Series solutions of second order linear differential equations

Transcription

1 Series solutions of second order linear differential equations We start with Definition 1. A function f of a complex variable z is called analytic at z = z 0 if there exists a convergent Taylor series for f at z = z 0 with positive radius of convergence: f(z) = for z z 0 < R with R > 0. a n (z z 0 ) n with a n = f (n) (z 0 ), n = 0, 1, 2,... n! Now we consider second order linear differential equations of the form y + p(z)y + q(z)y = 0 Then we have the following definition: or d 2 y dz 2 + p(z)dy + q(z)y = 0. (1) dz Definition 2. A point z 0 C is called a regular point of the differential equation (1) if both p(z) q(z) are analytic in z 0. Otherwise z 0 is called a singular point of (1). For singular points we distinguish between two cases: Definition 3. If z 0 C is a singular point of (1) both (z z 0 )p(z) (z z 0 ) 2 q(z) are analytic in z 0, then z 0 is called a regular singular point of (1). Otherwise z 0 is called an irregular singular point of (1). For regular points we have: Theorem 1. If z 0 C is a regular point of the differential equation (1), then p(z) q(z) are both analytic at z = z 0. This implies that there exist positive constants R 1 R 2 such that p(z) = p n (z z 0 ) n, z z 0 < R 1 q(z) = q n (z z 0 ) n, z z 0 < R 2. Then there exist solutions y of (1) which are analytic at z = z 0. In fact, if R = min(r 1, R 2 ), then the solution y can be written as y(z) = c n (z z 0 ) n for z z 0 < R. Moreover, c 0 = y(z 0 ) c 1 = y (z 0 ) can be chosen arbitrary. Then y(z) = c 0 y 1 (z) + c 1 y 2 (z) with {y 1 (z), y 2 (z)} linearly independent. For regular singular points we have Frobenius method: 1

2 Theorem 2. If z 0 C is a regular singular point of the differential equation (1), then (z z 0 )p(z) (z z 0 ) 2 q(z) are both analytic at z = z 0. This implies that there exist positive constants R 1 R 2 such that (z z 0 )p(z) = (z z 0 ) 2 q(z) = p n (z z 0 ) n, z z 0 < R 1 (2) q n (z z 0 ) n, z z 0 < R 2. (3) Then there exist solutions y of (1) which can be written in the form y(z) = (z z 0 ) r In fact, if R = min(r 1, R 2 ), then we have y(z) = (z z 0 ) r Here r C denotes a solution of the indicial equation c n (z z 0 ) n, c 0 0. (4) c n (z z 0 ) n for 0 < z z 0 < R. r(r 1) + p 0 r + q 0 = 0. (5) Moreover, if r 1, r 2 C denote the solutions of the indicial equation (5) with Re r 1 Re r 2, then we have: 1. If r 1 r 2 / {0, 1, 2,...}, then we have y 1 (z) = (z z 0 ) r 1 y 2 (z) = (z z 0 ) r 2 a n (z z 0 ) n, a 0 0 b n (z z 0 ) n, b 0 0 are two linearly independent solutions of the differential equation (1). 2. If r 1 r 2 = 0, then we have y 1 (z) = (z z 0 ) r 1 a n (z z 0 ) n, a 0 0 y 2 (z) = y 1 (z) ln(z z 0 ) + (z z 0 ) r 2 b n (z z 0 ) n are two linearly independent solutions of the differential equation (1). 2

3 3. If r 1 r 2 {1, 2, 3,...}, then we have y 1 (z) = (z z 0 ) r 1 a n (z z 0 ) n, a 0 0 y 2 (z) = Ay 1 (z) ln(z z 0 ) + (z z 0 ) r 2 b n (z z 0 ) n, b 0 0 are two linearly independent solutions of the differential equation (1), where A is a constant that might be zero or not. Some remarks: A series of the form (4) is called a generalized power series. Only for r {0, 1, 2,...} this generalized power series reduces to a normal power series. The solutions r 1 r 2 of the indicial equation (5) are called indices. In the second case it is always possible to choose b 0 = 0 since r 1 = r 2. In that case the second solution always has a logarithmic singularity at z = z 0. In the third case the second solution might have a logarithmic singularity at z = z 0 (in the case that A 0) or not (in the case that A = 0). The proof of this theorem due to Frobenius is constructive. equation (1) by (z z 0 ) 2 to obtain (z z 0 ) 2 y + (z z 0 ) 2 p(z)y + (z z 0 ) 2 q(z)y = 0. We multiply the differential Since (z z 0 )p(z) (z z 0 ) 2 q(z) are analytic at z = z 0 we use (2) (3) to find ( ) ( ) (z z 0 ) 2 y + (z z 0 ) p n (z z 0 ) n y + q n (z z 0 ) n y = 0. Now we set Then we have y(z) = (z z 0 ) r y (z) = y (z) = c n (z z 0 ) n = c n (z z 0 ) n+r, c 0 0. (n + r)c n (z z 0 ) n+r 1 (n + r)(n + r 1)c n (z z 0 ) n+r 2. Hence we have ( ) ( ) (n + r)(n + r 1)c n (z z 0 ) n+r + p n (z z 0 ) n (n + r)c n (z z 0 ) n+r ( ) ( ) + q n (z z 0 ) n c n (z z 0 ) n+r = 0. 3

