Differential Equations. Sequential differentiation method
Ordinary differential equations are equations that contain one or more derivatives of the desired function y = y (x)
F (x, y, y 1,…, y (n)) = 0, where x is the independent variable.
A solution to a differential equation is a function that, after substituting it into an equation, turns it into a triumph.
Some solution methods are known from the course of differential equations. For a number of first-order equations (with separable variables, homogeneous, linear, etc.), it is possible to obtain a solution in the form of formulas by analytical transformations.
In most cases, approximate methods are used to solve differential equations, which can be divided into two groups:
1) analytical methods that give a solution in the form of an analytical expression;
2) numerical methods that give an approximate solution in the form of a table.
Let's consider the listed methods in the form of the following examples.
8.1 Method of sequential differentiation.
Consider the equation:
with initial conditions, where 
- given numbers.
Suppose that the desired solution y = f (x) can be solved in a Taylor series in powers of the difference (x-x 0):
2 
n +….
Initial conditions (8.2) give us the values of y (k) (x 0) for k = 0,1,2, ..., (n-1). The values y (n) (x 0) are found from equation (8.1), substituting (x-x 0) and using the initial conditions (8.2):
y (n) (x 0) = f (x 0, y 0, y "0, ..., y 0 (n-1))
The values y (n + 1) (x 0), y (n + 2) (x 0) ... are successively determined by differentiating equation (8.1) and substituting x = x 0, y (k) (x 0) = y 0k (k - 0,1,2).
EXAMPLE: Find the first seven terms of the power series expansion of the solution y = y (x) of the equation y "" +0,1 (y ") 2 + (1 + 0,1x) y = 0 with the initial conditions y (0) = 1; y "(0) = 2.
SOLUTION: We are looking for the solution to the equation in the form of a series:
y (x) = y (0) + y "(0) x / 1! + y" "(0) x 2 /2!+...+y (n) (0) x n / n! ...
From the initial conditions, we have y (0) = 1, y "(0) = 2. To determine y" "(0), we solve this equation for y" ":
y "" (0) = - 0.1 (y ") 2 - (1 + 0.1x) y (8.3)
Using the initial conditions, we obtain
y "" (0) = –0.1 * 4 - 1 * 1 = –1.4
Differentiating with respect to x the left and right sides of equation (8.3)
y "" "= - 0.2 y" y "" - 0.1 (xy "+ y) - y",
y (4) = - 0.2 (y "y" "" + y "" 2) - 0.1 (xy "" + 2y ") - y" ",
y (5) = - 0.2 (y "y (4) + 3y" "y" "") - 0.1 (xy "" "+ 3y" ") - y" "",
y (6) = - 0.2 (y "y (5) + 4y" "y (4) + 3y" "" 2) - 0.1 (xy (4) + 4y "" "- y (4) )
Substituting the initial conditions and the value y "" (0), we find y "" "(0) = - 1.54;
y (4) (0) = - 1.224; y (5) (0) = 0.1768; y (6) (0) = - 0.7308. Thus, the desired approximate solution will be written in the form: y (x) ≈ 1 + 2x - 0.7x 2 - 0.2567x 3 + 0.051x 4 + 0.00147x 5 - 0.00101x 6.
8.2 Euler's method
The simplest of the numerical methods for solving differential equations is the Euler method, which is based on replacing the desired function with a polynomial of the first degree, i.e. linear extrapolation. We are talking about finding the values of the function at adjacent points of the argument x not between them.
Let us choose the step h small so that for all x between x 0 and x 1 = x 0 + h the value of the function y differs little from the linear function. Then on the indicated interval y = y 0 + (x - x 0) y "= y 0 + (x -
Continuing to determine the values of the function in the same way, we make sure that Euler's method is represented in the form of sequential execution of formulas:
∆y k = y "k h
y k + 1 = y k + ∆y k
EXAMPLE
Let us solve by the Euler method the equations y "= x - y with the initial condition x 0 = 0, y 0 = 0 on a segment with a step h = 0.1.
