Lecture Notes on PDE s: Separation of Variables and Orthogonality

Transcription

1 Lecture Notes on PDE s: Separation of Variables and Orthogonality Richard H. Rand Dept. Theoretical & Applied Mechanics Cornell University Ithaca NY version 15 Copyright 28 by Richard H. Rand 1

2 R.Rand Lecture Notes on PDE s 2 Contents 1 Three Problems 3 2 The Laplacian 2 in three coordinate systems 4 3 Solution to Problem A by Separation of Variables 5 4 Solving Problem B by Separation of Variables 7 5 Euler s Differential Equation 8 6 Power Series Solutions 9 7 The Method of Frobenius 11 8 Ordinary Points and Singular Points 13 9 Solving Problem B by Separation of Variables, continued 17 1 Orthogonality Sturm-Liouville Theory Solving Problem B by Separation of Variables, concluded Solving Problem C by Separation of Variables 27

3 R.Rand Lecture Notes on PDE s 3 1 Three Problems We will use the following three problems in steady state heat conduction to motivate our study of a variety of math methods: Problem A : Heat conduction in a cube 2 u = for <x<l, <y<l, <z<l (1) with the assumption that u = u(x, z,only) (that is, no y dependence)), and with the boundary conditions: Problem B : Heat conduction in a circular cylinder u = on x =,L (2) u = on z = (3) u =1 on z = L (4) 2 u = for <r<a, <z<l (5) with the assumption that u = u(r, z,only) (that is, no θ dependence), and with the boundary conditions: Problem C : Heat conduction in a sphere u = on r = a (6) u = on z = (7) u =1 on z = L (8) 2 u = for <ρ<a (9) with the assumption that u = u(ρ, φ,only) (that is, no θ dependence), and with the boundary conditions: Here φ is the colatitude and θ is the longitude. u = on r = a, π/2 φ π (1) u =1 on r = a, φ<π/2 (11)

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5 R.Rand Lecture Notes on PDE s 4 2 The Laplacian 2 in three coordinate systems Rectangular coordinates Circular cylindrical coordinates 2 u = 2 u x u y u z 2 (12) where and where Spherical coordinates where and where 2 u = 2 u r + 1 u 2 r r u r 2 θ + 2 u (13) 2 z 2 x = r cos θ, y = r sin θ, that is, r 2 = x 2 + y 2 (14) 2 u = 1 [ ( ρ 2 u ) + 1 ρ 2 ρ ρ sin φ φ θ<2π (15) ( ) u φ sin φ + 1 sin 2 φ 2 ] u θ 2 x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos θ, that is, ρ 2 = x 2 + y 2 + z 2 (17) (16) θ<2π, φ π (18)

6 R.Rand Lecture Notes on PDE s 5 3 Solution to Problem A by Separation of Variables In this section we solve Problem A by separation of variables. This is intended as a review of work that you have studied in a previous course. We seek a solution to the PDE (1) (see eq.(12)) in the form Substitution of (19) into (12) gives: u(x, z) =X(x)Z(z) (19) X Z + XZ = (2) where primes represent differentiation with respect to the argument, that is, X means dx/dx whereas Z means dz/dz. Separating variables, we obtain Z Z = X X = λ (21) where the two expressions have been set equal to the constant λ because they are functions of the independent variables x and z, and the only way these can be equal is if they are both constants. This yields two ODE s: X + λx = and Z λz = (22) Substituting the ansatz (19) into the boundary conditions (=B.C.) (2) and (3), we find: The general solution of the X equation in (22) is X() =, X(L) = and Z() = (23) X(x) =c 1 cos λx + c 2 sin λx (24) where c 1 and c 2 are arbitrary constants. The first B.C. of (23), X() =, gives c 1 =. The second B.C. of (23), X(L) =, gives λ = nπ/l, for n =1, 2, 3, : λ = n2 π 2 L, X(x) =c 2 2 sin nπx, n =1, 2, 3, (25) L The general solution of the Z equation in (22) can be written in either of the equivalent forms: Z(z) =c 3 cosh λz + c 4 sinh λz (26) or Z(z) =c 5 e λz + c 6 e λz (27) where λ = n2 π 2 L 2 and where cosh v = ev + e v 2 and sinh v = ev e v 2 (28)

7 R.Rand Lecture Notes on PDE s 6 Choosing the form (26), the third B.C. of (23), Z() =, gives c 3 =. Substituting the derived results into the assumed form of the solution (19), we have u(x, z) =X(x)Z(z) =a n sin nπx L nπz sinh, n =1, 2, 3, (29) L where a n =c 2 c 4 is an arbitrary constant. Since the PDE (1) is linear, we may superimpose solutions to obtain the form: u(x, z) = a n sin nπx nπz sinh (3) L L n=1 We still have to satisfy the B.C. (4), u(x, L)=1, which gives (from (3)): 1= a n sin nπx sinh nπ (31) L n=1 Eq.(31) is a Fourier series. We may obtain the values of the constants a n by using the orthogonality of the eigenfunctions sin nπx on the interval <x<l: L L sin nπx mπx sin dx = n m (32) L L where n and m are integers. Eq.(32) follows from the trig identity: sin nπx L mπx sin L = 1 (n + m)πx cos + 1 (n m)πx cos 2 L 2 L Substituting (33) into (32), the cosines integrate to sines and vanish at the upper and lower limits if n m. In the case that n=m, we have ( L 1 (n + m)πx cos 1 ) (n m)πx L ( cos dx = 1 2mπx cos 2 L 2 L 2 L + 1 ) dx = L (34) 2 2 We return now to eq.(31) and use (32) and (34) to obtain the coefficients a n. Multiplying (31) by sin mπx and integrating from to L, we obtain: L L sin mπx L dx = a L m sinh mπ (35) 2 Changing the index from m to n, this gives L L a n 2 sinh nπ = sin nπx L dx = [ L nπ ] nπx L { 2L cos = nπ L for n odd for n even whereupon we obtain the following solution u(x, z) from eqs.(3) and (36): (33) (36) u(x, z) = n=1,3,5, 4 nπ sinh nπz L sinh nπ sin nπx L (37)

