6.3 Linear System with Regular Singularities

6.3 Linear System with Regular Singularities A system of the form dx̄ 1 = A(t)x̄ dt t − t0 (6.3.1) where A(t) is analytic at t0 , is said to have a regular singularity at t0 . As before, we can assume that t0 = 0 and A(t) has series representation of the form A(t) = ∞ X A k tk (6.3.2) k=0 in the interval of the convergence |t| < r. Eq. (6.3.1) then reduces to the equation of the form dx̄ 1 = dt t ∞ X k=0 Ak t k ! x̄ (6.3.3) We prove the following theorem, giving the existence of a series solution of the equation given by Eq. (6.3.3) Theorem 6.3.1 The differential equation given by Eq. (6.3.3) has a series ∞ X solution of the form x̄(t) = tµ c̄k tk in the interval |t| < r, provided µ is an k=0 eigenvalue of A0 and no other eigenvalue of the form µ + n exists for A0 , where n is a positive integer. Proof : We introduce a new dependent variable ȳ(t) = tµ x̄(t). This gives dȳ dt = = 1 [A(t) − µI] ȳ(t) t" # ∞ 1 X k Ak t − µI ȳ(t) t k=0 Assuming that ȳ(t) = ∞ X k=0 c̄k tk , we get 6.3. LINEAR SYSTEM WITH REGULAR SINGULARITIES ∞ X k=1 c̄k ktk = ∞ X k=0 " k X Ak−l c̄l l=0 ! # − µc̄k tk . This gives us the recurrence relations (A0 − µI)c̄0 = 0 (A0 − (µ + I)) c̄1 = −A1 c̄0 .. . n−1 X (A0 − (µ + n)I)c̄n = − An−l c̄l (6.3.4) l=0 .. . Since |(A0 − (µ + n)I)c̄n | 6= 0 for all n ≥ 1, the above relations iteratively define ∞ X c̄0 , c̄1 , · · · , c̄n , · · · We shall now show that the series x(t) = tµ c̄k tk converges. k=0 As in Section 6.2, for some positive numbers M , and p < r we have kAk kpk ≤ M, k = 0, 1, 2, · · · (6.3.5) We first observe the following (i) |A0 − (µ + k)I| 6= 0 for all positive integer k (µ + k) I k→∞ k (A0 − µI) k = lim − I k→∞ k k =n (ii) lim A0 − In view of property (ii), it follows that there exists a positive quantity  > 0 such that kA0 − (µ + k)Ik > , k = 1, 2, 3, · · · k So, the recurrence relation given by Eq. (6.3.4) provides (k + 1)c̄k+1 = [A0 − (µ + k + 1)I] −1 (k + 1) k X Ak+1−l c̄l l=0 Using Eq. (6.3.5) and (ii) we get k (k + 1) kc̄k+1 k ≤ M (k + 1) X kc̄l k  pk+1−l l=0 Equivalently, k (k + 1) kc̄k+1 k pk+1 ≤ X M (k + 1) kc̄l kpl  l=0 Proceeding as in Theorem 6.2.1, we have the relation
M

M M   +1 ···  +k −1 kc̄k k ≤ kc̄0 k k!pk for all k ≥ 1. Applying this estimate to the series expansion for x̄(t), we get kx̄(t)k ≤ tµ ∞ X c̄k tk k=0 µ ≤ |t| kc̄0 k = ∞ X M  k=0
M 
+1 ··· k! M  +k−1
 |t| p k |t|µ kc̄0 k  M |t|  1− p in the interval of uniform convergence |t| ≤ p < r. This proves the theorem. Example 6.3.1 (Bessel’s Equation) d2 x dx +t + (t2 − µ2 )x(t) = 0 2 dt dt d2 x 1 dx µ2 ⇐⇒ 2 + + (1 − 2 )x(t) = 0 dt t dt t We use the substitution x1 = x(t), x2 = tẋ(t) to get t2 dx1 dt = 1 x2 t dx2 dt = dx1 d 2 x1 +t 2 dt dt = dx1 dx1 µ2 − − (t − )x1 (t) dt dt t 1 = − (t2 − µ2 )x1 (t) t This reduces Eq. (6.3.6) to dx̄ dt = =  1 t 0 1 µ2 − t 2 0  1 A0 + A2 t2 x̄ t   x̄ (6.3.6) 6.3. LINEAR SYSTEM WITH REGULAR SINGULARITIES     0 1 0 0 where A0 = , A2 = . It is clear that A is analytic in µ2 0 −1 0 (−∞, ∞) and A0 has eigenvalues ±µ. Case 1: µ is not an integer P∞ Then Eq. (6.3.6) has solution of the form x(t) = tµ k=0 c̄k tk in (−∞, ∞). For the eigenvalue µ > 0, plugging the representation x(t) = ∞ X cm xm+µ , m=0 d2 x dt2 = ∞ X m=0 ∞ X dx = cm (m + µ)tm+µ−1 dt m=0 cm (m + µ)(m + µ − 1)tm+µ−2 in Eq. (6.3.6) we get the recurrence relation c2m c2m+1 = − 1 c2m−2 , m = 1, 2 · · · 22 m(µ + m) = 0, m = 0, 1, 2, · · · This gives c2 c4 c0 22 (µ + 1) c2 c0 = − 2 = 4 2 2(µ + 2) 2 2!