Chapter 5
SPECIAL FUNCTIONS
Chapter 5 SPECIAL FUNCTIONS table of content
Chapter 5 Special Functions
5.1 Heaviside step function - filter function
5.2 Dirac delta function - modeling of impulse processes
5.3 Sine integral function - Gibbs phenomena
5.4 Error function
5.5 Gamma function
5.6 Bessel functions
1. Bessel equation of order ν (BE) 2. Singular points. Frobenius method 3. Indicial equation 4. First solution – Bessel function of the 1st kind 5. Second solution – Bessel function of the 2nd kind. General solution of Bessel equation 6. Bessel functions of half orders – spherical Bessel functions 7. Bessel function of the complex variable – Bessel function of the 3rd kind (Hankel functions) 8. Properties of Bessel functions: - oscillations - identities - differentiation - integration - addition theorem 9. Generating functions 10. Modified Bessel equation (MBE) - modified Bessel functions of the 1st and the 2nd kind 11. Equations solvable in terms of Bessel functions - Airy equation, Airy functions 12. Orthogonality of Bessel functions - self-adjoint form of Bessel equation - orthogonal sets in circular domain - orthogonal sets in annular fomain - Fourier-Bessel series
5.7 Legendre Functions
1. Legendre equation 2. Solution of Legendre equation – Legendre polynomials 3. Recurrence and Rodrigues’ formulae 4. Orthogonality of Legendre polynomials 5. Fourier-Legendre series 6. Integral transform
5.8 Exercises
Chapter 5 SPECIAL FUNCTIONS
Chapter 5
SPECIAL FUNCTIONS
Introduction
In this chapter we summarize information about several functions which are widely used for mathematical modeling in engineering. Some of them play a supplemental role, while the others, such as the Bessel and Legendre functions, are of primary importance. These functions appear as solutions of boundary value problems in physics and engineering.
The survey of special functions presented here is not complete – we focus only on functions which are needed in this class. We study how these functions are defined, their main properties and some applications.
Chapter 5 SPECIAL FUNCTIONS 5.1 Heaviside Step Function
5.1 Heaviside Function (unit step function)
The Heaviside step function has only two values: 0 and 1 with a jump at x = 0 :
0 x < 0 H ()x = (1) 1 x > 0
Graphically it can be shown as:
> plot(Heaviside(x),x=-3..3,discont=true);
H ( x)
Shifting of the step function along the x-axis: 0 x < a H ()x − a = (2) 1 x < a
> plot(Heaviside(x-2),x=-1..4,discont=true);
H ( x2− )
filter function The filter function can be constructed in terms of the step function:
0 x < a F()()()x,a,b = H x − a − H x − b = 1 a < x < b (3) 0 x > b
It cuts the values of functions to zero outside of the interval []a,b :
> F(x,1,3):=Heaviside(x-1)-Heaviside(x-3); > plot(g(x)*F(x,1,3),x=-1..5,discont=true);
g ( xFx,1,3)⋅ ( )
The Heaviside step function is used for the modeling of a sudden increase of some quantity in the system (for example, a unit voltage is suddenly introduced into an electric circuit) – we call this sudden increase a spontaneous source. Chapter 5 SPECIAL FUNCTIONS
5.2 Dirac Function (delta function)
The Dirac delta function δ (x) is not a function in the traditional sense – it is rather a distribution – a linear operator defined by two properties. The first describes its values to be zero everywhere except at x = 0 where the value is infinite:
∞ x = 0 δ ()x = (4) 0 x ≠ 0
The second property provides the unit area under the graph of the delta function:
b ∫δ ()x dx = 1 where a < 0 and b > 0 a
The delta function is vanishingly narrow at x = 0 but nevertheless encloses a finite area. It is also known as the unit impulse function.
The Dirac delta function can be treated as the limit of the sequence of the following functions:
a) rectangular functions:
H ( xh+ )(−− Hxh ) δ ()x== lim Sh () x lim h0→→ h0 2h
b) Gauss distribution functions:
x2 − 1 σ 2 δ ()x ==lim Gσ () x lim e σσ→→00σπ
c) triangle functions:
0x<− h −x1 +−≤
n δ ()x== lim Dn lim nn→∞ →∞ π ()1nx+ 22
e) sine functions:
sin nx δ ()x = lim n→∞ πx Chapter 5 SPECIAL FUNCTIONS
Properties 1) Extension of the interval of integration to all real numbers still keeps the unit area under the graph of the delta function: ∞ ∫δ ()x dx = 1 −∞ 2) The Dirac delta function is a generalized derivative of the Heaviside step function: dH (x) δ ()x = dx It can be obtained from the consideration of the integral from the definition of the delta function with variable upper limit. It is obvious, that x 0 x < 0 ∫δ ()t dt = = H (x ) −∞ 1 x > 0 Therefore, the step function is formally an antiderivative of the delta function which now can be interpreted as a derivative of a discontinuous function.
3) Shifting in x : ∞ x = a δ ()x − a = 0 x ≠ a a+c ∫δ ()x − a dx = 1, c > 0 a−c
4) Symmetry: δδ(−=x) ( x) δδ( x − aax) =−( )
5) Derivatives: 1 δ ′()x = − δ ()x x The derivative can be defined as a limit of triangle functions and interpreted as a pure torque in mechanics. The higher order derivatives of the delta function are: k! δ ()k ()x = (− 1 )k δ ()x k1,2,...= xk
6) Scaling: 1 δ ()ax = δ ()x for a ≠ 0 a
7) There are some important properties of the delta function which reflect its application to other functions. If f (x) is continuous at x = a , then
f (x)δ (x − a) = f (a)δ (x − a) c ∫ f ()(x δ x − a )dx = f ()a b < a < c b ∞ ∫ f ()(x δ x − a )dx = f ()a −∞
x ∫ f ()t δ (t − a )dt = f (a )H (x − a ) x0 ≤ a x0 Chapter 5 SPECIAL FUNCTIONS
Integration with derivatives of the delta function (integration by parts): ∞∞ ∞ fxδδ′ xdxfx=− x f′′ x δ xdx =− f 0 ∫∫() () () ()−∞ () () () −∞ −∞ ∞∞ ∞ f xδδ′′′′′′′′ xdx=− f x x f x δ xdx =−−= f 0′ f 0 ∫∫() () () ()−∞ () () () () −∞ −∞ 8) Laplace transform: ∞ L{}δ ()x = ∫ e−sxδ ()x dx = 1 0 ∞ L{}δ ()x − a = ∫ e−sxδ ()x − a dx = e−as a > 0 0 9) Fourier transform: ∞ F{}δ ()x − a = ∫ e−iωxδ ()x − a dx = e−iaω a > 0 −∞
Applications The delta function is applied for modeling of impulse processes.
For example, the unit volumetric heat source applied instantaneously at time t = 0 is described in the Heat Equation by the delta function: ∂u − k∇2u = δ ()t ∂t
If the unit impulse source is located at the point r = r0 and releases all
energy instantaneously at time t = t0 , then the Heat Equation has a source ∂u − k∇2u = δ ()()t − t δ r − r ∂t 0 0 Impulse models are used for calculation of the Green’s function for non- homogeneous DE.
The other interpretation of the delta function δ ()t − t0 as a force applied
instantaneously at time t = t0 yielding an impulse of unit magnitude.
Example Consider IVP: unit impulse is imposed on a dynamical system initially at rest at t = 5 : y′′ + 9 y = δ (t − 5) initial conditions: y(0) = 0 y′(0) = 0 Solution: Apply the Laplace transform to the given initial value problem (use the property of the Laplace transform): s 2Y + 9Y = e−5s Solve the algebraic equation forY : e−5s Y = s2 + 9 The inverse Laplace transform yields a solution of IVP: 1 y()t = H ()()t − 5 sin 3 t − 5 3
The graph of the solution shows that the system was at rest until the time t = 5 , when an impulse force was applied yielding periodic oscillations. Chapter 5 SPECIAL FUNCTIONS
5.3 Sine Integral Function The sine integral function is defined by the formula:
x sint Si( x ) = ∫ dt x ∈ ()− ∞,∞ (5) 0 t
The integrand can be expanded in Taylor series and then integrated term by term yielding a series representation of the sine integral function:
∞ (− 1)n x2n+1 Si()x = ∑ (6) n=0 ()()2n + 1 2n + 1 !
