Math 6730
Homework Exercises - Solutions
1
Regular Perturbation Theory
(due Sept. 22, 2015)
1. Find a two-term asymptotic expansion, for small ǫ, for all solutions of the following
equations:
(a)
ǫx3 − 3x + 1 = 0
(b)
x2+ǫ =
(1)
1
x + 2ǫ
(2)
(c)
xe−x = ǫ
(3)
x3 − x2 − ǫ = 0
(4)
(d)
2. Find a two-term asymptotic expansion, for small ǫ, of the solution of the following
boundary value problems:
(a)
y ′′ + y + ǫy 3 = 0,
y(0) = 0,
π
y( ) = 1.
2
(5)
(b)
y ′′ − y = 0,
y(0) = 0,
y(1 + ǫ) = 1
(6)
Hint: Use a change of variables to transform this to a problem on a fixed domain
(independent of ǫ).
3. Suppose that a one-dimensional cell that is 10 µm long and the voltage potential is
held fixed at both ends (so that Φ = 0 at the ends). Suppose that the cell contains
approximately equal amounts (say 400mM ) of sodium and chloride, but suppose there
is a slight excess of sodium. How much excess sodium is there if the total difference
between maximal and minimal potential in the cell is 0.1mV? Plot the (approximate)
distribution of potential and ion concentrations for both species.
Hint: Derive the Poisson-Boltzman equation as follows: According to the Poisson
equation,
X
ǫ∇2 φ = −qNa
zi ci
(7)
i
1
where ǫ is the dielectric constant of the medium, Na is Avagadro’s number, q is the
unit electric charge, ci and zi are the concentration and unit charge for the ith species.
According to the Nernst-Planck equation,
0 = ∇ci +
zi F
ci ∇φ.
RT
(8)
Use this information to derive the Poisson-Boltzmann equation in dimensionless units
X
∇2 Φ = −β
zi αi exp(−zi Φ),
(9)
i
where
L2 F NA
β=q
C0 ,
(10)
ǫRT
and C0 is a characteristic concentration, L is the length of a one dimensional domain.
What is αi ? For parameter values, use that ǫ = 6.938 × 10−10 C2 /Nm2 at 25C, q =
= 25.8mV. How large is β for this cell?
1.6 × 10−19 C(oulombs), RT
F
Use that the net charge
δ=
X
zi αi .
(11)
i
is a small dimensionless parameter to find the asymptotic solution of the problem.
4. Approximate the time it takes for a projectile whose terminal velocity is VT with height
y(t) above the earth, with y(0) = 0 and yt (0) = V0 , to return to earth, assuming that
V02
and VVT0 are small. What is the effect of drag on the return time?
gR
Hint: Use the equation
ytt = −
g
gR
yt .
y 2 −
(1 + R )
VT (R + y)
(12)
to account for viscous atmospheric drag.
1.1
Solutions
1. (a) There are three solutions,
x=
and
1
1 2
1
+ ǫ+
ǫ + O(ǫ3 ),
3 81
729
r
x=±
√
3 1
3√
− ∓
ǫ + O(ǫ).
ǫ 6
72
(13)
(14)
(b) Take the logarithm of both sides of the equation to rewrite the equation as
(3 + ǫ) ln x + ln(1 +
2
2ǫ
) + 2kπi.
x
(15)
)
A power series solution in ǫ of the form x = x0 +ǫx1 +ǫ2 x2 +· · · yields x0 = exp( ±iπ
3
x0
x1
2
1
2
or x = 1 and x1 = − 3 − 3 ln(x0 ), and x2 = 6x0 (3x1 + 4x1 + 4) − 3 .
For k = 0, this gives
2
2
(16)
x = 1 − ǫ + ǫ2 + O(ǫ3 ).
3
3
For k = ±1, the answer is tedious to write out.
(c) To find an asymptotic expansion for
x exp(−x) = ǫ,
(17)
−1
make a few changes of variables. Let γ = ln
and y = γx and set η = γ ln γ and
ǫ
the equation becomes
y = 1 − η + γ ln(y).
