1
Notes for Fourth Winter School in Applied Mathematics: Hong Kong
Strong Localized Perturbations and the Biharmonic
Problem
(Received 15 January 2013)
1 Biharmonic Equation with Hole
Problem 1: Consider the Biharmonic equation in the two-dimensional concentric annulus, formulated as
△2 u = 0 ,
u = 1,
x ∈ Ω\Ωε ,
ur = 0 ,
u = ur = 0 ,
on
(1.1 a)
r = 1,
r = ε.
(1.1 b)
(1.1 c)
Here Ω is the unit disk centered at the origin, containing a small hole of radius ε centered at x = 0, i.e. Ωε =
{x | |x| ≤ ε}. Calculate the exact solution, and from it determine an approximation to the solution in the outer
region |x| ≫ O(ε). Can you re-derive this result 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
We first find the exact solution of (1.1) and then expand it for ε → 0. Since the radially symmetric solutions to
(1.1 a) are linear combinations of {r2 , r2 log r, log r, 1}, we can write the solution to (1.1 a), which satisfies (1.1 c), as
u = A r2 − 1 + Br2 log r − (2A + B) log r + 1 ,
(1.2)
for any constants A and B. Then, imposing that u = ur = 0 on r = ε, we get two equations for A and B:
2A 1 − ε2 A + B 1 − ε2 − 2ε2 log ε = 0 ,
A 1 + 2 log ε − ε2 + B 1 − ε2 log ε = 1 .
(1.3 a)
(1.3 b)
Equation (1.3 a) gives
A=−
B
2
1−
2ε2 log ε
1 − ε2
.
(1.4)
2
Upon substituting this into (1.3 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 log ε Bε2 log ε 2ε2 B(log ε)2
1
B
+
+
+ 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
(1.5 a)
(1.5 b)
(1.5 c)
(1.5 d )
The last line of (1.5) determines B, while (1.4) determines A. In this way, we get
2
2
B ∼ −2 − 8ε2 (log ε) ,
A ∼ 1 + 4ε2 (log ε) .
(1.6)
Upon substituting (1.6) into (1.2), we obtain the following two-term expansion in the outer region r ≫ O(ε):
2
u ∼ u0 (r) + ε2 (log ε) u1 (r) + · · · ,
(1.7)
where u0 (r) and u1 (r) are defined by
u0 (r) = r2 − 2r2 log r ,
u1 = 4 r2 − 1 − 8r2 log r .
(1.8)
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
(1.2), and consequently u = 1 in the outer region.
Next, we show how to recover (1.7) from a matched asymptotic expansion analysis. In the outer region we expand
the solution as
u ∼ w0 + σw1 + · · · ,
(1.9)
where σ ≪ 1 is an unknown gauge function, and where w0 satisfies the following problem with a point constraint:
△ 2 w0 = 0 ,
0 < r < 1;
w0 (1) = 1 ,
w0r (1) = 0 ,
w0 (0) = 0 .
(1.10)
The solution is readily calculated as
w0 = r2 − 2r2 log r .
(1.11)
0 < r < 1;
(1.12)
The problem for w1 is
△ 2 w1 = 0 ,
w1 (1) = w1r (1) = 0 .
The solution to (1.12) is given in terms of unknown coefficients α1 and β1 as
w1 = α1 r2 − 1 + β1 r2 log r − (2α1 + β1 ) log r .
(1.13)
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 (1.11) that the terms of order O(ε2 log ε) and O(ε2 ) will be
Strong Localized Perturbations and the Biharmonic Problem
3
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) + · · · .
(1.14)
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 .
(1.15)
We substitute (1.15) into (1.14), and write the resulting expression in terms of the outer variable r = ερ. A short
calculation gives that the far-field behavior of (1.14) 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) . (1.16)
In contrast, the two-term outer solution from (1.9), (1.11), and (1.13), is
u ∼ r2 − 2r2 log r + σ α1 r2 − 1 + β1 r2 log r − (2α1 + β1 ) log r + · · · .
(1.17)
Upon comparing (1.17) with (1.16), we conclude that
B0 = 0 ,
B1 = A0 ,
A1 = 1 ,
B1 = −2 ,
2
σ = ε2 (log ε) .
(1.18)
This leaves the unmatched constant term −4ε2 (log ε)2 on the right-hand side of (1.16). 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 (1.17). This gives β1 = −8, and specifies the second-order term as
w1 = 4 r2 − 1 − 8r2 log r .
