- Email: [email protected]
A modification of the crandall-kikuta method
V. V.Ditkin
12
REFERENCES 1.
GAUTSCHI, W., Computational aspects of three-term recurrence relations. SIAM Rev., 1967,9,24-82.
2.
BLANCH, G., Numerical evaluation of continued fractions. SIAM Rev., 1964,4,383-421.
3.
MIKLOSKO, J., The numerical computation of three-term recurrence relations and the tridiagonal system of linear equations by the method of shooting. Zh. vj$risl. Ma. ma. Fiz., 1974,14, No. 6,1371--1377.
4.
MIKLOSKO, J., Numerioky vypooet mdiagonalnych systemov na pocitacoch s malou kapacitou pamate. Proc. Res. problems of techn. cybernetics. Bratishwa, Slovak Acad. Sci 1965,99-l 11.
5.
FADDEEV, D. K., FADDEEVA, V. N., Computational methods of linear algebra. M., Fizmatgiz, 1963.
AMODIFICATIONOFTHECRANDALL-KIKUTAMETHOD" V,V.DITKIN Moscow
(Received 14 Juiy 1975) A MODIFICATION of the Crandall-Kikuta method for finding the eigenelement and eigenvahre of the problem Arp==kp, is considered, which does not involve the solution of an ill-posed linear system. Furthe~ore, for the case when A is a differenti~ operator, a device is described whereby the pivotal condensation method can be used to solve the supplementary linear problems that arise. A method for finding the eigenelement and eigenvalue of the problem A cp=x(P, where A is an operator, specified in Hilbert space, was devised by Crandah and Kikuta in [I, 2] . In this method, the eigenelement and eigenvahre are found by an iterative process, which can be written as
where pn+l is a normalizing factor. A calculation from these expressions demands the solution of the equation. (il-h,,l) rp,,+i=p, +!cp>I almost on its eigenvalue, so that numerical realization of the Crandall-Kikuta method is extremely laborious. Clearly, the growth of II~,lrl II can be avoided by a suitable choice of the non&zing factor pn+t . However, the introduction of pn+l does not of itself improve the properties of the system for fmding q~,, ! ,. Below, we consider a modification of the Crandall-Kikuta method, such that there is no need to solve an ill-posed linear system. In addition, for the case when A is a differential operator, we outline a device for using the pivotal condensation method for solving the supplementary linear problems that arise. It seems likely that, on occasions, our method will offer a more convenient means for avoiding the above-mentioned difficulties than does the method of false perturbations [3,41. 1. Let H be a Hilbert space. We consider the eigenelement and eigenvaiue problem
XEH,
Ax-Ax,
Ilxll~0.
Here, A is a given non-self-adjoint operator, satisfying the following conditions.
*Zh. @hid. Mat. mat. Fir.,
lb,
4, 838-846,
1976.
(1.1)
A modification of the Crandall-Kikuta
method
13
Condition 1. The space H can be expanded in a direct sum of subspaces Ho and HI, invariant
under A, where the subspace Ho consists of the eigenvalues of A corresponding to the eigenvalue Ao (which is in general complex). This expansion of H in a direct sum of two subspaces defines a projector P onto Ho parallel to HI, where P is a linear bounded operator (see [ 51). Condition 2. We shall assume that Xo is not a point of the spectrum of the operator A,
considered in HI, i.e., the operator R (A,) = (A --&I) -’ is bounded in the subspace Hr. Put C=ll (A--L,Z)--‘llu,. w e want to find Au and the corresponding eigenelement. Consider the following iterative process. Given z,=H and h,. Let ~,=t,,fA,, GEH~, A,=H,; h,=L,+o,. We shall solve the problem ( A--h,Z)x,=p2s,, while imposing on the required element x2 the auxiliary condition zz=z,,+A2, A,EH ,. As a result, we find .Q, pz, L= (A&, G) / (z,, x1). We then solve the problem (~1-h,Z) X~=~LGTI:under the conditions that x:=xU+AJ, A+H,.We find.r3,~lz. AS= (Ax3. x3) / (x3, xJ) ,etc. We obtain a sequence of numbers h ,, . . . , h T1. . ; pi,. .,pr,... and a sequence of elements xl, . . . , xr, . . . . The iterative process just described has the following properties. Theorem 1
Let conditions 1 and 2 be satisfied, and let z,=~,fA, A,ER,, that IlA,ll/ll~,ll~~~‘/~, IS,IGq/(l+q)c, where q=4~(1i-~)
hh=
(A-hd)sh=~h~h-I,
where
x;,=x,+~~,
&,d,,
I*,,
hh=ho+6h
=ho-a
rz;)
k,
h,=hO% be given such Assume that G,, PI,, hk
k=2,3,.
