m A x Y m x m A m to she

BiiUGmm momm

" F " ™ '

i .jfT.i i or Preffeatittv

M l a o y P x e f m r o r

,7a Liu I Bireotpr oi 1sfa© Department of Mat3a©ffi&tiGS

MMBtasw

l«aa "of' tii© Graduate Sahool

AN APPROXIMATE SOLTJTIOBf TO THE

DIRICHLJBT PROBLEM

THESIS

Presented to the Graduate Council of the

lorth Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

MASTER Of SCIENCE

By

Edward William Redw±ne» £,S.

Denton, Texas

August, 1964

TABLE OP CONT3SKTS

Pag©

LIS® OF ILLUSTRATIONS. . iv

Chapter

I. IN TKODU CTION . » %

II, THE PHOCliJDUHE FOR THE APPROXIMATION 5

III. SOLUTION OP THE SET Of LINEAR EQUATIONS 34

BIBLIOGRAPHY 40

ill

LIST OF ILLUSTRATIONS

Diagraa Pag©

1* Lattice Points.. • * ....... 2

2# Lattice Point wit& Thy©© Adjacent Points......» 6

3« Lattice Point with. Two Adjacent Points*........ 7

4. Lattice Point with One Adjacent Point* 8

5* Lattice Point with Pour Adjacent Points*«...••• S

6 • Lattice Points on Polygon. 9

iv

CHAPTKH I

INTRODUCTION

In the category of mathematics called partial differ-

ential equations there is a particular type of problem

called the Dirichlet problem# Proof ie given in many

partial differential equation books that every Birichlet

problem has one and only one solution. The explicit

solution is very often not easily determined» so that a

method for approximating the solution at certain points

becomes desirable, The purpose of this paper ie to present

and investigate one such method, I'he Dirichlet problem can

be stated as follows.

Definition 1.1. Let H be a simply connected* bounded 2 *•

region in E whose boundary B is a contour. Let S * RUB#

If a gllren function g(x»y) is defined and continuous on B»

then the Dirichlet problem is that of determining a

function u « f(x,y) which is;

(a) defined and continuous on Hf

(b) harmonic on E» that is, a 2 fUty) + a

2f(».y) e 0 | 9x <9y

(c) identical with g(xty) on B.

Example. Let the function g(xty) « 3x ~ y + 2 for all

points (x,y) which satisfy x 2 + y2 » 4 be given. Then the

problem of determining a function u » f(x,y) such thatt

(a) f(x,y) is continuous for all points (x*y) for

which, + y®£ 4, aad

(b) f(x,y) is harmonic at all points (x,y) for

which x2 + y 2C 4,

is a Diriohlet problem.

Before proceeding to the method of approximation, a

description of lattice point® must be given#

Definition 1.2» Let E be a region in I 2 with bound-

ary B. Let (x,y) be a point of HUB, Let L^ be the set

of all points (x + mh, y + nh), m « 0, ~1, —2, «*• ,

n « 0, —1, ~2, , which are contained in H U B for fixed

h> 0. L^ is said to be a set of lattice points, defined

on HUB, and h is the mesh width.

(-3,0)

(0,3)

*2

•" 1

3 4 5 ©

7 8 9 10 II \

12 13 14 15 16 -NI

19 20

(o,o)

21 22 23 j

24 25 26 27 28

29 •

(3,0)

(0,-3)

diagram 1—Lattice Points

Example* Let B be the oirole whose equation is

s 9 and let E be its interior, let h « 1,

(x,y) » (0,0), then is the set of points (0,3)* (-2,2),

(-1,2), (0,2), (1,2), (2,2), (-2,1), (-1,1), (0,1), (1,1),

(2,1), (-5,0), (-2,0), (-1,0), (0,0), (1,0), (2,0), (3,0),

(-2,-1), (—1,—l), (0,—1), (1,—1)* (2,—1), (-2,-2), (-1,-2),

(0,-2), (1,-2), (2,-2), (0,-3). In Diagram 1, these

points have been assigned the numbers 1 - 29* respectively.

It is desirable now to divide into two disjoint

subsets Bjk and will be the set of lattice points

which in some manner lie near B and will be called the

lattice boundary, fhe remaining points of will make up

iijj and will be oailed the interior lattice points. The

following definitions describe how to determine and

exactly.

Definition 1.3. Let H be a region in with boundary

B and for fixed h>0 let be a set of lattice points

defined on HUB. Two points (x-j y )* (x2,y2) of are

adjacent if and only if the straight-line segment joining

them is contained in RUB and

(a) x1 * x2, jyg - y | « h, or

(b) y ^ es y^, |x2 — X x i » h .

Example. For the set of lattice points shown in Dia-

gram 1, the pairs of points 8,9; 10,16? 17*18$ 1,4 are

pairs of adjacent points, while the pairs 16,21$ 20,22$

1*5? 4,15 are not.

Definition 1.4, let H be a region in I 2 with boundary

B« If is a set of lattice points defined on R U B then

the interior of 1^ is the set of all points of which

have four adjacent points in X^.

Definition 1.5* Let tt he a region in S2 with boundary

B and for fixed h > 0 let be a set of lattice points

defined on RUB* She boundary of L^# also called the

lattice boundary, is defined by B^u m L^f **•&,

For the set of lattice point® displayed in

Diagram 1, consists of points 4, 8, 9* 10, 13# 14, 15#

16, 17# 20# 21, 22# 26, while consists of points 1, 2#

3, 5, 6, 7, 11, 12, 18, 19# 23, 24, 25, 27, 28, 29.

If u « f(x,y) is the unknown solution of a Diriohlet

problem on a plane point set H U B and is a set of lat-

tice points defined on H U B consisting of exactly M points#

then number these points in a one-to-one fashion with the

integers 1# 2, ••• , H. Denote the co-ordinates of the

point numbered k by (x^y^.) and the unknown function

u » f(x,y) at (Xfc#^) by uk, k «= 1, 2, ... , 1.

CHAPTER II

THE P OCJUURjJ ffGii THii

APPKOXIMATIQI

A method for approximating the solution to the

Dirichlet problem which is a variation of the Mebman-

Serschgorin-Gollatz approximation procedure will now he

described in four steps, fhis will be followed by a

consideration of uniqueness* convergence, and the solution

of a set of simultaneous linear equations which will give

the actual numerical values of the approximate solution

of the Dirichlet problem.

