Solutions and Behavior ofLattice Differential Equations
Brian E. Moore
A “Senior, But Still Kicking” Seminar at
Cha-Cha Days, College of Charleston
September 26, 2010
Sep 2010 – p.1/37
Contents1. Numerical reproduction of spinodal decomposition
Erik Van Vleck & Tony HumphriesCahn-Hilliard EquationNecessary conditions on lattice spacing
2. Numerical reproduction of conservation lawsSebastian ReichHamiltonian EquationsLattice spacing occasionally important
3. Derivation of exact solutions for an LDETony Humphries & Erik Van VleckNagumo EquationWave soltions with inhomogeneous diffusion coefficients
Sep 2010 – p.2/37
1.
Numerical reproduction of spinodal decomposition
Erik Van Vleck & Tony Humphries
Sep 2010 – p.3/37
Spinodal Decomposition
(a) (b) (c)
(a) Initial condition: well-mixed metals form a molten alloy
(b) Short time: phase separation, characteristic length scale
(c) Long time: metals are separated, decomposition is complete
Special thanks to David Landy and Linli Wang for the images which were posted on
http://www.personal.psu.edu/dal233/project3.html
Sep 2010 – p.4/37
Cahn-Hilliard Equation
ut = −∆(ǫ2∆u+ f(u)), f(u) = u− u3
u ∈ [−1, 1] represents concentrations of one metallic component.
f ′(m) > 0 ⇒ m ∈ (−1/√
3, 1/√
3) and m is unstable.
Almost all orbits starting close to m exit a neighborhood of thatpoint close to a strongly unstable invariant subspace.
These orbits
behave similarly to orbits of vt = −∆(ǫ2∆v + f ′(m)v)
exhibit characteristic pattern formation
Sep 2010 – p.5/37
Lattice Differential Equations
d2u
dx2≈ u(x+ h)− 2u(x) + u(x− h)
h2
d2u
dx2≈ 1
h2
. . . . . . . . .
1 −2 1
1 −2 1
1 −2 1. . . . . . . . .
...u−1
u0
u1
...
We call this matrix ∆h.It is an approximation of the Laplacian operator ∆ = d2
dx2 .
Sep 2010 – p.6/37
Linear Analysis
Define Aǫ = −∆(ǫ2∆ + f ′(m))
κj = j2π2 are eigenvalues of −∆, µj are eigenvalues of −∆h
So, eigenvalues of Aǫ are λj = κj(f′(m)− ǫ2κj) which are
positive when 0 < κj < f ′(m)/ǫ2
For 0 < ρ≪ 1, and µj < (2f ′(m))/ǫ2 for some j ∈ [0, . . . , n]
there exists a nρ > 0 such that for all n ≥ nρ/ǫ
|µj
κj− 1| < ρ for |µj − κj | <
3f ′(m)ρ
ǫ2
Sep 2010 – p.7/37
6
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1/ǫ
n n = n̂/ǫ n = nρ/ǫ
p p p p p p p p p p
p p p p p p p p p p p
p p p p p p p p p p p p
p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p p p p p p p
p p p p p p p p p p p p p p p p p p p p p p p p p p p p
Spinodal decomposition occurs in the shaded region.
Sep 2010 – p.8/37
2.
Numerical reproduction of conservation laws
Sebastian Reich
Sep 2010 – p.9/37
Hamiltonian Partial Differential Equations
According to Bridges (1997), these equations can be written
Kzt + Lzx = ∇zS(z)
K and L are skew-symmetric matrices
z = z(x, t) is the vector of state variables
S is smooth
Sep 2010 – p.10/37
Hamiltonian Partial Differential Equations
According to Bridges (1997), these equations can be written
Kzt + Lzx = ∇zS(z)
K and L are skew-symmetric matrices
z = z(x, t) is the vector of state variables
S is smooth
Examples include KdV, Boussinesq, Zakharov-Kuznetsov,nonlinear Schrödinger, nonlinear wave equations, etc.
Sep 2010 – p.10/37
Hamiltonian Partial Differential Equations
According to Bridges (1997), these equations can be written
Kzt + Lzx = ∇zS(z)
K and L are skew-symmetric matrices
z = z(x, t) is the vector of state variables
S is smooth
Examples include KdV, Boussinesq, Zakharov-Kuznetsov,nonlinear Schrödinger, nonlinear wave equations, etc.
