

Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 76, pp. 1–21.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

TRAVELING WAVES FOR UNBALANCED BISTABLE

EQUATIONS WITH DENSITY DEPENDENT DIFFUSION

PAVEL DRÁBEK, MICHAELA ZAHRADNÍKOVÁ

Abstract. We study the existence and qualitative properties of traveling

wave solutions for the unbalanced bistable reaction-diffusion equation with

a rather general density dependent diffusion coefficient. In particular, it al-
lows for singularities and/or degenerations as well as discontinuities of the first

kind at a finite number of points. The reaction term vanishes at equilibria and

it is a continuous, possibly non-Lipschitz function. We prove the existence
of a unique speed of propagation and a unique traveling wave profile (up to

translation) which is a non-smooth function in general. In the case of the

power-type behavior of the diffusion and reaction near equilibria we provide
detailed asymptotic analysis of the profile.

1. Introduction

We are concerned with the traveling wave solutions of the bistable equation

∂u

∂t
=

∂

∂x

(
d(u)

∣∣∂u
∂x

∣∣p−2 ∂u

∂x

)
+ g(u), (x, t) ∈ R× R+,

u(x, t) = U(x− ct)
(1.1)

for a speed of propagation c ∈ R. Here R+ := [0,+∞), 1 < p < +∞ and the
properties of the density dependent diffusion coefficient d = d(s) as well as the
reaction term g = g(s) will be specified later.

If p = 2, d ≡ 1 and g : [0, 1]→ R is a smooth function such that g(0) = g(s∗) =
g(1) = 0, g(s) < 0 for s ∈ (0, s∗), g(s) > 0 for s ∈ (s∗, 1), equation (1.1) is studied,
e.g., in [1, 2]. The authors of these articles explain how the mathematical modeling
of diploid individuals (homozygote and heterozygote) in population dynamics leads
to the bistable equation. If in addition the reaction term g satisfies the unbalanced
bistable condition ∫ 1

0

g(s) ds > 0, (1.2)

the mathematical model describes the so called heterozygote inferior case. If
g(s) ≤ 0 instead of g(s) < 0 for s ∈ (0, s∗), the bistable equation models the
flame propagation in chemical reactor theory. In contrast with the population dy-
namics model, where u denotes the relative density of the population of one allele,
in the combustion model u represents a normalized temperature and s∗ represents

2010 Mathematics Subject Classification. 35Q92, 35C07, 34A12, 35K92.
Key words and phrases. Density dependent diffusion; unbalanced bistable reaction term;

degenerate and singular diffusion; traveling wave; degenerate non-Lipschitz reaction.
c©2021. This work is licensed under a CC BY 4.0 license.

Submitted August 22, 2021. Published September 14, 2021.

1


2 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

a critical temperature at which an exothermic reaction starts (see, e.g., [10]). The
bistable equation with reaction term like above was also suggested in [4] as a model
for a nerve which has been treated with certain toxins. In [11] this equation serves
as a model for a bistable active transmission line. Other possible interpretations
may be found in [13].

In this paper we focus on more general, in particular non-smooth (even non-
Lipschitz) reaction term g as well as on the density dependent diffusion coefficient
d which can be singular and/or degenerate near the equilibria 0 and 1 and discon-
tinuous in a finite number of points in the interval (0, 1). We show the existence
of a traveling wave solution if the classical unbalanced bistable condition (1.2) is
replaced by the following more general one:∫ 1

0

(
d(s)

) 1
p−1 g(s) ds > 0. (1.3)

In our previous work [7] we proved that if equality holds in (1.3) then (1.1) possesses
nonconstant stationary solutions (also called standing wave solutions) “connecting”
the equilibria 0 and 1. We also studied qualitative properties of these stationary
solutions, in particular, the lack of smoothness and the asymptotic properties near
the equilibria 0 and 1. In contrast with the stationary case, inequality (1.3) renders
the time dependent traveling wave solutions of the form

u(x, t) = U(x− ct),
where U = U(z), z ∈ R, is the profile of the traveling wave and c ∈ R is the speed
of its propagation. Density dependent diffusion coefficient which is discontinuous
is studied, e.g., in [12] to model the temperature field in a wire of superconducting
material carrying an electrical current and immersed in a bath at constant temper-
ature. Convergence of the solution of the initial value problem for equation in (1.1)
with d ≡ 1 to a traveling wave is studied, e.g., in [9] for C1 reaction term g or in
[6] for reaction term g which is only one-sided Lipschitz continuous.

Our results concerning the existence and uniqueness of the profile U and speed
c extend and generalize those from [1, Theorem 4.2], [2, Theorem 4.1], [5, Theorem
3.1]. The asymptotic analysis of U near 0 and 1 extends that from [5, Section 6].

2. Preliminaries

Let g : [0, 1]→ R, g ∈ C[0, 1] be such that g(0) = g(s∗) = g(1) = 0 for s∗ ∈ (0, 1)
and

g(s) ≤ 0, s ∈ (0, s∗), g(s) > 0, s ∈ (s∗, 1).

The diffusion coefficient d : [0, 1] → R is supposed to be nonnegative lower semi-
continuous and d > 0 in (0, 1). There exist 0 = s0 < s1 < s2 < · · · < sn < sn+1 = 1
such that d

∣∣
(si,si+1)

∈ C(si, si+1), i = 0, . . . , n, and d has discontinuity of the first

kind (finite jump) at si, i = 1, . . . , n.
We introduce the moving coordinate z = x− ct and write u(x, t) = U(x− ct) =

U(z). For the sake of simplicity we write (·)′ instead of d
dz (·). Then (1.1) transforms

into (
d(U(z))|U ′(z)|p−2U ′(z)

)′
+ cU ′(z) + g(U(z)) = 0 . (2.1)

Let U : R→ [0, 1] be a monotone continuous function. We denote

MU := {z ∈ R : U(z) = si, i = 1, 2, . . . , n},


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 3

NU := {z ∈ R : U(z) = 0 or U(z) = 1}.
Then MU and NU are closed sets, MU is a union of a finite number of points or
intervals,

NU = (−∞, z0] ∪ [z1,+∞),

where −∞ ≤ z0 < z1 ≤ +∞ and we use the convention (−∞, z0] = ∅ if z0 = −∞
and [z1,+∞) = ∅ if z1 = +∞. Below we introduce the definition of a monotone
solution of (2.1).

Definition 2.1. A monotone continuous function U : R→ [0, 1] is called a solution
of (2.1) if

(a) For any z /∈ MU ∪ NU the derivative U ′(z) exists and it is finite. For
z ∈ intMU ∪ intNU we have U ′(z) = 0.

(b) For any z ∈ ∂MU there exist finite one sided derivatives U ′(z−), U ′(z+)
and

L(z) := |U ′(z−)|p−2U ′(z−) lim
ξ→z−

d(U(ξ)) = |U ′(z+)|p−2U ′(z+) lim
ξ→z+

d(U(ξ)).

(c) Function v : R→ R defined by

v(z) :=


d(U(z))|U ′(z)|p−2U ′(z), z /∈MU ∪NU ,
0, z ∈ NU ∪ intMU ,

L(z), z ∈ ∂MU

is continuous and for any z, ẑ ∈ R,

v(ẑ)− v(z) + c (U(ẑ)− U(z)) +

∫ ẑ

z

g(U(ξ)) dξ = 0 . (2.2)

Moreover, limz→±∞ v(z) = 0 if either
limz→−∞ U(z) = 1 and limz→+∞ U(z) = 0, or
limz→−∞ U(z) = 0 and limz→+∞ U(z) = 1.

