CYK\2010\PH402\Mathematical Physics\Tutorial Find two linearly independent power series solutions of the equation.

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1 CYK\010\PH40\Mathematical Physics\Tutorial 1. Find two linearly independent power series solutions of the equation For which values of x do the series converge?. Find a series solution for y xy + y = x )y + y = 0 about x = 0. Is P x) = 1 + x ) 1 analytic everywhere? 3. The equation 1 x )y xy + α y = 0 where α is a constant, is called the Chebyshev equation. a) Compute two linearly independent series solutions for x < 1. b) Show that for every non-negative integer α = n there is a polynomial solution of degree n. When appropriately normalized these are called Chebyshev polynomials. 4. The equation y xy + αy = 0 where α is a constant, is called Hermite equation. a) Find two linearly independant series solutions for < x <. b) Are these solutions convergent for all x? Find the behaviour of these solutions for large x. c) Show that there is a polynomial solution of degree n for every α = n non-negative integer. 5. Find the singular points of the following equations and determine those which are regular singular points: a) x y + x + x )y y = 0 b) 1 x ) y xy + ll + 1)y = 0 Legendre) c) x y 5y + 3x y = 0 d) x + x ) y + 3x + )y + x 1)y = 0 e) 1 x ) y xy + n y = 0 Chebyshev) f) y xy + αy = 0 Hermite) g) x y + xy + x n )y = 0 Bessel) h) xy + 1 x)y + ay = 0 Laguerre) 6. The equation xy + 1 x)y + ay = 0 where α is a constant, is called the Laguerre equation. solutions about x = 0. Check convergence. Find two linearly independent series 7. The interaction between two nucleons may be described by a mesonic potential V x) = Ae ax x where A and a are constants.find the rst few nonvanishing terms of the solution of 1D Schrodinger equation. 8. Find the series solutions of Legendre equation 1 x ) y xy + ll + 1)y = 0 about x = 1. Show that there are polynomial solutions for non-negative l. 1

2 Solutions 1. There are no singular points, thus let y = c k x k. Substitute in the Dierential equation kk 1)c k x k x kc k x k 1 + c k x k = 0 Adjusting the summation variable k in the rst term to k +, we get k + 1) k + ) c k+ k 1) c k x k = 0 Comparing the coecients of x k, k + 1) k + ) c k+ k 1) c k = 0 that is Explicitly, c k+ = k 1) k + 1) k + ) c k c = 1 1 c 0 = 1 c 0 c 3 = 0 3 c 1 = 0 c 4 = 1) 4 3 c = c 6 = Clearly, for odd k 3, c k = 0. For even k, c 4 = 1) 1 3 6! 1) 1) c 0 = 1 4! 4 c 0 c 0 = 1 40 c 0 The two solutions c k = y = c 0 + k=,4,... 1) 1 k 3) c 0 1) 1 k 3) x k + c 1 x = c x 1 4 x x6 + c 1 x. y 1 = 1 1 x 1 4 x x6 y = x are linearly independent and are convergent for all x. To see this, lim c k+ x = 0. c k. Substitute y = c k x k. 1 + x ) kk 1)c k x k + k k c k x k = 0

3 , recurrence relation is Explicitly, for even k, And for odd k, k + 1) k + ) c k+ + k k 1) + 1) c k x k = 0 k k k 1) + 1) c k+ = k + 1) k + ) c k. c = 1 1 c 0 = 1 c 0 c 4 = c = ! c 6 = 13 6! 6 5 c 4 = c 0 = 1 8 c 0 c 3 = 1 3 c 1 = 1 6 c 1 c 5 = c 3 = ! c 0 = c 0 y = c x x x6 + + c 1 x 1 6 x x5 +. The two solutions are linearly independent and lim C k+ x = x c k The series solutions converge only when x < 1. It is expected because 1 + x ) 1 is analytic only for x < The Chebyshev equation is analytic at x = 0. regular singular at ±1). substituting y = c k x k in the equation, we get recurrence relation ) k ) α c 1 c k = c k k k 1) = 1 k j α ) c 0 j=0, = 1 k j α ) c 0 j=1,3 even k odd k The two solutions are y 1 = 1 +,,... y = x + k=3,5, k j=0, k j=1,3 j α ) j α ) x k x k Clearly, series are convergent only for x < 1. And it is easy to see that if α = n some even integer then y 1 becomes a polynomial of degree n since c n+ = 0 and subsequently all further coecients are zero. If α = n+1, that is some odd integer, then y would reduce to a polynomial. 3