8 R.Rand Lecture Notes on PDE s 7 4 Solving Problem B by Separation of Variables Problem B has the PDE (see (5) and (13)): 2 u = 2 u r + 1 u 2 r r + 2 u z = (38) 2 Following the procedure we used on problem A, we seek a solution to the PDE (38) in the form u(r, z) =R(r)Z(z) (39) Substitution of (39) into (38) gives: R Z + 1 r R Z + RZ = (4) where again primes represent differentiation with respect to the argument. Separating variables, we obtain Z Z = + 1 R r R = λ (41) R This yields two ODE s: R + 1 r R + λr = and Z λz = (42) Substituting the ansatz (39) into the B.C. (6) and (7), we find: R(a) = and Z() = (43) Now if we were to continue to follow the procedure we used on problem A, we would solve the R equation of (42) and use the first B.C. of (43) to find λ, and so on. However, the R equation has a variable coefficient, namely in the 1 r R term. Thus we must digress and find out to how to solve such ODE s before we can continue with the solution of problem B.

9 R.Rand Lecture Notes on PDE s 8 5 Euler s Differential Equation The simplest case of a linear variable coefficient second order ODE is Euler s equation: ax 2 d2 y dy + bx + cy= (44) dx2 dx We look for a solution with the ansatz: y = x r (45) Substitution of (45) into (44) gives ar(r 1) + br + c = that is, ar 2 +(b a)r + c = (46) We may use the quadratic formula to obtain (in general) a pair of complex conjugate roots r 1 and r 2. Thus the general solution may be written y = c 1 x r 1 + c 2 x r 2 (47) If the roots are both real, then eq.(47) suffices. However in the general case in which r 1 and r 2 are complex, say r 1 = µ + iν, we obtain the form y = c 1 x µ+iν + c 2 x µ iν = x µ (c 1 x iν + c 2 x iν ) (48) Using Euler s formula, e iθ = cos θ + i sin θ, and the identity x = e log x, we obtain the real form y = x µ (c 1 e iν log x + c 2 e iν log x )=x µ (c 3 cos(ν log x)+c 4 sin(ν log x)) (49) where c 3 and c 4 are real arbitrary constants, and where log x stands for natural logarithms. In the case that the roots are repeated, r 1 =r 2 =r, the general solution to (44) is y = c 1 x r + c 2 x r log x (5)

10 6 Power Series Solutions R.Rand Lecture Notes on PDE s 9 So now we know how to solve Euler s equation. What about linear differential equations with variable coefficients which are not in the form of Euler s equation? A natural approach would be to look for the solution in the form of a power series: y = c + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x c n x n + (51) where the coefficients c i are to be found. The method is best illustrated with an example. Example 1 We substitute (51) into (52) d 2 y dx + x dy 2 dx + y = (52) 2c 2 +6c 3 x +12c 4 x x(c 1 +2c 2 x +3c 3 x 2 + )+c + c 1 x + c 2 x 2 + c 3 x 3 + = (53) and collect terms: 2c 2 + c + x(6c 3 +2c 1 )+x 2 (12c 4 +3c 2 )+x 3 (2c 5 +4c 3 ) (54) Next we require the coefficient of each power of x n to vanish, giving: 2c 2 + c = (55) 6c 3 +2c 1 = (56) 12c 4 +3c 2 = (57) 2c 5 +4c 3 = (58) Note that eqs.(55)-(58) can be abbreviated by the single recurrence relation: which may be written in the form (n + 2)(n +1)c n+2 +(n +1)c n =, n =, 1, 2, 3, (59) c n c n+2 =, n =, 1, 2, 3, (6) n +2 The nature of the recurrence relation (6) is that c and c 1 can be chosen arbitrarily, after which all the other c i s will be determined in terms of c and c 1. This leads to the following expression for the general solution of eq.(52): y = c f(x)+c 1 g(x) (61) where f(x) =1 x2 2 + x4 2 4 x x (62) g(x) =x x3 3 + x5 3 5 x x (63)

11 R.Rand Lecture Notes on PDE s 1 The method of power series has worked great on Example 1. Let s try it on Example 2: Example 2 We substitute (51) into (64) 2x 2 d2 y dy +3x +( 1+x)y = (64) dx2 dx 2(2c 2 x 2 +6c 3 x 3 +12c 4 x 4 + )+3x(c 1 +2c 2 x+3c 3 x 2 + )+( 1+x)(c +2c 1 x+2c 2 x 2 +2c 3 x 3 + )= (65) and collect terms: This time eqs.(67)-(69) have the recurrence relation: c = (66) 2c 1 + c = (67) 9c 2 + c 1 = (68) 2c 3 + c 2 = (69) c n c n+1 =, n =1, 2, 3, (7) (2n + 1)(n +2) But since eq.(66) requires that c =, we find from (7) that all the c i s must vanish. In other words, the method of power has failed to produce a solution for Example 2, eq.(64). This raises two questions: 1. How can we obtain a solution to eq.(64)?, and 2. For which class of equations will the method of power series work? We answer the first question in the next section by replacing the method of power series by a more general method called the method of Frobenius. We put off answering the second question until later.

