CUBATURE FORMULAS. APPLICATION TO THE NUMERICAL INTEGRATION OF SECOND-ORDER HYPERBOLIC PARTIAL DIFFERENTIAL EQUATIONS

In the first part of this work, we study the cubature formulas

	$\displaystyle\iint_{D}fdxdy=\frac{S}{4}\left[f\left(x_{1},y_{1}\right)+f\left(x_{1},y_{2}\right)+f\left(x_{2},y_{1}\right)+f\left(x_{2},y_{2}\right)\right]+R$		(1)
	$\displaystyle\iint_{D}fdxdy=Sf\left(x_{0},y_{0}\right)+R$		(2)

Or $D$ is the rectangle formed by the lines $x=x_{1},x=x_{2},y=y_{1},y=y_{2}$ , $\left(x_{0},y_{0}\right)$ are the coordinates of the center of the rectangle $D$ And $S$ is the area of the rectangle $D$ We focus primarily on determining the remainder of these formulas, which we put in the form

R=\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}}+\psi\frac{\partial^{2}f}{\partial

(3)

assuming that the function $f(x,y)$ has partial derivatives with respect to $x$ and to $y$ , of the first and second order continuous in $D$ .

We also determine the cubature formula

\iint_{D}fdxdy=\frac{S}{2}\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)\right]+R

(4)

and we put its remainder in the form (3). We will apply this formula in the second part of this work to the numerical integration of the second-order hyperbolic partial differential equation.

We obtained the cubature formulas (1), (2), (4) by extending the method given by J. Radon [1] for obtaining quadrature formulas, which we applied systematically [2]. We are continuing to develop this method and will publish further work on cubature formulas elsewhere.

In the second part of our work, we deal, as an application of the first part, with the numerical integration of the partial differential equation.

\frac{\partial^{2}z}{\partial x\partial y}=f(x,y,z,p,q)

(5)

Or $p=\frac{\partial z}{\partial x},q=\frac{\partial z}{\partial y}$ which admits a unique integral $z(x,y)$ in the rectangle $\Delta$ formed by the straight lines $x=0,x=\lambda,y=0,y=\mu$ and which is zero on the sides of this rectangle, located on the axes $Ox,Oy$ .

The problem we will address is the following: determining a network $\Gamma$ formed by the straight lines $x=x_{i},y=y_{k}$ where the points $x_{i}$ And $y_{k}$ share the intervals $(0, λ)$ And $(0,\mu)$ in $n$ And $m$ equal parts and find a calculation algorithm for the numbers $z_{ik}^{(v)},p_{ik}^{(v)},q_{ik}^{(v)}$ so that $ε$ given a positive number, the absolute values of the differences

z\left(x_{i},y_{k}\right)-z_{ik}^{(v)},p\left(x_{i},y_{k}\right)-p_{ik}^{(v)},q\left(x_{i},y_{k}\right)-q_{ik}^{(v)}

on the network nodes $\Gamma$ , are smaller than $2ε$
The results of this work were communicated at the Colloquium on the Theory of Partial Differential Equations held in Bucharest from September 21-26, 1959 .

I. CUBATURAL FORMULAS

§. First cubature formula

1.

Let's consider the rectangle $D$ defined by inequalities

x_{1}\leqslant x\leqslant x_{2},y_{1}\leqslant y\leqslant y_{2}

(1)

and a function $f(x,y)$ having first and second order partial derivatives in $D$ We are looking for a cubature formula for the double integral

\iint_{D}f(x,y)dxdy

(2)

relative to the rectangle $D$ . To this end, we will extend the method given by J. radon [1], and which we have applied in numerous works [2].

Let us designate by $\varphi(x,y),\psi(x,y),\theta(x,y)$ integrals of partial differential equations

\frac{\partial^{2}\varphi}{\partial x^{2}}=\alpha,\quad\frac{\partial^{2}\psi}{\partial x\partial y}=\beta,\quad\frac{\partial^{2}\theta}{\partial y^{2}}=\gamma,

(3)

defined in the rectangle $D$ and which satisfy boundary conditions that will be specified later. In the second members $\alpha,\beta,\gamma$ are constants that will be determined.

We are going to transform the integrals

\iint_{D}\frac{partial^{2}\varphi}{\partial

(4)

through suitable part-by-part integration.
$1^{\circ}$ We can write

\iint_{D}\frac{\partial^{2}\varphi}{\partial x^{2}}fdxdy=\int_{y_{1}}^{y_{2}}dy\int_{x_{1}}^{x_{2}}\frac{\partial}{\partial x}\right)dx+\iint_{D}\varphi\frac{\partial^{2}f}{\partial x^{2}}dxdy

and we will have

$\displaystyle\iint_{D}\frac{\partial^{2}\varphi}{\partial x^{2}}fdxdy$	$\displaystyle=\int_{y_{1}}^{y_{2}}\left[\frac{\partial\varphi}{\partial x}\left(x_{2},y\right)f\left(x_{2},y\right)-\varphi\left(x_{2},y\right)\frac{\partial f}{\partial x}\left(x_{2},y\right)\right]dy-$	(5)
	$\displaystyle-\int_{y_{1}}^{y_{2}}\left[\frac{\partial\varphi}{\partial x}\left(x_{1},y\right)f\left(x_{1},y\right)-\varphi\left(x_{1},y\right)\frac{\partial f}{\partial x}\left(x_{1},y\right)\right]dy+$
	$\displaystyle+\iint_{D}\varphi\frac{\partial^{2}f}{\partial x^{2}}dxdy$

$2^{\circ}$ We will also have

\iint_{D}\frac{\partial^{2}\theta}{\partial y^{2}}fdxdy=\int_{x_{1}}^{x_{2}}dx\int_{y_{1}}^{y_{2}}\frac{\partial}{\partial y}\left(\frac{\partial\theta}{\partial y}f-\theta\frac{\partial f}{\partial y}\right)dy+\iint_{D}\theta\frac{\partial^{2}f}{\partial y^{2}}dxdy

that's to say

$\displaystyle\iint_{D}\frac{\partial^{2}\theta}{\partial y^{2}}fdxdy$	$\displaystyle=\int_{x_{1}}^{x_{2}}\left[\frac{\partial^{\theta}}{\partial y}\left(x,y_{2}\right)f\left(x,y_{2}\right)-\theta\left(x,y_{2}\right)\frac{\partial f}{\partial y}\left(x,y_{2}\right)\right]dx-$	(6)
	$\displaystyle-\int_{x_{1}}^{x_{2}}\left[\frac{\partial\theta}{\partial y}\left(x,y_{1}\right)f\left(x,y_{1}\right)-\theta\left(x,y_{1}\right)\frac{\partial f}{\partial y}\left(x,y_{1}\right)\right]dx+$
	$\displaystyle+\iint_{D}\theta\frac{\partial^{2}f}{\partial y^{2}}dxdy$

$3^{\circ}$ We can write

	$\displaystyle 2\iint_{D}\frac{\partial^{2}\psi}{\partial x\partial y}fdxdy$	$\displaystyle=\int_{x_{1}}^{x_{2}}dx\int_{y_{1}}^{y_{2}}\frac{\partial}{\partial y}\left[\frac{\partial\psi}{\partial x}f-\psi\frac{\partial f}{\partial x}\right)dy+$
		$\displaystyle+\int_{y_{1}}^{y_{2}}dy\int_{x_{1}}^{x_{2}}\frac{\partial}{\partial x}\left(\frac{\partial\psi}{\partial y}f-\psi\frac{\partial f}{\partial y}\right)dx+2\iint_{\\|}\psi\frac{\partial^{2}f}{\partial x\partial y}dxdy$

that's to say

But

	$\displaystyle 2\iint_{D}\frac{\partial^{2}\psi}{\partial x\partial y}fdxdy$	$\displaystyle=\int_{x_{1}}^{x_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{2}\right)f\left(x,y_{2}\right)dx-\int_{x_{1}}^{x_{2}}\psi\left(x,y_{2}\right)\frac{\partial f}{\partial x}\left(x,y_{2}\right)dx-$
		$\displaystyle-\int_{x_{1}}^{x_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{1}\right)f\left(x,y_{1}\right)dx+\int_{x_{1}}^{x_{2}}\psi\left(x,y_{1}\right)\frac{\partial f}{\partial x}\left(x,y_{1}\right)dx+$
		$\displaystyle+\int_{y_{1}}^{y_{2}}\frac{\partial\psi}{\partial y}\left(x_{2},y\right)f\left(x_{2},y\right)dy-\int_{y_{1}}^{y_{2}}\psi\left(x_{2},y\right)\frac{\partial f}{\partial y}\left(x_{2},y\right)dy-$
		$\displaystyle-\int_{y_{1}}^{\frac{\partial\psi}{\partial y}}\left(x_{1},y\right)f\left(x_{1},y\right)dy+\int_{y_{1}}^{y_{2}}\psi\left(x_{1},y\right)\frac{\partial f}{\partial y^{\prime}}\left(x_{1},y\right)dy+$
		$\displaystyle+2\iint_{D}\psi\frac{\partial^{2}f}{\partial x\partial y}dxdy$

	$\displaystyle\int_{x_{1}}^{x_{2}}\psi\left(x,y_{2}\right)\frac{\partial f}{\partial x}\left(x,y_{2}\right)dx$	$\displaystyle=\psi\left(x_{2},y_{2}\right)f\left(x_{2},y_{2}\right)-\psi\left(x_{1},y_{2}\right)f\left(x_{1},y_{2}\right)-$
	$\displaystyle-$	$\displaystyle\int_{x_{1}}^{x_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{2}\right)f\left(x,y_{2}\right)dx$
	$\displaystyle\int_{x_{1}}^{x_{2}}\psi\left(x,y_{1}\right)\frac{\partial f}{\partial x}f\left(x,y_{1}\right)dx$	$\displaystyle=\psi\left(x_{2},y_{1}\right)f\left(x_{2},y_{1}\right)-\psi\left(x_{1},y_{1}\right)f\left(x_{1},y_{1}\right)-$
	$\displaystyle-$	$\displaystyle\int_{x_{1}}^{x_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{1}\right)f\left(x,y_{1}\right)dx$

\begin{gathered}\int_{y_{1}}^{y_{2}}\psi\left(x_{2},y\right)\frac{\partial f}{\partial y}\left(x_{2},y\right)dy=\psi\left(x_{2},y_{2}\right)f\left(x_{2},y_{2}\right)-\psi\left(x_{2},y_{1}\right)f\left(x_{2},y_{1}\right)-\\ -\int_{y_{1}}^{y_{2}}\frac{\partial\psi}{\partial y}\left(x_{2},y\right)f\left(x_{2},y\right)dy\\ \int_{y_{1}}^{y_{2}}\psi\left(x_{1},y\right)\frac{\partial f}{\partial y}\left(x_{1},y\right)dy=\psi\left(x_{1},y_{2}\right)f\left(x_{1},y_{2}\right)-\psi\left(x_{1},y_{1}\right)f\left(x_{1},y_{1}\right)-\\ -\int_{y_{1}}^{y_{2}}\frac{\partial\psi}{\partial y}\left(x_{1},y\right)f\left(x_{1},y\right)dy\end{gathered}

