Memoirs on Differential Equations and Mathematical Physics
Volume 42, 2007, 49–68
Sergo Kharibegashvili
ON THE GLOBAL SOLVABILITY OF
THE CAUCHY CHARACTERISTIC PROBLEM
FOR A NONLINEAR WAVE EQUATION
IN A LIGHT CONE OF THE FUTURE
Abstract. For a nonlinear wave equation occurring in relativistic quantum mechanics we investigate the problem on the global solvability of the
Cauchy characteristic problem in a light cone of the future.
2000 Mathematics Subject Classification. 35L05, 35L20, 35L70.
Key words and phrases. Nonlinear wave equations, the Cauchy characteristic problem, global solvability of the problem.
On the Global Solvability of the Cauchy Characteristic Problem
51
1. Statement of the Problem. Consider a nonlinear wave equation of
the type
3
( + m2 )u :=
∂ 2 u X ∂ 2u
2
3
−
2 + m u = −λu + F,
∂t2
∂x
i
i=1
(1)
where λ 6= 0 and m ≥ 0 are given real constants, F is a given and u is
an unknown real function. As is known, the equation (1) arises in the
relativistic quantum mechanics ([1]–[5]).
For the equation (1) we consider the Cauchy characteristic problem on
finding in a frustum of the light cone of the future DT : |x| < t < T ,
x = (x1 , x2 , x3 ), T = const > 0, a solution u(x, t) of that equation by the
boundary condition
(2)
uST = g,
where g is a given real function on the characteristic conic surface ST :
t = |x|, t ≤ T . When considering the case T = +∞, we assume that
D∞ : t > |x| and S∞ = ∂D∞ : t = |x|.
Note that the Cauchy characteristic problem for the equation (1) with
the boundary conditions u|t=0 = u0 , ut |t=0 = u1 has been investigated in
[1]–[5]. For more general semi-linear equations of type (1) the issues of
existence or nonexistence of a global solution of that problem have been
considered and studied in [6]–[21].
In the linear case, that is for λ = 0, the problem (1), (2) is posed correctly
and its global solvability takes place in the corresponding function spaces
([22]–[26]). In the nonlinear case, for the equation u+m2 u = −λ|u|p u+F ,
p = const 6= 0, coinciding with (1) for p = 2, the global solvability of the
Cauchy characteristic problem for λ > 0, m = 0 and 0 < p < 1 in the
Sobolev space W21 (DT ) has been proved in [27]. In the present paper we
prove the global solvability of the problem (1), (2) for λ > 0 in the Sobolev
space W22 (DT ). The uniqueness of a solution of that problem will also be
proved, and a result on the absence of the global solvability of the problem
(1), (2) will be given for the case λ < 0. In this direction the works [28]–[30]
are noteworthy.
2. A Priori Estimate of a Solution of the Problem (1), (2) for λ > 0.
For the sake of simplicity of our exposition, the boundary condition (2) will
be assumed to be homogeneous, i.e.,
u = 0.
(3)
ST
◦
Suppose W k2 (DT , ST ) = {u ∈ W2k (DT ) : u|ST = 0}, where W2k (DT ) is the
well-known Sobolev space, and the condition u|ST = 0 will be understood
in the sense of the trace theory ([31, pp. 56, 70]).
◦
Definition 1. Let F ∈ W 12 (DT , ST ). The function u = u(x, t) is said
to be a solution of the problem (1), (3) of the class W22 , if u ∈ W22 (DT ),
it satisfies both the equation (1) almost everywhere in the domain DT and
52
S. Kharibegashvili
the boundary condition (3) in the sense of the trace theory (and hence
◦
W 22 (DT , ST )).
◦
◦
Definition 2. Let F ∈ W 12 (DT , ST ). The function u ∈ W 22 (DT , ST ) is
said to be a strong generalized solution of the problem (1), (3) of the class
W22 , if there exists a sequence of functions un ∈ C ∞ (DT ) satisfying the
boundary condition (3) and
lim kun − ukW22 (DT ) = 0,
n→∞
lim kFn − F kW21 (DT ) = 0,
(4)
and
(5)
n→∞
where
Fn = un + m2 un + λu3n
supp Fn ∩ ST = ∅.
Remark 1. It can be easily seen that a strong generalized solution of the
problem (1), (3) of the class W22 in the sense of Definition 2 is also a solution
of the problem (1), (3) of the class W22 in the sense of Definition 1, since, as it
will be noted below, the first equality of (4) implies lim ku3n − u3 kL2 (DT ) =
n→∞
0. On the other hand, in the next section we will prove that the problem
(1), (3) is solvable in the sense of Definition 2, whereas the uniqueness of
a solution of that problem in the sense of Definition 1 will be proved in
Section 4. This obviously implies that a solution of the problem is unique
in the sense of Definition 2, and the above definitions are equivalent.
◦
Definition 3. Let F ∈ L2,loc (D∞ ) and F ∈ W 12 (DT , ST ) for any T > 0.
We say that the problem (1), (3) is globally solvable, if for any T > 0 this
problem has a solution of the class W22 in the domain DT in the sense of
Definition 1.
◦
Lemma 1. Let λ ≥ 0 and F ∈ W 12 (DT , ST ). Then for any strong
generalized solution u of the problem (1), (3) of the class W22 the following
a priori estimate
kukW22 (DT ) ≤ σ1 kF kL2 (DT ) + λσ2 kF k3L( DT ) +
h
i
+ σ3 kF kW21 (DT ) exp σ4 + λσ5 kF k2L2 (DT )
(6)
is valid with positive constants σi = σi (m, T ), i = 1, . . . , 5, not depending
on u and F .
