Spotlight on the Extended Method of Frobenius

Transcription

1 113 Spotlight on the Extended Method of Frobenius See Sections 11.1 and 11.2 for the model of an aging spring. Reference: Section 11.4 and SPOTLIGHT ON BESSEL FUNCTIONS. Bessel functions of the first kind were constructed earl in the nineteenth centur b the method of generalized power series, but, as noted in the SPOTLIGHT ON BESSEL FUNCTIONS, the method does not alwas produce a second, independent solution of Bessel s equation. The puzzle of finding a second solution remained unresolved until 1867, when C. G. Neumann ( ) published his pioneering work on solutions of Bessel s equation. Frobenius extended Neumann s methods to a complete treatment of all solutions near an regular singular point of an second-order linear ODE. The extended theor of Frobenius alwas allows us to construct a second solution even when the roots of the indicial polnomial differ b an integer. In this section we outline the extended Frobenius theor, define Bessel functions of the second kind, and express the solution of the problem of the aging spring in terms of Bessel functions of the first and second kinds. The Extended Method of Frobenius To review briefl, our goal is to find a pair of independent solutions near the regular singular point x = of the second-order linear ODE x 2 + xp(x) + q(x) = (1) The functions p(x) and q(x) are real analtic on an interval J centered at the origin: p(x) = The quadratic indicial polnomial is p n x n and q(x) = f (r) = r 2 + ( p 1)r + q q n x n and its indicial roots are r 1 and r 2. We assume here that the indicial roots are real and that r 2 r 1. The Method of Frobenius outlined in Section 11.4 alwas ields a nontrivial solution of ODE (1) of the form 1 = x r 1 a n x n However, the method ma not alwas ield a second solution when r 1 = r 2, or even when r 1 r 2 is a positive integer. In the former case, experience with Euler equations suggests that the second solution should involve a logarithm. The case where r 1 r 2 is a positive integer is somewhat msterious. What form do the second solutions have in that situation, and how can the be found? The following complete version of the Frobenius Theorem clears up these questions.

2 114 THEOREM 1 The Extended Theorem and Procedure of Frobenius Suppose that the power series p(x) = p n x n and q(x) = q nx n converge on the interval x < R. Then the ODE x 2 + xp(x) + q(x) = has a regular singular point at x =, and the Extended Method of Frobenius (described below) produces two independent solutions of the ODE if the indicial roots are real. 1. Find the indicial roots r 1 and r 2 of the indicial polnomial f (r) = r 2 + ( p 1)r + q and verif that the are real; index them so that r 2 r Construct the solution 1 (x) = x r 1 a nx n, x >, a = 1 b the Method of Frobenius described in the procedure of Section The recursion formula is f (r 1 + n)a n = n 1 k= [(k + r 1) p n k + q n k ]a k, n = 1, 2, If r 1 r 2 is not an integer, construct a second independent solution 2 (x) as described b the procedure in Section If r 1 r 2, then a second independent solution has the form 2 (x) = 1 (x) ln x + x r 1 c n x n, x > 5. If r 1 r 2 is a positive integer, then a second independent solution has the form 2 (x) = α 1 (x) ln x + x r 2 (1 + d n x n ), x > To determine the coefficients c n, α, and d n, substitute the appropriate form for 2 into the ODE and use the Identit Theorem to obtain recursion relations. The power series in the formulas for 2 (x) converge in an interval centered at zero. For formulas valid for x <, replace x b x in the terms x r 1, x r 2, and ln x. 1 The recursion formula in Step 2 shows just where the difficult lies if r 1 r 2 is an integer. If r 2 = r 1 and r 2 replaces r 1 in the recursion formula, then no new solution independent of 1 results. If r 1 r 2 is the positive integer m, then the recursion formula ma not produce a second solution independent of 1 when r 2 replaces r 1 because f (r 2 + m) = f (r 1 ) =, and so there ma be no wa to determine a m. The next two examples illustrate how to proceed when r 1 r 2 is an integer. EXAMPLE 1 The standard form of this ODE is x 2 + x + 2x =, so p = 1 and q =. The indicial polnomial is f (r) = r 2. Equal Indicial Roots The ODE x = has a regular singular point at zero. The indicial roots are r 1 = r 2 =. A basic solution

