Math 5120
Homework 8 Solutions
1. (a) The general reaction-diffusion equation in polar coordinates is given by
∂~u
D ∂
∂~u
D ∂ 2 ~u
= αf~(~u) +
R
+ 2 2.
∂t
R ∂R
∂R
R ∂φ
(b) To determine the conditions under which the growth of tentacles is possible, we need to
figure out when the spatial modes of the perturbed system will grow with time. Since we
can assume that there is no variation in the R direction, all derivatives in R are zero, leaving
us with the following equation:
D ∂ 2 ~u
∂~u
= αf~(~u) + 2 2 .
∂t
R ∂φ
Now, we want to determine the system governing perturbations to the spatially homogeneous
steady-state solution. Let ~v = ~u − ~u ∗ , where ~u ∗ is the homogeneous solution and ~v is the
perturbation to that solution. Substituting this expression into the reduced system gives a
new equation for changes in ~v :
D ∂ 2~v
∂~v
= αf~(~u ∗ + ~v ) + 2 2 .
∂t
R ∂φ
Linearizing about ~u ∗ (for ~v sufficiently small) gives
∂~v
D ∂ 2~v
= αJ~v + 2 2 ,
∂t
R ∂φ
where J is the matrix defined by
∂fi .
∂uj u~ =~u ∗
To solve this linearized system, separate variables, and assume a solution of the form
~v = f (φ) · g(t) = Fn (φ) exp(σn t)~cn , where Fn (φ) is an eigenfunction of the linear op2
∂
erator L = − ∂φ
2 that satisfies LFn (φ) = λn Fn (φ), subject to the appropariate boundary
conditions. Note that Fn is a function of φ only. Since φ ranges from 0 to 2π and the
problem is on an annulus, we require that the spatial modes be the same at the “ends” of
this interval, making the boundary condition Fn (0) = Fn (2π). We also require that Fn (φ)
have period 2π
n for each n to ensure that there are n peaks for each mode. Since both
sin(qn φ) and cos(qn φ) can satisfy these requirements, either would be acceptable choices for
Fn (φ). Suppose we choose Fn (φ) = cos(qn φ). Since the period must be 2π
n , qn = n, and so
Fn (φ) = cos(nφ). To determine λn , substitute this function into the eigenvalue equation.
The result is λn = n2 . Substituting the generic expression for ~v into the reduced system
yields
σn Fn (φ) exp(σn t)~cn = αFn (φ) exp(σn t)J~cn − λn
⇒
⇒
λn
σn~cn = αJ~cn − 2 D~cn
R n2
σn~cn = αJ − 2 D ~cn ,
a
1
D
Fn (φ) exp(σn t)~cn
R2
where a = R. Lastly, we could solve for σn and nontrivial ~cn by noting that the final equation
2
is an eigenvalue equation for the m × m matrix αJ − na2 D, where σn is a temporal eigenvalue
and ~cn is the corresponding eigenvector. σni are therefore the solutions to the equation
n2
det σn I − αJ + 2 D = 0,
a
for i = 1, ..., m. Further, since {σni } are the temporal eigenvalues corresponding to the nth
spatial mode, instability of perturbations to the spatially homogeneous solution will occur if
at least one σni satisfies ℜ(σni ) > 0 for some n and i. This condition makes the growth of
n tentacles possible.
2. (a) Without diffusion, the equation ~ut = J~ut has a unique steady state of ~u(x) = 0 since
det(J) 6= 0. To determine stability, we look at the eigenvalues of J, which are λ1,2 = ±2i.
Therefore, the steady state is not unstable.
u
= 0, we can approach the problem in
(b) Adding diffusion on the interval (0, π) with ∂~
∂x x=0,π
the same way as in problem 1. Since the original PDE is linear and the spatially homogenous
solution is 0, we can directly study the behavior of small ~u. Let ~u = Fn (x) exp(σn t)~cn ,
where Fn (x) is an eigenfunction satisfying −∇2 Fn (x) = λn Fn (x). To satisfy the assumed
homogeneous
Neumann boundary conditions, we must choose Fn (x) = cos(qn x). Solving
∂Fn =
0,
we get that qn = n. So Fn (x) = cos(nx) and λn = n2 . Substituting
∂x x=0,π
~u = Fn (x) exp(σn t)~cn into the original PDE gives
σn~cn = (J − n2 D)~cn .
Again, we have an eigenvalue equation for σn . An unstable spatial mode n will exist if there
is at least one σni associated with it such that ℜ(σni ) > 0. To find σn , we need to solve
0 = det(J − n2 D − σn I)
3 − n 2 − σn
13
= det
−1
−3 − 9n2 − σn .
√
Then σn1,2 = −5n2 ± 2 4n4 + 6n2 − 1. Note that since n2 > 0 always, an unstable spatial
mode will√only occur for real σn1 > 0. The only spatial mode that satisfies the inequality
−5n2 + 2 4n4 + 6n2 − 1 > 0 is n = 1 .
2