complex ODEs

Complex ODEs PhL 2.24.10 This is an oddball topic, just another that has me totally confused. On the one hand, one thinks of an ODE as being in a real variable x on some interval (a,b) of the real axis. There is nothing "complex" about it. On the other hand, there are regular singular points, branch points, cuts, Riemann sheets, and all the apparatus of the complex z plane. Usually the singular points are on the real axis, however. But to wrap around a branch point, you have to have the complex plane. Stakgold's ODE discussion has no complex aspects to it as I recall. In Vol I we do single real variable, then in Vol II we do multiple real variables. His book is not about "complex functions" really. Ahlfors has an ODE chapter at the end of his book which I have not read yet. Hypothesis #1. You start with any ODE in variable x. Lxu(x) = 0. Forget BC's for the moment. There are two independent solutions, call them u1(x) and u2(x). We then sort of "continue" everything to the complex z plane and we then magically have Lzu(z) = 0 with solutions u1(z) and u2(z) . Away from singularities, these ui(z) are magically analytic functions. One way to accept that is that the ui(x) had power series representations with real coefficients, and these then naturally become analytic functions when you replace x by z. In the complex ODE Lzu(z) = 0 the derivatives like ∂z are now really complex function derivatives which are "the same in all directions". For me, I think of f(z) as the complex function, and I think of f '(z) as that complex derivative. So according to this hypothesis, all ODE's can be thought of as "complex ODE's", an example being the Legendre ODE. When we do a problem like the oblate hyperboloid with boundary conditions in 3D space, each of the three variables is treated as a real variable, so this is back in the Stakgold or Smythe bailiwick. Example 1: –∂x2f(x) = k2f(x) // changing from u to f = u + iv. In the real ODE world, we have solutions e±ikx for fi(x). ( Perhaps I should have used sin(kx), cos(kx) here to be fully in the real world and not have fi(x) be complex. But this does show the intermediate world where fi(x) is a complex function of a real variable. ) Now let's jump to complex: –∂z2f(z) = k2f(z) f(z) = e±ikz = e±ik(x+iy) = e±ikx e∓iky How then do we compute a derivative ∂zf(z). You do it "in any direction" and Ahlfors on page 25 shows perhaps the simplest way to do it: f(z) = u(z) + iv(z) f(x,y) = u(x,y) + iv(x,y) ∂zf = ∂x u(x,y) + i ∂x v(x,y) This is "doing it in the x direction". So let's try it: f(z) = e±ikx e∓iky = cos(ikx) e∓iky ± i sin(ikx) e∓iky = u + iv ∂x u(x,y) = -iksin(ikx) e∓iky ∂x v(x,y) = ±ikcos(ikx) e∓iky ∂zf = ∂x u(x,y) + i ∂x v(x,y) = -iksin(ikx) e∓iky + i { ±ikcos(ikx) e∓iky } = ∓ kcos(ikx) e∓iky -iksin(ikx) e∓iky = ∓ k e∓iky[cos(ikx) ± i sin(ikx)] = ∓ k e∓iky e±ikx = ∓ k e±ikz Sort of the long way around, but the point is that you really can do it using the x and y notation only. Of course we would get the last result instantly just applying ∂z to f(z) = e±ikz . So we have then shown, using x and y notation only, that this complex function f(x,y) really does solve –∂z2f(z) = k2f(z) . Here is another way to view the above. Let f(x) = Σnanxn+r be the Frobenius solutions about point x = 0 which we assume is non-singular. Then an are just right so that –∂x2f(x) = k2f(x). The an are real. Then define the new complex function f(z) = Σnanzn+r . We know the rule for powers, and it is completely obvious that this thing must be a solution of –∂z2f(z) = k2f(z) . What about BC's ? Suppose we solve our Lxu(x) = 0 subject to some BC's. We just "continue" this equation to complex z in the same magical fashion away from singular points of the ODE. Then a BC has the form BC[ u(z1)] = 0 where it happens that z1 = real. Same idea for Lxu(x) = f(x). The notion of "continue" is just