(18)
Clearly y is nearly 1. A relatively easy way to proceed is iteratively, i.e., let
yn = F (yn−1 ) where F (y) = 1 − η + γ ln(y), so that
y0 = 1,
(19)
y1 = 1 − η,
(20)
y2 = 1 − η + γ ln(1 − η),
(21)
y3 = 1 − η + γ ln(1 − η + γ ln(1 − η)),
(22)
and so on. However, this is not an asymptotic series, so to make it into one we
now expand yn as a Taylor series in γ and find
3 !
ln(1 − η) 2
ln(1
−
η)
ln(1 − η) 1 (ln(1 − η))2
y ≈ −η+γ ln(1−η)+
γ 3 +O
γ +
−
γ4
(1 − η)
(1 − η)2
2 (1 − η)2
1−η
(23)
This is an asymptotic series, but a strange one.
(d) There are three solutions
and
x = 1 + ǫ − 2ǫ2 + O(ǫ3 ),
(24)
√
5 √
1
x = ±i ǫ − ǫ ∓ i ǫ3 .
2
8
(25)
2. (a)
1
1
3
2
y(x) = sin(x) + ǫ sin(x) − sin(x) cos2 (x) + x cos(x) + sin(x) .
4
8
8
8
(26)
(b)
where t =
t cosh(t) cosh(1) sinh(t) sinh(t)
+ O(ǫ2 ),
+ǫ
−
y(x) =
2
sinh(1)
sinh(1)
sinh (1)
x
.
1+ǫ
3
(27)
3. Take z1 = 1 and z2 = −1 for sodium and chloride, respectively, and then δ = α2 − α1 .
It follows that
∇2 Φ = −β(−α1 exp(−Φ) + (δ + α1 ) exp(Φ)).
(28)
Set Φ = δΦ1 + O(δ 2) and find that to leading order in δ,
∇2 Φ1 + 2βα1 Φ1 = −β
The solution of this equation is easy, being
√
cosh( 2βα1 (x − 12 )
1
√
1−
Φ1 = −
.
2α1
cosh( 12 2βα1 )
(29)
(30)
Now suppose that the maximum value of Φ is Φ∗ . This implies that
δ
= −Φ∗ .
2α1
Now by conservation, it must be that
Z 1
Z
1=
α2 exp(Φ)dx ≈
0
and since
1
α2 (1 + δΦ1 )dx,
1
(32)
0
√
tanh( 12 2βα1)
√
.
1−2
2βα1
0
√
tanh( 12 2βα1 )
1
√
1 = α2 1 − δ
1−2
2α1
2βα1
Z
(31)
1
Φ1 dx = −
2α1
Now we use that β is very large to get
1
1 = α2 1 − δ
= α2 (1 + Φ∗ ) .
2α1
Similarly, total sodium (in dimensionless units) is
1
1 + ∆N a = α1 1 + δ
= α1 (1 − Φ∗ )
2α1
(33)
(34)
(35)
(36)
We can now use the three equations (31,35,36) to find that
∆N a =
Convert to dimensional units: Φ∗ =
∆N a =
Φ∗2
.
(1 + Φ∗ )(1 − 2Φ∗ )
0.1
25.8
(37)
= 0.00387
Φ∗2
= 3.0 × 10−5 .
(1 + Φ∗ )(1 − 2Φ∗ )
(38)
This is the fractional charge deflection necessary. Thus, if the initial concentration is
400mM, then the charge deflection concentration is 12µM.
4
4. Begin by non-dimensionalizing the equation
ytt = −
gR
g
yt ,
y 2 −
(1 + R )
VT (R + y)
(39)
with initial data
y(0) = 0,
yt (0) = V0 .
(40)
Rescale time t = στ and length y = LY so that
LYτ τ = −
with initial data Yτ (0) =
and ǫ =
L
R
=
V02
.
gR
V0 σ
.
L
gσ 2
σg
−
LY 2
(1 + R )
VT (1 +
LY
R
)
LYτ ,
(41)
Pick L = V0 σ and gσ 2 = L so that σ =
V0
,
g
and L =
V02
,
g
The rescaled equation is
Yτ τ = −
1
V0
1
−
Yτ ,
2
(1 + ǫY )
VT (1 + ǫY )
(42)
with initial data Yτ (0) = 1.
