separation and transforms 2D chap 7

Separation and Transforms PhL 8.9.11 I have been working in Stak Chapter 7 where Stak is fiddling with polar coordinates in different ways, and I feel the need to generalize the discussion there somewhat. 0. Overview (written 12.9.11, 2.5 pages) 1 1. The PDE equation and BC's. 3 2. Doing the separation of the homo PDE and identifying the EV equations. 4 3. Expansions, projections and transforms associated with Lx and Ly 5 4. Double Expansion Solution for the inhomo equation Lu = f. 6 5. Going back to the homo equation Lu=0 8 6. What about the Green's Function equation Lu=δ? 9 7. Big Example: Let (x,y) = (r,θ), polar coordinates, with r in (0,a) and θ in (-π,π), and L = -2. 10 (a) The two Sturm-Liouville problems and their solutions 10 (b) Solve the Lu = f. 12 (c) Solve Lu = 0 13 (d) The Green's Function for a circle with point charge at (r0,θ0) 14 8. The Partial Eigenfunction Method of Analysis. 15 (a) Use the x eigenfunctions 15 (b) Compute the Green's Function for the y-equation. 17 (c) Use the y eigenfunctions 19 Comments. 20 9. Curvilinear Coordinates 20 (a) 2D systems 20 (b) 3D systems with solutions not dependent one of the variables 22 10. A review of Stak examples in light of this doc 22 ___________________________________________________________________________________ 0. Overview (written 12.9.11, 2.5 pages) Section 1 considers a 2D PDE having a certain simple form in hopes that it will be separable, L = gx(y)hx(x)Lx + gy(x) hy(y)Ly Lu = 0 homo Lu = f inhomo If the operators Lx and Ly are self-adjoint, then each offers the possibility of a 1D Sturm-Liouville problem. This 2D PDE need not be the 2D Laplace or Helmholtz equation. For that reason, and also since the coordinates are not necessarily "curvilinear coordinates", the Stackel theory (as I wrote about it) does not in general apply, though of course it does apply in the special case (see Appendix A). Section 2 then shows that the homo equation Lu=0 can indeed be separated with u = X(x)Y(y). The separated equations are these, written in standard Stakgold Sturm-Liouville form, Lx Xλx(x) = λx sx(x)Xλx(x) λx = ν sx(x) ≡ gy(x)/ hx(x) Ly Yλy(y) = λy sy(y)Yλy(y) λy = -ν sy(y) ≡ gx(y)/ hy(y) where Xλx(x) are EF's of Lx and Yλy(y) are EF's of Ly, and then it is verified that L Xλx(x) Yλy(y) = 0. Section 3 writes down some presumed S-L transforms for these X and Y systems. Section 4 then does a "full eigenfunction method" (double expansion) solution of inhomo Lu=f. In the "double diagonalized space" the solution is given by uλx,λy = Fλx,λy/ (λx + λy) where the F projections are obtained in a certain way from f as shown below, so the final solutions is u(x,y) = Σλx Σλy Xλx(x) Yλy(y) uλx,λy = Σλx Σλy Xλx(x) Yλy(y) Fλx,λy/ (λx+λy) where Fλx,λy = !Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) f(x,y)/ [ hx(x) hy(y) ] where it is understood that each spectrum could be discrete, continuous or mixed. Section 5 sets f=0 in Lu=f to reconsider the homo equation Lu=0 and one finds these equivalent atomic forms u(x,y) = Σλx uλx,-λx Xλx(x) Y-λx(y) u(x,y) = Σλy u-λy,λy X-λy(x) Yλy(y) Section 6 then solves the Lu=δ problem where δ = δ(x-x0) δ(y-y0) p(x0,y0) as a special case of Section 4 above and finds then that g(x,y| x0,y0) = { p(x0,y0) / [ hx(x0) hy(y0)} * Σλx Σλy [ Xλx(x) Yλy(y)] [Xλ(x0) Yλy(y0)]* / (λx +λy) = " Σλ φλ(r)φλ(r0)* /(λ) " which is a generic form for a green's function where here λ is the sum of λ's probably due to the simple PDE form we started with. Section 7 considers the example of polar coordinates and L = Laplace,. There are 4 parts: In part (a), everything is transcribed from x,y to r,θ and the two Sturm-Liouville problems are discussed in a typical BC situation which is periodicity in θ and u vanishing at r=a and interval (0,a). The associated transforms are Expo Fourier Series for θ and Mellin Sine for r. Then in part (b), the solution to Lu=f is transcribed from above and the appropriate helper g and h functions are inserted. In part (c), Lu=0 is solved and we have transcribed the atomic forms obtained earlier. These forms are then written just as I wrote them for 3D in a separate doc, and the atoms are shown to match Stak's 2D potential theory. Finally, part (d) sets f = δ for a point charge Green's problem with u vanishing at r=a and we get a well-defined solution. So all this work is using the full or double eigenfunction expansion idea. Section 8 considers the partial-eigenfunction method, and there are 3parts: In part (a) the expansion onto x eigenfunctions is u(x,y) = Σλx uλx(y) Xλx(x) and similarly for f/F, where here uλx(y) are the usual "coefficient functions" of the other variable. It is found that uλx(y) solves a certain ODE Ly uλx(y) + λx sy(y)uλx(y) = sy(y) Fλx(y) which one then has to solve. In part (b) the above equation is solved with RHS = δ(y-y0) as a first solution step. This is a standard 1D Green's function problem, and we consider the "left and right sides" of the Green's function in various situations such as regular or limit circle at both ends, then limit point at one end, then at both ends. This just controls whether each "side" has one or two terms. In the last case, g(y|y0) = α [Yλy(y<)] [ Yλy(y>)] for example, where the two Y functions shown are those of finite s-norm at their respective endpoints. I did not finish the task, but in theory then you can use this Green's Function to solve the above inhomo equation for uλx(y) and then, finally, you have your solution u(x,y). In part (c) we just restate the equations in the case the one uses the Y eigenfunctions. It is noted that the spectrum in this case of ν might be completely different from the X