1
Notes for Workshop: Asymptotic Methods in Fluid Mechanics: Survey and Recent Advances
Solutions to Problems in “Asymptotic Methods for PDE
Problems in Fluid Mechanics and Related Systems with
Strong Localized Perturbations in Two-Dimensional
Domains”
M. J. WARD, M. C. KROPINSKI
Michael J. Ward; Department of Mathematics, University of British Columbia, Vancouver, British Columbia, V6T 1Z2, Canada,
Mary Catherine Kropinski, Department of Mathematics, Simon Fraser University, Burnaby, British Columbia, V5A IS6, Canada
(Received 15 September 2009)
1 Problem 1
Problem 1: Consider a conventional infinite-order logarithmic expansion for the outer solution in the form
j
∞ X
−1
W0j (x) + σ(ε)W1 + · · · ,
W ∼
log(εd)
j=0
(1.1)
with σ(ε) ν k for any k > 0. By formulating a similar series for the inner solution, derive a recursive set of
problems for the W0j for j ≥ 0 from the asymptotic matching of the inner and outer solutions. Show that this series
can be summed and leads to the result in equation (2.17a) of the workshop notes.
Solution:
We consider the pipe flow problem of §2 of the notes, formulated as
4w = −β ,
x ∈ Ω\Ωε ,
(1.2 a)
w = 0,
x ∈ ∂Ω ,
(1.2 b)
w = 0,
x ∈ ∂Ωε .
(1.2 c)
In the outer region we expand the solution to (1.2) in an explicit infinite-order logarithmic expansion as
w(x; ε) = W0H (x) +
∞
X
j=1
ν j W0j (x) + · · · .
(1.3)
Here ν = O(1/ log ε) is a gauge function to be chosen. The smooth function W0H satisfies the unperturbed problem
in the unperturbed domain, given by
4W0H = −β ,
x ∈ Ω;
W0H = 0 ,
x ∈ ∂Ω .
(1.4)
By substituting (1.3) into (1.2 a) and (1.2 b), and letting Ωε → x0 as ε → 0, we get that W0j for j ≥ 1 satisfies
4W0j = 0 ,
W0j = 0 ,
W0j
x ∈ Ω\{x0 } ,
(1.5 a)
x ∈ ∂Ω ,
(1.5 b)
is singular as x → x0 .
(1.5 c)
2
M. J. Ward, M. C. Kropinski
The matching of the outer and inner expansions will determine a singularity behavior for W 0j as x → x0 for each
j ≥ 1.
In the inner region near Ωε we introduce the inner variables
y = ε−1 (x − x0 ) ,
v(y; ε) = W (x0 + εy; ε) .
(1.6)
We then pose the explicit infinite-order logarithmic inner expansion
v(y; ε) =
∞
X
γj ν j+1 vc (y) .
(1.7)
j=0
Here γj are ε-independent coefficients to be determined. Substituting (1.7) and (1.2 a) and (1.2 c), and allowing v c (y)
to grow logarithmically at infinity, we obtain that vc (y) satisfies
4y v c = 0 ,
y∈
/ Ω1 ;
vc ∼ log |y| ,
vc = 0 ,
y ∈ ∂Ω1 ,
as |y| → ∞ .
(1.8 a)
(1.8 b)
The unique solution to (1.8) has the following far-field asymptotic behavior:
vc (y) ∼ log |y| − log d + o(1) ,
as |y| → ∞ .
(1.8 c)
Upon using the far-field behavior (1.8 c) in (1.7), and writing the resulting expression in terms of the outer variable
x − x0 = εy, we obtain that
v ∼ γ0 +
∞
X
j=1
ν j [γj−1 log |x − x0 | + γj ] .
(1.9)
The matching condition between the infinite-order outer expansion (1.3) as x → x0 and the far-field behavior (1.9)
of the inner expansion is that
W0H (x0 ) +
∞
X
j=1
ν j W0j (x) ∼ γ0 +
∞
X
j=1
ν j [γj−1 log |x − x0 | + γj ] .
(1.10)
The leading-order match yields that
γ0 = W0H (x0 ) .
(1.11)
The higher-order matching condition, from (1.10), shows that the solution W0j to (1.5) must have the singularity
behavior
W0j ∼ γj−1 log |x − x0 | + γj ,
as x → x0 .
(1.12)
The unknown coefficients γj for j ≥ 1, starting with γ0 = W0H (x0 ) are determined recursively from the infinite
sequence of problems (1.5) and (1.12) for j ≥ 1. The explicit solution to (1.5) with W 0j ∼ γj−1 log |x − x0 | as x → x0
is given explicitly in terms of the Dirichlet Green’s function Gd (x; x0 ) by
W0j (x) = −2πγj−1 Gd (x; x0 ) ,
(1.13)
where Gd (x; x0 ) satisfies
4Gd = −δ(x − x0 ) , x ∈ Ω ;
Gd = 0 , x ∈ ∂Ω ,
1
log |x − x0 | + Rd (x0 ; x0 ) + o(1) , as x → x0 .
Gd (x; x0 ) = −
2π
(1.14 a)
(1.14 b)
Solutions to Problems
3
Next, we expand (1.13) as x → x0 and compare it with the required singularity structure (1.12). This yields
1
−2πγj−1 −
(1.15)
log |x − x0 | + Rd00 ∼ γj−1 log |x − x0 | + γj ,
2π
where Rd00 ≡ Rd (x0 ; x0 ). By comparing the non-singular parts of (1.15), we obtain a recursion relation for the γ j
given by
γj = −2πRd00 γj−1 ,
γ0 = W0H (x0 ) ,
(1.16)
which has the explicit solution
j
γj = [−2πRd00 ] W0H (x0 ) ,
j ≥ 0.
(1.17)
Finally, to obtain the outer solution we substitute (1.13) and (1.17) into (1.3) to obtain
w ∼ W0H (x) +
∞
X
ν j (−2πγj−1 ) Gd (x; x0 ) ,
j=1
∼ W0H (x) − 2πνGd (x; x0 )
∞
X
ν j γj
j=0
∼ W0H (x) − 2πνW0H (x0 )Gd (x; x0 )
∼ W0H (x0 ) −
∞
X
[−2πνRd00 ]
j
j=0
2πνW0H (x0 )
Gd (x0 ; x0 ) .
1 + 2πνRd00
(1.18)
The last expression (1.18) agrees with equation (2.17a) of the notes. Similarly, upon substituting (1.17) into the
infinite-order inner expansion (1.7), we obtain
v(y; ε) =
∞
X
γj ν j+1 vc (y) = νW0H (x0 )vc (y)
j=0
∞
X
j
[−2πRd00 ν] =
j=0
νW0H (x0 )
vc (y) ,
1 + 2πνRd00
(1.19)
which recovers equation (2.17b) of the notes.
2 Problem 2
Problem 2: Consider the following problem in an arbitrary two-dimensional domain with N small inclusions:
4u − m(x)u = 0 ,
u=f,
u = αj ,
x ∈ Ω\ ∪N
j=1 Ωεj ,
(2.1 a)
x ∈ ∂Ω .
(2.1 b)
x ∈ ∂Ωεj ,
j = 1, . . . , N ,
(2.1 c)
Here m(x) is an arbitrary smooth function with m(x) ≥ 0 in Ω, f is an arbitrary function on ∂Ω, and α j are
constants. Formulate a linear system, similar to equation (3.17) of the workshop notes, in terms of a certain Green’s
function, that effectively sums the infinite-order logarithmic series in the asymptotic expansion of the solution. Apply
your general theory to the unit disk Ω for the case N = 1, m ≡ 1, f ≡ 0, and α 1 = 1, and where there is an
arbitrarily-shaped hole centered at the origin of the unit disk.
