Example 23.2.4 Consider a series expansion about the origin of the equation w  + 1 − z z w  − 1 z 2 w = 0. The indicial equation is α 2 − 1 = 0 α = ±1. Substituting a Frobenius series into the differential equation, z 2 ∞  n=0 (n + α)(n + α − 1)a n z n−2 + (z −z 2 ) ∞  n=0 (n + α)a n z n−1 − ∞  n=0 a n z n = 0 ∞  n=0 (n + α)(n + α − 1)a n z n + ∞  n=0 (n + α)a n z n − ∞  n=1 (n + α − 1)a n−1 z n − ∞  n=0 a n z n = 0  α(α − 1) + α − 1  a 0 + ∞  n=1  n + α)(n + α − 1)a n + (n + α − 1)a n − (n + α − 1)a n−1  z n = 0. Equating powers of z to zero, a n (α) = a n−1 (α) n + α + 1 . We know that the first solution has the form w 1 = z ∞  n=0 a n z n . Setting α = 1 in the reccurence formula, a n = a n−1 n + 2 = 2a 0 (n + 2)! . 1214 Thus the first solution is w 1 = z ∞  n=0 2a 0 (n + 2)! z n = 2a 0 1 z ∞  n=0 z n+2 (n + 2)! = 2a 0 z  ∞  n=0 z n n! − 1 − z  = 2a 0 z ( e z −1 − z). Now to fin d the sec ond solution. Setting α = −1 in the reccurence formula, a n = a n−1 n = a 0 n! . We see that in this case there is no trouble i n defining a 2 (α 2 ). The second solution is w 2 = a 0 z ∞  n=0 z n n! = a 0 z e z . Thus we see that the general solution is w = c 1 z ( e z −1 − z) + c 2 z e z w = d 1 z e z +d 2  1 + 1 z  . 1215 23.3 Irregular Singular Points If a point z 0 of a differential equation is not ordinary or regular singular, then it is an irregular singular point. At least one of the solutions at an irregular singular point will not be of the Frobenius form. We will examine how to obtain series expansions about an irregular singular point in the chapter on asymptotic expansions. 23.4 The Point at Infinity If we want to determine the behavior of a function f(z) at infinity, we can make the transformation ζ = 1/z and examine the point ζ = 0. Example 23.4.1 Consider the behavior of f(z) = sin z at infinity. This is the same as considering the point ζ = 0 of sin(1/ζ), which has the series expansion sin  1 ζ  = ∞  n=0 (−1) n (2n + 1)!ζ 2n+1 . Thus we see that the point ζ = 0 is an esse ntial singularity of sin(1/ζ). Hence sin z has an essential singularity at z = ∞. Example 23.4.2 Consider the behavior at infinity of z e 1/z . We make the transformation ζ = 1/z. 1 ζ e ζ = 1 ζ ∞  n=0 ζ n n! Thus z e 1/z has a pole of order 1 at infi nity. 1216 In order to classify the point at infinity of a differential equation in w(z), we apply the transformation ζ = 1/z, u(ζ) = w(z). We write the derivatives with respect to z in terms of ζ. z = 1 ζ dz = − 1 ζ 2 dζ d dz = −ζ 2 d dζ d 2 dz 2 = −ζ 2 d dζ  −ζ 2 d dζ  = ζ 4 d 2 dζ 2 + 2ζ 3 d dζ Now we apply the transformation to the differential equation. w  + p(z)w  + q(z)w = 0 ζ 4 u  + 2ζ 3 u  + p(1/ζ)(−ζ 2 )u  + q(1/ζ)u = 0 u  +  2 ζ − p(1/ζ) ζ 2  u  + q(1/ζ) ζ 4 u = 0 Example 23.4.3 Classify the singular points of the differential equation w  + 1 z w  + 2w = 0. There is a regular singular point at z = 0. To examine the point at infinity we make the transformation ζ = 1/z, u(ζ) = w(z). u  +  2 ζ − 1 ζ  u  + 2 ζ 4 u = 0 u  + 1 ζ u  + 2 ζ 4 u = 0 1217 Thus we see that the differential equation for w(z) has an irregular singular point at infinity. 