Application of the method of successive approximations to the numerical integration of differential equations

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D.V. Ionescu
Institutul de Calcul

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D.V. Ionescu, L’application de la méthode des approximations successives à l’intégration numérique des équations différentielles. (French) Bull. Math. Soc. Sci. Math. Phys. R. P. Roumaine (N.S.) 3 (51) 1959 423–431.

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D.V. Ionescu, Application of the method of successive approximations to the numerical integration of differential equations, Acad. R. P. Romîne Fil. Cluj Stud. Cerc. Mat., 11 1960, 273–286.
[Aplicarea metodei aproximărilor succesive la integrarea numerică a ecuațiilor diferențiale]

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BULL. MATH. of the Society. Sci. Math. Phys. of the RPR
Volume 3 (53)

n^{\circ}

4, 1959

APPLICATION OF THE METHOD OF SUCCESSIVE APPROXIMATIONS TO THE NUMERICAL INTEGRATION OF DIFFERENTIAL EQUATIONS

BY

DV IONESCU (Cluj)

Let us consider the differential equation

y^{'} = f (x, y)

where the function

f (x, y)

is defined and has continuous first- and second-order partial derivatives in the rectangle

D

defined by inequalities

x_{0} ⩽ x ⩽ x_{0} + has, | y | ⩽ b .

Let y(x) denote the integral of this equation that satisfies the condition

y (x_{0}) = 0

and either

ε

a given positive number.

In this work, we will show, by applying the method of successive approximations, that we can determine on the interval

[x_{0}, x_{0} + has)

a network

Γ

knots

x_{0}, x_{1}, \dots, x_{n}

and a calculation algorithm for number calculations

y_{i}^{(z)}

Or

s = 0, 1, 2, \dots

, v so as to have

| y (x_{i}) - y^{(y)} | < 2 ε

on all network nodes

Γ

To fix the number v, we take into account the method of successive approximations, and for the choice of the number of nodes n and the calculation algorithm for the numbers

\sum_{i}^{(s)}, s = 0, 1, \dots

, v we will use the trapezoid quadrature formula.

This theorem was communicated at the Colloquium on Mechanics [1] held in Bucharest (October 25-29, 1959) and an extension to second-order hyperbolic partial differential equations was communicated at the Colloquium on the Theory of Partial Differential Equations [2] held in Bucharest (September 21-26, 1959)

§ 1. Dilential equations

Consider the differential equation
(1)

y^{'} = f (x, y)

where the function /

(x, y)

eal continue dane Ie metangle

D

defined by frog infegiltes
(2)

x_{b} ⩽ x ⩽ x_{b} + has . | y | ⩽ b

and satisfies the condition of LIPSCHITZ
(3)

| / (x, F) f (x, y) | \dots HAS | F y |

Or

HAS

is a constant.
Under these conditions, we know that the differential equation (1) has a unique integral of zero for

x

-

x_{0}

- kile cat definie wor l'intervalle

[x_{0} x_{0} + h_{1}]

where
(4)

h_{1} = min (has, \frac{b}{M})

the number

M

holding a saperiodic boundary of

| / (x, y) |

in the rectangle

D

The
complete collection

y (x)

can be obtained by the method of successive approximations. We construct the function map

{y^{(n)} (x)}

, where
(b)

\begin{aligned} y^{(m)} (x) - \int_{- 4}^{3} f [ξ, 0] d ξ \\ (6) & y^{(n)} (x) - \int_{α_{4}}^{π} f [ξ y^{(0 - 1)} (ξ)] d ξ . \end{aligned}

We know that the mother

y^{(1)} (x) + \sum_{i = 1}^{\infty} [y^{(1)} (x) - y^{(0 - 1)} (x)]

is absolutely and uniformly convergent on the interval

[x_{0}, x_{0} + h_{1}]

and that the sum of this series represents the integral of the differential equation (1) which satisfied the condition

y (x_{0}) - 0

We can read the complete text.

y (x)

in the form

y (x) = y^{(n)} (x) + \sum_{n = 1}^{\infty} [y^{(n)} (x) - y^{(n - 1)} (x)]

and we demonstrate the inequality

| y (x) - y^{(1)} (x) | < \frac{M}{HAS} e^{μ_{1}} \frac{{(HAS μ_{1})}^{γ + 3}}{(ν + 2)!} .

Given a positive number, we can then choose the smallest natural number v to have

\begin{matrix} (7) & \frac{M}{HAS} < \frac{{(HAS h_{1})}^{r + 1}}{(v + 2) T} < ϵ \end{matrix}

and we will have
(8)

| y (x) - y^{(y)} (x) | < c .

the natural number y once chosen, remain fixed and play an important role in the numerical integration of the differential equation (1).

