Electronic Journal of Differential Equations, Vol. 2004(2004), No. 77, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
BLOW-UP OF SOLUTIONS TO A NONLINEAR WAVE
EQUATION
SVETLIN GEORGIEV GEORGIEV
Abstract. We study the solutions to the the radial 2-dimensional wave equation
1
sinh 2χ
= g,
χtt − χr − χrr +
r
2r 2
γ
γ−1
χ(1, r) = χ◦ ∈ Ḣrad , χt (1, r) = χ1 ∈ Ḣrad
,
where r = |x| and x in R2 . We show that this Cauchy problem, with values into
a hyperbolic space, is ill posed in subcritical Sobolev spaces. In particular, we
construct a function g(t, r) in the space Lp ([0, 1]Lqrad ), with p1 + 2q = 3 − γ, 0 <
γ < 1, p ≥ 1, and 1 < q ≤ 2, for which the solution satisfies limt→0 kχ̄kḢ γ =
rad
∞. In doing so, we provide a counterexample to estimates in [1].
1. Introduction
In this paper, we study the properties of the solutions to the Cauchy problem
1
sinh(2χ)
χtt − χr − χrr +
= g(t, r),
r
2r2
γ
γ−1
χ(1, r) = χ◦ ∈ Ḣrad
, χt (1, r) = χ1 ∈ Ḣrad
,
(1.1)
(1.2)
where g ∈ Lp ([0, 1]Lqrad ), p1 + 2q = 3 − γ, 0 < γ < 1, p ≥ 1, 1 < q ≤ 2, r = |x|,
x ∈ R2 , and t ∈ [0, 1].
The homogeneous problem, i.e. (1.1) with g ≡ 0, has been investigated by several
authors; see for example [1, p. 89-141], [10], [11], and references therein. For the
non-homogeneous problem, we construct function g in Lp ([0, 1]Lqrad ), and a solution
χ̄ to (1.1)–(1.2) such that limt→0 kχ̄kḢ γ = ∞. This implies that (1.1)–(1.2) is an
rad
ill posed problem, and provides a counter example to [1, pages 3–4, Eq. (9)].
Equation (1.1) is obtained from the wave map equation with a source term when
x ∈ R2 and the target is the hyperboloid H2 : u21 + u22 − u23 = −1, H2 ,→ R3 . Let
us consider the equation
utt − 4u − (|ut |2 − |∇x u|2 )u = q(t, x),
2000 Mathematics Subject Classification. 35L10, 35L50.
Key words and phrases. Wave equation, blow-up, hyperbolic space.
c
2004
Texas State University - San Marcos.
Submitted March 16, 2004. Published May 26, 2004.
1
2
SVETLIN G. GEORGIEV
EJDE-2004/77
when x ∈ R2 and the target is the hyperboloid H2 : u21 + u22 − u23 = −1, H2 ,→ R3 .
Equation (1) is a wave map equation with a source term, where
|ut |2 = u21t + u22t − u23t ,
|∇x u|2 = |∇x1 u|2 + |∇x2 u|2 ,
|∇xi u|2 = u21xi + u22xi − u23xi ,
i = 1, 2.
Let q = (q1 , q2 , q3 ), with
sinh3 χ cos φ1
1
sinh
χ
cos
φ
+
,
1
r2
r2
1
sinh3 χ sin φ1
q2 = cosh χ sin φ1 (χtt − ∆χ) + 2 sinh χ sin φ1 +
,
r
r2
q3 = sinh χg.
q1 = cosh χ cos φ1 (χtt − ∆χ) +
For the hyperboloid H2 , we have the parametric representation
u = (u1 , u2 , u3 ),
(1.3)
u1 = sinh χ cos φ1 ,
(1.4)
u2 = sinh χ sin φ1 ,
(1.5)
χ ≥ 0,
(1.6)
u3 = cosh χ,
φ1 ∈ [0, 2π].
Let x1 = r cos φ1 , x2 = r sin φ1 . Then from (1), we get that χ satisfies (1.1).
Our main result is as follows.
Theorem 1.1. Let p1 + 2q = 3 − γ, p ≥ 1, 1 < q ≤ 2, 0 < γ < 1. Then there exist
function g ∈ Lp ([0, 1]Lqrad ) and solution χ̄ to (1.1)–(1.2) such that
lim kχ̄kḢ γ = ∞.
t→0
rad
Note that the case p = 1, q = 2 which is the energy case can not be reached by
the setting in this theorem.
2. Preliminary results
Let f be a real-valued function satisfying
(H1) f ∈ C 2 [0, ∞),
(H2) f (0) = 0, f (1) = f 0 (1) = 0.
