stakgold chap 7 exercise 7_38

Stakgold chap 7 exercise 7_38 PhL 12.28.11 I select this problem (am sick right now) because lots of the results are given so I will be able to verify my work. Page 291. Our "cylindrical" waveguide has some general cross section area R, not necessarily a disk. It is infinite in both +z and - z directions. This is just a generic Helmholtz problem where the BC is the Dirichlet one that u = 0 on the inner surface of the cylinder. We try to find a solution in the obvious factored form shown. Here is what happens if we stuff this form into the 3D Helmholtz equation (22 + ∂z2 + λ)u = 0 u = Z(z) φ(x,y) This is the usual sign relation Stak uses between 2 and parameter λ. So here we go (22 + ∂z2 + λ)[ Z(z) φ(x,y)] = 0 [Z(z)22 φ(x,y) + φ(x,y)∂z2 Z(z) + λ Z(z) φ(x,y)] = 0 22 φ(x,y)/φ+ Z"(z)/Z + λ = 0 22 φ(x,y)/φ + λ = - Z"(z)/Z LHS is function of x,y only, RHS is function of z only, so set to constant μ to get 22 φ(x,y)/φ + λ = μ => 22 φ + (λ-μ) φ = 0 OR 22 φ + ν φ = 0 - Z"(z)/Z = μ => -Z" = μZ => - 22 φ = νφ At this point, I have derived all results on page 291. Notice that λ is the original H parameter, but we end up now with a 2D H equation with a new parameter ν = λ-μ, same sign relation. Could have used λ' . What about BC's? The Z one is that things die off far away I presume, while the φ one is that we must have φ = 0 on the 2D boundary R. Part I: Solve the transverse problem first and use those φn as the partial eigenfunctions. In our first approach, imagine that we solve this Laplace EV problem: - 22 φn = νnφn with φn = 0 on R We now install a Green's point source inside the cylinder on the z=0 plane at (x0, y0). Our task is to compute the Green's function, fair enough. My first instinct is to use the "partial eigenfunction method". I will then write g(x,y,z| x0,y0, 0) = Σn Gn(z) φn(x,y) and try to find the Gn(z). Our Greens PDE is this -(22 + ∂z2 + λ) g(x,y,z| x0,y0, 0) = δ(x-x0)δ(y-y0)δ(z) Now insert the above expansion to get LHS = -(22 + ∂z2 + λ) g(x,y,z| x0,y0, 0) = -(22 + ∂z2 + λ) Σn Gn(z) φn(x,y) = - Σn (22 + ∂z2 + λ) Gn(z) φn(x,y) = - Σn (22 Gn(z) φn(x,y) + ∂z2 Gn(z) φn(x,y) + λ Gn(z) φn(x,y) ) = - Σn (Gn(z) 22 φn(x,y) + φn(x,y) ∂z2 Gn(z) + λ Gn(z) φn(x,y) ) But we know that 22 φn(x,y) = -νnφn so we then have LHS = - Σn (- Gn(z) νn φn(x,y) + φn(x,y) ∂z2 Gn(z) + λ Gn(z) φn(x,y) ) = - Σn [ - Gn(z) νn + ∂z2 Gn(z) + λ Gn(z) ] φn(x,y) = - Σn [ ∂z2 Gn(z) + (λ - νn) Gn(z) ] φn(x,y) If we now assume that the φn are discrete and normalized, we can then apply !Syntax Error, I dx dy φm*(x,y) to both sides of our equation to get !Syntax Error, I dx dy φm* (x,y) LHS = !Syntax Error, I dx dy φm*(x,y) RHS - Σn [ ∂z2 Gn(z) + (λ - νn) Gn(z) ] δn,m = φm*(x0,y0)δ(z) - [ ∂z2 Gm(z) + (λ - νm) Gm(z) ] = φm*(x0,y0)δ(z) - [ ∂z2 + (λ - νm) ] Gm(z) = φm*(x0,y0)δ(z) We can now regard this last equation as a 1D Green's problem in variable z. I am now inclined to define - k2 = (λ - νm) so - [ ∂z2 - k2 ] Gm(z) = φm*(x0,y0)δ(z) The homo solution for z > 0 will be Gm(z)homo+ = φm*(x0,y0)e-kz since ∂z2 on this produces +k2 and then [ ∂z2 - k2 ] = 0 as desired. So here are the left and right solutions: Gm(z)homo+ = A φm*(x0,y0)e-kz Gm(z)homo- = B φm*(x0,y0)e+kz Continuity of the solution at z = 0 requires B = A. We now integrate the ODE from z = -ε to ε ( I always do this from scratch to save the trouble of looking up the sign) -!Syntax