Untangling Complex Systems, page 60
→
Y
Assuming that the concentrations of A and B are kept constant by continuous matter sup-
ply, write the two partial differential equations (PDEs) for [ X] and [ Y] imaging to have
a non-stirred bi-dimensional squared reactor, whose sides have lengths r 1 = r 2 = L.
Nondimensionalize the system of PDEs.
9.12. Find the steady-state solution of the Schnackenberg model without considering diffusion.
Determine the relations among the parameters needed to have the possibility of observing
the formation of Turing patterns in the presence of diffusion.
9.13. The nondimensionalized partial differential equations of the Schnackenberg model are
∂ u
2
2
2
u u
= γ { a − u + u }
v + ∂
2 + ∂
2
∂τ
∂
x
∂
y
∂ v
2
2
2
v
v
= γ b
{ − u v}+ ∂
d 2 +
∂
d 2
∂τ
x
∂
y
∂
Solve them over time, on a square domain with a side included between [0, +1], and in four
conditions that differ only in the values of a and b.
a.
a = 0.07 and b = 1.61;
b.
a = 0.14 and b = 1.34;
c.
a = 0.02 and b = 1.77;
d.
a = 0.1 and b = 1.35
The values of d and γ are fixed to 20 and 10000, respectively. To integrate the partial differ-
ential equations, use the finite difference method with a spatial step size Δ x = Δ y = 0.01. For
more details on the computational method read the paper by Dufiet and Boissonade (1992).
9.14. The formation of the chemical gradients within a fertilized egg is a relevant process in the
embryological development. In fact, the inhomogeneous distribution of a morphogen can
induce the partition of an egg into distinct regions. Let us imagine that a morphogen, for
instance, protein P, is synthesized continuously in the anterior part of the egg (Figure 9.30).
Protein P diffuses into the rest of the egg, but it has a limited lifetime τ = (1 kd ). The kinetic
constant kd refers to a degradation process of P.
The Emergence of Order in Space
295
P
r
FIGURE 9.30 Sketch of a cell wherein protein P is synthesized in anterior cellular portion and diffuses towards the posterior cellular portion.
Which is the spatial distribution of the concentration of P along the spatial coordinate r,
i.e., [ P] = f ( r)?
9.15. This exercise has three main goals. First, we want to observe live chemical waves. Second,
we determine the dependence of the rate of chemical waves on the [H+][BrO−3] product.
Finally, we define the relationship between the period T of the oscillations and [H+][BrO−3]
product. Go to the wet laboratory, wear a white coat, gloves and safety glasses, and prepare
the following solutions using deionized water as the solvent:
• Solution A: KBrO 0.6 M in H SO 0.6 M.
3
2
4
• Solution B: CH (COOH) 0.48 M
2
2
• Solution C: KBr 0.97 M
• Solution D: Ferroin (tris(1,10-phenanthroline) iron(III) sulphate) 0.025 M.
Into a small Erlenmeyer flask, introduce 7 mL of A, 3.5 mL of B and 1 mL of C. Close
the container with a stopper and stir the solution with a magnetic stirrer. Bromate oxidizes
bromide to bromine, and the solution looks brown. The brown color slowly disappears
because bromine reacts with malonic acid to form bromomalonic acid. When the solu-
tion becomes transparent, add 0.5 mL of D and stir. Use a pipet to transfer 2.5 mL of the
mixture into a cuvette having a tiny stir bar and pour the rest into a Petri dish (about 10 cm
in diameter) to cover its surface uniformly. Place the cuvette in a UV-Visible spectropho-
tometer and maintain its solution under stirring. Record how the absorbance at 620 nm
changes over time (every three seconds). Measure the period of the oscillations. Place the
Petri dish on a sheet of millimeter graph paper and leave it quiescent for a while. Then,
chemical waves will appear; measure the change in the radial distance from the center of a
target pattern as the function of time. Also, measure the wavelength, which is the distance
between consecutive fronts. Swirl the Petri dish gently to remove bubbles of CO and after
2
leaving it quiescent for a while, observe the appearance of new patterns. Measure the speed
of the waves again. Does it remain constant? Repeat the experiment using the following
combinations of the solutions:
• 6 mL of A, 1 mL of deionized water, 3.5 mL of B, 1 mL of C, and finally 0.5 mL of D;
• 5 mL of A, 2 mL of deionized water, 3.5 mL of B, 1 mL of C, and finally 0.5 mL of D.
