sign in sign up
<<
Home , Limit of a sequence, Uniform convergence

54 1. THE THEORY OF CONVERGENCE

8. and differentiation Suppose that (i) a ∞

u(x)= uk(x) Xk=1 converges pointwise on some interval [a,b]; 0 (ii) the un(x) are continuous on [a,b]; 0 (iii) the series of the derivatives k un converges uniformly on [a,b]. P Let us show that under these hypotheses the sum u(x) is continuously differentiable on [a,b] and ∞ ∞ d dun(x) (8.1) u0(x)= u (x)= dx n dx Xn=1 Xn=1 In other words, the order of differentiation and summation can be changed if the series of derivatives converges uniformly. Since the series of derivatives converges uniformly and the deriva- tives are continuous, then there exists a continuous v(x) such that ∞ 0 v(x)= un(x) Xn=1 By the theorem about integration and uniform convergence, x ∞ x ∞ 0 v(t) dt = u (t) dt = un(x) un(a) Z Z n − a Xn=1 a Xn=1   = u(x) u(a) − for any x in [a,b], where the fundamental theorem of and the convergence of the series un were used. Since v is continuous, its is differentiable withP respect to the upper and, hence, u is differentiable. Moreover, taking the of both sides of the relation, x u(x)= u(a)+ v(t) dt Za it is concluded that u0(x)= v(x) which proves (8.1). It turns out that differentiability of u can also be established even 0 if the derivatives un(x) exist but are not necessarily continuous. In this 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 55 case, the fundamental theorem of calculus cannot be used to establish differentiability of u(x). Nevertheless, the result still holds7:

Theorem 8.1. Let un be a of differentiable functions on { } [a,b]. Suppose that the sequence un(x0) converges for some x0 [a,b] and the sequence of derivatives {u0 (x) }converges uniformly on∈[a,b]. { n } Then the sequence un converges uniformly to some u(x) on [a,b], the limit function u(x){is differentiable} on [a,b] and 0 0 0 u (x)= lim un(x) = lim un(x) . n→∞  n→∞ Naturally, a similar result holds for series if the above theorem is applied to the sequence of partial sums.

Example. The series ∞ sin2(nx) u(x)= 2+ n3 Xn=1 converges for all real x by the comparison test: 1 1 1 un(x) 3 3 and 3 < | |≤ 2+ n ≤ n X n ∞ In fact, it converges uniformly by the above argument. Each term of the series is a continuously differentiable function everywhere. The series of the derivatives converges uniformly everywhere because the series of upper bounds

0 n sin(2nx) n 1 1 un(x) = | 3 | 3 2 and 2 < | | 2+ n ≤ 2+ n ≤ n X n ∞ Therefore u(x) is continuously differentiable everywhere and ∞ ∞ n sin(2nx) u0(x)= u0 (x)= . n 2+ n3 Xn=1 Xn=1

8.1. Differentiability of a . The derivative of a function of a complex variable z is defined by f(z) f(w) f 0(z)= lim − w→z z w − In particular, (zn)0 = nzn−1

7see, e.g., W. Rudin, Principles of , Chapter 7 56 1. THE THEORY OF CONVERGENCE because zn wn =(z w)(zn−1 + zn−2w + + wn−2z + wn−1) − − ··· Let f(z) be defined by a power series ∞ n f(z)= anz , z < R | | Xn=0 where R> 0 is the . The series of derivatives ∞ ∞ n−1 n annz = an+1(n + 1)z Xn=1 Xn=0 is also a power series with the same radius of convergence:

n n n lim sup an+1 (n +1) = lim √n + 1 lim sup an+1 n→∞ n→∞ p| | n→∞ p| | n 1 = lim sup an = n→∞ R p| | Therefore it converges uniformly on any disk z b < R. Thus, f(z) is differentiable in the disk z < R and | |≤ | | ∞ 0 n−1 f (z)= annz , z < R | | Xn=1 The same argument can now be applied to the function g(z) = f 0(z) to conclude that ∞ 0 00 n−2 g (z)= f (z)= ann(n 1)z , z < R − | | Xn=2 In fact, it follows that a function defined by convergent power series is infinitely many times differentiable, its derivative are obtained term- by-term differentiation of the power series, ∞ (k) n−k f (z)= n(n 1) (n k)anz , z < R − ··· − | | Xn=k and these power series have the same radius of convergence.