eigenfunction case. Section 9 then considers the Laplace and Helmholtz equations in 2D curvilinear coordinates, 2 parts: In part (a) these 2D systems are considered. Basically for all such systems one finds that Q1 = Q2 and then 2f = (1/g11(12)) (∂21 + ∂22)f. For Laplace Lu=0, those looks just like the 2D Cartesian case and one gets the expected separated solutions with trivial functions. For Helmholtz, separation is only possible if g11(x,y) = a(x) + b(y). This only occurs in alternate polar, parabolic, and elliptic 2D coordinates. Most of M&S 2D systems are therefore not separable for Helmholtz. The conclusion reached here agrees with the conclusion of my N=2 Stackel doc section. In part (b) it is noted that these 2D systems effectively occur in 3D curvilinear coordinate work when a certain 3rd coordinate is symmetric and so does not enter the solution. Section 10 then reviews a lot of the Stak 2D Helmholtz applications in his Chapter 7, and I show how all these fit right into the mold of this document and this explains why "things always work". Whether you apply a partial EF "projector" to the PDE, or whether you expand the solution in those EF's, you always end up with an ODE for your coefficient functions of the other variable, and you never have any surviving integrals on the LHS of your ODE. This is what got me writing this doc! Here I show this for a very general class of PDE's, and all his examples are either polars or ρ-z in cylindricals. Appendix A considers the special "2D Helmholtz equation + curvilinear coordinates" case of our general 2D PDE form, and then makes the connection between our helper functions and the Stackel matrix functions. Comment: What led to this doc was Stak's "bandying around" with Helmholtz in polar coordinates ___________________________________________________________________________________ 1. The PDE equation and BC's. To that end, assume we have some PDE of the following form L u(x,y) = f(x,y) L = gx(y)hx(x)Lx + gy(x) hy(y)Ly [gx(y)hx(x)Lx + gy(x) hy(y)Ly] u(x,y) = f(x,y) where x and y are arbitrary coordinates (perhaps x=r and y=θ), and where f(x,y) is some driving function, perhaps a distribution such as "a ring of source". One hidden assumption here is that when we say Lx, we mean that Lx is a differential operator which is self-adjoint and does not involve the other variable y, and similarly for Ly. In our full L, we allow that Lx might be multiplied by some functions as shown, so this is a fairly restrictive form for L, but I think it covers our important cases. More generally we could assume L = qx(x,y)Lx + qy(x,y)Ly where the q functions don't factor, and this would no doubt be a much harder problem to deal with. In practice, we will find that these helper g and h functions are just scale factors encountered when changing to non-Cartesian coordinate systems, with possible constants to cause the resulting L operators to be standard ones you find in a book. 2. Doing the separation of the homo PDE and identifying the EV equations. Our usual starting point is to study the homo version of this equation [gx(y)hx(x)Lx + gy(x) hy(y)Ly] u(x,y) = 0 We shall now assume that there are some boundary conditions for our x and y systems which serve to determine the eigenfunctions for those systems. We assume that these BC's are not "interlinked" with each other, which means that for each coordinate, the boundaries lie along a surface of constant coordinate, which we sometimes call a "level surface". So maybe the interval for x is (ax, bx) and similarly for y. Sometimes a BC says u=0 at both ends of the interval, but sometimes (as for θ) the BC has the form that u and u' are continuous at a certain point (θ=±π). Comment added 12.9.11. The Stackel theory, as I present it in my paper, starts specifically with the Helmholtz equation (a homo equation) and specifically with some generic orthogonal coordinates ξn with scale factors hn. Here, I am not assuming Helmholtz, and I am assuming nothing about coordinates x and y, and below I will be seeking only "simple separation" meaning R = 1. I think one could apply the Stackel concept to the generic PDE above, and maybe that is what I am doing below. We are now ready to do our separation attempt. We are just searching for solutions of the separated form, and we are not claiming that this exhausts all possible solutions (though it probably does in the atomic sense). So u(x,y) = X(x)Y(y) Installing this into our homo equation gives [gx(y)hx(x)Lx + gy(x) hy(y)Ly]X(x)Y(y) = 0 [gx(y) hx(x)Lx X(x)] Y(y) + gy(x) hy(y)X(x) [Ly Y(y)] = 0 Divide both sides by X(x)Y(y) gx(y) gy(x) to get [ hx(x)Lx X(x)] / [gy(x) X(x)] = – hy(y)Ly Y(y)/[gy(y) Y(y)] The LHS is independent of y so it could be some t(x). The RHS is independent of x, so it could be some w(y). We then have t(x) = w(y). The only functions that satisfy this equation are that both are constants, and they are of course the same constant, and we shall call that ν, the separation constant. So we have [hx(x)Lx X(x)] /[ gy(x)X(x] = - [hy(y)Ly Y(y)]/ [ gx(y)Y(y) ] = ν hx(x)Lx X(x) = +ν gy(x)X(x) hy(y)Ly Y(y) = –ν gx(y)Y(y) and these then our the "separated equations". Rewrite them now as Lx X(x) = ν [gy(x)/ hx(x)] X(x) Ly Y(x) = - ν [gx(y)/ hy(y)] Y(y) Anticipating that ν will the ±λ, we put this into "standard EV form" by defining sx(x) ≡ gy(x)/ hx(x) => gy(x) = hx(x) sx(x) sy(y) ≡ gx(y)/ hy(y) => gx(y) = hy(y) sy(y) Then our separated equations are Lx X(x) = ν sx(x) X(x) Ly Y(x) = - ν sy(y) Y(y) If we now set ν = λx = - λy, these equations become Lx Xλx(x) = λx sx(x)Xλx(x) λx = ν Ly Yλy(y) = λy sy(y)Yλy(y) λy = -ν And these are our x and y EV equations now in "standard form". Meanwhile, we