@
Appl. Gen. Topol. 23, no. 2 (2022), 463-480

doi:10.4995/agt.2022.15963

© AGT, UPV, 2022

Classical solutions for the Euler equations of

compressible fluid dynamics: A new topological

approach

Dalila Boureni a , Svetlin Georgiev b , Arezki Kheloufi a and
Karima Mebarki c

a Laboratory of Applied Mathematics, Bejaia University, 06000 Bejaia, Algeria.

(dalila.boureni@univ-bejaia.dz; arezki.kheloufi@univ-bejaia.dz; arezkinet2000@yahoo.fr)
b Department of Differential Equations, Faculty of Mathematics and Informatics, University of

Sofia, Sofia, Bulgaria. (svetlingeorgiev1@gmail.com)
c Laboratory of Applied Mathematics, Faculty of Exact Sciences, Bejaia University, 06000 Bejaia,

Algeria. (karima.mebarki@univ-bejaia.dz; mebarqi karima@hotmail.fr)

Communicated by E. A. Sánchez-Pérez

Abstract

In this article we study a class of Euler equations of compressible fluid
dynamics. We give conditions under which the considered equations
have at least one and at least two classical solutions. To prove our
main results we propose a new approach based upon recent theoretical
results.

2020 MSC: 35Q31; 35A09; 35E15.

Keywords: Euler equations; classical solution; fixed point; initial value prob-
lem.

1. Introduction

In this paper, we investigate an initial value problem for Euler equations of
compressible fluid dynamics, see [6], [10], [21]. Namely, we are concerned with

Received 22 July 2021 – Accepted 26 June 2022

http://dx.doi.org/10.4995/agt.2022.15963
https://orcid.org/0000-0003-3227-0745
https://orcid.org/0000-0001-8015-4226
https://orcid.org/0000-0001-5584-1454
https://orcid.org/0000-0002-6679-5059


D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

the following system:

(1.1)

∂tρ + ∂x(ρu) = 0,

∂t(ρu) + ∂x
(
ρu2 + p(ρ)

)
= 0, t > 0, x ∈ R,

ρ(0,x) = ρ0(x), x ∈ R,

u(0,x) = u0(x), x ∈ R,

where

(H1): ρ0,u0 ∈ C1(R), 0 ≤ ρ0(x),u0(x) ≤ B, x ∈ R, with B is a given
positive constant.

Here the unknowns ρ = ρ(t,x) ≥ 0 and u = u(t,x) denote respectively, the
density and the velocity of the gas, while the pressure p = p(ρ) is a given
function so that

(H2): p ∈ C(R) is a nonnegative function for which p(z) ≤ Czq, z ≥ 0,
C is a positive constant, q ≥ 0.

Note that if

p(ρ) = Cρq, ρ ≥ 0, C > 0, q ≥ 1,
then, the fluid is called isentropic and isothermal when q > 1 and q = 1,
respectively. For other possibilities of the pressure function, readers may refer
to [5] and the references therein. Cauchy problem with bounded measurable
initial data for (1.1):

(ρ0,u0) ∈ L∞ ×L∞

where u0(x) and ρ0(x) ≥ 0 (6≡ 0) is studied in [4]. The authors established the
convergence of a second-order shock-capturing scheme. In [7], a convergence
result for the method of artificial viscosity applied to the isentropic equations
of gas dynamics is established. In [20], some properties for solutions of (1.1)
containing a portion of the t − x plane in which ρ = 0 called vacuum state,
were investigated. Conservation laws of the one-dimensional isentropic gas
dynamics equations in Lagrangian coordinates are obtained in [16]. In [19],
a 2 × 2 hyperbolic system of isentropic gas dynamics, in both Eulerian or
Lagrangian variables is considered.

Whereas local existence results for problems of type (1.1) were obtained,
see, for example, [2], [3], [13],[17], [18], [22], [23], the literature concerning
global existence of solutions for such kind of problems does not seem to be
very rich. The problem of the global in time existence of solutions of the
equations of fluid mechanics in one space dimension was treated by Glimm in
1965 [12]. The equations (1.1) was investigated in [11] for existence of global
periodic solutions. For Euler equations with damping, the global existence of
solutions can be found in [24], [27], [30] and the references therein. In [5], a class
of conditions for non-existence of global classical solutions is established for
the initial-boundary value problem of a three-dimensional compressible Euler

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 464



Classical solutions for the Euler equations

equations with (or without) time-dependent damping. We mention also the
works [15], [26] and [28].

The aim of this paper is to investigate the IVP (1.1) for existence of global
classical solutions. We call a solution a classical solution if it, along with its
derivatives that appear in the equations, is of class C([0,∞) ×R).

Our main result for existence of classical solutions of the IVP (1.1) is as
follows.