,
for any integer k > 2. Then,
~6,~cq’“-‘~6k-,~~q2*-‘-‘~6,~,
h--l,
~~A~ll~y’b-z~~Ah-,~l~q2K-‘-‘~~~~~~,
ka2.
Proof: Given any integer k > 2, we have Sk-,= xO+Ah_,, Ah_,=H,. in the form where AGH,; then (A-Ah-,/)
(x:,+11,)=,.ih
(ho-hk_,)xo+
Hence, since(h,--3L,_,)~p~H,, /.h=hO-h k--l, k>2. Further,
pAxo~H,,
(A-h&)
%N32
Xk
will be sought
(G+&-,)r
Ak=p~o+clJk-,.
(A-hk-II)Ah~H,r
Sk=-Gk-_i R (hr_i)xh-,=x0-&A
ptbk-,E~,y
(L-L) Ah-,,
since R (hh-,).zo=-&~~xo. In view of the fact that -ik-,R(hk-,) we find from Eq. (1.2) that Ak=--IS,,-,R (ha-,) Ah-,. It is easily shown that
a.,
Ah-,ER,,
we obtain
(1.2) while G,=s~+A~,
Y. K Ditkin
14
Consequently,
(AXA,Sk)
(Xl*4
= hr=ho
~o)+~~~-~~“(Ak-~~R(hk-i)Ak--L)
+ _~,_,(~,_,,~~)-_6~,(R(~~-_t)Ak-,, I
(x&,4 + s,_,S,“-, (R&+-i)&-+
(1.3)
R&--l) A,-l) ,
(&:k,3k)
k=2,3,.
..
We know that (D c
(Ak-ho)
mR”+i
(ho)
=fd
(hk),
k=l,
2,. . . ,
?u-0
if
~~,/=Jh~-hc/~lIIIR(ko)ll=l/C.
Byhypothesis,161jdq/C(l+q), With k = 2 we obtain from (1.3):
8, =
Hence
Hence
q~1.HencellR(L)ll~
(-8,) (Ai, zo) -6t2(R (hi) Ai, 20) (;zz,4
(Ifq)C,
~~~~llR(~~)ll~q~~.
A modification of the Crandall-Kikuta method
We thus have
p32I-=albl, q
llA211~IhI ll~(L)1lllA~ll~qllA,ll. In addition,! &I
llR(L) ll<61.
Then, in view of the fact that I iLi I -C . . . -=z~6,~~qlC(l+q), weget llR(L-i) I 6,1llR(~,,_i).ll
ll<(l+q)C,
we have IjR (hz) II-C(l+q)C.
Since162(<1611~q/(1+q)C,
Assume that, for k = n - 1, we have proved that
llA,ll~ I&,-i I IIR (L-1) IIIIAn-ill
IlA,,ll~q2n-211An--lll~q2”-1-illAillr
With k = n we obtain from (1.3): 6 =
-&,_,
(A,,-,, xo)
n
+
la,_,
-St-, (R L-I) An-i, 50) (G, 4
12(A,_,
zj (an_l) A,_,) +a-,a,“_,
(B
(k--i) An-i, R (in-i) An-,) ,
(%I, G) llznl12=llxol12-S,_i(xo, R(~L,-,)A,-,)-~,-~(R(~~-I)A,-,,
lo)
+16,_,12(R(h,-I)A,-,;R(h,-i)An--l), ll~~ll~~l120~12~21s,-,lll~(~,-,~ >llsol12 (i-2$+) x0
IIIIL-illll~oll
>llsoll”(i-2g)
> y
.
Hence
ISnI(.&h-ilIlA.-ill
[ll~oll+l6n-il
llfl(L-J
+l6,-ilIlR(h~-i)llllA,-~ll+16,-,l2lI~(h,-~~ll2llA,-~lll < q2”-‘-‘ls._.l~[ll.roll+llxoll+llAJl+llAilll
llll~oll
V. V. Ditkin
16 where q==4f,(f+g)<1.
Consequently, f&I<&+“/&,-,
j
Notes. 1. By the theorem, 16nl-0,IIA&+O, as k-f 00. 2. Let the conditions of Theorem 1 hold. Then, for any integer k > 2,
where
M=
IIR (h) II.
mu
I*--Li
For,byTheorem 1,wehave h~,(~(&_~~~... ( I& j for any integer k 2 2. Hence ah the hk lie inside the circle ) h.-ho 1G )Sit and IIR ( hA)IIGM for any k ;;S2. Since all the conditions of Theorem 1 are satisfied, we have
=
ISA-I
1II&-1119, IlAi\k--lllM,
AA=-~%+-IR(?v+--~)AA--~, IIAJ=S16k-1/
k&:2.