can rewrite L in a slightly more useful form: L = [ gx(y)hx(x)Lx + sx(x) hy(y)Ly] = [hy(y) sy(y)hx(x)Lx + hx(x) sx(x) hy(y)Ly] so L = hx(x) hy(y) [sy(y) Lx + sx(x) Ly] Notice what happens when we apply L to a 2D basis function: L Xλx(x) Yλy(y) = hx(x) hy(y) [sy(y) λxsx(x) + sx(x) λysy(y)] Xλx(x) Yλy(y) = hx(x) hy(y) sx(x) sy(y) [ λx+λy] Xλx(x) Yλy(y) = (λx+λy) { hx(x) hy(y) sx(x) sy(y)} Xλx(x) Yλy(y) = 0 // since by their definition, λx+λy = 0 [ stuff was here which is now Appendix A.] 3. Expansions, projections and transforms associated with Lx and Ly Since we can find complete eigenfunctions for each variable, assuming the BC's provide a clean SL problem in each, we know we can expand either separated function in them as shown here on the left: X(x) = Σλx aλxXλx(x) aλx = !Syntax Error, Idx sx(x) Xλx*(x) X(x) Y(y) = Σλy aλyYλy(y) aλy = !Syntax Error, Idy sy(y) Yλx*(y) Y(y) where the aλx are "coefficients" which are constants, and where the sums could be integrals or even combinations of sums and integrals. Since the eigenfunctions form a complete set, we know we can invert our expansions to get the projections as shown. The combination of expansion and projection is called a transform, so there are then two transforms associated with our problem, one in x and one in y. 4. Double Expansion Solution for the inhomo equation Lu = f. This discussion aligns with what Stak would call the "full eigenfunction method" of doing something, as opposed to the "partial EF method" which we shall do later. Since we are going to expand u in basis functions Xλx(x)Yλy(y), our solutions are appropriate for u(x,y) meeting the boundary conditions which are met by Xλx(x)Yλy(y). We start with Lu=f which reads hx(x) hy(y) [sy(y) Lx + sx(x) Ly] u(x,y) = f(x,y) Now since we are here doing a double expansion, we might as well write u(x,y) = Σλx Σλy Xλx(x) Yλy(y) uλx,λy Comparing to the above, we can identify uλx,λy = aλx aλy but here only the product will occur, so we will just use uλx,λy. Inserting our expansion we get hx(x) hy(y) [sy(y) Lx + sx(x) Ly] Σλx Σλy Xλx(x) Yλy(y) uλx,λy = f(x,y) Move in the operators and move out the sums and coefficients hx(x) hy(y) Σλx Σλy uλx,λy [sy(y) Lx + sx(x) Ly] Xλx(x) Yλy(y) = f(x,y) hx(x) hy(y) Σλx Σλy uλx,λy [sy(y) Lx Xλx(x) Yλy(y) + sx(x) Ly Xλx(x) Yλy(y)] = f(x,y) hx(x) hy(y) Σλx Σλy uλx,λy [sy(y) λx sx(x) Xλx(x) Yλy(y) + sx(x) λy sy(y) Xλx(x) Yλy(y)] = f(x,y) hx(x) hy(y) sx(x) sy(y) Σλx Σλy uλx,λy [λx Xλx(x) Yλy(y) + λy Xλx(x) Yλy(y)] = f(x,y) [ hx(x) hy(y) sx(x) sy(y) ] Σλx Σλy Xλx(x) Yλy(y) uλx,λy [λx + λy] = f(x,y) Σλx Σλy Xλx(x) Yλy(y) uλx,λy (λx + λy) = f(x,y)/ [ hx(x) hy(y) sx(x) sy(y) ] We now see that we want to define a new RHS function F(x,y) according to F(x,y) ≡ f(x,y)/ [ hx(x) hy(y) sx(x) sy(y) ] and then we want to do a double expansion of F F(x,y) = Σλx Σλy Xλx(x) Yλy(y) Fλx,λy When our equation Lu=f is fully diagonalized and we obtain uλx,λy (λx + λy) = Fλx,λy Thus we have replaced a complicated PDE in two variables with this one line of algebra which contains no integrals, derivatives, etc. I would call this a double diagonalization. This equation must be true for all pairs {λx, λy} where each λx runs over its complete spectrum, whatever that might be (determined by its SL problem). For any pair {λx, λy} which has the property (λx + λy)= 0, we shall need Fλx,λy = 0, and we would call this a "consistency condition" on the driving function, in line with Stak's vocabulary. If this property is not true, then our solution is uλx,λy = Fλx,λy/ (λx + λy) So all we have to do is compute the Fλx,λy using our transform projections stated above, Fλx,λy = !Syntax Error, Idx sx(x) Xλx*(x) !Syntax Error, Idy sy(y) Yλy*(y) F(x,y) = !Syntax Error, Idx sx(x) Xλx*(x) !Syntax Error, Idy sy(y) Yλy*(y) f(x,y)/ [ hx(x) hy(y) sx(x) sy(y) ] = !Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) f(x,y)/ [ hx(x) hy(y) ] Our solution is then uλx,λy = Fλx,λy/ (λx+λy) and in coordinate space we have u(x,y) = Σλx Σλy Xλx(x) Yλy(y) uλx,λy = Σλx Σλy Xλx(x) Yλy(y) Fλx,λy/ (λx+λy) and the problem is completely solved, although the solution appears as a double sum/integral which is not always what we would like. As a final check of this result, consider: L u(x,y) = Σλx Σλy L[Xλx(x) Yλy(y)] Fλx,λy/ (λx+λy) = Σλx Σλy L[Xλx(x) Yλy(y)] Fλx,λy/ (λx+λy) = Σλx Σλy (λx+λy) s(x,y) [Xλx(x) Yλy(y)] Fλx,λy/ (λx+λy) = s(x,y)Σλx Σλy Xλx(x) Yλy(y) Fλx,λy = s(x,y)F(x,y) = hx(x) hy(y) sx(x) sy(y) F(x,y) = f(x,y) 5. Going back to the homo equation Lu=0 We obtain uλx,λy (λx + λy) = 0 Now the only contributions to the solution u(x,y) come from pairs {λx, λy} such that (λx + λy) = 0. And for such pairs we can have uλx,λy be an arbitrary value. Thus, we can write our homo solution as u(x,y) = Σλx+λy=0 uλx,λy Xλx(x) Yλy(y) When we separated our PDE earlier, we had separation constant ν and we identified ν = λx = -λy => λx+λy = 0 all the time We could then write the above sum as u(x,y) = Σλx uλx,-λx Xλx(x) Y-λx(y) or u(x,y) = Σλy u-λy,λy X-λy(x) Yλy(y) where again the ua,b are arbitrary numbers. It is obvious that these are homo equation solutions since, for example using the first line above, Lu = hx(x) hy(y) [sy(y) Lx + sx(x) Ly] Σλx uλx,-λx Xλx(x) Y-λx(y) = hx(x) hy(y) Σλx uλx,-λx [sy(y) Lx + sx(x) Ly] Xλx(x) Y-λx(y) = hx(x) hy(y) Σλx uλx,-λx [sy(y) sx(x)λx+ sx(x)(-λx)s(y)] Xλx(x) Y-λx(y) = hx(x) hy(y) sx(x) sy(y) Σλx uλx,-λx [λx+ (-λx)] Xλx(x) Y-λx(y) = 0 I refer to the product Xλx(x) Y-λx(y) as an "atom", and point out that the factors of this atom are "cross linked" in that the label λx appears in both factors. The solution of the homo equation is a linear combination of atoms! I have a document that lists the triple atomic forms for various curvilinear coordinate systems in 3D, but here we happen to be working in 2D. 6. What about the Green's Function equation Lu=δ? We take our inhomo case now with f = δ. The solution will be g(x,y) = Σλx Σλy Xλx(x) Yλy(y) uλx,λy uλx,λy = Fλx,λy/ (λx +λy) Fλx,λy =!Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) f(x,y)/ [ hx(x) hy(y) ] The question now arises of what kind of δ you have in mind! To be flexible, let's allow some arbitrary function here and write (this would account for any curvilinear coordinates for example) δ = δ(x-x0) δ(y-y0) p(x0,y0) Then we get Fλx,λy =!Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) f(x,y)/ [ hx(x) hy(y) ] =!Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) δ(x-x0) δ(y-y0) p(x0,y0) / [ hx(x) hy(y) ] = Xλx*(x0) Yλy*(y0) p(x0,y0) / [ hx(x0) hy(y0) ] We then obtain g(x,y) = Σλx Σλy Xλx(x) Yλy(y) uλx,λy = Σλx Σλy Xλx(x) Yλy(y) Fλx,λy/ (λx +λy) = Σλx Σλy Xλx(x) Yλy(y) Xλx*(x0) Yλy*(y0) p(x0,y0) / [ hx(x0) hy(y0) ] / (λx +λy) = { p(x0,y0) / [ hx(x0) hy(y0)} Σλx Σλy Xλx(x) Yλy(y) Xλx*(x0) Yλy*(y0) / (λx +λy) = { p(x0,y0) / [ hx(x0) hy(y0)} * Σλx Σλy [ Xλx(x) Yλy(y)] [Xλ(x0) Yλy(y0)]* / (λx +λy) Recall from "review of the generic regular BV problem" that the 1D situation was this g(x|ξ;0) = Σn φn(x) Fn/(λn) = Σn φn(ξ)*φn(x) /(λn) Apart from our overall constant, which only controls scaling in a Green's function problem, we see how our 1D formula is generalized. We would regard this as the properly normalized EF φλx,λy(x,y) = Xλx(x) Yλy(y) and then we have g(x,y|x0,y0) = { p(x0,y0) / [ hx(x0) hy(y0)} Σλx,λy φλx,λy(x,y) φλx,λy*(x0,y0) / (λx +λy) g(x|x0) = Σλ φλ(x)*φλ(x0) /(λ) Notice that g(x,y|x0,y0) will have our usual symmetry only if p(x0,y0) = K hx(x0) hy(y0). Surely this is what happens when you work with Laplace or Helmholtz converted to some curvilinear coordinates where only two of the coordinates are active because it is a 2D problem or a 3D one with a symmetry. 7. Big Example: Let (x,y) = (r,θ), polar coordinates, with r in (0,a) and θ in (-π,π), and L = -2. (a) The two Sturm-Liouville problems and their solutions Consider -2u = f in polar coordinates so L = -2. This is our example Lu=f. L u(r,θ) = f(r,θ) L = gr(θ)hr(r)Lr + gθ(r)hθ(θ)Lθ = -2 [gr(θ)hr(r)Lr + gθ(r)hθ(θ)Lθ ]u(r,θ) = f(r,θ) [ -(1/r)∂r(r∂r) - (1/r)2∂θ2) u(r,θ) = f(r,θ) gr(θ) hr(r)Lr = -(1/r)∂r(r∂r) gθ(r) hθ(θ)Lθ = -(1/r)2∂θ2 And this point we make the choice that Lr = -∂r(r∂r) and Lθ = -∂θ2, and this then forces values onto the four g and h functions. We then have ( we select things to get "standard L's" ) Lr = -∂r(r∂r) gr(θ)= 1 hr(r) = 1/r sr(r) = 1/r Lθ = -∂θ2 gθ(r) = 1/r2 hθ(θ) = 1 sθ(θ) = 1 Notice that both L's are formally self-adjoint. We then find that sr(r) = gθ(r)/ hr(r) = (1/r2)/(1/r) = 1/r sθ(θ) = gr(θ)/ hθ(θ) = 1/1 Our two eigenfunction equations are then Lr Rλr(r) = λr (1/r)Rλr(r) λr = - λθ Lθ Θλθ(θ) = λθΘλθ(θ) Let's write them both out: -∂r(r∂r)Rλr(r) = λr (1/r)Rλr(r) -∂θ2 Θλθ(θ) = λθΘλθ(θ) The actual eigenvalues λr and λθ are determined by the details of the boundary conditions for our problem. If we have full θ azimuth (-π.π), for example, then we know λθ = n2 where n is an integer and Θλθ(θ)~ einθ. The corresponding transform is the Exponential Fourier Series Transform as in transforms.doc. The correctly normalized eigenfunctions are Θλθ(θ) = ()-1 e-inθ where λθ = n2. We often use fuzzy notation and say that Θn(θ) = ()-1 e-inθ even through λθ = n2. Proper normalization here means that !Syntax Error, I dθ Θn(θ) Θm(θ) = δn,m. The "measure" used in the eigenvalue summation is this: Σλθ = Σλθ = 0,1,4,9.. = Σn=0,1,2.. = Σn=0∞ Now consider instead the r equation say on an interval (0,a). It is easy to show that Rλr = r±iα with α = . If for example we want R to vanish at r = a, we get the situation I wrote up in "the Mellin Transforms" . You get the Mellin Sine Transform which says this: ( I refer to λr just as λ here) -∂r(r∂r)Rλr(r) = λ (1/r)Rλr(r) - (rRλr')' - λ (1/r)Rλr = 0 Rλr(r) = (π)-1/2 sin[ ln(r/a) ] // normalized eigenfunctions δ(r-r')r = (2/π)!Syntax Error, Idk sin[ k ln(r/a) ] sin[ k ln(r'/a) ] // completeness λ = k2 F(k) = !Syntax Error, Idr sin[ k ln(r/a) ] f(r) // the transform λ = k2 f(r) = (2/π) !Syntax Error, Idk sin[ k ln(r/a) ] F(k) The Rλ(r) shown form a complete set on the interval (0,a), the spectrum is the entire positive real λ axis, and we have λ = k2 as the relation between λ and the transform conjugate parameter k. The measure in our summation is now Σλr = !Syntax Error, Idλr = !Syntax Error, I[2krdkr] If we choose to express things in terms of λr = kr2, we note that dλr = 2krdkr which is the second form shown above. This is analogous to using n instead of λθ to enumerate the summation. So in this case we find that λr = all positive real numbers and it is not quantized the way λθ = n2 was. (b) Solve the Lu = f. Now let's assume these two sets of boundary conditions and solve Lu=f using our double eigenfunction expansion. We have, as just noted, Rλr(r) = (π)-1/2 sin[ ln(r/a) ] λr > 0 gr(θ)= 1 hr(r) = 1/r sr(r) = 1/r Θλθ(θ) = ()-1 e-inθ λθ = n2 , n=0,1,2... gθ(r) = 1/r2 hθ(θ) = 1 sθ(θ) = 1 We then do the following expansion for u(r,θ) u(r,θ) = Σn=0∞ !Syntax Error, Idλr Rλr(r) Θλθ(θ) uλr,λθ We then define F(r,θ) this way F(r,θ) ≡ f(r,θ)/ [ hr(r) hθ(θ) sr(r) sθ(θ) ] = f(r,θ)/[(1/r) 1 (1/r) ] = r2 f(r,θ) The double-diagonalized solution is uλr,λθ = Fλr,λθ/ (λr + λθ) and in coordinate space we have u(r,θ) = Σλr Σλθ Rλr(r) Θλθ(θ) Fλr,λθ/ (λr + λθ) = !Syntax Error, I[2krdkr] Σn=0∞ Rλr(r) Θλθ(θ) Fλr,λθ/ (kr2 + n2) λr = kr2 λθ= n2 where Θλθ(θ) = { ()-1 e-inθ } n = Rλr(r) = { (πkr)-1/2 sin[ kr ln(r/a) ] } kr = We