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2 The Dirichlet Problem
2
The Dirichlet Problem
This chapter is devoted to the simplest boundary value problem for the differential equation (1)
-y” + q(x)y
= ly,
0
4
x 5 1.
Here, l E C and q E L2 = Li[O, 11, the Hilbert space of all real valued, square integrable functions on [0, 11. We ask whether there are nontrivial solutions of equation (1) satisfying the Dirichlet boundary condition
y(0) = 0,
y(1) = 0.
This is the Dirichlet problem. A complex number l is called a Dirichlet eigenvalue of q if the Dirichlet problem can be solved. The corresponding nontrivial solutions are called eigenfunctions of q for I t . The collection of all eigenvalues of q is its Dirichlet spectrum. From a more abstract point of view, the Dirichlet problem belongs to the theory of self-adjoint operators. Precisely, the unbounded operator Q, ,defined by
25
Inverse Spectral Theory
26
for f in the dense subspace D of all functions in Hc[O, 11 that vanish at 0 and 1, is self-adjoint on the domain D.’ It can be shown that the spectrum of Q coincides with the Dirichlet spectrum of q, and so we may draw some conclusions about the Dirichlet problem from general facts about selfadjoint operators. For instance, the Dirichlet spectrum must be an unbounded sequence of real numbers. However, abstract operator theory is not helpful in answering many of the questions that interest us. For this reason, we shall follow Sturm’s classical approach [St] which may be called the “method of shooting. ” Intuitively speaking, the solution y2 is shot from the left end point of the interval with unit velocity. At x = 1, it reaches the height yz(1, A). The idea is to vary the parameter A so that the terminal height is 0. This point is illustrated in Figure 1. At the moment, the terminal velocityyz’(1,A) does not concern us. It becomes important in Chapter 3. The “shooting method” is easy to implement. If p is a zero ofyz( 1, A), then yz(x, p) is a nontrivial solution of equation (1) satisfying the Dirichlet boundary conditions. Hence, p is a Dirichlet eigenvalue of q. Conversely, suppose p is a Dirichlet eigenvalue of q with eigenfunction y(x). Then Y ( X ) = Y’(O)YZ(X,
Figure 1. ‘That is, for all f,g E D , (Qf,g) = ( f , Q g ) ,
and for all g
E
L;, the condition
f*;,:&-, l ( Q f y g ) l implies that g is in D. See [RS].
< O0
The Dirichlet Problem
21
by Corollary 1.1, hencey2( 1, p ) = 0 by the boundary conditions. We see that the Dirichlet spectrum of q is identical with the zero set of the entire function y2(l,A, q). From now on, we will therefore not distinguish between them. The spectrum of a finite dimensional linear map is the zero set of its characteristic polynomial. By analogy, we can consider y2( 1, A , q) as the “characteristic polynomial” of the linear operator - d2/dx2+ q with Dirichlet boundary conditions. For q = 0, the Dirichlet spectrum is the infinite sequence
n’, 4 n 2 ,9n2, ..., n 2 n 2 , ..., since y2( 1, A, 0) = sin fi/fi.The “Counting Lemma” below shows that there are infinitely many eigenvalues for every q. It also gives a first rough estimate of their location, to be refined later on. First, a simple technical lemma.
Lemma 1. If Iz - nnl 2 n/4 for all integers n, then
< 4lsin zl. Proof. Write z = x + iy with real x,y. Since JsinzJis even and periodic with period T I , it suffices to prove the lemma for 0 Ix I7112 and IzI 2 7114. We have lsin z12 = cosh’y - cos’x. For n/6
I
x
I
7712,
cos’x
I
$ I$ cosh’y
for all real y. For 0 Ix Id 6 , the assumption IzI 2 n/4 implies y 2 2 ( 7 ~ / 4 )-~ x2 2 &n2 2 and hence cosh’y
L
1
+ y 2 2 4 L 0 cos’x
as before. Thus, in both cases we have IsinzI2 2 $cosh2y > &e21Yl, and the result follows. The estimate of Lemma 1 is not optimal, but chosen for the sake of convenience. In general, there is a positive constant cg for each 6 > 0 such that
< calsinzl provided Iz - nnl > 6 for all n.
Inverse Spectral Theory
28
In the next lemma, we admit complex valued functions q. This is useful, for instance in the proof of Theorem 3.1. Lemma 2 (The Counting Lemma). Let q E Lf and N > 2e1I4l1be an integer. Then y2( 1, A, q) has exactly N roots, counted with multiplicities, in the open half plane ReA < (N + +)’n’, and f o r each n > N , exactly one simple root in the egg shaped region
There are no other roots.
and let K > N be another integer. Consider the Proof. Fix N > 2e11411, contours
+ i)n, R e f l = ( N + i)n, IflI
=
(K
and 71
Ifl-nnl=-, 2
n>N.
See Figure 2. By Lemma 1, the estimate eltrnfi1< 4lsinflI
holds on all of them. Therefore, by the Basic Estimate for y z , Yd1, A) - -
fl
also holds on them. It follows that yz(1, A) does not vanish on these contours. Hence, by Rouche’s theorem, yz(1, A) has as many roots, counted with multiplicities, as sin fl/flin each of the bounded regions and the remaining
29
The Dirichlet Problem
I-plane
I
1
I
1
Figure 2.
unbounded region. Since sin fi/fihas only the simple roots n2n2,n and since K > N can be chosen arbitrarily large, the lemma follows.
L
1,
Returning to real valued functions q , we have
Theorem 1. The Dirichlet spectrum of q in L i is an infinite sequence of real numbers, which is bounded from below and tends to + m.
Proof. We only have to prove reality, the rest follows from the Counting Lemma. Suppose A is a Dirichlet eigenvalue of q with eigenfunction y(x). Then -y”
+ q(x)y = Ay.
Conjugating the equation,
-p + q(x)P = xy, since q is real. Multiplying the first equation by 7, the second by y and taking the difference, we obtain [ y , jq’ = y p
- y”y= (A
- X)(yI2,
30
Inverse Spectral Theory
hence, by integration,
The left hand side vanishes, while the integral does not. Therefore - X = 0, that is, A is real. Of course, this is the standard argument, used to prove that eigenvalues of self-adjoint operators are real.
A
It has been shown that every eigenfunction for an eigenvalue A is a multiple of y2(x, A). The geometric multiplicity of A, which is the maximal number of linearly independent eigenfunctions for A , is therefore 1 . On the other hand, its algebraic multiplicity, defined as its order as a root of y2(l, A), might very well be larger. This, however, is not the case. We will often use the abbreviated notation = WdA.
-
Theorem 2.
s:
If A is a Dirichlet eigenvalue of q in L2, then 32(1, A ) ~ i ( lA) ,
=
&t,
= Ilu2(.,
A) d t
4112> 0.