(1.19)
This expression reproduces that obtained in (1.8) from the perturbation of the exact solution.
In Problem 9 below we elaborate on why it is impossible to match to an outer solution u0 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.
2 A Biharmonic Eigenvalue Problem
Problem 2: 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,
△2 u − λu = 0 ,
x ∈ Ω\Ωε ;
u = ∂n u = 0 ,
x ∈ ∂Ω ;
u = ∂n u = 0 ,
x ∈ ∂Ωε .
(2.1)
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:
△ 2 u ⋆ − λ⋆ u ⋆ = 0 ,
x ∈ Ω\{x0 } ;
u⋆ = ∂ n u⋆ = 0 ,
x ∈ ∂Ω ;
u⋆ (x0 ) = 0 .
(2.2)
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Finally, calculate the asymptotic behavior of the difference λ0ε − 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
(2.1) 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
△2 u − λu = 0 ,
with normalization condition
R
Ω
x ∈ Ω;
u = ∂n u = 0 ,
x ∈ ∂Ω ,
(2.3)
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 + · · · .
(2.4)
In the outer region we expand the outer eigenfunction as
uε (x) = u0 (x) + σu1 (x) + · · · .
(2.5)
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| .
(2.6)
Upon substituting (2.6) into (2.1), we obtain that v0 satisfies
△2 v0 = 0 ,
ρ = |y| ≥ 1 ;
v0 (1) = v0ρ (1) = 0 .
(2.7)
The general solution of this problem has the form
v0 = aρ2 + bρ2 log ρ + c log ρ + d ,
ρ ≥ 1.
(2.8)
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) + · · · .
(2.9)
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 v0 (1) = v0ρ (1) = 0, which is impossible.
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 constraint, given by (2.3). In other words, we claim that the limiting problem ε → 0 corresponds to the eigenvalue
problem (2.2) with point constraint. Point constraints are compatible with Biharmonic problems, but not with
Laplace’s equation.
Strong Localized Perturbations and the Biharmonic Problem
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We then look for an eigenvalue of (2.1) close to λ⋆0 . For some gauge function σ ≪ 1, we expand
λ0ε = λ⋆0 + σλ1 + · · · .
(2.10)
In the outer region, we expand the eigenfunction as
u0ε (x) = u⋆0 (x) + σu1 (x) + · · · .
(2.11)
Substituting (2.10) and (2.11) into (2.1), we obtain that λ1 and u1 (x) satisfy
△2 u1 − λ⋆0 u1 = λ1 u⋆0 ,
u1 = ∂ n u1 = 0 ,
x ∈ Ω\{x0 } ,
Z
x ∈ ∂Ω ;
u⋆0 u1 dx = 0 ,
(2.12 a)
(2.12 b)
Ω
u1
singular as x → x0 .
(2.12 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) .
(2.13)
In terms of the gauge function µ ≪ 1, we then expand
vε (y) = µv0 (y) ,
ρ = |y| .
(2.14)
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) + · · · .
(2.15)
∇u⋆0 ≡ ∇u⋆0 (x)|x=x0 .
(2.16)
Here we have defined
The problem for v0 is
△2 v0 = 0 ,
ρ = |y| ≥ 1 ;
v0 (1) = v0ρ (1) = 0 .
(2.17)
For any vector a, there is a solution to (2.17) of the form
v0 = A · eθ vc (ρ) ,
(2.18 a)
i
h
1
.
vc = ρ log ρ − ρ log e1/2 +
2ρ
(2.18 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.
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 .
(2.19)
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This far-field expression suggests that we define µ and ν by
ν=−
Then, the matching condition (2.15) becomes
1
,
log εe1/2
µ = εν
(2.20)
∇u⋆0 · (x − x0 ) + · · · + σu1 ∼ A · eθ |x − x0 | + A · eθ ν|x − x0 | log |x − x0 | + · · · .
(2.21)
Therefore, we conclude that
A = ∇u⋆0 ,
σ=ν
(2.22)
The matching condition (2.21) shows that the solution u1 to (2.12) must have the singularity behavior
u1 ∼ ∇u⋆0 · (x − x0 ) log |x − x0 | ,
as
x → x0 .
(2.23)
Finally, we apply the divergence theorem to (2.12) 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.
(2.24)
In summary, the principal eigenvalue of (2.1) 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 (2.2) with point constraint.
(2.25)