Theorem 1.1. Suppose (H1)-(H2). Then, the IVP (1.1) has at least one
nonnegative solution (ρ,u) ∈C1([0,∞) ×R) ×C1([0,∞) ×R).
Theorem 1.2. Suppose (H1)-(H2). Then, the IVP (1.1) has at least two
nonnegative solutions (ρ1,u1), (ρ2,u2) ∈C1([0,∞) ×R) ×C1([0,∞) ×R).

The strategy for the proof of Theorem 1.1 and Theorem 1.2 which we develop
in Section 2 uses the abstract theory of the sum of two operators. This basic and
new idea yields global existence theorems for many of the interesting equations
of mathematical physics.

The paper is organized as follows. In the next section, we give some auxiliary
results. In Section 3 we prove Theorem 1.1. In Section 4, we prove Theorem
1.2. In Section 5, we give an example to illustrate our main results.

2. Preliminaries and auxiliary results

2.1. Preliminaries. To prove our existence results we will use Theorem 2.1
and Theorem 2.8, that we will present and demonstrate in the sequel.

Theorem 2.1. Let � > 0, R > 0, E be a Banach space and

X = {x ∈ E : ‖x‖≤ R}.
Let also, Tx = −�x, x ∈ X, S : X → E is a continuous, (I −S)(X) resides in
a compact subset of E and

(2.1) {x ∈ E : x = λ(I −S)x, ‖x‖ = R} = ∅
for any λ ∈

(
0, 1

�

)
. Then, there exists x∗ ∈ X so that

Tx∗ + Sx∗ = x∗.

Proof. Define

r

(
−

1

�
x

)
=



−1
�
x if ‖x‖≤ R�

Rx
‖x‖ if ‖x‖ > R�.

Then, r
(
−1
�
(I −S)

)
: X → X is continuous and compact. Hence and the

Schauder fixed point theorem, it follows that there exists x∗ ∈ X so that

r

(
−

1

�
(I −S)x∗

)
= x∗.

Assume that −1
�
(I −S)x∗ 6∈ X. Then,∥∥∥(I −S)x∗∥∥∥ > R�, R

‖(I −S)x∗‖
<

1

�

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 465



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

and

x∗ =
R

‖(I −S)x∗‖
(I −S)x∗ = r

(
−

1

�
(I −S)x∗

)
and hence, ‖x∗‖ = R. This contradicts with (2.1). Therefore, −1

�
(I−S)x∗ ∈ X

and

x∗ = r

(
−

1

�
(I −S)x∗

)
= −

1

�
(I −S)x∗

or

−�x∗ + Sx∗ = x∗,
or

Tx∗ + Sx∗ = x∗.

This completes the proof. �

Let E be a real Banach space.

Definition 2.2. A closed, convex set P in E is said to be cone if
(1) αx ∈P for any α ≥ 0 and for any x ∈P,
(2) x,−x ∈P implies x = 0.

Every cone P defines a partial ordering ≤ in E defined by :

x ≤ y if and only if y −x ∈P.

Denote P∗ = P\{0}.

Definition 2.3. A mapping K : E → E is said to be completely continuous if
it is continuous and maps bounded sets into relatively compact sets.

In what follows, we give some results about the fixed point index theory for
perturbation of a completely continuous mapping by expansive one. First, we
recall the definition of an expansive mapping.

Definition 2.4. Let X and Y be real Banach spaces. A mapping K : X → Y
is said to be expansive if there exists a constant h > 1 such that

‖Kx−Ky‖Y ≥ h‖x−y‖X
for any x,y ∈ X.

In the following lemma, we present the key property of the expansive map-
pings which allows to extend the notion of the fixed point index in the case of
a completely continuous mapping perturbed by an expansive one.

Lemma 2.5. [29, Lemma 2.1] Let (X,‖.‖) be a linear normed space and D ⊂
X. Assume that the mapping T : D → X is expansive with constant h > 1.
Then, the inverse of I −T : D → (I −T)(D) exists and

‖(I −T)−1x− (I −T)−1y‖≤
1

h− 1
‖x−y‖, ∀x,y ∈ (I −T)(D).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 466



Classical solutions for the Euler equations

In the sequel, P will refer to a cone in a Banach space (E,‖.‖), Ω is a subset
of P, and U is a bounded open subset of P.

Assume that S : U → E is a completely continuous mapping and T : Ω → E
is a expansive one with constant h > 1. By Lemma 2.5, the operator (I−T)−1
is (h− 1)−1-Lipschtzian on (I −T)(Ω). Suppose that

(2.2) S(U) ⊂ (I −T)(Ω),

and

(2.3) x 6= Tx + Sx, for all x ∈ ∂U ∩ Ω.

Then, x 6= (I −T)−1Sx, for all x ∈ ∂U and the mapping (I −T)−1S : U →P
is a completely continuous. From [14, Theorem 2.3.1], the fixed point index
i ((I −T)−1S,U,P) is well defined. Thus we put

(2.4) i∗(T + S,U ∩ Ω,P) =
{
i (I −T)−1S,U,P), if U ∩ Ω 6= ∅
0, if U ∩ Ω = ∅.

