Stakgold Exercise 6_47

Stakgold Exercise 6.47 PhL 10.23.09 Here we are going to actually solve a real integral equation 6.190 with a log type kernel. Without peeking, my usual approach would be to diagonalize it with an appropriate transform. You need a set of basis functions that you know how to integrate against the log. You could differentiate to get rid of the log somehow. But I think Stak is going to give us an ad hoc method. He does a change of variables, and THEN he does a basis function expansion. There are lots of tricky jogs in his path that I would not have taken were I on my own. (1) I would have left sinβ in the integral, which messes things up. (2) taking sinβ into F(β), I would have done a sine transform on F(β). Both these wrong turns lead to death. (a) Derive 6.191 We start with: !Syntax Error, Idy f(y) ln |x-y| = g(x) First he wants to change variables this way x = cosα y = cosβ so we get y = -1 means β = π and y = +1 means β = 0 and also dy = -sinβ, so !Syntax Error, Idβ sinβ f(cosβ) ln | cosα - cosβ | = g(cosα) Now replace F(β) = f(cosβ) sinβ and G(α) = g(cosα), !Syntax Error, Idβ F(β) ln | cosα - cosβ | = G(α) // which is 6.191 (b) Derive 6.192 Let's start from the other end and see how 6.22 must look, Σn=1 (1/n) cos(nα)cos(nβ) = (1/2) Σn=1 (1/n) { cos(nα-nβ) + cos(nα+nβ) } = (1/2) Σn=1 (1/n) cos(nα-nβ) + (1/2) Σn=1 (1/n) cos(nα+nβ) Now set α' = 1 in 6.22 and we get, using 6.22 twice, = (1/2)(-1/2) ln[2-2cos(α-β)] + (1/2)(-1/2) ln[2-2cos(α+β)] = (-1/4) ln[2-2cos(α-β)] + (-1/4) ln[2-2cos(α+β)] Now set [2-2cos(α-β)] = 4 { [1-cos(α-β)]/2 } = 4 sin2[(α-β)/2] so we have ln[2-2cos(α-β)] = ln { 4 sin2[(α-β)/2] } = 2 ln2 + ln sin2[(α-β)/2] So we have now shown that Σn=1 (1/n) cos(nα)cos(nβ) = (-1/4) { 2 ln2 + ln sin2[(α+β)/2]} + (-1/4) { 2 ln2 + ln sin2[(α-β)/2]} = - ln2 + (-1/4) ln { sin2[(α+β)/2] sin2[(α-β)/2] } = - ln2 + (-1/2) ln | sin[(α+β)/2] sin[(α-β)/2] | = - ln2 + (-1/2) ln[ | cosα - cosβ |)/2 ] Schaum p 17 5.64 = - ln2 + (-1/2) ln | cosα - cosβ |) + (+1/2)ln2 = - (ln2)/2 + (-1/2) ln | cosα - cosβ |) Therefore we have ln | cosα - cosβ | = -2 Σn=1 (1/n) cos(nα)cos(nβ) - ln2 // which is 6.192 So this is going to be used to convert our kernel in 6.191 into a trig form. Notice the "separable" form of the kernel, although we have an infinite sum. Now go back to the above form Σn=1 (1/n) cos(nα)cos(nβ) = - (ln2)/2 + (-1/2) ln | cosα - cosβ |) (-1/π) Σn=1 (1/n) cos(nα)cos(nβ) = (+1/2π) [ln2 + ln | cosα - cosβ | ] ≡ m(α,β) and so we have also verified 6.188. (c) Derive 6.193. We are told to expand F and G on "cosine series". We have from above that F(β) = f(cosβ) sinβ 0,π G(β) = g(cosβ) 