Energy and momentum conservation laws are derived directlyfrom this equation
Sep 2010 – p.10/37
Conservation Laws
Kzt + Lzx = ∇zS(z)
Take the inner product of the equation with zt to get theEnergy Conservation law
Et + Fx = 0
for E = S(z) + (zTx Lz)/2 and F = (zT
Lzt)/2
Sep 2010 – p.11/37
Conservation Laws
Kzt + Lzx = ∇zS(z)
Take the inner product of the equation with zt to get theEnergy Conservation law
Et + Fx = 0
for E = S(z) + (zTx Lz)/2 and F = (zT
Lzt)/2
Take the inner product of the equation with zx to get theMomentum Conservation Law
Gx + It = 0
for G = S(z) + (zTt Kz)/2 and I = (zT
Kzx)/2
Sep 2010 – p.11/37
Spatial Discretizations
Implicit midpoint scheme in space gives
Kzn+1/2t + L
zn+1 − zn
h= ∇S(zn+1/2)
with
zn+1/2 =1
2
(
zn+1 + zn)
Symplectic Euler in space gives
Kznt + L+
zn+1 − zn
h+ L−
zn − zn−1
h= ∇S(zn)
withL = L+ + L−, L
T+ = −L−
Sep 2010 – p.12/37
Energy Conservation
Energy conservation laws satisfied exactly by lattice equations
Implicit Midpoint En+1/2t + F n+1
−F n
h = 0 for
En+1/2 = S(zn+1/2)− 1
2
(
zn+1 − zn
h
)T
Lzn+1/2
Fn =1
2(zn)T Lzn
t
Symplectic Euler Ent + F n+1
−F n
h = 0 for
En = S(zn)−(
zn − zn−1
h
)T
L+zn Fn = (zn)TL−z
n−1t
Sep 2010 – p.13/37
Modified Equations Represent the Lattice Equation
Taylor series implies
z(xn+1)− z(xn)
h= zx
(
xn+1/2
)
+h2
222!zxxx
(
xn+1/2
)
+ . . .
along with similar expansions gives the modified equation
Kzt + L
(
zx −h2
3 · 22zxxx + . . .
)
= ∇zS(z),
which can be rewritten
K̃z̃t + L̃z̃x = ∇z̃S̃(z̃).
This equation satisfies a momentum conservation law.
Sep 2010 – p.14/37
Residual in Momentum Conservation Law
h = ∆t = 0.01
0 5 10 15 20 25 30 35 400
0.5
1
1.5
2
t
r m
ρ =0ρ =1ρ =2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
x
r m
ρ =0ρ =1ρ =2
Sep 2010 – p.15/37
A Soliton Collision
Sep 2010 – p.16/37
Residual in Momentum Conservation Law
for a soliton collision, h = 0.02, ∆t = 0.01
0 2 4 6 8 100
2
4
6
r m
0 2 4 6 8 100
0.5
1
1.5
2
r m
0 2 4 6 8 100
2
4
6
x 10−3
t
r m
−10 −5 0 5 100
0.5
1
−10 −5 0 5 100
0.2
0.4
−10 −5 0 5 100
0.5
1
1.5x 10
−3
x
Sep 2010 – p.17/37
3.
Derivation of exact solutions for an LDE
Tony Humphries & Erik Van Vleck
Sep 2010 – p.18/37
Our Nervous System
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Na+
6sodium channel
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Na+
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myelin sheath
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Na+
The nervous cells live inside the “Hot Dog Buns” which arecalled myelin sheath.
The inrush of sodium (Na+) at the sodium channels causes theelectric impulse to jump to the next cell.
Multiple Sclerosis causes the destruction of myelin, which helpscarry electrical signals.
Sep 2010 – p.19/37
Bistable Equation with
Inhomogeneous Diffusion
u̇j = αj(uj+1 − uj)− αj−1(uj − uj−1)− f(uj)
with
αj =
{
αj −m ≤ j ≤ nα j < −m or j > n
m,n ∈ {0} ∪ N
The nonlinearity is the derivative of a double-well potential,
typically f(u) = u(u− a)(u− 1) with a ∈ (0, 1).
