stakgold Vol 2 Appendix B

Stakgold Vol 2 Appendix B PhL 4.6.11 I don't know where else you would see a section like this. Stak does it right. A. Getting the large t behavior of a Laplace-Transform-Like integral. 1 B. Special case of (B.1): Large s limit of the Laplace Transform 2 C. Large t limit of the Inverse Laplace Transform 3 ________________________________________________________________________________ A. Getting the large t behavior of a Laplace-Transform-Like integral. B.1 shows our integral type of interest where we want behavior for large t. I(t) = !Syntax Error, Idx e-h(x)t f(x) As noted, this is mostly going to come from region of small h(x) within the integration range (a,b). Suppose h(x) has a minimum in this middle of this range at some x = c. Then model as h(x) = h(c) + (1/2!) h"(c) (x-c)2 ` h"(c) > 0 cupping up we assume So the idea here is that h(x) is positive and thus we care about its minimum. Then get !Syntax Error, Idx f(x) exp[ -t {h(c) + (1/2!) h"(c) (x-c)2} ] = e-th(c) !Syntax Error, Idx f(x) exp[ -(t/2) h"(c) (x-c)2} ] ≈ e-th(c) f(c) !Syntax Error, Idx exp[ -(t/2) h"(c) (x-c)2} ] Change variables to y = x-c so this becomes ≈ e-th(c) f(c) !Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] = 2 e-th(c) f(c)!Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] ≈ 2 e-th(c) f(c) !Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] // since t large, makes little diff. = 2 e-th(c) f(c) (1/2) [(t/2) h"(c)]-1/2 // Maple = e-th(c) f(c) [(t/2) h"(c)]-1/2 = e-th(c) f(c) 1/ = e-th(c) f(c) // agrees with page 399 A. Then for this case in which h(x) is positive and has a single minimum at c inside interval (a,b) we get !Syntax Error, Idx f(x) e-th(x) ≈ e-th(c) f(c) Suppose we have a minimum instead at endpoint a and there is no interior minimum. This means that h'(a) ≥ 0 since h(x) has to be rising at the left edge if it is the intervals minimum. Then smallest h(x) is at x = a where h(x) = h(a) + (x-a) h'(a) I(t) ≈ !Syntax Error, Idx f(x) exp(-t [h(a) + (x-a) h'(a)]) ≈ !Syntax Error, Idx f(x) e-th(a) e-th'(a)(x-a) ≈ e-th(a) f(a) !Syntax Error, Idx e-th'(a) (x-a) Now let y = x-a to get = e-th(a) f(a) !Syntax Error, Idy e-th'(a)y ≈ e-th(a) f(a) !Syntax Error, Idy e-th'(a)y = e-th(a) f(a) [1/th'(a) ] // agrees with (B.3) Stak says that if h'(a) = 0 at this endpoint, then pretend it is the first case and add 1/2, I agree. B. Special case of (B.1): Large s limit of the Laplace Transform [ s = t here] Suppose h(x) = x and interval is (a,b) = (0,∞) as arises in Laplace transform. Then we have our second case and we get h'(x) = 1 so !Syntax Error, Idx f(x) e-tx ≈ f(0) /t // agrees with p 400 A. If f(x) is analytic about x = 0, expand in the usual Taylor series p 400B (when you expand around the point x = 0, it is called a McLaurin series for some reason). Then we go look up the Laplace of xn and find it to be [ Stak is using t in place of the usual Laplace variable s ] xn/n! → 1/tn+1 Thus the corrected B.5 should read f(x) = Σn=0∞ f(n)(0) xn/n! => L(t) = Σn=0∞ f(n)(0) / tn+1 // Stak has a typo in his B.5 I cannot find this on the web anywhere, no good search handle. Suppose instead that f(x) has an expansion in non-integer powers as in p 400 C (perhaps those powers are eigenvalues in some problem). Then ET I tells us that f(x) = Σn an xαn xαn → Γ(αn+1) t-αn-1 and then F(t) = Σn an Γ(αn+1) t-αn-1 = Σn an (αn)! t-αn-1 // agrees with B.6 and I checked that if αn = n, this gives my version of