Expansion into a complex Fourier series theory. Fourier series. Examples of solutions

Fourier series expansion of even and odd functions expansion of a function given on an interval into a series in sines or cosines Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series in general orthogonal systems of functions Fourier series in an orthogonal system Minimal property of Fourier coefficients Bessel’s inequality Equality Parseval Closed systems Completeness and closedness of systems

Fourier series expansion of even and odd functions A function f(x), defined on the interval \-1, where I > 0, is called even if the graph of the even function is symmetrical about the ordinate axis. A function f(x), defined on the segment J), where I > 0, is called odd if the graph of the odd function is symmetrical with respect to the origin. Example. a) The function is even on the interval |-jt, jt), since for all x e b) The function is odd, since Fourier series expansion of even and odd functions is expansion of a function given on an interval into a series in sines or cosines Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series for general orthogonal systems of functions Fourier series for an orthogonal system Minimal property of Fourier coefficients Bessel’s inequality Parseval’s equality Closed systems Completeness and closedness of systems c) Function f(x)=x2-x, where does not belong neither to even nor to odd functions, since Let the function f(x), satisfying the conditions of Theorem 1, be even on the interval x|. Then for everyone i.e. /(x) cos nx is an even function, and f(x) sinnx is an odd one. Therefore, the Fourier coefficients of an even function f(x) will be equal. Therefore, the Fourier series of an even function has the form 00 If f(x) - odd function on the interval [-тр, ir|, then the product f(x)cosnx will be an odd function, and the product f(x) sinпх will be an even function. Therefore, we will have Thus, the Fourier series of an odd function has the form Example 1. Expand the function 4 into a Fourier series on the interval -x ^ x ^ n Since this function is even and satisfies the conditions of Theorem 1, then its Fourier series has the form Find the Fourier coefficients. We have Applying integration by parts twice, we obtain that So, the Fourier series of this function looks like this: or, in expanded form, This equality is valid for any x €, since at the points x = ±ir the sum of the series coincides with the values ​​of the function f(x ) = x2, since the graphs of the function f(x) = x and the sum of the resulting series are given in Fig. Comment. This Fourier series allows us to find the sum of one of the convergent numerical series, namely, for x = 0 we obtain that Example 2. Expand the function /(x) = x into a Fourier series on the interval. The function /(x) satisfies the conditions of Theorem 1, therefore it can be expanded into a Fourier series, which, due to the oddness of this function, will have the form Integrating by parts, we find the Fourier coefficients. Therefore, the Fourier series of this function has the form This equality holds for all x B at points x - ±t the sum of the Fourier series does not coincide with the values ​​of the function /(x) = x, since it is equal to. Outside the interval [-*, i-] the sum of the series is a periodic continuation of the function /(x) = x; its graph is shown in Fig. 6. § 6. Expansion of a function given on an interval into a series in sines or cosines Let a bounded piecewise monotonic function / be given on the interval. The values ​​of this function on the interval 0| can be further defined in various ways. For example, you can define a function / on the segment tc] so that /. In this case they say that) “is extended to the segment 0] in an even manner”; its Fourier series will contain only cosines. If the function /(x) is defined on the interval [-l-, mc] so that /(, then the result is an odd function, and then they say that / is “extended to the interval [-*, 0] in an odd way”; in this In this case, the Fourier series will contain only sines. Thus, each bounded piecewise monotonic function /(x) defined on the interval can be expanded into a Fourier series in both sines and cosines. Example 1. Expand the function into a Fourier series: a) by cosines; b) by sines. M This function, with its even and odd continuations into the segment |-x,0) will be bounded and piecewise monotonic. a) Let's extend /(z) into the segment 0) a) Extend j\x) into the segment (-тр,0| in an even manner (Fig. 7), then its Fourier series i will have the form П=1 where the Fourier coefficients are equal, respectively for Therefore, b) Let us extend /(z) into the segment [-x,0] in an odd way (Fig. 8). Then its Fourier series §7. Fourier series for a function with an arbitrary period Let the function fix) be periodic with a period of 21.1 ^ 0. To expand it into a Fourier series on the interval where I > 0, we make a change of variable by setting x = jt. Then the function F(t) = / ^tj will be a periodic function of the argument t with period and it can be expanded on the segment into a Fourier series. Returning to the variable x, i.e., setting, we obtain All theorems valid for Fourier series of periodic functions with period 2π , remain valid for periodic functions with an arbitrary period 21. In particular, a sufficient criterion for the decomposability of a function in a Fourier series also remains valid. Example 1. Expand into a Fourier series a periodic function with a period of 21, given on the interval [-/,/] by the formula (Fig. 9). Because this function is even, then its Fourier series has the form Substituting the found values ​​of the Fourier coefficients into the Fourier series, we obtain Note one important property of periodic functions. Theorem 5. If a function has period T and is integrable, then for any number a the equality m holds. that is, the integral of a segment whose length is equal to the period T has the same value regardless of the position of this segment