ODEs with a k^2 type parameter

ODEs with a k2 type parameter. PhL 10.11.10 The purpose of this document is to investigate the implications of having a parameter appear in an ODE in a squared form only. The classic case is f "(x) = k2f(x) which has solutions [ ekx, e-kx] or [sh(kx), ch(kx)]. In the first pair we find that f(x; -k) gives "the other solution" and is independent of f(x; k), while for the second pair we find that f(x; -k) gives a multiple of f(x; k). Another issue is the possibility of having solutions which are even or odd in x or in k. Other cases of quadratic parameters are the m2 in the associated Legendre ODE as seen in Pnm(z), the n2 of the Bessel equation as in Jn(x), and the a2 and b2 which appear in the Lamé equation, this last case being the motivation for this digression. In Part A we show for f "(x) = k2f(x) that the indicial equation gives r = 0 and r = 1 (as for any ordinary point) with corresponding Frobenius solutions ch(kz) and sh(kz)/k and we note that both solutions are even in k when written in this manner. In Part B we convince ourselves that you can always find two independent solutions of an ODE containing k2 which are even in k. So these two solutions are solutions "of definite Frobenius index", whereas the e±bx solutions are not. A. The official Frobenius Method for f"(z) = k2f(z) 1 B. Question: Is it always possible, for an ODE which contains only k2, to select a pair of independent solutions which are both even in k? 4 Short answer: (added after the long answer was written) 4 Long answer: 4 1. The case that f1(z; -k) = C(k) f1(z;k) 5 2. The case that f1(z; -k) = independent of f1(z;k) 8 3. Summary 8 4. Examples 9 A. The official Frobenius Method for f"(z) = k2f(z) One of the main ODE tools we have is the Frobenius Method, so here I want to just follow it through for our prototype case. We try a solution of the form (where r and the an are to be determined) f(z) = zr Σn=0∞ anzn = Σn=0∞ anzn+r f'(z) = Σn=0∞ (n+r) anzn+r-1 f"(z) = Σn=0∞ (n+r)(n+r-1) anzn+r-2 Aside: We are going to insist that a0 ≠ 0, and will take it as 1. But suppose a0 = 0 and a1≠ 0. Then we would write, using n' = n-1, f(z) = Σn=1∞ anzn+r = Σn'=0∞ an'+1zn'+r+1 = Σn=0∞ an+1zn+r+1 = zs Σn=0∞ bnzn where s= r+1 and bn = an≈1 . So by letting a0 = 0, we end up with the exact same starting form for our assumed power series solution (just new names), and we would then assume b0 ≠ 0. So the point is that we might as well assume a0 ≠ 0 "without loss of generality". So, insert the above forms into the ODE to get Σn=0∞ (n+r)(n+r-1) anzn+r-2 = k2 Σn=0∞ anzn+r Multiply both sides by z2 (this really takes the ODE to standard Frobenius form which we should have done at the start) Σn=0∞ (n+r)(n+r-1) anzn+r = k2 Σn=0∞ anzn+r+2 Divide both sides by zr Σn=0∞ (n+r)(n+r-1) anzn = k2 Σn=0∞ anzn+2 Let n+2 = n' (then rename n' to be n) to write the second sum as { Σn=0∞ anzn+2} = { Σn'=2∞ an'-2zn'-2+2} = { Σn'=2∞ an'-2zn'} = { Σn=2∞ an-2zn} Break out the n=0 and n=1 terms in the first sum so we then have (0+r)(0+r-1) a0 + (1+r)(1+r-1) a1z + Σn=2∞ (n+r)(n+r-1) anzn -k2{ Σn=2∞ an-2zn} = 0 Now combine the two sums together r(r-1)a0 + r(1+r) a1z + Σn=2∞ [(n+r)(n+r-1) an - k2 an-2] zn = 0 If we set z = 0 and require a0 ≠ 0, the first term is all that is left and we have r(r-1) = 0 which is the indicial equation telling us there are only two choices, r = 0 and r = 1. For such