this: to the extent something is a power series with real coefficients, that is what you continue by replacing x by z. Example 2: Consider the string problem –∂x2f(x) = k2f(x) where f(0) = f(b) = 0. The solution we know is f(x) = A sin(kx). The complex solution is then f(z) = Asin(kz). plot3d(Re(sin(x+I*y)), x = -8..8, y = -3..3); Im over here ↓ The physical string lies along the center line of the two pictures. On the left we get sin(kx) along that line, and on the right we get 0 along that line. So the surfaces are what you get by "continuing" the string off the real axis. We show here THE solution to –∂z2f(z) = k2f(z) with f(0) = f(π/k) = 0. In all directions except the real axis, we get exponential growth. Question #1: Suppose f(z) is a solution to Lzf(z) = 0. Is f(z*) also a solution? Do we have Lzf(z*) = 0 ? Answer #1: If we assume that our solution f(z) is analytic in some mirrored region about the real axis (which might not be true if there are somehow complex branch points that make this not so) then when we say that Lzf(z) = 0 we mean this is true for all z in complex domain D. Then z* is also in that domain so definitely we have Lzf(z*) = 0. Answer #2: If we start with Lzf(z) = 0 and if we assume that Lz is "real", meaning that written as Lx the various functions inside L are real functions of real x, then we apply * to find that Lz* f(z*) = 0. To get this form, we just star all occurrences of z in Lz, and we assume that power series form for f(z) with real coefficients. So this shows that Lz* f(z*) = 0 , NOT that Lz f(z*) = 0 . So this answer is inconclusive and tends toward the negative. Example 3: –∂z2f(z) = k2f(z) We have our complex solutions of a complex variable: f(z) = e±ikz = e±ikx e∓iky. So we want to test out this function: f(z*) = e±ikz* = e±ikx e±iky. // sign change on y We can manually compute the derivative as we did above. Imagine writing all the steps as above, but then we edit everything and replace y → -y. Our derivatives were all x derivatives, so nothing changes except for this replacement. Our last step would then be: ∂zf (z*) == ∓ k e∓ik(-y) e±ikx = ∓ k e∓ik(-y) e±ikx = ∓ k e±ikx e±iky = ∓ k f(z*) It seems to work, and we seem then to have shown that Lz f(z*) = 0 and Answer #1 seems right. Again, our solution is for all z, so we just examine the solution at the particular point z = z*. This is a bit tricky. We always have f(z) being a function of z, not z*. So in some sense ∂z* f(z) = 0. Ahlfors talks this way sometimes, imagining functions F(z,z*) where these are two independent variables, replacing f(x,y). In this sense, we would say Lz f(z*) = [–∂z2 - k2] f(z*) = - k2 f(z*) ≠ 0. Similarly, Lz* g(z) = 0. I guess the key is that, when you declare a function, you have to clearly state what independent variables it is a function of: Is it x,y or is it z,z*. Example 4: Pnm(z) We know that Pnm(2 + i3) solves the Legendre ODE. That is to say, [Lz Pnm(z)]|2+i3 = 0. We loosely write this as Lz Pnm(2 + i3)] = 0. This is true for any z in the complex plane. So Lz Pnm(2 - i3)] = 0 as well. Theorem: In the sense of z = (x,y) as the variables, [Lzf(z)]z=z1 = 0 => [Lzf(z)]z=z1* = 0 or f(z) is a solution of Lzf(z) = 0 at point z = z1 => f(z) is a solution of Lzf(z) = 0 at point z = z1* or in very sloppy notation Lzf(z1) = 0 => Lzf(z1*) = 0 Question #2: Suppose [Lzf(z)] = 0 is an ODE and f(z) is a solution. Are either of the following also solutions of the ODE? (a) g(z) = A f(z) + B f(-z) (b) h(z) = A f(z) + B f(z*) Clarification: We have to make the distinction between a functional form and a function evaluation. Here is what we know: If [Lzf(z)] = 0, we say that f(z) is an ODE solution at any point z. In (a) or (b) above, if our meaning of the second term is "functional evaluation", then f(z) involves the point z, and f(-z) involves a completely different point -z. So the function g(z) is