Proof. By Definition 2 of strong generalized solution u of the problem
(1), (3) of the class W22 , there exists a sequence of functions un ∈ C ∞ (D T )
satisfying the conditions (3), (4) and (5), and hence
un + m2 un = −λu3n + Fn ,
un = 0.
un ∈ C ∞ (D T ),
ST
The proof of the lemma runs in several steps.
(7)
(8)
53
On the Global Solvability of the Cauchy Characteristic Problem
10 . Assuming Ωτ := D∞ ∩ {t = τ }, we first show that the a priori
estimate
2 X
2 #
Z "
Z
3 ∂un
∂un
2
+
un +
dx ≤ c Fn2 dx dt, 0 < t < T, (9)
∂t
∂xi
i=1
Ωt
Dt
is valid with a constant c not depending on un and Fn . Indeed, multiplying
n
both sides of the equation (7) by ∂u
∂t and integrating over the domain Dτ ,
0 < τ ≤ T , we obtain
2
Z
Z
Z
∂ ∂un
∂
∂un
m2
1
dx dt − ∆un
dx dt +
(un )2 dx dt+
2
∂t ∂t
∂t
2
∂t
Dτ
Dτ
Dτ
Z
Z
λ
∂un
∂
4
+
Fn
(un ) dx dt =
dx dt . (10)
4
∂t
∂t
Dτ
Dτ
By ν = (ν1 , ν2 , ν3 , ν0 ) we denote the unit vector of the outer normal to
ST \ {(0, 0, 0, 0)}. The integration by parts with regard for the equality (8)
and ν|Ωτ = (0, 0, 0, 1) results in
2
2
2
2
Z Z Z Z
∂un
∂un
∂un
∂ ∂un
dx dt =
ν0 ds =
dx +
ν0 ds,
∂t ∂t
∂t
∂t
∂t
Dτ
Ωτ
Sτ
Dτ
Z
Z
Z
∂
(un )2k ν0 ds = (un )2k dx, k = 1, 2,
(un )2k dx dt =
∂t
Dτ
Z
Dτ
Ωτ
∂Dτ
2
∂ un ∂un
dx dt =
∂x2i ∂t
=
Z
1
∂um ∂um
νi ds −
∂xi ∂t
2
1
2
Z ∂un
∂xi
∂Dτ
∂Dτ
−
Dτ
∂Dτ
1
∂un ∂un
νi ds −
∂xi ∂t
2
Z
Z ∂un
∂xi
Sτ
2
Z
ν0 ds −
1
2
Z 2
∂un
∂xi
Ωτ
ν0 ds =
2
∂ ∂un
dx dt =
∂t ∂xi
Z
∂un ∂un
νi ds−
∂xi ∂t
∂Dτ
2
i = 1, 2, 3,
dx,
whence by virtue of (10) we obtain
" 3 2
Z
Z
∂un
∂un
1 X ∂un
ν0 −
dx dt =
νi +
Fn
∂t
2ν0 i=1 ∂xi
∂t
Sτ
Dτ
+
∂un
∂t
2 ν02
−
3
X
j=1
νj2
#
1
ds +
2
Z "
∂un
∂t
Ωτ
+
λ
4
Z
Ωτ
2
+
2
3 X
∂un
i=1
u4n dx +
m2
2
∂xi
Z
Ωτ
#
dx+
u2n dx.
(11)
54
S. Kharibegashvili
Since Sτ is a characteristic surface, we have
3
X
2
2 ν0 −
νj = 0.
(12)
Sτ
j=1
∂
∂
− νi ∂t
), i = 1, 2, 3, is the interior difTaking into account that (ν0 ∂x
i
ferential operator on Sτ , owing to (8) we find
∂un
∂un
(13)
ν0 −
νi = 0, i = 1, 2, 3.
∂xi
∂t
Sτ
With regard for (12) and (13), the equation (11) yields
2 #
2 X
Z
Z
Z "
3 λ
∂un
∂un
4
2
u2n dx =
dx +
un dx + m
+
∂t
∂x
2
i
i=1
Ωτ
Ωτ
Ωτ
Z
∂un
dx dt .
= 2 Fn
∂t
(14)
Dτ
2
n
Since λ ≥ 0, m2 ≥ 0, by the Cauchy inequality 2Fn ∂u
∂t ≤ Fn +
it follows from (14) that
2 #
2 X
2
Z Z "
Z
3 ∂un
∂un
∂un
dx ≤
+
dx dt+ Fn2 dx dt .
∂t
∂xi
∂t
i=1
Ωτ
Rτ
|x|
Dτ
∂un 2
∂t
(15)
Dτ
Reasoning in a standard way, from the equalities un |Sτ = 0 and un (x, τ ) =
∂un (x,t)
∂t
dt, x ∈ Ωτ , 0 < τ ≤ T , we arrive at the inequality ([31, p. 63])
Z
u2n dx
≤T
Ωτ
Z ∂un
∂t
Dτ
2
(16)
dx dt.
Adding the inequalities (15) and (16), we obtain
2 #
2 X
Z "
3 ∂un
∂un
2
dx ≤
+
un +
∂t
∂xi
i=1
Ωτ
≤ (1 + T )
Z ∂un
∂t
Dτ
Introduce the notation
Z "
w(δ) =
Ωδ
u2n
+
∂un
∂t
2
+
2
dx dt +
Fn2 dx dt .
Dτ
2
3 X
∂un
i=1
Z
∂xi
#
dx.
(17)
55
On the Global Solvability of the Cauchy Characteristic Problem
Then by virtue of (17) we have
2 #
2 X
Z
Z "
3 ∂un
∂un
2
dx dt + Fn2 dx dt =
un +
w(δ) = (1 + T )
+
∂t
∂xi
i=1
Dδ
Dτ
= (1 + T )
Zδ
0
w(σ) dσ + kFn k2L( D ) ,
2
δ
0 < δ < T.