This integer is called the generalized fixed point index of the sum T + S on
U ∩ Ω with respect to the cone P.

The basic properties of the index i∗ are collected in the following lemma

Lemma 2.6 ([8, Theorem 2.3]). The fixed point index defined in (2.4) satisfies
the following properties:

(a) (Normalization). If U = Pr, 0 ∈ Ω, and Sx = z0 ∈B(−T0, (h−1)r)∩P for
all x ∈Pr, then,

i∗ (T + S,Pr ∩ Ω,P) = 1.
(b) (Additivity). For any pair of disjoint open subsets U1,U2 in U such that
T + S has no fixed point on (U \(U1 ∪U2)) ∩ Ω, we have

i∗ (T + S,U ∩ Ω,P) = i∗ (T + S,U1 ∩ Ω,P) + i∗ (T + S,U2 ∩ Ω,P),

where i∗ (T + S,Uj ∩ Ω,X) : = i∗ (T + S|Uj,Uj ∩ Ω,P), j = 1, 2.
(c) (Homotopy Invariance). The fixed point index i∗ (T + H(t, .),U ∩Ω,P) does
not depend on the parameter t ∈ [0, 1] whenever

(i) H : [0, 1] ×U → E be a completely continuous mapping,
(ii) H([0, 1] ×U) ⊂ (I −T)(Ω),
(iii) Tx + H(t,x) 6= x, for all t ∈ [0, 1] and x ∈ ∂U ∩ Ω.
(d) (Solvability). If i∗ (T + S,U ∩ Ω,P) 6= 0, Then, T + S has a fixed point in
U ∩ Ω.

Several considerations allowing computation of the index i∗ are shown in
[8]. The following result is an extension of [8, Proposition 2.11] in the case of
a completely continuous mapping perturbed by an expansive one.

Proposition 2.7. Let U be a bounded open subset of P with 0 ∈ U. Assume
that T : Ω ⊂ P → E is an expansive mapping, S : U → E is a completely
continuous one and S(U) ⊂ (I −T)(Ω).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 467



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

If T + S has no fixed point on ∂U ∩ Ω and there exists ε > 0 small enough
such that

Sx 6= (I −T)(λx) for all λ ≥ 1 + ε, x ∈ ∂U and λx ∈ Ω,

then, the fixed point index i∗ (T + S,U ∩ Ω,P) = 1.

Proof. The mapping (I − T)−1S : U → P is a completely continuous with-
out fixed point in the boundary ∂U and it is readily seen that the following
condition is satisfied

(I −T)−1Sx 6= λx for all x ∈ ∂U and λ ≥ 1 + ε.

Our claim then follows from the definition of i∗ and [1, Lemma 2.3]. �

Now we are able to present a multiple fixed point theorem. The proof rely
on Proposition 2.7 and [8, Proposition 2.16] producing the computation of the
index i∗. This result will be used to prove Theorem 1.2.

Theorem 2.8. Let U1,U2 and U3 three open bounded subsets of P such that
U1 ⊂ U2 ⊂ U3 and 0 ∈ U1. Assume that T : Ω ⊂ P → E is an expansive
mapping, S : U3 → E is a completely continuous one and S(U3) ⊂ (I−T)(Ω).
Suppose that (U2 \U1) ∩ Ω 6= ∅, (U3 \U2) ∩ Ω 6= ∅, and there exists v0 ∈ P∗
such that the following conditions hold:

(i): Sx 6= (I −T)(x−λv0), for all λ > 0 and x ∈ ∂U1 ∩ (Ω + λv0),
(ii): there exists ε > 0 small enough such that Sx 6= (I−T)(λx), for all
λ ≥ 1 + ε, x ∈ ∂U2, and λx ∈ Ω,

(iii): Sx 6= (I −T)(x−λv0), for all λ > 0 and x ∈ ∂U3 ∩ (Ω + λv0).
Then,T + S has at least two non-zero fixed points x1,x2 ∈P such that

x1 ∈ ∂U2 ∩ Ω and x2 ∈ (U3 \U2) ∩ Ω

or

x1 ∈ (U2 \U1) ∩ Ω and x2 ∈ (U3 \U2) ∩ Ω.

Proof. If Sx = (I−T)x for x ∈ ∂U2∩Ω, then we get a fixed point x1 ∈ ∂U2∩Ω
of the operator T + S. Suppose that Sx 6= (I − T)x for any x ∈ ∂U2 ∩ Ω.
Without loss of generality, assume that Tx+Sx 6= x on ∂U1∩Ω and Tx+Sx 6=
x on ∂U3 ∩ Ω, otherwise the conclusion has been proved. By Proposition 2.7
and [8, Proposition 2.16], we have

i∗ (T + S,U1 ∩Ω,P) = i∗ (T + S,U3 ∩Ω,P) = 0 and i∗ (T + S,U2 ∩Ω,P) = 1.