4 4. The Hermite equation is analytic at all x. Again, substituting y = c k x k, we get the recurrence relation k α) c k = c k. kk 1) For even k = m, And for odd k = m + 1 and Clearly, y 1 = 1 αx + c m = m m α) α) c 0 m)! c m+1 = m m 1 α) 1 α) c 1 m + 1)! y = x αα ) x 6 α 1) x c k αα )α 4) x α 1)α 3) x lim c k x = lim k α)x kk 1) = 0, both series converge for all x. To obtain asymptotic behaviour, terms with large k matter, thus Compare with The series behaves like e x. c m = m c m = 1 m! c 0 e x = 1 + x + 1 x m)! xm + It is easy to see that if α = n some integers either y 1 or y becomes a polynomial. 5. skipped. 6. Given xy + 1 x)y + ay = 0 Note that x = 0 is a regular singular point. The indicial equation rr 1) + r + 0 = 0 has only one solution given by r = 0. The rst solution is of the form y = k c kx k. Substitution gives us, c k+1 = = = k a) k + 1) c k k a) 1 a) a) k + 1)! c 0 Γk + 1 a) Γ a) k + 1)! c 0 y 1 = c 0 1 ax Γ a) a1 a) x 4 4

5 It is clear that the series converges for all x. The second solution has the form given by y = d k x k + log x) y 1 with d 0 = 0. Substituting in the dierential equation, k + 1) d k+1 + a k)d k x k = xy 1 = kc k x k, the second solution is d 1 = 0 4d = c 1 = d = a a4 3a) 9d 3 + a )d = 4c = d 3 = 18 y = a x a4 3a) x log x)y Potential is given to be The Schrodinger equation is given by V x) = Ae ax x m y + V x) E)y = 0 y + λ + Bx ) e ax y = 0 1) where λ = me/ and B = ma/. Here, x = 0 is a regular singular point. The solution has a form y = x r c k x k. Substitute in equation 1: r + k)r + k 1)c k x r+k + λ + B i=0 i=0 x i 1 Consider the double sum, put i + k = m and replace k sum by m sum c k m ) c m i i! xr+k+i 1 = i! And the for the rst sum put k = m, thenx rr 1)c 0 x r +r + 1)rc 1 + Bc 0 ) x r 1 Indicial equation is m=1 m=0 i=0 i! x m+r 1 c k x r+k = 0 r + m + 1)r + m)c m+1 + λ c m 1 + B rr 1) = 0. 5 m i=0 c m i i! ) x m+r 1 =

6 putting r = 1, we get the rst solution c 1 + Bc 0 = 0 = c 1 = B c 0 3 c + λ c 0 + B c 0 + c 1 ) = 0 = c 0 = 1 6 λ + B 1 B )) c 0 the solution is y = c 0 x 1 B x 1 λ + B 1 B )) x Legendre Equation 1 x) 1 + x) y xy + ll + 1)y = 0 Make a substitution z = 1 x. Then dy/dx = dy/dz. Legendre equation becomes z z) y + 1 z) y + l l + 1) y = 0 where the derivatives are wrt z. Clearly z = 0 and are regular singular points. Using Frobenius method with power series about z = 0. Let Indicial equation is y = z r c k z k. rr 1) + r = 0 = r = 0. The regular solution is given by y = c k z k. Substituting in the de, we get ll + 1) c 1 = c 0 kk + 1) ll + 1) c k+1 = k + 1) c k k > 0. for l = 0, c k = 0 for all k > 0 For l = 1, c 1 = c 0 and c k = 0 for all k > 1 y = c 0 For l =, c 1 = 3c 0, c = 3c 0 /, so y = c 0 1 z) = c x) = c 0 x y = c 0 1 3z + 3 z ) Finally, for l = 3, we get y = c 0 1 6z + 15 z 5 ) 3 z3 6

12d. Regular Singular Points

October 22, 2012 12d-1 12d. Regular Singular Points We have studied solutions to the linear second order differential equations of the form P (x)y + Q(x)y + R(x)y = 0 (1) in the cases with P, Q, R real