Now set
V0
VT
= γǫ, expand in ǫ, and find that
1
1
1
1
Y (τ ) = − τ 2 + t + ǫ(− τ 4 + ( (γ + 2))τ 3 − γτ 2 )
2
12
6
2
(43)
and the root is
2
4
τ = 2 + ǫ( − γ ).
3
3
Convert this back into dimensional parameters to find
t=2
(44)
V0 4 V03
2 V02
+
−
.
g
3 g 2 R 3 gVT
(45)
Notice that the correction due to drag decreases the time of flight.
2
Matched Asymptotic Expansions
(due Oct. 8, 2015)
1. Find a composite expansion for the solutions of each of the following (include a plot
of the approximate solution);
(a)
ǫy ′′ + y ′ = a(> 0),
y(0) = 0,
y(1) = 1,
(46)
ǫy ′′ + 2y ′ + y 3 = 0,
y(0) = 0,
y(1) =
1
2
(47)
(b)
5
(c)
ǫy ′′ − y ′ − y 2 = 1,
(d)
1
y(0) = ,
3
3
y(0) = ,
5
ǫy ′′ = yy ′ − y 3 ,
y(1) = 1
(48)
2
3
(49)
y(1) = −
(e)
ǫy ′′ = 9 − (y ′ )2 ,
2.1
y(0) = 0,
y(1) = 1
(50)
Solutions
1. (a)
x
y(x) = ax + (1 − a)(1 − e− ǫ ) + O(ǫ).
(b)
y(x) = √
1
2x
1
− √ e− ǫ + O(ǫ).
3+x
3
(51)
(52)
(c)
x−1
1
1
y(x) = tan(tan−1 ( ) − x) + (1 − tan(tan−1 ( ) − 1))e ǫ
3
3
(d) There are three solutions
i.
y1 (x) =
ii.
5
3
3 x − 1
1
3
4 3
− tanh
− .
+ tanh−1
2 ǫ
9
2
−x 2
y2 (x) =
iii.
y3 (x) = H(x−
(e)
3
2x
3 4x
− 2 tanh
− tanh−1 ( )
1 + 2x
ǫ
10
(53)
(54)
(55)
12 x − 7 7 12
7
12 12
1
1
12
)+H( −x)( 5
− )− tanh
)( − 1
(
)
12 13 2 + x
12
13
13
13
ǫ
−
x
3
(56)
1
3x − 1 y(x) = −1 + ǫ 2 ln cosh( 1 )
ǫ2
(57)
QSS Analysis
(due Nov. 12, 2015)
1. The calcium-activated potassium channel has two open states and two closed states
(see Fig. 1). The channel has two binding sites for calcium and can open when one of
the states is occupied and remains open when the second site is bound. The binding
process is a fast process, while the transition between open and closed states is slow.
Find the equation describing slow dynamics of the opening of the channel as a function
of calcium concentration.
6
(a)
(b)
0.6
1
0.8
0.5
0.6
0.4
0.4
y(x)
y(x)
0.2
0
0.3
-0.2
0.2
-0.4
-0.6
0.1
-0.8
-1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
x
0.5
0.6
0.7
0.8
0.9
1
x
(c)
(d)-1
1
1.5
0.8
1
0.6
0.5
0
0.2
y(x)
y(x)
0.4
0
-0.5
-0.2
-1
-0.4
-1.5
y 1(x)
y 2(x)
y 3(x)
-0.6
-0.8
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
-2
1
0
0.1
0.2
0.3
0.4
0.8
-50
0.6
-100
0.4
-150
0.2
y(x)
dy/dx
1
0
-200
0.7
0.8
0.9
1
0.6
0.7
0.8
0.9
1
0
-250
-0.2
-300
-0.4
-0.6
-350
-450
-2
0.6
(e)
(d)-2
50
-400
0.5
x
x
y 1(x)
y 2(x)
y 3(x)
-1.5
-0.8
-1
-0.5
0
0.5
1
-1
1.5
0
y(x)
0.1
0.2
0.3
0.4
0.5
x
7
Ca
2+
C1
Ca
C2
2+
O1
O2
Figure 1: Diagram of the four states for the calcium-activated potassium channel.