Solution: In the outer region, defined away from Ωεj for j = 1, . . . , N , we expand
u(x; ε) ∼ U0H (x) + U0 (x; ν) + σ(ε)U1 (x; ν) + · · · .
(2.2)
4
M. J. Ward, M. C. Kropinski
Here ν = (ν1 , . . . , νN ) is a set of logarithmic gauge functions to be determined and σ νjk as ε → 0 for j = 1, . . . , N .
In (2.2), U0H (x) is the smooth function satisfying the unperturbed problem in the unperturbed domain Ω
4U0H − m(x)U0H = 0 ,
x ∈ Ω;
U0H = f ,
x ∈ ∂Ω .
(2.3)
Substituting (2.2) into (2.1 a) and (2.1 b), and letting Ωεj → xj as ε → 0, we get that U0 satisfies
4U0 − m(x)U0 = 0 ,
x ∈ Ω\{x1 , . . . , xN } ,
U0 = 0 ,
U0
(2.4 a)
x ∈ ∂Ω ,
is singular as x → xj ,
(2.4 b)
j = 1, . . . , N .
(2.4 c)
The singularity behavior for U0 as x → xj will be found below by matching the outer solution to the far-field behavior
of the inner solution to be constructed near each Ωεj .
In the j th inner region near Ωεj we introduce the inner variables y and v(y; ε) by
y = ε−1 (x − xj ) ,
v(y; ε) = u(xj + εy; ε) .
(2.5)
v(y; ε) = αj + νj γj vcj (y) + µ0 (ε)V1j (y) + · · · ,
(2.6)
We then expand v(y; ε) as
where γj = γj (ν) is a constant to be determined. Here µ0 νjk as ε → 0 for any k > 0. In (2.6), the logarithmic
gauge function νj is defined by
νj = −1/ log(εdj ) ,
(2.7)
where dj is specified below. By substituting (2.5) and (2.6) into (2.1 a) and (2.1 c), we conclude that v cj (y) is the
unique solution to
4y vcj = 0 ,
y∈
/ Ωj ;
vcj = 0 ,
vcj (y) ∼ log |y| − log dj + o(1) ,
y ∈ ∂Ωj ,
(2.8 a)
as |y| → ∞ .
(2.8 b)
Here Ωj ≡ ε−1 Ωεj , and the logarithmic capacitance, dj , is determined by the shape of Ωj .
Writing (2.8 b) in outer variables and substituting the result into (2.6), we get that the far-field expansion of v
away from each Ωj is
v ∼ αj + γj + νj γj log |x − xj | ,
j = 1, . . . , N .
(2.9)
Then, by expanding the outer solution (2.2) as x → xj , we obtain the following matching condition between the
inner and outer solutions:
U0H (xj ) + U0 ∼ αj + γj + νj γj log |x − xj | ,
as x → xj ,
j = 1, . . . , N .
(2.10)
In this way, we obtain that U0 satisfies (2.4) subject to the singularity structure
U0 ∼ αj − U0H (xj ) + γj + νj γj log |x − xj | + o(1) ,
as x → xj ,
j = 1, . . . , N .
(2.11)
Observe that in (2.11) both the singular and regular parts of the singularity structure are specified. Therefore, (2.11)
will effectively lead to a linear system of algebraic equations for γj for j = 1, . . . , N .
The solution to (2.4 a) and (2.4 b), with U0 ∼ νj γj log |x − xj | as x → xj , can be written as
U0 (x; ν) = −2π
N
X
i=1
νi γi G(x; xi ) ,
(2.12)
Solutions to Problems
5
where G(x; xj ) is the Green’s function satisfying
4G − m(x)G = −δ(x − xj ) , x ∈ Ω ; G = 0 , x ∈ ∂Ω ,
1
G(x; xj ) ∼ −
log |x − xj | + R(xj ; xj ) + o(1) , as x → xj .
2π
(2.13 a)
(2.13 b)
Here Rjj ≡ R(xj ; xj ) is the regular part of G.
Finally, we expand (2.12) as x → xj and equate the resulting expression with the required singularity behavior
(2.11) to get
νj γj log |x − xj | − 2πνj γj Rjj − 2π
N
X
i=1
νi γi G(xj ; xi ) = αj − U0H (xj ) + γj + νj γj log |x − xj | ,
j = 1, . . . , N . (2.14)
i6=j
In this way, we get the following linear algebraic system for γj for j = 1, . . . , N :
−γj (1 + 2πνj Rjj ) − 2π
N
X
i=1
νi γi Gji = αj − U0H (xj ) ,
j = 1, . . . , N .
(2.15)
i6=j
Here Gji ≡ G(xj ; xi ) and νj = −1/ log(εdj ). We summarize the asymptotic construction as follows:
Principal Result: For ε 1, the outer expansion for (2.1) is
u ∼ U0H (x) − 2π
N
X
νi γi G(x; xi ) ,
i=1
for |x − xj | = O(1) .
(2.16 a)
The inner expansion near Ωεj with y = ε−1 (x − xj ), is
u ∼ αj + νj γj vcj (y) ,
for |x − xj | = O(ε) .
(2.16 b)
Here νj = −1/ log(εdj ), dj is defined in (2.8 b), vcj (y) satisfies (2.8), U0H satisfies the unperturbed problem (2.3),
while G(x; xj ) and R(xj ; xj ) satisfy (2.13). Finally, the constants γj for j = 1, . . . , N are obtained from the N
dimensional linear algebraic system (2.15).
To illustrate the theory, let Ω be the unit disk containing one arbitrarily-shaped hole centered at the origin. Suppose
that m(x) = 1 and f = 0. Then, U0H ≡ 0, and the Green’s function satisfying (2.13) is radially symmetric with a
singularity at the center of the disk. The explicit Green’s function is
1
K0 (1)
G(x; 0) =
K0 (r) −
I0 (r) ,
2π
I0 (1)
0 < r < 1,
(2.17)
where r ≡ |x|. Here I0 (r) and K0 (r) are the modified Bessel functions of the first and second kind, respectively,
of order zero. To identity the regular part of G at the origin, i.e. R(0; 0), we use the well-known asymptotics
K0 (r) ∼ − log r + log 2 − γe as r → 0, where γe is Euler’s constant. Then, from (2.17) and (2.13 b), we get that
K0 (1)
1
log 2 − γe −
.
(2.18)
R11 ≡ R(0; 0) =
2π
I0 (1)
For N = 1, U0H ≡ 0, and α1 = 1, the system (2.15) then determines γ1 in terms of R11 and ν = −1/ log(εd1 ) as
γ1 = − [1 + 2πν1 R11 ]−1 .
(2.19)
Therefore, γ1 is determined explicitly in terms of the logarithmic capacitance, d1 , of the arbitrarily-shaped hole
centered at the origin.
6
M. J. Ward, M. C. Kropinski
3 Problem 3
Problem 3: Consider the following problem in the disk Ω = {x | |x| ≤ 2} that contains three small holes:
x ∈ Ω\ ∪3j=1 Ωεj ,
4u = 0 ,
u = 4 cos(2θ) ,
u = αj ,
|x| = 2 .
x ∈ ∂Ωεj ,
j = 1, 2, 3 .
