Boundary-Value Problems: Part II

Transcription

1 Chapter 3 Boundary-Value Problems: Part II Problem Set #3: 3., 3.3, 3.7 Due Monday March. th 3. Spherical Coordinates Spherical coordinates are used when boundary conditions have spherical symmetry. The Laplace equation in spherical coordinates takes the following form, Φ= r r rφ + sin θ Φ Φ + r sin θ θ θ r sin =. 3. θ φ To use the separation of variables method we can plug in the following ansatz Φr, θ, φ = Ur P θqφ 3. r into 3. toyield P θqφ d r dr Ur+Ur d Qφ sin θ ddθ r r sin θ dθ P θ + Ur d P θ r r sin Qφ = θ dφ 3.3 r or after multiplying by 3 sin θ, UrP θqφ r sin θ d U dr U + sin θ d sin θ ddθ P dθ P = d Q. 3.4 Q dφ Since the right hand side depends only on φ and the left hand side does not there must be some constant m such that d Q dφ Q = m

2 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 5 and r sin θ d U dr U + sin θ d sin θ ddθ P dθ P = m 3.6 Equation 3.5 hassolutions Q = C m e imφ 3.7 where m must be an integer for Q to be single valued. Similarly we can separate variables θ and r in 3.6 toget r d U dr U = m sin θ d sin θ ddθ P sin θ dθ P 3.8 or r d U = ll U dr and m sin θ d sin θ ddθ P sin θ dθ P = ll + 3. for some constant ll +. These two equation can be rewritten as d dr U = ll +U 3. r and d sin θ ddθ m sin θ dθ P = sin ll + P 3. θ Equation 3. canbesolveusingpolynomialansatz where or U r α 3.3 αα = ll α = { l + Thus, the most general solution of 3. isgivenby l 3.5 U = A l r l+ + B l r l. 3.6 To determine l we switch to 3. rewrittenintermsofx =cosθ, d x d m dx dx P l = ll + P x l. 3.7 This is the generalized Legendre equation whose solutions are the associated Legendre functions.

3 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 6 3. Legendre polynomials For starters let us consider a special case m =corresponding to the ordinary Legendre equation d x d dx dx P l + ll +P l = 3.8 If we assume the following expansion of where a,thenweobtain + P l x =x α j= a j x j 3.9 [ α + jα + j aj x α+j α + jα + j + l l +a j x α+j] =3. j= αα a x α ++ααa x α + [α + j +α + j +a j+ α + jα + j + l l +a j ] x α+j =. j= Since the coefficient of all of the powers of x have to vanish separately, we find and a j+ = α α = for a α α + = for a 3. α + jα + j + l l + a j. 3. α + j +α + j + The two equation 3. areredundantandtheonlyconditionthatfollows from them is α =or. Then the expansion 3.9 haseitheroddfor α =orevenforα =powersofx. One can show that the series expansion converges for x =only if it terminates. Since α and j are positive integers or zero the relation 3. terminates only if l is a positive integer or zero. Then only one of the two series terminates when { j max + αj max + α + ll + j max = l for α = = j max + α +j max + α + j max = l for α = 3.3

4 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 7 Thus, { l/ j P l x = a jx j where a j+ = jj+ ll+ a j+j+ j for even l l / j a j x j+ where a j+ = j+j+ ll+ a j+3j+ j for odd l. 3.4 For example, if we set P l =, then P x = P x = x P x = +3x P 3 x = 3x +5x We note without proving that the Legendre Polynomial have the following properties P l x = P l+ x = P l x = P l x P l x = d l l l! dx l x l Rodrigues Formula 3.6 l + l + P lx l l + P l x Recurrence Relation # 3.7 d l +dx l+x P l x Recurrence Relation # 3.8 l + δ ll Orthogonality condition 3.9 The last property implies that the Legendre polynomials formacomplete set of orthogonal function on the interval, and any function can be expanded as fx = A l P l x where A l = l + fxp l xdx. 3.3 Then the general solution of the Laplace equation in spherical coordinates for the case of azimuthal symmetry i.e. m =isgivenby Φr, θ, φ = A l r l + B l r l P l cos θ Azimuthal symmetry Consider a sphere of radius a with potential V θ on its surface.to find the potential everywhere inside the sphere we note that the potential should not

