1
AARMS Course: Homework 1
Problem 1: Let Ω = Ω = {x | |x| ≤ 1} be the unit sphere in three-dimensions. Suppose that Ω contains four small
spherically-shaped non-overlapping spherical holes Ωεj of a common radius ǫ ≪ 1 centered at some xj for j = 1, . . . , 4.
Consider the following PDE where αj for j = 1, . . . , 4 are constants:
x ∈ Ω\ ∪4j=1 Ωεj ,
△u = 1 ,
u = 1,
|x| = 1 .
u = αj ,
x ∈ ∂Ωεj ,
j = 1, .., 4 .
(1) Derive a two-term asymptotic expansion for the solution to this problem in the form u = u0 + ǫu1 + · · · , when
ǫ ≪ 1. (Hint: your result will depend on the the Green’s function for the Laplacian in a sphere, which you can
calculate explicitly by the method of inages, and will involve an “unperturbed solution”, which can be readily
found.)
(2) Setting αj = 1 for j = 1, .., 4, use your asymptotic result to estimate u at x = (1/2, 0, 0), when the holes are
centered at the four symmetrically placed points x = (0, ±1/2, ±1/2) in the plane x = 0.
(3) Now suppose that each of the spherical holes is replaced by an ellipsodial shape with semi-axes ǫa, ǫb, and ǫc,
where a > 0, b > 0, and c > 0, are the same for each hole. Show how to estimate u at x = (1/2, 0, 0) now.
(Hint: Your answer here should be really brief).
(4) Finally, replace the condition u = 1 on |x| = 1 with the Neumann condition ∂r u = 0 on r = |x| = 1. Show how
to estimate u at x = (1/2, 0, 0) now.
Solution to Problem 1:
(1) In the outer region we expand
u = u0 + ǫu1 + · · · ,
where u0 is the unperturbed solution in the absence of any holes, and satisfies
(
∆u0 = 1
u0 = 1
for x ∈ Ω
for x ∈ ∂Ω .
(0.2)
where Ω is the unit sphere. We look for a radially symmetric solution to this problem and readily calculate
that
u0 = (r2 + 5)/6 ,
r = |x| .
(0.3)
2
In addition, u1 satisfies



∆u1 = 0
for x ∈ Ω\{x1 , . . . , xN }
u1 = 0
for x ∈ ∂Ω


u singular
1
(0.4)
as x → xj , j = 1, . . . , N .
Now in the inner region near x = xj we write y = ǫ−1 (x − xj ) and v(y; ǫ) = u(xj + ǫy, ǫ) = v0 (y) + · · · . The
matching condition yields that v0 → u0 (xj ) as |y| → ∞ so that


∆y v0 = 0

for y 6∈ Ωj = ε−1 Ωǫj
v 0 = αj


v → u (x )
0
0 j
The solution is decomposed as
for x ∈ ∂Ωj
(0.5)
as |y| → ∞ .
v0 = u0 (xj ) + (αj − u0 (xj )) vc (y) ,
(0.6)
where vc (y) satisfies



