LECTURE 1 Table of Contents Two special equations: Bessel s and Legendre s equations. p. 259-268. Fourier-Bessel and Fourier-Legendre series. p. 453-460. Boundary value problems in other coordinate system. Laplace s equation in polar coordinates. p. 504-507. Exercises: 11.5.1-3, 11.5.1-4, 11.5.1-5, 13.1-6 Two special equations: Bessel s and Legendre s equations The following linear, homogeneous differential equations of the 2nd order play an important role in mathematical physics x 2 y + xy + (x 2 ν 2 )y = 0, ν [0, ), x [0, ) (1) (1 x 2 )y 2xy + n(n + 1)y = 0, n = 1, 2,..., x ( 1, 1) (2) Eqs. (1) are named the Bessel differential equation and the Legendre differential equation, respectively. The solutions of Bessel s equations often appear for problems with rotational symmetry, which are formulated in polar coordinates. Similarly, solutions to Legendre s equation appear in spherical symmetric problems formulated in spherical coordinates. The general solution of (1) and (2) both have the form, cf. Theorem 4.2, p. 130 y(x) = c 1 y 1 (x) + c 2 y 2 (x) (3) y 1 (x) and y 2 (x) are linearly independent solutions to the homogeneous differential equations, and c 1 and c 2 are arbitrary constants. One should compare the Bessel equation (1) with the corresponding Cauchy-Euler equation, see p. 173 x 2 y + xy ν 2 y = 0 (4) The two differential equations are identical save the coefficient of y(x). However, for ν 0 even these factors become identical for x 0. Hence, the solutions of (1) and (4) for ν 0 should become asymptotically identical as x 0. For ν [0, ), ν 0, 1, 2,... (i.e. ν is not integer valued) a fundamental solution system to the Bessel equation can be shown to be 1
y 1 (x) = J ν (x) = y 2 (x) = J ν (x) = n=0 n=0 ( 1) n x ) 2n+ν (5) n! Γ(1 + ν + n)( 2 ( 1) n x ) 2n ν (6) n! Γ(1 ν + n)( 2 J ν (x) is known as the Bessel function of the first kind and order ν. Γ(x) signifies the so-called gamma function defined as, see App. 1. Γ(x) = 0 e t t x 1 dt (7) The gamma function fulfills the fundamental relation Γ(x) = (x 1)Γ(x 1), which can be proved by integration by parts of (7). Obvious, Γ(1) = 1. For n = 2, 3,... the following relation may then be derived recursively Γ(n) = (n 1)! (8) The series solutions (5) and (6) have been systematically derived in the textbook. Alternatively, the validity of these series may be proved simple by insertion into (1). Obviously, y 1 (x) and y 2 (x) become identical for ν = 0. Actually, for integer valued ν = n it can be shown that J n (x) and J n (x) become linearly dependent as follows J n (x) = cos(nπ)j n (x) = ( 1) n J n (x) (9) If ν = 0, 1, 2,... we can still use the solution (5). A linear independent solution can in this case be constructed from the limit passing Y n (x) = lim ν n cos(νπ)j ν (x) J ν (x) sin(νπ) (10) Throughout the limit passing the right hand side is a linear combination of the solutions J ν (x) and J ν (x). Hence, the Bessel equation is fulfilled at any time during the limit passing. Both the numerator and the denominator becomes zero in the limit as seen from (9). However, a finite limit value Y n (x) can be shown to exist, which also must be a solution to the Bessel equation. Y n (x) is known as the Bessel function of the second kind and order n. A series solution similar to (5) and (6) can also be derived for these functions. 2