5 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 8 blow up at r =and thus the B l s in 3.3 mustvanish. Thenourproblem reduces to finding the appropriate coefficients in Φr, θ, φ = A l r l P l cos θ 3.3 for boundary conditions V θ = A l a l P l cos θ Multiplying both sides of 3.33byP l cos θ sin θ and integrating over θ leads to or π V θp l cos θ sin θdθ = π A l a l P l cos θp l cos θ sin θdθ = A l a l A l = l + a l P l xp l xdx = A l a l l V θp l cos θ sin θdθ For instance the potential inside of two hemispherical shells of radius a held at different potentials, i.e. { +V for θ<π/ V θ = 3.36 V for π/ <θ π, can be estimated from [ 3 r Φr, θ =V a P cos θ 7 r 8 a 3 P3 cos θ+ r 5 P5 cos θ...]. 6 a 3.37 For the potential outside of the sphere r/a l would be replaced by r/a l+. In fact this solution could have been obtained from an alreadyknownsolution.9 forθ =.Since Φr, = A l r l + B l r l 3.38

6 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 9 and Φr, = V r a r = V r + a π we get Φr, θ = V π j= j= j j Γ j a j j! r 3.39 j j Γ j a j Pj cos θ 3.4 j! r which can be shown to be equivalent to Another important problem with azimuthal symmetry that can be solved by the same method is the potential due to a unit point charge at x placed anywhere on z axis. Then, the Taylor series expansion for x also on the z axis is given by x x = r r = r< l 3.4 r > where r < or r > isthesmallerorlargerofr and r,andthus, x x = r< l r> l+ 3.4 Fields in a Conical Hole r l > P l cos θ. 3.4 Consider a conical hole in a conductor that has angle β relative z axis. To find the potential inside the hole we are restricted to the region of interest in spherical coordinates: <θ<βand <φ<π. The problem has an azimuthal symmetry, but the solutions arenot given by Legendre polynomials since we can no longer expand around θ = π/ or x =in 3.8. For this particular problem it makes sense to look for an expansion around θ =or x =in 3.8, and thus it is convenient to change variables d dy in 3.8 torewritetheequationas y y d dy P ν d dy y = x 3.43 = d dx νν +P ν =. 3.45

7 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 3 We have switched from subscript l to ν to emphasize the fact that the functions P ν y are not Legendre polynomials. Just like before we can plug in a series expansion P ν y =y α j= a j y j 3.46 into 3.45 todeducethatvanishingofacoefficientofthelowestpowerof y requires α =and the recursion relation between successive coefficients is a j+ = j νj + ν + j + a j Moreover, if we choose a normalization corresponding to a then we obtain our final solution or P ν y =+ νν + y +!! νν + P ν x =+!! x ν ν +ν +ν + y !! ν ν +ν +ν + +!! x These are the so-called Legendre functions of the first kind and order ν, where ν sarenotnecessarilyintegers.theirvaluesaredeterminedfrom the boundary conditions Φθ = β = corresponding to and the most general solution is given by Φr, θ, φ =A + P ν cos β =, 3.5 A n r ν k P νk cos θ where P νk cos β =. 3.5 n= Deep inside the hole r ofagroundedconductora =onlythe leading term would dominate, i.e. Φr, θ, φ =Ar ν P ν cos θ, 3.5 where ν = {.45 β for small β ln for large β. π β 3.53

