On the Fredholm Solvability for a Class of Multidimensional Hyperbolic Problems

0.12 MB·English

by  Irina Kmit

A Fredholmness criterion with
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0 applications to hyperbolic PDEs
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I. Kmit
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Institute for Applied Problems of Mechanics and Mathematics,

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2 Ukrainian Academy of Sciences

Naukova St. 3b, 79060 Lviv, Ukraine
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E-mail: [email protected]
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Abstract
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[ We prove a criterion of Fredholmness for linear operators in Banach spaces

1 and apply it to some first-oder hyperbolic partial differential operators.
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1 Introduction
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2
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2 In [1] we developed an analytical approach for proving Fredholmness property for
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a class of linear hyperbolic problems. This subject is related to the following two
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0 topics, known results on which are far from being conclusive:
:
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• Correct posedness of initial-boundary value problems for hyperbolic PDEs;
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X
r • Local investigation of nonlinear hyperbolic equations, as smooth continua-
a
tion and local bifurcations (the Fredholmness property is crucial here).
The Fredholmness in the context of hyperbolic PDEs is related to propagating
singularities along special curves called characteristics. This entails two main
complications:
• Inverses to hyperbolic operators do not improve the regularity of right-
hand sides of differential equations so much as inverse elliptic or parabolic
operators do it. This causes a complication with using the well-known
classical fact that any Fredholm operator is bijective modulo a compact
operator. Namely, to prove the desired compactness property, we would
need more regularity than the hyperbolicity gives us;
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• To ensure an optimal regularity trade-off (needed for Fredholmness) be-
tween the spaces of solutions and the spaces of right hand sides, one hardly
could speak about the regularity of solutions separately in each indepen-
dent variable. This conclusion is based on the fact that the hyperbolicity
does not ensure uniform regularity of solutions in both characteristic and
non-characteristic directions.
To overcome these complications for a certain class of hyperbolic PDEs, we
here prove a new Fredholmness criterion for linear operators in Banach spaces
(Section 2, Theorem 1) and construct spaces of solutions by means of the corre-
sponding graph norms as suggested in [1] (Section 3 of the present paper). In the
case of n = 1 Theorem 1 gives a well-known classical result. If n = 2, it is a part