In particular, y2( 1, A) # 0. Thus, all roots of y2( 1 , A) are simple.
Proof. Let yields2
y2 =
yz(x, A). Differentiating equation (1) with respect to A
-32” + q(x132 = yz + A 3 2 . Multiplying this equation by y 2 , the original equation by 3 2 and taking the difference, we obtain 2 Y2 =
Y232 - 32Y2
= [32,y21’.
Hence
’To avoid the problem of interchanging x- and I-derivative, first assume q is continuous and argue as in the proof of Theorem 1.6.
The Dirichkt Problem
31
since yz(0, A) and )'2(0,A) vanish for all A, and yz(1, A) vanishes for a since y2 is real for Dirichlet eigenvalue I. The integral is equal to 11 y2( , real A.
-
Problem 1. Prove the identity of Theorem 2 by starting from
and using Theorem 1.6. It follows from the preceeding results that the Dirichlet spectrum of every q in L2is a sequence of real numbers p d q ) < P2(4) <
***
satisfying pn(q) = n2n2
+ O(n).
To each eigenvalue we associate a unique eigenfunction gn normalized by
= g n ( X , q)
By Theorem 2,
-
Y ~ XPn) ,
J h ( 1 , PnjYi(l9 P n ) '
where the argument q has been suppressed on the right hand side. In particular, gn(x,0) = *sin
nnx.
Let us investigate the Dirichlet eigenvalues as functions defined on L2.
Theorem 3. Pn, n 2 1 , is a compact, real analytic function on L2. Its gradient is
32
Inverse Spectral Theory
Since p,, is real analytic, g,, is also a real analytic function of q by the expression above and the Analyticity Properties of Chapter 1.
Proof. To verify compactness, suppose the sequence q m , m weakly to q. By the principle of uniform boundedness, 11q11 I SZP Ilq.mll IM
Let N > 2e‘,
E
L
1, converges
< 03.
> 0, and consider the intervals
I,,= [ A E
m: [ A - pn(q)l < E l ,
1 I n I N.
If E is sufficiently small, then these intervals are all disjoint and contained in the half line (- 00, ( N + $)’n2) by the Counting Lemma. Moreover, y2(1, A , q) changes sign on each of them, since pn(q) is a simple root. As m tends to infinity, the functions y2(l, A , q m ) converge to yz(1, A , q) uniformly on I, u ... u IN by Theorem 1.5. Hence, for sufficiently large m, they also change sign on I I , ..., IN, so they must all have at least one root in each of these intervals. But there are only N roots on the whole half line (- m, ( N + +)’n’) by the Counting Lemma. Therefore, yz(1, A , q m ) has exactly one root in each interval I,,, 1 I n IN , which must be the nth eigenvalue of qm. Consequently, lp(n(qm) - pun(q)J< E ,
1 In IN ,
for all sufficiently large m. It follows that M q m )
+
pun(4)
form + 00, since N a n d E > 0 were arbitrary. Thus, p,, is a compact function of q. To prove real analyticity, fix p in L2. By Theorem 2, Yz(l9
Pu,(P),P )
f
0.
So the implicit function theorem applies, and there exists a unique continuous function P,, ,defined on some small neighborhood U C L2 of p , such that Y2(1,P n ( 4 h
4) = 0,
Prim = Pn(P)
on U. Furthermore, f i n is real analytic. On the other hand, p,, is also a continuous function on U satisfying yz(1, p,,(q), q) = 0. Therefore,
Pn(d
= PrI(4)
on U by uniqueness, and so p,, is real analytic.
The Dirichlet Problem
33
To calculate the gradient observe that
since yz(1, pn) = 0 and y,(l)yl(l) = 1 by the Wronskian identity. Hence,
Here is another, perhaps more intuitive proof of the last identity. Differentiating both sides of the differential equation of gn in the direction u we obtain - d q g X ~ )+ qdqgn(tr) + ugn = pndqgn(u) + dqpn(V)gn.
If q is continuous, then gn is twice continuously differentiable, and we may interchange differentiation with respect to x and q to obtain - (dqgn(u))" + qdqgn(u) + vgn = pn dqgn(u)
+ dqpn(u)gn.
Multiplying both sides by gn and integrating we find (Qdqgn(u1,gn)
where Q
=
-d2/dx2
+
(d, u> = pn(dqgn(v), gn) + dqpUn(V)
+ q. The first term equals
(dqgn(v), Qgn) = pn(dqgn(u), gn).
Hence dqpn(u) = < g i , u > ,
and
However, both sides are continuous functions of q, and the continuous q are dense in Lz. So this identity holds in general.
34
Problem 2.
Inverse Spectral Theory
Show that Ipn(q) - pn(p)l 5 114 - Pll-9
n 2 1,
if q , p are in L", the space of all real valued, essentially bounded functions on [0, 11. [Hint: Write pn(q) - pn(P)
=
1;
g p n ( t q + (1 - t ) p )dt.1
We make a small digression to illustrate the usefulness of Theorem 3 . Consider the restriction of the function p n , n L 1, to the sphere 11q11 = 1 in L2. The ball 11q11 I 1 is compact in the weak topology of L2,3and p,,is a compact function on L2. Hence, pn attains its maximum and minimum on 11q11 5 1. However, we have pn(q
+ C)
= pn(q)
+c
for real c by the differential equation. It follows that the maximum and minimum must be attained on the unit sphere 11q11 = 1, for otherwise they would not be extremal. Consequently, p n has at least two critical points on the unit sphere in L2. Let q be a critical point of pn on 11q11 = 1. The gradient of p n must be proportional to the unit normal at q. That is,
where
By (*), q has exactly n + 1 double roots in [0, 11 including 0 and 1, and does not change sign. Moreover, q is infinitely often differentiable. For, g," is always continuous, hence so is q . But then g," is in C ' , hence so is q , ad in finitum. The critical point q also satisfies a differential equation. Apply the operator L of Theorem 1.7 to (*) to obtain -2pnq'
+ 3qq' - +q'" = 0.
'Recall that the sphere 11q11 = 1 is not compact.
35
The Dirichlet Problem
Integration, multiplication by q’ and another integration yields the identity -p(nq
2
+ +q3- $(q’l2= cq
with some constant c # 0. The substitution q differential equation
(**I
= 2u
+ fpn then leads to the
(u’)= ~ 4(u - el)(u - e2)(u - e3)
where e l , e2, e3 are complex numbers whose sum is zero. For distinct e l , e2, e3, (**) is the differential equation for the Weierstrass @-function. It is possible to carry this analysis further using properties of the Weierstrass @-function. But we shall not. The Counting Lemma gave us a first rough estimate of the location of the Dirichlet eigenvalues. This estimate is now refined. Let P2 denote the Hilbert space of all real sequences (a1,a2, ...) such that C a: is finite. More generally, for k L 0, Pi denotes the Hilbert space of all real sequences (a1,a2, ...) such that
C
n a 1
(nkanl2<
m.