(18)
From (18), taking into account the fact that the expression kFn k2L:2 (Dδ )
as a function of δ is nondecreasing, by the Gronwall lemma ([32, p. 13]) we
find that
w(δ) ≤ kFn k2L2 (Dδ ) exp(1 + T )δ ≤ kFn k2L2 (Dδ ) exp(1 + T )T,
whence for δ = t we obtain the inequality (9) with the constant c = exp(1 +
T )T .
20 . Since owing to (5) we have supp Fn ∩ ST = ∅, there exists a positive
number δn < T such that
supp Fn ⊂ DT,δn = {(x, t) ∈ DT : t > |x| + δn }.
In this section we will show that
un DT \D
T,δn
= 0.
(19)
(20)
Indeed, let (x0 , t0 ) ∈ DT \ Dt,δn . Introduce the domain Dx0 ,t0 = {(x, t) ∈
R : |x| < t < t0 − |x − x0 |} which is bounded from below by the surface ST
and from above by the boundary Sx−0 ,t0 = {(x, t) ∈ R4 : t = t0 − |x − x0 |}
4
0
0
of the light cone of the past G−
x0 ,t0 = {(x, t) ∈ R : t < t − |x − x |} with
0 0
the vertex at the point (x , t ). By virtue of (19), we have
Fn D 0 0 = 0, (x0 , t0 ) ∈ DT \ DT,δn .
(21)
4
x ,t
Assume Dx0 ,t0 ,τ := Dx0 ,t0 ∩ {t < τ }, Ωx0 ,t0 ,τ := Dx0 ,t0 ∩ {t = τ }, 0 <
τ < t0 . Then ∂Dx0 ,t0 ,τ = S1,τ ∪ S2,τ ∪ S3,τ , where S1,τ = ∂Dx0 ,t0 ,τ ∩ S∞ ,
S2,τ = ∂Dx0 ,t0 ,τ ∩ Sx−0 ,t0 , S3,τ = ∂Dx0 ,t0 ,τ ∩ Ωx0 ,t0 ,τ . Just in the same way
as when obtaining the equality (11), we multiply both sides of the equation
0
n
0 0
(7) by ∂u
∂t , integrate over the domain Dx ,t ,τ , 0 < τ < t , and taking into
account (8) and (21), we obtain
0=
Z
S1,τ ∪S2,τ
3 2 2 3
X
∂un
∂un
1 X ∂un
ν02 −
νj2 ds+
ν0 −
νi +
2ν0 i=1 ∂xi
∂t
∂t
j=1
+
Z
S2,τ ∪S3,τ
λ 4 m2 2
u +
u ν0 ds+
4 n
2 n
56
S. Kharibegashvili
1
+
2
Z "
∂un
∂t
S3,τ
2
+
2
3 X
∂un
i=1
∂xi
#
dx.
(22)
By (8) and (12), taking into account the fact that S2,τ is, just like S1,τ ,
3
P
a characteristic surface and hence ν02 −
= 0 and also that
νj2 S1,τ ∪S2,τ
j=1
we have
Z
S1,τ ∪S2,τ
1
1
ν0 S1,τ = − √ < 0, ν0 S2,τ = √ > 0, ν0 S3,τ = 1,
2
2
∂u
∂u
= 0,
ν0 −
νi ∂xi
∂t
S1,τ
2 ∂u
∂u
ν0 −
νi ≥ 0, i = 1, 2, 3,
∂xi
∂t
S2,τ
" 3 2 2 #
3
X
∂un
∂un
1 X ∂un
2
2
ν0 −
νj
ν0 −
νi +
ds ≥ 0. (23)
2ν0 i=1 ∂xi
∂t
∂t
j=1
Taking into account (23), from (22) we get
2 #
2 X
Z
Z "
3 ∂un
∂un
dx ≤ M
+
∂t
∂xi
i=1
u2n ds,
0 < τ < t0 . (24)
S2,τ ∪S3,τ
S3,τ
∞
Here, since un ∈ C (DT ) and |ν0 | ≤ 1, we can take
λ
(25)
kun k2C(DT ) < +∞
2
in the capacity of a nonnegative constant M independent of the parameter
τ . As far as un |ST = 0 and ST : t = |x|, t ≤ T , we will have
M = m2 +
un (x, t) =
Zt
∂un (x, σ)
dσ,
∂t
(x, t) ∈ S2,τ ∪ S3,τ .
(26)
|x|
Reasoning in a standard way ([31, p. 63]), from the equality (26) we get
that
2
Z
Z
∂un
2
0
un ds ≤ 2t
dx, 0 < τ < t0 .
(27)
∂t
S2,τ ∪S3,τ
Dx0 ,t0 ,τ
Assuming v(τ ) =
R
S3,τ
easily obtain
∂un 2
∂t
0
v(τ ) ≤ 2t M
+
Zτ
0
3
P
i=1
∂un 2
∂xi
v(δ) dδ,
dx, from (24) and (27) we
0 < τ < t0 ,
57
On the Global Solvability of the Cauchy Characteristic Problem
whence by virtue of (25) and the Gronwall lemma it immediately follows
∂un
∂un
∂un
n
that v(τ ) = 0, 0 < τ < t0 , and hence ∂u
∂t = ∂x1 = ∂x2 = ∂x3 = 0 in the
domain Dx0 ,t0 . Therefore un |Dx0 ,t0 = const, and taking into account the
homogeneous boundary condition (8), we find that un |Dx0 ,t0 = 0 ∀(x0 , t0 ) ∈
DT \ DT,δn . Thus the equality (20) is proved.
30 . We will now pass to proving the a priori estimate (6). By (20),
extending un from the domain DT to the layer ΣT = {(x, t) ∈ R4 : 0 < t <
T } as zero and preserving for it the same designation, we obtain
= 0.
(28)
u n ∈ C ∞ Σ T , u n P
T \D T,δn
In particular, from (28) it follows that un = 0 for |x| ≥ T .