Determine how the speed of the waves and the period of the oscillations depend on the
product [H+][BrO−3]. In other words, determine the exponents, n′ and n″, appearing in the
following two relations:
n′
v ∝
( +
−
H BrO
3 )
n′′
T ∝
( +
−
H
BrO3 )
296
Untangling Complex Systems
According to Field and Noyes (1974b), the wave velocity v ≥ kD +
−
(
[
][
]) /
4
1 2
H
BrO3
, where
D = 2× −
−
(
10 5
2
1
cm s ) is the diffusion coefficient of HBrO and
2
k is the kinetic constant of
the autocatalytic reaction, i.e.,
d[HBrO2] = k[H+][BrO−3][HBrO2]
dt
Assuming that v = kD +
−
(
[
][
]) /
4
1 2
H
BrO3
, determine k and compare your estimate with the
expected value of 20 M − 2s − 1 (Pojman et al. 1994).
9.16. Use the eikonal equation [9.46] to estimate the size of a spiral wave, knowing that the
velocity of a front wave is v
p = 0.0120 cm/s and D = 2 × 10−5 cm2 s−1.
9.17. A fire that spreads in a forest can be assumed to be an example of chemical waves. Why?
Is it a trigger or a phase wave? Which is the autocatalyst? How long is the refractory
period?
9.18. Compare the properties of chemical waves with those of physical waves, like light and
sound. Which features are peculiar to chemical waves?
9.19. In this exercise that requires roughly three weeks to be completed, you will observe the
phenomenon of periodic precipitation for Ag CrO in gelatin. The Liesegang patterns take
2
4
many days to form because ions must diffuse over long distances. Go to your wet labora-
tory, wear a white coat, gloves, and safety glasses. Dissolve 0.2 g of K Cr O in 200 mL of
2
2
7
distilled or deionized water, within a beaker. The solution appears yellow. Then, add 8 g
of gelatin and dissolve it by stirring and heating up to 50°C. Pour the hot solution into two
or more Petri dishes generating thin layers of gels, and within test tubes filling them up to
3 cm from the top.
Cover the Petri dishes with glass plates and the test tubes with parafilm. Let all of
them stand for, at least, 24 h at room temperature. Meanwhile, prepare two solutions of
AgNO at different concentrations. One by dissolving 1 g of AgNO in 10 mL of distilled or
3
3
deionized water; the other by dissolving 0.5 g of AgNO in 10 mL of distilled or deionized
3
water. After one day at ~25°C, the gel is formed.
Place a few drops of silver nitrate solution on the surface of the gels formed in Petri
dishes and cover them with glass plates to slow down evaporation.
Place 5 mL of the silver nitrate solutions on top of the gels contained in large test tubes
and 2 mL on top of small test tubes. Cover the test tubes, again, with parafilm. Let the
silver nitrate to diffuse within the gels.
Monitor the number of rings as a function of time, their spacing and the speed of spread-
ing of the most advanced ring.
Which is the effect of the AgNO concentration on the properties of the patterns? Are
3
the patterns easily reproducible? Does the size of the test tube affect the features of the
patterns?
Compare the Liesegang patterns with the transient structures obtained with the BZ
reaction in exercise 9.1 and 9.15. Which are the similarities and differences? Do we con-
firm the plausible presence of an autocatalytic process in the Liesegang phenomenon?
Verify the time law and apply equation [9.10] to determine the diffusion coefficient of
silver ions within the gel and the kinetic constant of the autocatalytic process of nuclei
association.
9.15 SOLUTIONS TO THE EXERCISES
9.1. (a) Soon after the addition of component E, the solution wherein there is the redox indica-
tor becomes blue. Quickly, it turns to red. Then, blue stripes originate spontaneously at
the bottom of the colored part of the solution (see Figure 9.31 that is in black and white).
The Emergence of Order in Space
297
Space
11 cm
0
Time
FIGURE 9.31 The spontaneous emergence of a spatial structure inside a test tube wherein the BZ reaction
takes place in unstirred conditions.
The stripes move slowly upward. After more than 30’, the colored portion of the solution
becomes covered by thin blue stripes over a red background (see Figure 9.31).
(b) The wavelength of the spatially periodic structure can be determined by using
a graph paper. The wavelength of the pattern corresponds to the distance between two
adjacent stripes. It is evident that the wavelength of the wave train is not constant, and it
increases moving towards the top of the solution. Figure 9.32 shows a schematic quantita-
tive representation of the pattern that can be obtained.