In analogy to the notation O ( l / n ) ,we use the notation $(n) for an arbitrary sequence of numbers which is an element of Pi’. For instance, a n
=
/jn
+ p2(n)
is equivalent to an = P n
Theorem 4.
+ I,,
C
n a 1
I: <
03.
For q in L2,
and
gXx, q ) = f i n n cos rrnx
+ 0(1).
These estimates hold uniformly on bounded subsets of [0, I ] x L2.
36
Inverse Spectral Theory
Proof. We “iterate” estimates on pn and g,, , thereby sharpening them. All estimates hold uniformly on bounded subsets of [O, 11 x Lz. Let p,, = p,,(q). By the Counting Lemma,
By the Basic Estimate for y2,
Using the identity 2 sin’ax = 1 - cos 2ax, one calculates
It follows that
Thus the preliminary estimate
is obtained. Our estimate of p,, may now be improved. By Theorem 3,
(3)
The Dirichlet Problem
31
Since (2) holds uniformly for tq, 0
It I1,
this implies
pn = n2n2 + 0(1)
or equivalently
a
=
nn
+
.(:),
and in turn gn(x) = *sin
nnx
+o
(3 -
.
Inserting the last estimate of gn into (3) and using the identity 2 sin’ux = 1 - cos 2ux again, we obtain pn -
n2n2
=
j: (1
-
cos2nnx
+0
1
= j o q ( t ) d f1
=
i o q ( t ) d t + t2(n),
since (cos 2nnx, q ) are the square summable Fourier cosine coefficients of q . Finally, we estimate gA . By the Basic Estimates,
= cos nnx
+ o(:).
This proves the theorem. It will also be important to have asymptotic estimates of the squares of the eigenfunctions and the products ~ ( x4,) = YI(X,P~)YZ(X, Pn),
n
1.
38
Inverse Spectral Theory
Corollary 1. For q in L2, 2
gn =
d -gi dx
=
1 - cos 2nnx
+O
2nn sin 2nnx
+ O(1)
and 1 an = -sin2nnx 2nn
d -an = cos 2nnx dx
+0
+0
All estimates hold uniformly on bounded subsets of [0, 11 x L2.
Proof. The estimates of gx and (d/dx)gi follow from Theorem 4 by straightforward calculations. Also, = nn + O(l/n) and the Basic Estimates for y1 and y2 yield
a
.YI(X, pn)
=
(3
+O
cos nnx
-
and
1
= -sin 2nnx
an = y1y2 h
2nn
= cos2nnx
(9
+0
-
.
+0
The Dirichlet Problem
Problem 3.
s:
(a) Show that the estimate of Theorem 4 can be improved to pn(q) =
(b)
39
n
71 2 2
+
q(t)dt - ( cos Z Z ~ X ,4)
+ P!(n).
The function q is odd on [0, 11, if g(l - x ) = - q ( x ) . Show that for
odd q, pn(q) =
n2n2 + ~ ? ( n ) .
(c) Show that for q in the Sobolev space H6 with
1; q(t) dt
= 0,
The characteristic polynomial of an n x n-matrix is equal to the product
where A,, ...,A n are the eigenvalues of the matrix. There is an analogous representation for yz(1, A , q). Theorem 5. For q in L2,
Proof. We havepn = n2n2 + O(1) by Theorem 4. It follows from Lemma E.2 that the infinite product
is an entire function of A, which satisfies
p(A)
=
(
sin dX 1 + 0 -(0; n))
-
a
uniformly on the circles 1A1 = rn = (n + +)'n2 for n large enough. Its roots are precisely the Dirichlet eigenvalues of q , so the quotient p(A)/yz(l, A) is also an entire function. By the Basic Estimate for y2 and Lemma 1, sin dX
Y 2 ( l , A) =-
dX
+
o('~;'p'>
Inverse Spectral Theory
40
uniformly for 111 = r n . Hence,
for 111
=
rn, that is,
as n + 03. If follows from the maximum principle that the difference vanishes identically. Hence, p(A) = yz(1, A). If the functions yz( 1, A, q) and yz(1, A, p ) are equal, then clearly P n ( q ) = p n ( p )for all n L 1. That is, q and p have the same Dirichlet spectrum. Now we know that the converse is also true. In other words, all the information in y2( 1, A) is already contained in the Dirichlet spectrum Pun, n L 1. Here are two useful consequences of the product formula. Corollary 2.
Proof. Part (a) follows from Lemma E.3. The first identity in (b) follows from (a), while the second one is a consequence of the first one and Theorem 2 . Problem 4. Prove Theorem 5 , using Hadamard's theorem [Ti]. [Hint: Observe that y2(l, A) is an entire function of order i, so that
41
The Dirichlet Problem
with
c1 = c o n
ms1
-.m2n2 Pm
Now divide by sin fi
fi and let I tend to
--03
A = n m2n2 m2n2
to obtain c1
-
msl
=
11.
The rest of this chapter is devoted to properties of the eigenfunctions. Theorem 6. (a) For every q in L2,the eigenfunction g n ,n 1 1, has exactIy n + 1 roots in [0, 11. They are aN simple. (b) If q is even, then g n is even when n is odd and odd when n is even.
4.
Even and odd mean even and odd about the point The task of counting the roots of gn is simplified by the following “deformation lemma”:
Lemma 3. Let h t , 0 It I1 , be a famiIy of real valued functions on a 5 x Ib, which is jointly continuously differentiable in t and x. Suppose that f o r every t , ht has a finite number of roots in [a, b ] , aI1 of which are simple, and has boundary values that are independent o f t . Then ho and hl have the same number of roots in [a, b ] . Intuitively speaking, the roots of hr in the interior of [a,b] move as t moves, but they can never collide or split, since they are all simple. Therefore, their number is independent of t.
Proof of Lemma 3. Suppose for convenience that ht(a) = 0 = ht(b),
0
It I1 .
Other cases are handled analogously. Fix t i n [0, 11. By assumption, the roots of ht are simple. Therefore, we can place an interval around each of them, so that ht changes sign once on the intervals in the interior, hi is bounded away from zero on all the intervals, and in addition, ht is bounded away from zero on the complementary intervals. See Figure 3.
42
Inverse Spectral Theory
Figure 3.