12 R.Rand Lecture Notes on PDE s 11 7 The Method of Frobenius The method of Frobenius generalizes the method of power series by seeking a solution in the form of a generalized power series : y = x r (c + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x c n x n + ) (71) Note that eq.(71) reduces to the power series (51) if r=. Let s try this method on Example 2: Example 2, continued 2x 2 d2 y dy +3x +( 1+x)y = (72) dx2 dx We substitute (71) into (72) and collect terms: (2r 2 + r 1)c x r + [(2r 2 +5r +2)c 1 + c ]x r+1 + [(2r 2 +9r +9)c 2 + c 1 ]x r+2 + = (73) Eqs.(75)-(77) have the recurrence relation: (2r 2 + r 1)c = (74) (2r 2 +5r +2)c 1 + c = (75) (2r 2 +9r +9)c 2 + c 1 = (76) [2r 2 +(4n +1)r +(n 2 + n 1)]c n+1 + c n = (77) c n c n+1 =, n =1, 2, 3, (78) (2r +2n 1)(r + n +1) In addition to (78), we also must satisfy, from (74), the indicial equation: 2r r 1= r = 1, 2 We consider each of these r-values separately, taking c =1 for each: For r= 1 we get: (79) For r= 1 2 y = f(x) = 1 x +1 x 2 + x2 18 x x4 126 x5 + (8) 684 we get: y = g(x) = ( x 1 x ) 5 + x2 7 x x x (81) The general solution of Example 2, eq.(72), is thus given by y = k 1 f(x)+k 2 g(x) (82)

13 R.Rand Lecture Notes on PDE s 12 where k 1 and k 2 are arbitrary constants. Next, let s try the method of Frobenius on Example 3: Example 3 x 3 d2 y dx 2 + y = (83) We substitute (71) into (83) and collect terms: c x r +[(r 2 r)c + c 1 ]x r+1 +[(r 2 + r)c 1 + c 2 ]x r+2 +[(r 2 +3r +2)c 2 + c 3 ]x r = (84) Eqs.(86)-(89) have the recurrence relation: c = (85) (r 2 r)c + c 1 = (86) (r 2 + r)c 1 + c 2 = (87) (r 2 +3r +2)c 2 + c 3 = (88) (r + n)(r + n 1)c n + c n+1 = (89) c n+1 = (r + n)(r + n 1)c n, n =1, 2, 3, (9) But since eq.(85) requires that c =, we find from (9) that all the c i s must vanish. In other words, the method of Frobenius has failed to produce a solution for Example 3, eq.(83). Summarizing, we have seen that for some equations the method of power series works (Example 1), whereas for other equations that method fails but the more general method of Frobenius works (Example 2). Now we have seen an example in which the method of Frobenius fails (Example 3). We need a classification scheme which will tell us which equations we can be guaranteed to solve using these two methods.

14 R.Rand Lecture Notes on PDE s 13 8 Ordinary Points and Singular Points Let s consider the general class of linear second order ODE s of the form: A(x) y + B(x) y + C(x) y = (91) in which A(x), B(x) and C(x) have power series expansions about x=: A(x) =A + A 1 x + A 2 x 2 + (92) B(x) =B + B 1 x + B 2 x 2 + (93) C(x) =C + C 1 x + C 2 x 2 + (94) This includes the possibility that A(x), B(x) and C(x) are polynomials. Definition: If A then x= is called an ordinary point. If A = and not both B and C are zero, then x= is called a singular point. Now suppose that x= is a singular point. Then let p(x) be defined by p(x) =x B(x) A(x) = B + B 1 x + B 2 x 2 + A 1 + A 2 x + A 3 x 2 + (95) and let q(x) be defined by q(x) =x 2C(x) A(x) = x C + C 1 x + C 2 x 2 + A 1 + A 2 x + A 3 x 2 + (96) Definition: Let x= be a singular point. If p(x) and q(x) don t blow up as x approaches zero, then x= is called a regular singular point. A singular point which is not regular is called an irregular singular point. These definitions will help us to determine whether the method of power series or the method of Frobenius will work to solve a given equation of the form (91). Rule 1: If x= is an ordinary point, then two linearly independent solutions can be obtained by the method of power series expansions about x=. Rule 2: If x= is a regular singular point, then at least one solution can be obtained by the method of Frobenius expanded about x=. As an example, consider Example 1, eq.(52): d 2 y dx + x dy 2 dx + y = (97) In this case A(x)=1 and x = is an ordinary point. According to Rule 1, two independent solutions can be obtained in the form of power series, see eqs.(62) and (63).

