Abstract: The propagation of ultra-short light pulses in silica optical fibers is modeled by the short pulse equation.
We introduce a nonlocal regularization of the problem and study the existence of solutions in a class of possibly
discontinuous functions.
Key-Words: Existence. Short pulse equation. Non-local formulation. Hyperbolc-elliptic system. Discontinuous
solutions. Cauchy problem.
Received: May 12, 2024. Revised: September 26, 2024. Accepted: October 18, 2024. Published: November 28, 2024.
1 Introduction
The short pulse equation reads
∂x∂tu+q∂xu3=bu, q, b ∈R,(1)
and has beed deduced in several context
• in [1] for the nonlinear propagation of optical
pulses of a few oscillations duration in dielectric
media,
• in [2] for the propagation of ultra-short light
pulses in silica optical fibers,
• in [3], [4], [5], [6], [7], [8] as non-slowly-varying
envelope approximation model that describes the
physics of few-cycle-pulse optical solitons,
• in [9], [10], [11] for pseudospherical surfaces,
• in [12] for the short pulse propagation in
nonlinear metamaterials characterized by a weak
Kerr-type nonlinearity in their dielectric
response,
• in [13], [14] in the context of plasma physics,
• in [15], [16] for the dynamics of radiating gases,
• in [17], [18], [19] for ultrafast pulse
propagation in a mode-locked laser cavity in the
few-femtosecond pulse regime.
The mathematical features of (1) have been widely
studied
• the wellposedness of the Cauchy problem in the
context of energy spaces can be found in [20],
[21], [22],
• the wellposedness of the Cauchy problem in the
context of entropy solution can be found in [23],
[24], [25],
• the wellposedness of the homogeneous initial
boundary value problem is in [26],
• the convergence of a finite difference numerical
scheme is studied in [27].
Here we regularize (1) with the following nonlocal
problem
∂tu+q∂xv=bP, t > 0, x ∈R,
∂xP=u, t > 0,x∈R,
α∂2
xv+β∂xv+γv =κu3, t > 0, x ∈R,
P(t, 0) = 0, t > 0,
u(0, x) = u0(x), x ∈R,
(2)
where q, b, α, β, γ, κ ∈R.
Nonlocal regularizations are widely used for
conservation laws
• in the context of traffic flow [28], [29], [30], [31],
Discontinuous Solutions for a non-local Regularization
of the Short Pulse Equation
GIUSEPPE MARIA COCLITE1, LORENZO DI RUVO2
1Dipartimento di Meccanica, Matematica e Management
Politecnico di Bari
via E. Orabona 4, 70125 Bari
ITALY
2Dipartimento di Matematica
Università di Bari
via E. Orabona 4, 70125 Bari
ITALY
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DOI: 10.37394/23206.2024.23.81
Giuseppe Maria Coclite, Lorenzo Di Ruvo
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Volume 23, 2024
• in the context of sedimentation [32],
• in the context of slow erosion [33], [34],
• in the context of the linearly polarized continuum
spectrum pulses in optical waveguides [35], [36].
Coherently with [23], [24], [37],[38].
• on the initial datum we assume
u0∈L1(R)∩L∞(R),ZR
u0(x)dx = 0; (3)
• on the function
P0(x) = Zx
−∞
u0(y)dy, (4)
we assume that
ZR
P0(x)dx
=ZRZx
−∞
u0(y)dydx = 0,
kP0k2
L2(R)
=ZRZx
−∞
u0(y)dy2
dx < ∞;
(5)
• on the constants q, b, α, β, γ, κ, we assume that
qβ
κ≥0, b =2qκ
γ, α, β, κ, γ 6= 0,(6)
or
b=2qκ
γ, α =−γ, β = 0, γ 6= 0.(7)
Since in (6) and (6) we assume α6= 0, it is not
restrictive to set it equal to 1. The assumptions (6)
and (7) are necessary to keep the solutions of (2) in
the energy space.
The main result of this paper is the following
theorem.
Theorem 1.1 Assume (3), (4), (5), and (6) or (7).
There exists a distributional solution (u, v, P )of (2)
such that
u∈L∞((0, T )×R)∩L∞(0, T ;L2(R)),
v∈H2((0, T )×R)∩L∞(0, T ;H2(R))∩
∩W1,∞((0, T )×R)∩
∩L∞(0, T ;W2,∞(R)),
∂tu∈L∞(0, T ;W1,∞(R))∩
∩L∞(0, T ;H1(R)),
∂t∂xv∈L∞(0, T ;L∞(R))∩
∩L∞(0, T ;L2(R)),
P∈L∞((0, T )×R)∩L∞(0, T ;L2(R)).
(8)
for every T > 0.
The well-posedness of (2) was proved in [39] and
[35] under the assumption:
u0∈L1(R)∩H2(R),(9)
and
u0∈L1(R)∩H1(R),(10)
respectively. Finally, we observe that the assumptions
(3), (4) and (5) are the ones used in [23], [24] in order
to prove the well-posedness of entropy solutions of
(1).
The remaining part of the manuscript is organized
as follows. Section 2 is dedicated to several a priori
estimates on a vanishing viscosity approximation of
(2). Those play a key role in the proof of our main
result, that is given in Section 3.
2 Vanishing Viscosity Approximation
Our existence argument is based on passing to the
limit in a vanishing viscosity approximation of (2).
Fix a small number 0<ε<1and let uε=
uε(t, x)be the unique classical solution of the
following mixed problem,[40]:
∂tuε+q∂xvε=bPε+ε∂2
xuε,
∂xPε=uε,
α∂2
xvε+β∂xvε+γvε=κu3
ε,
Pε(t, 0) = 0,
uε(0, x) = uε,0(x),
(11)
where t > 0,x∈Rand uε,0is aC∞approximation
of u0such that
kuε,0kL∞(R)≤ ku0kL∞(R),
kuε,0kL2(R)≤ ku0kL2(R),
ZR
uε,0(x)dx = 0,
kPε,0kL2(R)≤ kP0kL2(R),
ZR
Pε,0(x)dx = 0,
√εk∂xuε,0k2
L2(R)
+√ε
∂2
xuε,0
2
L2(R)≤C0
ε
∂3
xuε,0
2
L2(R)
+ε√ε
∂4
xuε,0
2
L2(R)≤C0,
√εk∂t∂xuε, 0kL2(R)≤C0,
uε, 0→u0(x)in Lp
loc(R)and a.e. in R,
(12)
and C0is a constant independent on ε.
