series

Appendix C PowerSeriesandSpecialFunctions A special function is a function — usually of a single variabl e — that arises in suffi- ciently many applications as to warrant its own name, an inve stigation of its basic proper- ties, and the development of special algorithms for computi ng its values. The first “special functions” one meets up with are the exponential function, t he natural logarithm, and the trigonometric functions. However, since these common func tions, along with polynomials, rational functions, algebraic functions (combinations of roots), and the hyperbolic func- tions appear much earlier in one’s mathematical education, they are usually referred to as elementary functions . Truespecial functions , such as the gamma function, Bessel func- tions, Legendre functions, Airyfunctions, hypergeometri c functions, andmany manymore, await a more advanced mathematical training. These special functions play a starring role in more advanced applications in physics, engineering and m athematics. They initially appear when one tries to solve linear partial differential eq uations in higher dimensions in non-rectangular coordinate systems. Application of the me thod of separation of variables method reduces the partial differential equation to an ordin ary differential equations of a non-elementary type. The solutions to these special ordina ry differential equations are the aforementioned special functions. In this appendix, we collect together the required results a bout a few of the most important classes of special functions, including a short p resentation of the series approach for solving non-elementary ordinary differential equation s. We assume that the reader is fluent in the very basics of infinite series, including the defi nitions of convergence and absolute convergence, and the basic convergence tests, spe cifically the comparison, ratio and root tests. Anby standard calculus text, e.g., [ 9,168,170], can be relied on for this. For more advanced treatment of special functions, the reade r can consult a variety of sources, including the handbooks [ 3,145], and the detailed texts [ 136,144]. C.1. Power Series. By definition, a power series has the form f(x) =c0+c1x+c2x2+···+cnxn+···=∞/summationdisplay k=0ckxk. (C.1) The coefficients cknd the variable xcan either be real or complex numbers. A power series clearly converges at x= 0, since then all terms but the first are zero, and so f(0) =c0. A simple application of the comparison test shows that if the series converges for a given 12/11/12 1265 c/ci∇cleco√y∇t2012 Peter J. Olver valuex=a, then is automatically converges, in fact absolutely, for a llxwith smaller modulus (absolute value): |x|<|a|. As a consequence, there are precisely three possibilites fo r the convergence of a power series: (a) The series converges for all x. (b) There is a positive number ρ >0, called the radius of convergence , such that the series converges absolutely whenever |x|< ρand diverges whenever |x|> ρ. The series may or may not converge at points on the interface |x|=ρ. (c) The series only converges, trivially, at x= 0. An example is the power series/summationtext nn!xn. In case ( a), we say ρ=∞, while in case ( c),ρ= 0. Thus, in the real case, a power series converges on a symmetric interval centered at the origin, in cluding, possibly, one or both endpoints. In the conplex category, the power series conver ges inside a disk or radius ρ centered at the origin, including possibly some of its bound ary points. A function f(x) represented by a convergent power series whose radius of co nvergence ρ >0 is called analytic. Applying the Weierstrass Mtest of Theorem 12.26, we deduce that the power series converges uniformly on any interval (disk) {/ba∇dblx/ba∇dbl ≤r}of radius 0 ≤r < ρ. Thus, Propo- sition 12.27 implies that the differentiated power series co nverges uniformly, on the same subset, to the derivative f′(x) =c1+2c2x+3c3x2+···+(n+1)cn+1xn+···=∞/summationdisplay k=0(k+1)ck+1xk.(C.2) In particular, the derivative of an analytic function exist s and is itself analytic. Setting x= 0 in (C.2) yields c1=f′(0). Clearly we can continue to differentiate, and hence ever y analytic function is infinitely differentiable. Interestin gly, the converse is true for complex functions — see Chapter 16 — but false for real functions, as n oted in Exercise 2.2.28. A straightforward induction shows that the nthderivative has the power series expansion f(n)(x) =n!cn+(n+1)!cn+1x+1 2(n+2)!cn+2x2+···. Again substituting x= 0 leads to the following key formula: cn=f(n)(0) n!, (C.3) for thenthcoefficient in the original power series as a multiple of the nthderivative of the function at the origin. Thus, replacing each coefficient by it s value, we conclude that f(x) =f(0)+f′(0)x+1 2f′′(0)x2+···=∞/summationdisplay k=0f(k)(0) k!xk. (C.4) You should instantly recognize the result as the Taylor series of the function f(x). Thus, a convergent†power series is the Taylor series for its sum ! †More accurately, a power series with a positive radius of convergence . 12/11/12 1266 c/ci∇cleco√y∇t2012 Peter J. Olver The self-same principles immediately extend to series f(x) =c0+c1(x−x0)+c2(x−x0)2+···=∞/summationdisplay k=0ck(x−x0)k(C.5) in powers of x−x0. Such a power series converges in the interval (disk) |x−x0|< ρ0 and diverge for |x−x0|> ρ0for some 0 ≤ρ0≤ ∞, the radius of convergence. Assuming ρ >0, the sum f(x) is, by definition, analytic at x=x0, and, by the same argument, (C.5) is the Taylor series for f(x) atx0: f(x) =f(x0)+f′(x0)(x−x0)+1 2f′′(x0)(x−x0)2+···=∞/summationdisplay k=0f(k)(x0) k!(x−x0)k.(C.6) With a little more work, it can be shown that f(x) is, in fact, analytic at all points x1such that|x1−x0|< ρ0. This means that f(x) can be represented, near the point x1, by a convergent series in powers of x−x1. Mmoreover, the radius of convergence of the resulting iseries is at least as large as ρ1=ρ0− |x1−x0|, which represents the distance from x1 to the points (circle) |x−x0|=ρrepresenting the interface between the convergent and non-convergent regions of the original series (C.5). If, in fact,ρ1> ρ0− |x1−x0|, then the new power series convrges at points not handled by the ori ginal power series, and the the result is an ana;lytic continuation of the function f(x). The process can often be continued, although may result in multiply-valued complex functions. See Chapter 16 for some additional dicsussion. The convergence of the Taylor series means that its partial s ums will provide polyno- mial approximations to the (analytic) function. We state th is result in a slightly altered notation that emphasizes the The scalar version of Taylor’s theorem with remainder. Theorem C.1. Letu(x)∈Cn. Then the nthorder Taylor expansion of u(x+h), for small h, is u(x+h) =u(x)+u′(x)h+u′′(x)h2 2+···+u(n)(x)hn n!