If we consider the following first five stripes, the average value of the wavelength is
λ a = 0 1
. 7
.
cm
(c) By using equation [9.10], we can determine the kinetic constant k of the autocatalytic
process, responsible for the periodic structure. We obtain
2.0
1.8
1.6
1.4
1.2
m)(c 1.0
Height 0.8
0.6
0.4
0.2
0.0
FIGURE 9.32 Schematic structure of the pattern shown in Figure 9.31.
298
Untangling Complex Systems
2
D
2
π
2 10 5
2
1
π
cm s
k
M
=
0 3
. 2
1
1
M s
b 2 =
× −
−
4
4
λ
(0.35 M )(0.002 L
=
− −
)
2
a
(0 1
. 7cm)
(0.0082 L)
(d) If we consider the formula [9.10], we expect that by reducing b, the wavelength of the
periodic structure increases. This relation is what we see by performing the experiment.
Since the concentration of KBrO is ten times smaller, the speed of the formation of the
3
periodic structure is definitely much smaller than before. It takes a lot of time to observe
the final structure. The final pattern has a wavelength λ that is more than three times lon-
d
ger according to the ratio λ d
( b) a
=
=
λ
10.
a
( b) d
9.2. The results of the average rates are listed in Table 9.3.
When we have perfect mixing and uniform distribution of u (it is the case represented by
trace 1), the rate is independent of x. On the other hand, when the mixing is imperfect, and
we have a gradient, the average value of the rate maintains equal to 1 only in the case of a
linear kinetic law. From the other values of the average rate, we also notice that the higher
the non-linearity of the kinetic law, the greater the difference between the uniform and
non-uniform distribution of u along x. This result means that non-linear reactions amplify
the effect of imperfect mixing.
9.3. If we think of equation [9.20] as a quadratic equation in K 2 of the form a( K 2)2 bK 2
+
+ c,
we calculate its derivative with respect to K 2 and we impose it equal to zero, to find its
minimum.
d ( det ( J′))
a K 2
2
b
( ) = ( )+ = 0
2
d K
It derives that
K 2 = −( b 2 a), i.e., K 2 = ( D ( )
)
X
Ry Y
D R X
2 D D
+
( )
. The latter coincides
s
Y
x
s
X
Y
with equation [9.24].
9.4. Two intuitive examples of how a “local self-activation and lateral inhibition” can give rise to
Turing structures have been proposed by Murray (1988, 2003). In both cases, the scenario
is that of dry forest, and the activator is a fire that tends to spread throughout the forest.
The inhibitor may be a team of firefighters (Murray 1988) or a large group of grasshoppers
(Murray 2003). The firefighters equipped with helicopters can disperse a fire-retardant
spray, whereas the grasshoppers can generate a lot of moisture by sweating when they
get warm. If the fire moves too quickly, then the outcome will be a uniform charred area.
TABLE 9.3
Average Rates for Different Kinetic Laws and the Three
Situations of u versus x Shown in Figure 9.29
Kinetic Law
Trace 1
Trace 2
Trace 3
u
1
1
1
u 2
1
1.4
2
u 3
1
2.2
4
e( u−1)
1
1.2
1.54
10( u 1−)
1
2.4
5.05
The Emergence of Order in Space
299
However, when the firefighters travel faster, or the grasshoppers sweat profusely and gen-
erate enough moisture, they prevent the fire spreading everywhere. Eventually, a stable
pattern of charred black and uncharred green regions will be established.
Other examples of “local self-activation and lateral inhibition” may be found in societ-
ies, ecosystems, et cetera. R ( )
x X
R X
( )
α
1
9.5. The Jacobian matrix is J
s
y
s
=
=
.
R
γ
β
( )
x Y
R Y
( )
s
y
s
The four conditions for having a Turing pattern are:
I.
α + β < 0;
II.
αβ − γ > 0;
III.
δ Dβ +αδ > 0;
IV.
δ Dβ +αδ > 2
δ
D 2 (αβ −γ )
1/2
1 α
β
The critical wavenumber is K =
+
.
2 Dδ
δ
Reminding that the wavelength is λ = ( π
2 K), the number of waves in the domain of
length L is given by the ratio ( L λ ) = ( LK π
2 ).