2. Suppose that an enzyme can bind two substrate molecules, so can exist in one of three
states, a free moleule S, a complex with one molecule bound C1 , and a complex with
two molecules bound C2 . The corresponding chemical reactions are
k1
S+E
−→
←−
k−1
k3
S+
C1 −→
←−
k−3
k
(58)
k
(59)
2
C1 −→
P +E
4
C2 −→
P + C1
Give a “proper” derivation of the corresponding Michaelis-Menten rate of production
of product P , using the assumption that the amount of substrate is much larger than
the amount of complex.
3. Consider the process of autocatalytic conversion of a substrate to an enzyme given by
k1
S+E
−→
←−
k−1
k
2
C −→
E+E
(60)
Find the (slow) rate of substrate conversion under the assumption that k2 ≪ k−1 .
4. Suppose a particle randomly switches between three states
k
k
−→
S−1 −→
←− S0 ←− S1 ,
k
(61)
k
and in states Sj it moves with velocity jV , j = −1, 1, while in state S0 the particle
diffuses with diffusion coefficient D. Suppose the switching rates k are quite large.
Find the effective diffusion coefficient for this particle.
3.1
Solutions
1. The differential equation system is
dpc1
dt
dpc2
dt
dpo1
dt
dpo2
dt
= −k1 pc1 + k−1 pc2 ,
(62)
= k1 pc1 − k−1 pc2 − k2 pc2 + k−2 po1 ,
(63)
= k2 pc2 − k−2 po1 − k3 po1 + k−3 po2 ,
(64)
= k3 po1 − k−3 po2 ,
(65)
8
where k1 and k3 are calcium dependent, k1 , k−1 , k3 , and k−3 are large compared to k2 ,
and k−2 . Set y = pc1 + pc2 . The qss approximation implies
0 = −k1 pc1 + k−1 pc2 ,
0 = k3 po1 − k−3 po2 ,
(66)
(67)
which implies that
k1 y
k−1 y
,
pc 2 =
,
k1 + k−1
k1 + k−1
k−3
k3
=
(1 − y),
po2 =
(1 − y).
k3 + k−3
k3 + k−3
pc 1 =
(68)
po1
(69)
This implies that
dy
= −k2 pc2 + k−2 po1 ,
dt
k−2 k−3
k2 k1
y+
(1 − y),
= −
k1 + k−1
k3 + k−3
(70)
(71)
which is the slow equation for the closed-open transition.
2. The differential equation system is
ds
dt
de
dt
dc1
dt
dc2
dt
dp
dt
Conservation implies
= −k1 es + k−1 c1 − k3 sc1 + k−3 c2 ,
(72)
= −k1 es + k−1 c1 + k2 c,
(73)
= k1 es − k−1 c1 − k2 c1 − k3 sc1 + k−3 c2 + k4 c2 ,
(74)
= k3 sc1 − k−3 c2 − k4 c2 ,
(75)
= k 2 c1 + k 4 c2 .
(76)
that e + c1 + c2 = E0 so that we eliminate e and find
ds
= −k1 s(E0 − c1 − c2 ) + k−1 c1 − k3 sc1 + k−3 c2 ,
dt
dc1
= k1 s(E0 − c1 − c2 ) − k−1 c1 − k2 c1 − k3 sc1 + k−3 c2 + k4 c2 ,
dt
dc2
= k3 sc1 − k−3 c2 − k4 c2 .
dt
Scale the variables using c1 = E0 u1 , c2 = E0 u2 , s = S0 σ and t =
S0
τ
k−1 E0
(78)
(79)
and find
dσ
= −κ1 σ(1 − u1 − u2 ) + u1 − κ3 σu1 + κ−3 u2 ,
dτ
du1
= κ1 σ(1 − u1 − u2 ) − u1 − κ2 u1 − κ3 σu1 + κ−3 u2 + κ4 u2 ,
ǫ
dτ
du2
= κ3 σu1 − κ−3 u2 − κ4 u2 ,
ǫ
dτ
9
(77)
(80)
(81)
(82)
k1
2
, κ2 = kk−1
, κ3 =
where κ1 = Sk0−1
quasi-steady approximation takes
S0 k 3
,
k−1
, κ−3 =
k−3
,κ
k−1 4
=
k4
,
k−1
and ǫ =
0 = κ1 σ(1 − u1 − u2 ) − u1 − κ2 u1 − κ3 σu1 + κ−3 u2 + κ4 u2 ,
0 = κ3 σu1 − κ−3 u2 − κ4 u2
E0
.
