Frobenius & Special Functions

Study Sheet

Frobenius & Special Functions

Singular points, indicial equations, Bessel and Legendre

Series at Singular Points

Tip
Regular singular points and the indicial equation

Where the leading coefficient vanishes (like x=0x = 0 in x2y′′+xy′+(x2−ν2)y=0x^2y'' + xy' + (x^2 - \nu^2)y = 0), plain power series can fail. Frobenius: try y=xr∑anxny = x^r\sum a_nx^n with an UNKNOWN starting power rr.

The lowest-order coefficient gives the INDICIAL equation — for Bessel, r2−ν2=0r^2 - \nu^2 = 0, roots ±ν\pm\nu — and each admissible rr seeds its own series. The pure case is Cauchy–Euler x2y′′+αxy′+βy=0x^2y'' + \alpha xy' + \beta y = 0, where y=xry = x^r alone works and the indicial equation r(r−1)+αr+β=0r(r-1) + \alpha r + \beta = 0 IS the whole story.

Tip
Bessel functions: the drumhead cosine

Bessel's equation governs radial vibrations of circular membranes and heat in cylinders. Its bounded solution J0(x)=1−x24+x464−⋯=∑(−1)k4k(k!)2x2kJ_0(x) = 1 - \tfrac{x^2}{4} + \tfrac{x^4}{64} - \cdots = \sum\tfrac{(-1)^k}{4^k(k!)^2}x^{2k} looks like a cosine whose amplitude slowly decays; its zeros (2.405, 5.520, …) set the drum's overtone frequencies — irregularly spaced, which is why a drum sounds like a thump and not a note. Half-integer orders collapse to elementary functions: J1/2(x)=2πxsin⁡xJ_{1/2}(x) = \sqrt{\tfrac{2}{\pi x}}\sin x.

Tip
Legendre polynomials: quantization by boundedness

Legendre's equation (1−x2)y′′−2xy′+n(n+1)y=0(1-x^2)y'' - 2xy' + n(n+1)y = 0 appears whenever a problem lives on a sphere. Series solutions generally blow up at x=±1x = \pm1 (the poles); demanding BOUNDEDNESS forces the parameter to be a whole number n(n+1)n(n+1), truncating the series to the polynomials P0=1P_0 = 1, P1=xP_1 = x, P2=3x2−12P_2 = \tfrac{3x^2-1}{2}, P3=5x3−3x2,…P_3 = \tfrac{5x^3-3x}{2}, \ldots — orthogonal on [−1,1][-1,1] and the angular half of spherical harmonics. The same mechanism that quantized Fourier eigenvalues, wearing spherical clothes.

Side note
Significance: the functions with their own names

Bessel, Legendre, Hermite, Laguerre — the "special functions" are the standard library of mathematical physics, and Frobenius is the compiler that builds them. Your phone's antenna design (cylindrical waves = Bessel) and every atomic orbital picture (Legendre in the angles) come from this machinery.

Try it
Try it: an indicial equation cold

For 2x2y′′+3xy′−y=02x^2y'' + 3xy' - y = 0, find the indicial roots. Work: substitute y=xry = x^r: 2r(r−1)+3r−1=2r2+r−1=(2r−1)(r+1)=02r(r-1) + 3r - 1 = 2r^2 + r - 1 = (2r - 1)(r + 1) = 0: r=12r = \tfrac12 or −1-1. General solution y=Ax+Bxy = A\sqrt x + \tfrac Bx — fractional powers, impossible for constant-coefficient equations, routine here.

Proofs & Why It Matters

Tip
Proof: the Cauchy–Euler indicial equation

Substitute y=xry = x^r into x2y′′+αxy′+βy=0x^2y'' + \alpha xy' + \beta y = 0: the derivatives give x2⋅r(r−1)xr−2+αx⋅rxr−1+βxr=[r(r−1)+αr+β]xrx^2\cdot r(r-1)x^{r-2} + \alpha x\cdot rx^{r-1} + \beta x^r = \left[r(r-1) + \alpha r + \beta\right]x^r.

