NATIONAL BUREAU OF STANDARDS REPORT

2332

ON THE ACCURACY OF THE NUMERICAL SOLUTION OF THE DIRtCHLET

PROBLEM BY FINITE DIFFERENCES

J. L. Walsh and David Young

U. S. DEPARTMENT OF COMMERCE

NATIONAL BUREAU OF STANDARDS
U. S. DEPARTMENT OF COMMERCE

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NATIONAL BUREAU OF STANDARDS REPORT

NBS PROJECT NBS REPORT

1101 -10-5100 March 10, 1953 2332

ON TOE ACCURACY OF TOE NUMERICAL SOLUTION OF TOE DIRICHLET

PROBLEM BY FINITE DIFFEREfCES

by

J a L« Walsh and David Yoong

Uni ve r sd t y,x JBia .Li f d r n i a .atrLo« nge^. * Harvard University

University of Maryland

The publication, reprli

unless permission Isob

25, D, C, Such permls

cally prepared If that

Approved for public release by the

Director of the National Institute of

Standards and Technology (NIST) on October 9, 2015

part. Is prohibited

lards, Washington

•t has been speclflrt for Its own use.

ON THE ACCURACY OF THE NUMERICAL SOLUTION OF THE DIRICHLET

PROBLEM BY FINITE DIFFERENCES 1

by

J®L® WALSH2 and DAVID YOUNG3

Harvard University

1 » Introduction

Although finite difference methods afford a powerful

tool for obtaining numerical solutions of partial differential

equations, little is known about the accuracy 0 It is the purpose of this paper to derive numerical bounds for the error,

in certain closed regions, of the difference analogue of the

Diriehlet problem® We shall be concerned only with the difference between the exact solution of the difference equation

and the solution of the Diriehlet problem* The error bounds

which we obtain involve quantities which can actually be computed such as the mesh size, and the oscillation and modulus

of continuity of the given function on the boundary* So far

1. This papeT^ was "presented to the American Mathematical Society

in September 1951; an abstract was published in Bull® Amer®

Math*Soc® 52, *+78(1951 )* The paper was prepared in part

while Professor Walsh was engaged as a consultant on a National Bureau of Standards contract with the University

of California at Los Angeles* The research was also supported by the Office of Naval Research under Contracts N5ori07834 and N5ori-76, Project 22, with Harvard University,

and by the Army Office of Ordnance Research Project TB2-0001

(*+07) with the University of Maryland®

2® Harvard University; Consultant, National Bureau of Standards

contract with the University of California at Los Angeles®

3® University of Maryland®
2

as method is concerned the chief novelty is the use of the

difference analogues of harmonic measure and the Schwarz

Alternating Process*,

k

Gerschgorin [h-] derived error bounds for boundary value

problems associated with elliptic partial differential equations*, These, and similar bounds derived by Gollatz [2], and

Mikeladz® [10], involve bounds, in the closed region, of

certain partial derivatives of the solution of the differential

equation*, However, the solution of the differential equation

itself ii not known, to say nothing of its derivatives, (although

approximate values for the derivatives may sometimes be found

by examining the corresponding difference quotients) » Also,

it may happen that although the derivatives in question are not

bounded in the closed region, the solution of the difference

equation may still converge to the solution of the differential

equation-:,

Rosenbloora [13], presented an error bound for the Dirichlet

problem which is closely related to Gerschgorin 8 ® but which

utilizes special properties ©f harmonic functions 0 By us® of

well-known inequalities giving bounds for the partial derivatives of a harmonic function at an interior point in terms of

the oscillation on the boundary and the distance from the

boundary, he obtains upper bounds for the derivatives in closed

subregions o Then, by solving the difference equation on subregions and, as the mesh size approaches zero, letting these

4 0 Numbers in brackets' ['^J7 refer to the bibliography at the

end of the paper*,
3

subregions approach the given region, Rosenbloom obtains an

error bound involving a/* (6), the modulus of continuity in the

closed region of the solution of the Dirichlet problem,. However, Rosenbloom does not discuss the question of finding

a> *(b) in terms o f to ( 6 ) , the known modulus of continuity of

the given function on the boundary 0 Furthermore in practical

numerical work one would not wish to change the boundary of the

network as the mesh size is decreased,.

In j 3 we sse the explicit solution of the difference analogue of the Dirichlet problem for the rectangle, obtained by

Le Roux [9] and by Phillips and Wiener [12], to derive an error

bound involving bounds for the derivatives of the given function

on the boundary 0 In $ 4- and § 5 harmonic measure and its finite difference analogue, discrete harmonic measure, are used

to obtain bounds for the moduli of continuity of harmonic and

discrete harmonic functions respectively in certain regions in

terms of aj(b) 0 Phillips and Wiener [12] showed the existence

of such bounds whereas we obtain suitable bounds in a precise

numerical form„ Then, in f 6, these upper bounds are used to

yield a uniform error estimate for the closed rectangle in

terms ©f the oscillation on the boundary and ^(b) a In § 7

the extension to regions made up of two or more overlapping

rectangles is discussed,,

/ 2o Discrete Harmonic Functions

Let h and k be arbitrary positive numbers, and let

L[h,k] denote the set of points (x,y) such that both x/h and
y/h are integers*, LetiX.be a simply connected closed region

with interior R such that the boundary S of XLeonsists of

straight lines each of whien is parallel to a coordinate axis

and contains a point of L[h 9 k]*> Let XX^ denote the subset

of points of L[h 9 k] contained inXX*, Two points (x^ ) and

(x 2? y 2

) ©f L[h,k] are ad lacent if

(2o 1 ) [(x

1 =x 2 )/hj 2 + [(y 1 =y2 )/k] i = 1„

A point of jfL.^ is an interior point ©f XX^ if the four adjacent

points belong toXX^o All othe^ points of XX^ are boundary

points*, We let R^ and denote respectively the set of interior and boundary points ofXX^o Evidently ? we have R^ — R

and S^c So

A function U(x 9 y) defined onXX^ is said to be disc rete

harmonic 9 (d.h*) in R^ if it satisfies the difference equation

(2»2) 6

2 [U(x 9 y) j = [2o~

2

/(1 + o2

) ][U(x+h ? y)+U(X“h 9 y)-=2U(x,y) ]

+ [2/(1+ o~ 2

) ][U(x,y+k) + U(x,y-k)-2U(x,y) ] = 0,

where

(2<>3) tr “ k/h 0

The finite difference analogue of the Dirichlet problem is

the following problems given a function f(x ? y) defined on S 9 to find a function U(x 9 y) defined ©nXX^? doh*, in R^ and coin5 « Heiibronn [5] introduced the term discrete harmonic function *) and studied the properties of these functions*,
5

ciding with f(x,y) on The existence and uniqueness of a solution of this problem for bounded regions is easy to prove;

see for instance Gerschgorin [4]„ The convergence to the solution of the Dirichlet problem has been proved using non-constructive methods by Le Roux [9]? Phillips and Wiener [12],

and others

o

§ 3« Error Estimate for the Rectangle under Differentiability

Assumptions

Let-Cl be bounded by the lines x=0, x=a, y=G , and y=b,

where a=Ah, and A and B are positive integers® Let f(x,y)

be defined and continuous on S and let u(x,y) and U(x,y) denote respectively the solution of the Dirichlet problem and

its finite difference analogue with boundary values determined

by f(x,y)o In this section we shall derive an upper bound for

the error U(x,y)-u(x,y) in the region XL^, under certain assumptions about the derivatives of f(x,y) 0 If f^(x,y) is defined

and continuous on S, we denote generieally by u^(x,y) and

U

A (x,y) the solutions respectively of the Dirichlet problem and

its finite difference analogue with boundary values f^(x,y).