15 R.Rand Lecture Notes on PDE s 14 Consider now Example 2, eq.(72): 2x 2 d2 y dy +3x +( 1+x)y = (98) dx2 dx In this case A = and C = 1 so that x= is a singular point. The quantities p(x) and q(x) are given by p(x) =x B(x) A(x) = 3 2 and q(x) =x 2C(x) A(x) = 1+x 2 and since neither p(x) nor q(x) blows up as x, we see that x= is a regular singular point. Then by Rule 2, at least one solution can be obtained by the method of Frobenius. In fact we found two such solutions, see eqs.(8) and (81). Next consider Example 3, eq.(83): x 3 d2 y dx + y = (1) 2 In this case A = and C =1 so that x= is a singular point. The quantities p(x) and q(x) are given by p(x) =x B(x) C(x) = and q(x) =x2 A(x) A(x) = 1 (11) x Since q(x) blows up as x, we see that x= is an irregular singular point and neither Rule 1 nor Rule 2 applies. In fact we saw that the method of Frobenius did not work on Example 3. You may have noticed that Rule 1 guarantees two linearly independent solutions in the case of an ordinary point, while Rule 2 only guarantees one solution in the case of a regular singular point. Nevertheless in the case of Example 2 (which has a regular singular point) we found two linearly independent solutions. The case where the method of Frobenius only yields one solution is illustrated by Example 4: Example 4 x 2 d2 y dy +4x +(2+x) y = (12) dx2 dx Here x= is a regular singular point. We apply the method of Frobenius by substituting the Frobenius series (71) into eq.(12) and collecting terms: (r 2 +3r +2)c x r +[(r 2 +5r +6)c 1 + c ]x r+1 +[(r 2 +7r + 12)c 2 + c 1 ]x r+2 + = (13) (99) (r 2 +3r +2)c = (14) (r 2 +5r +6)c 1 + c = (15) (r 2 +7r + 12)c 2 + c 1 = (16) [r 2 +(2n +5)r +(n 2 +5n + 6)]c n+1 + c n = (17)

16 R.Rand Lecture Notes on PDE s 15 Eqs.(15)-(17) have the recurrence relation: c n c n+1 =, n =1, 2, 3, (18) (r + n + 2)(r + n +3) In addition to (18), we also must satisfy, from (14), the indicial equation: r 2 +3r +2= r = 1, 2 (19) Let us consider each of these r-values separately: For r= 1 we get: ( 1 y = c x x 12 x x3 288 x x 5 ) For r= 2, eq.(15) requires that we take c =. Then we obtain the following solution: ( 1 y = c 1 x x 12 x x3 288 x x 5 ) (11) (111) Eqs.(11) and (111) are obviously not linearly independent. The method of Frobenius has obtained only one linearly independent solution. This case can be characterized by Rule 3: Rule 3: Ifx= is a regular singular point, and if the two indicial roots r 1 and r 2 are not identical and do not differ by an integer, then the method of Frobenius will yield two linearly independent solutions. In the case of Example 2, eq.(72), the indicial equation (79) gave r 1 = 1 and r 2 = 1. Since these 2 r-values do not differ by an integer, Rule 3 guarantees that the method of Frobenius will generate two linearly independent solutions. See eqs.(8) and (81). However, in the case of Example 4, eq.(12), the indicial roots were r 1 = 1 and r 2 = 2, see eq.(19). In this case Rule 3 does not apply and we have no guarantee that the method of Frobenius will generate two linearly independent solutions. A natural question to ask at this point is: How can we find the second linearly independent solution in cases like that of Example 4, where there is a regular singular point for which the indicial roots are repeated or differ by an integer? The answer is given by Rule 4: Rule 4: If x= is a regular singular point, and if the two indicial roots r 1 and r 2 are either identical or differ by an integer, then one solution can be obtained by the method of Frobenius. Let us refer to that solution as y = f(x). A second linearly independent solution can be obtained in the form y = Cf(x) log x + g(x) (112) where g(x) is a Frobenius series. The constant C may or may not be equal to zero. If C=, then both linearly independent solutions can be obtained in the form of Frobenius series. In the repeated root case r 1 =r 2, the constant C is not equal to zero.

17 R.Rand Lecture Notes on PDE s 16 In the case of repeated indicial roots, we should not be surprised to find the occurrence of log x in the solution, since we have seen that in Euler s equation, eq.(44), which has a regular singular point at x=, the presence of a repeated root implies the presence of log x in the solution, see eq.(5).

