Perturbation Theoryward/teaching/m605/every2_perturb.pdf · then perturbation theory says that for...
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Appendix D
Perturbation Theory
D.1 Simple Examples
Let
A =
1 0 0 00 1 0 00 0 2 00 0 0 3
, B =
0 1 10 10−1 0 10 1010 10 4 1010 10 10 6
. (D.1)
The eigenvalues of A are 1, 2 and 3, where λ = 1 has multiplicity 2. To find theeigenvalue of the perturbed matrix A + ǫB corresponding to the unperturbedeigenvalue λ0 = 2, we use non-degenerate perturbation theory, so
λ = λ0 + ǫλ1 + O(
ǫ2)
(D.2)
= 2 + ǫx∗T
0 Bx0
x∗T0 x0
+ O(
ǫ2)
(D.3)
= 2 + ǫ
[
0 0 1 0]
0 1 10 10−1 0 10 1010 10 4 1010 10 10 6
0010
[
0 0 1 0]
0010
+ O(
ǫ2)
(D.4)
= 2 + 4ǫ + O(
ǫ2)
, (D.5)
noting that A = AT , so x∗0, the eigenvector of AT with eigenvalue 2, is the same
as x0, the eigenvector of A with eigenvalue 2. We can do likewise for λ0 = 3 tofind
λ = 3 + 6ǫ + O(
ǫ2)
. (D.6)
203
For λ0 = 1, we must use degenerate perturbation theory, which says that
Bc = λ1Mc (D.7)
where
Bij ≡ v∗Ti Bvj Mij ≡ v∗T
i vj , (D.8)
so equation D.7 becomes[
0 1−1 0
]
c = λ1
[
1 00 1
]
c, (D.9)
hence λ1 is i or −i, so the perturbed eigenvalues for λ0 = 1 are
λ = 1 + ǫ
i−i
+ O(
ǫ2)
(D.10)
In summary, the eigenvalues of A + ǫB are
λ =
1 ± iǫ2 + 4ǫ3 + 6ǫ
+ O(
ǫ2)
(D.11)
These are compared to numerically calculated eigenvalues in figure D.1.If instead,
B =
1 0 10 100 −1 10 1010 10 4 1010 10 10 6
, (D.12)
then perturbation theory says that for ǫ small, the eigenvalues of A + ǫB are
λ =
1 ± 1ǫ2 + 4ǫ3 + 6ǫ
+ O(
ǫ2)
(D.13)
These are compared to numerically calculated eigenvalues in figure D.2.
D.2 Wilkinson’s Matrix
We now choose to look at an example where perturbation theory is very poorat approximating the perturbed eigenvalues. Following Wilkinson [31], we let
A =
20 2019 20 0
18 20. . .
. . .
0 2 201
. (D.14)
204
0 0.005 0.010.98
1
1.02Real Component of First Eigenvalue
Rea
l Com
pone
nt
Size of Perturbation (ε)0 0.005 0.01
−0.01
0
0.01Imag. Component of 1st Eigenvalue
Imag
. Com
pone
nt
Size of Perturbation (ε)
0 0.005 0.010.98
1
1.02Real Component of Second Eigenvalue
Rea
l Com
pone
nt
Size of Perturbation (ε)0 0.005 0.01
−0.01
0
0.01Imag. Component of 2nd Eigenvalue
Imag
. Com
pone
nt
Size of Perturbation (ε)
0 0.005 0.011.98
2
2.02
2.04
Real Component of Third Eigenvalue
Rea
l Com
pone
nt
Size of Perturbation (ε)0 0.005 0.01
−0.01
0
0.01Imag. Component of 3rd Eigenvalue
Imag
. Com
pone
nt
Size of Perturbation (ε)
0 0.005 0.01
3
3.05
3.1Real Component of Fourth Eigenvalue
Rea
l Com
pone
nt
Size of Perturbation (ε)0 0.005 0.01
−0.01
0
0.01Imag. Component of 4th Eigenvalue
Imag
. Com
pone
nt
Size of Perturbation (ε)
calculated by MATLAB
predicted by perturbation theory
Figure D.1: The real (first column) and imaginary (second column) componentsof the four eigenvalues of A + ǫB, plotted versus ǫ, with A and B defined byequation D.1.