Step 1. Choose a point (I,y)€SUB and an h> 0

and then construct a set of lattice points on RWB.

will consist of N distinct points. lumber these in a

one-to-one correspondence with the integers 1, 2» ••• » B«

Step 2. For each point of the lattice boundary Bh»

approximate the analytical solution of the Dirichlet

problem as follows. Let (xs,yg) be a point of and

suppose it has been assigned the number s, such that a is

an integer and l^s^H. If (x .y_) is also a point of 3 «$ o

then set

(1) ug » g(xB#ys).

Let the four points (xs + h,y8), (xs,yg + h), (x0 - h,yg)»

6

(x »y« - &) be numbered w» p, q., z$ not necessarily respec-S 0

tlYQljt iiacli of w, p, q.i z is among the first I integers

if the point which it represents is in 1^. If

but is not a point of B» then at least one of the four

points w, p, q.» z is either not a point of BUB or is a

point of RUB which is not adjacent to (xg,ys).

Oase X. Suppose only on© of the points w, pt q,, % is

like che one described above and* without loss of general-

ity, suppose it is (x„ + h,y ) and numbered w. Then there a H

is a point (xQ + &w»yB)f 0<dw<h« which lies on B and

should be numbered 2• » such that the straight-line segment

Joining (xa»ys) &&& Cxg + y g ) is contained in HUB.

low approximate u_ by

(2) (3^ + h)us - dwtip - d ^ - dwua » Ssug,.

Note that Ugt » g{xs + which can be calculated and

given a numerical value.

< w h >

(vh>)rJ 8 'tfs' q. fi_LL

xs+4w'ys

)

z

Diagram 2.—Lattice Poinx with 2hree Adjacent Points

Oase 2. Suppose two of the points w, p, q, z, say w and

p, are suoh that each one is either not a point of RUB or is

a point of HUB which is not adjacent to Also,

suppose w and p are (xQ + h,y0) and (x8#yg + &)» respectively.

Then there are two points (x0 + <V»y8) (xgtys * dp)»

0 <d <h and 0 d < h, which should toe numbered 2' and V» w P

respectively# Then calculate ug by

(3) ( 2 ^ ^ * M w + M p ) u g - dwdpuq. " dwdpuz 18 hdptt2' + M w u 3 f

Note that Ugt » g(x0 + dwfy0) and u^, « g(xsfys + dp) and

both can be given numerical ralu.es,

(x .y +d ) w s s p

/ , x 8 V x0+ dw»y 0)

( XB" h' yB )*i ' ^

(*B»ya-h)lz

Diagram 3.—Lattice Point with Two Adjacent Points

Case 5* Suppose three of the points p> c[f z$ say

w, p, q, are such that eaoh one is either not a point of

RUB or is a point of R^3 which is not adjacent to (xm,ya)9 3 »

Also, suppose arbitrarily that w, pf $ are (x_ + hjyJ*

(x@iyg + h)* (xs - k*ys)* respectively* Then there are

three points (x& + dw>y8), (xs#ys + dp), (xB - d<jty0)*

0 <d wCh t 0<dp<h, 0<d^<ht which should be numbered 2%

3*» 4 % respectively. Then calculate ug by

(4) Cd^dpd^ +• hdv/dp + hd^d^ *f hdpd^)Ug *» d dpdgUg, * hd^d^Ugi

+ hdwdqu5,

Ohoose h small enough so that there does not exist a lattice

point in I*k which oanmot be connected to all other lattice

points in by path® which move from one lattice point to

an adjacent one*

8.

(x„iy +d ) -8**8 P

(xsfys-h)Iz

Diagram 4»—lattice Point with One Ad^aoent Point

Step 3* Let (xs»yg) be a point of and suppose it

is numbered s* Then calculate u_ by

(5) 4u — u — u — u — u » 0* a w p q. a

Hot© that this implies that ug is the mean value of u^,

u • u • u • P <1 s;

z

(xg,ys+h)

s v <•vh-ys>

(3cB»ys-

h)

Diagram 5*—lattice Point with Four Adjacent Points

Step 4« Steps 1» 2, and 3 together with subscript

notation produce a system of B linear equations in I un-

knowns u » u2, ... t Construct the N x N matrix A

by using the coefficients of the equation used to calculate

u . If siN, as the nonzero elements of row s of A, But m

the coefficient of u^, 1 <t£:N, in column t of A. Also*

construct the column matrix € by using the constants, such

that the constant from the equation used to calculate ug,

1 < la the element in row s of 0, I'hus, the final

step is the solution of the linear algebraic system which

gives an approximation to the analytical solution at each

point of

(-1

C-i.-I)

17

12

18 19

13

8 ( W

3

14 15 16

20 21

10 II

(s[h,ilh)

m-i'h)

Di&grasi 6#—Lattice Points on Polygon

Example. Let B be the polygon with side® produced

toy connecting the points (-1, 2j|) to (3>|> 2jjr) to (5^*-2^)

to to (2|»-1) to (-1,-1} to (-1,2^) with straight

lines and let E he the interior. Let g(x,y) * x + y on

all sides. Set (x,y) » (0,0) and h = 1. As show* in Dia-

gram 6, the points of are (3,-2), (-1,-1), (0,-1),

(1,-1), (2,-1), (3,-1), (-1,0), (0,0), (1,0), (2,0),

(3,0), (-1,1), (0,1), (1,1), (2,1), (3,1), (-1,2), (0,2),

(1,2), (2,2), (5,2) and are numbered 1 - 21, respectively.

Consider each point in order and applying the appropriate

steps gives the twenty-one equations:

1 0

+ H - i + k'ia * - i ' i ' i u 6

- z ) + - 2 ) +

( 5 - 1 + 1 ) U 6

« t t g .