____

9. For the numerical integration of the differential equation (I), adjust the condition

y (x_{0}) = 0

, we will make new hypotheses about the function / (x, y), which are related to the numerical interpretation procedure that we will now give in this work.

We will assume that the function

f (z, g)

sit des derivatives participles par npport &

x

ot ay of the first and of the round order, continuous in the triangle D. Under these conditions the number A of Lipachitz's inequality (3) is an upper bound of

| \frac{\partial f}{\partial y} |

in the rectangle

D

.

We demonstrate that the gos functions (

x

) donsers by the formals (6) and (6) cat of the first and second order derivatives continuous by the interval

[x_{0}, x_{0} + h_{1}]

We can calculate upper bounds of

| \frac{d y^{(n)} (x)}{d x} | \cdot | \frac{d^{n} (x)}{d x^{2}} |, (x = 0, 1.2 \dots . v)

over the interval

[x_{0}, x_{0} + h_{1}]

by using the upper horns of

| / (x, y) | . | \frac{\partial f}{i x} | \cdot | \frac{\partial f}{\partial y} |

in your rectangle

D

It
follows that the functions

p^{2 n} (x) = / | x, y^{0 - 1} (x) |

an upper bound of

| \frac{P^{*} (x)}{P^{3}} |

For

s = 0, 1, \dots, v_{0}

on the interval

[x_{0}, x_{0} + h_{1}]

The number N will play an important role in the approximate calculation of the integrals (0) and (6) for

s = 0, 1, \dots, v

.

We will denote by h a positive number of 6́finite by
(9)

HAS = min (has \frac{b - δ}{M})

where & is a positive number, domá, awes small. I eat tvident que nons avous

h < h_{1}

.

With these hypotheses, everyone could proceed to the anmerical integration of the equation

over the interval

[x_{6}, x_{6} +

h

]

, in arithmetic programming and an algorithm for calculating numbers go for s

= 0, 1, \dots

, v so as to have on the nodes of the network I

\begin{matrix} (10) & | y^{(1)} (x_{1}) - y^{p} | < c . \end{matrix}

of your trapezoid quadrature formula

\begin{matrix} (11) & \int_{0}^{0} f (x) d x = \frac{β - α}{2} U (α) + / (β)) + R_{1} . \end{matrix}

on
(12)

R = \int_{0}^{0} \frac{(x - α) (x - β)}{2} f^{'} (x) d x .

If N, ext an upper bound of

| /^{\circ} (x) |

because the interval

[α, β]

we will have
(13)

| R | ⩽ \frac{(β - α)^{3}}{12} N_{1} .

4 Let's divide the interculle [

x_{0}, x + b

] that are equal parts by points

x_{1}, x_{2}, \dots, x_{n - 1}

. Northern islands

x_{3}, x_{1} \dots \dots x_{1}

Or

x_{1} = x_{6} + h

form a rewas

Γ

Now let's calculate the values of the functions

g^{(4)} (x)

, For

s \sim 0, 1, \dots

ver les meads do réwer Γ.

Name we have first

5^{\infty} (x_{1}) = \int_{z_{1}}^{z_{1}} / ∣ ξ, 0] d ξ .

È applying the quadrature formula (II) to each interval [

x_{a}

.

x_{1}

]

[x_{2}, x_{2}] \dots, [x_{i - 1}

.

x_{i}]

and by adding, we have the formula

\begin{matrix} (HA) & y^{(n)} (x_{i}) = y_{i}^{(n)} + R_{i}^{(n)} . \end{matrix}

ભ્યે
(Ib)
and

n^{p} = \frac{h}{2 n} [f (x_{0}, 0) + f (x_{i}, 0) + 2 \sum_{i = 1}^{i - 1} f (x_{i}, 0)]

| R^{p} | < \frac{H^{3}}{12 π^{3}} i N < \frac{n^{3}}{12 π^{3}} N

because

\frac{i}{n} < I

Let e
be a positive number, which we will determine later. We will choose the number n, the smallest natural number, such that

\begin{matrix} (16) & \frac{h^{3}}{12 n^{3}} N < c_{1} \end{matrix}

Since the number n is chosen in this way, it will remain fixed in the following, and we are sure of it in formula (14)
(17)

| R^{(n)} | < c

i. For In function y(1)(x). non nome

y^{11} (x_{1}) - \int_{x_{0}}^{J_{1}} ∣ ξ y_{1} (E_{1}) E_{0}

ei pur lea neuidn

x_{1}

we have

y^{(1)} (x_{1}) - \int_{x_{0}}^{x_{4}} f [ξ y_{5} (ξ)] ε_{6} .

We proceed as in no. 4, and apply the quadrature formula of the intervals

[x_{0}, x_{1}]

.