of our result from [1], where Fredholm Alternative is proved for periodic-Dirichlet
problem for linear one-dimensional hyperbolic systems.
A novelty of the paper consists in generalization of the Fredholmness criterion
obtained in [1] from n = 2 to an arbitrary n. Another result is a construction of
multidimensional hyperbolic partial differential operators for which the general-
ized criterion applies with n = 3 and does not with n < 3.
2 Fredholmness criterion
We here generalize the Fredholmness criterion proved in [1].
Theorem 1 Let W be a Banach space, I be the identity in W, and K ∈ L(W)
with Kn being compact for some n ∈ N. Then I −K is Fredholm.
Proof. Since
n−1
I −Kn = Ki(I −K)
i=0
X
and Kn is compact, we have
dimker(I −K) ≤ dimker(I −Kn) < ∞. (1)
Similarly dimker(I−K)∗ < ∞, hence codimim(I −K) < ∞. It remains to show
that im(I −K) is closed.
Take a sequence (wj)j∈N ⊂ W and an element w ∈ W such that
(I −K)w → w. (2)
j
We have to show that w ∈ im(I −K).
2
Because of (1) there exists a closed subspace V of W such that
W = ker(I −K)⊕V, (3)
Using the decomposition
w = u +v , u ∈ ker(I −K), v ∈ V,
j j j j j
we get from (2)
(I −K)v → w. (4)
j
First, we show that the sequence (vj)j∈N is bounded. If not, without loss of
generality we can assume that
lim kv k = ∞. (5)
j
j→∞
From (4) and (5) we get
v
j
(I −K) → 0, (6)
kv k
j
hence
v
(I −Kn) j → 0. (7)
kv k
j
On the other side, because Kn is compact, there exist v ∈ W and a subsequence
(vjk)k∈N such that v
Kn jk → v. (8)
kv k
jk
Inserting (8) into (7), we get
v
jk → v ∈ V. (9)
kv k
jk
Combining (9) with (6), we get (I−K)v = 0, i.e. v ∈ V ∩ker(I−K) and kvk = 1.
However, this contradicts (3).
Now we use the boundedness of (vj)j∈N to show that w ∈ im(I −K). As Kn
is compact, there exists v ∈ W and a subsequence (vjk)k∈N such that Knvjk → v.
n−1 n−1
On the other hand, (4) yields (I −Kn)v = Ki(I −K)v → Kiw. Hence
j j
i=0 i=0
n−1 P P
lim v = Kiw +v
k→∞ jk
i=0
X
and, therefore,
n−1
w = lim(I −K)v = (I −K) Kiw +v ∈ im(I −K)
k→∞ jk !
i=0
X
(cid:3)
as desired.
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3 Application to two-dimensional
hyperbolic PDEs
We here prove a well-posedness result (specifically, the Fredholmness) for two-
dimensional mixed hyperbolic problems of periodic-Dirichlet type. Consider the
following first-order hyperbolic system
n n
a (α ∂ u +∂ u +β ∂ u +γ (x,y,t)u )+ b (x,y,t)u = f (x,y,t),
ij i t j x j i y j i j ij j i
j=1 j=1
X i ≤ n,X(x,y,t) ∈ (0,1)×R×R,
(10)
supplemented with the periodic conditions in y and t
u (x,y +Y,t+T) = u (x,y,t), i ≤ n, (x,y,t) ∈ [0,1]×R×R, (11)
i i
and the trivial boundary conditions in x
u (0,y,t) = 0, i ≤ k, (y,t) ∈ R2,
i (12)
u (1,y,t) = 0, k +1 ≤ j ≤ n, (y,t) ∈ R2.
j
We will make the following assumptions: n ≥ 3, k is an integer in the range
2 ≤ k < n, the coefficients a , α , and β , are fixed real constants, the coefficients
ij i i
γ , b : [0,1]×R×R → R and the right hand sides f : [0,1]×R×R → R are
i ij i
known functions. Given l ∈ N in the range 1 ≤ l ≤ k −1, we will suppose also
that the matrix A = (a )n has the following diagonal-block structure
ij i,j=1
A 0 0
1
A = 0 A 0 (13)
2
 