Since xr≠0x^r \ne 0 for x>0x > 0, the bracket must vanish — a QUADRATIC in rr whose two roots give two power solutions (a double root contributes xrln⁡xx^r\ln x, precisely parallel to the xerxxe^{rx} of constant-coefficient theory, via the substitution x=etx = e^t which converts one theory into the other). ■\blacksquare

Tip
Proof: J₀ solves Bessel of order zero

Termwise, with y=∑ckx2ky = \sum c_kx^{2k}, ck=(−1)k4k(k!)2c_k = \tfrac{(-1)^k}{4^k(k!)^2}: compute xy′′+y′+xyxy'' + y' + xy (the ν=0\nu = 0 equation divided by xx).

The x2k+1x^{2k+1} coefficient is ck+1(2k+2)(2k+1)+ck+1(2k+2)+ck=ck+1(2k+2)2+ckc_{k+1}(2k+2)(2k+1) + c_{k+1}(2k+2) + c_k = c_{k+1}(2k+2)^2 + c_k, and indeed ck+1=−ck4(k+1)2c_{k+1} = \tfrac{-c_k}{4(k+1)^2} makes it vanish: −ck4(k+1)2⋅4(k+1)2+ck=0\tfrac{-c_k}{4(k+1)^2}\cdot4(k+1)^2 + c_k = 0. Every coefficient dies — the series solves the equation exactly. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: a Frobenius recurrence at a regular singular point

Find the first three nonzero terms of a Frobenius solution of 2xy′′+y′+y=02xy'' + y' + y = 0 for the larger indicial root.

Step 1 — try y=∑anxn+ry = \sum a_nx^{n+r}; the lowest-order terms give the indicial equation 2r(r−1)+r=r(2r−1)=02r(r-1) + r = r(2r - 1) = 0: r=0r = 0 or 12\tfrac12.

Step 2 — for r=12r = \tfrac12, matching the coefficient of xn+rx^{n+r} gives the recurrence an+1=−an(n+1)(2n+3)a_{n+1} = -\dfrac{a_n}{(n+1)(2n+3)}... (from 2(n+r+1)(n+r)an+1+(n+r+1)an+1+an=02(n+r+1)(n+r)a_{n+1} + (n+r+1)a_{n+1} + a_n = 0 with r=12r = \tfrac12).

Step 3 — run it from a0=1a_0 = 1: a1=−11⋅3=−13a_1 = -\tfrac{1}{1\cdot3} = -\tfrac13, a2=1/32⋅5=130a_2 = \tfrac{1/3}{2\cdot5} = \tfrac{1}{30}.

Step 4 — the solution: y=x1/2(1−x3+x230−⋯ )y = x^{1/2}\left(1 - \tfrac x3 + \tfrac{x^2}{30} - \cdots\right) — a square-root prefactor no ordinary power series could produce, and the recurrence generates as many terms as accuracy demands.

Example
Worked: a Cauchy–Euler initial-value problem

Solve x2y′′−3xy′+4y=0x^2y'' - 3xy' + 4y = 0 with y(1)=1y(1) = 1, y′(1)=0y'(1) = 0.

Step 1 — indicial equation: r(r−1)−3r+4=r2−4r+4=(r−2)2r(r-1) - 3r + 4 = r^2 - 4r + 4 = (r-2)^2: double root r=2r = 2.

Step 2 — double root means y=x2(A+Bln⁡x)y = x^2(A + B\ln x) (the ln⁡x\ln x plays the role tt plays for repeated constant-coefficient roots).

Step 3 — data: y(1)=A=1y(1) = A = 1; y′=2x(A+Bln⁡x)+Bxy' = 2x(A + B\ln x) + Bx, so y′(1)=2A+B=0y'(1) = 2A + B = 0, B=−2B = -2.

Step 4 — solution y=x2(1−2ln⁡x)y = x^2(1 - 2\ln x).

Step 5 — verify at x=ex = e: y=e2(1−2)=−e2y = e^2(1 - 2) = -e^2; direct substitution of y=x2−2x2ln⁡xy = x^2 - 2x^2\ln x into the equation collapses to 00 term by term ✓.