S0
The
(83)
(84)
which we solve to find
u1 =
u2 =
(κ−3 + κ4 )κ1 σ
,
+ κ1 σ(κ−3 + κ4 ) + (κ2 + 1)(κ4 + κ−3 )
(85)
κ1 κ3 σ 2
.
κ1 κ3 σ 2 + κ1 σ(κ−3 + κ4 ) + (κ2 + 1)(κ4 + κ−3 )
(86)
κ1 κ3
σ2
It follows that
dp
k2 (κ−3 + κ4 )σ + k4 κ3 σ 2
= E0 κ1
.
dt
κ1 κ3 σ 2 + κ1 σ(κ−3 + κ4 ) + (κ2 + 1)(κ4 + κ−3 )
(87)
3. The differential equation system is
ds
= −k1 es + k−1 c,
dt
de
= −k1 es + k−1 c + k2 c,
dt
dc
= k1 es − k−1 c − k2 c.
dt
(88)
(89)
(90)
Set e = E0 − c, so that
ds
= −k1 s(E0 − c) + k−1 c,
dt
dc
= k1 s(E0 − c) − k−1 c − k2 c.
dt
(91)
(92)
Under the assumption that k2 ≪ k−1 , we take s to be in quasi-equilibrium so that
c=
k1 E0 s
,
k1 s + k−1
(93)
and then the slow evolution is
or
where Kd =
to enzyme.
k−1
,
k1
d(s + c)
= −k2 c,
dt
(94)
E0 s
k2 E0 s
d
(s +
)=−
,
dt
s + Kd
s + Kd
(95)
which is a differential equation for the (slow) conversion of substrate
10
4. The deterministic part of the motion is
dy
=Vn
dt
(96)
where n has three states n = −1, 0, 1 with transitions
k
S−1 −→
←−
k
k
S0 −→
←−
S1
(97)
k
The master equations are
∂p−1
∂p−1
= kp0 − kp−1 + V
,
∂t
∂x
∂ 2 p0
∂p0
= kp−1 + kp1 − 2kp0 + D 2 ,
∂t
∂x
∂p1
∂p1
= kp0 − kp1 − V
.
∂t
∂x
We assume that k is large so introduce the parameter ǫ =
become
1
.
k
1
∂p−1
∂p−1
=
(p0 − p−1 ) + V
,
∂t
ǫ
∂x
∂p0
1
∂ 2 p0
=
(p−1 + p1 − 2p0 ) + D 2 ,
∂t
ǫ
∂x
1
∂p1
∂p1
=
(p0 − p1 ) − V
.
∂t
ǫ
∂x
The nullspace of the matrix is spanned by the vector
 
1
1
φ =  1 .
3
1
We set




p−1
w−1
p =  p0  = vφ +  w0  ,
p1
w1
(98)
(99)
(100)
Then, the equations
(101)
(102)
(103)
(104)
(105)
and find the equation for v by projecting
∂v
∂p−1
∂p1
∂ 2 p0
= V
−V
+D 2
∂t
∂x
∂x
∂x
1 ∂2
∂
(w−1 − w1 ) + D 2 (v + w0 ).
= V
∂x
3 ∂x
The vector w must satisfy the equation




∂v
−1 1
0
−V ∂x
1
1
1 −2 1  w =  0  .
ǫ
3
∂v
0
1 −1
V ∂x
11
(106)
(107)
(108)
The solution of this problem has
w−1 − w1 =
2ǫ ∂v
V
.
3 ∂x
(109)
Thus, the effective Fokker-Planck equation (ignoring higher order derivatives) is
∂v
∂2v
= Def f 2
∂t
∂x
(110)
where
1
2ǫ
Def f = D + V 2 ,
3
3
which, in case you are checking, has the correct units.