+Rn(x,h),(C.7) where the remainder goes to zero faster than Rn(x,h) hn−→0 as h−→0. There are various formulas for the error, of which the most re levant is Cauchy’s form Rn(x,h) =u(n+1)(ξ)hn+1 (n+1)!, (C.8) whereξis a point lying between†xandx+h. Cauchy’s version assumes that u∈Cn+1. This is a generalization of the well-known Mean Value Theore m †In the complex version, the point ξlies on the line segment connecting xtox+h. 12/11/12 1267 c/ci∇cleco√y∇t2012 Peter J. Olver Theorem C.2. Suppose f(u)is continuously differentiable. Then f(x)−f(a) =f′(ξ)(x−a) for some ξbetween xanda. (C .9) Similarly, the first order Taylor expansion takes the form f(x) =f(a)+f′(a)(x−a)+1 2f′′(ξ)(x−a)2for some ξbetween xanda. (C.10) Example C.3. Consider the function f(x) = (1+ x)r. Note that f(k)(x) =r(r−1)(r−2)···(r−k+1)(1+ x)r−k. Therefore, its Taylor expansion at x= 0 is (1+x)r= 1+rx+r(r−1) 2x2+r(r−1)(r−2) 6+···=∞/summationdisplay k=0/parenleftbiggr k/parenrightbigg xk.(C.11) The coefficients are called binomial coefficients , and denoted /parenleftbiggr k/parenrightbigg =r(r−1)(r−2)···(r−k+1) k(k−1)(k−2)···1. (C.12) A simple application of the ratio test proves that the series (C.11) converges for all xsuch that|x|<1, and diverges for |x|>1, and thus the radisu of copnvergence is ρ= 1. Ifr=nis a non-negative integer, then the Taylor expansion termin ates, and reduces to the well-known Binomial Formula (1+x)n= 1+nx+n(n−1) 2x2+···+nxn−1+xn=n/summationdisplay k=0/parenleftbiggn k/parenrightbigg xk.(C.13) In the vector-valued case, the first order Taylor expansion o f a vector-valued function at a point u⋆takes the form F(u) =F(u⋆)+∇F(u⋆)·(u−u⋆)+R(u−u⋆). (C.14) The remainder term depends quadratically on the distance fr omutou⋆, meaning that there is a positive constant C(depending on ε >0) such that† |R(u−u⋆)| ≤C/ba∇dblu−u⋆/ba∇dbl2whenever /ba∇dblu−u⋆/ba∇dbl< ε. (C.15) We will also have occasion to use the second order expansion F(u) =F(u⋆)+∇F(u⋆)·(u−u⋆)+1 2(u−u⋆)T∇2F(u⋆)(u−u⋆)+S(u−u⋆),(C.16) †One can use any convenient norm here. 12/11/12 1268 c/ci∇cleco√y∇t2012 Peter J. Olver where |S(u−u⋆)| ≤C3/ba∇dblu−u⋆/ba∇dbl3whenever /ba∇dblu−u⋆/ba∇dbl< ε. (C.17) Proposition C.4. A convergent power series u(x1,...,xn) =/summationdisplay JcJ J!xJ istheTaylorseries for theanalyticfunction u(x)attheorigin x0=0, andso itscoefficients cJ=∂Ju/∂xJ(0)are the partial derivatives of uat the origin. The proof is the same as the one-dimensional version Proposi tion C.4: differentiate and substitute. C.2. Special Functions. Very few differential equations can be solved explictly in cl osed form. Even for linear ordinary differential equations, once one tries to move beyo nd the simplest constant coeffi- cient equations, there are not very many examples with expli cit solutions. One important example that has been solved are the Euler equations (3.84) B ut many other fairly sim- ple second order equations, including the Bessel and Legend re equations that arose as a result of our separation of variables solution to partial di fferential equations, do not have elementary functions as solutions. These and other equatio ns that appear in a number of key applications lead to new types of “special functions” th at occur over and over again in applications. Just as the student learned to become familiar with exponent ial and trigonometric functions, thus, at a more advanced level, applications in p hysics, engineering and mathe- matics require gaining some familiaritywith the propertie s of these functions. The purpose of this section is to introduce the student to some basic prop erties of the most important special functions, including the gamma function, the Airy f unctions, the Legendre func- tions and, finally the Bessel functions. Lack of space will pr event us from introducing additional important special functions, such as hypergeom etric functions, confluent hyper- geometric functions, parabolic cylinder functions, the ze ta function, elliptic functions, and many others. The interested reader can consult more advance d texts, such as [ 144,190], and the handbook [ 3], as well as the soon to appear update [ 145] and its web site, for the latest information on this fascinating and very active field of mathematics and applica- tions. We should remark that there is no precise definition of the term “special function” — it merely designates a function that plays a distinguished role in a range of applications and whose properties and evaluation are therefore of partic ular interest. Most special functions arisemost naturallyassolutionsto second order linearordinary differential equations with variable coefficients. One metho d of gaining analytical insight into their properties is to formulate them as power series. T herefore, we will learn how to construct power series solutions to differential equatio ns when closed form solutions are not available. As we shall see, although computationally me ssy at times, the power series method is straightforward to implement in practice. When we are at a regular point for the differential equation, the solutions can be obtained as ordi nary power series. At so-called 12/11/12 1269 c/ci∇cleco√y∇t2012 Peter J. Olver regular singular points, a more general type of series known as a Frobenius expansion is required. More general singular points require more advanc ed techniques, and will not be discussed here. The Gamma Function The first special function that we shall treat does not, in fac t arise as the solution to a differential equation. Rather, it forms a generalizatio n of the factorial function from integers to arbitrary real and complex numbers. As such, it w ill often appear in power series solutions to differential equations when parameters take on non-integral values. First recall that the factorial of a non-negative integer nis defined inductively by the iterative formula n! =n·(n−1)!,starting with 0! = 1 . (C.18) Thus, ifnis a non-negative integer, the iteration based on the second formula terminates, and yields the familiar expression n! =n(n−1)(n−2)····3·2·1. (C.19) Ifnis not a non-negative integer, then the iteration will not te rminate, and we cannot use it to compute the factorial. Our goal is to circumvent thi s difficulty, and introduce a function f(x) that is defined for allvalues of x, and will play the role of such a factorial. The function should satisfy the functional equation f(x) =xf(x−1) (C .20) where defined. If, in addition, f(0) = 1, then we know f(n) =n! whenever nis a non- negative integer, and hence such a function will extend the d efinition of the factorial to more general real and complex numbers. A moment’s thought should convince the reader that there are many possible ways to construct such a function. The most important method relies on an integral formula, and leads to the definition of the gamma function, originally dis covered by Euler. Definition C.5. Thegamma function is defined by Γ(z) =/integraldisplay∞ 0e−ttz−1dt. (C.21) The