Taking these formulas into account, we will have

$\displaystyle\iint_{D}\frac{\partial^{2}\psi}{\partial x\partial y}fdxdy$	$\displaystyle=-\psi\left(x_{2},y_{2}\right)f\left(x_{2},y_{2}\right)+\psi\left(x_{1},y_{2}\right)f\left(x_{1},y_{2}\right)+$	(7)
	$\displaystyle+\psi\left(x_{2},y_{1}\right)f\left(x_{2},y_{1}\right)-\psi\left(x_{1},y_{1}\right)f\left(x_{1},y_{1}\right)+$
$\displaystyle+$	$\displaystyle\int_{x_{1}}^{x_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{2}\right)f\left(x,y_{2}\right)dx-\int_{x_{1}}^{y_{2}}\frac{\partial\psi}{\partial x}\left(x,y_{1}\right)f\left(x,y_{1}\right)dx+$
$\displaystyle+$	$\displaystyle\int_{y_{1}}^{y_{2}}\frac{\partial\psi}{\partial y}\left(x_{2},y\right)f\left(x_{2},y\right)dy-\int_{y_{1}}^{y_{2}}\frac{\partial\psi}{\partial y}\left(x_{1},y\right)f\left(x_{1},y\right)dy+$
$\displaystyle+$	$\displaystyle\iint_{D}\psi\frac{\partial^{2}f}{\partial x\partial y}dxdy$

The functions $\varphi(x,y),\psi(x,y),\theta(x,y)$ being the integrals of the partial differential equations (3), by adding formulas (5), (6) and (7) term by term we will have the formula

$\displaystyle(\alpha+\beta+\gamma)\iint_{D}fdxdy$	$\displaystyle=-\psi\left(x_{2},y_{2}\right)f\left(x_{2},y_{2}\right)+\psi\left(x_{1},y_{2}\right)f\left(x_{1},y_{2}\right)+$	(8)
	$\displaystyle+\psi\left(x_{2},y_{1}\right)f\left(x_{2},y_{1}\right)-\psi\left(x_{1},y_{1}\right)f\left(x_{1},y_{1}\right)-$
	$\displaystyle-\int_{y_{1}}^{y_{2}}\varphi\left(x_{2},y\right)\frac{\partial f}{\partial x}\left(x_{2},y\right)dy+\int_{y_{1}}^{x_{2}}\varphi\left(x_{1},y\right)\frac{\partial f}{\partial x}\left(x_{1},y\right)dy-$
	$\displaystyle-\int_{x_{1}}^{x_{2}}\theta\left(x,y_{2}\right)\frac{\partial f}{\partial y}\left(x,y_{2}\right)dx+\int_{x_{1}}^{x_{2}}\theta\left(x,y_{1}\right)\frac{\partial f}{\partial y}\left(x,y_{1}\right)dx+$

\begin{gathered}+\int_{x_{1}}^{x_{2}}\left[\frac{\partial\psi}{\partial x}\left(x,y_{2}\right)+\frac{\partial\theta}{\partial y}\left(x,y_{2}\right)\right]f\left(x,y_{2}\right)dx-\int_{x_{1}}^{x_{2}}\left[\frac{\partial\varphi}{\partial x}\left(x,y_{1}\right)+\frac{\partial\theta}{\partial y}\left(x,y_{1}\right)\right]f\left(x,y_{1}\right)dx+\\ +\int_{y_{1}}^{y_{2}}\left[\frac{\partial\psi}{\partial y}\left(x_{2},y\right)+\frac{\partial\varphi}{\partial x}\left(x_{2},y\right)\right]f\left(x_{2},y\right)dy-\int_{y_{1}}^{y_{2}}\left[\frac{\partial\psi}{\partial y}\left(x_{1},y\right)+\frac{\partial\varphi}{\partial x}\left(x_{1},y\right)\right]f\left(x_{1},y\right)dy+\\ +\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy.\end{gathered}

Formula (8) reduces to a cubature formula with its remainder if the functions $\varphi(x,y),\psi(x,y),\theta(x,y)$ satisfy the following boundary conditions

$\displaystyle\varphi\left(x_{2},y\right)=0,$	$\displaystyle\varphi\left(x_{1},y\right)=0$
$\displaystyle\theta\left(x,y_{2}\right)=0,$	$\displaystyle\theta\left(x,y_{1}\right)=0$
$\displaystyle\left.\frac{\partial\psi}{\partial x}\left(x,y_{2}\right)+\frac{\partial 0}{\partial y}x,y_{2}\right)=0,$	$\displaystyle\frac{\partial\psi}{\partial x}\left(x,y_{1}\right)+\frac{\partial\theta}{\partial y}\left(x,y_{1}\right)=0$	(9)
$\displaystyle\frac{\partial\psi}{\partial y}\left(x_{2},y\right)+\frac{\partial\varphi}{\partial x}\left(x_{2},y\right)=0,$	$\displaystyle\frac{\partial\psi}{\partial y}\left(x_{1},y\right)+\frac{\partial\varphi}{\partial x}\left(x_{1},y\right)=0.$

If it is possible to integrate the partial differential equations (3) with the boundary conditions (9), we are led to the cubature formula

	$\displaystyle(\alpha+\beta+\gamma)\iint_{D}fdxdy=$	$\displaystyle-\psi\left(x_{2},y_{2}\right)f\left(x_{2},y_{2}\right)+\psi\left(x_{1},y_{2}\right)f\left(x_{1},y_{2}\right)+$		(10)
		$\displaystyle+\psi\left(x_{2},y_{1}\right)f\left(x_{2},y_{1}\right)-\psi\left(x_{1},y_{1}\right)f\left(x_{1},y_{1}\right)+R$

with its remainder

R=\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy

(11)

The search for the cubature formula (10) is thus reduced to the integration of the partial differential equations (3), with the boundary conditions (9). The numbers $\alpha,\beta,\gamma$ will be determined in such a way that ^this problem is possible.
2. Let's determine the functions $\varphi(x,y),\psi(x,y),\theta(x,y)$ The first equation (3) shows us that

\varphi(x,y)=\frac{\alpha}{2}x^{2}+A_{1}(y)x+A_{2}(y)

and the conditions $\varphi\left(x_{1},y\right)=0,\varphi\left(x_{2},y\right)=0$ determine the functions $A_{1}(y)$ And $A_{2}(y)$ we will have

\varphi(x,y)=\frac{\alpha}{2}\left(x-x_{1}\right)\left(x-x_{2}\right)

(12)

Similarly, we will have

\theta(x,y)=\frac{Y}{2}\left(y-y_{1}\right)\left(y-y_{2}\right).

(13)

The other boundary conditions (9) become

\begin{array}[]{cc}\frac{\partial\psi}{\partial x}\left(x,y_{1}\right)=\frac{\gamma}{2}\left(y_{2}-y_{1}\right);&\frac{\partial\psi}{\partial y}\left(x_{1},y\right)=\frac{\alpha}{2}\left(x_{2}-x_{1}\right)\\ \frac{\partial\psi}{\partial x}\left(x,y_{2}\right)=-\frac{\gamma}{2}\left(y_{2}-y_{1}\right);&\frac{\partial\psi}{\partial y}\left(x_{2},y\right)=-\frac{\alpha}{2}\left(x_{2}-x_{1}\right).\end{array}

The second equation (3) gives us

\psi(x,y)=\beta xy+\psi_{1}(x)+\psi_{2}(y)

Or $\psi_{1}(x)$ And $\psi_{2}(y)$ are functions to be determined. By stating that conditions (14) are satisfied, we will have the equations

		$\displaystyle\beta y_{1}+\psi_{1}^{\prime}(x)=\frac{\gamma}{2}\left(y_{2}-y_{1}\right);\quad\beta x_{1}+\psi_{2}^{\prime}(y)=\frac{\alpha}{2}\left(x_{2}-x_{1}\right)$
		$\displaystyle\beta y_{2}+\psi_{1}^{\prime}(x)=-\frac{\gamma}{2}\left(y_{2}-y_{1}\right);\quad\beta x_{2}+\psi_{2}^{\prime}(y)=-\frac{\alpha}{2}\left(x_{2}-x_{1}\right),$

who will give us

		$\displaystyle\psi_{1}^{\prime}(x)=\frac{\gamma}{2}y_{2}-\left(\frac{\gamma}{2}+\beta\right)y_{1}=\frac{\gamma}{2}y_{1}-\left(\frac{\gamma}{2}+\beta\right)y_{2}$
		$\displaystyle\psi_{2}^{\prime}(y)=\frac{\alpha}{2}x_{2}-\left(\frac{\alpha}{2}+\beta\right)x_{1}=\frac{\alpha}{2}x_{1}-\left(\frac{\alpha}{2}+\beta\right)x_{2}.$

To make these equations possible, we choose the constants $\alpha,\beta,\gamma$ such as

\frac{\gamma}{2}=-\left(\frac{\gamma}{2}+\beta\right),\quad\frac{\alpha}{2}=-\left(\frac{\alpha}{2}+\beta\right),

that's to say

\alpha=-\beta,\quad\gamma=-\beta.

To determine completely $\alpha,\beta,\gamma$ let's add the equation

\alpha+\beta+\gamma=1

and we will have

\alpha=1,\quad\beta=-1,\quad\gamma=1,

(15)

from which it follows that

\psi_{1}^{\prime}(x)=\frac{1}{2}\left(y_{1}+y_{2}\right),\quad\psi_{2}^{\prime}(y)=\frac{1}{2}\left(x_{1}+x_{2}\right)

and consequently

\psi_{1}(x)+\psi_{2}(y)=\frac{\left(y_{1}+y_{2}\right)x+\left(x_{1}+x_{2}\right)y}{2}+\text{ const. }

So we will finally have

\psi(x,y)=-\frac{1}{2}\left[\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)\right]+C

(16)

Or $C$ is an arbitrary constant.
3. Functions $\varphi(x,y),\psi(x,y),\theta(x,y)$ being determined by formulas (12), (13), (16), let us return to the cubature formula (10). We have

\begin{array}[]{ll}\psi\left(x_{1},y_{1}\right)=C,&\psi\left(x_{1},y_{2}\right)=C+\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\\ \psi\left(x_{2},y_{2}\right)=C,&\psi\left(x_{2},y_{1}\right)=C+\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\end{array}

and formula (10) becomes

	$\displaystyle\iint_{U}fdxdy=-C\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)\right]+$		(17)
	$\displaystyle+\left[C+\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\right]\left[f\left(x_{1},y_{2}\right)+f\left(x_{2},y_{1}\right)\right]+R$

where the rest $R$ is given by the formula

R=\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial f^{2}}{\partial y^{2}}\right)dxdy

(18)

In formula (17) the coefficient of the arbitrary constant $C$ East

\iint_{D}\frac{\partial^{2}f}{\partial x\partial y}dxdy-\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)-f\left(x_{1},y_{2}\right)-f\left(x_{2},y_{1}\right)\right]

and it is zero, so this formula reduces to the cubature formula
with

	$\displaystyle\iint_{D}fdxdy=\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\left[f\left(x_{1},y_{2}\right)+f\left(x_{2},y_{1}\right)\right]+$			(19)
		$\displaystyle+\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy$

	$\displaystyle\varphi(x,y)=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right)$
	$\displaystyle\psi(x,y)=-\frac{1}{2}\left[\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)\right]$		(20)
	$\displaystyle\theta(x,y)=\frac{1}{2}\left(y-y_{1}\right)\left(y-y_{2}\right)$

But we can choose, in formula (17), the constant $C$ , so that this formula has other forms.