18 R.Rand Lecture Notes on PDE s 17 9 Solving Problem B by Separation of Variables, continued Now that we know how to solve linear variable coefficient ODE s, let us return to the problem which motivated our digression, namely the application of separation of variables to problem B. We wrote u(r, z) = R(r)Z(z), see eq.(39), and we found that the unkown function R(r) had to satisfy the following boundary value problem (see eqs.(42),(43)): d 2 R dr r dr dr + λr = with the B.C. R = when r = a (113) To begin with, let s stretch the r coordinate so as to absorb the separation constant λ and thereby make it disappear from the ODE. Let x = λr (114) and let us change notation from R(r) toy(x): dr dr = dy dx dx dr = dy d dx dr ( λr) = λ dy dx (115) where we have used the chain rule. Similarly, d2 R = λ d2 y. The differential equation in (113) dr 2 dx 2 becomes d 2 y dx + 1 dy 2 x dx + y = (116) Next let us multiply eq.(116) by x so as to put it in the form of eq.(91): x d2 y dx + dy + xy = (117) 2 dx Now we have to decide upon which method to use to solve eq.(117). Inspection of (117) shows that x= is not an ordinary point (since A, see eq.(92)). In fact x= is a regular singular point. To prove this we compute the quantities p(x) and q(x) defined in eqs.(95),(96): p(x) =x B(x) A(x) = 1 and q(x) =x2c(x) A(x) = x2 (118) and we note that neither p(x) nor q(x) blows up as x, telling us that x= is a regular singular point. Then by Rule 2 we know that at least one solution of (117) can be obtained by the method of Frobenius. We proceed with the method of Frobenius by substituting a Frobenius series (71) into eq.(117) and collecting terms. We find: r 2 c x r 1 +(r 2 +2r +1)c 1 x r +[(r 2 +4r +4)c 2 +c ]x r+1 +[(r 2 +6r +9)c 3 +c 1 ]x r+2 + = (119) r 2 c = (12) (r 2 +2r +1)c 1 = (121)

19 R.Rand Lecture Notes on PDE s 18 Eqs.(122)-(124) have the recurrence relation: (r 2 +4r +4)c 2 + c = (122) (r 2 +6r +9)c 3 + c 1 = (123) (r + n +2) 2 c n+2 + c n = (124) c n c n+2 =, n =, 1, 2, 3, (125) (r + n +2) 2 The indicial equation for this system is obtained from eq.(12): r 2 = r =, (repeated root) (126) So we are in the case of a regular singular point with a repeated indicial root. Rule 4 applies to this situation and tells us that the method of Frobenius will be able to generate only one solution, and that a second linearly independent solution will involve a log x term, see eq.(112). Note that once r= is chosen to satisfy eq.(12), we find from eq.(121) that c 1 =. This implies, from the recurrence relation (125) that all the c odd coefficients must vanish. Thus we obtain the following solution: ( y = c 1 x2 4 + x4 64 x x x 1 ) (127) The function so generated is very famous and is known as a Bessel function of order zero of the first kind. It is usually represented by the symbol J and can be written in the following form: ( 1) n x 2n J (x) = (128) n= 2 2n (n!) 2 We present without derivation the following expression for J (x) which is valid for large values of x: 2 J (x) (x πx cos π ) as x (129) 4 From Rule 4, a second linearly independent solution will take the form: y = CJ (x) log x + g(x) (13) where g(x) can be written as a power series. Let us abbreviate this second linearly independent solution by the symbol Y. Thus we have obtained the following expression for the general solution to eq.(117): y = k 1 J (x)+k 2 Y (x) (131)

20 R.Rand Lecture Notes on PDE s 19 Recall that we defined x in order to simplify the form of the R differential equation in (113). Now we return to the original independent variable r by setting x= λr (see eq.(114)). The general solution to the R differential equation in (113), R + R /r + λr = can be written: R(r) =k 1 J ( λr)+k 2 Y ( λr) (132) The next step is to apply the B.C. associated with this equation: R = when r = a (133) Before applying the B.C. (133), we note that the presence of the log term in Y will produce infinite temperatures at r=, and hence the Y part of the solution must be removed by choosing k 2 = in (132). The B.C. (133) then requires that R(a) =k 1 J ( λa) = (134) There is an infinite sequence of λ values, {λ i }, each of which satisfies eq.(134). These may be found by utilizing tables of the zeros of J (x). We shall represent the n th zero of J (x) byγ n. The first five zeros of J (x) are: Γ 1 =2.448, Γ 2 =5.521, Γ 3 =8.6537, Γ 4 = , Γ 5 =14.939, (135) The corresponding values of λ are of the form λ n =Γ 2 n/a 2 : λ 1 = ,λ a 2 2 = ,λ a 2 3 = ,λ a 2 4 = ,λ a 2 5 = , (136) a 2 And the corresponding R-eigenfunctions are: R n (r) =J ( ( ) r λ n r)=j Γ n (137) a Returning to the separation of variables solution of problem B, we must also solve the Z- equation (see eqs.(42),(43)): Z λz = with the B.C. Z = when z = (138) As in the separation of variables solution of problem A, the general solution of the Z equation (138) can be written in the form: The B.C. Z() = gives c 3 =. Z(z) =c 3 cosh λz + c 4 sinh λz (139) Substituting the derived results into the assumed form of the solution u(r, z) = R(r)Z(z), we have ( ) r u(r, z) =R(r)Z(z) =a n J Γ n sinh Γ nz, n =1, 2, 3, (14) a a

21 R.Rand Lecture Notes on PDE s 2 where a n is an arbitrary constant. Since the PDE (38) is linear, we may superimpose solutions to obtain the form: ( ) r u(r, z) = a n J Γ n sinh Γ nz (141) n=1 a a We still have to satisfy the B.C. (8), u(r, L)=1, which gives (from (141)): ( ) r 1= a n J Γ n sinh Γ nl (142) a a n=1 Our task now is to find the constants a n. When we were faced with this same task in solving problem A, we had a Fourier series (see eq.(31)). We obtained the values of the constants a n by using the orthogonality of the eigenfunctions sin nπx. In the case of problem B we will ( L ) r similarly use the orthogonality of the eigenfunctions J Γn a.

