# Spectral Properties of Elliptic Operators ```Spectral Properties of Elliptic Operators
In previous work we have replaced the strong version of an elliptic boundary value problem
L&szlig;u&Yacute;x&THORN;&agrave; = f&Yacute;x&THORN;
BC&szlig;u&Yacute;x&THORN;&agrave; = g&Yacute;x&THORN;
x5U
x5@
with the weak problem
find u 5 V such that
B&szlig;u, v&agrave; = f&Yacute;v&THORN;
-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
B&szlig;u, v&agrave; = &Yacute;L&szlig;u&agrave;, v&THORN; H + &Ouml;T 0 v, T 1 u&times;
-u, v 5 V.
If the bilinear form B is V ? elliptic,
B&szlig;u, u&agrave; &sup3; 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, B&szlig;u, u&agrave; &sup3; 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 &szlig;u, v&agrave; = f&Yacute;v&THORN; + W&Yacute;u, v&THORN; 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 &cedil; H is compact. That is,
L W u = f + Wu
?1
&eacute; u = L ?1
W f + WL W u &eacute; &Yacute;I ? K&THORN;u = 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 &Yacute;I ? K&THORN;u = 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
L&szlig;u&Yacute;x&THORN;&agrave; = ?div&Yacute;A&Yacute;x&THORN; 4u&Yacute;x&THORN;&THORN; + c&Yacute;x&THORN; u&Yacute;x&THORN;
x5U
1
is uniformly elliptic on U, and we will consider the eigenvalue problem of finding values of
the scalar V, such that
L&szlig;u&Yacute;x&THORN;&agrave; = V u&Yacute;x&THORN;
u&Yacute;x&THORN; = 0
x5U
x5@
has nontrivial solutions. Note that in the abscence of first order terms,
-d, f 5 C K0 &Yacute;U&THORN;
&Yacute;L&szlig;d&agrave;, f&THORN; 0 = &Yacute;d, L&szlig;f&agrave;&THORN; 0
and we say that L is symmetric. Then it follows that the associated bilinear form B is also
symmetric,
B&szlig;u, v&agrave; = B&szlig;v, u&agrave;
-u, v 5 V = H 10 &Yacute;U&THORN;.
Unless c&Yacute;x&THORN; &sup3; 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 &Yacute;U&THORN; &cedil; H 0 &Yacute;U&THORN; 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 &aacute;V k &acirc; of scalars such
that
L&szlig;u&Yacute;x&THORN;&agrave; = V u&Yacute;x&THORN;
u&Yacute;x&THORN; = 0
x5U
x5@
has nontrivial solutions. Moreover,
(a) the V k are all real and L &lt; V 1 &sup2; V 2 &sup2; ` &cedil; K
(b) Let N k = &aacute;u 5 H 10 &Yacute;U&THORN; : L&szlig;u&agrave; = V k u&acirc;. Then dim N k &lt; K all k and
&Yacute;u, v&THORN; 0 = 0
if
u 5 Nk, v 5 Nj
(c) The weak solutions &aacute;d k &acirc; , for
j&reg;k
L&szlig;d&Yacute;x&THORN;&agrave; = V d&Yacute;x&THORN;
d&Yacute;x&THORN; = 0
||d|| 0 = 1,
x5U
x5@
form an orthonormal basis for H 0 &Yacute;U&THORN;, and at the same time, an orthogonal
(but not orthonormal) basis for H 10 &Yacute;U&THORN;.
2
Proof- Let L denote the smallest value such that L L : H 10 &Yacute;U&THORN; &cedil; H 0 &Yacute;U&THORN; is an isomorphism.
Then for any W &sup3; L, there is a bounded linear operator S W : H 0 &Yacute;U&THORN; &cedil; H 10 &Yacute;U&THORN;, which is
inverse to L W ; i.e., L W &Yacute;S W f&THORN; = f for all f 5 H 0 &Yacute;U&THORN;. Let K = i E S W : H 0 &Yacute;U&THORN; &cedil; H 10 &Yacute;U&THORN; &ETH; H 0 &Yacute;U&THORN;.
Then K is a compact linear operator on H 0 &Yacute;U&THORN;. Note that
SWf = u
B W &szlig;u, v&agrave; = &Yacute;f, v&THORN; 0 -v 5 H 10 &Yacute;U&THORN;
&eacute;
&eacute;
LWu = f
For arbitrary f, g 5 H 0 &Yacute;U&THORN;, let Kf = u, Kg = v. Then
&Yacute;f, Kg&THORN; 0 = &Yacute;f, v&THORN; 0 = B W &szlig;u, v&agrave; = B W &szlig;v, u&agrave; = &Yacute;g, u&THORN; 0 = &Yacute;u, g&THORN; 0 = &Yacute;Kf, g&THORN; 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.