For example, to $C=-\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}$ we have the cubature formula

	$\displaystyle\iint_{D}fdxdy=\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)\right]+$		(21)
	$\displaystyle+\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y^{\prime}}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy$

with

		$\displaystyle\varphi(x,y)=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right)$
		$\displaystyle\psi(x,y)=-\frac{1}{2}\left[\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)+\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)\right]$
		$\displaystyle\theta(x,y)=\frac{1}{2}\left(y-y_{1}\right)\left(y-y_{2}\right)$

Similarly if $C=-\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{4}$ we have the cubature formula

	$\displaystyle\iint_{U}fdxdy=$	$\displaystyle\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{4}\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)+f\left(x_{1},y_{2}\right)+f\left(x_{2},y_{1}\right)\right]+$		(23)
		$\displaystyle+\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy$

with

\varphi(x,y)=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right)

	$\displaystyle\psi(x,y)=-\frac{1}{2}\left[\left[\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)+\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\right]\right.$		(24)
	$\displaystyle\theta(x,y)=\frac{1}{2}\left(y-y_{1}\right)\left(y-y_{2}\right)$

4.

The sign of the functions can be discussed. $\varphi(x,y),\psi(x,y),\theta(x,y)$ in the cubature formulas (19), (21) and (23) in $D$ .

In formula (19) we have $\varphi(x,y)<0,\theta(x,y)<0,\psi(x,y)>0$ In $D$ .

However, in formula (21), we have $\varphi(x,y)<0,\theta(x,y)<0$ , $\psi(x,y)<0$ In $D$ Indeed, if we make the change of variables in the second formula (22)

x=\frac{x_{1}+x_{2}}{2}+\xi,y=\frac{y_{1}+y_{2}}{2}+\eta,\lambda=\frac{x_{2}-x_{1}}{2},\quad\mu=\frac{y_{2}-y_{1}}{2}

we will have
we will have

\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)+\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)=2\left(\xi_{\eta}+\lambda_{\mu}\right)

and in the rectangle $D$ , We have $|\xi|<\lambda,|\eta|<\mu$ , so that the number $\xi\eta+\lambda\mu$ is positive and consequently $\psi(x,y)<0$ In $D$ .

In the cubature formula (23), the function $\psi(x,y)$ changes sign in the rectangle $D$ Indeed, according to the previous calculations, we have

\left(x-x_{1}\right)\left(y-y_{2}\right)+\left(x-x_{2}\right)\left(y-y_{1}\right)+\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}=2\xi_{\eta}

and consequently the function $\psi(x,y)$ changes sign in the rectangle $D$ 5.
In the second chapter, we will make an important application of the cubature formula (21). For this formula, we will give an evaluation of $|R|$ .

We first note that the functions $\varphi(x,y),\psi(x,y),\theta(x,y)$ of this formula being negative in $D$ , we can use the mean value theorem and we will have

R=\frac{\partial^{2}f}{\partial x^{2}}\left(P_{1}\right)\iint_{D}\varphi dxdy+\frac{\partial^{2}f}{\partial x\partial y}\left(P_{2}\right)\iint_{D}\psi dxdy+\frac{\partial^{2}f}{\partial y^{2}}\left(P_{3}\right)\iint_{D}\theta dxdy

Or $P_{1},P_{2},P_{3}$ are certain points of the rectangle $R$ .
We have

\begin{gathered}\iint_{D}\varphi dxdy=-\frac{\left(x_{2}-x_{1}\right)^{3}\left(y_{2}-y_{1}\right)}{12},\iint_{D}\psi dxdy=-\frac{\left(x_{2}-x_{1}\right)^{2}\left(y_{2}-y_{1}\right)^{2}}{4}\\ \iint_{D}\theta dxdy=-\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)^{3}}{12}\end{gathered}

and consequently the previous formula becomes
$R=-\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{12}\left[\left(x_{2}-x_{1}\right)^{2}\frac{\partial^{2}f}{\partial x^{2}}\left(P_{1}\right)+3\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)\frac{\partial^{2}f}{\partial x\partial y}\left(P_{2}\right)+\right.$

\left.+\left(y_{2}-y_{1}\right)^{2}\frac{\partial^{2}f}{\partial y^{2}}\left(P_{3}\right)\right]

(25)

If we designate by $M_{2}$ an upper limit of $\left|\frac{\partial^{2}f}{\partial x^{2}}\right|,\left|\frac{\partial^{2}f}{\partial x\partial y^{2}}\right|,\left|\frac{\partial^{2}f}{\partial y^{2}}\right|$ , In $D$ , we will have the following evaluation of the absolute value of the remainder $R$ of the cubature formula (21).

|R|\leqslant\frac{S}{12}\left[\left(x_{2}-x_{1}\right)^{2}+3\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)+\left(y_{2}-y_{1}\right)^{2}\right]M_{2}

(26)

Or $S$ is the area of the rectangle $D$ .

§ 2. Second cubature formula

(1)
6. Consider the rectangle $D$ defined by inequalities

x_{0}-h\leqslant x\leqslant x_{0}+h\quad,\quad y_{0}-k\leqslant y\leqslant y_{0}+k

We will determine the remainder of the cubature formula
(2)

\iint_{D}f(x,y)dxdy=4hkf\left(x_{0},y_{0}\right)+R

assuming that the function $f(x,y)$ has continuous first and second order partial derivatives in $D$ .

Let us designate by $D_{1},D_{2},D_{3},D_{4}$ , the rectangles defined by the inequalities
(3)
$\left(D_{1}\right)\quad x_{0}\leqslant x\leqslant x_{0}+h$ ,
$y_{0}\leqslant y\leqslant y_{0}+k$
$\left(D_{2}\right)\quad x_{0}-h\leqslant x\leqslant x_{0}\quad$ ,
$y_{0}\leqslant y\leqslant y_{0}+k$
$\left(D_{3}\right)\quad x_{0}-h\leqslant x\leqslant x_{0}$
$y_{0}-k\leqslant y\leqslant y_{0}$
$\left(D_{4}\right)\quad x_{0}\leqslant x\leqslant x_{0}+h$ ,
$y_{0}-k\leqslant y\leqslant y_{0}$ .

To these rectangles we attach the following functions and numbers

\begin{array}[]{llllll}\varphi_{1}(x,y),&\psi_{1}(x,y),&\theta_{1}(x,y);&\alpha_{1},&\beta_{1},&\gamma_{1}\\ \varphi_{2}(x,y),&\psi_{2}(x,y),&\theta_{2}(x,y);&\alpha_{2},&\beta_{2},&\gamma_{2}\\ \varphi_{3}(x,y),&\psi_{3}(x,y),&\theta_{3}(x,y);&\alpha_{3}.&\beta_{3},&\gamma_{3}\\ \varphi_{4}(x,y),&\psi_{4}(x,y),&\theta_{4}(x,y);&\alpha_{4},&\beta_{4},&\gamma_{4},\end{array}

such as

\begin{array}[]{lll}\frac{\partial^{2}\varphi_{1}}{\partial x^{2}}=\alpha_{1},&\frac{\partial^{2}\psi_{1}}{\partial x\partial y}=\beta_{1},&\frac{\partial^{2}\theta_{1}}{\partial y^{2}}=\gamma_{1}\\ \frac{\partial^{2}\varphi_{2}}{\partial x^{2}}=\alpha_{2},&\frac{\partial^{2}\psi_{2}}{\partial x\partial y}=\beta_{2},&\frac{\partial^{2}\theta_{2}}{\partial y^{2}}=\gamma_{2}\\ \frac{\partial^{2}\varphi_{3}}{\partial x^{2}}=\alpha_{3},&\frac{\partial^{2}\psi_{3}}{\partial x\partial y}=\beta_{3},&\frac{\partial^{2}\theta_{3}}{\partial y^{2}}=\gamma_{3}\\ \frac{\partial^{2}\varphi_{4}}{\partial x^{2}}=\alpha_{1},&\frac{\partial^{2}\psi_{4}}{\partial x\partial y}=\beta_{4},&\frac{\partial^{2}\theta_{4}}{\partial y^{2}}=\gamma_{4}\end{array}

Each rectangle $D_{1},D_{2},D_{3},D_{4}$ We apply formula (8) from § 1. We will have
(5)
$\left(\alpha_{1}+\beta_{1}+\gamma_{1}\right)\iint_{D_{1}}f(x,y)dxdy=-\psi_{1}\left(x_{0}+h,y_{0}+k\right)f\left(x_{0}+h,y_{0}+k\right)++\psi_{1}\left(x_{0},y_{0}+k\right)f\left(x_{0},y_{0}+k\right)+\psi_{1}\left(x_{0}+h,y_{0}\right)f\left(x_{0}+h,y_{0}\right)-\psi_{1}\left(x_{0},y_{0}\right)f\left(x_{0},y_{0}\right)-$

		$\displaystyle-\int_{y_{0}}^{y_{0}+k}\varphi_{1}\left(x_{0}+h,y\right)\frac{\partial f}{\partial x}\left(x_{0}+h,y\right)dy+\int_{y_{0}}^{y_{0}+k}\varphi_{1}\left(x_{0},y\right)\frac{\partial f}{\partial x}\left(x_{0},y\right)dy-$
		$\displaystyle-\int_{x_{0}}^{x_{0}+h}\theta_{1}\left(x,y_{0}+k\right)\frac{\partial f}{\partial x}\left(x,y_{0}+k\right)dx+\int_{x_{0}}^{x_{0}+h}\theta_{1}\left(x,y_{0}\right)\frac{\partial f}{\partial y}\left(x,y_{0}\right)dx+$

(6)

\begin{gathered}+\int_{x_{0}}^{x_{0}+h}\left[\frac{\partial\psi_{1}}{\partial x}\left(x,y_{0}+k\right)+\frac{\partial\theta_{1}}{\partial y}\left(x,y_{0}+k\right)\right]f\left(x,y_{0}+k\right)dx-\\ -\int_{x_{9}}^{x_{0}+h}\left[\frac{\partial\psi_{1}}{\partial x}\left(x,y_{0}\right)+\frac{\partial\theta_{1}}{\partial y}\left(x,y_{0}\right)\right]f\left(x,y_{0}\right)dx+\\ +\int_{y_{0}}^{y_{0}+k}\left[\frac{\partial\psi_{1}}{\partial y}\left(x_{0}+h,y\right)+\frac{\partial\varphi_{1}}{\partial x}\left(x_{0}+h,y\right)\right]f\left(x_{0}+h,y\right)dy-\\ -\int_{y_{0}}^{y_{0}+k}\left[\frac{\partial\psi_{1}}{\partial y}\left(x_{0},y\right)+\frac{\partial\varphi_{1}}{\partial x}\left(x_{0},y\right)\right]f\left(x_{0},y\right)dy+\\ +\iint_{D_{1}}\left(\varphi_{1}\frac{\partial^{2}f}{\partial x^{2}}+\psi_{1}\frac{\partial^{2}f}{\partial x\partial y}+\theta_{1}\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy\\ \left(\alpha_{2}+\beta_{2}+\gamma_{2}\right)\int_{D_{2}}f(x,y)=-\psi_{2}\left(x_{0},y_{0}+k\right)f\left(x_{0},y_{0}+k\right)+\\ +\psi_{2}\left(x_{0}-h,y_{0}+k\right)f\left(x_{0}-h,y_{0}+k\right)+\\ +\psi_{2}\left(x_{0},y_{0}\right)f\left(x_{0},y_{0}\right)-\psi_{2}\left(x_{0}-h,y_{0}\right)f\left(x_{0}-h,y_{0}\right)-\\ -\int_{y_{0}}^{y_{0}+k}\varphi_{2}\left(x_{0},y\right)\frac{\partial f}{\partial x}\left(x_{0},y\right)dy+\int_{y_{0}}^{x_{0}+k}\varphi_{2}\left(x_{0}-h,y\right)\frac{\partial f}{\partial x}\left(x_{0}-h,y\right)dy-\\ -\int_{x_{0}-h}^{x_{0}}\theta_{2}\left(x,y_{0}+k\right)\frac{\partial f_{2}}{\partial y}\left(x,y_{0}+k\right)dx+\int_{x_{0}-k}^{x_{0}}\theta_{2}\left(x,y_{0}\right)\frac{\partial f}{\partial y}\left(x,y_{0}\right)dx+\\ +\int_{x_{0}-h}\left[\frac{\partial\psi_{2}}{\partial x}\left(x,y_{0}+k\right)+\frac{\partial\theta_{2}}{\partial y}\left(x,y_{0}+k\right)\right]f\left(x,y_{0}+k\right)dx-\\ -\int_{x_{0}-h}^{x_{0}}\left[\frac{\partial\psi_{2}}{\partial x}\left(x,y_{0}\right)+\frac{\partial\theta_{2}}{\partial y}\left(x,y_{0}\right)\right]f\left(x,y_{0}\right)dx+\\ +\int_{y_{0}+k}\left[\frac{\partial\psi_{2}}{\partial x}\left(x_{0},y\right)+\frac{\partial\varphi_{2}}{\partial x}\left(x_{0},y\right)\right]f\left(x_{0},y\right)dy-\\ -\int_{y_{0}}^{y_{0}+k}\left[\frac{\partial\psi_{2}}{\partial y}\left(x_{0}-h,y\right)+\frac{\partial\varphi_{2}}{\partial x}\left(x_{0}-h,y\right)\right]f\left(x_{0}-h,y\right)dy+\end{gathered}

