Legendre\'s equation is the second-order differential equation (1 - t2)d2y/dt2 -

Question

Legendre's equation is the second-order differential equation (1 - t2)d2y/dt2 - 2tdy/dt + v(v + 1)y = 0, where v is a constant. Guess the Taylor series centered at t = 0 as a solution y(t), that is, y(t) = a0 + a1t + a2t2 + a3t3 + a4t4 + .... Compute the coefficients a2, a3, and a4 in terms of the initial conditions a0 = y(0) and a1 = y'(0). By choosing v to be a positive integer and using the initial conditions (y(0), y'(0)) = (1,0) or (y(0), y'(0)) = (0,1), show that there are polynomial solutions Pv(t) of Legendre's equation. Show that the first three of these polynomials are P0(t) = 1, P1(t) = t, and P2(t) = 1 - 3t2. Compute Pv(t) for v = 3,4, 5, and 6. Verify that kPv(t) is a solution of the same Legendre's equation as Pv(t) if k is a constant. (Consequently, Pv(t) is only determined up to a constant. For example, P2(t) is sometimes given as P2(t) = (3t2 - 1)/2).

Explanation / Answer

y(t) = ao + a1 t +a2*t^2 +a3*t^3+ .....................+a4*t^4+................

(1-t^2)d^2y/dt^2 -2tdy/dt + v(V+1)y =0 ...............(1)

a) y' = a1 + 2a2*t + 3a3*t^2 + 4a4*t^3 + ........

y'' = 2a2 + 6a3*t + 12a4*t^2 + ........

put he value of y ,y' and y'' in the equation (1)

(1-t^2)(2a2 + 6a3*t + 12a4*t^2) -2t(a1 + 2a2*t + 3a3*t^2 + 4a4*t^3 ) + V(V+1)(ao + a1 t +a2*t^2 +a3*t^3+ .....................+a4*t^4) =0

(2a2 +v(V+1)a0) +t[6a3- 2a1 +v(V+1)a1]+ t^2[-2a2+ 12a4-6a3)] + ...... =0

equating the constant term =0

then (2a2 +v(V+1)a0) = 0

a2 = -V(v+1)a0

a2 =-V(v+1)y(0)

equating the cofficient of t =0

[6a3- 2a1 +v(V+1)a1] =0

a3 = 2a1 -V(V+1)a1

= a1[2 -V(V+1)]

=y'(0) [2 -V(V+1)]

equating the cofficient of t^2 =0

-2a2+ 12a4-6a3) =0

a4 = (a2 +3a3)/6

a4 ={-V(v+1)y(0) +y'(0) [2 -V(V+1)]}/6

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b) given y(0) =1 , y'(0) =0

then a0 = 1

a1 = 0

a2 = -V(v+1)

a3 = 0

a4 = -V(v+1)/6

hence Pv(t) =

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