Decompose into a Fourier series on a segment. Fourier series: history and influence of the mathematical mechanism on the development of science
Fourier series are a representation of an arbitrary function with a specific period in the form of a series. IN general view this solution is called the expansion of the element in an orthogonal basis. Expansion of functions into Fourier series is a fairly powerful tool for solving various problems due to the properties of this transformation during integration, differentiation, as well as shifting expressions by argument and convolution.
A person who is not familiar with higher mathematics, as well as with the works of the French scientist Fourier, most likely will not understand what these “series” are and what they are needed for. Meanwhile, this transformation has become quite integrated into our lives. It is used not only by mathematicians, but also by physicists, chemists, doctors, astronomers, seismologists, oceanographers and many others. Let us also take a closer look at the works of the great French scientist who made a discovery that was ahead of its time.
Man and the Fourier transformFourier series are one of the methods (along with analysis and others). This process occurs every time a person hears a sound. Our ear automatically transforms elementary particles into elastic medium are arranged in rows (along the spectrum) of successive volume levels for tones different heights. Next, the brain turns this data into sounds that are familiar to us. All this happens without our desire or consciousness, on its own, but in order to understand these processes, it will take several years to study higher mathematics.
The Fourier transform can be carried out using analytical, numerical and other methods. Fourier series refer to the numerical method of decomposing any oscillatory processes - from ocean tides and light waves to cycles of solar (and other astronomical objects) activity. Using these mathematical techniques, you can analyze functions, representing any oscillatory processes as a series of sinusoidal components that move from minimum to maximum and back. The Fourier transform is a function that describes the phase and amplitude of sinusoids corresponding to a specific frequency. This process can be used to solve very complex equations that describe dynamic processes arising under the influence of thermal, light or electrical energy. Also, Fourier series make it possible to isolate constant components in complex oscillatory signals, making it possible to correctly interpret the experimental observations obtained in medicine, chemistry and astronomy.
The founding father of this theory is the French mathematician Jean Baptiste Joseph Fourier. This transformation was subsequently named after him. Initially, the scientist used his method to study and explain the mechanisms of thermal conductivity - the spread of heat in solids. Fourier suggested that the initial irregular distribution can be decomposed into simple sinusoids, each of which will have its own temperature minimum and maximum, as well as its own phase. In this case, each such component will be measured from minimum to maximum and vice versa. The mathematical function that describes the upper and lower peaks of the curve, as well as the phase of each of the harmonics, is called the Fourier transform of the temperature distribution expression. The author of the theory reduced the general distribution function, which is difficult to describe mathematically, to a very convenient series of cosine and sine, which together give the original distribution.
The principle of transformation and the views of contemporariesThe scientist's contemporaries - leading mathematicians of the early nineteenth century - did not accept this theory. The main objection was Fourier's assertion that a discontinuous function, describing a straight line or a discontinuous curve, can be represented as a sum of sinusoidal expressions that are continuous. As an example, consider the Heaviside step: its value is zero to the left of the discontinuity and one to the right. This function describes the dependency electric current from a temporary variable when the circuit is closed. Contemporaries of the theory at that time had never encountered a similar situation where a discontinuous expression would be described by a combination of continuous, ordinary functions such as exponential, sine, linear or quadratic.
After all, if the mathematician was right in his statements, then by summing the infinite trigonometric Fourier series, one can obtain an accurate representation of the step expression even if it has many similar steps. At the beginning of the nineteenth century, such a statement seemed absurd. But despite all the doubts, many mathematicians expanded the scope of study of this phenomenon, taking it beyond the study of thermal conductivity. However, most scientists continued to be tormented by the question: “Can the sum of a sinusoidal series converge to exact value discontinuous function?
Convergence of Fourier series: an exampleThe question of convergence arises whenever it is necessary to sum infinite series of numbers. To understand this phenomenon, consider a classic example. Will you ever be able to reach the wall if each subsequent step is half the size of the previous one? Let's say you're two meters from your target, the first step takes you to the halfway mark, the next one takes you to the three-quarters mark, and after the fifth you'll have covered almost 97 percent of the way. However, no matter how many steps you take, you will not achieve your intended goal in a strict mathematical sense. Using numerical calculations, it can be proven that it is eventually possible to get as close as a given distance. This proof is equivalent to demonstrating that the sum of one-half, one-fourth, etc. will tend to unity.
This issue was raised again at the end of the nineteenth century, when they tried to use Fourier series to predict the intensity of tides. At this time, Lord Kelvin invented an instrument, an analog computing device that allowed military and merchant marine sailors to monitor this natural phenomenon. This mechanism determined sets of phases and amplitudes from a table of tide heights and corresponding time points, carefully measured in a given harbor throughout the year. Each parameter was a sinusoidal component of the tide height expression and was one of the regular components. The results of the measurements were entered into Lord Kelvin's calculating instrument, which synthesized a curve that predicted the height of the water as a function of time over next year. Very soon similar curves were drawn up for all the harbors of the world.
