makes this a non{trivial problem, is the repulsion between electrons. The nu- clear kinetic ... In Chapter 7 we introduce a method which is essentially ... In particular, it was shown in Fe] that nuclei and electrons arrange themselves into atoms and ...... (5:11). Note that is increasing in . Lemma 5.5: Let u be any solution of u. 00.
LOWER BOUNDS FOR THE GROUND STATE ENERGY OF ATOMS
Luis A. Seco
A DISSERTATION PRESENTED TO THE FACULTY OF PRINCETON UNIVERSITY IN CANDIDACY FOR THE DEGREE OF DOCTOR OF PHILOSOPHY RECOMMENDED FOR ACCEPTANCE BY THE DEPARTMENT OF MATHEMATICS
October, 1989
Abstract The ground state energy of an atom is de ned to be lowest possible value of the energy Hamiltonian. This work describes an algorithm to obtain lower bounds for the ground state energy of atoms, together with its implementation. The results are within a few precent of the upper bounds given by Hartree{Fock's method.
1
Acknowledgements: It is a great pleasure to express my deep gratitude to my advisor Charles Fe erman; his wealth of ideas, inspiration and friendship made this work possible and fun; his inexhaustible patience and superb instruction are largely responsible for whatever worthwhile there may be in my mathematical education. I am very grateful to Rafael de la Llave, for innumerable conversations, numerous ideas, for showing me around, and for providing me with computer{assisted{ anything I wished. I want to express my special gratitude to my former teachers, A. Cordoba and J. C. Peral for their help and support. I want to thank also E. Lieb, J. P. Solovej and T. Spencer for useful conversations, and to D. Rana, for providing me with his interval arithmetic package. Last, but not least, I am specially grateful to Maite, for doing all the computations in Chapter 3, and too many other things to write here. If anyone nds anything worthwhile in this work, it is certainly due to some of the above.
2
Contents Abstract 1. The History : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 2. The Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9 3. The Potential
3.1 The De nition : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 15 3.2 The Formulas : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 19
4. The Computer
4.1 Interval Arithmetic : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28 4.2 Functional Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 30
5. The ODE
5.1 Initial Value Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 32 5.2 Eigenvalues : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 38 6. The Numerics : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 59
7. The Fuzz 7.1 7.2 7.3 7.4 7.5 7.6
The Set-up : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 68 Notation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 71 Pointwise Estimates : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 73 Fuzzed Estimates : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 81 Real Variables : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 89 The Implementation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 96 8. The Ball Packing : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 107 References : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114
3
Computer Programs 1. Numerical Analysis
1.1 Documentation 1.2 Listing
2. Rigorous Veri cation
2.1 Documentation 2.2 Listing
4
Chapter 1: The History Quantum Mechanics was introduced with the goal of understanding the atom. probably its rst success was to compute accurately the ground state energy of Hydrogen. However, it was realized very early that to do the same for the other atoms was no easy task. The formulation of the problem is as follows: The operator Z X
X HZ = ? jxZ j + 12 jx ?1 x j i j i=1 i6=j i is called the Hamiltonian for the atom of charge Z ; it acts on functions which ? are in L2 R3Z with some antisymmetry condition. The energy, E (Z ), corre? 21 xi
sponds to the in mum of its spectrum. The term ?i represents the kinetic energy of each atom, the term ?Z=jxi j represents the attraction each electron P feels towards the nucleus, and the term 21 i6=j jxi?1 xj j , which is the one that makes this a non{trivial problem, is the repulsion between electrons. The nuclear kinetic energy is not included in the formulation of the problem; this is a reasonable approximation, since the nucleus is very heavy. Some qualitative properties associated with the atomic Hamiltonian, such as essential self{adjointness (i.e. physical signi cance), qualitative properties of the spectrum, regularity results for the bound states (i.e. eigenfunctions), etc. we 5
soon established. See [K1], [K2] and [K3], for example. At the same time, a more computational motivation lead others to tackle the problem of computing atomic energies. The rst important result in this direction was the creation of Thomas-Fermi (TF) theory. They attempted to understand atoms using a di erent atomic model in which the object of study was not the wave function, but the electric density; this simpli es the problem in the sense that the nuclear charge does not enter as the dimension of the problem, but rather as a parameter in a one-dimensional problem. The main results according to this theory are: 1. ETF (Z ) = CTF Z = 7
3
2. Most of the electrons are at distance Z ? = from the nucleus. 1
3
3. The outermost electron (chemistry) is at a distance of the order of 1. 4. Atoms are neutral. 5. Molecules do not exist. It was clear empirically that Thomas-Fermi energy was quite accurate for large atoms (although not entirely satisfactory) and not so good for small ones, and with the exception of item 5, the other properties look obvious in nature. But a rigorous analysis had to wait until 1977, when it was proved that atoms behave as Thomas-Fermi theory predicts in the limit Z ! 1 (see [Li] and [LS]). Other people also considered the problem of atomic energies for large Z (in the limit as Z ! 1). Scott predicted a correction for Thomas-Fermi energy that recently has been proved correct (see [Hu] and [SW]), and other atomic models due to Dirac, von Weizsacker and others were created, in an attempt to capture certain aspects of the real problem neglected by TF theory. Also, with the creation of computers, other purely computational techniques were 6
introduced. A specially accurate method to compute upper bounds to atomic energies was developed by Hartree, Fock, Slater and others (although the rst Hartree papers were done by hand computation!). Their method consists in nding the ground{state energy for the same Hamiltonian restricted to a smaller class of functions (this gives upper bounds), namely the antisymmetric product of one{electron functions. This method leads to a collection of ODE's that can be solved numerically. It is believed that the results obtained by this method are very good for all values of Z . The present work will actually give a bound for the error in their approximation. Another method to compute both upper and lower bounds was introduced by A. Weinstein and his collaborators, the so{called method of \Intermediate Hamiltonians". It is basically a regularity result in which a particular orthogonal family of functions is constructed and it is proved that the ground{state eigenfunction has Fourier coecients that decrease very fast with respect to this orthogonal family; this allows to reduce the problem to a nite dimensional one, with a good control on the error. This method works well for small atoms and molecules. In particular, very good upper and lower bounds are known for the energy of Helium. See [W e]. Up to this point, an open problem is to produce good methods to compute lower bounds to atomic energies for atoms. An important result was obtained Hertel, Lieb and Thirring in [HLT]. In this work, we will present another approach to this problem that improves their bounds. There is one common feature in all these theories to approximate atomic energies: they are basically reductions to one{electron problems, which are much easier in general than many electron problems, even if you loose linearity or other nice features along the way. In Chapter 7 we introduce a method which is essentially many{electron analysis.
7
It is interesting to point out that good bounds for atomic energies are important not only quantitatively, but also because certain qualitative aspects of ordinary matter, such as binding and no-binding theorems, stability of matter, neutrality of atoms, etc., depend critically on whether the energy of atoms is within a certain range or not. See [Fe], [FS1] and [FS2] and [SSS] for examples of such results. In particular, it was shown in [Fe] that nuclei and electrons arrange themselves into atoms and molecules provided that the ground state energy of atoms is within a factor 2 |roughly speaking| of the constant in the inequality of the stability of matter (see [DL], [LT] and [Ll]). We also refer to [Fe] for a detailed description of some of these results.
8
Chapter 2: The Problem Consider the Hamiltonian for an atom of charge Z with Z electrons: HZ =
acting on functions
Z X i=1
? 12 xi
X ? jxZ j + 12 jx ?1 x j i j i6=j i ?
2 H = ^Zi=1 L2 (RZ ) C2
(2:1)
with \^" denoting antisymmetric product. The ground state energy is then de ned to be the in mum of the spectrum of HZ : E (Z ) = inf = inf 2H hHZ ; i 2Spec(H ) jj jj2 =1
It is known that this in mum is achieved, and the functions that give this in mum value are usually called \ground states". Upper bounds can be obtained by using trial functions. Lower bounds for E will be found as follows: If we have an inequality of the form: Z 1X 1 X 2 i6=j jxi ? xj j i=1 V (xi ) ? C
9
(2:2)
with V radially symmetric decreasing function vanishing at in nity, we have an operator inequality HZ
Z X i=1
? 12 xi
? jxZ j + V (x) i
? C def = HLB
Let now 1 < 2 < < 0 be the negative eigenvalues of
?
H 1 electron = ? 12 x ?
Z jxj + V (x)
acting on L2 R3 . Then, the eigenvalues of HLB =
Z X i=1
? 12 xi
? jxZ j + V (x) i
are then given by sums of the type i1 + i2 + + iZ
were the eigenfunctions are given by the tensor products of the corresponding eigenfunctions of each ik . The antisymmetry condition (2:1) implies that no more than two of these ik can be equal. Therefore,
8 Z=2 > X > i ? C 2 > > < i=1 E > X ? > > : 2 i=1 i +
Z even
Z 1 2
Z +1 2
? C Z odd
The way to determine the eigenvalues of H 1 electron is standard, and follows from decomposition into spherical harmonics; as a result, the eigenvalues of H 1 electron are the same as the eigenvalues of the ODE operator 10
? 12 u00 ? acting on
Z
l(l + 1) u? r 2r2 ? V (r) u
Hode = u : r?1 u(r) 2 L2(0; 1)
for l = 0; 1; 2; :::. Every eigenvalue of (2:3) corresponds to an eigenvalue of H 1 electron with multiplicity 2l + 1. Note that if u is a solution of (2:3), then u 2 Hode i u(0) = u(1) = 0. From now on, we will consider the ODE problem associated with
?u00 ? 2rZ u ? l(l r+2 1) ? 2V (r) u
(2:3)
and we will remember that there is a factor 2 a ecting the results. These eigenvalues can be worked out explicitly if V = 0, and they are given by the formula Z2 n = ? 2(n + l)2
(2:4)
but for general V the ODE cannot be solved explicitly. We will, still, however be able to estimate accurately its eigenvalues. For this we will use computer assisted techniques. This will be done roughly as follows: First, we will see how to obtain rigorous upper and lower bounds for the Taylor coecients of our function V , up to any nite order, as well as upper bounds for the high order terms. Next, we will use these upper and lower bounds (that we will refer simply as \intervals") to obtain upper and lower bounds (intervals) for the Taylor coecients of the solution of the ODE, and this, at least in principle, is trivial, by matching powers. This will give equations for the Taylor coecients of the solution |up to some nite order| in terms of those of the equation, and will also provide bounds the high order terms. 11
Thus, to obtain bounds for the solution of initial value problems is reduced to a nite number of manipulations with intervals. Chapter 4 explains how this can be done using the capabilities of a computer. To go from here to estimating the eigenvalues we will use comparison theorems for Sturm-Liouville problems. The next issue is is how to obtain good lower bounds for the energy of atoms with this method: First, it is important to obtain good functions V in inequality (2:2). Some care will be taken to optimize the choice of V . Next, for every choice of a function V , we will obtain a constant C , but it will not be optimal. It is also important to improve the inequality by choosing C as small as possible. This will lead to the problem of ball packings in R3. Finally, observe that what we really needed in the original problem is not the strong assertion Z 1X 1 X 2 i6=j jxi ? xj j i=1 V (xi ) ? C
but only
* X 1
1
2 i6=j jxi ? xj j
;
+ X Z
i=1
V (xi ) ;
?C
for equal to the ground state of the atom. Note that, even if Z X 1 X 1 E (x1 ; : : : ; xZ ) = 2 i6=j jxi ? xj j ? i=1 V (xi ) + C
is smallest possible pointwise, hE ; i might still be big. 12
(2:5)
Obviously, if E = 0 at some point, then there is no hope that hE ; i can be bounded from below with something strictly positive for all , since you can take more and more concentrated around that point. However, in this problem, this would cause the laplacian term to grow very much (roughly as the square of the inverse of the diameter of the support); this means that the that play a role are to some extent \smeared out" and cannot concentrate too much on the points were (2:2) is sharp; this argument will imply that hE ; i is in fact positive. As a consequence, if E~ (Z ) is the lower bound obtained using inequalities of type (2:2), then E (Z ) E~ (Z ) + hE ; i (2:6) for the ground state of the atom, which is a strictly better lower bound. Chapter 3 will be devoted to obtain all formulas involved in the calculation of V and C in (2:2). Chapter 4 will explain how the computer can be used to prove certain rigorous mathematical statements related to out problem. Chapter 5 contains the necessary results in theory of Ordinary Di erential Equations to reduce the sharp estimates for the eigenvalues of the ODE problem (2:3) to a nite number of computations. Chapter 6 discusses several techniques used to obtain good lower bounds for the energy, and lists the actual results that were obtained. Chapter 7 shows how the lower bounds can be obtained by leaving the realm of pointwise inequalities of type (2:2) and using inequalities of type (2:6) in two di erent ways: rst, by obtaining an a priori constant lower bounds for hE ; i and, second, by obtaining a lower bound as a sum of one particle potentials. Finally, Chapter 8 will take care of improving the constant C in (2:2). This analysis leads to ideas coming from the problem of ball packings in R3. 13
An announcement of our results appeared in [Se].