Graphically it can be shown as:
> plot(Si(x),x=-25..25);
Si( x)
The limiting values of the sine integral function are determined by the Dirichlet integral ∞ sinω π ∫ dω = 0 ω 2 which was obtained in Section 3.5 (Remark 3.7, p.227) as a particular case of the Fourier transformation of the step function.
In Chapter 3, we discussed connection of Gibbs phenomena in the Fourier series approximations of functions with jumps and the properties of sine integral function.
Chapter 5 SPECIAL FUNCTIONS
5.4 Error Function The error function is the integral of the Gauss density function
x 2 2 erf ( x ) = ∫ e−t dt x ∈ ()− ∞,∞ (7) π 0
> plot(erf(x),x=-4..4); erf ( x )
The complimentary error function is defined as
∞ 2 2 erfc( x ) = 1 − erf ()x = ∫e−t dt (8) π x
> plot(erfc(x),x=-4..4);
erfc( x )
Series expansion of the error function:
2 ∞ (− 1)n x2n+1 erf ()x = ∑ π n=0 n! ()2n + 1
Derivatives of the error function:
d 2 2 erf ()x = e −x dx π
2 d − 4x 2 erf ()x = e −x dx 2 π
Chapter 5 SPECIAL FUNCTIONS
5.5 Gamma Function
Definition The Gamma function appears in many integral or series representations of special functions. Gamma was introduced by Leonard Euler in 1729 who investigated the integral function 1 q ∫ x p ()1xdx− p,q∈ 0 which for natural values p,q∈ is equal to p!q!
()pq1!++ With some transformation of this integral and taking the limits, Euler ended up with the result 1 n ∫ ()−ln x dx=≡ n!Γ () n + 1 0 Later, the gamma function was defined by the improper integral which converges for all x except of 0 and negative integers (Euler, 1781):
∞ Γ ()x = ∫ e−tt x−1dt (9) 0
> plot(GAMMA(x),x=-5..4);
Γ ()x
Properties: a) Γ (x + 1) = xΓ (x) (10)
∞ Γ (x + 1) = ∫ e−tt ()x+1 −1dt 0 ∞ = ∫ e−tt xdt 0 ∞ = −∫t xde−t 0 ∞ ∞ = − t xe−t + e−t dt x (limt xe−t = 0 ) 0 ∫ t→∞ 0 ∞ = x∫ e−tt x−1dt 0 = xΓ (x) Chapter 5 SPECIAL FUNCTIONS
b) When x = n is a natural number then
Γ (n) = (n − 1)! n = 1,2,3,...
Γ (n + 1) = n! n = 0,1,2,... provided that 0! = 1 (11)
The gamma function is a generalization to real numbers of a factorial (which is defined only for non-negative integers).
Proof: Γ (1) = 1 then using property (a) Γ (2) = Γ (1 + 1)()= 1Γ 1 = 1 = 1! Γ (3) = Γ (2 + 1)()= 2Γ 2 = 2 ⋅1 = 2!
… (then by mathematical induction)
c) The gamma function does not exist at zero and negative integers.
d) The gamma function is differentiable everywhere except at x = 0,−1,−2,... . It is a differentiable extension of the factorial. The derivative of the gamma function is called the digamma function. It is denoted by
Ψ ( x) = Γ ′( x)
Ψ ( x)
e) Stirling formula (approximation for largex , x > 9 ) x x Γ ()x + 1 ≈ 2πx e
f) Calculation of gamma function: Lanczos’ approximation in Fortran or C++ Numerical recipes.
g) Binomial coefficients:
z z! Γ ( z1+ ) == (12) w w!() z− w !ΓΓ()( w+−+ 1 z w 1 )
Chapter 5 SPECIAL FUNCTIONS
5.6 Bessel Functions
1. Bessel Equation In the method of separation of variables applied to a PDE in cylindrical coordinates, the equation of the following form appears: r 2 R′′()r + rR′(r)+ (λ2r 2 −ν 2 )R(r) = 0 r ∈[0,∞) This equation is the second order linear ordinary differential equation with variable coefficients. It includes two parameters λ and ν . It is not of the Euler-Cauchy type. Can be solved by the Frobenius method.
Simplify equation by the change of variables to eliminate parameter λ : y()x = R(r) x = λr dR dy dx = = λy′ dr dx dr d 2 R d dR d dy d dy dx = = λ = λ = λ2 y′′ dr 2 dr dr dr dx dx dx dr Then the equation becomes
x2 y′′ + xy′ + (x2 −ν 2 )y = 0 (14)
Now the equation is written in traditional variables, and it includes only one parameter ν . This equation is called to be a Bessel Equation of order ν . Apply a power-series solution to this equation.
2. Singular Points Singular points of the differential equation with variable coefficients are the points x at which the first coefficient becomes zero: x2 = 0 ⇒ x = 0 is the only singular point of the Bessel Equation. Therefore, if we find a power-series solution around this point, it will be convergent for all real numbers. Determine the type of the singular point. Divide the equation by x2 to rewrite it in the normal form: 1 ν 2 y′′ + y′ + 1 − y = 0 2 x x Identify coefficients of the equation in normal form: 1 ν 2 P()x = and Q()x = 1 − . x x2 Check if the singularity is removable:
xP()x = 1 is analytic: p0 = 1 2 2 2 2 x Q()x = x −ν is analytic: q0 = −ν Therefore, x = 0 is a regular singular point, and the Frobenius theorem can be used for solution of the Bessel Equation.
3. Indicial Equation Substitute coefficients p0 and q0 into the indicial equation (Ch. 2.42): 2 r + ()p0 − 1 r + q0 = 0 r 2 −ν 2 = 0 There are two roots of this equation:
r1 =ν r2 = −ν (choose ν ≥ 0 for convenience, later we can abandon this assumption). The Frobenius approach depends on the form of the difference of roots of the indicial equation:
r1 − r2 =ν − (−ν ) = 2ν Two cases of the Frobenius theorem may be involved: Chapter 5 SPECIAL FUNCTIONS
a) r1 − r2 = 2ν ≠ integer
b) r1 − r2 = 2ν = integer
this case includes ν = n , n ∈ N0 (positive integers and zero) and 2n +1 ν = (half of the odd integer) 2
In both cases, the first solution, following the Frobenius theorem, has to be found in the form: ∞ ∞ n+r1 n+ν y1 = ∑ cn x = ∑ cn x , x > 0 (15) n=0 n=0 Proceed to this solution, and then we will analyze how it handles the abovementioned cases.