on the number axis. In fact, We make a change of variable in the second integral, assuming. This gives and therefore, Geometrically, this property means that in the case of the area shaded in Fig. 10 areas are equal to each other. In particular, for a function f(x) with a period we obtain at Expansion into a Fourier series of even and odd functions, expansion of a function given on an interval into a series in sines or cosines Fourier series for a function with an arbitrary period Complex notation of the Fourier series Fourier series in general orthogonal systems functions Fourier series in an orthogonal system Minimal property of Fourier coefficients Bessel’s inequality Parseval’s equality Closed systems Completeness and closedness of systems Example 2. The function x is periodic with a period Due to the oddness of this function, without calculating integrals, we can state that for any The proven property, in particular, shows that the Fourier coefficients of a periodic function f(x) with a period of 21 can be calculated using the formulas where a is an arbitrary real number (note that the functions cos - and sin have a period of 2/). Example 3. Expand into a Fourier series a function given on an interval with a period of 2x (Fig. 11). 4 Let's find the Fourier coefficients of this function. Putting in the formulas we find that for Therefore, the Fourier series will look like this: At the point x = jt (discontinuity point of the first kind) we have §8. Complex representation of the Fourier series This section uses some elements comprehensive analysis(see Chapter XXX, where all actions performed here with complex expressions are strictly justified). Let the function f(x) satisfy sufficient conditions for expansion into a Fourier series. Then on the segment x] it can be represented by a series of the form Using Euler’s formulas Substituting these expressions into series (1) instead of cos πx and sin φx we will have We introduce the following notation Then series (2) will take the form Thus, the Fourier series (1) is represented in complex form (3). Let's find expressions for the coefficients through integrals. We have Similarly, we find The final formulas for с„, с_п and с can be written as follows: . . The coefficients с„ are called the complex Fourier coefficients of the function. For a periodic function with a period), the complex form of the Fourier series will take the form where the coefficients Cn are calculated using the formulas. The convergence of series (3) and (4) is understood as follows: series (3) and (4) are called convergent for given value g, if there are limits Example. Expand the period function into a complex Fourier series. This function satisfies sufficient conditions for expansion into a Fourier series. Let us find the complex Fourier coefficients of this function. We have for odd for even n, or, in short. Substituting the values), we finally obtain Note that this series can also be written as follows: Fourier series for general orthogonal systems of functions 9.1. Orthogonal systems of functions Let us denote by the set of all (real) functions defined and integrable on the interval [a, 6] with a square, i.e., those for which an integral exists. In particular, all functions f(x) continuous on the interval [a , 6], belong to 6], and the values ​​of their Lebesgue integrals coincide with the values ​​of the Riemann integrals. Definition. A system of functions, where, is called orthogonal on the interval [a, b\, if Condition (1) assumes, in particular, that none of the functions is identically zero. The integral is understood in the Lebesgue sense. and we call the quantity the norm of the function. If in an orthogonal system for any n we have, then the system of functions is called orthonormal. If the system (y>„(x)) is orthogonal, then the system Example 1. The trigonometric system is orthogonal on a segment. The system of functions is an orthonormal system of functions on, Example 2. The cosine system and the sine system are orthonormal. Let us introduce the notation that they are orthogonal on the interval (0, f|, but not orthonormal (for I Ф- 2). Since their norms are COS Example 3. Polynomials defined by equality are called Legendre polynomials (polynomials). For n = 0 we have It can be proven , that the functions form an orthonormal system of functions on the interval. Let us show, for example, the orthogonality of the Legendre polynomials. Let m > n. In this case, integrating n times by parts, we find since for the function t/m = (z2 - I)m all derivatives up to order m - I inclusive vanish at the ends of the segment [-1,1). Definition. A system of functions (pn(x)) is called orthogonal on the interval (a, b) by an overhang p(x) if: 1) for all n = 1,2,... there are integrals. Here it is assumed that the weight function p(x) is defined and positive everywhere on the interval (a, b) with the possible exception of a finite number of points where p(x) can vanish. Having performed differentiation in formula (3), we find. It can be shown that the Chebyshev-Hermite polynomials are orthogonal on the interval Example 4. The system of Bessel functions (jL(pix)^ is orthogonal on the interval zeros of the Bessel function Example 5. Let us consider the Chebyshev-Hermite polynomials, which can be defined using equality. Fourier series in an orthogonal system Let there be an orthogonal system of functions in the interval (a, 6) and let the series (cj = const) converge on this interval to the function f(x): Multiplying both sides of the last equality by - fixed) and integrating over x from a to 6, due to the orthogonality of the system, we obtain that this operation has, generally speaking, a purely formal character. However, in some cases, for example, when the series (4) converges uniformly, all functions are continuous and the interval (a, 6) is finite, this operation is legal. But for us now it is the formal interpretation that is important. So, let a function be given. Let us form the numbers c* according to formula (5) and write. The series on the right side is called the Fourier series of the function f(x) with respect to the system (^n(i)). The numbers Cn are called the Fourier coefficients of the function f(x) with respect to this system. The sign ~ in formula (6) only means that the numbers Cn are related to the function f(x) by formula (5) (it is not assumed that the series on the right converges at all, much less converges to the function f(x)). Therefore, the question naturally arises: what are the properties of this series? In what sense does it “represent” the function f(x)? 