9.6. Based on the parameters listed in the text of the exercise, the differential equations become
∂ x
2
2
x
= α x(1− r 1 y ) + y + ( Dδ ) ∂
∂
Y
Assuming that the concentrations of A and B are kept constant by continuous matter sup-
ply, write the two partial differential equations (PDEs) for [ X] and [ Y] imaging to have
a non-stirred bi-dimensional squared reactor, whose sides have lengths r 1 = r 2 = L.
Nondimensionalize the system of PDEs.
9.12. Find the steady-state solution of the Schnackenberg model without considering diffusion.
Determine the relations among the parameters needed to have the possibility of observing
the formation of Turing patterns in the presence of diffusion.
9.13. The nondimensionalized partial differential equations of the Schnackenberg model are
∂ u
2
2
2
u u
= γ { a − u + u }
v + ∂
2 + ∂
2
∂τ
∂
x
∂
y
∂ v
2
2
2
v
v
= γ b
{ − u v}+ ∂
d 2 +
∂
d 2
∂τ
x
∂
y
∂
Solve them over time, on a square domain with a side included between [0, +1], and in four
conditions that differ only in the values of a and b.
a.
a = 0.07 and b = 1.61;
b.
a = 0.14 and b = 1.34;
c.
a = 0.02 and b = 1.77;
d.
a = 0.1 and b = 1.35
The values of d and γ are fixed to 20 and 10000, respectively. To integrate the partial differ-
ential equations, use the finite difference method with a spatial step size Δ x = Δ y = 0.01. For
more details on the computational method read the paper by Dufiet and Boissonade (1992).
9.14. The formation of the chemical gradients within a fertilized egg is a relevant process in the
embryological development. In fact, the inhomogeneous distribution of a morphogen can
induce the partition of an egg into distinct regions. Let us imagine that a morphogen, for
instance, protein P, is synthesized continuously in the anterior part of the egg (Figure 9.30).
Protein P diffuses into the rest of the egg, but it has a limited lifetime τ = (1 kd ). The kinetic
constant kd refers to a degradation process of P.
The Emergence of Order in Space
295
P
r
FIGURE 9.30 Sketch of a cell wherein protein P is synthesized in anterior cellular portion and diffuses towards the posterior cellular portion.
Which is the spatial distribution of the concentration of P along the spatial coordinate r,
i.e., [ P] = f ( r)?
9.15. This exercise has three main goals. First, we want to observe live chemical waves. Second,
we determine the dependence of the rate of chemical waves on the [H+][BrO−3] product.
Finally, we define the relationship between the period T of the oscillations and [H+][BrO−3]
product. Go to the wet laboratory, wear a white coat, gloves and safety glasses, and prepare
the following solutions using deionized water as the solvent:
• Solution A: KBrO 0.6 M in H SO 0.6 M.
3
2
4
• Solution B: CH (COOH) 0.48 M
2
2
• Solution C: KBr 0.97 M
• Solution D: Ferroin (tris(1,10-phenanthroline) iron(III) sulphate) 0.025 M.
Into a small Erlenmeyer flask, introduce 7 mL of A, 3.5 mL of B and 1 mL of C. Close
the container with a stopper and stir the solution with a magnetic stirrer. Bromate oxidizes
bromide to bromine, and the solution looks brown. The brown color slowly disappears
because bromine reacts with malonic acid to form bromomalonic acid. When the solu-
tion becomes transparent, add 0.5 mL of D and stir. Use a pipet to transfer 2.5 mL of the
mixture into a cuvette having a tiny stir bar and pour the rest into a Petri dish (about 10 cm
in diameter) to cover its surface uniformly. Place the cuvette in a UV-Visible spectropho-
tometer and maintain its solution under stirring. Record how the absorbance at 620 nm
changes over time (every three seconds). Measure the period of the oscillations. Place the
Petri dish on a sheet of millimeter graph paper and leave it quiescent for a while. Then,
chemical waves will appear; measure the change in the radial distance from the center of a
target pattern as the function of time. Also, measure the wavelength, which is the distance
between consecutive fronts. Swirl the Petri dish gently to remove bubbles of CO and after
2
leaving it quiescent for a while, observe the appearance of new patterns. Measure the speed
of the waves again. Does it remain constant? Repeat the experiment using the following
combinations of the solutions:
• 6 mL of A, 1 mL of deionized water, 3.5 mL of B, 1 mL of C, and finally 0.5 mL of D;
• 5 mL of A, 2 mL of deionized water, 3.5 mL of B, 1 mL of C, and finally 0.5 mL of D.