**10

( 3 - i + D a , , -" X win 11*11'

4 U 1 5

4 u 1 4

4 U X 5

( 5 - | + 1 ) ^ -

( 3 * | ? + l ) u ^ g -

( 5 ' | + D t ^ g "

( 5 - f + 1 ) « 2 0 -

U g ** —1 "*1

u , = 0 - 1

tt4 - 1 - 1

a 2 — 1

K - K - K i •

® **X + 0

3 ~ " 7 ~ " 9 " " 1 3

1 - < 3 | - 1 )

u

l 4 - « g - U^Q - « l 4 « 0

~ ^ — H *» 9 " " U " * 1 5 - 0

$ " i 6 " i b i o " be " 1 - ( 5 I + 0 )

•"X + X U 1 2

~ u 1 2 ~ ^ 8 " ^ 4 *" *aX8 88 0

" U X 3 " n9 ~ ®X5 * * X 9 " 0

" U X 4 ~ ~ ^ 6 ~ * 2 0 m 0

• K i - K 5 - K i • 1 - ( 3 t + 1 )

ttX7 * ~ 1 + 2

' 1 * 1 7 * l ° 1 3 ~ ^ 1 9 m 1 ( 0 + 4 *

I t ^ X S ~ f ^ X # " § * 2 0 " i ' C 1 + 2 § )

' l h * i q *• ^ c " i f t o ' i « X • { 2 + 2 ^ ) C X? a, Xy c CX C.

11

< 2 > H + l - i + l ,i ) uzi - f K o - l - K 6

m * 2) 4- X*^(5 + 2^) &* *Y *T 4B*>

The following is the matrix form for these equations, such

that the first twenty-one columns make up A and the last column

is 0.

if 0 0 0 0 =1

M 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ° n

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -2

0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

*T| 0 0 0 =1 1Z 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1

0 0 -1 0 0 0 -1 4 ""1 0 0 0 -1 0 0 0 0 0 0 0 0 0

0 0 0 -1 0 0 0 *•1 •»1 0 0 0 "*1 0 0 0 0 0 0 0 0

0 0 0 0 •1 0 0 0 *•1 4 -1 0 0 0 -1 0 0 0 0 0 0 0

0 0 0 0 0 -1 0 0 0 si 11 4 4 0 0 0 0 s| 0 0 0 0 o 1|

0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 -1 0 0 0 -1 4 -1 0 0 0 -1 0 0 0 0

0 0 0 0 0 0 0 0 •1 0 0 0 *1 0 0 0 -1 0 0 0

0 0 0 0 0 0 0 0 0 -1 0 0 0 -1 4 -1 0 0 0 -1 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 si 11 0 0 0 0 z l 3L7 mj.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1

0 0 0 0 0 0 0 0 0 0 0 0 =1 0 0 0 dk i ^ L W mmW

0 - 1

0 0 0 0 0 0 0 0 0 0 0 0 0 z

0 0 0 •»1 9 -%L

0 ^

0 0 0 0 0 0 0 0 0 0 0 0 0 0 »1 ~f 0 0 fx k Jit JSt — — ym mm— M

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - * 0 0 0 rnrnm 1 4

12

It will now be shown that the method described above

is a reasonable technique of approximation in that

(a) if the points {xs>y0)# (xQ + h»y0)f (xg#yg + h),

(xg - h,y0), (x0,y8 - h)f numbered s, w, pf 2,

respectively, are elements of R, then (5) can be

derived within an order of h,

(b) the resulting linear algebraic system always has

one and only one solution, end

(c) the approximate solution converges to the analyt-

ical solution as the mesh width h converges to

aero#

X.et the five points UQ,y@), (xg + h*yg), (x0,ys • h),

(x0 - h,y@)t (xfi»y8 • h) be numbered s, wt p, qf z, respec-

tively, as in Diagram 5. Define D as follows*

(6) D * (aufl + bu^ + ctip + du^ + euz)

P2*(*a»ya) ^2f(x ,yfl)

d X e)y^ 0

such that a, b, c, d, e are constants and u « f(x,y) is the

analytical solution of the Diriohlet problem* Since all

five points are elements of R and u * f(x,y) is analytic

on R, faylor*s formula with a remainder can be used to gives

<7) ^ - u(x..y.) + h ^ 8 + §- — - J •* + o(h?),

13

(8) up • u(x0tys5 + h df(xtt,yj „2 S^CXstya)

0(h3)

(9) - «(r.,y.) - hJ^(*«'y«) • | 2 9 jY*"* • °<»>5>

(10) u_ m m(xfl>7_) - h. 11 '" +

Ox 6 i>x

a*c*..y.> »2 a2*<*«.y„>

- -*~sF*g, « fl)y

Substituting (7) through (10) into (6) gives

(11)

+ o(h5).

2) = u_(a + b + o + d + «) + (hi* - M ) 8 dx

+ t £ l i ^ ( h c . h 9 ) +

c>7 2

a2f(*.f7_) j»2

c)x S p £ (f * 1)

JLI^lA (|% +12.. i)

c>y

+ b0(h3) + c0(h3) + dO(h?) + «0(h3) •

She right side of (6) is equal to the right side of (11)

for any values of a, b, c, d, and e. Let a « ~4/h » , o

b s d a d s M 1/h » so that

a + f e + c + d + © « 0

hb » hd m 0

bo ** he w 0

.2 -2 | i j + | d - i s O 2 h2 0 + 1 8 - 1 * 0 ,

From (6) and (11) then

-*«» * °w • *i> + aq + u. -j. 3 f(xg>ya) c) f(xs>ys)

£>x« 3y' 0(h)

14

or

~^ub + u v + ur, + ua + "a + o ( h ) „ + ^ V I a i

jbr c)x o>y

She harmonic property of f(x,y) at (xg,yfl) implies

-»4u + u + u + u„ + u„ S SL_JB J L - — S + o{h) « 0.

h 2

Discarding the terns 0(h) gives the approximation

-4u +• n, + + u„ + u,_ SL * ^ P — a & m o

and

4u — u -» u — u — 11 « 0 s w p q z

and (5) been derived.

farther investigation require® careful distinction

between the analytical solution of the Dirichlet problem

and the approximate solution under consideration* Froa

this point on, let u « f(xty) represent the analytical

solution of 'the problem and let U « U(x,y) "be the approxi-

mate solution. Therefore, U = U(x,y) is defined only on

L^. If ( x8#y s)

e s* then (1) implies

(12) U(xB,y0) • g(xs,ys).

If a n d ie a type of Case 1, Step 2, then (2)

implies

(13) (3&y. + h)U(xB,ys) - V,^3C8,y» + &) - dwIJ(xs - h,yg)

-VJ(x8,ys - h) » hg(xs + dw,yg).