4 This can only be true if the coefficients of all powers of z z 0 are equal to zero. The coefficient of the lowest power of z z 0, id est (z z 0 ) r, is {r(r 1) + p 0 r + q 0 } c 0. Since c 0 0 this leads to the indicial equation (5). More details are left out here. Frobenius method is based on the theory of Euler s differential equation: x 2 y + αxy + βy = 0, α, β R. (6) By using the change of variables x = e t or t = ln x we find that This implies that dy dx = dy dt dt dx = 1 x dy dt d 2 y dx 2 = d ( 1 dx x dy ) = 1 dt x d2 y dt 2 dt dx 1 x 2 dy dt = 1 x 2 d2 y dt 2 1 x 2 dy dt. x dy dx = dy x 2 d2 y dt dx 2 = d2 y dt 2 dy dt. Hence, with y(x) = Y (t) Euler s differential equation (6) turns into Y Y + αy + βy = 0 Y + (α 1)Y + βy = 0, which is a second order linear differential equation with constant coefficients. The characteristic equation for this differential equation is r 2 + (α 1)r + β = 0 r(r 1) + αr + β = 0. (7) If r 1 r 2 denote the roots of this characteristic equation, then we have 1. If r 1, r 2 R r 1 r 2, then Y 1 (t) = e r 1t Y 2 (t) = e r 2t are two linearly independent solutions. This leads to y 1 (x) = x r 1 y 2 (x) = x r 2 being two linearly independent solutions of Euler s differential equation (6). 2. If r 1, r 2 R r 1 = r 2, then Y 1 (t) = e r 1t Y 2 (t) = te r 1t are two linearly independent solutions. This leads to y 1 (x) = x r 1 y 2 (x) = x r 2 ln x being two linearly independent solutions of Euler s differential equation (6). 3. If r 1, r 2 / R, say r 1,2 = λ ± µi with λ, µ R µ 0, then Y 1 (t) = e λt cos µt Y 2 (t) = e λt sin µt are two linearly independent solutions. This leads to y 1 (x) = x λ cos(µ ln x) y 2 (x) = x λ sin(µ ln x) being two linearly independent solutions of Euler s differential equation (6). This can also be obtained by looking for solutions of the form y(x) = x r of Euler s differential equation (6) directly. Then we have y (x) = rx r 1 y (x) = r(r 1)x r 2 which leads to r(r 1)x r + αrx r + βx r = 0 {r(r 1) + αr + β} x r = 0. This leads to the characteristic equation (7). 4

5 Examples 1. The hypergeometric differential equation is given by z(1 z)y + {c (a + b + 1)z} y aby = 0, a, b, c C. (8) It is clear that z = 0 z = 1 are the only singular points of this differential equation. Since c (a + b + 1)z zp(z) = z 2 q(z) = abz 1 z 1 z are both analytic at z = 0, we conclude that z = 0 is a regular singular point of (8). Note that p 0 = c q 0 = 0 which leads to the indicial equation We also have that r(r 1) + cr = 0 r(r 1 + c) = 0. c (a + b + 1)z (z 1)p(z) = z (z 1) 2 q(z) = ab(z 1) z are both analytic at z = 1, which implies that z = 1 is a regular singular point of (8) too. In that case we have p 0 = a + b c + 1 q 0 = 0 which leads to the indicial equation r(r 1) + (a + b c + 1)r = 0 r(r + a + b c) = 0. This implies that there exists solutions of the form y 1 (z) = a n z n y 2 (z) = z 1 c b n z n for the hypergeometric differential equation (8). These two solutions are linearly independent at least for c / Z. We also have solutions of the form y 3 (z) = c n (z 1) n y 4 (z) = (z 1) c a b d n (z 1) n for the hypergeometric differential equation (8). These two solutions are linearly independent at least for c a b / Z. Note that we have y(z) = a n z n = y (z) = na n z n 1 y (z) = Substitution into the differential equation (8) leads to n(n 1)a n z n 1 n(n 1)a n z n n(n 1)a n z n 2. + c na n z n 1 (a + b + 1) na n z n ab a n z n = 0. 5