Differentiating the equality (7) with respect to the variable xi , we obtain
un,xi + m2 un,xi = −3λu2n un,xi + Fn,xi ,
where un,xi =
∂un
∂xi ,
3
E(τ ) =
1X
2 i=1
Fn,xi =
Z ∂Fn
∂xi .
u2n,xi t +
3
X
(29)
Suppose
u2n,xi xk
k=1
Ωτ
i = 1, 2, 3,
dx,
Ωτ = D∞ ∩ {t = τ }.
(30)
By virtue of (28), in the right-hand side of (30) we can take instead of Ωτ
a three-dimensional ball B(0, T ) : |x| < T in the plane t = τ .
Differentiating the equality (30) with respect to the variable τ and then
integrating by parts, with regard for (7), (28) and (29) we find that
3
3 Z X
X
0
un,xi xk un,xi xk t dx =
un,xi t un,xi tt +
E (τ ) =
i=1 Ω
=
i=1 Ω
=
τ
t
Z
3
X
i=1
un,xi tt un,xi t −
3
X
k=1
un,xi xk xk un,xi t dx =
un,xi un,xi t dx =
τ
3 Z
X
i=1 Ω
=
3 Z
X
i=1 Ω
=
k=1
τ
3 Z
X
− m2 un,xi − 3λu2n un,xi + Fn,xi un,xi t dx =
|x|<T,
t=τ
− m2 un,xi − 3λu2n un,xi + Fn,xi un,xi t dx.
By Hölder’s inequality ([33, p. 134])
Z
f1 f2 f3 dx ≤ kf1 kLp1 kf2 kLp2 kf3 kLp3
(31)
58
S. Kharibegashvili
1
1
1
+
+
= 1,
p1
p2
p3
and by the Cauchy inequality, for the right-hand side of (31) we have the
estimate
X
Z
3
2
2
− m un,xi − 3λun un,xi + Fn,xi un,xi t dx ≤
I=
for p1 = 3, p2 = 6, p3 = 2,
i=1
≤
|x|<T,
t=τ
Z
3
2 X
m
2
i=1
3
1X
+
2 i=1
≤
m2 + 1
2
3
+
1X
2 i=1
u2n,xi dx +
|x|<T,
t=τ
Z
u2n,xi t
|x|<T,
t=τ
3
X
i=1
Z
3
m2 X
2 i=1
3
X
dx + 3λ
u2n,xi dx +
2
Fn,x
dx + 3λ
i
3
X
i=1
|x|<T,
t=τ
Z
i=1
1X
2 i=1
u2n un,xi un,xi t
|x|<T,
t=τ
Z
3
2 X
m
2
|x|<T,
t=τ
u2n,xi t dx +
|x|<T,
t=τ
i=1
Z
3
Z
Z
2
dx+
Fn,x
i
|x|<T,
t=τ
dx ≤
u2n,xi t dx+
|x|<T,
t=τ
ku2n kL3 ((|x|<T, t=τ )) ×
× kun,xi kL6 ((|x|<T, t=τ )) kun,xit kL2 ((|x|<T, t=τ )) . (32)
l
By the well-known theorem of imbedding of the space Wm
(Ω) into Lp (Ω)
for m = 2, l = 1, p = 6 ([31, p. 84]; [34, p. 111]), there takes place the
inequality
kvkL6 ((|x|<T )) ≤ c1 kvk ◦ 1
W 2 ((|x|<t))
◦
∀v ∈ W 12 ((|x| < T ))
with a positive constant c1 not depending on v.
We also have ([31, p. 117])
Z
Z
n
3
X
X
◦
vx2i xk dx ∀v ∈ W 22 ((|x| < T ))
vx2i dx ≤ c2
i=1
|x|<T
(33)
(34)
i,j=1
|x|<T
with a positive constant c2 not depending on v.
Applying the inequality (33) to the functions un and un,xi which by virtue
◦
of (28) belong to the space W 12 ((|x| < T )) for the fixed t = τ , we obtain
kun kL6 ((kxk<T, t=τ )) ≤ c1 kun k ◦ 1
W 2 ((kxk<T, t=τ ))
kun,xi kL6 ((kxk<T, t=τ )) ≤ c1 kun,xi k ◦ 2
,
W 2 ((kxk<T, t=τ ))
.
By (9), (30), (33) and (35), we have
ku2n kL3 ((kxk<T, t=τ )) kun,xi kL6 ((kxk<T, t=τ )) kun,xi t kL2 ((kxk<T, t=τ )) ≤
(35)
59
On the Global Solvability of the Cauchy Characteristic Problem
1
≤ kun k2L6 ((kxk<T, t=τ )) c1 kun k ◦ 2
W 2 ((kxk<T, t=τ ))
≤
1
1
c1 ckFn k2L2 (Dτ ) c1 [2E(τ )] 2 [2E(τ )] 2
[2E(τ )] 2 ≤
= 2cc21 kFn k2L2 (Dτ ) E(τ ).
(36)
By (9), (32), (34) and (36), we have
I≤
1
m2 + 1
c2 2E(τ ) + m2 E(τ ) + kFn k2W 1 ((kxk<T, t=τ )) +
2
2
2
2
2
+ 6λcc1 kFn kL2 (Dτ ) E(τ ).
(37)
By (31) and (37), we find that
E 0 (τ ) ≤ α(τ )E(τ ) + β(τ ) ≤ α(T )E(τ ) + β(τ ),
τ ≤ T.
(38)
Here
α(τ ) = (m2 + 1)c2 + m2 + 6λcc21 kFn k2L2 (Dτ ) ,
(39)
1
β(τ ) = kFn k2W 1 ((kxk<T, t=τ )) .