9.3. Convergence on average Definition. A sequence converges to the element ] on average if the norm is in the space Theorem 6. If a sequence ) converges uniformly, then it converges on average. M Let the sequence ()) converge uniformly on the interval [a, b] to the function /(x). This means that for everyone, for all sufficiently large n, we have Therefore, from which our statement follows. The converse is not true: the sequence () may converge on average to /(x), but not be uniformly convergent. Example. Consider the sequence nx. It is easy to see that But this convergence is not uniform: there exists e, for example, such that, no matter how large n is, on the interval cosines Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series for general orthogonal systems of functions Fourier series for an orthogonal system Minimal property of Fourier coefficients Bessel’s inequality Parseval’s equality Closed systems Completeness and closedness of systems and let We denote by c* the Fourier coefficients of the function /(x ) by an orthonormal system b Consider a linear combination where n ^ 1 is a fixed integer, and find the values ​​of the constants at which the integral takes a minimum value. Let us write it in more detail. Integrating term by term, due to the orthonormality of the system, we obtain. The first two terms on the right side of equality (7) are independent, and the third term is non-negative. Therefore, the integral (*) takes a minimum value at ak = sk. The integral is called the mean square approximation of the function /(x) by a linear combination of Tn(x). Thus, the root mean square approximation of the function /\ takes a minimum value when. when Tn(x) is the 71st partial sum of the Fourier series of the function f(x) over the system (. Setting ak = sk, from (7) we obtain Equality (9) is called the Bessel identity. Since its left side is non-negative, then Bessel’s inequality follows from it. Since I am arbitrary here, Bessel’s inequality can be represented in a strengthened form, i.e., for any function / the series of squared Fourier coefficients of this function in an orthonormal system ) converges. Since the system is orthonormal on the interval [-x, m], then inequality (10) translated into the usual notation of the trigonometric Fourier series gives the relation do that is valid for any function /(x) with an integrable square. If f2(x) is integrable, then due to necessary condition convergence of the series on the left side of inequality (11), we obtain that. Parseval's equality For some systems (^„(x)), the inequality sign in formula (10) can be replaced (for all functions f(x) 6 ×) by an equal sign. The resulting equality is called the Parseval-Steklov equality (completeness condition). Bessel's identity (9) allows us to write condition (12) in an equivalent form. Thus, the fulfillment of the completeness condition means that the partial sums Sn(x) of the Fourier series of the function /(x) converge to the function /(x) on average, i.e. according to the norm of space 6]. Definition. An orthonormal system ( is called complete in b2[аy b] if every function can be approximated with any accuracy on average by a linear combination of the form with a sufficiently large number of terms, i.e. if for any function /(x) ∈ b2[a, b\ and for any e > 0 there is natural number nq and numbers a\, a2y..., such that No From the above reasoning follows Theorem 7. If by orthonormalization the system ) is complete in space, the Fourier series of any function / over this system converges to f(x) on average, i.e. by norm It can be shown that the trigonometric system is complete in space. This implies the statement. Theorem 8. If a function /o its trigonometric Fourier series converges to it in average. 9.5. Closed systems. Completeness and closedness of systems Definition. An orthonormal system of functions \ is called closed if in the space Li\a, b) there is no nonzero function orthogonal to all functions. In the space L2\a, b\, the concepts of completeness and closedness of orthonormal systems coincide. Exercises 1. Expand the function 2 into a Fourier series in the interval (-i-, x) 2. Expand the function into a Fourier series in the interval (-tr, tr) 3. Expand the function 4 into a Fourier series in the interval (-tr, tr) into a Fourier series in the interval (-jt, tr) function 5. Expand the function f(x) = x + x into a Fourier series in the interval (-tr, tr). 6. Expand the function n into a Fourier series in the interval (-jt, tr) 7. Expand the function /(x) = sin2 x into a Fourier series in the interval (-tr, x). 8. Expand the function f(x) = y into a Fourier series in the interval (-tr, jt) 9. Expand the function f(x) = | sin x|. 10. Expand the function f(x) = § into a Fourier series in the interval (-π-, π). 11. Expand the function f(x) = sin § into a Fourier series in the interval (-tr, tr). 12. Expand the function f(x) = n -2x, given in the interval (0, x), into a Fourier series, extending it into the interval (-x, 0): a) in an even manner; b) in an odd way. 13. Expand the function /(x) = x2, given in the interval (0, x), into a Fourier series in sines. 14. Expand the function /(x) = 3, given in the interval (-2,2), into a Fourier series. 15. Expand the function f(x) = |x|, given in the interval (-1,1), into a Fourier series. 16. Expand the function f(x) = 2x, specified in the interval (0,1), into a Fourier series in sines.