First, one can verify directly that the function u^ (x,y)

defined by

(3*1) u.j(x,y) = f (0,0) + [ f (a ,0) - f(0,Q)J (x/a)

+ [f (0 ,b) - f (0 ,0) ] (y/b)

+ [g(a,b) + g (0 ,0) - g(a,G) - g(0,b)j (x/a)(y/b),

which is linear on every line parallel to either coordinate
6

axis s is both harmonic in R and d„h 0 in R^o Thus if f^(x 9 y)

is defined on S and equal to (x,y) on S, then u

1 (x,y)

= U

1 (x,y) ini\ L

o Next 9 if f 2 (x»y) - f(x,y) - f.j(x,y) on S, then we have by

linearity of Laplace B s equation and (3°1)°

(3»2) U(x 9 y) - u(x 9 y) = [U2

(x,y)+U.| (x,y) ]-[u2 (x,y)+u

1 (x,y) ]

= U

2 (x 9 y) - u 2 (x ? y) 0

Evidently both U

2 (x ? y) and u 2 (x ? y) vanish at all the corners

of XL o We study arbitrary boundary values f(x 9 y) by studying in

turn four functions each of which has the boundary values

on 0ne of XL and values identically zero ©n the

other three sides,, By the symmetry of the situation it is sufficient t© study in detail only ©ne of the latter functions,,

Let us set f^(x 9 y) = f

2 (x ? 0) when y=0 and f^(x ? y) = 0 elsewhere ©e So Since f 2

(x ? y) is continuous and vanishes at the

corners @ fXL ? £^(x ? y) is continuous on So

It can be verified direc tly^that if the Fourier series

of f^(x 9 0) converges to f^(x 9 0) 9 then we have

(3.3) uj( x,y) = £

A

n sin(n^/a)

^

J

n=1

where

6* f*r [dj pages 95=96,
7

(3.4)

and that 7

A

n - (2/a) / f^(t ? 0) sin (n7rt/a)dt ? '0

(3.5) U

3 (x,y)= £ A* -i-W.) ^n^

s a{nU///b) ] n=1

where

A-1

(3-6) A* - (2h/a) "T~ sin(n7rjh/a)f -A jh,0)

n 4—r ^

J=1

and where m and n satisfy the relation

(3.7) sinh (m7rk/2a ) = C~ sin(nirh/2a) .

Now let g(x) = f-^(x 9 0) o We now prove the following theorem?

Theorem 3. 1 » Let g(x) satisfy the following conditions ;

(a) g(x) and its first (s-2) derivatives are continuous ,

0 < x < a

( s— 1 ) (b) g

y (x) fails to exist or fails to be continuous

for at most a finite number of points in the interval 0 < x < a 5 and g

v y (x) is uniformly continuous in each open interval in

which it exists and is continuous

(c) g

v '(x) exists, except possibly for a finite number

of points, and is bounded , 0 < x < a 0 Then uniformly for all (x,y) £ A. we have

(3.8) U:j(x,y) - u

3 (x,y)l ^ (2+C) D

2 (h/a),

if s = 2, and

(3.9) U^(x ? y) - u^(x 9 y)| ^ (1+Ce) D^(h/a)

if s = 3? where

7 o See for instance Le Roux [9] or Phillips and Wiener [12]
8

(3*10) C = (1/2!+)ir2 (l+ cr 2 )coth(c6 7rb/a)/c6e

(3^11) fv> 2 - 2(a/Tr) 2[2J

1 M1 /a + Mg]

|0 3 = 2(a/ir) 3[2(J2

+1 )M2/a+M

3

]

( 3 * 12 ) C8 = (2/7TCT) Sinh 1 (CT)

and J . , i=

? 2^ denotes the number of points in the open interval °atOTO” 3L / O \ 0 < x < a at which g

1

( x ) does not exist or is not continuous „ Here NL , i= 1

,

9 3 ? denotes the least upper bound of the modulus

of g^(x).

Proof * Lemma 3*1 » If ra and n satisfy (3*7)? then

(3*13) 0 < n-m ^ (1/24)(1+ cr 2

)Tr

2

n 3 (h/a) 2 ?(n=1 , 2, . . „ , A-1 )

Proof . Let p = mtrk/2a, q = nTrh/2a. Then (3.7) takes the form

(3® 1 4) sinh p = CT sin q e We now study the

terval 0 ^ q ^ tt/ 2*

p

1 (0)= 0"? and

function p(q) defined by

One can verify directly

p"(q)

(3.+) in the in * that p(0)=0,

p

,?

(q) ; 2 2

•2 <T sin g)eos q

^2 . 2 . 5/2

( 1 + CT sxn q)

y

where primes denote differentiation with respect to q. By the

Extended Mean Value Theorem, since p

n (q) ^ 0, we have p crq»

Therefore m ^ n 9 On the other hand, we also have

P

" a ( q

)

o~(i + ( 2 2 -20“ sin '

(1+ (T sin q)

^ cro+O

since | (1 -2 (T~sin^q) (1+ (J“ 2 sin 2

q)

2

j <

Extended Mean Value Theorem we get

P(q) - CTq=( 1 /6 )q 3 p«’ ! (])

and

Again using the

(0 < ^

< q)
9

| p(q) ^crq| < (l/6)q^ cr (1+C“ 2

) and the lemma fellows

®

Lemma 3*2 ® If m and n satisfy (3o7) and if && is given

by (3*12) , then

(3.1 5) ra > n ? ( n = 1 ? 2 , . o • § A) .

Proof 3 Defining p and q by (3«1*0 have

(p(q)/q-[p ! (q)q~p(q) J/q

2

.

By the Mean Value Theorem we have

p(q) = q p

8 (|) (0 < f

< q).

Sinee 9 as shown in the proof ©f Lemma 1 f p

,9 (q) < 0, we have

p

s (q) < p

B (|) and p’(q)q=p(q) < 0® Therefore the ratio

p(q )/q is a non-inereasing function ©f q , (0 < q < ^/2).

Its minimum value in the interval 0 ,< q < ir/2 is assumed when

q = ir/2 0 Therefore

p(q)/q > p(^/2)/(tt/2) = (aAJsinh”1 (fir)

,

and the lemma fellows®

Now for convenience let us define the function

( 3 , 16 ) p (y) = £ilhI lm^b/aJi1 ry/b)J

m

sinhCntfrb/®)

We shall study its behavior as a function of both m and y ? where ra is assumed to be a continuous variable®

Lemma 3®3 ® If m >. 0. then for 0 < y < b we have

0 — dm P ra ^ > -(* y/«)coth(npr b/a)exp[-itfr y/a)].

Proof® Differentiating (3®l6) with respect to m we obtain

1m P m

(y) = (r b/a)

£

(1 ~y/b)cesh[ (m>f b/a) (l=y/b) ]sinh(mFb/a)
10

-cosh(m?r b/a)§inh[ (m*r b/a) (1 -y/b) ] j

sinh“2

(mir b/a)

= ('ir b/a) Pm (y) ^(l-y/b)coth[ (mT b/a) (1 -y/b) ]-c«th(nfir b/a )^

We note that for K > 0 ? x > 0 ?

4- (x e©th(Kx) )=coth(Kx)-=Kx csch 2 (Kx) = A

dx 2 sinh2

(Kx)

But since sioh(2Kx) > 2Kx ? the last expression is non^negativ*

therefore by the Mean Value Theorem we have

(1 =y/b)eoth[ (iwr b/a)(1-y/b)j < cothCnpr b/a) , (0<y<b)

and

35 rB

( y) < 0.

On the other hand

p ( v ) = (tt b/a) 5 (i-v/b) gpsh[(mir b/a)0Ty/bjJ _ dm I m

{yj k D/aJ y/D; sinh (inpr b/a) /.

-coth(mir b/a) Pm (y)J

= (, b/a)c9th(mir b/a) [(1 -y/b)

= (IT b/a)coth(rair b/a) £(1-y/b) sinh(S

( b/afcosh(mir b/a)

- (y/b)Pm (y)J.

For 0 < y <_ b the terms in the brackets are respectively nonnegative and non-positive 9 whence

l^mCy) 1= (

fr y/a)coth(i®r b/a)Pm

(y) s
11

Finally? w© note that

Pm(y) = exp(-nflr y/a)

1 -exp[-2(m<r b/a) (1 ~y/b) 1 1-expL-2«ir b/a)”~

< exp ("DPT y/a)

provided

Lemma

0 < y < b, The lemma now follows®

3 o^ o If m and n satisfy (3° 7)? then

(3 ' 17)

| Hn(y)

I -

[c8th(ot,r b/a)/ote]n 1

(1 <. n <, A) 5 (m < < n) ,

where oc is given by (3*12).

Proof . It is easily verified that if 0 < t, 0 <_ K, then

It e“

Kt

| < (Ke)”1 o Therefore

y expC-eflt y/a) < [ (npr/a) ©]“**

.