As an example of a function satisfying (H1)-(H2), we have
f (x) = (1 − x)2 x.
(2.1)
Certainly f ∈ C 2 [0, ∞); therefore, (H1) holds. Note that f (0) = 0 and f 0 (x) =
−2(1 − x)x + (1 − x)2 thus f (1) = f 0 (1) = 0; therefore, (H2) holds.
In addition, assume that:
(H3) p1 + 2q = 3 − γ, 0 < γ < 1, p ≥ 1, q > 1
(H4) 0 < α ≤ 2 − q, β > 0
β
q
< 2p(q−1)
or β < α with
(H5) Either β > α with α
β
α
Note that when (H3) holds, q ≤ 2; because if q > 2 then
from where γ > 1 which contradicts 0 < γ < 1.
q(2p−1)
.
2p
3 − γ = p1 + 2q
>
< 1 + 1 = 2,
EJDE-2004/77
BLOW-UP OF SOLUTIONS
Let z 1/α =
r
.
tβ/α
Then z 2/α =
3
r2
,
t2β/α
∂z 2/α
2β 1 2/α
∂ 2 z 2/α
2β(α + 2β) 1 2/α
=−
z ,
=
z ,
2
∂t
α t
∂t
α2
t2
∂z 2/α
2 1
2
∂ 2 z 2/α
= β zα,
= 2β .
2
∂r
∂r
α
t
tα
Let f be a function satisfying (H1)–(H2) and let
(
f (r2 ) for r ≤ 1,
χ◦ =
0
for r ≥ 1,
(
2 0 2
− 2β
α r f (r ) for r ≤ 1,
χ1 =
0
for r ≥ 1,
4β 2 4/α 00 2/α 2β(α + 2β) 2/α 0 2/α z f z
+
z f z
,
α2
α2
B2 = z 2/α f 00 (z 2/α ) + f 0 (z 2/α ),
(
sinh(2χ̄)
B1
4
for r ≤ tβ/α ,
2 − t2β/α B2 +
2r 2
g= t
0
for r ≥ tβ/α ,
B1 =
(2.2)
(2.3)
(2.4)
(2.5)
(2.6)
and let
(
f (z 2/α ) for r ≤ tβ/α
χ̄ =
0
for r ≥ tβ/α .
(2.7)
Note that χ̄ is a solution of (1.1)–(1.2). Indeed, for z ≤ 1 we have
2β 2/α 0 2/α
z f (z ),
tα
4β 2 1 4
2β(α + 2β) z 2/α 0 2/α
f (z ),
χ̄tt = 2 2 z α f 00 (z 2/α ) +
α t
α2
t2
2 1
χ̄r = β/α z α f 0 (z 2/α ),
t
2
4 2/α 00 2/α
χ̄rr = 2β z f (z ) + 2β f 0 (z 2/α ).
tα
tα
Then, for z ≤ 1, we have
χ̄t = −
sinh(2χ̄)
1
sinh 2χ̄
B1
4
χ̄tt − χ̄r − χ̄rr +
= 2 − 2β B2 +
≡ g,
2
r
2r
t
2r2
tα
Lemma 2.1. Under Assumptions (H1)-(H5), the function g defined by (2.6) is in
the space Lp ([0, 1]Lqrad ).
Proof. Note that
kgkqLq
rad
Z
tβ/α
q
Z
|g| r dr =
=
0
tβ/α sinh(2χ̄) q
4
B1
2 − 2β B2 +
r dr.
t
2r2
tα
0
1
By making the change of variable r = z α tβ/α ,
1 1
dr = tβ/α z α −1 dz,
α
2β
r dr = t α
1 2 −1
z α dz
α
4
SVETLIN G. GEORGIEV
EJDE-2004/77
(with t fixed) and
2β
kgkqLq
rad
Z 1
4
sinh(2χ̄) q 2 −1
1
B1 2 − 2β B2 + 2β
z α dz
t
tα
2t α z 2/α
0
2β
2β
Z
Z
2
2
t α −2q 1
t α (1−q) 1
q α
−1
≤ c1
|B1 | z
dz + c2
|B2 |q z α −1 dz
α
α
0
0
2β
(1−q) Z 1 q
α
t
sinh(2χ̄) α2 −1
dz.
+ c3
z
α
2z 2/α
0
=
tα
α
Let
Z
I1 =
1
Z
2
|B1 |q z α −1 dz,
1
0
2
|B2 |q z α −1 dz,
I2 =
I3 =
0
Z 1
sinh(2χ̄) q α2 −1
dz.
z
2z 2/α
0
From the definition of B1 and B2 and since (H1) and (H4) hold (we note that
2
2
α ∈ (0, 1) because 1 < q ≤ 2), we have |B1 |q z α −1 ∈ C[0, 1], |B2 |q z α −1 ∈ C[0, 1].