do have to compute our Fλr,λθ coefficients this way Fλx,λy = !Syntax Error, Idx Xλx*(x) !Syntax Error, Idy Yλy*(y) f(x,y)/ [ hx(x) hy(y) ] Fλr,λθ = !Syntax Error, Idr Rλr*(r) !Syntax Error, Idθ Θλθ*(θ) f(r,θ)/ [ hr(r) hθ(θ) ] = !Syntax Error, Idr Rλr*(r) !Syntax Error, Idθ Θλθ*(θ) f(r,θ)/ [ (1/r) ] = !Syntax Error, Idr r Rλr*(r) !Syntax Error, Idθ Θλθ*(θ) f(r,θ) So one could try out some sample f(r,θ) and carry this all through. (c) Solve Lu = 0 As for the homo equation Lu=f , we have two options for writing the solution u(x,y) = Σλx uλx,-λx Xλx(x) Y-λx(y) u(x,y) = Σλy u-λy,λy X-λy(x) Yλy(y) which now become u(r,θ) = Σλr uλr,-λr Rλr(r) Θ-λr(θ) u(r,θ) = Σλθ u-λθ,λθ R-λθ(r) Θλθ(θ) Here we have to pay attention more then normal: Θλθ(θ) = { ()-1 exp(-inθ) } = { ()-1 exp(-iθ) } Θ-λr(θ) = { ()-1 exp(-iθ) } = { ()-1 exp(-krθ) } We think of our λr integration variable as running along the "top of the cut", so -λr is below the real axis on the left side, and therefore = -i and therefore exp(-iθ) } = exp(-θ) = exp(-krθ). Now consider Rλr(r) = { (πkr)-1/2 sin[ kr ln(r/a) ] } kr = If we want R-λθ(r), we need λr = -λθ = -n2. If we think of λθ as just above the cut (though no cut, just in the upper half plane), then -λθ is lower left at phase -π so = -i n so then = kr = = -in. Then we have R-λθ(r) = { (-inπ)-1/2 sin[ -in ln(r/a) ] } = { - (-inπkr)-1/2 sinh[n ln(r/a)] } So our two forms for the homo solution were u(r,θ) = Σλr uλr,-λr Rλr(r) Θ-λr(θ) u(r,θ) = Σλθ u-λθ,λθ R-λθ(r) Θλθ(θ) and we insert to get u(r,θ) = !Syntax Error, I[2krdkr] uλr,-λr { (πkr)-1/2 sin[ kr ln(r/a) ] } { ()-1 exp(-krθ) } u(r,θ) = Σn=0∞ u-λθ,λθ { - (-inπ)-1/2 sinh[n ln(r/a)] } { ()-1 exp(-inθ) } where the coefficients can be arbitrary numbers. Basically L acting on either {..} gives 0, so the coefficients don't matter. Now if we think of the potential inside a disk with u=0 on the perimeter r=a, then yes, all the coefficients will be 0. But these "atomic expansions" would be appropriate for a situation where we have charges inside, but we are away from those charges, such as in a Green's Function situation. Then the coefficients are not 0 and have to be adjusted to match the charges somehow. We can identify the basic "atoms" here as osc expo (1/) sin[ kr ln(r/a) ] e-krθ λr = kr2 = positive real r in (0,a) expo osc (1/) sinh[ n ln(r/a) ] e-inθ n2 = λθ = 1,4,9... n = 1,2.. θ in (-π,π) [ ln(r/a), 1] 1 n=0 where in the second case we have to treat n=0 as a special case with different atoms. Notice that a set of atoms like this is specific to the boundary conditions. In this example, our r BC was really that Rλr(r=a) = 0, and for θ was the usual continuity stuff. Had we chosen Rλr'(r=a) = 0, both atomic forms would be different. So any linear combination of the above atoms solves the Laplace equation for the inside of a grounded circle of radius a. In the second case above, notice that sinh[ n ln(r/a) ] = (1/2)[ enln(r/a) - e-nln(r/a)] = (1/2)[ (r/a)n - (r/a)-n] and one could more generally think of the atoms as (r/a)±n e-inθ which is how things appear in Stak p 128. (d) The Green's Function for a circle with point charge at (r0,θ0) We now have u = g and f(r,θ) = δ(r-r0) = δ(θ-θ0) δ(r-r0)/r0 . According to work above, we know that δ = δ(x-x0) δ(y-y0) p(x0,y0) g(x,y) = { p(x0,y0) / [ hx(x0) hy(y0)} * Σλx Σλy [ Xλx(x) Yλy(y)] [Xλx(x0) Yλy(y0)]* / (λx +λy) g(r,θ) = { p(r0,θ0) / [ hr(r0) hθ(θ0)} * Σλr Σλθ [ Rλr(r) Θλθ(θ)] [Rλr(r0) Θλθ(θ0)]* / (λr +λθ) = {[1/r0] / [1/r0] } * Σλr Σλθ [ Rλr(r) Θλθ(θ)] [Rλr(r0) Θλθ(θ0)]* / (λr +λθ) = Σλr Σλθ [ Rλr(r) Θλθ(θ)] [Rλr(r0) Θλθ(θ0)]* / (λr +λθ) g(r,θ|r0,θ0) = !Syntax Error, I[2krdkr] Σn=0∞ [ Rλr(r) Θλθ(θ)] [Rλr(r0) Θλθ(θ0)]* / (kr2 + n2) g(x|x0) = Σλ φλ(x)*φλ(x0) /(λ) where Θλθ(θ) = { ()-1 e-inθ } Rλr(r) = { (πkr)-1/2 sin[ kr ln(r/a) ] } and there is our final answer. The double sum answers are never very convenient of course, and we know other more compact expressions of g(r,θ) such as from the image method. Remember this is g inside a grounded disk, it is not the free space g. Since g(x|x0) = -ln(1/R) in 2D, we can regard the above as just one of our many "1/R expansions" but in 2D we get the extra log 8. The Partial Eigenfunction Method of Analysis. We first did the double expansion stuff shown above mainly just to get it out of the way. In practice, we are going to "work a problem" by using the complete set of eigenfunctions in either x or y. (a) Use the x eigenfunctions When we "separated variables" writing u(x,y) = X(x)Y(y), we got Lx X(x) = ν sx(x) X(x) Ly Y(x) = - ν sy(y) Y(y) If we now set ν = λx = - λy, these equations become Lx Xλx(x) = λx sx(x)Xλx(x) λx = ν Ly Yλy(y) = λy sy(y)Yλy(y) λy = -ν Suppose we ignore the second equation for the moment, and solve the first equation for its EF's and its EV's, We then obtain some set of legal values ν = {λx} and we know the Xλx(x) functions. So consider our inhomo equation Lu=f hx(x) hy(y) [sy(y) Lx + sx(x) Ly] u(x,y) = f(x,y) We as before write f(x,y) ≡ F(x,y) [ hx(x) hy(y) sx(x) sy(y) ] u(x,y) = Σλx uλx(y) Xλx(x) F(x,y) = Σλx Fλx(y) Xλx(x) and we obtain for the LHS of Lu=f hx(x) hy(y) [sy(y) Lx + sx(x) Ly] Σλx uλx(y) Xλx(x) hx(x) hy(y) Σλx [sy(y) Lx uλx(y) Xλx(x)+ sx(x) Ly uλx(y) Xλx(x)] hx(x) hy(y) Σλx [sy(y) λx sx(x) uλx(y) Xλx(x)+ sx(x) (Ly uλx(y)) Xλx(x)] hx(x) hy(y) sx(x) sy(y) Σλx Xλx(x) [λx uλx(y) + {(Ly uλx(y))/ sy(y)}] We now set this to the RHS which is f(x,y) ≡ [ hx(x) hy(y) sx(x) sy(y) ] F(x,y) = hx(x) hy(y) sx(x) sy(y) Σλx Fλx(y) Xλx(x) We then cancel the four helper functions and use completeness of the Xλx(x) to get λx uλx(y) + {(Ly uλx(y))/ sy(y)} = Fλx(y) Ly uλx(y) + λx sy(y)uλx(y) = sy(y) Fλx(y) // KEY RESULT! This is then a 2nd order 