23 Elements of analytic ODE theory. Bessel s functions
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1 23 Elements of analytic ODE theory. Bessel s functions Recall I am changing the variables) that we need to solve the so-called Bessel s equation 23. Elements of analytic ODE theory Let x 2 u + xu + x 2 m 2 )u =, m =,, 2,... px)u + qx)u + rx)u = 23.) be a second order linear homogeneous ODE with non-constant coefficients. Recall that function f is called analytic at x if it can be represented in some neighborhood of x by a convergent power series: The coefficients can be found by fx) = u + u x x ) + u 2 x x ) 2 + u 3 x x ) u k = f k) x ), k! where k! = k. From the calculus course, for example, we know that exponent, sine and cosine are analytic anywhere in R: e x = + x + x2 2! + x3 3! +..., sin x = x x3 3! + x5 5! x7 7! +..., cos x = x2 2! + x4 4! x6 6! +... Theorem 23.. Consider problem 23.) and assume p, q, r are analytic at x and px ). Then problem 23.) with the initial conditions ux ) = u, u x ) = u has a unique analytic solution. This theorem actually gives us a way to work through the problem. All we need to do is to look for the solution in the form and determine u 2, u 3,.... Since the general solution to 23.) is given by ux) = u + u x x ) + u 2 x x ) ux) = Aûx) + Bǔx), where A, B are two arbitrary constants and û and ǔ are two linearly independent solutions, we can always use our power series method with two different and linearly independent) initial conditions, e.g., we can take ûx ) =, û x ) =, Math 483/683: Partial Differential Equations by Artem Novozhilov artem.novozhilov@ndsu.edu. Spring 26
2 and ǔx ) =, ǔ x ) =. In one of the previous lectures we already saw this method applied to the equation u + ωu =. Consider another example. Example 23.2 Airy equation). Consider and take the initial conditions u xu =, u) = = u, u ) = = u. The stated theorem obviously works in this case since p is constant and rx) = x. I take and hence ux) = u + u x + u 2 x 2 + u 3 x 3 + u 4 x u x) = 2u u 3 x + 4 3u 4 x Plugging the obtained expressions into my equation I find 2u u 3 x + 4 3u 4 x u 3 x 3... = u x + u x 2 + u 2 x 3 + u 3 x 4 + u 4 x Two convergent power series equal only if the coefficients at the same powers are equal, that is Using the initial conditions I find u 3k = 2u 2 =, 6u 3 = u, 2u 4 = u, 2u 5 = u 2, 3u 6 = u 3,... n + )n + 2)u n+2 = u n,... u 3k 3, k =, 2, 3,... 3k3k ) and all other u i are zero. The last expression is enough to write that my first linearly independent solution to the Airy equation is ûx) = u 3k x 3k, k= which, as can be proved, converges for any x R. I will leave it as an exercise to show that for the initial conditions u) =, u ) = the solution is ǔx) = k= u 3k+ x 3k+, u 3k+ = u 3k 2 3k + )3k. 2
3 23.2 Solving Bessel s equation Unfortunately, the same method will not work for Bessel s equation, if I d like to build a power series solution around. The reason is that x = is not a regular point, meaning that p) =. For Bessel s equation x = is a singular point, but fortunately for as a regular singular point a point x is called a regular singular point if the equation can be written as x x ) 2 ax)u + qx)u + rx)u =, which is obviously holds for our equation with x = and ax) = ). In this case it turns out the Frobenius method will work. Frobenius method says that in this case a solution can be sought in the form ux) = x x ) ν where ν does not to be integer or positive. I have, assuming that u =, The coefficient at x ν must be n= ux) = x ν + u x ν+ + u 2 x ν xux) = x ν+ + u x ν+2 + u 2 x ν x 2 ux) = x ν+2 + u x ν+3 + u 2 x ν u n x x ) n, u x) = νx ν + ν + )u x ν + ν + 2)u 2 x ν u x) = νν )x ν 2 + ν + )νu x ν + ν + 2)ν + )u 2 x ν +... νν ) + ν m 2 =, which is true only if ν = ±m, if m. For the ν + n degree I have, replacing m 2 with ν 2, x ν+n : [ ν + n) 2 ν 2] u n + u n 2 = u n = n2ν + n) u n 2, n = 2, 3, 4,... Starting with u =, u = I get that all the odd indices are zero, whereas for even n = 2k and hence my solution is u 2k 2 u 2k = 4kk + ν) =... = ux) = k= ) k 2 2k k!ν + k)ν + k )... ν + ) ) k x ν+2k 2 2k k!ν + k)ν + k )... ν + ). In general this is not necessary, but in our case m is an integer, and if ν = m then the denominator in the series above vanishes. Hence we only found one solution to Bessel s equation: ux) = k= ) k x m+2k 2 2k k!m + k)m + k )... m + ). 3