1 MINISTRY OF EDUCATION AND SCIENCE OF THE RF NOVOSIBIRSK STATE UNIVERSITY FACULTY OF PHYSICS R. K. Belkheeva FOURIER SERIES IN EXAMPLES AND PROBLEMS Textbook Novosibirsk 211

2 UDC BBK V161 B44 B44 Belkheeva R.K. Fourier series in examples and problems: Textbook / Novosibirsk. state univ. Novosibirsk, s. ISBN B textbook basic information about Fourier series is presented, examples are given for each topic studied. An example of the application of the Fourier method to solving the problem of transverse vibrations of a string is analyzed in detail. Illustrative material is provided. There are tasks for independent decision. Designed for students and teachers Faculty of Physics NSU. Published by decision of the methodological commission of the Faculty of Physics of NSU. Reviewer Dr. physics and mathematics Sci. V. A. Aleksandrov The manual was prepared as part of the implementation of the NRU-NSU Development Program for the years. ISBN from Novosibirsk State University, 211 c Belkheeva R.K., 211

3 1. Expansion of a 2π-periodic function into a Fourier series Definition. The Fourier series of the function f(x) is the functional series a 2 + (a n cosnx + b n sin nx), (1) where the coefficients a n, b n are calculated using the formulas: a n = 1 π b n = 1 π f(x) cosnxdx, n = , 1,..., (2) f(x) sin nxdx, n = 1, 2,.... (3) Formulas (2) (3) are called Euler Fourier formulas. The fact that the function f(x) corresponds to the Fourier series (1) is written as the formula f(x) a 2 + (a n cosnx + b n sin nx) (4) and we say that the right side of formula (4) is a formal series Fourier function f(x). In other words, formula (4) only means that the coefficients a n, b n were found using formulas (2), (3). 3

4 Definition. A 2π-periodic function f(x) is called piecewise smooth if there are a finite number of points = x in the interval [, π]< x 1 . Рассмотрим два условия: а) f(l x) = f(x); б) f(l + x) = f(x), x [, l/2]. С геометрической точки зрения условие (а) означает, что график функции f(x) симметричен относительно вертикальной прямой x = l/2, а условие (б) что график f(x) центрально симметричен относительно точки (l/2;) на оси абсцисс. Тогда справедливы следующие утверждения: 1) если функция f(x) четная и выполнено условие (а), то b 1 = b 2 = b 3 =... =, a 1 = a 3 = a 5 =... = ; 2) если функция f(x) четная и выполнено условие (б), то b 1 = b 2 = b 3 =... =, a = a 2 = a 4 =... = ; 3) если функция f(x) нечетная и выполнено условие (а), то a = a 1 = a 2 =... =, b 2 = b 4 = b 6 =... = ; 4) если функция f(x) нечетная и выполнено условие (б), то a = a 1 = a 2 =... =, b 1 = b 3 = b 5 =... =. ЗАДАЧИ В задачах 1 7 нарисуйте графики и найдите ряды Фурье для функций, { предполагая, что они имеют период 2π:, если < x a cosx + a2 В задачах найдите ряды Фурье в комплексной форме для функций. 26. f(x) = sgn x, π < x < π. 27. f(x) = ln(1 2a cosx + a 2), a < 1. 1 a cosx 28. f(x) = 1 2a cosx + a2, a < Докажите, что функция f, определенная в промежутке [, π], вещественнозначна, если и только если коэффициенты c n ее complex series Fourier relations are related by c n = c n, n =, ±1, ±2, Prove that the function f defined in the interval [, π] is even (i.e., satisfies the relation f(x) = f(x)), if and only if the coefficients c n of its complex Fourier series are related by the relations c n = c n, n = ±1, ±2, Prove that the function f defined in the interval [, π] is odd (i.e., satisfies the relation f(x ) = f(x)), if and only if the coefficients c n of its complex Fourier series are related by the relations c n = c n, n =, ±1, ±2,.... Answers 1 2π 24. a n a π a n i + e 2inx, where it is implied that the term corresponding to n = is omitted. π n n= a n n cosnx. 28. a n cosnx. n= 46