By continuity, h, and h: behave in exactly the same way on these intervals, if s is sufficiently close to t. Consequently, h, has exactly the same number of roots as h r , when s is close to t . This argument applies to all t in [0, 11, so the number of roots is independent of t. In particular, it is the same for ho and hl . H
Proof of Theorem 6. (a) Since g,(x, q) does not vanish identically, its roots are all simple by Corollary 1.1. Their number is finite, since otherwise they would cluster at a multiple root. Now consider the deformation hr of gn given by h r ( ~= ) gn(X, tq),
0 It, x I1.
Clearly, for each I , hr vanishes at 0 and 1 and has only a finite number of roots in [0, 11 all of which are simple. Thus we may apply Lemma 3, and it follows that hl = gn has as many roots in [0, 11 as ho = a s i n nnx. Hence, g, has exactly n + 1 roots in [0, 11. (b) Let q be even. That is, q( 1 - x) = q(x) for 0 I x I 1. Substituting 1 - x for x in -S i
+ q(x)gn = pngn
9
we see that gn( 1 - x) is another eigenfunction of q forp,, with norm 1. Hence, gn(x) = - sgn gXl)gn(l -
XI.
Since sgn gXl) = (- 1)" by Corollary 2, the claim follows.
H
One of the main objectives of classical Sturm-Liouville theory is to generalize the Fourier sine expansion. The idea is to replace the sines by eigenfunctions of a boundary value problem for a second order equation.
The Dirichlet Problem
43
This was a precursor of the abstract spectral theorem for self-adjoint operators. For the Dirichlet problem, the result is
Theorem 7. For each q in Lz, the sequence gn(x, d,
n2 1
is an orthonormal basis for L2.
=
0.
Since pm # pn for m # n, this implies (gmvgn) = 0,
m
f
n.
Since also (gn, gn) = 1 by construction, the functions gn, n 2 1, are orthonormal. It remains to show that this system is complete in Lz. We do this by comparing the vectors gn with the vectors
en = \/z sin nnx,
n
L
1,
which do form an orthonormal basis of Lz. Consider the linear operator A on Lz defined by
Af=
C
n r 1
(f,en)gn.
A is well defined, since ( f , e n ) , n L 1, is an Pz-sequence and gn, n 2 1, is an orthonormal basis. A is an isometry, since IIAfII’ =
1
n.? 1
I(f,
=
IIfII’
by the orthogonality of the gn and Parseval’s identity. In particular, A is one-to-one. Also, by Theorem 4,
Inverse Spectral Theory
44
a,
so A - I is Hilbert-Schmidt and hence compact by Theorem D. 1. It follows
from the Fredholm Alternative, Theorem D.2, that A is onto L2 and has a bounded inverse. Therefore, gn, n 1 1, is an orthonormal basis for L2. There are more elementary proofs of the completeness of the Dirichlet eigenfunctions, which do not appeal directly to the Fredholm Alternative. See for instance Birkhoff and Rota [BR], who also make a number of interesting historical remarks. The method used here, however, is easy to generalize. Roughly speaking, any sequence of vectors, which is linearly independent and “sufficiently close” to some orthonormal basis, is itself a basis. Theorem D.3 makes this statement precise. It will be very useful in the sequel. Actually, we will never need Theorem 7, but include it as a prototype for other completeness theorems. For us, it will be much more important to settle independence and completeness questions about squares of eigenfunctions, for example in the proof of Theorem 3.2. The next theorem is essential for this and many other purposes. Recall that
Theorem 8. For m , n 2 1,
It is only a slight exaggeration to say that Theorem 8 is the basis of almost everything else we are going to do.
The Dirichlet Problem
45
Proof. (a) Integration by parts yields
lo 1
=
This clearly vanishes for m
=
gmgn[gm gnl dx.
n. If m # n, then p m # p,, and we can use
[gm, gnl’ = ( p m - punlgmgn to obtain
(b) Again, by partial integration,
If rn
=
n, then y2 is a multiple of g,, hence [ y z ,g,]
=
1,
using the Wronskian identity. (c) is proven in the same manner.
=
0, and we get
Inverse Spectral Theory
46
For q
=
0, the vectors 1,
n L 1, are a basis for L2[0,11. Theorem 8 and the asymptotic estimates of Corollary 1 make it possible to prove the same statement for every q . First, let us specify the notions of linear independence and basis. A sequence of vectors u1 , u 2 , ... in a Hilbert space is linearly independent, if none of the vectors un , n 1 1, is contained in the closed linear span of all the other vectors urn, m # n. One might try to define a stronger notion of linear independence by requiring U n to lie outside the weak closure of the span of urn, rn # n. However, if a sequence of vectors converges weakly, then the arithmetic means of some subsequence converge strongly to the same limit by the theorem of Banach-Saks [Ba]. Thus, the weak and strong closure of a linear space are the same, and nothing is gained. A sequence of vectors u l , u 2 , ... is a basis for a Hilbert space H, if there , ...) such that the correexists a Hilbert space A of sequences a = ( a ] a2, spondence a+
C
n z1
anvn
is a linear isomorphism between A and H.4 Theorem 9. A t every point in L2, the vectors 2
1,g, - 1,
n
2 1
are linearly independent. So are the vectors
The two sequences are mutually perpendicular, and together are a basis for L' . Precisely, ((,v)+
d
C t n z d + V O + nCa 1 Vn(g,'
n z1
is a linear isomorphism between pf x R x
P
- 1)
and L2.
4 A linear isomorphism between Banach spaces is an isomorphism between vector spaces, which is continuous and has a continuous inverse.
41
The Dirichlet Problem
It follows from Theorem 6 that at an even point in L2,the vectors 1, g,' - 1, n 2 1, are a basis for the subspace of even functions in L2, and the vectors (d/dx)g,', n 1 1, are a basis for the subspace of odd functions.
Proof. The vector g,' - 1, n 2 1, is not in the closed linear span of the vectors 1, g& - 1, m # n, since
but
by Theorem 8. Similar arguments apply to all other vectors. The two sequences are mutually perpendicular, since
for all m ,n again by Theorem 8. By Corollary 1, 2
gn - 1 =
- ~ 0 ~ 2 n n+x 0
1=1 1 d -g,' = sin27rnx 27rn dx
+o
We have shown that the vectors on the left are linearly independent. The vectors on the right without the error terms are (up to an irrelevant factor an orthonormal basis of L2, and the error terms are square summable. Therefore, Theorem D.3 applies. It follows that the vectors on the left are a basis with coefficients in x R x ?. From this, the last statement of the theorem follows.
a)
It is useful to write Theorem 9 in a slightly different form.
Inverse Spectral Theory
48
Corollary 3. A t every point q in L2, the subspaces
are perpendicular and closed. Their direct sum is all of L2.In particular, they are the odd and even subspaces respectively when q is even.
Problem 5. (a) Show that log N Conclude that 1 is in the closed linear span of gn', n L 1. (b) Show that the vectors gn', (d/dx)gn',n 1 1, span L2 (that is, finite linear combinations are dense) and are linearly independent, but not a basis. The basic facts about the Dirichlet eigenvalues and eigenfunctions are now proven. We are ready to approach the inverse Dirichlet problem.
