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IndianJ.pureapplMath,27(3):313-322,March1996 THE SOLUTION FOR MIXED BOUNDARY VALUE PROBLEMS OF TWO-DIMENSIONAL POTENTIAL THEORY FUQIAN YANG! AND RONG Yao University ofRochester, Rochester, NY 14627, U.S.A (Received 10August 1994; after revision 20October 1995; ‘accepted 1November 1995) Using Green's functions, theDitichlet_mixed boundary value problems for10 ‘dimensional potemtal theory aretransformed tosingular integral equations, andsolved exacly 1,INTRODUCTION Mixed boundary value problems forcomputing potential anddiffusion rates arise inmathematical physics, mechanics andengineering": °.These problems have been studied extensively :Mills etal, Mushkelishilit and Sneddon! have reviewed the available solution procedures. These techniques were classified asfollows :integral transforms; integral equation techniques; complex variables and numerical calculation, For the mixed boundary value problems oftwo dimensional potential theory, several works have recently been published. Ranger® considered thetwodimensional potential problem ofaplate charged toanarbitrary potential situated symmetrically between and parallel toearthed planes. Srivastav had earlier considered thecase where the plate isperpendicular tothe planes. The problem was reduced todual series equations andthen solved'. The stated problem inRanger’ was reduced, by successive integral transformations, toatwo part boundary value problem fora harmonic function andsolved. Rose andDeHoog® andTait andMoodie? using the complex variables, andEjike', Singh etal’, Yang andLi! andDavidson" using Fourier transforms and integral equation methods solved asetofmixed boundary value problems forthetwo dimensional potential problem. ‘Asshown inFig. 1,theproblem tobeconsidered here isthat offinding the potential distribution associated with twoparallel planes; oneplane hasacoplanar infinite strip 0.<|X|<a, ¥=0with potential g(X), and|X|za, ¥=0isinsulated, and Y=Ltheother boundary either isearthed ofinsulated. The problems may be "DeparmentofMechanicalEngineering Deparment ofElectrical Engineering 318 FUQIANYANGANDRONGYAO reduced using Green's function totheintegral equations with logarithmic kernels, and solved exactly. avy avoY V=g(X) ay! ~a 0 a > LVorwv=0ay Y Fro, 1.Physical configuration 2.THE MIXED BOUNDARY VALUE PROBLEM I 2.1, Formulation ofthe Mixed Boundary Value Problem I Consider two parallel planes, one has a coplanar infinite strip, 0s|X|sa, Y=0with potential g(X) and|X|=a,Y=0isinsulated, andthe other boundary Y=Lisearthed. We want tofind thepotential Vatany point between theplanes. ‘This isequivalent tofinding thesolution V(X, Y)ofLaplace’s equation vVUK,Y) =0 a inthedomain 0<Y<L,subject totheboundary conditions a. ~@Ve 2a,¥=0aa for|X|=a,¥ Vix,0)=g(%),for|X]sa,¥=0 @) Vor,0)=0, atY=. ~@) Further V(X, Y)should tend tozero as|X|+2. Before solving theabove boundary value problem, wefirst introduce the following dimensional parameters x= Xa, y=Yio. -(6) Substituting theabove dimensionless parameters intoeqns. (1)-(4), weobtain VFVix,y) =0 6) ‘TWO-DIMENSIONAL POTENTIAL THEORY 31S inthedomain 0<y<L/a, subject totheboundary conditions pa for|xJzly=0 ~O Vo, y)=8), for|x]s1, y=0 ..