47 5. Lyapunov’s equality Theorem (Lyapunov’s equality). Let the function f: [, π] R be such that f 2 (x) dx< +, и пусть a n, b n ее коэффициенты Фурье. Тогда справедливо равенство, a (a 2 n + b2 n) = 1 π называемое равенством Ляпунова. f 2 (x) dx, ПРИМЕР 13. Напишем равенство Ляпунова для функции { 1, если x < a, f(x) =, если a < x < π и найдем с его помощью суммы числовых рядов + sin 2 na n 2 и + Решение. Очевидно, 1 (2n 1) 2. 1 π f 2 (x) dx = 1 π a a dx = 2a π. Так как f(x) четная функция, то для всех n имеем b n =, a = 2 π f(x) dx = 2 π a dx = 2a π, 47

48 a n = 2 π f(x) cosnxdx = 2 π a cos nxdx = 2 sin na πn. Therefore, the Lyapunov equality for the function f(x) takes the form: 2 a 2 π + 4 sin 2 na = 2a 2 π 2 n 2 π. From the last equality for a π we find sin 2 na n 2 = a(π a) 2 Setting a = π 2, we obtain sin2 na = 1 for n = 2k 1 and sin 2 na = for n = 2k. Consequently, k=1 1 (2k 1) 2 = = π2 8. EXAMPLE 14. Let’s write Lyapunov’s equality for the function f(x) = x cosx, x [, π], and use it to find the sum of the number series (4n 2 + 1) 2 (4n 2 1) 4. 1 π Solution. Direct calculations give = π π f 2 (x) dx = 1 π x 2 cos 2 xdx = 1 π x sin 2xdx = π π x cos x = π x 21 + cos 2x dx = 2 π 1 4π cos 2xdx =

49 Since f(x) is an even function, then for all n we have b n =, a n = 2 π = 1 π 1 = π(n + 1) = f(x) cosnxdx = 2 π 1 cos(n + 1)x π (n + 1) 2 x cosxcosnxdx = x (cos(n + 1)x + cos(n 1)x) dx = 1 π sin(n + 1)xdx sin(n 1)xdx = π(n 1) π π 1 + cos(n 1)x = π(n 1) 2 1 (= (1) (n+1) 1) 1 (+ (1) (n+1) 1) = π(n + 1) 2 π(n 1) 2 () = (1)(n+1) 1 1 π (n + 1) + 1 = 2 (n 1) 2 = 2 (1)(n+1) 1 n k π (n 2 1) = π (4k 2 1) 2 if n = 2k, 2 if n = 2k + 1. The coefficient a 1 must be calculated separately, since in the general formula for n = 1 the denominator of the fraction goes to zero. = 1 π a 1 = 2 π f(x) cosxdx = 2 π x(1 + cos 2x)dx = π 2 1 2π 49 x cos 2 xdx = sin 2xdx = π 2.