The calculations are shown in the table.
The first row in columns 1 and 2 is filled with initial data. Then y is calculated by given equation(in column 4), then ∆y = y "h - in column (4).
Column (5) contains a table of values of the exact solution to the given equation.
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The table shows that for x = 1 the relative error of the Euler method is δ = 0.37 - 0.35 / 0.37 * 100% ≈5.4% |
REFINED EULER'S METHOD
With the same amount of computational work, it gives a higher accuracy.
Previously, we considered the integrand to be constant, equal to its value f (x k, y k) at the left end of the segment. A more accurate value will be obtained if we assume f (x, y (x)) equal to the value in the center of the plot. To do this, you need to take a double section (x k-1, x k + 1), replacing the formula
y k + 1 = y k + ∆y k on y k + 1 = y k-1 + 2hy "k (8.5)
It is this formula that expresses the refined Euler method. But in this case, you must adhere to the following sequence of actions:
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EXAMPLE For comparison, consider the same equation y "= x - y with the initial conditions x 0 = 0, y 0 = 0. The refined method, as can be seen from the table, gives a higher accuracy relative error at x = 1, y = 0.370, and y 0.368. |
If the equation has the form We have a difference in the Taylor series Let us investigate the convergence of the resulting series, into which we substitute the initial conditions. The series can be used to solve algebraic equations. View. The solution of such equations is carried out by the method of indeterminate coefficient and subsequent differentiation.
51. Periodic functions. Trigonometric. Determination of coefficients by the Euler-Fourier method.
A periodic function with a period 2П, satisfying the Dirichlet conditions on the interval (-П, П), can be represented by the Fourier series:
The coefficients of which are found by the formulas
At the points of continuity of the function f (x), the Fourier series converges to f (), and at the points of discontinuity, to. The Fourier series expansion of a periodic function f (x) with period 2l has the form where
53 Orthogonal systems of functions. Fourier series for an arbitrary orthogonal system of functions. Definition 1. An infinite system of functions f 1 (x), f 2 (x) .. fn (x) (1) is called orthogonal on the interval [a, b] if, for any n ≠ k, the equality (x) ϕ k ( x) dx = 0 (2) It is assumed that dx ≠ 0 Let the function ϕ (x), defined on the interval [a, b], be such that it is represented by a series in functions of the orthogonal system (1), which converges to the given functions on [a, b]: f (x) = (x) (6). Let us define the coefficients with p. Suppose that the series obtained after multiplying the series (6) by any ϕ k (x) admits term-by-term integration. We multiply both sides of equality (6) by ϕ k (x) and integrate within the limits from a to b. Taking into account equalities (2), we obtain (x) ϕ k (x) dx = c k whence (7) Coefficients with к, calculated by formulas (7), are called 5 Fourier coefficients of the function f (x) by the system of orthogonal functions (1). Series (6) is called the Fourier series in the system of functions (1).
54. Dirichlet conditions. A sufficient condition for the representation of a function in a Fourier series. The function f (x) is definite and continuous in some range of values of x, is called non-decreasing (non-increasing) if from the condition x 2> x 1; f (x 2) ≥f (x 1) - non-decreasing f (x 2) ≤f (x 1) - non-increasing The function f (x) is called piecewise monotone on a segment if this segment can be divided into a finite number of points х 1, x 2, x 3 ... .. x n -1 into intervals so that on each of the intervals the function is monotone, that is, either does not decrease or does not increase, it follows from this that if the function f (x) is piecewise monotone and is limited to segments, then it may have break points of the 1st kind. x = c = f (c-0) = f (c + 0); f (c-0) f (c + 0). T. Dirikhlet. If a function f (x) with period 2π is piecewise monotone and bounded on a closed interval x [-π; π], then the Fourier series built on this function converges at all points the sum of the obtained series S (x) is equal to the value of f (x) at the points of continuity of this function, at the points of discontinuity of the function f (x) the sum of the series is the mean arithmetic side of the function f (x) on the right and on the left S (c) = (f (c-0) + f (c + 0)) / 2. The conditions of this theorem are called the Dirichlet conditions.