first fact is that the gamma function integral converges w henever Re z >0; otherwise the singularity of tz−1is too severe to permit convergence of the improper integral at t= 0. The key property that turns the gamma function into a subs titute for the factorial function relies on an elementary integration by parts: Γ(z+1) =/integraldisplay∞ 0e−ttzdt=−e−ttz/vextendsingle/vextendsingle/vextendsingle∞ t=0+z/integraldisplay∞ 0e−ttz−1dt. The boundary terms vanish whenever Re z >0, while the final integral is merely Γ( z). Therefore, the gamma function satisfies the recurrence rela tion Γ(z+1) =zΓ(z) provided Re z >0. (C.22) 12/11/12 1270 c/ci∇cleco√y∇t2012 Peter J. Olver Ifweset f(x) = Γ(x+1), then(C.22)isthesameas(C.20). Moreover, by directin tegration Γ(1) =/integraldisplay∞ 0e−tdt= 1. Combining this with the recurrence relation (C.22), we dedu ce that Γ(n+1) =n! (C .23) whenever n≥0 is a non-negative integer. Therefore, we can identify x! with the value Γ(x+1) whenever x >−1 isanyreal number. Remark: The reader may legitimately ask why not replace tz−1bytzin the definition of Γ(z), which would avoid the n−1 in (C.23). There is no simple answer; we are merely following a well-established precedent set originally by E uler. Thus, at integer values of z, the gamma function agrees with the elementary factorial. A few other values can be computed exactly. One important cas e is when z=1 2. Using the substitution t=x2, withdt= 2xdx, we find Γ/parenleftbig1 2/parenrightbig =/integraldisplay∞ 0e−tt−1/2dt=/integraldisplay∞ 02e−x2dx=√π, (C.24) where the final integral was evaluated earlier, (Gaussint ). Thus, using the identification with the factorial function, we identify this value with/parenleftbig −1 2/parenrightbig ! =√π. The recurrence relation (C.22) will then fix the value of the gamma function a t all half-integers1 2,3 2,5 2,.... For example, Γ/parenleftbig3 2/parenrightbig =1 2Γ/parenleftbig1 2/parenrightbig =1 2√π, (C.25) and hence1 2! =1 2√π. Further properties of the gamma function are outlined in th e exercises. A graph of the gamma function appear in Figure C.1 . Note the appearance of singularities at negative integer values of x=−1,−2,.... One of the most useful formulas involving the gamma function isStirling’s Formula , Γ(n+1) =n!∼√ 2πnnn en, n −→ ∞, (C.26) which gives the asympotitic values of the factorial functio n for large n. A proof is outlined in the exercises. C.3. Series Solutions of Ordinary Differential Equations. When confronted with a novel differential equation, there ar e a few standard options for making progress in solving and understanding the soluti ons. One of these is the “look- up”method, thatreliesonpublishedcollectionsofdifferen tialequationsandtheirsolutions. One of the most useful references that collects together man y solved differential equations is the classic German compendium written by Kamke, [ 112]. Two more recent English- language handbooks are [ 196,198]. Of course, numerical integration — see Chapter 20 for a prese ntation of basic methods — is always an option for approximating the solution. Numeri cal methods do, however, 12/11/12 1271 c/ci∇cleco√y∇t2012 Peter J. Olver Figure C.1. The Gamma Function. have their limitations, and are best accompanied by some und erstanding of the underlying theory, coupled with qualitative or quantitative expectat ions of how the solutions should behave. Furthermore, numericalmethodsprovidelessthana dequateinsightintothenature of the special functions that appear as solutions of the part icular differential equations arising in separation of variables. A numerical approximat ion cannot, in itself, be used to establish rigorous mathematical properties of the solutio ns of the differential equation. A more classical means of constructing and approximating th e solutions of differential equations is based on their power series or Taylor series exp ansions. The Taylor expansion ofasolutionatapoint x0isfoundbysubstitutingageneralpowerseriesintothediffe rential equation and equating coefficients of the various powers of x−x0. The initial conditions at x0serve to uniquely determine the coefficients and hence the der ivatives of the solution at the initial point. The Taylor expansion of a special functio n can be used to deduce many of the key properties of the solution, as well as provide reason able numerical approximations to its values within the radius of convergence of the series. (Howeer, serious numerical computations more often rely on non-convergent asymptotic expansions, [ 144].) Example C.6. Before developing the general computational machinery, a n a¨ ıve computation of a simple equation will be enlightening. Cons ider the initial value problem d2x du2+u= 0, u (0) = 1, u′(0) = 0. Let’s see how to construct a soloution in the form of a power se ries u(x) =u0+u1x+u2x2+u3x3+···=∞/summationdisplay n=0unxn 12/11/12 1272 c/ci∇cleco√y∇t2012 Peter J. Olver for the solution Term-by-term differentiation yields the se ries expansions† u′′(x) =u2+6u3x+12u4x2+20u5x3+···=∞/summationdisplay n=0(n+1)(n+2)un+2xn,(C.27) for its derivatives. Substituting into the equation, and th en equating the various powers of xto 0leadstothefollowingrecurrence relationsrelatingth ecoefficients ofourpower series. The left column indicates the power of x, while the right column tells us the recurrence relation:1 2 u2+u0= 0, x 6u3+u1= 0, x212u4+u2= 0, x320u5+u3= 0, x430u6+u4= 0, ...... xn(n+1)(n+2)un+2+un= 0. The initial conditions serve to prescribe the first two coeffic ients; u0=u(0) = 1, u1=u′(0) = 0. We solve the recurrence relations in order. The first equatio n determines u2=−1 2u0= −1 2. The second prescribes u3=−1 6u0= 0. Next we find u4=−1 12u2=1 24. Next, u5=−1 20u3= 0. Then u6=−1 30u4=−1 720. In general, it is not hard to see that u2k=(−1)k (2k)!, u2k+1= 0, and hence the resulting series solution is u(x) = 1−1 2x2+1 24x3−1 720x6+···=∞/summationdisplay k=0(−1)k k!x2k, which is merely the well-known Taylor series for cos x, the solution to the initial value problem. Note that the generation of the Taylor series does n ot rely on any a priori knowl- edge of trigonometric functions or alternatuve solution me thods for ordinary differential equations. In this section, we provide a brief introduction to the basic series solution techniques for ordinary differential equations, concentrating on seco nd order linear differential equa- tions, since these form by far the most important class of exa mples arising in applications. Whenx0is a regular point, the method will construct a standard Tayl or expansion for the †If we choose to work with the series in summation form, we need to re- index appropriately in order to display the term of degree n. 12/11/12 1273 c/ci∇cleco√y∇t2012 Peter J. Olver solution, whileso-calledregular singularpointsrequire aslightlymoregeneral seriesexpan- sion. Generalizations to higher order equations, nonlinea r equations, and even (nonlinear) systems are left to other more detailed texts, including [ 107]. Regular Points We shall concentrate on solving a homogeneous linear differe ntial equation of the form p(x)d2u dx2+q(x)du dx+r(x)u= 0. (C.28) The coefficients p(x),q(x),r(x) are assumed to be analytic functions where defined. This means that, at a point x0, they admit convergent power series expansions p(x) =p0+p1(x−x0)+p2(x−x0)2+···, q(x) =q0+q1(x−x0)+q2(x−x0)2+···, r(x) =r0+r1(x−x0)+r2(x−x0)2+···.