Let us prove some a priori estimates on uε,Pεand
vε. We denote with C0the constants which depend
only on the initial data, and with C(T), the constants
which depend also on T.
Following [5, Lemma 2.1], we prove the following
result.
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Lemma 2.1 For each t > 0, we have that
Pε(t, −∞) = Pε(t, ∞) = 0,
Z0
−∞
uε(t, x)dx =Z∞
0
uε(t, x)dx = 0,
ZR
uε(t, x)dx = 0.
(13)
Remark 2.1 In light of (13), we have that
Pε(t, x) = Zx
0
uε(t, y)dy =Zx
−∞
uε(t, y)dy. (14)
Proof of Lemma 2.1. We begin by proving
Pε(t, −∞) = 0.(15)
Thanks to the smoothness of uε, from the first
equation of (11), we have
lim
x→−∞ (∂tuε+q∂xvε) = bPε(t, −∞) = 0,(16)
that is (15).
In a similar way, we can prove that
Pε(t, ∞) = 0.(17)
(15) and (17) give (13).
We prove (13). Integrating the second equation of
(11) (0, x), again by (11), we have
Pε(t, x) = Zx
0
uε(t, y)dy. (18)
(13) follows from (13) and (18).
Finally, we prove (13). We begin by observing
that, by (13),
Z0
−∞
uε(t, x)dx = 0.(19)
Therefore, by (13) and (19),
Z0
−∞
uε(t, x)dx +Z∞
0
uε(t, x)dx
=ZR
uε(t, x)dx = 0,
(20)
that is (13). ♠
Following [35, Lemma 2.5], we have the
following result.
Lemma 2.2 For each t≥0, we have that
Z−∞
0
Pε(t, x)dx =−1
b∂tPε(t, 0)
−q
bvε(t, 0) + ε
b∂xuε(t, 0),
Z∞
0
Pε(t, x)d=−1
b∂tPε(t, 0)x
−q
bvε(t, 0) + ε
b∂xuε(t, 0),
ZR
Pε(t, x)dx = 0.
(21)
Proof. We begin by proving (21). Integrating the first
equation on (0, x), we have that
Zx
0
∂tuε(t, y)dy +qvε(t, x)−qvε(t, 0)
−ε∂xuε(t, x) + ε∂xuε(t, 0)
=bZx
0
Pε(t, y)dy.
(22)
By (13), we obtain that
d
dt Z−∞
0
uε(t, x)dx
=Z−∞
0
∂tuε(t, x)dx = 0.
(23)
Moreover, the regularity of uεand vεgive
lim
x→−∞ (qvε(t, x)−ε∂xuε(t, x)) = 0.(24)
Therefore, by (22), (23) and (24),
bZx
0
Pε(t, y)dy =−qvε(t, 0) + ε∂xuε(t, 0),(25)
which gives (21).
We prove (21). Observe that by (13),
d
dt Z∞
0
uε(t, x)dx =Z∞
0
∂tuε(t, x)dx = 0,(26)
while, thanks to the regularity of uε, vε,
lim
x→∞ (qvε(t, x)−ε∂xuε(t, x)) = 0.(27)
(21) follows from (22), (26) and (27).
Finally, (21) and (21) give (21). ♠
Arguing as in [39, Lemma 2.2], we have the following
result.
Lemma 2.3 We have that
ZR
u3
ε∂xvεdx
=
β
κk∂xvε(t, ·)k2
L2(R),if (6) holds,
0,if (7) holds.
(28)
We continue with some L2type estimates of the
solution.
Lemma 2.4 Let T > 0. If (6) or (7) hold
kuε(t, ·)kL4(R)≤C(T),
kPε(t, ·)kL2(R)≤C(T),
kuε(t, ·)kL2(R)≤C(T),
εZt
0k(uε∂xuε)(s, ·)k2
L2(R)ds ≤C(T),
εZt
0kuε(s, ·)k2
L2(R)ds ≤C(T),
εZt
0k∂xuε(s, ·)k2
L2(R)ds ≤C(T),
(29)
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for every 0≤t≤T. In particular, if (6) holds, we
have
Zt
0k∂xvε(s, ·)k2
L2(R)ds ≤C(T),
kPεkL∞((0,T )×R)≤C(T),
(30)
for every 0≤t≤T.
Proof. We begin by observing that, thanks to (21), we
can consider the following function:
Fε(t, x) = Zx
−∞
Pε(t, y)dy. (31)
Integrating the second equation of (11) on (−∞, x),
thanks (31) and Remark 2.1, we have the following
equation:
∂tPε(t, x) + qvε(t, x)
=bFε(t, x) + ε∂xuε(t, x).(32)
Therefore, arguing as in [39, Lemma 2.3], we have
(29), (30) and (30). ♠
A key role in our compactness argument is played
by the following a priori estimates.
Lemma 2.5 Assume (6) or (7). Let T > 0. We have
that
k∂xvε(t, ·)kL∞(R)≤C(T),
,k∂xvε(t, ·)kL2(R)≤C(T),
kvε(t, ·)kL∞(R),kvε(t, ·)kL2(R)≤C(T),
(33)
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by observing that,
thanks to (29) and the Young inequality, we have that
κu3
ε(t, ·)∈L1(R),0≤t≤T. (34)
Therefore, by [35, Lemma 2.1], (33) holds. ♠
Arguing as in [5, Lemma 2.6] and [35, Lemma
2.8], we have the following result.
Lemma 2.6 Assume (6) or (7).We have that
kuεkL∞((0,T )×R)≤C(T),
∂2
xvε
L∞((0,T )×R)≤C(T),
∂2
xvε(t, ·)
L2(R)≤C(T),
(35)
for every 0≤t≤T.
In the next lemma we prove an H1energy
estimate.