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23 1 Orthogonality R.Rand Lecture Notes on PDE s 21 In order to complete our treatment of problem B, we need to use orthogonality of the associated eigenfunctions. In the case of problem A, the orthogonal eigenfunctions X(x) satisfied the following boundary value problem (see eqs.(22) and (23)): X + λx = with the B.C. X() =, X(L) = (143) In order to prove orthogonality, we solved this problem to get X n (x) =sinλ n x = sin nπx (144) L and then we used the properties of trig functions to prove that L sin nπx mπx sin dx = n m (145) L L See eqs.(32)-(34). Unfortunately, we can t use a comparable method to prove the orthogonality of Bessel functions. Instead we will use a different method called Sturm-Liouville theory. We demonstrate the method by using it on eq.(143). Let X n (x) satisfy the eqs: and let X m (x) satisfy the eqs: We want to prove that X n + λ n X n = with the B.C. X n () =, X n (L) = (146) X m + λ m X m = with the B.C. X m () =, X m (L) = (147) We multiply (146) by X m and (147) by X n, and subtract, giving L X m X n dx = for n m (148) X m X n X n X m +(λ n λ m )X n X m = (149) Next we integrate (149) from to L: L L (X m X n X n X m) dx +(λ n λ m ) X n X m dx = (15) The idea now is to get the first term to vanish by using integration by parts and the B.C. in (143). Consider the first part of the first term: L X m X n dx =[X m X n] L L X mx n dx (151) The integrated terms vanish due to the B.C. X m ()=X m (L)=. Thus L L X m X n dx = X m X n dx (152)

24 R.Rand Lecture Notes on PDE s 22 Similarly, Therefore eq.(15) becomes L X n X m dx = L X n X m dx (153) L L (λ n λ m ) X n X m dx = X n X m dx = if n m (154) We have thus proven the orthogonality of the eigenfunctions X n (x) without constructing X n (x). This method is quite different from the method we used in eqs.(32)-(34), where we solved for X n (x) = sin nπx and used various trig identities. L Now in the case of problem B, the eigenfunctions R n (r) satisfy the boundary value problem (113): d 2 R n + 1 dr n dr 2 r dr + λ nr n = with the B.C. R n = when r = a (155) Based on the procedure given above in eqs.(146)-(154), it would seem like the right way to proceed with eq.(155) would be to multiply by R m, subtract the same equation with m and n interchanged, and integrate by parts to eliminate the terms which are not part of the orthogonality condition. Try it: it doesn t work! The correct procedure is to multiply eq.(155) by r and write it in the form: ( d r dr ) n + λ n rr n = (156) dr dr Now we proceed as before, multiplying (156) by R m and subtracting the same equation with m and n interchanged, giving: ( d R m r dr ) ( n d R n r dr ) m +(λ n λ m )rr n R m = (157) dr dr dr dr Now we integrate (157) from to a: ( ( a d R m r dr ) ( n d R n r dr )) a m dr +(λ n λ m ) rr n R m dr = (158) dr dr dr dr As before, the idea is to get the first term to vanish by using integration by parts and the B.C. in (155). Consider the first part of the first term: a ( d R m r dr ) [ ] a n dr a n dr = rr m r dr n dr dr dr dr dr m dr (159) dr The integrated terms vanish at the upper limit because R m = at r=a. At the lower limit, the presence of the factor r kills the term, assuming that R and dr/dr remain bounded as r. So we have a ( d R m r dr ) a n dr = r dr n dr dr dr dr m dr (16) dr

25 R.Rand Lecture Notes on PDE s 23 Similarly, we find that a ( d R n r dr ) a m dr = r dr m dr dr dr dr n dr (161) dr so that eq.(158) becomes a (λ n λ m ) rr n R m dr = a rr n R m dr = if n m (162) Since we showed in eq.(137) R n (r) =J ( ( ) r λ n r)=j Γ n (163) a we have the following orthogonality condition for Bessel functions: a ( ) ( ) r r rj Γ n J Γ m dr = if n m (164) a a where as a reminder we note that Γ n is the n th zero of J (x), see eq.(135).

26 R.Rand Lecture Notes on PDE s Sturm-Liouville Theory In this section we present some results which generalize the orthogonality results in the previous section. A linear second order ODE is said to be in Sturm-Liouville form if it can be written as follows: [ d p(x) dy ] +(q(x)+ λw(x))y = (165) dx dx We suppose that this equation is defined on an interval a x b and that B.C. on y(x) are given at the endpoints x=a and x=b. [In the case of problem B, eq.(156) is in Sturm-Liouville form with R(r) corresponding to y(x), and where p(x)=x, q(x)=, and w(x)=x. The interval r a corresponds to a x b.] Let {λ n } be a set of eigenvalues satisfying the ODE (165) and the B.C., and let {y n (x)} be the corresponding eigenfunctions. The orthogonality condition for this system is b Here w(x) is called the weight function. a In order for eq.(166) to hold, the following terms must vanish: w(x)y n (x)y m (x)dx = if n m (166) [p(x)(y m (x)y n(x) y n (x)y m(x))] b a = (167) A variety of B.C. will satisfy eq.(167), including y(x) vanishing at both endpoints x=a and x=b. As we have seen in the case of problem B, another way to satisfy (167) is if p(x)= at one endpoint and y(x)= at the other endpoint. A related question is given a second order ODE which is not in Sturm-Liouville form: How can it be put into Sturm-Liouville form? A(x)y + B(x)y + C(x)y + λd(x)y = (168) Recall that this was the situation in problem B, where the original R-equation, (155), had to be multiplied by r to get it into the Sturm-Liouville form (156). The procedure is to first calculate the following expression for p(x): p(x) =e (B(x)/A(x))dx (169) Then multiply eq.(168) by p(x)/a(x). The result may then be written in the Sturm-Liouville form (165), where q(x)= p(x)c(x) (17) A(x)