\begin{gathered}+\int_{x_{0}}^{x_{0}+h}\left[\frac{\partial\psi_{4}}{\partial x}\left(x,y_{0}\right)+\frac{\partial\theta_{4}}{\partial y}\left(x,y_{0}\right)\right]f\left(x,y_{0}\right)dx-\\ -\int_{y_{0}}^{x_{0}+h}\left[\frac{\partial\psi_{4}}{\partial x}\left(x,y_{0}-k\right)+\frac{\partial\theta_{4}}{\partial y}\left(x,y_{0}-k\right)\right]f\left(x,y_{0}-k\right)dx-\\ -\int_{y_{0}-k}^{y_{0}}\left[\frac{\partial\psi_{1}}{\partial y}\left(x_{0}+h,y\right)+\frac{\partial\varphi_{4}}{\partial x}\left(x_{0}+h,y\right)\right]f\left(x_{0}+h,y\right)dy+\\ +\int_{y_{0}-k}^{y_{0}-k}\left[\frac{\partial\psi_{4}}{\partial y}\left(x_{0},y\right)+\frac{\partial\varphi_{4}}{\partial y}\left(x_{0},y\right)\right]f\left(x_{0},y\right)dy+\\ +\iint_{I_{4}}\left(\varphi_{4}\frac{\partial^{2}f}{\partial x^{2}}+\psi_{4}\frac{\partial^{2}f}{\partial x\partial y}+\theta_{4}\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy\end{gathered}

Let's add formulas (5), (6), (7), and (8) term by term and impose boundary conditions so that the new formula is of the form
(2). We first assume that the constants $\alpha_{i},\beta_{i},\gamma_{i}$ are such that
(9) $\alpha_{1}+\beta_{1}+\gamma_{1}=\alpha_{2}+\beta_{2}+\gamma_{2}=\alpha_{3}+\beta_{3}+\gamma_{3}=\alpha_{4}+\beta_{4}+\alpha_{4}=\lambda$

The boundary conditions will be as follows

$\displaystyle\psi_{1}\left(x_{0}+h,y_{0}+k\right)=0,$	$\displaystyle\psi_{2}\left(x_{0}-h,y_{0}+k\right)=0$	(1)
$\displaystyle\psi_{3}\left(x_{0}-h,y_{0}-k\right)=0,$	$\displaystyle\psi_{4}\left(x_{0}+h,y_{0}-k\right)=0$
$\displaystyle\psi_{1}\left(x_{0},y_{0}+k\right)-\psi_{2}\left(x_{0},y_{0}+k\right)=0,$	$\displaystyle\psi_{2}\left(x_{0}-h,y_{0}\right)-\psi_{3}\left(x_{0}-h,y_{0}\right)=0$	(2)
$\displaystyle\psi_{3}\left(x_{0},y_{0}-k\right)-\psi_{4}\left(x_{0},y_{0}-k\right)=0,$	$\displaystyle\psi_{4}\left(x_{0}+h,y_{0}\right)-\psi_{1}\left(x_{0}+h,y_{0}\right)=0$
$\displaystyle\varphi_{1}\left(x_{0}+h,y\right)=0,$	$\displaystyle\theta_{1}\left(x,y_{0}+k\right)=0$
$\displaystyle\varphi_{2}\left(x_{0}-h,y\right)=0,$	$\displaystyle\theta_{2}\left(x,y_{0}+k\right)=0$	(3)
$\displaystyle\varphi_{3}\left(x_{0}-h,y\right)=0,$	$\displaystyle\theta_{3}\left(x,y_{0}-k\right)=0$
$\displaystyle\varphi_{4}\left(x_{0}+h,y\right)=0,$	$\displaystyle\theta_{4}\left(x,y_{0}-k\right)=0$
$\displaystyle\varphi_{1}\left(x_{0},y\right)-\varphi_{2}\left(x_{0},y\right)=0,$	$\displaystyle\theta_{1}\left(x,y_{0}\right)-\theta_{4}\left(x,y_{0}\right)=0$	(4)
$\displaystyle\theta_{2}\left(x,y_{0}\right)-\theta_{3}\left(x,y_{0}\right)=0,$	$\displaystyle\varphi_{3}\left(x_{0},y\right)-\varphi_{4}\left(x_{0},y\right)=0$
$\displaystyle\frac{\partial\psi_{1}}{\partial x}\left(x,y_{0}+k\right)+\frac{\partial\theta_{1}}{\partial y}\left(x,y_{0}+k\right)=0,$	$\displaystyle\frac{\partial\psi_{2}}{\partial x}\left(x,y_{0}+k\right)+\frac{\partial\theta_{2}}{\partial y}\left(x,y_{0}+k\right)=0$
$\displaystyle\frac{\partial\psi_{2}}{\partial y}\left(x_{0}-h,y\right)+\frac{\partial\varphi_{2}}{\partial x}\left(x_{0}-h,y\right)=0,$	$\displaystyle\frac{\partial\psi_{3}}{\partial y}\left(x_{0}-h,y\right)+\frac{\partial\varphi_{3}}{\partial x}\left(x_{0}-h,y\right)=0$	(5)
$\displaystyle\frac{\partial\psi_{3}}{\partial x}\left(x,y_{0}-k\right)+\frac{\partial\theta_{3}}{\partial y}\left(x,y_{0}-k\right)=0,$	$\displaystyle\frac{\partial\psi_{4}}{\partial x}\left(x,y_{0}-k\right)+\frac{\partial\theta_{4}}{\partial y}\left(x,y_{0}-k\right)=0$
$\displaystyle\frac{\partial\psi_{4}}{\partial y}\left(x_{0}+h,y\right)+\frac{\partial\varphi_{4}}{\partial x}\left(x_{0}+h,y\right)=0,$	$\displaystyle\frac{\partial\psi_{1}}{\partial y}\left(x_{0}+h,y\right)+\frac{\partial\varphi_{1}}{\partial x}\left(x_{0}+h,y\right)=0$

		$\displaystyle-\frac{\partial\psi_{1}}{\partial y}\left(x_{0},y\right)-\frac{\partial\varphi_{1}}{\partial x}\left(x_{0},y\right)+\frac{\partial\psi_{2}}{\partial y}\left(x_{0},y\right)+\frac{\partial\varphi_{2}}{\partial x}\left(x_{0},y\right)=0$
		$\displaystyle-\frac{\partial\psi_{1}}{\partial x}\left(x,y_{0}\right)-\frac{\partial\theta_{1}}{\partial y}\left(x,y_{0}\right)+\frac{\partial\psi_{4}}{\partial x}\left(x,y_{0}\right)+\frac{\partial\theta_{4}}{\partial y}\left(x,y_{0}\right)=0$
		$\displaystyle-\frac{\partial\psi_{2}}{\partial x}\left(x,y_{0}\right)-\frac{\partial\theta_{2}}{\partial y}\left(x,y_{0}\right)+\frac{\partial\psi_{3}}{\partial x}\left(x,y_{0}\right)+\frac{\partial\theta_{3}}{\partial y}\left(x,y_{0}\right)=0$
		$\displaystyle-\frac{\partial\psi_{4}}{\partial y^{\prime}}\left(x_{0},y\right)-\frac{\partial\varphi_{4}}{\partial x}\left(x_{0},y\right)+\frac{\partial\psi_{3}}{\partial y}\left(x_{0},y\right)+\frac{\partial\varphi_{3}}{yx}\left(x_{0},y\right)=0$

This gives us the cubature formula

	$\displaystyle\lambda\iint_{D}f(x,y)dxdy=$		(11)
	$\displaystyle=\left[-\psi_{1}\left(x_{0},y_{0}\right)+\psi_{2}\left(x_{0},y_{0}\right)-\psi_{3}\left(x_{0},y_{0}\right)+\psi_{4}\left(x_{0},y_{0}\right)\right]f\left(x_{0},y_{0}\right)+R,$

where the rest $R$ is given by the formula

R=\iint_{U}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+0\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy

(12)

the functions $\varphi,\psi,0$ being equal to $\varphi_{1},\psi_{1},\theta_{1}$ In $D_{1}$ has $\varphi_{2},\psi_{2},\theta_{2}$ In $D_{2}$ , has $\varphi_{3},\psi_{3},\theta_{3}$ In $D_{3}$ , and to $\varphi_{4},\psi_{4},\theta_{4}$ In $D_{4}$ .

The search for the cubature formula (11) is thus reduced to the integration of the partial differential equations (4) with the boundary conditions (10). The constants $\alpha_{i},\beta_{i},\gamma_{i}$ will be determined in such a way that this problem is possible.
7. We will now determine the functions $\varphi_{i},\psi_{i},\theta_{i}$ and the numbers $\alpha_{i},\beta_{i},\gamma_{i}$ .