Lemma 2.7 Assume (6) or (7). Fix T > 0. We have
that
√εk∂xuε(t, ·)k2
L2(R)
+2ε√εetZt
0
e−s
∂2
xuε(s, ·)
2
L2(R)ds
≤C(T),
(36)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by
−2√ε∂2
xuε, an integration on Rgives
√εd
dt k∂xuε(t, ·)k2
L2(R)
+2ε√ε
∂2
xuε(t, ·)
2
L2(R)
= 2√εb ZR
Pε∂2
xuεdx
−2q√εZR
∂2
xuε∂xvεdx.
(37)
Observe that, by (11) and (13),
2bZR
Pε∂2
xuεd
=−2bZR
∂xPε∂xuεdx
=−2bZR
uε∂xuεdx = 0.
(38)
Moreover,
2q√εZR
∂2
xuε∂xvεdx
=−2q√εZR
∂xuε∂2
xvεdx.
(39)
Consequently, by (37),
√εd
dt k∂xuε(t, ·)k2
L2(R)
+2ε√ε
∂2
xuε(t, ·)
2
L2(R)
=−2q√εZR
∂xuε∂2
xvεdx.
(40)
Since 0< ε < 1, thanks to (35) and the Young
inequality,
2√ε|q|ZR|∂xuε||∂xvε|dx
= 2√εZR|∂xuε||q∂2
xvε|dx
≤√εk∂xuε(t, ·)k2
L2(R)
+√εq2
∂2
xvε(t, ·)
2
L2(R)
≤√εk∂xuε(t, ·)k2
L2(R)
+q2
∂2
xvε(t, ·)
2
L2(R)
≤√εk∂xuε(t, ·)k2
L2(R)+C(T).
(41)
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Therefore, by (40),
√εd
dt k∂xuε(t, ·)k2
L2(R)
+ε√ε
∂2
xuε(t, ·)
2
L2(R)
≤√εk∂xuε(t, ·)k2
L2(R)+C(T).
(42)
The Gronwall Lemma and (12) give
√εk∂xuε(t, ·)k2
L2(R)
+2ε√εetZt
0
e−s
∂2
xuε(s, ·)
2
L2(R)ds
≤C0et+C(T)etZt
0
e−sds ≤C(T),
(43)
which gives (36). ♠
The following lemma gives an estimate on the
blow-up of the H3norm of the solution.
Lemma 2.8 Assume (6) or (7). We have that
√ε
∂3
xvε(t, ·)
2
L2(R)≤C(T),(44)
for every 0≤t≤T.
Proof. Differentiating the third equation of (11) with
respect to x, we have
α∂3
xvε+β∂2
xvε+γ∂xvε= 3κu2
ε∂xuε.(45)
Since
uε(t, ±∞) = ∂xuε(t, ±∞)
=vε(t, ±∞) = ∂xvε(t, ±∞)
=∂2
xvε(t, ±∞) = 0,
(46)
then
∂3
xvε(t, ±∞) = 0.(47)
Multiplying (46) by 2αε∂3
xvεan integration on Rof
(45 gives
2√εα2
∂3
xvε(t, ·)
2
L2(R)
= 6√εακ ZR
u2
ε∂xuε∂3
xvεdx
−2√εαβ ZR
∂2
xvε∂3
xvεdx
−2√εαγ ZR
∂xvε∂3
xvεdx.
(48)
Observe that, by (47),
−2√εαβ ZR
∂2
xvε∂3
xvεdx
=√εαβ ZR
∂x(∂2
xvε)2= 0,
(49)
and
−2√εαγ ZR
∂xvε∂3
xvεdx
= 2√εαγ
∂2
xvε(t, ·)
2
L2(R).
(50)
Consequently, since 0< ε < 1, by (35) and (48),
2√εα2
∂3
xvε(t, ·)
2
L2(R)
≤6√ε|α||κ|ZR
u2
ε|∂xuε||∂3
xvε|dx
+2√ε|γα|
∂2
xvε(t, ·)
2
L2(R)
≤6√ε|α||κ|ZR
u2
ε|∂xuε||∂3
xvε|dx
+C(T).
(51)
Due to (35), (36) and the Young inequality,
6√ε|α||κ|ZR
u2
ε|∂xuε||∂3
xvε|dx
= 2√εZR|3κu2
ε∂xuε||α∂3
xvε|dx
≤9√εκ2ZR
u4
ε(∂xuε)2dx
+√εα2
∂3
xvε(t, ·)
2
L2(R)
≤9√εκ2kuεk4
L∞((0,T )×R)×
×k∂xuε(t, ·)k2
L2(R)
+√εα2
∂3
xvε(t, ·)
2
L2(R)
≤C(T) + √εα2
∂3
xvε(t, ·)
2
L2(R).
(52)
It follows from (51) that
√εα2
∂3
xvε(t, ·)
2
L2(R)≤C(T),(53)
which gives (44). ♠
In the next lemma we prove an H2energy
estimate.
Lemma 2.9 Assume (6) or (7). Fix T > 0. We have
that
√ε
∂2
xuε(t, ·)
2
L2(R)
+2ε√εetZt
0
e−s
∂3
xuε(s, ·)
2
L2(R)ds
≤C(T),
(54)
and
4
√εk∂xuεkL∞((0,T )×R)≤C(T),(55)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by
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2√ε∂4
xuε, it follows from integration on Rthat
√εd
dt
∂2
xuε(t, ·)
2
L2(R)
+2ε√ε
∂3
xuε(t, ·)
2
L2(R)
= 2√εb ZR
Pε∂4
xuεdx
−2q√εZR
∂4
xuε∂xvεdx.
(56)
Observe that, by (11) and (13),
2b√εZR
Pε∂4
xuεdx
=−2bZR
∂xPε∂3
xuεdx
=−2√εZR
uε∂3
xuεdx
= 2bZR
∂xuε∂2
xuεdx = 0,
(57)
and
−2q√εZR
∂4
xuε∂xvεdx
= 2√εq ZR
∂3
xuε∂2
xvεdx
=−2qZR
∂2
xuε∂3
xvεdx.
(58)
It follows from (44), (56) and the Young inequality
that
√εd
dt
∂2
xuε(t, ·)
2
L2(R)
+2ε√ε
∂3
xuε(t, ·)
2
L2(R)
=−2√εq ZR
∂2
xuε∂3
xvεdx
≤2√ε|q|ZR|∂2
xuε||∂3
xvε|dx
≤√ε
∂2
xuε(t, ·)
2
L2(R)
+q2√ε
∂3
xvε(t, ·)
2
L2(R)
≤√ε
∂2
xuε(t, ·)
2
L2(R)+C(T).