27 R.Rand Lecture Notes on PDE s 25 w(x) = p(x)d(x) A(x) For example, in the case of eq.(155), which we may write in the form: (171) y + 1 x y + λy = (172) we have A(x) =1, B(x) = 1, C(x) =, D(x) = 1 (173) x Then eqs.(169)-(171) give p(x) =e (B(x)/A(x))dx = e (1/x)dx = e log x = x (174) q(x)= p(x)c(x) = A(x) (175) w(x) = p(x)d(x) = x A(x) (176) which is to say that if eq.(172) is multiplied by p(x)/a(x)=x, it can be written in the Sturm- Liouville form: ( d x dy ) + λxy = (177) dx dx

28 R.Rand Lecture Notes on PDE s Solving Problem B by Separation of Variables, concluded Now that we know about the orthogonality of Bessel functions, let s return to solving problem B by separation of variables. We previously obtained the following expression for u(r, z) (see eq.(141): ( ) r u(r, z) = a n J Γ n sinh Γ nz (178) n=1 a a We still have to satisfy the B.C. (8), u(r, L)=1, which gives (from (178)): ( ) r 1= a n J Γ n sinh Γ nl (179) a a n=1 Now we can use the orthogonality result (164) a ( ) ( ) r r rj Γ n J Γ m dr = a a if n m (18) ( ) r We multiply (179) by rj Γm a and integrate from to a. The orthogonality condition (18) zaps everything on the RHS of the resulting equation except for the n=m term: a ( ) r rj Γ m dr = a m sinh Γ ml a ( )] r 2 r [J Γ m dr a a a (181) Solving (181) for a m, we obtain: a m = I 1 (m) I 2 (m) sinh ΓmL a (182) where we have used the notation I 1 and I 2 to abbreviate the following integrals: a ( ) r a ( )] r 2 I 1 (m) = rj Γ m dr and I 2 (m) = r [J Γ m dr a a (183) Using this notation, the expression (178) for u(r, z) becomes: ( ) I 1 (n) u(r, z) = n=1 I 2 (n) J r sinh Γ nz a Γ n a sinh ΓnL a (184) Now it turns out that the integrals I 1 and I 2 can be evaluated in terms of a tabulated function J 1 (x), which is the notation for a Bessel function of order 1 of the first kind. Skipping all the details, the result is: I 1 (n) I 2 (n) = 2 1 Γ n J 1 (Γ n ) (185) Using (185) in (184), we obtain the final solution of problem B : ( ) r 2 J Γn a sinh Γnz a u(r, z) = n=1 Γ n J 1 (Γ n ) sinh ΓnL a (186)

29 R.Rand Lecture Notes on PDE s Solving Problem C by Separation of Variables Problem C has the PDE (see (9) and (16)): 2 u = 1 [ ( ) ρ 2 u + 1 ρ 2 ρ ρ sinφ φ ( )] u φ sinφ = (187) Following the procedure we used on problems A and B, we seek a solution to the PDE (187) in the form u(ρ, φ) = R(ρ)Φ(φ) (188) Substitution of (188) into (187) gives: [ 1 (ρ 2 R ) Φ + 1 ] ρ 2 sinφ (Φ sinφ) R = (189) where primes represent differentiation with respect to the argument. Separating variables, we obtain This yields two ODE s: ρ 2 R + 2ρR R = 1 Φ sinφ + Φ cos φ = λ (19) sinφ Φ ρ 2 R + 2ρR λr = and Φ + Φ cotφ + λφ = (191) Let s begin with the Φ-equation. This equation can be simplified by changing the independent variable from the colatitude φ to x=cosφ. In order to accomplish this step, we will use the chain rule, and we will write Φ(φ) as y(x) instead of Φ(x) to avoid confusion. We have dφ dφ = dy dx dx dφ = sinφ dy dx (192) d 2 Φ dφ = d ( sinφ dy ) = cos φ dy 2 dφ dx dx + sin2 φ d2 y dx = xdy 2 dx + (1 x2 ) d2 y (193) dx 2 Using eqs.(192) and (193), the Φ-equation of (191) becomes: x dy ( dx + (1 x2 ) d2 y dx + cotφ sinφ dy ) + λy = (194) 2 dx Eq.(194) can be simplified by using cot φ sin φ=cos φ=x, giving: (1 x 2 ) d2 y dx 2xdy + λy = (195) 2 dx Note that x= is an ordinary point of eq.(195). By Rule 1, we can use the method of power series expansions about x= to obtain two linearly independent solutions. (See the section on Ordinary Points and Singular Points.) We look for a solution in the form of a power series: y = c + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x c n x n + (196)