8 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II Spherical Harmonics Let us now consider the Laplace equation in spherical coordinates without azimuthal symmetry which boils down to solving the generalized Legendre equation for m d dx x d dx P l m = ll + x P l It can be shown that for the equation to be finite between and the parameter l must be positive integer and Then the solution of 3.54 aregivenby m = { l, l +,..., l,l} P m l x = { m x m/ dm P dx m l x for m m l m! l+m! lx for m< 3.56 Another definition of P m l s is obtained from the Rodrigues formula 3.6 and is valid for positive as well as negative m, P m l x = m l l! x m/ dl+m dx l+m x l The associated Legendre function form a set of orthogonal functions on the interval,, Pl m xpl m xdx = δ ll l + l + m! l m! It is now convenient to define a set of orthonormal functions l +l m! Y lm θ, φ = 4π l + m! P l m cos θe imφ 3.59 with Y l, m θ, φ = m Ylm θ, φ. 3.6 which also satisfy the orthogonality and normalization conditions π π dφ dθ sin θy l m θ, φylm θ, φ =δ ll δ mm 3.6

9 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 3 and the completeness relation l m= l Y lmθ,φ Y lm θ, φ =δφ φ δcos θ cos θ. 3.6 These function are known as spherical harmonics as they represent the angulardependent part of the solutions of the Laplace equation in spherical coordinates θ, φ. For example, Y = 4π 3 Y = sin θeiφ 8π 3 Y = 8π cos θ 3 Y, = 8π sin θe iφ, 3.63 Note that for m =the spherical harmonic reduce to the Legendre polynomials l + Y l θ, φ = 4π P lcos θ 3.64 and for m = l Y mm θ, φ = m m +m! e imφ sin m θ m m! 4π Since the spherical harmonics form a complete orthonormal set of function, an arbitrary square-integrable function on a unit sphere can be expanded as, l gθ, φ = A lm Y lm θ, φ, 3.66 m= l where the coefficients are given by A lm = dωylmθ, φgθ, φ, 3.67 and the most general solution to the Laplace equation can also beexpanded in spherical harmonics as Φr, θ, φ = l Alm r l + B lm r l+ Y lm θ, φ m= l

10 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 33 It will be useful to prove the so-called addition theorem for spherical harmonics which expresses a Legendre polynomial of order l and angle γ in terms of spherical harmonics of angles θ, φ and θ,φ where cos γ =cosθ cos θ + sin θ sin θ cosφ φ Clearly, the expansion 3.66 implies that P l cos γ = l = m= l A l mθ,φ Y l mθ, φ. 3.7 If θ =,correspondingtoθ,φ lying on the z axis, then cos γ =cosθ and then P l cos γ must satisfy d sin θ d sin θ dθ dθ P lcos γ + ll +P l cos γ = 3.7 and, since the Laplacian operator ψ = d r sin θ dθ sin θ d ψ is a scalar which dθ must be invariant under rotations, we have P l cos γ+ ll + r P l cos γ = 3.7 for arbitrary θ.thereforep l cos γ must be expressible in terms of spherical harmonics of order l, i.e. with coefficients A m θ,φ = P l cos γ = l m= l dω Y lm θ, φp lcos γ = A m θ,φ Y lm θ, φ π l + Y lm θ,φ in agreement with the addition theorem for spherical harmonics P l cos γ = 4π l + l m= l Y lm θ,φ Y lm θ, φ This result can, for example, be used to express the potential atx due to a unit point charge at x.bysubstituting3.75 into3.4 weobtain x x =4π l m= l l + r< l r> l+ Y lm θ,φ Y lm θ, φ. 3.76

11 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II Bessel functions The cylindrical coordinates ρ, φ, z are used when the boundary surface has a cylindrical symmetry, but the boundary conditions are not uniform along the z axis. The Laplace equation in cylindrical coordinates takes thefollowing form Φ= ρ ρ ρ ρ Φ + Φ ρ φ + Φ = z As usual we apply the separation of variables technique and substitute Φρ, φ, z =RρQφZz 3.78 to obtain the three ordinary differential equations d R dρ + dr ρ dρ + d Z dz k Z = 3.79 d Q dφ + ν Q = 3.8 R =. 3.8 k ν ρ Solutions of the first two equations 3.79 and3.8 aregivenby { Zz = A { A + B z for k = k e kz + B k e kz for k Qφ = C + D φ for ν = C ν e iνφ + D ν e iνφ for ν and the solutions of 3.8 willbederivedusingseriesexpansion. Aswewill see shortly that by choosing a real k the solutions will be given in terms of Bessel function, when a chose of imaginary k would have led to solutions in terms of modified Bessel functions. In terms of variable x = kρ equation 3.8 takestheformofthebessel equation d x dx If we substitute expansion x drx + dx ν x Rx = Rx = a j x j+α, 3.85 j=