55. Expansion of even / odd functions in a Fourier series.
It follows from the definition of an even and odd function that if ψ (x) is an even function, then Indeed
Since, by the definition of an even function, ψ (-x) = ψ (x).
Similarly, we can prove that if φ (x) is an odd function, then If an odd function f (x) is expanded in a Fourier series, then the product f (x) cos (kx) is also an odd function, and f (x) sin (kx) - even; hence, the Fourier series of an odd function contains “only sines”
If an even function is expanded in a Fourier series, then the product f (x) sin (kx) is an odd function, and f (x) cos (kx) is even, hence
That is, the Fourier series of an even function contains “only cosines”. The obtained formulas make it possible to simplify calculations when searching for the Fourier coefficients in cases where the given function is even or odd. Obviously, not every periodic function is even or odd.
Izvestia
OF THE TOMSK ORDER OF THE OCTOBER REVOLUTION AND THE ORDER OF LABOR RED BANNER OF THE POLYTECHNICAL INSTITUTE named after S.M.KIROV
APPLICATION OF THE SEQUENTIAL METHOD
DIFFERENTIATION IN THE CALCULATION OF TRANSIENT PROCESSES OF ELECTRIC MACHINE SOURCES
IMPULSES
A. V. LOOS
(Presented by the scientific seminar of the departments of electrical machines and general electrical engineering)
Transient processes of electric machine sources of impulses, for example, single-phase shock generators, valve pulse generators, etc., are described by systems of differential equations with periodic coefficients, which cannot be eliminated by any transformations. Investigations of transient processes of electrical machines in the general case of asymmetry are based on the use of the principle of constancy of flux linkage, the use of integral equations, approximate methods of solution, etc. etc.
In some cases, the equations of transient processes of electric machine pulsed energy sources can be reduced to equations with constant coefficients, however, the need to consider the case of two or more winding systems on the rotor requires solving a cubic equation or characteristic equations of higher degrees with complex coefficients, which is impossible in an algebraic form. ... The need to take into account the saturation of the magnetic circuit and changes in the rotor speed further complicates the solution of such problems. In these cases, the most acceptable is the use of analytical methods for approximate solutions.
Among the analytical methods for the approximate integration of systems of differential equations, integration using power series by the method of sequential differentiation is very widespread. This method is applicable both for solving systems of linear differential equations with constant and variable coefficients, and for solving nonlinear problems. The sought-for particular solution is represented in the form of an expansion in a Taylor series. The effectiveness of the application of the method to a large extent depends on the ability of the researcher to use a priori information about physical nature the problem to be solved.
Indeed, if we compose a system of differential equations for an electric machine source of pulses, taking currents as unknown functions, then it is known in advance that the solutions will represent rapidly oscillating functions. Obviously, to represent them in the form of a Taylor series, a large number of terms will be required, i.e., the solution will be extremely cumbersome. Differential Equations transient processes are more profitable to compose not for currents, but for flow-couplings. This is due to the fact that the flux linkage of the windings changes
the number I in time is much smaller, since they are, as a rule, monotonically changing functions, for a sufficiently accurate representation of which in the form of an expansion in a Taylor series, only a few terms are required. After determining the flux linkages, the currents are found by solving ordinary algebraic equations.
As an example, consider the use of the sequential differentiation method to calculate the transient processes of a valve pulse generator.
The calculation of the load current of the valve generator can, however, be performed according to the envelope curve of the phase currents obtained when the synchronous generator is suddenly switched on to a symmetrical three-phase active load. The value of the equivalent symmetrical resistive load is determined by the ratio R3 - 2 / sRh. Thus, to calculate the load current curve and phase currents, it is necessary to solve the complete system of differential equations of the synchronous generator when connected to a symmetrical resistive load.