(8) Mix, y)=0, aty =Lia ~ 1VQ,y)|<, forfx]+e. .-(0) The solution forV(x, y)interms oftheGreen’s functiom can beexpressed as 1 ay=fFGlx,y, x,0)2 .a MandefGten%TsIyeat® co) where G(x,y,Xp,Yo)istheGreen's function andisdefined belowas V?G(x,¥,x0,Y0)=8(xo,Yo) ne(12) inthedomain 0<y<L/a, subject totheboundary conditions AGU,ys40.Ya)_ .ey =0,fory=0 o»(13) G(x, ¥,X,Yo)=0, fory=Lia. we(14) 2.2.Solution forGreen's Function G(x,y,XqYo) Using theFourier series, theGreen’s function G(x,y,xo,Yo)canbeexpressed as i 1)a; Gtastor ¥soeos((2+3) 7") 3) satisfying theboundary conditions (13) and(14). Substituting eqn. (15) intoeqn. (12), wwe obtain a 2 2 28m _((,,1)an l\ay). 3[SEAer2)f) el(2) F)-9 ws(16) which inturn gives =eos(ne3)P| 7) where 316 FUOIAN YANG AND RONG YAO Phy 1) ax)ao((nea) 2)noe. o-(18) Using eqn. (18) and thecondition Ihgl<@% for|x|, ww(19) we obtain L (1\ax Inospeian7y %?(-("43) Tbs) =(20) which inturn gives Ce i\e-rowd=— 3estan (-(*42) 7sl) Using eqn. (21), weobtain Geran0)Re[In(E=1 | on(22) where ris roexo|-Tx-nl-0)). @3) 2.3. Solution for the Potential V(x, y) ‘Simplifying eqn. (22) and using eqns. (8)and (11), weobtain ate)=1ffx)in|tanh(ante—19)/AL)|do ~»Q4) which isasingular integral equation, where flx,) isdefined as a,fea)i ~@5) Tosolve this equation, weconsider thefollowing two cases. (a)g(x) isaneven function inx—Since g(x) iseven inxand lx) iseven inx,weobtain xg(x)=fflxg)In|tanh(a(x—x0)/4L)tanh(a(x+.x0)/4L)|dro, a forOs<1. ..(26) ‘TWO-DIMENSIONAL POTENTIAL THEORY 317 ‘Taking thederivative inx,weobtain 12a sinh(arx/2L)cosh(anxo/2L)T£49cosh(anx/L)~cosh(anay/L) #7#1for08951, ~(27) which isidentical to @pa) cosh (anxy/2L)L{‘F00)San(anx/2L) ~sink?(amxy/2L) “ =—80) 28)sinh (axx/2L) Using thesolution given byCooke", thesolution ofeqn. (28) can beexpressed as fon, # (Sin (a70%9/L) deg 2Lsinh(axx7L)deJ[sinh?(aruxp/2L) sinh?(anx/2L)}? * ,xfiF(&)sinh(ax&/L) d&4,(sink? (axg/2L) ~sinh?(an /2L)]'? a a0) * [sinh(an/2L)— sinh?(axx/2L))"? a where Aisaconstant tobedetermined by0"), which isthecharge density onthe strip 0°s.xs1 forasymmetric potential distribution. Substituting eqns. (29) and (22) into eqn. (11), wefind that thepotential Vix, y)isgiven by 1eye!ffin|VSBaRxD SiaRVZALYVosy)xl![cosh(ax|x=x9|/2L)+cos(axy/2L) Vinh?