4. First Solution Using assumed form of solution (2), calculate the derivatives ∞ n+ν y1 = ∑ cn x n=0 ∞ n+ν −1 y1′ = ∑()n +ν cn x n=0 ∞ n+ν −2 y1′′= ∑ ()()n +ν n +ν − 1 cn x n=0 and substitute them into the Bessel Equation (1): ∞ ∞ ∞ ∞ n+ν n+ν n+ν +2 2 n+ν ∑()()n +ν n +ν − 1 cn x + ∑ ()n +ν cn x + ∑cn x −ν ∑cn x = 0 n=0 n=0 n=0 n=0
Divide the equation by xν and collect the terms ∞ ∞ n n+2 ∑n()n + 2ν cn x + ∑cn x = 0 n=0 n=0 Rename indices: in the first sum m = n ; in the second sum n = m + 2 ∞ ∞ m m ∑ m()m + 2ν cm x + ∑ cm−2 x = 0 m=0 m=2 Combine both series: ∞ m 0 ⋅ c0 + ()1 + 2ν c1 x + ∑[]m (m + 2ν )cm + cm−2 x = 0 (16) m=2 Applying the Identity Theorem 2.6 to the term with summation, we obtain a recurrence relationship: −c c = m−2 for m ≥ 2 m m()m + 2ν
Using this relationship and also the first two terms of the equation, we get:
m = 0 0 ⋅ c0 = 0 ⇒ c0 = arbitrary
m = 1 ()1 + 2ν c1 = 0 ⇒ c1 = 0 (by assumption,ν > 0 ) −c −c −c m = 2 c = 0 = 0 ⇒ c = 0 2 2()2 + 2ν 2 ⋅ 2()1 +ν 2 2 ⋅ 2()1 +ν −c m = 3 c = 1 = 0 ⇒ c = 0 3 3()3 + 2ν 3 −c c c m = 4 c = 2 = 0 ⇒ c = 0 4 4()4 + 2ν 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2()()1 +ν 2 +ν 4 24 ⋅ 2()()1 +ν 2 +ν Chapter 5 SPECIAL FUNCTIONS
−c m = 5 c = 3 = 0 ⇒ c = 0 5 5()5 + 2ν 5 −c −c −c m = 6 c = 4 = 4 ⇒ c = 0 6 6()6 + 2ν 2 ⋅ 2 ⋅ 3()3 +ν 6 24 ⋅ 3!()()()1 +ν 2 +ν 3 +ν … All coefficients with odd indices are equal to zero. Recognizing the pattern, we can determine the coefficients with even indices: (− 1)k c c = 0 k = 0,1,2,... 2k 22k ⋅ k! ()()()1 +ν 2 +ν k +ν This expression may be written in more elegant form if the gamma function is used. Multiply and divide the expression by Γ ()1 +ν : (− 1)k c Γ (1 +ν ) c = 0 2k 22k ⋅ k! Γ ()()()()1 +ν 1 +ν 2 +ν k +ν Repeatedly using the property (a) of the gamma function, we squeeze the product in the denominator: Γ ()1 +ν (1 +ν )(2 +ν ) (k +ν ) = Γ (2 +ν )(2 +ν )(3 +ν )( k +ν )= Γ (k +ν + 1) Then the expression for the coefficients becomes: (− 1)k c Γ (ν + 1) c = 0 2k 22k ⋅ k! Γ ()ν + k + 1 1 Choose the value for the arbitrary coefficient c = , then 0 2ν Γ ()ν + 1 (− 1)k c = 2k 22k+ν ⋅ k! Γ ()ν + k + 1 Then the solution becomes 2k+ν k x k 2k+ν ()− 1 ∞ ()− 1 x ∞ 2 y1 ()x = ∑ 2k+ν = ∑ k=0 2 ⋅ k! Γ ()ν + k + 1 k=0 k! Γ ()ν + k + 1 This series solution converges absolutely for all x ≥ 0 because there are no other singular points. The function represented by this power-series solution is called to be a Bessel function of the 1st kind of order ν and it is denoted by Bessel function of the 1st kind 2k+ν k x ()− 1 ∞ 2 Jν ()x = ∑ (17) k=0 k! Γ ()ν + k + 1
This formula is valid for any real ν ≥ 0 (including integers ν = n and 2n +1 half of the odd integers ν = ). 2 If ν is integer (let ν = n ), then the gamma function is replaced by the factorial Γ ()n + k + 1 = (n + k)! and the solution simplifies to:
2k+n k x ()− 1 ∞ 2 J n ()x = ∑ (18) k=0 k! ()n + k !
This is a Bessel function of the 1st kind of integer order (including zero).
Chapter 5 SPECIAL FUNCTIONS
5. Second Solution Case 1) r2 = −ν ≠ int eger 2n +1 (including halfs of the odd integers ν = ) 2 Because x appears squared in the equation, the second solution may be obtained from the first by a simple replacement of ν by −ν :
2k−ν k x ()−1 ∞ 2 J −ν ()x = ∑ (19) k=0 k!Γ()−ν + k +1
Functions Jν (x) and J −ν (x) are linearly independent. It can be shown that the Wronskian of Bessel functions (17) and (19) is:
Jν (x) J −ν (x) 2 sinνπ W []Jν ()x , J −ν ()x = = − (20) Jν′ ()x J −′ν ()x πx
If ν ≠ n is not integer, then the Wronskian is not zero and the Bessel
functions Jν (x) and J −ν (x) are linearly independent. Then the general solution of the Bessel Equation may be written as
y(x) = c1Jν (x)+ c2 J −ν (x) (21)
Case 2) ν = n When ν is integer, the Wronskian (20) is equal to zero
for any x > 0 , therefore, Bessel functions of integer orders J n (x) and
J −n (x) are linearly dependent.
We can show that in this case functions J n ()x and J −n (x) are just multiples of each other. Indeed, write a Bessel function of negative integer order replacing ν by n in equation (19): 2k−n k x ()−1 ∞ 2 J −n ()x = ∑ (22) k=0 k!Γ()− n + k +1 and change the index of summation k by s to make substitution in the exponentiation 2k − n = 2s + n then k = s + n , and equation (22) becomes 2s+n s+n x ()−1 ∞ 2 J −n ()x = ∑ (23) s=−n ()()s + n !Γ s +1 Consider a factor in the denominator Γ()s +1 : when s +1 ≤ 0 (non- positive integer), the gamma function is unbounded, therefore, the first n terms from s = −n to s = −1 in the summation (23) are equal to zero, then taking that into account for integers, Γ()s +1 = s! , we obtain 2s+n 2s+n s+n x s x −1 −1 ∞ () ∞ () 2 n 2 J −n ()x = ∑ == ()−1 ∑ s=0 ()s + n !s! s=0 ()s + n !s! or n J −n (x) = (−1) J n (x) (24)
So, function J −n (x) is the function J n (x) up to the sign. We need to find the second linearly independent solution. According to the Frobenius Theorem 2.11, it can be found in the form: Chapter 5 SPECIAL FUNCTIONS
∞ k−ν J −ν ()x = ∑ dk x + cJν ()x ln x k=0 or we can use the reduction formula (Part I, 2.2.9 3, p.131) to find the second solution: dx J x = J x −ν ()ν ()∫ 2 xJν ()x where long division and the Cauchy product should be used (which is tedious but manageable). Bessel function of the 2nd kind But, instead, traditionally, the second independent solution is introduced by the definition of the Bessel function of the 2nd kind of orderν :
cosνπJ (x)− J (x) Y ()x = ν −ν for ν ≠ n not integer (25) ν sinνπ
and for integers, as the limit
cosνπJν (x)− J −ν (x) Yn ()x = limYν ()x = (26) ν →n sinνπ
which appear to exist for all n = 0,±1,±2,... (or n ∈ Z ). Can be derived: 2 x Yn ()x = ln + γ J n ()x − ...... (27) π 2 1 1 1 γ = lim1+ + + + − ln m = 0.5772... Euler’s constant m→∞ 2 3 m
Functions Yn (x) have a logarithmic singularity at zero, while functions
J n ()x are finite at zero, that leads to their linear independence. It can be shown that the Wronskian for these functions is given by
Jν (x) Yν (x) 2 W []Jν ()x ,Yν ()x = = (28) Jν′ ()x Yν′ ()x πx
General solution of Bessel equation Functions Jν (x) and Yν (x) are linearly independent for all ν (including integers), and can be used for construction of the general solution of the Bessel equation:
y(x) = c1Jν (x)+ c2Yν (x) (29)
When the order of the Bessel equation is not integer, the complete solution may be also given only in the terms of Bessel functions of the first kind:
y(x) = c1Jν (x)+ c2 J −ν (x) (30)
The second solution was derived mostly for integer roots, so, we emphasize it by the following statement: the complete solution of the Bessel equation of integer order n is given by:
y(x) = c1J n (x)+ c2Yn (x) (31)
1 6. Bessel functions of half orders It is happens that functions of orders ν = n ± can be expressed in terms 2 1 of elementary functions. Show it forν = ± . Consider BE (14). 2 Chapter 5 SPECIAL FUNCTIONS
1 u − Use substitution y = = x 2 u x 3 1 1 − − du y′ = − x 2 u + x 2 2 dx 5 3 1 3 − − du − du y′ = x 2 u − x 2 + x 2 4 dx dx then the equation becomes
1 2 −ν u′′ + 1+ 4 u = 0 2 x 1 Forν = ± , this equation reduces to a linear ODE with constant 2 coefficients u′′ + u = 0 the general solution of which is given by
y = c1 cos x + c2 sin x Apply the back substitution u = y x , then the solution becomes cos x sin x y = c1 + c2 x x 2 If we choose for constants c = c = to be consistent with definition 1 2 π (17), we obtain that Bessel functions of half orders are given by: 2 cos x J 1 ()x = (32) 2 π x 2 sin x J 1 ()x = (33) − 2 π x 3 It can be verified that Bessel functions of ν = ± are: 2 sin x − cos x 2 x J 3 ()x = (34) 2 π x cos x + sin x 2 x J 3 ()x = (35) − 2 π x Other Bessel functions of half of odd integer orders also can be expressed in terms of elementary functions. These functions are used for construction of spherical Bessel functions
J 1 (x) n n+ π 2 n 1 d sin x jn ()x = = x − (36) 2 x x dx x
Y 1 (x) n n+ π 2 n 1 d cos x yn ()x = = −x − (37) 2 x x dx x which are solutions of the Bessel equation x2 y′′ + 2xy′ + [x2 − n(n +1)]y = 0 n = 0,±1,±2,... (38) This equation appears as one of the ODE in the separation of variables of the Laplacian in spherical coordinates [Abramowitz and Stegun, p.437]
Chapter 5 SPECIAL FUNCTIONS
7. Bessel functions of the 3rd kind A linear combination in the general solution (29) assumes that coefficients
c1 and c2 are real numbers. We also considered the variable x to be a real number too. But the obtained equations and functions are valid also for complex numbers. Two particular combinations of Bessel functions
J v ()z and Yν (z) with complex coefficients lead to the introduction of the complex version of Bessel functions, which also are the solutions of the Bessel equation but in the field of complex numbers z ∈ Z :
We define two Bessel functions of the 3rd kind of order ν ( they are also called Hankel functions) as
(1) Hν (z) = Jν (z)+ iYν (z) (39)
(2) Hν (z) = Jν (z)− iYν (z) (40)
or we can express them in terms of the Bessel function Jν ()z only if
function Yν (z) is replaced in this definition by a complex version of equation (25): J (z) − e−νπi J (z) H ()1 ()z = −ν ν (41) ν isinνπ J (z)− eνπi J (z) H ()1 ()z = −ν ν (42) ν − i sinνπ Definitions (41) and (42) are for v ≠ integer . For ν = n , n = 0,1,2,... , we take the limits: −νπi ()1 J −ν (x) − e Jν (x) H n ()z = lim (43) ν →n isinνπ νπi ()2 J −ν (x)− e Jν (x) H n ()z = lim (44) ν →n − i sinνπ (1) (2) Because Hankel functions Hν (z) and Hν (z) are linear combinations of Bessel functions of the 1st and the 2nd type, they have the similar properties. (1) (2) The Wronskian of Hankel functions Hν (z) and Hν ()z is 4i W [H ()1 ()z , H ()2 ()z ]= − ν ν πz (1) (2) therefore functions Hν (z) and Hν (z) form a fundamental set for the Bessel equation; and the general solution of the Bessel equation can be written as (1) (2) y(x) = c1Hν (z)+ c2 Hν (z) for any order of the Bessel equation ν (including integers).