&Yacute;Kf, f&THORN; 0 = &Yacute;u, f&THORN; 0 = B W &szlig;u, u&agrave; &sup3; c 0 ||u|| 21 ,
so K is a compact self adjoint, positive operator on the Hilbert space H = H 0 &Yacute;U&THORN;. 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, &aacute;S k &acirc;, accumulating at zero; i.e.,
||K|| L&Yacute;H&THORN; &sup3; S 1 &sup3; S 2 &sup3; ` &cedil; 0
(b) Let N k = &aacute;u 5 H : K&szlig;u&agrave; = S k u&acirc;. Then dim N k &lt; K all k and
&Yacute;u, v&THORN; H = 0
if
u 5 Nk, v 5 Nj
j&reg;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
&eacute;
L W &szlig;Sd&agrave; = d
Kd = S d
&eacute;
B W &szlig;Sd, v&agrave; = &Yacute;d, v&THORN; 0 -v 5 V = H 10 &Yacute;U&THORN;
hence
L W &szlig;Sd&agrave; = d
&eacute;
L W &szlig;d&agrave; = 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|| L&Yacute;H&THORN; &sup3; S 1 &sup3; S 2 &sup3; ` &cedil; 0
implies
0&lt;
1
&lt; J 1 &sup2; J 2 &sup2; ` &cedil; K.
||K|| L&Yacute;H&THORN;
L W &szlig;d k &agrave; = J k d k &eacute; L&szlig;d k &agrave; = &Yacute;J k ? W&THORN; d k = V k d k
with
0&lt;
1
? W &lt; V 1 &sup2; V 2 &sup2; ` &cedil; K.
||K|| L&Yacute;H&THORN;
If &aacute;d k &acirc; denote the weak solutions of
L W &szlig;d&Yacute;x&THORN;&agrave; = J d&Yacute;x&THORN;
d&Yacute;x&THORN; = 0
||d|| 0 = 1,
x5U
x5@
u = &gt; k&gt;0 &Yacute;u, d k &THORN; 0 d k = &gt; k&gt;0 u k d k
then for u 5 V = H 10 &Yacute;U&THORN;, write
Define
&Yacute;&Yacute;u, v&THORN;&THORN; V = B W &szlig;u, v&agrave; + &Yacute;u, v&THORN; H
Note that
&Yacute;&Yacute;u, v&THORN;&THORN; V = B W &gt; k&gt;0 u k d k , &gt; j&gt;0 v j d j + &gt; k&gt;0 u k d k , &gt; j&gt;0 v j d j
-u, v 5 V.
H
= &gt; k &gt; j u k v j B W &szlig;d j , d k &agrave; + &gt; k u k v k .
B W &szlig;d j , d k &agrave; = &Yacute;L W d j , d k &THORN; H = J j &Yacute;d j , d k &THORN; H = V j N jk
But
&Yacute;&Yacute;u, v&THORN;&THORN; V = &gt; k &Yacute;J k + 1&THORN;u k v k .
hence
and, in particular,
&Yacute;&Yacute;d j , d k &THORN;&THORN; V = &Yacute;J k + 1&THORN; N jk .
To see that the eigenfunctions form a complete family for the inner product on V, suppose
&Yacute;&Yacute;u, d k &THORN;&THORN; V = 0
Then
-k.
0 = &Yacute;&Yacute;u, d k &THORN;&THORN; V = B W &szlig;u, d k &agrave; + &Yacute;u, d k &THORN; H = J k &Yacute;u, d k &THORN; H + &Yacute;u, d k &THORN; H
i.e.,
&Yacute;J k + 1&THORN;&Yacute;u, d k &THORN; H = 0
But
Jk &sup3;
hence
1
&gt;0
||K|| L&Yacute;H&THORN;
&Yacute;u, d k &THORN; H = 0
-k.
-k
-k and since the family is complete in H, u = 0.n
4
u 5 V = H 10 &Yacute;U&THORN; iff ||u|| 2V = &Yacute;&Yacute;u, u&THORN;&THORN; V = &gt; k &Yacute;J k + 1&THORN;u 2k &lt; K
Note that
u 5 H = H 0 &Yacute;U&THORN; iff ||u|| 2H = &Yacute;u, u&THORN; 0 = &gt; k u 2k &lt; K
Similarly,
For F 5 D v &Yacute;U&THORN;, f 5 D&Yacute;U&THORN; &ETH; H,
F&Yacute;f&THORN; = F&Yacute;&gt; k f k d k &THORN; = &gt; k f k F&Yacute;d k &THORN; = &gt; k f k F k
and we have F 5 V v iff
|F&Yacute;f&THORN;| = &gt; k f k F k = &gt; k &Yacute;1 + J k &THORN; 1/2 f k &Yacute;1 + J k &THORN; ?1/2 F k
&sup2; &gt; k &Yacute;1 + J k &THORN; ?1 F 2k
u 5 H iff
||f|| 2V
&Yacute;1 + J k &THORN; ?1/2 F k
F 5 V v iff
i.e.,
1/2
5 &sect;2.