It is not possible in practice to solve the problem
&E-H,,
ka2,
(1.4)
inasmuch as information is then needed about the subspaces Ho and HI. Let us modify the problem (1.4), under the assumption that ah the conditions of Theorem 1 hold with respect to this problem. For this, we consider the numbers wh=(&, u&f -', k= 1, 2, . . . , where ‘&lf is a sequence of known elements of H, convergent as k -t m to the known element a~7 such that (x0, a) +O. Obviously, lim k-m
I
Wh--
1 I
=O.
We write problem (1.4) in the form (A-l,_,Z)Z
= ~&*,
17
A modification of the Crandall-Kikuta method
cfk, Gt)=l.
(A-h*_,z)fk=phfk-,,
(1.5)
Since &=u~~,,=u~~~,,+~u~A,, and ask + 0~
we have
-& IIf,- ____ c%a>\I I II- (x0,a) II+o. 50
d
1Ilxoll+lal
Wh
IIAAI,
&
Xk---
In addition, since ~=L-L.l,
we have
We have thus proved: Theorem 2 I.& the conditions of Theorem 1 be satisfied. Assume that, by solving problem (1.5), we have found 5% @, hk~ (A&., C)/( &, &) fork~2,We~enhave,askjm,
II&-& II-+a
Ihn-ho I “0,
I&l-+0.
Note 3. The following may be suggested as a possible method for selecting the sequence {akf . Assume that a,, sl=zl, (&, Q#O are known. Asak we take tie element C--II(ZI, Z”H), where &_ 1 is obtained when solving problem (1.5) at the previous step, k > 2. In this case, obviously, for any k > 2 the problem (A-
hk_,f)2”h=Gkzk--l,(z”h, F&-l)=
(31, L-f),
(1.6)
hk = Wkr z”k) ,
will be solved, and since 1
(21,Sk-f)
(i,, 2"A-1) =
WAS-=
(Zkrak)
(rkr %-l)
(Zkrzk-1)
V. V. D&kin
IS then
(54,x0)
lim wk= (50,fo) k-ta, and hence
2. Consider problem (1.6). We put L1==h. Zi-,==j,L=r. ps=p. (Zt, &) ==x,where are known numbers. We have to find the element xE:H and the number ~1,
fis a knownelement of H, and h, x
It was shown in Paragraph 1 that, given a specific choice of initial approximation (i.e., of a pair ht ,x1), we have I(f--~+,/l<& where 6 is a smafl quantity and Y# 0. We shall show that the problem (A-hZ)s=yf, is well posed with respect to its ~~t-h~d problem
(2-I)
side. For this, we consider, along with problem (2.1), the
(A -hZ) 2=$qY-h, where &zH
(X?f, ==%
(3, /! =X+F,
(24
and the number E are small perturbations.
We shall show that, if h and e are small, then’;;,; are close to x, 1. We put f-x=6x, We obtain
a-p==Ap
and subtract the corresponding equations (2.1) from (2.2).
(A-M)
Ax=Apf+h,
(Ax, f) =e,
(2.3)
where Ax=H, Ap is a number. Let us write Ax, J h as Ax=Ax”+Ax’==ax,3_dr’,
f=f”+f’=p&+f’,
(2.4)
h=h”i-h’==ys,-kh’, where As’, f”, h%H,,; Ax’, f’, IzkH,; 01, @, y are numbers. These representations are unique. Since ~/~--~xO~~~~,then p + u as 6 -+ 0. For, consider the element !4=f---Y~0=rj~o+f’-W0= @--~)~~+l’. ~b~ou~y, Py= ~~~-~7~~o+~~= @v ) xor where I-’is the operator ~tr~uced in Paragraph 1. Hence we fmd that II @-v)~oll= i.e.,
II Pit I@--Vi
G ---I% llr Ii
I p-%q Ilzoll=II~yll~llPllJiyl~~IIPll~,
A modification of the Crandall-Kikuta method
In addition, since h is small and fis close to “0, then h’=H,, in (2.3), we get A (ax,+Ax’) -h (ax,+Ax’) !ax,+Ax’, fix@+/“) =E. ~Ax~---~~z~~H~,
Since
19
7 are also small. Substitut~g (2.4)
=Ap( /3xo+l’) +yxoih’,
Apfixoi-~xOEHo, AA?-LAx’EH,,
Apf’+h’EH,,
we have
AAx’-iLAx’=A&V-h’, aAx,--c&xo=A~~x,+yx,, a6 (x0, x0) +a(xo, f’) +p (Ax’, x0) + (Ax’, f’} ==E, Noting that Ax,=h,xO,
we obtain (A-U)
(ho-h) a=ApB+y, a~ll~~~~‘+a(x~, f’) +B (Ax’, x0)+ (Ax*, f’) =E. Ax’=A.CLfi+hl,
In order to find hx and Ap, we have to solve this system of three equations with three unknowns Ax’, A v, a. It has already been remarked that the norm of the operator R(X) in the subspace HI is bounded by a constant M, where M=
max IIR(h) II. lb--b/
Hence
The determinant D of the system of two equations is equal to
Although problem (2.1) is solved almost on the eigenvalue of the problem (1 .l), i.e., (ho - A) is a small quantity, in view of the fact that in this case Ilf-v.q/I is also small, the term 1fi ~2~~xo~~z will differ considerably from zero. Hence D is not close to zero. On solving the system and estimating the qu~tities Act, AX, we obtain
Since 1~1, 1e 1 and Ilh’Jj are small, 1Ap 1, IIAxll are also small. We have thus shown that problem (2.1) is well posed with respect to its right-hand side. 3. As H we take the space L2(0, l), in which
V. V. Ditkin
20
(a,
b)=fu(t)T&e. 0
We consider in this space the problem
Lx=hx,
B,x=O,
octc1,
1=0, 1, *.
