On a certain type of nonlinear hyperbolic equations derived from astrophysical problems

0.21 MB·

by  Tetu Makino

On a certain type of nonlinear hyperbolic
6 equations derived from astrophysical problems
1
0
2 Tetu Makino ∗

n
a
February 2, 2016
J

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3

] Abstract
P
A Investigationsofsphericallysymmetricmotionsofself-gravitatinggaseous
starsgovernedbythenon-relativisticNewtoniangravitationtheoryorby
.
h the general relativistic theory lead us to a certain type of non-linear hy-
t perbolic equations defined on a finite interval of the space variable. The
a
m linearized principal part has regular singularities at the both ends of the
interval of the space variable. But the regularity loss caused by the sin-
[
gularities at the boundaries requires application of the Nash-Moser tech-

1 nique. An abstract unified treatment of theproblem is presented.
v
8 KeyWordsandPhrases. Non-linearhyperbolicequations,Nash-Moser
6 theory,Singular boundary,Time periodic solutions, Cauchy problems.
1 2010MathematicsSubjectClassificationNumbers. 35L70,35Q31,35Q85.
0
0
. 1 Introduction
2
0
6 We have investigated spherically symmetric motions of gaseous stars governed
1
by the non-relativistic Euler-Poisson equations in [5] and by the relativistic
:
v Einstein-Euler equations in [6]. See also [7] on the problem governed by the
i Einstein-Euler-de Sitter equations with the cosmologicalconstant. In this arti-
X
cle a unified formulation of the discussions developed in the previous works is
r
a presented. This treatment using the Nash-Moser theorem is based on the work
[4]. Let us begin by giving an abstract settlement of the problem.
The set of equations we consider is
∂y
J(x,y,z)v =0, (1a)
∂t −
∂v
+H (x,y,z,v) y+H (x,y,z,v,w)=0 (1b)
1 2
∂t L
with
∗ProfessorEmeritusatYamaguchiUniversity,Japan/e-mail: [email protected]
1
d2 5 N d d
= x(1 x) (1 x) x +ℓ (x)x(1 x) +L (x). (2)
L − − dx2 − 2 − − 2 dx 1 − dx 0
(cid:16) (cid:17)
∂y ∂v
Herexrunsonthe interval]0,1[andz =x ,w =x . N isaparameterand
∂x ∂x
we put the assumption
(B0): N >4.
Let us denote by A the set of all functions defined and analytic on a neigh-
borhood of [0,1]. Let us denote by AQ(Up) the set of all analytic functions
f(x,y , ,y )ofxinaneighborhoodof[0,1]andy ,y inaneighborhood
1 p 1 p
··· ···
U of 0 such that
f(x,y , ,y )= a (x)yk1 ykp.
1 ··· p k1···kp 1 ··· p
k1+··X·+kp≥Q
We put
(B1): We suppose ℓ ,L A and there is a neighborhood U of 0 such that
1 0
∈
J A0(U2), H A0(U3), H A2(U4).
1 2
∈ ∈ ∈
LetusfixT >0arbitrarilyandfixfunctionsy ,v C ([0,T] [0,1])such
∗ ∗ ∞
∈ ×
that y (t,x),z (t,x) = x∂y /∂x,v (t,x),w (t,x) = x∂v /∂x U for 0 t
∗ ∗ ∗ ∗ ∗ ∗
∈ ≤ ∀ ≤
T,0 x 1. We seek a solution y,v C ([0,T] [0,1]) of (1a)(1b) of the
∞
≤ ∀ ≤ ∈ ×
form
y =y∗+y˜, v =v∗+v˜ (3)
which satisfies
y˜ =0, v˜ =0. (4)
t=0 t=0
| |
We suppose the following assumptions:
(B2): We have
H (x,0,0,0)=J(x,0,0) 1 (5)
1 −
and there is a constant C >1 such that
1
<J(x,0,0)<C (6)
C
for x U .
0
∀ ∈
(B3): We have
∂ J 0, (∂ H ) y+∂ H 0, ∂ H 0 (7)
z z 1 z 2 w 2
≡ L ≡ ≡
as x 1.
→
2
Here a function f defined and analytic on a neighborhood of (1,0, ,0) is
···
said to satisfy f 0 as x 1 if there is a function Ω defined and analytic on a
≡ →
neighborhood of (1,0, ,0) such that
···
f(x,y , ,y )=(1 x)Ω(x,y , ,y ).