2
2
Since according to (28) we have E(0) = 0, multiplying both sides of the
inequality (38) by exp[−α(T )τ ] and integrating in a standard way, we obtain
Zτ
Zτ
α(T )τ
−α(T )σ
α(T )τ
E(τ ) ≤ e
e
β(σ) dσ ≤ e
β(σ) dσ =
0
=
1 α(T )τ
e
2
0
Zτ
0
kFn k2W 1 ((kxk<T, t=σ)) dσ =
2
1
≤ eα(T )T kFn k2W 1 (DT ) ,
2
2
By the equality (7),
1 α(T )τ
e
kFn k2W 1 (Dτ ) ≤
2
2
0 ≤ τ ≤ T.
(40)
un,tt = ∆un − m2 un − λu3n + F.
Squaring both parts of the equality (41) and using the inequality
n
n
P
i=1
i=1
a2i
for n = 4, we will have
Z
u2n,tt dx ≤ 4λ2
Ωτ
Z
u6n dx + 4
|x|<T,
t=τ
Z
|x|<T,
t=τ
find that
Z
u2n,tt dx ≤ 4λ2 c61 kun k2◦ 1
W 2 ((kxk<T, t=τ ))
(41)
2
ai ≤
(∆un )2 + m4 u2n + Fn2 dx,
whence by virtue of (9), (35) and the fact that (∆un )2 ≤ 3
Ωτ
P
n
3
P
i=1
u2n,xi xi , we
+ 24E(τ ) + 4m2 ckFn k2L2 (Dτ ) +
+ 4kFn k2L2 ((kxk<T, t=τ )) ≤ 4λ2 c61 kFn k6L2 (Dτ ) + 4m4 ckFn k2L2 (Dτ ) +
60
S. Kharibegashvili
+ 4kFn k2L2 ((kxk<T, t=τ )) + 24E(τ ).
(42)
Due to (40), it follows from (42) that
Z
u2n,tt
dx dt =
ZT
dτ
0
DT
+ 4m
4
Z
u2n,tt dx ≤ 4λ2 c61 T kFn k6L2 (DT ) +
Ωτ
cT kFn k2L2 (DT )
+ 4kFn k2L2 (DT ) + 12eα(T )T kFn k2W 1 (DT ) .
(43)
2
From (9), (30), (40) and (43) we now find that
kun k2W 2 (DT )
2
+
3
X
=
ZT
dτ
0
Z ≤
≤
0
+
Ωτ
u2n,xi t +
i=1
ZT
u2n
3
X
i,k=1
∂un
∂t
2
+
2
3 X
∂un
∂xi
i=1
+ u2n,tt +
u2n,xi xk dx ≤
Z
ZT
Z
2
2
un,tt dx dt + 2E(τ ) dτ ≤
c Fn dx dt dτ +
Dτ
cT kFn k2L2 (DT )
+
DT
2 6
4λ c1 T kFn k6L2 (DT )
0
+ 4m4 cT kFn k2L2 (DT ) +
+ 4kFn k2L2 (DT ) + 12eα(T )T kFn k2W 1 (DT ) + T eα(T )T kFn k2W 1 (DT ) =
2
4
= (cT + 4m cT +
+ (12 + T )e
4)kFn k2L2 (DT )
α(T )T
+
kFn k2W 1 (DT ) .
2
2
4λ2 c61 T kFn k6L2 (DT ) +
(44)
P
3
1
2
3
P
|ai | 2 and
kun kW22 (DT ) ≤ σ1 kFn kL2 (DT ) + λσ2 kFn k3L2 (DT ) +
+ σ3 kFn k2W 1 (DT ) exp σ4 + λσ5 kFn k2L2 (DT ) ,
(45)
Taking into account the obvious inequality
i=1
the equality (39), from (44) we obtain
|ai |
≤
i=1
1
2
1
1
1
where σ1 = (cT + 4m4 cT + 4) 2 , σ2 = 2c31 T 2 , σ3 = (12 + T ) 2 , σ4 =
1
2
2
2
2 ((m + 1)c2 + m ), σ5 = 3cc1 . By (4), passing the in inequality (45) to
the limit as n → ∞, we obtain the a priori estimate (6). Thus the proof of
Lemma 1 is complete.
3. The Global Solvability of the Problem (1), (3) for λ > 0.
Remark 2. Before we proceed to considering the question on the solvability of the nonlinear problem (1), (3), let us consider the same question
for the linear case in the form needed for us, when in the equation (1) the
parameter λ = 0, i.e., for the problem
Lu(x, t) = F (x, t),
u(x, t) = 0,
(x, t) ∈ DT
(x, t) ∈ ST .
(L := + m2 ),
(46)
(47)
61
On the Global Solvability of the Cauchy Characteristic Problem
◦
In this case for F ∈ W 12 (DT , ST ) we analogously introduce the notion of a
◦
strong generalized solution u ∈ W 22 (DT , ST ) of the problem (46), (47) for
which there exists a sequence of the functions un ∈ C ∞ (DT ) satisfying the
boundary condition (47) and
lim kun − ukW22 (DT ) = 0,
n→∞
lim kLun − F kW21 (DT ) = 0.
n→∞
(48)
As it follows from the a priori estimate (6), for that solution for λ = 0
we have the estimate
kukW22 (DT ) ≤ c0 kF kW21 (DT )
(49)
with a positive constant c0 not depending on u and F .
