Electronic Journal of Differential Equations, Vol. 2006(2006), No. 133, pp. 1–6.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
ON EXACT SOLUTIONS OF A DEGENERATE QUASILINEAR
WAVE EQUATION WITH SOURCE TERM
NOUR-EDDINE AMROUN, ABBÈS BENAISSA
Abstract. In this paper, we construct exact solutions of the Cauchy problem
of degenerate quasilinear wave equation of Kirchhoff type, and study the effect
of the nonlinear terms on the existence of solutions. We construct solutions
that exist globally for some initial data, and that blow up in a finite time for
some other initial data.
1. Introduction
In this paper, we construct exact solutions of the following initial-value problem of a quasilinear wave equation with nonlinear source term, and study their
asymptotic behaviour concerning the time variable,
(|u0 |l−2 u0 )0 − M (kux k22 )uxx = γ|u|p−1 u,
in R × [0, +∞[,
(1.1)
u(x, 0) = u0 (x), u (x, 0) = u1 (x)in R.
R∞
p−1
where M (r) = r 2 , kux k22 = −∞ |ux (x, t)|2 dx, l > 2, p > l − 1 and γ > 0 are
constants.
For problem (1.1), when l > 2 and M ≡ 1 without source term and with nonlinear
dissipation, Benaissa and Mimouni [3] determined suitable relations between l and
p, so that the energy decays exponentially or alternatively polynomially. More
precisely, they showed that the energy of the solutions decays with exponentially if
l + 1 ≥ p and decays polynomially if l + 1 < p.
For problem (1.1) (in the case of bounded domain), when l > 2 and M is not
a constant function, with nonlinear dissipation, Benaissa and Messaoudi [2] have
investigated the blowup of solutions. They show that, for suitably chosen initial
data and a relation between l and p, any classical solution blows up in finite time.
When l = 2, the equation without a source term is often called the wave equation
of Kirchhoff type which has been introduced for studying nonlinear vibrations of an
elastic string by Kirchhoff [9]. The existence of local and global solutions in Sobolev
and Gevrey classes was investigated by many authors; see for example [7, 6, 8].
When l = 2, Ebihara, Hoshino and Kurokiba [4] constructed one of the solutions
for (1.1) and studied their behavior in t. They have shown that there exists a
solution which blows up at a finite time under some initial condition and in the other
0
2000 Mathematics Subject Classification. 35C05, 35L70.
Key words and phrases. Kirchhoff equation; exact solutions.
c
2006
Texas State University - San Marcos.
Submitted June 21, 2006. Published October 19, 2006.
1
2
N.-E. AMROUN, A. BENAISSA
EJDE-2006/133
2
case there exists a global solution which decays with the order O(t− p−1 )(t → ∞).
We think that the interaction of the term (|u0 |l−2 u0 )0 (l > 2) and the source term
|u|p−1 u and the velocity M (r) have an effect on the result of [4].
We construct the solution of the form u(x, t) = v(x)ϕ(t) for (1.1) when M (r) =
p−1
r 2 , p > 1, l > 2 and p + 3 = 2l and we study the behavior of the solutions as the
time t increases.
2. Separation of variables
Let u(x, t) = v(x)ϕ(t) and substitute in (1.1). Then (1.1) is changed to
Z ∞
p−1
2
|v|l−2 v(|ϕ0 |l−2 ϕ0 )0 −
|vx (x, t)|2 dx
−∞
× |ϕ|p−1 ϕvxx − γ|v|p−1 v|ϕ|p−1 ϕ = 0,
x ∈ R,
u(x, 0) = v(x)ϕ(0) = ϕ0 v(x),
x ∈ R,
ut (x, 0) = v(x)ϕt (0) = ϕ1 v(x),
t > 0,
(2.1)
x ∈ R.