1 p 1 p
··· − ···
Of course it is the case if f(1,y , ,y ) = 0 for y , , y . In the assump-
1 p 1 p
··· ∀ ··· ∀
tion (B3) the functions in (7) are regardedas functions of x,y,Dy,D2y,v,Dv.
Here D stands for ∂/∂x.
Under the above situation, we are going to prove the following
Theorem 1 There is a small positive number δ(T) and a large number K such
that, if
j+mkaxKk∂tj∂xk(y∗,v∗)kL∞ ≤δ(T), (8)
≤
there exists a solution (y,v) of (1a)(1b)(3)(4).
2 Frame-work to apply the Nash-Moser(-Hamilton)
theorem
The equations for w:=(y˜,v˜)T turn out to be
∂y˜
Jv˜ (∆J)v =c , (9a)
∗ 1
∂t − −
∂v˜
+H y˜+(∆H ) y +∆H =c , (9b)
1 1 ∗ 2 2
∂t L L
where
∂y˜
J =J(x,y +y˜,z +z˜) with z˜=x
∗ ∗
∂x
∆J =J(x,y +y˜,z +z˜) J(x,y ,z ),
∗ ∗ ∗ ∗
−
∂y
∗
c1 = +J(x,y∗,z∗)v∗,
− ∂t
∂v˜
H1 =H1(x,y∗+y˜,z∗+z˜,v∗+v˜,w∗+w˜) with w˜ =x ,
∂x
∆H1 =H1(x,y∗+y˜,z∗+z˜,v∗+v˜) H1(x,y∗,z∗,v∗),
−
∆H =H (x,y +y˜,z +z˜,v +v˜,w +w˜) H (x,y ,z ,v ,w ),
2 2 ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗
−
∂v
∗
c = H (x,y ,z ,v ) y H (x,y ,z ,v ,w ).
2 1 ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗
− ∂t − L −
We write the equations (9a)(9b) as
P(w)=c, (10)
3
wherec=(c ,c )T. The domainofthe nonlinearmapping P is U,the setofall
1 2
functions w=(y˜,v˜)T E E such that
0 0
∈ ×
y˜ + xDy˜ + v˜ + xDv˜ <ǫ , (11)
0
| | | | | | | |
where D = ∂/∂x. Here ǫ is so small that (11) implies y(t,x) = y (t,x)+
0 ∗
y˜(t,x),z = xDy,v,w = xDv U for t [0,T], x U . We have defined
0
∈ ∀ ∈ ∀ ∈
E=C ([0,T] [0,1]) and E = φ E φ =0 .
∞ 0 t=0
× { ∈ | | }
We are going to apply the Nash-Moser(-Hamilton) theorem ([1, p. 171,
III.1.1.1]).
First we introduce gradings of norms on E to make it a tame space in the
sense of Hamilton [1].
To do so, we use a cut off function ω C (R) such that ω(x) = 1 for
∞
∈
x 1/3, 0 < ω(x) < 1 for 1/3 < x < 2/3 and ω(x) = 0 for 2/3 x. For a
≤ ≤
function u of x [0,1], we define
∈
u[0](x):=ω(x)u(x), u[1](x):=(1 ω(x))u(x). (12)
−
We consider the functional spaces
E = u E u=0 for 5/6 x 1 , (13a)
[0]
{ ∈ | ≤ ≤ }
E = u E u=0 for 0 x 1/6 (13b)
[1]
{ ∈ | ≤ ≤ }
endowed with the gradings (k·|([µ∞]n))n,(k·k([µ2)]n)n, µ=0,1, defined as follows.
Put
d2 5 d
:=x + ,
△[0] dx2 2dx
and
d2 N d
:=X + with X =1 x,
△[1] dX2 2 dX −
and put
kuk([µ∞]n) = sup k(−∂t2)j(−△[µ])kukL∞,
j+k n
≤
T 1/2
u (2) = ( ∂2)j( )ku 2 dt ,
k k[µ]n k − t −△[µ] k[µ]
(cid:16)j+Xk≤nZ0 (cid:17)
where
1 1/2
u = u2x3/2dx ,
[0]
k k | |
(cid:16)Z0 (cid:17)
1 1/2
u = u2(1 x)N/2 1dx .
[1] −
k k | | −
(cid:16)Z0 (cid:17)
4
Proposition 1 Let µ = 0 or = 1. The gradings (k · |([µ∞]n))n,(k · k([µ2)]n)n are
equivalent and make E a tame space.
[µ]
Proof. Let us consider E and denote X =1 x by x and by
[1] [1]
− △
d2 N d
=x + .