The Dirichlet Problem
This chapter is devoted to the simplest boundary value problem for the differential equation (1)
-y” + q(x)y
= ly,
0
4
x 5 1.
Here, l E C and q E L2 = Li[O, 11, the Hilbert space of all real valued, square integrable functions on [0, 11. We ask whether there are nontrivial solutions of equation (1) satisfying the Dirichlet boundary condition
y(0) = 0,
y(1) = 0.
This is the Dirichlet problem. A complex number l is called a Dirichlet eigenvalue of q if the Dirichlet problem can be solved. The corresponding nontrivial solutions are called eigenfunctions of q for I t . The collection of all eigenvalues of q is its Dirichlet spectrum. From a more abstract point of view, the Dirichlet problem belongs to the theory of self-adjoint operators. Precisely, the unbounded operator Q, ,defined by
25
Inverse Spectral Theory
26
for f in the dense subspace D of all functions in Hc[O, 11 that vanish at 0 and 1, is self-adjoint on the domain D.’ It can be shown that the spectrum of Q coincides with the Dirichlet spectrum of q, and so we may draw some conclusions about the Dirichlet problem from general facts about selfadjoint operators. For instance, the Dirichlet spectrum must be an unbounded sequence of real numbers. However, abstract operator theory is not helpful in answering many of the questions that interest us. For this reason, we shall follow Sturm’s classical approach [St] which may be called the “method of shooting. ” Intuitively speaking, the solution y2 is shot from the left end point of the interval with unit velocity. At x = 1, it reaches the height yz(1, A). The idea is to vary the parameter A so that the terminal height is 0. This point is illustrated in Figure 1. At the moment, the terminal velocityyz’(1,A) does not concern us. It becomes important in Chapter 3. The “shooting method” is easy to implement. If p is a zero ofyz( 1, A), then yz(x, p) is a nontrivial solution of equation (1) satisfying the Dirichlet boundary conditions. Hence, p is a Dirichlet eigenvalue of q. Conversely, suppose p is a Dirichlet eigenvalue of q with eigenfunction y(x). Then Y ( X ) = Y’(O)YZ(X,
Figure 1. ‘That is, for all f,g E D , (Qf,g) = ( f , Q g ) ,
and for all g
E
L;, the condition
f*;,:&-, l ( Q f y g ) l implies that g is in D. See [RS].
< O0
The Dirichlet Problem
21
by Corollary 1.1, hencey2( 1, p ) = 0 by the boundary conditions. We see that the Dirichlet spectrum of q is identical with the zero set of the entire function y2(l,A, q). From now on, we will therefore not distinguish between them. The spectrum of a finite dimensional linear map is the zero set of its characteristic polynomial. By analogy, we can consider y2( 1, A , q) as the “characteristic polynomial” of the linear operator - d2/dx2+ q with Dirichlet boundary conditions. For q = 0, the Dirichlet spectrum is the infinite sequence
n’, 4 n 2 ,9n2, ..., n 2 n 2 , ..., since y2( 1, A, 0) = sin fi/fi.The “Counting Lemma” below shows that there are infinitely many eigenvalues for every q. It also gives a first rough estimate of their location, to be refined later on. First, a simple technical lemma.
Lemma 1. If Iz - nnl 2 n/4 for all integers n, then
< 4lsin zl. Proof. Write z = x + iy with real x,y. Since JsinzJis even and periodic with period T I , it suffices to prove the lemma for 0 Ix I7112 and IzI 2 7114. We have lsin z12 = cosh’y - cos’x. For n/6
I
x
I
7712,
cos’x
I
$ I$ cosh’y
for all real y. For 0 Ix Id 6 , the assumption IzI 2 n/4 implies y 2 2 ( 7 ~ / 4 )-~ x2 2 &n2 2 and hence cosh’y
L
1
+ y 2 2 4 L 0 cos’x
as before. Thus, in both cases we have IsinzI2 2 $cosh2y > &e21Yl, and the result follows. The estimate of Lemma 1 is not optimal, but chosen for the sake of convenience. In general, there is a positive constant cg for each 6 > 0 such that
< calsinzl provided Iz - nnl > 6 for all n.
Inverse Spectral Theory
28
In the next lemma, we admit complex valued functions q. This is useful, for instance in the proof of Theorem 3.1. Lemma 2 (The Counting Lemma). Let q E Lf and N > 2e1I4l1be an integer. Then y2( 1, A, q) has exactly N roots, counted with multiplicities, in the open half plane ReA < (N + +)’n’, and f o r each n > N , exactly one simple root in the egg shaped region
There are no other roots.
and let K > N be another integer. Consider the Proof. Fix N > 2e11411, contours
+ i)n, R e f l = ( N + i)n, IflI
=
(K
and 71
Ifl-nnl=-, 2
n>N.
See Figure 2. By Lemma 1, the estimate eltrnfi1< 4lsinflI
holds on all of them. Therefore, by the Basic Estimate for y z , Yd1, A) - -
fl
also holds on them. It follows that yz(1, A) does not vanish on these contours. Hence, by Rouche’s theorem, yz(1, A) has as many roots, counted with multiplicities, as sin fl/flin each of the bounded regions and the remaining
29
The Dirichlet Problem
I-plane
I
1
I
1
Figure 2.
unbounded region. Since sin fi/fihas only the simple roots n2n2,n and since K > N can be chosen arbitrarily large, the lemma follows.
L
1,
Returning to real valued functions q , we have
Theorem 1. The Dirichlet spectrum of q in L i is an infinite sequence of real numbers, which is bounded from below and tends to + m.
Proof. We only have to prove reality, the rest follows from the Counting Lemma. Suppose A is a Dirichlet eigenvalue of q with eigenfunction y(x). Then -y”
+ q(x)y = Ay.
Conjugating the equation,
-p + q(x)P = xy, since q is real. Multiplying the first equation by 7, the second by y and taking the difference, we obtain [ y , jq’ = y p
- y”y= (A
- X)(yI2,
30
Inverse Spectral Theory
hence, by integration,
The left hand side vanishes, while the integral does not. Therefore - X = 0, that is, A is real. Of course, this is the standard argument, used to prove that eigenvalues of self-adjoint operators are real.
A
It has been shown that every eigenfunction for an eigenvalue A is a multiple of y2(x, A). The geometric multiplicity of A, which is the maximal number of linearly independent eigenfunctions for A , is therefore 1 . On the other hand, its algebraic multiplicity, defined as its order as a root of y2(l, A), might very well be larger. This, however, is not the case. We will often use the abbreviated notation = WdA.
-
Theorem 2.
s:
If A is a Dirichlet eigenvalue of q in L2, then 32(1, A ) ~ i ( lA) ,
=
&t,
= Ilu2(.,
A) d t
4112> 0.