Sep 2010 – p.20/37
McKean’s Caricature of the Cubic
0 0.5 1
−0.2
−0.1
0
0.1
0.2
φ
f(φ)
a = 1/2
0 0.5 1
−0.2
−0.1
0
0.1
0.2
φ
f(φ)
a = 3/4
Solid blue line: f(u) = u(u− a)(u− 1)
Dashed red line:
f(u) = u− h(u− a) h(x) =
1 x > 0
[0,1] x = 0
0 x < 0
Sep 2010 – p.21/37
Numerical Simulations
for the Evolution Equation
For the case of a single defect αj =
{
0.6 j = 30
1 j 6= 30
10 20 30 40 50 600
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
j
u j(t)
a = .79
10 20 30 40 50 600
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ju j(t
)
a = .78
A slightly slower wave is stopped by the defect.
Sep 2010 – p.22/37
Steady State Solutions
Definition: The range of a values that yield standing waves iscalled the interval of propagation failure.
Standing waves are solutions of u̇j = 0 or equivalently
αj(uj+1 − uj)− αj−1(uj − uj−1) = f(uj)
limj→∞
uj = 1 limj→−∞
uj = 0
where f(uj) = uj − h(uj − a).
Note: Solutions are not translationally invariant.
Sep 2010 – p.23/37
Families of Standing Waves
Define ξ∗ ∈ R a parameter that determines the position of thewave relative to the defect region.
Sep 2010 – p.24/37
Families of Standing Waves
Define ξ∗ ∈ R a parameter that determines the position of thewave relative to the defect region.
Define j∗ = ⌊ξ∗⌋ so that
uj∗ = a or uj∗ < a < uj∗+1
Sep 2010 – p.24/37
Families of Standing Waves
Define ξ∗ ∈ R a parameter that determines the position of thewave relative to the defect region.
Define j∗ = ⌊ξ∗⌋ so that
uj∗ = a or uj∗ < a < uj∗+1
We seek solutions that satisfy
uj < a for j < ξ∗ and uj > a for j > ξ∗.
Sep 2010 – p.24/37
Families of Standing Waves
Define ξ∗ ∈ R a parameter that determines the position of thewave relative to the defect region.
Define j∗ = ⌊ξ∗⌋ so that
uj∗ = a or uj∗ < a < uj∗+1
We seek solutions that satisfy
uj < a for j < ξ∗ and uj > a for j > ξ∗.
=⇒ f(uj) = uj − h(uj − a) = uj − h(j − ξ∗)
Sep 2010 – p.24/37
Derivation of Solutions
We use Jacobi operator theory (Teschl 2000) to solve thedifference equation
αj(uj+1 − uj)− αj−1(uj − uj−1)− uj = −hj
for
hj =
1 j > j∗
hj∗ j = j∗
0 j < j∗
where
hj∗ =
{
[0,1] ξ∗ = j∗
0 j∗ < ξ∗ < j∗ + 1
Sep 2010 – p.25/37
Standing Wave Solutions
General Solution = Homogeneous Solution + Particular Solution
uj = uj∗ρj + uj∗+1σj +
−∑jk=j∗+1
hk
αkσj−k j > j∗
0 j = j∗
hj∗
αj∗σj−j∗ j < j∗
ρj and σj are fundamental solutions and may be constructedrecursively using
αj(uj+1 − uj)− αj−1(uj − uj−1)− uj = 0
withρj∗ = 1, ρj∗+1 = 0, σj∗ = 0, σj∗+1 = 1.
The particular solution can be found in Teschl (2000).
Sep 2010 – p.26/37
Interval of Propagation Failure
Theorem (Humphries, Moore, Van Vleck 2010)
If a yields a traveling wave for α0 = α, then
Either a ∈ (0, 1/(λ + 2)) or a ∈ ((λ+ 1)/(λ+ 2), 1) withλ = (1 +
√1 + 4α)/2α
There are no corresponding standing waves for α0 < α andξ∗ /∈ (0, 1), nor for α0 > α and ξ∗ ∈ (0, 1).
There exist standing waves for α0 < α and ξ∗ ∈ (0, 1), andfor α0 > α and ξ∗ /∈ (0, 1), provided
a ∈[
α0/α
λ+ 2(α0/α),λ+ α0/α
λ+ 2(α0/α)
]
.