where Lx=~‘“)+a~(t)s~“-“+. . . -I-a,_,(t)x(“+a,(t)r, complex functions in 0 4 t Q 1, t
al(t),
.(
n-l,
. . . , u,,(t)
(3.1) are continuous
n-i
I= 0,1,...,k - 1,
‘Cbilx(*) (O), ,=o
B,x = ’
n--i 1= k, k + 1, . . . , n - 1,
b,GP (1))
and bil are complex constants, i, Z-O, 1, . . . , n- 1. Allowing for the boundary conditions, we can write problem (3.1) as Ax=hx,
(3.2)
and we shah assume that everything said about the properties of the operator A in Paragraph 1 likewise holds with respect to problem (3.2). While a single equation is considered for simplicity, the following discussion applies equally for a system of equations. We arrive at the problem (L4Z)x=pf,
0<1<1,
(x,f)=j2(t)iiij;lt=x.
B,z=O,’
(3.3)
z-o,
1,. . , , n-l,
where Ais a known number, sufficiently close to the eigenvahre of problem (1. l), and fis a known function belonging to L2(0, 1). We require to find the function 5 (t) EL’ (0, 1) and the number ~1. The pivotal condensation method is well known to be very useful for solving a system of linear ordinary differential equations (see e.g., [6]). However, in our case, direct application of this method for solving problem (3.3), i.e., elimination of x from the first equation, and the process of finding x/p= ( (L-AZ) -‘f, f), would lead to difficulties of the type outlined earlier. The following device makes it possible to use the pivotal condensation method to solve (3.3). We introduce the auxiliary functions
xn+i = p,
x,+2
=
s h)f(z)d~ Xl
A modification of the Crandall-Kikuta method
21
and rewrite (3.3) as a system of differenti~ equations, using the relations (z (t) =x, (t) ) : I 5, ===xz, .., .
x;*-i = xx, J,,‘-bi (t)
.
.
.
.
.
.
.
.
.
.
x,,+ . . . +a,(t)xI-hx,-x,+if=O,
I
x,,+, = 0,
z:+z - x,f = 0, while adding to the n given boundary conditions B,x=O, Z=O, I, ..., n-1, the two extra conditionsz, +,(O)=O, Z,+Z(1)=x. In this way, the well-posed problem (3.3) reduces to a system of n + 2 linear differential equations with n + 2 boundary conditions, of which k + 1 are specified on the left-hand end, and n - k + 1 on the ~~t-h~d end. In view of what was said above, this boundary value problem is numerically stable and can be solved without difficulty by the method of transferring the boundary conditions (see [7,8] ). The author sincerely thanks A. A. Abramov for his guidance. TF~sl~ted by D. E. Brown REFERENCES 1.
CRANDALL, S. H., Iterativeproceduresrelated to relaxation methods for eigenvalues problems, Sot., A207,No. 1090,416-424,195l.
2.
KMUTA, T., Convergence of iterative methods, Progr. Theorer. Phys., 10,No. 6,653-6?2,1953.
3.
GAVURIN, M., A method of false perturbations ISI-170,196l.
4.
GAVURIN, M. K., The solution of “ahost 151-154,196O.
5.
KATO, T., Perturbation theor for linear operators, Springer. 1966.
6.
BAKHVALOV, N. S., Numerical methods (Chislennye metody), Nauka, Moscow,1973,
7.
ABRAMOV, A. A., A version of the pivotal condensation method, Zh. vjichisl.Mat. mat, Fiz., 1, No. 2, 349-351.1961.
8.
ABRAMOV, A. A., On transfer of the boundary conditions for systems of linear ordinary differential equations (a variant of the pivotal condensation method), Zh. vj%hisZ. Mat. mat. Fiz., 1, No. 3, 542~545,196l.
Roe.
Roy.
for finding eigenvalues, Zh. vJchi.sl.Mat. mat. Fiz., 1, No. 5,
singular” operator equations, Usp. matem. Nauk, 15, No. 5,
12
REFERENCES 1.