When determining the armature current, the external active resistance can be added to the active resistance of the stator r = R3 + rc. The equations of transient processes of the synchronous generator in the d, q axes are as follows:
pYd = - Ud - (ü ^ q -rld, (1)
р - - Uq + with W6 riq, (2)
P ^ f = Uf - rfif, (3)
P ^ Dd - - rodiDcb (4)
PXVD :( = - rDq ioq, (5)
XfXDd - X2ag | m Xad (XDd-XaH) Tf. xad (Xj - Хпн) w
D "d ri" d Tßd 9
, * _ x ° q w „xaq / 7)
q ~ "Ä7 ™ q q"
XdXDd ~~ x "ad ig xad (xDd" ~ "xad) m Xad (xd Xad) -CG f ^ -D- 1 ~~" - ~ D- d "---- d" * "
XdXf X2ad yep xad (xf ~~ xari) m xad (xd ~ xad) w / n \ iDd = - ~ q- ^ Dd - D- Td --d - M »w)
D - XdXfXDd ^ 2x3ad - x2ad (xd + xr -f X [) d), (11)
A "= XqXDq - X2aq. (12)
There is no general analytical solution to the system of equations (1- ^ 12). An attempt to obtain design ratios for the currents of a synchronous generator in the presence of active resistances in the stator circuit was made in. However, the author made a mistake physically related to the inadmissibility of assuming the constancy of flux linkages along the longitudinal and transverse axes in a rotating machine in the presence of active resistance in the stator circuit. This error was pointed out in, where an exact solution was obtained for the case of one winding system on the rotor and the impossibility of using conventional solution methods when considering two or more winding systems on the rotor was shown. Therefore, the example considered here is of considerable interest.
Substituting (6-10) into (1-5) and taking into account that Ud = Uq =: 0, we obtain the equations of transient processes written with respect to flux linkages in the normal form of Kosh and:
[(x (x1) c1 - x. ^ H ^ - xa (1 (x0 (1 - x ^ H ^ _
3 d7 ~ (xOo (H ^ x, 1 (] H ^)
P ^ = bmr - ^ [(xc] x0c1 - x2aa) H * (- Xa (1 (XO (1 - xa<1№
Ha<1 (хс! - Х^Ч^] ,
P = --- X2a (1) ¥ 141 - hi (x (- x ^ H ^
Hayo (Xs1 - has1) ¥ (],
p CHTs = ^ -¿g (xh H ^ - xach H ^).
Suppose that before switching on the load, the synchronous generator was idling with the excitation current, then the initial conditions are at 1 = 0.
H ^ o = * Gox = Mb ^ H "o = 1 Goxa (b ChTs0 - O, ¥ C (0 = 0.
With the initial conditions adopted, the solution for ^, Ъa, ^, Ь can be represented in the form of an expansion in the Maclaurin series
Similarly for the flux linkages Ch ^, Ch ^, Tm, Ch ^. The initial values of the derivatives of flux linkages in equations of the form (18) are easy to find under known initial conditions by successive differentiation of equations (13-17). After substituting the initial values of flux linkages and their derivatives into equations of the form (18), we obtain:
(3 = 1Gohas1
XrX ^ - x ^ \
^ = Cho has1 N
1 GHop "+2 1 ^ - 4 G --- 7- W X
2 A "(x2ochg + x2achGoch)
X? 1 g (xaH (Hoa - Xls1) ®2
syo ~ 1 goal (1
1__GR (1 xyas1 (x (- khas!) S ° 2
L X2ad Year
(20) (21) (22) (23)
The convergence of solutions for,,, Ъч can be determined by studying the remainder terms of expansions in the Maclaurin series (19-23)
KnNo) = - ^ mt P (n + 1) ^ (H), (24)
where 0
Similarly for "Moat, According to the found values of the flux
using equations (6-10), it is easy to find the fluxes 1r »a. Using the formulas of linear transformations, we determine the phase currents:
1a = ¡c) coe co 1 - ¡d et co 1 (25) 1b = 1st sob 1 --- 1h e1n ^ -> (26)
"-c = - 1a -> b- (27)
The load current of the valve pulse generator is found as the sum of the instantaneous values of the phase currents 1a, 1b, ¡from one sign.