a linear combination of solutions to the ODE evaluated at different points in the z plane! So in this sense, we cannot say that we are linearly combining solutions at the same point, so the whole thing makes no sense. Therefore, when we write things like (a) and (b), we have to be talking about the "functional form" interpretation of the second terms. Question #3. Suppose we write f(z) = Σnanzn . How do we express our two different meanings for f(-z)? functional form: f(-z) = Σnfn(-z)n = Σnfn(-1)n zn = Σn Fn zn = F(z), some new function of z. function evaluation: f(-z) = Σnfn(-z)n , evaluation of f(z) at the point z = -z. Example 5. So we can ask about the validity of this equation: –∂z2F(z) = k2F(z) where F(z) ≡ f(-z) in the functional sense. For this particular ODE, it turns out that Lz = L-z since d/dz d/dz = d/d(-z) d/d(-z) so we can write: –∂z2F(z) = –∂[-z]2 F(z) = –∂[-z]2 f(-z) = –∂w2 f(w) w= -z, new complex variable = k2f(w) = k2f(-z) Therefore: –∂z2f(-z) = k2f(-z) so that Lz f(-z) = k2 f(-z) in functional sense, meaning really that LzF(z) = k2F(z). This would also be true if the ODE had a term z∂z since z d/dz = (-z)d/d(-z). But it would not be true if the ODE had a term 5 ∂z . So for this example, the linear combination g(z) = A f(z) + B F(z) would in fact also be a solution of the ODE at the point z. We are no longer talking about functions at two different points. Both f(z) and F(z) are evaluated at the point z. So this is an example of the (a) case above. What about the (b) case? Let's try to mimic the previous example. Example 6. So we can ask about the validity of this equation: –∂z2G(z) = k2G(z) where G(z) ≡ f(z*) in the functional sense. Let's paste and then edit stuff from above: functional form: f(z*) = Σnfn(z*)n ≠ Σn Gn zn = G(z), some new function of z. // Note the NOT equal. function evaluation: f(z*) = Σnfn(z*)n , evaluation of f(z) at the point z = z*. So things are now quite different. There exists no G(z) that we can even talk about! So we cannot really ask whether LzG(z) = 0 since there is no G(z). We can say all these things, however: Lzf(z) = 0 Lwf(w) = 0 Lz*f(z*) = 0 But we cannot claim that Lz = Lz* . For example, we do not have d/dz d/dz = d/dz* d/dz* . This is because we do not have (dz)2 = (dz*)2 . That is to say, (dz)2 = (dx + idy)2 ≠ (dx – idy)2 Therefore, if we write h(z) = A f(z) + B f(z*) this cannot be represented in a functional form sense as h(z) = A f(z) + B G(z) so we cannot even talk about this being a linear combination of two ODE solutions at the same point. All we can say is that it is the sum of solutions at two different points. Example 7: Pnm(z). The Legendre ODE has the property that Lz = L-z so we can then make these claims: F(z) ≡ Pnm(-z) is a solution of LzF(z) = 0 at z, the Legendre ODE g(z) = APnm(z) + Pnm(-z) is a solution of Lzg(z) = 0 at z, the Legendre ODE In contrast, we cannot say that F(z) ≡ Pnm(z*) is a solution of LzG(z) = 0 at z, the Legendre ODE h(z) = APnm(z) + Pnm(z*) is a solution of Lzh(z) = 0 at z, the Legendre ODE Combining solutions on different Riemann sheets. Let f(z) be an analytic function that has multiple sheets. Then f(z)n refers to the solution on the nth sheet, and perhaps n = 0 is the principle sheet. If there are three branch points we might write f(z)n1,n2,n3 to fully describe the function, but we just say f(z)n as a symbolic notation: The ODE Sheet Theorem: if Lzf(z)n1 = 0, then if Lzf(z)n2 = 0 for any sheet n2 . I am not quite sure how to prove this, but I know it is true. The ODE is a local operator object that acts on a tiny patch of the z plane somewhere. It does not know which "sheet" that patch lies on, so it is valid for any sheet. The sheet structure is a global property, not a local property. So that is my first argument. The second is that I know that when you start with some f(z)0 and cycle around a branch