∆y vc = 0
vc = 1


 v c ∼ Cj
for y 6∈ Ωj
(0.7)
for y ∈ Ωj
as |y| → ∞ .
|y|
However, since each Ωεj is a sphere of radius ε, or equivalently Ωj is a sphere of radius 1 in the y-variable, we
calculate exactly that
vc = 1/|y| ,
|y| ≥ 1 ,
so C = 1 .
(0.8)
In general if we have that all holes have the same shape, but non-spherical, then we would use vc ∼ C/|y| for
the far-field behavior of each hole (see part (3) below)
This yields the far-field behavior
v0 ∼ u0 (xj ) + (αj − u0 (xj ))
1
,
|y|
as
|y| → ∞ .
(0.9)
Now the matching condition is simply
u0 (xj + ǫu1 + · · · ∼ u0 (xj ) + (αj − u0 (xj ))
ǫ
+ εv1 + · · · .
|x − xj |
(0.10)
This yields that
u1 ∼ (αj − u0 (xj ))
1
|x − xj |
as
x → xj .
Therefore, the problem for u1 is simply
(
PN
∆u1 = −4π j=1 (αj − u0 (xj )) δ(x − xj ) ,
u1 = 0
for x ∈ Ω
for x ∈ ∂Ω .
The solution is given by
u1 = 4π
N
X
j=1
(αj − u0 (xj )) .G(x; xj ) ,
(0.11)
(0.12)
3
where G(x; xj ) is the Dirichlet Green’s function satisfying
(
∆G = −δ(x − xj ) for x ∈ Ω
G=0
(0.13)
for x ∈ ∂Ω .
This gives the two-term asymptotic expansion in the outer region given by
4 X
|x|2 + 5
(|xj |2 + 5)
u∼
G(x; xj )
αj −
+ 4πε
6
6
j=1
(0.14 a)
For the unit sphere, the Green’s function as obtained by the method of images is (this is well-known in potential
theory)
G(x; ξ) = −
1
1
+
.
4π|x − ξ| 4π|x||ξ − x/|x|2 |
(0.14 b)
(2) We calculate the terms in (0.14) for the data provided. We get |xj |2 = 1/2 for j = 1 . . . , 4, and |x − xj | =
√
3/2,
and |x − xj /|xj |2 | = 3/2 for all j. Therefore, we get from (0.14) that
4 7
4
11 X
2
u∼ +ε 1−
,
−√ +
8
12 j=1
3 3
√ 7
7 ε
4 − 2 3 ≈ + 0.05954ε ,
u∼ +
8 9
8
which gives the desired estimate.
(3) The solution is to simply replace ǫ in (0.14) by ǫC, where C is the capacitance of the ellipsoid. This capacitance
−1
R∞
is given by C = 2 0 (a2 + η)−1/2 (b2 + η)−1/2 (c2 + η)−1/2 dη
from the earlier notes.
(4) Upon replacing u = 1 on r = 1 with ur = 0 on r = 1, we observe that there is no solution to the unperturbed
problem. Thus, as similar to the calculation of the MFPT in the notes, we must expand u = u0 /ε + u1 + · · ·
and the same for the inner region v = v0 /ε + v1 + · · · near each xj . This is for the same reason as explained
in the MFPT notes, in that the Neumann eigenvalue problem has principal eigenvalue λ0 = O(ε).
We assume for simplicity below that Ωǫj = {x| |x − xj | = ǫ} so that we have N -small spheres of radius ǫ.
This simplification will allow us to readily calculate a two-term expansion for the solution.
In the outer region, we expand
u0
+ u1 + · · · .
ǫ
We obtain that ∆u0 = 0 with ∂n u0 = 0 on ∂Ω, and so u0 is a constant. The problem for u1 is


for x ∈ Ω\{x1 , . . . , xN }

∆u1 = 1
u=
∂ n u1 = 0


u singular
1
while u2 satisfies



∆u2 = 0
∂ n u2 = 0


u singular
2
for x ∈ ∂Ω
(0.15)
(0.16)
as x → xj , j = 1, . . . , N ,
for x ∈ Ω\{x1 , . . . , xN }
for x ∈ ∂Ω
as x → xj , j = 1, . . . , N .
(0.17)
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Now in the inner region we let y = ǫ−1 (x − xj ), and we expand the inner solution as
v=
v0
+ v1 + ǫv2 + · · · .
ǫ
(0.18)
We obtain, upon using the matching condition v0 → u0 as |y| → ∞ that



∆y v0 = 0 for |y| ≥ 1
v0 = 0
for |y| = 1


v → u
as |y| → ∞ .
0
(0.19)
0
Notice that at this order, we do not see the αj term. It occurs only at one higher order.
The solution is written as v0 = u0 (1 − vc ), where vc = 1/|y|. The matching condition for x → xj becomes
u0
v0
ε
u0
1−
+ v1 + · · · .
(0.20)
+ u1 + ǫu2 + · · · ∼
+ v1 + · · · =
ǫ
|x − xj |
{z
} |ǫ
{z
}
|ǫ
x→xj
y→∞
Therefore, we obtain
u1 → −
The problem for u1 is simply
u0
,
|x − xj |