(3.1 a)
(3.1 b)
(3.1 c)
Suppose that each of the holes has an elliptical shape with semi-axes ε and 2ε. Apply the theory for summing infinite
logarithmic expansions to first derive and then numerically solve a linear system for the source strengths. In your
implementation assume that the holes are centered at x1 = (1/2, 1/2), x2 = (1/2, 0) and x3 = (−1/4, 0). The
boundary values on the holes are to be taken as α1 = 1, α2 = 0 and α3 = 2.
Solution: This is just a simple application of the theory in Problem 2 for the special case of a disk of radius 2 with
m(x) ≡ 0 and f = 4 cos(2θ) = 4(cos2 θ − sin2 θ) = x2 − y 2 on (x2 + y 2 )1/2 = 4.
For this problem, the solution to the unperturbed problem (2.3) is simply
U0H (x, y) = x2 − y 2 .
(3.2)
Next, the Green’s function satisfying (2.13) of Problem 2 with m(x) ≡ 0 and its regular part are calculated from the
method of images as
1
G(x; xj ) = −
log
2π
2|x − xj |
|x − x0j ||xj |
!
,
Rjj
#
"
1
2
≡ R(xj ; xj ) = −
.
log
2π
|xj − x0j ||xj |
(3.3)
Here x0j is the image point of xj in the unit disk of radius two..
Next, we note that since each of the holes has an elliptic shape with semi-axes ε and 2ε, then from Table 1 of the
notes their common logarithmic capacitance is d = 3/2. The holes are assumed to be centered at x 1 = (1/2, 1/2),
x2 = (1/2, 0) and x3 = (−1/4, 0), and have the constant boundary values α1 = 1, α2 = 0 and α3 = 2.
Therefore, upon defining ν = −1/ log (3ε/2) we obtain from (2.15) of Problem 2 that γ j for j = 1, . . . , 3 is the
solution of the linear system
−γ1 [1 + 2πνR11 ] − 2πν [γ2 G(x1 ; x2 ) + γ3 G(x1 ; x3 )] = 1 ,
(3.4 a)
−γ2 [1 + 2πνR22 ] − 2πν [γ1 G(x2 ; x1 ) + γ3 G(x2 ; x3 )] = −1/4 ,
(3.4 b)
−γ3 [1 + 2πνR33 ] − 2πν [γ1 G(x3 ; x1 ) + γ2 G(x3 ; x2 )] = 31/16 .
(3.4 c)
Here Rjj and G(xj ; xi ) are to be evaluated from (3.3).
We solve this linear system numerically for γj as a function of ε. The curves γj (ε) as a function of ε are plotted in
Fig. 1. We observe that the leading-order approximation to (3.4), valid for ν 1, is simply γ 1 = −1, γ2 = 1/4 and
γ3 = −31/16. From Fig. 1 we observe that this approximation, which neglects interaction effects between the holes,
is rather inaccurate unless ε is very small.
4 Problem 4
Problem 4: Beginning with the steady, incompressible, Navier-Stokes equations in velocity-pressure form, derive the
problem in equation (4.1) of the notes for the streamfunction for steady viscous flow over an infinite cylinder.
Solutions to Problems
7
4
γ
1
γ
2
γ3
3
2
γ
1
0
−1
−2
−3
0
0.05
0.1
0.15
ε
0.2
0.25
Figure 1. Plot of γj = γj () for j = 1, 2, 3 obtained from the numerical solution to (3.4).
Solution: We begin with the steady-state incompressible dimensionless Navier-Stokes equations in two space dimensions for the velocity u(x) and the pressure p(x), with x = (x1 , x2 , 0), given by
∇· u = 0,
ε (u · ∇) u = −∇p + 4u .
(4.1)
Lengths are made dimensionless with respect to the radius a of the cross-section of the infinitely-long cylinder, the
dynamic viscosity ν, and the free-stream speed u∞ in the x1 direction at infinity. The parameter ε = u∞ a/ν is the
Reynolds number, and it is assumed to be small. As |x| → ∞, we have u → î, indicating a uniform flow in the x1
direction. The no slip condition u = 0 is to hold on the body.
To eliminate the pressure, we take the curl of the momentum equation and then use ∇ × ∇p = 0 to get
(u · ∇) (∇ × u) − [(∇ × u) · u] =
1
4 (∇ × u) .
ε
(4.2)
Next, since the flow is two dimensional, we introduce the scalar Lagrangian stream function ψ(ρ, θ) defined in terms
of the velocity components u = uρ r̂ + uθ θ̂ in the radial and tangential directions by
uρ =
1 ∂ψ
,
ρ ∂θ
uθ =
∂ψ
.
∂ρ
(4.3)
With this choice, the divergence-free condition ∇ · u = 0 holds automatically. Then, by calculating the cross-products
in cylindrical coordinates, (4.3) readily transforms to
ε
42 ψ = − [∂ρ ψ∂θ (4ψ) − ∂θ ψ∂ρ (4ψ)] .
ρ
(4.4)
which agrees with equation (3.1) of the notes.
Next, we take ψ = 0 to correspond to the boundary of the cross-section of the cylinder. Next, if u = 0 on the
body, then this corresponds from (4.3) to ∂n ψ = 0, where ∂n is the outward normal derivative on the body. Finally,
if we put ψ ∼ ρ sin θ as ρ → ∞, then from (4.3), we get ur → cos θ and uθ → sin θ as r → ∞. In terms of cartesian
coordinates this corresponds to uniform flow in the x direction at infinity.
8
M. J. Ward, M. C. Kropinski
5 Problem 5
Problem 5: Consider the Biharmonic equation in the two-dimensional concentric annulus, formulated as
42 u = 0 ,
u=f,
x ∈ Ω\Ωε ,
ur = 0 ,
u = ur = 0 ,
(5.1 a)
on r = 1 ,
(5.1 b)
r = ε.
(5.1 c)
Here Ω is the unit disk centered at the origin, containing a small hole of radius ε centered at x = 0, i.e. Ω ε =
{x | |x| ≤ ε}. Consider the following two choices for f : Case I: f = 1. Case II: f = sin θ. For each of these
two cases calculate the exact solution, and from it determine an approximation to the solution in the outer region
|x| O(ε). Can you re-derive these results from singular perturbation theory in the limit ε → 0? (Hint: the leading-
order outer problem for Case I is different from what you might expect).
Solution:
Case I: We consider the perturbed problem
42 u = 0 ,
u = 1,
ε < r < 1,
ur = 0 ,
u = ur = 0 ,
(5.2 a)
on r = 1 ,
(5.2 b)
on r = ε .
(5.2 c)
We first find the exact solution of (5.2) and then expand it for ε → 0. Since the radially symmetric solutions to
(5.2 a) are linear combinations of {r 2 , r2 log r, log r, 1}, we can write the solution to (5.2 a), which satisfies (5.2 b), as
u = A r2 − 1 + Br2 log r − (2A + B) log r + 1 ,
(5.3)
for any constants A and B. Then, imposing that u = ur = 0 on r = ε, we get two equations for A and B:
Equation (5.4 a) gives
2A 1 − ε2 A + B 1 − ε2 − 2ε2 log ε = 0 ,
A 1 + 2 log ε − ε2 + B 1 − ε2 log ε = 1 .
(5.4 a)
(5.4 b)
2ε2 log ε
B
1−
.
