Find L for Lr(x)=d(x) for Love case

Is there an Lx for the Love Equation? PhL 1.10.11 Overview (1/2 page, written 1.12.11) 1 1. Where this kernel appears. 1 2. The related Fourier Transform and Fourier Cosine Transform. 2 3. Proof using the Fourier Transform. 3 4. Proof using the Cosine Transform. 5 5. Solution to Lxg(x) = δ(x) for C∞ functions g(x) 7 6. Is the Love kernel Hilbert-Schmidt? 10 _________________________________________________________________________________ Overview (1/2 page, written 1.12.11) I show that, for the Love integral equation's kernel, r(x) = (d/π) / [ d2 +x2 ] there exists no differential operator Lx with constant coefficients (even if we allow infinite order) such that Lx r(x) = δ(x). I show this two different ways and arrive at the same conclusion twice. It follows that there is no Lx such that Lx r(x-y) = δ(x-y) and thus it follows that the Love integral equation cannot be represented as a differential equation of the form Lx[f(x) - 1] = ∓ !Syntax Error, Idt Lx r(x-t) f(t) = ∓ !Syntax Error, Idt Lx δ(x-t) f(t) = ∓ f(x) θ(x>-1)θ(x<+1) where Lx has the form just stated. _________________________________________________________________________________ 1. Where this kernel appears. Sneddon gives the following form for the Love integral equation, f(x) - 1 = ∓!Syntax Error, Idt r(x-t) f(t) r(x) = (d/π) / [ d2 +x2], where the two parallel disks of r = 1 are located at z = 0 (with V = V0) and z = d (with V = ±V0), so it is the lower sign which corresponds to the capacitor problem. 2. The related Fourier Transform and Fourier Cosine Transform. We can look up this Fourier Transform (d2+x2)-1 = !Syntax Error, Idk e-ikx F(k) F(k) = (1/2π) !Syntax Error, Idx e+ikx (d2+x2)-1 = e-d|k| /2d where in the second line I put |k| to allow that k might be negative (it is in half the k integral! ) Let's consider this transform a little more carefully. We first write F(k) = (1/2π) !Syntax Error, Idx e+ikx (d2+x2)-1 = (1/2π) !Syntax Error, Idx[cos(kx)+isin(kx)] (d2+x2)-1 The sin(kx) term is odd in x, so contributes nothing to the integral, so we have F(k) = (1/2π) !Syntax Error, Idx cos(kx) (d2+x2)-1 There are no poles or issues with the integral, so F(k) is real, and it is even in k. Furthermore, the integrand is even in x, so we can fold it to get F(k) = (1/π) !Syntax Error, Idx cos(kx) (d2+x2)-1 and we can regard this if we like as a Fourier Cosine projection. Now we turn to the other direction: (d2+x2)-1 = !Syntax Error, Idk e-ikx F(k) = !Syntax Error, Idk [cos(kx)-isin(kx)] F(k) Since F(k) is even, the sine term gives nothing and we get (d2+x2)-1 = !Syntax Error, Idk cos(kx)F(k) = 2 !Syntax Error, Idk cos(kx)F(k) Since the integrand is even in k, we can write this as shown on the right above. So we can summarize our transform if we like in this manner, which is the famous Fourier Cosine Integral Transform valid for "even" functions f(x), [ even means it would have only even powers if expanded in a power series in x ] (d2+x2)-1 = 2 !Syntax Error, Idk cos(kx)F(k) F(k) = (1/π) !Syntax Error, Idx cos(kx) (d2+x2)-1 = e-d|k| /2d = e-dk /2d Since only k > 0 is involved in this version of the transform, we can remove the || in the last step 2nd line. Since r(x) = (d/π) / [ d2 +x2] we could express the above transform as r(x) = 2 (d/π)!Syntax Error, Idk cos(kx)F(k) = (2d/π)!Syntax Error, Idk cos(kx)F(k) F(k) = (1/π)(π/d) !Syntax Error, Idx cos(kx) r(x) = (1/d) !Syntax Error, Idx cos(kx) r(x) = e-dk /2d On the other hand, we can still use the full Fourier Integral Transform, which above said (d/π) (d2+x2)-1 = (d/π)!Syntax Error, Idk e-ikx F(k) F(k) = (1/2π)(π/d) !Syntax Error, Idx e+ikx (d/π) (d2+x2)-1 = e-d|k| /2d which again we write as r(x) = (d/π)!Syntax Error, Idk e-ikx F(k) F(k) = (1/2π)(π/d) !Syntax Error, Idx e+ikx r(x) = e-d|k| /2d 3. Proof using the Fourier Transform. Now let us consider this problem which we want to solve for Lx Lx r(x) = δ(x) We shall now expand both sides using the full Fourier Integral Transform: Lx (d/π)!Syntax Error, Idk e-ikx F(k) = (1/2π) !Syntax Error, Idk e-ikx We then move Lx into the integral with the usual caveat to get (d/π)!Syntax Error, Idk [ Lx e-ikx ] F(k) = (1/2π) !Syntax Error, Idk e-ikx which we rewrite inserting our known result for F(k) (d/π)!Syntax Error, Idk [ Lx e-ikx ] e-d|k| /2d = (1/2π) !Syntax Error, Idk e-ikx Now let's make the ansatz that we can represent Lx in this manner Lx = Σn=0∞ an ∂xn an = constants! We then get (d/π)!Syntax Error, Idk [Σn=0∞ an ∂xn e-ikx ] e-d|k| /2d = (1/2π) !Syntax Error, Idk e-ikx (d/π)!Syntax Error, Idk [Σn=0∞ an(-ik) n e-ikx ] e-d|k| /2d = (1/2π) !Syntax Error, Idk e-ikx (2d)!Syntax Error, Idk e-ikx { [Σn=0∞ an(-ik) n ] e-d|k| /2d} = !Syntax Error, Idk e-ikx !Syntax Error, Idk e-ikx { [Σn=0∞ an(-ik) n ] e-d|k|} = !Syntax Error, Idk e-ikx We now appeal to the completeness of the basis set e-ikx and conclude that [Σn=0∞ an(-ik) n ] e-d|k| = 1 [Σn=0∞ an(-ik) n ] = e+d|k| where we are careful to keep the |k| since k can be either sign in the Fourier Integral Transform. We now want to solve this equation for the an coefficients. We can do this, expanding e+d|k|, Σn=0∞ an(-ik) n = Σn=0∞ dn |k|n / n! Now we can write |k| = sign(k) k to get Σn=0∞ an(-i)n kn = Σn=0∞ dn (sign(k))n kn / n! We write this as Σn=0∞ bn kn = 0 bn = [an(-i)n - dn (sign(k))n/n! ] This must be valid for all k in (-∞,∞). But even if it were valid only on some finite interval, we would still conclude that bn must be 0, see Power Series doc on this subject. Therefore we must have an(-i)n = (sign(k))n dn/ n! But this says that an is a function k, but we assumed that an were just constants like 2.5, and cannot be functions of anything other than n. We cannot just consider k > 0 because our Fourier integrals shown above all involved both signs of k. The conclusion is that no an exists and that therefore our ansatz Lx = Σn=0∞ an ∂xn has failed! Note: When I did this analysis earlier, I omitted the |k| absolute values and got an = (id)n/n! and this in turn implied that Lx = Σn=0∞ an ∂xn = Σn=0∞ (id)n/n! ∂xn = exp(id∂x). But that was incorrect! 