205
0 0.005 0.010.98
1
1.02First Eigenvalue
Eig
enva
lue
Size of Perturbation (ε)
0 0.005 0.010.98
1
1.02Second Eigenvalue
Eig
enva
lue
Size of Perturbation (ε)
0 0.005 0.011.98
2
2.02
2.04
Third Eigenvalue
Eig
enva
lue
Size of Perturbation (ε)
0 0.005 0.01
3
3.05
3.1Fourth Eigenvalue
Eig
enva
lue
Size of Perturbation (ε)
calculated by MATLAB
predicted by perturbation theory
Figure D.2: The four eigenvalues (all real) of A+ ǫB, plotted versus ǫ, where Bis defined by equation D.12.
206
This has eigenvalues 1, 2, . . . , 20 and the eigenvector corresponding to eigenvalueλ is
(−20)20−λ
(20−λ)!
...(−20)2
2!−2010...0
λ − 1 zeroes
(D.15)
The eigenvector of AT corresponding to eigenvalue λ is
0...0120
(20)2
2!...
(20)λ−1
(λ−1)!
20 − λ zeroes
(D.16)
D.2.1 Eigenvalues
Let
B =
0 0... . .
.
0 01 0 · · · 0
. (D.17)
Then perturbation theory says that the eigenvalues of A+ǫB are given, to orderǫ, by
λn = λ0n + ǫλ1n (D.18)
= n + ǫx∗T
0 Bx0
x∗T0 x0
(D.19)
= n + ǫ2019 (−1)
n
(20 − n)! (n − 1)!(D.20)
or, if we generalise A and B to be m by m matrices,
λn = n + ǫmm−1 (−1)
m−n
(m − n)! (n − 1)!(D.21)
207
10−15
10−10
10−5
10
10.1
10.2
10.3
10.4
10.5
10.6
Size of Perturbation (ε)
Rea
l Com
pone
nt o
f Eig
enva
lue
Real Component of Tenth Eigenvalue
predicted range of validity of perturbation theorycalculated by MATLABpredicted by perturbation theory
Figure D.3: The 10th eigenvalue in the m = 20 case, as a function of ǫ, ascomputed numerically by MATLAB. The eigenvalue predicted by perturbationtheory is also shown. The ‘predicted range’ gives the number η for which wemust have ǫ ≪ η in order for perturbation theory to be valid, as in equationD.22.
where this is valid for
ǫmm−1
(m − n)! (n − 1)!≪ 1. (D.22)
For example, when m = 50, perturbation theory predicts
λ8 = 8 + ǫ · 2.5085 × 1028 (D.23)
so we expect this to be valid for ǫ ≪ 10−28.The eigenvalues in the m = 20 case, as a function of ǫ, were computed
numerically in MATLAB and are shown in figures D.4–D.9. Also, the 10theigenvalues in the m = 20 case, as a function of ǫ, is compared to that predictedby perturbation theory. This is shown in figure D.3.
Error in Computation
Since the Wilkinson matrix has such poorly behaved eigenvalues, we expectthat numerical computations of its perturbed eigenvalues will be difficult. Toillustrate this, we compute the eigenvalues of the perturbed Wilkinson matrixfor different values of m and ǫ using two different programs, MATLAB andMathematica. The calculations were performed to 15 digits. The result is shown
208
10−15
10−10
10−5
−5
0
5
10
15
20
25Real Component of Eigenvalues
Rea
l Com
pone
nt o
f Eig
enva
lues
Size of Perturbation (ε)0 10 20
−10
−5
0
5
10Eigenvalues
Imag
inar
y C
ompo
nent
of E
igen
valu
es
Real Component of Eigenvalues
10−15
10−10
10−5
−10
−5
0
5
10Imaginary Component of Eigenvalues
Imag
inar
y C
ompo
nent
of E
igen
valu
es
Size of Perturbation (ε)
predicted range of validity of perturbation theorycalculated by MATLAB
Figure D.4: The eigenvalues in the m = 20 case, as a function of ǫ, as computednumerically by MATLAB. The ‘predicted range’ gives the number η, as a func-tion of λ, for which we must have ǫ ≪ η in order for perturbation theory to bevalid, as in equation D.22.
209
−100
1020
30
−10−5
05
1010
−15
10−10
10−5
Real Component
Eigenvalues
Imaginary Component
Siz
e of
Per
turb
atio
n (ε)
Figure D.5: A 3D representation of the eigenvalues in the m = 20 case, as afunction of ǫ, as computed by MATLAB.