Let us first integrate the equations
where
(12)
L The sob n

		$\displaystyle\frac{\partial^{2}\varphi_{1}}{\partial x^{2}}=\alpha_{1},\quad\frac{\partial^{2}\varphi_{2}}{\partial x^{2}}=\alpha_{2}$
		taking into account the boundary conditions
		$\displaystyle\varphi_{1}\left(x_{0}+h,y\right)=0,\varphi_{2}\left(x_{0}-h,y\right)=0$
		$\displaystyle\varphi_{1}\left(x_{0},y\right)-\varphi_{2}\left(x_{0},y\right)=0$
		$\displaystyle\text{ Nous aurons }\quad\begin{aligned} \varphi_{1}(x,y)&=\frac{\alpha_{1}}{2}x^{2}+A_{1}(y)x+B_{1}(y)\\ \varphi_{2}(x,y)&=\frac{\alpha_{2}}{2}x^{2}+A_{2}(y)x+B_{2}(y)\\ 0&=\frac{\alpha_{1}}{2}\left(x_{0}+h\right)^{2}+A_{1}(y)\left(x_{0}+h\right)+B_{1}(y)\end{aligned}$

0=\frac{\alpha_{2}}{2}\left(x_{0}-h\right)^{2}+A_{2}(y)\left(x_{0}-h\right)+B_{2}(y),

from which it follows that

		$\displaystyle\varphi_{1}(x,y)=\left(x-x_{0}-h\right)\left[\frac{\alpha_{1}}{2}\left(x+x_{0}+h\right)+A_{1}(y)\right]$
		$\displaystyle\varphi_{2}(x,y)=\left(x-x_{0}+h\right)\left[\frac{\alpha_{2}}{2}\left(x+x_{0}-h\right)+A_{2}(y)\right].$

The condition $\varphi_{1}\left(x_{0},y\right)-\varphi_{2}\left(x_{0},y\right)=0$ , gives us the equation

\left(\alpha_{1}+\alpha_{2}\right)x_{0}+\frac{h}{2}\left(\alpha_{1}-\alpha_{2}\right)+A_{1}(y)+A_{2}(y)=0

(15)

who determines $A_{2}(y)$ we will have

A_{2}(y)=-\left(\alpha_{1}+\alpha_{2}\right)x_{0}-\frac{h}{2}\left(\alpha_{1}-\alpha_{2}\right)-A_{1}(y)

and consequently

	$\displaystyle\varphi_{1}(x,y)=\left(x-x_{0}-h\right)\left[\frac{\alpha_{1}}{2}\left(x+x_{0}+h\right)+A_{1}(y)\right]$		(13)
	$\displaystyle\varphi_{2}(x,y)=\left(x-x_{0}+h\right)\left[\frac{\alpha_{2}}{2}\left(x-x_{0}\right)-\alpha_{1}x_{0}-\frac{h}{2}\alpha_{1}-A_{1}(y)\right]$

Or $A_{1}(y)$ is an arbitrary function.
Similarly, we find

	$\displaystyle\varphi_{3}(x,y)=\left(x-x_{0}+h\right)\left[\frac{\alpha_{3}}{2}\left(x-x_{0}\right)-\alpha_{4}x_{0}-\frac{h}{2}\alpha_{4}-A_{4}(y)\right]$		( $\prime$ )
	$\displaystyle\varphi_{4}(x,y)=\left(x-x_{0}-h\right)\left[\frac{\alpha_{4}}{2}\left(x+x_{0}+h\right)+A_{4}(y)\right]$

Or $A_{4}(y)$ is an arbitrary function.
By a similar process, we find

	$\displaystyle\theta_{1}(x,y)=\left(y-y_{0}-k\right)\left[\frac{\gamma_{1}}{2}\left(y+y_{0}+k\right)+C_{1}(x)\right]$		(17)
	$\displaystyle\theta_{2}(x,y)=\left(y-y_{0}-k\right)\left[\frac{\gamma_{2}}{2}\left(y+y_{0}+k\right)+C_{2}(x)\right]$		(14)
	$\displaystyle\theta_{3}(x,y)=\left(y-y_{0}+k\right)\left[\frac{\gamma_{3}}{2}\left(y-y_{0}\right)-\gamma_{2}y_{0}-\frac{k}{2}\gamma_{2}-C_{2}(x)\right]$
	$\displaystyle\theta_{4}(x,y)=\left(y-y_{0}+k\right)\left[\frac{\gamma_{4}}{2}\left(y-y_{0}\right)-\gamma_{1}y_{0}-\frac{k}{2}\gamma_{1}-C_{1}(x)\right]$

Or $C_{1}(x)$ And $C_{2}(x)$ are arbitrary functions.
With formulas (13), (13') and (14) the boundary conditions ( $10_{3}$ ) And $\left(10_{4}\right)$ are satisfied.

Let's move on to determining the function $\psi_{1}(x,y)$ Boundary conditions $\left(10_{5}\right)$ give us

		$\displaystyle\frac{\partial\psi_{1}}{\partial x}\left(x,y_{0}+k\right)=-\frac{\partial\theta_{1}}{\partial y}\left(x,y_{0}+k\right)$
		$\displaystyle\frac{\partial\psi_{1}}{\partial y}\left(x_{0}+h,y\right)=--\frac{\partial\varphi_{1}}{\partial x}\left(x_{0}+h,y\right)$

and formulas (13) and (14) give us

		$\displaystyle\frac{\partial\psi_{1}}{\partial y}\left(x,y_{0}+k\right)=-\gamma_{1}\left(y_{0}+k\right)-C_{1}(x)$
		$\displaystyle\frac{\partial\psi_{1}}{\partial y}\left(x_{0}+h,y\right)=-\alpha_{1}\left(x_{0}+h\right)-A_{1}(y)$

The next step is to integrate the equation.

\frac{\partial^{2}\psi_{1}}{\partial x\partial y}=\beta_{1}

with conditions (15) and the first condition (101).
We will have

\frac{\partial\psi_{1}}{\partial x}=\beta_{1}\left(y-y_{0}-k\right)-\gamma_{1}\left(y_{0}+k\right)-C_{1}(x)

and by integrating with respect to $x$ , we find

	$\displaystyle\psi_{1}(x,y)$	$\displaystyle=\beta_{1}\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)-\gamma_{1}\left(y_{0}+k\right)\left(x-x_{0}-h\right)$		(16)
		$\displaystyle-\alpha_{1}\left(x_{0}+h\right)\left(y-y_{0}-k\right)-\int_{x_{0}+h}^{x}C_{1}(s)ds-\int_{y_{0}+k}^{y}A_{1}(t)dt$

The function $\psi_{1}(x,y)$ thus determined satisfies conditions (15) and condition $\psi_{1}\left(x_{0}+h,y_{0}+k\right)=0$ .

By introducing the functions

\begin{array}[]{ll}P_{1}(x)=\int_{x_{0}+h}^{x}C_{1}(s)ds,&Q_{1}(y)=\int_{y_{0}+k}^{y}A_{1}(t)dt\\ P_{2}(x)=\int_{x_{0}-h}^{x}C_{2}(s)ds,&Q_{2}(y)=\int_{y_{0}-k}^{v}A_{4}(t)dt\end{array}

we will have the formulas

\begin{gathered}\psi_{1}(x,y)=\beta_{1}\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)-\gamma_{1}\left(y_{0}+k\right)\left(x-x_{0}-h\right)-\\ -\alpha_{1}\left(x_{0}+h\right)\left(y-y_{0}-k\right)-P_{1}(x)-Q_{1}(y)\end{gathered}

	$\displaystyle\psi_{2}(x,y)=$	$\displaystyle\beta_{2}\left(x-x_{0}+h\right)\left(y-y_{0}-k\right)+\left(\alpha_{1}x_{0}+\frac{\alpha_{1}+\alpha_{2}}{2}h\right)\left(y-y_{0}-k\right)-$		(18)
		$\displaystyle-\gamma_{2}\left(y_{0}+k\right)\left(x-x_{0}+h\right)-P_{2}(x)+Q_{1}(y)$

\begin{gathered}\psi_{3}(x,y)=\beta_{3}\left(x-x_{0}+h\right)\left(y-y_{0}+k\right)+\left(\gamma_{2}y_{0}+\frac{\gamma_{2}+\gamma_{3}}{2}k\right)\left(x-x_{0}+h\right)+\\ +\left(\alpha_{4}x_{0}+\frac{\alpha_{3}+\alpha_{4}}{2}h\right)\left(y-y_{0}+k\right)+P_{2}(x)+Q_{2}(y)\\ \psi_{4}(x,y)=\beta_{4}\left(x-x_{0}-h\right)\left(y-y_{0}+k\right)+\left(\gamma_{1}y_{0}+\frac{\gamma_{1}+\gamma_{4}}{2}k\right)\left(x-x_{0}-h\right)-\\ -\alpha_{4}\left(x_{0}+h\right)\left(y-y_{0}+k\right)+P_{1}(x)-Q_{2}(y)\end{gathered}

With formulas (18) we satisfy the conditions ( $10_{1}$ ) And ( $10_{5}$ ).
Let's now write that the conditions $\left(10_{2}\right)$ are satisfied.
We will have the equations

\begin{gathered}\left(\gamma_{1}+\gamma_{2}\right)\left(y_{0}+k\right)h+P_{2}\left(x_{0}\right)-P_{1}\left(x_{0}\right)=0\\ \left(\alpha_{1}+\alpha_{4}\right)\left(x_{0}+h\right)k+Q_{2}\left(y_{0}\right)-Q_{1}\left(y_{0}\right)=0\\ {\left[\left(\gamma_{1}+\gamma_{2}\right)y_{0}+\frac{\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}}{2}k\right]h+P_{2}\left(x_{0}\right)-P_{1}\left(x_{0}\right)=0}\\ {\left[\left(\alpha_{1}+\alpha_{4}\right)x_{0}+\frac{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}{2}h\right]k+Q_{2}\left(y_{0}\right)-Q_{1}\left(y_{0}\right)=0}\end{gathered}

which give us

	$\displaystyle\gamma_{1}+\gamma_{2}=\gamma_{3}+\gamma_{4}=-\frac{P_{2}\left(x_{0}\right)-P_{1}\left(x_{0}\right)}{\left(y_{0}+k\right)h}$		(19)
	$\displaystyle\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{3}=-\frac{Q_{2}\left(y_{0}\right)-Q_{1}\left(y_{0}\right)}{\left(x_{0}+h\right)k}$

By writing that the conditions ( $10_{6}$ are also satisfied, we will have the following equations

	$\displaystyle\alpha_{1}+\alpha_{2}+\beta_{1}+\beta_{2}=0$
	$\displaystyle\beta_{1}+\beta_{4}+\gamma_{1}+\gamma_{4}=0$		(20)
	$\displaystyle\gamma_{3}+\gamma_{2}+\beta_{2}+\beta_{3}=0$
	$\displaystyle\alpha_{3}+\alpha_{4}+\beta_{3}+\beta_{4}=0$

8.

We still need to determine the numbers. $\alpha_{i},\beta_{i},\gamma_{i}$ satisfying relations (9), (19) and (20).

Let us introduce the notations
(21)

\frac{P_{1}\left(x_{0}\right)-P_{2}\left(x_{0}\right)}{2h\left(y_{0}+k\right)}=I,\quad\frac{Q_{1}\left(y_{0}\right)-Q_{2}\left(y_{0}\right)}{2k\left(x_{0}+h\right)}=J

and the parameters $t,u,v,z$ through relationships

\alpha_{1}-\alpha_{4}=2t,\quad\alpha_{3}-\alpha_{2}=2u,\quad\gamma_{1}-\gamma_{2}=2v,\quad\gamma_{3}-\gamma_{4}=2z

The equations

\begin{array}[]{lll}\gamma_{1}+\gamma_{2}=2I;&\gamma_{3}+\gamma_{4}=2I;&\alpha_{1}+\alpha_{4}=2J;\\ \gamma_{1}-\gamma_{3}=2v;&\gamma_{3}-\gamma_{4}=2z;&\alpha_{1}-\alpha_{4}=2t;\\ \alpha_{3}-\alpha_{2}=2u\end{array}

determine $\alpha_{i}$ And $\gamma_{i}$ and we will have

\begin{array}[]{ll}\alpha_{1}=J+t,&\gamma_{1}=I+v\\ \alpha_{2}=J-u,&\gamma_{2}=I-v\\ \alpha_{3}=J+u,&\gamma_{3}=I+z\\ \alpha_{1}=J-t,&\gamma_{4}=I-z.\end{array}

Finally, equations (9) where $\lambda$ is supposedly known, they give us

		$\displaystyle\beta_{1}=\lambda-I-J-v-t$
		$\displaystyle\beta_{2}=\lambda-I-J+v+u$
		$\displaystyle\beta_{3}=\lambda-I-J-z-u$
		$\displaystyle\beta_{4}=\lambda-I-J+z+t$

By stating that equations (20) are satisfied, we will have the conditions

\lambda=I=J.