Determine how the speed of the waves and the period of the oscillations depend on the
product [H+][BrO−3]. In other words, determine the exponents, n′ and n″, appearing in the
following two relations:
n′
v ∝
( +
−
H BrO
3 )
n′′
T ∝
( +
−
H
BrO3 )
296
Untangling Complex Systems
According to Field and Noyes (1974b), the wave velocity v ≥ kD +
−
(
[
][
]) /
4
1 2
H
BrO3
, where
D = 2× −
−
(
10 5
2
1
cm s ) is the diffusion coefficient of HBrO and
2
k is the kinetic constant of
the autocatalytic reaction, i.e.,
d[HBrO2] = k[H+][BrO−3][HBrO2]
dt
Assuming that v = kD +
−
(
[
][
]) /
4
1 2
H
BrO3
, determine k and compare your estimate with the
expected value of 20 M − 2s − 1 (Pojman et al. 1994).
9.16. Use the eikonal equation [9.46] to estimate the size of a spiral wave, knowing that the
velocity of a front wave is v
p = 0.0120 cm/s and D = 2 × 10−5 cm2 s−1.
9.17. A fire that spreads in a forest can be assumed to be an example of chemical waves. Why?
Is it a trigger or a phase wave? Which is the autocatalyst? How long is the refractory
period?
9.18. Compare the properties of chemical waves with those of physical waves, like light and
sound. Which features are peculiar to chemical waves?
9.19. In this exercise that requires roughly three weeks to be completed, you will observe the
phenomenon of periodic precipitation for Ag CrO in gelatin. The Liesegang patterns take
2
4
many days to form because ions must diffuse over long distances. Go to your wet labora-
tory, wear a white coat, gloves, and safety glasses. Dissolve 0.2 g of K Cr O in 200 mL of
2
2
7
distilled or deionized water, within a beaker. The solution appears yellow. Then, add 8 g
of gelatin and dissolve it by stirring and heating up to 50°C. Pour the hot solution into two
or more Petri dishes generating thin layers of gels, and within test tubes filling them up to
3 cm from the top.
Cover the Petri dishes with glass plates and the test tubes with parafilm. Let all of
them stand for, at least, 24 h at room temperature. Meanwhile, prepare two solutions of
AgNO at different concentrations. One by dissolving 1 g of AgNO in 10 mL of distilled or
3
3
deionized water; the other by dissolving 0.5 g of AgNO in 10 mL of distilled or deionized
3
water. After one day at ~25°C, the gel is formed.
Place a few drops of silver nitrate solution on the surface of the gels formed in Petri
dishes and cover them with glass plates to slow down evaporation.
Place 5 mL of the silver nitrate solutions on top of the gels contained in large test tubes
and 2 mL on top of small test tubes. Cover the test tubes, again, with parafilm. Let the
silver nitrate to diffuse within the gels.
Monitor the number of rings as a function of time, their spacing and the speed of spread-
ing of the most advanced ring.
Which is the effect of the AgNO concentration on the properties of the patterns? Are
3
the patterns easily reproducible? Does the size of the test tube affect the features of the
patterns?
Compare the Liesegang patterns with the transient structures obtained with the BZ
reaction in exercise 9.1 and 9.15. Which are the similarities and differences? Do we con-
firm the plausible presence of an autocatalytic process in the Liesegang phenomenon?
Verify the time law and apply equation [9.10] to determine the diffusion coefficient of
silver ions within the gel and the kinetic constant of the autocatalytic process of nuclei
association.
9.15 SOLUTIONS TO THE EXERCISES
9.1. (a) Soon after the addition of component E, the solution wherein there is the redox indica-
tor becomes blue. Quickly, it turns to red. Then, blue stripes originate spontaneously at
the bottom of the colored part of the solution (see Figure 9.31 that is in black and white).
The Emergence of Order in Space
297
Space
11 cm
0
Time
FIGURE 9.31 The spontaneous emergence of a spatial structure inside a test tube wherein the BZ reaction
takes place in unstirred conditions.
The stripes move slowly upward. After more than 30’, the colored portion of the solution
becomes covered by thin blue stripes over a red background (see Figure 9.31).
(b) The wavelength of the spatially periodic structure can be determined by using
a graph paper. The wavelength of the pattern corresponds to the distance between two
adjacent stripes. It is evident that the wavelength of the wave train is not constant, and it
increases moving towards the top of the solution. Figure 9.32 shows a schematic quantita-
tive representation of the pattern that can be obtained.