210
Size of Perturbation (ε)
Siz
e of
Mat
rix
10−50
10−40
10−30
10−20
10−10
10
15
20
25
30
35
40
45
50
1e−14
1e−12
1e−10
1e−08
1e−06
1e−04
1e−02
1
1e+02
Difference Between Mathematica and MATLAB EigenvaluesRegion Where Eigenvalues Become All Real
Figure D.6: The difference between the computed eigenvalues of A+ ǫB accord-ing to MATLAB and Mathematica, as a function of m and ǫ. The dashed linesindicate the region in which the eigenvalues, according to Mathematica, becomeentirely real.
211
Size of Perturbation (ε)
Siz
e of
Mat
rix
10−50
10−40
10−30
10−20
10−10
10
15
20
25
30
35
40
45
50
1e−14
1e−12
1e−10
1e−08
1e−06
1e−04
1e−02
1
1e+02
Difference Between Mathematica and Perturbation Theory EigenvaluesRegion Where Eigenvalues Become All Real
Figure D.7: The difference between the computed eigenvalues of A+ ǫB accord-ing to Mathematica and the predicted eigenvalues from perturbation theory, asa function of m and ǫ. The dashed lines indicate the region in which the eigen-values, according to Mathematica, become entirely real. This also correspondsto the approximate boundary of the region of validity of perturbation theory.
212
Size of Perturbation (ε)
Siz
e of
Mat
rix
10−50
10−40
10−30
10−20
10−10
10
15
20
25
30
35
40
45
50
1e−14
1e−12
1e−10
1e−08
1e−06
1e−04
1e−02
1
1e+02
Difference Between MATLAB and Perturbation Theory EigenvaluesRegion Where Eigenvalues Become All Real
Figure D.8: The difference between the computed eigenvalues of A+ ǫB accord-ing to MATLAB and the predicted eigenvalues from perturbation theory, as afunction of m and ǫ. The dashed lines indicate the region in which the eigen-values, according to Mathematica, become entirely real. This also correspondsto the approximate boundary of the region of validity of perturbation theory.
213
in figure D.6. More precisely, the plot is of the maximum of the absolute valueof the complex difference between the eigenvalues as computed by MATLABand Mathematica. To determine which is most accurate, we also plot each ascompared with perturbation theory (figures D.7 and D.8). Observe that, withinthe region of validity of perturbation theory, Mathematica agrees very well withperturbation theory. We assume then that the error depicted in figure D.7 isdue to errors from perturbation theory, not Mathematica.
D.2.2 Sherman-Woodbury-Morrison Formula
If we wish to solve the related problem Ax + ǫBx = b, with A and B as above,then perturbation theory predicts
x =(
A−1 − A−1ǫBA−1)
b. (D.24)
Since for our choice of B, we can write B = uvT , then we can apply the Sherman-Woodbury-Morrison formula which says that the exact solution is
x =
(
A−1 − A−1ǫBA−1
1 + ǫvT A−1u
)
b, (D.25)
so we expect perturbation theory to be valid for ǫvT A−1u ≪ 1.Now
A−1 =
19!20! (−20) 18!
20! (−20)2 17!20! · · · (−20)19 0!
20!
18!19! (−20) 17!
19!
17!18!
0. . .
0!1!
(D.26)
so
vT A−1u =(−20)19
20!(D.27)
for m = 20. We can generalise this for any m to obtain
vT A−1u =(−m)m−1
m!. (D.28)
We then expect perturbation theory to be valid for
ǫ ≪(
vT A−1u)−1
(D.29)
ǫ ≪(
(−m)m−1
m!
)−1
. (D.30)
For example, when m = 50, this means that perturbation theory is valid forǫ ≪ 2 × 10−19.
214
−100
1020
30
−10−5
05
1010
−15
10−10
10−5
100
Real Component
Imaginary Component
Siz
e of
Per
turb
atio
n
pseudospectrumeigenvalues of perturbed matrix
Figure D.9: A 3D version of the pseudospectrum for the unperturbed matrix(the surface), in the m = 20 case, in comparison with the eigenvalues of theperturbed matrix (the lines), as computed by MATLAB. The lines in this figureare identical to those in figure D.5.