△ dx2 2 dx
( E can be treated similarly.) Then u E means u C ([0,1]) and
[0] [1] ∞
∈ ∈
u(x)=0 for x 5/6.
≥
Now, we can define the Fourier transformation Fu(ξ) of a function u(x) of
x [0,+ ) by
∈ ∞
Fu(ξ):= ∞K(ξx)u(x)xN/2 1dx,
−
Z0
where K(X) is an entire function of X C given by
∈
∞ ( 4X)k
K(X)=2(√X)−N/2+1JN/2−1(4√X)=2N/2 k!Γ(−N/2+k),
k=0
X
J being the Bessel function of order N/2 1. Note that
N/2 1
− −
F( u)(ξ)=4ξ Fu(ξ)
−△ ·
and the inverse of F is F itself. Then we see that E endowed with ( )
[1] k·k∞[1]n n
is a tame direct summand of the tame space
F=L (R [0,+ ),dτ ξN/2 1dξ,log(1+τ2+4ξ))
∞1 × ∞ ⊗ −
through the transformation
1
u(τ,ξ)= e √ 1τtFu(t, )(ξ)dt
− −
F √2π ·
Z
and its inverse applied to the space
E˜ :=C (( 2T,2T) [0,2)),
0∞ − ×
intowhichtheextensionoffunctionsofE multipliedbyafixedcutofffunction
[1]
χ(t) which vanishes for t T can be imbedded, and the space
≤−
E˙ :=C˙ (R [0,+ )):= u j k lim sup ( ∂2)j( )ku =0 ,
∞ × ∞ { | ∀ ∀ L→∞L≤|t|,L≤x| − t −△ | }
for which functions of E are restrictions. In fact, if we denote by e:E E˜
[1] [1]
→
the extension operator, and by r : E˙ E the restriction operator, then the
[1]
→
mappings e : E F and r : F E are tame and the composition
[1] [1]
F → F →
(r ) ( e) is IdE . For the details, see the proof of [1, p.137, II.1.3.6.]. This
F ◦ F [1]
implies that E[1] is tame with respect to (k·k∞[1])n)n. On the other hand
N
2 kuk[1] ≤kukL∞ ≤Csupk(−△)juk[1]
r j σ
≤
5
bytheSobolevimbeddingtheorem(see[4,Appendix]),providedthat2σ >N/2.
(Recall ( u )2 = 1 u2xN/2 1dx.) The derivative with respect to t can be
k k[1] 0 | | −
treated more simply. Thus we see the equivalence
R
1
u (2) u (∞) C u (2)
Ck k[1]n ≤k k[1]n ≤ k k[1],n+s
with 2s>1+N/2. (cid:4)
We put
kuk(n∞) := sup kuk([µ∞]n),
µ=0,1
1/2
u (2) := ( u (2) ) .
k kn k k[µ]n
(cid:16)µX=0,1 (cid:17)
Proposition 2 The gradings ( (n∞))n, ( (n2))n are equivalent and the space
k·k k·k
E becomes a tame space by these gradings.
Proof. In fact E is a tame direct summand of the Cartesian product E
[0]
×
E , which is a tame space, (see [1, p.136, 1.3.1. and 1.3.4]), by the linear
[1]
mapping L : E E E : u (u[0],u[1]) and M : E E E :
[0] [1] [0] [1]
→ × 7→ × →
(u0,u1) u0+u1. Clearly L is tame and ML = IdE. To verify the tameness
7→
ofM, we note that, if the supportof a function u is included in [1/6,5/6],then
k△m[µ]ukL∞ ≤C k△k[1−µ]ukL∞.
0 k m
≤X≤
A proof can be found in [5, Appendix B]. If h E for µ = 0,1, then
µ [µ]
∈
h=M(h ,h )=h +h and h[0] =ω h +ω h . Then by [4, Proposition 4]
0 1 0 1 0 1
· ·
we have
k△m[0]h[0]kL∞ ≤C k△k[0]h0kL∞ +k△m[0](ω·h1)kL∞
k m
X≤
≤C k△k[0]h0kL∞ +C′ k△k[1](ω·h1)kL∞
k m k m
X≤ X≤
≤C k△k[0]h0kL∞ +C′′ k△k[1]h1kL∞,
k m k m
X≤ X≤
sincethe supportofω h isincludedin[1/6,2/3]. Similarlywe getanestimate
1
·
|d△irm[e1c]ht[s1u]kmL∞m.anTd.he(cid:4)se estimates imply the tameness of M. Thus E is a tame
Then E E is a tame space as the Cartesian product of the tame space E
×
and its copy. Let us denote
kwk(n∞) =max(ky˜k(n∞),kv˜k(n∞)),
1/2
w (2) = ( y˜ (2))2+( v˜ (2))2 .
k kn k kn k kn
(cid:16) (cid:17)
6
Then we can claim
Proposition 3 If 2s>1+max(N,5)/2, then we have
1
w (2) w ( ) C w (2) .