GAUTSCHI, W., Computational aspects of three-term recurrence relations. SIAM Rev., 1967,9,24-82.
2.
BLANCH, G., Numerical evaluation of continued fractions. SIAM Rev., 1964,4,383-421.
3.
MIKLOSKO, J., The numerical computation of three-term recurrence relations and the tridiagonal system of linear equations by the method of shooting. Zh. vj$risl. Ma. ma. Fiz., 1974,14, No. 6,1371--1377.
4.
MIKLOSKO, J., Numerioky vypooet mdiagonalnych systemov na pocitacoch s malou kapacitou pamate. Proc. Res. problems of techn. cybernetics. Bratishwa, Slovak Acad. Sci 1965,99-l 11.
5.
FADDEEV, D. K., FADDEEVA, V. N., Computational methods of linear algebra. M., Fizmatgiz, 1963.
AMODIFICATIONOFTHECRANDALL-KIKUTAMETHOD" V,V.DITKIN Moscow
(Received 14 Juiy 1975) A MODIFICATION of the Crandall-Kikuta method for finding the eigenelement and eigenvahre of the problem Arp==kp, is considered, which does not involve the solution of an ill-posed linear system. Furthe~ore, for the case when A is a differenti~ operator, a device is described whereby the pivotal condensation method can be used to solve the supplementary linear problems that arise. A method for finding the eigenelement and eigenvalue of the problem A cp=x(P, where A is an operator, specified in Hilbert space, was devised by Crandah and Kikuta in [I, 2] . In this method, the eigenelement and eigenvahre are found by an iterative process, which can be written as
where pn+l is a normalizing factor. A calculation from these expressions demands the solution of the equation. (il-h,,l) rp,,+i=p, +!cp>I almost on its eigenvalue, so that numerical realization of the Crandall-Kikuta method is extremely laborious. Clearly, the growth of II~,lrl II can be avoided by a suitable choice of the non&zing factor pn+t . However, the introduction of pn+l does not of itself improve the properties of the system for fmding q~,, ! ,. Below, we consider a modification of the Crandall-Kikuta method, such that there is no need to solve an ill-posed linear system. In addition, for the case when A is a differential operator, we outline a device for using the pivotal condensation method for solving the supplementary linear problems that arise. It seems likely that, on occasions, our method will offer a more convenient means for avoiding the above-mentioned difficulties than does the method of false perturbations [3,41. 1. Let H be a Hilbert space. We consider the eigenelement and eigenvaiue problem
XEH,
Ax-Ax,
Ilxll~0.
Here, A is a given non-self-adjoint operator, satisfying the following conditions.
*Zh. @hid. Mat. mat. Fir.,
lb,
4, 838-846,
1976.
(1.1)
A modification of the Crandall-Kikuta
method
13
Condition 1. The space H can be expanded in a direct sum of subspaces Ho and HI, invariant
under A, where the subspace Ho consists of the eigenvalues of A corresponding to the eigenvalue Ao (which is in general complex). This expansion of H in a direct sum of two subspaces defines a projector P onto Ho parallel to HI, where P is a linear bounded operator (see [ 51). Condition 2. We shall assume that Xo is not a point of the spectrum of the operator A,
considered in HI, i.e., the operator R (A,) = (A --&I) -’ is bounded in the subspace Hr. Put C=ll (A--L,Z)--‘llu,. w e want to find Au and the corresponding eigenelement. Consider the following iterative process. Given z,=H and h,. Let ~,=t,,fA,, GEH~, A,=H,; h,=L,+o,. We shall solve the problem ( A--h,Z)x,=p2s,, while imposing on the required element x2 the auxiliary condition zz=z,,+A2, A,EH ,. As a result, we find .Q, pz, L= (A&, G) / (z,, x1). We then solve the problem (~1-h,Z) X~=~LGTI:under the conditions that x:=xU+AJ, A+H,.We find.r3,~lz. AS= (Ax3. x3) / (x3, xJ) ,etc. We obtain a sequence of numbers h ,, . . . , h T1. . ; pi,. .,pr,... and a sequence of elements xl, . . . , xr, . . . . The iterative process just described has the following properties. Theorem 1
Let conditions 1 and 2 be satisfied, and let z,=~,fA, A,ER,, that IlA,ll/ll~,ll~~~‘/~, IS,IGq/(l+q)c, where q=4~(1i-~)
hh=
(A-hd)sh=~h~h-I,
where
x;,=x,+~~,
&,d,,
I*,,
hh=ho+6h
=ho-a
rz;)
k,
h,=hO% be given such Assume that G,, PI,, hk
k=2,3,.