8. Properties of Bessel functions
a) Functions Jν (x) and Yν (x) are both oscillatory; they have infinitely many roots for x > 0 .
x2 x4 x6 x8 J ()x = 1− + − + + 0 4 64 2304 147456 x x3 x5 x7 J ()x = − + − + 1 2 16 384 18432 x2 x4 x6 x8 J ()x = − + − + 2 8 96 3072 184320 Chapter 5 SPECIAL FUNCTIONS
2ν Identities J ()x = J ()x − J ()x (45) ν +1 x ν ν −1 2ν J ()x = J ()x − J ()x (46) ν −1 x ν ν +1
d Differential identities [xν J ()x ]= xν J ()x J ′ (x)()= −J x (47) dx ν ν −1 0 1 d [x −ν J ()x ]= −x −ν J ()x (48) dx ν ν +1
ν xJ ′ (x)+νJ (x) = xJ (x) J ′ ()x = J ()x − J ()x (49) ν ν ν −1 ν ν −1 x ν ν xJ ′ (x)−νJ (x) = −xJ (x) J ′ ()x = −J ()x + J ()x (50) ν ν ν +1 ν ν +1 x ν
ν +1 ν +1 Integral identities ∫ x Jν (x)dx = x Jν +1 (x)+ c (51) −ν +1 −ν +1 ∫ x Jν (x)dx = −x Jν −1 (x)+ c (52)
∞ Addition theorem J n ()x + y = ∑ J k ()()x J n−k y (53) k =−∞
The same properties also hold for Yν (x) .
b) Functions Iν (x) and Kν (x) are not oscillatory
x2 x4 x6 x8 I ()x = 1+ + + + + 0 4 64 2304 147456 x x3 x5 x7 I ()x = + + + + 1 2 16 384 18432 x2 x4 x6 x8 I ()x = + + + + 2 8 96 3072 184320
1 x t − ∞ t 2 n 9. Generating functions e = ∑ J n ()x t (55) n=−∞
cos x = J 0 (x)− 2J 2 (x)+ 2J 4 (x)− 2J 6 (x)+
sin x = 2J1 (x)− 2J 3 (x)+ 2J 5 (x)+
(What can be obtained for x=1 ?)
n Expansions of x 1 = J x (x)+ 2J 2 (x)+ 2J 4 (x)+ + 2J 2k (x)+
1 x = J ()x + 3J ()x + 5J ()x + + (2k +1 )J ()x + 2 1 3 5 2k +1
1 x = J ()x + 4J ()x + 9J ()x + + k 2 J ()x + 8 2 4 6 2k Chapter 5 SPECIAL FUNCTIONS
10. Modified Bessel equation The modified Bessel equation is given by x 2 y′′ + xy′ − (x 2 +ν 2 )y = 0 (56) which can be written in the form of Bessel equation (14) with the second parameter λ = i : x 2 y′′ + xy′ + (i 2 x 2 −ν 2 )y = 0 (57) which has a general solution given by equation (29) with x replaced by ix:
y(x) = c1 Jν (ix)+ c2Yν (ix) (58) Equation (58) provides a solution of the modified Bessel equation (91) in complex form. But it is desirable to have a real form of solution. Consider 2k +ν 2k +ν k ix x −1 ∞ () ∞ 2 ν 2 Jν ()ix = ∑ = i ∑ (59) k =0 k!Γ()ν + k +1 k =0 k!Γ()ν + k +1 because: k (−1 )k ()i 2k +ν = (−1 )k ()i 2k iν = (−1 )k (i 2 ) iν = ()()−1 k −1 k iν = ()−1 2k iν = iν . Then define a function which is called the modified Bessel function of the 1st kind of order ν :
2k +ν x ∞ −ν 2 Iν ()x = i Jν (ix )= ∑ (60) k =0 k!Γ()ν + k +1
which is a real function and which is a solution of the modified Bessel equation (56). Notation I for this function reflects the method of its definition, and it means “the function of imaginary argument”. For negative values of parameter −ν , define a second solution of the modified Bessel equation as
2k −ν x ∞ ν 2 I −ν ()x = i J −ν (ix )= ∑ (61) k =0 k!Γ()−ν + k +1
The Wronskian of functions Iν (x) and I −ν (x) can be calculated as 2sinνπ W[]I ()x , I ()x = − ≠ 0 for v ≠ integer ν −ν πx therefore, functions Iν (x) and I −ν (x) are linearly independent and form a fundamental set, then the general solution of the modified Bessel equation of non-integer order is given by
y(x) = c1 Iν (x)+ c2 I −ν (x) v ≠ integer (62)
In the case of integer orders, function I −n (x) is the same as function
I n ()x . Indeed, when ν is changed for n in equation (61) n n n n n n −n I −n ()x = i J −n (ix) = i (−1) J n (ix) = i (−1)()i [i J n ix ]
2 n n n n = ()i ()−1 I n ()()()()x = −1 −1 I n x = I n ()x For integer orders ν = n , n = 0,1,2,3,… , the second solution of the modified Bessel equation is defined with the help of the modified Bessel function of the 2nd kind of order ν : Chapter 5 SPECIAL FUNCTIONS
π I (x)− I (x) K ()x = −ν ν (63) ν 2 sinνπ
as the limit of
K n (x) = lim Kν (x) (64) ν →n
11. Equations solvable in terms of Bessel functions
Consider some generalizations of the Bessel equation which also can be solved in terms of the Bessel function. a) Consider a generalized Bessel differential equation of the form
1− 2m α()2m −1 m 2 − p 2ν 2 y′′ + − 2α y′ + p 2 a 2 x 2 p−2 +α 2 + + y = 0 2 x x x
if y = y(x) is any solution of the Bessel equation, then the function y = x m eαx y(ax p ) is a solution of the generalized equation. For instance, for any real ν ≥ 0 (including integers), the general solution can be written as m αx p p y = x e [c1 Jν (ax )+ c2Yν (ax )] or for non-integer orders, a general solution can be written as m αx p p y = x e [c1 Jν (ax )+ c2 J −ν (ax )] Proof of this statement can be made by the appropriate change of variables and by reduction of the differential equation to the Bessel equation.