A=−
2
1 − ε2
Upon substituting this into (5.4 b), we obtain that B satisfies
B
2ε2 log ε
2 log ε
1
−
1−
1+
+ B log ε =
2
1 − ε2
1 − ε2
1 − ε2
B
B log ε Bε2 log ε 2ε2 B(log ε)2
1
− −
+
+
+ B log ε =
2
1 − ε2
1 − ε2
(1 − ε2 )2
1 − ε2
B
− − B(1 + ε2 ) log ε + Bε2 (1 + ε2 ) log ε + 2ε2 (log ε)2 B + B log ε ∼ 1 + ε2
2
B
− + 2ε2 (log ε)2 B ∼ 1 + O(ε2 ) .
2
(5.5)
(5.6 a)
(5.6 b)
(5.6 c)
(5.6 d )
The last line of (5.6) determines B, while (5.5) determines A. In this way, we get
2
B ∼ −2 − 8ε2 (log ε) ,
2
A ∼ 1 + 4ε2 (log ε) .
(5.7)
Solutions to Problems
9
Upon substituting (5.7) into (5.3), we obtain the following two-term expansion in the outer region r O(ε):
2
u ∼ u0 (r) + ε2 (log ε) u1 (r) + · · · ,
(5.8)
where u0 (r) and u1 (r) are defined by
u0 (r) = r2 − 2r2 log r ,
u1 = 4 r2 − 1 − 8r2 log r .
(5.9)
It is interesting to note that the leading-order outer solution u0 (r) is not a C 2 smooth function, but that it does
satisfy the point constraint u0 (0) = 0. Hence, in the limit of small hole radius the ε-dependent solution does not tend
to the unperturbed solution in the absence of the hole. This unperturbed solution would have B = 0 and A = 0 in
(5.3), and consequently u = 1 in the outer region.
Next, we show how to recover (5.8) from a matched asymptotic expansion analysis. In the outer region we expand
the solution as
u ∼ w0 + σw1 + · · · ,
(5.10)
where σ 1 is an unknown gauge function, and where w0 satisfies the following problem with a point constraint:
42 w 0 = 0 ,
0 < r < 1;
w0 (1) = 1 ,
w0r (1) = 0 ,
w0 (0) = 0 .
(5.11)
The solution is readily calculated as
w0 = r2 − 2r2 log r .
(5.12)
0 < r < 1;
w1 (1) = w1r (1) = 0 .
(5.13)
The solution to (5.13) is given in terms of unknown coefficients α1 and β1 as
w1 = α1 r2 − 1 + β1 r2 log r − (2α1 + β1 ) log r .
(5.14)
The problem for w1 is
42 w 1 = 0 ,
The behavior of w1 as r → 0, as found below by matching to the inner solution, will determine α1 and β1 .
In the inner region we set r = ερ and obtain from (5.12) that the terms of order O(ε 2 log ε) and O(ε2 ) will be
generated in the inner region. Therefore, this suggests that in the inner region we expand the solution as
v(ρ) = u(ερ) = ε2 log ε v0 (ρ) + ε2 v1 (ρ) + · · · .
The functions v0 and v1 must satisfy vj (1) = vjρ (1) = 0. Therefore, we obtain for j = 0, 1 that
vj = Aj ρ2 − 1 + Bj ρ2 log ρ − (2Aj + Bj ) log ρ .
(5.15)
(5.16)
We substitute (5.16) into (5.15), and write the resulting expression in terms of the outer variable r = ερ. A short
calculation gives that the far-field behavior of (5.15) is
2
2
v ∼ − (log ε) B0 r2 + (log ε) (A0 − B1 )r2 + B0 r2 log r + A1 r2 + B1 r2 log r + 2A0 ε2 (log ε) + O(ε2 log ε) . (5.17)
In contrast, the two-term outer solution from (5.10), (5.12), and (5.14), is
u ∼ r2 − 2r2 log r + σ α1 r2 − 1 + β1 r2 log r − (2α1 + β1 ) log r + · · · .
(5.18)
Upon comparing (5.18) with (5.17), we conclude that
B0 = 0 ,
B 1 = A0 ,
A1 = 1 ,
B1 = −2 ,
2
σ = ε2 (log ε) .
(5.19)
10
M. J. Ward, M. C. Kropinski
This leaves the unmatched constant term −4ε2 (log ε)2 on the right-hand side of (5.17). Consequently, it follows that
the outer correction w1 is bounded as r → 0 and has the point value w1 (0) = −4. Consequently, 2α1 + β1 = 0 and
α1 = 4 in (5.18). This gives β1 = −8, and specifies the second-order term as
w1 = 4 r2 − 1 − 8r2 log r .
(5.20)
This expression reproduces that obtained in (5.9) from the perturbation of the exact solution.
In Problem 9 below we elaborate on why it is impossible to match to an outer solution u 0 that does not satisfy
u0 (0) = 0. In addition, we further remark that point constraints are possible with the Biharmomic operator, since
the free-space Green’s function has singularity O |x − x0 |2 log |x − x0 | as x → x0 . However, with a point constraint
we will not have C 2 smoothness.
Case II: Next, we consider the perturbed problem
42 u = 0 ,
u = sin θ ,
ε < r < 1,
ur = 0 ,
u = ur = 0 ,
(5.21 a)
on r = 1 ,
(5.21 b)
on r = ε .
(5.21 c)
We first find the exact solution of (5.21) and then expand it for ε → 0. Since the solutions to (5.21 a) proportional
to sin θ are linear combinations of {r 3 , r log r, r, r−1 } sin θ, then we can write the solution to (5.21 a), which satisfies
(5.21 b), as
1
B 1
1 B
r+
sin θ ,
+A+
u = Ar + Br log r + −2A + −
2
2
2
2 r
for any constants A and B. Then, imposing that u = ur = 0 on r = ε, we get two equations for A and B:
1 B
B
1
3
Aε + Bε log ε + −2A + −
ε+
ε−1 = 0 ,
+A+
2
2
2
2
1 B
1
B
2
3Aε + B + B log ε + −2A + −
−
ε−2 = 0 .
+A+
2
2
2
2
3
(5.22)
(5.23 a)
(5.23 b)
By comparing the O(ε−1 ) and O(ε−2 ) terms in (5.23), it follows that
B
1
+A+
= κε2 ,
2
2
(5.24)
where κ is an O(1) constant to be found. Substituting (5.24) into (5.23), and neglecting the higher order Aε 3 and
3Aε2 terms in (5.23), we obtain the approximate system
1 B
≈ −κ ,
B log ε + −2A + −
2
2
1 B
B + B log ε + −2A + −
≈ κ.
2
2
(5.25)
By adding the two equations above to eliminate κ, we obtain that
B + 2B log ε + (−4A + 1 − B) = 0 .
From (5.26), together with A ∼ −(1 + B)/2 from (5.24), we obtain that
B∼
3ν
,
2−ν
A=1−
3
,
2−ν
where ν ≡
(5.26)
−1
.
log εe1/2
Finally, substituting (5.27) into (5.22), we obtain that the outer solution has the asymptotics
u ∼ (1 − Ã)r3 + ν Ãr log r + Ãr sin θ , r O(ε) .
(5.27)
(5.28 a)
Solutions to Problems
11
where à is defined by
à ≡
3
,
2−ν
ν≡
−1
.
log εe1/2
(5.28 b)
We remark that (5.28) is an infinite-order logarithmic series approximation to the exact solution. However, it does
not contain transcendentally small terms of algebraic order in ε as ε → 0.
Next, we show how to derive (5.28) by employing the hybrid formulation used in the low Reynolds number flow
problem of §4.