12 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 35 where a,into3.84 then a j x j+α j + α ν + a j x j+α = j= Since coefficient of every power of x must vanish independently we get and a recursive relation It follows that and If we choose j= a α ν = 3.87 a + α ν = 3.88 a j+ = j ++α ν a j α = ±ν 3.9 a j+ = 3.9 a j = 4jj + α a j. 3.9 α = α Γα + then the recursion relation leads to th following solutions J ν x = J ν x = x ν j= x ν j= 3.93 j x j 3.94 j!γj + ν + j x j 3.95 j!γj ν + known as Bessel functions of the first kind of order ±ν. Thesetwosolutions are linearly independent only when ν is not an integer and otherwise J ν x = ν J ν x In either case one can construct another linearly independent solutions using the Neumann function or Bessel function of the second kind N ν x J νxcosνπ J ν x sinνx

13 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 36 Alternatively, one can use the Hankel functions or Bessel function of the third kind defined as linear combinations of Bessel and Neumann functions, H ν x J ν x+in ν x 3.98 H ν x J ν x in ν x 3.99 which also form a fundamental set of solutions to the Bessel equation We note that all of these functions, i.e. Ω ν = J ν,n ν,h ν or H ν satisfy very useful recursion formulas, Ω ν x+ω ν+ x = ν x Ω νx 3. Ω ν x Ω ν+ x = d dx Ω νx. 3. One can use Bessel function to expand an arbitrary function on theinterval ρ a into Fourier-Bessel series where x νn are the roots of and A νn = fρ = n= a J ν+ x νn A νn J ν x νn ρ a 3. J ν x = 3.3 a ρ ρfρj ν x νn. 3.4 a Although the expansion in terms of ρ ρj ν x νn is very useful for homogeneous Dirichlet boundary conditions an alternative expansion in terms of a ρ ρjν y νn,wherey a νn are the roots of d dx J νx =, 3.5 is useful for homogeneous Neumann boundary conditions. Some otherex- pansions are given by a n J ν+n x = Neumann series 3.6 n= a n J ν+n ν + nx = Kapteyn series 3.7 n= a n J ν n + x = Schlomilch series. 3.8 n=

14 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II Cylindrical Coordinates By summing over all of the solutions to the Laplace equation of theform 3.78 weobtain Φρ, φ, z = A + B zc + D φe + F log ρ ν A ν + B ν zc ν e iνφ + D ν e iνφ E ν ρ ν + F ν ρ ν + + k A k e kz + B k e kz C k + D k φe k J kρ+f k N kρ + + k A νk e kz + B νk e kz C νk e iνφ + D νk e iνφ E νk J ν kρ+f νk N ν kρ. ν As an example we can consider a cylinder with radius a and height L with boundary conditions for ρ = a Φρ, φ, z = for z = 3. V ρ, φ for z = L. If we are to solve the Laplace equation for the interior of the cylinder, then the solutions which blow up at ρ =must vanish. Moreover the solutions must be matched at φ =and φ =π. Thisleadsto Φρ, φ, z = A + B z ν A ν + B ν zc ν e iνφ + D ν e iνφ ρ ν + + k A k e kz + B k e kz C k + D k φj kρ+ + k A νk e kz + B νk e kz C νk e iνφ + D νk e iνφ J ν kρ. ν The boundary condition Φρ, φ, z ==implies that = A ν A ν C ν e iνφ + D ν e iνφ ρ ν + + k A k + B k C k + D k φj kρ+ + k A νk + B νk C νk e iνφ + D νk e iνφ J ν kρ, ν