14
Chapter 3: The Potential
3.1 The De nition The method to obtain inequalities (2:2) has its origin in the elementary formula rst used in [FL]:
Z Z dz dR 1 B(z;R) (x) B(z;R) (y ) 5 = jx ? yj R>0 z2R R 1
3
that implies
dzdR 1X 1 = 1 Z Z N (N ? 1) 5 2 i6=j jxi ? xj j 2 R>0 z2R R 3
where N is the number of xi that belong to B (z; R): N = N (x1 ; : : : ; xZ ; z; R) =
Observe that
Z X i=1
B(z;R) (xi )
N (N ? 1) = (N ? N )2 + (2N ? 1)N ? N 2
15
for any number N = N (z; R); and, since (N ? N )2 0 we get 1 X 1 1 ZZ (2N ? 1)N dz dR ? 1 ZZ N 2 dz dR 2 i6=j jxi ? xj j 2 R5 2 R5
ZZ dz dR Z ZZ X dz dR 1 1 (2N ? 1) xi 2B(z;R) R5 ? 2 N 2 5 = 2 R i=1
=
Z X i=1
V (xi ) ? C:
with
ZZ 1 (z; R) ? 1) dz dR V (xi ) = 2( N 2 jz?R>xj RZ?1(z)
In practice we will take Ri (z) to be piecewise linear, for that closely approximates the functions (3:1). Observe that this does not a ect the correctness of our estimates. The following Lemma proves useful
Lemma 3.1: f (jxj), we have
Let (x) be a radial function de ned on R3 ; then, if (x) =
Z R?x R2 ? (x ? y )2 dy + 4 y 2 f (y ) dy (x) dx = yf (y ) x 0 B(z;R) jR?xj where jzj = x, and the second integral does not appear if x > R. Z
Z R+x
Corollary:
8 0 > > > 4 > > < 3 M3 jB (0; M ) \ B (x; R)j = > 4 > R3 > 3 > > : H (x; R; M ) 17
if M < jxj ? R if R > M + jxj if M > R + jxj otherwise
for
1 H (x; R; M ) = ? (M 2 ? R2 )2 + 4 R3 + M 3 ? jxj R2 + M 2 + 1 jxj3 4x 3 2 2 12
(3:2)
The following Lemma lists some of the properties of the functions Ri .
Lemma 3.2a:
Let Ri be de ned as in (3:1), for integrable, radially symmetric, decreasing function. Thus, Ri will take values in R1 from now on. 1. Ri is a radially symmetric, continuous increasing function. 2. If we de ne bi to be
Z 1Z 1 Z 1 bi
Then
3. Ri(x) x + Ri (0).
?1 ?1
(x1 ; x2 ; x3 ) dx1 dx2 dx3 = i
lim R (x) ? (x ? bi ) = 0 jxj!1 i
It will also be useful to consider the inverse functions
?1 Mi (R) = Ri (R) if R Ri (0) 0 if R < Ri (0)
The properties for Ri then translate into properties for Mi as follows.
Lemma 3.2b:
Let Mi be de ned as before. 1. Mi is a radially symmetric, continuous increasing function. 2. limjxj!1 Mi (x) ? (x + bi ) = 0 3. Mi(R) R ? Ri (0) for all R.
18
(3:3)
Note that the freedom in choosing N translates in the freedom to choose any set of functions Ri , or Mi . This will be of importance later.
3.2 The Formulas Given functions Mi (R), we will describe all the calculations involved in the computation of V (x) and C :
ZZ ?2N (z; R) ? 1 dz dR 1 V (x) = 2 jz?xj R (z) 21 2
and call
Clearly,
ZZ dz dR 1 (2 Ni (z; R) ? 1) 5 Vi (x) = 2ZZ R dz dR = 1 R5 R>Ri (z)
V (x) =
since
i
if R < Ri (z)
N (z; R) =
ZX ?1 i=1 ZX ?1 i=1
Vi (x) Ni (z; R)
If we do the same with the constant C , then C=
ZX ?1 i=1
19
i Ci
for
ZZ ? dz dR 1 Ni2 ? 41 Ci = 2ZZ R5 dz dR = 1 R5 R>Ri (z)
In order to see this, we have to check that ZX ?1 j =1
? j Nj2 ? 41 = Ni2 ? 41
or, equivalently ZX ?1 j =1
Since we have
?
j Nj2 ? 14 = i2 + i
Nj2 ? 41 ZX ?1 j =1
for Ri(z) < R < Ri+1(z)
0 for j i + 1 = 2 for j i
Xi ? 2 1 j N ? = 2 j = i2 + i j
4
j =1
as claimed Thus, it is enough to calculate Vi (x) and Ci . Let's put Mi (R) = M (R). Observe that
2 if jzj < M (R) 2Ni ? 1 =
0 if jzj M (R)
therefore, 20
ZZ dz dR 1 2 Vi (x) = j z j jxj jxj+R x x +R>M (R) R x x R x >M (R)
4 M 3 (R) dR 3 R5
where H is as in (3:2). At this point, a diculty appears when trying to solve the equations
jxj + R = M (R);
R ? jxj = M (R);
jxj ? R = M (R)
(3:4)
For this reason, since we are free to choose M as we please, we will take functions Mi (R) that are piecewise linear, i.e.: Pick numbers ai , bi , n for 0 i n, such that Ri (0) = r0 < r1 < < rn a0 > a 1 > > a n = 1 b0 < b 1 < < b n
(3:5)
and take M (R) = ai R + bi for ri < R < ri+1 . We will limit our attention to convex piecewise linear Mi , to ensure that equations (3:4) have unique solutions, which simpli es the problem. Call jxj simply x. Let's consider two cases: x < Ri (0) and x > Ri (0).
case I: x < Ri(0) The straight line R ? x will cross M (R) at a point R where ri0 R ri0 +1
21
In this case, R =
x + b i0 1 ? ai0
Similarly, R + x will cross M (R) for R = ?1x?+abii for ri R < ri +1. Therefore, we have: 1
1
Vi (x) =
Z
x+bi0 a i0
?
1
1
1
4 M 3(R) dR + Z ?ai H (x; R; M (R)) dR + Z 1 4 R3 dR ?x bi 3 R5 x bi 3 R5 R5 ?ai ?ai ?x+bi1 1
1
+ + 1 0 Ri (0) 1 1 0 1 Z 1x?+abii00 4 iX 0 ?1 Z rk+1 4 dR dR = ( a k R + b k )3 5 + ( ai0 R + bi0 )3 5 R 3 R ri0 k=0 Zrk 3 ri0 +1 + x+bi0 H (x; R; ai0 R + bi0 ) dR R5
?ai0
1
+ +
iX 1 ?1
Zr
k+1
k=i0 +1 rk
Z
?x+bi1 1?ai 1
ri1
H (x; R; ak R + bk )
dR R5
Z 1 4 dR dR H (x; R; ai1 R + bi1 ) 5 + ?x+bi R3 5 : 1 3 R R 1?ai 1
Now, de ne functions:
4 a3 a2b 4 ab2 1 b3 Z 4 dR 3 G1 (a; b; R) = (aR + b) R5 = ? 3 R + 2 R2 + 3 R3 + 3 R4 : 3 Z 4 dr ?4 G2 (R) =
G3 (a; b; x; R) =
Z
3 R2 = 3R : dR H (x; R; aR + b) 5 : R
All integrals are taken to be inde nite integrals. So, 22
Vi (x) =
iX 0 ?1h k=0
G1 (ak ; bk ; rk+1 ) ? G1 (ak ; bk ; rk )
+ G1 ai0 ; bi0 ; 1x ?+ abi0 i0
i
? G1 (ai ; bi ; ri )
0
0
0
+ G3 (ai0 ; bi0 ; x; ri0+1 ) ? G3 ai0 ; bi0 ; x; 1x ?+ abi0 i0 iX ? 1 1 h i + G3 (ak ; bk ; x; rk+1 ) ? G3 (ak ; bk ; x; rk ) k=i0 +1
?x + b ? x + bi ? G3 (ai ; bi ; x; ri ) ? G2 1 ? a i : + G3 ai ; bi ; x; 1 ? a i i
1
1
1
1
1
1
1
1
1
Since all these functions are always evaluated at points of the form x + , the functions we really need are F1 (a; b; ; ; x) = G1 (a; b; x + ) 4 a3 a2 b 4 ab2 1 b3 = ? 3 ( x + ) + 2 ( x + )2 + 3 ( x + )3 + 3 ( x + )4 d d2 d3 d4 1 = ( x + ) + ( x + )2 + ( x + )3 + ( x + )4 : (3:6) F2 ( ; ; x) = G2 ( x + ) =
?4
3( x + ) :
F3 (a; b; ; ; x) = G3 (a; b; x; x + ) p1 (x) = ( xp4+(x ) )4 + ( xp3+(x ) )3 + ( xp2+(x ) )2 + ( x + ) + p0(x) log( x + ):
where 23
(3:7)
(3:8)
p0 (x) =
?(a2 ? 1)2
(3:9a)
4x
ab(a2 ? 1) 2 ? 3 (1 + a3) x b2 (1 ? 3a2 ) 2 (1 + a2 ) p2 (x) = ?4x ? a b + 4 x ab3 2 2 ab p3 (x) = 3x ? 3 ab + 3 x b4 b3 b2 1 p4 (x) = ? + x ? x3 16x 6 8 48
(3:9b)
p1 (x) =
(3:9c) (3:9d) (3:9e)
So, Vi (x) =
iX 0 ?1h k=0
F1 (ak ; bk ; 0; rk+1; x) ? F1 (ak ; bk ; 0; rk ; x)
i
+ F1 ai ; bi ; 1 ?1a ; 1 ?bi a ; x i i ? F1 (ai ; bi ; 0; ri ; x) + F3 (ai ; bi ; 0; ri +1 ; x) ? F3 ai ; bi ; 1 ?1a ; 1 ?bi a ; x i i 0
0
0
0
0
+
iX 1 ?1
h
k=i0 +1
0
0
0
0
0
0
0
0
0
0
0
F3 (ak ; bk ; 0; rk+1 ; x) ? F3 (ak ; bk ; 0; rk ; x)
i
bi ? 1 + F3 ai ; bi ; 1 ? a ; 1 ? a ; x ? F3 (ai ; bi ; 0; ri ; x) ?1 b i i (3:10) ? F2 1 ? a ; 1 ?i a ; x : i i
1
1
1
1
1
1
1
1
1
1
1
Given a; b; ; all three functions F1 ; F2 and F3 are analytic functions in x in some xed common neighborhood of some x0. All singularities come from terms of the form x( x1+ )k or log( x + ); therefore the radius of convergence of Fi around x0 is min x; x + : 24
The values for are either bi or ?bi ; thus, if either bi or bi are very close to x, the radius of convergence of V will be very small; this will happen if the piecewise linear pieces M (R) is made of, have slopes very close to 1. 0
1
0
1
case 2: x > Ri(0) As in the previous case, consider the intersection of M (R) with x ? R and R + x. The crossing points are respectively x ? bi ?x + bi y0 = and y1 = 1 + ai 1 ? ai We thus have 0
1
0
1
Z 1 4 dR dR R3 5 Vi (x) = H (x; R; M (R)) 5 + R 3 R y1 (x) y0 (x) Z ri0+1 1 ?1 Z rk +1 dR iX dR = H (x; R; M (R)) 5 + H (x; R; M (R)) 5 R k=i0 +1 rk R y0 (x) Z 1 4 dR Z y1(x) dR R3 5 + H (x; R; M (R)) 5 + R y1 (x) 3 R ri1 ? bi0 1 = ?F3 ai0 ; bi0 ; 1 + a ; 1 + a ; x) + F3 (ai0 ; bi0 ; 0; ri0 ; x i0 i0 iX ? 1 1 h i + F3 (ak ; bk ; 0; rk+1 ; x) ? F3 (ak ; bk ; 0; rk ; x) Z y (x) 1
k=i0 +1 ? F3 (ai1 ; bi1 ; 0; ri1 ; x)
bi ? 1 + F3 ai ; bi ; 1 ? a ; 1 ? a ; x ?1 b i i ? F2 1 ? a ; 1 ?i a ; x : i i 1
1
1
1
1
1
1
1
(5:15)
The considerations for the radius of convergence in this case are similar to those of the previous section. We point out that all the radii of convergence of V are bounded below by R(0) min a ? 1 ; x n?1 25
because x + , being the intersection point of M (R) and R + x, has to be at least R(0), and is at most an?1 ?1 , since fai g is a decreasing sequence. However, the singularity at 0 is unavoidable, and this is inconvenient. What we will do is pick a small number, , and for x < just set V (x) = V (); since V is decreasing, this will still give a lower bound to the correct potential; moreover, it is not wasteful, for two reasons: rst, V 0 (0) = 0, therefore V is close to a constant around 0; second, since the solution of the ODE vanishes at 0, rst order perturbation theory says that the e ect of the potential around 0 is not important. 1
Therefore, the ODE (2:3) takes the form a l(l + 1) 00 u + + +c u=0 r
r2
(3:12)
around zero Also, if x > 2rn + bn then
1 ? 1 + 4 3 x x ? bn and so, we may take (since we are only looking for lower bounds) 3x ? b c Vi (x) = for c = min 1; 3(x ? bn ) x n and in particular, given > 0 and l 0, we can nd a positive integer V (x) =
k = k() =
Z ? P c
i +1
(where [ ] denotes \largest integer smaller than"), such that X Z ? k l(l + 1) ? 2r2 Vi (x) r i
(3:13)
(3:14) (3:15)
This will be useful in order to bound our ODE (2:3) by means of the easier to study ODE's 2k 00 u + (3:16) ? 2 u = 0 r 26
and at in nity.