In order to sum the infinite logarithmic series we formulate a hybrid method by following equations (4.13)–(4.15)
of the workshop notes. In the inner region, with inner variable ρ ≡ ε−1 r, we look for an inner solution in the form
(see equations (4.14) and (4.21) of the notes)
1
ρ
v(ρ, θ) = u(ερ, θ) ∼ εν Ã(ν) ρ log ρ − +
2 2ρ
sin θ .
(5.29)
Here ν ≡ −1/ log εe1/2 and à ≡ Ã(ν) is a function of ν to be found. The extra factor of ε in (5.29) is needed since
the solution in the outer region is not algebraically large as ε → 0. Now letting ρ → ∞, and writing (5.29) in terms
of the outer variable r = ερ, we obtain that the far-field form of (5.29) is
v ∼ Ãνr log r + Ãr sin θ .
(5.30)
Therefore, the approximate outer hybrid solution wH to (5.21) that sums all the logarithmic terms must satisfy
42 w H = 0 ,
0 < r < 1,
wH = sin θ , wHr = 0 , on r = 1 ,
wH ∼ Ãνr log r + Ãr sin θ , as r → 0 .
The solution to (5.31 a) and (5.31 b) has the explicit given in
1 β
1
β 1
wH = αr3 + βr log r + −2α + −
r+
sin θ .
+α+
2
2
2
2 r
(5.31 a)
(5.31 b)
(5.31 c)
(5.32)
The condition (5.31 c) then yields the three equations
β = Ãν ,
−2α +
1 β
− = Ã ,
2
2
1
β
+α + = 0,
2
2
(5.33)
for α, β, and Ã. We solve this system to obtain
β = Ãν ,
à =
3
,
2−ν
α = 1 − Ã .
(5.34)
Upon substituting (5.34) into (5.32), we obtain that the resulting expression agrees exactly with the result (5.28)
obtain from the asymptotics of the exact solution.
This simple example of Case II has shown explicitly, without numerical methods, that the hybrid asymptotic
numerical method for summing infinite logarithmic expansions agrees with the results that can be obtained from the
exact solution.
12
M. J. Ward, M. C. Kropinski
6 Problem 6
Problem 6: Consider the following convection-diffusion equation for T (X), with X = (X 1 , X2 ) posed outside two
circular disks Ωj for j = 1, 2 of a common radius a, and with a center-to-center separation 2L between the two disks:
X ∈ R2 \ ∪2j=1 Ωj ,
κ4T = U · ∇T ,
T = Tj ,
X ∈ ∂Ωj ,
T ∼ T∞ ,
j = 1, 2 ,
(6.1 a)
(6.1 b)
|X| → ∞ .
(6.1 c)
Here κ > 0 is constant, Tj for j = 1, 2 and T∞ are constants, and U = U(X) is a given bounded flow field with
U(X) → (U∞ , 0) as |X| → ∞, where U∞ is constant.
• Non-dimensionalize (6.1) in terms of U∞ and the length-scale γ = κ/U∞ to derive a convection-diffusion equation
outside of two circular disks of radii ε ≡ U∞ a/κ, with inter-disk separation 2Lε/a. Here ε is the Peclet number.
• In the low Peclet number limit ε → 0 show how a hybrid asymptotic-numerical solution can be implemented to
sum the infinite logarithmic expansions for two different distinguished limits: Case 1: L/a = O(1). Case 2:
L/a = O(ε−1 ). For Case 1, we require an explicit formula for the logarithmic capacitance, d, of two disks of a
common radius, a, and with a center-to-center separation of 2l. By using bipolar coordinates, the result for d is
(see the workshop notes for the precise reference)
∞
X
e−mξc
ξc
+
,
log d = log (2β) −
2
m cosh(mξc )
m=1
(6.2)
where β and ξc are determined in terms of a and l by
β=
p

l
ξc = log  +
a
l 2 − a2 ;

s 2
l
− 1 .
a
(6.3)
• For a uniform flow with U = (U∞ , 0) for X ∈ R2 , determine the required Green’s function and its regular part.
Solution:
We introduce the dimensionless variables x, u(x), and w(x) by
x = X/γ ,
T = T∞ w ,
u(x) = U(γx)/U∞ ,
γ ≡ κ/U∞ .
(6.4)
We define the dimensionless centers of the two circular disks by xj for j = 1, 2, and their constant boundary
temperatures αj for j = 1, 2, by
xj = Xj /γ ,
αj = wj /T∞ ,
j = 1, 2 .
(6.5)
Then, (6.1) transforms in dimensionless form to
4w = u · ∇w ,
w = αj ,
w ∼ 1,
x ∈ R2 \ ∪2j=1 Dεj ,
x ∈ ∂Dεj ,
|x| → ∞ .
j = 1, 2 ,
(6.6 a)
(6.6 b)
(6.6 c)
Solutions to Problems
13
Here Dεj = {x | |x − xj | ≤ ε} is the circular disk of radius ε centered at xj . The center-to-center separation is
|x2 − x1 | = 2lε ,
l ≡ L/a .
(6.7)
The dimensionless flow has limiting behavior u ∼ (1, 0) as |x| → ∞.
Case 1: We assume that l = O(1) as ε → 0, so that |x2 − x1 | = O(ε). This is the case where the bodies are close
together. It leads below to a new inner problem, not considered previously in the notes.
We assume without loss of generality that x1 + x2 = 0. We then introduce the inner variables y and v(y) by
y = ε−1 x ,
v(y) = w(εy) .
(6.8)
Then, we obtain that (6.6 a) and (6.6 b) transform to
y ∈ R2 \ ∪2j=1 Dj ,
4y v = εu0 · ∇y v ,
v = αj ,
y ∈ ∂Dj ,
j = 1, 2 ,
(6.9 a)
(6.9 b)
Here Dj = {y | |y − yj | ≤ 1} is the circular disk centered at yj = xj /ε of radius one, and u0 ≡ u(0). The inter-disk
separation is
|y2 − y1 | = 2l .
(6.10)
We then look for a solution to (6.9) in the form
v = v0 + νAvc ,
(6.11)
where ν = O(−1/ log ε) and A = A(ν) is to be found. Here v0 is the solution to
4y v 0 = 0 ,
y ∈ R2 \ ∪2j=1 Dj ,
v0 = α j ,
y ∈ ∂Dj ,
j = 1, 2 ,
v0 bounded as |y| → ∞ .
(6.12 a)
(6.12 b)
(6.12 c)
Moreover, vc (y) is the solution to
4y v c = 0 ,
vc = 0 ,
y ∈ R2 \ ∪2j=1 Dj ,
(6.13 a)
y ∈ ∂Dj ,
(6.13 b)
vc ∼ log |y| ,
j = 1, 2 ,
as |y| → ∞ .
(6.13 c)
Since Dj for j = 1, 2 are non-overlapping circular disks, the problem (6.12) can be solved explicitly using conformal
mapping and the introduction of symmetric points. In this way, we can derive that
v0 ∼ v0∞ + o(1) ,
as |y| → ∞ .
(6.14)
The simple calculation of v0∞ is omitted. When α1 = α2 = αc , then clearly v0∞ = α1 . Next, we can solve (6.13)
exactly by introducing bipolar coordinates as suggested in the hint, which was motivated by Appendix B of Reference
[10] of the workshop notes. In this way, we calculate that
vc (y) ∼ log |y| − log d + o(1) ,
|y| → ∞ ,
(6.15)
where d is given by setting a = 1 in (6.2) and (6.3). Therefore, in this analysis we have neglected the transcendentally
small O(ε) term in (6.9), representing a weak drift in the inner region.