D.2.3 Pseudospectrum
Following Embree and Trefethen [7], we define the pseudospectrum of A to be
Λǫ(A) = z ∈ C : z ∈ Λ (A + E) for some E with ‖E‖ ≤ ǫ (D.31)
where Λ (A + E) is the set of eigenvalues of the matrix (A + E) and ‖ · ‖ is amatrix norm induced by a vector norm. We choose ‖ · ‖ to be the 2-norm. Since‖ǫB‖ = ǫ, then the eigenvalues of the perturbed matrix A + ǫB are elementsof Λǫ(A). The pseudospectrum of A in the m = 20 case is shown in figuresD.9 and D.10. Note that the pseudospectrum of A agrees very well with theeigenvalues of A + ǫB as computed by MATLAB, although the eigenvalues ofA + ǫB are all strictly within the boundary predicted by the pseudospectrum.This is especially visible in figure D.10. This suggests that there exist somematrices C with ‖C‖ = 1 such that the eigenvalues of A + ǫC are slightlyfurther away from the unperturbed eigenvalues than the eigenvalues of A + ǫB.
215
Real Component
Imag
inar
y C
ompo
nent
0 5 10 15 20−5
−4
−3
−2
−1
0
1
2
3
4
5
eigenvalues of unperturbed matrixeigenvalues of perturbed matrixcontour plot of pseudospectrum
−13
−12
−11
−10
−9
−8
−7
−6
−5
−4
Figure D.10: The pseudospectrum for the unperturbed matrix, in the m = 20case.
216
D.3 Domain Perturbations
Consider now the first order corrections to the eigenvalues of a differential equa-tion where the domain has been perturbed by some perturbation of order ǫ. Welook in specific at the problem
∆φ + λφ = 0 in Ωǫ (D.32a)
φ = 0 on δΩǫ (D.32b)
whereδΩǫ : R = 1 + ǫf (θ) (D.33)
and
f (θ) =
∞∑
n=−∞
aneinθ, a−n = an. (D.34)
We will now find the first order corrections to the eigenvalues by two differentmethods, and show that they produce the same result.
D.3.1 General Method
For the first method, we start by writing
φ = φ0 + ǫφ1 + ǫ2φ2 + . . . (D.35)
λ = λ0 + ǫλ1 + ǫ2λ2 + . . . (D.36)
and equating terms of equal order from equation D.32a, we obtain that
O (1) : ∆φ0 + λ0φ0 = 0 (D.37)
O (ǫ) : ∆φ1 + λ0φ1 = −λ1φ0 (D.38)
L (φ) = −λ1φ0. (D.39)
Performing a Taylor expansion of the boundary conditions,
0 = φ (1 + ǫf (θ) , θ) (D.40)
= φ (1, θ) + ǫf (θ) φ,r (1, θ) +ǫ2f2 (θ)
2φ,rr (1, θ) (D.41)
and equating terms of equal order,
O (1) : 0 = φ0 (1, θ) (D.42)
O (ǫ) : 0 = φ1 (1, θ) + f (θ) φ0,r (1, θ) (D.43)
217
First Eigenvalue
The unperturbed eigenfunction of the first eigenvalue is
φ0 = J0 (j01r) (D.44)
where jmk is the kth positive zero of the Bessel function Jm. Now we know that
(φ0,L (φ1)) − (φ1,L (φ0)) =
∫
∂D
(φ0φ1,r − φ1φ0,r) ds (D.45)
where
(f, g) ≡∫
D
fg dx. (D.46)
Using equations D.37, D.39, D.42 and D.43, equation D.45 becomes
−λ1 (φ0, φ0) =
∫
∂D
f (θ) (φ0,r)2
ds (D.47)
so
λ1 = −∫
∂Df (θ) (φ0,r)
2ds
∫
D(φ0)
2dx
(D.48)
= −∫ θ=2π
θ=0
∑∞n=−∞
(
aneinθ)
(J ′0 (j01) j01)
2dθ
∫ θ=2π
θ=0dθ∫ r=1
r=0r (J0 (j01r))
2dθ
(D.49)
= − (J ′0 (j01))
2j201
∫ θ=2π
θ=0
(
a0 +∑∞
n=1
(
aneinθ + a−ne−inθ))
dθ
2π 12 (J ′
0 (j01))2 (D.50)
= −j201
2
∫ θ=2π
θ=0
(
a0 +∞∑
n=1
(2Re (an) cos (nθ) +
2 Im (an) sin (nθ))
)
dθ (D.51)
= −2j201a0. (D.52)
The first eigenvalue is then
λ = j201 − ǫ2j2
01a0 + O(
ǫ2)
. (D.53)
Second and Third Eigenvalues
The second and third eigenvalues of the unperturbed system are the same, sowe have
φ0 = c1v1 + c2v2 (D.54)
where c1 and c2 are arbitrary constants and
v1 = J1 (j11r) cos (θ) , v2 = J1 (j11r) sin (θ) . (D.55)
218
As before, we can use
(v1,L (φ1)) − (φ1,L (v1)) =
∫
∂D
(v1φ1,r − φ1v1,r) ds, (D.56)
which, upon applying equations D.37, D.39, D.42 and D.43, becomes
−λ1 (v1, φ0) =
∫
∂D
f (θ) φ0,rv1,r ds (D.57)
λ1c1 (v2, v1) − λ1c2 (v2, v2) = c1
∫
∂D
f · v21,r ds + c2
∫
∂D
f · v1,rv2,r ds. (D.58)
We can do likewise, with v1 in equation D.56 replaced with v2 to obtain
λ1c2 (v2, v1) − λ1c2 (v2, v2) = c1
∫
∂D
f · v1,rv2,r ds + c2
∫
∂D
f · v22,r ds. (D.59)
We can express these two equations together as
[ ∫
f · v21,r
∫
f · v1,rv2,r∫
f · v1,rv2,r
∫
f · v22,r
] [
c1
c2
]
= −λ1
[
(v1, v1) (v1, v2)(v1, v2) (v2, v2)
] [
c1
c2
]
.