The additivity property of the index i∗ yields

i∗ (T + S, (U2 \U1) ∩ Ω,P) = 1 and i∗ (T + S, (U3 \U2) ∩ Ω,P) = −1.

Consequently, by the existence property of the index i∗, T + S has at least two
fixed points x1 ∈ (U2 \U1) ∩ Ω and x2 ∈ (U3 \U2) ∩ Ω. �

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 468



Classical solutions for the Euler equations

2.2. Auxiliary results. In this subsection, we give some properties of solu-
tions of IVP (1.1). Let X1 = C1([0,∞) ×R) be endowed with the norm

‖u‖X1 = max
{

sup
(t,x) ∈ [0,∞) ×R

|u(t,x)|, sup
(t,x) ∈ [0,∞) ×R

|ut(t,x)|,

sup
(t,x) ∈ [0,∞) ×R

|ux(t,x)|
}
,

provided it exists. Let X2 = X1 ×X1 be endowed with the norm
‖(ρ,u)‖X2 = max{‖ρ‖X1, ‖u‖X1}, (ρ,u) ∈ X2,

provided it exists. For (ρ,u) ∈ X2, we will write (ρ,u) ≥ 0 if ρ(t,x) ≥ 0,
u(t,x) ≥ 0 for any (t,x) ∈ [0,∞) ×R. For (ρ,u) ∈ X2, define the operators

S11 (ρ,u)(t,x) =

∫ x
0

(ρ(t,x1) −ρ0(x1)) dx1 +
∫ t

0

ρ(t1,x)u(t1,x)dt1,

S21 (ρ,u)(t,x) =

∫ x
0

(ρ(t,x1)u(t,x1) −ρ0(x1)u0(x1)) dx1

+

∫ t
0

(
ρ(t1,x)(u(t1,x))

2 + p(ρ(t1,x))
)
dt1,

S1(ρ,u)(t,x) =
(
S11 (ρ,u)(t,x),S

2
1 (ρ,u)(t,x)

)
, (t,x) ∈ [0,∞) ×R.

Lemma 2.9. Suppose (H1) and p ∈C(R). If (ρ,u) ∈ X2 satisfies the equation

(2.5) S1(ρ,u)(t,x) = 0, (t,x) ∈ [0,∞) ×R,
then it is a solution of the IVP (1.1).

Proof. Let (ρ,u) ∈ X2 is a solution to the equation (2.5). Then
(2.6) S11 (ρ,u)(t,x) = 0, S

2
1 (ρ,u)(t,x) = 0, (t,x) ∈ [0,∞) ×R.

We differentiate the first equation of (2.6) with respect to t and x and we find

ρt(t,x) + (ρu)x(t,x) = 0, (t,x) ∈ [0,∞) ×R.
We put t = 0 in the first equation of (2.6) and we arrive at∫ x

0

(ρ(0,x1) −ρ0(x1)) dx1 = 0, x ∈ R,

which we differentiate with respect to x and we find

ρ(0,x) = ρ0(x), x ∈ R.
Now, we differentiate the second equation of (2.6) with respect to t and x and
we find

(ρu)t(t,x) + (ρu
2 + p(ρ))x(t,x) = 0, (t,x) ∈ [0,∞) ×R.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 469



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

We put t = 0 in the second equation of (2.6) and we get∫ x
0

(ρ(0,x1)u(0,x1) −ρ0(x1)u0(x1)) dx1 = 0, x ∈ R,

which we differentiate with respect to x and we obtain

ρ(0,x)u(0,x) −ρ0(x)u0(x) = 0, x ∈ R,

whereupon

u(0,x) = u0(x), x ∈ R.

Thus, (ρ,u) is a solution to the IVP (1.1). This completes the proof. �

Lemma 2.10. Suppose (H1) and let h ∈C([0,∞) × R) be a positive function
almost everywhere on [0,∞) ×R. If (ρ,u) ∈ X2 satisfies the following integral
equations:∫ t

0

∫ x
0

(t− t1)2(x−x1)2h(t1,x1)S11 (ρ,u)(t1,x1)dx1dt1 = 0, (t,x) ∈ [0,∞) ×R

and∫ t
0

∫ x
0

(t− t1)2(x−x1)2h(t1,x1)S21 (ρ,u)(t1,x1)dx1dt1 = 0, (t,x) ∈ [0,∞) ×R,

then, (ρ,u) is a solution to the IVP (1.1).

Proof. We differentiate three times with respect to t and three times with
respect to x the integral equations of Lemma 2.10 and we find

h(t,x)S1(ρ,u)(t,x) = 0, (t,x) ∈ [0,∞) ×R,

whereupon

S1(ρ,u)(t,x) = 0, (t,x) ∈ [0,∞) ×R.