4. Proof using the Cosine Transform. Let's repeat the above section using Cosine Transforms instead and see if we can be more successful. Maybe if we only deal with k > 0 things will be somehow better: Now let us consider this problem which we want to solve for Lx Lx r(x) = δ(x) We shall now expand both sides using the full Cosine Integral Transform. Recall from Section x that we had r(x) = 2 (d/π)!Syntax Error, Idk cos(kx)F(k) = (2d/π)!Syntax Error, Idk cos(kx)F(k) F(k) = (1/π)(π/d) !Syntax Error, Idx cos(kx) r(x) = (1/d) !Syntax Error, Idx cos(kx) r(x) = e-d|k| /2d so we then get LHS = Lx [(2d/π)!Syntax Error, Idk cos(kx)F(k)] Now what is δ(x) expressed in Fourier Cosine language? Well start with δ(x) = (1/2π) !Syntax Error, Idk e-ikx = (1/2π) !Syntax Error, Idk [cos(kx)-isin(kx)] I argue now that δ(x) is even in x, so the sin(kx) contribution must vanish. Another argument for this same conclusion is that we think of δ(x) as being real. We then have δ(x) = (1/2π) !Syntax Error, Idk cos(kx) = (1/π) !Syntax Error, Idk cos(kx) (*) Now just out of curiosity, suppose we start with our generic Fourier Cosine Transform f(x) = 2 !Syntax Error, Idk cos(kx)F(k) F(k) = (1/π) !Syntax Error, Idx cos(kx) f(x) which I have just carefully shown is consistent with my Summary Case 2 in transforms.doc. If we were to insert f(x) = δ(x) into the second integral here, what do we get? F(k) = (1/π) !Syntax Error, Idx cos(kx) δ(x) = ? Since our Cosine Transform deals only with x ≥ 0, we pick up "half the delta function" here to get F(k) = (1/π) !Syntax Error, Idx cos(kx) δ(x) = (1/2π) Then we get δ(x) = 2 !Syntax Error, Idk cos(kx)F(k) = 2 !Syntax Error, Idk cos(kx) (1/2π) = (1/π) !Syntax Error, Idk cos(kx) in agreement with (*) above. See "disk green attempt2" section 4a for more detail on this idea of picking up half the delta function. Our equation with both sides expanded as above is now Lx (2d/π)!Syntax Error, Idk cos(kx)F(k) = (1/π) !Syntax Error, Idk cos(kx) x ≥ 0 We now move Lx inside the integral to get (2d)!Syntax Error, Idk [ Lx cos(kx)] F(k) = !Syntax Error, Idk cos(kx) x ≥ 0 I keep saying x ≥ 0 because our Cosine Transform is only defined on x ≥ 0. Now let's assume the following form for Lx Lx = Σn=0,2∞ an ∂xn Here I assume only even powers because odd powers will generate sin(kx) terms which are odd in x, but the RHS is even in x. ( by saying g(x) is "even" for x ≥ 0 I mean that if expanded in an xn power series, only even powers would appear. ) So we get, [(2d)!Syntax Error, Idk [Σn=0,2∞ an ∂xn cos(kx)] F(k) = !Syntax Error, Idk cos(kx) all x Now for even n we know that ∂x2 cos(kx) = -k2 cos(kx) ∂x4 cos(kx) = k4 cos(kx) .... ∂xn cos(kx) = (-1)n/2 kn cos(kx) So we then have (2d)!Syntax Error, Idk [Σn=0,2∞ an (-1)n/2 kn cos(kx)] F(k) = !Syntax Error, Idk cos(kx) all x Now let's install our F(k). Since we are only thinking k ≥ 0 now we use F(k) = e-dk /2d to get !Syntax Error, Idk [Σn=0,2∞ an (-1)n/2 kn cos(kx)]e-dk = !Syntax Error, Idk cos(kx) all x !Syntax Error, Idk cos(kx) [Σn=0,2∞ an (-1)n/2 kn]e-dk = !Syntax Error, Idk cos(kx) all x We now appeal to cos(kx) being complete for even functions of x to conclude that Σn=0,2∞ an (-1)n/2 kn = edk = cosh(dk) + sinh(dk) or Σn=0,2∞ an (-1)n/2 kn - cosh(dk) = sinh(dk) No matter what an we choose, the LHS has only even powers of k and the RHS has only odd powers of k, so these cannot be equal! Therefore, no an exists for the assumed form of our Lx. This is the same conclusion we reached in the previous section! 