50 Thus, Lyapunov’s equality for the function f(x) has the form: 8 π + π (4n 2 + 1) 2 π 2 (4n 2 1) = π, from where we find the sum of the number series (4n 2 + 1) 2 (4n 2 1) = π π PROBLEMS 32. Write the Lyapunov equality for the function ( x f(x) = 2 πx, if x< π, x 2 πx, если π < x. 33. Напишите равенства Ляпунова для функций f(x) = cos ax и g(x) = sin ax, x [, π]. 34. Используя результат предыдущей задачи и предполагая, что a не является целым числом, выведите следующие классические разложения функций πctgaπ и (π/ sin aπ) 2 по rational functions: πctgaπ = 1 a + + 2a a 2 n 2, (π) = sin aπ (a n) 2. n= 35. Derive the complex form of the generalized Lyapunov equality. 36. Show that the complex form of the Lyapunov equality is valid not only for real-valued functions, but also for complex-valued functions. 5

51 π (2n + 1) = π sin 2απ 2απ = 2sin2 απ α 2 π 2 Answers + 4 sin2 απ π 2 α 2 (α 2 n 2) 2; sin 2απ 1 2απ = απ n 2 4sin2 π 2 (α 2 n 2) 2. 1 π 35. f(x)g(x) dx= c n d n, where c n is the Fourier coefficient 2π of the function f(x), and d n is the Fourier coefficient functions g(x). 6. Differentiation of Fourier series Let f: R R be a continuously differentiable 2π-periodic function. Its Fourier series has the form: f(x) = a 2 + (a n cos nx + b n sin nx). The derivative f (x) of this function will be a continuous and 2π-periodic function, for which we can write a formal Fourier series: f (x) a 2 + (a n cos nx + b n sin nx), where a, a n, b n, n = 1 , 2,... Fourier coefficients of the function f (x). 51

52 Theorem (on term-by-term differentiation of Fourier series). Under the above assumptions, the equalities a =, a n = nb n, b n = na n, n 1 are valid. EXAMPLE 15. Let the piecewise smooth function f(x) be continuous in the interval [, π]. Let us prove that if the condition f(x)dx = is satisfied, the inequality 2 dx 2 dx, called Steklov’s inequality, holds, and we will make sure that equality in it holds only for functions of the form f(x) = A cosx. In other words, Steklov’s inequality gives conditions under which the smallness of the derivative (in the mean square) implies the smallness of the function (in the mean square). Solution. Let us extend the function f(x) to the interval [, ] in an even manner. Let us denote the extended function by the same symbol f(x). Then the extended function will be continuous and piecewise smooth on the interval [, π]. Since the function f(x) is continuous, then f 2 (x) is continuous on the interval and 2 dx< +, следовательно, можно применить теорему Ляпунова, согласно которой имеет место равенство 1 π 2 dx = a () a 2 n + b 2 n. 52

53 Since the continued function is even, then b n =, a = by condition. Consequently, Lyapunov’s equality takes the form 1 π 2 dx = a 2 π n. (17) Let us make sure that for f (x) the conclusion of the theorem on term-by-term differentiation of the Fourier series is satisfied, that is, that a =, a n = nb n, b n = na n, n 1. Let the derivative f (x) undergo kinks at points x 1, x 2,..., x N in the interval [, π]. Let us denote x =, x N+1 = π. Let us divide the integration interval [, π] into N +1 intervals (x, x 1),..., (x N, x N+1), on each of which f(x) is continuously differentiable. Then, using the property of additivity of the integral, and then integrating by parts, we obtain: b n = 1 π = 1 π = 1 π f (x) sin nxdx = 1 π N f(x) sin nx j= N f(x) sin nx j= x j+1 x j x j+1 x j n n π N j= x j+1 x j x j+1 x j f (x) sin nxdx = f(x) cosnxdx = f(x) cosnxdx = = 1 π [(f(x 1) sin nx 1 f(x) sin nx) + + (f(x 2) sinnx 2 f(x 1) sin nx 1)

54 + (f(x N+1) sin nx N+1 f(x N) sin nx N)] na n = = 1 π na n = = 1 π na n = na n. x j+1 a = 1 f (x)dx = 1 N f (x)dx = π π j= x j = 1 N x j+1 f(x) π = 1 (f(π) f()) = . x j π j= The last equality occurs due to the fact that the function f(x) was continued in an even way, which means f(π) = f(). Similarly we obtain a n = nb n. We have shown that the theorem on term-by-term differentiation of Fourier series for a continuous piecewise smooth 2π-periodic function whose derivative in the interval [, π] undergoes discontinuities of the first kind is correct. This means f (x) a 2 + (a n cosnx + b n sin nx) = (na n)sin nx, since a =, a n = nb n =, b n = na n, n = 1, 2,.... Since 2 dx< +, то по равенству Ляпунова 1 π 2 dx = 54 n 2 a 2 n. (18)