Sep 2010 – p.27/37
Interval of Propagation Failure
α = 1
−3 −2 −1 0 1 2 3 40
0.2
0.4
0.6
0.8
1
−→−→−→
←− ←− ←−
ξ∗
a
α0 = 0.2
Sep 2010 – p.28/37
Interval of Propagation Failure
α = 1
−3 −2 −1 0 1 2 3 40
0.2
0.4
0.6
0.8
1
−→−→
←− ←−
ξ∗
a
α0 = 5
Sep 2010 – p.29/37
Interval of Propagation Failure
α = 1, αdefect = 0.2
−3 −2 −1 0 1 2 30.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
ξ∗
a
1 defect2 defects3 defects
Sep 2010 – p.30/37
Non-zero Wave Speed SolutionsWe make a ‘traveling wave’ ansatz
uj(t) = ϕ(ξj ; ξ∗) ξj =
{
j − cj(t), j ∈ Rj − ct, j /∈ R
and define
R = {j ∈ Z : cj(t) 6= ct},S = R = {j ∈ Z : j ∈ R, j + 1 ∈ R, or j − 1 ∈ R},T = {j ∈ Z : αj 6= α}.
−c′jϕ′(ξj) = αj(ϕ(ξj+1)− ϕ(ξj)) + αj−1(ϕ(ξj−1)− ϕ(ξj))
−ϕ(ξj) + h(ξj − ξ∗)
Sep 2010 – p.31/37
Non-Zero Wave Speed Solution: Part 1
ϕ(ξ; ξ∗) = ψ(ξ; ξ∗) + χ(ξ; ξ∗),
where
ψ(ξ; ξ∗) =1
2+
1
π
∫
∞
0
A(s) sin(s(ξ − ξ∗))s (A(s)2 + c2s2)
ds+c
π
∫
∞
0
cos(s(ξ − ξ∗))A(s)2 + c2s2
ds
forA(s) = 1 + 2α(1− cos(s)),
ψ is the solution to the equation with αj = α, ∀ j ∈ Z .
Sep 2010 – p.32/37
Non-Zero Wave Speed Solution: Part 2
χ(ξ; ξ∗) =∑
j∈R
bjFj(ξ)Bj(ξ∗) + α
∑
j∈S
Fj(ξ)Cj(ξ∗)
+∑
j∈T
γj (Fj(ξ)− Fj+1(ξ))Dj(ξ∗)
Fj(ξ) =1
π
∫
∞
0
A(s) cos(s(ξ − ξj))− cs sin(s(ξ − ξj))A(s)2 + c2s2
ds.
Bj(ξ∗) = αj (ϕ(ξj+1; ξ∗) − ϕ(ξj ; ξ
∗)) + αj−1 (ϕ(ξj−1; ξ∗) − ϕ(ξj ; ξ∗))
−ϕ(ξj ; ξ∗) + h(ξj − ξ∗).
Cj(ξ∗) = ϕ(ξj+1; ξ∗) − ϕ(ξj + 1; ξ∗) + ϕ(ξj−1; ξ∗) − ϕ(ξj − 1; ξ∗),
Dj(ξ∗) = ϕ(ξj+1; ξ∗) − ϕ(ξj ; ξ
∗),
Sep 2010 – p.33/37
Approximation of Wave Speeds
8 10 12 14 16 18 20
0.6
0.7
0.8
0.9
1
1.1
1.2
t
d
α0 = 0.9
α0 = 0.8
α0 = 0.7
Sep 2010 – p.34/37
Computation of Traveling Front
−5 −4 −3 −2 −1 0 1 2 3 4−0.2
−0.1
0
j
χ
d0 = 0.5552 and ξ* = 0.4794
−5 −4 −3 −2 −1 0 1 2 3 40
0.5
1
j
φ
−5 −4 −3 −2 −1 0 1 2 3 40
0.02
0.04
j
|u −
φ|
Sep 2010 – p.35/37
Computation of Traveling Front
−5 −4 −3 −2 −1 0 1 2 3 40
0.01
0.02
0.03
j
χ
d−1
= 1.1392 and ξ* = −0.6029
−5 −4 −3 −2 −1 0 1 2 3 40
0.5
1
j
φ
−5 −4 −3 −2 −1 0 1 2 3 40
0.01
0.02
0.03
j
|u −
φ|
Sep 2010 – p.36/37
Conclusions
Go to seminars. It could change your life.
LDEs are the coolest.
Sep 2010 – p.37/37