0 0 A
3
 
where A , A , and A are l×l, (k−l)×(k−l), and (n−k)×(n−k)-matrices,
1 2 3
respectively, while 0 denotes the null matrices of respective sizes. Moreover, the
matrix B = (b )n is assumed to be one of the following two kinds:
ij i,j=1
0 0 B
1
A = B 0 0 (14)
2
 
0 B 0
3
 
or
0 B˜ 0
1
A = 0 0 B˜ , (15)
2
 
B˜ 0 0
3
 
4
where B ,B ,B ,B˜ ,B˜ , and B˜ are, respectively, l ×(n−k), (k −l)×l, (n−
1 2 3 1 2 3
k) × (k − l), l × (k − l), (k − l) × (n − k), and (n − k) × l-matrices. Without
loss of generality we can consider the matrix (14) (similar argument works also
for (15)).
Our aim is to prove the Fredholmness of the operator generated by problem
(10)–(12) between the two spaces of continuous functions, namely, between the
space of the right-hand sides
W := (C ([0,1]×R×R))n
per
and the space of solutions
V := u ∈ W : u (0,y,t) = 0 for i ≤ k,u (1,y,t) = 0 for k +1 ≤ i ≤ n,
i i
n
n (16)
a (α ∂ u +∂ u +β ∂ u ) ∈ W for i ≤ n .
ij i t j x j i y j
Xj=1 o
Here u = (u ,...,u ), C ([0,1]×R×R) denotes the space of continuous on
1 n per
[0,1]×R×R functions being Y-periodic in y and T-periodic in t. The function
spaces W and V are endowed with the norms
n
kuk := max |u |
W i
[0,1]×[0,Y]×[0,T]
i=1
X
and
n
kuk := kuk + a (α ∂ u +∂ u +β ∂ u ) . (17)
V W ij i t j x j i y j
(cid:13) (cid:13)
(cid:13)Xi=1 (cid:13)W
(cid:13) (cid:13)
(cid:13) (cid:13)
Lemma 2 The space V is com(cid:13)plete. (cid:13)
Proof. Let (um)m∈N be a fundamental sequence in V. Then (um)m∈N and
n n
a (α ∂ um +∂ um +β ∂ um),..., a (α ∂ um +∂ um +β ∂ um)
1j 1 t j x j 1 y j nj n t j x j n y j
!
Xj=1 Xj=1 m∈N
are fundamental sequences in W. Dueto completeness of W, there exist u,v ∈ W
such that
n n
um → u and a (α ∂ um +∂ um +β ∂ um) → v
ij i t x i y
!
Xj=1 i=1
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in W as m → ∞. It remains to show that, given i ≤ n, n a (α ∂ u +∂ u +
j=1 ij i t j x j
β ∂ u ) = v inthesenseofgeneralizedderivatives, wherev arethecomponentsof
i y j i i
P
vector v. But thisis obvious: Take a smoothfunction ϕ : (0,1)×(0,Y)×(0,T) →
R with compact support. Then
1 Y T
n
a u (α ∂ +∂ +β ∂ )ϕdxdydt
ij j i t x i y
Z Z Z j=1
0 0 0 X
1 Y T
n
= lim a um(α ∂ +∂ +β ∂ )ϕdxdydt
ij j i t x i y
m→∞
Z Z Z j=1
0 0 0 X
1 Y T 1 Y T
n
= − lim a (α ∂ +∂ +β ∂ )umϕdxdydt = − v ϕdxdydt
ij i t x i y j i
m→∞
Z Z Z j=1 Z Z Z
0 0 0 X 0 0 0
as desired. Since i is an arbitrary natural number in the range i ≤ n, the proof
(cid:3)
is complete
To formulate our result let us introduce linear operators: Given a ,α ,β ∈ R
ij i i
and γ ,b ∈ C ([0,1]×R×R), we define
i ij per
n n
Cu := a (α ∂ u +∂ u +β ∂ u +γ (x,y,t)u ) ,
ij i t j x j i y j i j
!
Xj=1 i=1
n n
Du := b (x,y,t)u .
ij j
!
Xj=1 i=1
Moreover, we will need a modified notation for α and β , namely,
i i
0 if b ≡ 0 for all 1 ≤ j ≤ n,
α˜ = ij
i α otherwise ,
i
(cid:26)
0 if b ≡ 0 for all 1 ≤ j ≤ n,
β˜ = ij
i β otherwise .
i
(cid:26)
Theorem 3 Assume that γ ,b ∈ C1 ([0,1]×R×R), det(a )n 6= 0, and
i ij per ij i,j=1
(β˜ −β˜ )(α˜ −α˜ )−(β˜ −β˜ )(α˜ −α˜ ) 6= 0 (18)
i j j s j s i j
for all i,j,s ∈ {1,...,n} with 1 ≤ i ≤ l, l + 1 ≤ j ≤ k, k + 1 ≤ s ≤ n unless
α˜ = α˜ = 0. Then the following is true:
i j
(i) The operator C is an isomorphism from V onto W.
(ii) The operator C +D is Fredholm from V into W.
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Proof. To prove (i), we have to show that, given f ∈ W, there exists a unique
u ∈ V such that
n
a (α ∂ u +∂ u +β ∂ u +γ (x,y,t)u ) = f (x,y,t), i ≤ n. (19)
ij i t j x j i y j i j i
j=1
X
Rewrite (19) as
n
d
+γ a u (ξ,y+β (ξ −x),t+α (ξ −x)) = f (x,y,t), i ≤ n.
i ij j i i i
dξ
" # ξ=x
(cid:18) (cid:19) Xj=1 (cid:12)
(cid:12)
(cid:12)
Set
A 0
A = 1 .
0 0 A
2
(cid:18) (cid:19)
On the account of (13) and the assumption that det(a )n 6= 0, system (19)
ij i,j=1
has a unique solution in V explicitely given by the formula
k
1
u (x,y,t) = A0 ad
i detA ji
0
j=1
X(cid:0) (cid:1)
x ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
 j 1 j 1 j 1 1
Z0 Zx 
×f (ξ,y +β (ξ −x),t+α (ξ −x))dξ, i ≤ k;
j j j
 