55 Since each term in the series in (18) is greater than or equal to the corresponding term in the series in (17), then 2 dx 2 dx. Recalling that f(x) is an even continuation of the original function, we have 2 dx 2 dx. Which proves Steklov’s equality. Now we examine for which functions equality holds in Steklov’s inequality. If for at least one n 2, the coefficient a n is different from zero, then a 2 n< na 2 n. Следовательно, равенство a 2 n = n 2 a 2 n возможно только если a n = для n 2. При этом a 1 = A может быть произвольным. Значит в неравенстве Стеклова равенство достигается только на функциях вида f(x) = A cosx. Отметим, что условие πa = f(x)dx = (19) существенно для выполнения неравенства Стеклова, ведь если условие (19) нарушено, то неравенство примет вид: a a 2 n n 2 a 2 n, а это не может быть верно при произвольном a. 55

56 PROBLEMS 37. Let the piecewise smooth function f(x) be continuous in the interval [, π]. Prove that when the condition f() = f(π) = is satisfied, the inequality 2 dx 2 dx, also called the Steklov inequality, holds, and make sure that equality in it holds only for functions of the form f(x) = B sin x. 38. Let the function f be continuous in the interval [, π] and have in it (except perhaps a finite number of points) a derivative f (x) that is square integrable. Prove that if the conditions f() = f(π) and f(x) dx = are satisfied, then the inequality 2 dx 2 dx, called the Wirtinger inequality, holds, and equality in it holds only for functions of the form f(x ) = A cosx + B sin x. 56

57 7. Application of Fourier series for solving partial differential equations When studying a real object (natural phenomena, production process, control systems, etc.) two factors turn out to be significant: the level of accumulated knowledge about the object under study and the degree of development of the mathematical apparatus. On modern stage scientific research The following chain has been developed: phenomenon physical model mathematical model. The physical formulation (model) of the problem is as follows: the conditions for the development of the process and the main factors influencing it are identified. The mathematical formulation (model) consists of describing the factors and conditions selected in the physical formulation in the form of a system of equations (algebraic, differential, integral, etc.). A problem is called well-posed if in a certain functional space a solution to the problem exists, uniquely and continuously depends on the initial and boundary conditions. Mathematical model is not identical to the object under consideration, but is an approximate description of it. Derivation of the equation for free small transverse vibrations of a string. We will follow the textbook. Let the ends of the string be secured and the string itself stretched taut. If you move a string from its equilibrium position (for example, pull it back or hit it), then the string will begin to 57

58 hesitate. We will assume that all points of the string move perpendicular to its equilibrium position (transverse vibrations), and at each moment of time the string lies in the same plane. Let us take a system of rectangular coordinates xou in this plane. Then, if at the initial moment of time t = the string was located along the Ox axis, then u will mean the deviation of the string from the equilibrium position, that is, the position of the point of the string with the abscissa x at an arbitrary moment of time t corresponds to the value of the function u(x, t). For each fixed value of t, the graph of the function u(x, t) represents the shape of the vibrating string at time t (Fig. 32). At a constant value of x, the function u(x, t) gives the law of motion of a point with abscissa x along a straight line parallel to the Ou axis, the derivative u t is the speed of this movement, and the second derivative is 2 u t 2 acceleration. Rice. 32. Forces applied to an infinitesimal section of a string Let’s create an equation that the function u(x, t) must satisfy. To do this, we will make a few more simplifying assumptions. We will consider the string to be absolutely flexible - 58

59 koy, that is, we will assume that the string does not resist bending; this means that the stresses arising in the string are always directed tangentially to its instantaneous profile. The string is assumed to be elastic and subject to Hooke's law; this means that the change in the magnitude of the tension force is proportional to the change in the length of the string. Let us assume that the string is homogeneous; this means that its linear density ρ is constant. We neglect external forces. This means that we are considering free vibrations. We will study only small vibrations of the string. If we denote by ϕ(x, t) the angle between the abscissa axis and the tangent to the string at the point with the abscissa x at time t, then the condition for small oscillations is that the value ϕ 2 (x, t) can be neglected in comparison with ϕ (x, t), i.e. ϕ 2. Since the angle ϕ is small, then cosϕ 1, ϕ sin ϕ tan ϕ u therefore, the value (u x x,) 2 can also be neglected. It immediately follows that during the vibration process we can neglect the change in the length of any section of the string. Indeed, the length of a piece of string M 1 M 2, projected into the interval of the abscissa axis, where x 2 = x 1 + x, is equal to l = x 2 x () 2 u dx x. x Let us show that, under our assumptions, the magnitude of the tension force T will be constant along the entire string. To do this, let’s take any section of the string M 1 M 2 (Fig. 32) at time t and replace the action of the discarded sections - 59