B.5. C. Large t limit of the Inverse Laplace Transform Good grief. Usual Laplace uses t = time and s = Laplace. In the above, Stak uses x = time and t = Laplace. Now in page 401 a we are back to the usual t = time and s = Laplace. So p 401A is just the usual Laplace inversion formula. He shows a sample case where there are two poles and two cuts and drags the contour to the left and assumes the left side great half circle vanishes. This gives equation B which goes with the figure. A pole of order m. [402] He actually intends s2 to have a pole of order m, not just order 1. This means that the function F(s)(s-s1)m is analytic in s, so F(s) could have for example a term (s-s1)-m+1. Thus we get the strange looking power series of p 402 A where the largest negative power is (s-s1)-m . P 402 B is correct, but I will do this my way instead, We have f(t) = (1/2πi) ds Σk=-m∞ ak (s-s2)k ets = Σk=-m∞ ak (1/2πi) ds (s-s2)k ets But only the negative powers of k can make contributions, so = Σk=-m-1 ak (1/2πi) ds (s-s2)k ets = Σk=1m a-k (1/2πi) ds (s-s2)-k ets = Σk=1m a-k (1/k-1)!) ∂t(k-1)(es2t) = Σk=1m a-k (tk-1/k-1)!) (es2t) = es2t Σk=1m a-k tk-1/(k-1)! // agrees with B.7 So this would be a contribution from a "pole of order m" and of course you need all the a-k coefficients in order to really know what you've got. He notes that we don't care about the s4 pole since it is to the left and will give a contribution exp(s4t) ~ exp(-|Re(s4)| t) and is down by large ratio from the s2 term, it is down by an exponential exp(-tΔs) that is to say, so we ignore s4. A Cut. Now we have to face the cut and the one with branch point s1 dominates. Now we will do the discontinuity integral. I agree with modeling F(s) as shown in B.8 since this is a Frobenius form for a solution of an ODE in s near a singular point, though I question the α range. We then have for example term = -(1/2πi) ∫C2 est F(s) ds = -(1/2πi) !Syntax Error, Ids est disc(F(s)) // a is Re(s1) Now R = |s-s1| = s1-s for s on the cut, disc(s-s1)α = Rα ( eiπa - e-iπα) = Rα 2isin(πα) R = s1- s > 0 on the integral disk(F(s)) = (s1-s)α2isin(πα) Σn=0∞bn(s-s1)n so get ( assume that Im(s1) = r and Re(s1) = a, this latter used in the page 401 picture ) term = -(1/2πi) !Syntax Error, Ids est (s1-s)α2isin(πα) Σn=0∞bn(s-s1)n = - (1/π) sin(πα) Σn=0∞ bn (-1)n!Syntax Error, Ids est(s1-s)α+n Now, let s' ≡ s-ir be a complex variable which runs along the cut with Im(s') = 0. Then ds = ds' and est = et(s'+ir) and (s1-s) = Re(s1-s) = a - s' and endpoints are then -∞ and a, so we have = - (1/π) sin(πα)eirt Σn=0∞ bn (-1)n!Syntax Error, Ids' es't (a - s')α+n But of course s' is now just a real variable, so let's first replace it with -y to get = - (1/π) sin(πα)eirt Σn=0∞ bn (-1)n!Syntax Error, Idy e-yt (a+y)α+n Now let x = y+a and we then get = - (1/π) sin(πα)eirt Σn=0∞ bn (-1)n!Syntax Error, Idx e-(x-a)t xα+n = - (1/π) sin(πα)eirt eat Σn=0∞ bn (-1)n!Syntax Error, Idx e-xt xα+n Now notice that s1 = a + ir so we really have = - (1/π) sin(πα)ets1 Σn=0∞ bn (-1)n!Syntax Error, Idy e-xt xα+n Now consider this integral L(t) = !Syntax Error, Idy e-xt xα+n Think of this as one term in the series p 400C where αn = α+n and an = 1. We can then claim L(t) ≈ (α+n)! t-α-n-1 so our result is now term = - (1/π) sin(πα)ets1 Σn=0∞ bn (-1)n (α+n)! t-α-n-1 // agrees with (B.9)' and that certainly took a long time! Once again, this "term" is what you get in the inverse Laplace transform if you have a cut of this Frobenius