According to the method under consideration, the transient processes of the valve pulse generator were calculated with the parameters:
X (1 = = Xos! = Hvch = 1.05; ha (1 = has, = 1; x (= 1.2; rc = r - !! = goa = = 0.02; Yn = 0.05 ...
In fig. 1 shows the calculated curves of the phase currents \ b, ¡c and the load current ¡c. Comparison of analytical calculations with the results obtained on the AVM MN-14 when investigating the complete system of equations gives
Rice. 1. Design curves tokos without generator and load
good convergence. An estimate of the convergence of the solution by studying the remainder of the Maclaurin series expansion (24) also shows that the maximum calculation error does not exceed 5 - = - 7%.
The method of sequential differentiation can be used to analyze transient processes of electrical machine sources of impulses, the equations of which contain variable coefficients. The study of transient processes described by nonlinear differential equations also does not encounter fundamental difficulties when using this method, but its application in this case can lead to cumbersome expressions. For the correct choice of the form of the initial system of differential equations, it is necessary in all cases to use a priori information about the physical picture of the processes, which greatly simplifies the solution.
LITERATURE
1.I.I. Treshchev. Machine research methods alternating current... "Energy", 1969.
2. A. I. In azhio V. Fundamentals of the theory of transient processes of a synchronous machine. Gosenergoizdat, 1960.
3. Ch. Konkord and a. Synchronous machines. Gosenergoizdat, 1959.
4. E. Ya. Kazovskii. Transient processes in AC electric machines. Publishing house of the Academy of Sciences of the USSR, 1962.
5.L.E. Elsgolts. Differential equations and calculus of variations. "Science", 1969.
6. G. A. Sipaylov, A. V. Los, and Yu. I. Ryabchikov. Research of transient processes of the valve pulse generator. Izv. TPI. This collection.
Theorem.
Given:
If the right side of the remote control, i.e. function 
, is an analytic function of its arguments in some neighborhood of the point 
, then for values close enough to, there is a unique solution to the Cauchy problem, which can be represented as a power series (Taylor series).
Consider the above Cauchy problem. We will seek a solution to the Cauchy problem for nth order DE in the form of a Taylor series in powers in a neighborhood of a point.
The coefficients of the series are the derivatives of the function calculated at the point.
Let's find them:
1) From the initial conditions, we determine the first n expansion coefficients:
;
2) The value of the (n + 1) -th coefficient is determined by substituting the values in the DU:
3) To find all subsequent coefficients, we will sequentially differentiate the left and right sides of the original DE and calculate the values of the coefficients using the initial conditions and all the coefficients already obtained.
Comment. If the conditions of the existence and uniqueness theorem for the solution are satisfied, then the partial sum of the resulting Taylor series will be an approximate solution of the Cauchy problem posed.
Algorithm of the method of sequential differentiation
1. Write the solution y (x) in the form of an infinite power series in powers:
, where
2. Determine the values of the first n coefficients (here n is the order of the original equation) using the initial conditions.
3. Express the highest derivative from the DE. Calculate its value at the starting point using the initial conditions. Calculate the coefficient.
4. Differentiating the expression for the highest derivative from item 3 with respect to x, find the n + 1 derivative of the function. Calculate its value at the starting point using the initial conditions and the value of the highest derivative calculated in step 3. Calculate the coefficient.
5. The rest of the coefficients are calculated similarly to the procedure described in item 4.