special values of r we have r(1+r) a1z + Σn=2∞ [(n+r)(n+r-1) an - k2 an-2] zn = 0 Divide by z r(1+r) a1 + Σn=2∞ [(n+r)(n+r-1) an - k2 an-2] zn-1 = 0 We must have the first term vanish AND the bracket [] in the second term must vanish for all n: r = 0: a1 = any [n(n-1) an - k2 an-2] = 0 n=2... r = 1: a1 = 0 [n(n+1) an - k2 an-2] = 0 n=2... For the r = 0 case we arbitrarily assume a1 = 0 which kills off all odd an (we are certainly allowed to do that searching for a solution) and for the even n the recursion relation says 2*1 a2 = k2 a0 = k2 => a2 = k2/2! 4*3 a4 = k2a2 = k4/2! => a4 = k4/ 4! 6*5a6 = k2a4 = k6/4! => a6 = k6/6! => an = kn/n! n = 2,4,.... and a0 = 1 Our solution is then f(z) = Σn=0∞ anzn = Σn=0,2,4 (kz)n/n! = 1 + (kz)2/2! + (kz)4/4! = ch(kz) // r=0 sol For the r = 1 case, the recursion relation says 2*3 a2 = k2 a0 = k2 => a2 = k2/3! 4*5 a4 = k2a2 = k4/3! => a4 = k4/ 5! 6*7a6 = k2a4 = k6/5! => a6 = k6/7! => an = kn/(n+1)! n = 2,4,.... and a0 = 1 Our solution is then (remember r = 1 now) f(z) = Σn=0∞ anzn+1 Let n' = n+1 and write this as f(z) = Σn'=1∞ an'-1zn'-1+1 = Σn'=1∞ an'-1zn' = Σn=1∞ an-1zn where from above we then have an-1 = kn-1 /n! so we get f(z) = Σn=1∞ (kn-1 /n!)zn = (1/k) Σn=1∞ (kz) /n! = (1/k) [ kz + (kz)3/3! + ...] = (1/k) sh(kz) So with the assumption that a0 = 1 we find these two Frobenius solutions to our ODE: r = 0 : f0(z) = ch(kz) r = 1 : f1(z) = (1/k) sh(kz) In both solutions, we get only even an coefficients so Σn=0∞ anzn is even. Our first solution then is even, and our second solution is z times an even function, which is of course an odd function. We could think of this as f(z) = z [ (1/(kz) sh(z)] where the bracket is even. [ We are talking even and off in z here.] Note that both solutions as shown are even in k. So for either of these solutions we have fi(z; -k) = fi(z; k) So in this example, we might note that the solutions [ ekx, e-kx] are not Frobenius solutions, but are linear combinations of same. ________________________________________________________________________________ B. Question: Is it always possible, for an ODE which contains only k2, to select a pair of independent solutions which are both even in k? My tentative answer to this question is "yes", but I don't have a bulletproof proof. Short answer: (added after the long answer was written) Let fi(x; k) be two independent solutions. Then construct: F1(x; k) = f1(x; k) + f1(x; -k) Whether or not f1(x; -k) is independent of f1(x; k), this is an even solution. If f1(x; -k) = - f1(x; k), this solution would vanish. In this case, construct this alternative solution F1alt(x; k) = k f1(x; k) which is again even in k. So here is our suggestion: F1(x; k) = k f1(x; k) if f1(x; k) is odd in k = f1(x; k) + f1(x; -k) otherwise F2(x; k) = k f2(x; k) if f1(x; k) is odd in k = f2(x; k) + f2(x; -k) otherwise How do we know that F1 and F2 so selected will be independent? They might be the same, as in the case where f1 = ekx and f2 = e-kx. In this case, use F2(x; k) = k [f2(x; k) – f2(x; -k)] as the second even function. Long answer: This seems to be a pretty simple question. For our case above we can pick ch(kz) and k sh(kz) which are both even in k. But showing this for a general ODE is not easy. We begin: Suppose we have some solution f1(z; k) to an ODE in which only k2 appears. We know that f1(z; -k) is therefore also a solution: L(k2) f1(z; k) = 0 => L([-k]2) f1(z; -k) = 0 => L(k2) f1(z; -k) = 0 Either f1(z; -k) = C(k) f1(z;k) in which case f1(z; -k) is not an independent solution along with f1(z; k), or f1(z; -k) can be taken as an independent solution. 