Spectral Properties of Elliptic Operators

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Spectral Properties of Elliptic Operators
In previous work we have replaced the strong version of an elliptic boundary value problem
LßuÝxÞà = fÝxÞ
BCßuÝxÞà = gÝxÞ
x5U
[email protected]
with the weak problem
find u 5 V such that
Bßu, và = fÝvÞ
-v 5 V.
The connection between theweak problem and the strong problem, (and in particular the
boundary conditions), was established by making use of the abstract Green’s formula
Bßu, và = ÝLßuà, vÞ H + ÖT 0 v, T 1 u×
-u, v 5 V.
If the bilinear form B is V ? elliptic,
Bßu, uà ³ c 0 || u|| 2V -u 5 V. then it follows immediately
from the Lax-Milgram lemma that the weak problem has a unique solution for every f. This
is the same thing as saying that L is an isomorphism from V onto H.
On the other hand, if B is only V ? H ? coercive, Bßu, uà ³ c 0 || u|| 2V ? c 1 ||u|| 2H -u 5 V, then
we are not able to apply this lemma directly to the weak problem. Instead, we replace the
weak problem with
find u 5 V such that
B W ßu, và = fÝvÞ + WÝu, vÞ H
-v 5 V.
For W sufficiently large, B W is V-elliptic and the Lax Milgram lemma implies the existence of
a solution to this perturbed problem. Then L W is an isomorphism from V onto H, hence
there is a bounded inverse, L ?1
W from H to V. Then, since the embedding of V into H is
compact by the Rellich lemma, we have that i E L ?1
W : H ¸ H is compact. That is,
L W u = f + Wu
?1
é u = L ?1
W f + WL W u é ÝI ? KÞu = F
?1
where Ku = WL ?1
W u, and F = L W f. The operator I ? K is an operator of Fredholm type for
which there is a theorem, known as the Fredholm alternative theorem, stating conditions
under which there is a unique solution to ÝI ? KÞu = F. By translating this theorem into the
notation and terminology of the weak boundary value problem, we obtain a similar
existence-uniqueness theorem for the weak boundary value problem.
Now we are going to apply a similar approach in considering the eigenvalue problem for an
elliptic operator. We suppose that
LßuÝxÞà = ?divÝAÝxÞ 4uÝxÞÞ + cÝxÞ uÝxÞ
x5U
1
is uniformly elliptic on U, and we will consider the eigenvalue problem of finding values of
the scalar V, such that
LßuÝxÞà = V uÝxÞ
uÝxÞ = 0
x5U
[email protected]
has nontrivial solutions. Note that in the abscence of first order terms,
-d, f 5 C K0 ÝUÞ
ÝLßdà, fÞ 0 = Ýd, LßfàÞ 0
and we say that L is symmetric. Then it follows that the associated bilinear form B is also
symmetric,
Bßu, và = Bßv, uà
-u, v 5 V = H 10 ÝUÞ.
Unless cÝxÞ ³ 0, the bilinear form B is not V-elliptic, but it is V-H-coercive, which is to say,
B W is V-elliptic, for W sufficiently large. Then we recall that L W : H 10 ÝUÞ ¸ H 0 ÝUÞ is an
isomorphism for W sufficiently large. Then, instead of considering the eigenvalue problem
for L, we will consider the eigenvalue problem for L ?1
W which is a compact operator for which
there is a classical theorem. By translating the conclusions of the classical theorem into the
terminology of our elliptic operator, we obtain the following theorem.
Theorem Under the assumptions on L, there is a countable collection áV k â of scalars such
that
LßuÝxÞà = V uÝxÞ
uÝxÞ = 0
x5U
[email protected]
has nontrivial solutions. Moreover,
(a) the V k are all real and L < V 1 ² V 2 ² ` ¸ K
(b) Let N k = áu 5 H 10 ÝUÞ : Lßuà = V k uâ. Then dim N k < K all k and
Ýu, vÞ 0 = 0
if
u 5 Nk, v 5 Nj
(c) The weak solutions ád k â , for
j®k
LßdÝxÞà = V dÝxÞ
dÝxÞ = 0
||d|| 0 = 1,
x5U
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form an orthonormal basis for H 0 ÝUÞ, and at the same time, an orthogonal
(but not orthonormal) basis for H 10 ÝUÞ.