If we consider the following first five stripes, the average value of the wavelength is
λ a = 0 1
. 7
.
cm
(c) By using equation [9.10], we can determine the kinetic constant k of the autocatalytic
process, responsible for the periodic structure. We obtain
2.0
1.8
1.6
1.4
1.2
m)(c 1.0
Height 0.8
0.6
0.4
0.2
0.0
FIGURE 9.32 Schematic structure of the pattern shown in Figure 9.31.
298
Untangling Complex Systems
2
D
2
π
2 10 5
2
1
π
cm s
k
M
=
0 3
. 2
1
1
M s
b 2 =
× −
−
4
4
λ
(0.35 M )(0.002 L
=
− −
)
2
a
(0 1
. 7cm)
(0.0082 L)
(d) If we consider the formula [9.10], we expect that by reducing b, the wavelength of the
periodic structure increases. This relation is what we see by performing the experiment.
Since the concentration of KBrO is ten times smaller, the speed of the formation of the
3
periodic structure is definitely much smaller than before. It takes a lot of time to observe
the final structure. The final pattern has a wavelength λ that is more than three times lon-
d
ger according to the ratio λ d
( b) a
=
=
λ
10.
a
( b) d
9.2. The results of the average rates are listed in Table 9.3.
When we have perfect mixing and uniform distribution of u (it is the case represented by
trace 1), the rate is independent of x. On the other hand, when the mixing is imperfect, and
we have a gradient, the average value of the rate maintains equal to 1 only in the case of a
linear kinetic law. From the other values of the average rate, we also notice that the higher
the non-linearity of the kinetic law, the greater the difference between the uniform and
non-uniform distribution of u along x. This result means that non-linear reactions amplify
the effect of imperfect mixing.
9.3. If we think of equation [9.20] as a quadratic equation in K 2 of the form a( K 2)2 bK 2
+
+ c,
we calculate its derivative with respect to K 2 and we impose it equal to zero, to find its
minimum.
d ( det ( J′))
a K 2
2
b
( ) = ( )+ = 0
2
d K
It derives that
K 2 = −( b 2 a), i.e., K 2 = ( D ( )
)
X
Ry Y
D R X
2 D D
+
( )
. The latter coincides
s
Y
x
s
X
Y
with equation [9.24].
9.4. Two intuitive examples of how a “local self-activation and lateral inhibition” can give rise to
Turing structures have been proposed by Murray (1988, 2003). In both cases, the scenario
is that of dry forest, and the activator is a fire that tends to spread throughout the forest.
The inhibitor may be a team of firefighters (Murray 1988) or a large group of grasshoppers
(Murray 2003). The firefighters equipped with helicopters can disperse a fire-retardant
spray, whereas the grasshoppers can generate a lot of moisture by sweating when they
get warm. If the fire moves too quickly, then the outcome will be a uniform charred area.
TABLE 9.3
Average Rates for Different Kinetic Laws and the Three
Situations of u versus x Shown in Figure 9.29
Kinetic Law
Trace 1
Trace 2
Trace 3
u
1
1
1
u 2
1
1.4
2
u 3
1
2.2
4
e( u−1)
1
1.2
1.54
10( u 1−)
1
2.4
5.05
The Emergence of Order in Space
299
However, when the firefighters travel faster, or the grasshoppers sweat profusely and gen-
erate enough moisture, they prevent the fire spreading everywhere. Eventually, a stable
pattern of charred black and uncharred green regions will be established.
Other examples of “local self-activation and lateral inhibition” may be found in societ-
ies, ecosystems, et cetera. R ( )
x X
R X
( )
α
1
9.5. The Jacobian matrix is J
s
y
s
=
=
.
R
γ
β
( )
x Y
R Y
( )
s
y
s
The four conditions for having a Turing pattern are:
I.
α + β < 0;
II.
αβ − γ > 0;
III.
δ Dβ +αδ > 0;
IV.
δ Dβ +αδ > 2
δ
D 2 (αβ −γ )
1/2
1 α
β
The critical wavenumber is K =
+
.
2 Dδ
δ
Reminding that the wavelength is λ = ( π
2 K), the number of waves in the domain of
length L is given by the ratio ( L λ ) = ( LK π
2 ).
9.6. Based on the parameters listed in the text of the exercise, the differential equations become
∂ x
2
2
x
= α x(1− r 1 y ) + y + ( Dδ ) ∂
∂
