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# Partial Differential Equations I MATH 545

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Laplace s Equation 1 Equilibrium Phenomena Consider a general conservation statement fora region U in Rquot containing a material which is being transported through U by a flux field F Ft Let u mil denote the scalar concentration field for the material u equals the concentration at 20 Note that u is a scalar valued function while Ft is a vector valued function whose value at each it is a vector whose direction is the direction of the material flow at it and whose magnitude is proportional to the speed of the flow at it In addition suppose there is a scalar source density field denoted by st This value of this scalar at it indicates the rate at which material is being created or destroyed at it If B denotes an arbitrary ball inside U then for any time interval t1t2 conservation of material requires that B ul2dx B ul1dx j j 88 ml d5xdl j Bsldxdl Now gt gt t gt B uxt2dx B uxt1dx 18 Btuxt 1de and t t it 88 Fxt nxdSxdt if B dlvFxt 1de hence I I2 3min diVFt sldxdl 0 for all B c U and all m2 11 t1 B Since the integrand here is assumed to be continuous it follows that limit divFQJ sl 0 for all X E U and all t 12 Equation 11 is the integral form of the conservation statement while 12 is the differential form of the same statement This conservation statement describes a large number of physical processes We consider now a few special cases a Transport u ut Ft uxt V where V constant sl 0 In this case the equation becomes atut gradut 0 b Steady Diffusion u M Ft KVux where K constant gt 0 s In this case the equation becomes K divgradu SQ 0r KV2u This is the equation that governs steady state diffusion of the contaminant through the region U The equation is called Poisson s equation if 32 0 and Laplace s equation when s 0 These are the equations we will study in this section Another situation which leads to Laplace s equation involves a steady state vector field having the property that div x 0 When denotes the velocity field for an incompressible fluid the vanishing divergence expresses that conserves mass When denotes the magnetic force field in a magnetostatic field the vanishing divergence asserts that there are no magnetic sources In the case that represents the vector field of electric force the equation is the statement that U contains no electric charges ln addition to the equation div Vx 0 it may happen that Vsatisfies the equation curl Vx 0 This condition asserts that the field Vis conservative energy conserving Moreover it is a standard result in vector calculus that curl Vx 0 implies that V grad uG for some scalar field it MG Then the pair of equations div 0 and curl 0 taken together imply that WW 0 and i grad MG We say that the conservative field is derivable from the potential it MGquot To say that u is a potential is to say that it satisfies Laplace s equation The unifying feature of all of these physical models that lead to Laplace s equation is the fact that they are all in a state of equilibrium Whatever forces are acting in each model they have come to a state of equilibrium so that the state of the system remains constant in time lfthe balance of the system is disturbed then it will have to go through another transient process until the forces once again all balance each other and the system is in a new equilibrium state 2 Harmonic Functions A function u ux is said to be harmonic in U C Rquot if i u e C2U ie u together with all its derivatives of order 5 2 is continuous in U ii V2u 0 at each point in U Note that in Cartesian coordinates BuBxl divVuQ BBx1Bano E azuBxf azuBx BitBxquot aTauG V2u It is clear from this that all linear functions are harmonic A function depending on x only through the radial variable r ix x is said to be a radial function If u is a radial function then BuBxi u r BrBxi and BrBxi x72xi xir uBx2 u rBrBxi2 u r r xir u rBrBxi2 u r and VZMG 21 