VTU Edusat Programme 16

Subject : Engineering Mathematics Sub Code: 10MAT41

UNIT 5: Special Functions

Dr. K.S.Basavarajappa Professor & Head

Department of Mathematics Bapuji Institute of Engineering and of Technology

Davangere-577004 Email: ksbraju@hotmail.com

SPECIAL FUNCTIONS

The special function deals with the solution of two important ordinary differential equations. Two special functions are Bessels function and Legendre polynomial. Bessels function deals with solving the boundary value problem pertaining to axial symmetry. Legendre function deals with problems pertaining to spherical symmetry. Many differential equations arising from physical problems are linear but have variable coefficients and do not permit a general solution in terms of known functions. Such equations can be solved by numerical methods, but in many cases it is easier to find a solution in the form of an infinite convergent series.

Series solution of certain differential equations give rise to special functions such as Bessels function, Legendres polynomial, Lagurres polynomial, Hermite polynomial, Chebyshev Polynomial Strun-Lioville problem based o the orthogonality of functions is also included which show that Bessels Legendres and other equations can be considered from a common point of view. These special functions have many application Engineering.

Solution of Laplaces equation in cylindrical polar coordinates (Equation reducible to Bessels differential equation)

Let (, , ) denote cylindrical polar coordinates, then the Laplacian of a function = (, , ) is given by = + + + . (1) This is a linear second order partial differential equation; we solve this equation by method of separation of variables

Assume the tentative solution as

(, , ) = () () () ..(2) or

= Then = , it is called the laplaces equation

Substituting the derivatives of (2) in (1) then,

+ 1 + 1 + = 0

Divide by then + 1 + 1 + = 0 Then + + = Choose the common solution as -p2 1 + 1 ! + 1 1 ! = 1 = p Taking 2nd and 3rd term then

1 = p p = 0 Auxiliary equation is

For (# $) = 0 % $ = 0 % = $ = '()* + '(+)* Also 1 + 1 ! + 1 1 ! = p + 1 ! + 1 = pr

Then we group the terms of r in L.H.S and the terms of on R.H.S then + 1 ! + pr = 1 L.H.S is a differential equation with r as the independent variable and independent of

and R.H.S is a differential equation with as the independent variable and independent of r then this type of equation holds when each side is a constant. Then, we assume the above equation holds only if it is equal to some n2.

Then taking two equations to be solved separately as follows Let R.H.S as

1R R = n

= 0 + 0 = 0 (# + 0) = 0 Auxiliary equation is % + 0 = 0

M = i n

Then = 1'230 + 143500 Also we have LHS as

+ 1 ! + $ = 0 Multiply R1

6 + 1 7 + $ 0 = 0 + + $ 0 = 0

Setting x = p r, put R1 = y then, r = x/p

= 88 =898 =

898: 6

8:87 = $ 6

898:7

Where p = dx/dr.

= 88 =

898 =

88 6

8987 =

88: 6

8987 6

8:87 =

88: 6$ 6

898:77 $

= ;;?< = $ @;

BESSEL FUNCTIONS

Definition: The second order differential equation

x dydx + xdydx + x ny = 0

Where n is a constant, is called Bessels equation of order n. It is one of the most important differential equations in applied mathematics. Its particular solutions are Bessel functions.

Series solution of Bessels Equation: The Bessels equation of order n is

x L

Equating the coefficient of xm+1 to zero, we have

Y% + 1 0Y = 0 (5) To get the relation between the successive as, we equate the coefficient of xm+r-1 to zero.

Therefore Y?+X% + 1 0[ + Y?+ = 0 i.e., Y?+ = XZU?+

y = AiH(F) + Bi+H(F) where A and B are arbitrary constants.

Expansion for J0(x) and J1(x): When n is an integer, (0 + 1) = 0! Hence iH(F) becomes e^ (:) = (+)b(^U?U)! ?! @BC^U?h?kQ (1) = @BC^ [ ^! !(^U)! @BC + !(^U)! @BC4 . Putting n = 0 and 1 in turn in (1), we have

eQ(:) = !! @BC + !! @BC4 (2) e(:) = B [ ! !! @BC + !! @BC4 (3) The series representation (2) and (3) show that J0(x) is an even function like cosx and J1(x) is an odd function like sinx as shown in graphs.

Orthogonality of Bessel functions:

To prove that:l m no(pm)no(qm)rm = s , p q [noU(m)[ , p = q t Where , are the roots of Jx(x) = 0.

Proof:

Consider the Bessels equation as x L

xu @{NC + xu @{NC + (x n)u @{NC = 0 (2)1 xv @|NC + xv @|NC + (x n)v @|NC = 0 .(3)1 Subtracting between them (2)1 - (3)1 x[u v u v[ + [uv uv[ + ( )x u v = 0 ddx [x}uv uv~[ = ( )x u v Integrating both sides from 0 to 1

( ) x u v dxQ = ddx [x}uv uv~[ dx

Q

= uv uv When u = Jx(x) > u1 = Jx(x) v = Jx(x) > v1 = Jx(x) Therefore

( ) x Jx(x)Jx(x)dx = Jx(x) Jx(x) Jx(x) Jx(x) Q at x = 1 on RHS we get

l x Jx(x)Jx(x)dx = > () ()+ () > ()

Put = , then x Jx(x)dx = Jx() Jx()2

Q

= 12 [ JxU()[ Therefore l m [ no(pm)[ = [ noU(p)[

SOLUTION OF LAPLACE EQUATION IN SPHERICAL POLAR COORDINATES

Let (r, , ) denote the spherical polar coordinates. Then the expression for the Laplacian of a function = (r, , ), is given by, =

r + 2r + + cot = 0 1 [r + 2r[ = 1 [ + cot[ Choose K2 as the common solution then,

1 [r + 2r[ = 1 [ + cot[ = K (say) > [r + 2r[ = K ; < [ + cot[ = K

r + 2r K = 0

This is a Cauchys type ordinary homogeneous linear differential equation of second order.