Here we consider the case ϕ(t) > 0, so that the first equation of (2.1) is equivalent
to
αp−1 vxx + γ|v|p−1 v
(|ϕ0 |l−2 ϕ0 )0
=
= λ,
(2.2)
l−2
|v| v
ϕp
R∞
where α2 = −∞ |vx (x)|2 dx and λ is a positive constant. Therefore, we obtain two
problems from (2.2): First problem
αp−1 vxx − λ|v|l−2 v + γ|v|p−1 v = 0,
Z ∞
2
α =
|vx |2 dx < +∞ ,
x ∈ R,
(2.3)
−∞
and second problem
(|ϕ0 |l−2 ϕ0 )0 = λϕp ,
ϕ(0) = ϕ0 ,
ϕ(t) ≥ 0,
t ≥ 0,
ϕt (0) = ϕ1 ,
(2.4)
t ≥ 0.
3. First Problem
In this section, we construct a positive solution v(x) of (2.3) with lim|x|→∞ v(x) =
0. For this purpose we study the problem
αp−1 vxx − λ|v|l−2 v + γ|v|p−1 v = 0,
x ∈ R,
v(0) = A (a positive constant),
vx (0) = 0,
lim v(x) = 0,
(3.1)
|x|→∞
α2
=
2
Z
∞
|vx (x)|2 dx < +∞,
0
where α is a fixed positive number. If v(x) is a solution of (3.1), then we can solve
(2.3) by setting v(−x) = v(x) for x > 0, because of the second equation in (3.1).
EJDE-2006/133
ON EXACT SOLUTIONS
3
Multiplying the first equation of (3.1) by 2vx and integrating from 0 to x, we obtain
2λ l
2γ p+1 2λ l
2γ
v −
v
−
A +
Ap+1 .
l
p+1
l
p+1
If we choose A > 0 such that
1
(p + 1)λ p+1−l
A=
,
γl
αp−1 vx2 =
(3.2)
(3.3)
then (3.2) implies
s
p−1
2
α
vx = ∓
2λ l
2γ p+1
v −
v
.
l
p+1
(3.4)
Here we consider the case where v is positive and vx < 0, so that we treat the
following equation which is derived from (3.4):
p−1
vx
q
= −α− 2 .
(3.5)
l
2γ
2λ
p+1−l
v2
l − p+1 v
If we integrate (3.5) from c to v, then we obtain
Z c
p−1
dz
q
= α− 2 x.
l
2γ p+1−l
2λ
v z2
l − p+1 z
(3.6)
If there exists x∗ ∈ (0, ∞) such that v(x∗ ) = 0, then we get
Z c
p−1
dz
q
= α− 2 x∗ .
l
2γ
2λ
p+1−l
0 z2
l − p+1 z
But one can easily show
Rc
dz
0 z 2l
q
2γ
2λ
p+1−l
l − p+1 z
= ∞ with use of (3.3) and α−
p−1
2
x∗ <
+∞, thus v(x) is monotone decreasing and v(x) > 0. And if limx→∞ v(x) = k > 0,
then from (3.6), we obtain
Z c
p−1
dz
q
= lim α− 2 x (= ∞).
l
x→∞
2γ
2λ
p+1−l
k z2
l − p+1 z
However,
Rc
dz
k z 2l
q
2γ
2λ
p+1−l
l − p+1 z
By putting y =
(3.6), we see that
−
2γ p+1−l
p+1 z
Z
dz
I=
v
2λ
l
is finite, so we deduce that limx→∞ v(x) = 0.
q
c
Z
q
l
z2
q
2λ
l
We suppose that
q
Z
I=
0
−
2γ p+1−l
p+1 z
2p−l
2(p+1−l)
2γ
2λ
l−2
l − p+1 z
in order to calculate the left hand side of
2γ
2λ
p+1−l
l − p+1 z
=
2p−l
l−2
γ
2 2(p+1−l) ( (p+1)
) 2(p+1−l)
2p−l
2 2(p+1−l)
(p + 1 − l)( 2λ
l −y )
0
= 32 . We obtain
3
1
γ
2 2 ( (p+1)
)2
(l − 2)( 2λ
l
l
3 dy =
2
λ(l − 2)
− y )2
q
2λ
l
−
v
2γ l−2
p+1 v
l−2
2
.
dy.