(59)
The Gronwall Lemma and (12) give
√ε
∂2
xuε(t, ·)
2
L2(R)
+2ε√εetZt
0
e−s
∂3
xuε(s, ·)
2
L2(R)ds
≤C0+C(T)etZt
0
e−sds ≤C(T),
(60)
which gives (54).
Finally, we prove (55). Thanks to (36), (54) and
the Hölder inequality,
(∂xuε(t, x))2
= 2 Zx
−∞
∂xuε∂2
xuεdy
≤2ZR|∂xuε||∂2
xuε|dx
≤2k∂xuε(t, ·)kL2(R)
∂2
xuε(t, ·)
L2(R)
≤C(T)
√ε.
(61)
Hence,
√εk∂xuεk2
L∞((0,T )×R)≤C(T),(62)
which gives (55). ♠
The following lemma gives a bound on the time
derivative of the solution.
Lemma 2.10 Assume (6) or (7). We have that
k∂tuε(t, ·)kL2(R)≤C(T),(63)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by 2∂tuε,
an integration on Rgives
2k∂tuε(t, ·)k2
L2(R)
= 2bZR
∂tuεPεdx
+2εZR
∂tuε∂2
xuεdx
−2qZR
∂tuε∂xvεdx.
(64)
Since 0<ε<1, thanks to (29), (33), (36) and the
Young inequality,
2|b|ZR|∂tuε||Pε|dx
=ZR|∂tuε||2bPε|dx
≤1
2k∂tuε(t, ·)k2
L2(R)
+2b2kPε(t, ·)k2
L2(R)
≤1
2k∂tuε(t, ·)k2
L2(R)+C(T),
(65)
and
2εZR|∂tuε||∂2
xuε|dx
=ZR|∂tuε||2ε∂2
xuε|dx
≤1
2k∂tuε(t, ·)k2
L2(R)
+ε2
2
∂2
xuε(t, ·)
2
L2(R)
≤1
2k∂tuε(t, ·)k2
L2(R)
+√ε
2
∂2
xuε(t, ·)
2
L2(R)
≤1
2k∂tuε(t, ·)k2
L2(R)+C(T),
(66)
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and
2|q|ZR|∂tuε||∂xvε|dx
=ZR|∂tuε||2q∂xvε|dx
≤1
2k∂tuε(t, ·)k2
L2(R)
+2q2k∂xvε(t, ·)k2
L2(R)
≤1
2k∂tuε(t, ·)k2
L2(R)+C(T).
(67)
Therefore, by (64), we have that
1
2k∂tuε(t, ·)k2
L2(R)≤C(T),(68)
which gives (63). ♠
We continue with some estimates of the high order
derivatives of mixed type.
Lemma 2.11 Assume (6) or (7). We have that
k∂t∂xvε(t, ·)kL∞(R)≤C(T),
k∂t∂xvε(t, ·)kL2(R)≤C(T),
k∂tvε(t, ·)kL∞(R)≤C(T),
k∂tvε(t, ·)kL2(R)≤C(T),
(69)
for every 0≤t≤T.
Proof. Differentiating the third equation of (11) with
respect to t, we have that
α∂t∂2
xvε+β∂t∂xvε+γ∂tvε= 3κu2
ε∂tuε.(70)
We begin by observing that, thanks to (29), (63) and
the Young inequality, we have that
3κu2
ε(t, ·)∂tuε(t, ·)
L1(R)≤C(T),(71)
for every 0≤t≤T. Therefore, by [35, Lemma 2.1],
(69) holds. ♠
We continue with blow-up rate of the H4norm.
Lemma 2.12 Assume (6) or (7). We have that
ε
∂4
xvε(t, ·)
L2(R)≤C(T),(72)
for every 0≤t≤T.
Proof. Differentiating (45) with respect to x, we have
that
α∂4
xvε+β∂3
xvε+γ∂2
xvε
= 6κuε(∂xuε)2+ 3κu2
ε∂2
xuε.(73)
Observe that, since
∂2
xuε(t, ±∞) = 0,(74)
by (46) and (47),
∂4
xvε(t, ±∞) = 0.(75)
Multiplying (73 by 2εα∂4
xvε, an integration on R
gives
2α2ε
∂4
xvε(t, ·)
2
L2(R)
= 12ακε ZR
uε(∂xuε)2∂4
xvεdx
+6ακε ZR
u2
ε∂2
xuε∂4
xvεdx
−2αβε ZR
∂3
xvε∂4
xvεdx
−2αγε ZR
∂2
xvε∂4
xvεdx.
(76)
Observe that, by (47) and (75),
−2αβε ZR
∂3
xvε∂4
xvεdx
=−αβε ZR
∂x((∂3
xvε)2dx = 0,
(77)
and
−2αγε ZR
∂2
xvε∂4
xvεdx
= 2αγε
∂3
xvε(t, ·)
2
L2(R).
(78)
Consequently, since 0< ε < 1, by (44) and (76),
2α2ε
∂4
xvε(t, ·)
2
L2(R)
≤12|ακ|εZR|uε|(∂xuε)2|∂4
xvε|dx
+6|ακ|εZR
u2
ε|∂2
xuε||∂4
xvε|dx
+2|αγ|ε
∂3
xvε(t, ·)
2
L2(R)
≤12|ακ|εZR|uε|(∂xuε)2|∂4
xvε|dx
+6|ακ|εZR
u2
ε|∂2
xuε||∂4
xvε|dx
+2|αγ|√ε
∂3
xvε(t, ·)
2
L2(R)
≤12|ακ|εZR|uε|(∂xuε)2|∂4
xvε|dx
+6|ακ|εZR
u2
ε|∂2
xuε||∂4
xvε|dx
+C(T).