2(k + 1) ? 2 u = 0 u00 + r
This completes the description of Vi (x). The upshot is that for every x0 > 0 we can write the potential
x ? x n N X 0 V (x) = an +H r
n=0
where r is the radius of convergence and is bounded below. H is a function with small norm in a sense to be made precise later; it contains all the high order terms. The values of an and estimates for j H j can be obtained from x0 by a nite computation. Vi , as de ned by this procedure, is a piecewise analytic function; that is: there are points fxi g (where in fact each xi is of the form either ri + M (ri ) or ri ? M (ri ), such that Vi has a power series expansion around every point 0 < x < 1, and it agrees with this power series except at the xi , where it only agrees to the left of xi . Vi is continuous everywhere, and has continuous rst derivatives except at the smallest and largest of the fxi g.
The fact that V has so many singularities has as a consequence that a regularity theory analysis of the ODE (2:3) (for example, reducing the problem to a nite dimensional one via a clever orthonormal family) will not be appropriate. Also, the fact that V introduces a perturbation which is unbounded as Z grows implies that you can expect trouble if you try to analyse the ODE as a perturbation of | for instance| the Hydrogen atom (V 0). It seems then that the ODE problem has to be dealt with using the powerful machinery of computing the solution of the ODE everywhere with very good bounds. 27
Chapter 4: The Computer In this chapter, we explain the basic techniques used in proving rigorous results with a computer. All of this is becoming quite standard. For similar and more detailed analysis see [Mo], [KM], [EW], [EKW] and [Ra].
4.1 Arithmetic Let R be the set of \representable numbers" in a computer, that is those numbers that the computer can represent exactly. Depending on the speci c machine, they are usually real numbers with some nite binary expansion. It is well known that computers can only perform arithmetic in an approximate way: the addition |for example| of two representable numbers is another representable number that will probably be close to the true sum, but is not exactly the true sum. The idea to perform rigorous arithmetic is to instruct the computer on how to produce upper and lower bounds to the true results of arithmetic operations between representable numbers; in other words, we work with intervals with endpoints in R, and we implement arithmetic operations on intervals in such a way that given two intervals, the computer will produce a third that is guaranteed 28
to contain the result of all arithmetic operations between points in the initial intervals. This is usually called \interval arithmetic". As an example, let's consider one possible interval arithmetic implementation of the sum. This is fact the one that was used in our programs. The computer manufacturer provides two operations on representables, a1 and a2 with the property that
8r; s 2 Ra1 (r; s) r + s a2(r; s) Then, given two intervals, I1 = [r1; s1] and I2 = [r2 ; s2], the sum of these two intervals is de ned to be I3 = [a1 (r1 ; r2); a2 (s1 ; s2 )]
It is then clear that for all real numbers x 2 I1 and y 2 I2 we have that x + y 2 I3 Another possible implementation, that was also occasionally used in our programs, is as follows: We construct two functions on representables up and dn, with the property that up (r ) sinf fs > rg 2R dn (r ) sinf fs < rg 2R
The computer manufacturer again provides us with a function a, which is in fact
oating point addition for the computer we used, with the property that a(r; s) is the closest representable number to r + s. This implies that, again, given two intervals, I1 = [r1; s1] and I2 = [r2; s2], if we de ne the sum of these two intervals to be I3 = [r3 ; s3)] with r3 = dn (a(r1; r2)) and s3 = up (a(s1; s2)) then for all real numbers x 2 I1 and y 2 I2 we have that x + y 2 I3
29
4.2 Functional Analysis A convenient Banach space to use in this theory is the space of piecewise analytic functions, with a lower bound on the size of the domains of analyticity. The purpose of this section is to formalize de nitions and set up the framework for computer assisted analysis in function space. Consider the Banach space
(
H 1 = f (z )jf (z ) =
1 X n=0
with norm
jfj =
an z n ;
1 X n=0
1 X n=0
jan j < 1
)
jan j
This is a subspace of the set of analytic functions in the unit disk. It becomes a Banach Algebra with j j . We consider a neighborhood basis for the topology induced by j j consisting of sets U (I1; : : : ; IN ; C ) of the form
f (z ) =
1 X n=0
an z n an 2 In ; 0 n N;
1 X n=N +1
jan j C
(4:1)
where C is a positive real number and In are intervals in the real line. For the computer implementation, C will run over the set of computer-representable numbers, and the intervals will be those with representable endpoints. The reason why this is a convenient space to work in is because elementary operations, such as addition, product, integration, di erentiation (composed with a slightly contracting dilation), evaluation at a point and integration of initial value problems in ordinary di erential equations can be conveniently bounded by elementary formulas in terms of this set of neighborhoods. 30
Observe that if we have a function f (z) which is analytic in some disk, jz ?z0 j < r, then, for any r~ < r, it is clear that if we de ne the normalized version of f f~(~z ) = f (
z~ ? z0 ) r~
then f~ 2 H 1. In the real analytic case, H 1 [a; b] will denote H 1 of the disc with center a and radius jb ? aj. In the previous section, we saw that we will have to deal with functions with are sums of rational expressions. It is possible to produce neighborhoods of type (4:1) that contain those functions locally.
31
Chapter 5: The ODE In this chapter, we will see how our ODE problem (2:3) can be dealt with the functional analysis introduced in the previous chapter, and therefore, how one can obtain computer assisted results about ODE's. The discussion will be divided into two parts: the solution of initial value problems and the localization of eigenvalues. The presentation is taylored to deal with our special problem, but it can be modi ed, at the expense of complication, to deal with more general problems.
5.1 Initial Value Problems Lemma 5:1 below takes care of the solutions of an IVP with analytic coecients. Lemmas 5:2 and 5:3 take care of the expansion of the solutions at the singularities of the ODE, around 0 and 1.
Lemma 5.1:
Consider the ODE: 9 u00 + qu = 0 > = u(0) = u0 > u0 (0) = u1 ;
32
(5:1)
where q(x) 2 U (q0 ; ; qN ; ). Then, u 2 U (u0 ; ; uN +2 ; C ) where n X 1 un+2 = ? (n + 2)(n + 1) i=0 ui qn?i
0nN
(5:2)
and NX +2n
C
i=0
N X
jqk j
o
jqk j
!
jui j + ( k + i + 2)( k + i + 1) ( N + 3 + i )( N + 2 + i ) k=N +1?i
1?
X N
+ k=0 (k + N + 5)(k + N + 4) (2N + 6)(2N + 5)
(5:3)
+
Operations are performed in the set{theoretic sense.
Proof: For simplicity in the formulas, put q?1 , q?2 , = 0.
The rst part can be obtained by matching powers. For the estimate for C , note that
X n>N +2
jun j 33
n XX jui qn?i j
n>N i=0 (n + 2)(n + 1) +2 n X NX X j ui qn?ij j ui qn?i j = + ( n + 2)( n + 1) n>N i=0 i>N +2 (n + 2)(n + 1) NX +2 X jui qn?i j + X X juiqn?i j = i=0 n>N (n + 2)(n + 1) i>N +2 ni (n + 2)(n + 1) NX +2 1 NX +2 iX +N X jqn?i j j qn?i j + j ui j = jui j i=1 n=i+N +1 (n + 2)(n + 1) n=N +1 (n + 2)(n + 1) i=0
+ =
NX +2
i>N +2
jui j
jui j N X
X
jqn?i j
ni (n + 2)(n + 1)
jqk j
k=N +1?i (k + i + 2)(k + i + 1) 1 NX +2 X jqk j + jui j i=0 k=N +1 (k + i + 2)(k + i + 1)
i=1
+
X
X
jui j
i>N +2 NX +2n X N i=0
+
X
jqn?i j
ni (n + 2)(n + 1)
o
jqk j
jui j + ( k + i + 2)( k + i + 1) ( N + 3 + i )( N + 2 + i ) k=N +1?i
X N
jqk j
k=0 (k + N + 5)(k + N + 4)
This yields the result.
+ (2N + 6)(2N + 5)
X
i>N +2
jui j
QED
Note that if q is analytic in a disk D(z; R) other that the unit disk, it is possible to apply the lemma to its normalized version, q~, with initial conditions u0 and R u1 , and the solution obtained is, then, u~, the normalized version of the true solution u. Also note that by doing this, the coecients qi are multiplied by a 34
factor shrinking with the radius, thus improving the estimate for C ; in particular, note that the denominator in the expression for C , equation (5:3), may very well be negative (in which case the lemma does not claim any result, although one clearly exists); however, this denominator becomes positive and if fact goes to one when you take smaller and smaller radius for q. In other words, it is unavoidable (in this formulation) to take small steps when solving the ODE. We point out, for the sake of completeness, an alternative argument to solve the problem in Lemma 5.1: A solution to ODE (5:1) is characterized as the xed point in H 1 of the functional
T [f ](x) = u0 +
Z x 0
u1 ?