uBx2 212th u r MO ur MW n 1u r r We see from this computation that the radial function u unr is harmonic for various n if n 1 ul r 0 ie u1r Arl B n 2 uz r u r ru 2r 0 ie u2r Cln r n gt 2 71 rl quotr kl 0 unr Crz quot Note also that since V2auBx BBxiV2u for any i it follows that every derivative of a harmonic function is itself harmonic Of course this presupposes that the derivative exists but it will be shown that every harmonic function is automatically infinitely differentiable so every derivative exists and is therefore harmonic It is interesting to note that if u and u2 are both harmonic then u must be constant To see this write V2u2 divgradu2 div2uVu 2Vu Vu Zuvzu Zqul2 Then V2u2 0 implies qul2 0 which is to say u is constant Evidently then the product of harmonic functions need not be harmonic It is easy to see that any linear combination of harmonic functions is harmonic so the harmonic functions form a linear space It is also easy to see that if u ux is harmonic on Rquot then for any 2 e Rquot the translate vx ux z is harmonic as is the scaled function w wlx for all scalars 1 Finally V2 is invariant under orthogonal transformations To see this suppose coordinates x and y are related by Q11x1 anxn Q E mel anxn Then Vx BBx1Ban and Bx BylBxi8y aynaxQByn Qli 9y1 QmBy ith row of QVy ie Vx QTVy and V QTVyT VQ Then V3 V V VQQT Vy V Vy for QQT I A transformation Q with this property QQT I is said to be an orthogonal transformation Such transformations include rotations and reflections Problem 6 Suppose u and v are both harmonic on R3 Show that in general the product of u times v is not harmonic Give one or more examples of a special case where the product does turn out to be harmonic 3 Integral Identities Let U denote a bounded open connected set in Rquot having a smooth boundary BU This is sufficient in order forthe divergence theorem to be valid on U That is if EYE denotes a smooth vector field over U ie F e CU C1U and if 2x denotes the outward unit normal to 8U at x e BU then the divergence theorem asserts that IUdivF dx jaUF dsoc 31 Considerthe integral identity 31 in the special case that 131x Vux for u e C1U nC2U Then div xx divVux V2ux and FT 2 Vu E BNux the normal derivative of u Then 31 becomes IU V2uxdx IaUaNuxdSx 32 The identity 32 is known as Green s first identity lf functions u and v both belong to 010 n C2U and if Fx vxVux then div xx div vxVux vxV2ux Vu Vv and 712 wow2 vaNux and with this choice for 31 becomes Green s second identity I vxV2ux Vu Vv dx I vxaNuxdSx 33 U 8U Finally writing 33 with u and v reversed and subtracting the result from 33 we obtain Green s symmetric identity IUvxV2ux uxV2vx dx I8UvxBNux uxanx dSx 34 Problem 7 Let u uxyz be a smooth function on R3 and let A denote a 3 by 3 matrix whose entries are all smooth functions on R3 Let 1 AVu If U denotes a bounded open set in R3 having smooth boundary BU then find a surface integral over the boundary whose value equals the integral of the divergence off over U If v vxyz is also a smooth function on R3 then write the integral of vdivFT over U as the sum of 2 integrals one of which is a surface integral over BU 4 The Mean Value Theorem for Harmonic Functions We begin by introducing some notation Ba x e Rquot lx al lt r the open ball of radius r with center at xa Ba x e Rquot lx al 5 r the closed ball of radius r with center at xa Sa x e Rquot lx al r the surface of the ball of radius r with center at xa LetAn denote the ndimensional volume ofBl0 Then A2 7rA3 4713 and in general Aquot 7rquot2 quotn2 1 Then the volume of the nball of radius r is WAquot Also let Squot denote the area of the n1dimensional surface of 310 in Rquot ie Squot is the area of 8310Then Squot 11Aquot and the area of BB0 is equal to nAnr H In particular Szr 27V S3V2 47rr2 etc We will also find it convenient to introduce the notation A 1 IB39agtfxdx Anrn IB39agtfxdx average value af x over Ba A 1 IaB39agtfxdSx Snrml IaB39agtfxdSx average value af x over BBa Recall that it