∆u1 = 1
∂ n u1 = 0


u1 ∼ − u0
(
x → xj .
(0.21)
for x ∈ Ω\{x1 , . . . , xN }
for x ∈ ∂Ω
∆u1 = 1 + 4πu0
∂ n u1 = 0
(0.22)
as x → xj , j = 1, . . . , N ,
|x−xj |
which is equivalent to
as
PN
j=1
δ(x − xj )
for x ∈ Ω
for x ∈ ∂Ω .
Upon using the divergence theorem, we obtain that
(0.23)
|Ω| + 4πu0 N = 0 .
This yields the leading-order outer solution as
u∼
u0
,
ǫ
where
u0 = −
|Ω|
.
4πN
Now we proceed to one higher order in the asymptotic construction. To do so, we must solve for u1 explicitly.
This is done by introducing the Neumann Green’s function G(x; xj ) defined uniquely by the solution to
∆G =
1
− δ(x − xj ) ,
|Ω|
x ∈ Ω,
∂n G = 0 , x ∈ ∂Ω ,
Z
G dx = 0 .
(0.24 a)
(0.24 b)
(0.24 c)
Ω
We notice that G(x; xj ) exists since
R Ω
1
|Ω|
R
− δ(x − xj ) dx = 0, and the condition Ω G dx = 0 specifies G
uniquely. In addition, the local behavior of G(x; xj ) is
G(x; xj ) ∼
1
+ R(xj ) ,
4π|x − xj |
as
x → xj ,
(0.24 d )
5
where R(xj ) is the regular (smooth) part of the Neumann Green’s function.
Now we write the solution to the problem for u1 as
u1 = −4πu0
N
X
G(x; xi ) + ū1 .
(0.25)
i=1
Here ū1 is a constant to be found.
Now we expand the solution in (0.25) as x → xj for each j = 1, . . . , N to obtain
u1 ∼ −4πu0
X
N
G(xj ; xi ) +
i6=j
1
+ Rj
4π|x − xj |
+ ū1 .
We write this expression as
u1 ∼ Bj + ū1 −
where Bj is defined by
u0
,
|x − xj |
as
x → xj ,
N
X
G(xj ; xi ) .
Bj = −4πu0 Rj +
i6=j
Then, the matching condition is
v0
u0
+ u1 + ǫu2 + · · · ∼
+ v1 + · · · .
ǫ
ǫ
Writing this condition out in detail we have
u0
ǫ
u0
u0
1−
+ v1 .
+ ū1 + Bj −
+ ǫu2 ∼
ǫ
|x − xj |
ǫ
|x − xj |
This implies that for each j = 1, . . . , N , v1 must satisfy


for |y| ≥ 1

∆y v1 = 0
The solution is given explicitly by
v 1 = αj


v ∼ ū + B
1
1
j
for |y| = 1
as |y| → ∞ .
v1 = (ū1 + Bj ) − [(ū1 + Bj ) − αj ]
1
.
|y|
1
Therefore, v1 ∼ (ū1 + Bj ) − [(ū1 + Bj ) − αj ] |y|
as |y| → ∞. This implies from the matching condition that u2
must satisfy