Remark 2.2. Constant functions U ≡ k, where k is such that g(k) = 0, are
solutions of (2.1). In particular, U ≡ 0, U ≡ 1 and U ≡ s∗ are solutions.

Remark 2.3. Let z /∈MU ∪NU , ẑ = z+h, h 6= 0. Divide (2.2) by h and let h→ 0.
Then, by Definition 2.1 (a), the derivative U ′(z) exists and

v′(z) + cU ′(z) + g(U(z)) = 0. (2.3)

In particular, v is differentiable in z.

Remark 2.4. Let U be a solution of (2.1) in the sense of Definition 2.1 such
that either U(z) → 1 as z → −∞ and U(z) → 0 as z → +∞ or U(z) → 0 as
z → −∞ and U(z) → 1 as z → +∞. If d is not continuous in (0, 1) then MU 6= ∅
and either MU = ∂MU (i.e., intMU = ∅) or else intMU 6= ∅. In the former case
MU = {ξ1, . . . , ξn}, where U(ξi) = si, i = 1, . . . , n. In the latter case there exist
1 ≤ k ≤ n and 1 ≤ i1 < i2 < · · · < ik ≤ n such that U ≡ sij on some interval
[aij , bij ], aij < bij , j = 1, . . . , k, and intMU = ∪nj=1(aij , bij ). The equation (2.1) is
satisfied pointwise in intMU and it follows from the continuity of v that if aij > −∞
or bij < +∞ we have U ′(aij ) = 0 or U ′(bij ) = 0, j = 1, . . . , k, respectively, because
d(s) > 0, s ∈ (0, 1).

If z0 > −∞ then U ′(z0−) = 0 and U ′(z0+) exists finite or infinite. Similarly, if
z1 < +∞ then U ′(z1+) = 0 and U ′(z1−) exists finite or infinite.


4 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

Remark 2.5. Let p = 2, d ≡ 1 and g ∈ C1[0, 1]. Let U = U(z) be a solution in
the sense of Definition 2.1. Then MU = ∅, NU = ∅, and (2.1) holds pointwise, i.e.,
U ∈ C2(R) and it is a classical solution. For more general d we have to employ the
first integral (2.2) because of the lack of differentiability of a solution U .

3. Equivalent first order ODE

We will look for monotone traveling waves U = U(z) satisfying boundary condi-
tions

lim
z→−∞

U(z) = 1 and lim
z→+∞

U(z) = 0. (3.1)

Let U : R → [0, 1] be a monotone nonincreasing solution of the BVP (2.1),
(3.1) such that U is strictly decreasing at any point z ∈ R where U(z) ∈ (0, 1).
Then there exist −∞ ≤ z0 < z1 ≤ +∞ such that U(z) = 1, z ∈ (−∞, z0] and
U(z) = 0, z ∈ [z1,+∞). Moreover, MU = {ξ1, ξ2, . . . , ξn} where U(ξi) = si, i =
1, 2, . . . , n. In particular, intMU = ∅ and MU = ∂MU , see Remark 2.4. For
all z /∈ MU ∪ NU we have U ′(z) < 0 and for all z ∈ MU we have U ′(z−) < 0
and U ′(z+) < 0. The function U is continuous and piecewise C1 in the sense
that U

∣∣
(ξi,ξi+1)

∈ C1(ξi, ξi+1). Therefore, there exists strictly decreasing inverse

function U−1 : (0, 1)→ (z0, z1), z = U−1(U), such that U−1
∣∣
(si,si+1)

∈ C1(si, si+1),

i = 0, 1, . . . , n and the limits

lim
U→si−

d

dU
U−1(U), lim

U→si+

d

dU
U−1(U)

exist and are finite for i = 1, 2, . . . , n. Hence, we make the change of variables (cf.
[7, 8])

w(U) = v(U−1(U)), U ∈ (0, 1). (3.2)

It follows from Remark 2.3 that w = w(U) is a piecewise C1-function in (0, 1),

w
∣∣
(si,si+1)

∈ C1(si, si+1), i = 0, 1, . . . , n,

with finite limits limU→si− w
′(U), limU→si+ w

′(U), i = 1, 2, . . . , n. Therefore, for
any z ∈ (ξi, ξi+1) and U ∈ (si, si+1), i = 0, 1, . . . , n, we have

d

dz
v(z) =

d

dz
w(U(z)) =

dw

dU
(U(z))U ′(z). (3.3)

From v(z) = −d(U(z))|U ′(z)|p−1 we deduce that

U ′(z) = −
∣∣∣ v(z)

d(U(z))

∣∣∣p′−1

, p′ =
p

p− 1
. (3.4)

From (3.2), (3.3) and (3.4),

dv

dz
= −dw

dU
(U(z))

∣∣∣ v(z)

d(U(z))

∣∣∣p′−1

= −dw

dU

∣∣∣w(U)

d(U)

∣∣∣p′−1

.

Therefore, equation (2.3), namely

v′(z) + cU ′(z) + g(U(z)) = 0, z ∈ (ξi, ξi+1),

becomes

−dw

dU

∣∣∣w(U)

d(U)

∣∣∣p′−1

− c
∣∣∣w(U)

d(U)

∣∣∣p′−1

+ g(U) = 0, U ∈ (si, si+1),


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 5

i = 0, 1, . . . , n. This is equivalent to

|w|p′−1 dw

dU
= −c|w|p′−1 + (d(U))

p′−1
g(U), (3.5)

or
1

p′
d

dU
|w|p′ = c|w|p′−1 − (d(U))

p′−1
g(U). (3.6)

Set f(U) = (d(U))
1
p−1 g(U) and write t instead of U , and y(t) = |w(t)|p′ . Then

(3.6) becomes

y′(t) = p′
[
c (y(t))

1/p − f(t)
]
, t ∈ (0, 1) \ ∪ni=1{si}. (3.7)

From (3.1) and Definition 2.1(c) we deduce that v(z)→ 0 as z → z0+ or z → z1−
which is equivalent to limU→0+ w(U) = limU→1− w(U) = 0. Therefore, y = y(t)
satisfies the boundary conditions

y(0) = y(1) = 0. (3.8)

On the other hand, let us suppose that y = y(t) is a positive solution of (3.7),

(3.8). Set w(s) := −(y(s))1/p′ . Then w satisfies (3.5) and (3.6). For U ∈ (0, 1) set

z(U) = z(s∗)−
∫ U

s∗

∣∣ d(s)

w(s)

∣∣ 1
p−1 ds. (3.9)

Then the function z = z(U) is continuous strictly decreasing and maps the interval
(0, 1) onto (z0, z1), where −∞ ≤ z0 < z1 ≤ +∞. Let us denote by U : (z0, z1) →
(0, 1) the inverse function to z = z(U). Then U(z(s∗)) = s∗, U is continuous strictly
decreasing,

lim
z→z0+

U(z) = 1 and lim
z→z1−

U(z) = 0.