The lemma follows from Lemmas 3 e 2 and 3^3 and from the fact that

coth(x) is a decreasing function of x? for x > Q®

Lemma 3 . 5 . If m and n satisfy (3^7) ? then

| pm

(y) - pn (y)| < C(h/a) 2

n 2

, (0 < y < b) , (1 < n < A)

,

where C and c/ are given by (3*10) and (3»12)j respec tively «,

Proof o By the Mean Value Theorem we have

I rra <y> -r„(y) I r m

(y)]Ml m=n

do This inequality was proved by Phillips and Wiener [12]
12

The lemma follows from ( 3 o 1 3 ) and (3°17)»

Lemma 3 q6 0 If A^ is defines b y (3,^), then

(3.13) I AJ < D„/n!

where 1 D is defined by (3.11).

S

Proof o The proof involves the use of repeated integration

by parts of the integral expression for A

R

, and is similar to

that given by Jackson [8], pages 13”1^? for the case s = 2.

We omit the details

„

We now define the function f 4 (x,y )

fi.(x,y) = 0 unless y = 0 and

A

fh(x ? 0) = A sin(nTT x/a).

n-1 n

For s and 0 <, x <, a , we have

| 6® , W

f

+ (x ? 0)=g(x) I = [^“ A

n sin(nfr x/a) | < D

§ J

~s

n=A+1

< D

g J^Vs dt = D

s A

1 “S /(s-l)

or

(3.19) |i'i+ (x»0) - g(x) | < [D / (s-1 ) ] (h/a) s_1

On the other hand if we replace

£^(x,0) by f^CxjO) in (3»^) and (3*8) we get

A — A^ (n — 1 ^ 2 ^ * ® <> a A=1 ) o

Therefore for (x,y) £ L[h ? k], we have, since sinCA^ x/a) = 0,

A-1

(3.20) (x ? y)“U^(x 9 y) = ^

A^sininir x/a) [ Pra (y)-

n (y) ] n=1
to accompany

National Bureau of Standards Report 2332

On the Accuracy oi the Numerical Solution of the Dirichlet

Problem by Finite Differences

by JoL« Walsh and David Young

(a) The following paragraphs should be inserted on page 13?

line 9o

Evidently (3»22) could be used to obtain an error bound

for the case s=3, and this was done in the original manuscript

of the present paper,, Since then, however, a paper by Wasow,

[15], has appeared which contains a stronger result for the

case s=3 than the original form of Theorem 3o1« A combination of his methods and the original ones now yields an improvement (the present Theorem 3«1 ) of Wasow's result in this

case since we allow g^^(x) and g^ 2 ^(x) to fail to exist for

a finite number of points 0 Thus 9 following [15], we obtain from Lemmas 3- 1 ,3°2,3o3?

and the Mean Value Theorem

)|P (y) _ "P (

y

)

| Coi e(h/a) 2 (7ry/a)n^exp(-ocn7ry/a) 0 n n

By (3o20) and Lemma 3°6 we have

I i o _2P

(3o24)

I Ui+ (x,y)-u1+ (x,y)

j < C D^oCeCh/a) (7ry/a) 2P. exp (-o(.mry/a )

« Ce D

3

(h/a) 2

, oo r

— 1 —1

exp(-ooi^y/a)^[exp(oC7ry/a)-1 ] ^

(c&Try/ a ) ^ n=1

13

and

( 3 - 21 ) l l^.(x,y)-u4

(x,y)| £ X I An

l I P^y) ~ Pn (y) I n=1

A-1

< CD

§

(h/a) 2

21 “

by Lemma 3«5 and (3°l8)® It can be shewn that

{3* 2* 2*) A-1 p _ /-A p „

( 3^23-) 21 « <1+1 t

2-s dt

n=1 ~"I

f A = (h/a)'1

, (s-2)

<

j

1 |l ®g A°1 -°l eg (h/2) , (s =3K

Therefore.'*/ /2"' 2~

(3. 2-3) | j (3-<r22-) lUj+Cxjy) - u^Cxjy)! <

^ ifl-43^9^-® Since d 0 h„ functions possess

mean value properties they assume their maximum and minimum

values on the boundary as do harmonic functions*. Since the

functions [u^(x 9 y)-u1+ (x ? y) J and [ U^(x ,y)-Ui+ (x,y) ] are

harmonic and d«h-» respectively we have

1 u 3(x?y)-u1+ (x,y) < Max.,

0<x<a

g(x) “f^(x,0) |

U^CxjyJ-U^Cxjy) < Max

0<x<a

g(x) -f^CxjO))

Therefore in

I u

3 (x»y)~u 3

( x ?y)! 1 ! U^(x,y)-U1+ (x,y)| + | U^Cx ,y)-u1+ (x,y)

+ |u

1+ (x 9 y)-u^(x 9 y)| £ |u^ (x,y)-u1+ (x 9 y) |

+ 2 Max | g(x)-fh (x,0)|

„

0<x<a ^
I

1*

(3,2.3J (3.2-V),

By (3o19) 03^20 ("3o2~3) 9 the theorem follows® 0®E®D®

§

h 0 Harmonic Measure

Let JlLdenote a simply connected region with interior R

and boundary So Let S

5 denote a subset of S consisting of a finite number @£ connected subsets of S® We define the h

rmorale measure ^ (homJ 9 H[(x 9 y) 9 S

8 9 -fij as the unique function

which is harmonic and bounded in R 9 is continuous inXLexcept

perhaps at a finite number ©f points ©f S 9 and equals unity on

S

9 and zero on S=S 8 « The properties ©f harmonic measure have

been studied in considerable detail^ see for instance Nevanliona

[11] Chapter IIIo

By analogy we define discrete harmonic measure -, (doh®iSo ) 9 for regions ©f the type described in §2^ as follows?

H

L

t (x,y) , is a function d®h® and bounded in R^ 9 equal

t© unity ©n S^AS 0 and to s@r© on (i=S 9 )/l S^®

F©r bounded regions the existence and uniqueness of h.io,

is well kn@wn ? see for instance [ 1 1 ] 0 The existence and uniqueness for the half plane can be proved by the use of conformal raapping 0 The existence and uniqueness ©f d®h®m® for bounded regions follows from the existence and uniqueness ©f the solution

of the difference analogue of the Dirichlet problem,, Later we

shall prove existence and uniqueness ©f d®hom® for a half

plane and for certain other unbounded regions®
15

We list some elementary properties of d„h*m. which are

analogues of well known properties of h

»

(*f.1) 0 < H

L[(x s y ) 9 S' ,-fI

L

] < 1 for all S

8 £ S »

(4.2} If S'CS"CS, then

H

L[(x ? y) 9 S

9 9 il

L

] < H

L[(x,y),S«,^L

].

(4-. 3) If S

1 is included in the boundary of bothi!^ andlTX^

where n$nL

and if (x ? y)^XXL?then

H

L[(x,y), S

1 ,-Ql

] < H

L[(x,y),S'

The first property follows at once from the fact that the

maximum and minimum values of d,h. functions are assumed on

the boundary* The second follows from this fact and from the

fact that the expression (x ,y) ,S " ,-0.^] - H^[ (x ? y) ,S ! J

is non-megative on S^« The third follows since by the maximum

and minimum principles H^CCx^y), S

5 J > 0 and hence

[ (x ? y) 5 S

8 ] - H

l

[ (x,y) , S * r^L ]| > 0 for any point

(x 9 y) on the boundary ofjQ^. This is the so-called principle

©f qebietseweiterung . We shall us® h«m B and doh®ra® for two purposes® In this

section and in §

5

lower bounds will be derived to enable us to

obtain upper bounds for the modulus of continuity of harmonic

and d. h® functions in a closed region in terms of their

moduli of continuity on the boundary. In § 7? upper bounds

will be derived which will enable us t® use the Schwarz Alternating Process and its difference analogue for overlapping
1

rec tangles

„

Let jf}. denote the rectangle of § 3 and let S' denote the

side contained in the line y

= 0 . Theorem 4.1 . If (x*-a/2) 2 +

O O

y < ^ then

(4.4) H[ (x,y) ,S ' ,-&] > 11 - \> (b/a )(6/a)

(4.5) H

l

[

x,y) ,S > 1 - V ( b/a ) ( 6/a

)

where

(4.6) V ( b/a )

== [4 + \T

l"coth^(w b/a)

j

1

Proof . Let u(x,y), U(x ? y) be harmonic and d»ho functions

respectively in R vanishing on S=S ‘ and equal to sin(7r x/a) on

S’. By (3°3)“(3<>7) we have

u(x,y) = sin(<r x/a)

and

U(x ,y)

where

sin Or x/a) ginh[(mir b/a)( 1 -y/b)]

sinh(mir b/aj

sinh(mfJ- k/2a) = crsinOr h/2a)„

By Lemma 3»1 we have m <, 1 » Also by Lemma 3»3 we have

Hm TM ° 8 Therefore Pm

(y) _> P^y) and we get

(*+•7) U(x,y) >, u(x 9 y)o

Now, for 0 < © £

ir /2 we have'* sin © >. ( 2/0© and hence

for 0 < x < a/2 we have sin(ir x/a) > 2x/a = 1 -( 2/a) (a/2-x)

But since sin Or/2) (a/2 +J») = sin(ir/2) ( 2- 3/0) for all y> it

follows that

9. This is Jordan's inequality^ see for instance Copson [3]

page 136 *
17

sin (if x/a) > 1 - (2/a)

Also

sinh[(JT b/a) (1 -y/b) ]

and by the Mean Value Theorem

a/2-x J 9 (0 < x < a ) »

(7r/a)cosh[ (fr b/a) (1 -y/b)]

< (^/a)c©sh('T b/a)

sinh[(fl" b/a)(1-y/b)] ^ sinh(7T b/a)-(«" y/a)cosh(«' b/a).