(µ + 1)(µ + 2) = − .. . c2m c2m+1 (−1)m c0 , + 1)(µ + 2) · · · (µ + m) m = 1, 2, · · · = 0, m = 0, 1, 2, 3, · · · = 22m m!(µ (6.3.7) Now define the gamma function Γ(µ) as Z ∞ Γ(µ) = e−t tµ−1 dt, µ > 0 0 which is a convergent integral. Γ(µ) satisfies the property that Γ(µ + 1) = µΓ(µ) 1 and Γ(n + 1) = n! If we put c0 = µ Γ(µ + 1), then c2m is given by 2 c2m = (−1)m 22m+µ m!(µ + 1)(µ + 2) · · · (µ + m)Γ(µ + 1) = (−1)m 22m+µ m!Γ(µ + m + 1) This gives us the first solution Jµ (t), called Bessl’s function of order µ of the differential equation given by Eq. (6.3.6), as Jµ (t) = tµ ∞ X (−1)m t2m 22m+µ m!Γ(µ + m + 1) m=0 Extending gamma function for µ < 0, we get the representation for the second solution J−µ (t) as J−µ (t) = t−µ ∞ X m=0 (−1)m t2m + m + 1) 22m−µ m!Γ(−µ Jµ (t) and Jµ (t) are linearly independent. Case 2: µ = n integer We have Jn (t) = tn One can easily derive that ∞ X (−1)m t2m 22m+n m!(n + m) m=0 J−n (t) = (−1)n Jn (t) This gives J0 (t) = ∞ X (−1)m t2m 22m (m!)2 m=0 =1− J1 (t) = ∞ X t2 t4 t6 + − +··· 22 (1!)2 24 (2!)2 26 (3!)2 (−1)m t2m+1 22m+1 (k!)(k + 1)! m=0 =t− t3 t5 t7 + − +··· 23 (1!)(2!) 25 (2!)(3!) 27 (3!)(4!) J0 looks similar to cosine function and J1 looks similar to sine function. The zeros of these functions are not completely regularly spaced and also oscillations are damped as we see in the following graphics. 6.3. LINEAR SYSTEM WITH REGULAR SINGULARITIES 1 0.8 J0 0.6 0.4 J1 0.2 5 10 15 20 t -0.2 -0.4 Figure 6.3.1: Sketch of J0 and J1 Example 6.3.2 (Bessel’s equation of order zero) Consider the equation x d2 x dx + +x =0 dt2 dt This is equivalent to the first order system of the form   dx̄ 1 0 1 =− x̄ −t2 0 dt t where x1 = x, x2 = t (6.3.8) dx . dt As we have seen before, a solution of this system is given by x1 (t) x2 (t) = = ∞ X (−1)k t2k k=0 ∞ X k=0 4k (k!)2 (−1)k t2k 4k (k!)2 (6.3.9) (6.3.10) To get another solution ȳ(t) linearly independent of x̄(t) and satisfying Eq. (6.3.8), we set   1 x1 . ȳ(t) = ψ(t)z̄(t) where ψ(t) =  0 x2 This gives dȳ dφ dz̄ = z̄(t) + ψ(t) = A(t)ψ(t)x̄(t) dt dt dt That is      1 1 1 x1 x2 0 x 2    t t  dz =    z̄ +    z̄(t) dt 0 x 2 0 −tx1 −t −tx1  Solving for 0 dz we get dt  and hence  1 x1 0 x2   dz =  dt  dz  =  dt So, we need to solve dz1 dt dz2 dt  t 0 −t 0 x1 x2 0 t x2 0 − 0   z̄(t)    z̄(t)  x1 x˙2 z1 = − z1 x2 x2 t = − z1 x2 = t (6.3.11a) (6.3.11b) Integrating Eq. (6.3.11(a)) - Eq. (6.3.11(b)) we get Z c1 t z1 = , z2 = −c1 dt 2 x2 x2 (t) Hence ȳ(t) = ψ(t)z̄(t) is given by Z  c tdt 1 − c 1 x1  x2 x22 (t)  ȳ(t) =   R t −c1 x2 dt x22 (t) That is y1 (t)       Z  c1 −x˙2 /x1 − c 1 x1 dt x2 x22 Z x˙2 = −c1 x1 dt x2 x21 Z 1 = −c1 x1 dt tx21 = 6.3. LINEAR SYSTEM WITH REGULAR SINGULARITIES y2 (t) = −c1 x2 Z t dt x22 Using the represntations of Eq. (6.3.9) - Eq. (6.3.10), we get x1 (t) x2 (t) t2 +··· 4 t2 t4 = − + +··· 2 16 = 1− This gives x1 Z dt tx21 = x1 Z 1+ t2 2 +··· dt t t2 = x1 lnt + x1 ( + · · · ) 4 and x2 Z  Z t 1+ t2 4 +··· t dt = 4x2 x22 t4 −2x2 = + x2 lnt + · · · t2  dt Hence, it follows that the second solution ȳ(t), linearly independent of x(t), is given by   t2 y(t) = x(t)lnt + x(t) 1 + + · · · 4 To be more precise, one can show that ∞ X (−1)m hm t2m y(t) = J0 (t)lnt + 22m (m!)2 m=1 where hm = 1 + 1 2 +···+ 1 m The function Y0 (t) defined as
. Y0 (t) = ay(t) + bJ0 (t), a = 2 , b = r − ln2 π
1 where r is the Euler’s constant limn→∞ 1 + 12 + · · · + m − lnn is called the Bessel’s function of the second kind (order zero). Hence, it is given by " #   X ∞ 2 t (−1)m hm t2m Y0 (t) = J0 (t) ln + r + π 2 22m (m!)2 m=1