Ck kn ≤k kn∞ ≤ k kn+s
SinceP(w)isasmoothfunctionoft,x,y˜[0],y˜[1],Dy˜[0],Dy˜[1], y˜[0], y˜[1],
[0] [1]
△ △
v˜[0],v˜[1],Dv˜[0],Dv˜[1], ∂ y˜[0],∂ y˜[1],∂ v˜[0],∂ v˜[1],thanksto [4,Proposition3,4,5]
t t t t
we see
kP(w)k(n∞) ≤C(1+kwk(n∞+)1),
that is, P is a tame mapping from U E E into E E.
0 0
⊂ × ×
Besides these gradings, we shall use another gradings. Namely, we put
( )mu for ℓ=2m
u := k △[µ] k[µ]
h i[µ]ℓ ( D˙[µ]( [µ])mu [µ] for ℓ=2m+1,
k △ k
where µ=0,1, and
d d
D˙ =√x , D˙ =√X with X =1 x,
[0] dx [1] dX −
and for h=(h,k)T we put
1/2
h := ( h )2+( k )2 ,
[µ]k [µ],ℓ+1 [µ]ℓ
k k h i h i
(cid:16)0≤Xℓ≤k (cid:17)
khk[hµτ]in := sup k∂tjh(t,·)k[µ]k,
0 t τ
≤ ≤ j+Xk≤n
T 1/2
h := ( ∂jh )2dt ,
|k k|[µ]n k t k[µ]k
(cid:16)j+Xk≤nZ0 (cid:17)
1/2
h := ( h[0] )2+( h[1] )2 ,
k [0]k [1]k
k k k k k k
(cid:16) (cid:17)1/2
khknhτi := (kh[0]kh[0τ]in)2+(kh[1]kh[1τ]in)2 ,
(cid:16) (cid:17) 1/2
h := ( h[0] )2+( h[1] )2 .
n [0]n [1]n
|k k| |k k| |k k|
(cid:16) (cid:17)
Thanks to [4, Proposition 7] and the Sobolev’s imbedding with respect to t we
can claim
Proposition 4 For any non-negative integer m, we have
1
h h T C h
C|k k|n ≤k khn i ≤ |k k|n+1
and
1
h (2) h C h (2).
Ck km ≤|k k|2m ≤ k km
7
Moreover we put
|h|n :=j+kmna,µx=0,1k∂tj(D˙[µ])kh[µ]kL∞,
≤
h τ := sup h(t) .
| |nh i | |n
0 t τ
≤ ≤
By interpolations, we can claim
Proposition 5 For any non-negative integer m we have
1
Ckhk(m∞) ≤|h|h2Tmi ≤Ckhk(m∞).
3 Analysis of the Fr´echet derivative I
We have to analyze the Fr´echetderivative DP(w) of the mapping P at a given
w=(y˜,v˜)T U. For h=(h,k)T we have DP(w)h=((DP) ,(DP) )T, where
1 2
∈
∂h ∂
(DP) = Jk (∂ J)v+(∂ J)vx h,
1 y z
∂t − − ∂x
∂k (cid:16) (cid:17)
(DP) = +H h+
2 1
∂t L
∂
+ (∂ H ) y+∂ H +((∂ H ) y+∂ H )x h+
y 1 y 2 z 1 z 2
L L ∂x
(cid:16) ∂ (cid:17)
+ (∂ H ) y+∂ H +(∂ H )x k.