Therefore

1-u(x ? y) < 1- -£[l-(2/a) | a/2-x| J[ 1 -Or y/a)c©th(w b/a)]}

(2/a) |a/2-x| + (it y/a)c©th(ir b/a) 0

By the Schwarz Inequality the last expression does not exceed

(l/a)[ (a/2-x) 2 + y

2

] 1y/2 V (b/a) <L (6/a)V (b/a)

provided (x-a/2) 2 + y

2

< 6

2

0 We now observe that

H[ (x,y) ,S ? ,-fl] u(x ? y)

H

Lt(x,y), S',^\] > U(x ? y) 0

The theorem now follows from (4- 0 7)»

Now letJfl^ denote the subset of L[h ? k] such that y >. 0

and let and denote the interior and boundary off^ respec tively

.

Theorem 4 0 2 o For the regionf!^ doh a m 0 exists and is

unique « Proof » Consider the sequence jf*Hj (x ,y)^

where
18

H

L

(n) (x ? y) = H

L[(x 9 y),S«AS L

(n)

9 H

L

(n)

]

and where.Og^ 0 ''* denotes the rectangle with vertices (nh,0),

(-nh 9 0 ) ? (nh 9 nk) and (-nh ? nk)® Evidently H

L

^

(x,y) exists

since doh®m® exists for bounded regions,, By (*+®3) it follows

that the sequence (x 9 y)J

is non-decreasing and bounded

above by unity® Therefore a unique limit exists which we

denote by H^(x 9 y)®

Since H

L ^(x 9 y) is d®h 0 it satisfies (2®2)® Taking limits

of both sides of (2 0 2) we find that H^(x 9 y) satisfies (2*2)

and is therefore doh® Evidently H^(x ? y) < 1 , H^(x ? y) = 1

on S

8/\ S L and H| (x ? y ) = 0 on S

9 -S^o Hence H^(x 9 y) has all the

properties required for doh®m®

In order to establish uniqueness we prove

Lemma 4- 0 1 „ If U(x ? y) is doh„ and bounded in R^ and vanishes

on S L

then U(x 9 y) vanishes in R^«

Proof o By hypothesis there exists a constant P such that

for (x,y)e/l, we have |u(x,y)| <_ P,

Given any (x ? y)£ and any ^ > 0 ? let a and b be chosen

so that (a ? b) € L[h,k] and

a > P(x 2 + y

2

)

1//2 V (l/2)/2 €

b > a®

Let denote the rectangle with vertices at (a ? b) , (-a ? b)

(a ? 0) and (-a 9 0) and let I denote the interval (-a < x < a)® $

Since |u(x,y)| < P on and vanishes on I 9 we have

a |u(x,y)| < P H

Lt(x,y), S*-I

g
19

By Theorem 4*1 we have, replacing a by 2a,

H

L[(x,y),Ia,HL J > 1 - V (b/2a)(6/2a)

2 2 2 provided x + y < 6^* But by hypothesis we have

(x

2 + y2

) 1//2 < 2 €• a/PV (1/2).

Substituting we get

Since b > a we have V (b/2a) < V 0/2); hence

and

This proves the lemma*

The uniqueness can now be proved by assuming two d.h.oio 's

and showing that their difference vanishes identically

The proof of Theorem 4*2 is complete*

We note that the existence of a unique bounded solution of

the difference analogue ©f the Dirichlet problem for-^1^ with

bounded boundary values is almost immediate* If the boundary

values are determined by g(x), then the limit of the absolutely and uniformly convergent series

solution exists* The uniqueness follows at once from Lemma

4* 1 *

is d*h* in R^, bounded in£X^ and equals g(x) on S^* Thus a
20

We next consider the region x >. 0 ? y >. 0* To find d.h»m«

for subsets of the line y = 0 we perform a sign-changing reflection about the line x = 0 and use Theorem Since the

d«hoin» is zero on the line x = 0 ? this can be done in such a way that the new function will be d«h e in Similarly to

find d.hofflo for subsets of the line y = 0 we reflect in the

line x = 0.

By similar methods the existence and uniqueness of the d.h.m*

can be established for other regions such as the semi-infinite

strip 0 < x < a, y > 0 o

$ 5» Modulus of Continuity

The modulus of continuity of a function f(x,y) in a closed regional is defined by

oj (b) = LUB | f (x

1 ,y 1 )-f (x 2

,y2

)|

where (x^y-j), (x2? y2 )€D.and (x

1 -x2

> 2

+ (y 1 "y2

^ 2

£

&2 ° Evi~

dently if 6 is not less than the diameter of XI then oj ( 6 ) equals

the oscillation of f(x ? y) in XI « For harmonic functions we prove the following theeremfe

Theorem 5<>1 o Let XI denote a bounded simply connected

closed region with interior R and whose boundary S is a closed

Jordan curve with the following propcrty s there exist constants

r0

> 0 and Q > 0 such that for any point P of S there exists

a circular sector with vertex at P, with radius Tq and included angle & containing no point of R„ Let u(x,y) be
21

harmonic in R , continuous in Cl with modulus of continuity

go (5) oj) S » If for some positive number D we have

6 < Max ,D,D( Tq/D) ^ ^ir

then the modulus of continuity of u (x,y) in XI satisfies the

(5.1) u,*(&) < u> [D(6/D),r/(,r+W >(‘+M/^)(6/D)'

r/(r+ '>)

where

= Max(2ir - ©,jt)

and

(5*2) M = Max [u(x,y>] - Min [u(x,y>]

(x,y) € S (x,y)es

is the oscillation of u(x,y) ©n S„

Proof

.

Lemma . 5»

1

» Let I denote the interval -r < x < r, y

= 0

M. 2 2 2 and let denote the region x + y , y > 0. If

(x,y) € C/> j then H[ (x,y) , I

r

, y > 0] > 1-2!/> Ar,

Proof . It is easy to show that if 0 is the angle at

(•=r,0) from I to the circle C through the points (-r ,0) , ( r ,0)

and (x,y) 9 then

H[(x,y), I p9 y > 0] = 1 - 0A.

Now if

-^/r=(1-cos 0)/sin 0

then Qjz is contained in the region bounded by C and the line

y

= 0* Hence, by the minimum principle for harmonic functions
22

we have for (x,y) €

H[(x,y), I

r ,y > 0] > 1 -(2/r) tan

-1

(>>/r) > 1-VAr

and the lemma is proved,,

2 2 2 Lemma 5o2 0 Let C„ .. denote the region x +y < r ,

° r , ip — —— —

0 £ tan

1 (y/x)< if < 2^ and let

^ denote the bounding

r radii ~ ^ ^ ^ of C ... » (If W = 2^ then B_ ... denotes the line " r,^ — 1 ” r ? <f

" " 0 < x < r, y = 0) o If

^ < r and if (x*y)

C„ , then “ ?T

H[(x,y),B

r ^,Cr ^ ] > l-O+AXvVrM'.

2 2 ? Proof o We first consider the case if = *r 0 if x +y =r^

and y > 0, then as in Lemma 1 we have

H[(x,y), I

r

, y > 0] = 1/2.

Hence, as we verily at once,

H[(x,y), I

r?