15

If (x ,y ,)6B, is a type of Gas® 2, Step 2, then (3) 8 & "

implies

(14) (2dwap + tidw + Mp)0(xB,ys) - d„<ipB(*a - h,ya)

-Vp U ( ls , yB "

h ) " hdp«(xB + + V '

if (xs.r8)£\ i» a type of Case 3, Step 2, then (4)

implies

(15) < V p d q + " V p + h V 4 +

- V p V ( x « " s " h ) * hdpdqg(l8 + dw'ye1

+ »yg + dp) + dwdp®^xs ~ d4,ya^*

I* (xs»ys)CHh, then (5) implies

(16) W(xs,ys) - V(xa + h,yg) - U(xfifys + h)

-U(xs - htjB) - (x0»ys - h) « 0.

Theorem 2*1. The system of linear equations result-

ing from the method of approximating the solution to the

Dirichlet problem always has a solution and this solution

is unique.

Proof* The constants of the equations are all that

is affected by ohaaging g(x,y). from matrix theory* if

a system of linear equations has one and only one solution,

then the same equations with the constants changed will

have on® and only one solution. Therefore, if the system

of equations has one and only one solution for g(x,y) « 0,

then the system of equations with g(x,y) equal to any

other function will have a unique solution.

16

With g(x,y) ss 0, tia® constants are all equal to

zero. One solution to this system is the on® with all

£©ros» To prove the uniqueness» assume there is a solution

U(xfy)» not a l l zeros, which satisfies the equations. Then

U(xfy) t 0 at some point Assume 11(^7^)<0*

If U(x »y]fc)>0» the proof is similar, Let ~M he the minimum

value of U(xi»y) on and (%»?^) aaif point o f ^ suc&

that

(17) * -«<<>•

Therefore, for all Cx»y)Cl» >

(18) tfGtjJx) = -M£U(x,y).

If (x^ti^CB^t then *

(a) (12) iaplies uC«xt7x^ * 0 wiliok contradicts (17) >

(fc) (13) implies that at least one of U(x ,yx +

- h,^), TJ(x ,y - h) is leas than tKx^y^)

which contridicts (18)f

(o) (14) implies that at least one of TJ(x - h,y^) and

11( ,7 • h) is less than contradicting

(18)i and

(d) (l'S) implies uGc^y^ - h) < u (%>?!) which contra-

dicts (3.8).

Therefore, is not an element of This leaves

*1* 1 Sh ^or there are only two possibilities to

consider:

(a*) if U(x, + h.y,) , + h), U(E. - h j , ) , and

- J>) are all e4ual, ttom fro. (16)

17

m 1 $ 7 X ) . y ( % + h9rx). fiiis can be continued

a finite number of times m to give

«»JC m ss U(*^ +• Bh)^) ,

wiser® (x^ + m h t y 3 1 ) ^ H o w e v e r , as shown above,

E(x,y) cannot equal »M at any point of and a

con traduction has been reached.

(b*) suppose that one of + h,^), 8(3^,3?^ + h),

v a x ~ 1 ^ ) , u(xlfyx - h) is not equal to at

least one of the other three. Ihen there le one

which 1® less than or equal to the others. With-

out loss of generality, assume that U(x^ + h,y^)

is that one* So that

^(*1*^1 + **) * + k»y^) +

U("x - h,y^) as + &g

DfxjL.?! - h) rn viXi + lijj) + a,

such that a 1 # ag, a^ are nonnegative and at least

one 1® positive. Substituting; these three equa-

tions into (16) gives

u ( % + h , ^ ) <tl(I1,y1),

contradicting the fact that UCx^y^) was the mini-

m m #

As each possible case was considered, the assumption that

there was a solution not all aseros led to a contradiction

and the theorem Is proved.

Consider now the convergence of the approximate solu-

tion to the analytical solution as h approaches zero.

18

Definition 2.1* In the case for which each of the Indi-

cated function values exists, let

(19) o[jr(x,y)J « 2£-4v(x#y) + v(x + h,y) + v(xty + Jx)

+ r(x - h,y) + v(x,y - h)J •

iOtUJLl* " «i MA c2 are «Wtr«y constants and all

tie indicated function values exist, then

sjV^tx.y) + c2T2(z,y)j . O1ofr1(x,7)] + OjOfVjCx.y)].

Proof* From Definition 2.1

o [ O 1 T 1 ( X , J) + o2r2(x.y)j - lg [ojTjU.y) + o^ix.j)]

+ [ c ^ x + h>y) -I- 0 2 ^ 2 • ii#y)J +

+ ©2v2(x,y + fe)j + f°iviCx - h,y) + «2T2*X ~ h'*H

+ f°lvl(x»y - h) + o2T2(x,y - h ) ] J rn o1[-AT1U,y)

+ Tx(x + h»y) + T^x^y + li) + r^x - hfy) + Tjtetjr - &)}

+ ° 2 C * » y ) + v2(x • h,y) + Vg(x,y + h) + r2(x - h,y)

+ r2(xty • h)j « + ©2$£v2(xty)J.

Lemma 2.2. Let E be a region with boundary S and

be a set of lattice points defined on RUB. If v(x,y) i®

defined on 1^ and sjV(x»y)J^O on and r(x,y)£0 on B&f

then T(xfy)>0 on 1^.

Proof. First, assume the opposite of the conclusion,

that v(x,y) <0 at some point of Since contains only

a finite number of points, there is a point (x^y^) CR^ at

which v(x,y) is a minimum, therefore,

(20) T(x1,y1) <0

19

and for each (xty^H^

(21) •(x1,y,)<Y(xty).

Using (19) and tie assumption that s[vU,y)J:£o leads to

(22) *(x»y)> $ f r { * + h, y) + v(x»y + &) + v<* " *'y) + T<x'* " h)3

for all (x,y)C K^,

But tli® equation© (20) to (22) imply by teolm.iq.ue8

used in (a*) and (b*) of the proof of fheorem 2»lf that

v(x,y)< 0 at some point of Since this ie a contradic-

tion, the proof is complete.