6 Hence [{n(n + 1) + c(n + 1)} a n+1 {n(n 1) + (a + b + 1)n + ab} a n ] z n = 0. This leads to the recurrence relation (n + 1)(n + c)a n+1 = (n + a)(n + b)a n, n = 0, 1, 2,... for the coefficients {a n }. The solution can be written as Hence Similarly, for we have y (z) = y(z) = a n = (a) n(b) n (c) n n! a 0, n = 0, 1, 2,.... a n z n = a 0 y(z) = z 1 c (a) n (b) n (c) n b n z n = (n + 1 c)b n z n c y (z) = z n ( ) a, b n! = a 0 2F 1 c ; z. b n z n+1 c (n + 1 c)(n c)b n z n 1 c. Substitution into the differential equation (8) leads to Hence (n + 1 c)(n c)b n z n c (n + 1 c)(n c)b n z n+1 c + c (n + 1 c)b n z n c (a + b + 1) (n + 1 c)b n z n+1 c ab b n z n+1 c = 0. [{(n + 2 c)(n + 1 c) + c(n + 2 c)} b n+1 {(n + 1 c)(n c) + (a + b + 1)(n + 1 c) + ab} b n ] z n+1 c = 0. This leads to the recurrence relation (n + 1)(n + 2 c)b n+1 = (n + a + 1 c)(n + b + 1 c)b n, n = 0, 1, 2,... for the coefficients {b n }. The solution can be written as b n = (a + 1 c) n(b + 1 c) n b 0, n = 0, 1, 2,.... (2 c) n n! 6

7 Hence y(z) = z 1 c b n z n = b 0 z 1 c = b 0 z 1 c 2F 1 ( a + 1 c, b + 1 c 2 c (a + 1 c) n (b + 1 c) n z n (2 c) n n! ) ; z. 2. The confluent hypergeometric differential equation is given by zy + (c z)y ay = 0, a, c C. (9) Note that z = 0 is the only singular point of this differential equation. Further we have in this case zp(z) = c z z 2 q(z) = az, which are both analytic at z = 0. Note that p 0 = c q 0 = 0, which implies that z = 0 is a regular singular point of the differential equation (9) with indicial equation r(r 1) + cr = 0 r(r 1 + c) = 0. Hence, the indices are r 1 = 0 r 2 = 1 c. This implies that for c / Z we have two linearly independent solutions of the form If we substitute y(z) = y 1 (z) = a n z n y 2 (z) = z 1 c b n z n. a n z n, y (z) = na n z n 1 y (z) = into the differential equation (9) we obtain or equivalently n(n 1)a n z n 2 n(n 1)a n z n 1 + c na n z n 1 na n z n a a n z n = 0 [{(n + 1)n + c(n + 1)} a n+1 (n + a)a n ] z n = 0. This leads to the recurrence relation (n + 1)(n + c)a n+1 = (n + a)a n, n = 0, 1, 2,... with solution a n = (a) n (c) n n! a 0, n = 0, 1, 2,.... 7

8 Hence we have a solution of the form If we substitute y (z) = y(z) = a n z n = a 0 y(z) = z 1 c (a) n z n ( a ) (c) n n! = a 0 1F 1 c ; z. b n z n = b n z n+1 c, (n + 1 c)b n z n c y (z) = into the differential equation (9) we obtain or equivalently (n + 1 c)(n c)b n z n 1 c (n + 1 c)(n c)b n z n c + c (n + 1 c)b n z n c (n + 1 c)b n z n+1 c a b n z n+1 c = 0 [{(n + 2 c)(n + 1 c) + c(n + 2 c)} b n+1 (n + a + 1 c)b n ] z n+1 c = 0. This leads to the recurrence relation with solution (n + 2 c)(n + 1)b n+1 = (n + a + 1 c)b n, n = 0, 1, 2,... Hence we have a solution of the form y(z) = z 1 c b n = (a + 1 c) n (2 c) n n! b 0, n = 0, 1, 2,.... b n z n = b 0 z 1 c 3. The Bessel differential equation is given by (a + 1 c) n z n ( ) a + 1 c (2 c) n n! = b 0 z 1 c 1F 1 ; z. 2 c z 2 y + zy + (z 2 ν 2 )y = 0, ν 0. (10) In this case z = 0 is the only singular point of the differential equation. Note that we have zp(z) = 1 z 2 q(z) = z 2 ν 2, which are both analytic at z = 0. This implies, since p 0 = 1 q 0 = ν 2, that z = 0 is a regular singular point of the Bessel differential equation (10) with indicial equation r(r 1) + r ν 2 r 2 ν 2 = 0. 8