4
(111)
Multiscale/Averaging
(due Dec. 3, 2015)
1. Using your choice of method (multiscale analysis or averaging), find approximate solutions for the equations
(a)
u′′ + ǫ(u′ )3 + u = 0
(112)
u′′ − ǫ cos t(u′ )2 + u = 0
(113)
u′′ + ǫ cos(ǫt)(u′ )3 + u = 0
(114)
u′′ + ǫ(ǫ + u′ )u′ + u = 0
(115)
u′′ + ǫ(ǫ + (u′ )2 )u′ + u = 0
(116)
(b)
(c)
(d)
(e)
Remark: For this problem, use higher order averaging. Why is this method
preferable for this problem?
4.1
Solutions
1. (a) Using multiscaling, the equation is rewritten as
2
∂2u
∂2u
∂u
∂u 3
2∂ u
+
2ǫ
+
ǫ
+
ǫ(
+
ǫ
) +u=0
∂τ 2
∂τ ∂s
∂s2
∂τ
∂s
(117)
u = A(s)(eiτ +iφ(s) + e−iτ −iφ(s) + ǫu1 (τ, s) + O(ǫ2 ),
(118)
Setting
12
we require
∂
∂ 2 u1
+ u1 + 2 (A(s)(ieiτ +iφ(s) − ie−iτ −iφ(s) ) + A3 (s)(ieiτ +iφ(s) − ie−iτ −iφ(s) )3 = 0.
2
∂τ
∂s
(119)
The solvability condition is
2A′ i − 2Aφ′ + A3 i = 0
(120)
which, after separating into real and imaginery parts, gives
2A′ + A3 = 0,
φ′ = 0
(121)
Notice that this is not structurally stable but that is the nature of the problem
and nothing can be done to rectify this.
(b) Use multiscaling to write the equation as
2
∂2u
∂2u
∂u
∂u 2
2∂ u
+
2ǫ
+
ǫ
+
ǫ
cos(τ
)(
+
ǫ
) +u=0
∂τ 2
∂τ ∂s
∂s2
∂τ
∂s
(122)
where τ = t, σ = ǫt. With u = A(σ) cos(τ + φ(σ)) + ǫu1 we find at O(ǫ):
∂ 2 u1
dA
dφ
+ u1 − 2
sin(τ + φ) − 2A cos(τ + φ) + A2 cos(τ ) cos2 (τ + φ) = 0 (123)
2
∂τ
ds
ds
Thus, we require
and
dA 1 2
− A sin φ = 0
ds
8
(124)
dφ 3
− A cos φ = 0
dt
8
(125)
One cannot find
(c) Use multiscaling to write the equation as
2
∂2u
∂2u
∂u
∂u 3
2∂ u
+
2ǫ
+
ǫ
+
ǫ
cos(s)(
+
ǫ
) +u=0
∂τ 2
∂τ ∂s
∂s2
∂τ
∂s
(126)
Set u = A(s) cos(t + φ(s) + ǫu1 and find at O(ǫ):
∂ 2 u1
dA
dφ
+ u1 − 2
sin(τ + φ) − 2A cos(τ + φ) + A3 cos(s) sin3 (τ + φ) = 0 (127)
2
∂τ
ds
ds
and then require
dA
3
dφ
= A3 cos(s),
=0
(128)
ds
8
ds
(d) For this problem, one can use either averaging or multiscaling, but in either case,
one must go beyond the first order to get a correct approximation. Using a
multiscale approach, choose a slow time scale s = ǫ2 t and write the equation as
2
2
∂u
∂u ∂u
∂u
∂2u
2 ∂ u
4∂ u
+
2ǫ
+
ǫ
+ ǫ(ǫ +
+ ǫ2 )(
+ ǫ2 ) + u = 0
2
2
∂τ
∂τ ∂s
∂s
∂τ
∂s ∂τ
∂s
13
(129)
and then the approximation u = u0 + ǫu1 + ǫ2 u2 yields
∂u0 2
∂ 2 u1
+ u1 + (
) =0
2
∂τ
∂τ
(130)
and
∂ 2 u0
∂u0
∂u0 ∂u1
∂ 2 u2
+
u
+
2
+
+2
=0
(131)
2
2
∂τ
∂τ ∂s
∂τ
∂τ ∂τ
Take u0 = A(s) cos(t + φ(s)) and then u1 = A2 ( 61 cos(2τ + 2φ) − 21 ), and the
equation for u2 becomes
∂ 2 u2
dA
dφ
1
+u2 −2
sin(τ +φ)−2A cos(t+φ)+A sin(τ +φ)+ A3 sin(τ +φ) sin(2τ +2φ) = 0
2
∂τ
ds
ds
3
(132)
(e) To use averaging, first introduce polar coordinates u = R cos θ, u′ = R sin θ.