,
for any integer k > 2. Then,
~6,~cq’“-‘~6k-,~~q2*-‘-‘~6,~,
h--l,
~~A~ll~y’b-z~~Ah-,~l~q2K-‘-‘~~~~~~,
ka2.
Proof: Given any integer k > 2, we have Sk-,= xO+Ah_,, Ah_,=H,. in the form where AGH,; then (A-Ah-,/)
(x:,+11,)=,.ih
(ho-hk_,)xo+
Hence, since(h,--3L,_,)~p~H,, /.h=hO-h k--l, k>2. Further,
pAxo~H,,
(A-h&)
%N32
Xk
will be sought
(G+&-,)r
Ak=p~o+clJk-,.
(A-hk-II)Ah~H,r
Sk=-Gk-_i R (hr_i)xh-,=x0-&A
ptbk-,E~,y
(L-L) Ah-,,
since R (hh-,).zo=-&~~xo. In view of the fact that -ik-,R(hk-,) we find from Eq. (1.2) that Ak=--IS,,-,R (ha-,) Ah-,. It is easily shown that
a.,
Ah-,ER,,
we obtain
(1.2) while G,=s~+A~,
Y. K Ditkin
14
Consequently,
(AXA,Sk)
(Xl*4
= hr=ho
~o)+~~~-~~“(Ak-~~R(hk-i)Ak--L)
+ _~,_,(~,_,,~~)-_6~,(R(~~-_t)Ak-,, I
(x&,4 + s,_,S,“-, (R&+-i)&-+
(1.3)
R&--l) A,-l) ,
(&:k,3k)
k=2,3,.
..
We know that (D c
(Ak-ho)
mR”+i
(ho)
=fd
(hk),
k=l,
2,. . . ,
?u-0
if
~~,/=Jh~-hc/~lIIIR(ko)ll=l/C.
Byhypothesis,161jdq/C(l+q), With k = 2 we obtain from (1.3):
8, =
Hence
Hence
q~1.HencellR(L)ll~
(-8,) (Ai, zo) -6t2(R (hi) Ai, 20) (;zz,4
(Ifq)C,
~~~~llR(~~)ll~q~~.
A modification of the Crandall-Kikuta method
We thus have
p32I-=albl, q
llA211~IhI ll~(L)1lllA~ll~qllA,ll. In addition,! &I
llR(L) ll<61.
Then, in view of the fact that I iLi I -C . . . -=z~6,~~qlC(l+q), weget llR(L-i) I 6,1llR(~,,_i).ll
ll<(l+q)C,
we have IjR (hz) II-C(l+q)C.
Since162(<1611~q/(1+q)C,
Assume that, for k = n - 1, we have proved that
llA,ll~ I&,-i I IIR (L-1) IIIIAn-ill
IlA,,ll~q2n-211An--lll~q2”-1-illAillr
With k = n we obtain from (1.3): 6 =
-&,_,
(A,,-,, xo)
n
+
la,_,
-St-, (R L-I) An-i, 50) (G, 4
12(A,_,
zj (an_l) A,_,) +a-,a,“_,
(B
(k--i) An-i, R (in-i) An-,) ,
(%I, G) llznl12=llxol12-S,_i(xo, R(~L,-,)A,-,)-~,-~(R(~~-I)A,-,,
lo)
+16,_,12(R(h,-I)A,-,;R(h,-i)An--l), ll~~ll~~l120~12~21s,-,lll~(~,-,~ >llsol12 (i-2$+) x0
IIIIL-illll~oll
>llsoll”(i-2g)
> y
.
Hence
ISnI(.&h-ilIlA.-ill
[ll~oll+l6n-il
llfl(L-J
+l6,-ilIlR(h~-i)llllA,-~ll+16,-,l2lI~(h,-~~ll2llA,-~lll < q2”-‘-‘ls._.l~[ll.roll+llxoll+llAJl+llAilll
llll~oll
V. V. Ditkin
16 where q==4f,(f+g)<1.
Consequently, f&I<&+“/&,-,
j
Notes. 1. By the theorem, 16nl-0,IIA&+O, as k-f 00. 2. Let the conditions of Theorem 1 hold. Then, for any integer k > 2,
where
M=
IIR (h) II.
mu
I*--Li
For,byTheorem 1,wehave h~,(~(&_~~~... ( I& j for any integer k 2 2. Hence ah the hk lie inside the circle ) h.-ho 1G )Sit and IIR ( hA)IIGM for any k ;;S2. Since all the conditions of Theorem 1 are satisfied, we have
=
ISA-I
1II&-1119, IlAi\k--lllM,
AA=-~%+-IR(?v+--~)AA--~, IIAJ=S16k-1/
k&:2.