1218 23.5 Exercises Exercise 23.1 (mathematica/ode/series/series.nb) f(x) satisfies the Hermite eq uation d 2 f dx 2 − 2x df dx + 2λf = 0. Construct two linearly independent solutions of the equation as Taylor series about x = 0. For what values of x do the series converge? Show that for certain values of λ, called eigenvalues, one of the solutions is a polynomial, called an eigenfunction. Calculate the first four eigenfunctions H 0 (x), H 1 (x), H 2 (x), H 3 (x), ordered by degree. Hint, Solution Exercise 23.2 Consider the Legendre equation (1 − x 2 )y  − 2xy  + α(α + 1)y = 0. 1. Find two linearly independent solutions in the form of power series about x = 0. 2. Compute the radius of convergence of the series. Explain why it is possible to predict the radius of convergence without actually deriving the series. 3. Show that if α = 2n, with n an integer and n ≥ 0, the series for one of the solutions reduces to an even polynomial of degree 2n. 4. Show that if α = 2n + 1, with n an integer and n ≥ 0, the series for one of the solutions reduces to an odd polynomial of degree 2n + 1. 5. Show that the first 4 polynomial solutions P n (x) (known as Legendre polynomials) ordered by their degree and normalized so that P n (1) = 1 are P 0 = 1 P 1 = x P 2 = 1 2 (3x 2 − 1) P 4 = 1 2 (5x 3 − 3x) 1219 6. Show that the Legendre equation can also be written as ((1 − x 2 )y  )  = −α(α + 1)y. Note that two Legendre polynomials P n (x) and P m (x) must satisfy this relation for α = n and α = m respectively. By multiplying the first relation by P m (x) and the second by P n (x) and integrating by parts show that Legendre polynomials satisfy the orthogonality relation  1 −1 P n (x)P m (x) dx = 0 if n = m. If n = m, it can be shown that the value of the integral is 2/(2n + 1). Verify this for the first three polynomials (but you needn’t prove it in general). Hint, Solution Exercise 23.3 Find the forms of two linearly independent series expansions about the point z = 0 for the differential equation w  + 1 sin z w  + 1 − z z 2 w = 0, such that the series are real-valued on the positive real axis. Do not calculate the coefficients in the expansions. Hint, Solution Exercise 23.4 Classify the singular points of the equation w  + w  z −1 + 2w = 0. Hint, Solution 1220 Exercise 23.5 Find the series expansions about z = 0 for w  + 5 4z w  + z −1 8z 2 w = 0. Hint, Solution Exercise 23.6 Find the series expansions about z = 0 of the fundamental solutions of w  + zw  + w = 0. Hint, Solution Exercise 23.7 Find the series expansions about z = 0 of the two linearly independent solutions of w  + 1 2z w  + 1 z w = 0. Hint, Solution Exercise 23.8 Classify the singularity at in finity of the differential equation w  +  2 z + 3 z 2  w  + 1 z 2 w = 0. Find the forms of the series solutions of the differential equation about infinity that are real-valued when z is real-valued and positive. Do not calculate the coefficients in the expansions. Hint, Solution 1221 Exercise 23.9 Consider the second order differential equation x d 2 y dx 2 + (b − x) dy dx − ay = 0, where a, b are real constants. 