3 115 set { 1, 2 } consists of a function 1 (x) given b a Frobenius series 1 = a n x n and a function 2 (x) with a logarithmic term 2 = 1 ln x + c n x n Use the Method of Power Series to find the values of a n in 1 : x n(n 1)a n x n 2 + na n x n a n x n = [n(n 1)a n + na n + 2a n 1 ]x n 1 = This leads to the recursion relation n 2 a n + 2a n 1 =, so For a solution 1 (x), set a = 1: a n = 2a n 1 = 4a n 2 n 2 n 2 (n 1) = = ( 1)n 2 n a 2 (n!) 2 1 = 1 + ( 1) n 2 n x n (n!) 2, all x (2) Turn now to the calculation of 2 for x > : 2 = 1 ln x + c nx n. Insert 2 into the ODE and rearrange terms: ( x 1 ln x x 1 x 2 ) ( + 1 ln x + ) ln x + x Collect the terms that include ln x, and rewrite the expression above: (x ) ln x = = (n 2 c n + 2c n 1 )x n 1 = (n 2 c n + 2c n 1 )x n 1 (n 2 c n + 2c n 1 )x n 1 (since x = ) 1=1 ( 1) n 2 n+1 x n 1 (n 1)!n! + (n 2 c n + 2c n 1 )x n 1 = We used the series for 1 to obtain the first summation on the line above. The recursion relation is n 2 c n + 2c n 1 + ( 1)n 2 n+1 =, n = 1, 2,... (n 1)!n!

4 116 It is hard to solve the recursion relation to find c n in terms of c. Instead, use the recursion relation to find the coefficients c 1, c 2, and c 3 in terms of c : c 1 + 2c 4 =, so c 1 = 2(2 c ) 4c 2 + 2c =, so c 2 = 3 + c 9c 3 + 2c 2 4/3 =, so c 3 = 22/27 2c /9 For simplicit, set c = and obtain c 1 = 4, c 2 = 3, and c 3 = 22/27. Then 2 = 1 ln x + c n x n = 1 (x) ln x + 4x 3x x3 +, x > (3) where 1 (x) is given b (2). For x <, replace x b x in the logarithm. For all practical purposes, an unsolved recursion relation is just as useful as a solved relation. If ou use a computer to calculate the coefficients, the unsolved form is preferable. Initialize with c and then iterate using the recursion relation. The following example illustrates the procedure when r 1 r 2 is a positive integer. EXAMPLE 2 The standard form of this ODE is x 2 x =, so p = and q =. The indicial polnomial is f (r) = r 2 r. Indicial Roots Differ b a Positive Integer The ODE x = has a regular singular point at zero. The indicial polnomial has roots r 1 = 1 and r 2 =. The equation has a Frobenius series solution of the form 1 = x a n x n = a n x n+1, a Find the a n s b the Method of Power Series with a = 1: a n = 1/((n + 1)!n!). So x n 1 = x (n + 1)!n! = x n+1 (n + 1)!n! = x x x3 + From Theorem 1, a second solution has the form 2 = α 1 ln x + d n x n, d = 1, x > (4) Insert the expression for 2 into the ODE x 2 2 =. Since x 1 1 = : ( α(x 1 1) ln x + α = α The recursion relation is x 2n + 1 (n + 1)!n! xn + ) + [n(n + 1)d n+1 d n ]x n [n(n + 1)d n+1 d n ]x n n(n + 1)d n+1 d n = α(2n + 1) (n + 1)!n!