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N.-E. AMROUN, A. BENAISSA
EJDE-2006/133
Then (3.6) becomes
q
l
λ(l − 2)
2λ
l
−
v
2γ l−2
p+1 v
l−2
2
= α−
p−1
2
x,
which implies
v(x) =
λ(l−2) 2
l
We put µ =
λ(m−2) 2 −(p−1)
α
m
vx (x) = −
1/(l−2)
l/(2λ)
α−(p−1) x2 +
2γ
p+1
.
(3.7)
and differentiate (3.7), to obtain
1
2
l l−2
x
µ
1 .
2γ 1+ l−2
m − 2 2λ
µ x2 + p+1
Thus we have
Z ∞
Z ∞
1 2 m l−2
2
x2
2
2
µ
dx.
|vx (x)| dx =
2
2γ 2+ l−2
l − 2 2λ
0
0
µ x2 + p+1
p
R∞
x2
When we put H = 0
2+ 2 dx and let z = µ/a x, where a =
l−2
2γ
µ x2 + p+1
2γ
p+1 ,
we have
Z
1
∞
z2
dz
2
a
µ 0 (1 + z 2 )2+ l−2
Z ∞
l−2
1
1
=
dz
1
2
2
3
+
2l a 2 l−2 µ 2 0 (1 + z 2 )1+ l−2
H=
=
1
2
2 + l−2
3
2
1
l−2
K
2
3
2l a 12 + l−2
µ2
(K < +∞).
Therefore,
Z
∞
2
|vx (x)| dx =
0
If we take α such as
α=
1
2 m l−2
l − 2 2λ
2
1
√ l−2
µ
1
2 K .
+
2l a 2 l−2
√
2
4 2 l(l − 1) γ 12 + l−2
1/l 3
√
K
λ 2l ,
2
λ
l
then (3.1)(4) is satisfied. Therefore, we have the following theorem.
Theorem 3.1. If l > 2 and p = 2l − 3, then the solution v(x) of (3.1) such that
1
λ l−2
v(0) = 2(l−2)
and vx (0) = 0, is given by
l
γ
1/(l−2)
1/(2λ)
v(x) =
,
2
λ(l−2)
−(p−1) x2 + 2γ
α
l
p+1
where
√
2
4 2 l(l − 1) γ 12 + l−2
1/l 3
α= √
K
λ 2l ,
2
λ
l
Z
K=
This solution can be extended to a solution of (2.3).
∞
1
2
0
(1 + z 2 )1+ l−2
.
EJDE-2006/133
ON EXACT SOLUTIONS
5
4. Second Problem
In this section, we consider Problem
(|ϕ0 |l−2 ϕ0 )0 = λϕp ,
ϕ(0) = ϕ0 ,
t ≥ 0,
ϕt (0) = ϕ1 ,
ϕ(t) ≥ 0,
(4.1)
t≥0
where λ is the constant from the previous section. If we multiply the equation
(4.1)(1) by ϕt (t) and integrate from 0 to t as in section 2, then we have
1/l
lλ
lλ
,
(4.2)
ϕt (t) = ±
v p+1 + |ϕ1 |l −
ϕp+1
(l − 1)(p + 1)
(l − 1)(p + 1) 0
because of (4.1)(2) and (4.1)(3). In order that we construct the solution of (4.1)
from (4.2), the following condition has to be satisfied
G(ϕ) ≡
lλ
lλ
v p+1 + |ϕ1 |l −
ϕp+1 ≥ 0.
(l − 1)(p + 1)
(l − 1)(p + 1) 0
We consider the following three cases.