Example 1 Check that the modified Bessel equation x 2 y′′ + xy′ − (x 2 +ν 2 )y = 0 is a particular case of the generalized equation. Rewrite it in the form of the generalized equation: 1 ν 2 y′′ + y′ + −1− y = 0 2 x x 1− 2⋅0 0 −1⋅ν 2 y′′ + − 2⋅0 y′ + 1 −1 x 2⋅1−2 + y = 0 ()( ) 2 x x from which we can identify m = 0 , α = 0 , p = 1 , and a 2 = −1 and, therefore, solutions of the modified Bessel equation should include functions
Jν (ix), J −ν (ix) , and Yν ()ix what we know from Section 10.
Airy equation Example 2 Consider the Airy equation y′′ − xy = 0 which is the simplest case of the linear 2nd order ODE with variable coefficients. This equation has applications in dynamics (oscillation of an aging spring), quantum mechanics and optics. Chapter 5 SPECIAL FUNCTIONS
Rewrite the Airy equation in the form of the generalized equation
1 1 9 2 1− 2⋅ 3 − ν 2 9 4 2⋅ −2 4 4 y′′ + − 2⋅0 y′ + ⋅− x 2 + 0 + 0 + y = 0 x 4 9 x from which we can identify 1 3 4 1 m = , α = 0 , p = , a 2 = − , and ν 2 = 2 2 9 9 from the last equation we can determine the order of the equation 1 ν = ± 3 Then solutions of the Airy equation can be written as 1 2 3 y ()x = x 2 J ix 2 1 1 3 3 1 2 3 2 2 y 2 ()x = x J 1 ix − 3 3 If we rewrite Bessel functions of the 1st kind of complex arguments in terms of modified Bessel functions using equation (95), then the solutions become 2 3 y ()x = xI x 2 1 1 3 3 2 3 2 y 2 ()x = xI 1 x − 3 3 These two linearly independent solutions (note, that order of modified Bessel functions is not integer) may be used for construction of the traditional form of solutions x 2 3 2 3 2 2 Ai()x = I 1 x − I 1 x 3 − 3 3 3 3 x 2 3 2 3 2 2 Bi()x = I 1 x + I 1 x 3 − 3 3 3 3 which are called Airy functions. The next plot shows the graph of Airy functions
Bi( x)
Aix( )
Chapter 5 SPECIAL FUNCTIONS
b) This form of equation is a particular case of the previous equation, but it is more convenient for applications in simpler cases: y′′ + a 2 x p−2 y = 0 solutions of this equation have the forms 2a p y()x = xJ x 2 1 p p
2a p y()x = xY x 2 1 p p Solutions of the Airy equation can be obtained in this case much faster.
c) Equation y′′ + (a 2 e 2x − p 2 )y = 0 has solutions x y(x) = J p (ae ) x y(x) = J − p (ae ) x y(x) = Y p (ae )
Chapter 5 SPECIAL FUNCTIONS
12. Orthogonality of Bessel functions We know from Sturm-Liouville theory that solutions of the self-adjoint differential equation satisfying homogeneous boundary conditions form a complete set of functions orthogonal with some weight function (Sturm- Liouville theorem). Consider application of this theory to solutions of BVP for BE.
Self-adjoint form of BE The Bessel equation of order ν with parameter λ x 2 y′′ + xy′ + (λ2 x 2 −ν 2 )y = 0 can be reduced to a self-adjoint form with the help of a multiplying factor a x 1 dx dx 1 ∫ a 1 ∫ 2 1 1 µ x = e 0 = e x = e ln x = () 2 2 a0 x x x 1 After multiplication of BE by it can be reduced to self-adjoint form x ν 2 ′ 2 []xy′ + − + λ xy = 0 (identify p()x = x ) x
Then, the Sturm-Liouville Problem in the interval x ∈[L,L12] produces
infinitely many values of the parameter λn (eigenvalues) for which there
exist non-trivial solutions yn (x) (eigenfunctions). According to the Sturm-Liouville theorem, the obtained eigenfunctions are orthogonal with the weight function p(x) = x :
L2 ∫ xyxyxdx0nm() () = for n ≠ m L1
Orthogonal sets for circular domain Consider BE in the finite circle 0 ≤ x ≤ L . The general solution is given by
0 ≤ x ≤ L y(x) = c1Jν (λx)+ c2Yν (λx)
The physical sense of solution of classical PDE requires a finite value of
solution in all points of [0,L]. Bessel functions of the second kind Yν (x) are unbounded at x = 0 , therefore, to satisfy this condition we have to put 0 L x an arbitrary constant c2 equal to zero. Then solution of BE becomes
y(x) = c1 Jν (λx)
boundary conditions: Consider the homogeneous boundary conditions at x = L :
I Dirichlet y x = 0 ( ) x=L
II Neumann y′ x = 0 ( ) x=L
III Robin [y′(x)+ Hy(x)]x=L = 0
Chapter 5 SPECIAL FUNCTIONS
We are looking for the values of the parameter λ which provide non-
trivial solutions y(x) = c1 Jν (λx) satisfying boundary conditions I-III. These values can be found by substitution of the solution into the boundary conditions I-III as the positive roots of the following equations:
Equations for eigenvalues I Jν (λL) = 0
ν II − λJ ()λL + J ()λL = 0 ν +1 L ν
( λ0 = 0 is also an eigenvalue when ν = 0 ) d J ()λx = 0 λJ′ (λx) = 0 dx ν ν ν λ− Jν +1 ()λx + Jν ()λx = 0 λx
ν III − λJν +1 ()λL + H + Jν ()λL = 0 L
Proof: y′(x)+ Hy(x) d = J ()λx + HJ ()λx (= λJ′ ()λx + HJ (λx) , Oz) dx ν ν ν ν ν = λ− Jν +1 ()λx + Jν ()λx + HJν ()λx λx ν = −λJν +1 ()λx + H + Jν ()λx x
In the particular case, when the Bessel function is of zero order, ν = 0 , equations for eigenvalues are:
I J 0 (λL) = 0
II J 1 (λL) = 0 (and λ0 = 0 is also eigenvalue)
III −λJ 1 (λL)+ HJ 0 (λL) = 0
Orthogonality Obtained equations generate infinitely many eigenvalues λn , n = 1,2,3,... For which the corresponding eigenfunctions are:
yxnn( ) = { Jν (λ x)}
The corresponding set of solutions {Jν (λn x)} is orthogonal with respect to the weight function p(x) = x :
L 0 n ≠ m xJ λ x J λ x dx = ∫ ν ()()n ν m 2 0 N n = m
where the squared norm of eigenfunctions is determined as (Ozisik N. Heat Transfer, p.133; McLachlan Bessel Functions for Engineers, p.110): Chapter 5 SPECIAL FUNCTIONS
L L x 2 ν 2 Norm of eigenfunctions N 2 = xJ 2 ()λ x dx = J ′ 2 ()λ x + 1− J 2 ()λ x ν ,n ∫ ν n ν n 2 2 ν n 0 2 λn x 0 L2 ν 2 = J ′2 ()λ L + 1− J 2 ()λ L 2 ν n 2 2 ν n λn L or integrating with Maple: L L 2 2 2 x 2 Nν ,n = xJν ()λn x dx =− Jννν()λλλn1n1nxJ−+ () xJ () x ∫ 2 0 0 The derivative can be expressed as (use chain rule and identity in sec. 6) d J ′ ()λ x = J ()λ x ν n dx ν n ν = λn − Jν +1 ()λn x + Jν ()λn x λn x ν = −λ J ()λ x + J ()λ x n ν +1 n x ν n or if we use the other identity for lower order then d J ′ ()λ x = J ()λ x ν n dx ν n ν =−λλn1nJνν− ()xJx() λ n λn x ν =−λλJ ()xJx() λ n1nνν− x n
Then taking into account that eigenvalues λn satisfy equations I-III (that simplifies expressions), the squared norm for specific boundary conditions is given:
2 2 2 L 2 L 2 I N ,n = J 1 ()λn L or = J ()λ L ν 2 ν + 2 ν −1n L2 ν 2 L2 II N 2 = 1 − J 2 λ L N 2 = ν ,n 2 2 ν ()n 0,0 2 λn L 2 L2 H 2 ν 2 III N 2 = + 1 − J 2 λ L ν ,n 2 2 2 ν ()n 2 λn λn L
Bessel-Fourier Series The obtained orthogonal systems can be used for constructing the function expansion in a generalized Fourier series
∞ fx()= ∑ aJnnν ()λ x n1=
where coefficients cn are determined from the equation L L xJxfxdxλ ∫ xJxfxdxν ()λn () ∫ ν ()n () 0 0 an = L = 2 2 Nν ,n ∫ xJxdxν ()λn 0
Chapter 5 SPECIAL FUNCTIONS
Example 3 (I Dirichlet boundary condition)
yL( ) = 0 Consider an orthogonal set obtained as a solution of a Dirichlet problem yxJ()ν x with Bessel functions of zero order { nn()= ν ()λ } where
0 L x eigenvalues λn are positive roots of equation Jν ()λL0= . The squared norm of eigenfunctions can be calculated as L2 N 2 = J 2 ()λ L ν ,n 2 ν +1n Then expansion of function f (x) in Fourier-Bessel series has the form L xJxfxdxλ ∞ ∫ ν ()n () 2 0 fx()= ∑ aJn0()λ n x, where an = 22 n1= LJLν +1n()λ (it is also known as the Hankel series (1869)).