(C.29) We expect that the solutions to the differential equation wil l also be analytic functions. This expectation is justified provided that the equation is regularat the point x0, in the following sense. Definition C.7. A pointx=x0is aregular point of a second order linear ordinary differential equation (C.28) provided the leading coefficien t does not vanish there: p0=p(x0)/ne}ationslash= 0. A point where p(x0) = 0 is known as a singular point . In short, a regular point iswhere thesecond order derivativ etermsdoes not disappear, and so the equation is “genuinely” second order. Remark: The definition of a singular point assumes that the other coe fficients do not both vanish there, i.e., either q(x0)/ne}ationslash= 0 orr(x0)/ne}ationslash= 0. If all three functions happen to vanish at x0, we would factor out a common factor ( x−x0)k, and hence, without loss of generality, can assume at least one of the coefficients is nonz ero atx0. The basic existence theorem for differential equations at re gular points follows. See, e.g., [100,107] for a proof. Theorem C.8. Letx0be a regular point for the second order homogeneous linear ordinary differential equation (C.28). Then there existsa unique solution u(x)totheinitial value problem u(x0) =a, u′(x0) =b. (C.30) Moreover, the solution u(x)is an analytic function for xsufficiently close to x0. Remark: It can be proved, [ 100], that the radius of convergence of any analytic solution u(x) is equal to the distance from the regular point to the neares t singular point in the complex plane. 12/11/12 1274 c/ci∇cleco√y∇t2012 Peter J. Olver At any regular point, the second order differential equation (C.28) admits two linearly independent analytic solutions, which we denote by u(x) and/tildewideu(x). The general solution can be written as a linear combination of the two basis soluti ons: u(x) =au(x)+b/tildewideu(x). (C.31) A standard choice for the two basis solutions is to take the fir st to satisfy the initial conditions u(x0) = 1, u′(x0) = 0, (C.32) and the second to satisfy /tildewideu(x0) = 0, /tildewideu′(x0) = 1, (C.33) although other choices may be used depending upon particula r circumstances. With this choice, the linear combination automatically satisfies the initial conditions (C.30). Therefore, every solution to an analytic differential equat ion at a regular point can be expanded in an ordinary power series u(x) =u0+u1(x−x0)+u2(x−x0)2+···=∞/summationdisplay n=0un(x−x0)n(C.34) at the point. Since the power series coincides with the Taylo r series for u(x), its the coefficients un=u(n)(x0) n! are multiples of the derivatives of the function at the point x0. (Some authors prefer to keep the n!’s in the power series; this is purely a matter of taste.) In p articular, the first two coefficients u0=u(x0) =a, u1=u′(x0) =b. (C.35) are prescribed by the initial conditions. Once the initial c onditions have been specified, the remaining coefficients must be uniquely prescribed since there is only one solution to the initial value problem. The basic computational technique for constructing the pow er series solution to the initial value problem is quite straightforward. One substi tutes the known power series (C.29) for the coefficient functions and the unknown power ser ies (C.34) for the solution into the differential equation (C.28). Multiplying out the f ormulae will result in a (compli- cated) power series thatmust beequated tozero. Atthispoin t, oneanalyzesthe individual coefficients. We rely on the basic observation that Two power series are equal if and only if their individual coe fficients are equal, generalizing the standard test for equality of polynomials . In particular, a power series represents the zero function if and only if all its coefficient s are 0. Thus, thepower seriessolutionmethodcontinues byequatin g, inorder, thecoefficients of the resulting power series to zero, starting with the lowe st order (constant) and working 12/11/12 1275 c/ci∇cleco√y∇t2012 Peter J. Olver upwards. The lowest order terms are multiples of ( x−x0)0= 1, i.e., the constant terms in the differential equation, lead to a linear recurrence rel ation u2=R2(u0,u1) =R2(a,b) that prescribes the coefficient u2in terms of the initial data. The coefficients of ( x−x0) lead to a linear recurrence relation u3=R3(u0,u1,u2) =R3(a,b,R2(a,b)) that prescribes the coefficient u3in terms of the initial data and the prviously computed coefficient u2. And so on. At the nthstage of the procedure, the coefficients of ( x−x0)n lead to the nthlinearrecurrence relation un+2=Rn(u0,u1,...,un+1), n = 0,1,2,... , (C.36) that will prescribe the ( n+ 2)ndorder coefficient in terms of the previous ones. Once the coefficients u0andu1have been specified by the initial conditions, the remaining coefficients u2,u3,u4,...are successively fixed by the recurrence relations (C.36). I n this fashion, we easily deduce the existence of a formal power ser ies solution to the differential equation at a regular point. The one remaining issue is wheth er the resulting power series actually converges. This can be proved with a detailed analy sis, [107], and will serve to complete the proof of the general existence Theorem C.8. Rather than continue in generality, the best way to learn the method is to investigate simple examples. The Airy Equation A particularly easy case to analyze is the Airy equation u′′=xu. (C.37) This second order ordinary differential equation arises in o ptics, dispersive waves, caustics (focusing of light waves as with a magnifying glass) and diffr action, and rainbows. It was first derived by the English mathematician George Airy in 183 9, [5]. In Exercise , we saw how it arises in a separation of variables solution to the Tricomi equation arising in supersonic fluid motion. The solutions to the Airy equation are known as Airy functions . While Airy functions cannot be written in terms of the standard elementary functi ons, it is relatively straight- forward to determine their power series expansion. Since th e leading coefficient p(x)≡1 is constant, and every point x0is a regular point of the Airy equation. For simplicity, we only treat the case x0= 0, and therefore consider a power series u(x) =u0+u1x+u2x2+u3x3+···=∞/summationdisplay n=0unxn 12/11/12 1276 c/ci∇cleco√y∇t2012 Peter J. Olver for the solution. Term-by-term differentiation yields the s eries