n
1
u (x,y,t) = A3 ad (20)
i detA ji
3
j=k+1
X (cid:0) (cid:1)
1 ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
 j 1 j 1 j 1 1
Zx Zx 
×f (ξ,y +β (ξ −x),t+α (ξ −x))dξ, k +1 ≤ i ≤ n.
j j j
 
Here andsubsequently, (As )ad stands for theadjoint matrix toA . Assertion
ij i,j s
(i) is therewith proved.
(cid:8) (cid:9)
To prove Assertion (ii), note that C + D is Fredholm from V into W iff
I + DC−1 is Fredholm from W into W, where I is the identity in W. Set
K = DC−1 and apply Theorem 1 with n = 3. Our task is therefore reduced to
showing that K3 is compact.
Take a bounded set N ⊂ W and let M be its image under (DC−1)3. To show
that M is precompact in W, we use Arzela-Ascoli precompactness criterion in
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C([0,1]×[0,Y]×[0,T]). As (DC−1)3 is a bounded operator on W, the set M is
uniformly bounded in W. It remains to check the equicontinuity property of M
in W. Given u ∈ W, set u˜ := DC−1Du. On the account of the representation
(19) for C−1, we get
n n
1
u˜ (x,y,t) = b (x,y,t) A3 ad
i ij detA rj
3
j=k+1 r=k+1
X X (cid:0) (cid:1)
1 ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
r 1 r 1 r 1 1
 
Zx Zx 
k
 
× (b u )(ξ,y+β (ξ −x),t+α (ξ −x))dξ, i ≤ l;
rq q r r
q=l+1
X
l l
1
u˜ (x,y,t) = b (x,y,t) A1 ad
i ij detA rj
1
j=1 r=1
X X(cid:0) (cid:1)
x ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ (21)
 r 1 r 1 r 1 1
Z0 Zx 
n
× (bu )(ξ,y+β (ξ −x),t+α (ξ −x))dξ, l+1 ≤ i ≤ k;
rq q r r
q=k+1
X
k k
1
u˜ (x,y,t) = b (x,y,t) A2 ad
i ij detA rj
2
j=l+1 r=l+1
X X (cid:0) (cid:1)
x ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
r 1 r 1 r 1 1
 
Z0 Zx 
l
 
× (b u )(ξ,y+β (ξ −x),t+α (ξ −x))dξ, k +1 ≤ i ≤ n.
rq q r r
q=1
X
Obviously, given f ∈ W, we have u˜ = (DC−1)3f iff u˜ is defined by (21) with
u = C−1DC−1f explicitely given by
l
1
u (x,y,t) = A1 ad
i detA ji
1
j=1
X(cid:0) (cid:1)
x ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
j 1 j 1 j 1 1
 
Z0 Zx 
 8 
n n
1
× b (ξ,y+β (ξ −x),t+α (ξ −x))dξ A3 ad
jr j j detA qr
3
r=k+1 q=k+1
X X (cid:0) (cid:1)
1 ξ1
× exp γ (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ
q 2 j q 2 j q 2 2
 