The graphs of the lower order Bessel functions J 0 (x), J 1 (x), Y 0 (x), Y 1 (x) have been shown on Fig. 1, along with the asymptotic form of the series solutions valid in the vicinity of x = 0 The following recursive formulas can be derived, see p. 264-265 J ν(x) = ν x J ν(x) J ν+1 (x) Y n(x) = J 0(x) = J 1 (x) Y 0(x) = Y 1 (x) n x Y n(x) Y n+1 (x) (11) Next, x is replaced with λx and dx with λdx in (1). This leads to the so-called parametric Bessel equation of order ν x 2 y + xy + (λ 2 x 2 ν 2 )y = 0 (12) For ν = 0, 1, 2,... the solution becomes y(x) = c 1 J n (λx) + c 2 Y n (λx) (13) (12) can be written in the self-adjoint form d ( xy (x) ) + (λ 2 x n2 ) y(x) = 0 (14) dx x where ν = n has been assumed. (14) is recognized as a differential equation for a Sturm-Liouville problem with r(x) = x, p(x) = x and q(x) = n 2 /x, see Box 1. The differential equation (14) is solved in the interval (0, b). Since, r(0) = 0 the problem is singular at x = 0. Assume the following boundary condition of Sturm-Liouville type is prescribed for solutions of (14) at x = b A 2 y(b) + B 2 y (b) = 0 (15) It then follows from (20) that the eigenfunctions y i (x) of (14), satisfying the boundary condition (15), fulfills the orthogonality condition (21) in Box 1 (with a = 0). Of the two solution J n (λ i x) and Y n (λ i x) only J n (λ i x) is bounded at x = 0. Here and in what follows we shall only consider problems for which the eigenfunctions are of bounded variations, so y i (x) = J n (λ i x). Then, J n (λ i x) are seen to be orthogonal with respect to the function p(x) = x over the interval [0, b] as follows 3
0 xj n (λ i x)j n (λ j x)dx = 0, λ i λ j (16) The boundary condition (15) attains the form A 2 J n (λb) + B 2 λj n(λb) = 0 (17) In (17), A 2 and B 2 are given from the physics of the problem. Further, the chain rule has been d used, dx J n(λx) = J n(λx) d dx (λx) = λj n(λx). Here and in what follows J n( ) denotes differentiation with respect to the argument. Eigenvalues λ i are determined as solutions of λ fulfilling (17). For i = j the integral (16) can be evaluated analytically, based on simple manipulations of the differential equation (14), see p. 456. With y(x) = J n (λx), the result becomes 0 x Jn(λx) 2 dx = 1 ) ) 2 2 b2( J n(λb) 1 ( + (b 2 n2 ) 2 2 λ 2 J n (λb) (18) a) b) ( x ) 2 J 0 (x) = 1 + 2 J 1 (x) = x 2 1 ( x ) 3 + 2 2 Fig.1: a) Bessel functions of the first kind. b) Bessel functions of the second kind. Y 0 (x) = 2 π ln x 2 + Y 1 (x) = 2 π 1 x + Fig. 2: Legendre polynomials. 4
Box 1: Sturm-Liouville Problems, p. 449 A Sturm-Liouville problem is defined by the two-point boundary value problem d ( ) ( ) r(x)y + q(x) + λp(x) y(x) = 0, x (a, b) dx A 1 y(a) + B 1 y (a) = 0 A 2 y(b) + B 2 y (b) = 0 (19) where p, q, r and r are continuous functions on the closed interval [a, b], and r(x) > 0 and p(x) > 0 in the open interval (a, b). Obviously, (19) has the trivial solution y(x) 0. Non-trivial solutions y 1 (x), y 2 (x),... exist for discrete values λ 1, λ 2,... of the parameter λ. ( λ i, y i (x) ) represent the eigensolutions of the problem. Based on (19) the following identity may be proved for two different eigensolutions ( λ i, y i (x) ), ( λ j, y j (x) ), cf. p.450 ( ) λi λ j p(x)y i (x)y j (x)dx = a ( r(b) y i (b)y j(b) y j (b)y i(b) ) ( ) r(a) y i (a)y j(a) y j (a)y i(a) (20) The terms within the brackets on the right hand side of (20) vanish due to the boundary conditions in (19). Then, the following orthogonality condition of the eigenfunctions prevail a p(x)y i (x)y j (x)dx = 0 (21) Hence, the eigenfunctions of the Sturm-Liouville problem (19) are orthogonal over the interval of definition weighted with the function p(x). If r(a) > 0 and r(b) > 0 we talk about a regular Sturm-Liouville problem. In this case both the boundary conditions in (19) are necessary in order to prove the orthogonality condition (21). However, if r(a) = 0 or r(b) = 0 it becomes indifferent whether the terms within the brackets on the right hand side of (20) vanish. Hence, if r(a) = 0 or r(b) = 0 the corresponding boundary condition at x = a or x = b needs not be fulfilled of eigenfunctions of the differential equation, in order that the orthogonality condition (21) is valid. Such problems are referred to as singular Sturm-Liouville problems. 5