8.2. Differentiation of a trigonometric . Let f(x)bea2π periodic function whose derivatives are continuous (up to some order p).− By Fej´er’s theorem, its Fourier series converges to f(x) pointwise every- where: ∞ π ikx 1 −ikx f(x)= Ake ,Ak = f(x)e dx 2π Z k=X−∞ −π 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 57

Let us investigate how fast the Fourier coefficients Ak decay with in- creasing k . Using the p times | | π π π 1 ikx 1 ikx 1 0 ikx Ak = f(x)e dx = f(x)e f (x)e dx 2π Z−π 2πik −π − 2πik Z−π π 1 = f 0(x)eikxdx −2πik Z−π 1 p 1 π = f (p)(x)eikxdx −ik  2π Z−π Since, by assumption, the derivative f (p)(x) is continuous and periodic, it is bounded on [ π,π] and it follows that − Mp (p) Ak p , Mp = sup f | |≤ k [−π,π] | | Thus, the Fourier coefficients of a function that has at least p continu- ous derivatives fall off at least as k−p with increasing k. In particular, if the function has infinitely many derivatives, then its Fourier coeffi- cients decay faster than any reciprocal power. This observation allows to conclude that the Fourier series of a function that is infinitely many times differentiable can be differenti- ated term-by-term any number of times and (p) p ikx f (x)= (ik) Ake Xk because the series converges uniformly for any p 0. Furthermore, suppose that f(x) is a real continuous,≥ 2π periodic function, and f 0(x) satisfies the hypotheses of Fej´er’s theorem− (see Sec- 0 tion 5.2). If Bk are the Fourier coefficients of f (x), then by integration by parts (as shown above)

Bk Ak = Bk = ikAk , k = 0 − ik ⇒ − 6 π 0 Assuming that B0 = 0 (or −π f (x)dx = 0), it is concluded by Fej´er’s theorem that the series obtainedR by term-by-term differentiation of the Fourier series for f(x) converges pointwise:

ikx ikx 1 0 0 ikAke = Bke = lim f (t)+ lim f (t) 2 t→x+ t→x−  Xk Xk So, if the derivative f 0 is not continuous (and satisfies the above hy- potheses), the Fourier series obtained by term-by-term differentiation of the Fourier series for f converges to f 0 at each point where f 0 is continuous and to the mid-point of each jump discontinuity of f 0. 58 1. THE THEORY OF CONVERGENCE

8.3. Solutions to partial differential equations represented by Fourier series. Solutions to partial differential equations (PDEs) are often sought as series over some sets of functions (e.g., orthogonal sets of functions). In order to be a solution to a PDE, the sum of such a series should have continuous partial derivatives of sufficiently high order. For example, a solution to a wave equation 1 ∂2u ∂2u = 0 c2 ∂2t − ∂2x in two variables (x,t) should be at least twice continuously differen- tiable. If a solution is represented by a series of smooth functions, then differentiability of its sum is not obvious and requires an investigation. Suppose that periodic solutions to the above wave equation are sought: u(x + 2π,t)= u(x) The equation describes small vibrations of a circular string. Think of a circle of radius 1 as an equilibrium configuration of the string. The string can vibrate in the plane in which the circle lie and each point of the string can move only in the radial direction. In this case, u(x,t) is the shape of the vibrating string at a time moment t. By Fej´er’s theorem any continuous 2π periodic function can be represented by its trigonometric Fourier series−

∞ π ikt 1 −ikx u(x,t)= vk(t)e , vk(t)= u(x,t)e dx 2π Z k=X−∞ −π

In particular, the sequence of its partial sums un(x,t) converges point- wise n ikt u(x,t)= lim un(x,t)= lim vk(t)e n→∞ n→∞ kX=−n for every real t. Let us find vk(t) for which a partial sum un(x,t) satisfies the wave equation. A physical solution is real. Owing to the linearity of the wave equation, a real solution can be obtained by taking either real or imaginary parts of a complex solution. A formulation using com- plex valued functions is technically simpler. Naturally, the periodicity ikx condition is fulfilled by un(x,t). Since the functions e are linearly independent, the coefficients vk(t) must satisfy the equation 1 v00(t)+ k2v (t) = 0 c2 k k 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 59

Its general solution reads ikct −ikct vk(t)= ake + bke , k = 0 , v (t)= a + tb 6 0 0 0 where ak and bk are (complex) constants so that n ik(x+ct) ik(x−ct) un(x,t)= ake + bke kX=−n  