follows from Green s first identity that if ux is harmonic in U then for any ball Ba contained in U we have BNux dSx I Ba j V2uxdx 0 8801 This simple observation is the key to the proof of the following theorem Theorem 41 Mean Value Theorem for Harmonic Functions Suppose u e C2U and V2ux 0 for everyx in the bounded open set U in Rquot Then for every Bx c U um i umdsky i new 41 ie 41 asserts that for every x in U and r gt 0 sufficiently small that Bx is contained in U ux is equal to the average value of u overthe surface BBx and ux is also equal to the average value of u over the entire ball Bx A function with the property asserted by 41 is said to have the mean value property 934x 800 Proof Fix a point x in U and an r gt 0 such that Bx is contained in the open set U Let gr IaB39xgtuyd y 8810gtux rzd z Here we used the change of variable y x r2 or z y xr so as y ranges over BBx z ranges over 8310 Then g r 8810gtVux rz Z 152 I Vqu yx r 98m It is evident that as y ranges over BBx ly xl r hence y xr is just the outward unit normal to the surface BBx which means that x my y 6mm Then g r I aNuydS39y B V2uyd9 0 since u is harmonic in U Now g r 0 implies that gr constant which leads to 934x x gr limo gt1imt gt0 j ux tzd z ux ie ux I Notice that this result also implies LWumdy j 881Xuyd5ydt IuxSnt Hdt uxAnrquot 9810 88xgtuyd y for all r gt 0 such that Bx c U or um lmuowy iuodyA which completes the proof of the theorem The converse of theorem 41 is also true Theorem 42 Suppose U is a bounded open connected set in Rquot and u e C2U has the mean value property ie for every x in U and for each r gt 0 such that Bx C U abs iultyds y Then V2ux 0 in U Proof If it is not the case that V2ux 0 throughout U then there is some Bx C Usuch that V2ux is say positive on Bx Then for gr as in the proof of theorem 41 o gm j 3Nu d y gjmyzumdy gt o This contradiction shows there can be no Bx C Uon which V2ux gt 0 and hence no point in U where V2ux is different from zero 634x For M uxy a smooth function of two variables we have a uxy z ux hy 2uxy ux hyh2 8Wuxy z uxy h 2uxy uxy hh2 hence h2V2uxy z 4uxy ux hy ux hy uxy h uxy h Then the equation V2uxy 0 in U is approximated by the equation uxy ux hy ux hy uxy h uxy h4 The expression on the right side of this equation is recognizable as an approximation for law blow y Thus in the discrete setting the connection between the property of being harmonic and the mean value property is more immediate 5 Maximumminimum Principles The following theorem known as the strong maximumminimum principle is an immediate consequence of the mean value property Theorem 51 strong maximum minimum principle Suppose U is a bounded open connected set in Rquot and u is harmonic in U and continuous on U the closure of U Let M and m denote respectively the maximum and minimum values of u on BU Then either ux is constant on Uso then ux m M or else for every x in U we have m lt ux lt M Proof Let M denote the maximum value of ux on Uand suppose ux0 M lf x0 is inside U then there exists an r gt 0 such that Bx0 C Uand ux 5 ux0 for all x e Bx0 Suppose there is some yo in Bxo such that quo lt uxo But this contradicts the mean value property since it implies M 060 IBMU It follows that ux ux0 for all x in Bx0 Similarly for any other point yo 6 U the assumption that quo lt uxo leads to a contradiction of the mean value property Then if x0 is an interior point of U we a force to conclude that ux is identically equal to M on U and by continuity on the closure U On the other hand if u is not constant on U then x0 must lie on the boundary of U uyd9 lt M Note that if u uxy satisfies the discrete Laplace equation uxy W lay W lay uxy h uxy h4 on a square grid then u can have neither a max nor a min at an interior point of the grid since at such a point the left side of the equation could not equal the right side At an interior maximum the left side would be greater than all four of the values on the right side preventing equality A similar situation