(79)
Since 0<ε<1, by (35), (36), (54), (55) and the
WSEAS TRANSACTIONS on MATHEMATICS
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Young inequality,
12|ακ|εZR|uε|(∂xuε)2|∂4
xvε|
=εZR|12κuε(∂xuε)2||α∂4
xvε|dx
≤72κ2εZR
u2
ε(∂xuε)4dx
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤72κ2εkuεk2
L∞((0,T )×R)ZR
(∂xuε)4dx
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤C(T)εZR
(∂xuε)4dx
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤C(T)εk∂xuεk2
L∞((0,T )×R)×
×k∂xuε(t, ·)k2
L2(R)
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤C(T) + α2ε
2
∂4
xvε(t, ·)
2
L2(R),
(80)
and
6|ακ|εZR
u2
ε|∂2
xuε||∂4
xvε|dx
=εZR|6κu2
ε∂2
xuε||α∂4
xvε|dx
≤18κ2εZR
u4
ε(∂2
xuε)2dx
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤18κ2√εkuεk4
L∞((0,T )×R)×
×
∂2
xuε(t, ·)
2
L2(R)
+α2ε
2
∂4
xvε(t, ·)
2
L2(R)
≤C(T) + α2ε
2
∂4
xvε(t, ·)
2
L2(R).
(81)
It follows from (79) that
α2ε
∂4
xvε(t, ·)
2
L2(R)≤C(T),(82)
which gives (72). ♠
In the next lemma we prove an H3energy
estimate.
Lemma 2.13 Assume (6) or (7). We have that
ε
∂3
xuε(t, ·)
2
L2(R)
+2ε2etZR
e−s
∂4
xuε(s, ·)
2
L2(R)ds
≤C(T),
(83)
and
8
√ε3
∂3
xuε
L∞((0,T )×R)≤C(T),(84)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by
−2ε∂6
xuε, we have that
−2ε∂6
xuε∂tuε
=−2bεPε∂6
xuε−2ε2∂2
xuε∂6
xuε
−2qε∂6
xuε∂xvε.
(85)
Observe that, thanks the second equation of (11) and
(13),
−2bε ZR
Pε∂6
xuεdx
= 2bε ZR
∂xPε∂5
xuεdx
= 2bε ZR
uε∂5
xuεdx
=−2bε ZR
∂xuε∂4
xuεdx
= 2bε ZR
∂2
xuε∂3
xuε= 0.
(86)
Moveover,
−2εZR
∂6
xuε∂tuεdx
= 2εZR
∂5
xuε∂t∂xuεdx
=−2εZR
∂4
xuε∂t∂2
xuεdx
= 2εZR
∂3
xuε∂t∂3
xuεdx
=εd
dt
∂3
xuε(t, ·)
2
L2(R),
(87)
and
−2ε2ZR
∂2
xuε∂6
xuεdx
= 2ε2ZR
∂5
xuε∂3
xuε
=−2ε2
∂4
xuε(t, ·)
2
L2(R),
(88)
and
−2qε ZR
∂6
xuε∂xvεdx
= 2qε ZR
∂5
xuε∂2
xvεdx
=−2qε ZR
∂4
xuε∂3
xvεdx
= 2qε ZR
∂3
xuε∂4
xvεdx.
(89)
It follows from (86), (88) and an integration of (85)
on Rthat
ε
∂3
xuε(t, ·)
2
L2(R)
+2ε2
∂4
xuε(t, ·)
2
L2(R)
= 2qε ZR
∂3
xuε∂4
xvεdx.
(90)
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Due to (72) and the Young inequality,
2|q|εZR|∂3
xuε||∂4
xvε|dx
≤ε
∂3
xuε(t, ·)
2
L2(R)
+q2ε
∂4
xvε(t, ·)
2
L2(R)
≤ε
∂3
xuε(t, ·)
2
L2(R)+C(T).
(91)
Therefore, by (90),
ε
∂3
xuε(t, ·)
2
L2(R)
+2ε2
∂4
xuε(t, ·)
2
L2(R)
≤ε
∂3
xuε(t, ·)
2
L2(R)+C(T).
(92)
The Gronwall Lemma and (12) give (83).
Finally, we prove (84). Thanks to (54), (83) and
the Hölder inequality,
(∂2
xuε(t, x))2
= 2 Zx
−∞
∂2
xuε∂3
xuεdy
≤2ZR|∂2
xuε||∂3
x|dx
≤
∂2
xuε(t, ·)
2
L2(R)
∂3
xuε(t, ·)
2
L2(R)
≤C(T)
4
√ε3.
(93)
Hence,
4
√ε3
∂2
xuε
2
L2((0,T )×R)≤C(T),(94)
which gives (84). ♠
We prove an uniform L∞bound on the time
derivative.
Lemma 2.14 Assume (6) or (7). We have that
k∂tuεkL∞((0,T )×R)≤C(T).(95)
Proof. By the first equation of (11), (30) and (33), we
have
|∂tuε|
=|bPε−q∂xvε+ε∂2
xuε|
≤ |b||Pε|+|q|+ε|∂2
xuε|
≤ |b|kPεkL∞((0,T )×R)
+|q|k∂xvεkL∞((0,T )×R)
+ε
∂2
xuε
L∞((0,T )×R)
≤C(T) + ε
∂2
xuε
L∞((0,T )×R).
(96)
Since 0< ε < 1, thanks to (84),
ε
∂2
xuε
L∞((0,T )×R)
=8
√ε58
√ε3
∂2
xuε
L∞((0,T )×R)
≤8
√ε3
∂2
xuε
L∞((0,T )×R)≤C(T).
(97)
It follows from (96) and (97) that
|∂tuε| ≤ C(T),(98)
which gives (95). ♠
We prove an uniform L2bound on the mixed time-
space second derivative.
Lemma 2.15 Assume (6) or (7). We have that
k∂t∂xuε(t, ·)kL2(R)≤C(T),(99)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by
−2∂t∂2
xuε, we have that
−2∂t∂2
xuε∂tuε
=−2b∂t∂2
xuεPε+ 2q∂t∂2
xuε∂xvε
−2ε∂t∂2
x∂2
xuε.
(100)
Observe that by the second equation of (11),
−2bZR
∂t∂2
xuεPεdx
= 2bZR
∂xPε∂t∂xuεdx
= 2bZR
uε∂t∂xuεdx.
