Traveling Wave Solutions in a Reaction-Diffusion Model for Criminal ...

Lemma 4 (Base Case). There exists a monotonically decreasing solution, sL0 (x), to (5) with
L = L0, such that
lim sL0 (x) = 1 lim sL0 (x) = 0.
x→−∞ x→∞
Proof. We prove this lemma in three steps. The first step is to construct a suitable supersolution
and subsolution. We then build a solutions by using the supersolution and subsolution as
barriers. The first solution constructed is not necessarily monotone, but it is bounded above by
b for all x > L. Using this upper bound we prove there also exists monotone solution that is
below the first solution we constructed.
Step 1: (Super/subsolution) Let ψ(x) be the unique solution to Cauchy problem
−u ′′ = f(u) x ∈ [L0, ∞)
u(L0) = b, u ′ (L0) = 0.
Such a solution exists and in fact following the arguments from the beginning of §3.4 we can
show that ψ ′ (x) < 0 and limx→∞ ψ(x) = 0. Now, define φ+(x) by
⎧
⎨ 1
x ≤ 0
b L0 φ+(x) =
⎩ 2 e e−x −L0 + e ex 0 ≤ x ≤ L0
ψ(x) x > L0.
The function φ+(x) is a supersolution as it satisfies
lim
x→ξ− φ′ +(x) ≥ lim
x→ξ + φ′ +(x),
for ξ = 0 and ξ = L0. Similarly, we build a subsolution, φ−(x). Let ψ− now be the solution to
the Cauchy problem
−u ′′ = f(u) x ∈ [−∞, 0)
u(−∞) = 1, u(0) = 0.
As before, we know that such a solution exists, and we define the subsolution
ψ−(x) x < 0
φ− =
0 x > 0.
Furthermore, s ′
(0) = − 2 1
0 f(θ)d θ, which is bounded away from zero and so the condition
on the derivative at x = 0 for a subsolution is satisfied.
Step 2: (Barrier Method) Having constructed a supersolution and a subsolution we use the
standard barrier method (see for example [18]) to show that there exists a solution to (23). This
steps follow those of the proof existence of the entire solutions in Theorem 3 (see Step 3 of the
proof), so we leave out the details.
Step 3: If the solution is monotone then we are done. If it is not monotone then the lack of
monotonicity happens in the gap. Indeed, for x > L we know that sL(x) ≤ b and then
− 1
2 (s′ L(x)) 2 =
sL(x)
0
f(θ) dθ < 0.
Since, the solution must be concave up in the gap then it reaches a minimum at a unique
point x ⋆ ∈ (0, L). If sL(x ⋆ ) ≥ a we construct a supersolution by letting s+(x) = sL(x) for x ≤ x ⋆ ,
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