fourierseries（6页）.doc

4.2 The Method of Frobenius: Solutions about Regular Singular Points Consider the equation L[y] = x2y” + xb(x)y’ + c(x)y = 0 (4.2-1) where b(x) = , c(x) = are analytic at x = 0 Assume a power series solution y = (4.2-2) Differentiating the series solution (4.2-2) yields y’ = , y” = Substitution of the series expressions above into Eq. (4.2-1) gives, in expanded form, L[y] = [r(r – 1)a0xr + (r + 1)ra1xr+1 + …] + [(b0 + b1x + …)( ra0xr + (r + 1)a1xr+1 + …] + [(c0 + c1x + …)( a0xr + a1xr+1 + …] (4.2-3) Collect like powers of x to obtain L[y] = [r(r – 1) + b0r + c0]a0xr + (4.2-4) Set the coefficients of xr? to zero: r(r – 1) + b0r + c0 = 0 (4.2-5) Eq. (4.2-5) is called the indicial equation. It has two roots. Set the coefficients of xm+r? to zero: = 0 (4.2-6) Eq. (4.2-6) is called the recurrence relation. Let r1 and r2 be the roots of the indicial equation. Then we have the following three cases. Case 1. Distinct roots not differing by an integer. Case 2. Double roots r1 = r2 = r. Case 3. Roots differing by an integer Consider the equation L[y] = x2y” + xb(x)y’ + c(x)y = 0 (4.2-7) Any solution y(x) of (4.2-7) satisfies the equation L[y] = 0. Choose am(r) to satisfies the recursion relation but not necessary the indicial relation, y(x,r) satisfies L[y(x,r)] = a0[r2 + (b0 – 1)r +c0]xr (4.2-8) Case 2. If the indicial relation has double roots then L[y(x,r)] = a0(r - r1)2 xr (4.2-9) If r = r1, then a0(r - r1)2 xr = 0, y(x,r1) is a first independent series solution of L[y] = 0. If we differentiate both sides of (4.2-9) with respect to r and let r = r1 {L[y(x,r)]} = L[y(x,r)] = {a0(r - r1)2 xr} {a0(r - r1)2 xr} = a0[2(r - r1) xr + (r - r1)2 xr ln x] = 0 Thus y(x,r)at r = r1 is a second independent series solution of (4.2-7) because L[y(x,r)] = 0. Case 3. Roots of indicial relation different by an integer (r2 = r1 + j, j = positive integer) L[y(x,r)] = a0(r - r1)(r - r2) xr (4.2-10) If r