point to an adjacent sheet, the result is some linear combination of the two independent solutions of the ODE, and we write this as f(z)1. We generally don't have f(z)1 = f(z)0. For example, ≠ - to pick a very simple example. One more try: The ODE's Lzf(z)n1 = 0 and Lzf(z)n2 = 0 look exactly the same! We are just talking two different sheets. The solution to each equation is some linear combination of some set of two independent solutions that you select. Notice that our theorem is not telling us how to find f(z)n2 if we are given f(z)n1 , it is just saying that whatever f(z)n2 comes out to be, in will be an ODE solution. But I will go on to show below how to actually find f(z)n2 if we are given f(z)n1. The solution to Lzf(z)0 = 0 has a Frobenius form Σn an (z-a)n+r where a is a branch point and r is a fixed Frobenius exponent or index. If we wind once around this branch point we get f(z)0 = Σn an (z-a)n+r → f(z)1 = Σn an [(z-a)e2πi] n+r = Σn an e2πi(n+r) (z-a)n+r = Σn an e2πir (z-a)n+r = e2πir Σn(z-a)n+r = e2πir f(z)0 But suppose we started with some linear combination: F(z)0 = A Σn an (z-a)n+r1 + B Σn bn (z-a)n+r2 then when we go around the branch point once we get F(z)1 = A e2πir1 Σn an (z-a)n+r1 + B e2πir2 Σn bn (z-a)n+r2 = A e2πir1 Σn an (z-a)n+r1 + B [ e2πir2 - e2πir1 + e2πir1 ] Σn bn (z-a)n+r2 = e2πir1 [A Σn an (z-a)n+r1 + B Σn bn (z-a)n+r2 ] + B [ e2πir2 - e2πir1] Σn bn (z-a)n+r2 = e2πir1 F(z)0 + B [ e2πir2 - e2πir1] Σn bn (z-a)n+r2 So this is what I wanted! Here I show that when you wrap around a branch point, you get a multiple of the original function, plus a new term which is some linear combination of whatever two independent solutions you picked. If we happen to pick the exact Frobenius solutions as the two independent solutions, then the wrap causes us to pick up a multiple of the second solution. I don't know if this is a theorem with someone's name on it. I don't know where to find this result even stated, though I have read it perhaps in Messiah or elsewhere. So this pretty much firms up my ODE Sheet Theorem stated above. As you go to a nearby sheet, the result you get is still an ODE solution. This is what "analytic continuation" to nearby sheets means. Of course if the exponents are just right, we have to deal with the log case and all that, but I think the point is clearly made without getting involved in those details. When we wrap a log branch point, we pick up an additive term, but I know it will still be an ODE solution. For example, when we start with Q1(z) on its principle sheet, we pick up an additive term Az and we know z = P1(z) which is an ODE solution. In fact, recall these facts from ODE notes 1 doc, for the case of a branch point at z = 0, "(6) What happens if r1 = r2 + s with s = positive integer ? In this case the solutions are p 195 B: w1(z) = zr1 [ 1 + Σn=1∞ Anzn] w2(z) = zr2 [ - 1/s + Σn=1∞ hnzn ] + gs w1(z) ln(z) " So you see if z wraps, we pick up a multiple of w1(z) which is a solution. Corollary #1 : If if Lzf(z)n1 = 0, then Lz { A f(z)n1 + B f(z)n2 } = 0. This says that you can linear combine solutions at the same point on different sheets and what you get will still be an ODE solution. Corollary #2 : If if Lzf(x+iε) = 0, then Lz { A f(x+iε) + B f(x-iε) } = 0 Here we might imagine that f(x+iε) = f(x+iε)0 where z = x + iε is on Sheet 0. We are of course imagining that we are on a cut. So f(x-iε) means f(x+iε)1 . So we are just applying Corollary #2 to the particular location z = x + iε. What we really mean by the above is this: Lz {A f(z)0 }|z = x+iε + Lz {B f(z)1 }|z = x+iε = 0 or Lz [ A f(z)0 + B f(z)1 ] |z = x+iε = 0 Of course it will also be true in general that Lz [ A f(z)0 + B f(z)1 ] = 0 for all z on the complex z plane. Notice please that our Corollary does not say Lz { A f(z) + B f(z*) } = 0 for general z. We have already shown above how this makes no sense.