would apply at an interior minimum Unless u is constant on the grid the only possible location for an extreme value is at a boundary point of the grid There is a weaker version of theorem 51 that is based on simple calculus arguments Theorem 52 Weak Maximumminimum principle Suppose U is a bounded open connected set in Rquot and u e CU C2U Let M and m denote respectively the maximum and minimum values of u on BU Then a V2ux S 0 in U implies ux SM for all x E U b V2ux 2 0 in U implies ux 2 m for all x E U c V2ux 0 in U implies m S ux S M for all x E U Proof of a The argument we plan to use can not be applied directly to ux lnstead let vx ux 8lxl2 forx e U and note that V2vx V2ux 2118 lt 0 for all x in U It follows that vx can have no interior maximum since at such a point x0 we would have BvBxi 0 and yBx2 S 0 for 1 S i lt n x x0 This is in contradiction to the previous inequality since it implies V2vx 2 0 This allows us to conclude that vx has no interior max and vx must therefore assume its maximum value at a point on the boundary of U Now U is bounded so for some R sufficiently large we have U C BR0 and this implies the following bound on maxst vx max vx S maxvx S M slxlz S M 8R2 XEU anU Finally we have ux 5 vx 5 M 2R2 for all x in U and all s gt 0 Since this holds for all s gt 0 it follows that ux M for all x in U Statement b can be proved by a similar argument or by applying a to u Then c follows from a and bl In the special case n 1 it is easy to see why theorem 52 holds In that case U ab and Vzu u x and the figure illustrates a b and c a ux S M b ux 2 m c m S ux S M The following figure illustrates why it is necessary to have both of the hypotheses u E CU and u E C2U ue CU u CU but u C2U but u E C2U If U is not bounded then the maxmin principle fails in general For example if U denotes the unbounded wedge xy y gt lxl in R2 then uxy y2 x2 is harmonic in U equals zero on the boundary of U but is not the zero function inside U An extended version of the maxmin principle due to E Hopf is frequently useful Theorem 53 Suppose U is a bounded open connected set in Rquot and u e CU C2U Suppose also that V2ux 0 in U and that u is not constant Finally suppose U is such that for each point y on the boundary of U there is a ball contained in U with y lying on the boundary of the ball If uy M then BNuy gt 0 and if uy m then 31mm lt 0 ie at a point on the boundary of U where ux assumes an extreme value the normal derivative does not vanish Problem 8 Let ux be harmonic on U and let vx quxl2 Show that vx S maXXEaU vx forx E U Hint compute Vzv and show that it is nonnegative on U 6 Consequences of the Mean Value Theorem and Mm Principles Throughout this section U is assumed to be a bounded open connected set in Rquot We list now several consequences of the results of the previous two sections It is a standard result in elementary real analysis that if a sequence of continuous functions um converges uniformly to a limit u on a compact set K then u is also continuous Moreover for any open subset W in K we have 1mm Wumdx Wudx Lemma 61 Suppose umx is a sequence of functions which are harmonic in U and which converge uniformly on U Then u limmuo u is harmonic in U Proof Since each u is harmonic in U theorem 41 implies that for every ball Bx C U we have umoc i ummd o i ummdy The uniform convergence of the sequence on U implies that 634x x umx ux lawnmows oala39uyd lty W ammdywl uydy hence 3906 W l uygtds lty I umdy But this says M has the mean value property and so by theorem 42 u is harmonicl 83m m Lemma 62 Suppose u e CU C2U satisfies the conditions V2ux 0 in U and ux 0 0n BU Then ux 0 for all x in U Proof The hypotheses u e CU C2U and V2ux 0 in U imply that m 5 ux M in U Then ux 0 0n BU implies m M 0 Lemma 62 asserts that the so called Dirichlet boundary value problem V2ux Fx x E U and ux gx x E BU has at most one solution in the class CU C2U Solutions having this degree of smoothness are called classical solutions of the Dirichlet