9 This implies that the indices are r 1 = ν r 2 = ν. Hence, for 2ν / {0, 1, 2,...} we have two linearly independent solutions of the form y 1 (z) = z ν a n z n y 2 (z) = z ν b n z n. For ν = 0 we have two linearly independent solutions of the form y 1 (z) = a n z n y 2 (z) = y 1 (z) ln z + b n z n. Finally, for 2ν {1, 2, 3,...} we have two linearly independent solutions of the form y 1 (z) = z ν where the constant A might be zero or not. a n z n y 2 (z) = Ay 1 (z) ln z + z ν b n z n, For instance, in the case ν 2 = 1/4 the indices are r 1 = 1/2 r 2 = 1/2. In that case we have two linearly independent solutions y 1 (z) = z 1/2 ( 1) n z 2n (2n + 1)! = sin z z y 2 (z) = z 1/2 ( 1) n z 2n (2n)! = cos z z. This can easily be seen by setting y(z) = z 1/2 u(z) into the differential equation (10) with ν 2 = 1/4. Then we have y (z) = z 1/2 u (z) 1 2 z 3/2 u(z) y (z) = z 1/2 u (z) z 3/2 u (z) z 5/2 u(z), which leads to z 3/2 u (z) z 1/2 u (z)+ 3 4 z 1/2 u(z)+z 1/2 u (z) 1 2 z 1/2 u(z)+z 3/2 u(z) 1 4 z 1/2 u(z) = 0 or z 3/2 [ u (z) + u(z) ] = 0 = u (z) + u(z) = 0. This leads to the linearly independent solutions u 1 (z) = sin z u 2 (z) = cos z. In the case ν = 0 we substitute y(z) = a n z n, y (z) = na n z n y (z) = into the differential equation (10) with ν = 0 to find which leads to a 1 z + n(n 1)a n z n 2 n(n 1)a n z n + na n z n + a n z n+2 = 0, [{(n + 2)(n + 1) + (n + 2)} a n+2 + a n ] z n+2 = 0. 9

10 This implies that The solution can be written as a 1 = 0 (n + 2) 2 a n+2 = a n, n = 0, 1, 2,.... This leads to the solution y(z) = a n z n = a 0 a 2k+1 = 0 a 2k = ( 1)k a 0 2 2k, k = 0, 1, 2,.... (k!) 2 k=0 ( 1) k z 2k ( ) 2 2k (k!) 2 = a 0 0F 1 1 ; z2 = a 0 J 0 (z), 4 where J 0 (z) denotes the Bessel function of the first kind of order zero. In the general case with ν > 0 we substitute y(z) = z ν a n z n = y (z) = a n z n+ν, y (z) = (n + ν)a n z n+ν 1 (n + ν)(n + ν 1)a n z n+ν 2 into the differential equation (10) to obtain [ ν(ν 1) + ν ν 2 ] a 0 z ν + [ (ν + 1)ν + (ν + 1) ν 2] a 1 z ν+1 + [{ (n + ν + 2)(n + ν + 1) + (n + ν + 2) ν 2 } ] a n+2 + a n z n+ν+2 = 0. This leads to the recurrence relation (2ν + 1)a 1 = 0 [ (n + ν + 2) 2 ν 2] a n+2 = a n, n = 0, 1, 2,... or equivalently, since ν > 0, a 1 = 0 (n + 2)(n + 2ν + 2)a n+2 = a n, n = 0, 1, 2,.... The solution can be written as This leads to the solution y(z) = z ν a 2k+1 = 0 a 2k = a n z n = a 0 z ν k=0 ( 1) k a 0 2 2k, k = 0, 1, 2,.... (ν + 1) k k! ( 1) k z 2k ( ) 2 2k (ν + 1) k k! = a 0 z ν 0F 1 ν + 1 ; z2. 4 The Bessel function J ν (z) of the first kind of order ν is defined by ( ) J ν (z) = (z/2)ν Γ(ν + 1) 0 F 1 ν + 1 ; z