15 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 38 or A = 3.3 A v = A k = B k A νk = B νk and Φρ, φ, z = B z zc ν e iνφ + D ν e iνφ ρ ν + ν + k e kz e kz C k + D k φj kρ+ + e kz e kz C νk e iνφ + D νk e iνφ J ν kρ k ν = zc ν e iνφ + D ν e iνφ ρ ν ν sinhkzc νk e iνφ + D νk e iνφ J ν kρ k ν The boundary condition Φρ = a, φ, z =implies that or = zc ν e iνφ + D ν e iνφ a ν 3.5 ν + sinhkzc νk e iνφ + D νk e iνφ J ν ka k ν C ν = 3.6 D ν = and if J m k mn a= 3.7 defines k mn s from its roots k mn a then Φρ, φ, z = sinhkza mn sinmφ+b mn cosmφj ν k mn ρ. m= n= 3.8

16 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 39 The remaining boundary condition fixes the coefficients A mn and B mn using.4 and3.4, A mn = B mn = sinhk mn Lπa J m+k mn a sinhk mn Lπa J m+k mn a π π dφ dφ a a with a correction B n B n due to coefficient in.4. How would the solution 3.8 changeifa? dρ ρv ρ, φj m k mn ρ sinmφ 3.9 dρ ρv ρ, φj m k mn ρcosmφ Spherical expansion of Green s functions If the problem involves the distributions of charges AND boundary conditions, then it is important to be able to express the Green functions using expansion appropriate for the boundary conditions. For example, consider asphereofradiusa, withaspecifiedpotentialv φ, θ at its surface, and a point charge. This problem was already solved using the method of Green s functions, Φx = 4πɛ ρx G D d 3 x 4π such that G D =at the surface of the sphere, i.e. rˆx r ˆx = G D x, x = It was already shown 3.76 that { 4π l m= l 4π l m= l Φ n G Dda, 3. rˆx r ˆx r r ˆx aˆx a. 3. r l Y l+ r l+ r l Y l+ r l+ lm θ,φ Y lm θ, φ for r<r lm θ,φ Y lm θ, φ for r <r 3.3 and the expansion of 3. insphericalharmonicsisgiveby, 4π l r l G D x, x m= l al+ Y l+ r = l+ rr l+ lm θ,φ Y lm θ, φ for r<r 4π l r l m= l al+ Y l+ r l+ rr l+ lm θ,φ Y lm θ, φ for r <r, 3.4 for the exterior of the sphere and 4π l r l G D x, x m= l rr l Y l+ r = l+ a l+ lm θ,φ Y lm θ, φ for r<r 4π l r l m= l rr l Y l+ r l+ a l+ lm θ,φ Y lm θ, φ for r <r. 3.5

17 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 4 for the interior of the sphere. For illustration let us find potential inside a sphere of radius b at potential V θ, φ and a charge Q on a uniformly charged ring of radius a<bin the xy plane with center at x = y =,i.e. ρx = Q πa δr aδcos θ. 3.6 From 3.4 G D n = [ ] GD r r =b = 4π and the second term in 3. is Φ l 4π n G Dda = m= l l m= l r l b l+ Y lm θ,φ Y lm θ, φ 3.7 r l b Y lmθ, φ l+ dω V θ,φ Y lm θ,φ. 3.8 To calculate the first term in 3. wealsousetheexpansion3.4 and 3.6 4πɛ ρx G D d 3 x = Q πa ɛ π dφ l m= l l + r< l r r >r < l l+ > b l+ 3.9 where r < or r > isthesmallerorlargerofr and a. But because of the azimuthal symmetry of the potential due to the ring only m =terms would survive, 4πɛ Y lm π/,φ Y lm θ, φ ρx G D d 3 x = = dφ r< l l + r r >r < l Y l+ > b l+ l π/,φ Y l θ, φ Q r< l = 4πa ɛ r r >r < l P l+ > b l+ l P l cos θ 3.3 and the final answer is Φx = 3.3 Q r< l = 4πa ɛ r r >r < l P l+ > b l+ l P l cos θ+ l r l + b Y lmθ, φ dω V θ,φ Y l+ lm θ,φ. 3.3 m= l