boundary value problem The partial differential equation is satisfied at each point of U and the boundary condition is satisfied at each point of the boundary Later we are going to consider solutions in a wider sense Lemma 63 For anyF e CU andg e CBU there exists at most one u e CU C2U satisfying V2ux F in U and ux g on BU Proof Suppose M1 M2 6 CU C2U both satisfy the conditions of the boundary value problem Then w m M2 satisfies the hypotheses of lemma 62 and is therefore zero on the closure of U Then u1 uz on the closure of U Lemma 64 Suppose u e CU C2U satisfies V2ux 0 in U and ux g on BU where gx 2 0 If gx0 gt 0 at some point x0 6 BU then ux gt 0 at every x e U Proof First gx 2 0 implies that m 0 Then gx0 gt 0 at some point x0 6 8U implies M gt 0 It follows now from the strong M m principle that 0 lt ux lt M at every x e U Note that lemma 64 asserts that if a harmonic function that is nonnegative on the boundary of its domain is positive at some point of the boundary then it must be positive at every point inside the domain Le a local stimulus applied to the quotskinquot of the body produces a global response felt everywhere inside the body This could be referred to as the organic behavior of harmonic functions This mathematical behavior is related to the fact that Laplace s equation models physical systems that are in a state of equilibrium lfthe boundary state of a system in equilibrium is disturbed even if the disturbance is very local then the system must readjust itself at each point inside the boundary to achieve a new state of equilibrium This is the physical interpretation of quotorganic behaviorquot Lemma 65 ForF e CU andg e CBU suppose u e CU C2U satisfies V2ux Fx x E U and ux gx x E BU Then maXXEUluxl S Cg MCF where Cg maxxgaU lgxl CF maXXEU lFxl M a constant depending on U Proof The estimate asserts that Cg MCF 5 ux 5 Cg MCF forx e U First let M abs ixi2 Then V2vx V2ux CF Fx CF 5 0 in U and vx S maXXEaU lxl2 for x E U Since U is bounded there exists some R gt 0 such that lxlz 5 R2 for x e U Then vx S Cg R2 and ux S vx S Cg MCF forx E 7 Similarly let wx W lxl2 and show that ux 2 wx 2 Cg MCF for x E 7 If we define a mapping S CU x CBU gt CU C2U that associates the data pair Fg forthe boundary value problem of lemma 65 to the solution ux then we would write u SFg Evidently lemma 65 asserts that the mapping 8 is continuous To make this statement precise we must explain how to measure distance between data pairs F1g1F2g2 in the data space CU x CBU and between solutions u1u2 in the solution space CU Although we know that the solutions belong to the space CU C2U this is a subspace of the larger space CU so we are entitled to view the solutions as belonging to this larger space We are using the term quotspacequot to mean a linear space of functions that is a set that is closed underthe operation of forming linear combinations Define the distance between u1u2 in the solution space CU as follows llul zllcm maxer u1x 200l Similarly define the distance from F1g1 to F2g2 in the data space CU x CBU by llF1gl F2582llca7gtxcaugt maXxEU lF1x F2xl maXxeaU lg1x g2xl Each of these quotdistance functionsquot defines what is called a norm on the linear space where it has been defined In order to be called a norm the functions have to satisfy the following conditions 139 llaull M Hull for all scalars a and for all functions M ii lluvll 5 Hull llvll for all functions uv iii Hull 2 0 for all u and Hull 0 if and only if u 0 One can check that the distance functions defined above both satisfy all three of these conditions and they therefore qualify as norms on the spaces where they have been defined Now the estimate of lemma 65 asserts that if Mj solves the boundary value problem with data Fjgj j 12 then maXXEU lu1x u2xl S maXXEaU lg1x g2xl MmaXXEU lF1x F2xl iey llul uZHCW E maX1M llF1g1 F25g2llca7gtxcaugt Evidently if the data pairs are close in the data space then the solutions are correspondingly close in the solution space This is