(an[x+xq72L)+sin®(axy72L) +Hf‘cosh(ax|x+x|/2L)+cos(axy/2L)idavo e»30) (©)g(x) isanodd function inx—Since g(x) isodd inxandfixo) isodd inx, ‘eqn. (26) ischanged to 1tanh(ax(x-29)/4L) eatffedtn|aaT) [doforOsa], GI) which isidentical to 38 FUQIAN YANG AND RONG YAO ' sinh(axx/2L)+sinh(axceo/2L) xa)=~ffix)in|‘inh(axx/2L)—sinh(anxp/2L)|# ‘Using thesolution given byCooke", weobtain joe4(Sin lorae/L)droDaef[Sinb?(azcxy/2L) —sinh?(anx/2L)]"2 * . fe heJ[sinh(arxg/2L) -sinh?(ax/2)}"* sh (ax/2L)eanh(ae/2L) ~BOT[sinh?(an/2L)=sinh?(axx/2L)]"> &) which isthecharge density onthestrip 0*sx<1 for.theantisymmetric case. ‘Substituting eqns. (33) and(22) intoeqn. (11), wefind that thepotential V(x, y)for y>0isgiven by 1 1Meydaf(|cosh(az|x=x9|/2L) +e0s(axy/2L) i? (a[+¥q72L)+Sin”(0a72L -wf‘cosh(ax|x+x9|/2L)+00s(axy/2L) fedi aG4) 3. THE MIXED BOUNDARY VALUE PROBLEM II 3.1. Formulation ofthe Mixed Boundary Value Problem Ii Inthe above section, amixed boundary value problem leading toasingular integral equation issolved. Here weconsider another mixed boundary value problem: twoparallel infinite planes, oneplane hasainfinite stripe onit,O=|X|<a, ¥=0 with potential g(X), theother, |X|=a,Y=Oandtheotherplane¥=Zisinsulated Wewant tofind thepotential Vatanypoint between theplanes. ‘This isequivalent tofinding thesolution V(X, Y)ofLaplaces’s equation VVX.Y) =0, »-(35) inthedomain 0<Y<L,subject totheboundary conditions we <|X}ue, Ye a36)¥20, for|X|2=a,¥=0, Vx, 0)=g(%), for|X]<a, Y= 0, ~-G2) ‘TWO-DIMENSIONAL POTENTIAL THEORY 319 avas. = «»- (38)ay0, atY=L. (38) Using thedimensionless parameters given ineqn. (5),weobtain eqns. (6)-(8), (10) and a we : 89¥20, aty=Lia, G9) [Meyb<o% for|x|re .-(40) Similarly, thesolution forV(x, y)byGreen's function can beexpressed aseqn. (11). ‘The Green’s function G(x, y,x,Yo)isdefined below V?Gtx,¥sx0Yo)=B%0:Yo)s 41) inthedomain 0<y<L/a, subject totheboundary conditions G(x,y,X0Yo) =For) a0, ay=0, (42) peo, aty=Lia. ..(43) 3.2. Solution for Green's Function G(x,,XoYo) Using the same method asinthe previous section, the Green’s function G(x,y,x0,yo)canbeexpressed as 3 Geasezord=aple—ral-geP|—“Tlel] xcos(Ty)eos(E0}. Using eqn. (44), weobtain a 1 Glx,y,x0,0) =37[a0]#7Re[In(1) ~(45) where risdefined ineqn. (23). 3.3. Solution for the Potential V(x, y) Simplifying, eqn.(45)andusingeqns.(8)and(11),weobtain 1ata)=2 JAso)In|2sinh(ax(x—%9)/2L)|dr, ow(46) 4 which isasingular integral equation, andwhere ffx) isdefined ineqn.(25). To solve this equation, weconsider thefollowing two cases. 320 FUQIANYANGANDRONGYAO (a)g(x) isaneven function inx—Since g(x) iseven inxandfix) iseven in2g,weobtain 1 rgx)=ffl)In|4sinh(ax(x—%)/2L) ° sinh(ax(x+2%)/2L)|drq for0x51, .0) which isidentical to 1 glx)=ffiz)In|2.cosh (axx/L) —2cosh(arty/L) dro. -(48) ° Using thesolution given byCooke'?, thesolution ofeqn. (48) is a 1 ra $2)"Tosh (anx/L)~1)"? de (Sith (axx/L)dpS(esata Seahwry * i cosh(anE/L)— 1 , xS|coax conei|r@a) ota (eme/b) ‘LT?