Error, I dz [ ∂z2 - k2 ] Gm(z) = φm*(x0,y0) !Syntax Error, I dz δ(z) -!Syntax Error, I dz d [ ∂zGm(z)] = φm*(x0,y0) ∂zGm(z)+ – ∂zGm(z)– = – φm*(x0,y0) Installing our homo± solutions from above gives A φm*(x0,y0) ∂z e-kz – A φm*(x0,y0) ∂z e+kz = – φm*(x0,y0) -kA φm*(x0,y0) e-kz - k A φm*(x0,y0) e+kz = – φm*(x0,y0) -kA - k A = – 1 => 2kA = 1 => A = 1/2k Therefore our 1D Green's function is this: Gm(z) = (1/2k) φm*(x0,y0) e-k|z| Now recall from above that g(x,y,z| x0,y0, 0) = Σn Gn(z) φn(x,y) = Σn (1/2k) e-k|z|φn*(x0,y0) φn(x,y) k = and this agrees with 7.227. Now suppose λ < νn for all n. Then k is real, and we damp in both z directions, and we get no propagation in the waveguide! In particular, this is true for λ = 0 which is the Laplace Green's problem and we then just have k = . So let's write down this problem and its solution: Consider the Laplace problem λ=0 in the cylinder. The Green's function is given by g(x,y,z| x0,y0, 0) = Σn (1/2k) e-k|z|φn*(x0,y0) φn(x,y) where k = where φn are the transverse eigenfunctions and νn the corresponding EV's, - 22 φn = νnφn Thus we have a solution to an electrostatics problem I don't think we ever did in Chapter 6. A key fact that Stak does not mention much is that in the general λ case, if λ > some νi, then you will get propagation of your wave down the waveguide! So only a finite number of modes will be activated for any finite λ. As you increase λ, you get more modes going, carrying power down the guide. Part II: Now we are supposed to redo this entire problem using Zn eigenfunctions instead of φn ones. Our equation is -Z" = μZ which has solution Z = e-ikz so that Z" = - k2Z and k2 = μ and we have a continuous spectrum of oscillatory eigenfunctions ( see later for why I add the 2π) Zk(z) = e-ikz /2π k = in (-∞,∞) So now we want to expand G on this set of partial eigenfunctions, g(x,y,z| x0,y0, 0) = !Syntax Error, Idk Zk(z) Gk(x,y) = (1/2π)!Syntax Error, Idk e-ikz Gk(x,y) and this is the form Stak shows in page 292. But this is precisely a Fourier Transform and he uses α instead of k as the integration variable, and he has a constant 1/2π added in. Once again, here is how Stak likes to do his FT's: (page 23) f(x) = (1/2π) !Syntax Error, I du e-iux f^(u) f^(u) = !Syntax Error, Idx f(x) eiux f(x) = (1/2π) !Syntax Error, I dk e-ikx f^(k) f^(k) = !Syntax Error, Idx f(x) eikx f(z) = (1/2π) !Syntax Error, I dα e-iαz f^(α) f^(α) = !Syntax Error, Idz f(z) eiαz So taking the last line with f = g, we can then identify Gk(x,y) = g^(k; x,y) and g(x,y,z| x0,y0, 0) = (1/2π)!Syntax Error, Idα e-iαz g^(α; x,y) // agrees with p 292 B. OK fine. Now we have a continuous spectrum μ = α2 whereas earlier we had a discrete spectrum doing things the other way with n as the discrete label. So as we did before, we want to jam this expansion above into the full Green's PDE which was -(22 + ∂z2 + λ) g(x,y,z| x0,y0, 0) = δ(x-x0)δ(y-y0)δ(z) -(22 + ∂z2 + λ) (1/2π)!Syntax Error, Idα e-iαz g^(α; x,y) = δ(x-x0)δ(y-y0)δ(z) Now write ∂z2 e-iαz = (-iα)2 e-iαz = -α2 e-iαz so that - (1/2π)!Syntax Error, Idα (22 + ∂z2 + λ) e-iαz g^(α; x,y) = δ(x-x0)δ(y-y0)δ(z) - (1/2π)!Syntax Error, Idα (22 - α2 + λ) e-iαz g^(α; x,y) = δ(x-x0)δ(y-y0)δ(z) - (1/2π)!Syntax Error, Idα e-iαz (22 - α2 + λ) g^(α; x,y) = δ(x-x0)δ(y-y0)δ(z) Now my inclination is to write δ(z) = (1/2π)!Syntax Error, Idα e-iαz on the RHS to get (1/2π)!Syntax Error, Idα e-iαz { - (22 - α2 + λ) g^(α; x,y) } = (1/2π)!Syntax Error, Idα e-iαz δ(x-x0)δ(y-y0) and then we use completeness of the e-iαz to conclude that - (22 - α2 + λ) g^(α; x,y) = δ(x-x0)δ(y-y0) and this agrees with p 292 C. Now we need some way to express this 2D Green's function g^(α; x,y). In the previous part of this problem we determined that - 22 φn = νnφn with φn = 0 on R determined the set of EF's φn and EV's νn and let's write (22 + νn)φn = 0 So think of νn = λ-α2 and then we can compose our Green's function by the "general formula" for a Helmholtstatics Green's Function. But now I have to locate this thing somewhere. Again, Stak was so busy in his Helmholtz Section in this chapter that he neglected to produce this thing. I think I can conjecture the general result in n dimensions. _____________________________________________________________________ Digression: Find the general bilinear form for a Helmholtz Green's function: Assume the problem is this (where I try to use the usual signs) (-2 - k2)g(x|ξ) = δ(x-ξ) and assume that this is our set of Laplace EF's -2 φn = λn φn Then try g(x|ξ) = - Σn φn*(ξ) φn(x) / (k2-λn) To verify this conjecture, apply operator to both sides, where means x , -(2 + k2)g(x|ξ) = (2 + k2)Σn φn*(ξ) φn(x) / (k2-λn) = Σn (k2-λn)-1 φn*(ξ) (2 + k2) φn(x) = Σn (k2-λn)-1 φn*(ξ) (-λn + k2) φn(x) = Σn φn*(ξ) φn(x) = δ(x-ξ) // completeness with std normalization so this verifies our little formula. ________________________________________________________________________ Now we want to apply this digression result to this equation, - (22 - α2 + λ) g^(α; x,y) = δ(x-x0)δ(y-y0) where n=2 and we then identify k2 = λ-α2 and also λn = νn and we then get g^(α; x,y | x0,y0) = - Σn φn*(ξ) φn(x) / (k2-λn) = - Σn φn*(x0,y0) φn(x,y) / (λ-α2-νn) = Σn φn*(x0,y0) φn(x,y) / (-λ+α2+νn) = Σn φn*(x0,y0) φn(x,y) / (νn+α2-λ) Now insert this into our continuous z expansion: g(x,y,z| x0,y0, 0) = (1/2π)!Syntax Error, Idα e-iαz g^(α; x,y) = (1/2π)!Syntax Error, Idα e-iαz Σn φn*(x0,y0) φn(x,y) / (νn+α2-λ) = (1/2π) Σn !Syntax Error, Idα e-iαz φn*(x0,y0) φn(x,y) / (νn+α2-λ) and indeed this agrees with Stak's presented result on p 292 D. We can now compare the two solutions to this problem. The first is a single sum on n, while the second is a "double expansion" as he calls it. Now to make them look the same, we need to show that !Syntax Error, Idα e-iαz / (νn+α2-λ) = π exp(-|z|) / So I will do this little "extra credit" integral now. Let νn- λ = a2. Then I need to show that !Syntax Error, Idα e-iαz / (a2+α2) = π exp(-a|z|) / a (*) This is just an inverse FT one can look up, but I can do it directly faster this way LHS = !Syntax Error, Idα e-iαz / [(α+ia)(α-ia)] If z > 0 we close the contour down since then e-iαz = e-i(-100i)z = e-100z. We then pick up the residue of the pole at σ = -ia so (wrong way makes -) (1/2πi) LHS = (1/2πi) !Syntax Error, Idα e-iαz / [(α+ia)(α-ia)] = – Res{e-iαz / [(α+ia)(α-ia)] }α=-ia = – e-i(-ia)z / (-ia-ia) = e-az/2ai Therefore LHS = 2πi e-az/2ai = π e-az/a // agrees for z > 0 If z < 0, close contour up and then get (right way makes +) (1/2πi) LHS = (1/2πi) !Syntax Error, Idα e-iαz / [(α+ia)(α-ia)] = + Res{e-iαz / [(α+ia)(α-ia)] }α=+ia = e-i(ia)z / (ia+ia) = eaz/2ai Therefore LHS = 2πi eaz/2ai = π e-a|z|/a and then the two results can be combined and we have shown (*) above. It certainly seems better to solve this problem the Part I way.