[x_{1}, x_{1}] \dots, [x_{1}, \dots, x_{1}]

none nurons
(18)
where
(10)

g^{(1)} (x_{1}) - [g^{(1)}] + η^{(1)}

(II)

| r_{1}^{(n)} | < c_{1}

and where
(20)

[y^{(1)}] - \frac{h}{2 n} | f [x_{0}, y^{(n)} (x_{0})] + f [x_{1}, y^{(n)} (x_{1})] + 2 \sum_{i = 1}^{L - 1} | [x_{1}, y^{n} (x_{1})] |

Introduce the numbers yp

^{(1)}

by formula
(21)

y^{(1)} - \frac{h}{2 n} {f (x_{0}, 0) + f (x_{1}, y_{1}^{(n)}) + 2 \sum_{i = 1}^{i - 1} f (x_{1}, y_{i}^{(n)})}

.

We will have
(22)

[y^{(1)}] - y^{(1)} + p^{(1)}

Or

p^{(1)} - \frac{h}{2 n} {f (x_{1}, y (x_{1}^{(n)})) - f (x_{1}, y_{1}^{(n)}) + 2 \sum_{- 1}^{4} U (x_{1}, y^{(n)} (x_{1})) - l (x_{1}, y^{n})}

Finally, taking into account Lipsciirz inequality (3) and the integlities (17), we have

| p_{i}^{(1)} | < \frac{1}{2 n} (2 i - 1) d c_{1}

that is to say
(29)

| p^{(1)} | < Δ h c_{1}

Returning to formula (18) we can berire
(24)

y^{(1)} (x_{1}) = y^{v} + B^{v}

where, according to the inequalities (19) and (23) we have
that is to say
(26)

\begin{aligned} | R^{(1)} | ⩽ ε_{1} + d A ε_{1} \\ | R^{(1)} | ⩽ (1 + R) ε_{1} \end{aligned}

od nons avons note
(26)

K - A h

Let us move on to the general case. Suppose that nons avone demonstrates that
(27)

| g^{(t - 1)} (x_{j}) - y_{j}^{f - 4)} | < (1 + K + \dots + K^{- 1}) ε_{1}

for jI. 2.... n, od̀

y^{f - 1)} - \frac{1}{2 n} | f (x_{0}, 0) + f (x_{1}, y_{1}^{f - m}) + 2 \sum_{m = 1}^{i - 1} | (x_{1}, y_{0}^{f - m_{1}}) |

and we demonstrate that by introducing the number yso by the formula
(28)

y^{(1)} - \frac{h}{2 n} {f (x_{0}, 0) + f (x_{0}, x^{(0 - 1)}) + 2 \sum_{j = 1}^{j - 1} f (x_{1}, y_{j}^{(0 - 1)})}

ones avons aresi
(29)

| y^{(1)} (x_{1}) - y^{(2)} | ⩽ (1 + K + \dots + K^{+}) r_{1},

ED, indeed, mons have

y^{(1)} (x_{1}) = \int_{x_{0}}^{\infty} \int [ξ y^{(0 - 1)} (ξ)] d ξ .

By proceeding as in No. O, we have

\begin{matrix} (30) & y^{(0)} (x_{i}) = [y^{(0)}] + η^{(0)} \end{matrix}

where
(31)

| γ^{(0)} | < r_{1}

and
(32)

[y^{(i)}] = \frac{h}{2 π} {[x_{0}, y^{(k - 1)} (x_{0})] + / [x_{1}, y^{(0 - 1)} (x_{1})] + 2 \sum_{j = 1}^{j - 1} / [x_{1}, y^{(0 - 1)} (x_{j})]}

Formulas (28) and (32) showed that

\begin{matrix} [y^{(i)}] = y^{(i)} + p^{(0)} \\ {\hat{p}}^{(i)} = \frac{1}{2 n} {/ [x_{i}, y^{(o - 1)} (x_{i})] - / [x_{i}, y_{i}^{(o - 1)}] + 2 \sum_{j = 1}^{i = 1} U (x_{i}, y^{(o - 1)} (x_{i})) - / (x_{i}, y_{i}^{(o - 1)})]} . \end{matrix}

Taking into account the lapschitz inequality and inequalities (27), we have

| p (p) | < \frac{h}{2 n} (2 i - 1) A (1 + K + \dots + K^{2 - 1}) r_{1}

"
13)

| {\hat{p}}^{(i)} | ⩽ K + K^{2} + \dots + K^{0}) c_{1} .

II rAulte Alon que nons pouvais Ariur

y^{(n)} (x_{1}) - y_{0}^{n} + k_{1}^{n} .

Or

R^{(n)} - r^{n} + {\hat{p}}^{n} .

Tonant compo des integaliten (31) et (33). il réwulte que nous avons

| R^{(n)} | ⩽ {(1 + K + \dots + K^{'})}_{n_{1}}

This proves that inequality (29) is demonstrated.
7. We have therefore found an algorithm for calculating the numbers s₀ using formulas (16) and (28). It now remains for us to determine the number ₀. We will first take the number

ϵ_{1}

in such a way that the middle member of Pinégalité (20) for s - y is smaller than c. We will therefore prove

c_{1} ⩽ \frac{ε}{1 + K + \dots + K} .