Let z ∈ (ξi, ξi+1), i = 0, 1, . . . , n, where U(ξi) = si, i = 0, 1, . . . , n, n + 1. Then
from (3.9) we deduce

dU

dz
=

1
dz(U)

dU

= −
∣∣∣w(U)

d(U)

∣∣∣ 1
p−1

, U ∈ (si, si+1),

i.e., U ∈ C1(ξi, ξi+1), U ′(z) < 0 and

− d(U(z))
∣∣∣dU(z)

dz

∣∣∣p−1

= w(U(z)) =: v(z), (3.10)

i.e.,
d

dz

[
d(U(z))

∣∣∣dU
dz

∣∣∣p−2 dU

dz

]
=

d

dz
w(U(z)) =

dw

dU

dU(z)

dz
. (3.11)

From (3.5), (3.10) we deduce that

dw

dU
= −|w(U)|−(p′−1)

(
− c|w(U)|p′−1 + (d(U))

p′−1
g(U)

)
= −c+ |w(U)|−(p′−1) (d(U))

p′−1
g(U)

= −c+ (d(U(z)))
−(p′−1)

∣∣∣dU(z)

dz

∣∣∣−(p−1)(p′−1)

(d(U(z)))
p′−1

g(U(z))

= −c+
∣∣∣dU(z)

dz

∣∣∣−1

g(U(z)).


6 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

Let us substitute this into (3.11):

d

dz

[
d(U(z))

∣∣dU
dz

∣∣p−2 dU

dz

]
=
[
− c+

∣∣dU(z)

dz

∣∣−1
g(U(z))

]dU(z)

dz

= −cdU(z)

dz
− g(U(z)),

i.e.,

d

dz

[
d(U(z))

∣∣dU
dz

∣∣p−2 dU

dz

]
+ c

dU(z)

dz
+ g(U(z)) = 0, z ∈ (ξi, ξi+1),

i = 0, 1, . . . , n. It follows from (3.10) and the continuity of U that

lim
z→z0+

d(U(z))|U ′(z)|p−2U ′(z) = lim
z→z1−

d(U(z))|U ′(z)|p−2U ′(z) = 0

and the following one sided limits are finite

lim
z→ξi−

d(U(z))|U ′(z)|p−2U ′(z) = lim
z→ξi+

d(U(z))|U ′(z)|p−2U ′(z),

i = 1, 2, . . . , n. Since U is monotone decreasing in (z0, z1), we have

lim
z→ξi−

d(U(z)) = lim
s→si+

d(s) and lim
z→ξi+

d(U(z)) = lim
s→si−

d(s).

Therefore, U satisfies the transition condition

|U ′(ξi−)|p−2
U ′(ξi−) lim

s→si+
d(s) = |U ′(ξi+)|p−2

U ′(ξi+) lim
s→si−

d(s).

We summarize the above reasoning in the following equivalence.

Proposition 3.1. A function U : R→ [0, 1] is a monotone non-increasing solution
of (2.1), (3.1) which is strictly decreasing at any point z ∈ R where U(z) ∈ (0, 1) if
and only if y : [0, 1]→ R is a positive solution of (3.7), (3.8).

Thanks to this proposition we can study the first order problem (3.7), (3.8)
and derive the existence result for (2.1), (3.1). Let us recall that there are “two
unknowns” in the first order problem. Indeed, besides the positive solution y = y(t)
we also look for unknown speed of propagation c ∈ R.

Lemma 3.2. Let us assume that (1.3) holds and BVP (3.7), (3.8) has a positive
solution. Then c > 0.

Proof. Let y(t) > 0, t ∈ (0, 1) be a positive solution of (3.7), (3.8). Integrating (3.7)
and using (3.8) we obtain

0 = y(1)− y(0) =

∫ 1

0

y′(t) dt = p′
[
c

∫ 1

0

y(t) dt−
∫ 1

0

f(t) dt
]
.

Hence

c =

∫ 1

0
(d(t))

1
p−1 g(t) dt∫ 1

0
y(t) dt

> 0. �

Remark 3.3. Let d and g be as in Section 2 and the following balanced condition
holds ∫ 1

0

(d(s))
1
p−1 g(s) ds = 0.

Then

y(t) = −p′
∫ t

0

(d(s))
1
p−1 g(s) ds, t ∈ (0, 1) (3.12)


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 7

is a unique positive solution of (3.7), (3.8) with c = 0. The solution given by (3.12)
leads to the standing wave. Its profile U = U(x) satisfies(

d(U(x))|U ′(x)|p−2U ′(x)
)′

+ g(U(x)) = 0, x ∈ R.

Detailed analysis of these solutions for g satisfying g < 0 in (0, s∗) is given in [7].

4. Existence result

In this section we first concentrate on the existence result for the first order
BVP (3.7), (3.8). More precisely, we prove that under appropriate assumptions on
f = f(t) there exists a unique real number c > 0 and an absolutely continuous
function y = y(t) such that y(t) > 0, t ∈ (0, 1), (3.8) holds and the equation (3.7)
is satisfied in the sense of Carathéodory (see [3, Chapter 2]).

The following result is interesting on its own. In combination with Proposition
3.1 it is a tool to prove the existence and uniqueness of the traveling speed c and
monotone decreasing traveling wave profile U .

Theorem 4.1. Let f ∈ L1(0, 1), f(t) ≤ 0, t ∈ (0, s∗), f(t) > 0, t ∈ (s∗, 1), and∫ 1

0

f(s) ds > 0. (4.1)

Then there is a unique number c > 0 and an absolutely continuous function y = y(t),
t ∈ [0, 1], such that y(0) = y(1) = 0, y(t) > 0, t ∈ (0, 1), and

y′(t) = p′
[
c
(
y+(t)

)1/p − f(t)
]

for a.a. t ∈ (0, 1). Here y+ = max{y, 0} denotes the positive part of y.

We prove Theorem 4.1 using the concept of solution of the first order ODE in
the sense of Carathéodory. For (t, y, c) ∈ [0, 1]× R2 and f = f(t) we set

h(t, y, c) := p′
[
c
(
y+
)1/p − f(t)

]
and consider the following two initial value problems which depend on a parameter
c ∈ R:

y′(t) = h(t, y(t), c), y(0) = 0 (4.2)

and

y′(t) = h(t, y(t), c), y(1) = 0. (4.3)

In both cases we look for a solution y = y(t), t ∈ [0, 1]. Therefore, (4.2) is re-
ferred to as a forward initial value problem, while (4.3) is referred to as a backward
initial value problem. Note that f ∈ L1(0, 1) implies that h = h(t, y, c) satisfies
Carathéodory’s conditions, i.e., for a.e. t ∈ [0, 1] fixed, h(t, ·, ·) is continuous with
respect to y and c and for every y ∈ R and c ∈ R fixed, h(·, y, c) is measurable with
respect to t. In what follows, for any fixed c ∈ R, yc = yc(t) denotes the solution in
the sense of Carathéodory of the forward and backward initial value problem (4.2)
and (4.3), respectively. In particular, yc is absolutely continuous in [0, 1] and the
equation holds a.e. in [0, 1]. Let us mention the following global existence result.

Lemma 4.2. Let f ∈ L1(0, 1), c ∈ R. Then there exists at least one global solution
yc = yc(t) of both (4.2) and (4.3) defined on the entire interval [0, 1].


8 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

Proof. For any fixed K > 0 and c ∈ R there exists mc,K ∈ L1(0, 1) such that for
any y ∈ [−K,K] we have |h(t, y, c)| ≤ mc,K(t) for a.a. t ∈ [0, 1]. This fact follows
from the definition of h. But then according to [14, Theorem 10.XX] there exist
solutions yc = yc(t) of both (4.2) and (4.3) which are defined for all t ∈ [0, 1]. �

Remark 4.3. The uniqueness of the solution in the above lemma does not hold in

general due to the fact that the function y 7→ c (y+)
1/p

, y ∈ R, does not satisfy the
Lipschitz condition at 0. However, it is nondecreasing for c > 0 and non-increasing
for c < 0. Therefore, it satisfies one-sided Lipschitz condition in either case and
we have the following uniqueness results separately for the forward and backward
initial value problems, depending on the sign of c.