Zt 0
q (s) f (s) ds dt
By using the recursion formula (5:2) for the coecients of u up to any nite order we want, we can obtain a polynomial u] that we believe to be close to the true solution u. It is possible to compute a neighborhood of type (4:1) containing T [u] ], and, therefore, we can compute an upper bound for T [u] ] ? u] H , that if our guess was good, will be small. Also, it is elementary to compute upper bounds to the Lipschitz norm of T , j T j . All this upper bounds have to be rigorous; they can be implemented using interval arithmetic. 1
1
Finally, the contraction theorem in Banach Spaces tells us that that if j T j < 1, then there is a xed point u of T , and thus a solution of ODE (5:1), and moreover (this is the important thing) 1
u] ? u T [u]] ? u] H 1 ? jT j 1
(5:4)
1
and this gives us very good control over the solution u. Note the close resemblance of this estimate with estimate (5:3). However, there is one di erence which is that in our approach, all the approximation error goes into the \high order terms", 35
whereas in this method, it goes into what is called \general error" (we don't know which Taylor coecients it hits), and this makes it more convenient to work with a slightly di erent type of neighborhoods in H 1 that control this general error. These procedures based on the contraction theorem is unavoidable (as far as I know) to study questions in which existence is not given. Note that this method proves existence, as a consequence of the xed point theorem. This was used in fact to study very deep questions in [LL], [EW], [EW] and [EKW]. Also, in other problems (see [FL]), in which one has to study dependence on parameters, the contraction method is better since the recursive computation of the coecients leads to explosive growth of the intervals. We now deal with the singularities of the ODE at zero and in nity.
Lemma 5.2:
Consider the ODE 2k 00 ? 2 u = 0 u + r
for k a positive integer, > 0, then, the only solution of the ODE that vanishes at 1 is k X ? r u=e an rn (5:5) n=0
where ak is an arbitrary constant, and n(n + 1) an = an+1 2(n ? k)
nk?1
Proof: Clearly (5:5) is a solution of the ODE. The lemma follows by observing that the solution that is linearly independent from it is k X r a~n r?n u=e n=0
36
where a~n+1 = ?a~n+1
n(n + 1) 2(k ? n ? 1)
nk
and from here, any solution which is not (5:5) will grow exponentially at 1. QED
Lemma 5.3:
Consider the ODE
? b n(n + 1) u = 0 u00 + a + ? r r2 for n a positive integer. Then, the only solution of the ODE that vanishes at 0 belongs to U (u0 ; ; uN +2; C ) for un+1 any constant, and kn
uk = 0
b un+1 (n + 2)(n + 1) ? n(n + 1) b uk+1 ? a uk uk+2 = ? (k + 2)(k + 1) ? n(n + 1)
un+2 = ?
k n+1
ja uN +2 j jb uN +2 j + ja uN +1 j + (N + 3)(N + 2) ? n(n + 1) (N + 4)(N + 3) ? n(n + 1) C j bj j aj 1 ? (N + 4)(N + 3) ? n(n + 1) ? (N + 5)(N + 4) ? n(n + 1) + If a and b are sets of numbers, operations are performed in the set{theoretic sense.
Proof: The rst part of the lemma follows from the theory of regular singular 37
points. For the estimate for C , note that the recursion formula gives
X k>N +2
juk j (N + 3)(N +jb2)j ? n(n + 1) juN +2 j jbj
+ (N + 4)(N + 3) ? n(n + 1)
X k>N +2
juk j
+ (N + 3)(N +ja2)j ? n(n + 1) juN +1 j
+ (N + 4)(jNaj + 3) ? n juN +2 j
jaj
+ (N + 5)(N + 4) ? n(n + 1)
X k>N +2
juk j
QED
So, the result follows.
From these lemmas, and from the remarks of Chapter 4, we can solve initial value problems and evaluate the solution and its derivative at any point within its domain of analyticity. For its derivative it turns out that it convenient to use the formula u0 (x) = u1 ?
Zx 0
q (t) u(t) dt
This formula allows us to evaluate the derivative up to including the boundary of the circle inside which q is in H 1 (we do not need to contract it a little).
5.2 Eigenvalues A crucial device in the study of eigenvalue problems is the \match" function, M () associated with the ODE operator ?u00 + q u acting on Hode . 38
Here we will in fact consider general one{parameter boundary value problems
9
u00 (x) + p(x; ) u(x) = 0 > = u(0) = 0 > u0 (1) = 0 ;
for which we will say that ?2 is an eigenvalue i the previous equation has a solution for this value of . For the case p(x; ) = q (x) ? 2
the two de nitions of eigenvalue agree. Recall that our boundary condition u 2 Hode was equivalent to u(0) = u(1) = 0, by virtue of Lemmas 5.2 and 5.3. We de ne the match function then as follows: Take any point 0 < y < 1, and consider on the one hand, u0, one solution of the ODE which vanishes at 0, and on the other hand, u1 , one solution vanishing at 1. Then, de ne M () =
q
u00 (y ) u1 (y ) ? u0 (y ) u01 (y )
u20 (y ) + u00 2 (y )
Then, ?2 is an eigenvalue i M () = 0.
q
2 1 (y ) + u01 (y )
u2
:
In our analysis we will be considering two di erent \eigenvalue" problems, associated with the functions p(x; ) given respectively by Z l(l + 1) ? x2 ? V (x) ? 2 x 8 Z l(l + 1) > ? ? V (x) ? 2 for x x0 > < 2 x x pk (x; ) = > 2k ? 2 > : for x x0 x They will give rise to match functions M0 and M respectively. p0 (x; ) =
39
(5:6) (5:6up)
The reason is that, for k = k() as in (3:16), we have pk() (x; ) p0 (x; ) (5:7) In fact, when the subscript k is missing we will understand it is this p that we are talking about. (5:7) will have as a consequence that M will act as a maximal problem for M0, and it is easier to study by virtue of Lemma 5.2. Observe that the problem associated with p is not even a generalized eigenvalue problem since it is not guaranteed that @p(x; ) 0 @
and, even worse, it is not even continuous, since k() isn't. This can be remedied as follows: We will only estimates for M () for a nite number of values of , 1 > > n . De ne k( ) ? k(i+1 ) Gi = ? i i ? i+1 and Z ? V (x0 ) k() = Recall that k() was de ned to be the smallest integer above k(). Since the graph of k is convex, given > 0 we can nd a di erentiable function k () such that 1. k() k(), what implies that pk (x; ) p(x; ) 2. k(i ) = k(i ) for all i. 3. dkd () < Gi + for i+1 i . We then de ne M to be the match function associated with pk . The two de nitions of M agree at those i and as long as Gi i + k(i+1 ) 0 M(i ) < 0 i = 1; 3; : : : dn M(up i ) < 0 M(i ) > 0 i = 2; 4; : : :
(5:17)
and that, in the case M is not a generalized eigenvalue problem, we impose the condition that k does not jump in each interval i . If we can prove (5:17), (5:15) and the fact that M is continuous would prove up that each interval i = (dn i ; i ) contains at least one eigenvalue. If we fail to prove this we look for other choices of iup;dn . Note that here we are not using condition (5:6) to full power, but only the fact that sign M (r) = sign M(r) for r 2 R where the sign of an interval is de ned to be +1 if the interval is completely to the right of 0, ?1 if it is completely to the left and is left unde ned if it contains the value 0. The reason why we impose condition (5:17) is because expect M to change sign between zeros. The fact that we assume that
M(up i )>0 has to do with the fact that we de ned
M = det (u ; u1 ) 0
and not
M = det (u1 ; u ) 0
50
and with Lemma 5.9. It is interesting to point out that to expand the phase in time does not work if we substitute the phase by the vector (u(x); u0(x)); the reason is that interval arithmetic estimates are far to conservative and the bounds we obtain are very bad after a few steps. The reason why the previous algorithm in fact manages to prove good bounds for the eigenvalues lies in the fact that the solution of this particular ODE is of the form e?x with a factor out front small compared to the exponential. This has as a consequence that the normalizing factor in the phase has a contractive e ect that provides stability of the bounds: that's why we expand from 1 towards 0. In other words, if you look at the time ow Tt (u (x)) = u (x + t)
it turns out for the particular ODE we are working with, the phases of the eigenfunctions are almost always in the unstable manifold, and expansion of the phase backward in time gives good answers. If the phase happened to be in the stable manifold, the previous procedure would also if we expand forward in time, for the same reason. In the general case that solutions to the ODE can have signi cant projections in both the stable and unstable manifolds, one sided shooting is not enough: you need to consider these two projections separately. For this important generalization see [LR]. All this tells us that, for the maximal problem M , there are negative eigenvalues ? 2 , with i 2 i , i = 1; ; k, with eigenfunctions uk ; but in order to proof that these eigenvalues are lower bounds to the eigenvalues for the true problem (2:3) we proceed as follows: If M is a true eigenvalue problem, then it would suce to prove that the solution to the ODE corresponding to up k has k ? 1 zeros in (0; 1), with k of the right parity, in view of Lemma 5.10. 51
In the case M is not a generalized eigenvalue problem, we need to show that uk has k ? 1 zeros in (0; 1), in view of Lemma 5.8. In order to see this, we use the following Lemma:
Lemma 5.11:
Assume
dn k(up ) + 2 > up k(dn )
and that the solution of u00 (x) + pk(
up
)+2 (x; d n) u(x)u = 0
)
u(1) = 0
(5:18)
with pk( )+2 (x; dn) as in (5:6up), has k ? 1 zeros. Then, for any dn up , the solution of 9 2 k() = ? u(x) = 0 > u(x)00 + up
x
>
u(1) = 0 ;
has at most k ? 1 zeros.
Proof: Since k() is decreasing in , the rst hypothesis in the lemma simply
says that
p (x; ) pk(up )+2 (x; dn )
8 2 (dn ; up)
and the rest just follows by the usual argument.
QED
m Call um (x) the solution of (5:18), or u1 k . Note that u is the solution of one of those ODE's we can solve via the lemmas in the IVP section. In particular, we know intervals that contain the values of the solution at certain points. We will now see that under very mild conditions on the thickness of the partition x0 ; ; xn , the number of zeros of um is essentially the same as the number of sign changes in um (x0 ); ; um (xn ) up
52
Lemma 5.13:
Consider any solution of l(l + 1) Z 00 2 u ? V (x) + u=0 2 ? + x
x
on [a; b], with 0 < a < b < 1, and V decreasing. De ne Z k(a; b) = ? V (a) b
Then, the following holds: 1. if
? 1 b ? a max Z ?1 ; k(a; b) ?+ =2
then u cannot have two zeros in [a; b]. 2. If u(0) = 0, u 6= 0 for
0 < x < Z ?1
3. If u(1) = 0, then u 6= 0
for
2k < x < 1
where k is de ned as in (3:15), and x0 is the largest point at which V is singular.
Proof: Consider the equation
9
b v 00 + v = 0 = x v (0) = 0 ;
By Lemma 5.3 we know that
(5:19)
6b x2 (b < 6) 12 ? 2b which implies that v(x) > 0 for 0 < x < 126?b2b . Thus, if w(x) is a solution of v >x?