0,π Our angle range of interest is (0,π), so we are allowed to do "as we like" outside this range. Suppose then we consider the range (-π,π) but we force F to be even F(β) = f(cosβ) | sinβ | = f(cosβ) (-π,π) G(β) = g(cosβ) (-π,π) Now we can do a "cosine series" expansion on both these functions, to wit [ Schaum p 131 2L = 2π ] F(β) = Σm=1 fm cos(mβ) + f0/2 fm = (1/π)!Syntax Error, IF(β) cos(mβ) dβ = (2/π)!Syntax Error, IF(β) cos(mβ) dβ G(β) = Σm=1 gm cos(mβ) + g0/2 gm = (1/π)!Syntax Error, IG(β) cos(mβ) dβ = (2/π)!Syntax Error, IG(β) cos(mβ) dβ Now here is our equation of interest, where on the second line we use our previous identity: !Syntax Error, Idβ ln | cosα - cosβ| F(β) = G(α) !Syntax Error, Idβ {-2 Σn=1 (1/n) cos(nα)cos(nβ) - ln2} F(β) = G(α) We are hoping this thing will "diagonalize". Let's then apply (2/π)!Syntax Error, Idα cos(mα) to both sides: gm = (2/π)!Syntax Error, Idα cos(mα) !Syntax Error, Idβ {-2 Σn=1 (1/n) cos(nα)cos(nβ) - ln2} F(β) Now in the ln2 term we have all by itself !Syntax Error, Idα cos(mα) = (1/m)sin(mα)|π0 = 0 m ≠ 0 = π m =0 = π δm,0 so this term in gm has this value gm (the ln2 term) = (2/π) π δm,0 (-ln2) !Syntax Error, Idβ F(β) = (2/π) π δm,0 (-ln2) (π f0/2) = π δm,0 (-ln2) ( f0) = -π ln2 f0 δm,0 So this ln2 term will only affect the g0 equation. Now the other term in gm is (2/π)!Syntax Error, Idα cos(mα) !Syntax Error, Idβ {-2 Σn=1 (1/n) cos(nα)cos(nβ)} F(β) = (-4/π) Σn=1 (1/n) !Syntax Error, Idβ cos(nβ) F(β)!Syntax Error, Idα cos(mα) cos(nα) = (-4/π) Σn=1 (1/n) !Syntax Error, Idβ cos(nβ) F(β)( πδm,n/2) = 0 if m = 0, else: = (-2) (1/m) !Syntax Error, Idβ cos(nβ) F(β) = (-2/m) πfm/2 = (- πfm/m) We have now found that g0 = -π ln2 f0 gm= (- πfm/m) m = 1,2... which we can solve to get f0 = - (1/πln2) g0 fm = -(m/π)gm Now we can then write: F(β) = Σm=1 fm cos(mβ) + f0/2 = - Σm=1 cos(mβ) (m/π)gm - (1/2) (1/πln2) g0 = - Σm=1 cos(mβ) (m/π) (2/π)!Syntax Error, IG(α) cos(mα) dα - (1/2) (1/πln2) (2/π)!Syntax Error, IG(α) dα = - (1/π2ln2) !Syntax Error, IG(α) dα - (2/π2) Σm=1 m cos(mβ) !Syntax Error, IG(α) cos(mα) dα which at last is 6.193. (c) Derive 6.194. Start with B, derive C: F(β) = (-1/π2ln2) !Syntax Error, Idα G(α) - (2/π2)Σn=1 n cos(nβ) !Syntax Error, Idα G(α) cos(nα) Let's try the trick he used earlier to say -n cos(nβ) = ∂β2 cos(nβ)/n so the above becomes F(β) = (-1/π2ln2) !Syntax Error, Idα G(α) + (2/π2) ∂β2 Σn=1 (1/n) cos(nβ) !Syntax Error, Idα G(α) cos(nα) = (-1/π2ln2) !Syntax Error, Idα G(α) + (2/π2) ∂β2 !Syntax Error, Idα G(α) {Σn=1 (1/n) cos(nβ) cos(nα)} Now we can use our identity Σn=1 (1/n) cos(nα)cos(nβ) = - ln2 + (-1/2) ln | cosα - cosβ | to