It is not possible in practice to solve the problem
&E-H,,
ka2,
(1.4)
inasmuch as information is then needed about the subspaces Ho and HI. Let us modify the problem (1.4), under the assumption that ah the conditions of Theorem 1 hold with respect to this problem. For this, we consider the numbers wh=(&, u&f -', k= 1, 2, . . . , where ‘&lf is a sequence of known elements of H, convergent as k -t m to the known element a~7 such that (x0, a) +O. Obviously, lim k-m
I
Wh--
1 I
=O.
We write problem (1.4) in the form (A-l,_,Z)Z
= ~&*,
17
A modification of the Crandall-Kikuta method
cfk, Gt)=l.
(A-h*_,z)fk=phfk-,,
(1.5)
Since &=u~~,,=u~~~,,+~u~A,, and ask + 0~
we have
-& IIf,- ____ c%a>\I I II- (x0,a) II+o. 50
d
1Ilxoll+lal
Wh
IIAAI,
&
Xk---
In addition, since ~=L-L.l,
we have
We have thus proved: Theorem 2 I.& the conditions of Theorem 1 be satisfied. Assume that, by solving problem (1.5), we have found 5% @, hk~ (A&., C)/( &, &) fork~2,We~enhave,askjm,
II&-& II-+a
Ihn-ho I “0,
I&l-+0.
Note 3. The following may be suggested as a possible method for selecting the sequence {akf . Assume that a,, sl=zl, (&, Q#O are known. Asak we take tie element C--II(ZI, Z”H), where &_ 1 is obtained when solving problem (1.5) at the previous step, k > 2. In this case, obviously, for any k > 2 the problem (A-
hk_,f)2”h=Gkzk--l,(z”h, F&-l)=
(31, L-f),
(1.6)
hk = Wkr z”k) ,
will be solved, and since 1
(21,Sk-f)
(i,, 2"A-1) =
WAS-=
(Zkrak)
(rkr %-l)
(Zkrzk-1)
V. V. D&kin
IS then
(54,x0)
lim wk= (50,fo) k-ta, and hence
2. Consider problem (1.6). We put L1==h. Zi-,==j,L=r. ps=p. (Zt, &) ==x,where are known numbers. We have to find the element xE:H and the number ~1,
fis a knownelement of H, and h, x
It was shown in Paragraph 1 that, given a specific choice of initial approximation (i.e., of a pair ht ,x1), we have I(f--~+,/l<& where 6 is a smafl quantity and Y# 0. We shall show that the problem (A-hZ)s=yf, is well posed with respect to its ~~t-h~d problem
(2-I)
side. For this, we consider, along with problem (2.1), the
(A -hZ) 2=$qY-h, where &zH
(X?f, ==%
(3, /! =X+F,
(24
and the number E are small perturbations.
We shall show that, if h and e are small, then’;;,; are close to x, 1. We put f-x=6x, We obtain
a-p==Ap
and subtract the corresponding equations (2.1) from (2.2).
(A-M)
Ax=Apf+h,
(Ax, f) =e,
(2.3)
where Ax=H, Ap is a number. Let us write Ax, J h as Ax=Ax”+Ax’==ax,3_dr’,
f=f”+f’=p&+f’,
(2.4)
h=h”i-h’==ys,-kh’, where As’, f”, h%H,,; Ax’, f’, IzkH,; 01, @, y are numbers. These representations are unique. Since ~/~--~xO~~~~,then p + u as 6 -+ 0. For, consider the element !4=f---Y~0=rj~o+f’-W0= @--~)~~+l’. ~b~ou~y, Py= ~~~-~7~~o+~~= @v ) xor where I-’is the operator ~tr~uced in Paragraph 1. Hence we fmd that II @-v)~oll= i.e.,
II Pit I@--Vi
G ---I% llr Ii
I p-%q Ilzoll=II~yll~llPllJiyl~~IIPll~,
A modification of the Crandall-Kikuta method
In addition, since h is small and fis close to “0, then h’=H,, in (2.3), we get A (ax,+Ax’) -h (ax,+Ax’) !ax,+Ax’, fix@+/“) =E. ~Ax~---~~z~~H~,
Since
19
7 are also small. Substitut~g (2.4)
=Ap( /3xo+l’) +yxoih’,
Apfixoi-~xOEHo, AA?-LAx’EH,,
Apf’+h’EH,,
we have
AAx’-iLAx’=A&V-h’, aAx,--c&xo=A~~x,+yx,, a6 (x0, x0) +a(xo, f’) +p (Ax’, x0) + (Ax’, f’} ==E, Noting that Ax,=h,xO,
we obtain (A-U)
(ho-h) a=ApB+y, a~ll~~~~‘+a(x~, f’) +B (Ax’, x0)+ (Ax*, f’) =E. Ax’=A.CLfi+hl,
In order to find hx and Ap, we have to solve this system of three equations with three unknowns Ax’, A v, a. It has already been remarked that the norm of the operator R(X) in the subspace HI is bounded by a constant M, where M=
max IIR(h) II. lb--b/
Hence
The determinant D of the system of two equations is equal to
Although problem (2.1) is solved almost on the eigenvalue of the problem (1 .l), i.e., (ho - A) is a small quantity, in view of the fact that in this case Ilf-v.q/I is also small, the term 1fi ~2~~xo~~z will differ considerably from zero. Hence D is not close to zero. On solving the system and estimating the qu~tities Act, AX, we obtain
Since 1~1, 1e 1 and Ilh’Jj are small, 1Ap 1, IIAxll are also small. We have thus shown that problem (2.1) is well posed with respect to its right-hand side. 3. As H we take the space L2(0, l), in which
V. V. Ditkin
20
(a,
b)=fu(t)T&e. 0
We consider in this space the problem
Lx=hx,
B,x=O,
octc1,
1=0, 1, *.