Then, require
R′ cos θ − Rθ′ sinθ − R sin θ = 0
(133)
and
R′ sin θ + Rθ′ cos θ + ǫ(ǫ + (R sin θ)2 )R sin θ + R cos θ = 0
(134)
Combining these the appropriate way, we find
R′ + ǫ(ǫ + (R sin θ)2 )R sin2 θ = 0
(135)
θ′ + 1 + ǫ(ǫ + (R sin θ)2 ) sin θ cos θ = 0
(136)
and
Finally, set θ = ψ − t so that
R′ = −ǫ(ǫ + R2 sin2 (ψ − t))R sin2 (ψ − t)
ψ ′ = −ǫ(ǫ + R2 sin2 (ψ − t)) sin(ψ − t) cos(ψ − t)
(137)
(138)
Now we seek a change of variables
R = ρ + ǫR1 (ρ, φ, t) + ǫ2 R2 (ρ, φ, t) · · · ,
ψ = φ + ǫP1 (ρ, φ, t) + ǫ2 P2 (ρ, φ, t) + · · ·
(139)
(140)
so that the equations have the form
ρ′ = ǫH0 (ρ, φ) + ǫ2 H1 (ρ, φ) + · · ·
φ′ = ǫG0 (ρ, φ) + ǫ2 G(ρ, φ)
(141)
(142)
This means
O(ǫ) : H0 + R1t = −ρ3 sin4 (ψ −t),
G0 + P1t = −ρ3 sin3 (φ −t) cos(φ −t) (143)
so that
3
H0 = − ρ3 ,
8
1
5
R1 = − sin(φ−t) cos3 (φ−t)ρ3 + ρ3 cos(φ−t) sin(φ−t) (144)
4
8
14
and
1
P1 = ρ3 sin4 (φ − t)
4
To next order the equations are complicated, but the answer is
G0 = 0,
1
5 6
3 5
H1 = − ρ,
G1 =
ρ +
ρ
2
128
512
This gives us the averaged equations we want, namely
1
3
ρ′ = −ǫ ρ3 − ǫ2 ρ.
8
2
The solution of this equation is
s
4ǫ
ρ(t) =
A exp(ǫ2 t) − 3
(145)
(146)
(147)
(148)
where A = 3 + ρ24ǫ(0) > 3, which behaves much differently than the leading order
solution, which decays algebraically.
5
Homogenization
(due Dec. 17, 2013)
1. A laminate material consists of many thin layers of alternating materials with different
thermal conductivities. Suppose the layers have thermal conductivities Di and thicknesses li , i = 1, 2. Find the effective conductivity tensor for the composite material.
2. Find the effective diffusion tensor for a periodic medium in which the diffusion coefficient is rapidly varying in space, according to
x
y D(x, y) = exp α( ) + β( ) ,
(149)
ǫ
ǫ
where α(x) and β(y) are periodic functions with period a and b, respectively.
3. Find the coupled phase equations for a system of coupled lambda-omega (λ − ω)
systems, of the form
P
d
g
xi
λ(ri ) −ω
xi
ij
j
,
(150)
=
+ǫ
ω
λ(ri )
yi
0
dt yi
i = 1, · · · , N, where ri = x2i + yi2, ǫ ≪ 1, λ(1) = 0, λ′ (1) < 0, and
(a)
(diffusive coupling)
gij = cij (xj − xi ),
(151)
gij = cij xj
(152)
(b)
(synaptic coupling).