Case 1. If ϕ0 ≥ 0 and ϕ1 > 0, then G(ϕ(0)) = G(ϕ0 ) > 0. Hence ϕt (t) =
(G(ϕ(t)))1/l > 0 for sufficiently small t > 0 since (G(ϕ(t)))1/l is monotone increasing function,
we see that ϕt > 0 for all t > 0 where ϕ(t) exist. Since one
R∞
can show ϕ0 (1\(G(ϕ))1/l ) dϕ < +∞, we see that ϕ → +∞ as t → T ∗ , where
R∞
T ∗ = ϕ0 (1\(G(ϕ))1/l ) dϕ.
lλ
Case 2. In the case of ϕ0 ≥ 0 and |ϕ1 |l − (l−1)(p+1)
ϕp+1
= 0, by solving (4.2), we
0
obtain
1/l
λl
l
− p+1−l
p+1−l
(p + 1 − l) (l−1)(p+1)
− l
∓
t
,
ϕ± (t) = ϕ0
l
l
with double signs in same order. Obviously ϕ+ decays with the order O(t)− p+1−l
p+1−l
−
(l−1)(p+1) 1/l
l
as t → +∞ and ϕ− blows up at ϕ0 l p+1−l
.
λl
Case 3. If ϕ0 ≥ 0, ϕ1 < 0 and |ϕ1 |l −
p+1
lλ
(l−1)(p+1) ϕ0
> 0, we have ϕt (t) =
1/l
−(G(ϕ(t))) < 0 locally. Then, since G(ϕ(t)) is monotone increasing function, we
see that ϕt (t) < 0 for all t > 0 where ϕ(t) exist. From (4.2),
Z 0
dϕ
≡ T ∗∗ < +∞.
−
(G(ϕ))1/l
ϕ0
Therefore, T ∗∗ exists such that ϕ(t) → 0 as t → T ∗∗ . From (4.2), we can get
Z 0
1
dϕ
− ∗∗
= 1.
(4.3)
1/l
T
−
t
(G(ϕ))
ϕ(t)
Let s = (T ∗∗ − t)r in (4.3) and let t → T ∗∗ , we have
1/l
lλ
lim∗∗ ϕ(t)(T ∗∗ − t)−1 = |ϕ1 |l −
ϕp+1
.
t→T
(l − 1)(p + 1) 0
(4.4)
From (4.2) and (4.4), we have
lim∗∗ ϕ(t)ϕ0 (t)(T ∗∗ − t)−1 = − |ϕ1 |l −
t→T
Then we have the following theorem.
2/l
lλ
ϕp+1
.
0
(l − 1)(p + 1)
(4.5)
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N.-E. AMROUN, A. BENAISSA
EJDE-2006/133
Theorem 4.1. (1) If ϕ0 ≥ 0 and ϕ1 > 0, then by putting
Z ∞
(1\(G(ϕ))1/l ) dϕ,
T∗ =
ϕ0
the solution ϕ(t) of (4.1) blows up at t = T ∗ , that is ϕ(t) → +∞ as t → T ∗ .
lλ
(2) If ϕ0 ≥ 0 and |ϕ1 |l − (l−1)(p+1)
= 0, then the solution ϕ(t) of (4.1) is
ϕp+1
0
given by
1/l
λl
l
p+1−l
− p+1−l
(p + 1 − l) (l−1)(p+1)
− l
ϕ± (t) = ϕ0
,
∓
t
l
with double signs in the same order.
lλ
(3) If ϕ0 ≥ 0, ϕ1 < 0 and |ϕ1 |l − (l−1)(p+1)
ϕp+1
> 0, then by putting
0
Z 0
dϕ
−
≡ T ∗∗ ,
1/l
(G(ϕ))
ϕ0
the solution ϕ(t) vanishes at t = T ∗∗ and satisfies (4.4) and (4.5).
Depending on the choice of the constants ϕi (i = 0, 1), we obtain a solution
which blows up in a finite time and another case we have a solution which decays
l
with order O(t)− p+1−l as t → ∞.
References
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Université Djillali Liabès, Faculty des Sciences, Departement de mathematiques, B.
P. 89, Sidi Bel Abbès 22000, Algeria
E-mail address, N.-E. Amroun: amroun nour@yahoo.com
E-mail address, A. Benaissa: benaissa abbes@yahoo.com