For Legendre s equation with the parameter n one solution can be shown to be an nth degree polynomial y 1 (x) = P n (x) = a n x n + a n 1 x n 1 + + a 1 x + a 0 (22) The coefficients a n, a n 1,..., a 1, a 0 are determined so Legendre s differential equation is fulfilled. Additional, another solution y 2 (x) in terms of an infinite polynomial series may be derived. The systematic derivation of y 1 (x) and y 2 (x) has been performed in the textbook, but is not that interesting. P n (x) is denoted the Legendre polynomial of order n. The lower order Legendre polynomials are P 0 (x) = 1, P 1 (x) = x P 2 (x) = 1 ( 3x 2 1 ), P 3 (x) = 1 5x 2 2( 3 3x ) P 4 (x) = 1 ( 35x 4 30x 2 + 3 ), P 5 (x) = 1 63x 8 8( 5 70x 3 + 15x ) (23) The graph of the functions have been shown on Fig. 2. As seen all polynomials pass through the point (1, 1). Further, Legendre polynomials of even order n are symmetric around the y axis, whereas the polynomials of odd order are all skew symmetric. The validity of the results (23) can be proved by insertion of y(x) = P n (x) into the relevant version of (2). The first few of these are n = 0 : (1 x 2 )P 0 2xP 0 = 0 n = 1 : (1 x 2 )P 1 2xP 1 + 2P 1 = 0 n = 2 : (1 x 2 )P 2 2xP 2 + 6P 2 = 0 n = 3 : (1 x 2 )P 3 2xP 3 + 12P 3 = 0 (24) It is not necessary to know the explicit expression for the Legendre polynomials in order to evaluate function values. These are more efficiently obtained from the recurrence relation, p. 267 P k+1 (x) = 2k + 1 k + 1 xp k(x) k k + 1 P k 1(x), k = 1, 2,... (25) As an example, (25) provides for k = 1: P 2 (x) = 3 2 xp 1(x) 1 2 P 0(x) = 3 2 x x 1 2 1 (26) Eq. (2) can be written in the following self-adjoint form 6
d ( (1 x 2 )y ) + n(n + 1)y = 0, x ( 1, 1) (27) dx The equivalence of (2) and (27) follows by carrying out the differentiation in (27). (27) is recognized as a differential equation for a Sturm-Liouville problem with r(x) = (1 x 2 ), p(x) = 1, q(x) = 0 and λ 2 = n(n + 1), see Box 1. Since, r( 1) = r(1) = 0 the problem is singular at both the boundaries x = 1 and x = 1. Then, the orthogonality condition (21) with a = 1, b = 1 and p = 1 is fulfilled for any non-trivial solution y n (x) to (27), whatever boundary conditions these may fulfill at x = ±1. Hence, the Legendre polynomials P n (x) are orthogonal with respect to the weight function p(x) = 1 over the interval [ 1, 1], i.e. 1 1 P m (x)p n (x) dx = 0, m n (28) For m = n the integral (28) can be evaluated analytically, p. 459 1 1 P 2 n(x) dx = 2 2n + 1 (29) Fourier-Bessel and Fourier-Legendre series Bessel functions of the first kind J n (λ i x), and the Legendre polynomials P n (x) have been shown to be eigenfunctions to special, singular Sturm-Liouville eigenvalue problems. Then, it follows that it is possible to expand a function defined on the interval of the said eigenvalue problem into a generalized Fourier series after these eigenfunctions. Fourier-Bessel series We want to expand the function f(x), x [0, b] after the function system {J n (λ 1 x), J n (λ 2 x),...}, where the order n is common for all functions. These deviates by the eigenvalues, which are determined as solutions to (17). The expansion reads f(x) = c j J n (λ j x) (30) j=1 (30) is multiplied with r(x)j n (λ i x) = xj n (λ i x), followed by an integration over the interval [0, b]. From the orthogonality condition (16) follows 7