Lemma 4.4. Let f ∈ L1(0, 1). If c ≤ 0 then (4.2) has exactly one solution yc =
yc(t), t ∈ [0, 1]. If c ≥ 0 then (4.3) has exactly one solution yc = yc(t), t ∈ [0, 1].

The proof of the above lemma follows from combination of Theorems 9.X and
10.XX in [14]. Thanks to the uniqueness result we also have continuous dependence
of solutions on the parameter c.

Lemma 4.5. Let f ∈ L1(0, 1), c0 ≥ 0. Then c → c0 6= 0 (c → 0+ if c0 = 0)
implies that solutions yc = yc(t) of the backward initial value problem (4.3) converge
uniformly in [0, 1] (i.e., in the topology of C[0, 1]) to yc0 . Similar result holds for
c0 ≤ 0 and the forward initial value problem (4.2).

The proof of the above lemma follows from the uniqueness result in Lemma 4.4
and [3, Theorems 4.1 and 4.2].

As we already observed in the proof of Lemma 3.2, the assumption (4.1) yields
c > 0. For this reason, we further focus on parameters c ∈ [0,+∞) and the
backward initial value problem (4.3). We know that for any c ∈ [0,+∞) there is
a unique solution of (4.3), yc = yc(t), t ∈ [0, 1]. Our goal is to show that there is
unique c∗ > 0 such that yc∗(t) > 0, t ∈ (0, 1), yc∗(0) = 0. To this end we have to
investigate in more detail the dependence of the solution yc = yc(t) of the backward
initial value problem (4.3) on the parameter c.

Let us introduce the notion of the defect Pcϕ of a function ϕ = ϕ(t) with respect
to the differential equation y′ = h(t, y, c), see e.g. [14, §9.II]:

Pcϕ := ϕ′ − h(t, ϕ, c).

The following comparison argument is one of our basic tools.

Lemma 4.6. Let f ∈ L1(%, 1),0 ≤ % < 1, c ≥ 0, ϕ(1) ≤ ψ(1), Pcϕ ≥ Pcψ a.e. in
[%, 1]. Then either ϕ < ψ in [%, 1] or there exists ξ ∈ [%, 1] such that ϕ = ψ in [ξ, 1]
and ϕ < ψ in (%, ξ]. In particular, ϕ ≤ ψ in [%, 1].

The proof of the above lemma follows from combination of Theorems 9.X and
10.XXI in [14].

Corollary 4.7. Let f be as in Theorem 4.1, and 0 ≤ c1 < c2. Then

yc1(t) > yc2(t), t ∈ (0, 1).

In particular, we have the weak comparison at the terminal value 0: yc1(0) ≥ yc2(0).

Proof. We have

Pc2yc1 = y′c1 − h(t, yc1 , c2) = y′c1 − h(t, yc1 , c1)︸ ︷︷ ︸
=0

+h(t, yc1 , c1)− h(t, yc1 , c2)


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 9

= p′(c1 − c2)
(
y+
c1

)1/p ≤ 0 = y′c2 − h(t, yc2 , c2) = Pc2yc2 .

Then Lemma 4.6 with % = 0 yields that either yc1 > yc2 in (0, 1) or there exists
ξ ∈ [0, 1] such that yc1 = yc2 in [ξ, 1] and yc1 > yc2 in (0, ξ). Since both yc1 and
yc2 solve the backward initial value problem, we subtract the equation in (4.3) for
c = c2 from that for c = c1 and obtain

p′(c1 − c2)
(
y+
c1

)1/p
= 0 in [ξ, 1],

i.e., yc1 = yc2 ≤ 0 in [ξ, 1]. But from (4.1) and (4.3) we then obtain

yc1(ξ) = yc2(ξ) = p′
∫ 1

ξ

f(σ) dσ > 0

if ξ < 1. Therefore, ξ = 1 and yc1 > yc2 in (0, 1). Using Lemma 4.2 and extending
yc1 , yc2 continuously to 0, we end up with yc1(0) ≥ yc2(0). �

Remark 4.8. Unfortunately, the comparison argument above does not allow us
to conclude the strict inequality yc1(0) > yc2(0) in general. For this reason the
proof of uniqueness of c∗ is more involved and requires more detailed analysis of
the solution of the backward initial value problem (4.3) at the terminal value 0.

Corollary 4.9. Let f ∈ L1(0, 1), f(t) ≤ 0, t ∈ (0, s∗), f(t) > 0, t ∈ (s∗, 1) and

f̃(t) = 0, t ∈ (0, s∗), f̃(t) = f(t), t ∈ (s∗, 1). Let c ≥ 0 and ỹc = ỹ(t), t ∈ [0, 1] be a

solution of (4.3) with f replaced by f̃ . Then yc ≤ ỹc in (0, 1].

Proof. Set h̃(t, y, c) := p′
[
c(y+(t)1/p)− f̃(t)

]
. Then h̃ ≤ h and so

Pcỹc = ỹ′c − h(t, ỹc, c) = ỹ′c − h̃(t, ỹc, c)︸ ︷︷ ︸
=0

+h̃(t, ỹc, c)− h(t, ỹc, c)

≤ 0 = y′c − h(t, yc, c) = Pcyc a.e. in (0, 1).

It then follows from Lemma 4.6 with % = 0 that yc ≤ ỹc in [0, 1]. �

Corollary 4.10. Let f ∈ L1(0, 1), f(t) ≤ 0, t ∈ (0, s∗), f(t) > 0, t ∈ (s∗, 1). Then
there exists c# > 0 such that yc#(0) < 0.

Proof. Let f̃ be as in Corollary 4.9. Since f̃ = 0 in (0, s∗), we have

ỹ′c = p′c
(
ỹ+
c (t)

)1/p
a.e. in (0, s∗). (4.4)

Assume that there exist cn → +∞ such that ỹcn ≥ 0 in [0, 1]. Then ỹ+
c = ỹc and

separating variables in (4.4) yields

(ỹcn(t))
1/p′

= (ỹcn(s∗))
1/p′

+ cn(t− s∗), t ∈ [0, s∗), (4.5)

and

ỹ′cn(t) = p′
[
cn (ỹcn(t))

1/p − f(t)
]
, t ∈ [s∗, 1),

with ỹcn(1) = 0. Therefore

ỹcn(s∗) ≤ p′
∫ 1

s∗

f(σ) dσ < +∞. (4.6)

Then we conclude from (4.5), (4.6) that there exists n0 ∈ N such that for all n ≥ n0

we must have ỹcn(0) < 0. But Corollary 4.9 yields that ycn(0) ≤ ỹcn(0) < 0.
Therefore, we may set c# = cn, n ≥ n0. �


10 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

Corollary 4.11. Let f ∈ L1(0, 1), f(t) ≤ 0, t ∈ (0, s∗), f(t) > 0, t ∈ (s∗, 1)
and yc = yc(t) be a solution of (4.3) with c > 0. If yc(0) ≥ 0 then yc(t) > 0 for
t ∈ (0, 1).

Proof. We have

Pc0 = 0− h(t, 0, c) = p′f(t) ≥ 0 = y′c − h(t, yc, c) = Pcyc a.e. in [s∗, 1].