Consider now expansion of the function f ()x = 1− H (x − 1 ), x ∈[0,3] in the Hankel series of order ν = 0 . Coefficients are 3 ∫ xJ0n()λ x dx 0 JJ1n()λλ12 1n() an ===2222 NNLJL0,nλn 0,nλλ n 1() n and the expansion becomes
∞ 2 J1n0n(λλ) Jx( ) fx()= ∑ 2 9 n1= λλn1J3() n
This example can be illustrated with a Maple presentation (SF-1.mws)
Example 4 (II Neumann boundary condition)
Consider an orthogonal set obtained as a solution of the Neumann ′ yL()= 0 (ν ) problem with Bessel functions y n (x)()= Jν λn x where eigenvalues
λn are positive roots of the equation ν 0 L x − λJ ()λL + J ()λL = 0 ( λ = 0 is also an eigenvalue for ν = 0 ) ν +1 L ν 0 The squared norm of eigenfunctions can be calculated as L2 ν 2 L2 N 2 = 1− J 2 λ L and N 2 = ν ,n 2 2 ν ()n 0,0 2 λnL 2 Fourier-Bessel series: L xJxfxdxλ ∞ ∫ 0n()() ()0 0 ν = 0 fx()=+ a0nn∑ ay() x, where an = 2 n1= Nν ,n 2 L axfxdx= 0 2 ∫ () L 0 L axJxfxdx= λ ∞ nn∫ ν ()() ()ν 0 ν > 0 fx()= ∑ aynn () x, where 2 n1= Nν ,n
Chapter 5 SPECIAL FUNCTIONS
Consider now expansion of the function f (x)()= 1− H x − 1 , x ∈[0,3], ν = 0 :
3 ∫ xdx 0 1 an ==2 N0,0 9
3 xJλ x dx ∫ 0n() 2 0 L2JJ1n()λλ 1n() an ==222 = NJLJ30,n29λ n 0()λλλ n n 0() n
Maple solutions: ν = 0 SF-1-2-0.mws ν = 1 SF-1-2-1.mws
Example 5 (III Robin boundary condition)
Consider an orthogonal set obtained as a solution of the Robin problem yL′()+= HyL () 0 ()ν with Bessel functions of zero order yxJnn()= ν ()λ x where eigenvalues
λn are positive roots of equation
0 L x ν −++=λλJνν+1 ()LH JL0() λ L
The squared norm of eigenfunctions can be calculated as L2 H 2 ν 2 N 2 = + 1 − J 2 λ L ν ,n 2 2 2 ν ()n 2 λn λn L
Fourier-Bessel series: L xJxfxdxλ ∞ ∫ 0n()() ()ν 0 f ()x = ∑ aynn () x where an = 2 n1= Nν ,n
Consider expansion of function f (x) = 1− H (x − 1), x ∈[]0,3 , H = 2
Maple solution (for ν = 1 ) SF-1-3.mws
Chapter 5 SPECIAL FUNCTIONS
SF-1.mws Example 3 Fourier-Bessel series I Dirichlet boundary condition ν = order of Bessel Functions > nu:=0; ν := 0 > L:=3; L := 3 > f(x):=1-Heaviside(x-1); f()x := 1() − Heaviside x − 1 Characteristic equation: > w(x):=BesselJ(nu,x*L); w()x := BesselJ()03 , x > plot(w(x),x=0..10);
Eigenvalues: > lambda:=array(1..200); λ := array()1.. 200 , [ ] > n:=1: for m from 0 to 50 do y:=fsolve(w(x)=0,x=m/2..(m+1)/2): if type(y,float) then lambda[n]:=y: n:=n+1 fi od: > for i to 2 do lambda[i] od; 0.8016085192 1.840026037 > N:=n-1;n:='n':i:='i':m:='m':y:='y':x:='x': N := 24 Eigenfunctions: ¾ y[n]:=BesselJ(nu,lambda[n]*x); yn := BesselJ()0, λn x Squared Norm: >NY[n]:=int(x*y[n]^2,x=0..L): NY[n]:=subs(BesselJ(nu,L*lambda[n])=0,NY[n]):
Fourier-Bessel coefficients: ¾ a[n]:=int(x*y[n]*f(x),x=0..L)/NY[n];
2 BesselJ ()1, λ a := n n 9 2 λn BesselJ()13 , λn Fourier-Bessel series: > u(x):=Sum(a[n]*y[n],n=1..N); 24 BesselJ ()1, λ BesselJ ()0, λ x 2 n n u()x := ∑ 2 n = 1 9 λ BesselJ()13 , λ n n > u(x):=sum(a[n]*y[n],n=1..N): > plot({f(x),u(x)},x=0..L,discont=true,color=black);
Chapter 5 SPECIAL FUNCTIONS
Orthogonal sets for annular domain Consider BE in the finite circle 2 2 2 2 LxL12≤≤ x y′′ + xy′ + (λ x −ν )y = 0 , x ∈ (L1 ,L2 ) with homogeneous boundary conditions: dy h − k + h y = 0 H = 1 1 dx 1 1 k x=L1 1 0 L L dy h 12 k + h y = 0 H = 2 2 dx 2 2 k x=L2 2 The general solution is given by
y(x) = c1Jν (λx)+ c2Yν (λx) A BVP for BE in the finite domain according to the Sturm-Liouville
theorem generates an infinite set of eigenvalues λn and corresponding
eigenfunctions yn (x) orthogonal with the weight function p()x = x . A particular form of the orthogonal set depends on the type of boundary conditions. Consider a case when both boundary conditions are of Dirichlet type:
Example 6 (Dirichlet-Dirichlet boundary conditions)
Boundary conditions: y = 0 x=L1 y = 0 x=L2 Apply boundary conditions to the general solution of BE:
c1Jν (λL1 )+ c2Yν (λL1 ) = 0
c1Jν (λL2 )+ c2Yν (λL2 ) = 0
This is a homogeneous system of two linear algebraic equations for c1
and c2 . Rewrite it in the matrix form
Jν (λL1 ) Yν (λL1 )c1 0 = Jν ()λL2 Yν ()λL2 c2 0 We are looking for non-trivial solution of BVP, i.e. both coefficients in general solution cannot be zero
c1 0 ≠ c2 0 A homogeneous linear system has a non-trivial solution only if the determinant of the system matrix is equal to zero:
Jν (λL1 ) Yν (λL1 ) Equation for eigenvalues λn : det = Jν (λL1 )Yν (λL2 )()()− Jν λL2 Yν λL1 = 0 Jν ()λL2 Yν ()λL2
The roots of this equation yield the eigenvalues λn for which BVP has
non-trivial solutions yn (x) (eigenfunctions). Oscillatory property of
Bessel functions provides an infinite set of eigenvalues λn and corresponding eigenfunctions are
yn (x) = c1,n Jν (λn x)+ c2,nYν (λn x)
Determine now the coefficients c1,n and c2,n from a system where eigenvalues are substituted Chapter 5 SPECIAL FUNCTIONS
Jν (λn L1 ) Yν (λn L1 )c1,n 0 = Jν ()()λn L2 Yν λn L2 c2,n 0
Because a linear system has a singular matrix, solutions for c1,n and c2,n are linearly dependent and can be determined just from one equation, let it be the second one
c1,n Jν (λn L2 )+ c2,nYν (λn L2 ) = 0 one of the unknowns in this equation is a free parameter, choose 1 −1 c1,n = , then c2,n = Jν ()λn L2 Yν ()λn L2 Then eigenfunctions have the form:
Jν (λn x) Yν (λn x) Eigenfunctions yn ()x = − Jν ()λn L2 Yν ()λn L2
The norm of eigenfunctions is given by:
L2 2 2 Nν ,n = ∫ xyn ()x dx L1 2 L2 Jν ()λn x Yν ()λn x = x − dx ∫ J λ L Y λ L L1 ν ()n 2 ν ()n 2 1 L2 1 L2 = xJ 2 ()λ x dx + xY 2 ()λ x dx − J 2 λ L ∫ ν n Y 2 λ L ∫ ν n ν ()n 2 L1 ν ()n 2 L1 1 L2 − xJ ()()λ x Y λ x dx J λ L Y λ L ∫ ν n ν n ν ()()n 2 ν n 2 L1
= ... express in terms of Jν +1 , …