Then I1 < ∞, I2 < ∞. Now we consider I3 . Let B be constant for which
B − 2 cosh(2χ̄)χ̄0 ≥ 0.
Such a constant B exists because χ̄(z 2/α ) and χ̄0 (z 2/α ) are bounded functions for
z ∈ [0, 1] and supp χ̄ ⊂ [0, 1], supp χ̄0 ⊂ [0, 1]. Let
p(z) = Bz 2/α − sinh(2χ̄).
Then
2 2 −1
z α (B − 2 cosh(2χ̄)χ̄0 ) ≥ 0 ∀z ∈ [0, 1].
α
Consequently, p(z) is an increasing function for z ∈ [0, 1]. Therefore, p(z) ≥ p(0) =
0 for all z ∈ [0, 1] or Bz 2/α ≥ sinh(2χ̄) for z ∈ [0, 1]. Then
Z 1 2/α q
2
Bz
I3 ≤
2/α z α −1 dz ≡ const < ∞.
2z
0
Consequently,
2β
2β
kgkqLq ≤ c̄1 t α −2q + c̄2 t α (1−q)
p0 (z) =
rad
and from here
2β
kgkpLp ([0,1]Lq
rad
2β
2β
kgkLqrad ≤ c̃1 t αq −2 + c̃2 t αq − α ,
Z 1
Z 1
2βp
2βp
2βp
αq −2p dt + c
≤
c
t
t αq − α dt < ∞
1
2
)
0
0
because (H5) holds. Here c1 , c2 , c3 , c̄1 , c̄2 , c̃1 , c̃2 , c̄1 , c̄2 are positive constants. As special notation we have
s
Ḣ s (Rn ) ≡ Ḟ2,2
(Rn ),
0<s<∞
where (see [1, p. 94, def. 2])
s
s
Ḟ2,2
(Rn ) = Ḃ2,2
(Rn ),
−∞ < s < ∞.
As in [5, p. 30-31], when f (r) is a function with compact support in [0, 1], we have
Z 1
1/2
kf kḢ γ :=
h−1−2γ k∆h f k2L2 dh
,
rad
0
where ∆h f = f (r + h) − f (r), 0 < γ < 1.
rad
EJDE-2004/77
BLOW-UP OF SOLUTIONS
5
Lemma 2.2. Under assumptions (H1)–(H5), the function χ◦ defined by (2.2) is
γ
in the space Ḣrad
.
Proof. By definition of the norm,
Z 1
Z
kχ◦ k2Ḣ γ =
h−(1+2γ)
rad
0
Z
|f ((r + h)2 ) − f (r2 )|2 r dr dh
0
0
Using the Mean Value Theorem,
Z 1
Z
kχ◦ k2Ḣ γ =
h−(1+2γ)
rad
1
1
|f 0 (ξ)|2 [(r + h)2 − r2 ]2 r dr dh
0
1
=
h−(1+2γ)
0
Z
1
|f 0 (ξ)|2 (h2 + 2rh)2 r dr dh
0
1
Z
≤ Q1
h
−(1+2γ)
0
Z
Z
1
(h4 r + 4r2 h3 + 4r3 h2 ) dr dh
0
1
h4
4
+ h3 + h2 )dh
2
3
0
4
1
1
1
+ Q1
+ Q1
,
= Q1
2(4 − 2γ) 3 3 − 2γ
2 − 2γ
= Q1
h−(1+2γ)
where Q1 is positive constant. Therefore, kχ◦ kḢ γ ≤ Q2 for some constant Q2 and
γ
χ◦ ∈ Ḣrad
.
rad
Lemma 2.3. Under Assumptions (H1)-(H5), the function χ1 defined by (2.3) is
γ−1
in the space Ḣrad
.
γ−1
Proof. We have that L2rad ,→ Ḣrad
. On the other hand
Z
Z
2β 2 1 4 0 2 2 2
4β 2 1 5 0 2 2
r |f (r )| dr = 2
r |f (r )| dr < ∞
kχ1 k2L2 = 2
rad
α
α
0
0
γ−1
because f 0 (r2 ) ∈ C[0, 1]. Consequently, χ1 ∈ Ḣrad
.