C

r

J = 2H [(x,y),I

r

, y > 0] -1

If (x,y) € Cjo ir , then by Lemma 5»1 we have

H[(x,y), I

r?

C

r ] > 1 -(4/tt ) (2

0/ r) „

The lemma can now be verified for the general case by mapping

C

r

jj. onto ^

by means of the conformal transformation

(w/r) = (z/r)

We omit the details 0 Now by the maximum principle for harmonic functions, the

maximum value of |u(x

1 ,y 1 )~U(x2 ,y2 >| for

^

-x 2

)

2+ (y.j =y2

)

2<6 2

occurs when either (x^ ,y^ ) or (x 2

,y 2

) beion 9 s t@ s °

( X

1 ?y1 ) € S then for all r such that 0 < r <_ r^ there exists
23

a circular sector C ... containing at least one point of R

r 5 x with vertex at (x^ ,y^ ) with radius r >, b and angle such

that B

r the union of the bounding radii is disjoint from R»

Since the theorem is trivially true if (^2*^2^ ^ S , we assume

(x 2

? y2

) 6 R„ Le t-0» 1 denote the closure of the connected

component of (x 2

? y2

) for the region /O C

r

. Evidently

_nt

c c

v 9 tjr

Since for (x,y )6 O R, we have

H[(x,y),B

r y

, ,C J < 1 = H[(x,y),il

i nR,-n.1 ]

and for (x,y) contained in^Q^ and on the arc of C

p ^

, we have

H[(x,y), B

r>v, , C

rjy,j = 0,

then for all (x,y)£ J"!., it follows that

H[(x,y),B

r ^

jC^] < H[ (x ? y) ,-fl.j /H R,!^ }•

But by Lemma 5<>2 we have

H[(x2

,y2),B

ril#, ,C

r > 1-(4A)(6/r

since the lemma is obviously true if we rotate the sector*

Therefore, we have

I u(x2

,y2)=u(x

1 ,y1 )| <a)(r)H[(*2

,y2 ),C

p

S 9 -flj

+ MH [ (x2

,y 2

) , S—

]

< 0>(r) + (^M/TT)( 6/r)

ir/V

Now by the assumptions on 6 we have

D( 6/D)

ff//(ir+ ^ < D(r

Q/D) = r o

*

Hence, we can choose r=D( 6/D)

7r// ^ 7r+ ^ <_ r and the theorem

follows
2MWe new consider the case where XL is an arbitrary simply

connected region, We first prove the following general theorem,

which is essentially a formulation for harmonic functions

which is equivalent to Carath^odory 5 s form of the theorem of

Milieux for analytic functions?

Theorem 5 » 2 , Let G be a Jordan subregion of |z| <1 whose

boundary consists of a Jordan arc od-, which passes through

z = 0 plus an a rc cx^ ©j[ P? |z| = 1 . Let u(z) be harmonic and

bounded in G, continuous in the corresponding closed region

except at the end points of and 06} * equal to unity in the

interior points of ^ and to zero in the interior points of

<^2» Then in every point z of G we have

(5-3) u(z) > 1~(Vjr )tan" 1 |z|

1y/2

»

Proof , Let v(z) be conjugate to u(z) in G, and let f(z) =

®xp[“-u(z)-i v(z)]« Except perhaps at the end points of the

arcs, on o4

j we have jf (z)| = e" 1

, and on we have |f(z)| = 1;

on these open arcs |f(z)j is continuous in the two-dimensional

sense. Then by the form of Milieux" s theorem presented by

Caratheodory , [l], £ 358, we have for z € G

log |£(z)| = -u(z) < -1 + (h/ir ) tan”1 |zj

as we were to prove.

Theorem 5 » 3 <> Let jQ. denote a bounded simply-connected closed

region with interior R and boundary S« Let u(x,y) be harmonic

in R, continuous in j(~i. and having modulus of continuity ^j(b) on S

If D is any positive constant, then the modulus of continuity
25

of u(x ? y) in ,0, satisfies the inequality

u>* (6) < OJ [D(6/D) ,/3

j + (‘m/ir)(6/D) l/3

for all 6 < Do

Proof a As in the proof of the previous theorem it is

sufficient to show that if C

r is a circle with center (x^ ,y^

)

on S and radius r and if (

x

? y p ^ an<^

(x-j-Xg)*1 + (y

1 “=y 2

> 2

X X r2

? then

(5oM H[(x2? y2

) ? C

r ^ S ,il] ^ 1-(Vr)(6/r) l/2

.

The theorem would then follow if we let r=D(6/D) ^ ^

since

6 < D and hence D(6/D)^^ > 6.

Now, if XL is a Jordan region, ( 5 »

1+) follows at once from

Theorem 5»2 since tan |z| £ |z| for |z| < 1, We

indicate the modifications necessary to include the case where

£L is an arbitrary simply connected region,.

Let 0 be a boundary point ofjfL, let C

r be the circle |z |- r,

and let z^ be a point interior to-fland to C^. If no point of

XL lies exterior to C

r

?

then a function v(z) harmonic in R

and continuous in XLequal to unity on S is identically unity

in ilso we have v(z

@

) = 1„ We proceed to study the contrary

case 0 If points of R lie exterior to such points can be

joined to by a Jordan arc lying wholly in R, so at least

one arc of C

r lies in R„ Let the totality of mutually dis-

«

7 o * * o Denote by R the sub- * Q joint arcs of in R be ,A2
26

region of R interior to containing and by R^ the sura of

R its reflection (inverse) in € otS and the arcs which form

part of the boundary of R

o

; if an arc of is part of the

boundary of R, it does not belong to R^ „ We modify R^ by

adjoining to R^ the interior of each Jordan curve that can be

drawn in R. and to which O and i. are exterior,. Denote this

1 o new region by R

2

<> Then R

2

contains is simply connected,

and has O as either an exterior or a boundary pointo The

subregion R^ of H- interior to C

p

is also simply connected, and

part of its boundary is an open arc A

Q

of of which every

point is accessible^ we choose A

o as the largest such arc,

so that every boundary point of R^ not on A

q

is either a boundary point of R interior to or is a point ©f not contained

in an arc of consisting wholly of accessible boundary points

of R^o By a conformal map of R^ onto the interior of a circle

y

(in which AQ necessarily corresponds t© an arc of % ) it

is clear that H(z,A

0 ,R^) exists and is unique^ this function

takes the boundary value unity at every point of A

Q

and the

boundary value zero at every boundary point of R^ not on the

closure of A » Caratheodory J § proof of Milloux's theorem is

valid for the region R^, the distance from to the boundary

of R^ is not greater than

j z

q | , so (5©*+) is valid for an

arbitrary simply-connected region, and the theorem follow#

„

The modulus of continuity of a function F(x,y) on any

subsetXXj^ of L[h,k] is defined by

MJ l

( 5) = LUB

| F ( Xl 9 y.,) - F(x2

,y2 )|
27

o 0 0 where (x^y^, (x 2? y2 )£Xl, L

ao<3 where (x^x^) +(y

1 =-y 2

) ^ & •

For d 0 h 0 functions in a rectangle we prove the following

theorems

Theorem 5 <>4 o Let .0. denote the rectangle 0 < x < a,

0 < y < b, where (a ? b) £ ^[hjkjo Let Q.^ denote the subset

of L[h 9 k] contained in Xl » If U(x 9 y) is doho in the interior of JlL » and has modulus of continuity &a>( 6) on S^ ? the

boundary of XI j_ ? then for 6 ,< r = ( a b ) 7 the modulus of continuity of U(x 9 y) inXi L

satisfies the inequality

(5.5) W* (6) < ui

L

[2 l/2

{

(r&) l/2

+ h+k}]+(M/2)M(l/2)(6/r) l/2

where M and V (l/2) are defined by (5« 2) and (4<,6) 9 respectively , Proof o As in the proof of Theorem 5®1 the maximum value

2 2 2

of |U(x^ ,y^ )-U(x 2 >y2

)| $ or (x^-x2

) + (y^ -y2

) < & is assumed

when either (x^ f y^) or (x

2 ?y 2

) belongs to S^ 0 Let us assume

that (x^,y^) € S L

o Now if (x 2? y2

) also belongs to S L

the

theorem is trivial since 5 < r„ We therefore assume

(x2? yp

) C R^o

Let p denote a straight line containing (x^ 9 y^) and ineluding one of the sides of the rectangle,. Let C denote the

closed interior of a semi-circle with center at (x^ ? y^) 9 with

radius 6^ = 2^^ ^

( 6 r )

"* h+kj, with bounding diameter included in p and containing at least one point of R 9 0 Since

6 < r we have 5^ > 21/2 S Evidently there exists a rectangle T included in C with one side contained in p and with

vertices contained in L[h 9 k] such that the sides perpendicular
to p are at least half as long as those parallel to p and

1 /2 such that the latter sides have length at least 2(6r) o

By (4-o 2) and (*+o3) we have

H

L[(x,y),Cns L ,HL

] > H

L[(x,y), T A S L ,il L

J

> Hj(x,y), Tr\S L ,Xl L r\T].