15
5.1
Solutions
1. Use the answer to Problem 2 to calculate the effective diffusion tensor. We find, in the
direction normal to the layers, the diffusion coefficient is
Def f =
D1 D2 (l1 + l2 )
.
l 1 D2 + l 2 D1
(153)
In the direction parallel to the lamination, the diffusion coefficient is
Def f =
l 1 D1 + l 2 D2
.
l1 + l2
(154)
2. Follow standard arguments to conclude that
Z
1
Def f =
D(I + ∇W )dV,
VΩ Ω
(155)
where W is the vector that satisfies
∇ · (D∇W ) = −∇ · (DI),
(156)
where I is the identity matrix.
For this problem with D = exp(α(x) + β(y)), W must satisfy
∇ · (exp(α(x) + β(y))∇W ) = −∇ · (exp(α(x) + β(y))I),
(157)
or
(exp(α(x))wx1 )x = −(exp(α(x))x ,
(exp(β(y))wy2)y = −(exp(β(y))y ,
(158)
exp(β(y))wy2 = K2 − exp(β(y),
(159)
wy2 = K2 exp(−β(y)) − 1,
(160)
which integrates once to
exp(α(x))wx1 = K1 − exp(α(x)),
wx1 = K1 exp(−α(x)) − 1,
where
K1 =
1 Z
a
a
0
−1
exp(−α(x))dx
,
K2 =
1 Z
b
b
−1
(161)
.
(162)
exp(−β(y))dy
0
It follows from (155) that
Def f =
K1 1b
Rb
0
exp(β(y))dy
Ra 0
0
K2 a1 0 exp(α(x))dx
16
3. The multiscaled equations are
P
d
d
xi
λ(ri ) −ω
xi
xi
j gij
,
+ǫ
+ǫ
=
yi
ω
λ(ri )
0
dt yi
dσ yi
(163)
so setting xi = x0i + ǫx1i + · · ·, yi = yi0 + ǫyi1 + · · ·, we find
0 0 d
λ(ri0 ) −ω
xi
xi
=
0
0
yi0
ω
λ(ri )
dt yi
(164)
and
L
x1i
yi1
d
≡
dt
d
= −
dσ
x1i
yi1
−
x0i
yi0
(x0i )2
r0
x0 y 0
′
+ λ (r0 ) ir0 i
P 0 j gij
λ′ (r0 )
+
ω
0
−ω + λ′ (r0 )
λ′ (r0 )
x0i yi0
r0
(yi0 )2
r0
!
,
(x0i )2
r0
x0 y 0
′
λ (r0 ) ir0 i
λ′ (r0 )
−ω +
x0i yi0
r0
(yi0 )2
′
λ (r0 ) r0
ω + λ′ (r0 )
(168)
!
u
v
It follows that the solvability condition for the order ǫ system is
Z 2π
X
ω
ω
dφi
sin(ωt + φi )
gij0 .
=−
dσ
2π 0
j
0
2π
ω
0
=
Z
= −
sin(ωt + φi )gij (x0j − αx0i )dt
(169)
(170)
(171)
(172)
2π
ω
sin(ωt + φi )cij (cos(ωt + φj ) − α cos(ωt + φi ))dt (173)
0
1
=
ω
(165)
(167)
and one can directly verify that the nullspace is spanned by
u
− sin(ωt + φi )
=
.
v
cos(ωt + φi )
We calculate
Z 2π
Z
ω
0
sin(ωt + φi )gij dt =
(166)
The leading order periodic solution is given by
0 xi
cos(ωt + φi )
.
=
sin(ωt + φi )
yi0
x
has a nullspace
It is easy to see that the linearized operator L
y
− sin(ωt + φi )
cos(ωt + φi )
The adjoint operator is
d
u
u
∗
−
=−
L
v
dt v
xi
yi
Z
2π
0
sin(τ )cij (cos(τ + φj − φi ) − α cos(t)dt
π
sin(φj − φi )
ω
17
(174)
(175)
It follows that the phase equations are
1X
dφi
cij sin(φj − φi ).
=
dσ
2 j
18
(176)