∆u2 = 0
∂ n u2 = 0


u2 ∼ [αj − (ū1 + Bj )]
for x ∈ Ω\{x1 , . . . , xN }
1
|x−xj |
for x ∈ ∂Ω
as x → xj
j = 1, . . . , N .
Therefore, we can write the problem for u2 as
∆u2 = −4π
N
X
[αj − (ū1 + Bj )]δ(x − xj )
j=1
x ∈ Ω;
∂ n u2 = 0 ,
on
∂Ω .
6
Finally, we determine ū1 by the divergence theorem. We calculate
N
X
j=1
[αj − (ū1 + Bj )] = 0 ,
so that
ū1 =
where
N
1 X
(αj − Bj ) ,
N j=1
Bj = −4πu0 Rj +
N
X
(0.26 a)
G(xj ; xi ) .
i6=j
(0.26 b)
In summary, a two-term outer expansion is given by
u0
|Ω|
+ u1 + · · · , with u0 = −
ǫ
4πN
PN
where u1 is given by u1 = ū1 − 4πu0 i=1 G(x; xi ). Finally, the constant term ū1 is given in (0.26).
u∼
Problem 2: In a 3-D domain, calculate a two-term asymptotic expansion for the principal eigenvalue of the
Neumann problem:
∆u + λu = 0 ,
∂n u = 0 ,
u = 0,
x ∈ Ω\Ωa ;
Z
u2 dx = 1 ,
(0.27 a)
Ω\Ωa
x ∈ ∂Ω ,
(0.27 b)
x ∈ ∂Ωa ≡ ∪N
j=1 ∂Ωεj .
(0.27 c)
Here Ωεj is an arbitrarily shaped trap of “radius” ǫ for which Ωεj → xj as ǫ → 0. The result you find should be the
same as given in Principal Result 2.1 of the Day 2 lecture notes.
Solution to Problem 2:
We now use the method of matched asymptotic expansions to derive a two-term expansion for the principal
eigenvalue λ(ε) of (0.27) as ε → 0. For the problem with no traps, λ0 = 0 and u0 = |Ω|−1/2 is the unperturbed
eigenfunction, where |Ω| denotes the volume of Ω. We expand the principal eigenvalue for (0.27) as
λ = ελ1 + ε2 λ2 + · · · .
(0.28)
In the outer region away from an O(ε) neighborhood of xj , we expand the outer solution as
u = u0 + εu1 + ε2 u2 + · · · .
(0.29)
Upon substituting (0.28) and (0.29) into (0.27 a) and (0.27 b), we obtain that u1 and u2 satisfy
Z
∆u1 = −λ1 u0 , x ∈ Ω\{x1 , . . . xN } ;
∂n u1 = 0 , x ∈ ∂Ω ;
u1 dx = 0 .
(0.30)
Ω
∆u2 = −λ2 u0 − λ1 u1 ,
x ∈ Ω\{x1 , . . . xN } ;
∂ n u2 = 0 ,
x ∈ ∂Ω ;
Z
Ω
u2 dx = −
|Ω|1/2
2
Z
u21 dx .
(0.31)
Ω
The matching of u1 and u2 to inner solutions defined in an O(ε) neighborhood of each trap will yield singularity
conditions for u1 and u2 as x → xj for j = 1, . . . , N . .
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In the inner region near the j th trap we introduce the local variables y and w(y) by
y = ε−1 (x − xj ) ,
w(y) = u(xj + εy, ε) .
(0.32)
Upon substituting (0.32) into (0.27 a) and (0.27 c), we obtain that ∆y w = −ε2 λw, where ∆y denotes the Laplacian
in the y variable. We expand the inner solution as
w = w0 + εw1 + ε2 w2 + · · · ,
(0.33)
and then use λ = O(ε) to obtain the following inner problems for k = 0, 1, 2:
∆y wk = 0 ,
y 6∈ Ωj ;
wk = 0 ,
y ∈ ∂Ωj .
(0.34)
Here Ωj denotes an O(ε−1 ) magnification of Ωεj so that Ωj = ε−1 Ωεj . The appropriate far-field boundary condition
for (0.34) is determined by matching w to the outer asymptotic expansion of the eigenfunction.
The matching condition is that the near-field behavior of the outer eigenfunction as x → xj must agree asymptot-
ically with the far-field behavior of the inner eigenfunction as |y| = ε−1 |x − xj | → ∞, so that
u0 + εu1 + ε2 u2 + · · · ∼ w0 + εw1 + ε2 w2 + · · · .
(0.35)
Since u0 = |Ω|−1/2 , the first matching condition is that w0 ∼ |Ω|−1/2 as |y| → ∞. We then introduce wc by
w0 =
1
(1 − wc ) ,
|Ω|1/2
(0.36)
so that from (0.34) with k = 0, we get that wc satisfies
∆y wc = 0 ,
y 6∈ Ωj ;
wc = 1 ,
y ∈ ∂Ωj ;
wc → 0 as
|y| → ∞ .
(0.37 a)
This is a classic problem in electrostatics, and it is well-known that the far-field behavior of wc is
wc ∼
Cj
Pj · y
+
+ ···
|y|
|y|3
as
|y| → ∞ .
(0.37 b)
Here Cj is the capacitance of Ωj and Pj denotes the dipole vector, both determined by the shape of Ωj . These intrinsic
quantities can be found explicitly for different trap shapes such as spheres, ellipsoids, etc..
Upon substituting (0.37 b) into (0.35), we obtain that the matching condition becomes
ε2 Pj · (x − xj )
εCj
1
1
2
+ εw1 + ε2 w2 + · · · .
−
1−
+ εu1 + ε u2 + · · · ∼
|x − xj |
|x − xj |3
|Ω|1/2
|Ω|1/2
(0.38)
Therefore, we require that u1 has the singular behavior u1 ∼ −|Ω|−1/2 Cj /|x − xj | as x → xj for j = 1, . . . , N . The
problem (0.30) for u1 with this singularity behavior can be written in Ω in terms of the Dirac distribution as
∆u1 = −λ1 u0 +
with
R
Ω
N
4π X
Cj δ(x − xj ) ,
|Ω|1/2 j=1
x ∈ Ω;
∂ n u1 = 0 ,
x ∈ ∂Ω ,
(0.39)
u1 dx = 0. Upon using the divergence theorem, and recalling that u0 = |Ω|−1/2 , we determine λ1 as
N
λ1 =
4π X
Cj .
|Ω| j=1
(0.40)
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This leading order result does not depend on the trap locations it does not indicate how to optimize λ. As such, we
must extend the calculation to one higher order.
To solve (0.39), we introduce the Neumann Green’s function G(x; ξ), which satisfies
1
− δ(x − ξ) , x ∈ Ω ;
∂n G = 0 , x ∈ ∂Ω ,
|Ω|
Z
1
G(x; ξ) ∼
+ R(ξ) , as x → ξ ;
G(x; ξ) dx = 0 .
4π|x − ξ|
Ω
∆G =
(0.41 a)
(0.41 b)
Here R(ξ) is called the regular part of G(x; ξ). In terms of G, the unique solution to (0.39), which satisfies
is simply
u1 = −
R
Ω
u1 dx = 0,
N
4π X
Ck G(x; xk ) .
|Ω|1/2 k=1
(0.42)
Next, we expand u1 in (0.42) as x → xj . Upon using (0.41 b) to obtain the local behavior of G, we obtain