1. Show that x = 0 is a regular singular point. Determine the location of any additional singular points and classify them. Incl ude the point at infinity. 2. Compute the indicial equation for the point x = 0. 3. By solving an appropriate recursion relation, show that one solution has the form y 1 (x) = 1 + ax b + (a) 2 x 2 (b) 2 2! + ··· + (a) n x n (b) n n! + ··· where the notation (a) n is defined by (a) n = a(a + 1)(a + 2) ···(a + n −1), (a) 0 = 1. Assume throughout this problem that b = n where n is a non-negative integer. 4. Show that when a = −m, where m is a non-negative integer, that there are polynomial solutions to this equation. Compute the radius of convergence of the series above when a = −m. Verify that the result you get is in accord with the Frobenius theory. 5. Show that if b = n + 1 where n = 0, 1, 2, . . ., then the second solution of this equation has logarithmic terms. Indicate the form of the second solution in this case. You need not compute any coefficients. Hint, Solution 1222 [...]... the Frobenius form y1 = x + a2 x2 + a3 x3 + O(x4 ) We substitute y1 into the differential equation and equate coefficients of powers of x xy + 2xy + 6 ex y = 0 x(2a2 + 6a3 x + O(x2 )) + 2x(1 + 2a2 x + 3a3 x2 + O(x3 )) + 6(1 + x + x2 /2 + O(x3 ))(x + a2 x2 + a3 x3 + O(x4 )) = 0 (2a2 x + 6a3 x2 ) + (2x + 4a2 x2 ) + (6x + 6(1 + a2 )x2 ) = O(x3 ) = 0 17 a2 = 4, a3 = 3 17 y1 = x − 4x2 + x3 + O(x4 ) 3 Now we... , n ≥ 0 n +2 n≥0 a0 and a1 are arbitrary We determine the rest of the coefficients from the recurrence relation We consider the cases for even and odd n separately a2n 2 2n a2n 4 = (2n)(2n − 2) a2n = − a0 (2n)(2n − 2) · · · 4 · 2 a0 = (−1)n n , n≥0 m=1 2m = (−1)n a2n−1 2n + 1 a2n−3 = (2n + 1)(2n − 1) a2n+1 = − a1 (2n + 1)(2n − 1) · · · 5 · 3 a1 = (−1)n n , n≥0 m=1 (2m + 1) = (−1)n If {w1 , w2 } is the... (x)P0 (x) dx = −1 1 dx = 2 −1 1 1 x3 P1 (x)P1 (x) dx = x dx = 3 −1 −1 1 2 1 1 1 1 9x4 − 6x2 + 1 dx = 4 4 P2 (x)P2 (x) dx = −1 −1 1 1 P3 (x)P3 (x) dx = −1 −1 1 1 25 x6 − 30x4 + 9x2 dx = 4 4 = −1 5 2 3 9x − 2x3 + x 5 7 1 = −1 25 x − 6x5 + 3x3 7 2 5 1 = −1 2 7 Solution 23 .3 The indicial equation for this problem is 2 + 1 = 0 Since the two roots α1 = i and 2 = −i are distinct and do not differ by an integer,... = − 2 + 2( n + α) − 1 8(n + α) The First Solution Setting α = 1 /4 in the recurrence formula, an−1 an (α1 ) = − 2 + 2( n + 1 /4) − 1 8(n + 1 /4) an−1 an (α1 ) = − 2n(4n + 3) Thus the first solution is ∞ w1 = z 1 /4 an (α1 )z n = a0 z 1 /4 1 − n=0 1 1 2 z+ z + ··· 14 616 The Second Solution Setting α = −1 /2 in the recurrence formula, an−1 an = − 2 + 2( n − 1 /2) − 1 8(n − 1 /2) an−1 an = − 2n(4n − 3) 12 34 an... = − 2 dt t d d = −t2 dz dt w + d2 d d = −t2 −t2 2 dz dt dt 2 d d = t4 2 + 2t3 dt dt 1 24 0 The equation for u is