(D.60)Using the definition of v1 and v2 from equation D.55 and the result that
∫ r=1
r=0
r (J1 (j11r))2dr = −J0 (j11) J2 (j11)
2=
1
2(J0 (j11))
2, (D.61)
this simplifies to
[ ∫
f · v21,r
∫
f · v1,rv2,r∫
f · v1,rv2,r
∫
f · v22,r
] [
c1
c2
]
= −π
2(J0 (j11))
2λ1
[
c1
c2
]
. (D.62)
If we re-write this as[
A BB C
] [
c1
c2
]
= Q
[
c1
c2
]
(D.63)
then the eigenvalues are given by
Q =A + C ±
√A2 + C2 − 2AC + 4B2
2, (D.64)
219
where
A =
∫
∂D
f (θ) (v1,r (1, θ))2
ds (D.65)
=
∫ θ=2π
θ=0
(
∞∑
n=−∞
aneinθ
)
(J ′1 (j11) j11 cos (θ))
2dθ (D.66)
= (J ′1 (j11))
2j211
∫ θ=2π
θ=0
(
a0 +
∞∑
n=1
(
aneinθ + a−ne−inθ)
)
cos2 (θ) dθ (D.67)
= (J ′′0 (j11))
2j211
∫ θ=2π
θ=0
(
a0 +
∞∑
n=1
(2Re (an) cos (nθ)
+2 Im (an) sin (nθ))
)
cos2 (θ) dθ (D.68)
= (J ′′0 (j11))
2j211π (Re (a2) + a0) (D.69)
≡ A0 + ξ. (D.70)
We likewise obtain that
C =
∫
∂D
f (θ) (v2,r (1, θ))2
ds (D.71)
=
∫ θ=2π
θ=0
(
∞∑
n=−∞
aneinθ
)
(J ′1 (j11) j11 sin (θ))
2dθ (D.72)
= (J ′′0 (j11))
2j211π (−Re (a2) + a0) (D.73)
= −A0 + ξ (D.74)
and
B =
∫
∂D
f (θ) v1,r (1, θ) v2,r (1, θ) ds (D.75)
=
∫ θ=2π
θ=0
(
∞∑
n=−∞
aneinθ
)
(J ′1 (j11) j11)
2sin (θ) cos (θ) dθ (D.76)
= − (J ′′0 (j11))
2j211π Im (a2) , (D.77)
so that
Q =A + C ±
√A2 + C2 − 2AC + 4B2
2(D.78)
= ξ ±√
A20 + B2. (D.79)
220
From the definition of Q, we obtain that
−λ1π
2(J0 (j11))
2= Q (D.80)
−λ1π
2(J0 (j11))
2= (J ′′
0 (j11))2j211a0π ± (J ′′
0 (j11))2j211π |a2| (D.81)
λ1 = 2j211 (± |a2| − a0) (D.82)
So the second eigenvalue is
λ = j211 + ǫ2j2
11 (− |a2| − a0) + O(
ǫ2)
(D.83)
and the third isλ = j2
11 + ǫ2j211 (|a2| − a0) + O
(
ǫ2)
. (D.84)
D.3.2 Method of Assumed Solution
For this method, we follow Wolf and Keller [32]. We start by assuming that thesolution is of the form
φ (r, θ, ǫ) =∞∑
n=−∞
An (ǫ) Jn (kr) einθ (D.85)
where
An (ǫ) = δ|n|mαn + ǫβn + ǫ2γn + . . . (D.86)
R (θ, ǫ) = 1 + ǫ∑
n
aneinθ + ǫ2∑
n
bneinθ + O(
ǫ3)
(D.87)
√λ ≡ k = k0 + ǫk1 + ǫ2k2 + . . . (D.88)
Applying boundary conditions,
φ (R (θ, ǫ) , θ, ǫ) =∑
n
An (ǫ) Jn (kR (θ, ǫ)) einθ = 0. (D.89)
Expanding this in a Taylor series, and writing kR = k0 + (k − k0) + k (R − 1),we obtain
∑
n
An
Jn (k0) + J ′n (k0) (k − k0 + k (R − 1)) +
1
2J ′′
n (k0) (k − k0 + k (R − 1))2
+ . . .