expansions† u′(x) =u1+2u2x+3u3x2+4u4x3+···=∞/summationdisplay n=0(n+1)un+1xn, u′′(x) =u2+6u3x+12u4x2+20u5x3+···=∞/summationdisplay n=0(n+1)(n+2)un+2xn,(C.38) for its derivatives. On the other hand, xu(x) =u0x+u1x2+u2x3+···=∞/summationdisplay n=1un−1xn. (C.39) Equating this power series to that of u′′(x) leads to the following recurrence relations relating the coefficients of our power series. The left column indicates the power of x, while the right column displays the recurrence relation: 1 u2= 0, x 6u3=u0, x212u4=u1, x320u5=u2, x430u6=u3, ...... xn(n+1)(n+2)un+2=un−1. We solve the recurrence relations in order. The first equatio n determines u2. The second prescribes u3=1 6u0in terms of u0. Next we find u4=1 12u1in terms of u1. Next, u5=1 20u2= 0. Then u6=1 30u3=1 180u0is first given in terms of u3, but we already know the latter in terms of u0. And so on. At the nthstage of the recursion, stage we determine un+2using our previously tabulated formula for un−1. The only coefficients that are not determined by this procedur e are the first two, u0 andu1. These correspond to the value of the solution and its deriva tive at the initial point x0= 0, as in (C.30). Let us construct the two basis solutions. The first uses the in itial conditions u0=u(0) = 1, u1=u′(0) = 0. The recurrence relations then show that the only nonvanishi ng coefficients cnare when n= 3kis a multiple of 3; all others are zero. Moreover, c3k=c3k−3 3k(3k−1) †If we choose to work with the series in summation form, we need to re- index appropriately in order to display the term of degree n. 12/11/12 1277 c/ci∇cleco√y∇t2012 Peter J. Olver A straightforward induction proves that c3k=1 3k(3k−1)(3k−3)(3k−4)···6·5·3·2. The resulting solution is u1(x) = 1+1 6x3+1 180x6+···=∞/summationdisplay k=1x3k 3k(3k−1)(3k−3)(3k−4)···6·5·3·2.(C.40) Note that the denominator is similar to a factorial, except e very third term is omitted. Similarly, starting with the initial conditions u0=u(0) = 0, u1=u′(0) = 1, we find that the only nonvanishing coefficients cnare when n= 3k+1 leaves a remainder of 1 when divided by 3. The recurrence relation c3k+1=c3k−2 (3k+1)(3k)yields c3k+1=1 (3k+1)(3k)(3k−2)(3k−3)···7·6·4·3. The resulting solution is u2(x) =x+1 12x4+1 504x7+···=∞/summationdisplay k=1x3k+1 (3k+1)(3k)(3k−2)(3k−3)···7·6·4·3.(C.41) Again, the denominator skips every third term in the product . Every solution to the Airy equation can be written as a linear combination u(x) =au1(x)+bu2(x),where a=u(0), b=u′(0) correspond to the initial conditions of u(x) atx= 0. The power series (C.40,41), con- verge quite rapidly for all values of x, and so the first few terms provide a reasonable approximation to the two solutions for moderate values of x. The solutions (C.40,41), while easiest to derive using powe r series techniques, are not the most useful for applications in mathematics and physics . The most important solution is theAiry function of the first kind Ai(x) =1 π/integraldisplay∞ 0cos/parenleftbig tx+1 3t3/parenrightbig dt. (C.42) An independent solution is provided by the Airy function of the second kind Bi(x) =1 π/integraldisplay∞ 0/bracketleftbig exp/parenleftbig tx−1 3t3/parenrightbig +sin/parenleftbig tx+1 3t3/parenrightbig/bracketrightbig dt. (C.43) They have the initial values Ai(0) =1 32/3Γ/parenleftbig2 3/parenrightbig, Ai′(0) =−1 31/3Γ/parenleftbig1 3/parenrightbig,Bi(0) =1 31/6Γ/parenleftbig2 3/parenrightbig, Bi′(0) =31/6 Γ/parenleftbig1 3/parenrightbig.(C.44) 12/11/12 1278 c/ci∇cleco√y∇t2012 Peter J. Olver -10-7.5 -5-2.5 2.5 5 -0.4-0.20.20.4 Ai(x)-12-10 -8-6-4-2 2 -0.50.511.522.53 Bi(x) Figure C.2. The Airy Functions. Graphs of the two Airy functions appear in Figure C.2. Both fu nctions oscillate for negative values of x, with a slowly decreasing amplitude. An intuitive explanat ion is that whenx <0 the Airy equation (C.37) corresponds to a constant coefficie nt differential equation of the form u′′=−k2u, which has oscillatory trigonometric solutions. On the other hand, when x >0, the Airy equation is more like a constant coefficient differe ntial equation of the form u′′= +k2u, whose basis solutions ekxande−kxare, respectively, exponentially growing and exponentially decaying. Indeed , asx→ ∞, the first Airy functionAi( x)decaysveryrapidly,whereasthesecondBi( x)growsevenmoredramatically. Actually, the growth/decay rates are faster than exponenti al. It can be shown that Ai(x)∼  e−2x3/2/3 2√π x1/4, x →+∞, sin/parenleftbigg2 3(−x)3/2+π 4/parenrightbigg √π(−x)1/4, x→ −∞, Bi(x)∼  e2x3/2/3 2√π x1/4, x →+∞, cos/parenleftbigg2 3(−x)3/2+π 4/parenrightbigg √π(−x)1/4, x→ −∞. Every solution to the Airy equation can be written as a linear combination u(x) =aAi(x)+bBi(x). Detailed investigations into the properties and numerical computation of the Airy functions can be found in [ 3,145,144]. The Legendre Equation A particularly important example is the Legendre equation (1−t2)2d2P dt2−2t(1−t2)dP dt+/bracketleftbig λ(1−t2)−m2/bracketrightbig P= 0. (C.45) 12/11/12 1279 c/ci∇cleco√y∇t2012 Peter J. Olver The parameter mgoverns the orderof the Legendre equation, which in the cases of in- terest to us, is an integer, while λplays the role of an eigenvalue. As we learned in the preceding sections, this differential equation arises in th e solutions to a wide variety of partial differential equations in spherical coordinates. T he boundary conditions that serve to specify the eigenvalues are that the solution remain boun ded at the two singular points t=±1, leading to |P(−1)|<∞, |P(+1)|<∞. (C.46) The point t= 0 is a regular point of the Legendre equation. Indeed, the on ly singular points are the boundary points t=±1. Therefore, we can determine the solutions to the Legendre equation by the method of power series based at t0= 0. However, the general recurrence relations are rather complicated to solve in clo sed form, and we use some tricks to get a handle on the solutions. Consider first the case m= 0. The Legendre equation of order 0 is (1−t2)d2P dt2−2tdP dt+λP= 0. (C.47) As we noted above, the eigenfunctions are the Legendre polynomials Pn(t) =1 2nn!dn dtn(t2−1)n. They clearly satisfy the boundary conditions (C.46). To ver ify that they are indeed solu- tions to the differential equation (C.47), we let qn(t) = (t2−1)n. By the chain rule, the derivative of qn(t) is q′ n= 2nt(t2−1)n−1and hence ( t2−1)q′ n= 2nt(t2−1)n= 2ntqn. Differentiating the latter formula, (t2−1)q′′ n+2tq′ n= 2ntq′ n+2nqn,or (t2−1)q′′ n= 2(n−1)tq′ n+2nqn. A simple induction proves that the kthorder derivative q(k) n(t) =dkqn/dtksatisfies (t2−1)q(k+2) n= 2(n−k−1)tq(k+1) n+2[n+(n−1)+···+(n−k)]q(k) n = 2(n−k−1)tq(k+1) n+(k+1)(2n−k)q(k) n.