1D ODE in the variable y. If we can somehow solve this equation, somehow paying attention to the boundary conditions, then we would have the solution to our entire problem as : u(x,y) = Σλx uλx(y) Xλx(x) and we would then have completely avoided finding the Yλy(y)eigenfunctions and eigenvalues {λy}. Notice that it is an eigenvalue λx of the X problem which is sitting quietly in our 1D uλx(y) y equation as a parameter, and of course uλx(y) will thus depend on λx, as our notation indicates. Notice also that our ODE in y for the function uλx(y) is NOT the ODE Ly Yλy(y) = λy sy(y)Yλy(y). Let's compare Ly uλx(y) + λx sy(y)uλx(y) = sy(y) Fλx(y) (*) Ly Yλy(y) - λy sy(y)Yλy(y) = 0 If f(x,y) = 0, these two equations are the same apart from the eigenvalue sign and name. But if we have the inhomo case with some f, then the equations are not the same. (b) Compute the Green's Function for the y-equation. One modus operandi is to compute the Green's function for (*) and then use that to find the solution to the equation with the full RHS. So we now shall compute that Green's function for g(ay) = g(by) = 0 (assume Dirichlet). We start with Ly uλx(y) + λx sy(y)uλx(y) = δ(y-y0) This is a Standard Issue 1D Green's function problem which we know how to solve. In the rest of this section, we shall not assume that Yλy(y) are the Y eigenvalue functions as described above, but instead they are just "some" generic solution of the EV equation without concern for its BC's. We will continue to use the λy label to avoid a bunch of minus signs. If the endpoints are not limit-point singular, we know that at each endpoint we are free to add in a mixture of a second independent solution which we might call Yλy(y). If we are limit-point at only one endpoint, there is only one possible s-norm solution at that end point, which we can again call Yλy(y). So our general plan is to write (assuming not limit point) g(y|y0) = [A Yλy(y<) + B Yλy(y<)] [A' Yλy(y>) + B' Yλy(y>)] Note that a form like this is automatically continuous at y=y0. At that point we have g(y0|y0) = [A Yλy(y0) + B Yλy(y0)] [A' Yλy(y0) + B' Yλy(y0)] whether we let y→y0 from the left or right. Our (assumed here as Dirichlet type) end conditions are A Yλy(ay) + B Yλy(ay) = 0 A' Yλy(by) + B' Yλy(by) = 0 Finally we require that the jump be equal to 1/a0 , whatever that is for Ly Ly = a0(y) ∂y2 + etc The jump is given by ∂yg+ = [A Yλy(y0) + B Yλy(y0)] [A' Yλy'(y0) + B' Yλy'(y0)] ∂yg- = [A Yλy'(y) + B Yλy'(y)] [A' Yλy(y0) + B' Yλy(y0)] jump = [A Yλy(y0) + B Yλy(y0)] [A' Yλy'(y0) + B' Yλy'(y0)] - [A Yλy'(y0) + B Yλy'(y0) ] [A' Yλy(y0) + B' Yλy(y0)] = AA' [ Yλy(y0) Yλy'(y0)- Yλy'(y0) Yλy(y0) ] + AB' [Yλy(y0) Yλy'(y0)- Yλy'(y0) Yλy(y0) ] + A'B [Yλy(y0) Yλy'(y0) - Yλy'(y0) Yλy(y0) ] + A'B'[Yλy(y0) Yλy'(y0) - Yλy'(y0) Yλy(y0)] = AB' W[Yλy(y0), Yλy(y0)] + A'B W[Yλy(y0), Yλy(y0)] = (AB' - A'B) W[Yλy(y0), Yλy(y0)] The Wronskian is always some simple expression, and we then have our third condition which we then follow by our first two conditions (AB' - A'B) W[Yλy(y0), Yλy(y0)] = 1/a0(y0) A Yλy(ay) + B Yλy(ay) = 0 A' Yλy(by) + B' Yλy(by) = 0 How can we have four coefficients and only 3 conditions? Well think of the 3 conditions this way (AB' - A'B) = α A = βB A' = β'B' Insert into the first to get βBB' - β'B'B = α => BB'(β-β') = α BB' = α/(β-β') So we know BB'. Now multiply 2nd equation by B' AB' = βBB' = β α/'(β-β') So we know AB'. Now multiply 3rd equation by A: AA' = β'AB' = β2 α/'(β-β') So we know AA'. Now multiply 3rd equation by B BA' = β'BB' = = β' α/(β-β') So we know BA'. Thus we know all four coefficients appearing in g g(y|y0) = [A Yλy(y<) + B Yλy(y<)] [A' Yλy(y>) + B' Yλy(y>)] = AA' Yλy(y<) Yλy(y>) + BA' Yλy(y<) Yλy(y>) + etc It is true that we don't know the individual coefficients so this is the degree of freedom that is missing by having only 3 equations on 4 unknowns. Fine, we have what we need. If we are limit point at the right end, say, then we will have g(y|y0) = [A Yλy(y<) + B Yλy(y<)] [ Yλy(y>)] which is as if A'=0 and B'=1. Note that this Yλy(y) is now a specific function, whereas before it could be any function independent of Yλy(y). We no longer have a right-end g=0 condition so here are our equations this time left end: A Yλy(ay) + B Yλy(ay) = 0 jump: (A) W[Yλy(y0), Yλy(y0)] = 1/a0(y0) and this lets us determine A and B. If we are limit point at both ends, then we know that g(y|y0) = α [Yλy(y<)] [ Yλy(y>)] where Y is the only solution that works at the left end and Y is the only one that works at the right end. Now we have no endpoint conditions and only the jump condition and that determines α. If we think of this case as A=α, B=0, A' = 0, B'=1 then the jump condition can be read from above, (AB' - A'B) W[Yλy(y0), Yλy(y0)] = 1/a0(y0) α W[Yλy(y0), Yλy(y0)] = 1/a0(y0) So notice that we have successfully found the Green's Function without computing the actual Y eigenfunctions. We only needed to have two independent solutions of the EV equation ignoring BC's. So a big point is that we avoided that detailed Y eigenfunction work, and we were probably able to select our generic Y and Y functions in some simple way to naturally meet our BC's at the individual end. (c) Use the y eigenfunctions We just repeat everything stated above with x↔y. This time we have to compute the Y eigenfunctions, but we don't have to do the X ones. Here are some of the equations: f(x,y) ≡ F(x,y) [ hx(x) hy(y) sx(x) sy(y) ] u(x,y) = Σλy uλy(x) Yλy(y) F(x,y) = Σλy Fλy(x) Yλy(y) We end up with a 1D equation in x, where now λy = -λx but we write it as λy, and we compare Lx uλy(x) + λy sx(x)uλy(x) = sx(x) Fλy(x) Lx Xλy(x) - λx sx(x)Xλy(x) = 0 // the EV equation for X We can compute