Now, for (x,y) € S^At we have

H

L[(x,y), T/Ap, T] < 1 = HL[(x,y), TAS^Hl].

Moreover, since the intersection of q and the boundary of

tA-Q.l

is contained in 1/0$^ we ^ ave

H

L[(x,y), T/^P, T] = 0

for all points of the boundary ofXl^/OT not contained in

S^/^To Therefore, for all points of the boundary of t/LTL

we have

H

L[(x,y), TnSL ,il L /^T] > H

L[(x,y), T A P, T].

It follows that for all (x,y) € t nnk we have

H

L[(x,y), T nsL ,n L AT] > H

L[(x,y), lAp, T].

Now since

2)

^ T C\ jQ. l we have by Theorem 4„1

H

L[(x 2

,y2

), T AP,T] > 1 -(1/2) V (l/2) (6/r) l/2

.

Hence

H

L[(x2

,y2), S l

-(S l AC),-Q l

] < (1/2) v(l/2)(6/r) l/2

.

Therefore

|U(x

1 ,y 1 )-U(x2 ,y2 )j < ta> L

[2l/2

{

(r6) l/2

+ h + k] ]

1/2 + (M/2) V (l/2) (6/r)
29

and the theorem follows * In the above proof if h = k, one could have obtained a rectangle with the desired properties by letting C have radius

2**

( ro) ^ hjo Therefore we have

Corollary o If h = k ? then

u> L*(c) <o> L

[2 1/2 [(r£>) 1/2+h^+(H/2) V (1/2)(6/r) l/2

.

§ 6 . Error Estimate for the Rectangle in Terms of the Modulus

of Continuity on the Boundary

In this section we obtain an error bound for the rectangle

of i 3 under the assumption that the function f(x 9 y), which

determines the boundary values , has modulus of continuity

u> (6) on So The function f^(x,y) defined in / 3 is also continuous and we denote by co^(6) its modulus of continuity for

0 < x < a ? y-0o

Let U^(x ? y) and «^(x 9 y) be given by (3^3) and (3«6) respectively,, W© define the function f^(x) by the partial sum

A

(6*1) fcr(x) = ^ sin(n*rx/a)

where the A

fi are the Fourier coefficients for f^(x,0) = g(x)

as defined by (3»4) and where the coefficients are summation coefficients defined by Jackson [7] page 9® Let

u^(x,y) and U^(x 5 y) denote respectively the solutions of the

Dirichlet problem and its difference analogue vanishing on

S except for y = 0 „ If y

= 0 the values are determined by

fcj(x)* We now prove

Theorem b , 1 » If (x ? y)6„Q.L and 0<x<a ? €<y<b,
30

then

( 6 . 2 )

where

(6-3)

JU^(x 9 y)~o^(x 9 y) | < 8&F’C TV (h/a) 2 (€/*)'

r = OC^TT^ Q^Tf^e^

M* -- Max jg(x)|

0<x<a 1 1

and where C and o/ are determined by (3,10) and (3»12) respec

t

-

ivel]

(6 .k)

(6.5)

Proof, It can be verified that

A

u^(x ? y) = 2 dA n A

n sin ( n

' ,rx/«> P„(y)

U^Cx.y) = «A

n=1

A

dA

u A

n

s in ( n^x/a) Pm

(y)

n=1 ?n

where m and n are related by (3.7) and where P (y) is defined

m

by ( 3 »

1

6 ) o Evidently we have

(6.6) I U

5

(x,y)-u

5 (x,y)| < jr |dA>n A

n J |pm

(y)- Pn (y> j n®3 s By Lemma 3*3 we have

(6,7) d n

dm m

< (

7Jr y/a)coth(m/rb/a)exp(“in(ry/a)

(0 < y < b), (m > 0).

By Lemma 3,2 we have m J> ocn for 1 < n < A; and

(6 * 8) 1 35 rm

(y) < (

jr y/a)coth (oc jrb/a)exp(- o<. n^y/a).

Using Lemma 3*1 we get ? as in the proof of Lemma 3*5
31

| rm

(y>- Pn

( y ) | ^ C <^e(h/a) 2 (ry/a )n^exp(-o< n^y/a)

Therefore we have

A 00 2 |Hm (y)-rn (y)| <2, |r.(»-Pn

(y)| n— i n—

i

p

oo

p < C oCe(h/a) C^y/a) rPexpC-oinry/a)

o=1

which is convergent for y > 0 o Now the following statement can be verified easily? if

G(x) is a continuous non-negative function for x >_ 0, nondecreasing for 0 < x < X 9 and non-increasing for X <. x, then

oo G(n)

c

J g( t; It + G(X)

n= Therefore since for V > 0« we have

Max [ x^exp(- V x) j = (3/y»e)^

x>0

it follows that

oo

n= exp (=od nr y/a ) < j t^exp(-cx<ry t/a)dt v-yo + (3a/c>4 If ye)TM*

Ur N 3

< 6 (a/octry) + (3a/cx rye

0

Therefore we get

(6.9) X |rm(y)- rn (y)| < >C(h/a) 2 (y/a)-3

< "A C(h/a) 2

( S/a)"3

since y > € ° For the Fourier coefficients, since |g(t)| < M , we have
32

( 6 . 10 ) (2/a) j g(t)sin(n^t/a)dt I ©

< (2/a) [ |g(t)j dt < 2M*

Jackson [6] proved that

I 1 " dA,n I < 3»/A < 3

hence

(6.11 ) | d

A, J »

1+0

From (6.6), (6.9)? (6.10) and (6.11) the theorem now follows.

Corollary . For (x?y)€“ XL ^ and 0<x<a?6<y<.b

we have

(6.12) | U^(x,y)-u^(x,y) | < 8m*C 7- (h/a) 2

( € /a)”^

+ 2K ( 2h

)

1 o where K is a positive absolute constant less than 3*

Proof. I U^(x,y)»u^(x,y) | < I U^(x,y)=U^(x,y)

|

+ lu^(x,y)-u^(x,y)| + |u^(x,y)-u^(x,y)|

.

By the maximum and minimum principles for harmonic and d.h.

functions we have for (x?y)6Xl^

I Uo(x?y)“U^(x?y)| , |u^(x,y)-uc(x,y)| < Max | f *(x)-g(x)|

.

0<x<a

10. See’ Jackson L7J page 8
33

Using a theorem of Jackson [7] page 7 we obt in

0<x<a

Max I f r'(x)-g(x)j < K to ^ (2a/A)

where K is a positive absolute constant less than 3® The corollary now follows from the theorem,,

We shall now derive an error bound for the original problem where we no longer assume that the boundary values vanish

at the corners or along the sides of the given rec tangle

®

Theorem 6„2» For all (x ? y )£ _(1. ^ we have

where

6/7 (a/b )9/l\ h/a) 2/7

( 6 . 1 *+)

(6.15)

( 6 , 16 ) = (2/ircr)sinh 1 (<r) , ft =(2cr/ir)sinh

-

^ (&•

1 )

6e 27/2

(6.17)

M = Max

(x ? y) € S

u(x ? y)-Min u(x ? y)

(x,y) 6 S

(6*1 8)
3*t

(6.19) «l

*( 6) = CO [2 1/2

C

(r6) l/2

+ h+k£ ]+M V (1/2) (6/r) l/2

and w(6) is the modulus of continuity of u(x 9 y) on S* The

function ^ ( b/a ) is defined by (4*6) and V (1/2) equals 3 <>97

approximately * i g.ainWnftn POM —— ———B — i—i 1 Proof * Lemma 6*1 * If u^(x 9 y) is defined by (3d) ? then for all