14
M. J. Ward, M. C. Kropinski
Upon substituting (6.14) and (6.15) into (6.11), and writing y = ε−1 x, we obtain in terms of outer variables that
the far-field behavior of v is
−1
.
(6.16)
log (εd)
The behavior (6.16) is the singularity behavior for the infinite-logarithmic series approximation V 0 (x; µ) to the
v ∼ v0∞ + A + νA log |x| ,
ν≡
outer solution as x → 0. This approximation satisfies
4V0 = u · ∇V0 ,
x ∈ R2 \{0} ;
V0 ∼ 1 ,
|x| → ∞ ,
(6.17)
with singularity behavior (6.16) as x → 0.
To solve this problem we introduce the Green’s function G(x; ξ) satisfying
4G = u · ∇G − δ(x − ξ) , x ∈ R2 ,
1
G(x; ξ) ∼ −
log |x − ξ| + R(ξ; ξ) + o(1) , x → ξ ,
2π
(6.18 a)
(6.18 b)
with G(x; ξ) → 0 as |x| → ∞. Here R(ξ; ξ) is the regular part of this Green’s function at x = ξ.
The solution to (6.17) with singular behavior V0 ∼ νA log |x| as x → 0 is
V0 = 1 − 2πνAG(x; 0) .
(6.19)
By expanding (6.19) as x → 0, and equating the regular part of the resulting expression with that in (6.16), we get
1 − 2πνAR00 = A + v0∞ . This determines A = A(ν) by
A=
1 − v0∞
,
1 + 2πνR00
ν≡
−1
,
log(εd)
(6.20)
where R00 ≡ R(0; 0). The outer and inner solutions are then given in terms of A. Finally, one can calculate the
Nusselt number, representing the average heat flux across the bodies, by using the divergence theorem together with
the form (6.16) of the far-field behavior in the inner region.
Case 2: We assume that l = O(ε−1 ) as ε → 0, and define l = l0 /ε with l0 = O(1), so that |x2 − x1 | = 2l0 . This
is the case where the small disks of radius ε are separated by O(1) distances in (6.6). In the analysis there are two
distinct inner regions; one near x1 and the other at an O(1) distance away centered at x2 . Since each separated disk
is a circle of radius ε, it has a logarithmic capacitance d = 1. Therefore, the infinite-logarithmic series approximation
V0 (x; µ) to the outer solution satisfies
4V0 = u · ∇V0 ,
x ∈ R2 \{0} ;
V0 ∼ 1 ,
V0 ∼ αj + Aj + νAj log |x − xj | ,
|x| → ∞ ,
−1
.
ν≡
log ε
(6.21 a)
(6.21 b)
The solution to (6.21) is given explicitly by
V0 = 1 − 2πν
2
X
Ai G(x; xi ) .
(6.22)
i=1
We then let x → xj for j = 1, 2 in (6.22) and equate the nonsingular part of the resulting expression with the regular
part of the singularity structure in (6.21 b). This yields that A1 and A2 satisfy the linear algebraic system
A1 (1 + 2πνR11 ) + 2πνA2 G12 = 1 − α1 ;
A2 (1 + 2πνR22 ) + 2πνA1 G21 = 1 − α2 .
Here Gij = G(xj ; xi ) and Rjj = R(xj ; xj ) are the Green’s function and its regular part as defined by (6.18).
(6.23)
Solutions to Problems
15
Finally, we remark that for the case of a uniform flow where u = (1, 0), then the explicit solution to (6.18) is
1
x1 − ξ 1
K0 (|x − ξ|) ,
(6.24 a)
exp
G(x; ξ) =
2π
2
where x = (x1 , x2 ) and ξ = (ξ1 , ξ2 ). By letting x → ξ, and using K0 (r) ∼ − log r + log 2 − γe , as r → 0+ , where γe
is Euler’s constant, we readily calculate that
R(ξ, ξ) =
1
(log 2 − γe ) .
2π
(6.24 b)
These results for G and its regular part can be used in the results of either (6.20) or (6.23) for Case I or Case II,
respectively.
7 Problem 7
Problem 7: Let Ω be the unit disk containing one arbitrarily-shaped hole Ω ε centered at the origin. Our goal is to
calculate the principal eigenvalue of
Z
u2 dx = 1 ,
(7.1 a)
∆u + λu = 0 , x ∈ Ω\Ωε ;
Ω\Ωε
u = 0 , x ∈ ∂Ω ;
u = 0 , x ∈ ∂Ωε .
(7.1 b)
This eigenvalue problem is slightly different from the problem (5.1) of the notes in that here we pose the Dirichlet
condition u = 0 on ∂Ω. For (7.1), derive an explicit transcendental equation for the infinite-order logarithmic series
approximation to the principal eigenvalue.
Solution:
The analysis in §5 of the notes can be repeated, and we readily obtain equation (5.12) of the notes. Thus, the
infinite-order logarithmic series approximation λ? to the principal eigenvalue λ satisfies the transcendental equation
Rh (x0 ; x0 , λ∗ ) = −
1
,
2πν
ν=−
1
,
log(εd)
(7.2)
where Rh (x0 ; x0 , λ∗ ) is the regular part of the Helmholtz Green’s function, satisfying
∆Gh + λ∗ Gh = −δ(x − x0 ) , x ∈ Ω ; Gh = 0 , x ∈ ∂Ω ,
1
Gh (x; x0 , λ∗ ) ∼ −
log |x − x0 | + Rh (x0 ; x0 , λ∗ ) + o(1) , as x → x0 .
2π
(7.3 a)
(7.3 b)
Notice that Gh = 0 on ∂Ω. Since the hole is centered at the origin then x0 = 0.
When Ω is the unit disk with a hole centered at the origin, then (7.3) becomes a radially symmetric problem whose
solution can be found explicitly. A simple calculation gives
√ 

?
λ
Y
√
0
√
1
√ J0
λ? r −
λ? r  ,
G = −  Y0
4
J
λ?
0 < r < 1,
(7.4)
0
where r = |x|. Here J0 (z) and Y0 (z) are the Bessel functions of the first and second kind, of order zero. By using the
well-known asymptotic behavior Y0 (z) ∼ 2π −1 [log z − log 2 + γe + o(1)] and J0 (z) ∼ 1 + o(1) as z → 0+ , we obtain
16
M. J. Ward, M. C. Kropinski
from (7.4) that the local behavior for G as x → 0 is given by
1
log |x| + Rh + o(1) , as x → 0 ,
2π
 √ 
λ?
√
1
1 Y0
− log 2 + γe + log
λ? +  √  ,
Rh ≡ −
2π
4 J0
λ?
G(x; 0) ∼ −
(7.5 a)
(7.5 b)
where γe is Euler’s constant. Finally, upon substituting (7.5 b) for Rh into (7.2), we conclude that λ? (εd) is the first
root of the transcendental equation
 √ 
√ π Y0
λ?
1
log 2 − γe − log
λ? +  √  = − = log(εd) .
2 J0
ν
?
λ
(7.6)
Here d is the logarithmic capacitance of the arbitrarily-shaped hole centered at the origin of the unit disk.
It is interesting to note that the result (7.6) can also be obtained by first finding the exact eigenvalue relation for
the concentric annulus ε < |x| < 1 with u = 0 on |x| = ε and on |x| = 1, and then letting ε → 0 in this resulting
expression. The eigenfunction is proportional to
√ 

√ √ J0
λ
√ Y0
λr −
λr  ,
u =  J0
Y0
λ
0 < r < 1,
(7.7)
and upon setting u = 0 at r = ε, we get the eigenvalue relation
Y0
√
λε = J0
√
√ λ
√ .