Hence and Lemma 2.9, we conclude that (ρ,u) is a solution to the IVP (1.1).
This completes the proof. �

Let

B1 = max{B,B2,B3,CBq}.

Lemma 2.11. Suppose (H1) and (H2). For (ρ,u) ∈ X2 with ‖(ρ,u)‖X2 ≤ B,
we have

|S11 (ρ,u)(t,x)| ≤ 2B1(1 + t)(1 + |x|),

|S21 (ρ,u)(t,x)| ≤ 2B1(1 + t)(1 + |x|), (t,x) ∈ [0,∞) ×R.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 470



Classical solutions for the Euler equations

Proof. We have

|S11 (ρ,u)(t,x)| =
∣∣∣∫ x

0

(ρ(t,x1) −ρ0(x1)) dx1 +
∫ t

0

ρ(t1,x)u(t1,x)dt1

∣∣∣
≤

∣∣∣∫ x
0

(ρ(t,x1) −ρ0(x1)) dx1
∣∣∣ + ∣∣∣∫ t

0

ρ(t1,x)u(t1,x)dt1

∣∣∣
≤

∣∣∣∫ x
0

(|ρ(t,x1)| + ρ0(x1)) dx1
∣∣∣ + ∫ t

0

|ρ(t1,x)||u(t1,x)|dt1

≤ 2B|x| + B2t

≤ 2B1|x| + B1t

≤ 2B1(1 + |x|)(1 + t), (t,x) ∈ [0,∞) ×R,

and

∣∣∣S21 (ρ,u)(t,x)∣∣∣ = ∣∣∣∫ x
0

(ρ(t,x1)u(t,x1) −ρ0(x1)u0(x1)) dx1

+

∫ t
0

(
ρ(t1,x)(u(t1,x))

2 + p(ρ(t1,x))
)
dt1

∣∣∣
≤

∣∣∣∫ x
0

(ρ(t,x1)u(t,x1) −ρ0(x1)u0(x1)) dx1
∣∣∣

+
∣∣∣∫ t

0

(
ρ(t1,x)(u(t1,x))

2 + p(ρ(t1,x))
)
dt1

∣∣∣
≤

∣∣∣∫ x
0

(|ρ(t,x1)||u(t,x1)| + ρ0(x1)u0(x1)) dx1
∣∣∣

+

∫ t
0

(
|ρ(t1,x)|(u(t1,x))2 + C(ρ(t1,x))q

)
dt1

≤ 2B2|x| + B3t + CBqt

≤ 2B1|x| + 2B1t

≤ 2B1(1 + |x|)(1 + t), (t,x) ∈ [0,∞) ×R.

This completes the proof. �

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 471



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

3. Proof of Theorem 1.1

(A1): Let A be a positive constant such that A ≤ 1 and g ∈C([0,∞)×R)
is a nonnegative function such that

16(1 + t)
(
1 + t + t2

)
(1 + |x|)

(
1 + |x| + x2

)∫ t
0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣∣dt1 ≤ A,
(t,x) ∈ [0,∞) ×R.

In the last section, we will give an example for a function g that satisfies (A1).
For (ρ,u) ∈ X2, define the operators

S12 (ρ,u)(t,x) =

∫ t
0

∫ x
0

(t− t1)2(x−x1)2g(t1,x1)S11 (ρ,u)(t1,x1)dx1dt1,

S22 (ρ,u)(t,x) =

∫ t
0

∫ x
0

(t− t1)2(x−x1)2g(t1,x1)S21 (ρ,u)(t1,x1)dx1dt1,

S2(ρ,u)(t,x) =
(
S12 (ρ,u)(t,x),S

2
2 (ρ,u)(t,x)

)
, (t,x) ∈ [0,∞) ×R.

Lemma 3.1. Suppose (H1)-(H2). For (ρ,u) ∈ X2, ‖(ρ,u)‖X2 ≤ B, we have

‖S2(ρ,u)‖X2 ≤ AB1,

where

B1 = max{B,B2,B3,CBq}.

Proof. We have

|S12 (ρ,u)(t,x)| =
∣∣∣∣
∫ t

0

∫ x
0

(t− t1)2(x−x1)2g(t1,x1)S11 (ρ,u)(t1,x1)dx1dt1
∣∣∣∣

≤
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)2(x−x1)2g(t1,x1)|S11 (ρ,u)(t1,x1)|dx1
∣∣∣∣dt1

≤ 2B1
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)2(x−x1)2g(t1,x1)(1 + t1)(1 + |x1|)dx1
∣∣∣∣dt1

≤ 8B1(1 + t)t2(1 + |x|)x2
∫ t

0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ 16B1(1 + t)(1 + t + t2)(1 + |x|)(1 + |x| + x2)

∫ t
0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ AB1, (t,x) ∈ [0,∞) ×R,