what is meant by continuous dependence of the solution on the data Note that if we were to change the definition of the norm in one orthe other or both of the spaces the solution might no longer depend continuously on the data Consider the solution for the following boundary value problem V2uxy 0 for 0 lt x lt 71 y gt 0 ux0 0 Byux0 gx sin rm 0 lt x lt 71 u0y u7ry 0 y gt 0 For any integer n the solution is given by uxy i sin mc sinh ny Evidently the distance between g and zero in the data space is llgx 0HCR maxm sin ml 5 a while the distance between uxy and zero in the solution space is 1 N W n Zsmmc s1nh ny quot2 H x930 0HClt0ltxltJI ygt0 maX0ltxltnygt0 This means that the data can be made arbitrarily close to zero by choosing n large while the solution can simultaneously be made as far from zero as we like by choosing y gt 0 large Then the solution to this problem does not depend continuously on the data since arbitrarily small data errors could lead to arbitrarily large solution errors This problem is said to be quotnot well posedquot 7 Uniqueness from Integral Identities lntegral identities can be used to prove that various boundary value problems cannot have more than one solution For example consider the following boundary value problem V2ux Fx x E U BNux gx x E BU This is known as the Neumann boundary value problem for Poisson s equation Green s first identity leads to UFxdx Uvzumdx aUaNuxdSx aUgxdSx Then a necessary condition forthe existence of a solution to this problem is that the data F g satisfies IUFxdx IaUgxdSx lfthis condition is satisfied and if u1u2 denote two solutions to the problem then w m uz satisfies the problem with F g 0 Then we have 0 Uwvzwdx IanBNwdSx jUVw oVde lele2dx But this implies that lel 0 which is to say w is constant in U Then the solutions to this boundary value problem may differ by a constant they are not unique We should point out that in order for the equation and the boundary condition to have meaning in the classical sense we must assume that the solutions to this problem belong to the class 010 n C2U On the other hand considerthe problem V2ux Fx x E U ux g1x x E BUl BNux g2x x E BUZ where BU is composed of two distinct pieces BUl and BUz Now if u1u2 denote two solutions to the problem and w m m then we have as before 0 Uwvzwdx IanBNwdSx jUVw oVde IaUleNwdSx jaU2wandSx levwl2 In this case w 0 on BU1 and an 0 on BUZ so we again reach the conclusion that w is constant in U Since w e C1U C2U it follows that if w 0 0n BUl then w 0 on U Then the solution to this problem is unique Finally consider the Neumann problem forthe so called Helmholtz equation V2ux cxux Fx x E U ux gx x E BU where we suppose that Cx 2 C0 gt 0 forx e U We can use integral identities to show that this problem has at most one smooth solution As usual we begin by supposing the problem has two solutions and we let wx denote their difference Then V2wx cxwx 0 x E U wx 0 x E BU and 0 Uwx V2wx cxwxdx IanBNwdSx va dexl Ucxwx2dx Since w 0 on BU it follows that quwlz cxwx2dx 2 COJU wx2dx 0 and this implies that wx vanishes at every point of U Notice that this proof of uniqueness doesn t work if we don t know that the coefficient Cx is nonnegative How would the proof have to be modified if we knew only that Cx 2 0 Problem 9 Prove that the following problem has at most one smooth solution V2ux Fx x E U and ux gx x E BU Use first the Green s identity approach and then use the result in lemma 65 Note that this result was already established by means of the Mm principle Problem 10 Prove that the following problem has at most one smooth solution V2ux Fx in U and ux BNux gx 0n BU Eigenvalues for the Laplacian The eigenvalues for the Dirichlet problem forthe Laplace operator are any scalars A for which there exist nontrivial solutions to the Dirichlet boundary value problem V2ux lux x E U ux 0 x E BU Note that if ux 0 then any choice of A will satisfy the conditions of the problem Therefore we allow only nontrivial solutions