the Green's Function as before, but now in the x world. g(x|x0) = [A Xλx(x<) + B Xλx(x<)] [A' Xλx(x>) + B' Xλx(x>)] // regular or limit circle g(x|x0) = [A Xλx(x<) + B Xλx(x<)] [Xλx(x>)] // right end limit point g(x|x0) = α [Xλx(x<)] [Xλx(x>)] // both ends limit point Comments. (a) Notice that in our first method, we used ν = λx as our spectrum (since we used the X eigenfunctions). In the second method we used ν = -λy and in this is a completely different spectrum! (b) If we did the Helmholtz equation, we would have an extra parameter "λ" floating around everywhere which is separate from our λx and λy here. 9. Curvilinear Coordinates (a) 2D systems The general Laplacian formula is this for orthogonal 2D systems 2f = (1/) { ∂1 [g'11 (∂1f) ] + ∂2 [g'22 (∂2f) ] } = = Q1Q2 g'ii = Qi-2 So we can write this out a bit as 2f = (1/Q1Q2) { ∂1 [(Q2/Q1) (∂1f) ] + ∂2 [(Q1/Q2) (∂2f) ] } In general, 2f = (1/Q1Q2) { ∂1 [(Q2/Q1) (∂1f) ] + ∂2 [(Q1/Q2) (∂2f) ] } Now all the usual 2D systems listed in M&S Chapter 2 (including polars treated in the special way), have Q1 = Q2 due to their conformal mapping origin, so the above simplifies quite a bit 2f = (1/g11(12)) (∂21 + ∂22)f where the g11 functions are listed off on page 52-55, and in general one sees that g11(12) does not factor in to a form a(1)b(2). For the Laplace equation, we would then have (1/g11(x,y)) [∂x2+∂y2] u(x,y) = 0 => [∂x2+∂y2] u(x,y) = 0 and then we are back into the mold of this document (starting with Section 1 above). So we know we can still find separation in the sense u(x,y) = X(x)Y(y). Helmholtz is a different story: (1/g11(x,y)) [∂x2+∂y2] + κ2 ) X(x)Y(y) = 0 (1/g11(x,y)) [∂x2 X(x)Y(y)+∂y2 X(x)Y(y)] + κ2 X(x)Y(y) ) = 0 (1/g11(x,y)) [(∂x2 X(x))Y(y)+ X(x) (∂y2 Y(y))] + κ2 X(x)Y(y) ) = 0 [(∂x2 X(x))/X(x)+ (∂y2 Y(y)/Y(y)] + κ2 g11(x,y) ) = 0 How can this possibly separate? Well the trick is that g11(x,y) must have the form g11(x,y) = a(x) + b(y) Looking through the tables p 52-55 we see this is true only for: (1) E1 special polar coordinates [ a(x) only ]; (2) P2 which is 2D parabolic coordinates; (3) H1 (Fig 214 p 69) elliptic coordinates. This agrees with the conclusions of my N=2 Stackel paper section. Looking at one of these three cases, for elliptic coordinates, if one writes (as on page 53) x = a cosh u cosv g11= (acoshu)2 – (a cosv)2 = a(x) + b(y) = Q12 = Q22 y = a sinh u sinv So if the separation shown above into a(x) + b(y) then we DO get EV problems. [(∂x2 X(x))/X(x)+ (∂y2 Y(y)/Y(y)] + κ2a(x) + κ2b(y) ) = 0 (∂x2 X(x))/X(x) + κ2a(x) = - (∂y2 Y(y)/Y(y)] - κ2b(y) = ν ∂x2 X(x) + κ2a(x)X(x) - ν X(x) = 0 ∂y2 Y(y) + κ2b(y)Y(y) + ν Y(y) = 0 Lx X(x) = ν X(x) Lx = ∂x2 + κ2a(x) Ly Y(y)= - ν Y(a) Ly = ∂y2 + κ2b(y) (b) 3D systems with solutions not dependent one of the variables If the 3D system was formed by extruding or rotating a 2D system, and if it is that special 3rd variable that the solution does not depend on, then I think there is nothing new to talk about. For example, M&S talk about elliptic-cylinder coordinates on page 20, and z is the ignored variable, then this is the same as elliptical coordinates under section (a) above. It happens that for these a(x) and b(y) functions we then have Lx X(x) = ν X(x) Lx = ∂x2 + κ2a2cosh2x Ly Y(y)= - ν Y(a) Ly = ∂y2 – κ2a2 cos2y and you see these equations on page 20. The eigenfunctions happen to be some horrible cem class functions which are Mathieu functions about which I known nothing whatsoever. Comment: When you include the z coordinate dependence in this last case, you get the tangle of two separation constants for the η and ψ equations as shown page 19. If you fix one separation constant α2, you might think of it as an EV problem for the other separation constant α1. But then your spectrum will be some α1 = f(α2). For each α2, you might get a spectrum {α1}, but all the points will "move" as you change α2. This means the spectrum is really a set of trajectories in α1 space, α1i = fi(α2). So this is they very far from my doc here. But still you can consider one of the separated tangled equations with its BC on level surface, and the BC will then give you some relation between α1 and α2. I think the Lamé functions case is typical of this separation constant tangle. I cannot remember enough to make sense of it right now. There are two quantum numbers, but what are the BC's that make them quantize? I quote my Hobson notes: "We are made aware that the En,p(μ)En,p(ν) situation is a 2D Sturm Liouville problem which quantizes the n and p parameters all at once, and cannot be decomposed into a product of 1D systems. " I think the {n,p} eigenvalues are both determined by finiteness requirements, the way the n in Pn(cosθ) is quantized to integer. Comment: None of the 2D systems considered in this paper is separable with R ≠ constant, that is to say, they are all "simple separable". 