N
X
4π 
Cj

+ Aj
as x → xj ;
Aj = − 1/2 Cj Rj +
Ck Gj,k  .
(0.43)
u1 ∼ − 1/2
|Ω| |x − xj |
|Ω|
k=1
k6=i
Here we have defined Rj ≡ R(xj ) and Gj,k ≡ G(xj ; xk ). Upon substituting this expression into the matching condition
(0.38), we obtain
εCj
1
Cj
ε2 Pj · (x − xj )
1
2
1−
+ε − 1/2
+ Aj +ε u2 +· · · ∼
+εw1 +ε2 w2 +· · · . (0.44)
−
|x − xj |
|x − xj |3
|Ω|1/2
|Ω| |x − xj |
|Ω|1/2
We then conclude that w1 ∼ Aj as |y| → ∞. The solution w1 to (0.34) is
Cj
w1 = Aj (1 − wc ) ∼ Aj 1 −
+ ···
as
|y|
|y| → ∞ ,
(0.45)
where wc is the solution to (0.37). Next, we write the far-field behavior in (0.45) in outer variables and substitute
the resulting expression into the right-hand side of the matching condition (0.44) to identify the terms of O(ε2 ). In
this way, we obtain that the outer eigenfunction u2 must have the following singularity behavior as x → xj :
u2 ∼ −
Aj Cj
Pj · (x − xj )
−
|x − xj |
|x − xj |3
as
x → xj ,
j = 1, . . . , N .
(0.46)
The problem (0.31) for u2 , together with singularity behavior (0.46), can be written in Ω in terms of the Dirac
distribution as
∆u2 = −λ2 u0 − λ1 u1 + 4π
N
X
j=1
Aj Cj δ(x − xj ) − 4π
N
X
j=1
Pj · ∇δ(x − xj ) ,
with ∂n u2 = 0 for x ∈ ∂Ω. Then, applying the divergence theorem to (0.47), and using
λ2 =
N
4π X
Aj Cj .
|Ω|1/2 j=1
R
Ω
x ∈ Ω,
(0.47)
u1 dx = 0, we get
(0.48)
We remark that this eigenvalue correction λ2 does not depend on the dipole vector Pj defined in (0.37 b).
Next, it is convenient to introduce the capacitance vector c and the symmetric Neumann Green’s matrix G with
diagonal entries R(xj ), and off-diagonal entries G(xi ; xj ). Upon substituting (0.40) and (0.48) into (0.28), we obtain
9
the main result
16π 2 ε2
4πεN
C̄ −
pc (x1 , . . . , xN ) .
(0.49)
|Ω|
|Ω|
Here C̄ ≡ N −1 (C1 + . . . + CN ) and pc (x1 , . . . , xN ) is the discrete sum defined in terms of the entries Gi,j of the
λ0 (ε) ∼
Green’s matrix G, as defined by
pc (x1 , . . . , xN ) ≡ cT Gc =
N
N X
X
i=1 j=1
Ci Cj Gi,j .
(0.50)