30 R.Rand Lecture Notes on PDE s 28 Substituting (196) into (195) and collecting terms gives c λ +2c 2 +(c 1 (λ 2) + 6 c 3 ) x +(c 2 (λ 6) + 12 c 4 ) x 2 +(c 3 (λ 12) + 2 c 5 ) x 3 + (197) Requiring the coefficient of each power of x n to vanish gives: c λ +2c 2 = (198) c 1 (λ 2) + 6 c 3 = (199) c 2 (λ 6) + 12 c 4 = (2) c 3 (λ 12) + 2 c 5 = (21) which gives the following recurrence relation: ( ) n(n +1) λ c n+2 = c n (22) (n + 2)(n +1) Here c and c 1 can be chosen arbitrarily, giving rise to two linearly independent solutions f(x) and g(x): y = c f(x)+c 1 g(x) (23) where f(x) =1 λx2 λ(λ 6) x4 + + (24) 2 24 (λ 2) x3 (λ 12) (λ 2) x5 g(x) =x + + (25) 6 12 It turns out that these two series diverge at x=±1, that is, at cos φ=±1, that is, at φ= and φ=π. Physically this represents unbounded temperatures along the polar axis of the sphere. No choice of the arbitrary constants c and c 1 will eliminate this problem, as it did in the case of the cylinder, compare with eqs.(132),(134). However, if λ is taken equal to n(n + 1), for n an integer, then one of the two infinite series will terminate, that is, will become a polynomial, and the convergence problem will disappear. We therefore take λ=n(n + 1), n=, 1, 2, 3, and we obtain a sequence of polynomial solutions called Legendre polynomials, y = P n (x): P (x) = 1 (26) P 1 (x) = x = cos φ (27) P 2 (x) = 1 2 (3x2 1) = 1 2 (3 cos2 φ 1) (28) P 3 (x) = 1 2 (5x3 3x) = 1 2 (5 cos3 φ 3 cos φ) (29) Rodrigues s formula gives an expression for P n (x): P n (x) = 1 n!2 n d n dx n (x2 1) n (21)

31 R.Rand Lecture Notes on PDE s 29 Let us turn now to the R(ρ)-equation in (191): ρ 2 R +2ρR λr =, λ = n(n + 1) (211) This is an Euler equation, so we look for a solution in the form R(ρ) =ρ r (see eq.(45)). Substitution into (211) gives: r(r 1) + 2r n(n +1)= r = n and r = (n + 1) (212) Thus eq.(211) has the general solution: R(ρ) =k 1 ρ n 1 + k 2 (213) ρ n+1 For bounded temperatures at the sphere s center, ρ=, we require that k 2 =: R(ρ) =k 1 ρ n (214) Substituting the derived results into the assumed form of the solution (188), we have u(ρ, φ) =R(ρ)Φ(φ) =b n P n (cos φ)ρ n, n =, 1, 2, 3, (215) where b n is an arbitrary constant. Since the PDE (187) is linear, we may superimpose solutions to obtain the form: u(ρ, φ) = b n P n (cos φ)ρ n (216) n= Our task now is to choose the constants b n so as to satisfy the B.C. on the outside of the sphere, ρ = a, see eqs.(1),(11): u(a, φ) = b n a n P n (cos φ) = n= or, switching to x=cos φ, { 1 for φ<π/2 for π/2 φ π (217) u(a, x) = b n a n P n (x) = n= { 1 for x 1 for 1 x< (218) As in the case of problems A and B, we need to use orthogonality of the eigenfunctions to find the b n coefficients. We can write Legendre s equation (195) in Sturm-Liouville form (165) as follows: [ d (1 x 2 ) dy ] + λy = (219) dx dx Comparison of (219) with the Sturm-Liouville form (165) gives the weight function w(x)=1, from which we obtain the following orthogonality condition: 1 1 P n (x)p m (x)dx = if n m (22)

32 R.Rand Lecture Notes on PDE s 3 We multiply eq.(217) by P m (x) and integrate from x= 1 tox=1. The orthogonality condition (22) eliminates all terms in the infinite series except for the n=m term: 1 b m a m [P m (x)] 2 dx = 1 1 P m (x)dx (221) Solving (221) for b m, we obtain: b m = I 3(m) a m I 4 (m) where we have used the notation I 3 and I 4 to abbreviate the following integrals: (222) I 3 (m) = 1 P m (x)dx and I 4 (m) = 1 1 [P m (x)] 2 dx (223) Using this notation, the expression (216) for u(ρ, φ) becomes: ( ) I 3 (n) ρ n u(ρ, φ) = I 4 (n) P n(cos φ) (224) a n= Now it turns out that the integrals I 3 and I 4 can be evaluated in closed form. Here are the results, without derivation: 1 I 4 (n) = [P n (x)] 2 dx = 2 1 2n +1 (225) 1 if n= 1 1/2 if n=1 I 3 (n) = P n (x)dx = if n=2,4,6, if n=3,5,7, (226) where ( 1) n 1 2 (n 2)!! (n+1)!! { k(k 2)(k 4) for k even k!! = k(k 2)(k 4) for k odd (227)