Since the space C0∞ (D T , ST ) = {F ∈ C ∞ (DT ) : supp F ∩ ST = ∅} of the
infinitely differentiable in D T functions vanishing in some bounded neigh◦
borhood (its own for each function) of the set ST is dense in W 12 (DT , ST ),
◦
for a given F ∈ W 12 (DT , ST ) there exists a sequence of functions Fn ∈
C0∞ (D T , ST ) such that lim kFn − F kW21 (DT ) = 0. For the fixed n, exn→∞
tending the function Fn from the domain DT to the layer ΣT = {(x, t) ∈
R4 : 0 < t < T } as zero and preserving the same notation, we will have
Fn ∈ C ∞ (ΣT ) with the carrier supp Fn ⊂ D∞ : t > |x|. Denote by un the
n
solution of the Cauchy linear problem: Lun = Fn , un |t=0 = 0, ∂u
∂t t=0 = 0
in the layer ΣT , which, as is known, exists, is unique and belongs to the
∞
space
C (ΣT ) ([35, p. 192]). Moreover, since supp Fn ⊂ D∞ , un |t=0 = 0,
∂un ∂t t=0 = 0, taking into account the geometry of the domain of dependence
of a solution of the linear wave equation, we will have supp un ⊂ D∞ ([35,
p. 191]). Retaining for the restriction of the function un to the domain DT
the same designation, we can easily see that un ∈ C ∞ (D T ), un |ST = 0, and
because of (49),
kun − ukW22 (DT ) ≤ c0 kFn − F kW21 (DT ) .
(50)
◦
Since the sequence {Fn } is fundamental in W 12 (DT , ST ), by (50) and
the fact that un |ST = 0, the sequence {un } is likewise fundamental in the
◦
whole space W 22 (DT , ST ) = {u ∈ W22 (DT ) : u|ST = 0}. Thus there exists
◦
a function u ∈ W 22 (DT , ST ) such that lim kun − ukW22 (DT ) = 0, and since
n→∞
by the condition Lun = Fn → F in the space W21 (DT ), the function u will,
by Remark 2, be a strong generalized solution of the problem (46), (47)
◦
from the space W 22 (DT , ST ). The uniqueness of the solution from the space
◦
W 22 (DT , ST ) follows from the a priori estimate (49). Consequently, for the
solution u of the problem (46), (47) we can write u = L−1 F , where L−1 :
◦
◦
W 12 (DT , ST ) → W 22 (DT , ST ) is a linear continuous operator, whose norm
62
S. Kharibegashvili
admits, by virtue of (49), the estimate
kL−1 k ◦ 1
◦
W 2 (DT ,ST )→W 22 (DT ,ST )
◦
≤ c0 .
(51)
Remark 3. Below, we will show that the Nemytski operator N :
◦
W 22 (DT , ST ) → W 12 (DT , ST ) acting by the formula N u = −λu3 is continuous and compact.
◦
Indeed, first we note that if u ∈ W 22 (DT , ST ), then since DT is a bounded
domain from R4 , therefore u ∈ Lq (DT ) for any q ≥ 1, and moreover, the
◦
imbedding operator I1 : W 22 (DT , ST ) → Lq (DT ) is continuous and compact
([31, p. 84]). Note also that the imbedding operator I2 : W21 (DT ) →
Lp (DT ) is a linear continuous one for 1 < p < 4 ([31, p. 83]). Therefore,
◦
∂u
if u ∈ W 22 (DT , ST ), then u2 , uk ∈ L6 (DT ), and since ∂x
∈ W21 (DT ), we
i
∂u
have ∂xi ∈ L3 (DT ). As is known, if fi ∈ Lpi (DT ), i = 1, 2, p11 + p12 = r1 ,
pi > 1, r > 1, then f1 f2 ∈ Lr (DT ) ([4, p. 45]). For p1 = 6, p2 = 3, r = 2,
∂u
u
2 ∂u
, we obtain ∂N
( 16 + 31 = 12 ), f1 = u2 , f2 = ∂x
∂xi = −3λu ∂xi ∈ L2 (DT ),
i
∂N u
i = 1, 2, 3. Analogously, we have ∂t ∈ L2 (DT ).
◦
Let X be a bounded set in W 22 (DT , ST ), and let {un } be a sequence
◦
taken arbitrarily in X. Since the space W 22 (DT , ST ) is compact-imbedded
◦
into the space W 12 (DT , ST ) ([34, p. 183]), there exist a subsequence {unk }
◦
and a function u ∈ W 12 (DT , ST ) such that
∂unk
∂u =
−
lim kunk − ukL2 (DT ) = lim k→∞
k→∞
∂t
∂t L2 (DT )
∂unk
∂u = lim = 0.
−
k→∞ ∂xi
∂xi L2 (DT )
(52)
On the other hand, according to the above-said, there exists a subsequence
of the sequence {unk }, for which we will preserve the same designation, such
that
∂unk
−
v
= 0, i = 1, 2, 3,
lim ku2nk −v0 kL6 (DT ) = 0, lim i
k→∞
k→∞
∂xi
L3 (DT )
(53)
∂unk
3
lim kunk −vkL2 (DT ) = 0, lim = 0,
−v4 k→∞
k→∞
∂t
L3 (DT )
where v0 , v, vi , i = 1, 2, 3, 4, are some functions from the corresponding
spaces: L6 (DT ), L2 (DT ) for v0 , v, and L3 (DT ) for the remaining vi . Reasoning in a standard manner and using the notion of generalized Sobolev’s
derivative, from (52) and (53) we find that
v0 = u 2 ,
v = u3 ,
vi =
∂u
, i = 1, 2, 3,
∂xi
v4 =
∂u
.
∂t
(54)
63
On the Global Solvability of the Cauchy Characteristic Problem
Let us now show that
∂N unk
∂N u lim −
= 0, i = 1, 2, 3,
k→∞ ∂xi
∂xi L2 (DT )
∂N unk
∂N u = 0.