cut form: F(s) = (s-s1)α Σn=0∞ bn(s-s1)n Now what about the little half turn around the branch point? term' = -(1/2πi) ∫halfturnccw est F(s) ds The usual thing is to set s-s1 = Reiθ and then we have ds = iReiθdθ so term' = -(1/2π) !Syntax Error, I dθ Reiθ F(s1+Reiθ) But F(1+Reiθ) = (Reiθ)α Σn=0∞ bn(Reiθ)n and the lowest power of R is with n = 0 and as R→0 and this power is Rα. So for small R we have term' ≈ -(1/2π) !Syntax Error, I dθ Reiθ Rα eiαθ b0 and this vanishes as long as R1+α → 0 which means 1+α > 0 which means α > -1, that is, Re(α) > -1. This is how I would state things. He puts an upper limit on Re(α) of 0 which seems unnecessarily restrictive on this result. Summary of the cut result: Assume there is a branch point at s1 and that in a disk centered at this branch point we have F(s) = (s-s1)α Σn=0∞ bn(s-s1)n where Re(α) > -1 True, if α were 1 + 1/3, say, we could absorb the integer part into the series and say b0 = 0. Now given this branch point form for F(s), its contribution to the inverse Laplace transform is this: f(t) = - (1/π) sin(πα)ets1 Σn=0∞ bn (-1)n (α+n)! t-α-n-1 Application: Suppose our F(s) has a branch point at s = s1 = 0 with α = -1/2 and at the same time suppose it has a pole of order 1 also at s = 0, as shown in page 402 D. Then the pole gives a-1 and the cut gives fcut(t) = + (1/π) Σn=0∞ bn (-1)n (n-1/2)! t-n-1/2 sin(πα) = -1 I guess this example is going to arise somewhere. Comment: Notice that we assume the form shown F(s) just near the branch point. This is not going to be the form for F(s) all along the infinite cut. But just as in other cases, we know that most of the contribution is going to come from the integration region "just to the left of the branch point". We are then allowed to extend this region to -∞ without making too much difference. A. Redo Part A above in the case that h(x) is complex. B.1 shows our integral type of interest where we want behavior for large t. I(t) = !Syntax Error, Idx e-h(x)t f(x) As noted, this is mostly going to come from region of small Re(h(x)) within the integration range (a,b). Suppose Re(h(x)) has a minimum in this middle of this range at some x = c. Then model as h(x) = h(c) + (1/2!) h"(c) (x-c)2 Re{h(c)}" > 0 cupping up we assume Note in passing that h(x) = u(x) + iv(x) so h"(x) = u"(x) + iv"(x) etc. So the idea here is that Reh(x) is positive and thus we care about its minimum. Then get !Syntax Error, Idx f(x) exp[ -t {h(c) + (1/2!) h"(c) (x-c)2} ] = e-th(c) !Syntax Error, Idx f(x) exp[ -(t/2) h"(c) (x-c)2} ] ≈ e-th(c) f(c) !Syntax Error, Idx exp[ -(t/2) h"(c) (x-c)2} ] Change variables to y = x-c so this becomes ≈ e-th(c) f(c) !Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] = 2 e-th(c) f(c)!Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] ≈ 2 e-th(c) f(c) !Syntax Error, Idy exp[ -(t/2) h"(c) y2} ] // since t large, makes little diff. = 2 e-th(c) f(c) (1/2) [(t/2) h"(c)]-1/2 // Maple = e-th(c) f(c) [(t/2) h"(c)]-1/2 = e-th(c) f(c) 1/ = e-th(c) f(c) // agrees with page 399 A. Then for this case in which h(x) is positive and has a single minimum at c inside interval (a,b) we get !Syntax Error, Idx f(x) e-th(x) ≈ e-th(c) f(c) Suppose we have a minimum instead at endpoint a and there is no interior Re(h) minimum. This means that Reh(a)' ≥ 0 since Reh(x) has to be rising at the left edge if it is the intervals minimum. Then smallest h(x) is at x = a where h(x) = h(a) + (x-a) h'(a) => Reh(x) = Reh(a) + (x-a) Reh'(a) I(t) ≈ !Syntax Error, Idx f(x) exp(-t [h(a) + (x-a) h'(a)]) ≈ !Syntax Error, Idx f(x) e-th(a) e-th'(a)(x-a) ≈ e-th(a) f(a) !Syntax Error, Idx e-th'(a) (x-a) Now let y = x-a to get = e-th(a) f(a) !Syntax Error, Idy e-th'(a)y ≈ e-th(a) f(a) !Syntax Error, Idy e-th'(a)y = e-th(a) f(a) [1/th'(a) ] // agrees with (B.3) Stak says that if h'(a) = 0 at this endpoint, then pretend it is the first case and add 1/2, I agree. B. Redo the single endpoint case with a full power series and see what happens We start with I(t) = !Syntax Error, Idx e-h(x)t f(x) Assume the min of h(x) occurs at endpoint a, and that one of two situations arises, both of which are legal for a minimum at this left endpoint h'(a) > 0 h'(a) = 0 and h"(a) > 0 etc. The claim is that for very large t, only the region very close to the endpoint contributes. We then integrate over a to a+δ, and then we replace δ by ∞ saying that by adding a part far from the endpoint, we don't make any difference. So I would argue that we want to model h(x) as best we can in the region of the endpoint. So let's put in a full series like this h(x) = h(a) + Σn=1∞ h(n)(a)/n! * (x-a)n Then we have I(t) = !Syntax Error, Idx e-h(x)t f(x) ≈ f(a) !Syntax Error, Idx exp( -t [h(a) + Σn=1∞ h(n)(a)/n! * (x-a)n ] ) ≈ exp( -t h(a) )f(a) !Syntax Error, Idx exp( -t Σn=1∞ h(n)(a)/n! * (x-a)n ) Now take z = x-a to get = exp( -t h(a) )f(a) !Syntax Error, Idz exp( -t Σn=1∞ h(n)(a)/n! * zn ) The problem with keeping multiple terms is that this integral is a bit ugly. If we try to keep just two terms, for example, we get this (see Maple below) !Syntax Error, Idz exp( -t [Az+Bz2 ) = (1/2) exp( tA2/[4B]) erfc(t1/2A/[2]) where we now define α so that α = A/[2] A = h'(a) B = h"(a)/2 α2 = A2/[4B] α = h'(a)/ [ 2 ] = h'(a) / α2 = h'(a)2/ [2 h"(a)] and then !Syntax Error, Idz exp( -t [Az+Bz2 ) = (1/2) exp(tα2) erfc(α t1/2) In this case, to find the large t limit, we are forced to study the error function. So we need to know about the erf for large argument, and we call upon new AS 2010 to get so that erfc(α t1/2) ~ [ exp(-α2t)/(αt1/2)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m Now above we can write !Syntax Error, Idz exp( -t [Az+Bz2 ) = (1/2) exp(tα2) [ exp(-α2t)/(αt1/2)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m = (1/2) [ /(αt1/2)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m But [ /(αt1/2)] = [ /(αt1/2)] = t-1/2B-1/2 α-1 t-1/2 But α = A/[2] = A 2-1 B-1/2 => α-1 = A-1 21 B1/2 so that [ /(αt1/2)] = t-1/2B-1/2 [A-1 21 B1/2] t-1/2 = t-1 A-121 = 2/(At) so we then have !Syntax Error, Idz exp( -t [Az+Bz2 ) = (1/2) [2/(At)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m = [1/(At)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m Then our single endpoint result keeping just these two terms is I(t) = exp( -t h(a) )f(a) [1/(At)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m ≈ exp( -t h(a) )f(a) [1/(At)] [ 1 - 3/(2α2) t-1 + ...] ≈ exp( -t h(a) )f(a) [1/(t h'(a))] [ 1 - 3/(2α2) t-1 + ...] Then 3/(2α2) = 3 2-1 α-2 = 3 2-1 2 h"(a)/ h'(a)2 = (3 h"(a)/ h'(a)2) so our result is then I(t) = exp( -t h(a) )f(a) [1/(t h'(a))] [ 1 - (3 h"(a)/ h'(a)2) t-1 + ...] Now is this consistent with Stak's B.3 for example? Yes it is! (finally). But now I have a second term as well. Conclusion: if we keep just the linear and quadratic terms in our model for h(x) near its minimum, we get the last result above as our approximation for the integral I(t) for large t. If we try to keep 3 terms, the dz