einθ = 0. (D.90)
Applying equations D.86–D.88 and equating terms of equal order, we obtainthat
O (1) :(
αmeimθ + α−me−imθ)
Jm (k0) = 0 (D.91)
O (ǫ) :(
α−mJ ′−me−imθ + αmJ ′
meimθ)
(
k1 + k0
∑
l
aleilθ
)
+
∑
n
βnJneinθ = 0. (D.92)
221
From equation D.91, we obtain that k0 = jmp, so m = 0, p = 1 corresponds tothe first eigenvalue and m = 1, p = 1 will give the second and third eigenvalues.
First Eigenvalue
Under the condition that m = 0, equation D.92 becomes
0 = 2α0J′0
(
k1 + k0
∑
l
aleilθ
)
+∑
n
βnJneinθ. (D.93)
Equating terms that have θ = 0, we obtain
0 = 2α0J′0 (j01) (k1 + k0a0) + β0J0 (j01) (D.94)
0 = 2α0J′0 (j01) (k1 + k0a0) (D.95)
0 = k1 + k0a0, (D.96)
sok1
k0= −a0 (D.97)
and
λ = k2 (D.98)
= k20 + ǫ2k0k1 + O
(
ǫ2)
(D.99)
= j201 − ǫ2j2
01a0 + O(
ǫ2)
, (D.100)
which agrees with equation D.53.
Second and Third Eigenvalues
Again following Wolf and Keller [32], we equate the coefficients of eimθ in equa-tion D.92 to obtain
0 = α−mJ ′−m (k0) k0a2m + αmJ ′
m (k0) (k1 + k0a0) + βmJm (k0) (D.101)
0 = α−mJ ′−m (k0) k0a2m + αmJ ′
m (k0) (k1 + k0a0) (D.102)
0 = α−mk0a2m + αm (k1 + k0a0) (D.103)
andk1
k0= −αma2m
αm− a0. (D.104)
Alternately, equating the coefficients of e−imθ in equation D.92, we obtain
k1
k0= −αma2m
αm− a0. (D.105)
Equating equations D.104 and D.105, we obtain that∣
∣
∣
∣
αma2m
αm
∣
∣
∣
∣
eiω =
∣
∣
∣
∣
αma2m
αm
∣
∣
∣
∣
e−iω (D.106)
222
where
ω = arg
(
αma2m
αm
)
(D.107)
= − arg
(
αma2m
αm
)
(D.108)
so that
eiω = e−iω (D.109)
eiω = ±1. (D.110)
It then follows that
k1
k0= ±
∣
∣
∣
∣
αma2m
αm
∣
∣
∣
∣
− a0 (D.111)
= ± |a2m| − a0, (D.112)
so the second and third eigenvalues are
λ = j211 + ǫ2j2
11 (± |a2| − a0) + O(
ǫ2)
(D.113)
which agrees with equations D.83 and D.84.
D.3.3 Result
Similarly, Wolf and Keller [32] compute the second order correction to the firsteigenvalue. Then under the condition that the area of the domain does notchange under the perturbation (which causes a0 = 0, among other things), theyobtain that
AλII = πj211−
π1/2j211
j201
[
AλI − πj201
1 + j01J ′2 (j01) /J2 (j01)
]1/2
+O[
AλI − πj201
]
. (D.114)
where A is the area of the domain, and λI and λII are the first and secondeigenvalues, respectively. Normalising A to unity, we obtain the plot shown infigure D.11.
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0 20 40 60 80 1000
20
40
60
80
First Eigenvalue (λI)
Sec
ond
Eig
enva
lue
(λ II
)
Figure D.11: The first and second eigenvalues for perturbations to a circle.When there are no perturbations, λI is at a minimum.
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