What if the process is disrupted by a discontinuous function?At that time it seemed obvious that a tidal wave prediction instrument with big amount account elements can calculate a large number of phases and amplitudes and thus provide more accurate predictions. However, it turned out that this pattern is not observed in cases where the tidal expression that should be synthesized contained a sharp jump, that is, it was discontinuous. If data from a table of time moments is entered into the device, it calculates several Fourier coefficients. The original function is restored thanks to the sinusoidal components (in accordance with the found coefficients). The discrepancy between the original and reconstructed expression can be measured at any point. When carrying out repeated calculations and comparisons, it is clear that the value of the largest error does not decrease. However, they are localized in the region corresponding to the discontinuity point, and at any other point they tend to zero. In 1899, this result was theoretically confirmed by Joshua Willard Gibbs of Yale University.
Fourier analysis is not applicable to expressions containing an infinite number of spikes over a certain interval. In general, Fourier series, if the original function is represented by the result of the real physical dimension, always converge. Questions about the convergence of this process for specific classes of functions led to the emergence of new branches in mathematics, for example, the theory of generalized functions. She is associated with such names as L. Schwartz, J. Mikusinski and J. Temple. Within the framework of this theory, a clear and precise theoretical basis under such expressions as the Dirac delta function (it describes a region of a single area concentrated in an infinitesimal neighborhood of a point) and the Heaviside “step”. Thanks to this work, Fourier series became applicable to solving equations and problems involving intuitive concepts: point charge, point mass, magnetic dipoles, and concentrated load on a beam.
Fourier methodFourier series, in accordance with the principles of interference, begin with the decomposition of complex forms into simpler ones. For example, a change in heat flow is explained by its passage through various obstacles made of heat-insulating material irregular shape or a change in the surface of the earth - an earthquake, a change in the orbit of a celestial body - the influence of planets. As a rule, such equations that describe simple classical systems can be easily solved for each individual wave. Fourier showed that simple solutions can also be summed to obtain a solution of more complex tasks. In mathematical terms, Fourier series are a technique for representing an expression as a sum of harmonics - cosine and sine. Therefore, this analysis is also known as “harmonic analysis”.
Fourier series - an ideal technique before the “computer age”Before the creation of computer technology, the Fourier technique was best weapon in the arsenal of scientists when working with the wave nature of our world. The Fourier series in complex form makes it possible to solve not only simple tasks, which are amenable to the direct application of Newton's laws of mechanics, but also fundamental equations. Most of the discoveries of Newtonian science in the nineteenth century were made possible only by Fourier's technique.
With the development of computers, Fourier transforms have risen to a qualitatively new level. This technique is firmly established in almost all areas of science and technology. An example is digital audio and video. Its implementation became possible only thanks to a theory developed by a French mathematician at the beginning of the nineteenth century. Thus, the Fourier series in a complex form made it possible to make a breakthrough in the study of outer space. In addition, it influenced the study of the physics of semiconductor materials and plasma, microwave acoustics, oceanography, radar, and seismology.
Trigonometric Fourier seriesIn mathematics, a Fourier series is a way of representing arbitrary complex functions the sum of simpler ones. In general cases, the number of such expressions can be infinite. Moreover, the more their number is taken into account in the calculation, the more accurate the final result is. Most often, trigonometric functions of cosine or sine are used as the simplest ones. In this case, Fourier series are called trigonometric, and the solution of such expressions is called harmonic expansion. This method plays an important role in mathematics. First of all, the trigonometric series provides a means for depicting and also studying functions; it is the main apparatus of the theory. In addition, it allows you to solve a number of problems in mathematical physics. Finally, this theory contributed to the development of a number of very important branches of mathematical science (the theory of integrals, the theory of periodic functions). In addition, it served as the starting point for the development of the following functions of a real variable, and also laid the foundation for harmonic analysis.
Transcript1 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. Given illustrative material. There are tasks for independent solution. Intended for students and teachers of the Faculty of Physics of NSU. Published by decision of the methodological commission of the Faculty of Physics of NSU. Reviewer: Dr. Phys.-Math. 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 комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является четной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n = ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является нечетной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2,.... Ответы 1 2π 24. a n a π a n i + e 2inx, где подразумевается, что слагаемое, соответствующее n =, пропущено. 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. Therefore, 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 general formula at n = 1 the denominator of the fraction becomes 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 рациональным функциям: πctgaπ = 1 a + + 2a a 2 n 2, (π) = sin aπ (a n) 2. n= 35. Выведите комплексную форму обобщенного равенства Ляпунова. 36. Покажите, что complex form 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. A 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 general meaning T 1 = T 2 through T. Now let's calculate the sum of projections F u of the same forces onto 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 is a 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 have obtained a 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