Learnta 2 >3, Let E be a region with boundary B and let

be a set of lattice point® defined on HUB. If r^(xiy)t

Tg(x»y) are defined on and - lafrl(x*y)]l - a[T2^*»yO on

\ |^(x»y)| ,Tg(x»y) on B^* then (T^(xty)| Vg(xty)

on Bjj.

jP$go£. From the assumptions, y2(x ,y) * ^ ( x t y J ^ O on

B^. From Lemma 2„1

s/v2(x,y) - v1(x»y )] . a[rs(x,3rl] - ajVjUy)]

on \ * Since o[r2(x#y)J 4- fofcUty)]! £ Q on \ $ then

°[r2(x,y)] - ajr^xiy)] oJV2(x,y)J + ^^(x^yjJ) < 0

®u Bcfuo i |

ajr2(x,y) - r^Xty)] < 0

on By Lemma 2*2, then

v2(x*y) - ^(xtyJ^O

on R. . A

low using assumptions made.

20

r2(x,y) + r1(x,y)>v2(x,y) - Jv1(x,y)|>0

on Bh# and

a(V2<x,y) + Tj_(x, r>] * G{V2(x,y)J + ^ 0

on Hence, from Lemma 2.2,

v2(x,y) + •1(x,y>> 0

on Since v2(x,y) + v^(x»y)>Q arid v2(xty) - v Cxty)**: 0

on then

T2(x,y)2/v1(x»y)|

on and the proof Is complete.

Lemma 2.4. Let E be a region with, boundary B and L^ "be

a set of lattice point® defined on HUB. If r(x,y) is de-

fined on I»h and |g|v(xtyj]f < A1 on Jv(x»y)J^ Ag on

and r is the radius of any circle which contains HUB in

its interior,- then

on R^.

Proof. Let the equation of the circle with center

(a,lb) and radius r which contains SUB in its interior be

(x - a)2 + (y - b)2 « r2.

Define w by

w(x,y) « Lr2|l - fe" ^ ~ b?2

T + a 2 .

By direct calculation

o{w(x,y)J « -A^.

Since for any" (x,y)£RUB

21

2 . ^ 2 (* - a) ,+, (y - Z<xt

then w(x»y)> k2 on Since

on and

la[rU,y)JI •c Aj,

®|w(xiyj| cs «»A f

then

|&£r(x,y)) J ~&{w(x,y)J

on Also sine©

|v(x,y)|^A2

on and

w(x,y)> Ag

on S^f then

w(x,y)> I v(x,y) |

on B^. low, ~l&ljr(xfyjJl > ® jw(xfy)jon and w(xty)> j v(xfy) |

on which together with Lemma 2«3 implies

w(x»y)> jv(x,y)|

on 1^, or

|T(x,y)| w(x»y)^^A1r + Ag

on which concludes the proof.

Lemma 2.5. for the Dirichlet problem let u « f(x,y) be

the analytical solution and lot U(x»y) bo th© approxiiiat®

solution. If f(xiy)€£0^* on RUB and (x^»y^) is any point

of then

(2?) |t?ixify ) — f(x^>y^)|^ e + »

22

where

(24) e « max |u(x»y) -* f(x»y)| ,

lub HUB

m&.T? ( rr% ox M W M M M &

5 ?

and r is the radius of any clrole whieh. contains HUB in

its interior#

Proof. Sinoe tf(x»y) exists on only a finite number of

point® and f(x,y) is continuous on HUB, e does exist. Due

to the assumption that f(x,y)£c4 on HUB, also exists.

Let (x^y^CR^ and define Q as

Q - sji^.yj)].

Us© of finite Taylor expansions gives

(25) Gff(*1,yi)] * (xi,yi^ + $(*±97$) - h

2 D2t(x» ,y4) y3 d?t{x*»y4) 04f (E, »y,) t'd j-r^x1,y1; ^ ynx**!* mgm III I m i III! I i j F > ! Ml I l i f e Till mm < 8 p M * iiiiniinoimiini I mi i f lg i i i i « i i i A

» JhJ T* 5 ? Ox wil l inff lki I . A n n

dt(Xi>y±) hz 32 f (x±,y i ) h3 a3 f (x1 >y1 ) + f(*1t7i) - b. '."•i..nftii,,. .ft., + ii ^ 11"

OfCx^y*) h2 32f(%f3T4]

+ + f (x^y^) + h ' |

+ yj. '*tii" + ^ + f(x1 #y1) + it *—*-

* fr 3y* + fr * + fr 3 I 1

where x ^ B ^ x ^ + '^i^ E2^yi + il*xi ~ 'k<1&34<'xt*

23

yA - h<2^<y£# Since f(x,y) Is hamonie, that isf

+ ? 2X,(«^ . o, 6>x c>y

(25) becomes

S^f (B.. ty. ) 3 *f (at* fBj%) 9Or<xi»'iO • tr

i ^ L . + 3 y

<0*f(&,,7^ + 94f(3^,B4)

c) x 5 y

therefore, 2

(26) |Q| • | vCtixi.yiUM-g-* .

By direct calculation,, O^U(x^>y^)J • 0 and by use of

Lemma 2.1,

|&[u(xi,y1) • f(xi»yi)]| * lG[v(Xirf±)] - ^Cr(3c±#y1>J/

» ^[^(ac^ty^jjl * |Q{£ g t ,

low, by (24), on

(27) l^(x»y) - f(x,y)| e*

Herns®, (26) and (27) together with Lemma 2.4 imply that

' u ( xi^i > - '("l-'i)! - • +

on fi^, where r is the radius of any circle which contains

S C B in its interior and this concludes the proof,

JLtit* For the Dirichlet problem let u « f(x#y)

be the analytical solution and let U(x,y) be the approx-

imate solution. If f(x»y)£IO^ on B U B and (x^y^) is any

24

point of then

r2h2M, (28) lv(xlfy±) - f(x1,y1)| < 1 - ™ ^ + 6h

2M2,

where

(29) m lub HUB

# j «s 2t4 ^(ISI'CT and r is the radius of any circle which contains HUB in

its interior.

Prooj;, first» consider a point of (x ,y )« There

are four general oases for whioh (x.,y4 )£B. . J i) JBL

Case 1. Suppose (x^y^e^flB), then

(30) |u(xr^) - f(xyy3)j m leUyy^) - g U ^ y ^ j » 0.