4 I am allowed to multiply my solution by any constant, and I choose, by convention, to multiply the series above by /2 m m!), in this case I have J m x) = k= ) k x m+2k 2 2k+m k!m + k)!, Bessel s function of the first kind of the m-th order. An application of the ratio test yields that the series converges for any x R and hence J m is analytic anywhere in R or even in C). Just to get first idea on the Bessel s functions, note that J ) =, J m ) =, m =, 2,... So, I found one independent solution and in need of another one. Abel s formula see Math 266) tells me that for two linearly independent solutions to u + ax)u + bx)u = I must have [ ] u u det 2 u u = Ce ax) dx. 2 In my case ax) = /x and hence I end up with or, after some rearranging Using u x) = J m x) I have Therefore, u u 2 u u 2 = C x, u2 u ˆ u 2 x) J m x) = ) = C ˆ u 2 x) = J m x) xu 2. C J 2 mx) dx. C J 2 mx) dx is the second linearly independent solution, which is actually called Neumann s function of the second kind of the m-th order Some facts about solutions to Bessel s equation The full analysis of the solutions to Bessel s equation is beyond the scope of this course. I, however, would like to show how at least some of the important results can be obtained and proved. I will start with Bessel s function of the first kind of order zero: J x) = x x x x Since the ratio of two consecutive terms is x2 2k) 2, 4
5 which approaches zero as k for any fixed x then this series converges absolutely and uniformly, and hence J and all its derivatives are continuous. For J I have J x) = x 2 x x x , which immediately implies that cf. with cos x = sin x). Analogously, dj dx x) = J x) d dx xj x)) = xj x). Let me use my formula for the second independent solution to find Neumann s function of the second kind of zero order. I have Using the fact that prove it) and integrating by terms I find ˆ N x) = J x) dx xj 2 x). xj 2 x) = x + x 2 + 5x N x) = J x) ) log x + x2 4 3x The most important fact here is that N is not defined at zero and approaches it behaves like log for small x). Using Neumann s function of the second kind I can define Bessel s function of the second kind of zero order as a special linear combination of J and N : Y x) = 2 π N x) log 2 γ)j x)), where γ is the Euler constant, γ = lim n n i= ) n log n) Hence the general solution to Bessel s equation of zero order can be written this is the most standard form) as ux) = AJ x) + BY x). Recall that we are mostly interested in solutions to R + ) r R + λ m2 R =. By above and generalizing I showed that the general solution to this equation is given by r 2 Rr) = AJ m λr) + BY m λr). 5
6 Now let me denote vx) = J αx) and wx) = J βx), where α and β are some constants. Due to the above I have that v and w solve xv + v + α 2 xv =, xw + w + β 2 xw =. If I multiply the first equation by w, second by v and subtract then I get, after simplifications, xv w vw ) ) = β 2 α 2 )xvw. By integrating from to I proved that β 2 α 2 ) ˆ xj αx)j βx) dx = αj α)j β) βj β)j α). Similarly, by multiplying the equation for v by 2xv I can show exercise) that ˆ xj 2 αx) dx = 2 J 2 α) + J 2 α)). An important corollary is as follows: if α and β are two roots of J then ˆ xj αx)j βx) dx = and ˆ xj 2 αx) dx = 2 J 2 α). The question is, of course, do we have any roots at all? To see that there are always infinitely many roots, let me make the change of variables vx) = ux) x in Bessel s equation of the order zero. Then, after straightforward manipulations, I find that v = + ) 4x 2 v, that is, when x is large, then the equation is approximately v + v =, and hence for large x Bessel s equation has an approximate solution ux) = A cosx ϕ) x, for some constants A and ϕ, which indicates that Bessel s functions approach zero as x and that Bessel s functions have infinitely many real positive roots. Let me introduce the inner product f, g B = ˆ xfx)gx) dx. 6
7 Note that the functions J ζ k x), k =, 2, 3,... are orthogonal on [, ] with respect to this inner product. Here ζ k is the k-th root of J x). It can be proved that any nice function f can be represented as a convergent series fx) = c J ζ x) + c 2 J ζ 2 x) + c 3 J ζ 3 x) This expansion is called the Fourier Bessel expansion, and the coefficients can be found, from the proved formulas above, as c k = f, J ζ k x) B = 2 J ζ k x), J ζ k x) B J 2ζ k) 23.4 Summary about Bessel s functions ˆ xfx)j ζ k x) dx. In a way similar to the above one can show that the following theorem holds. Theorem Consider Bessel s ODE of the order m, where m =,, 2,...: x 2 u + xu + x 2 m 2 )u =. The general solution to this equation can be written as ux) = AJ m x) + BY m x), where J m is Bessel s function of the first kind of order m and Y m is Bessel s function of the second kind of order m. J ) =, J m ) =, m =, 2,.... Bessel s functions of the second kind have a singularity at x = for any m. In particular, lim x + Y m x) =. Both J m and Y m approach zero as x, both J m and Y m have infinitely many positive roots. Let ζ k,m denote the k-th root of J m. Then ˆ {, l k, xj m ζ k,m x)j m ζ l,m x) dx = 2 J m+ 2 ζ k,m), l = k. Any sufficiently nice function f can be represented as Fourier-Bessel series fx) = c k J m ζ k,m x), where the explicit form of the coefficients can be inferred from the relation above. k= 7
8 Figure : Graphs of several Bessel s functions of the the first kind Figure 2: Graphs of several Bessel s functions of the the second kind 8
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