5 117 For n =, 1, 2, the recursion relation reduces to d = α, 2d 2 d 1 = 3α 2, 6d 3 d 2 = 5α 12 Since d = 1 is alread built into Step 5, α = 1. Thus, α = d = 1, d 2 = d 1 /2 3/4, d 3 = d 1 /12 7/36 Let d 1 = (or an other convenient value) and use the recursion to determine d n, n = 2, 3,.... Use formula (4) for 2 (x) with α = 1, d = 1, d 1 =, d 2 = 3/4, d 3 = 7/36. Use recursion for d n, n 4. Then is a second solution independent of 1. 2 = 1 ln x x x3 + Although α is not zero in formula (4) of Example 2, it can be zero. for another ODE. Then, the second solution is a power series, and there is no logarithmic term. Bessel Functions of the Second Kind of Integer Order We now appl Theorem 1 to the problem of finding a second solution of Bessel s equation of order n x 2 + x + (x 2 n 2 ) = (5) where n is a nonnegative integer. We found one solution, J n (x), in the SPOTLIGHT ON BESSEL FUNCTIONS. The difference of the two roots n and n of the indicial polnomial f (r) = r 2 n 2 is an integer, so Frobenius Theorem in Section 11.4 does not appl in our search for a second solution independent of J n. There are two cases. Case 1: n =. Since r 1 = r 2 =, Theorem 1 tells us that a second solution of Bessel s equation of order zero independent of J has the form 2 = 1 ln x + c n x n (6) and we can determine the c n s using the Method of Power Series. This process requires ingenuit and patience. In practice, we take the second solution to be a certain linear combination of J (x) and the function 2 of (6). This is the Bessel function of the second kind (or Weber function) of order, Y (x): Y (x) = 2 ( γ + ln x ) J (x) 2 π 2 π k= ( 1) k h(k) ( x ) 2k (k!) 2 2

6 118 Y, Y1, Y2 Y Y 1 Y2 FIGURE 1 Graphs of Bessel functions of the second kind: Y, Y 1, Y 2. x where h(k) = 1 + 1/ /k. The number γ is Euler s constant, 2 ( 1 γ = 1 + n + ln n 1 ) = lim [h(k) ln k] n k n=2 Case 2: n > The second solution is Y n (x) = 2 ( γ + ln x ) J n (x) 1 x π 2 π 2 1 x π 2 n k= (n k 1)! ( x 2k k! 2) k= ( x ) (7) 2k 2 n n 1 ( 1) k [h(k) + h(k + n)] k!(n + k)! It is the Bessel function of the second kind of order n (or a Weber function). Although the first and third terms in the expression for Y n remain bounded near zero, the second does not, and Y n (x) as x +. See Figure 1 for the graphs of the first three functions of integer order. We describe the zeros and the oscillator character of Y n, and indeed of an nontrivial solution of Bessel s equation, in the SPOT- LIGHT ON BESSEL FUNCTIONS. The functions Y n (x) satisf the same Recursion Formulas given in Theorem 1 in the SPOTLIGHT ON BESSEL FUNCTIONS for J n (x) and have the same oscillator behavior and structure of the set of zeros as J n. The most notable visual difference between the 2 Euler s constant γ is It is not known whether γ is rational, algebraic, or transcendental.

7 e t/1 = = J (1e t/2 ) + e t =, () = 5, () = 1 FIGURE 2 Spring stretches to a finite length. t FIGURE 3 A rapidl aging spring stretches. t graphs of J n (x) and Y n (x) is the behavior at the singularit x = : J n () is 1 (if n = ) or (if n > ), but lim x + Y n (x) = for ever nonnegative n. The unbounded behavior of Y n (x) as x + plas a role in the problem of the aging spring. The Aging Spring and Bessel Functions m See Example for a power series solution of an aging spring IVP. We introduced the ODE of the aging spring model (t) + k m e εt (t) = (8) in Section Here is a new measure of time, s = αe βt, where α and β are constants to be chosen later. (The exponential form of the elastic coefficient ke εt in (8) suggests an exponential measure of time). Let s see what happens when we change from t-variables to s-variables in equation (8). Use the formulas ds/dt = βαe βt = βs, (s/α) ε/β = e εt, and the chain rule: d dt = d ds ds dt = d ds βs d 2 dt = d ( ) d = d 2 dt dt ds ( ) d ds dt dt = d ( ) d ds ds ds βs dt = d2 ds 2 (βs)2 + d ds β2 s Substitute into ODE (8): d 2 dt + k 2 m e εt = β 2 s 2 d2 ds + 2 β2 s d ds + k ( s ) ε/β = m α Divide the above ODE in and s b β 2.Then make the strategic choices for α and β: β = ε 2, α = 2 ε k m