Proof of main result
Let (H1)-(H5) hold. By (2.1), f (z 2/α ) = (1 − z 2/α )2 z 2/α . Let g be defined
by (2.6) and χ̄ by (2.7). From Lemma 2.1, g ∈ Lp ([0, 1]Lqrad ). Also, we have
γ
χ̄(1, r) = χ◦ , χ̄t (1, r) = χ1 . From Lemma 2.2, χ̄(1, r) ∈ Ḣrad
, and from Lemma
γ−1
2.3, χ̄t (1, r) ∈ Ḣrad . Therefore, χ̄ is solution of (1.1)–(1.2). Let θ = tβ/α which is
in (0, 1]. Then
Z 1
Z θ r2 2
r + h 2 kχ̄k2Ḣ γ =
h−(1+2γ)
− f 2 r dr dh =: I.
(2.8)
f
rad
θ
θ
0
0
Then
Z 1
Z √θ r2 2
3
r + h 2 I≥ √
h−(1+2γ)
− f 2 r dr dh.
f
1
θ
θ
θ
( 2− 2 )θ
2
and
r + h 2 r2 h
(r + h)2 i2 (r + h)2 r 2 2 r 2
f
−f 2 = 1−
−
1
−
θ
θ
θ2
θ2
θ2 θ2
2
2 2
2
2
[θ − (r + h) ] (r + h) − (θ − r2 )2 r2
=
.
θ6
6
SVETLIN G. GEORGIEV
EJDE-2004/77
Note that the numerator of the above expression is
L := [θ2 − (r + h)2 ]2 (r + h)2 − (θ2 − r2 )2 r2
= (θ2 − r2 )(2rh + h2 )(θ2 − 3r2 ) + (3r2 − 2θ2 )(2rh + h2 )2 + (2rh + h2 )3 .
For r ∈ ( θ2 , √θ3 ) we have
L ≥ (3r2 − 2θ2 )(2rh + h2 )2 + (2rh + h2 )3
= (2rh + h2 )2 (3r2 − 2θ2 + 2rh + h2 )
= (2rh + h2 )2 [2r2 + (r + h)2 − 2θ2 ].
√
For h ∈ (( 2 − 12 )θ, 1) and r ∈ ( θ2 , √θ3 ),
r+h≥
√
1
θ √
+ 2θ − θ = 2θ,
2
2
(r + h)2 ≥ 2θ2 .
√
Therefore, for h ∈ (( 2 − 12 )θ, 1) and r ∈ ( θ2 , √θ3 ),
L ≥ (2rh + h2 )2 2r2 ≥ h4 r2 ,
r2 2
h8 r 4
r + h 2 − f 2 ≥ 12 .
f
θ
θ
θ
Consequently
Z 1
I≥ √
θ
√
3
h8 r 5
dr dh
θ
θ12
( 2− 12 )θ
2
Z 1
θ6
θ6 1
h7−2γ
−
dh
= 12 √
27 64
6θ
( 2− 12 )θ
h
√
37
1 8−2γ 8−2γ i
=
1
−
(
2
−
)
θ
6 × 27 × 64θ6 .(8 − 2γ)
2
37
37
√
=
−
θ2−2γ
6 × 27 × 64θ6 .(8 − 2γ) 6 × 27 × 64(8 − 2γ)( 2 − 12 )8−2γ
h
−(1+2γ)
Z
which approcahes zero as θ → 0. From this statement and (2.8), we obtain
lim kχ̄kḢ γ = ∞.
t→0
rad
2.1. Comments. Let (H1)-(H5) hold. By (2.1), f (z 2/α ) = (1 − z 2/α )2 z 2/α . Then
the function χ̄ defined by (2.7) is a solution to the Cauchy problem
1
χtt − χr − χrr = g1 (t, r),
(2.9)
r
γ
γ−1
χ(1, r) = χ◦ ∈ Ḣrad
, χt (1, r) = χ1 ∈ Ḣrad
,
(2.10)
where χ◦ and χ1 are the functions defined with (2.2) and (2.3),
(
B1
4
for r ≤ tβ/α
2 − t2β/α B2
g1 = t
0
for r ≥ tβ/α .
Then because of (H5),
kg1 kpLp ([0,1]Lq )
rad
Z
≤ l1
1
t
0
2βp
αq −2p
Z
dt + l2
1
2βp
t αq
0
− 2βp
α
dt < ∞,
EJDE-2004/77
BLOW-UP OF SOLUTIONS
7
where l1 , l2 are positive constants. In [2, Corollary 1.3] it is proved that
kχ̄kC([0,1]Ḣ γ
rad )
Therefore, χ̄ is in
≤ kχ◦ kḢ γ + kχ1 kḢ γ−1 + kg1 kLp ([0,1]Lqrad ) .
rad
rad
γ
C([0, 1]Ḣrad
).
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University of Sofia, Faculty of Mathematics and Informatics, Department of Differential Equations, Bulgaria
E-mail address: sgg2000bg@yahoo.com