(C.48) In particular, when k=n, this reduces to (t2−1)q(n+2) n=−2tq(n+1) n+n(n+1)q(n) n= 0, and sovn=q(n) nsatisfies (1−t2)v′′ n−2tv′ n+n(n+1)vn= 0, which is precisely the order 0 Legendre equation (C.47) with eigenvalue parameter λ= n(n+ 1). The Legendre polynomial Pnis a constant multiple of vn, and hence it too 12/11/12 1280 c/ci∇cleco√y∇t2012 Peter J. Olver satisfies the order 0 Legendre equation and hence forms an eig enfunction for the Legendre boundary value problem (C.45–46). While it is not immediate ly apparent that the Leg- endre polynomials form a complete system of eigenfunctions , this is the case. This is the result of a general theory of eigenfunctions of Sturm–Liouv ille boundary value problems, [42], or, more particlarly, the theory of orthogonal polynomia ls, [145]. Indeed, the orthog- onality of the Legendre polynomials that was noted in Chapte r 5 is, in fact, a consequence of the fact that they are eigenfunctions for this self-adjoi nt boundary value problem. More generally, if we substitute k=m+nin (C.48), we have (1−t2)w′′ n−2(m+1)tw′ n+(m+n+1)(n−m)wn= 0, (C.49) wherewn=q(m+n) n. This is notthe order mLegendre equation, but can be converted into it by setting wn= (1−t2)−m/2zn. Differentiating, we find w′ n= (1−t2)−m/2z′ n−mt(1−t2)−m/2−1zn, w′′ n= (1−t2)−m/2z′′ n−2mt(1−t2)−m/2−1z′ n+(m+m(m+1)t2)(1−t2)−m/2−2zn. Therefore, after a little algebra, equation (C.49) takes th e alternative form (1−t2)−m/2+1z′′ n−2t(1−t2)−m/2z′ n+(n(n+1)(1−t2)−m2)(1−t2)−m/2−1zn= 0, which, when multiplied by (1 −t2)m/2+1, is precisely the order mLegendre equation (C.45) with eigenvalue parameter λ=n(n+1). We conclude that zn(t) = (1−t2)m/2wn(t) = (1−t2)m/2dn+m dtn+m(t2−1)n is a solution to the order mLegendre equation. Moreover, zn(±1) = 0, and hence zn(t) is an eigenfunction for the order mLegendre boundary value problem. Indeed, zn(t) is a constant multiple of the associated Legendre function Pm n(t), as defined in (18.28). With some more work, it can be proved that the associated Legendre functions form a complete system of eigenfunctions for the the order mLegendre boundary value problem. Regular Singular Points Inalargerangeofapplications, oneisarticularlyinteres tedinthebehaviorofsolutions to a differential equation near a singular point. Usually, a p ower series expansion (C.34) fails to produce a solution at a singular point. As before, we write the differential equation as p(x)d2u dx2+q(x)du dx+r(x)u= 0. (C.50) Here, we assume that the functions p,q,rare analytic at x0, where now we assume that p(x0) = 0, but at least one of q(x0),r(x0) is non-zero. If the singular point is not too “wild”, onecanconstruct solutionsusingarelativelysimp lemodificationofthebasicpower series. 12/11/12 1281 c/ci∇cleco√y∇t2012 Peter J. Olver In order to formulate the key definition, we rewrite the differ ential equation in solved form d2u dx2=g(x)du dx+h(x)uwhere g(x) =−q(x) p(x), h(x) =−r(x) p(x). Ifp(x0) = 0, then, typically, the functions g(x),h(x) will have singularities at x=x0, and we need to ensure that these singularities are not too bad. Definition C.9. A singular point x0is called a regular singular point if g(x) =k(x) x−x0, h (x) =ℓ(x) (x−x0)2, (C.51) wherek(x) andℓ(x) are analytic at x=x0. Thus, in the language of complex analysis, the point x0is a regular singular point provided g(x) has a pole of order at most 1, while h(x) has a pole of order at most 2 at x=x0. In terms of the original coefficients, the regularity condit ions (C.51) require that we can write the differential equation in the form (x−x0)2a(x)d2u dx2+(x−x0)b(x)du dx+c(x)u= 0, (C.52) wherea(x),b(x) andc(x) are analytic at x=x0and, moreover, a(x0)/ne}ationslash= 0. Fortunately, almost all ordinary differential equations ar ising in applicationshave only regular singular points. The irregular singular points are much harder to deal with, and must be relegated to an advanced treatment, e.g., [ 100,107]. Thesimplestexampleofanequationwitharegularsingularp ointisthe Euler equation ax2u′′+bxu′+cu= 0, (C.53) wherea/ne}ationslash= 0,b,care constants. The point x= 0 is a regular singular point; indeed, the solved form of the Euler equation is u′′=−b axu′−c ax2u, andhence satisfies(C.51). Allotherpoints x0/ne}ationslash= 0areregularpointsfortheEulerequation. As discussed in Example 7.35, Euler equations are solved by s ubstituting the power ansatzu(x) =xrinto the equation. As a result, the exponent ris determined by the associated characteristic equation (7.53), namely ar(r−1)+br+c= 0. Ifthisquadraticequationhastwodistinctroots r1/ne}ationslash=r2, weobtaintwolinearlyindependent (possibly complex) solutions u(x) =xr1and/tildewideu(x) =xr2. The general solution u(x) = c1xr1+c2xr2is a linear combination of these two basis solutions. Note th at unless r1and r2are non-negative integers, the solutions have a singularit y — either a pole or branch point — at the singular point x= 0. A repeated root, r1=r2, requires an additional 12/11/12 1282 c/ci∇cleco√y∇t2012 Peter J. Olver logarithmic term, /tildewideu(x) =xr1logx, in the second solution, and the general solution has the form u(x) =c1xr1+c2xr1logx. The series solution method at more general regular singular points is modeled on the simple example of the Euler equation. One now seeks a solutio nthat has a series expansion of the form u(x) = (x−x0)r∞/summationdisplay n=0un(x−x0)n=u0(x−x0)r+u1(x−x0)r+1+u2(x−x0)r+2+···.(C.54) The full theory was established by the German mathematician Georg Frobenius in the late 1800’s, and the series are sometimes known as Frobenius expansions . The exponent ris known as the indexof the expansion. Remark: If the index r=−nis a negative integer, then (C.54) has the form of a Laurent series expansion, [ 4] Butrcan be non-integral, or even complex, and the resulting expansion is known in complex analysis as a Puiseux expansion , [95]. We can assume, without any loss of generality, that the leadi ng coefficient u0/ne}ationslash= 0. Indeed, if uk/ne}ationslash= 0 is the first non-zero coefficient, then the series begins wit huk(x−x0)r+k, and we replace rbyr+kto write it in the preceding form. Moreover, since any scalar multiple of a solution is a solution, we can divide by u0and assume that u0= 1 or any other convenient non-zero value, as desired. Warning : Unlike ordinary power series expansions, the coefficients u0andu1arenot prescribed by the initial conditions at the point x0. Indeed, as we learned in our study of the Bessel and Legendre equations, one cannot typically i mpose specific initial values for the solutions at a singular point. Often, mere boundedne ss will suffice to distinguish a solution. Here, the solution is usually completely detrem ined by the index rand the leading coefficient u0. The Frobenius solution method proceeds by substituting the series (C.54) into the differential equation (C.52). Since u(x) = (x−x0)r+u1(x−x0)r+1+···, (x−x0)u′(x) =r(x−x0)r+(r+1)u1(x−x0)r+1+···, (x−x0)u′′(x) =r(r−1)(x−x0)r+(r+1)ru1(x−x0)r+1+···, the lowest order terms are multiples of ( x−x0)r. Equating this particular coefficient to zero leads to a quadratic equation of the form s0r(r−1)+t0r+r0= 0, (C.55) where s0=s(x0) =1 2p′′(x0), t0=t(x0) =q′(x0), r0=r(x0), are the leading coefficients in the power series expansions of the coefficients of the differ- ential equation. The quadratic equation (C.55) is known as t heindicial equation , since it determines the possible indices rin the Frobenius expansion of a solution. 