−
lim
k→∞ ∂t
∂t L2 (DT )
(55)
Indeed, using Hölder’s inequality for p = 3, q = 32 ( p1 + 1q = 1), we obtain
2
2
Z ∂N unk
∂N u 2 ∂u
2
2 ∂unk
dx dt =
−
−
u
=
9λ
u
nk
∂xi
∂xi L2 (DT )
∂xi
∂xi
DT
= 9λ
2
Z (u2nk
∂unk
−u )
+ u2
∂xi
2
DT
≤ 18λ2
Z
(u2nk − u2 )2
DT
∂unk
∂xi
2
∂unk
∂u
−
∂xi
∂xi
dx dt + 18λ2
Z
DT
2
u4
dx dt ≤
∂unk
∂u
−
∂xi
∂xi
2
dx dt
∂unk 2 2
2
2 2
+
≤ 18λ k(unk − u ) kL3 (DT ) ∂xi
L 3 (DT )
2
2 ∂unk
∂u + 18λ2 ku4 kL3 (DT ) =
−
∂xi
∂xi L 3 (DT )
2
2
∂unk = 18λ2 ku2nk − u2 k2L6 (DT ) +
∂xi L3 (DT )
2
∂unk
∂u −
+ 18λ2 ku2 k2L6 (DT ) .
(56)
∂xi
∂xi L3 (DT )
n
o
∂u 2
By virtue of (53), the sequence ∂xnik is bounded. Hence from
L3 (DT )
(56), by virtue of (53) and (54), we obtain the first three equalities from (55)
for i = 1, 2, 3. The last equality from (55) is proved analogously. Finally,
the fact that N unk → N u in L2 (DT ) follows directly from (53) and (54).
Thus the statement of Remark 3 is proved.
Remark 4. When writing the equalities
lim ku2nk − v0 kL6 (DT ) = 0,
k→∞
from (53), we used the following:
◦
W 12 (DT , ST )
lim ku3nk − vkL2 (DT ) = 0
k→∞
(1) the imbedding operator I1 :
→ Lq (DT ) for any q ≥ 1 is continuous and compact; (2) if
u ∈ Lpr (DT ), p ≥ 1, r ≥ 1, then |u|p ∈ Lr (DT ), and from the fact that
un → u in Lpr (DT ) it follows that |un |p → |u|p in Lr (DT ) (analogously,
if un → u in L3r (DT ), then u3n → u3 in Lr (DT )), which is a consequence
of the following theorem ([34, p. 66]): the nonlinear Nemytski operator H,
64
S. Kharibegashvili
acting by the formula u → h(x, u), where the function h = h(x, ξ) possesses
the Carathéodory property, acts continuously from the space Lp (DT ) to
p
Lr (DT ), p ≥ 1, r ≥ 1, if and only if |h(x, ξ)| ≤ d(x) + α|ξ| r ∀ξ ∈ (−∞, ∞),
where d ∈ Lr (DT ) and δ = const ≥ 0. In our case, h(x, ξ) = |ξ|p ,
i.e., Hu = |u|p . Note also that we have for the present proved that the
◦
operator N from Remark 3 is compact from the space W 22 (DT , ST ) to
W21 (DT ). In its turn, this implies that the operator is continuous as well,
since the above-mentioned spaces, being Hilbert ones, are also reflexive ([34,
◦
p. 182]). Finally, the fact that the image N (W 22 (DT , ST )) is, in fact, a sub◦
space of the space W 12 (DT , ST ) follows from the following reasoning. If
◦
◦
u ∈ W 22 (DT , ST ), then there exists a sequence un ∈ C 2 (D T , ST ) = {u ∈
◦
C 2 (DT ) : u|ST = 0} such that un → u in the space W 22 (DT , ST ). But
according to the above-said, N un → N u in the space W21 (DT ), and since
◦
◦
N un = −λu3n ∈ C 2 (DT , ST ) ⊂ W 12 (DT , ST ), owing to the completeness of
◦
◦
◦
the space W 12 (DT , ST ), we finally obtain N (W 22 (DT , ST )) ⊂ W 12 (DT , ST ),
◦
◦
and hence the operator N : W 22 (DT , ST ) → W 12 (DT , ST ) is continuous and
compact.
Remark 5. As is said in Remark 1 of Section 2, the first equality of (4)
implies that lim ku3n − u3 kL2 (DT ) = 0. The latter is a direct consequence
n→∞
of the statement in Remark 3. From the above remarks it immediately
◦
◦
follows that if F ∈ W 12 (DT , ST ), then the function u ∈ W 22 (DT , ST ) is a
strong generalized solution of the problem (1), (3) of the class W22 if and
only if this function is, in view of (51), a solution of the following functional
equation
u = L−1 (−λu3 + F )
(57)
◦
in the space W 22 (DT , ST ).
We rewrite the equation (57) in the form
u = Au := L−1 (N u + F ),
◦
(58)
◦
where, according to Remark 3, the operator N : W 22 (DT , ST ) → W 12 (DT , ST )
is continuous and compact. Consequently, owing to (51), the operator A :
◦
◦
W 22 (DT , ST ) → W 22 (DT , ST ) is likewise continuous and compact. At the
same time, by Lemma 1, for any parameter τ ∈ [0, 1] and any solution
◦
u ∈ W 22 (DT , ST ) of the equation with the parameter u = τ Au the following
a priori estimate is valid:
kukW22 (DT ) ≤ σ1 τ kF kL2 (DT ) + τ λσ2 τ 3 kF k3L2(DT ) +
2
2
+ σ3 τ kF kW21 (DT ) exp σ4 + τ λσ5 τ kF kL2 (DT ) ≤ σ1 kF kL2(DT ) +
65
On the Global Solvability of the Cauchy Characteristic Problem
+ λσ2 kF k3L2 (DT ) + σ3 kF kW21 (DT ) exp σ4 + λσ5 kF k2L2 (DT ) = C0 (λ, σi , F ),
where C0 = C0 (λ, σi , F ) is a positive constant not depending on u and on
the parameter τ .
Therefore, by the Leray–Schauder theorem ([36, p. 375]), the equation
(58) and hence the problem (1), (3) has at least one strong generalized solution of the class W22 in the domain DT . Thus taking into account Remark
1 and Definitions 1, 2 and 3 of Section 2, the following theorem is valid.