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 472



Classical solutions for the Euler equations

and∣∣∣∣ ∂∂tS12 (ρ,u)(t,x)
∣∣∣∣ = 2

∣∣∣∣
∫ t

0

∫ x
0

(t− t1)(x−x1)2g(t1,x1)S11 (ρ,u)(t1,x1)dx1dt1
∣∣∣∣

≤ 2
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)(x−x1)2g(t1,x1)|S11 (ρ,u)(t1,x1)|dx1
∣∣∣∣dt1

≤ 4B1
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)(x−x1)2g(t1,x1)(1 + t1)(1 + |x1|)dx1
∣∣∣∣dt1

≤ 16B1(1 + t)t(1 + |x|)x2
∫ t

0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ 16B1(1 + t)(1 + t + t2)(1 + |x|)(1 + |x| + x2)

∫ t
0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ AB1, (t,x) ∈ [0,∞) ×R,

and∣∣∣∣ ∂∂xS12 (ρ,u)(t,x)
∣∣∣∣ = 2

∣∣∣∣
∫ t

0

∫ x
0

(t− t1)2(x−x1)g(t1,x1)S11 (ρ,u)(t1,x1)dx1dt1
∣∣∣∣

≤ 2
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)2|x−x1|g(t1,x1)|S11 (ρ,u)(t1,x1)|dx1
∣∣∣∣dt1

≤ 4B1
∫ t

0

∣∣∣∣
∫ x

0

(t− t1)2|x−x1|g(t1,x1)(1 + t1)(1 + |x1|)dx1
∣∣∣∣dt1

≤ 8B1(1 + t)t2(1 + |x|)|x|
∫ t

0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ 16B1(1 + t)(1 + t + t2)(1 + |x|)(1 + |x| + x2)

∫ t
0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣dt1
≤ AB1, (t,x) ∈ [0,∞) ×R.

As above,∣∣S22 (ρ,u)(t,x)∣∣ ,
∣∣∣∣ ∂∂tS22 (ρ,u)(t,x)

∣∣∣∣ ,
∣∣∣∣ ∂∂xS22 (ρ,u)(t,x)

∣∣∣∣ ≤ AB1,
(t,x) ∈ [0,∞) ×R. Therefore,

‖S2(ρ,u)‖X2 ≤ AB1.
This completes the proof. �

Below, let

(A2): � ∈ (0, 1), A, B, B1 and q satisfy the inequalities �B1(1 + A) < 1
and AB1 < B.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 473



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

Let Ỹ denotes the union of the set {(ρ0,u0)} and the closure of the set of all
equi-continuous families in X2 with respect to the norm ‖ ·‖X2 . Let also,

Y = {(ρ,u) ∈ Ỹ : (ρ,u) ≥ 0, ‖(ρ,u)‖X2 ≤ B}.

Note that Y is a compact set in X2. For (ρ,u) ∈ X2, define the operators

T(ρ,u)(t,x) = −�(ρ,u)(t,x),

S(ρ,u)(t,x) = (ρ,u)(t,x) + �(ρ,u)(t,x) + �S2(ρ,u)(t,x), (t,x) ∈ [0,∞) ×R.

For (ρ,u) ∈ Y , using Lemma 3.1, we have

‖(I −S)(ρ,u)‖X2 = ‖�(ρ,u) + �S2(ρ,u)‖X2

≤ �‖(ρ,u)‖X2 + �‖S2(ρ,u)‖X2

≤ �B1 + �AB1

= �B1(1 + A)

< B.

Thus, S : Y → X2 is continuous and (I −S)(Y ) resides in a compact subset of
X2. Now, suppose that there is a (ρ,u) ∈ X2 so that ‖(ρ,u)‖X2 = B and

(ρ,u) = λ(I −S)(ρ,u)

or
1

λ
(ρ,u) = (I −S)(ρ,u) = −�(ρ,u) − �S2(ρ,u),

or (
1

λ
+ �

)
(ρ,u) = −�S2(ρ,u)

for some λ ∈
(
0, 1

�

)
. Hence, ‖S2(ρ,u)‖X2 ≤ AB1 < B,

�B <

(
1

λ
+ �

)
B =

(
1

λ
+ �

)
‖(ρ,u)‖X2 = �‖S2(ρ,u)‖X2 < �B,

which is a contradiction. Hence and Theorem 2.1, it follows that the operator
T + S has a fixed point (ρ∗,u∗) ∈ Y . Therefore,

(ρ∗,u∗)(t,x) = T(ρ∗,u∗)(t,x) + S(ρ∗,u∗)(t,x)

= −�(ρ∗,u∗)(t,x) + (ρ∗,u∗)(t,x) + �(ρ∗,u∗)(t,x) + �S2(ρ∗,u∗)(t,x),

(t,x) ∈ [0,∞) ×R, whereupon

0 = S2(ρ
∗,u∗)(t,x), (t,x) ∈ [0,∞) ×R.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 474



Classical solutions for the Euler equations

From here and from Lemma 2.10, it follows that (ρ∗,u∗) is a solution to the
IVP (1.1). This completes the proof.