If ak and bk are such that the corresponding Fourier series converges, then it defines a function u(x,t). For example, if

ak < , bk < X | | ∞ X | | ∞ then the series converges uniformly on the plane spanned by (x,t) and, hence, the sum u(x,t) is continuous everywhere. It can be written in the form u(x,t)= a(x + ct)+ b(x ct) − where a and b are continuous functions of a single variable being the sums of the two Fourier series with coefficients ak and bk, respectively. If, in addition, a and b are twice continuously differentiable, then u(x,t) is a general periodic solution to the wave equation 8. The following question arises. There is a sequence of functions with terms satisfying a PDE and the sequence converges. Does the limit function satisfy the PDE? In general, the answer is negative because the limit function may not even be differentiable everywhere in its domain, and, hence, cannot be a solution. In the case of the wave equation, a sufficient condition for the limit function to be a solution is the uniform convergence of the sequence of second partial derivatives. Indeed 1 ∂2u ∂2u 1 ∂2 ∂2 = lim un lim un c2 ∂t2 − ∂x2 c2 ∂t2 n→∞ − ∂x2 n→∞ 2 2 (1) 1 ∂ u ∂ u = lim n lim n n→∞ c2 ∂t2 − n→∞ ∂x2 2 2 (2) 1 ∂ u ∂ u = lim n n = 0 , (x,t) D n→∞ c2 ∂t2 − ∂x2  ∀ ∈

8The function a(x + ct) describes a wave traveling in the direction of increas- ing x because the amplitude remains constant if x + ct is constant which is true if x is decreasing with rate c when the physical time t is increasing. Similarly, b(x ct) describes a wave traveling with the speed c in the direction of increasing x. So,− a general solution is a superposition of two waves traveling clockwise and counterclockwise on the circular string. 60 1. THE THEORY OF CONVERGENCE

Here (1) is true if the of the second partial derivatives of un converge uniformly on D, and (2) holds because each term of the sequence satisfies the wave equation. So, that the order of taking the limit and partial derivatives can be changed (thanks to the uniform convergence) is sufficient for the limit function to satisfy the equation. In particular, a sufficient condition for the twice continuous differ- entiability of the sums a and b is the uniform convergence of the series of second derivatives

2 iky+ 2 iky− k ake , k bke , y± = x ct X X ± A sufficient condition for the latter is provided, for example, by the Weierstrass test for :

2 2 k ak < , k bk < X | | ∞ X | | ∞ −p −p which requires that ak = O(k ) and bk = O(k ) where p > 3, whereas the uniform| convergence| of the| Fourier| series for u(x,t) is guaranteed if p> 1. So, the mere convergence of the sequence of partial sums with terms satisfying the wave equation is not sufficient for that the sum of the series satisfies the wave equation. This is the reason why a function represented by a convergent Fourier series whose partial sums satisfy the wave equation is called a formal or generalized solution. Whether or not that this function has continuous partial derivatives up to second order satisfying the wave equation has yet to be determined. If the answer is affirmative, then the function is called a classical solution. Clearly, the answer depends on how fast the coefficients ak and bk fall off with increasing k . In mechanics, it is known that a motion of a dynamical system| | is uniquely determined (by Newton’s law of motion) by the initial config- uration and velocities. The above wave equation is nothing but New- ton’s law of motion written for small vibrations of an elastic uniform string that has a uniform internal tension. The initial configuration and velocity are given by ∂u u = w0(x) , = w1(x) t=0 ∂t t=0

These functions can be expanded into Fourier series which converges pointwise if the initial data are continuous and 2π periodic: − ikx ikx w0(x)= Ake , w1(x)= Bke X X Suppose that the sequences of partial sums un(x,t) (satisfying the wave equation) and its partial derivatives ∂un/∂t converge uniformly. Then 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 61 by the theorem about uniform convergence and continuity and by the theorem about uniform convergence and differentiation

ikx u = lim u(x,t)= a0 + (ak + bk)e t=0 t→0 Xk=06 ∂u ∂ ∂ = lim lim un(x,t) = lim lim un(x,t) ∂t t=0 t→0 ∂t n→0 t→0 n→0 ∂t

∂ ikx = lim lim un(x,t)= b0 + ikc(ak bk)e n→0 t→0 ∂t − Xk=06 A comparison of these series with the Fourier series of the initial data show that

1 iBk 1 iBk ak = Ak , bk = Ak + , k = 0 2  − ck  2  ck  6 and a0 = A0, b0 = B0. So, the initial data determine how fast ak and bk fall off with increasing k . In particular, using the result from Section 8.2, it is concluded that| for| sufficiently smooth initial data, the solution to the initial value problem for the wave equation can indeed be represented by a Fourier series. It will be shown later that even in the case when the initial data are not smooth enough, the Fourier series defines a distributional solu- tion which is a good approximation (in some sense to be defined) to a classical solution. Thus, generalized or distributional solutions are just as important in applications as the classical ones. Yet, distributional solutions are often easier to obtain (their Fourier coefficients are de- fined by computationally simpler as compared to the classical solutions).