60 by the tension forces T 1 and T 2. Since, according to the condition, all points of the string move parallel to the Ou axis and there are no external forces, the sum of the projections of the tension forces on the Ox axis must be equal to zero: T 1 cosϕ(x 1, t) + T 2 cosϕ(x 2, t) =. Hence, due to the smallness of the angles ϕ 1 = ϕ(x 1, t) and ϕ 2 = ϕ(x 2, t), we conclude that T 1 = T 2. Let us denote the total value of T 1 = T 2 by T. Now we calculate the sum of the projections F u of the same forces on the Ou axis: F u = T sin ϕ(x 2, t) T sin ϕ(x 1, t). (2) Since for small angles sin ϕ(x, t) tan ϕ(x, t), and tan ϕ(x, t) u(x, t)/ x, then equation (2) can be rewritten as F u T (tg ϕ(x 2, t) tan ϕ(x 1, t)) (u T x (x 2, t) u) x (x 1, t) x x T 2 u x 2(x 1, t) x . Since the point x 1 is chosen arbitrarily, then F u T 2 u x2(x, t) x. After all the forces acting on the section M 1 M 2 have been found, we apply Newton’s second law to it, according to which the product of mass and acceleration is equal to the sum of all active forces. The mass of a piece of string M 1 M 2 is equal to m = ρ l ρ x, and the acceleration is equal to 2 u(x, t). Newton's t 2 equation takes the form: 2 u t (x, t) x = u 2 α2 2 x2(x, t) x, where α 2 = T ρ constant positive number. 6

61 Reducing by x, we get 2 u t (x, t) = u 2 α2 2 x2(x, t). (21) As a result, we obtained a linear homogeneous second-order partial differential equation with constant coefficients. It is called the string vibration equation or the one-dimensional wave equation. Equation (21) is essentially a reformulation of Newton's law and describes the motion of the string. But in the physical formulation of the problem there were requirements that the ends of the string are fixed and the position of the string at some point in time is known. We will write these conditions as equations as follows: a) we will assume that the ends of the string are fixed at points x = and x = l, i.e. we will assume that for all t the relations u(, t) =, u(l, t ) = ; (22) b) we will assume that at time t = the position of the string coincides with the graph of the function f(x), i.e. we will assume that for all x [, l] the equality u(x,) = f( x); (23) c) we will assume that at the moment t = the point of the string with the abscissa x is given speed g(x), i.e. we will assume that u (x,) = g(x). (24) t Relations (22) are called boundary conditions, and relations (23) and (24) are called initial conditions. Mathematical model of free small transverses 61

62 string oscillations is that it is necessary to solve equation (21) with boundary conditions (22) and initial conditions (23) and (24) Solving the equation of free small transverse string oscillations by the Fourier method Solving equation (21) in the region x l,< t . Подставляя (25) в (21), получим: X T = α 2 X T, (26) или T (t) α 2 T(t) = X (x) X(x). (27) Говорят, что произошло разделение переменных. Так как x и t не зависят друг от друга, то левая часть в (27) не зависит от x, а правая от t и общая величина этих отношений 62

63 must be a constant, which we denote by λ: T (t) α 2 T(t) = X (x) X(x) = λ. From here we get two ordinary differential equations: X (x) λx(x) =, (28) T (t) α 2 λt(t) =. (29) In this case, the boundary conditions (22) will take the form X()T(t) = and X(l)T(t) =. Since they must be satisfied for all t, t >, then X() = X(l) =. (3) Let us find solutions to equation (28) that satisfy boundary conditions (3). Let's consider three cases. Case 1: λ >. Let us denote λ = β 2. Equation (28) takes the form X (x) β 2 X(x) =. Its characteristic equation k 2 β 2 = has roots k = ±β. Hence, common decision equation (28) has the form X(x) = C e βx + De βx. We must select the constants C and D so that the boundary conditions (3) are met, i.e. X() = C + D =, X(l) = C e βl + De βl =. Since β, this system of equations has a unique solution C = D =. Therefore, X(x) and 63

64 u(x, t). Thus, in case 1 we got trivial solution, which we will not consider further. Case 2: λ =. Then equation (28) takes the form X (x) = and its solution is obviously given by the formula: X(x) = C x+d. Substituting this solution into the boundary conditions (3), we obtain X() = D = and X(l) = Cl =, which means C = D =. Therefore, X(x) and u(x, t), and we again have a trivial solution. Case 3: λ