0 c i = x J n (λ i x) f(x) dx = c j x J n (λ i x)j n (λ j x) dx j=1 0 x J n(λ i x) f(x) dx 0 x J = n(λ 2 i x) dx 0 2 0 x J n(λ i x) f(x) dx b 2( J n(λ i b) ) ( ) 2 (Jn + b 2 n2 (λ i b) ) 2 λ 2 i (31) where (18) has been used. From (17) follows that J n(λ i b) = h bλ i J n (λ i b), h = ba 2 B 2 (32) Then (31) can be written as c i = 2λ2 i 0 x J n(λ i x) f(x) dx ( )( h 2 n 2 + b 2 λ 2 i Jn (λ i b) ) 2 (33) The boundary condition (32) with the result (33) for the Fourier-Bessel expansion coefficients is referred to as Case II, p. 457. Assume that B 2 = 0 in (17), corresponding to the boundary condition J n (λ i b) = 0 (34) Then, from the recurrence relation (11) follows that J n(λ i b) = n λ i b J n(λ i b) J n+1 (λ i b) = J n+1 (λ i b) (35) By the use of (34) and (35), (31) reduces to c i = 2 0 x J n(λ i x) f(x) dx b 2( J n(λ i b) ) 2 = 2 0 x J n(λ i x) f(x) dx b 2( J n+1 (λ i b) ) 2 (36) The boundary condition (34) with the result (36) for the Fourier-Bessel expansion coefficients is referred to as Case I, p. 456. Finally, assume that A 2 = 0 and n = 0 in (17), corresponding to the boundary condition J 0(λ i b) = 0 J 1 (λ i b) = 0 (37) 8
where (11) has been used. Since, J 1 (0) = 0 it follows that λ 1 = 0 is an eigenvalue to the boundary condition (37). For λ 2, λ 3,..., which are all non-zero, (31) reduces to c i = 2 0 x J 0(λ i x) f(x) dx b 2( J 0 (λ i b) ) 2, i = 2, 3,... (38) For i = 1 the first statement in (31) for n = 0 and λ 1 = 0 provides c 1 = 0 x J 0(0 x) f(x) dx 0 x J 0 2(0 x) dx = 0 x f(x) dx 0 x dx = 2 b 2 0 x f(x) dx (39) where the result J 0 (λ 1 x) = J 0 (0 x) 1 has been used. The boundary condition (37) with the results (38), (39) for the Fourier-Bessel expansion coefficients is referred to as Case III, p. 457. In this case the Fourier-Bessel series (30) becomes f(x) = c 1 + c j J 0 (λ j x) (40) j=2 Fourier-Legendre series We want to expand the function f(x), x [ 1, 1] after the function system {P 0 (x), P 1 (x), P 2 (x),...}. The expansion reads f(x) = c n P n (x) (41) n=0 (41) is multiplied with P m (x), followed by an integration over the interval [ 1, 1]. From the orthogonality condition (28) follows 1 1 c m = P m (x) f(x) dx = 1 c n P m (x)p n (x) dx n=0 1 1 P m(x) f(x) dx 1 1 P 2 m(x) dx 1 = 2m + 1 2 1 1 P m (x) f(x) dx, m = 0, 1, 2,... (42) where (29) has been used. (41) is a power series quite similar to an truncated Taylor expansion. The difference is that (41) exists even for a non-analytical function, for which a Taylor expansion pr. definition does not exist. 9
Fig. 3: Fourier-Legendre approximation to non-analytical functions. Fig. 3 shows a Fourier-Legendre approximation function truncated after the 5th term to a nonanalytical (different Taylor expansions exist for the two branches). Theorem 11.5: Conditions for Convergence If f (x) and f 0 (x) are piecewise continuous on the open intervals (0, b) or ( 1, 1), then the Fourier-Bessel series or the Fourier-Legendre series converge to f (x) at any point, where f (x) is continuous, and to the average 21 [f (x ) + f (x+ )] at a point, where f (x) is discontinuous. 10
Box 2: Gradient vector and Laplace operator in polar coordinates Fig. 4: Differentials in polar and Cartesian coordinates. By use of simple projections the following relations between the polar differentials dr, rdθ and the Cartesian differentials dx and dy can be established } dr = cos θdx + sin θdy rdθ = sin θdx + cos θdy r x = cos θ, r y = sin θ θ x = 1 r sin θ, θ y = 1 (43) r cos θ u = u(r, θ) is considered a function of r and θ. By the use of the chain rule of partial differentiation the following result is obtained for the components of the gradient