4. Proof of Theorem 1.2

Let X2 be the space used in the previous section. Let

(A3): m > 0 be large enough and A, B, r, L, R1 be positive constants
that satisfy the following conditions

r < L < R1 ≤ B, � > 0, R1 +
A

m
B1 +

L

5m
>

(
2

5m
+ 1

)
L,

AB1 <
L

5
.

Let

P̃ = {(ρ,u) ∈ X2 : (ρ,u) ≥ 0 on [0,∞) ×R}.

With P we will denote the set of all equi-continuous families in P̃ . For (ρ,v) ∈
X2, define the operators

T1(ρ,v)(t,x) = (1 + m�)(ρ,v)(t,x) −
(
�
L

10
,�
L

10

)
,

S3(ρ,v)(t,x) = −�S2(ρ,v)(t,x) −m�(ρ,v)(t,x) −
(
�
L

10
,�
L

10

)
,

(t,x) ∈ [0,∞) × R. Note that any fixed point (ρ,v) ∈ X2 of the operator
T1 + S3 is a solution to the IVP (1.1). Define

U1 = Pr = {(ρ,v) ∈P : ‖(ρ,v)‖X2 < r},

U2 = PL = {(ρ,v) ∈P : ‖(ρ,v)‖X2 < L},

U3 = PR1 = {(ρ,v) ∈P : ‖(ρ,v)‖X2 < R1},

R2 = R1 +
A

m
B1 +

L

5m
,

Ω = PR2 = {(ρ,v) ∈P : ‖(ρ,v)‖X2 ≤ R2}.

(1) For (ρ1,v1), (ρ2,v2) ∈ Ω, we have

‖T1(ρ1,v1) −T1(ρ2,v2)‖X2 = (1 + m�)‖(ρ1,v1) − (ρ2,v2)‖X2,

whereupon T1 : Ω → X2 is an expansive operator with a constant
h = 1 + m� > 1.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 475



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

(2) For (ρ,v) ∈PR1 , we get

‖S3(ρ,v)‖X2 ≤ �‖S2(ρ,v)‖X2 + m�‖(ρ,v)‖X2 + �
L

10

≤ �
(
AB1 + mR1 +

L

10

)
.

Therefore, S3(PR1 ) is uniformly bounded. Since S3 : PR1 → X2 is
continuous, we have that S3(PR1 ) is equi-continuous. Consequently,
S3 : PR1 → X2 is completely continuous.

(3) Let (ρ1,v1) ∈PR1 . Set

(ρ2,v2) = (ρ1,v1) +
1

m
S2(ρ1,v1) +

(
L

5m
,
L

5m

)
.

Note that S12 (ρ1,v1) +
L
5
≥ 0, S22 (ρ1,v1) +

L
5
≥ 0 on [0,∞) × R. We

have ρ2,v2 ≥ 0 on [0,∞) ×R and

‖(ρ2,v2)‖X2 ≤ ‖(ρ1,v1)‖X2 +
1

m
‖S2(ρ1,v1)‖X2 +

L

5m

≤ R1 +
A

m
B1 +

L

5m

= R2.

Therefore, (ρ2,v2) ∈ Ω and

−εm(ρ2,v2) = −εm(ρ1,v1) −εS2(ρ1,v1) −ε
(
L

10
,
L

10

)
−ε

(
L

10
,
L

10

)
or

(I −T1)(ρ2,v2) = −εm(ρ2,v2) + ε
(
L

10
,
L

10

)
= S3(ρ1,v1).

Consequently, S3(PR1 ) ⊂ (I −T1)(Ω).
(4) Assume that for any (ρ1,u1) ∈ P∗ there exist λ ≥ 0 and (ρ,v) ∈

∂Pr ∩ (Ω + λ(ρ1,u1)) or (ρ,v) ∈ ∂PR1 ∩ (Ω + λ(ρ1,u1)) such that

S3(ρ,v) = (I −T1)((ρ,v) −λ(ρ1,u1)).

Then

−�S2(ρ,v) −m�(ρ,v) − �
(
L

10
,
L

10

)
= −m�((ρ,v) −λ(ρ1,u1)) + �

(
L

10
,
L

10

)
or

−S2(ρ,v) = λm(ρ1,u1) +
(
L

5
,
L

5

)
.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 476



Classical solutions for the Euler equations

Hence,

‖S2(ρ,v)‖X2 =
∥∥∥∥λm(ρ1,u1) +

(
L

5
,
L

5

)∥∥∥∥
X2

>
L

5
.

This is a contradiction.
(5) Let ε1 =

2
5m
. Assume that there exist a (ρ1,v1) ∈ ∂PL and λ1 ≥ 1 + ε1

such that λ1(ρ1,v1) ∈PR2 and

(4.1) S3(ρ1,v1) = (I −T1)(λ1(ρ1,v1)).