Then put = e log r = t ,then we get, r = D r = D(D 1) Then,

(D + D K) = 0 A. E m = 12 1 + K2 = Ce+>

Setting = y , x = cos , K = n(n + 1) , Differentiating w. r. t Then

dd = dyd = dydx dxd = (sin) dydx L

LEGENDRE FUNCTION

The differential equation of the form

(1 x) L

f(m k)(m k 1)a xO++ hkQ f}(m k n)~[(m k) + n + 1[axO+ = 0h

kQ (4)

Equating the coefficient of highest power of x i.e., xm, we get the indicial equation

(by taking k = 0)

(m n)(m + n + 1) aQ = 0 Since aQ 0, then m n = 0 m = n m + n + 1 = 0 m = (n + 1) Equating the coefficients of the next lowest power of x i.e., xm-1, we get (taking k=1)

(m + n)(m n 1) a = 0, a = 0 For m = n or m = -(n+1) the coefficients of a are not zero then a = 0

Equating to zero the coefficients of xm-k, we get the recurrence relation,

(m (k 2))(m (k 2) 1)a+ }(m k n)~[(m k) + n + 1[a = 0 a = (m k + 2)(m k + 1)}(n m + k)~[n + m k + 1[ a+ Since a = 0, we get a = a = a = = 0 When m= n, the recurrence relation reduces to

a = +(x+U)(x+U)[x+U[ a+ Putting k = 2, 4, 6 .. we get

a = n(n 1)2(2n 1) aQ a4 = (n 2)(n 3)4(2n 3) a = n(n 1)(n 2)(n 3)2 4(2n 1)(2n 3) aQ etc. For m = n, we get y = y1

y = aQ [xx n(n 1)2(2n 1) xx+ + n(n 1)(n 2)(n 3)2 4(2n 1)(2n 3) xx+4 . . [ When m = - (n+1), the recurrence relation reduces to,

a = (xU+)(xU) [xUU[ a+ Putting k = 2, 4, 6 .. we get

a = (n + 1)(n + 2)2(2n + 3) aQ a4 = (n + 3)(n + 4)4(2n + 5) a = (n + 1)(n + 2)(n + 3)(n + 4)2 4(2n + 3)(2n + 5) aQ etc. For m = - (n+1) , we get y = y2

y = aQ [x+x+ + (n + 1)(n + 2)2(2n + 3) x+x+ + (n + 1)(n + 2)(n + 3)(n + 4)2 4(2n + 3)(2n + 5) x+x+ + [ When n is a positive integer and

aQ = 1 3 5 (2n 1)n! Therefore y = 1 3 5 (2n 1)n! [xx n(n 1)2(2n 1) xx+ + . . [ y1 = Pn(x),

Pn(x) is called the Legendres function of the first kind.

Pn(x) is a terminating series and gives what are called Legendres polynomials for different

values of n such that Pn(1) = 1

We can write Pn(x) as,

Px(x) = f (1) (2n 2k)!2xk! (n k)! (n 2k)! xx+

kQ

Where N = x , if n is evenx+ , if n is odd

When aQ is taken as, aQ = n!1 3 5 . (2n + 1) Then

= n!1 3 5 . (2n + 1) x+x+ + (n + 1)(n + 2)2(2n + 3) x+x+ + = o(m), Qx(x) is called the Legendres function of second kind. Therefore General solution of equation is

= o(m) + o(m) ,where C1 and C2 are arbitrary constants.

Rodrigues Formula And Legendres Polynomials;

The relation,

Px(x) = x! LLN (x 1)x is known as Rodrigues formula Proof: Let v = (x 1)x v = L{LN = n(x 1)x+2x Multiply by (x 1) on both sides (x 1)v = 2nx(x 1)x (x 1)v = 2nxv

(x 1)v 2nxv = 0 Using Liebnitz formula, taking U = v1 and v = 1 - x2

(1 x)vxU + 0C (2x)vx + nC(2)vx+ 2nxvx nC2nvx+ = 0 Replacing n by (n + 1), we get

(1 x) ddx (vx) 2x ddx (vx) + n(n + 1)vx = 0 Which is Legendres equation and vn is its solution. But Px(x) and Qx(x)are the solutions of Legendres equation.

Since vx = L

and so on.

Example:

Express f(x) = x4 + 3x x + 5x 2 in terms of Legendre polynomials. Solution:

Seperating from Legendre polynomials for 1, x, x2, x3 and x4, in terms of p0(x) , p1(x), p2(x),

p3(x), p4(x) further replacing in the given problem

f(x) = 835 P4(x) + 3 23 P(x) + 35 x! 17 23 P(x) + 13! + 5x 73 35 f(x) = P4(x) + ` P(x) P(x) + 4 P(x) 4Q PQ(x)

Problems for practice:

1. f(x) = 5x3 + x 2. f(x) = 4x3 2x2 -3 x+ 8 3. f(x) = x4 + 2x3 +2 x2 -x 3