10. A review of Stak examples in light of this doc Problems "in more than one dimension" is Section 7.9 starting page 253, but it is not really until Section 7.12 on Helmholtz that things start moving Page 268 (7.160) Lu=0. Here we have our classic polar separated equations for the first time I think at least in this chapter. On page 269 he considers Lu=δ(r-r0)δ(φ) for the point charge, an example of our Section 6 above. Stak uses φ expo eigenfunctions in his "partial EF analysis" and obtains the 1D Green's in variable r which he then solves. This is done with no finite BC's,. so you end up with a fancy way to write the E2 free-space Helmholtz solution, and this then leads to Bessel function sum rules. Now in his analysis he applies the φ projector to both sides of the PDE and I kept wondering: how do you know this is going to separate? How do you know you won't end up with some integral on the LHS that you cannot get rid of? Well, I show in Section 8 (a) above that, since this PDE has the simple form considered in this doc, you really do get an ODE for the coefficient function uλx(y), so this shows that things must separate in this sense, and there are no "integrals" in this ODE. Then the Green's function is just a special case of equation (*) at the end of Section 8 (a). Alternatively, we know it separates because polar coordinates separate according to Stackel theory or just according to the fact that the classical cylindrical system separates for Helmholtz. Page 272, 2D Green's function inside a wedge. We already did this earlier with Green's 7.173 as the partial EF solution where sines are the eigenfunctions leaving a 1D Green's in r. On page 273, he instead does the partial EF solution using r as the EF's and this is Kantorovich with the Kiγ functions. So this is like switching from x EF's to y EF's as I did above in Section 8 (c). Again, we know ahead of time that everything is going to separate "with no integrals on the LHS side". Page 276: 3D Helmholtz with ring source in cylindrical coords, no φ dependence, so only ρ and z present. We know that even when all three dependencies are present, things here will be simple-separable, so clearly that will still be true when you are dealing only with ρ and z. Here he first does the problem using the r EF's J0(γr) (Hankel transform) with coefficient functions UH(z), and he obtains a 1D Greens ODE for UH(z). Again, no integrals on the LHS. We know it has to work because we know things separate. Just as I did this in Section 8 (a) in "the general case". Then on page 278 he does it all over again using z EF's (Fourier Transform) Page 281: We are back for more! This is the point source near grounded half-wire in 2D. He expands on the sin φ type EF's and gets a 1D Green's problem in r and solves it and replicates an earlier result. Page 290 Exercises: 3D point charge on z axis on φ dependence, spherical coordinates. So here our two coordinates of interest are r and θ. But again, we know Helmholtz separates in r,θ,φ so it surely separates in this r,θ only problem and we get another sum rule this time involving Legendre functions. This same stuff appears in more of these exercises, they are all special cases of my assumed form above, and of course they are all Helmholtz in known-separable systems! ________________________________________________________________________________ Appendix A: stuff moved out of the main text flow above, where marked. Comment added 12.9.11. If our PDE really was the Helmholtz equation, and if the variables really were some orthogonal curvilinears, then I would just USE the Stackel theory, find the matrix Φ, and know the ODE's for the two coordinates. I would not use this document at all. So I guess therefore I don't now see the point of the following digression. I am therefore right now demoting this section to an Appendix A. New Section Added 8.30.11. How does the above relate to 2D Stackel separation? We know in that case that if we are starting with the Helmholtz equation, we get this, where κ12 = K12, LnXn = (1/fn)∂n[fn(∂nXn)] + [ K12Φn1(n) + k22Φn2(n)]Xn = 0 LnXn = (1/fn)∂n[fn(∂nXn)] + [ K12Φn1(n) + k22Φn2(n)]Xn = 0 or LxX = (1/fx)∂x[fx(∂xX)] + [ K12Φx1(x) + k22Φx2(x)]X = 0 LyY = (1/fy)∂y[fy(∂yX)] + [ K12Φy1(y) + k22Φy2(y)]Y = 0 which we can compare to our forms above, for which I will now use script L Lx X(x) = ν sx(x) X(x) Ly Y(x) = - ν sy(y) Y(y) so rewrite the forms above as (1/fx)∂x[fx(∂xX)] + [ K12Φx1(x) ]X = - k22Φx2(x)]X (1/fy)∂y[fy(∂yX)] + [ K12Φy1(y) ]Y = - k22Φy2(y) Y {(1/fx)∂x[fx(∂x)] + K12Φx1(x) } X = - k22Φx2(x)]X {(1/fy)∂y[fy(∂y)] + K12Φy1(y) }Y = - k22Φy2(y) Y We can then make the connection by saying Lx = {(1/fx)∂x[fx(∂x)] + K12Φx1(x) } ν = -k22 Φx2(x) = +s(x) Ly = {(1/fy)∂y[fy(∂y)] + K12Φy1(y) } ν = -k22 Φy2(y) = -s(y) Now we would like to somehow "add these" to get some total L object. Stackel (3.10) says (1/ψ)[Lψ] = Σn(1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + k12/Q + K12 = 0 (3.10) and for simple-separation we set k12 = 0, so we have (1/ψ)[Lψ] = Σn(1/[hn2Xn]) (1/fn)∂n[fn(∂nXn)] + K12 = (1/[hx2X])(1/fx) ∂x[fx(∂xX)] + (1/[hy2Y])(1/fy) ∂y[fy(∂yY)] + K12 = (1/[hx2X]){ Lx – K12Φx1(x)}X + (1/[hy2Y]) { Ly – K12Φy1(y)}Y + K12 Now multiply each term by ψ = XY to get [Lψ] = (1/[hx2]){ Lx – K12Φx1(x)}XY + (1/[hy2]) { Ly – K12Φy1(y)}XY + K12 XY [Lψ] = (1/[hx2]){ Lx – K12Φx1(x)}ψ + (1/[hy2]) { Ly – K12Φy1(y)}ψ + K12 ψ Finally we can identify L = (1/[hx2]){ Lx – K12Φx1(x)} + (1/[hy2]) { Ly – K12Φy1(y)} + K12 L = (1/hx2) Lx + (1/hy2) Ly – { (1/hx2) K12Φx1(x) + (1/hy2) K12Φy1(y)} + K12 L = (1/hx2) Lx + (1/hy2) Ly – { (1/hx2) K12Φx1(x) + (1/hy2) K12Φy1(y) – K12 } L = (1/hx2) Lx + (1/hy2) Ly – K12 { (1/hx2)Φx1(x) + (1/hy2)Φy1(y) - 1 } This is the first time I have ever written L as a sum of Lx and Ly . The surprise is that there are "extra terms" which don't fit my starting mold! I should write this down at once for polar coordinates. Meanwhile, we know that the full Helmholtz equation has this form H-1{ ∂1[(H/h12)(∂1ψ)] + ∂2[(H/h22)(∂2ψ)]}ψ + K12ψ = 0 H ≡ h1h2 H-1{ ∂x[(H/hx2)(∂xψ)] + ∂y[(H/hy2)(∂yψ)]}ψ + K12ψ = 0 H ≡ h1h2 Now we have been assuming simple separation in our ψ = XY above, so set R = Q = 1, and now also assume (3.5) which says (H/[hx2]) = fx(x)gx(y) (H/[hy2]) = fy(y)gy(x) Then our Helmholtz above reads H-1{ ∂x[fx(x)gx(y) (∂xψ)] + ∂y[fy(y)gy(x) (∂yψ)]}ψ + K12ψ = 0 or H-1{ gx(y)∂x[fx(x) (∂xψ)] + gy(x) ∂y[fy(y) (∂yψ)]}ψ + K12ψ = 0 _______________________________________________________________________________