Case 2. Suppose hut and

is the type discussed is Case 1, Step 2. Then using finite

Baylor expansions (5i) f(x3.y;j) -

f(x3.y3 + h) - h ~U y r 3 + h )

if aftlx^E}) + ^ ^

and for y^< \<y^ + h» a

(32) r(x3,yj - h) - f(xry3 + h) -c>y

. (2h)2 5 f(x1# )

^ ~~o?

for y - h < < y^ + h, From (32)

(33) - >. * h ) f(x1'y-1 - h ) + h> Py 2 2

25

. ( | ^ a 2 f < * r V

Substituting (33) into (31) gives

(34) f(x1»y1 + h) fixity* «* h)

f(xj,y^) « "" J \A + !i11"

2 d 2 a2£(xltB2) — #• — h """ f—

«>y c? y*

Howt

(35) f(*3.yj) - f(*3 + \,j }) - a v3 f ( x ^ V * ! *

5o2f(B3,r1)

p x

for x < x^ and

(36) f(xj - it#3T|) * f(x^ + a^jj) - (& + d^)

(b + O 2 aZf(B4,yi) + 'g J" ' "'" '%"" f Ox*

for x^ - h <B^<Xj + from (36)

„ „ - . / ' " i , ; W . . ^ „ , i , v t i ,

N V + a*) a 2 f < V i > + r r ^ f<*j - "»yj> - * a * - j p P - -

Substituting (37) into (35) gives

(38) (&. + f(xj»y^) » h f(x^ + d^y^) + d ^ x ^ - bty^)

ag(h + <*») a^fBj.y;,) ^(h + a,)2 a2f(E4,yi)

+ * . g " Px*

Use of (34) and (38) gives

Px'

26

(39) (3&w + h)f(x^y^) • + b) + V^ x3* yJ - h)

/ / 2 0 4- la £(x^ + + dwf^xj - hty^) + V 1 —T~$—

c/ y

v2w_ w \ ,2/^ . „ v ->2. 2 3' £(x1#E2) d£(h • d ^ 3

. ^ • — , ^ H -

+ d,,)2 2f(BA,yi) **» *w* (nmluilimiiuiil i«gw iii«W«*i««i|iiiia«iNM« wJfcw W .

* e)a?

from (13), since g(x^ + d^y^) « t{x^ + dw#y^)#

<«) u(*jty3) - D ( xj - ^-yj' + 3 3 7 T E u ( x3

d + w w

w+ h ® < v * j -

h ) + 3 3 7 T T f ( i i + V ] ) -

$&©», from (39) and (40)

OCZj.yj) - f(i ry 3) - j g ^ . f[0(xji7j + h)

- f(x^fyj + h)J + [d(3E3 #y^ - h) - f(xj»yj - h)J

+ [u(x^ - hty^) - £(x^ - h,y^)j - | fib2 f^ x| , 3V

~ y

2 02f(x.,S2) 32X(2,»yJ - 43i 111"» 11111»" + d (h. + d ) £ A

ay"i w w 3 i 2

c>x

so that

;J} I ^ V j 5 - f ( x r y a ) ' ^ ? s ^ H [la(xy'} + h )

f(*yyj * h>l +M xyy^ - k) - f(x 3 #y^ - h)(

+ I (x^ - Hty^) - F(x^ - h,yj)lJ

27

+ 2 ( 3 ^ tt) (2J>2 + ^ + V + ^ + h 2 + 2 V l + 4 ' »

and i t can easily be shows, that

(41) )u(Xj,y-j) - ^ [ j u ( ^ . y^ + h) - f(x^ ty^ + h)|

+ - *) - - * ) | + |u(x^ - w 3 ) . -j 3h2M2

- f(*2 • h#y^)|J + g 6

where ML is defined by (29)» ijCs

Cage 3. Suppose (x^vy^)CB^ but ( x ^ y ^ ^ ^ / I B and

is the type discussed in Case 2, Step 2. TJsiag methods

similar to those used to get (38) gives

(42) dp(h + dw) f(x^»y^) » dpfcf(xj + d ^ y ^

( (h + O l , , d J 32 f(E5 ,y.) «• d^dpfCx^ - h,y^) + " jp " ' d'Jj""' """"

(ii + a_)2<Ldv 3 2 f (Eg.y j 2 ^ 7

for Xj <Bg<Xj + d^ and x^ - h < ^ < X j j + and

(45) «,<& + dp) ' (X j . y j ) - i^fitCXyTj + 4p>

(£d (fc + d ) aZf(*1,B_) + i v a p t ( x y y i - h) + ; S ' j '

a ^ t e .+ tp)2 3 2 f (x j .E 8 ) 5 y i

for y3 <^<.3^ + dp and y^ - h<Eg<y^ > dp. So that

adding (42) and (43) giTes

(4*) [ap(ii + 4 ^ + a^tn + dp)] f ( i 3 , y 3 )

" < 2 V p + h av +

28

+ d p ) +• dwdpf(x.j - h ,y^) + d w d p f ( x j , y j - h)

( k + A w ) dP

d v <h + y 2 d A o 2 f < s g » y i )

Ox* * diF

d ^ C h + dD) ^ 2 f ( x , , E 7 ) d ^ h + d B ) 2 ^ 2 f < x , f B 8 )

* '2 0 y 2 2 0 y 2 #

Proa ( 1 4 ) , sine® g(x^ + *v»Tj) » f ( x ^ + ^ y ^ ) and

g(3tj»y^ + <lp) » *(xyyj + dp ) » t i l® :a

(45) D t e j . y j ) - K > f ( l 3 + a » ' * 3 )

+ lid^£(x^»y^ + d p ) + dwdpU(x^ - 3i,y^) + d^d^irCx^y^ - h)J

t h e r e f o r e , us ing (44) and (45) >

" ( X j . y j ) - f ( * 3 , y 3 ) « i V p + A r * h A p { & ( x 3 " h ' y 3 }

- £(x^ - b»yj)J + £u(x^ty^ - k) • f(x^#y^ - h)J

(& + O d ^ 0 2 f ( B 5 # y 1 ) (h + 4 ) 2 9 2 l ( B g , y J * # iiwriniiiiiimiiiii IN UI . ITTF". . Z „ | | L » * J * n i M i n i N i. I I R M I W M I I , R - P IT • i j | r I i M I R F I

2 Ox2 2 a x2

d p ( f e ^ d p ) 0 2 t i x ^ ) ^ (h + d p ) 2 3 2 f (X |»B Q ) J

Qy ( ) y J

| u ( x 3 , y j ) - 5 3 ^ 4 ^ — n s a j C f f e j - h «* 3 >

"* ^(*|j "* ^»yjj)l + ju(Xj»y^ — h.) «•» £ ( X j f y j — ix)|

+ (fcd , + d£ + h? + 2iidw, + d | + hd

+ + h z + 2 M p + a * ) ]

29

and it is easily shown that

(46) lUCx^y^) - ^ (|u(xj - h,y^) - f(x^ - h.y^l

3h2K9 + |U(x^,y^ - h) - - *0!) + —gr» .