v 1 v 2 w 2
L ∂x
(cid:16) (cid:17)
Thankstotheassumption(B3)thereareanalyticfunctionsa ,a ,a ,a ,a ,a
01 00 11 10 21 20
ofx,y(=y +y˜),Dy,D2y,v(=v +v˜),Dv,whereD =∂/∂x,onaneighborhood
∗ ∗
of [0,1] 0 0 such that
×{ }×···×{ }
∂h
(DP) = Jk+(a x(1 x)D+a )h, (14a)
1 01 00
∂t − −
∂k
(DP) = +H h+(a x(1 x)D+a )h+
2 1 11 10
∂t L −
+(a x(1 x)D+a )k. (14b)
21 20
−
We put
X:=L2([0,1];x3/2(1 x)N/2 1dx),
−
−
dφ
X1 := φ X D˙φ:= x(1 x) X ,
{ ∈ | − dx ∈ }
X2 := φ X1 ΛφpX ,
{ ∈ | − ∈ }
with
d2 5 N d
Λ=x(1 x) + (1 x) , (15)
− dx2 2 − − 2 dx
(cid:16) (cid:17)
We claim
8
Proposition 6 Given g C([0,T],X1 X), the equation DP(w)h=g admits
∈ ×
a unique solution h C([0,T],X2 X1) such that h(t=0)=(0,0)T.
∈ ×
Proof. We can rewrite
∂k
(DP) = H Λh+b Dˇh+b h+a Dˇk+a k,
2 1 1 0 21 20
∂t −
where
Dˇ :=x(1 x)D,
−
b :=H ℓ +a ,
1 1 1 11
b :=H L +a .
0 1 0 10
Of course b ,b are also analytic on a neighborhood of [0,1] 0 . Then
1 0
×{ }×···
we can write the equation DP(w)h=g(=(g ,g )T) as
1 2
∂ h a J h g
+ 1 − = 1 . (16)
∂t k a2 k g2
(cid:20) (cid:21) (cid:20)A (cid:21)(cid:20) (cid:21) (cid:20) (cid:21)
Here
a =a Dˇ +a ,
1 01 00
a :=a Dˇ +a ,
2 21 20
:= H Λ+b Dˇ +b .
1 1 0
A −
The standard calculation gives
1 d
k 2+((H /J)D˙hD˙h) +
1
2dt k k |
+(β(cid:16)D˙hD˙h)+(β D˙hh)+((cid:17)β D˙hk)++(β hk)+(β k k)=
1 2 3 4 5
| | | | |
=((H /J)D˙hD˙g )+(k g ), (17)
1 1 2
| |
where D˙ = x(1 x)D and
−
1 p 1∂(H /J)
β = (3+(N +3)x+2Dˇ)(H /J)a 1 +(H /J)(Dˇa +a ),
1 1 01 1 01 00
−4 − 2 ∂t
β =(H /J)D˙a ,
2 1 00
β = (H /J)D˙J +D˙H + x(1 x)(b +a ),
3 1 1 1 21
− −
β4 =b0, β5 =a20. p
Ofcourse ( ) and standfor the inner product andthe normof the Hilbert
·|· k·k
space X. We have used the following formulas:
Formula 1: If φ X2,ψ X1,α C1([0,1]), then
∈ ∈ ∈
( αΛφψ)=(αD˙φD˙ψ)+((Dα)Dˇφψ). (18)
− | | |
9
Formula 2: If φ X2 and α C1([0,1]), then
∈ ∈
(αD˙φD˙Dˇφ)=(α D˙φD˙φ), (19)
∗
| |
with
1
α∗ = (3+(N +3)x+2Dˇ)α.
−4
Since w is confined to U, we can assume
1 1
<J <M , <H <M
0 1 0
M M
0 0
witha constantM independentofw thanks to the assumption(B2). Now the
0
energy
:= k 2+((H /J)D˙hD˙h)
1
E k k |
enjoys the inequality
1d
E M( h 2H+ h H g H),
2 dt ≤ k k k k k k
where H=X1 X and
×
(φ,ψ)T 2 = φ 2 + ψ 2 = φ 2+ D˙φ 2+ ψ 2
k kH k kX1 k kX k k k k k k
and
5
M = βj L∞ +(M0)2+1.
k k
j=1
X
Since is equivalent to k 2+ D˙h 2, the Gronwall argument and application
E k k k k
of the Kato’s theory ([2]) deduce the conclusion. Here h should be estimated
k k
by as follows: The first component of (16) implies
E
t
h(t)= ( a Dˇh a h+Jk+g )(t)dt,
01 00 1 ′ ′
− −
Z0
therefore
t t
h(t) C h(t) dt + (C (t )+ g (t) )dt,
′ ′ ′ 1 ′ ′
k k≤ k k E k k
Z0 Z0
whereC =max(ka00kL∞,ka01kL∞M02+M0),whichimplies,throughthe Gron-
wall’s argument,
t
h(t) (eC(t t′) 1)(C (t)+ g (t) )dt.
− ′ 1 ′ ′
k k≤ − E k k
Z0
10