2
Proof- Let L denote the smallest value such that L L : H 10 ÝUÞ ¸ H 0 ÝUÞ is an isomorphism.
Then for any W ³ L, there is a bounded linear operator S W : H 0 ÝUÞ ¸ H 10 ÝUÞ, which is
inverse to L W ; i.e., L W ÝS W fÞ = f for all f 5 H 0 ÝUÞ. Let K = i E S W : H 0 ÝUÞ ¸ H 10 ÝUÞ Ð H 0 ÝUÞ.
Then K is a compact linear operator on H 0 ÝUÞ. Note that
SWf = u
B W ßu, và = Ýf, vÞ 0 -v 5 H 10 ÝUÞ
é
é
LWu = f
For arbitrary f, g 5 H 0 ÝUÞ, let Kf = u, Kg = v. Then
Ýf, KgÞ 0 = Ýf, vÞ 0 = B W ßu, và = B W ßv, uà = Ýg, uÞ 0 = Ýu, gÞ 0 = ÝKf, gÞ 0 .
Thus the symmetry of the bilinear form implies that K is symmetric, and since K is bounded,
in fact compact, it follows that K is self adjoint.
In addition,
ÝKf, fÞ 0 = Ýu, fÞ 0 = B W ßu, uà ³ c 0 ||u|| 21 ,
so K is a compact self adjoint, positive operator on the Hilbert space H = H 0 ÝUÞ. We have
the following theorem about such operators,
Theorem (Fredholm-Riesz-Schauder) For K a compact self adjoint, positive operator on the
Hilbert space H,
(a) K has a countable collection of positive eigenvalues, áS k â, accumulating at zero; i.e.,
||K|| LÝHÞ ³ S 1 ³ S 2 ³ ` ¸ 0
(b) Let N k = áu 5 H : Kßuà = S k uâ. Then dim N k < K all k and
Ýu, vÞ H = 0
if
u 5 Nk, v 5 Nj
j®k
(c) The normalized eigenfunctions for K form an orthonormal basis for H.
The compact self adjoint, positive operators on the Hilbert space H are the analogues of
symmetric positive definite matrices on R n .
Now
é
L W ßSdà = d
Kd = S d
é
B W ßSd, và = Ýd, vÞ 0 -v 5 V = H 10 ÝUÞ
hence
L W ßSdà = d
é
L W ßdà = 1 d
S
so the eigenfunctions of K are the eigenfunctions of L W but the eigenvalues of L W are the
reciprocals of the eigenvalues of K. Thus
3
||K|| LÝHÞ ³ S 1 ³ S 2 ³ ` ¸ 0
implies
0<
1
< J 1 ² J 2 ² ` ¸ K.
||K|| LÝHÞ
L W ßd k à = J k d k é Lßd k à = ÝJ k ? WÞ d k = V k d k
In addition,
with
0<
1
? W < V 1 ² V 2 ² ` ¸ K.
||K|| LÝHÞ
If ád k â denote the weak solutions of
L W ßdÝxÞà = J dÝxÞ
dÝxÞ = 0
||d|| 0 = 1,
x5U
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u = > k>0 Ýu, d k Þ 0 d k = > k>0 u k d k
then for u 5 V = H 10 ÝUÞ, write
Define
ÝÝu, vÞÞ V = B W ßu, và + Ýu, vÞ H
Note that
ÝÝu, vÞÞ V = B W > k>0 u k d k , > j>0 v j d j + > k>0 u k d k , > j>0 v j d j
-u, v 5 V.
H
= > k > j u k v j B W ßd j , d k à + > k u k v k .
B W ßd j , d k à = ÝL W d j , d k Þ H = J j Ýd j , d k Þ H = V j N jk
But
ÝÝu, vÞÞ V = > k ÝJ k + 1Þu k v k .
hence
and, in particular,
ÝÝd j , d k ÞÞ V = ÝJ k + 1Þ N jk .
To see that the eigenfunctions form a complete family for the inner product on V, suppose
ÝÝu, d k ÞÞ V = 0
Then
-k.
0 = ÝÝu, d k ÞÞ V = B W ßu, d k à + Ýu, d k Þ H = J k Ýu, d k Þ H + Ýu, d k Þ H
i.e.,
ÝJ k + 1ÞÝu, d k Þ H = 0
But
Jk ³
hence
1
>0
||K|| LÝHÞ
Ýu, d k Þ H = 0
-k.