◦
Theorem 1. Let λ > 0, F ∈ L2,loc (D∞ ) and F ∈ W 12 (DT , ST ) for any
T > 0. Then the problem (1), (3) is globally solvable, i.e., for any T > 0
this problem has a solution of the class W22 in the domain DT in the sense
of Definition 1.
Suppose
◦
◦
W k2,loc (D∞ , S∞ ) = {v ∈ L2,loc (D∞ ) : v|DT ∈ W k2 (DT , ST ) ∀T > 0}.
In the next section we will prove the uniqueness of a solution of the
problem (1), (3) of the class W22 in the sense of Definition 1. This and
Theorem 1 allow us to conclude that the theorem below is valid.
◦
Theorem 2. Let λ > 0 and F ∈ W 12,loc (D∞ , S∞ ). Then the problem
(1), (3) has in the light cone of the future D∞ a unique global solution u from
◦
the space W 22,loc (D∞ , S∞ ) which satisfies the equation (1) almost everywhere
in the domain D∞ and the boundary condition (3) in the sense of the trace
theory.
4. The Uniqueness of a Solution of the Problem (1), (3) of the
Class W22 .
Lemma 2. The problem (1), (3) fails to have more than one solution of
the class W22 in the sense of Definition 1.
Proof. Let u1 and u2 be two solutions of the problem (1), (3) of the class
W22 in the domain DT in the sense of Definition 1. Then for the difference
u = u2 − u1 we have
( + m2 )u = −3λu32 + 3λu31 ,
(59)
u, u1 , u2 ∈ W 22 (DT , ST ).
(60)
◦
Multiplying both parts of the equation (59) by ut and integrating over the
domain Dτ just in the same manner as when writing (14), we obtain
Z
Z 3
X
ut +
u2xi dx + m2 u2 dx =
Ωτ
i=1
Ωτ
= −6λ
Z
Dτ
(u32 − u31 )ut dx dt, 0 < τ ≤ T. (61)
66
S. Kharibegashvili
Estimate the first part of the equality (61). We have
Zτ
Z
Z
− 6λ (u32 − u31 )ut dx dt = 6|λ| dσ (u32 − u31 )ut dx dt ≤
0
Dτ
≤ 6|λ|
Zτ
dσ
≤ 9|λ|
Zτ
dσ
0
0
Z
Ωσ
Z
Ωσ
|u2 − u1 ||u22 + u2 u1 + u21 | |ut | dx dt ≤
(u22 + u21 )|u| |ut | dx dt.
(62)
Ωσ
Using Hölder’s inequality for p1 = 6, p2 = 3, p3 = 2 ( 16 + 13 + 12 = 1), we
obtain
Z
(u22 +u21 )|u| |ut | dx dt ≤ ku22 kL3 (Ωσ )+ku21 kL3 (Ωσ ) kukL6 (Ωσ ) kut kL2 (Ωσ ) =
Ωσ
= ku2 k2L6 (Ωσ ) + ku1 k2L6 (Ωσ ) kukL6(Ωσ ) kut kL2 (Ωσ ) ,
0 < σ ≤ T. (63)
By the imbedding theorem, we have ([31, pp. 69, 78])
kv|Ωσ k ◦ 1
W 2 (Ωσ )
≤ C(T )kvkW22 (DT )
kv|Ωσ kL6 (Ωσ ) ≤ βkv|Ωσ k ◦ 1
W 2 (Ωσ )
(dim Ωσ = 3, dim DT = 4),
(64)
≤ βC(T )kvkW22 (DT ) ,
where the positive constants C(T ) and β do not depend on the parameter
σ ∈ (0, T ] and on the function v.
By virtue of (60), it follows from (63) and (64) that
Z
(u22 + u21 )|u| |ut | dx ≤ 2M kuk ◦ 1
kut kL2 (Ωσ ) ≤
W 2 (Ωσ )
Ωσ
2
≤ M kuk ◦ 1
W 2 (Ωσ )
+
kut k2L2 (Ωσ )
=M
Z Ωτ
u2t
+
3
X
i=1
u2xi
dx,
(65)
where
M = β 3 C 2 (T ) max ku1 k2W 2 (DT ) , ku2 k2W 2 (DT ) < +∞.
2
2
R
P
3
Assuming w(τ ) = Ωτ u2t + i=1 u2xi dx, by (62)–(65) we have
w(τ ) ≤ 9|λ|M
Zτ
w(σ) dσ,
0
whence by the Gronwall lemma we find that w = 0, i.e., ut = ux1 = ux2 =
ux3 = 0. Consequently, u = const, and since by the condition of the lemma
u|ST = 0, we have u = 0 and hence u2 = u1 , which proves the lemma. On the Global Solvability of the Cauchy Characteristic Problem
67
5. The Absence of the Global Solvability of the Problem (1), (3) for
λ < 0. The following statement is valid: let λ < 0, F = µF0 , F0 ∈ C(D∞ ),
supp F0 ∩S∞ = ∅, F0 ≥ 0, F0 6≡ 0, µ = const > 0. Then for any T > 0 there
exists a number µ0 = µ0 (T ) > 0 such that for µ ≥ µ0 the problem (1), (3)
◦
fails to have a classical solution u ∈ C 2 (DT , ST ) = {u ∈ C 2 (DT ) : u|ST = 0}
in the domain DT .
We omit here the proof of that statement because it repeats the proof
of an analogous result for the wave equation with the power nonlinearity
with two spatial variables [27] and relies essentially on the method of test
functions [14].
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(Received 9.07.2007)
Author’s address:
A. Razmadze Mathematical Institute
1, M. Aleksidze St., Tbilisi 0193, Georgia
E-mail: kharibegashvili@yahoo.com