Case 4« Suppose (x^y^)^!^ but (x^y^J^B^nB and

is the type discussed in Gas® 3» Step 2. Methods similar

to those used to get (38) give

(47) v V * 1 * *V f^x3ty3^ * + dp)

„ V n 4 ^ 1 * + 32«(*h.Bo) + dwd^dpf(x^,y^ - h) +

v q p ^ ' li—2~ Oy'

dwaaa0(ii + a 0)2a 2f(x ,e10)

'2 O y 2

f•03? ^ ^ y^ dp and y^ <•» h ^ +• dp i and

(48) MJ)(dw + d(1) X(x^fy^) » hdpdqf(xj + dw,y^)

, _, , . M «^(dw + a ) a^tE, ,y ) + - d^y^) + — p w ^ ^ — ~ J r —

_ M » a * ( d w + da)2«>2f<Ei2»yj)^

X2 '

for x j < \ x < Xj + dw and x^ - + < V M d i » S

(47) and (48) gives

(49) [ V a ( h + dp) 4- hdp(dw + d^jJ^x^y^)

« (d^dpd^ + hdpd^ + hdwdp + hdwd^) f(x^,y^)

« ^ p V ( a : i + * M w d p f ( x 3 - V V

4- hdwdqtf(x3,yd + dp) • V p V ( x r y 3 " h )

30

bdpdw^dw + "V o2f(4i.yJ) h < V V 4 w + 4„)2 a2i(2il2»y1)

- 2 — 3 7 " 2 —JP 4,1,^(1 + 40) dwdpd0(la + dc)

2

+ 2 " 3yZ " 2 Jy2

i'roi (15)» since

g(Xj + dw>y^) = f(Xj + dw,yj),

gCx^,;^ + dp) * + dp),

g(*j - dq>y^) = f(xj - dqty^)t then

(50) Utij.yj)

= V p d , + M p d 4 * + Mw dq LMPV<X;J + W + hd»dqf(xi,yj + V + Mwdpf'xj - V y j '

+ dwdpdqD(xj'yj " h>]-

Subtracting (49) from (50) gives

U(x^,y^) - f(x^,y^)

1 s d! d d + lid d! + hd d +• lid 'ci w p q P q. w p V q [ v P

4 # V ' j - «

• " y S f r + V

ha»dp(*« * V * - W * i h * *»> °gf(xrIiq'

^ y

+ ^vdpda + f > ,<JioO

2 Oy2 J'

31

low

|V(xj,jj) - £(xyXj)l

- W l + " A * H r S + K , 4 , ( V A ' ^ V j - h )

- - *>l + I 2 <M pa£ + + ha£dp + 2 * ^ ^

+ +

+ * V p 4 * + + V > , ' ]

and it is easily shown that

(51) ItfUj./j) - fUj.y^liJlo^.yj - h)

- f(xj,y, - h)l + —j2 .

Henee, Oases 1, 2, 3» and 4, above, imply by results

(30), (41)» (#6), and (51) that if (x^y^^B^, then

(52) Iu(x^fyj) - tUyfj)! < (hKxyjj + h) - f(xyj^ + h)|

+ ) (*jty-j - &) - f(*^y^ - 3a)l + lu(x3 - hfyj)

/ xTi 3h%U - f(x^ - .

Consider now the three points (x^fy^ + h), (x^,y^ «• h), and

(xj - la»y )« If gill three points are elements of then

(52) implies

I^Cx^yj) - f(x^ty^)| £-2j| + — y S -

If two points are elements of and one is an element of

\ then (52) and (23) imply

32

, \ , , * , . 0 r 2 h 2 3 h 2 M 0

(53) |u (x^ ,y^ ) - f ( x j » y ^ ) | ^ - | + ^ + —55— + g ^ •

I f on© po in t I s an element of and two are elements o f

Rh t then (52) and (23) imply

t ^ A l ^h%€ (54) | U ( X j , y j ) - f ( x j # y 3 ) | + | + yg £ + — j - l «

I f a l l three po in ts are elements o f then (23) impl ies

tha t

( 5 5> * 2 ^ • » * ! « .

I n any case, (53)» (54) , and (55) imply tha t

, . £ , lrv, . . v, > 3« 3r2h2KJL ItoPvu (56) M x y ^ ) - f ( x r y 3 ) | i -2| + ^ 4 + ^

But (X|»y^) i s an a r b i t r a r y po in t o f @0 tha t (56)

imp l ies

3 r V M . 3 f c V * 5S + — j r "

and

(5?) + .

Subs t i t u t i ng (57) i n t o (23) gives

!u(3t1»y i) - f ( x 1 , y i ) | £ 4- 6h2K2

f o r and the lemma i s proved#

JfePyfR ,£.•,£• I f u a f ( x f y ) i s the so l u t i on of the

M r i o h l e t problem and U(x f y ) i s the approximate so lu t i on

and f ( x , y)CC^ on HUB, then tT(Xfy) converge© to f ( x # y )

as h converges t o zero*

33

Proo£» At points of the proof follows directly from

(28). At points of the proof follows directly from (23)

and (57).

Hote that if |U(x»y) - f(x,y)| is called the error in

the approximate solution U(x,y)# then (28) is an error

bound. Howeverf (28) has very little value as an error

bound since M2 and are known to exist, but, in general,

cannot be evaluated.