(22)

No, we will finally have

\begin{array}[]{lll}\alpha_{1}=J+t,&\beta_{1}=-J-v-t,&\gamma_{1}=J+v\\ \alpha_{2}=J-u,&\beta_{2}=-J+v+u,&\gamma_{2}=J-v\\ \alpha_{3}=J+u,&\beta_{3}=-J-z-u,&\gamma_{3}=J+z\\ \alpha_{4}=J-t,&\beta_{4}=-J+z+t,&\gamma_{4}=J-z\end{array}

Thus the problem on partial differential equations (4) with boundary conditions $\left(10_{1}\right),\ldots,\left(10_{6}\right)$ is resolved. The functions $\varphi_{i},\psi_{i},\theta_{i}$ are given by formulas (13), (14), (18) where $A(y),A_{4}(y),C_{1}(x),C_{2}(x)$ are arbitrary functions, which, through functions (17), are linked by the relation

\frac{P_{1}\left(x_{0}\right)-P_{2}\left(x_{0}\right)}{2h\left(y_{0}+k\right)}=\frac{Q_{1}\left(y_{0}\right)-Q_{2}\left(y_{0}\right)}{2k\left(x_{0}+h\right)}

(24)

9.

Let's now calculate the coefficient of $f\left(x_{0},y_{0}\right)$ in the cubature formula (11). We have

\begin{gathered}-\psi_{1}\left(x_{0},y_{0}\right)+\psi_{2}\left(x_{0},y_{0}\right)-\psi_{3}\left(x_{0},y_{0}\right)+\psi_{4}\left(x_{0},y_{0}\right)=\\ =-\left(\beta_{1}+\beta_{2}+\beta_{3}+\beta_{4}\right)hk-\left(\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}\right)\frac{hk}{2}-\left(\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}\right)\frac{hk}{2}-\\ -\left(\gamma_{1}+\gamma_{2}\right)h\left(y_{0}+k\right)-\left(\gamma_{1}+\gamma_{2}\right)hy_{0}-\left(\alpha_{1}+\alpha_{4}\right)k\left(x_{0}+h\right)-\left(\alpha_{1}+\alpha_{4}\right)kx_{0}+\\ +2\left[P_{1}\left(x_{0}\right)-P_{2}\left(x_{0}\right)\right]+2\left[Q_{1}\left(y_{0}\right)-Q_{2}\left(y_{0}\right)\right].\end{gathered}

Taking into account formulas (21), (22), (23), (24) we find

-\psi_{1}\left(x_{0},y_{0}\right)+\psi_{2}\left(x_{0},y_{0}\right)-\psi_{3}\left(x_{0},y_{0}\right)+\psi_{4}\left(x_{0},y_{0}\right)=4Jhk.

17 - Mathematica

The cubature formula (11) therefore becomes

\iint_{D}f(x,y)dxdy=4hkf\left(x_{0},y_{0}\right)+\frac{R}{I}

(25)

10.

Let's now look at the role of arbitrary constants. $t,u,v,z$ and arbitrary functions $A_{1}(y),A_{4}(y),C_{1}(x),C_{2}(x)$ which enter into the expression of the functions $\varphi_{i},\psi_{i},\theta_{i}$ , in the expression of the rest $R$ of the cubature formula (25).

The coefficient of $t$ in the expression of the rest $R$ , according to formulas (13), (14), (18) and (23) is

		$\displaystyle T=\frac{1}{2}\iint_{D_{1}}\left(x-x_{0}-h\right)\left(x+x_{0}+h\right)\frac{\partial^{2}f}{\partial x^{2}}dxdy-$
		$\displaystyle-\left(x_{0}+\frac{h}{2}\right)\iint_{J_{2}}\left(x-x_{0}+h\right)\frac{\partial^{2}f}{\partial x^{2}}dxdy+$
		$\displaystyle+\left(\left(x_{0}+\frac{h}{2}\right)\iint_{D_{3}}\left(x-x_{0}+h\right)\frac{\partial^{2}f}{\partial x^{2}}dxdy-\right.$
		$\displaystyle-\frac{1}{2}\iint_{D_{4}}\left(x-x_{0}-h\right)\left(x+x_{0}+h\right)\frac{\partial^{2}f}{\partial x^{2}}dxdy-$
		$\displaystyle-\left(x_{0}+h\right)\iint_{D_{1}}\left(y-y_{0}-k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy-$
		$\displaystyle-\iint_{D_{1}}\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy+$
		$\displaystyle+\left(x_{0}+\frac{h}{2}\right)\iint_{D_{2}}\left(y-y_{0}-k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy-$
		$\displaystyle-\left(x_{0}+\frac{h}{2}\right)\iint_{D_{3}}\left(y-y_{0}+k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy+$
		$\displaystyle+\left(x_{0}+h\right)\iint_{D_{4}}\left(y-y_{0}+k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy+$
		$\displaystyle+\iint_{D_{4}}\left(x-x_{0}-h\right)\left(y-y_{0}+k\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy$

Through properly conducted part-by-part integration, it can be demonstrated that $T=0$ This is demonstrated in the same way as in the expression $R$ of the remainder of the cubature formula, the coefficients of $u,v,z$ are worthless.

It follows that we can do it in formulas (23) $t=u=v=z=0$ and then these formulas reduce to
(26)

		$\displaystyle\alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha_{4}=J$
		$\displaystyle\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=-J$
		$\displaystyle\gamma_{1}=\gamma_{2}=\gamma_{3}=\gamma_{4}=J$

Formulas (13), (14), and (18) become

\begin{gathered}\varphi_{1}(x,y)=\left(x-x_{0}-h\right)\left[\frac{J}{2}\left(x+x_{0}+h\right)+A_{1}(y)\right]\\ \varphi_{2}(x,y)=\left(x-x_{0}+h\right)\left[\frac{J}{2}\left(x-x_{0}\right)-J\left(x_{0}+\frac{h}{2}\right)-A_{1}(y)\right]\\ \varphi_{3}(x,y)=\left(x-x_{0}+h\right)\left[\frac{J}{2}\left(x-x_{0}\right)-J\left(x_{0}+\frac{h}{2}\right)-A_{4}(y)\right]\\ \rho_{4}(x,y)=\left(x-x_{0}-h\right)\left[\frac{J}{2}\left(x+x_{0}+h\right)+A_{4}(y)\right]\\ \theta_{1}(x,y)=\left(y-y_{0}-k\right)\left[\frac{J}{2}\left(y+y_{0}+k\right)+C_{1}(x)\right]\\ \theta_{2}(x,y)=\left(y-y_{0}-k\right)\left[\frac{J}{2}\left(y+y_{0}+k\right)+C_{2}(x)\right]\\ \theta_{3}(x,y)=\left(y-y_{0}+k\right)\left[\frac{J}{2}\left(y-y_{0}\right)-J\left(y_{0}+\frac{k}{2}\right)-C_{2}(x)\right]\\ \theta_{4}(x,y)=\left(y-y_{0}+k\right)\left[\frac{J}{2}\left(y-y_{0}\right)-J\left(y_{0}+\frac{k}{2}\right)-C_{1}(x)\right]\\ \begin{array}[]{c}\psi_{1}(x,y)=-J\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)-J\left(y_{0}+k\right)\left(x-x_{0}-h\right)-\\ x\\ -J\left(x_{0}+h\right)\left(y-y_{0}-k\right)-\int_{x_{0}+h}^{y}C_{1}(s)ds-\int_{y_{0}+k}A_{1}(t)dt\\ \psi_{2}(x,y)=-J\left(x-x_{0}+h\right)\left(y-y_{0}-k\right)+J\left(x_{0}+h\right)\left(y-y_{0}-k\right)-\\ -J\left(y_{0}+k\right)\left(x-x_{0}+h\right)-\int_{x_{0}-h}^{x}C_{2}(s)ds+\int_{y_{0}+k}^{y}A_{1}(l)dt\\ \begin{array}[]{c}\psi_{3}(x,y)=-J\left(x-x_{0}+h\right)\left(y-y_{0}+k\right)+J\left(y_{0}+k\right)\left(x-x_{0}+h\right)+\\ x\end{array}C_{0}^{y}A_{4}(t)dt.\end{array}\end{gathered}

	$\displaystyle\psi_{4}(x,y)=$	$\displaystyle-J\left(x-x_{0}-h\right)\left(y-y_{0}+k\right)+J\left(y_{0}+k\right)\left(x-x_{0}-h\right)-$
		$\displaystyle-J\left(x_{0}+h\right)\left(y-y_{0}+k\right)+\int_{x_{0}+h}^{x}C_{1}(s)ds-\int_{y_{0}-k}^{y}A_{4}(t)dt$

The functions $\psi_{1}(x,y),\psi_{2}(x,y),\psi_{3}(x,y),\psi_{4}(x,y)$ can also be written in the following form

	$\displaystyle\psi_{1}(x,y)=-J\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)-\int_{x_{0}+h}^{x}\left[C_{1}(s)+J\left(y_{0}+k\right)\right]ds-$	$\displaystyle\int_{y_{0}+k}^{y}\left[A_{1}(t)+J\left(x_{0}+h\right)\right]dt$
	$\displaystyle\begin{array}[]{l}\psi_{2}(x,y)=-J\left(x-x_{0}+h\right)\left(y-y_{0}-k\right)-\int_{x_{0}-h}^{x}\left[C_{2}(s)+J\left(y_{0}+k\right)\right]ds+\\ +\int_{y_{0}+k}^{y}\left[A_{1}(t)+J\left(x_{0}+h\right)\right]dt\end{array}$	$\displaystyle\begin{array}[]{c}x_{3}(x,y)=-J(x-h)\left(y-y_{0}+k\right)+\int_{x_{0}-h}^{x}\left[C_{2}(s)+J\left(y_{0}+k\right)\right]ds+\\ +\int_{y_{0}-k}^{y}\left[A_{4}(t)+J\left(x_{0}+h\right)\right]dt\\ \psi_{4}(x,y)=-J\left(x-x_{0}-h\right)\left(y-y_{0}+k\right)+\int_{x_{0}+h}^{x}\left[C_{1}(s)+J\left(y_{0}+k\right)\right]ds-\\ -\int_{y_{0}-k}^{y}\left[A_{4}(t)+J\left(x_{0}+h\right)\right]dt\end{array}$