and we refer to these as eigenfunctions lf ux is an eigenfunction for this problem corresponding to an eigenvalue 1 then lqux2dx quxV2uxdx jaUuaWdSx lUqul2dx Then 1 satisfies 1 qulzdx U gt IUux2dx Note that qul 0 since this would lead to u 0 which is not allowed ifu is an eigenfunction We have shown that all eigenvalues of the Dirichlet problem for the Laplace operator are strictly positive Problem 11 Show that the Neumann problem V2ux lux x E U BNux 0 x E BU has a zero eigenvalue which has the corresponding eigenfunction ux constant Problem 12 Under what conditions on the function ax does the boundary value problem V2ux lux x E U axux BNux 0 x E BU have only positive eigenvalues Problem 13 Show that for each of the eigenvalue problems considered here if ux is an eigenfunction corresponding to an eigenvalue 1 then for any nonzero constant k vx kux is also an eigenfunction corresponding to the eigenvalue 1 8 Fundamental Solutions for the Laplacian Let 6x denote the quotfunctionquot with the property that for any continuous functionfx I n5xfxdx f0 or equivalently max yfydy fx Of course this is a purely formal definition since there is no function 6x which could have this property Later we will see that 6x can be given a rigorous consistent meaning in the context of generalized functions However using the delta in this formal way we can give a formal definition of a fundamental solution for the negative Laplacian as the solution of v Eltx y so y w e Rquot 81 Formally this definition implies v3 l Eoc yfydy i W yfydy fltx Then the solution of the equation V2ux fx x E Rquot is given by abs JREltx yfydy 82 Although these steps are only formal they can be made rigorous Note that since there are no side conditions imposed on Ex or on ux neither of these functions is unique For example any harmonic function could be added to either of them and the resulting function would still satisfy the same equation Since 6x and V2 are both radially symmetric it seems reasonable to assume that Ex is radially symmetric as well ie Ex Er for r le x Then a definition for Ex which does not make use of 6x can be stated as follows Enx is a fundamental solution for V2 on Rquot if 139 Enr e C2Rquot0 ii VZEHU 0 for r gt 0 83 m umM 8850 BNEnxdSx 1 The properties i and ii in the definition imply that VZEMU EX n 1 0 for r gt 0 ie En rEr n 1r 10g n 110g r C Crl quot En Czlog r ifn 2 I Cn rz quot ifn gt 2 The constant Cquot can be determined from part iii of the definition It is this part of the definition that causes V2Enx to behave like 6x For n 2 we have 2 2n 8850BNEnxdSx 0 aCz1og mare C2 0 8610 2702 Then um 0 j BNE2xdSx 27m 1 8850 SO C2 127r and E2r 21 logr When n 3 we have 8850gt8NEnxdSx j BC3r82da C3 8850 jszda 4703 Then limguo I 8850 8850 BNE3xdSx 4775C3 1 so C3 147r and E3r 147rr We will now show that condition 83iii really does produce the 6 behavior for V2E Of course we can t try to show that V2En 6x since are not allowed to refer to 6x Instead we will show equivalently that V2ux fx for u given by 82 Here we suppose thatfx is continuous together with all its derivatives of order less than or equal to 2 and we suppose further thatfx has compact support ie for some positive K x vanishes for lxl gt K The notation for this class of functions is C Rquot Theorem 81 Let Enr denote a fundamental solution for V2 on Rquot Then for any f e C Rquot um IREnxyfydy satisfies u E C2Rquot V2ux fx for any x E Rquot Proof The smoothness of f implies the smoothness of u ie for i 12n altam limbo ux hellZ ux fx hei fx 2 dz fGl hEi d fG z h umM jREz Now converges uniformly to a axi and it follows that for each i BuBxi IR Ez BXL x z dy Similarly 32uxaxiaxj exists for each i and j since the corresponding derivatives of f all exist To show the second assertion write V ux 1R EnzV fx zdy j Enzvgfx 2 dz Since Enz tends to infinity as lzl tends to zero we treat this as an improper integral IR EnzV x 2 dz I8 EnzV x 2 dz IR EnzV fx 2 dz First note that 50 quotBE0 IBE0gtEnzV x 2 