In particular, y2( 1, A) # 0. Thus, all roots of y2( 1 , A) are simple.
Proof. Let yields2
y2 =
yz(x, A). Differentiating equation (1) with respect to A
-32” + q(x132 = yz + A 3 2 . Multiplying this equation by y 2 , the original equation by 3 2 and taking the difference, we obtain 2 Y2 =
Y232 - 32Y2
= [32,y21’.
Hence
’To avoid the problem of interchanging x- and I-derivative, first assume q is continuous and argue as in the proof of Theorem 1.6.
The Dirichkt Problem
31
since yz(0, A) and )'2(0,A) vanish for all A, and yz(1, A) vanishes for a since y2 is real for Dirichlet eigenvalue I. The integral is equal to 11 y2( , real A.
-
Problem 1. Prove the identity of Theorem 2 by starting from
and using Theorem 1.6. It follows from the preceeding results that the Dirichlet spectrum of every q in L2is a sequence of real numbers p d q ) < P2(4) <
***
satisfying pn(q) = n2n2
+ O(n).
To each eigenvalue we associate a unique eigenfunction gn normalized by
= g n ( X , q)
By Theorem 2,
-
Y ~ XPn) ,
J h ( 1 , PnjYi(l9 P n ) '
where the argument q has been suppressed on the right hand side. In particular, gn(x,0) = *sin
nnx.
Let us investigate the Dirichlet eigenvalues as functions defined on L2.
Theorem 3. Pn, n 2 1 , is a compact, real analytic function on L2. Its gradient is
32
Inverse Spectral Theory
Since p,, is real analytic, g,, is also a real analytic function of q by the expression above and the Analyticity Properties of Chapter 1.
Proof. To verify compactness, suppose the sequence q m , m weakly to q. By the principle of uniform boundedness, 11q11 I SZP Ilq.mll IM
Let N > 2e‘,
E
L
1, converges
< 03.
> 0, and consider the intervals
I,,= [ A E
m: [ A - pn(q)l < E l ,
1 I n I N.
If E is sufficiently small, then these intervals are all disjoint and contained in the half line (- 00, ( N + $)’n2) by the Counting Lemma. Moreover, y2(1, A , q) changes sign on each of them, since pn(q) is a simple root. As m tends to infinity, the functions y2(l, A , q m ) converge to yz(1, A , q) uniformly on I, u ... u IN by Theorem 1.5. Hence, for sufficiently large m, they also change sign on I I , ..., IN, so they must all have at least one root in each of these intervals. But there are only N roots on the whole half line (- m, ( N + +)’n’) by the Counting Lemma. Therefore, yz(1, A , q m ) has exactly one root in each interval I,,, 1 I n IN , which must be the nth eigenvalue of qm. Consequently, lp(n(qm) - pun(q)J< E ,
1 In IN ,
for all sufficiently large m. It follows that M q m )
+
pun(4)
form + 00, since N a n d E > 0 were arbitrary. Thus, p,, is a compact function of q. To prove real analyticity, fix p in L2. By Theorem 2, Yz(l9
Pu,(P),P )
f
0.
So the implicit function theorem applies, and there exists a unique continuous function P,, ,defined on some small neighborhood U C L2 of p , such that Y2(1,P n ( 4 h
4) = 0,
Prim = Pn(P)
on U. Furthermore, f i n is real analytic. On the other hand, p,, is also a continuous function on U satisfying yz(1, p,,(q), q) = 0. Therefore,
Pn(d
= PrI(4)
on U by uniqueness, and so p,, is real analytic.
The Dirichlet Problem
33
To calculate the gradient observe that
since yz(1, pn) = 0 and y,(l)yl(l) = 1 by the Wronskian identity. Hence,
Here is another, perhaps more intuitive proof of the last identity. Differentiating both sides of the differential equation of gn in the direction u we obtain - d q g X ~ )+ qdqgn(tr) + ugn = pndqgn(u) + dqpn(V)gn.
If q is continuous, then gn is twice continuously differentiable, and we may interchange differentiation with respect to x and q to obtain - (dqgn(u))" + qdqgn(u) + vgn = pn dqgn(u)
+ dqpn(u)gn.
Multiplying both sides by gn and integrating we find (Qdqgn(u1,gn)
where Q
=
-d2/dx2
+
(d, u> = pn(dqgn(v), gn) + dqpUn(V)
+ q. The first term equals
(dqgn(v), Qgn) = pn(dqgn(u), gn).
Hence dqpn(u) = < g i , u > ,
and
However, both sides are continuous functions of q, and the continuous q are dense in Lz. So this identity holds in general.
34
Problem 2.
Inverse Spectral Theory
Show that Ipn(q) - pn(p)l 5 114 - Pll-9
n 2 1,
if q , p are in L", the space of all real valued, essentially bounded functions on [0, 11. [Hint: Write pn(q) - pn(P)
=
1;
g p n ( t q + (1 - t ) p )dt.1
We make a small digression to illustrate the usefulness of Theorem 3 . Consider the restriction of the function p n , n L 1, to the sphere 11q11 = 1 in L2. The ball 11q11 I 1 is compact in the weak topology of L2,3and p,,is a compact function on L2. Hence, pn attains its maximum and minimum on 11q11 5 1. However, we have pn(q
+ C)
= pn(q)
+c
for real c by the differential equation. It follows that the maximum and minimum must be attained on the unit sphere 11q11 = 1, for otherwise they would not be extremal. Consequently, p n has at least two critical points on the unit sphere in L2. Let q be a critical point of pn on 11q11 = 1. The gradient of p n must be proportional to the unit normal at q. That is,
where
By (*), q has exactly n + 1 double roots in [0, 11 including 0 and 1, and does not change sign. Moreover, q is infinitely often differentiable. For, g," is always continuous, hence so is q . But then g," is in C ' , hence so is q , ad in finitum. The critical point q also satisfies a differential equation. Apply the operator L of Theorem 1.7 to (*) to obtain -2pnq'
+ 3qq' - +q'" = 0.
'Recall that the sphere 11q11 = 1 is not compact.
35
The Dirichlet Problem
Integration, multiplication by q’ and another integration yields the identity -p(nq
2
+ +q3- $(q’l2= cq
with some constant c # 0. The substitution q differential equation
(**I
= 2u
+ fpn then leads to the
(u’)= ~ 4(u - el)(u - e2)(u - e3)
where e l , e2, e3 are complex numbers whose sum is zero. For distinct e l , e2, e3, (**) is the differential equation for the Weierstrass @-function. It is possible to carry this analysis further using properties of the Weierstrass @-function. But we shall not. The Counting Lemma gave us a first rough estimate of the location of the Dirichlet eigenvalues. This estimate is now refined. Let P2 denote the Hilbert space of all real sequences (a1,a2, ...) such that C a: is finite. More generally, for k L 0, Pi denotes the Hilbert space of all real sequences (a1,a2, ...) such that
C
n a 1
(nkanl2<
m.