Zξ Zξ 
×f (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ , i ≤ l,
q 1  j q 1 j q 1 1 
k
1
u (x,y,t) = A2 ad
i detA ji
2
j=l+1
X (cid:0) (cid:1)
x ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
 j 1 j 1 j 1 1
Z0 Zx 
l l
1
× b (ξ,y+β (ξ −x),t+α (ξ −x))dξ A1 ad
jr j j detA qr
1
r=1 q=1
X X(cid:0) (cid:1)
ξ ξ1
× exp γ (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ
q 2 j q 2 j q 2 2
 
Z0 Zξ 
×f (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ , l+1 ≤ i ≤ k,
q 1  j q 1 j q 1 1 
n
1
u (x,y,t) = A3 ad
i detA ji
3
j=k+1
X (cid:0) (cid:1)
1 ξ
× exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
j 1 j 1 j 1 1
 
Zx Zx 
k k
1
× b(ξ,y +β (ξ −x),t+α (ξ −x))dξ  A2 ad
jr j j detA qr
2
r=l+1 q=l+1
X X (cid:0) (cid:1)
ξ ξ1
× exp γ (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ
 q 2 j q 2 j q 2 2
Z0 Zξ 
×f (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ , k +1 ≤ i ≤ n.
q 1  j q 1 j q 1 1 
To prove the desired equicontinuity property we have to show the existence of a
function α : R → R with α(p) → 0 as p → 0 for which we have
+
ku˜(x+h ,y +h ,t+h )−u˜(x,y,t)k ≤ α(|h|)kfk , (22)
1 2 3 W
the estimate being uniform in f ∈ W and h = (h ,h ,h ) ∈ R3. To achieve this
1 2 3
9
property we transform the expression for u˜ to a form appropriate for our pur-
poses. We make calculations only for one summand contributing into u˜ (similar
argument works for all other summands as well). Specifically, we will consider
the following summand (up to a multiplicative constant):
x ξ
b (x,y,t) exp γ (ξ ,y +β (ξ −x),t+α (ξ −x))dξ
ij r 1 r 1 r 1 1
 
Z0 Zx 
×b (ξ,y+β (ξ −x),t+α (ξ −x))dξ
rq r r
 
1 ξ1
× exp γ (ξ ,y +β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ
p 2 r p 2 r p 2 2
 
Zξ Zξ 
×b (ξ ,y+β (ξ −x)+β (ξ −ξ),t+α (ξ −x)+α (ξ −ξ))dξ (23)
ps 1  r p 1 r p 1 1 
ξ1 ξ2
× exp γ (ξ ,y +β (ξ −x)+β (ξ −ξ)+β (ξ −ξ ),
m 3 r p 1 m 3 1
Z0 nξZ1
t+α (ξ −x)+α (ξ −ξ)+α (ξ −ξ ))dξ
r p 1 m 3 1 3
×f (ξ ,y +β (ξ −x)+β (ξ −ξ)+β (ξ −oξ ),
m 2 r p 1 m 2 1
t+α (ξ −x)+α (ξ −ξ)+α (ξ −ξ ))dξ .
r p 1 m 2 1 2
Changing the order of integration, we have
x 1 ξ1 x ξ 1 1 1
dξ dξ dξ = dξ dξ dξ + dξ dξ
1 2 2 1 2 1
 
Z Z Z Z Z Z Z Z
0 ξ 0 0 0 ξ ξ ξ2
 
x x 1 x ξ2 1 1 x 1
= dξ dξ dξ + dξ dξ dξ + dξ dξ dξ . (24)
2 1 2 1 2 1
Z Z Z Z Z Z Z Z Z
0 ξ2 ξ 0 0 ξ2 x 0 ξ
Moreover, we introduce new variables µ and η (instead of ξ and ξ ) by
1
µ := y −β x+ξ(β −β )+ξ (β −β )
r r p 1 p m (25)
η := t−α x+ξ(α −α )+ξ (α −α ).
r r p 1 p m
Owing to (14), the integers r,p,m belong to three different sets {1,...,l}, {l +
1,...,k}, and {k +1,...,n}. Thus, on the account of (18), the mapping (25) is
non-generate. By means of (24) and (25) the summand (23) can be transformed
10