Summary: For an annular domain with boundary conditions: y = 0 x=L1 y = 0 x=L2
Eigenvalues λn are positive roots of the characteristic equation
Jν (λL1 )Yν (λL2 )()()− Jν λL2 Yν λL1 = 0 The eigenfunctions are
Jν (λn x) Yν ()λn x yn ()x = − Jν ()λn L2 Yν ()λn L2 Fourier-Bessel series: ∞ f ()x = ∑ an yn ()x n=1 L L ∫ xyn ()()x f x dx ∫ xJν ()()λn x f x dx 0 0 where an = L = 2 2 Nν ,n ∫ xyn ()x dx 0
Maple examples: ν = 0 SF-AD-1-0.mws ν = 1 SF-AD-1-1.mws
L1 = 2 , L2 = 5 f (x)()= 1 − H x − 3 Chapter 5 SPECIAL FUNCTIONS
Consider a case when both boundary conditions are of Robin type:
Example 7 (Robin-Robin boundary conditions) SF-5.mws
Consider BE in the annular domain 2 2 2 2 x y′′ + xy′ + (λ x −ν )y = 0 , x ∈ (L1 ,L2 ) with homogeneous boundary conditions:
dy h1 − k1 + h1 y = 0 H1 = dx x=L k1 0 L12 L 1 dy h k + h y = 0 H = 2 2 dx 2 2 k x=L2 2 The general solution is given by
y(x) = c1Jν (λx)+ c2Yν (λx) The derivative of the general solution (use chain rule and differential identities) d νν yx()=− c11λλ Jνν++ ( x ) + J() λλλ x +− c 21 Y νν () x + Y() λ x dx λλ x x Substitute into boundary conditions:
ν ν x = L1 − c1λ− Jν +1 ()λL1 + Jν ()λL1 − c2λ− Yν +1 ()λL1 + Yν ()λL1 + c1H1Jν (λL1 )()+ c2 H1Yν λL1 = 0 λL1 λL1 ν ν x = L2 c1λ− Jν +1 ()λL2 + Jν ()λL2 + c2λ− Yν +1 ()λL2 + Yν ()λL2 + c1H 2 Jν (λL2 )()+ c2 H 2Yν λL2 = 0 λL2 λL2 Collect terms ν ν x = L1 c1 λJν +1 ()λL1 + H1 − Jν ()λL1 + c2 λYν +1 ()λL1 + H1 − Yν ()λL1 = 0 L1 L1 ν ν x = L2 c1 − λJν +1 ()λL2 + H 2 + Jν ()λL2 + c2 − λYν +1 ()λL2 + H 2 − Yν ()λL2 = 0 L2 L2 Denote: ν a11 = λJν +1 ()λL1 + H1 − Jν ()λL1 L1 ν a12 = λYν +1 ()λL1 + H1 − Yν ()λL1 L1 ν a21 = − λJν +1 ()λL2 + H 2 + Jν ()λL2 L2 ν a22 = − λYν +1 ()λL2 + H 2 − Yν ()λL2 L2 Then a system for coefficients has the following matrix form:
a11 a12 c1 0 = a21 a22 c2 0 A necessary condition for a system to have a non-trivial solution is
a11 a12 det = 0 a21 a22 it yields a characteristic equation for values of the parameter λ for which the BVP has a non-trivial solution: Chapter 5 SPECIAL FUNCTIONS
Equation for eigenvalues λn : ν ν λJν +1 ()λL1 + H1 − Jν ()λL1 − λYν +1()λL2 + H2 + Yν ()λL2 L1 L2
ν ν − λYν +1 ()λL1 + H1 − Yν ()λL1 − λJν +1 ()λL2 + H 2 − Jν ()λL2 = 0 L1 L2
The positive roots of this equation provide an infinite set of eigenvalues
λn . Then for the determined eigenvalues λn , coefficients c1,n and c 2, n can be found from one of the equations of the system (choose the second one):
a21c1 + a22c2 = 0 One of the coefficients can be taken as a free parameter, choose 1 1 c1 = , then c2 = a21 a22
With determined coefficients, solutions of the BVP yn (x) (eigenfunctions) have the form:
Jν (λn x) Yν (λn x) Eigenfunctions: yn (x) = − a21,n a22,n
J (λ x) = ν n ν − λn Jν +1 ()λn L2 + H 2 + Jν ()λn L2 L2 Y (λ x) − ν n ν − λnYν +1 ()λn L2 + H 2 − Yν ()λn L2 L2
The norm of the eigenfunctions is determined by the integral
L2 2 2 Nν ,n = ∫ xyn ()x dx L1
Fourier-Bessel series: ∞ f ()x = ∑ an yn ()x n=1
L2 L2 2 ∫ xyn ()()x f x dx ∫ xyn ()()x f x dx L1 L1 where an = = L2 2 2 Nν ,n ∫ xyn ()x dx L1
Maple example ν = 0 SF-AD-9-0.mws ν = 1 SF-AD-9-1.mws
L1 = 2 , L2 = 5
H1 = 2 , H 2 = 3 f (x) = 1 − H (x − 3) Chapter 5 SPECIAL FUNCTIONS
5.7 Legendre Functions
1. Legendre Equation Separation of variables of the Laplacian in a spherical coordinate system yields a group of ODE one of which has the form ′ m 2 1− x 2 y′ + n n + 1 − y = 0 []() () 2 1− x where m and n are separation constants. This equation is called Legendre’s associated differential equation. Solution of this equation includes Legendre’s associated functions of st nd m m degree n and of order m of the 1 and the 2 kind Pn ()x and Qn (x). When m = 0 (in a case when the Laplacian does not depend on the variable φ ), equation is called the Legendre’s differential equation ′ [(1− x 2 )y′] + n()n + 1 y = 0 Solution of this equation include Legendre’s functions of degree n of the st nd 1 and the 2 kind Pn (x) and Qn (x).
2. Solution of Legendre Equation Consider the Legendre differential equation rewritten in standard form (1− x 2 )y′′ − 2xy′ + n(n + 1)y = 0 x ∈ R This equation has two singular points x = ±1, all other points are ordinary points. We will apply a power-series solution method around the ordinary point x = 0 (the interval of convergence for this solution is (−1,1) ). Assume that the solution is represented by a power series ∞ k y = ∑ ak x k =0 then derivatives of the solution are ∞ k −1 y′ = ∑ kak x k =1 ∞ k −2 y′′ = ∑ k()k − 1 ak x k =2 Substitute them into equation
∞∞∞ 2k2k1k−− ()1x−−−∑∑∑ kk1ax()kk 2xkax ++= nn1 () ax k 0 k2=== k1 k0 ∞∞∞∞ k2− k k k ∑∑∑∑kk()−−−−++= 1axkkkk kk () 1ax 2kax nn () 1 ax 0 k2==== k2 k1 k0 ∞∞∞∞ kkkk (change kk2=− in the first term) ∑∑∑∑()()k1k2ax++k2+ − kk1ax () − k − 2kaxnn1 k ++ () ax0 k = k0==== k2 k1 k0
∞ 2 ⋅1⋅a2 + 2 ⋅3⋅a3 x − 2 ⋅ a1 x + n(n + 1)()a0 + n n + 1 a1 x + ∑{}ak []− k()k − 1 − 2k + n ()n + 1 + ()()k + 1 k + 2 ak +2 = 0 k =2 ∞ nn()++⋅⋅+⋅⋅++− 1a02 12a{} 23a 3 nn () 1 2a 1 x +∑{}ak []n()()n + 1 − k k + 1 + ()()k + 1 k + 2 ak +2 = 0 k =2
Using the comparison theorem, determine the relation for coefficients: −n(n + 1) a = a 2 1⋅2 0 2 − n(n + 1) (n −1)(n + 2) a = a = a 3 2⋅3 1 2⋅3 1 Chapter 5 SPECIAL FUNCTIONS
k(k + 1)− n(n + 1) −(n − k)(n + k + 1) a = a = a k = 2,3,... k +2 ()()k + 1 ⋅ k + 2 k ()()k + 1 ⋅ k + 2 k
Coefficients a0 and a1 are arbitrary, consider them to be the parameters for the general solution and collect the terms corresponding to these coefficients, then the power series solution of the Legendre Equation becomes
nn1()+24 nn2n1n3() −++( )( ) nn2n4n1n3n5( −−+++)( )( )( )( ) 6 y()x =−a10 x + x − x + ... 2! 4! 6!