where Lx=~‘“)+a~(t)s~“-“+. . . -I-a,_,(t)x(“+a,(t)r, complex functions in 0 4 t Q 1, t
al(t),
.(
n-l,
. . . , u,,(t)
(3.1) are continuous
n-i
I= 0,1,...,k - 1,
‘Cbilx(*) (O), ,=o
B,x = ’
n--i 1= k, k + 1, . . . , n - 1,
b,GP (1))
and bil are complex constants, i, Z-O, 1, . . . , n- 1. Allowing for the boundary conditions, we can write problem (3.1) as Ax=hx,
(3.2)
and we shah assume that everything said about the properties of the operator A in Paragraph 1 likewise holds with respect to problem (3.2). While a single equation is considered for simplicity, the following discussion applies equally for a system of equations. We arrive at the problem (L4Z)x=pf,
0<1<1,
(x,f)=j2(t)iiij;lt=x.
B,z=O,’
(3.3)
z-o,
1,. . , , n-l,
where Ais a known number, sufficiently close to the eigenvahre of problem (1. l), and fis a known function belonging to L2(0, 1). We require to find the function 5 (t) EL’ (0, 1) and the number ~1. The pivotal condensation method is well known to be very useful for solving a system of linear ordinary differential equations (see e.g., [6]). However, in our case, direct application of this method for solving problem (3.3), i.e., elimination of x from the first equation, and the process of finding x/p= ( (L-AZ) -‘f, f), would lead to difficulties of the type outlined earlier. The following device makes it possible to use the pivotal condensation method to solve (3.3). We introduce the auxiliary functions
xn+i = p,
x,+2
=
s h)f(z)d~ Xl
A modification of the Crandall-Kikuta method
21
and rewrite (3.3) as a system of differenti~ equations, using the relations (z (t) =x, (t) ) : I 5, ===xz, .., .
x;*-i = xx, J,,‘-bi (t)
.
.
.
.
.
.
.
.
.
.
x,,+ . . . +a,(t)xI-hx,-x,+if=O,
I
x,,+, = 0,
z:+z - x,f = 0, while adding to the n given boundary conditions B,x=O, Z=O, I, ..., n-1, the two extra conditionsz, +,(O)=O, Z,+Z(1)=x. In this way, the well-posed problem (3.3) reduces to a system of n + 2 linear differential equations with n + 2 boundary conditions, of which k + 1 are specified on the left-hand end, and n - k + 1 on the ~~t-h~d end. In view of what was said above, this boundary value problem is numerically stable and can be solved without difficulty by the method of transferring the boundary conditions (see [7,8] ). The author sincerely thanks A. A. Abramov for his guidance. TF~sl~ted by D. E. Brown REFERENCES 1.
CRANDALL, S. H., Iterativeproceduresrelated to relaxation methods for eigenvalues problems, Sot., A207,No. 1090,416-424,195l.
2.
KMUTA, T., Convergence of iterative methods, Progr. Theorer. Phys., 10,No. 6,653-6?2,1953.
3.
GAVURIN, M., A method of false perturbations ISI-170,196l.
4.
GAVURIN, M. K., The solution of “ahost 151-154,196O.
5.
KATO, T., Perturbation theor for linear operators, Springer. 1966.
6.
BAKHVALOV, N. S., Numerical methods (Chislennye metody), Nauka, Moscow,1973,
7.
ABRAMOV, A. A., A version of the pivotal condensation method, Zh. vjichisl.Mat. mat, Fiz., 1, No. 2, 349-351.1961.
8.
ABRAMOV, A. A., On transfer of the boundary conditions for systems of linear ordinary differential equations (a variant of the pivotal condensation method), Zh. vj%hisZ. Mat. mat. Fiz., 1, No. 3, 542~545,196l.
Roe.
Roy.
for finding eigenvalues, Zh. vJchi.sl.Mat. mat. Fiz., 1, No. 5,
singular” operator equations, Usp. matem. Nauk, 15, No. 5,