This invites us to ask

\begin{array}[]{ll}\bar{A}_{1}(y)=A_{1}(y)+J\left(x_{0}+h\right),&\bar{C}_{1}(x)=C_{1}(x)+J\left(y_{0}+k\right)\\ \bar{A}_{4}(y)=A_{4}(y)+J\left(x_{0}+h\right),&\bar{C}_{2}(x)=C_{2}(x)+J\left(y_{0}+k\right)\end{array}

and the previous formulas become

\begin{gathered}\varphi_{1}(x,y)=\left(x-x_{0}-h\right)\left[\frac{J}{2}\left(x-x_{0}-h\right)+\bar{A}_{1}(y)\right]\\ \varphi_{2}(x,y)=\left(x-x_{0}+h\right)\left[\frac{J}{2}\left(x-x_{0}+h\right)-\bar{A}_{1}(y)\right]\\ \varphi_{3}(x,y)=\left(x-x_{0}+h\right)\left[\frac{J}{2}\left(x-x_{0}+h\right)-\bar{A}_{4}(y)\right]\\ \varphi_{4}(x,y)=\left(x-x_{0}-h\right)\left[\frac{J}{2}\left(x-x_{0}-h\right)+\bar{A}_{4}(y)\right]\\ \theta_{1}(x,y)=\left(y-y_{0}-k\right)\left[\frac{J}{2}\left(y-y_{0}-k\right)+\bar{C}_{1}(x)\right]\\ \theta_{2}(x,y)=\left(y-y_{0}-k\right)\left[\frac{J}{2}\left(y-y_{0}-k\right)+\bar{C}_{2}(x)\right]\\ \theta_{3}(x,y)=\left(y-y_{0}+k\right)\left[\frac{J}{2}\left(y-y_{0}+k\right)-\bar{C}_{2}(x)\right]\\ \theta_{4}(x,y)=\left(y-y_{0}+k\right)\left[\frac{J}{2}\left(y-y_{0}+k\right)-\bar{C}_{1}(x)\right]\\ \psi_{1}(x,y)=-J\left(x-x_{0}-h\right)\left(y-y_{0}-k\right)-\int_{x_{0}+h}^{x}\bar{C}_{1}(s)ds-\int_{y_{0}+k}^{y}\bar{A}_{1}(l)dt\\ \psi_{2}(x,y)=-J\left(x-x_{0}+h\right)\left(y-y_{0}-k\right)-\int_{x_{0}-h}^{x}\bar{C}_{2}(s)ds+\int_{y_{0}+k}^{y}\bar{A}_{1}(l)dt\\ \psi_{3}(x,y)=-J\left(x-x_{0}+h\right)\left(y-y_{0}+k\right)+\int_{x_{0}-h}^{x}\bar{C}_{2}(s)ds+\int_{y_{0}-k}^{y}\bar{A}_{4}(l)dl\\ \psi_{4}(x,y)=-J\left(x-x_{0}-h\right)\left(y-y_{0}+k\right)+\int_{x_{0}+h}^{x}\bar{C}_{1}(s)ds-\int_{y_{0}-k}^{y}\bar{A}_{4}(t)dt.\end{gathered}

With the new features (27), we have

	$\displaystyle P_{1}\left(x_{0}\right)-P_{2}\left(x_{0}\right)$	$\displaystyle=\int_{x_{0}+h}^{x_{0}}\overline{C_{1}}(s)ds-\int_{x_{0}-h}^{x_{0}}\overline{C_{2}}(s)ds+2Jh\left(y_{0}+k\right)$
	$\displaystyle Q_{1}\left(y_{0}\right)-Q_{2}\left(y_{0}\right)$	$\displaystyle=\int_{y_{0}+k}^{y_{0}}\overline{A_{1}}(l)dl-\int_{y_{0}-k}^{y_{0}}\overline{A_{4}}(l)dl+2Jk\left(x_{0}+h\right)$

and condition (24) becomes

\frac{\int_{x_{0}+h}^{x_{0}}\overline{C_{1}}(s)ds-\int_{x_{0}-h}^{x_{0}}\overline{C_{2}(s)}ds}{2h\left(y_{0}+k\right)}=\frac{\int_{y_{0}+k}^{y_{0}}\overline{A_{1}(t)}dt-\int_{y_{0}-k}^{y_{0}}\overline{A_{1}(t)}dt}{2k\left(x_{0}+h\right)}

(29)

The contribution of the functions $\bar{A}_{1}(y),\bar{A}_{4}(y),\bar{C}_{1}(x),\bar{C}_{2}(x)$ to the expression of the rest $R$ is the following

-\iint_{D_{3}}\bar{C}_{2}(x)\left(y-y_{0}+k\right)\frac{\partial^{2}f}{\partial y^{2}}dxdy+\iint_{D_{3}x_{0}-h}^{r}\left(\bar{C}_{2}(s)ds\right)\frac{\partial^{2}f}{\partial x\partial y}dxdy

Through part integrations, we demonstrate that

\left(30^{\prime}\right)\quad Q=\left[\int_{y_{0}+k}^{y_{0}}\bar{A}_{1}(t)dt-\int_{y_{0}-k}^{v_{0}}\overline{A_{4}}(t)dt\right]\left[f\left(x_{0}-h,y_{0}\right)-2f\left(x_{0},y_{0}\right)+f\left(x_{0}+h,y_{0}\right)\right]+

(

\prime

)

+\left[\int_{x_{0}+h}^{x_{0}}\overline{C_{1}}(s)ds-\int_{x_{0}-h}^{x_{0}}\overline{C_{2}}(s)ds\right]\left[f\left(x_{0},y_{0}-k\right)-2f\left(x_{0},y_{0}\right)+f\left(x_{0},y_{0}+k\right)\right]

Thus, the expressions of the functions $\varphi_{i},\psi_{i},0_{i}$ are given by formulas (28), the arbitrary functions $\bar{A}_{1}(y),\bar{A}_{4}(y),\bar{C}_{1}(x),\bar{C}_{2}(x)$ being obliged to satisfy condition (29). The contribution of the functions $\bar{A}_{1}(y)$ , $\bar{A}_{4}(y),\bar{C}_{1}(x),\overline{C_{2}}(x)$ to the expression of the rest $R$ is given by formulas (30) and (30'). If we take $\bar{A}_{1}(y)=\bar{A}_{4}(y)=\bar{C}_{1}(x)=\bar{C}_{2}(x)=0$ , condition (29) is satisfied, and the contribution of $\bar{A}_{1}(y),\bar{A}_{4}(y),\bar{C}_{1}(x),\bar{C}_{2}(x)$ to the expression of the rest $R$ is zero.

Done, we can do it in formulas (28) $\bar{A}_{1}(y)=\bar{A}_{4}(y)==\bar{C}_{1}(x)=\bar{C}_{2}(x)=0$ and we will finally have the cubature formula

\int_{E}\int_{E}f(x,y)dxdy=4hkf\left(x_{0},y_{0}\right)+\iint_{D}\left(\varphi\frac{\partial^{2}f}{\partial x^{2}}+\psi\frac{\partial^{2}f}{\partial x\partial y}+\theta\frac{\partial^{2}f}{\partial y^{2}}\right)dxdy

(31)

where the functions $\varphi,\psi,0$ coincide within the rectangles $D_{i}$ with the functions $\varphi_{i},\psi_{i},\theta_{i}$ Or $i^{\prime}=1,2,3,4$ , given by the following formulas

	$\displaystyle\varphi_{1}(x,y)=\frac{1}{2}\left(x-x_{0}-h\right)^{2},\quad\psi_{1}(x,y)=-\left(x-x_{0}-h\right)\left(y-y_{0}-k\right),$
	$\displaystyle\theta_{1}(x,y)=\frac{1}{2}\left(y-y_{0}-k\right)^{2}$
	$\displaystyle\varphi_{2}(x,y)=\frac{1}{2}\left(x-x_{0}+h\right)^{2},\quad\psi_{2}(x,y)=-\left(x-x_{0}+h\right)\left(y-y_{2}-k\right),$		(32)
	$\displaystyle\theta_{2}(x,y)=\frac{1}{2}\left(y-y_{0}-k\right)^{2}$		(32)
	$\displaystyle\varphi_{3}(x,y)=\frac{1}{2}\left(x-x_{0}+h\right)^{2},\quad\psi_{3}(x,y)=-\left(x-x_{0}+h\right)\left(y-y_{0}+k\right),$
	$\displaystyle\theta_{3}(x,y)=\frac{1}{2}\left(y-y_{0}+k\right)^{2}.$

\begin{gathered}\varphi_{4}(x,y)=\frac{1}{2}\left(x-x_{0}-h\right)^{2},\quad\psi_{4}(x,y)=-\left(x-x_{0}-h\right)\left(y-y_{0}+h\right)\\ \theta_{4}(x,y)=\frac{1}{2}\left(y-y_{0}+k\right)^{2}.\end{gathered}

11.

We can give an assessment of the absolute value of the remainder $R$ of the cubature formula (31). By designating $M_{2}$ an upper limit of $\left|\frac{\partial^{2}f}{\partial x^{2}}\right|,\left|\frac{\partial^{2}f}{\partial x\partial y}\right|,\left|\frac{\partial^{2}f}{\partial y^{2}}\right|$ In $D$ , we can write

	$\displaystyle\|R\|$	$\displaystyle\leqslant\frac{M_{2}}{2}\left\{\iint_{D_{1}}\left(x-x_{0}-h\right)^{2}dxdy+\ldots\right\}$
		$\displaystyle+M_{2}\left\{\iint_{D_{1}}\left(x_{0}+h-x\right)\left(y_{0}+k-y\right)dxdy+\ldots\right\}$
		$\displaystyle+\frac{M_{2}}{2}\left\{\iint_{D_{1}}\left(y-y_{0}-k\right)^{2}dxdy+\ldots\right\}$

We have

\begin{gathered}\iint_{D_{1}}\left(x-x_{0}-h\right)^{2}dxdy=\frac{h^{3}k}{3},\ldots\\ \iint_{D_{1}}\left(x_{0}+h-x\right)\left(y_{0}+k-y\right)dxdy=\frac{h^{2}k^{2}}{4},\ldots\\ \iint_{D_{1}}\left(y-y_{0}-k\right)^{2}dxdy=\frac{hk^{3}}{3},\ldots\end{gathered}

from which it follows that, that
is to say

R\left\lvert\,\leqslant\frac{M_{2}}{2}\left(\frac{4h^{3}k}{3}+\frac{4hk^{3}}{3}\right)+M_{2}h^{2}k^{2}\right.

|R|\leqslant\frac{S}{12}\left[2\left(h^{2}+k^{2}\right)+3hk\right]M_{2}

(33)

Or $S$ is the area of the rectangle $D,S=4hk$ 12.
We can compare the two cubature formulas demonstrated in this chapter

	$\displaystyle\iint_{D}fdxdy=\frac{\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)}{2}\left[f\left(x_{1},y_{1}\right)+f\left(x_{2},y_{2}\right)\right]+R_{1}$		( $21,\penalty 10000\ \S 1$ )
	$\displaystyle\iint_{D}fdxdy=4hkf\left(x_{0},y_{0}\right)+R_{2}$		( $31,\penalty 10000\ \S 2$ )

The second formula is preferable to the first. First, the second formula uses only one node. $\left(x_{0},y_{0}\right)$ the center of the rectangle $D$ , while the second uses two nodes, the opposite vertices $\left(x_{1},y_{1}\right)$ And $\left(x_{2},y_{2}\right)$ Regarding the remains $R_{1}$ And $R_{2}$ , we have the formulas ( $26,\S 1$ ) and (33, § 2) that is to say

\left|R_{1}\right|\leqslant\varrho_{1}\frac{SM_{2}}{12},\quad\left|R_{2}\right|\leqslant\varrho_{2}\frac{SM_{2}}{12}

\begin{gathered}Q_{1}=\left(x_{2}-x_{1}\right)^{2}+3\left(x_{2}-x_{1}\right)\left(y_{2}-y_{1}\right)+\left(y_{2}-y_{1}\right)^{2}\\ Q_{2}=2\left(h^{2}+k^{2}\right)+3hk.\end{gathered}

By substituting in the expression $\varrho_{1},x_{2}-x_{1}$ And $y_{2}-y_{1}$ by $2h$ And $2k$ , We have

\varrho_{1}-\varrho_{2}=2\left(h^{2}+k^{2}\right)+9hk>0

that's to say $\varrho_{1}>\varrho_{2}$ In
the case of a square, $h=k$ , We have

\frac{\rho_{1}}{\rho_{2}}=\frac{20}{7}=2,8\ldots

$. 3. Approximate calculation of a double integral in a rectangle $D$

13.