1. The case that f1(z; -k) = C(k) f1(z;k) In the first case f1(z;k) is almost even in k to begin with. Now, if it is true that f1(z; -k) = C(k) f1(z;k), then if we apply this twice we find that C(k) C(-k) = 1 Side Question: If someone tells you that the real function f(x) has the property f(x)f(-x) = 1, what can you say about the solution f(x) ? I had to go off and write a whole doc to answer this question! The answer is this: pick any fodd(z) you like, then and most general solution is f(z) = fodd(z) + Thus, C(k) cannot be "just any function", it must be one of this form C(k) = Codd(k) + where Codd(k) is an arbitrary odd function of k. Now, our goal is to define F1(z; k) ≡ A(k) f1(z;k) such that F1 is even in k. This will then require that A(k) f1(z;k) = A(-k) f1(z;-k) = A(-k) C(k) f1(z;k) We must then select an A(k) such that this is true: A(k) = A(-k) C(k) or A(k)/A(-k) = C(k) In order to find a suitable A(k), decompose it into even and odd parts, so [Aeven(k) + Aodd(k)] = [Aeven(k) - Aodd(k)]C(k) = [Aeven(k) - Aodd(k)] [Codd(k) + ] Thus we have this requirement: [Aeven(k) + Aodd(k)] = [Aeven(k) - Aodd(k)] [Codd(k) + ] = Aeven(k) Codd(k) – Aodd(k) Codd(k) + Aeven(k) - Aodd(k) We can imagine decomposing each side of this equation into its even piece and its odd piece: Aeven(k) = – Aodd(k) Codd(k) + Aeven(k) // even Aodd(k) = Aeven(k) Codd(k) – Aodd(k) // odd Aeven(k) [ 1 - ] = – Aodd(k) Codd(k) Aodd(k) [ 1 + ] = Aeven(k) Codd(k) (*) Insert the first into the second and you find Aodd(k) [ 1 + ] = {Aeven(k)} Codd(k) = {– Aodd(k) Codd(k)/ [ 1 - ]} Codd(k) = - Aodd(k) C2odd(k)/ [ 1 - ]} But this will be true for any Aodd we select, because cancelling it we get [ 1 + ] = - C2odd(k)/ [ 1 - ] [ 1 + ] [ 1 - ] = - C2odd(k) 1 - { 1 + [Codd(k)]2} = -C2odd(k) which is true. This seems to say that any old Aodd(k) is OK. So imagine that we select some Aodd(k). Then we need to have (from (*)) Aeven(k) = Aodd(k) [ 1 + ]/ Codd(k) The simplest Aodd(k) I can think of is Aodd(k)= k, so we have Aeven(k) = k [ 1 + ]/ Codd(k) Then we have A(k) = k{ 1 + [ 1 + ]/ Codd(k) } A(-k) = k{ -1 + [ 1 + ]/ Codd(k) } A(k)/A(-k) = { 1 + [ 1 + ]/ Codd(k) } / { -1 + [ 1 + ]/ Codd(k)} = { Codd(k) + [ 1 + ] } / { - Codd(k) + [ 1 + ]} = { 1 + Codd(k) + ] } / { 1 - Codd(k) + ]} We want to show this is equal to C(k) = [Codd(k) + ] . So we want to show this is true": [Codd(k) + ] = { 1 + Codd(k) + ] } / { 1 - Codd(k) + ]} which means [Codd(k) + ]{ 1 - Codd(k) + ]} = { 1 + Codd(k) + ] } The LHS we can multiply out to get LHS = Codd(k) - Codd2(k) + Codd(k) + - Codd(k) + [ 1 + Codd2(k)] = Codd(k) + Codd(k) + - Codd(k) + 1 = Codd(k) + + 1 and this is indeed the RHS. So I have verified my solution for A(k) shown above. So going way back, we were considering the situation where f1(z; -k) = C(k) f1(z;k). We have shown several things: (1) C(k) cannot be arbitrary, but must have the form C(k) = Codd(k) + (2) The function F1(z; k) ≡ A(k) f1(z;k) will then be an even function in k, provided we make this selection for A(k) : A(k) = k{ 1 + [ 1 + ]/ Codd(k) } So all this mess was just to show that we can find multiple of f1 which is even in k in this case. 2. The case that f1(z; -k) = independent of f1(z;k) In this case We