()()n1n2−+35() n1n3n2n4 −−++( )( )( ) ( n1n3n5n2n4n6 −−−++)( )( )( )( )( +) 7 +−ax1 x + x − x + ... 3! 5! 7!
= a0 Ln,1 ()x + a1 Ln,2 ()x Choose a sequence of non-negative values of n = 0,1,2,... then corresponding solutions are (note, that in the solution all terms except for finite number alternatingly disappear: if n=2k is even then in the first series all terms with multiple (n − 2k) disappear, if n=2k+1 is odd then in the second series all terms with multiple ()n − 2k − 1 disappear, and they become the finite polynomials)
L1,n (x) L2,n ()x
n = 0 a0 a1 L2,0 ()x
n = 1 a0 L1,1 (x) a1 x 2 n = 2 a0 (1− 3x ) a1 L2,2 ()x
5 3 n = 3 a0 L1,3 (x) a1 x − x 3 … n n! n−1 (n + 1)! Choose a = − 1 2 a = − 1 2 0 () 2 1 () 2 2 n n − 1 n + 1 2 n ! 2 n ! ! 2 2 2 Then Legendre functions of the 1st kind for different values of parameter n generate the following set of polynomials
P0 (x) = 1
P1 (x) = x 3 1 P ()x = x 2 − 2 2 2 5 3 P ()x = x 3 − x 3 3 2 35 15 3 P ()x = x 4 − x 2 + 4 8 4 8 … which are called Legendre polynomials. Because Legendre polynomials are solutions of the separated Laplace equation in spherical coordinates, they are also called spherical harmonics (and the method of solution in terms of Legendre functions is called correspondingly the Method of Spherical Harmonics). Recall that this system of polynomials up to scalar multiple was also obtained from orthogonalization of the linear independent set of monoms {1,x,x23 ,x ,...} on the interval [−1,1] . Chapter 5 SPECIAL FUNCTIONS
Recurrence formula (n + 1)Pn+1 (x) = (2n + 1)xPn (x)− nPn−1 (x)
n 1 d n Rodrigues’ formula P ()x = ()x 2 − 1 n 2 n n! dx 2
Orthogonality of Legendre polynomials Legendre polynomials are orthogonal in the interval [−1,1] with the weight function p(x) = 1 0mn≠ 1 ∫ PxPxdxmn() () = −1 2 mn= 2n+ 1
Fourier-Legendre series Legendre polynomials can be used for expansion of the function f (x), x ∈[]−1,1 in the Fourier-Legendre series:
∞ f ()x = ∑ cn Pn ()x n=0
where expansion coefficients are
1 f x P x dx ∫ ()n () 1 1 c = −1 = n + f x P x dx n 1 ∫ ()n () 2 2 −1 ∫ Pn ()x dx −1
Example 8 (expansion in Fourier-Legendre series (spherical harmonics))
f ()x = H (x) x ∈[−1,1]
Maple Solution: SF-8.mws
∞ 1 1 f ()x = ∑ n + ∫ Pn ()x dxPn ()x n=0 2 0
Integral transform The integral transform based on the Fourier-Legendre expansion 1 f n = ∫ f ()x K n ()x dx −1 with inverse transform ∞ f ()x = ∑ f n K n ()x n=0 where the kernel of the integral transform is defined as a normalized Legendre function 1 K ()x = n + P ()x n 2 n Chapter 5 SPECIAL FUNCTIONS
SF-8.mws Example 8 Fourier-Legendre Series
> restart; > with(orthopoly);
[]G,,,,,H L P T U > for n from 0 to 6 do P(n,x) od; 1 x 1 3 x2 − + 2 2 5 3 x3 − x 2 2 3 35 15 + x4 − x2 8 8 4 63 35 15 x5 − x3 + x 8 4 8 5 231 315 105 − + x6 − x4 + x2 16 16 16 16 > f(x):=Heaviside(x);
f()x := Heaviside ()x Fourier-Legendre coefficients: > c[n]:=(n+1/2)*int(f(x)*P(n,x),x=-1..1); 1 1 c := n + ⌠ P,()nxd x n 2 ⌡0 Fourier-Legendre series: > u(x):=sum(c[n]*P(n,x),n=0..10);
1 218295 315315 1702701 984555 1616615 u()x := + x − x3 + x5 − x7 + x9 2 65536 16384 32768 16384 65536 > plot({f(x),u(x)},x=-1..1);
> u(x):=sum(c[n]*P(n,x),n=0..100): > plot({f(x),u(x)},x=-1..1);
Chapter 5 SPECIAL FUNCTIONS
The Best Approximation by Polynomials
Consider a vector space of square integrable functions L1,12 [− ] . The span of all
polynomials of order n is a subspace of L1,12 [− ] . Call it π n .
Let fx()∈π n and let
n fxnnn()= ∑ cPx () k1= be the nth partial sum of the Fourier-Legendre expansion of the function f ( x) in [−1,1] .
th Then fn ()x provides the best approximation of the function f ()x by the n
order polynomials, i.e. function fn ( x) is the closest to the function f ()x among
the functions in π n in the sense that it minimizes the distance
1 2 fx()−=−−=− pnnnn () x()() f p,fp∫ f p dx −1
Chapter 5 SPECIAL FUNCTIONS
Graphs of Legendre polynomials
Legendre-1.mws Legendre polynomials
> restart; > with(orthopoly):
> plot({P(0,x),P(1,x),P(2,x)},x=-1..1);
P0
P1 P2
> plot({P(3,x),P(4,x),P(5,x)},x=-1..1);
P3 P5
P4
> plot({P(6,x),P(7,x),P(8,x)},x=-1..1);
P7 P8 P6
Chapter 5 SPECIAL FUNCTIONS
5.8 Exercises:
1) Show
sin xδ (x − nπ ) = 0 sin xδ ′(x − nπ ) = (− 1)n+1δ (x − nπ )
(by multiplying both sides by an arbitrary differentiable function and integrating)
2) Show ∞ f (− 2) ∫δ ()()3x + 6 f x dx = −∞ 3
In general ∞ f (x0 ) ∫δ []g()x f ()x dx = where g(x0 ) = 0 −∞ g′()x0
3) Solve the IVP and sketch the solution curves (use Maple and Laplace transform):
y′′ − y = (t − 2)H (t − 2) y(0) = 0 y′()0 = 0
y′′ + 2y′ + y = 3δ (t − 2) y(0) = 1 y′()0 = 1
4) Sign function is defined as
1 x > 0 sgn()x = −1 x < 0 a) Express sgn(x) in terms of Heaviside step function H (x) b) Express Heaviside step function H (x) in terms of sgn(x) d c) Calculate sgn()x dx d) Sketch the graph of sgn(x − 3)
5) Investigate convergence of the Bessel function Jν ()x using the ratio test (see supplemental materials)
6) Use term-by-term differentiation (why can we do it?) to show
d J ()x = −J ()x dx 0 1
Chapter 5 SPECIAL FUNCTIONS
7) Solve in terms of Bessel functions
x 2 y′′ + (x − 2)y = 0
8) Finish Examples 4 and 5 in section 6.12 Orthogonality of Bessel Functions 9) Hermite’s differential equation with parameter λ is
y2xyy0′′′− +=λ x ∈−∞∞(), , λ ∈ (HE)
a) Solve the HE by the power series method
b) Consider two linearly independent solutions
y1 ( x...) =
yx2 ( ) = ... which include parameter λ
c) If λ is a non-negative even integer, λ = 0,2,4,...,2n,..., then the series
terminates, and one obtains, alternating for y1 and y2 , polynomials of
degree n , which are called Hermitian polynomials H n ()x . Write them in traditional form in which the coefficient of xn is equal to 2n (the second solution is not polynomial).
d) Rewrite HE in self-adjoint form and determine the weight function wx( )
e) Check if the HP are orthogonal with the weight function wx( ) over (−∞∞, ) : ∞ ∫ Hmn()xH ()() xwxdx0= if mn≠ −∞
f) Give an example of function representation into Fourier-Hermite series
Chapter 5 SPECIAL FUNCTIONS