Then by Lemma 4.6 with % = s∗ either yc > 0 in [s∗, 1) or there exists ξ ∈ [s∗, 1]
such that yc = 0 in [ξ, 1] and yc > 0 in [s∗, ξ). In the latter case

0 = yc(ξ) = p′
[
c

∫ ξ

1

(
y+
c (t)

)1/p
dt−

∫ ξ

1

f(t) dt
]

= p′
∫ 1

ξ

f(t) dt

forces ξ = 1 (note that f > 0 in (s∗, 1)). Therefore, yc > 0 in [s∗, 1). Assume that
yc vanishes at (0, 1) and η ∈ (0, s∗) be its largest zero. Then

y′c(t) = p′
[
c(y+

c (t))1/p − f(t)
]
≥ p′c(y+

c (t))1/p, for a.e. t ∈ [0, η].

Separating variables and integrating over [t, η],

(y+
c (t))1/p′ ≤ −c(η − t), t ∈ [0, η].

In particular, we have (
y+
c (0)

)1/p′ ≤ −cη < 0,

a contradiction with yc(0) ≥ 0. �

Proof of Theorem 4.1. It follows from the assumptions on f that

y0(t) = p′
∫ 1

t

f(σ) dσ > 0

for all t ∈ [0, 1). In particular, y0(0) > 0. On the other hand, from Corollary
4.10 there exists c# > 0 such that yc#(0) < 0. The continuous dependence on
parameter c in Lemma 4.5, intermediate value theorem and the monotonicity of
function S : c 7→ yc(0) in Corollary 4.7 imply that there exist 0 < c1 ≤ c2 < c#
such that S(c) = 0 for all c ∈ [c1, c2], S(c) > 0, c < c1 and S(c) < 0, c > c2. Below
we derive the strong comparison argument which shows that, actually, c1 = c2.
Indeed, let c1 < c2. By Corollaries 4.7 and 4.11 we have

yc1(t) > yc2(t) > 0 for t ∈ (0, 1). (4.7)

Notice that for c ∈ (c1, c2) we also have

y′c1(t) = p′
[
c1 (yc1(t))

1/p − f(t)
]
≤ p′

[
c (yc1(t))

1/p − f(t)
]
, (4.8)

y′c2(t) = p′
[
c2 (yc2(t))

1/p − f(t)
]
≥ p′

[
c (yc2(t))

1/p − f(t)
]
, (4.9)

for a.e. t ∈ (0, 1). Set z1 = (yc1)1/p′ > 0, z2 = (yc2)1/p′ > 0. Then z1 > z2 in (0, 1)
and it follows from (4.8) and (4.9) that

z′1(t) ≤ c− f(t)

(z1(t))
1
p−1

, (4.10)

z′2(t) ≥ c− f(t)

(z2(t))
1
p−1

, (4.11)


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 11

for a.e. t ∈ (0, 1). Let us subtract (4.11) from (4.10) and restrict on the interval
(0, s∗). Then

(z1(t)− z2(t))
′ ≤ −f(t)

( 1

(z1(t))
1
p−1

− 1

(z2(t))
1
p−1

)
and

(z1(t)− z2(t)) (z1(t)− z2(t))
′

≤ −f(t)
( 1

(z1(t))
1
p−1

− 1

(z2(t))
1
p−1

)
(z1(t)− z2(t)) ≤ 0

for a.e. t ∈ (0, s∗) (notice that f(t) ≤ 0 in (0, s∗)). Hence

1

2

d

dt
(z1(t)− z2(t))

2 ≤ 0 for a.e. t ∈ (0, s∗). (4.12)

Since z1(0) = z2(0) = 0, it follows from (4.12) that z1(t) = z2(t), t ∈ (0, s∗), i.e.,
yc1(t) = yc2(t), t ∈ (0, s∗). However, this contradicts (4.7). Therefore, c1 = c2 and
c∗ = c1 = c2 is the unique value of c for which yc(0) = 0. As mentioned above,
yc∗(t) > 0, t ∈ (0, 1), and yc∗ is the unique solution of the backward initial value
problem (4.3). The proof is complete. �

Theorem 4.12. Let d and g be as in Section 2 and (1.3) holds. Then there is a
unique value of c = c∗ and unique non-increasing traveling wave profile U = U(z),
z ∈ R, such that U solves the BVP (2.1), (3.1). Furthermore, c∗ > 0 and

(i) there exist −∞ ≤ z0 < 0 < z1 ≤ +∞ such that U(z) = 1 for z ∈ (−∞, z0],
U(z) = 0 for z ∈ [z1,+∞);

(ii) U is strictly decreasing in (z0, z1), U(0) = s∗;
(iii) for i = 0, 1, 2, . . . , n, n + 1 let ξi ∈ [z0, z1] be such that U(ξi) = si, then U

is a piecewise C1- function in the sense that U is continuous,

U
∣∣
(ξi,ξi+1)

∈ C1(ξi, ξi+1), i = 0, 1, . . . , n,

and the limits U ′(ξi−) := limz→ξi− U
′(z), U ′(ξi+) := limz→ξi+ U

′(z) exist
are finite for all i = 1, 2, . . . , n;

(iv) for any i = 1, 2, . . . , n the following transition condition holds:

|U ′(ξi−)|p−2
U ′(ξi−) lim

s→si+
d(s) = |U ′(ξi+)|p−2

U ′(ξi+) lim
s→si−

d(s).

Proof. The existence and uniqueness of c∗ and U follow directly from Theorem 4.1
and Proposition 3.1. The properties of U are derived in the reasoning preceding
the statement of Proposition 3.1. �

Remark 4.13. The inequality “>” in (1.3) was motivated by modeling heterzygote
inferior case. The assumption∫ 1

0

(d(s))
1
p−1 g(s) ds < 0 (4.13)

leads to negative traveling speed of propagation c∗ < 0 and it can be treated in a
similar way. However, in this case the main tool is a shooting argument applied
to the forward initial value problem and the strong comparison argument must be
derived at the terminal value 1.


12 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

We can also prove similar results for increasing traveling wave profile U satisfying

lim
z→−∞

U(z) = 0 and lim
z→+∞

U(z) = 1.

In this case the assumption (1.3) leads to c∗ < 0 while (4.13) leads to c∗ > 0,
respectively.

Remark 4.14. Notice that condition U(0) = s∗ has just a normalizing character.
Indeed, since the equation (2.1) is autonomous then given any ξ ∈ R the translation
V (z) = U(z − ξ), z ∈ R, is also a solution of (2.1) which satisfies V (ξ) = s∗.

5. Asymptotic analysis of the traveling wave profile

In this section we focus on the asymptotic behavior of the traveling wave profile
U = U(z) as z → ±∞. Similar asymptotic analysis for standing waves (c = 0)
was done in [7]. Even though the main idea is the same also in the case c 6= 0, the
analysis is much more involved and not so precise because the solution of equation
(3.7) for c 6= 0 cannot be obtained in a closed form by simple integration of f = f(t)
as in the stationary case (c = 0), cf. [7].

For the sake of brevity, for t0 ∈ R we write

h1(t) ∼ h2(t) as t→ t0 if and only if lim
t→t0

h1(t)

h2(t)
∈ (0,+∞).

5.1. Asymptotics near 1. Let us assume that g(t) ∼ (1− t)γ and d(t) ∼ (1− t)δ
as t→ 1− for some γ > 0 and δ ∈ R. Then, formally,

f(t) = (d(t))
1
p−1 g(t) ∼ (1− t)γ+ δ

p−1 as t→ 1− .