Mnis there is still one condition for the number a. For formula (28) to make sense for s - v, the point with coordinates (

x_{i}

.

{\hat{y}}^{- 1}

) is found in the rectangular D. Of identity

y^{(n - 1)} - - [y^{(n - 1)} (x_{k}) - y^{(- 1)}] + y^{(n - 1)} (x_{l})

and from the definition of the number h, we do not deduce that

| ℜ^{(- 1)} | ⩽ | R^{(- 1)} | + b - δ < (1 + K + \dots + K^{-}} c_{1} + b - δ

or even
郎

| m^{(p - 1)} | < (1 + K + \dots + K) c_{1} + b - c

To have

| y^{(p - 1)} | ⩽ b

, it is necessary that

c_{1}

checks the condition

c_{1} ⩽ \frac{3}{1 + K + \dots + K}

So, we choose the number ca using the formula

\begin{matrix} (34) & ε_{1} - min (\frac{ε}{1 + K + \dots + K}, \frac{λ}{1 + K + \dots + K}) \end{matrix}

lo nombro

c_{1}

building ainai prevish, the coordinate point

(x_{1}, y^{(n)})

Or

3 = 0, 1, \dots, v - 1

is located in the rectangle

D

Indeed, by proceeding as above, we have

| y^{(j)} | < | k^{(k)} | + b - δ < (1 + K + \dots + K^{'}) c_{1} + b - b

or enooro

| y^{(0)} | < (1 + K + \dots + K^{'}) ε_{1} + b - δ < b .

Tranat accounts for inequalities and identity

! g (x) - y^{(n)} (x) \leq e . y^{(n)} (x_{1}) - y^{(n)} ∣< 0

(38)

η (x_{1}) - ξ^{''}! < 2 ε .

pine priser pue ic.
9. Dansua auter uravil [3] nons avons montró que si la fonction

f (x, y)

a des deviver partiedes par rapport ì

x

and to the left, of order greater than two, continuous in the rectangle D. It is practical to choose the network I in another way, using another quadrature formula. We have already treated in detail the integration and differential equation (I), using the method of mechanical approximations of the quadrature formula of K. Petr. [4,6]. We have also treated the quadrature formula of K. Petr. [4,6].

Buntim eax partial didvies of the secend exdre of hyperbelic type

IR. Now aroa makes an extension of the previous method of numerical integration of differential equations (1), to partial differential equations of the second order of hyperbolic type [2]. This work will soon appear in Marbrastica Toure ? (5). We have first established by an extension of the method of J. Radon [6]. the cubature formula
(36)

\iint_{D} f (x, y) d z d y = \frac{(x_{1} - x_{1}) (y_{1} - y_{1})}{2} U (x_{1}, y_{1}) + f (y_{2}, y_{0})] + R

,
od

D

ent the rectangle defined by the inequalities

x_{1} ⩽ x ⩽ x_{2}, y_{1} ⩽ y ⩽ y_{2}

and where the revte

R

is done by the formula

\begin{matrix} (37) & R - \iint_{D} (ρ \frac{\partial^{2} f}{\partial x^{2}} + ψ \frac{\partial^{2} f}{\partial x \partial y} + θ \frac{\partial^{2} f}{\partial y^{2}}) d x d y \end{matrix}

arec

\begin{aligned} φ (x, y) - \frac{1}{2} (x - x_{1}) (x - x_{1}), \\ \dot{ψ} (x, y) - - \frac{1}{2} [(x - x_{1}) (y - y_{2}) + (x - x_{2}) (y - y_{1}) + (x_{3} - x_{1}) (y_{2} - y_{1})), \\ 0 (x, y) - \frac{1}{2} (y - y_{1}) (y - y_{1}) . \end{aligned}

We now apply the mucoxiver approximation method and the cubature formula (3si) to the numerical integration of the equation anx derivative parlinles

\frac{\partial^{a} s}{\partial x \partial y} - f (x, y, z, p, q) .

Or

p - \frac{\partial z}{\partial z} \cdot q \dots \frac{\partial z}{\partial y}

, with the conditions

z (x, 0) - 0, z (0, y) - 0

, in the rectangle

Δ

formed by your rights

x - 0, x - x, y - 0, y - μ

We have shown that we can determine us rewan T tormé by the lines

x - x_{1}, y - y_{2}

where the points

x_{1}

And

y_{2}

share the intervals (

0, λ

) And (

0, μ

) in not m equal parts and seek a calculation algorithm for the numbers: Sy, P&. (') such that c being a positive number doan6, the values sbolues of the difference