In analogy to the notation O ( l / n ) ,we use the notation $(n) for an arbitrary sequence of numbers which is an element of Pi’. For instance, a n
=
/jn
+ p2(n)
is equivalent to an = P n
Theorem 4.
+ I,,
C
n a 1
I: <
03.
For q in L2,
and
gXx, q ) = f i n n cos rrnx
+ 0(1).
These estimates hold uniformly on bounded subsets of [0, I ] x L2.
36
Inverse Spectral Theory
Proof. We “iterate” estimates on pn and g,, , thereby sharpening them. All estimates hold uniformly on bounded subsets of [O, 11 x Lz. Let p,, = p,,(q). By the Counting Lemma,
By the Basic Estimate for y2,
Using the identity 2 sin’ax = 1 - cos 2ax, one calculates
It follows that
Thus the preliminary estimate
is obtained. Our estimate of p,, may now be improved. By Theorem 3,
(3)
The Dirichlet Problem
31
Since (2) holds uniformly for tq, 0
It I1,
this implies
pn = n2n2 + 0(1)
or equivalently
a
=
nn
+
.(:),
and in turn gn(x) = *sin
nnx
+o
(3 -
.
Inserting the last estimate of gn into (3) and using the identity 2 sin’ux = 1 - cos 2ux again, we obtain pn -
n2n2
=
j: (1
-
cos2nnx
+0
1
= j o q ( t ) d f1
=
i o q ( t ) d t + t2(n),
since (cos 2nnx, q ) are the square summable Fourier cosine coefficients of q . Finally, we estimate gA . By the Basic Estimates,
= cos nnx
+ o(:).
This proves the theorem. It will also be important to have asymptotic estimates of the squares of the eigenfunctions and the products ~ ( x4,) = YI(X,P~)YZ(X, Pn),
n
1.
38
Inverse Spectral Theory
Corollary 1. For q in L2, 2
gn =
d -gi dx
=
1 - cos 2nnx
+O
2nn sin 2nnx
+ O(1)
and 1 an = -sin2nnx 2nn
d -an = cos 2nnx dx
+0
+0
All estimates hold uniformly on bounded subsets of [0, 11 x L2.
Proof. The estimates of gx and (d/dx)gi follow from Theorem 4 by straightforward calculations. Also, = nn + O(l/n) and the Basic Estimates for y1 and y2 yield
a
.YI(X, pn)
=
(3
+O
cos nnx
-
and
1
= -sin 2nnx
an = y1y2 h
2nn
= cos2nnx
(9
+0
-
.
+0
The Dirichlet Problem
Problem 3.
s:
(a) Show that the estimate of Theorem 4 can be improved to pn(q) =
(b)
39
n
71 2 2
+
q(t)dt - ( cos Z Z ~ X ,4)
+ P!(n).
The function q is odd on [0, 11, if g(l - x ) = - q ( x ) . Show that for
odd q, pn(q) =
n2n2 + ~ ? ( n ) .
(c) Show that for q in the Sobolev space H6 with
1; q(t) dt
= 0,
The characteristic polynomial of an n x n-matrix is equal to the product
where A,, ...,A n are the eigenvalues of the matrix. There is an analogous representation for yz(1, A , q). Theorem 5. For q in L2,
Proof. We havepn = n2n2 + O(1) by Theorem 4. It follows from Lemma E.2 that the infinite product
is an entire function of A, which satisfies
p(A)
=
(
sin dX 1 + 0 -(0; n))
-
a
uniformly on the circles 1A1 = rn = (n + +)'n2 for n large enough. Its roots are precisely the Dirichlet eigenvalues of q , so the quotient p(A)/yz(l, A) is also an entire function. By the Basic Estimate for y2 and Lemma 1, sin dX
Y 2 ( l , A) =-
dX
+
o('~;'p'>
Inverse Spectral Theory
40
uniformly for 111 = r n . Hence,
for 111
=
rn, that is,
as n + 03. If follows from the maximum principle that the difference vanishes identically. Hence, p(A) = yz(1, A). If the functions yz( 1, A, q) and yz(1, A, p ) are equal, then clearly P n ( q ) = p n ( p )for all n L 1. That is, q and p have the same Dirichlet spectrum. Now we know that the converse is also true. In other words, all the information in y2( 1, A) is already contained in the Dirichlet spectrum Pun, n L 1. Here are two useful consequences of the product formula. Corollary 2.
Proof. Part (a) follows from Lemma E.3. The first identity in (b) follows from (a), while the second one is a consequence of the first one and Theorem 2 . Problem 4. Prove Theorem 5 , using Hadamard's theorem [Ti]. [Hint: Observe that y2(l, A) is an entire function of order i, so that
41
The Dirichlet Problem
with
c1 = c o n
ms1
-.m2n2 Pm
Now divide by sin fi
fi and let I tend to
--03
A = n m2n2 m2n2
to obtain c1
-
msl
=
11.
The rest of this chapter is devoted to properties of the eigenfunctions. Theorem 6. (a) For every q in L2,the eigenfunction g n ,n 1 1, has exactIy n + 1 roots in [0, 11. They are aN simple. (b) If q is even, then g n is even when n is odd and odd when n is even.
4.
Even and odd mean even and odd about the point The task of counting the roots of gn is simplified by the following “deformation lemma”:
Lemma 3. Let h t , 0 It I1 , be a famiIy of real valued functions on a 5 x Ib, which is jointly continuously differentiable in t and x. Suppose that f o r every t , ht has a finite number of roots in [a, b ] , aI1 of which are simple, and has boundary values that are independent o f t . Then ho and hl have the same number of roots in [a, b ] . Intuitively speaking, the roots of hr in the interior of [a,b] move as t moves, but they can never collide or split, since they are all simple. Therefore, their number is independent of t.
Proof of Lemma 3. Suppose for convenience that ht(a) = 0 = ht(b),
0
It I1 .
Other cases are handled analogously. Fix t i n [0, 11. By assumption, the roots of ht are simple. Therefore, we can place an interval around each of them, so that ht changes sign once on the intervals in the interior, hi is bounded away from zero on all the intervals, and in addition, ht is bounded away from zero on the complementary intervals. See Figure 3.
42
Inverse Spectral Theory
Figure 3.