get F(β) == (-1/π2ln2) !Syntax Error, Idα G(α) + (2/π2) ∂β2 !Syntax Error, Idα G(α) {- ln2 + (-1/2) ln | cosα - cosβ | } Now the ln2 term gives nothing under ∂β2 so have now F(β) = (-1/π2ln2) !Syntax Error, Idα G(α) + (2/π2) (-1/2) ∂β2 !Syntax Error, Idα G(α) ln | cosα - cosβ | = (-1/π2ln2) !Syntax Error, Idα G(α) - (1/π2) ∂β2 !Syntax Error, Idα G(α) ln | cosα - cosβ | and this is his result C. (d) Derive 6.195. Suppose g(x) = 1. Then G(β) = 1 and we have ( we just did this integral above) gm = (2/π)!Syntax Error, I1 cos(mβ) dβ = (2/π ) π δm,0 = 2 δm,0 Then we have f0 = - (1/πln2) g0 = - (2/πln2) fm = -(m/π)gm = 0 and then F(β) = Σm=1 fm cos(mβ) + f0/2 = f0/2 = - (1/πln2) But from above we had F(β) = f(cosβ) | sinβ | = f(cosβ) (-π,π) Therefore f(cosβ) = - (1/πln2) f(x) = - (1/πln2) f(x) = - (1/πln2)/ which is 6.195 Comments on this problem: We know that we can diagonalize any equation of this form: a(g1) = ∫dg b(g1g-1) c(g) where dg is the invariant Haar measure of the group of interest. I think we need ∫dg = 1 but I forget why. Here is the "proof" of the diagonalization: aσ ≡ ∫dg Dσ(g)a(g) // the items on the left and D are all square matrices bσ ≡ ∫dg Dσ(g)b(g) cσ ≡ ∫dg Dσ(g)c(g) a(g1) = ∫dg b(g1g-1) c(g) aσ = ∫dg1 Dσ(g1)a(g1) = ∫dg1 Dσ(g1){ ∫dg b(g1g-1) c(g)} = ∫dg { ∫dg1 Dσ(g1) b(g1g-1) } c(g) = ∫dg { ∫dg1 Dσ(g1g-1) Dσ(g) b(g1g-1) } c(g) // D is a group representation = ∫dg { ∫d(g1g-1) ( Dσ(g1g-1) Dσ(g) b(g1g-1) } c(g) // rearrangement theorem = ∫dg { ∫dg2 ( Dσ(g2) Dσ(g) b(g2) } c(g) = ∫dg2 Dσ(g2) b(g2) ∫dg Dσ(g) c(g) = bσ cσ so we end up with this matrix diagonalization where σ is a rep label aσ = bσ cσ The D functions have to be properly normalized. For the abelian SO(2) group we have (σ = integer) only one dimensional representations, Dσ(g) = eiθσ Dσ(g)Dσ(g-1) = eiθσ ei(-θσ) = 1 aσ = ∫dg Dσ(g)a(g) = (1/2π) !Syntax Error, Idθ eiσθ a(θ) If a,b,c are all real and even functions of θ, then we have aσ = (1/2π) !Syntax Error, Idθ eiσθ a(θ) = (1/2π) !Syntax Error, Idθ cos(σθ) a(θ) = (1/π)!Syntax Error, Idθ cos(σθ) a(θ) So our convolution equation is this and its diagonalization is on the next line a(α) = (1/2π) ∫dβ g(α-β)c(β) aσ = gσ cσ // diagonalization In our application we had a = G, c = F and according to the following, b = {πln | cosα - cosβ |} : G(α) = (1/2)!Syntax Error, Idβ ln | cosα - cosβ | F(β) = (1/2π)!Syntax Error, Idβ {πln | cosα - cosβ |} F(β) Our result was this: g0 = (-π ln2) f0 gm= (- π/m) fm m = 1,2... I think we can conclude that b0 = (-π ln2) bm = (- π/m) Somewhere from 30+ years ago I have good notes on this, Stakgold did not get involved in group interpretations of things.