λε
J0
λ
Y0
(7.8)
Next, in (7.8) we use the small argument expansions of Y0 (z) and J0 (z) as z → 0+ , and then, finally, replace ε
by εd in the resulting expression by recalling Kaplun’s equivalence principle. In this way, we readily recover the
transcendental equation (7.6) for the approximation λ? to λ.
8 Problem 8
Problem 8: For the eigenvalue problem (5.1) of the notes, consider the special case of K holes that have a common
logarithmic capacitance d = d1 = . . . , dK . By introducing two-term expansions directly in equation (5.1) of the notes
for the eigenvalue and the outer and inner approximations to the eigenfunction, re-derive the two-term approximation
in equation (5.27) of the Corollary.
Solution: We write the eigenvalue problem as
4u + λu = 0 ,
∂n u = 0 ,
u = 0,
x ∈ Ω\Ωp ;
Ωp ≡ ∪ K
j=1 Ωεj ,
(8.1 a)
u2 dx = 1
(8.1 b)
x ∈ ∂Ω ;
Z
x ∈ ∂Ωεj ,
j = 1, . . . , N .
Ω\Ωp
(8.1 c)
We assume that each hole Ωεj is centered at xj ∈ Ω and has the same logarithmic capacitance d.
We look for a two-term expansion for the principal eigenvalue λ0 (ε) as
λ0 (ε) = λ1 ν + λ2 ν 2 + · · · ,
ν = −1/ log(εd) .
(8.2)
Solutions to Problems
17
In the outer region, away from O(ε) neighborhoods of the holes, we expand the outer solution for u as
u = u0 + νu1 + ν 2 u2 + · · · .
(8.3)
u0 = |Ω|−1/2 ,
(8.4)
The leading-order term is
where |Ω| is the area of Ω. Upon substituting (8.2) and (8.3) into (8.1 a) and (8.1 b), and collecting powers of ν, we
obtain that u1 satisfies
4u1 = −λ1 u0 ,
∂ n u1 = 0 ,
Z
x ∈ Ω\{x1 , . . . , xK } ;
x ∈ ∂Ω ;
u1 singular as x → xj ,
u1 dx = 0 ,
(8.5 a)
j = 1, . . . , K ,
(8.5 b)
Ω
while u2 satisfies
4u2 = −λ2 u0 − λ1 u1 ,
∂ n u2 = 0 ,
Z
x ∈ Ω\{x1 , . . . , xK } ;
x ∈ ∂Ω ;
Ω
u2 singular as x → xj ,
u21 + 2u0 u2 dx = 0 ,
j = 1, . . . , K .
(8.6 a)
(8.6 b)
Now in the j th inner region we introduce the new variables by
y = ε−1 (x − xj ) ,
v(y) = u(xj + εy) .
(8.7)
We then expand the inner solution as
v(y) = νA0j vcj (y) + ν 2 A1j vcj (y) + · · · .
(8.8)
Upon substituting (8.7) and (8.8) into (8.1 a) and (8.1 c), we obtain that vcj satisfies
4y vcj = 0 ,
y∈
/ Ωj ;
vcj = 0 ,
vcj (y) ∼ log |y| − log d + o(1) ,
y ∈ ∂Ωj ,
as |y| → ∞ .
(8.9 a)
(8.9 b)
Here 4y is the Laplacian in the y variable, and Ωj ≡ ε−1 Ωεj . We consider the special case where d is independent
of j.
Upon using the far-field form (8.9 b) in (8.8), and writing the resulting expression in outer variables, we get
v = A0j + ν [A0j log |x − xj | + A1j ] + ν 2 [A1j log |x − xj | + A2j ] + · · · .
(8.10)
The far-field behavior (8.10) must agree with the local behavior of the outer expansion (8.3). Therefore, we obtain
that
A0j = u0 = |Ω|−1/2 ,
j = 1, . . . K ,
u1 ∼ u0 log |x − xj | + A1j ,
u2 ∼ A1j log |x − xj | + A2j ,
as x → xj ,
as x → xj ,
(8.11 a)
j = 1, . . . , K ,
j = 1, . . . , K .
(8.11 b)
(8.11 c)
Equations (8.11 b) and (8.11 c) give the required singularity structure for u 1 and u2 in (8.5) and (8.6), respectively.
18
M. J. Ward, M. C. Kropinski
The problem for u1 with singular behavior (8.11 b) can be written in terms of the delta function as
Z
K
X
u1 dx = 0 ,
δ(x − xj ) , x ∈ Ω ;
4u1 = −λ1 u0 + 2πA0
(8.12 a)
Ω
j=1
∂ n u1 = 0 ,
Upon using the divergence theorem we obtain that −λ1 u0
we get
x ∈ ∂Ω .
R
Ω
1dx + 2πA0 K = 0, so that with u0 = A0 from (8.11 a),
2πK
.
|Ω|
The solution to (8.12) can be written in terms of the Neumann Green’s function as
λ1 =
u1 = −2πu0
where the Neumann Green’s function GN (x; ξ) satisfies
K
X
(8.12 b)
(8.13)
GN (x; xi ) ,
(8.14)
i=1
1
− δ(x − ξ) , x ∈ Ω ; ∂n GN = 0 , x ∈ ∂Ω ,
|Ω|
Z
1
GN (x; ξ) ∼ −
log |x − ξ| + RN (ξ; ξ) + o(1) , as x → ξ ;
GN (x; ξ) dx = 0 .
2π
Ω
∆GN =
(8.15 a)
(8.15 b)
The constant RN (ξ; ξ) is the regular part of GN at the singularity. Since GN has a zero spatial average, it follows
R
from (8.14) that Ω u1 dx = 0, as required in (8.12 a).
Next, we expand u1 as x → xj . We use the local behavior for GN , given in (8.15 b), to obtain from (8.14) that


K
X


u1 ∼ u0 log |x − xj | − 2πu0 RN jj +
GN ij  , x → xj ,
(8.16)
i=1
i6=j
where GN ji = GN (xj ; xi ) and RN jj = RN (xj ; xj ). Comparing (8.16) and the required singularity behavior (8.11 b),
we obtain that

K

X


A1j = −2πu0 RN jj +
GN ij  ,
j = 1, . . . , N .
(8.17)
i=1
i6=j
Next, we write the problem (8.6) in Ω as
4u2 = −λ2 u0 − λ1 u1 + 2π
K
X
j=1
A1j δ(x − xj ) ,
x ∈ Ω;
∂ n u2 = 0 ,
x ∈ ∂Ω .
(8.18)
R
Since Ω u1 dx = 0 and u0 = |Ω|−1/2 , the divergence theorem applied to (8.18) determines λ2 as λ2 u0 |Ω| =
P
2π j=1 A1j . Finally, we use (8.17) for A1j , we get


K
N
X
X
4π 2

(8.19)
GN ji  .
p(x1 , . . . , xK ) ,
p(x1 , . . . , xK ) ≡
λ2 = −
RN jj +
|Ω|
i=1
j=1
i6=j
Combining (8.2) with (8.13) and (8.19) we get the two-term expansion given in equations (5.27) and (5.28) of the
Corollary in §5 of the workshop notes given by
λ0 (ε) ∼
4νπ 2
2πνK
−
p(x1 , . . . , xK ) + · · · ,
|Ω|
|Ω|
ν = −1/ log(εd) .