integral becomes difficult. For example, Maple cannot do it, nor can Wolfram. In any event, our first term agrees exactly with B.3. What if at the other endpoint? We just copy and edit from above, and I show changes in red except as all a→b, I won't make all the b's red. We start with I(t) = !Syntax Error, Idx e-h(x)t f(x) Assume the min of h(x) occurs at endpoint b, and that one of two situations arises, both of which are legal for a minimum at this left endpoint h'(b) < 0 h'(b) = 0 and h"(b) > 0 etc. The claim is that for very large t, only the region very close to the endpoint contributes. We then integrate over b-δ to b, and then we replace δ by ∞ saying that by adding a part far from the endpoint, we don't make any difference. So I would argue that we want to model h(x) as best we can in the region of the endpoint. So let's put in a full series like this h(x) = h(b) + Σn=1∞ h(n)(b)/n! * (x-b)n Then we have I(t) = !Syntax Error, Idx e-h(x)t f(x) ≈ f(b) !Syntax Error, Idx exp( -t [h(b) + Σn=1∞ h(n)(b)/n! * (x-b)n ] ) ≈ exp( -t h(b) )f(b) !Syntax Error, Idx exp( -t Σn=1∞ h(n)(b)/n! * (x-b)n ) Now take z = x-b to get = exp( -t h(b) )f(b) !Syntax Error, Idz exp( -t Σn=1∞ h(n)(b)/n! * zn ) and then take z→-z to get = exp( -t h(b) )f(b) !Syntax Error, Idz exp( -t Σn=1∞ h(n)(b)/n! * (-z)n ) So the changes are a→b and z→-z. In our 2-term calculation above, we accomplish this by A→ -A (but α2 2 = A2/[4B] stays the same). We then get this intermediate result !Syntax Error, Idz exp( -t [-Az+Bz2 ) = - [1/(At)] Σm=0∞ (-1)m (2m-1)!! / (2α2t)m which is just an overall sign change. Then we get I(t) = - exp( -t h(b) )f(b) [1/(t h'(b))] [ 1 - (3 h"(b)/ h'(b)2) t-1 + ...] and we can in effect say that we take our "a" result and take a→b and h'(a)→ - h'(b) to get the "b" result. C. Redo Part A above in the case that h(x) leading endpoint terms cancel Assume that the minimum of h(x) occurs at the endpoints and is the same at both endpoints. So there is a max of h(x) out in the interval somewhere. In this case, we want to add the two minima contributions. I will assume for the moment that the endpoint minima are not quite equal. I(t) = !Syntax Error, Idx e-h(x)t f(x) Expand the exponent h(x) = h(a) + h'(a)(x-a) + (1/2)h"(a)(x-a)2 at endpoint a h(x) = h(b) + h'(a)(x-b) + (1/2)h"(b)(x-b)2 at endpoint b We have solved this problem above and we know the contribution from each endpoint: I(t) = exp( -t h(a) )f(a) [1/(t h'(a))] [ 1 - (3 h"(a)/ h'(a)2) t-1 + ...] // from endpoint a I(t) = - exp( -t h(b) )f(b) [1/(t h'(b))] [ 1 - (3 h"(b)/ h'(b)2) t-1 + ...] // from endpoint b If h(a) ≠ h(b), one of these two terms dominates in the expo factor, so we forget the other term. But suppose h(a) = h(b). Then we have to add the two terms. Example: Consider this integral, where y > 0 I(t) = !Syntax Error, Idx e-ytcosx h(x) = ycos(x) Here we have an example of two equal minimums, but in this case the integrand is even in x so we can write I(t) = 2 !Syntax Error, Idx e-ytcosx and now we have only one minimum at endpoint b = π/2. Our large t approximation is then I(t) = - exp( -t h(b) )f(b) [1/(t h'(b))] [ 1 - (3 h"(b)/ h'(b)2) t-1 + ...] We compute (note that f(x) = 1 ) h(x) = y cos(x) h(b) = h(π/2) = 0 h'(x) = -ysin(x) h'(b) = -y h"(x) = -ycos(x) h"(b) = 0 so that I(t) = - exp( -t h(b) )f(b) [1/(t h'(b))] [ 1 - (3 h"(b)/ h'(b)2) t-1 + ...] = + f1 [1/(t y] [ 1 - 0 t-1 + O(t-2) ] So we conclude that !Syntax Error, Idx e-ytcosx ≈ (2/y)t-1 as t → ∞