18 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II Cylindrical expansion of Green s functions For the boundary value problems with cylindrical symmetry it is useful to expand the Green s function in cylindrical coordinates. This corresponds to solving the Laplace equation in cylindrical coordinates, i.e. Gx, x = 4π ρ δρ ρ δφ φ δz z, 3.33 where δz z = dk e ikz z = dk coskz z 3.34 π π δφ φ = e imφ φ π m= The Green s function can be expanded as Gx, x = π m= dk e imφ φ coskz z g m k, ρ, ρ, 3.36 where the functions g m k, ρ, ρ must satisfy d ρ dg m k + m = 4π ρ dρ dρ ρ ρ δρ ρ 3.37 For ρ ρ the solutions of 3.37 arethemodifiedbesselfunctions I m kρ i m J m ikρ 3.38 K m kρ π iν+ H m ikρ Then the most general solution is a linear combination of 3.38 and3.39, ψ kρ = ai m kρ+bk m kρ satisfies b.c. for ρ<ρ 3.4 ψ kρ = a I m kρ+b K m kρ satisfies b.c. for ρ <ρ 3.4 but due to symmetry ρ ρ wemusthave g m k, ρ, ρ =ψ kρ < ψ kρ >. 3.4

19 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 4 Then 3.37 givesus ρ +ɛ d ρ dg m dρ ρ ɛ ρ +ɛ ρ ɛ [ ] dgm dρ ρ=ρ +ɛ k ρ + m [ ρ dg m dρ [ dgm dρ ρ ] ρ=ρ +ɛ ] ρ=ρ ɛ ρ=ρ ɛ = 4π = 4π = 4π ρ k ψ kρ ψ kρ ψ kρ ψ kρ = 4π ρ ρ +ɛ ρ ɛ δ ρ ρ dρ kw[ψ kρ,ψ kρ ] = 4π ρ 3.43 where W [ψ,ψ ] is the Woronskian of ψ and ψ. As an aside, consider the Sturm-Liouville equation [ d px dyx ] +qx+λrx yx = 3.44 dx dx with two solutions y x and y x from some λ. Then, and thus d dx [pxw y,y ] = d dx [pxy xy x y xy x] = 3.45 = y xpxy x y xpxy x =3.46 In our problem p = ρ and W y,y px W ψ kρ,ψ kρ = 4π kρ 3.48 for all ρ.for the solution g m k, ρ, ρ to be finite at ρ =and to vanish at ρ = ψ kρ = AI m kρ 3.49 ψ kρ = K m kρ 3.5 and the constant A =4π determined from 3.48 since W I m,k m = x. 3.5

20 CHAPTER 3. BOUNDARY-VALUE PROBLEMS: PART II 43 Therefore the expansion of the Green s function in cylindrical coordinates is given by = dk e imφ φ coskz z I x x m kρ < K m kρ > 3.5 π m= = 4 dk coskz z π I kρ < K kρ > + cosmφ φ I m kρ < K m kρ > If we let x, thenonlym =term is finite and ρ + z = π m= coskzk kρdk 3.53 If we let z then the Green s function can be obtain by two methods. We can either replace ρ R = ρ + ρ ρρ cosφ φ in 3.53 or we can derive it directly from 3.5. By equating the integrands of both expressions we obtain, K k ρ + ρ ρρ cosφ φ = I kρ < K kρ > + cosmφ φ I m kρ < K m kρ > 3.54 which must hold for all z. Then one can show by expanding around k that in polar coordinates the Green s function is given by log ρ + ρ ρρ cosφ φ = log ρ > + m= m= m ρ< ρ > m cosmφ φ For example, the potential of a line of charge with linear charge density λ is Φx = ρ e x Gx, x d 3 x = π + λδρ dφ dz dρ 4πɛ 4πɛ πρ λ + = dz dk coskz z π ɛ I K kρ λ + = dk dz coskz K π kρ ɛ λ = dk δk K kρ πɛ λ kρ = lim log k πɛ = Gx, x λ log ρ πɛ + const. 3.56