The fact that f ∈ L1(0, 1) then implies the following necessary condition for pa-
rameters γ and δ:

γ +
δ

p− 1
> −1. (5.1)

It follows from (3.9) that the inverse function to a profile U = U(z) corresponding
to the speed c > 0 and normalized by U(0) = s∗ is given by

z(U) = −
∫ U

s∗

(d(t))
1
p−1

(yc∗(t))
1
p

dt, U ∈ (0, 1), (5.2)

where yc∗ = yc∗(t) is the unique positive solution of (3.7), (3.8). In order to find
the asymptotic behavior of z = z(U) as U → 1− from (5.2) we need to establish the
asymptotics of yc∗ = yc∗(t) as t → 1−. The asymptotics of U = U(z) as z → −∞
then would follow applying the inverse point of view to z = z(U) for U → 1−.

From our assumptions it follows that there exists θ > 0 (small enough) such that
both d and g are continuous in (1− θ, 1). Therefore, f = f(t) is also continuous in

(1− θ, 1) and hence f(t) ∼ (1− t)γ+ δ
p−1 is equivalent to

f(t) = η(t)(1− t)γ+ δ
p−1 , t ∈ (1− θ, 1),

where η = η(t) is a continuous function in (1− θ, 1), limt→1− η(t) ∈ (0,+∞).
In what follows we discuss different cases with respect to parameters γ, δ and p.


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 13

A. Let γ+ δ
p−1 ≤ 1

p−1 . Then for κ > 0 we set yκ(t) = κ(1− t)γ+ δ
p−1 +1, t ∈ (1−θ, 1)

and calculate the defect

Pc∗yκ = y′κ − p′[c∗ (yκ)
1
p − f(t)]

= −κ
(
γ +

δ

p− 1
+ 1
)

(1− t)γ+ δ
p−1

− p′
[
c∗κ

1
p (1− t)

γ+ δ
p−1

+1

p − η(t)(1− t)γ+ δ
p−1

]
= (1− t)γ+ δ

p−1

[
− κ
(
γ +

δ

p− 1
+ 1
)

+ p′η(t)
]

− (1− t)
γ+ δ

p−1
+1

p p′c∗κ
1
p ,

(5.3)

t ∈ (1− θ, 1). Our assumption γ + δ
p−1 ≤ 1

p−1 implies

γ +
δ

p− 1
≤
γ + δ

p−1 + 1

p
,

and therefore the power (1 − t)γ+ δ
p−1 dominates the power (1 − t)

γ+ δ
p−1

+1

p near 1.
It then follows from (5.3) that we may distinguish between two cases:

A1. There exists κ� 1 so small that Pc∗yκ > 0 = Pc∗yc∗ a.e. in (1− θ, 1).
A2. There exists κ̄� 1 so large that Pc∗yκ̄ < 0 = Pc∗yc∗ a.e. in (1− θ, 1).

Case A1. Let γ−δ+1
p < 1. It follows from Lemma 4.6 with % = 1− θ that

yc∗(t) ≥ yκ(t) in (1− θ, 1). (5.4)

From (5.2) and (5.4) we conclude that there exists c1 > 0 such that

z0 = lim
U→1−

z(U) = −
∫ 1

s∗

(d(t))
1
p−1

(yc∗(t))
1
p

dt

≥ −
∫ 1

s∗

(d(t))
1
p−1

(yκ(t))
1
p

dt

≥ −c1
∫ 1

s∗

(1− t) δ
p−1

(1− t)
γ+ δ

p−1
+1

p

dt

= −c1
∫ 1

s∗

dt

(1− t)
γ+ δ

p−1
+1

p − δ
p−1

= −c1
∫ 1

s∗

dt

(1− t) γ−δ+1
p

> −∞.

Case A2. Let γ−δ+1
p ≥ 1. It follows from Lemma 4.6 with % = 1− θ that

yc∗(t) ≤ yκ̄(t) in (1− θ, 1). (5.5)

From (5.2) and (5.5) we conclude that there exists c2 > 0 such that

z0 = −
∫ 1

s∗

(d(t))
1
p−1

(yc∗(t))
1
p

dt ≤ −
∫ 1

s∗

(d(t))
1
p−1

(yκ̄(t))
1
p

dt


14 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

≤ −c2
∫ 1

s∗

(1− t) δ
p−1

(1− t)
γ+ δ

p−1
+1

p

dt

= −c2
∫ 1

s∗

dt

(1− t) γ−δ+1
p

= −∞.

Taking into account (5.1), we may summarize these two cases as follows.

Theorem 5.1. Let us assume γ > 0,

−1 < γ +
δ

p− 1
≤ 1

p− 1
, (5.6)

γ − δ + 1

p
< 1, (5.7)

Then z0 > −∞. Let us assume γ > 0, (5.6), and

γ − δ + 1

p
≥ 1. (5.8)

Then z0 = −∞.

B. Let γ+ δ
p−1 >

1
p−1 . Then for κ > 0 we set yκ(t) = κ(1− t)p(γ+ δ

p−1 ), t ∈ (1−θ, 1)

and calculate

Pc∗yκ = y′κ − p′[c∗ (yκ)
1
p − f(t)]

= −κp
(
γ +

δ

p− 1

)
(1− t)p(γ+ δ

p−1 )−1

− p′
[
c∗κ

1
p (1− t)γ+ δ

p−1 − η(t)(1− t)γ+ δ
p−1
]

= −κp
(
γ +

δ

p− 1

)
(1− t)p(γ+ δ

p−1 )−1

− p′
[
c∗κ

1
p − η(t)

]
(1− t)γ+ δ

p−1 ,

(5.9)

for t ∈ (1− θ, 1). Our assumption γ(p− 1) + δ > 1 implies

γ +
δ

p− 1
< p

(
γ +

δ

p− 1

)
− 1,

and the power (1 − t)tγ+ δ
p−1 dominates the power (1 − t)p(γ+ δ

p−1 )−1 near 1. It
follows from (5.9) that we may distinguish between two cases:

B1. There exists κ� 1 so small that Pc∗yκ > 0 = Pc∗yc∗ a.e. in (1− θ, 1).
B2. There exists κ̄� 1 so large that Pc∗yκ̄ < 0 = Pc∗yc∗ a.e. in (1− θ, 1).

Case B1. Let γ < 1. From Lemma 4.6 with % = 1− θ we obtain

yc∗(t) ≥ yκ(t) in (1− θ, 1)

and therefore, similarly as in Case A1 we conclude that there exists c3 > 0 such
that

z0 ≥ −c3
∫ 1

s∗

(1− t) δ
p−1

(1− t)γ+ δ
p−1

dt = −c3
∫ 1

s∗

dt

(1− t)γ > −∞.

Case B2. Let γ ≥ 1. From Lemma 4.6 with % = 1− θ we obtain

yc∗(t) ≤ yκ̄(t) in (1− θ, 1)


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 15

and as in Case A2 we conclude that there exists c4 > 0 such that

z0 ≤ −c4
∫ 1

s∗

(1− t) δ
p−1

(1− t)γ+ δ
p−1

dt = −c4
∫ 1

s∗

dt

(1− t)γ = −∞.

We summarize these two cases as follows.

Theorem 5.2. Let us assume γ > 0,

γ +
δ

p− 1
>

1

p− 1
, (5.10)

γ < 1. (5.11)

Then z0 > −∞. Let us assume γ > 0, (5.10) and

γ ≥ 1. (5.12)

Then z0 = −∞.

Remark 5.3. To visualize conditions (5.6)–(5.8) and (5.10)–(5.12), we introduce
the following sets:

M1
1 := {(γ, δ) ∈ R2 : γ > 0,−1 < γ +

δ

p− 1
≤ 1

p− 1
, γ − δ + 1 < p},

M2
1 := {(γ, δ) ∈ R2 : γ > 0,−1 < γ +

δ

p− 1
≤ 1

p− 1
, γ − δ + 1 ≥ p},

M3
1 := {(γ, δ) ∈ R2 : γ > 0, γ +

δ

p− 1
>

1

p− 1
, γ < 1},

M4
1 := {(γ, δ) ∈ R2 : γ > 0, γ +

δ

p− 1
>

1

p− 1
, γ ≥ 1}.