8 e t/1 = = J (1e t/2 ) + Y (1e t/2 ) +.25e t/1 = = J (1e t/2 ) Y (1e t/2 ) FIGURE 4 A slowl aging spring stretches. t FIGURE 5 A slowl aging spring compresses. t and obtain the Bessel equation of order : The general solution of (9) is s 2 d2 ds 2 + s d ds + s2 = (9) = C 1 J (s) + C 2 Y (s) (1) where C 1 and C 2 are arbitrar constants. In terms of the time variable t, = C 1 J ( αe εt/2 ) + C 2 Y ( αe εt/2 ), where α = 2 ε Formula (11) is the general solution of the ODE + (k/m)e εt =. k m (11) EXAMPLE 3 Recall that Y (s) as s +. The Aging Spring Stretches As time t increases, the argument αe εt/2 of J and of Y in (11) tends to zero, so the behavior of solutions of Bessel s equation of order zero near the regular singularit at the origin models the asmptotic behavior of the aging spring. There are two distinct possibilities. First, if C 2 =, then as t +, the displacement from the rest position (modeled b (11)) asmptoticall approaches the value C 1 since J () = 1 (see Figure 2 where C 1 = 1). It is more likel, however, that C 2 is nonzero and that the displacement tends to C 1 J () + C 2 Y (). The limiting value is depending, respectivel, on whether C 2 is positive or negative. If C 2 is positive, then the spring oscillates with increasing amplitude and eventuall begins to stretch without bound. In realit, the spring stretches beond its elastic limit and snaps, or else behaves in a completel inelastic manner that ODE (8) cannot model. Figures 2 5 illustrate some of the possibilities. In Figure 2, k/m =.25, ε = 1/1, α = 1, C 1 = 1, and C 2 = in formula (11). In Figure 3, k/m = 1, ε = 1, α = 2,

9 121 C 1 and C 2 are positive. In Figure 4, k/m =.25, ε = 1/1, α = 1 and C 2 = C 1 = 1, but in Figure 5 we reverse the values of C 1 and C 2. Figure 5 illustrates the case C 2 <, which corresponds to the unreal situation of approach to infinite compression ( + ). So, model ODE (8) is onl valid over a restricted domain of compression and extension. PROBLEMS Extended Theorem of Frobenius. Check that zero is a regular singular point of each equation and find a basic solution set on the interval (, ). [Hint: in Problems 1 3, the solution 1 is a polnomial. Then use the Wronskian Reduction Method of Problem 14 in Section 3.7 to find a second independent solution.] 1. x + (1 + x) + = 2. x 2 + x(x 1) + (1 x) = 3. x x + = 4. x x 2 + = Equations Reducible to Bessel s Equation. Several second-order, variable coefficient linear ODEs are equivalent to Bessel s equation. Here are some. 5. Suppose that w(s) is the general solution of the Bessel equation of order n, s 2 w + sw + (s 2 n 2 )w =. Show that (x) = e ax w(bx) is the general solution of the equation x 2 + x(1 2ax) + [(a 2 + b 2 )x 2 ax n 2 ] = where a and b are real constants, b. 6. Find the general solution of x 2 + x(1 2x) + (2x 2 x 1) =. [Hint: see Problem 5.] 7. Suppose that (x) is a solution of Bessel s equation of order n. Show that w(z) = z c (az b ) is a solution of z 2 w + (2c + 1)zw + [a 2 b 2 z 2b + (c 2 n 2 b 2 )]w = 8. Find the general solution of + x 4 =. [Hint: see Problem 7.] Modeling Problems: Aging Springs. 9. Suppose that = J (6e t ) is the solution of a model of an aging spring. Plot as a function of t. What happens as t +? 1. Repeat Problem 9 with J (6e t ) Y (6e t ). 11. Summarize what has been shown in this chapter about the problem of the aging spring. Critique the model s validit. Propose and analze other models. In particular, consider the model + (b 2 + a 2 e εt ) = where a, b, and ε are constants. Critique this model. Plot solutions for various values of a, b, ε, and interpret these solutions in terms of motions of the aging spring.