Since (ρ1,v1) ∈ ∂PL and λ1(ρ1,v1) ∈PR2 , it follows that(
2

5m
+ 1

)
L < λ1L = λ1‖(ρ1,v1)‖X2 ≤ R1 +

A

m
B1 +

L

5m
.

Moreover,

−�S2(ρ1,v1) −m�(ρ1,v1) − �
(
L

10
,
L

10

)
= −λ1m�(ρ1,v1) + �

(
L

10
,
L

10

)
,

or

S2(ρ1,v1) +

(
L

5
,
L

5

)
= (λ1 − 1)m(ρ1,v1).

From here,

2
L

5
>

∥∥∥∥S2(ρ1,v1) +
(
L

5
,
L

5

)∥∥∥∥
X2

= (λ1 − 1)m‖(ρ1,v1)‖X2 = (λ1 − 1)mL

and
2

5m
+ 1 > λ1,

which is a contradiction.

Therefore, all conditions of Theorem 2.8 hold. Hence, the IVP (1.1) has at
least two solutions (ρ1,u1) and (ρ2,u2) so that

‖(ρ1,u1)‖X2 = L < ‖(ρ2,u2)‖X2 ≤ R1
or

r ≤‖(ρ1,u1)‖X2 < L < ‖(ρ2,u2)‖X2 ≤ R1.

5. An Example

Below, we will illustrate our main results. Let q = 2, C = 1 and

R1 = B = 10, L = 5, r = 4, m = 10
50, A = � =

1

104
.

Then

B1 = max
{

10, 103
}

= 103

and

AB1 =
1

104
· 103 < B, �B1(1 + A) =

1

104
· 103

(
1 +

1

104

)
< B,

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 477



D. Boureni, S. Georgiev, A. Kheloufi and K. Mebarki

i.e., (A2) holds. Next,

r < L < R1 ≤ B, � > 0, R1 >
(

2

5m
+ 1

)
L, AB1 <

L

5
.

i.e., (A3) holds. Take

h(s) = log
1 + s11

√
2 + s22

1 −s11
√

2 + s22
, l(s) = arctan

s11
√

2

1 −s22
, s ∈ R, s 6= ±1.

Then

h′(s) =
22
√

2s10(1 −s22)
(1 −s11

√
2 + s22)(1 + s11

√
2 + s22)

,

l′(s) =
11
√

2s10(1 + s22)

1 + s44
, s ∈ R, s 6= ±1.

Therefore,

−∞ < lim
s→±∞

(1 + s + s2 + s3 + s4 + s5 + s6)h(s) < ∞,

−∞ < lim
s→±∞

(1 + s + s2 + s3 + s4 + s5 + s6)l(s) < ∞.

Hence, there exists a positive constant C1 so that

(1 + s + s2 + s3 + s4 + s5 + s6)

(
1

44
√

2
log

1 + s11
√

2 + s22

1 −s11
√

2 + s22
+

1

22
√

2
arctan

s11
√

2

1 −s22

)
≤ C1,

s ∈ R. Note that lim
s→±1

l(s) = π
2

and by [25] (pp. 707, Integral 79), we have

∫
dz

1 + z4
=

1

4
√

2
log

1 + z
√

2 + z2

1 −z
√

2 + z2
+

1

2
√

2
arctan

z
√

2

1 −z2
.

Let

Q(s) =
s10

(1 + s44)(1 + s + s2)2
, s ∈ R,

and

g1(t,x) = Q(t)Q(x), t ∈ [0,∞), x ∈ R.
Then, there exists a constant C2 > 0 such that

16(1 + t)
(
1 + t + t2

)
(1 + |x|)

(
1 + |x| + |x|2

)
∫ t

0

∣∣∣∣
∫ x

0

g1(t1,x1)dx1

∣∣∣∣∣dt1 ≤ C2, (t,x) ∈ [0,∞) ×R.
Let

g(t,x) =
A

C2
g1(t,x), (t,x) ∈ [0,∞) ×R.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 478



Classical solutions for the Euler equations

Then

16(1 + t)
(
1 + t + t2

)
(1 + |x|)

(
1 + |x| + |x|2

)
∫ t

0

∣∣∣∣
∫ x

0

g(t1,x1)dx1

∣∣∣∣∣dt1 ≤ A, (t,x) ∈ [0,∞) ×R,
i.e., (A1) holds. Therefore, for the IVP

∂tρ + ∂x(ρu) = 0,

∂t(ρu) + ∂x
(
ρu2 + ρ2

)
= 0, t > 0, x ∈ R,

ρ(0,x) = u(0,x) = 1
1+x8

, x ∈ R,

are fulfilled all conditions of Theorem 1.1 and Theorem 1.2.

Acknowledgements. The authors D. Boureni, A. Kheloufi and K. Mebarki
acknowledge support of ”Direction Générale de la Recherche Scientifique et du
Développement Technologique (DGRSDT)”, MESRS, Algeria.

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