Then z0 > −∞ if and only if (γ, δ) ∈ M1
1 ∪ M3

1 and z0 = −∞ if and only if
(γ, δ) ∈M2

1∪M4
1. See Figure 1 for geometric interpretation. Our results generalize

those from [5, Section 6].

γ

δ

−1 1

−1

1

γ
+
δ
= −

1

γ
+
δ
=
1

γ
− δ

=
1

M1
1

M2
1

M3
1

M4
1

Figure 1. Visualization of the sets M1
1, M2

1, M3
1 and M4

1 for p = 2


16 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

5.2. Asymptotics near 0. Let us assume that g(t) ∼ −tα and d(t) ∼ tβ as

t → 0+ for some α > 0 and β ∈ R. Then, formally, f(t) ∼ −tα+ β
p−1 as t → 0+.

The assumption f ∈ L1(0, 1) yields necessary condition for parameters α and β:

α+
β

p− 1
> −1. (5.13)

The main idea to find the asymptotics of U = U(z) as z → +∞ is now based on
the investigation of the asymptotics of its inverse z = z(U) as U → 0+. For this
purpose we employ the formula (5.2) and, in particular, its limit for U → 0+:

z1 =

∫ s∗

0

(d(t))
1
p−1

(yc∗(t))
1
p

dt. (5.14)

The main difference between this and previous case (asymptotics near 1) consists in
the fact that we cannot use the comparison argument based on Lemma 4.6 due to
the lack of uniqueness for the forward initial value problem (4.2). However, special
form of our equation allows for the uniqueness result for this problem if we restrict
on the set of positive solutions in a neighbourhood of 0. We will explain this idea
below.

Lemma 5.4. Let f be as in Theorem 4.1. Then the forward initial value problem
(4.2) with c > 0 has a unique positive solution in (0, s∗).

Proof. Let y = y(t), t ∈ (0, s∗), be a solution of the forward initial value problem
(4.2) with c > 0, cf. Lemma 4.2. Then

y′(t) = p′
[
c
(
y+(t)

) 1
p − f(t)

]
≥ 0, t ∈ (0, s∗)

and therefore

y(t) = y(0) +

∫ t

0

y′(σ) dσ ≥ 0, t ∈ (0, s∗).

Assume that there are two positive solutions y1 = y1(t), y2 = y2(t), t ∈ (0, s∗) of

(4.2). Then z1 = (y1)1/p′ > 0, z2 = (y2)1/p′ > 0 solve the forward initial value
problem

z′i(t) = c− f(t)

(zi(t))
1
p−1

for a.e. t ∈ (0, s∗),

zi(0) = 0

for i = 1, 2. It then follows that

(z1(t)− z2(t))
′

= −f(t)
( 1

(z1(t))
1
p−1

− 1

(z2(t))
1
p−1

)
,

(z1(t)− z2(t))
+

(z1(t)− z2(t))
′

= −f(t)
( 1

(z1(t))
1
p−1

− 1

(z2(t))
1
p−1

)
(z1(t)− z2(t))

+

for a.e. t ∈ (0, s∗). Since f(t) ≤ 0, t ∈ (0, s∗), it follows from here that

1

2

d

dt

[
(z1(t)− z2(t))

+ ]2 ≤ 0, a.e. in (0, s∗). (5.15)

But z1(0) = z2(0) = 0 and (5.15) imply z1(t) ≤ z2(t). Similarly, we prove that
z2(t) ≤ z1(t). Therefore, y1(t) = y2(t) for t ∈ (0, s∗). �


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 17

Remark 5.5. It follows from Lemma 5.4 that the restriction of the unique positive
solution yc∗ = yc∗(t), t ∈ [0, 1], of the boundary value problem (3.7), (3.8) to the
interval (0, s∗) is also the unique solution of the forward initial value problem (4.2)
with c = c∗ on (0, s∗).

With the uniqueness result from Lemma 5.4 in hand, we can use the following
comparison argument which is our tool for the asymptotic analysis near 0.

Lemma 5.6. Let f ∈ L1(0, 1) be as in Theorem 4.1, 0 < θ < s∗, ϕ = ϕ(t), ψ =
ψ(t), t ∈ [0, θ] satisfy ϕ(0) = ψ(0) = 0, ϕ′(t) ≤ h(t, ϕ(t), c∗), ψ

′(t) ≥ h(t, ψ(t), c∗)
for a.e. t ∈ [0, θ], and let yc∗ = yc∗(t), t ∈ [0, 1], be the unique solution of (3.7),
(3.8). Then

ϕ(t) ≤ yc∗(t) ≤ ψ(t), t ∈ [0, θ].

Proof. The proof follows directly from [14, §10.XXII] combined with the uniqueness
result in Lemma 5.4 and Remark 5.5. �

The assumptions on d and g imply that for θ such that 0 < θ < min{s∗, s1} the

function f = f(t) is continuous in (0, θ) and f(t) ∼ −tα+ β
p−1 is equivalent to

f(t) = −η(t)(1− t)α+ β
p−1 , t ∈ (0, θ),

where η = η(t) is a continuous function in (0, θ), limt→0+ η(t) ∈ (0,+∞).
In what follows we discuss different cases with respect to parameters α, β and

p.

A. Let α+ β
p−1 ≤ 1

p−1 . For κ > 0 we set yκ(t) = κtα+ β
p−1 +1, t ∈ [0, θ]. Then

y′κ − p′[c∗ (yκ)
1
p − f(t)]

= κ
(
α+

β

p− 1
+ 1
)
tα+ β

p−1 − p′
[
c∗κ

1
p t

α+
β
p−1

+1

p + η(t)tα+ β
p−1
]

= tα+ β
p−1
[
κ
(
α+

β

p− 1
+ 1
)
− p′η(t)

]
− t

α+
β
p−1

+1

p p′c∗κ
1
p ,

(5.16)

for a.e. t ∈ [0, θ]. The assumption α+ β
p−1 ≤ 1

p−1 implies

α+
β

p− 1
≤
α+ β

p−1 + 1

p

and therefore the power tα+ β
p−1 dominates the power t

α+
β
p−1

+1

p near 0.

A1. There exists κ � 1 so small that y′κ(t) ≤ p′[c∗(yκ(t)
1
p ) − f(t)] for a.e.

t ∈ [0, θ].

A2. There exists κ̄ � 1 so large that y′κ̄(t) ≥ p′[c∗(yκ̄(t)
1
p ) − f(t)] for a.e.

t ∈ [0, θ].

It follows from Lemma 5.6 that solution yc∗ = yc∗(t) of the BVP (3.7), (3.8) must
satisfy

yκ(t) ≤ yc∗(t) ≤ yκ̄(t), t ∈ [0, θ].

Case A1. Let α−β+1
p < 1. Then there exists c1 > 0 such that

z1 =

∫ s∗

0

(d(t))
1
p−1

(yc∗(t))
1
p

dt ≤
∫ s∗

0

(d(t))
1
p−1

(yκ(t))
1
p

dt


18 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

≤ c1
∫ s∗

0

t
β
p−1

t
α+

β
p−1

+1

p

dt = c1

∫ s∗

0

dt

t
α−β+1

p

< +∞.