&aacute;u k &acirc; 5 &sect; 2 .
&Yacute;1 + J k &THORN; 1/2 u k
u 5 V iff
5 &sect;2.
and we can define, more generally,
u 5 H s &Yacute;U&THORN; iff
&Yacute;1 + J k &THORN; s/2 u k
5 &sect;2,
and we have, with this definition, &Yacute;H s &Yacute;U&THORN;&THORN; v = H ?s &Yacute;U&THORN;. This scale of Hilbert spaces, each with
the inner product
&Yacute;&Yacute;u, v&THORN;&THORN; s = &gt; k &Yacute;1 + J k &THORN; s u k v k
is completely analogous to the Fourier spaces H s &Yacute;R n &THORN; we have defined previously.
Consider now the following examples.
1)
L&szlig;u&Yacute;x&THORN;&agrave; = ?u”&Yacute;x&THORN;,
u&Yacute;0&THORN; = u&Yacute;^&THORN; = 0
0&lt;x&lt;^
Here c&Yacute;x&THORN; = 0 so W = 0 and J k = V k . For this operator, it is easy to compute the eigenvalues
and eigenfunctions
d k &Yacute;x&THORN; =
2
2
^ sin&Yacute;kx&THORN; and V k = k
k = 1, 2, u
5
Then the function
2x
^
x
2 1? ^
u&Yacute;x&THORN; =
has
u k = &Yacute;u, d k &THORN; 0 = ? 4 2 sin
^k
&Yacute;1 + V k &THORN; 1/2 ~ k
Now
i.e.
so
k^
2
0 &lt; x &lt; ^/2
^/2 &lt; x &lt; ^
.
&Yacute;1 + V k &THORN; s/2 u k ~ k s?2 5 &sect; 2
for s &sup2; 1;
u 5 H 1 = H 10 &Yacute;U&THORN;.
v&Yacute;x&THORN; = 1, 0 &lt; x &lt; ^,
On the other hand, for the function
vk =
we find
1 ? &Yacute;?1&THORN; k
5 &sect;2
k
but
&Yacute;1 + V k &THORN; s/2 v k ~ k s?1 5 &sect; 2
i.e.,
v 5 H 0 = L 2 &Yacute;U&THORN;.
w&Yacute;x&THORN; = N&Yacute;x ? x 0 &THORN;
Finally, for
wk =
and
wk 6 &sect;2
for s &gt; 1/2;
N 5 H ?1 &Yacute;U&THORN;.
L&szlig;u&Yacute;x, y&THORN;&agrave; = ?4 2 u&Yacute;x, y&THORN;
u=0
2)
x 0 5 &Yacute;0, ^&THORN;
2
^ sin&Yacute;kx 0 &THORN; 5 &sect; K
&Yacute;1 + V k &THORN; ?s/2 w k ~ k ?s 5 &sect; 2
In particular,
for s &sup2; 0;
in U = &aacute;0 &lt; x, y &lt; ^&acirc;
on @
Then, again, c = 0 so J = V, and
2 sin&Yacute;jx&THORN; sin&Yacute;ky&THORN;
d j,k &Yacute;x, y&THORN; = ^
In this case, for
V j,k = &Yacute;j 2 + k 2 &THORN;, j, k = 1, 2, u
w&Yacute;x, y&THORN; = N&Yacute;x ? x 0 &THORN; N&Yacute;y ? y 0 &THORN; &Yacute;x 0 , y 0 &THORN; 5 U
we have
2 sin&Yacute;jx 0 &THORN; sin&Yacute;ky 0 &THORN;
w j,k = ^
and
&Yacute;1 + V j,k &THORN; ?s/2 w k ~ &Yacute;1 + j 2 + k 2 &THORN; ?s/2 5 &sect; 2
for s &gt; 1
6
Then
w 6 H ?1 &Yacute;U&THORN;.
For the function
u&Yacute;x, y&THORN; =
we have
u j,k =
0 &lt; x &lt; ^/2, 0 &lt; y &lt; ^
^/2 &lt; x &lt; ^, 0 &lt; y &lt; ^
c j,k
5 &sect;2
j k2
&Yacute;1 + j 2 + k 2 &THORN; s/2 u j,k ~
but
2x
^
x
2 1? ^
&Yacute;1 + j 2 + k 2 &THORN; s/2
6 &sect; 2 if s = 1
j k2
Then u 5 H but u 6 V.
.
7
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