(x 9 y)gH we have

|u(x,y) - u

1 (x,y)| < M*

Proof * Since u(x 9 y) and u^ (x,y) are harmonic, the maximum

of |u(x ,y)-u^ (x 9 y) | is assumed on S, We can assume, without

loss of generality that the maximum occurs at a point whose

ordinate is zero* By (3»1) we have

u.Cx.O) = (1-x/a)u(0,0) + (x/a)u(a.O)

and

u(x ? 0)=u^ (x 9

< < 1 1 —x/a | u(x,0)=u(0 9 0)

+ |x/a u(x 9 0)~u (a 9 0) | *

But

| u(x 9 0)-u(0 9 0) u(x 9 0)»u < Max u(x 9 0)~Min u(x 9 0)<M9 0<x<a 0<x<a

and the lemma follows*

The above result cannot be improved as one can show by considering the case where u^ (x 9 y) is identically zero and

u(x 9 y) > 0*

From Lemma 6*1 we conclude that

|u^(x,y)| < M*
In particular we have

|g(x)| < M§

hence

(6„20) M* < Mo

Lemma 6o2 a

(6*21) < to (6) + 6 M/a c

Proof o Evidently we have

uj ^(b) < uj ( 6) + (6)

where co^ (b) denotes the modulus of continuity of u^ (x,0)

considered as a function of x, But since

u

1 (x + 6 9 0)-u

1 (x,0)=(6/a)[u(a,0)-u(0,0)

]

we have

(b) < b H/a

and the lemma follows

»

Lemma 6 o3 q If € < x < a-€ 9 € < y < b- € , (x,y) € XI

then

(6.22) |U(x,y)-u(x,y)| < 1 6m {*NaC

a (h/a) 2 (€/a)“3

>

bCb (k/b) 2 (^/b)”3j

+ ^K[co(2h)+ co(2k)

+ 2M(h/a+k-^b) ]

«

Proof o We observe that
36

| U(x,y)-u(x 9 y) j = |u2 (x 9 y)-u2 (x,y) |

.

The function [U2 (x,y)“U2 (x 9 y> ] is the sum of [U^(x 9 y)-u^(x 9 y)]

and three other terms of similar type. The lemma follows from

(6.12), from (6.20) and from Lemma 6.2. (We note that from the

hypothesis we have €/a, €/b < l/2).

Lemma 6.3 affords us an error bound for those points of

_0_ k which are at a distance of not less than €. from the boundary. In order to obtain a uniform error bound we now consider

those points which are within € of the boundary.

Since h/w <v, (l/2)/2 it follows from Theorem 5»1 9 (with

D=r) 9 and Theorem 5 oh that the moduli of continuity of U(x,y)

and u(x 9 y) do not exceed the upper bound for oj^*( 6) given by

Theorem 5 A? provided b < (ab)

1//2

. If (x,y)€jQ.^ and if (x 9 y)

is within 6 of a point of S 9 then there exists a point of S

L

at a distance not greater than € from (x 9 y)© We remark that

every point of _0- Lis within (ab) 1//2 of some point of S^.

Since U(x 9 y) = u(x 9 y) on we have |u(x 9 y)-u(x 9 y)| ,<2tj* L (6).

If I denotes the right member of (6.22) 9 then we have

(6.23) |U(x 9 y) - u(x 9 y)| <_ Max £l; 2w[ (O

In order that the error in should approach zero with

the highest power of Y- (hk)^2 whenever u>(6)/6 is bounded

as a function of 6 9 we choose

6 = r(hk/r2

) 2///\
37

Evidently we have

«/. = ^(h/o^b/a)3714

e/b = <r-2/?

( k/b) 4/7(a/b)3/l\

The theorem now follows from (6»23)? Lemma 6»3? and Theorem

5o4o

For the special case of the unit square Theorem 6*2 and

Theorem 5o*+? Corollary? give

Corollary o Ifa=b=cT = 1 then we have

|U(x?y)-u(x?y)| < Max

where

J

1 = 32 M 1 C h

2//? + 8K [ to (2h) + *+Mh]

J

2 = 2 co [2

1//2 (h

2^7 + h ) ] + M V (l/2)h

2// ^

C= CT ^/1 2 o4 e ) coth(o4-r) 0 536

c< = (2/*r)sinh“^ (1 ) a ^6b — - 2Z/2 oo 3,512

o<:3jr3 oc^ir^e 2

Substituting numerical values we obtain

J

1 < 61 M h

2//? + 2b cu (2h) + 96m h

J2

< 2u>[21//2 (h

2^7 + h)J + bu h

2^7

The above expressions for and J

2 represent a slight

improvement in the formulas previously given by the authors

[1^.
38

J 7 ® Other Regions

In this section we consider regions of the type

described in f 2 other than rectangles*. The case of two

overlapping rectangles is studied in detail as an illustration of the method which can be extended to more complicated regions o We first prove a theorem valid for two overlapping regions whether rectangles or not.

Theorem 7«1 o Let Cl ~X1° U XI” where R® R

n is not empty

and where neither region includes the other . Let u(x 9 y) and

U(x 9 y) denote respectively the solution of the Dirichlet problem and its finite difference analogue for Xlo LetXl^j^L

and HL denote the points of L[

h

9

k. ] belonging toS2,_5X.° and XL!1

respec tively 0 Let

and let

= Max HL[(x,y),Tr,-^- ]

(x,y)£T' L

L L

jj. 2 = Max

(x,y) € T"

H

L[(x ? y) ? T* l? XL' l

J

Next let U* and U” denote functions doh„ in X2 8 ^ and XT'l?

respec tively , and egual to u on S

1 ^ and S*8 ^ respec tively ,, Then for (x 1 y)gXl^ we have

| U(x,y)-u(x,y)| < Max(A ? B)

where
39

A = E'(1 +>“

1 -A1 A2) + £

"

1 “^1 A2

E”(1 + + E'

1

E’= Uax I U* (x,y)-u(x,y)

I

* (x,y) € R

1 L

E"= Max

(x,y) € R” l

|u"(x,y)-u(x,y)|

.

Proof

.

Let

F

1 = Max |u(x,y)-u(x ,y) | (x,y) 6 T’ l

F

M = Max |U(x,y)-u(x,y) I • (x,y)6T»*L

Clearly, for (x,y) € T' L

I U(x,y)-u(x,y) | < | U(x,y)-U" (x,y) | + | U"(x,y)-u(x,y) |

Max | U(x,y)-U"(x,y)j

(x,y)€T» L

+ |uM (x,y)-u(x,y)J .

Hence by the maximum and minimum principles for deh«, functions

Similarly

Therefore

F' < F" + E" „

F " </*- 2 F ' + E ' .

yu. E' + E"

' 1 1 -'/^|/«2 * F

1 <
40

E

f{ + E’

F

n <

i1 A

2

By the maximum and minimum principles

Max |U(x 9 y)-u(x,y)| < E

1 + F8 (x,y) € H' L

Max |u(x,y)~u(x,y)| £ E

88 + F"

e Si L

and the theorem follows

»

The proof of this theorem was, of course, motivated by

consideration of the difference analogue of the Schwarz Alternating Processo Indeed, in order to solve the difference equa

tion forsi, one might guess values for U(x,y) on S^Z^R'^

and then solve the difference equation in obtaining in

particular values on R'^o Next one solves the difference equation for u ®i n 9 the computed values on

S" L /0 R

8 l

? an^ °t> ta i ns new values for S This process can be repeated and the successive values thus obtained

converge to the exact solution of the difference equation,,

Moreover, the rapidity of convergence can be estimated if/*^

and

2 are known 0 In fact one can show that after a complete

iteration the maximum error is reduced by a factor of ( 1 )

If for andj^8 the quantities E

8 and E

88 are known whenever the modulus of continuity and oscillation of u(x,y) are

known on S

B and S

88 , then by using Theorem 5»1 or Theorem

E

! and E

88 can be computed provided the modulus of continuity
of u(x,y) on S is known. If, moreover, and/or/c^ are known,

an error bound for the composite region can be obtained.

We consider now the case of two overlapping rectangles

-Q-l and_Q2

with sides parallel to the coordinate axes. If the

intersection of the interiors is not empty and if neither region is a subset of the other, then the following cases can

occur.

Ca.se A Case £> Case- C—

The composite regions of Cases C and D are of the same type

as those of Cases A and B respectively. The problem of determining^^ and ^/<2 became problems in d.h.m* for the rectangle

of the following types:

FT]

1 1

i ;I r i

31 "Typetypes £ Type, h

Type. A

Figure 7«2
Upper bounds for dohoffio are required in points on the

closed dotted lines for the open arcs on the boundary indicated by heavy lines

«

By (4- o 3 ) 9 these upper bounds are not greater than the

upper bounds for the following problems?