8.4. Continuous nowhere differentiable functions. One might get an im- pression that a continuous sum of a functional series may not have the derivative only at some points. This is not so. A function defined as or the sum of series may be continuous everywhere but nowhere differentiable9! Let f(x)= x for x 1. Let us extend f to R by periodicity: | | | |≤ f(x +2) = f(x) Evidently, f(x) is continuous everywhere and bounded: f(x) 1 , x R . | |≤ ∀ ∈ 9In physics, trajectories of particles participating in a Brownian motion (e.g., molecules in the air) are modeled by such functions. 62 1. THE THEORY OF CONVERGENCE

Define the function u(x) on R by the series

∞ 3 n u(x)= f(4nx) . 4 Xn=0 This series of continuous terms converges uniformly on R because the series of upper bounds converges:

∞ 3 n 3 n 3 n f(4nx) = 4 < 4 ≤ 4 ⇒ 4 ∞ Xn=0

as a . Therefore the sum u(x) is continuous on R. However u(x) is differentiable nowhere. By definition

u(x + δ) u(x) u0(x) = lim − δ→0 δ This limit does not exist for any x. If this limit existed, then for any sequence δm converging to 0 as m , the limit →∞

u(x + δm) u(x) lim − m→∞ δm

0 1 −m must exist and coincide with u (x). Put δm = 24 where the sign is ± m m chosen so that no between lies between 4 x and 4 (x + δm)= m 1 4 x . Clearly, δm 0 as m . By the choice of δm: ± 2 → →∞ m m m 1 m 1 1 f(4 (x + δm)) f(4 x) = f(4 x ) f(4 x) = = | − | | ± 2 − | ± 2 2 because the function f has a constant , 1 or 1, between any two m m − , and no integer lies between 4 x and 4 (x + δm). Put

n n f(4 (x + δm)) f(4 x) γn = − δm Then it follows from the above property of f that

1 2 m γm = = 4 . | | δm | | n If n>m, the number 4 δm is even, and by periodicity of f

γn = 0 , n>m 8. UNIFORM CONVERGENCE AND DIFFERENTIATION 63

Then ∞ n ∞ n u(x + δm) u(x) 1 3 n 1 3 n − = f(4 (x + δm)) f(4 x) δm δm 4 − δm 4 Xn=0 Xn=0 ∞ n n n 3 f(4 (x + δm)) f(4 x) = − 4 δm Xn=0 ∞ n m n m−1 n 3 3 m 3 = γn = γn = 3 γn 4 4 − 4 Xn=0 Xn=0 Xn=1 This sequence diverges as m . Indeed, using the inequality →∞ A B A B | − |≥ | |−| |

m−1 n m−1 n m 3 m 3 3 γn 3 γn − 4 ≥ − 4 | | Xn=0 Xn=0 m−1 1 3m 3m = (3m 1) ≥ − 2 − →∞ Xn=0 as m . So, the lower bound diverges as m and this implies that the→∞ limit that defines the derivative u0(x)→∞ does not exist. Thus, u0(x) does not exist for any x R. ∈ 8.5. Exercises.

1. Put x u (x)= , n = 1, 2,... n 1+ n2x (i) Show that the sequence un converges uniformly to a function u(x); (ii) Show that the equation{ } 0 0 u (x)= lim un(x) n→∞ is correct if x = 0, but false if x = 0. 6 2 1 . Let f(x) = 2 x if π,x < π, f( π) = 0, and f(x) is defined by the periodic extension−f(x + 2π) =−f(x) for all other x. Sketch the graph f(x). Use Fej´er’s theorem to show that ∞ sin(nx) f(x)= ( 1)n+1 . − n Xn=1 Sketch the graph f 0(x) for all x for which it exists. Can the derivative f 0(x) be obtained by term-by-term differentiation of the Fourier series? 64 1. THE THEORY OF CONVERGENCE

3. Show that partial sums of the Fourier series ∞ −k2t ikx u(x,t)= ake e k=X−∞ satisfy the heat equation on a ∂u ∂2u = , u(x + 2π,t)= u(x,t) ∂t ∂x2 Show that: (i) the sum of the series d continuous partial derivatives of any order for all x and t> 0 if

ak < X | | ∞ and, hence, satisfies the heat equation if t> 0; (ii) the following relation holds lim u(x,t)= u(x, 0) t→0+ (iii) u(x,t) is continuous for all x and t 0 and, hence, is a classical solution to the initial value problem for the≥ heat equation.