Case A2. Let α−β+1
p ≥ 1. Then there exists c2 > 0 such that

z1 =

∫ s∗

0

(d(t))
1
p−1

(yc∗(t))
1
p

dt ≥
∫ s∗

0

(d(t))
1
p−1

(yκ̄(t))
1
p

dt

≥ c2
∫ s∗

0

t
β
p−1

t
α+

β
p−1

+1

p

dt = c2

∫ s∗

0

dt

t
α−β+1

p

= +∞.

We can summarize these two cases as follows.

Theorem 5.7. Let us assume α > 0,

−1 < α+
β

p− 1
≤ 1

p− 1
, (5.17)

α− β + 1

p
< 1. (5.18)

Then z1 < +∞. Let us assume α > 0, (5.17) and

α− β + 1

p
≥ 1. (5.19)

Then z1 = +∞.

B. Let α+ β
p−1 >

1
p−1 . For κ > 0 we set yκ(t) = κtp

′
, t ∈ [0, θ]. Then

y′κ − p′[c∗ (yκ)
1
p − f(t)] = κp′tp

′−1 − p′[c∗κ
1
p t

p′
p + η(t)tα+ β

p−1 ]

=
(
κp′ − p′c∗κ

1
p
)
t

1
p−1 − p′η(t)tα+ β

p−1 ,
(5.20)

for a.e. t ∈ [0, θ]. The assumption α + β
p−1 > 1

p−1 implies that the power t
1
p−1

dominates tα+ β
p−1 near 0.

B1. There exists κ � 1 so small that y′κ(t) ≤ p′[c∗(yκ(t)
1
p ) − f(t)] for a.e.

t ∈ [0, θ].

B2. There exists κ̄ � 1 so large that y′κ̄(t) ≥ p′[c∗(yκ̄(t)
1
p ) − f(t)] for a.e.

t ∈ [0, θ].

From Lemma 5.6 we conclude that solution yc∗ = yc∗(t) of the BVP (3.7), (3.8)
must satisfy

yκ(t) ≤ yc∗(t) ≤ yκ̄(t), t ∈ [0, θ].

Case B1. Let β > p− 2. Then there exists c3 > 0 such that

z1 =

∫ s∗

0

(d(t))
1
p−1

(yc∗(t))
1
p

dt ≤
∫ s∗

0

(d(t))
1
p−1

(yκ(t))
1
p

dt ≤ c3
∫ s∗

0

t
β
p−1

t
p′
p

dt = c3

∫ s∗

0

dt

t
1−β
p−1

< +∞.

Case B2. Let β ≤ p− 2. Then there exists c4 > 0 such that

z1 =

∫ s∗

0

(d(t))
1
p−1

(yc∗(t))
1
p

dt ≥
∫ s∗

0

(d(t))
1
p−1

(yκ̄(t))
1
p

dt ≥ c4
∫ s∗

0

t
β
p−1

t
p′
p

dt = c4

∫ s∗

0

dt

t
1−β
p−1

= +∞.

We summarize these two cases as follows.


EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 19

Theorem 5.8. Let us assume α > 0,

α+
β

p− 1
>

1

p− 1
, (5.21)

β > 2− p. (5.22)

Then z1 < +∞. Let us assume α > 0, (5.21) and

β ≤ 2− p. (5.23)

Then z1 = +∞.

Remark 5.9. To visualize conditions (5.17)–(5.19) and (5.21)–(5.23), we introduce
the sets:

M1
0 := {(α, β) ∈ R2 : α > 0,−1 < α+

β

p− 1
≤ 1

p− 1
, α− β + 1 < p},

M2
0 := {(α, β) ∈ R2 : α > 0,−1 < α+

β

p− 1
≤ 1

p− 1
, α− β + 1 ≥ p},

M3
0 := {(α, β) ∈ R2 : α > 0, α+

β

p− 1
>

1

p− 1
, β > 2− p},

M4
0 := {(α, β) ∈ R2 : α > 0, α+

β

p− 1
>

1

p− 1
, β ≤ 2− p}.

Then z1 < +∞ if and only if (α, β) ∈ M1
0 ∪ M3

0 and z1 = +∞ if and only if
(α, β) ∈M2

0∪M4
0. The reader is invited to see Figure 2 for geometric interpretation

and compare the sets M1
0, M2

0, M3
0, M4

0 and M1
1, M2

1, M3
1, M4

1.

α

β

−1 1

−1

1

α
+
β
= −

1

α
+
β
=
1

α
− β

=
1

M1
0

M2
0

M3
0

M4
0

U ′(z1−) = −∞

0 > U ′(z1−) ≥ −∞

Figure 2. Visualization of the sets M1
0, M2

0, M3
0 and M4

0 for p = 2

Remark 5.10. Let us assume z0 > −∞, i.e., (γ, δ) ∈M1
1∪M3

1. Then U ′(z0−) = 0
and it follows from Definition 2.1 that U ′(z0+) exists finite or infinite, see Remark
2.4. Since U is a monotone decreasing function, we have −∞ ≤ U ′(z0+) ≤ 0.
If z1 < +∞, i.e., (α, β) ∈ M1

0 ∪ M3
0 then by similar reasons U ′(z1+) = 0 and

−∞ ≤ U ′(z1−) ≤ 0.


20 P. DRÁBEK, M. ZAHRADNÍKOVÁ EJDE-2021/76

In the case (α, β) ∈ M3
0 our one-sided estimates on z1 allow for more precise

information about the smoothness of U at z1. Indeed, in this case we have

0 ≥ z′(0+) = lim
U→0+

z′(U)

= − lim
U→0+

(d(U))
1
p−1

(yc∗(U))
1
p

≥ − lim
U→0+

(d(U))
1
p−1(

yκ(U)
) 1
p

≥ −c3 lim
U→0+

U
β−1
p−1 .

(5.24)

We distinguish the following two cases:

1. If β > 1 then from (5.24) we obtain z′(0+) = 0 and therefore U ′(z1−) =
−∞.

2. If β = 1 then we deduce from (5.24) that 0 ≥ z′(0+) ≥ −c3 and therefore
0 > U ′(z1−) ≥ −∞.

In either case the traveling wave profile U is “sharp” in the sense that U ′ has a
jump at z1 (finite or infinite).

In other cases (α, β) ∈M1
0 and (γ, δ) ∈M1

1 ∪M3
1 our one-sided estimates on z1

and z0, respectively, do not provide analogous information as above. This is a big
difference between the traveling wave and standing wave, see [7].

Acknowledgements. Michaela Zahradńıková was supported by the project SGS-
2019-010 of the University of West Bohemia in Pilsen.

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EJDE-2021/76 TRAVELING WAVES FOR BISTABLE EQUATIONS 21

[14] W. Walter; Ordinary differential equations, Graduate Texts in Mathematics, Vol. 182,

Springer-Verlag, New York, 1998.

Pavel Drábek

Department of Mathematics, Faculty of Applied Sciences, University of West Bohemia,
Univerzitńı 8, 30100 Plzeň, Czech Republic

Email address: pdrabek@kma.zcu.cz

Michaela Zahradńıková

Department of Mathematics, Faculty of Applied Sciences, University of West Bohemia,

Univerzitńı 8, 30100 Plzeň, Czech Republic
Email address: mzahrad@kma.zcu.cz


	1. Introduction
	2. Preliminaries
	3. Equivalent first order ODE
	4. Existence result
	5. Asymptotic analysis of the traveling wave profile
	5.1. Asymptotics near 1
	5.2. Asymptotics near 0
	Acknowledgements

	References