9
Z w00 + w = 0 = x w(0) = 0 ;
53
then v(x) = w(x) is a solution of (5:19) with b = Z ; thus v (x) 6= 0
what implies w(x) 6= 0
12 ? 2Z for x 2 0; 6Z
12 ? 2Z for x 2 0; 6Z (Z < 6)
Taking the limit as ! 0, we get w(x) 6= 0
2 for x 2 0; Z
Since p(x) > Z=x, we have that u cannot vanish in 0 < x < 2Z ?1 , thus proving 2. This argument implies that u cannot vanish twice between two zeros of w anywhere, and since Z=x is decreasing, the zeros of w never get any closer than Z ?1 , which proves the rst half in 1. For the other statement in 1., note that Z l(l + 1) 2 p(x) < ? V (a) ? ? < k(a; b) b a2
for a < x < b
If k(a; b) < 0, the distance between neighboring zeros of u is bigger that for the solution of u00 + k(a; b) u = 0 (5:20)
p which happens to be e ?k(a;b)x and has no zeros. So, u cannot vanish twice in (a; b).
p If k(a; b) > 0, then the solution of (5:20) is sin k(a; b)x , whose zeros are at a
distance
1 k(a; b)? =2
which proves 1. 54
For 3., since by (3:15)
2k ? 2
for x x0 the condition says that p is negative in the range x x0 and x 2k=. Assume then that u(t) = 0 and, say, u0 (t) = 0; then u is positive a little to the right of t and since u00 (x) = ?p(x) u(x) u grows and cannot vanish at in nity. QED p(x) =
x
We assume that the partition x0 ; ; xn introduced at the end of Chapter 3 is ne enough so that every interval (xi ; xi?1) satis es the hypothesis of this lemma. This can always be arranged from the start since our potential (5:6) is bounded above. Note that property 3. in the lemma allows us to count the number of zeros to the right of x0 (which is useful if x0 happens to be less that 2k=); simply recall that we know the solution is a polynomial with a non-zero factor, so we only have to count the number of zeros of this polynomial between x0 and 2k=. Since these zeros cannot get any closer that ? = 2k ? = 2 (k(x ; 1)) = ? 0
+
1
1
2
x0
2
+
we can take points at a distance less that half this amount from each other and evaluate intervals containing the value of the polynomial at these points; make the assumption that none of these intervals contains the value 0; then the number of zeros of u in (x0; 1) is exactly the number of sign changes of these intervals. Note that this does not involve to solve any ODE, except at x0 . In this way, everything is reduced to counting the number of sign changes in um (xn ); ; um (x0 ) and adding up the zeros the previous procedure gives. The following argument takes care of the fact that we only know an interval bound to the true values of um at those points. 55
Lemmas 5:1 and 5:2 give us intervals fIig that contain um(xi ). We now make the assumption (otherwise we quit the proof) that 0 2 Ii )
( Ii?1 < 0 < Ii+1 or
Ii?1 > 0 > Ii+1
in which case we remove Ii from the collection. Therefore, we have a collection of intervals fIi g all of which are always either above or below 0. Let's now count the number of pairs (Ii; Ii+1) such that Ii Ii+1 < 0. It is clear that every such pair is associated with one zero of um . Note also that there can be no more; this can be seen as follows: 1. If Ii and Ii+1 were neighbors before we removed intervals, then a. If Ii Ii+1 < 0, because um is continuous, there is one zero, and only one, by Lemma 5:4, and this is the one we accounted for. b. If Ii Ii+1 > 0, then we are counting no zeros. Now, um must have an even number of zeros in (xi+1 ; xi ), and by Lemma 5:4 can have at most one. So it has no zeros in this interval. 2. If there was some interval between them that we removed, then they have di erent sign, and therefore we are accounting for one zero. Let y be the point xj corresponding to the interval we removed: a. If um (y) > 0, by Lemma 5:4 we have no zeros in (y; xi), and an odd number less than two of them in (xi+1 ; y). So, we have one zero in (xi+1 ; xi ). If um (y) < 0 argue the same way. b. If um (y) = 0, by Lemma 5:4 we have no other zeros either in (xi+1 ; y) or in (y; xi), so again, one zero in total. This proves that the number of zeros of um is the same as the number of sign changes of the intervals fIi g, which is trivial to compute. 56
It is clear that with the control we have over the solutions of ODE's it is possible to bound the number of zeros without using Sturm-Liouville theory, but comparison theorems simplify the algorithm enormously since everything is reduced to counting sign changes in a sequence of intervals. With the previous procedure, given l, we can obtain lower bounds for the k smallest eigenvalues of H 1 electron : E1;l < E2;l < < Ek;l
Recall that each eigenvalue has to be taken with multiplicity 2(2l + 1). But remember that what we really needed was the smallest eigenvalues among all these En;l for all values of l = 0; 1; : : :. Although it is trivial to arrange this once we know enough En;l , since calculating eigenvalues turns out to be quite time{consuming, it is important to have a good procedure. One possibility is as follows: Chemistry predicts that for 1 Z 86, the eigenvalues En;l will be ordered as follows E1;0 E2;0 E1;1 E3;0 E2;1 E4;0 E1;2 E3;1 E5;0 E2;2 E4;1 E6;0 E1;3 E3;2 E5;1
and transitions occur at the following values of Z Z = 2; 4; 10; 12; 18; 20; 30; 36; 38; 48; 54; 56; 70; 86
Each En;l corresponds to the orbital ks, kp, kd, kf,... depending on the value of l, for k = l + n. Since En;l En;l+1 , in order to check that the chemical predictions are correct, it is enough to do as follows: 57
Given 1 Z 86, let En;l be the largest eigenvalue corresponding to a non{ empty shell predicted by chemistry. For each l, let Enl;l be the largest eigenvalue that appears before En;l ; let l0 be the rst l such that E0;l does not appear before En;l ; then 0
check 1. En;l Enl+1;l check 2. En;l E0;l 0
Occasionally, these checks will not be satis ed: electrons will arrange themselves in shells di erently from chemical predictions; in this case, we simply replace all unknown eigenvalues by the smallest one known. In practice this only happened for the outermost shells, which correspond to the largest eigenvalues, which happen to be very similar from each other and therefore didn't have any noticeable change in the results.
58
Chapter 6: The Numerics As remarked before, how good a lower bound for the energy we obtain depends on how good a \charge density guess" we make, or equivalently, how good functions Ri we select. Equivalently, all the parameters we need for carrying out the algorithm described in Chapters 2|5 are given by the functions Ri that; since they are chosen to be piecewise linear, we are left with only a nite number of parameters, = ( 1; ; n ), which are essentially the numbers is (3:5). Thus, we write the potential as V (x) and we write the lower bound for the energy that we obtain with our algorithm as E ( ; Z ); we then have E ( ; Z ) E (Z )
8
and in this way, the problem of nding a good lower bound is reduced to the purely numerical problem of nding a good . This was done as follows: Following [HLT], we use as (x) an empirical approximation to TF density introduced in [Ti] ? 6 29 = (x) = ? 9 = 4 4jxj jxj + 2 R (note that = 1), and then for each Z , take 2
3
1
Z (x) = Z 2 Z =3 x
59
1
3
With this charge density, apply the algorithm from the previous chapters. This gives a lower bound for the energy, and also, it provides us with a new charge guess in the following way: Let fn;l the solution to our ODE (2:3) for each value of l, corresponding to the n'th eigenvalue, normalized so that it has L2 norm 1. As a consequence, a ground state for H 1 electron is given by lb (x1; ; xZ ) =
^
n;l;m
fn;l (ri ) Yl;m (i )
(6:9)
where xi = ri i , with 0 ri 1, and i a point in the unit sphere in R3; Yl;m () is a normalized spherical harmonic of degree l, the antisymmetric product contains Z factors and by the normalization of the fn;l has L2 norm 1. This means that for atoms with an incomplete orbital we make a random choice of which Yl;m we take into the product. This ground state has a density
Z
j (x; x2; ; xZ )j2 dx2 dxZ X 2 2 ( ) = fn;l (r) Ym;l
(x) =
l;m;n
For incomplete orbitals, then, is not radially symmetric. It will be radial, however, if we as ground state the average over all possible choices of Ym;l (the average of a ground state is still a ground state). Thus, since
X m
we can take
2 ( ) = 2l + 1 Ym;l
(x) =
X n;l
2 (r ) w(n; l)fn;l
where the weight w(n; l) is equal to 2(2l +1) except in the last orbital n; l, where it is X w(n; l) = Z ? 2(2l + 1) (n;l)0 z2R R5 3
(7:3)
where N is the number of xi that fall inside B (z; R). In Chapter 3 we used the estimate E 0 and we simply threw it away. What we want to prove here is the following Sobolev{type inequality. 68
Theorem 7.1: Let
be a function satisfying
j r j 2 CKE Z = : 7
3
for some constant CKE . Then, there exists a constant CE such that
hE ; i CE Z = 5
3
And this theorem has as a consequence the result stated in the introduction
Corollary:
Let
E~ (Z ) = inf 2H hHLB ;
i
j jj2 =1
Then
Proof: Let
know that
E (Z ) E~ (Z ) + CE Z =3 5
Z be a normalized ground state for HZ . By the virial theorem, we
h? Z ; Z i = ?E (Z ) where ? means the laplacian in R3Z .
From TF theory, we know (see [Li] or [LS]) that
E (Z ) = CTF Z =3 + o Z =3 7
7
which implies that Z satis es the hypotheses of Theorem 7.1; therefore, E (Z ) = hHZ Z ; Z i = hHLB Z ; Z i + hE Z ; Z i E~ (Z ) + CE Z 5=3
69
In order to prove Theorem 7.1, rst note that
Z dR 1 V (R) 5 E= 2 R>0 R where V (R) (that has nothing to do with the potential V of Chapter 3) is de ned as Z ? V (R) = (N ? N )2 ? 14 dz z2R and has the following geometrical interpretation: 3
de ne
9
k (R) = fz : N (x1; : : : ; xZ ; z; R) = kg > >
i (R) = fz : N (x1; : : : ; xZ ; z; R) 6= i; i ? 1g > > > [ k = =
(7:4)
(N ? i + 21 )2 ? 41 = 0 for z 2 i (R)
(7:5)
> > > > > ;
k6=i;i?1 [Z
(R) = ( i (R) \ Si (R)) i=1
Note that Therefore,
Z dR 1 V (R) 5 E= 2 R
with V (R) =
=
Z Z X i=1 Si (R) Z X Z Z X
?(N ? i + 1 )2 ? 1 dz 2
4
?(k ? i + 1 )2 ? 1 dz
2 4 i=1 k=1 Si (R)\ k (R) Z X Z X ? ? = Vol Si (R) \ k (R) (k ? i)2 + (k ? i) i=1 k=1
70
(7:6)
It will also be of importance to consider the simpler version of V (R) and E , de ned as V~ (R) = E~ =
Z X
Vol (Si (R) \ i (R)) = Vol
i=1 Z
1 2
V~ (R)
dR R5
It is clear by (7:4) and (7:6) that V (R) V~ (R) E E~
A moment's thought reveals that V (R) is strictly positive, independently of the set of points xi , due to the geometry of balls in R3 (although the corresponding function may be zero in R2!); to study this is the purpose of Chapter 8. Here, we will use a di erent approach that works even if E attains the value 0 on a thin set. As a result, the inequality is sharper than the one you can obtain by pointwise estimates.
7.2 Notation Let H (x; y; R) denote the volume of the intersection of the ball with center x and radius R, and the ball of center y and same radius. Recall that
8 4 3 < 3 R ? jx ? yjR2 + 12 jx ? yj3 if R jx?2 yj H (x; y; R) = : 0 if R < jx?yj 2
71
Given points x1 ; :::; xZ , de ne (xi ) = min jxi ? xj j i= 6 j
and call x(i) to one of its closest neighbors. Also, RZ = R Z (0) 3 4
the radius of a ball that we expect contains a xed percentage of all the electrons, and R0 = R1 (x) for jxj = RZ Note by (7:3) that R Z (0) CZ ? = and RZ CZ Z ? = . The following Lemma shows that R0 cZ ? = . 1
5 8
1
3
2
Lemma 7.2: If jxj cZ ? = 1
3
then
R1 (x) = C Z =3 jxj
and in particular
5
= 1
3
3
3
(1 + o (() Z ))
R1 (x) C0 Z ? =3 2
Proof: Let
~(x) jxj with ~(x) analytic. Assume that R1(x) jxj otherwise there is nothing to prove. Then, by de nition of R1 we have 0 (x) =
1 = Z2
Z
Z B(x;R (x))
0 Z ? =3 x dx 1
1
=Z
B(Z x;Z R1 (x)) Z Z1=3 (x+R1(x)) 1= 3
1= 3
jzj?1 ~(z) dz
Z =3 R1 (x)2 ? (y ? Z =3 x)2 ~(y ) = Z 1= dy Z 1=3 jxj Z 3 (x?R1 (x)) 2
1
72
ZZ
(x+R1 (x))
=3 2 (y ? Z =3 x)2 (~(0) + O (y)) Z R1(x) ? = Z 1= dy Z 1=3 jxj Z 3 (x?R1 (x)) 1 3 5 = 35 Z =3 ~(0) Rj1x(jx) 1 + O Z ? =3 1= 3
Thus
2
1
1=3 5 ? = 3 R1 (x) C Z jxj QED
and the result follows.