12/11/12 1283 c/ci∇cleco√y∇t2012 Peter J. Olver Therefore, just as in the Euler equation, it turns out that (t ypically) there are two allowable indices, say r1andr2, which are the roots of the quadratic indicial equation. If the indices are distinct, then one expects to find two differ ent Frobenius expansions. Usually, this assumption is valid, but there is an important exception, which occurs when the roots differ by an integer. The general result is summariz ed in the following list. (i) Ifr2−r1is not an integer, then there are two linearly independent so lutionsu(x) and /tildewideu(x), each having a convergent Frobenius expansions of the form (C.54). (ii) Ifr1=r2, then there is only one solution with a convergent Frobenius expansion. (iii) Finally, if r2=r1+k, wherek >0 is a positive integer, then there is a solution with a convergent Frobenius expansion corresponding to the smal ler index r1. The solution associated with the larger index r2may or may not have a convergent Frobenius expansion. Thus, in every case the differential equation has at least one solution with a Frobenius expansion. When the leading coefficient u0= 1 is fixed, then the remaining coefficients u1,u2,...are uniquely prescribed by the recurrence relations stemmi ng from substitution oftheexpansionintothedifferentialequation. Ifthesecon dsolutiondoesnothaveaFrobe- nius expansion, then it has an additional logarithmic term, as with the Euler equation, of a well-prescribed form. Details appear in the exercises. Ra ther than try to develop the theory in any more detail here, we suffice with consideration o f some particular examples. Example C.10. Consider the second order ordinary differential equation u′′+/parenleftbigg1 x+x 2/parenrightbigg u′+u= 0 (C .56) that we needed to solve for finding the fundamental solution t o the heat equation; see (rheatode ). We look for series solutions based at x= 0. Since the coefficient of u′has a simple pole, the point x= 0 is a regular singular point, and thus we can work with a Frobenius expansion as in (C.58). Substituting into the diff erential equation, we find that the coefficients of xrlead to the indicial equation r2= 0. There is only one root, r= 0, and hence even though we are at a singular point, we are dealing with an ordinary power series. The next term tells us thatu1= 0. Since r= 0, the general recurrence relation is (n+2)2un+2+1 2(n+2)un= 0, and hence un+2=−un 2(n+2). Therefore, the odd coefficients u2k+1= 0 are all zero, while the even ones are u2k=−u2k−2 4k=u2k−4 4k(4k−4)=−u2k−6 4k(4k−4)(4k−8)=···=(−1)k 4kk!since u0= 1. 12/11/12 1284 c/ci∇cleco√y∇t2012 Peter J. Olver The resulting power series takes a familiar form: u(x) =∞/summationdisplay k=1u2kx2k=∞/summationdisplay k=11 k!/parenleftbigg −x2 4/parenrightbiggk =e−x2/4. The second solution will require a logarithmic term. Howeve r, it can be found directly by a general reduction method. Once we know one solution to a s econd order ordinary differential equation, the second solution can be found by su bstituting the ansatz /tildewideu(x) =u(x)v(x) =e−x2/4v(x) into the equation. Thus, /tildewideu′′+/parenleftbigg1 x+x 2/parenrightbigg /tildewideu′+/tildewideu=/bracketleftbigg u′′+/parenleftbigg1 x+x 2/parenrightbigg u′+u/bracketrightbigg v+uv′′+2u′v′+/parenleftbigg1 x+x 2/parenrightbigg uv′ =e−x2/4/parenleftbigg v′′+1 xv′/parenrightbigg . Therefore, v′satisfies a linear first order ordinary differential equation : v′′+v′ x= 0,and hence v′=c1 x, v=clogx+d. The general solution to the original differential equation i s /tildewideu(x) =u(x)v(x) =e−x2/4(clogx+d). Bessel’s Equation Perhaps the most important “non-elementary” ordinary diffe rential equation is x2u′′+xu′+(x2−m2)u= 0, (C.57) known as Bessel’s equation of order m. We assume here that the order m≥0 is a non- negative real number; see Exercise for the Bessel equation of imaginary order. As we have seen, the Bessel equation arises from separation of var iables in a remarkable number of partial differential equations, including the Laplace, h eat and wave equations on a disk, a cylinder, and a spherical ball. Interestingly, the soluti ons to the Bessel equation were first discovered by the German mathematician Bessel in a comp letely different context: the study of celestial mechanics, i.e., the Newtonian theor y of planets orbiting around a central sun; see (19.16). The Bessel equation cannot (except in a few particular insta nces) be solved in terms of elementary functions, and so the use of power series is nat ural. The leading coefficient p(x) =x2is nonzero exceptwhenx= 0, and so all points except the origin are regular points. Therefore, at all nonzero points x0/ne}ationslash= 0, the standard power series construction can be used to produce the appropriate power series solutions of the Bessel equation. However, the recurrence relations for the coefficients are not particu larly easy to solve in clsoed form. 12/11/12 1285 c/ci∇cleco√y∇t2012 Peter J. Olver Moreover, applications tend to demand understanding the be havior of the solutions to the Bessel equation at the singular point x0= 0. Writing the Bessel equation in solved form u′′=−1 xu′+/parenleftbiggm2 x2−1/parenrightbigg u, we immediately see that x= 0 satisfies the conditions to qualify as a regular singular point. Consequently, we are led to seek a solution in the form of a Frobenius expansion. We first compute the expressions for the first two derivatives u(x) =xr+u1xr+1+u2xr+2+··· u′(x) =rxr−1+(r+1)u1xr+(r+2)u2xr+1+··· u′′(x) =r(r−1)xr−2+(r+1)ru1xr−1+(r+2)(r+1)u2xr+···,(C.58) of our purported solution. Substituting these expressions into (C.57), we find /bracketleftbig r(r−1)xr+(r+1)ru1xr+1+(r+2)(r+1)u2xr+2+···/bracketrightbig + +/bracketleftbig rxr+(r+1)u1xr+1+(r+2)u2xr+2+···/bracketrightbig + +/bracketleftbig xr+2+u1xr+3+u2xr+4+···/bracketrightbig −/bracketleftbig m2xr+m2u1xr+1+m2u2xr+2+···/bracketrightbig = 0, We equate the coefficients of the various powers of xto zero. The coefficient of the lowest order power, xr, is the indicial equation r(r−1)+r−m2=r2−m2= 0. There are two solutions to the indicial equation, r=±m, unless m= 0 in which case there is only one possible index r= 0. The higher powers of xlead to recurrence relations for the successive coefficients un. If we replace m2byr2, we find the following constraints: xr+1:/bracketleftbig (r+1)2−r2/bracketrightbig u1= (2r+1)u1= 0, u1= 0, xr+2:/bracketleftbig (r+2)2−r2/bracketrightbig u2+1 = (4r+4)u2+1 = 0, u2=−1 4r+4, xr+3:/bracketleftbig (r+3)2−r2/bracketrightbig u3+u1= (6r+9)u3+u1= 0, u3=−u1 6r+9= 0, and, in general, xr+n:/bracketleftbig (r+n)2−r2/bracketrightbig un+un−2=n(2r+n)un+un−2= 0. Thus, the basic recurrence relation is un=−1 n(2r+n)un−2, n = 2,3,4,... . (C.59) Starting with u0= 1,u1= 0, it is easy to deduce that all un= 0 for all odd n= 2k+1, while for even n= 2k, u2k=−u2k−2 4k(k+r)=u2k−4 16k(k−1)(r+k)(r+k−1)=··· =(−1)k 22kk(k−1)···3·2(r+k)(r+k−1)···(r+2)(r+1). 