dz 5 maXBE0lV x zl 850 lEnzl dz But 5 271 127r logr rdrd0 C8210g8 n 2 l iEnzi dz 010 850 Cquot I0 I rz quotrquot 1drda C82 n gt 2 hence 2 Ilmgqo IBE0EnzVz x zle 0 Next 2 o IRMEltOgtEnzVZfx z dz 8Rm50gtgtEnzBNfx zdSz j WVE2 Vz x 2 dz and l lama Ema aN x 2 dSzl maXzeaBltogtl3Nfx Zl 8850 lEnz l dszgt 12njj i1ogsigd0 Czllogsls if n 2 ansz quotsquot 1da C38 if n gt 2 We used the fact that 8RquotVBS0 BBS0 Finally since Enz is harmonic in RquotBg0 VEnz Vz x 2 dz j BNEnzfx 2 152 j VZEnzfx 2 dz IImam 7880 BNEnZfx 2 152 BE0 I1950 Now we can write BNEnzfx 2 152 j BNEnz fx 2 fx 152 17 and note that becausefx is continuous BNEnZ fx 2 fx dSZ S C mangagzo lfx Z fxl gt 0 as 8 gt 0 43350 3350 8850 BNEnzfx 152 850 In addition BNEnzdSz j BNEnzdSz a 1 as a a 0 J23190 830 because of 83139139139 and then it follows that v2ux mugs0 788 0 BNEnzfx z 152 fx Vx e Rn We remark again that since no side conditions have been imposed on ux this solution is not unique Any harmonic function could be added to ux and the sum would also satisfy V2ux 9 Green s Functions for the Laplacian Throughout this section U is assumed to be a bounded open connected set in Rquot whose boundary BU is sufficiently smooth that the divergence theorem holds Consider the Dirichlet boundary value problem for Poisson s equation V2ux Fx for x E U and ux gx for x E BU 91 We know that W iEnltx yFltygtdy satisfies the partial differential equation but this function does not in general satisfy the Dirichlet boundary condition In order to find a function which satisfies both the equation and the boundary condition recall that for smooth functions ux and vx IUvyV uy uwmn dy IaUlvaNqu no 6mm also 92 For x in U fixed but arbitrary let vy Enx y y in 92 where denotes a yet to be specified function that is harmonic in U Then since Enx y is a fundamental solution and 5 is harmonic in U Humvlvm Union Vino y 0 dy no Since ux solves the Dirichlet problem 92 becomes now abs lvltyvluygt dy laUlvaNqu uyr7Nvy also lUvygtFltydy laUgowNvmdso lwvmaNuWSo If the values of BNugz were known on BU then this would be an expression for the solution ux in terms of the data in the problem Since 31mm on the boundary is not given we instead choose the harmonic function 5 in such a way as to make the integral containing this term disappear Let 5 be the solution of the following Dirichlet problem V5150 0 fory E U W Enxy fory E 3U where we recall that x denotes some fixed but arbitrary point in U Then vy Enx y 501 0 on the boundary and the previous expression for ux reduces to um lUGxyfltydy laUaNGltxygodSo 93 where Gxy Enx y y Formally Gxy solves V2Gxy V2Ex y o 5x y for xy e U 94 Gx y o forxe U ye at and Gxy is known as the Green s function forthe Dirichlet problem for the Laplacian or alternatively as the Green s function of the first kind Note that if there are two Green s functions then their difference satisfies a completely homogeneous Dirichlet problem This would seem to imply uniqueness for the Green s function except for the fact that the uniqueness proofs were for the class of functions C2U CU and it is not known that Gxy is in this class This point will be cleared up later It can be shown rigorously that Gxy Gyx for all xy e U However a formal demonstration based on 94 proceeds as follows For xz e U be careful to note that x and y are points in Rquot apply 92 with uy Gyz and vy Gyx IUuyV Vy V V uy dy IUG35 3 GGRWG 3dZ lanuo 8Nvy memo also l 6166 8m wawdndso 0 The last integral vanishes because Gyz Gyz 0 for y e BU Then 92 implies 0 IUGZ5 x Gyx5y Zdz Gxz Gzx for all xz E U This proof will become rigorous when we have developed the generalized function framework in which this argument has meaning Example 91 Let U x1x2 e R2 x2 gt 0 The half space is the simplest example of a set having a boundary ie the boundary of the half space is the x1 axis x2 0 and we will be able to construct the Green s function of the first kind forthis simple set Note that 18

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