7.3 Pointwise Estimates The aim of this section is to prove that if electrons are not arranged as you expect, then E is big pointwise. Roughly, electrons are expected to be at a distance Z ? = of the closest neighbor and most of them are within Z ? = of the nucleus. This is what Thomas-Fermi theory predicts. This implies that we only have to analyze the smeared out potential over a very precise con guration: elsewhere its big pointwise. 2
1
Lemma 7.3:
If for some i, (xi ) < R10(0) , then: 1
E>
C
R1 (0)
Moreover, if there are k such points, then E>
C k (xi )
73
3
3
Proof:
Observe that when R < R1(0), since Mi(R) = 0 for all R, S1 (R) = R3 and thus V (R) =
Z X
?
i=1
= 21
X
i6=j Z 1X
2
?
Vol k1 (R) k2 ? k
i=1
Vol (B (xi; R) \ B (xj ; R)) H (xi ; x(i) ; R)
Let x and y be one such pair of closest neighbors. Note that H (x; y; R) increases with R; in particular for R > 23 jx ? yj,
11 jx ? yj3 2 H (x; y; R) H x; y; jx ? y j = 3 324 therefore, 11 E 648
Z X i=1 R (0) (xi )< 110
11 = 2592 11 1296
C
jxi ? x(i)
Z X
i=1 R (0) (xi )< 110
j3
81
Z R (0)
dR 2 jx ?x R5 i (i) j 3 1
16jxi ? x(i) j
j xi ? x(i) j3 ?
X 810 10?3 16(xi) ? R1(0) (x ) X ?1
(xi )
R1 (0)4
R1 (0)
i
10
R1 (0)
(xi )
10
QED
74
Lemma 7.4:
If for some xi , (xi ) R 10(xi ) , then 1
C R1 (xi )
E
and in general
E
X
C R1 (xi )
(xi ) R110(xi )
Proof: Let R1(x)=2 R R1(x). Note that z 2 B (x; R) \ B (y; R) implies x; y 2 B (z; R).
Note also that
B x;
R ), since for z 2 B (x; 10
R
10 B (x; R) \ B (y; R)
jz ? yj jz ? xj + jx ? yj 310R R ), x; y 2 B (z; R). Thus, for all z 2 B (x; 10 If jzj jxj, we have R1 (z) R1(x), since R1 is increasing. Thus, since R R1(x) we have N (z; R; x1; : : : ; xZ ) = 21
and since N (z; R) 2
Z R (x)
Z 1 E B x; R
1
jzjj10xj
)
(
R1 (x) 2
dz dR R5
R C(x) 1
If there are N disjoint pairs of closest neighbors, (xi ; x(i) ), with (xi ) R 10(xi ) , then N (z; R) 2 for 1
R1 (xi )
z 2 B xi ; 10
(jzj jxij) 75
and
R1 (x)
2
R R1(xi )
for some i, and since
(N ? 21 )2 ? 41 N N 2
we have that
XZ 1 E R B xi ;
X
i
Z R (x ) 1
R1 (xi )
jzjjx10i j (
i
)
2
dz dR R5
C i R1 (xi )
QED
Lemma 7.5: Consider x1; :::; xZ ordered in such a way that jx1j jxZ j. For each xi , (1 i Z ? 1), let's call Rxi the solution of the equation
jxi j = R + Mi (R) If, for some i Z ? 1
(7:7)
Rxi > 10Ri (0)
then
E>
C
Ri (0)
:
Observe that this allows one of the xi to go o to 1.
Proof: For all R < Rxi we have that B (xj ; R) does not intersect B (0; Mi(R)) for j i. Thus, all points in Si+1 (R); ; SZ (R) belong to at most i ? 1 of the Bk , that is, one less Bk than they should, and so
V (R) [Zj=i+1 Sj (R) = jB (0; Mi(R)j
76
From Lemma 3.2b we know that Mi (R) > R ? Ri(0) for R Ri(0); therefore, we have: E
ZR
xi
V (R)
Ri (0) Z Rxi
C
C
ZRR(0) i
(Mi (R))3 dR R5
xi
2Ri (0)
CRi
dR R5
(0)3
(R ? Ri (0))3 dR R5
ZR
dR 5 2Ri (0) R xi
1 ? 1 CRi 4 Rx4 i 1 16Ri(0) ?4 C 16R (0) ? R10(0) i i (0)3
RC(0) i
The previous lemma seems to indicate that it is hard to pull out an innermost electron, but not so hard to pull out an outer electron. The way to measure how far xi is by solving the equation jxi j = R + Mi(R); this may seem a little cavalier; note that if i > Z=2, then Mi(R) < R and so Rxi > 10Ri (0) if jxi j > 20Ri (0), and thus this strange condition can be replaced by simply jxj > 20Ri (0). As before, the lemma can be applied to the case in which di erent xi that are too far out, and add up the conclusions. But in this case much more is true, taking in to account the multiplicity in the error.
77
Lemma 7.6:
As before, let's have jxi j arranged in increasing order. Assume that for some n < Z ? 1, and for some p > 0,
jRxn?p j > 10 Rn(0) Then, E>C
(p + 1)2 Rn (0)
Proof: For each i between n ? p and n, every point in B (0; Mi(R)) belongs to p + 1 less of the Bk than it should. The rest is like the previous lemma.
Lemma 7.7: a. Consider B (0; 20RZ ) If 41 Z + k points are outside of this ball, then
( k ? 1)2 E>C RZ
b. If B (0; R Z (0) contains ( + )Z electrons then 1 10
5 8
5 8
E
CZ R 58Z (0)
The proof is again straightforward. This last lemma in particular implies that if cZ points leave a ball of radius Z = , with c > 41 , then E will be of the order of Z = . 1
5
3
3
Lemmas 7.3 and 7.4 told us that electrons cannot have very close neighbors. Let us now prove that most of them cannot have very distant neighbors either. 78
Set
~(xi ) = (xi )Z =3 2
We would like ~(xi ) to be bounded from above and below for all xi unless E gets very big. That it is bounded below, we saw already; that it is bounded above is only true for most of the electrons. The following lemma states it precisely.
Lemma 7.8:
Assume that there are N 2 electrons in B (0; 20RZ ). Then, N 3 1X ~(xi ) C 0 Z N ?1 N i=1
Proof: Let Bi = B (xi; (2xi) ). These balls do not intersect. Since N 2, we know that (xi ) 10RZ , therefore N [ i=1
what implies that
B (xi ;
(xi )
2 ) B (0; 25RZ )
N (x ) X B xi; i C 0RZ3 2 i=1
or, since RZ CZ Z ? = , 1
3
N ~(x ) ! X i C 0 Z ?1 = Z i=1
2
3
QED
Now, divide by N and we are done.
79
Lemma 7.9:
At least 1% the points xi will satisfy C1 ~(xi ) C2 jxij < C2 Z ?1=3
or else
)
E (x1 ; : : : ; xZ ) C7:9 Z =3 5
Proof: By Lemma 7.7, we can assume that there are 23 Z electrons inside
B (0; 20RZ ) and B (0; 101 R 58Z (0)) contains less than ( 85 + 481 )Z electrons; so, we ? can assume that at least 32 ? ( 85 + 481 ) Z = 481 Z electrons are in B (0; 20RZ ) ? Z electrons satisfying B (0; 101 R 58Z (0)); therefore, there are at least 48 1 2 CZ ? =3 jxj C 0 Z ? =3
Lemma 7.2 implies that Ri (xi ) Z ? = for these electrons. 2
3
Z electrons have closest neighbors at a distance less than R (xi ) , If 10% of these 48 10 Lemma 7.4 says that E cZ = . 1
5
Therefore, there are at least for a constant C1 .
3
3 160 Z
electrons inside B (0; 20RZ ) with ~(xi) C1
Assume that 1% of these satisfy ~(xi ) C2 , for C2 to be picked in a moment; then 3 X ~(x )3 3 X ~(x )3 i 2Z 2Z ~(x )C i i 54 C 3 6000 2 2
which contradicts Lemma 7:8 for C2 large enough. Therefore, at least electrons satisfy the conclusion of the lemma. 80
297 16000 Z
7.4 Fuzzed Estimates In this section we will develop the theory that unless E is big pointwise by the previous section, when you average E over a small ball of radius like Z ? = , you get something strictly positive. 2
3
For this section we will not need the multiplicity in the de nition of V (R), and therefore it will be easier to work with V~ . Recall that V~ (R) represents the volume of some set . Consider now a ball of any center z and radius R = 10C2 Z ? =3
(7:8)
2
and de ne
Vz (R) = volume of ( \ B (z; R ))
Z dR 1 Vz (R) 5 Ez = 2 R>0 R
and Clearly
Z C ~ E= (R)3 R Ez dz This is convenient because, by the pointwise estimates, unless E is very big, Vz depends only on a xed nite number of variables. One more modi cation. De ne 3
?
Vz (R) = min Vz (R); C3 2 R
8 1 Z 10 R dR > R all i < Vz (R) 5 if (xi ) > 100 2 R 10? R Ez = > : 00 2 2 6
0
6
0
otherwise
C Z
for C3 to be picked later. 81
Clearly,
E
Z
C Ez dz 3 R R3
This is more convenient because the supremum of Ez will be comparable with its minimum smeared out value independently of , and this will allow us to use Sobolev estimates. The purpose of this section is to prove that averaging Ez(x1; : : : ; xZ ) over two of the electrons taking values over a ball, you get something which is bounded below by Z = . We will however need this result when we average only over a certain certain cone of directions, because we will like to apply this result to Ez restricted to a subset of R3Z , which has nice geometric features but, averaging over balls, would cause Ez to take values outside of this set: this would be bad because it would cause an overlapping which would be unbounded as Z ! 1. 2
3
De nition: We say that a set U V R6 has property P , if given any two points x 2 U and y 2 V , there exist cones ?x and ?y contained in U and V respectively, centered at x and y, with solid angle at least some xed constant, height . Note that, given any (u; v) 2 U V
f(x; y) : (u; v) 2 ? g C6 x;y
Lemma 7.10:
Given x and y belonging to a set U that satis es P , with x; y 2 B (z; 101 R ), we have 1 Z E (x0; y0; x ; :::; x ) dx0 dy0 C 2 Z 2 3 n 7:10 j?x;y j ?x;y z for suciently small, where the cones are like in the de nition of P .