(cosh (axx/L) -1)(cosh (ax/L)-cosh(axx/L))}" en In[cosh (an/L)— 1]-In2 f{(cosh(ax§/L) —1)(cosh(ax/L)—cosh(ax§/L))]"”? *«) whichisthechargedensityonthestrip0*=x<1forasymmetric potentialdistribu-tion. Substituting eqns. (49) and (45) into eqn. (11), we find that thepotential Vix, y)isgiven by 1 Mar)3gf(In4+neonaxx—r/L)—cosh (9/2)° +In[cosh(ax|x+.x9/L)-cos(omy/t)fi) =»(50) (b)g(x)isanoddfunctioninx—Sinceg(x)isoddinxandffx)isoddin eqn. (46) ischanged to ‘TWO-DIMENSIONAL POTENTIAL THEORY 321 y sinh(ax(x-x9)/2L)rete)=ffain|TREE aR|temforOs291, «(51 which isidentical t0 ytanh(axx/2L)+tanh(axxo/2L) nats)=~ffe)|tanh(axx/2L)~tanh(arxy/2L) | © Using thesolution given byCooke", weobtain . faye p(—————_tanarag/ 20)dig TaeSco(g/L)Uta(arag/2)—tan(wx)? 6 xf——— #8Oef(tanh?(axx9/2L) -tanh?(anE/2L)}"? 2a__tanh (ax/2L)sinh(anx/L) ~807 [tanh?(ax/2L)—tanh?(anx/2L)]"? ~63) which isthecharge density onthestrip 0*=x<1forantisymmetric case. Substituting eqns. (53) and (45) into eqn. (11), wefind that thepotential Vix, y)fory>0is given by 11cosh(ax|x-x9|/L)—08(any/L) Mey)55fin|cosh(an|x+ap/L)—cos(any/L)|1%)ato.~(54) 4, SPECIAL CASES We first consider the symmetric case ofahalf space (ie. L-+%) due toan infinite strip 0<|X|<a, Y¥=0being charged to2unifrom voltage Vo,Since L—* =,we have amo||ao inh(222)-SH s!wan") randsn(5°) .(85) For problem 1,substituting eqn. (29) into eqn. (26) ,eqn. (29) becomes Vo ..(36)f)--n3 vies which isthepotential distribution onthestrip |x|<1,y=0,andisthesame as that given byBjike', Substitution ofeqn. (56) into eqn. (30) yields thepotential distribution v(x,y)fory>0. ForproblemIf,sincecosh(=)=1forf+,eqn.(49)gives 32 FUQIAN YANG AND RONG YAO fix) =0 -(57) which indicates thatthere isanuniform field inside thehalfspace. Fortheantisymmetric case ofahalf space duetoaninfinite strip 0<|X|=a, ¥= 0being charged toauniform voltage V,sgn(x),we,combining eqns. (55), (33) and (53) obtain 2_Mo__ =(58) WmVI which isthesame asthat obtained bydual integral equation method*, Substitution ofeqn. (58) into eqn. (54) ‘yields thepotential distribution 4x, y)fory>0. ACKNOWLEDGEMENT The author would like tothank Professor Alfred Clark Jr.for his discussion. REFERENCES 1.LN. Sneddon, Mixed Boundary Value Problems inPotential Theory, North Holland Pub. Co. Amsterdam, 1966. 2Douglas Ruth, Cand. J.Chem. Engng. 68(1990), 230. 3.PL Milly§.S.LaiandM.P.Dudukovic, Ind.Engng.Chem.Fundam.24(1985),64 4.N.L Masthelshvll, Singular Integral Equations, Noordhoff, Groninges, theNetherlands, 1953,S.K.B.Ranger,In.J.Engng.Sci.12(1974),853.6LRFRoseandF.R.DeHoog.Q.J!Mech.Appl.Math,36(1983),419. 7.RJ. Tait and T.B.Moodie, IJ.Engng, Sci 19(1981), 221 8.Uwadiegwu B.C. 0.Bike, IncJ.Engng. Sci 19(1981), 471.9.B.M.Singh,T.B.MoodieandJ.B.Haddow,ActaMech,38(1981),99 10,agian Yang andJ.C.M.Li,J.Appl Phys. 74(1993), 4382. 11, Stuart Davidson, Engng. Fracture Mech. 46(193) 727. 12. 5.C.Cooke, Glasgow Math. J.11(1970), 9