By continuity, h, and h: behave in exactly the same way on these intervals, if s is sufficiently close to t. Consequently, h, has exactly the same number of roots as h r , when s is close to t . This argument applies to all t in [0, 11, so the number of roots is independent of t. In particular, it is the same for ho and hl . H
Proof of Theorem 6. (a) Since g,(x, q) does not vanish identically, its roots are all simple by Corollary 1.1. Their number is finite, since otherwise they would cluster at a multiple root. Now consider the deformation hr of gn given by h r ( ~= ) gn(X, tq),
0 It, x I1.
Clearly, for each I , hr vanishes at 0 and 1 and has only a finite number of roots in [0, 11 all of which are simple. Thus we may apply Lemma 3, and it follows that hl = gn has as many roots in [0, 11 as ho = a s i n nnx. Hence, g, has exactly n + 1 roots in [0, 11. (b) Let q be even. That is, q( 1 - x) = q(x) for 0 I x I 1. Substituting 1 - x for x in -S i
+ q(x)gn = pngn
9
we see that gn( 1 - x) is another eigenfunction of q forp,, with norm 1. Hence, gn(x) = - sgn gXl)gn(l -
XI.
Since sgn gXl) = (- 1)" by Corollary 2, the claim follows.
H
One of the main objectives of classical Sturm-Liouville theory is to generalize the Fourier sine expansion. The idea is to replace the sines by eigenfunctions of a boundary value problem for a second order equation.
The Dirichlet Problem
43
This was a precursor of the abstract spectral theorem for self-adjoint operators. For the Dirichlet problem, the result is
Theorem 7. For each q in Lz, the sequence gn(x, d,
n2 1
is an orthonormal basis for L2.
=
0.
Since pm # pn for m # n, this implies (gmvgn) = 0,
m
f
n.
Since also (gn, gn) = 1 by construction, the functions gn, n 2 1, are orthonormal. It remains to show that this system is complete in Lz. We do this by comparing the vectors gn with the vectors
en = \/z sin nnx,
n
L
1,
which do form an orthonormal basis of Lz. Consider the linear operator A on Lz defined by
Af=
C
n r 1
(f,en)gn.
A is well defined, since ( f , e n ) , n L 1, is an Pz-sequence and gn, n 2 1, is an orthonormal basis. A is an isometry, since IIAfII’ =
1
n.? 1
I(f,
=
IIfII’
by the orthogonality of the gn and Parseval’s identity. In particular, A is one-to-one. Also, by Theorem 4,
Inverse Spectral Theory
44
a,
so A - I is Hilbert-Schmidt and hence compact by Theorem D. 1. It follows
from the Fredholm Alternative, Theorem D.2, that A is onto L2 and has a bounded inverse. Therefore, gn, n 1 1, is an orthonormal basis for L2. There are more elementary proofs of the completeness of the Dirichlet eigenfunctions, which do not appeal directly to the Fredholm Alternative. See for instance Birkhoff and Rota [BR], who also make a number of interesting historical remarks. The method used here, however, is easy to generalize. Roughly speaking, any sequence of vectors, which is linearly independent and “sufficiently close” to some orthonormal basis, is itself a basis. Theorem D.3 makes this statement precise. It will be very useful in the sequel. Actually, we will never need Theorem 7, but include it as a prototype for other completeness theorems. For us, it will be much more important to settle independence and completeness questions about squares of eigenfunctions, for example in the proof of Theorem 3.2. The next theorem is essential for this and many other purposes. Recall that
Theorem 8. For m , n 2 1,
It is only a slight exaggeration to say that Theorem 8 is the basis of almost everything else we are going to do.
The Dirichlet Problem
45
Proof. (a) Integration by parts yields
lo 1
=
This clearly vanishes for m
=
gmgn[gm gnl dx.
n. If m # n, then p m # p,, and we can use
[gm, gnl’ = ( p m - punlgmgn to obtain
(b) Again, by partial integration,
If rn
=
n, then y2 is a multiple of g,, hence [ y z ,g,]
=
1,
using the Wronskian identity. (c) is proven in the same manner.
=
0, and we get
Inverse Spectral Theory
46
For q
=
0, the vectors 1,
n L 1, are a basis for L2[0,11. Theorem 8 and the asymptotic estimates of Corollary 1 make it possible to prove the same statement for every q . First, let us specify the notions of linear independence and basis. A sequence of vectors u1 , u 2 , ... in a Hilbert space is linearly independent, if none of the vectors un , n 1 1, is contained in the closed linear span of all the other vectors urn, m # n. One might try to define a stronger notion of linear independence by requiring U n to lie outside the weak closure of the span of urn, rn # n. However, if a sequence of vectors converges weakly, then the arithmetic means of some subsequence converge strongly to the same limit by the theorem of Banach-Saks [Ba]. Thus, the weak and strong closure of a linear space are the same, and nothing is gained. A sequence of vectors u l , u 2 , ... is a basis for a Hilbert space H, if there , ...) such that the correexists a Hilbert space A of sequences a = ( a ] a2, spondence a+
C
n z1
anvn
is a linear isomorphism between A and H.4 Theorem 9. A t every point in L2, the vectors 2
1,g, - 1,
n
2 1
are linearly independent. So are the vectors
The two sequences are mutually perpendicular, and together are a basis for L' . Precisely, ((,v)+
d
C t n z d + V O + nCa 1 Vn(g,'
n z1
is a linear isomorphism between pf x R x
P
- 1)
and L2.
4 A linear isomorphism between Banach spaces is an isomorphism between vector spaces, which is continuous and has a continuous inverse.
41
The Dirichlet Problem
It follows from Theorem 6 that at an even point in L2,the vectors 1, g,' - 1, n 2 1, are a basis for the subspace of even functions in L2, and the vectors (d/dx)g,', n 1 1, are a basis for the subspace of odd functions.
Proof. The vector g,' - 1, n 2 1, is not in the closed linear span of the vectors 1, g& - 1, m # n, since
but
by Theorem 8. Similar arguments apply to all other vectors. The two sequences are mutually perpendicular, since
for all m ,n again by Theorem 8. By Corollary 1, 2
gn - 1 =
- ~ 0 ~ 2 n n+x 0
1=1 1 d -g,' = sin27rnx 27rn dx
+o
We have shown that the vectors on the left are linearly independent. The vectors on the right without the error terms are (up to an irrelevant factor an orthonormal basis of L2, and the error terms are square summable. Therefore, Theorem D.3 applies. It follows that the vectors on the left are a basis with coefficients in x R x ?. From this, the last statement of the theorem follows.
a)
It is useful to write Theorem 9 in a slightly different form.
Inverse Spectral Theory
48
Corollary 3. A t every point q in L2, the subspaces
are perpendicular and closed. Their direct sum is all of L2.In particular, they are the odd and even subspaces respectively when q is even.
Problem 5. (a) Show that log N Conclude that 1 is in the closed linear span of gn', n L 1. (b) Show that the vectors gn', (d/dx)gn',n 1 1, span L2 (that is, finite linear combinations are dense) and are linearly independent, but not a basis. The basic facts about the Dirichlet eigenvalues and eigenfunctions are now proven. We are ready to approach the inverse Dirichlet problem.