then t4 u + 2t3 u + (2t + 3t2 )(−t2 )u + t2 u = 0 1 u + −3u + 2 u = 0 t We see that t = 0 is a regular singular point To find the indicial equation, we substitute u = tα + O(tα+1 ) into the differential equation α(α − 1)tα 2 − 3αtα−1 + tα 2 = O(tα−1 ) Equating the coefficients of the tα 2 terms,... + 1 /2) (j + 1) (−1)n an = a0 n! n−1 j=0 1 j + 1 /2 Now let’s find the second solution Differentiating w2 , ∞ (n + 1 /2) bn z n−1 /2 w2 = n=0 ∞ (n + 1 /2) (n − 1 /2) bn z n−3 /2 w2 = n=0 Substituting these expansions into the differential equation, ∞ (n + 1 /2) (n − 1 /2) bn z n−3 /2 + n=0 1 2 ∞ ∞ (n + 1 /2) bn z n−3 /2 + n=0 bn−1 z n−3 /2 = 0 n=1 Equating the coefficient of the z −3 /2 term, 1 2 − 1 2 b0 + 11 b0 = 0, 22 we... 1 /2) (n − 1 /2) bn + (n + 1 /2) bn + bn−1 = 0 2 bn−1 bn = − n(n + 1 /2) 123 9 Calculating the bn ’s, b0 1· 3 2 b0 b2 = 3 5 1 2 2 · 2 (−1)n 2n b0 bn = n! · 3 · 5 · · · (2n + 1) b1 = − Thus the second solution is ∞ w2 = b0 z 1 /2 n=0 (−1)n 2n z n n! 3 · 5 · · · (2n + 1) Solution 23 .8 3 1 2 + 2 w + 2 w = 0 z z z In order to analyze the behavior at infinity we make the change of variables t = 1/z, u(t) = w(z) and. .. positive odd integer, then the second solution, y2 , is a polynomial of order n For λ = 0, 1, 2, 3, we have H0 (x) = 1 H1 (x) = x H2 (x) = 1 − 2x2 2 H3 (x) = x − x3 3 122 6 Solution 23 .2 1 First we write the differential equation in the standard form 1 − x2 y − 2xy + α(α + 1)y = 0 (23 .2) α(α + 1) 2x y + y = 0 (23 .3) 2 1−x 1 − x2 Since the coefficients of y and y are analytic in a neighborhood of x = 0, We...Exercise 23 .10 Consider the equation xy + 2xy + 6 ex y = 0 Find the first three non-zero terms in each of two linearly independent series solutions about x = 0 Hint, Solution 122 3 23 .6 Hints Hint 23 .1 Hint 23 .2 Hint 23 .3 Hint 23 .4 Hint 23 .5 Hint 23 .6 Hint 23 .7 Hint 23 .8 Hint 23 .9 Hint 23 .10 12 24 23 .7 Solutions Solution 23 .1 f (x) is a Taylor series about x = 0 ∞ an... following form: y2 = y1 ln(x) + a0 + a2 x2 + O(x3 ) The first three terms in the solution are y2 = a0 + x ln x − 4x2 ln x + O(x2 ) We calculate the derivatives of y2 y2 = ln(x) + O(1) 1 y2 = + O(ln(x)) x We substitute y2 into the differential equation and equate coefficients xy + 2xy + 6 ex y = 0 (1 + O(x ln x)) + 2 (O(x ln x)) + 6 (a0 + O(x ln x)) = 0 1 + 6a0 = 0 1 y2 = − + x ln x − 4x2 ln x + O(x2 ) 6 1 24 7 23 .8 . 0. Hint, Solution 122 3 23 .6 Hints Hint 23 .1 Hint 23 .2 Hint 23 .3 Hint 23 .4 Hint 23 .5 Hint 23 .6 Hint 23 .7 Hint 23 .8 Hint 23 .9 Hint 23 .10 12 24 23 .7 Solutions Solution 23 .1 f(x) is a Taylor series about. solution a 2n = 2 n  n k=0 (2( n − k) −λ) (2n)! , a 2n+1 = 0. The difference equation for y 2 is a n +2 = 2 n − λ (n + 1)(n + 2) a n , a 0 = 0, a 1 = 1, which has the solution a 2n = 0, a 2n+1 = 2 n  n−1 k=0 (2( n. 2 ∞  n=0 a n x n Equating coefficients gives us a difference equation for a n : (n + 2) (n + 1)a n +2 − 2na n + 2 a n = 0 a n +2 = 2 n − λ (n + 1)(n + 2) a n . 122 5 The first two coefficients, a 0 and