vector in polar coordinates u x = u r r x + u θ θ x u y = u r r y + u θ θ y u = cos θ r 1 u sin θ r θ u = sin θ r + 1 u cos θ r θ Partial differentiation of the gradient components in (44) with respect to x and y and use of the chain rule provides 2 u x 2 = 1 r sin2 θ u r + cos2 θ 2 u r 2 1 r cos θ sin θ 2 u r θ + 2 u cos θ sin θ r2 θ 1 r cos θ sin θ 2 u r θ + 1 r 2 sin2 θ 2 u θ 2 (45) 2 u y 2 = 1 r cos2 θ u r + sin2 θ 2 u r 2 + 1 r cos θ sin θ 2 u r θ 2 u cos θ sin θ r2 θ + 1 r cos θ sin θ 2 u r θ + 1 r 2 cos2 θ 2 u θ 2 finally, addition of the components in (45) provides the following result for the Laplace operator in polar coordinates 2 u = 2 u x 2 + 2 u y 2 = 2 u r 2 + 1 u r r + 1 2 u r 2 θ 2 (46) (44) 11
Boundary value problems in other coordinate system. Laplace s equation in polar coordinates Fig. 5: Laplace boundary value problems in Cartesian and polar coordinates. In Section 12.2 we learned how to solve Laplace s differential equation for a rectangular domain by means of the separation method. The method require, that homogeneous boundary conditions (either Dirichlet of Neumann conditions) are prescribed on two opposite sides. In this lecture we shall learn how to solve the same equation within a segment or the entire of a ring shaped domain with inner radius r 0 and outer radius r 1. It turns out to be favorable to solve the problem in polar coordinates. Then we consider u = u(r, θ) as a function of the polar coordinates r and θ. Hence, the first problem is to express the Laplace operator in these coordinates. The derivations is described in Box 2, and the result becomes 2 u = 2 u x 2 + 2 u y 2 = 2 u r 2 + 1 u r r + 1 2 u r 2 θ 2 = 0 (47) Next, the boundary value problem defined in Fig. 5 can be solved by the separation method. We search for product solutions to (47) of the type u(r, θ) = R(r)Θ(θ) (48) Insertion in (47) provides 2 u = 2 u r 2 + 1 u r r + 1 2 u r 2 θ 2 = R (r)θ(θ) + 1 r R (r)θ(θ) + 1 r 2 R(r)Θ (θ) = 0 r 2 R (r) + rr (r) R(r) = Θ (θ) Θ(θ) = λ2 (49) In the last statement of (49) the left hand side of the equation is a function of r, whereas the right hand side is a function of θ. This is only possible if both sides are equal to a constant λ 2, the so-called separation constant, which here at first is assumed to be positive real. Then, (49) is equivalent to the following ordinary differential equations 12
r 2 R (r) + rr (r) λ 2 R(r) = 0 (50) Θ (θ) + λ 2 Θ(θ) = 0 (51) (51) has the solution Θ(θ) = c 1 cos(λθ) + c 2 sin(λθ) (52) (50) is a differential equation of Cauchy-Euler s type, see p. 173. The solution is given as R(r) = c 3 r λ + c 4 r λ (53) The eigenvalues λ 1, λ 2,... of the parameter λ is determined by (52) by insertion into the homogeneous boundary conditions at θ = 0 and θ = θ 0. By use of the superposition principle the general solution can be written u(r, θ) = n=1 ( ) c 1,n cos(λ n θ) + c 2,n sin(λ n θ) )(c 3,n r λn + c 4,n r λn (54) we shall make a distinction between the following cases: 1 : Normal case : r 0 > 0, r 1 <, θ 0 < 2π a) r 0 = 0 c 4 = 0 b) r 1 = c 3 = 0 2 : Degenerated cases : (r λ, r ) (r λ, r 0) c) θ 0 = 2π λ n = n ( u(r, θ) = u(r, θ + 2π) ) 13
Example: Dirichlet boundary values The normal case is considered. u(r, 0) = 0, u(r, θ 0 ) = 0 implies Θ(0) = 0, Θ(θ 0 ) = 0. Insertion of the solution (52) into the indicated boundary conditions provides c 1 cos(λ 0) + c 2 sin(λ 0) = 0 c 1 = 0 c 2 sin(λθ 0 ) = 0 λ n θ 0 = nπ λ n = nπ θ 0 (55) Then, the solution (54) reduces to u(r, θ) = n=1 (A n r nπ θ 0 ) ) + B n r nπ θ 0 sin (nπ θθ0 (56) A n = c 2 c 3 and B n = c 2 c 4 are determined from the non-homogeneous boundary conditions at r = r 0 and r = r 1. 14