(8.20)
Solutions to Problems
19
9 Problem 9
Problem 9: Consider the following Biharmonic eigenvalue problem in a two-dimensional bounded domain Ω containing a small circular hole Ωε of radius ε centered at x0 ∈ Ω,
42 u − λu = 0 ,
x ∈ Ω\Ωε ;
u = ∂n u = 0 ,
x ∈ ∂Ω ;
u = ∂n u = 0 ,
x ∈ ∂Ωε .
(9.1)
Let λ0ε denote the first positive eigenvalue of this problem. Let λ0 be the first eigenvalue of the unperturbed problem
with no hole, with corresponding eigenfunction u0 (x). Assume that u0 (x0 ) 6= 0. By using a matched asymptotic
expansion argument, show that λ0ε does not approach λ0 as ε → 0, in contrast to that for Laplacian eigenvalue
problems in perforated domains. Instead, show that λ0ε → λ?0 as ε → 0, where λ?0 is the first eigenvalue of the
following problem with a point constraint:
4 2 u? − λ ? u? = 0 ,
x ∈ Ω\{x0 } ;
u ? = ∂ n u? = 0 ,
x ∈ ∂Ω ;
u? (x0 ) = 0 .
(9.2)
Finally, calculate the asymptotic behavior of the difference λ 0ε − λ?0 as ε → 0 using a matched asymptotic analysis.
Solution: Let λ0ε and u0ε (x) be the principal eigenvalue of the Biharmonic eigenvalue problem with a hole, given by
R
(9.1) with normalization condition Ω\Ω u20ε dx = 1. Next, let λ0 and u0 (x) be the first eigenpair of the unperturbed
ε
problem with no hole
42 u − λu = 0 ,
with normalization condition
R
Ω
x ∈ Ω;
u = ∂n u = 0 ,
x ∈ ∂Ω ,
(9.3)
u2 dx = 1.
We now show that λ0ε does not tend to λ0 as ε → 0. To show this, suppose to the contrary that for some σ 1
we have
λ0ε = λ0 + σλ1 + · · · .
(9.4)
In the outer region we expand the outer eigenfunction as
uε (x) = u0 (x) + σu1 (x) + · · · .
(9.5)
Now at x = x0 , we assume that u0 (x0 ) 6= 0.
In the inner region we introduce the new variables y = ε−1 (x − x0 ) and vε (y) = uε (x0 + εy). Then, for some
gauge function µ, we put
vε (y) = µv0 (ρ) ,
ρ ≡ |y| .
(9.6)
Upon substituting (9.6) into (9.1), we obtain that v0 satisfies
42 v 0 = 0 ,
ρ = |y| ≥ 1 ;
v0 (1) = v0ρ (1) = 0 .
(9.7)
The general solution of this problem has the form
v0 = aρ2 + bρ2 log ρ + c log ρ + d ,
ρ ≥ 1.
(9.8)
The matching condition is that the outer solution as x → x0 must agree with the inner expansion as ρ = |y| → ∞.
Therefore,
u0 (x0 ) + · · · + σu1 ∼ µv0 (ρ) + · · · .
(9.9)
The only possibility for matching is that a = b = 0, and that c = u0 (x0 ) with µ = −1/ log ε. However, this choice
leaves only one free parameter d to satisfy the two boundary conditions v 0 (1) = v0ρ (1) = 0, which is impossible.
20
M. J. Ward, M. C. Kropinski
Therefore, we conclude that if we assume that the perturbed eigenfunction is close to the unperturbed eigenfunction
with no hole in the outer region, then asymptotic matching is impossible. This suggests that this assumption must
be modified, and that the limiting problem as ε → 0 is not the problem with no hole.
Instead, we let λ?0 and u?0 (x) be the principal eigenpair of the Biharmonic eigenvalue problem with a point con-
straint, given by (9.3). In other words, we claim that the limiting problem ε → 0 corresponds to the eigenvalue
problem (9.2) with point constraint. Point constraints are compatible with Biharmonic problems, but not with
Laplace’s equation.
We then look for an eigenvalue of (9.1) close to λ?0 . For some gauge function σ 1, we expand
λ0ε = λ?0 + σλ1 + · · · .
(9.10)
In the outer region, we expand the eigenfunction as
u0ε (x) = u?0 (x) + σu1 (x) + · · · .
(9.11)
Substituting (9.10) and (9.11) into (9.1), we obtain that λ1 and u1 (x) satisfy
42 u1 − λ?0 u1 = λ1 u?0 ,
u1 = ∂ n u1 = 0 ,
x ∈ Ω\{x0 } ,
Z
x ∈ ∂Ω ;
u?0 u1 dx = 0 ,
(9.12 a)
(9.12 b)
Ω
u1
singular as x → x0 .
(9.12 c)
Next, we must derive a singularity condition for u1 as x → x0 .
In the inner region, we introduce the new variables
y = ε−1 (x − x0 ) ,
v(y) = u(x0 + εy) .
(9.13)
In terms of the gauge function µ 1, we then expand
vε (y) = µv0 (y) ,
ρ = |y| .
(9.14)
Since u0 (x0 ) = 0, the matching condition is that the outer expansion of the eigenfunction as x → x0 must agree
with the far-field form of the inner expansion as y → ∞,
∇u?0 · (x − x0 ) + · · · + σu1 ∼ µv0 (y) + · · · .
(9.15)
∇u?0 ≡ ∇u?0 (x)|x=x0 .
(9.16)
Here we have defined
The problem for v0 is
42 v 0 = 0 ,
ρ = |y| ≥ 1 ;
v0 (1) = v0ρ (1) = 0 .
(9.17)
For any vector a, there is a solution to (9.17) of the form
v0 = A · eθ vc (ρ) ,
(9.18 a)
h
i
1
.
vc = ρ log ρ − ρ log e1/2 +
2ρ
(9.18 b)
where eθ ≡ (cos θ, sin θ) and vc (ρ) is given by
Notice that this is the Stokes solution given in equation (4.21) of the workshop notes.
Solutions to Problems
21
We then write the far-field expansion of the inner solution in terms of the outer variables as
i
h
µv0 (y) ∼ ε−1 µA · eθ |x − x0 | log |x − x0 | − log εe1/2 .
(9.19)
This far-field expression suggests that we define µ and ν by
ν=−
Then, the matching condition (9.15) becomes
1
,
log εe1/2
µ = εν
∇u?0 · (x − x0 ) + · · · + σu1 ∼ A · eθ |x − x0 | + A · eθ ν|x − x0 | log |x − x0 | + · · · .
(9.20)
(9.21)
Therefore, we conclude that
A = ∇u?0 ,
σ=ν
(9.22)
The matching condition (9.21) shows that the solution u1 to (9.12) must have the singularity behavior
u1 ∼ ∇u?0 · (x − x0 ) log |x − x0 | ,
as x → x0 .
(9.23)
Finally, we apply the divergence theorem to (9.12) over Ω0 , where Ω0 ≡ Ω\Ωγ , and Ωγ is a small disk of radius
γ 1, centered at x0 . In this way, we get
λ1 = −4π|∇u?0 |2 ,
σ =ν.
(9.24)
In summary, the principal eigenvalue of (9.1) has the two-term asymptotic expansion
λ0ε ∼ λ?0 − 4πν|∇u?0 |2 + · · · ,
ν=−
1
.
log εe1/2
Here u?0 and λ?0 are the principal eigenpair of the problem (9.2) with point constraint.
(9.25)