5. Solution to Lxg(x) = δ(x) for C∞ functions g(x) In order to have Lxg(x) = δ(x), it just seems to me that something dramatic needs to happen with f(x) at the point x = 0. For the string problem we know that the derivative takes a jump at x = 0. The Green's Function f(x) is a sawtooth that peaks at x = 0 (in this case), and so it has zero curvature away from x = 0 and something dramatic happens at x = 0. Similarly for the case of 1/r in 2D or 3D, this function does something dramatic at r = 0. Now suppose g(x) is a boring function at x = 0, such as our Love kernel g(x) = 1/(x2+a2) a = fixed If you plot this close to x = 0, you get a nice smooth parabola as shown on the right below, All derivatives exist and are finite at x = 0, so I suppose this thing is C∞ . So now can some Lx find something dramatic at x = 0 so Lxg(x) = δ(x) ? If you try Lx with some coefficient functions an(x), you get this ∂ng(x) = Cn pn(x) / (x2+ a2)n+1 pn(x) = poly of degree n Cn = a number So then Lxg(x) = Σn=0∞ an(x) ∂ng(x) = Σn=0∞ an(x) Cn pn(x) / (x2+ a2)n+1 If the an(x) are polynomials of some degree, then each term is some poly over (x2+ a2)n+1 which is analytic at x = 0. How would you make this sum vanish near x = 0 but not at x = 0 ? In order to have any kind of kink at x = 0, it seems to me you need to use some complete basis functions. We have δ(x-x') = Σn=0∞ φn(x)* φn(x') δ(x) = Σn=0∞ φn(x)* φn(0) So we need to arrange for this to happen Σn=0∞ an(x) Cn pn(x) / (x2+ a2)n+1 = Σn=0∞ φn(x)* φn(0) for some set of basis functions. So that says an(x) Cn pn(x) / (x2+ a2)n+1 = φn(x)* φn(0) an(x) = φn(x)* φn(0) (x2+ a2)n+1/ [ Cnpn(x) ] For example, we could select normalized Legendre Polynomials, Σn=|m|∞(1/Knm) Pnm(z') Pnm(z) = δ(z'-z) Knm = (n+1/2)-1 f(n,m) Σn=0∞(1/Kn) Pn(z')Pn(z) = δ(z'-z) Kn = (n+1/2)-1 φn(x) = Pn(z)/ = Pn(z) Then a solution would be an(x) = (n+1/2) Pn(x)Pn(0) φn(0) (x2+ a2)n+1/ [ Cnpn(x) ] So here is my candidate Lx Lx = Σn=0∞ (n+1/2) Pn(x)Pn(0) φn(0) (x2+ a2)n+1/ [ Cnpn(x) ] ∂xn So maybe I can generalize this as follows: δ(x) = Σn=0∞ φn(x)* φn(0) Lx = Σn=0∞ φn(x)* φn(0) [ ∂xng(x) ]-1 ∂xn Then Lx g(x) = Σn=0∞ φn(x)* φn(0) [ ∂xng(x) ]-1 ∂xn g(x) = Σn=0∞ φn(x)* φn(0) = δ(x) so this seems to give an answer for any g(x) you want as long as it is C∞ so all those derivatives exist. They have to all exist so you can write Lx as shown with no divide by zeros. Theorem(?): If g(x) is C∞ in an interval of interest surrounding the origin x = 0, then a solution to the posed question: Lx g(x) = δ(x) is this: Lx = Σn=0∞ φn(x)* φn(0) [ ∂xng(x) ]-1 ∂xn where φn(x) is any orthonormalized complete set of basis functions on that interval. This theorem is applicable to the Love kernel, but of course it is a pretty ugly Lx and of infinite order. 6. Is the Love kernel Hilbert-Schmidt? Here is our Love integral equation f(x) - 1 = ∓!Syntax Error, Idt r(x-t) f(t) r(x) = (d/π) / [ d2 +x2], So the question is whether or not the following integral is finite: !Syntax Error, Idx!Syntax Error, Idy | (d/π) / [ d2 +(x-y)2] |2 < ∞ ? or I ≡ !Syntax Error, Idy!Syntax Error, Idx 1 / [ d2 +(x-y)2] 2 < ∞ ? Define new variable z = x-y to replace x and dx = dz so have I = !Syntax Error, Idy!Syntax Error, Idz 1 / (d2 +z2)2 Things are pretty messy from this point on, so here is Maple on the subject: Therefore we find that !Syntax Error, Idx!Syntax Error, Idy | (d/π) / [ d2 +(x-y)2] |2 = (d/π) I = (d/π) *(2/d3) tan-1(2/d) = (2/πd2)tan-1(2/d) and as long as d > 0, this is finite. Therefore the Love kernel is indeed a Hilbert-Schmidt kernel.