(101)
Moreover,
−2ZR
∂t∂2
xuε∂tuεdx
= 2 k∂t∂xuε(t, ·)k2
L2(R),
2qZR
∂t∂2
xuε∂xvεdx
=−2qZR
∂t∂xuε∂2
xvεdx,
−2εZR
∂t∂2
xuε∂2
xuεdx
= 2εZR
∂t∂xuε∂3
xuεdx.
(102)
An integration of (100) on R, (101) and (102) give
2k∂t∂xuε(t, ·)k2
L2(R)
= 2bZR
uε∂t∂xuεdx
−2qZR
∂t∂xuε∂2
xvεdx
+2εZR
∂t∂xuε∂3
xuεdx.
(103)
Since 0< ε < 1, due to (29), (35), (83) and the Young
inequality,
2|b|ZR|uε||∂t∂xuε|dx
≤b2kuε(t, ·)k2
L2(R)
+k∂t∂xuε(t, ·)k2
L2(R)
≤C(T) + k∂t∂xuε(t, ·)k2
L2(R),
(104)
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and
2εZR|∂t∂xuε||∂3
xuε|dx
=ZR|∂t∂xuε||2ε∂3
xuε|dx
≤1
2k∂t∂xuε(t, ·)k2
L2(R)
+2ε2
∂3
xuε(t, ·)
2
L2(R)
≤1
2k∂t∂xuε(t, ·)k2
L2(R)
+2ε
∂3
xuε(t, ·)
2
L2(R)
≤1
2k∂t∂xuε(t, ·)k2
L2(R)+C(T),
(105)
and
2|q|ZR|∂t∂xuε||∂2
xvε|dx
=ZR
√2∂t∂2
xuε
√3
√6q∂2
xvεdx
dx
≤1
3k∂t∂xuε(t, ·)k2
L2(R)
+3q2
∂2
xvε(t, ·)
2
L2(R)
≤1
3k∂t∂xuε(t, ·)k2
L2(R)+C(T).
(106)
Therefore, by (103),
1
6k∂t∂xuε(t, ·)k2
L2(R)≤C(T),(107)
which gives (99). ♠
We continue with the blow-up rate of the H5norm
of the solution.
Lemma 2.16 Assume (6) or (7). We have that
ε√ε
∂5
xvε(t, ·)
L2(R)≤C(T),(108)
for every 0≤t≤T.
Proof. Differentiating (73) with respect to x, we have
α∂5
xvε+β∂4
xvε+γ∂3
xvε
= 6κ(∂xuε)3+ 18κuε∂xuε∂2
xuε
+3κu2
ε∂3
xuε.
(109)
Observe that, since ∂3
xuε(t, ±)=0, by (46), (47),
(74) and (75),
∂5
xvε(t, ±∞) = 0.(110)
Multiplying (109) by 2ε√εα∂5
xvε, an integration on
Rgives
2ε√εα2
∂5
xvε(t, ·)
2
L2(R)
= 12ε√εακ ZR
(∂xuε)3∂5
xvεdx
+32ε√εακ ZR
uε∂xuε∂2
xuε∂5
xvεdx
+6ε√εακ ZR
u2
ε∂3
xuε∂5
xvεdx
−2ε√εαβ ZR
∂4
xvε∂5
xvεdx
−2ε√εαγ ZR
∂3
xvε∂5
xvεdx.
(111)
Since 0< ε < 1, thanks to (47), (75) and (110),
−2ε√εαβ ZR
∂4
xvε∂5
xvεdx
=−ε√εZR
∂x((∂4
xvε))2dx = 0,
(112)
and
−2ε√εαγ ZR
∂3
xvε∂5
xvεdx
= 2ε√εαγ
∂4
xvε(t, ·)
2
L2(R)
≤2ε|αγ|
∂4
xvε(t, ·)
2
L2(R).
(113)
Therefore, by (111),
2ε√εα2
∂5
xvε(t, ·)
2
L2(R)
≤12ε√ε|ακ|ZR|∂xuε|3|∂5
xvε|dx
+32ε√ε|ακ|ZR|uε∂xuε∂2
xuε|×
|∂5
xvε|dx
+6ε√ε|ακ|ZR|u2
ε∂3
xuε||∂5
xvε|dx
+C(T).
(114)
Since 0<ε<1, due to (35), (36), (54), (55), (83)
and the Young inequality,
12ε√ε|ακ|ZR|∂xuε|3|∂5
xvε|dx
=ε√εZR|12κ(∂xuε)3||α∂5
xvε|dx
≤72κ2ε√εZR
(∂xuε)6dx
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤72κ2ε√εk∂xuεk4
L∞((0,T )×R)×
×k∂xuε(t, ·)k2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤√εC(T)k∂xuε(t, ·)k2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T) + ε√εα2
2
∂5
xvε(t, ·)
2
L2(R),
(115)
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and
32ε√ε|ακ|ZR|uε∂xuε∂2
xuε||∂5
xvε|dx
=ε√εZR|32κuε∂xuε∂2
xuε||α∂5
xvε|dx
≤512κ2ε√εZR
u2
ε(∂xuε)2(∂2
xuε)2dx
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤512κ2ε√εkuεk2
L∞((0,T )×R)×
×ZR
(∂xuε)2(∂2
xuε)2dx
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T)ε√εk∂xuεk2
L∞((0,T )×R)×
×
∂2
xuε(t, ·)
2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T) + ε√εα2
2
∂5
xvε(t, ·)
2
L2(R),
(116)
and
6ε√ε|ακ|ZR|u2
ε∂3
xuε||∂5
xvε|dx
=ε√εZR|6κu2
ε∂3
xuε||α∂5
xvε|dx
≤18κ2ε√εZR
u4
ε(∂3
xuε)2dx
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤18κ2kuεk4
L∞((0,T )×R)×
×
∂3
xuε(t, ·)
2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T)ε√ε
∂3
xuε(t, ·)
2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T)ε
∂3
xuε(t, ·)
2
L2(R)
+ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)
≤C(T) + ε√εα2
2
∂5
xvε(t, ·)
2
L2(R).
(117)
Consequently, by (114),
ε√εα2
2
∂5
xvε(t, ·)
2
L2(R)≤C(T),(118)
which gives (108). ♠
We continue by proving an H4energy type
estimate.