82
R , for otherwise jx0 ? y 0 j < R for all Proof: We can assume that jx ? yj > 1000 100 (x0 ; y0) 2 ?x;y and so we have directly 0
0
Ez (x0 ; y 0; x3 ; ; xZ ) CZ 2 2
Although, as remarked before, it is impossible, let's start assuming that Vz = 0. This means that all points inside B (z; R) belong to the correct number of Bk . If this is so, consider real numbers r and s, smaller than 21 . Then, Vz (x + r(y ? x); y + s(x ? y ); x3; :::; xn; R) 41 rsjx ? y j3
or equivalently, Vz (x + u; y + v; x3 ; :::; xn; R) 41 juj jv j jx ? y j
for u and v vectors in the direction between x and y with norm less than jx ? yj=2. This is so because, when you move x and y closer, all those points in the new intersection that were not in the old union W (x; y; r; s; R) = B (x + r(y ? x); R) \ B (y + s(x ? y ); R)
?B (x; R) [ B (y; R)
increased their quota of Bk by two!. Therefore, we count the volume of all those points, and say that Vz (R) is bigger than that. Now, this volume is exactly
jW (x; y; r; s; R)j = H (x + r(y ? x); y + s(x ? y); R) ? H (x + r(y ? x); y; R) ? H (x; y + s(x ? y); R) + H (x; y; R) ? = 121 (1 ? r ? s)3 + 1 ? (1 ? r)3 ? (1 ? s)3 jx ? yj3 = 41 srjx ? yj3 because juj; jvj < 21 . Note that if you just fuzz one the electrons, then you don't gain anything; the quota is modi ed only by one, and since points are allowed to belong to two 83
di erent number of balls, they could just be shifting from one to the other. In other words, it seems that there is very little one{electron information in E , and you have to use essentially two{electron analysis. If, instead, we translate x and y by arbitrary vectors, u and v, with
juj < 21 jx ? yj
and
jvj < 21 jx ? yj
(cond: 1)
almost the same thing as before is still true. We can see it as follows: Consider a coordinate system centered at x, with basis vectors e1 ; e2 ; e3 orthogonal to each other and e1 = jxy??xyj . Consider only those vectors u, v with coordinates: (r; +; +); (?s; +; +) r; s 0 (7:9) then, we claim that all those points w = (w1 ; w2 ; w3 ) 2 W (x; y; r; s; R); with w2 ; w3 0
are also in W (x; y; u; v; R) = B (x + u; R) [ H (y + v; R) ? B (x; R) \ B (y; R)
For that we only have to check that w 2 B (x + u; R) \ B (y + v; R)
since it already is not in their intersection. This will follow if
jw ? (x + u)j jw ? (x + re1 )j and
jw ? (y + v)j jw ? (y ? se1 )j
In order to see this, we will write u = (r; u2 ; u3 ); with u2 ; u3 0
84
(7:10)
v = (?s; v2 ; v3 ); with v2 ; v3 0
So, the condition to be checked amounts to
j(w1 ? r; w2 ? u2; w3 ? u3 )j j(w1 ? r; w2; w3)j and
j(w1 ? jx ? yj + s; w2 ? u2; w3 ? u3 )j j(w1 ? jx ? yj + s; w2; w3)j which are trivial, since by (7:10), w2 , w3 0. )j ; this implies The volume of points represented by (7:10) is exactly jW (x;y;r;s;R 4
jW (x; y; u; v; R)j 161 rsjx ? yj
for (u; v) 2 ?x;y
and so, we have for this subset of directions u and v Vol ( (x + u; y + v; x3; ; xZ ; R)) 161 rsjx ? yj where r and s are the projection of u and v in the direction x ? y. Assuming
B (x; R + ) [ B (y; R + ) B (z; R )
(cond: 2) all these points whose volume we are estimating are inside B (z; R), and thus Vz (x + u; y + v; x3 ; :::; xZ ; R)
1 16 rsjx ? y j
and if C3 161 , this proves Vz (x + u; y + v; x3 ; :::; xZ ; R)
1 16 rsjx ? y j
Now, if the cones ?x;y are contained in this set of directions (7:9), we have Z 1 Z V (x0 ; y0; x ; :::; x ; R) dx0 dy0 C jx ? yj?6 Z rs du dv 3 n j?x;y j ?x;y z u2?x v2?y
C2 jx ? yj
85
Otherwise, pick (x0 ; y0) 2 ?x;y such that B (x0 ; c ) ?x
B (y0 ; c ) ?y
and
where c is a constant that depends only on the aperture of the cones; since x0 and y0 then have cones in every possible direction and aperture contained in our set U , we can apply the previous result to ?c x ;y to obtain 1
0
Z
0
Vz (x0 ; y 0; x3 ; :::; xn; R) dx0 dy 0 j?x;y j ?x;y Z C0 j?c j c Vz (x0; y0; x3; :::; xn; R) dx0 dy0 x0 ;y0 ?x0 ;y0
C 00 jx ? yj?6 C4 2jx ? yj
Z
Z
u2
?c x0
v2
?c y0
(7.11)
rs du dv
If Vz (R) = , for in the range 0 < < 10?6C4 2 jx ? yj
(7:12)
note that the previous argument says that when you move x and y by vectors u and v as in (7:9), there are points whose total volume is at least 161 rsjx ? yj that change their quota of Bk by two. Thus, making C3 > C4 , (7:12) implies Vol ( (x; y; x3; ; xZ )) = and therefore
1
Vol ( (x + u; y + v; x3; ; xZ )) 16 rsjx ? yj ? what implies, if C3 < 161 , that Vz (x + u; y + v; x3 ; :::; xZ ; R)
86
? 1 rsjx ? yj ? 16
+
+
And arguing as before, and using the fact that Av(f )+ (Avf )+ we see by (7:12) and (7:11) that
Z
(u; v; x3; ; xZ ; R) du dv ?C4 2 jx ? y j ? C4 2 jx ? y j V z + j?x;y j ?x;y 2
1
If however > 10?6C4 2 jx ? yj, then, if there is a pair of vectors 2 (u0; v0) 2 ?= x;y
such that
Vz (u0 ; v0 ; x3 ; ; xZ ; R) < 10?6 C4 2 jx ? y j
apply the previous estimate to any of the cones 2 ?= u ;v ?x;y 0
0
to obtain 1
j?x;y j
Z
Vz (x0 ; y 0 ; x3 ; :::; xn; R) dx0 dy 0 0 0 (x ;y )2?x;y
Z 1 j? j 0 0 = Vz (x0 ; y0; x3; :::; xn; R) dx0 dy0 x;y (x ;y )2?u ;v =2 ?u ;v
2 0 0
j? j C4 2 jx ? yj ? 10?6 C4 2 jx ? yj x;y 0
which, since
0
=2 1 ?u ;v 10 ?x;y 0
implies (7:11) also in this case
2
0
87
If no such pair of u0 and v0 exist, then 2 =2 Vz (x + u; y + v; x3 ; ; xZ ; R) 10?6 C4 2 jx ? y j for u 2 ?= x ; v 2 ?y which implies directly that (7:11) holds in all cases, probably with a di erent constant C4 , as long as conditions (cond: 1) and (cond: 2) are satis ed. This will be the case if B (x; R + u) [ B (y; R + v ) B (z; R) and 10 < jx ? yj < 2R which holds by hypothesis and whenever 5R > R > jx?2 yj Integrating (7:11) over this set of R0 s, (note that jx?2 yj 10?6R0 ) we get:
Z
R
5 Av Ez jx?yj AvVz (R) dR R5 2
C2 jx ? yj 2 C Rjx 4? yj
16 625 jx ? yj4 ? R
C 2 R0 4
(R) Recalling that both R and R0 are bounded above and below by constant multiples of Z ? = , we get the conclusion of the lemma. 2
3
will be like Z ? =3 ; this will make AvEz of the order of Z =3 . Later, we will be able to superimpose the estimates for Ezi for a number of zi of the order of Z , what will give that the smeared out error is of the order of Z 5=3 . 2
2
Also, note that, by the way Ez was de ned, it is trivial that
j Ezj 1 C1 Z 2 2 88
(7:13)
7.4 Real Variables Recall
R = 10C2 Z ? =3 = C Z ? =3 2
2
We will chose how small C has to be according to our needs. Starting with the cube Q0 with center the origin and diameter C2 Z ? = , consider all dyadic subcubes of size to be determined later, Q . Put 1
Z
R3Z
3
E (x1; :::; xZ )j (x1; :::; xZ )j2 dx1 ; :::; xZ =
=
XZ
Q1 :::QZ
E (x1 ; :::; xZ )j (x1 ; :::; xZ )j2 dx
Let's put EQ ;;QZ equal to E restricted to this product of cubes. Given one Q , we call Q its only dyadic ancestor with size C7 times bigger, with 1
?
C7 > max C1 ; 107
Lemma 7.11:
(7:14)
Either
min EQ ;;QZ C7:9 Z = 5
3
1
or, there exist c1 Z of these cubes fQi g, that we call fQ~ g, with c a small constant, such that 1. Q~ \ Q~ 0 = ;, for 6= 0 . 2. Every Q~ has a neighbor among the rest of the Q : call it Q( ). By neighbor we mean that it is either equal to itself or one of the 26 cubes adjacent to it.
89
Proof: Assume that there are (x1; : : : ; xZ ) such that E (x1; : : : ; xZ ) < C7:9 Z = . By Lemma 7.9, there are N of these xi , (call them fx~ g), with N > Z=100, such 5
that
3
)
C1 Z ? =3 (~x ) C2 Z ? =3 (7:15) jx~ j < C2 Z ?1=3 Assume that no one of the xi belongs to any of the sides of any dyadic subcube of Q0 . This is OK since E is continuous. Perform then a dyadic subdivision to Q0 such that 2
2
1. If at some point, one subcube contains no x~ , we throw it away. 2. If for one subcube Q, we have that Q contains less that C8 of these x~ , we store it away. Note that by (7:15), this is a nite procedure. Call fQ]g the set of cubes we stored. There are at least N=C8 of these cubes. Note also that fQ] g has a xed nite overlapping, since X Q] C8
so we can extract a maximal disjoint subfamily that contains at least N=C8 cubes, and take the corresponding family fQ] g. Note that by property 2 of these cubes, if 2Q] is its dyadic predecessor, (2Q]) contains at least C8 of the x~ 's, so by (7:15), ] 4 ? 3 ] (2C7 ) Vol Q = Vol 2Q 3 C8 (C1 =2)3 Z ?2 what implies diam Q] 10?6 (C1 =C7 ) C8= Z ? = Now we pick the size of the Q to be largest dyadic size smaller than the size in the previous expression. Thus, each Q] contains one of the initial Q , and statement 1 of the lemma follows by property 1 of the fQ]g and the properties of dyadic subdivisions. 1
3
2
3
For statement 2, simply realize that by (7:15) again, points x~ have a close neighbor at a distaince C1 Z ? = . By picking C8 an integer so that ? = 10?6 C1 C = Z ? = =C7 < C1 Z ? = 1 (7:16) 8 4 C1 Z 2
2
3
3
1
3
90
2
3
2
3
(such an integer exists by (7:14)) these selected cubes (or any other containing x~ 's) has a neighbor in the collection. Let's consider now one such product of cubes that satis es the second alternative in the previous Lemma. Let's select one such pair of neighboring cubes, say Q and Q . Make x = (x1 ; x2) and w = (x3 ; :::; xZ ), and de ne 1
2
Q = Q1;2 = Q1 QZ W = W1;2 = Q3 QZ
Note that Q satis es property P . Consider also centers z satisfying
jz ? c j < 1
Note that by (7:16) and by (7:8)
C1 Z ? =3 2
(7:17)
10
diam Q 10C1Z ? = 2
R Q B z;
(We can take C2 > 10C1 in Lemma 7.9). Then, argue as follows:
3
10
(7:18)
We know already that, by (7:18) and Lemma 7.10,
Z Z
2 j (x; w)j2 dx dw C7:10 3 R W S Z Z
Z 1 2 0 0 j Ez (x ; w) j (x; w)j dx dx dw (7.19) j ? W S x ?x The last term in the previous inequality di ers from Z Z 1 Z 0 0 2 0 (7:20) j Ez (x ; w)j (x ; w)j dx dx dw j ? W S x ?x 91
by at most
Z Z 1 Z W
Q j?x j ? Z Z Z ? 6 j (x; w)j2 ? j (x0; w)j2 dx0 dx dw C j Ez j 1 Ez (x0 ; w) j (x; w)j2 ? j (x0 ; w)j2 dx0 dx dw x
W Q ?x
= C?6 j Ezj 1
Z Z Z (x; w) + (x0; w) (x; w) ? (x0; w) dx0 dx dw W Q ? = Z Z Z 2 0 0 ? 6 (x; w) + (x ; w) dx dx dw C j Ez j 1 W Q ? = Z Z Z 2 0 0 (x; w) ? (x ; w) dx dx dw x
1
2
x
1
W Q ?x C?6 j Ezj 1
C 3 j
j L (QW ) + 2
Z Z Z Z 1
W Q ?x 0 C?6 j Ezj 1
Z Z Z W Q ?x
(x0; w)j2 dx0 dx dw
2
1
2
1
W Q B(x0 ;)
Z
1
2
2
!= 1
2
=
!= 1 j (x0; w)j2 dx dx0 dw A
jrx (u; w)j du dy dt dw W 0 B(0;) Q C j Ezj 1 j j L2 (QW ) j rx j L2(QW )
= !
jrx (x + t(x0 ? x); w)j2 jx ? x0 j2 dt dx0 dx dw
0 Z Z Z 3 @C j j L (QW ) + Z Z 1Z
j
2
2
and by (7:13) this is less than CZ 2 3 j j L2 (QW ) j rx j L2 (QW )
92
(7:21)
Note that
Z Z 1 Z W
Q j?x j
?x
Ez (x0 ; w)j (x0; w)j2 dx0 dx dw
C?6 C0
Z Z Z
W Q B(0;)
Z Z
W Q
Ez (x0 ; w)j (x0; w)j2 dx dx0 dw
Ez (x; w)j (x; w)j2 dx dw
which, with (7:19) and (7:21) implies that
Z Z W Q
Ez (x; w)j (x; w)j2 dx dw C7:10 Z 2 2
Z Z
j (x; w)j2 dx dw W Q 2 ? C10Z 3 j j L (QW ) j r j L (QW ) 2
2
for some constant C10 . So, if
j rx j L (QW ) C9 Z = j j 2L (QW ) 2
2
for C9 = 4
Q1 QZ
(7:21)
2
C = KE
1
2
(7:23)
c1
we have
Z
3
Ez (x1 ; ; xZ )j (x1; ; xZ )j2 dx1 dxZ
Z 2 2 (C7:10 ? C C10 C9 ) j j 2L (Q QZ ) (7.24) 2
1
and, taking C small enough, we get
Z
Q1 QZ
Ez (x1 ; ; xZ )j (x1; ; xZ )j2 dx1 dxZ
C11 Z = j j 2L (Q QZ ) (7.25) 2
3
2
1
for another constant C11, provided (7:21) holds for these cubes. 93
?= Recall that this holds for jz ? c j < C Z10 ; thus, integrating (7:25) over this set of z's, we get 1
1
Z
Z
Q1 QZ jz?c1 j
c T (z) 0 1 Gz; (x; ) =
x jr Gz; (x; )j c3 T2 (z) det j@x Gz; (x; )j c43
if jx ? zj < c1 T1 (z)
Fix a constant c5 < 1.