can construct F2(z; k) = f1(z; k) + f1(z; -k) as our second solution. We have ruled out f1(z; -k) = - f1(z; k) because we have said f1(z; -k) is independent, so F2 ≠ 0 and is of course even in k. 3. Summary We have an ODE in which only k2 appears, and we have two independent solutions fi(z; k). We know that fi(z; -k) are both solutions of the ODE. We first work with f1 : If f1(z; -k) is not independent of f1(z; k), then we have f1(z; -k) = C(k) f1(z;k) where C(k) is some multiple which must have a certain special form C(k) = Codd(k) + . In this case, we can construct an even solution function F11(z; k) ≡ A(k) f1(z;k) where A(k) has the form shown above. If f1(z; -k) is independent of f1(z; k), then we can construct an even solution function F12(z; k) = f1(z; k) + f1(z; -k) Whichever case it is, we have obtained an even solution function. So F11(z; k) ≡ A(k) f1(z;k) F12(z; k) = f1(z; k) + f1(z; -k) Now we can repeat this entire whizbang for solution f2(z; k) and we will end up with an even solution function which is one of these F21(z; k) ≡ A(k) f2(z;k) F22(z; k) = f2(z; k) + f2(z; -k) If it happens that both solutions have independent fi(z; k) so we get F12 and F22 as our even functions, and if it happens that F12 = F22, then we take as our second even function F23 = k[f2(z; k) - f2(z; -k)] 4. Examples Example 1: Suppose f1(z; k) = ch(kz). Then C(k) = 1. We then have Codd(k) = 0. And we then have A(k) = k{ 1 + [ 1 + 1]/ Codd(k) } ≈ 2 k / Codd(k) before final limit. But A(-k) = A(k) so we then have A(k)/A(-k) = 1 = C(k). Our even solution is then F11(z; k) ≡ A(k) f1(z;k) = [2 k / Codd(k)] ch(kz) but we can just take it then to be ch(kz). [ Codd(k) = [ C(k)2 - 1] / [2 C(k)] = (1-1)/2*1 = 0 ] Example 2: Suppose f1(z; k) = sh(kz) sp C(k) = -1. Then Codd(k) = 0 again, but we take the other sign of the square root to get C(k) = -1. In this case we find that A(k) = k{ 1 + [ 1 - 1]/ Codd(k) } = k and then we have A(k)/A(-k) = k / (-k) = -1 = C(k). Our even solution is then F11(z; k) ≡ A(k) f1(z;k) = k sh(kz). Example 3: Suppose f1(z; k) = ezk. Then F12 = ekx + e-kz = 2 ch(kz) as our even function. Example 4: Suppose in Example 1, we also had f2(z; k) = ezk. By example 3, this would lead us to F22 = 2 ch(kz) which is the same basically as F11, so we use the escape hatch to get F23 = k 2 sh(kz) . So our method for generating two even-in-k independent solutions seems to work for our prototype case, but what about some more general case? Example 5: How about associated Legendres? In the case that m ≠ integer, we have Pn-m not a multiple of Pnm and then our one solution would be just Pn-m + Pnm. Maybe this same thing with Q's would be the second even-in-m solution. Example 6: In the case that m is an integer, we know that Pn-m = f(n,-m) Pnm so C(m) = f(n,-m). As expected, we then have C(m)C(-m) = 1. But what is A(m) in this case? First, what is Codd(m) ? C(m) = f(n,-m) = Γ(n-m+1)/ Γ(n+m+1) Then Codd(m) = [ C(m)2 - 1] / [2 C(m)] = [f2(n,-m) - 1] / [2 f(n,-m)] = f(n,-m)[f(n,-m) - f(n,m)] / [2 f(n,-m)] = [f(n,-m) - f(n,m)]/2 = odd, yup A(m) = m{ 1 + [ 1 + ]/ Codd(m) } = m{ 1 + [ 1 + ]/ [f(n,-m) - f(n,m)]/2 } = m{ 1 + [ 1 + ]/ [f(n,-m) - f(n,m)]/2 } = m { 1 + } = m { 1 + } Notice that the second term in {} is odd in m and the first even, so A(m) has neither symmetry. But presumably this is our resulting even-in-m solution pnm(z) = m { 1 + } Pnm(z) I am too lazy to check it.