Lemma 2.17 Assume (6) or (7). We have that
ε√ε
∂4
xuε(t, ·)
2
L2(R)
+2ε2√εet
2Zt
0
e−s
∂5
xuε(s, ·)
2
L2(R)ds
≤C(T),
(119)
and
8
√ε5
∂3
xuε
L∞((0,T )×R)≤C(T),(120)
for every 0≤t≤T.
Proof. Multiplying the first equation of (11) by
2ε√ε∂8
xuε, we get
2ε√ε∂8
xuε∂tuε
= 2bε√εPε∂8
xuε
+2ε2√ε∂2
xuε∂8
xuε
−2qε√ε∂8
xuε∂xvε.
(121)
Observe that by (13) and the second equation of (11),
2bε√εZR
Pε∂8
xuεdx
= 2bε√εZR
∂xPε∂7
xuεdx
= 2bε√εZR
uε∂7
xuεdx
= 2bε√εZR
∂xuε∂6
xuεdx
= 2bε√εZR
∂2
xuε∂5
xuεdx
= 2bZR
∂3
xuε∂4
xuεdx = 0.
(122)
Moreover,
2ε√εZR
∂8
xuε∂tuεdx
=−2ε√εZR
∂7
xuε∂t∂xuεdx
= 2ε√εZR
∂6
xuε∂t∂2
xuεdx
=−2ε√εZR
∂5
xuε∂t∂3
xuεdx
=ε√εd
dt
∂4
xuε(t, ·)
2
L2(R),
(123)
and
2ε2√εZR
∂2
xuε∂8
xuεdx
=−2ε2√εZR
∂3
xuε∂7
xuεdx
= 2ε2√εZR
∂4
xuε∂6
xuεdx
=−2ε2√ε
∂5
xuε(t, ·)
2
L2(R),
(124)
and
−2qε√εZR
∂8
xuε∂xvεdx
= 2qε√εZR
∂7
xuε∂2
xvεdx
=−2qε√εZR
∂6
xuε∂3
xvεdx
= 2qε√εZR
∂5
xuε∂4
xvεdx
=−2qε√εZR
∂4
xuε∂5
xvεdx.
(125)
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Integrating (121), by (122) and (124), we have that
ε√εd
dt
∂4
xuε(t, ·)
2
L2(R)
+2ε2√ε
∂5
xuε(t, ·)
2
L2(R)
=−2qε√εZR
∂4
xuε∂5
xvεdx.
(126)
Due to (108) and the Young inequality,
2|q|ε√εZR|∂4
xuε||∂5
xvε|dx
≤ε√ε
∂4
xuε(t, ·)
2
L2(R)
+q2ε√ε
∂4
xuε(t, ·)
2
L2(R)
≤ε√ε
∂4
xuε(t, ·)
2
L2(R)+C(T).
(127)
It follows from (126) that
ε√εd
dt
∂4
xuε(t, ·)
2
L2(R)
+2ε2√ε
∂5
xuε(t, ·)
2
L2(R)
≤ε√ε
∂4
xuε(t, ·)
2
L2(R)+C(T).
(128)
The Gronwall Lemma and (12) gives (119).
Finally, we prove (120). Thanks to (83), (119) and
the Hölder inequality,
(∂3
xuε(t, x))2
= 2 Zx
−∞
∂3
xuε∂4
xuεdx
≤2ZR|∂3
xuε|∂4
xuε|dx
≤
∂3
xuε(t, ·)
L2(R)
∂3
xuε(t, ·)
L2(R)
≤C(T)
4
√ε5.
(129)
Hence,
4
√ε5
∂3
xuε
2
L∞((0,T )×R)≤C(T),(130)
which gives (120). ♠
We show an uniform L∞bound on the second
order mixed derivative.
Lemma 2.18 Assume (6) or (7). We have that
k∂t∂xuεkL∞((0,T )×R)≤C(T).(131)
Proof. Differentiating the first equation of (11) with
respect to, thanks to the second one of (11), we have
that
∂t∂xuε=buε+ε∂3
xuε−q∂3
xvε.(132)
Due to (35) and (35),
|∂t∂xuε|
=|buε−q∂3
xvε+ε∂3
xuε|
≤ |b||uε|+|q||∂2
xvε|+ε|∂3
xuε|
≤ |b|kuεkL∞((0,T )×R)
+|q|
∂2
xvε
L∞((0,T )×R)
+ε
∂3
xuε
L∞((0,T )×R)
≤C(T) + ε
∂3
xuε
L∞((0,T )×R).
(133)
Observe that, since 0< ε < 1, thanks to (120),
ε
∂3
xuε
L∞((0,T )×R)
=8
√ε38
√ε5
∂3
xuε
L∞((0,T )×R)
≤8
√ε5
∂3
xuε
L∞((0,T )×R)≤C(T).
(134)
(131) follows from (133) and (134). ♠
Consider the fast decaying function
χ(x) = e−|x|, x ∈R,(135)
that satisfies
0≤χ≤1,|χ0|=χ. (136)
We prove the following result
Lemma 2.19 Assume (6) or (7). We have that
εZR
(∂t∂xuε)2χdx
+Zt
0ZR
(∂2
tuε)2χdtdx ≤C(T),
(137)
for every 0≤t≤T.
Proof. Differentiating the first equation of (11) with
respect to t, we have
∂2
tuε=b∂tPε+ε∂t∂2
xuε−q∂t∂xvε.(138)
Multiplying (138) by 2∂2
tuεχ, and integration on R
give,
2ZR
(∂2
tuε)2χdx
= 2bZR
∂tPε∂2
tuεχdx
+2εZR
∂t∂2
xuε∂2
tuεχdx
−2qZR
∂t∂xvε∂2
tuεχdx.
(139)
Observe that
2εZR
∂t∂2
xuε∂2
tuεχdx
=−εd
dt ZR
χ(∂tuε)2dx
−2εZR
∂t∂xuε∂2
tuεχ0dx.
(140)
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Consequently, by (139),
εd
dt ZR
χ(∂tuε)2dx
+2 ZR
(∂2
tuε)2χdx
= 2bZR
∂tPε∂2
tuεχdx
−2εZR
∂t∂xuε∂2
tuεχ0dx
−2qZR
∂t∂xvε∂2
tuεχdx.