Problem _ZT

( For

Problem JZZT

( F~of Tytpe. B/ J

Problem M- Prekl&m TV~

( Fer Types 3, f) (

^

Figure 7 „3

For problem X, since the harmonic measure of the open

arc is required, l/2 is an upper bound on the dotted line,

by symmetry 0 Theorem 7o2 0 Let _Q, ^ be bounded in part by a segment of

a line / 0 Let contain no point of Jl » Let denote a line perpendicular to H and containing a point of X f\ S^o

Let S

s denote any subset of Ji f\ contained in one of the

open half planes bounded by £

0 If (x,y) is any point of

r

l

t ^ien
**•3

H

L[(x,y), S’,XIL

J < 1/2,

Proof. The theorem follows easily by (4.2), (4.3), and

Problem I.

As a corollary we have an upper bound of l/2 for Problem

II. Also, for Problem III if we extend the lines whose

harmonic measure is required to the right, we get by symmetry

as in Problem I, 1/2 as an upper bound for the d.h.m. on the

left dotted line. Similarly 1/2 is an upper bound for the

d.h.m. on the right dotted line.

For Problem IV the situation is somewhat more complicated.

We prove

Theorem 7.3. LetXldenote the

region: x > 0, y > 0, and letfl^

denote the subset of L[h,k] contained inXX .

1

Pt ' Ft

1

| Let 1^,1^ denote respectively, the regions

+ !

!

1 0<x<a, y=0 and 0<y<b, x=0, e jC(a>)

where the point (a,b) belongs to L[h,k]« T\

!

i

We have

«Lb i

i * - >. Figure 7*4

(7.1) H

L[(x,b),I

a L7Ib ,nL

] < Max [1|£; 1 -

* o

where

(7.2) w = ( 2A)i0 g ] iirp
Here y Q

is the smallest, number not less than a/2oC such that

yo/k is an integer s The quantities o<l and o~ are defined by

( 3 . 12 ) and ( 2 a 3 ) respec tively o Proof o Let I*

a

? 1*^ denote respectively the regions

x > a ? y = 0 and y > b ? x = 0 o We first prove

Lemma 7 ol . If 0 < x, 0 < y, then

( 7°3) H^[ (x 3 b) , — ^kL(xjb)j

and

(7.^) H

L [(a,y), > H

L[(a,y),

Proof o By Theorem 7°2 we have

H

l

[ (x ? b) , y > b 71 x = 0 ? x 2 > Hj[ (x f b) ? y < b/lx = 0 ,x >. 0 ]„

Hence we certainly have

H

L[(x ? b), [y > bAx = Oj U [=b < y < 0A x = 0J , x > 0]

2 H^[ (x ? b ) 3 [0 < y < bA x = 0 j U [ y 1. “bA x = ojj 3 x > 0 ] o

But by a sign-=ehanging reflection about the line y = 0 we can

show that the right member of the above expression minus the

left member equals

H

l

[ (x,b) ? I*

b?=Q, L

] ~ H^[(x 3 b) ? 3 0 °

This proves (7<>3)o Evidently (7o^) can be proved by the

same methodo

From Lemma 7.1 we conclude that

(7°5) H^[ (x j b ) 3 j 2 (x 9 b) 5 X^

J

+ (l/2)HL

[ (x,b) , 0 < y/^ x=0,^L

]o
^5

But by (7.*+) we have

(l/2)HL[(a,y),0 < yfl x = 0,S^

L

] + H[ (a,y) ,I

g

,-^

L

] < 1/2.

Therefore the d.h. function

(7.6) W(x,y) = H

l

[ (x,y) ,I

a ,^L ]+(l/2)H

L

[ (x,y) ,0 < yr\x=0,i^

L

J

satisfies the conditions

f W(x,0) = 1 , (0 < x < b)

(7.7) < W(0,y) = 1/2, (0 < y)

.W(a,y) < 1/2, (0 < y).

Let5^* denote the region 0 < x < a/ly > 0, and let52.*L

denote the corresponding subset of L[h,k]„ Evidently

(7.8)

where

W(x,y) < (x,y)

(7.9)

We now prove

[(a) (x,y) is doh» and bounded in R*^

< (b) (x,0) =1 5 (0 < x < a)

1(c) 1^(0, y) = llj (a ,y) = 1/2, (y > 0).

Lemma 7.2.

and such that

, Let U^(x«y) be d.h. and bounded for (x,y)6 R* L

U

2 (x,0) = 1 , (0 < x < 1

)

We have

U

2 (0,y) = U2 (a,y) =0, (y > 0).

(7.10) U

2 (x,c) £ Max[w; 1 -(1 -w)c/yo

J

.

Proof . Substituting in (3*5) and (3-6) and taking the

limit as b becomes infinite we get
46

A

U9 (x? y) = 2T (2/A)cot (n^/2N)sin(nWx/a)exp(=m<Ty/a)

^ n=1

n odd

where Ah = a Q

By Lemma 3o2 we have m > cK Also 9 since cot[nn"/2A] <

2A/n*r for n < A we have

A

U

2 (*?y) 21 (4A n)exp(~^o< yn/a)

n=1

n odd

GO < 2^ (4A n) exp (-it <X yn/a)

n odd

= (2/r) log

^^p(-f.<Ky/a)

^

1 - exp y/a)

Evidently if y > a/2o( , we have

(7° 1 1 ) U

2 (x ? y) < Wo

Now since 0 < w < 1

9 the function

(7o12) U^(x,y) = 1=(1=w)y/y

o

is doho in R* L and is not less than l^Cx^y) for S*^y < yQ

and for y = yg

o Therefore 9 for y < y Q

we have

(7»13) U

2 (x,y) < 1-(1-w)y/y

o

and the lemma follows 0 Now U

2 (x ? y) = 2U^(x ? y) - 1, We also have by (7«5)» (7»6)

and (7 *8)

H

L[(x,b),I

a l7Ib ,r2

L

] < W(x,b) < U, (x,b)

and the theorem now follows 0
In applying the theorem to Problem IV we note that we can

assume the value zero at the point (0,0) without changing the

d o h oiiis Thus an error bound can be obtained for the case of two

overlapping rectangles. The above methods can also be used

to find error bounds for other regions of the type considered
48

BIBLIOGRAPHY

1 o C,

2o L<

3»

4. So

5 o Ho

6 , Do

7 O Do

8« Do

9. Jo

10o So

1 1 o Ro

1 2o Ho

13 ° Po

Cara theodory, Funktionentheorie , Vol® II ? Verlag Birkhauser Basel, 1950,

Collatz, Bemerkunqen zur Fehlerabschatzunq fur das

Dif f erenzenverf ahren bei partiellen Differentialqleiefaunqen , ZeitSo fo angew. Math 0 u D Mech,,

13, 56-570933).

Copson, Functions oi a Complex Variable , Oxford, 1935.

Gerschgorin, Fehlerabscha tzunq fur das Dif f erenzenver- fahren zur Losunq partieller Dif f erentialgieichunqen , ZeitSo fo angewo Math® u, Meeho, 10, 373~382, (1 930 )

Heilbronn, On Discrete Harmonic Functions , Proc 0 Cambo

Philo Soc o 7 45, 194=206 (1949)o

Jackson, On the Degree of Convergence of S turm-Liouville

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Jackson, The Theory of

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Jackson, Fourier Series and Orthogonal Polynomials , Mathematical Association of America Ca ms Monograph, No,

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1 94l

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Le Roux, Sur le Probleme de Dirichlet , Jour 0 de Math,,

10, 1»9-230, (T9i4) . Mikeladze, tl Uber die numerische Losunq der Differential- lleichungen von Laplace und Poisson , Bull, Ac ad * Sci,

IJ 0 S 0 S 0 R 0 , (Sero Math,), No, 2, 271-290 (1938)-

Nevanlinna, Eindeutiqe analytische Funktionen, Berlin,

1936o ~ ^

Phillips and N, Wiener, Nets and the Dirichlet Problem , Jo Matho and PhySo,“ 2, 1 05-1 24 { 1 923) 7

Co Rosenbloom, On the Difference Equation Method for

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Applications of Conformal Maps, Institute for Numerical Analysis, Los Angeles, 1949®

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15® W » W a s ow , On the Truncation Error in the Solution of

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"-TM-- ——

Res o

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