variational formulations of electromagnetic problems.
Problem 2. This problem involves integration by parts for differential operators which arise in the theory of linearized elasticity. Let u be a vector
field on C 2 (Ω̄)3 and φ be a vector field in C 1 (Ω̄)3 where Ω ⊂ R3 (we think
of these as column vectors). The “strain” tensor ǫ(u) is the 3 × 3 matrix
defined on each point of Ω given by
(ǫ(u)ij )(x) =
ui,j (x) + uj,i (x)
,
2
1
i, j = 1, 2, 3.
2
Here u = (u1 , u2 , u3 )t and
∂ui
.
∂xj
Here we use the convention that indices to the left of the comma indicate
components while those to the right indicate differentiation. Given a tensor
T̃ (T̃ : Ω → R3×3 ), define the (row-wise) divergence operator div(T̃ ) to be
the vector field given by
ui,j ≡
(div(T̃ ))i ≡
3
X
T̃ij,j =
j=1
3
X
∂ T̃ij
j=1
∂xj
,
i = 1, 2, 3.
For µ and λ scalars and u a vector field, we define a linearized “stress”
tensor,
Tij (u) = 2µǫ(u)ij + λδij (∇ · u), i, j = 1, 2, 3.
Here δij denotes the Kronecker-delta function. Finally, we consider the differential operator (for u ∈ (C 2 (Ω̄))3
Lu ≡ −div(T (u)).
It is clear from the definitions that Lu is a vector field on Ω.
(a) Derive the following integration by parts formula: For φ and u as above,
X
Z
Z
3 Z
ǫ(u)ij φi,j dx
(T (u)n) · φ ds + 2µ
(Lu) · φ dx = −
Ω
∂Ω
+λ
i,j=1 Ω
Z
(∇ · u)(∇ · φ) dx.
Ω
(b) It follows from the symmetry of ǫ(u)(x) that
ǫ(u)ij φi,j + ǫ(u)ji φj,i = ǫ(u)ij ǫ(φ)ij + ǫ(u)ji ǫ(φ)ji
for each x. Use this to show that the double sum (in the large parenthesis
above) equals
X
3 Z
ǫ(u)ij ǫ(φ)ij dx .
i,j=1 Ω
Problem 3. Consider the vector field u(x, y) = (x, x + y)t . Let Ω be the
quarter circle consisting of points in the first quadrant with magnitude less
than one.
3
(a) Compute
Z
u · n ds
∂Ω
by computing the contribution from the three segments of ∂Ω, i.e., the
edge along y = 0, the edge along x = 0 and the arc θ ∈ [0, π/2] of radius
one.
(b) Compute
Z
∇ · u da
Ω
The results of (a) and (b) above should be the same!