(141)
Since 0< ε < 1, thanks to (69), (99) and the Young
inequality,
2εZR|∂t∂xuε||∂2
tuε||χ0|dx
≤2C0ZR|∂t∂xuε||∂2
tuε|χdx
≤C0ZR
χ(∂t∂xuε)2dx
+1
2ZR
(∂2
tuε)2χdx
≤C0k∂t∂xuε(t, ·)k2
L2(R)
+1
2ZR
(∂2
tuε)2χdx
≤C(T) + 1
2ZR
(∂2
tuε)2χdx,
2|b|ZR|∂tPε||∂2
tuε|χdx
≤2b2ZR
(∂tPε)2χdx
+1
2ZR
(∂2
tuε)2χdx,
2|q|ZR|∂t∂xvε||∂2
tuε|χdx
≤2q2ZR
χ(∂t∂xvε)2dx
+1
2ZR
(∂2
tuε)2χdx
≤C0k∂t∂xvε(t, ·)k2
L2(R)
+1
2ZR
(∂2
tuε)2χdx
≤C(T) + 1
2ZR
(∂2
tuε)2χdx.
(142)
It follows from (141) that
εd
dt ZR
χ(∂tuε)2dx
+1
2ZR
(∂2
tuε)2χdx
≤C(T)+2b2ZR
(∂tPε)2χdx.
(143)
Observe that, by the second equation of (11),
∂tPε=Zx
0
∂tuε(t, y)dy. (144)
Therefore, by (144), (63) and the Jensen inequality
2b2ZR
(∂tPε)2χdx
=ZR
χZx
0
∂tuε(t, y)dy2
dx
≤ZR
χ|x|
Zx
0
(∂tuε)2dy
dx
≤ k∂tuε(t, ·)k2
L2(R)ZR
χ|x| ≤ C(T).
(145)
Thus, by (143), we have
εd
dt ZR
χ(∂tuε)2dx +1
2ZR
(∂2
tuε)2χdx
≤C(T).
(146)
Integrating on (0, t), by (12), we get
εZR
χ(∂tuε)2dx
+1
2Zt
0ZR
(∂2
tuε)2χdsdx
≤C0+C(T)t≤C(T)
(147)
that is (137). ♠
3 Proof of Theorem 1.1
This section is devoted to the proof of Theorem 1.1.
Proof of Theorem 1.1. Thanks to Lemmas 2.4, 2.5,
2.6, 2.10, 2.11, 2.15, (95), (131), and (2.19),
{∂tuε}ε>0is bounded in H1
loc((0,∞)×R)(148)
Consequentially, there exists w∈H1
loc((0,∞)×R)
such that
∂tuε* w in H1
loc((0,∞)×R),
∂tuε→win Lp
Loc((0,∞)×R),
1≤p < ∞and a.e. in (0,∞)×R.
(149)
We define the following functon:
u(t, x) = Zt
0
w(s, x)ds +u0(x).(150)
We prove that
uε→uin Lp
Loc((0,∞)×R),
1≤p < ∞and a.e. in (0,∞)×R.(151)
Observe that
uε(t, x) = Zt
0
uε(s, x)ds +uε, 0(x).(152)
consequentially, we have that
uε(t, x)−u(t, x)
=Zt
0
(∂tuε(s, x)−w(s, x))ds
+uε, 0(x)−u0(x).
(153)
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Therefore, by (149),
ZT
0ZR
−R|uε(t, x)−u(t, x)|dtdx
≤ZT
0ZR
−R
Zt
0|∂tuε(s, x)−w(t, x)|dsdtdx
+TZR
−R|uε, 0(x)−u0(x)|dx →0,
(154)
which gives (151).
By (151), we have that
Pεκ→Pin Lp
loc((0, T ); W1, p(R))
1≤p < ∞, and a.e. in (0,∞)×R,(155)
where
P(t, x) = Zx
0
u(t, y)dy, t > 0, x ∈R.(156)
Moreover, thanks to Lemmas 2.4, 2.5, 2.6, 2.10, 2.11,
2.15 , (95), (131), and (2.19),
{vε}ε>0is bounded in H1
loc((0,∞)×R)(157)
Consequentially, there exists v∈H1
loc((0,∞)×R)
such that
vε* v in H1
loc((0,∞)×R),
vε→vin Lp
Loc((0,∞)×R),
1≤p < ∞and a.e. in (0,∞)×R.
(158)
Therefore, the triple (u, v, P )is a distributional
solution of (2) and (8) hold. ♠
4 Conclusion
We consider the short pulse equations that is a second
order evolutive PDE that appear in the
modeling of several physical and mathematical phe-
nomena. Moreover, if can rewritten in the form of an
hyperbolic equation of the first order with a
nonlocal source term. Here we consider a nonlocal
regularization fo the flux and studied the existence
of possibly discontinuous solutions using a vanishing
viscosity type argument and energy type estimates.
Acknowledgment:
GMC is member of the Gruppo Nazionale per
l’Analisi Matematica, la Probabilità e le loro
Applicazioni (GNAMPA) of the Istituto Nazionale di
Alta Matematica (INdAM).
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Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
The authors equally contributed in the present
research, at all stages from the formulation of
the problem to the final findings and solution.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
GMC has been partially supported by the Project
funded under the National Recovery and Resilience
Plan (NRRP), Mission 4 Component 2 Investment 1.4
-Call for tender No. 3138 of 16/12/2021 of Italian
Ministry of University and Research funded by the
European Union -NextGenerationEUoAward
Number: CN000023, Concession Decree No. 1033
of 17/06/2022 adopted by the Italian Ministry of
University and Research, CUP: D93C22000410001,
Centro Nazionale per la Mobilità Sostenibile, the
Italian Ministry of Education, University and
Research under the Programme Department of
Excellence Legge 232/2016 (Grant No. CUP -
D93C23000100001), and the Research Project of
National Relevance “Evolution problems involving
interacting scales” granted by the Italian Ministry
of Education, University and Research (MIUR Prin
2022, project code 2022M9BKBC, Grant No. CUP
D53D23005880006).
Conflicts of Interest
The authors have no conflicts of interest to
declare that are relevant to the content of this
article.
Creative Commons Attribution License 4.0
(Attribution 4.0 International , CC BY 4.0)
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Creative Commons Attribution License 4.0
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