Lemma 7.12:
If x2 2 B (z; T1 (z )) x1 2 B (z; T2 (z )) ? B (z; T1 (z )) def = S (z; )
then
Z1 c5 2
W (z; ; Gz; (x1 ; ); x2; ; xZ ) d
98
1 ? c5 2
Proof:
Note that for < 2?1 (z)
jz ? G(x1 ; )j jz ? (z ? (x1 ? z)j T2 (z) < T1 (z )
so
G(x1 ; ) 2 B (z; T1 )
and so
W (z; ; G(xi ; ); x2; ; xZ ) 1
thus
Z1 c5 2
W (z; ; G(x1 ; ); x2; ; xZ ) d
Z
1
2 c5 2
W (z; ; G(x1 ; ); x2; ; xZ ) d
1 ?(zc)5 2
QED
Lemma 7.13: F h? ;
There exist a function w(z; R) and a constant F such that
i + hE ; i
ZZ
?
w(z; R) N (z; R) ? 1 ;
99
dz dR R5
Proof: For the moment, we x z and and we drop the dependence of param-
eters on (z; ). Let
Ei;j = (x1 ; : : : ; xZ ) j xi 2 B (z; T2 ) ? B (z; T1 (z )) xj 2 B (z; T1 (z )); xk 2= B (z; T2 ) k 6= i; j
and
E = (x1 ; : : : ; xZ ) j NB(z; T2 ) > 2 :
Then
Z
W (z; ; x1 ; : : : ; xz ) j (x1 ; : : : ; xZ )j2 dx1 dxZ
Z
E
+
W (x1 ; : : : ; xZ ) j (x1 ; : : : ; xZ )j dx1 dxZ
XZ
i;j Ei;j
(7.28)
W (x1 ; : : : ; xZ ) j (x1 ; : : : ; xZ )j dx1 dxZ
Let's deal with the term in the previous sum corresponding to E1;2. The other terms are treated in the same manner. Let's introduce the following notation: u = u(x; ) = (u1 ; : : : ; uZ ) = (G(x1 ; ); x2; ; xZ ) ? x = x(u; ) = (x1 ; : : : ; xZ ) = G?1 (u1 ; ); u2; ; uZ U ( ) = fu j x(u; ) 2 E1;2 g S = fx 2 E1;2 j x 6= u(x; ) for 6= 1g ? (u; ) = det(Jx1 G) G?1 (u; );
where J denotes Jacobian matrix. 100
By the previous Lemma, we know that
Z
Av? c 1W (z; ; G(xi ; ); xk6=i) j (x1; : : : ; xZ )j2 dx E; Z 1 ? c5 j j2 (x1; : : : ; xZ ) dx1 dxZ (7.29) ?c 1 2
5 2
2
5
E1;2
Changing variables and applying Fubini's theorem, we get that the left hand side of (7:29) is equal to
ZZ ?x(u; ) 2 du d c5 ?1 W ( z; ; u ) 1? U ( ) (u; ) 2 c5 1 2
We proceed as in the previous section; the integral in this last expression di ers from ZZ du d (7:30) W (z; ; u) j (u)j2 U (u; ) c5 2
( )
1
by at most (note that W 1)
ZZ 2 du d j (u)j2 ? j (x(u; ))j ( u; ) ZZ =
6
u=x(u; ) U ( )
(ZZ +
j
du d j (u) ? (x(u; ))j j (u) + (x(u; ))j ( u; )
1 du d =2 (u)j (u; ) +
ZZ
2
j
ZZ u6=x2E1;2
1=2 ) du d (x(u; ))j2 (u; )
j (u) ?
du d (x(u; ))j2 ( u; )
!= 1
2
Since the range of u we are integrating over is included in E1;2 for any 1, we can expand the domain in the rst integral in the last expression to E1;2, and 101
changing variables back to x in the second and third integrals we obtain
9 8 ! = ZZ Z1 = = < d + j ( x)j2 dx d :j j L (E ) ; inf u (u; ) != ZZ 2 1
2
1;2
1
2
c5 2
1
u6=x2E1;2
j (u(x; )) ? (x)j du d
2
Note that (x) ? (u(x; )) = (u(x; 1)) ? (u(x; )) =? =? =?
Z1d
(7:31) j j L (E ; 2
1 2
Z1
Z1
S c5 2?1 2
1 2
c =
1?
0 Z 1 @ )
5 2
1
rx (u(x; t)) r G(x1; t) dt
2
Z Z1 S c5 2?1
!= 1
inf u (u; )
j r G(x1; )j 1
Z Z1 Z1
j j L (E ;
1
d
c5 2
sup
rx (u(x; t)) dtd u1(x1 ; t) dt 1
0 Z 1 @ )
c5 2?1 1
(u(x; t)) dt
dt
Therefore, we have
2
2
c =
+ 1?
!
5 2
jrx (u(x; t))j dt d dx d
inf u (u; )
c5 2
sup
c5 2?1 1
!= 1
2
j r G(x1; )j 1
!=
jrx (u(x; t))j dx dt 102
!
2
2
c =
+ 1?
1
1
2
1 A
!= 1
1
1
5 2
1
2
1 A
(7:31)
j j L (E ; 2
1 2
0 Z 1 @ ) 1
1?
5 2
Z1
jrx
x(u;)2S c5 2?1
j j L (E ; 2
2 1
0 Z 1 @ c5 2
0 Z 1 @ )
inf u (u; ) sup
d
!=
c =
1
2
1?
!
2
1 A
c =
1 A
5 2
2
2
!= 1
2
+ 1? 1
1
5 2
!
!=
1
d
c5 2?1 1
1
c =
+ 1?
dx d (u)j ( u; )
inf u (u; )
c5 2
2
j r G(x1; )j 1
c5 2?1 1
Z
inf u (u; )
sup
2
1
d
c5 2
c =
!=
1
5 2
1 A
2
j r G(x1 ; )j 1 j rx j L (S) 2
1
since u(S ; ) S . Let's introduce the following notation " = "(2 ; c5 ) = 1 ? c25 p = p(G; c5 ; 2 ) = = (G; c5 ; 2 ) =
Thus, we have
Z1
d
inf u (u; ) sup j r G(x1 ; )j 1 ?
c5 2
c5 2 1 1
(7:31) " = + p = (p") = j j L (E ; ) j rx j L (S) 21 " = + p = (p") = M ?1 j j 2L (E ; ) + M j rx j 2L (S) for any M . Note also that 1
1
2
1
2
2
1
2
1
2
2
1
2
(7:30) p (G; c5 ; 2)
1
1 2
2
Z
E1;2
103
2
1 2
W (z; ; x) j (x)j2 dx
1
2
to conclude that p (G; c5 ; 2 )
Z
W (z; ; x1 ; : : : ; xZ ) j (x1; : : : ; xZ )j2 dx1 dxZ
"
Z
E1;2
j j2 (x1 ; : : : ; xZ ) dx1 dxZ
? 21 " = + p = (p") =
1
2
M ?1 j
1
2
1
2
j L (E ; ) + M j rx j L (S) 2
1
1 2
2
Doing the same for all Ei;j and adding the results, we obtain
Z [Ei;j
W (z; ; x1 ; : : : ; xZ ) j (x1; : : : ; xZ )j2 dx1 dxZ
1 1 "p?1 ? 21 M ?1 " =2 + p =2
Z
? ?1 = p " 1
2
j j2 (x1 ; : : : ; xZ ) dx1 dxZ
[Ei;j 1=2 X 1 1 = = 1 2 2 ? 2 M " + p (p") j rxi j L2 (Si;j ) i;j
Note that the fact that G is a contraction, implies that p > . Since W (x1 ; : : : ; xZ ) 1
for (x1; : : : ; xZ ) 2 E
the last inequality and (7:28) give
Z R3Z
W (z; ; x1; : : : ; xZ ) j (x1 ; : : : ; xZ )j2 dx1 dxZ
104
? "p?1 ? 12 M ?1 " = + p = p?1 " = 1
Z
1
2
1
2
2
j j2 (x1 ; : : : ; xZ ) dx1 dxZ
E [Ei;j 1 X ? 21 M "1=2 + p1=2 (p") =2 j rxi j L2 (Si;j ) i;j ?1 1 ?1 1=2 1=2 ? ?1 1=2 "p ? M " + p p "
2
Z ? N (z; T2 (z ) ? 1 j j2 (x1 ; : : : ; xZ ) dx1 dxZ R = X = = 1 3Z
? 2M " + p 1
1
2
2
(p") 1
2
i;j
j rxi j L (Si;j ) 2
Z ? = w1 (z; ; G; c5; 2) N (z; T2 (z )) ? 1 j j2 (x1 ; : : : ; xZ ) dx1 dxZ R X 3Z
? w2 (z; ; G; c5; 2 )
i;j
j rxi j L (Si;j ) 2
Recall that all parameters are taken to be functions of the speci c ball we are studying. If we multiply the previous estimate by T2 (z)?4 and integrate against ?1 ?5 dzd
we obtain
?
dz d T ?4 (z ) 5 ZZ X ? w2 (z; ) j rxi j L2 (Si;j (z; T (z))) T ?dz4 (d z ) 5 i;j ZZ
? w1 (z; R) N (z; R) ? 1 ; dzRdR 5 ? F h? ;
hE ; i
for
ZZ
w1 (z; ) N (z; T2 (z )) ? 1 ;
F = sup
X ZZ
x1 ;:::;xZ i;j
(x1 ;:::;xZ )2Si;j (z;R)
w2 (z; R)
i
dz dR R5
QED
what proves the Lemma.
105
Note that the lower bound in this Lemma can be trivially improved to
ZZ
D?
w(z; R) N (z; R) ? 1
+
;
E dz dR R5
but although this is a better estimate, it does not give rise to one{electron potentials, which is our goal. The inequality that the Lemma proves depends on a large number of parameters. In this case, like in the case of the potential V in Chapter 3, some optimization is called for. Crudely speaking, the lower bound obtained in the Lemma will be like
?1 ? c Z = 5
5
2
3
whereas the price we pay, F h? ; i, since F represents a volume, will be like
?1 ? c 3 Z = 2
5
2
3
This implies that for an appropriate choice of the parameters, this method will give a contribution to the energy of cZ = . 5
3
Note also that this in this section, we are fuzzing with respect to the group of dilations, whereas in the previous ones we were dealing with translations. This is because before, the integration dz was done rst, and translations are natural for this measure. This was useful to superimpose estimates for di erent balls. In this section however, we are only looking for lower bounds as one{particle potentials, and we let the ODE solver take care of extracting any estimates that they yield, and thus it is more convenient to do the dR R integration rst. 5
106
Chapter 8: The Ball Packing Lemma 8.1:
Assume there are at least two disjoint balls Bk = B (ck ; Rk ) with ci in a ball S R3 , with the property that R 1 i 0