-k
-k and since the family is complete in H, u = 0.n
4
u 5 V = H 10 ÝUÞ iff ||u|| 2V = ÝÝu, uÞÞ V = > k ÝJ k + 1Þu 2k < K
Note that
u 5 H = H 0 ÝUÞ iff ||u|| 2H = Ýu, uÞ 0 = > k u 2k < K
Similarly,
For F 5 D v ÝUÞ, f 5 DÝUÞ Ð H,
FÝfÞ = FÝ> k f k d k Þ = > k f k FÝd k Þ = > k f k F k
and we have F 5 V v iff
|FÝfÞ| = > k f k F k = > k Ý1 + J k Þ 1/2 f k Ý1 + J k Þ ?1/2 F k
² > k Ý1 + J k Þ ?1 F 2k
In addition,
u 5 H iff
||f|| 2V
Ý1 + J k Þ ?1/2 F k
F 5 V v iff
i.e.,
1/2
5 §2.
áu k â 5 § 2 .
Ý1 + J k Þ 1/2 u k
u 5 V iff
5 §2.
and we can define, more generally,
u 5 H s ÝUÞ iff
Ý1 + J k Þ s/2 u k
5 §2,
and we have, with this definition, ÝH s ÝUÞÞ v = H ?s ÝUÞ. This scale of Hilbert spaces, each with
the inner product
ÝÝu, vÞÞ s = > k Ý1 + J k Þ s u k v k
is completely analogous to the Fourier spaces H s ÝR n Þ we have defined previously.
Consider now the following examples.
1)
LßuÝxÞà = ?u”ÝxÞ,
uÝ0Þ = uÝ^Þ = 0
0<x<^
Here cÝxÞ = 0 so W = 0 and J k = V k . For this operator, it is easy to compute the eigenvalues
and eigenfunctions
d k ÝxÞ =
2
2
^ sinÝkxÞ and V k = k
k = 1, 2, u
5
Then the function
2x
^
x
2 1? ^
uÝxÞ =
has
u k = Ýu, d k Þ 0 = ? 4 2 sin
^k
Ý1 + V k Þ 1/2 ~ k
Now
i.e.
so
k^
2
0 < x < ^/2
^/2 < x < ^
.
Ý1 + V k Þ s/2 u k ~ k s?2 5 § 2
for s ² 1;
u 5 H 1 = H 10 ÝUÞ.
vÝxÞ = 1, 0 < x < ^,
On the other hand, for the function
vk =
we find
1 ? Ý?1Þ k
5 §2
k
but
Ý1 + V k Þ s/2 v k ~ k s?1 5 § 2
i.e.,
v 5 H 0 = L 2 ÝUÞ.
wÝxÞ = NÝx ? x 0 Þ
Finally, for
wk =
and
wk 6 §2
for s > 1/2;
N 5 H ?1 ÝUÞ.
LßuÝx, yÞà = ?4 2 uÝx, yÞ
u=0
2)
x 0 5 Ý0, ^Þ
2
^ sinÝkx 0 Þ 5 § K
Ý1 + V k Þ ?s/2 w k ~ k ?s 5 § 2
In particular,
for s ² 0;
in U = á0 < x, y < ^â
on @
Then, again, c = 0 so J = V, and
2 sinÝjxÞ sinÝkyÞ
d j,k Ýx, yÞ = ^
In this case, for
V j,k = Ýj 2 + k 2 Þ, j, k = 1, 2, u
wÝx, yÞ = NÝx ? x 0 Þ NÝy ? y 0 Þ Ýx 0 , y 0 Þ 5 U
we have
2 sinÝjx 0 Þ sinÝky 0 Þ
w j,k = ^
and
Ý1 + V j,k Þ ?s/2 w k ~ Ý1 + j 2 + k 2 Þ ?s/2 5 § 2
for s > 1
6
Then
w 6 H ?1 ÝUÞ.
For the function
uÝx, yÞ =
we have
u j,k =
0 < x < ^/2, 0 < y < ^
^/2 < x < ^, 0 < y < ^
c j,k
5 §2
j k2
Ý1 + j 2 + k 2 Þ s/2 u j,k ~
but
2x
^
x
2 1? ^
Ý1 + j 2 + k 2 Þ s/2
6 § 2 if s = 1
j k2
Then u 5 H but u 6 V.
.
7
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