CHAPTER III

SOLUTION OP THE SET OF

LINEAR EQUATIONS

The final step in the procedure for finding the approx-

imate solution to a Diriohlet problem is the solution of

the set of linear equations which can he accomplished by

many methods. These methods are usually divided into direct

and iterative. The matrix notation for this set of linear

equations is AX « 0, where A * (a^»a^) is the I x I matrix

composed of the coefficients of the H unknowns in the system

I of linear equations 2 5 * x4 « e., 1^i£lf, X is the

3*1 a*3 3 i

column vector of the 1 unknowns x^, Xg, • «. » x^ and C is

the oolumn vector of the constants, c1# c2, • , c^, of

the equations. Direct methods, of which Gaussian elimination

together with its many variants are typical, yield the exact

answer in a finite? number of operations if there is no

round-off error. Usually the procedure for a direct method

is complicated and nonrepetitious. Iterative methods con-

sist of the repetition of a simple procedure, hut usually

only give the exact answer as a limit of a sequence, even

in the absence of round-off error.

Iterative methods are preferred for solving large

spars© systems AX » C since they usually take full advantage

34

35

of the numerous zeros of A in storage and in operation.

They minimise round-off error trouble because they tend to

be self-correcting, which is not ordinarily a characteristic

of direct methods#

In any iterative method, to solve the nonsingular sys-m

tern AX * 0 (the solution of which is A C ) , a sequence of

column vectors ••• is defined with the

hop® that as The Gauss-Seidel method is one such method. In this method the termt composed

(V) (V)

of p x2 » ••• » » of the sequence of column vectors

is calculated with the use of the following equation.

» - s - JcE If i £H. 1 J»1 ai,i 3 j«i+l ai,i * aifi

fhis indicates that a^^ for l£l£N cannot he zero. She

first column vector in the sequence is a guess to the

solution made by the user* 9?he sequence produced by the

0auss~Seidel method does not always converge. Proof will

now be considered that every set of linear equations asso-

ciated with the procedure for the approximate solution to

a Mriohlet problem can be solved by the Gauss-Seidel

method.

Definition 3.1. The S x I matrix A « (&±,a^) has

diagonal dominance if

I*1-1' -

5*1

36

for all liiil with strict inequality for at least one i.

Definition 3.2. 3!he I x If matrix A m ( a ^ ) is

reducible if the set of integers £~1» 2, ... 9 bJ is the

" t o 0 f t w 0 8 e t B 3 8,111 1 B»011 t h a t ai,j - 0 f o r

all i in S and 3 in 3?*

Ufa® following theorem can be found in most book© which

diecuss iterative methods for solving linear simultaneous

equations.

Theorem 3.1. If the M x $ matrix A has diagonal

dominance and is not reducible, then the Gauss-Seidel

method converges for any initial vector approximation X ^ #

Sheorem 5.2. The G-auss-Seidel method converges for

any set of linear equations produced by the procedure of

approximating the solution of the Dirichlet problem with

any initial vector approximation .

Proof, First» A must be shown to have diagonal domi-

nance. Consider row i of A which is associated with the

equation used to determine u^ as described in Steps 2 and

3 of Chapter II* If then four cases exist,

fhe manner in which the matrix A was constructed causes

the coefficient of uA to be the diagonal element ai

and the coefficients of u^, up, u f uB, if they exist, are ai,w* ai#p*

ai»q» ai»z* respectively# From Case 1, Step 2,

(34w + h>ui - dwup - 4 A - V z - hu2-

so that the only nonzero elements of row i of A are

37

(3dw + h), - d^, - dw, and - dw for which

+ n| > | - I + I - dw I + / - dw |

or

K , l l >

m

]?rom Case 2, Step 2,

(2d d + hd + hd )Ui - d d a - d d u » hd u«, + hd u«« v w p w p' i w p q. w p z p 2' Y 3 1

so that the only nonzero elements of row i of A ar«

(2d.,dr. + hd__ + hd„)f - d-d. and - d-.d for whioh

w p w p w p w p

I 2 V p + + Mpl>l- V p l + V p 1

or

l ai.i J >

am

From Case 3, Step 2,

( V p d q + hiwdp + " V * + M p V u i " V p V k

• hdpdqu2, + hdv/dqu3, + hdwdpu4,

so that the only nongero elements of row i of A are

( V p 1 ! + h V p + " V , + & < W m a " W i f o r whio11

1 W o . + h V p + hV(i + Mp dq 1 ? | - V p d , 1

01*

? j5'al.a1-

}*i

38

If is on B» then

ui m 8^xi,yi^

so that the only nonzero element of row i is 1 and

|l|>0 or

H

•ai,i' >

3**

if "^en

4ni - "w - up - uq - u2 " 0

so that the only nonzero ©laments of row i of A are 4, - 1,

- 1, - 1# and - 1 for which

|#| l| + 1 * it + I » It + I* 11

or

If

K . J - j g K . j ) -

m

There must he at least one lattice point which is an

element of B^, so that strict inequality does hold for

at least one i and therefore A has diagonal dominance.

Consider now whether A is reducible or not# By the

way A was constructed all diagonal elements are nonzero,

ai*i ^ 8 0 "^at n o ©lament i of 3 can also he an element

of f if A is reducible. If two lattice points (x^»y^) and

(x^»y^) are adjacent then u± is used in the calculation of

u^ and is used in the calculation of u^» So that row i

has a nonzero element in column j and row j has a nonzero

39

element in column i. Then if A is reducible, one of S and

$ must contain both i and J since # 0 and * 0»

flie assumption was made earlier that any two lattice points

can be connected by a path moving along lattice points from

on® point to an adjacent point. Shua the above procedure

can be repeated a finite number of times to show that any

two numbers i»3 must both be in on© of the two set® 8 and

T. therefore, if one of the integers £l, 2, ... , is in

8, then all of them must bt in 3 and likewise for f« Thus#

A is not reducible since 3 and f must both be nonempty for

A to be reducible# By Theorem 3*1 the §suss-Seidel method

converges for matrix A and the proof is complete#

Proof is given in many books that the maximum and

minimum value® of the solution of the Diriohlet problem

occur on the boundary B* Thus, an initial vector approxi-

mation can be best mad# by considering values on 3.

BXBLIGGMPHir

Porsythe, George E, and Wasow, Wolfgang a,» iouatioftSi

Greenspan, Donald,

Varga.^Hichard^S,, i S t i M S s I e w Jersey,

40