12/11/12 1286 c/ci∇cleco√y∇t2012 Peter J. Olver Therefore, the series solution is u(x) =∞/summationdisplay k=0u2kxm+2k=∞/summationdisplay k=0(−1)kxm+2k 22kk(k−1)···3·2(r+k)(r+k−1)···(r+2)(r+1). (C.60) So far, we not paid attention to the precise values of the indi cesr=±m, or whether oursolutiontotherecurrencerelationsisvalid. Inordert ocontinuetherecurrence, weneed to ensure that therecurrence relation(C.59)islegitimate , meaning thatthe denominator is never 0. Since n >0, this will notbe the case if and only if 2 r+n= 0, which requires that r=−1 2nbe either a negative integer −1,−2,−3,..., or half-integer, −1 2,−3 2,−5 2,.... These cases occur when the order m=−r=1 2nis either an integer or a half-integer. Indeed, these cases are precisely the cases when the two indi ces, namely r1=−mand r2=m, differ by an integer, r2−r1=n, and so we are in the tricky case ( iii) of the Frobenius method. There is, in fact, a key distinction between the integral and the half integral cases. Recall that the odd coefficients u2k+1= 0 in the Frobenius series automatically vanish, and so we only have to worry about the recurrence relation (C. 59) forevenvalues of n. Thus, for even n= 2k, the factor 2 r+n= 2(r+k) = 0 vanishes only when r=−kis a negative integer; the half integral values do not, in fact ca use problems. Therefore, if the orderm≥0 isnota non-negative integer, then the Bessel equation of order madmits two linearly independent Frobenius solutions, given by the exp ansions (C.60) with exponents r= +mandr=−m. Ifmis an integer, however, there is only one Frobeius solution, namely the expansion (C.60) with r= +mgiven by the positive exponent. The second independent solution has an additional logarithmic term in its formula; details appear in Exercise . By convention, the standard Bessel function of order mis obtained by multiplying this solution by 1 2mm!or, rather,1 2mΓ(m+1), (C.61) where the first factorial form can be used if mis a non-negative integer, while the more general gamma function expression must be employed for non- integral values of m. The result is Jm(x) =∞/summationdisplay k=0(−1)kxm+2k 22k+mk!Γ(m+k+1), (C.62) The series is well-defined for all†mexcept when m=−1,−2,−3,...is a negative integer. We conclude that Theorem C.11. Ifm >0isnotan integer, then the two linearly independent solutions to the Bessel equation of order mare the Bessel functions Jm(x)andJ−m(x). †Actually, if mis a negative integer, the first 2 m+ 1 terms in the series vanish because Γ(−n) =∞at negative integer values. The series J−m(x) =Jm(x) then actually coincides with its positive sibling. 12/11/12 1287 c/ci∇cleco√y∇t2012 Peter J. Olver Ifm= 0,1,2,3,...is an integer, the the Bessel function Jm(x)is a solution to the Bessel equation. The second solution, traditionally denoted Ym(x), can be found as a limiting case Ym(x) = lim ν→mYν(x) = lim ν→mcosνπ Jν(x)−J−ν(x) sinνπ(C.63) of a certain linear combination of Bessel functions of non-i ntegral order ν. The justification of the last statement of the theorem can be f ound in Exercise . We note that for ν/ne}ationslash=m, the linear combination of Bessel functions in the limiting expression is a solution to the Bessel equation of order νwhich is indepedent from Jν(x). It can be proved that this continues to hold in the limit. The series formula for Ym(x) is quite complicated, [ 144,186], and its derivation is left to a more advanced course. Example C.12. Consider the particular case when m=1 2. There are two indices, r=±1 2, for the Bessel equation of order m=1 2, leading to two solutions J1/2(x) and J−1/2(x) obtained by the Frobenius method. For the first, with r=1 2, the recurrence relation (C.59) takes the form un=−1 (n+1)nun−2. Starting with u0= 1 and u1= 0, the general formula is easily found to be un=  (−1)k (n+1)!, n= 2keven, 0 n= 2k+1 odd. Therefore, the resulting solution is u(x) =√x∞/summationdisplay k=0(−1)k (2k+1)!x2k=1√x∞/summationdisplay k=0(−1)k (2k+1)!x2k+1=sinx√x. According to (C.61), the Bessel function of order1 2is obtained by dividing this function by √ 2 Γ/parenleftbig3 2/parenrightbig =/radicalbiggπ 2, where we used (C.25) to evaluate the gamma function at3 2. Therefore, J1/2(x) =/radicalbigg 2 πxsinx. (C.64) Similarly, for the other index r=−1 2, the recurrence relation un=−1 n(n−1)un−2 12/11/12 1288 c/ci∇cleco√y∇t2012 Peter J. Olver leads to the formula un=  (−1)k n!, n= 2keven, 0 n= 2k+1 odd, for the coefficients, corresponding to the solution u(x) =x−1/2∞/summationdisplay k=0(−1)k (2k)!x2k=cosx√x. Therefore, using (C.61) and (C.24), the Bessel function of o rder−1 2is J−1/2(x) =√ 2 Γ/parenleftbig1 2/parenrightbigcosx√x=/radicalbigg 2 πxcosx. (C.65) Remark: Ifwenowsubstitute(C.64)intothedefining formula(18.90 )forthespherical Bessel functions, we prove our earlier elementary formula ( 18.91) for the spherical Bessel function of order 0. Finally, we demonstrate how Bessel functions of different or ders are related by an important recurrence relation. Proposition C.13. The Bessel functions are interconnected by the following re cur- rence formulae: dJm dx+m xJm(x) =Jm−1(x), −dJm dx+m xJm(x) =Jm+1(x).(C.66) Proof: Let us differentiate the power series xmJm(x) =∞/summationdisplay k=0(−1)kx2m+2k 22k+mk!(m+k)!. We find d dx[xmJm(x)] =∞/summationdisplay k=0(−1)k2(m+k)x2m+2k−1 22k+mk!(m+k)! =xm∞/summationdisplay k=0(−1)kxm−1+2k 22k+m−1k!(m−1+k)!=xmJm−1(x). Expansion of the left hand side of this formula leads to xmdJm dx+mxm−1Jm(x) =d dx[xmJm(x)] =xmJm−1(x), which proves the first recurrence formula (C.66). The second formula is proved by a similar manipulation involving differentiation of x−mJm(x). Q.E.D. 12/11/12 1289 c/ci∇cleco√y∇t2012 Peter J. Olver Example C.14. Forinstance, wecanuse(C.66)tofindthecorresponding recu rrence formulae for the spherical Bessel functions Sn(x) =/radicalbiggπ 2xJn+1/2(x). Differentiating and using the second recurrence relation, w e find dSn dx=/radicalbiggπ 2xdJn+1/2 dx−1 2/radicalbiggπ 21 x3/2Jn+1/2(x) =−/radicalbiggπ 2x/parenleftbigg Jn+3/2(x)+n+1 2 xJn+1/2(x)/parenrightbigg −1 2/radicalbiggπ 21 x3/2Jn+1/2(x) =−/radicalbiggπ 2xJn+3/2(x)+n x/radicalbiggπ 2xJn+1/2(x) =−Sn+1(x)+n xSn(x). This completes the proof of the spherical Bessel recurrence formula (18.92). With this, we conclude our brief introduction to the method o f Frobenius and the theory of Bessel functions. The reader interested in furthe r delving into either the general method, or the host of additional properties of Bessel funct ions is encouraged to consult the texts [ 190,144,107,186]. 12/11/12 1290 c/ci∇cleco√y∇t2012 Peter J. Olver