We will apply the volume formulas established in this chapter to the approximate calculation of a double integral in the rectangle $D$ .

Let us consider the double integral

\iint_{D}f(x,y)dxdy=\int_{a}^{b}dx\int_{c}^{d}f(x,y)dy

and suppose that the function $f(x,y)$ has continuous first- and second-order partial derivatives in $D$ .

Let's share the interval $(a,b)$ in $n$ equal parts by points $x_{1},x_{2},\ldots,x_{n-1}$ and the interval ( $c,d$ ) in $m$ equal parts by points $y_{1},y_{2},\ldots,y_{m-1}$ Let us designate by $D_{ik}$ the rectangle formed by the lines $x=x_{i-1},\quad x=x_{i},\quad y=y_{k-1},\quad y=y_{k}$ Or $i=1,\ldots,n,\quad k=1,\ldots,m,\quad$ with $x_{0}=a,x_{n}=b,y_{0}=c,y_{m}=d$ .

Let's apply to each rectangle $D_{ik}$ the first cubature formula. We will have

\iint_{D_{ik}}f(x,y)dxdx=\frac{(b-a)(d-c)}{2nm}\left[f\left(x_{i},y_{k}\right)+f\left(x_{i+1},y_{k+1}\right)\right]+R_{ik}

\left|R_{ik}\right|\leqslant\frac{(b-a)(d-c)}{12mn}\left[\frac{(b-a)^{2}}{n^{2}}+3\frac{(b-a)(d-c)}{nm}+\frac{(d-c)^{2}}{m^{2}}\right]M_{2}

$M_{2}$ being an upper bound of $\left|\frac{\partial^{2}f}{\partial x^{2}}\right|,\left|\frac{\partial^{2}f}{\partial x\partial y}\right|,\left|\frac{\partial^{2}f}{\partial y^{2}}\right|$ In $D$ By
summing the double integrals in the rectangles $D_{ik}$ we will have the approximate calculation formula

	$\displaystyle\iint_{B}f(x,y)dxdy=$	$\displaystyle\frac{(b-a)(d-c)}{2nm}\left\{\sum_{i=0}^{n-1}f\left(x_{i},c\right)+\sum_{i=1}^{n}f\left(x_{i},d\right)+\sum_{k=1}^{m-1}f\left(b,y_{k}\right)+\right.$		(1)
		$\displaystyle\left.+\sum_{k=1}^{m-1}f\left(a,y_{k}\right)+2\sum_{i=1}^{n-1}\sum_{k=1}^{m-1}f\left(x_{i},y_{k}\right)\right\}+R$

and we have the rest

|R|\leqslant\sum_{i=1}^{n}\sum_{k=1}^{m}\left|R_{ik}\right|

that's to say

|R|\leqslant\frac{(b-a)(d-c)}{12}\left[\frac{(b-a)^{2}}{n^{2}}+3\frac{(b-a)(d-c)}{mn}+\frac{(d-c)^{2}}{m^{2}}\right]M_{2}.

(2)

\frac{(b-a)(d-c)}{12}\left[\frac{(b-a)^{2}}{n^{2}}+3\frac{(b-a)(d-c)}{nm}+\frac{(d-c)^{2}}{m^{2}}\right]M_{2}<\varepsilon

With this choice of $n$ And $m$ we will have in the approximate calculation formula (1), $|R|<\varepsilon$ .
14. The second cubature formula can also be applied to the approximate calculation of a double integral.

We will have

\iint_{D_{ik}}f(x,y)dxdy=\frac{(b-a)(d-c)}{nm}f\left(\xi_{i},\eta_{k}\right)+R_{ik}^{\prime}

Or ( $\xi_{i},r_{ik}$ ) are the coordinates of the center of the rectangle $D_{ik}$ and we have

\left|R_{ik}^{\prime}\right|\leqslant\frac{(b-a)(d-c)}{12nm}\left[\frac{(b-a)^{2}}{2n^{2}}+3\frac{(b-a)(d-c)}{4nm}+\frac{(d-c)^{2}}{2m^{2}}\right]M_{2}

This results in the approximate calculation formula

\iint_{D}f(x,y)dxdy=\frac{(b-a)(d-c)}{nm}\sum_{i=1}^{n}\sum_{k=1}^{m}f\left(\zeta_{i},\eta_{k}\right)+R^{\prime}

(3)

with

\left|R^{\prime}\right|\leqslant\frac{(b-a)(d-c)}{12}\left[\frac{(b-a)^{2}}{2n^{2}}+3\frac{(b-a)(d-c)}{4nm}+\frac{(d-c)^{2}}{2m^{2}}\right]M_{2}

(4)

Given a positive number e, we can choose the natural numbers $n$ And $m$ large enough, so that

\frac{(b-a)(d-c)}{12}\left[\frac{(b-a)^{2}}{2n^{2}}+3\frac{(b-a)(d-c)}{4nm}+\frac{(d-c)^{2}}{2m^{2}}\right]M_{2}<\varepsilon

We will then have in the approximate calculation formula (3), $|R|<\varepsilon$ We
will apply the approximate calculation formula (1) to the numerical integration of second-order hyperbolic partial differential equations.

II. NUMERICAL INTEGIATION OF THE SECOND-ORDER HYPERBOLIC PARTHELIC DIFFERENTIAL EQUATION

s. 1. General considerations and assumptions

15.

Consider the partial differential equation
(1)

\frac{\partial^{2}z}{\partial x\partial y}=f(x,y,z,p,q)

where the function $f(x,y,z,p,q)$ is continuous in the domain $\dot{D}$ defined by inequalities

0\leqslant x\leqslant a,\quad 0\leqslant y\leqslant b,\quad|z|\leqslant\alpha,\quad|p|\leqslant\beta,\quad|q|\leqslant\gamma

(2)

and satisfies Lipschitz's condition

|f(x,y,Z,P,Q)-f(x,y,z,p,q)|\leqslant A|Z-z|+B|P-p|+C|Q-q|.

(3)

We know that under these conditions the partial differential equation (1) has a unique integral $z(x,y)$ zero on the axes $Ox,Oy$ It is defined within the rectangle. $\Delta_{1}$ formed by the straight lines $x=0,x=\lambda_{1},y=0,y=\mu_{1}$ , Or $\lambda_{1}$ And $\mu_{1}$ are positive numbers satisfying the conditions

\lambda_{1}=\min\left(a,\frac{\gamma}{M}\right),\quad\mu_{1}=\min\left(b,\frac{\beta}{M}\right),\quad\lambda_{1}\mu_{1}\leqslant\frac{\alpha}{M}

Or $M$ is an upper bound of $|f(x,y,z,p,q)|$ in the field $D$ We
can obtain the integral $z(x,y)$ by the method of successive approximations, by integrating the equations

	$\displaystyle\frac{\partial^{2}z^{(0)}}{\partial x\partial y^{\prime}}=f(x,y,0,0,0)$		(4)
	$\displaystyle\frac{\partial^{2}z^{(s)}}{\partial x\partial y}=f\left(x,y,z^{(s-1)},p^{(s-1)},q^{(s-1)}\right)$

Or $s=1,2,\ldots$ with the conditions $z^{(s)}(x,0)=z^{(s)}(0,y)=0$ , Or $s==0,1,\ldots$ .

The series

z^{(0)}+\sum_{s=1}^{\infty}\left[z^{(s)}-z^{(s-1)}\right],\quad p^{(0)}+\sum_{s=1}^{\infty}\left[p^{(s)}-p^{(s-1)}\right],\quad q^{(0)}+\sum_{s=1}^{\infty}\left[q^{(s)}-q^{(s-1)}\right]

are absolutely and uniformly convergent in the rectangle $\Delta_{1}$ we have

		$\displaystyle\left\|z^{(s)}-z^{(s-1)}\right\|\leqslant ML^{s}\frac{\left(\lambda_{1}+\mu_{1}\right)^{s+2}}{(s+2)!}$
		$\displaystyle\left\|p^{(s)}-p^{(s-1)}\right\|\leqslant ML^{s}\frac{\left(\lambda_{1}+\mu_{1}\right)^{s+1}}{(s+1)!}$
		$\displaystyle\left\|q^{(s)}-q^{(s-1)}\right\|\leqslant ML^{s}\frac{\left(\lambda_{1}+\mu_{1}\right)^{s+1}}{(s+1)!}$

So we have

\begin{gathered}\left.\sim(x,y)=z^{(0)}+\sum_{s=1}^{\infty}z^{(s)}-z^{(s-1)}\right]\\ p(x,y)=p^{(0)}+\sum_{s=1}^{\infty}\left[p^{(s)}-p^{(s-1)}\right],\quad q\left(x,y^{\prime}\right)=q^{(0)}+\sum_{s=1}^{\infty}\left[q^{(s)}-q^{(s-1)}\right]\end{gathered}

and we write

		$\displaystyle z(x,y)-z^{(\nu)}(x,y)=\sum_{s=\nu}^{\infty}\left[z^{(s+1)}-z^{(s)}\right]$
		$\displaystyle p(x,y)-p^{(\nu)}(x,y)=\sum_{s=\nu}^{\infty}\left[p^{(s+1)}-p^{(s)}\right]$
		$\displaystyle q(x,y)-q^{(\nu)}(x,y)=\sum_{s=\nu}^{\infty}\left[q^{(s+1)}-q^{(s)}\right]$

It is demonstrated that

		$\displaystyle\left\|z(x,y)-z^{(\nu)}(x,y)\right\|\leqslant ML^{\nu+1}e^{L\left(\lambda_{1}+\mu_{1}\right)}\frac{\left(\lambda_{1}+\mu_{1}\right)^{\nu+3}}{(\nu+3)!}$
		$\displaystyle\left\|p(x,y)-p^{(\nu)}(x,y)\right\|\leqslant ML^{\nu+1}e^{L\left(\lambda_{1}+\mu_{1}\right)}\frac{\left(\lambda_{1}+\mu_{1}\right)^{\nu+2}}{(\nu+2)!}$
		$\displaystyle\left\|q(x,y)-q^{(\nu)}(x,y)\right\|\leqslant ML^{\nu+1}e^{L\left(\lambda_{1}+\mu_{1}\right)}\frac{\left(\lambda_{1}+\mu_{1}\right)^{\nu+2}}{(\nu+2)!}$

Given a positive number, we can choose the natural number $\nu$ , such that the second members of the preceding inequalities are smaller than $\varepsilon$ Thus we will have for the smallest number $\nu$ satisfying these conditions

	$\displaystyle\left\|z(x,y)-z^{(\nu)}(x,y)\right\|\leqslant\varepsilon$
	$\displaystyle\left\|p(x,y)-p^{(\nu)}(x,y)\right\|\leqslant\varepsilon$		(5)
	$\displaystyle\left\|q(x,y)-q^{(\nu)}(x,y)\right\|\leqslant\varepsilon$

The number $\nu$ as specified, will remain fixed thereafter and will play an important role in the numerical integration of the partial differential equation (1).
16. For the numerical integration of the partial differential equation (1), we will make new assumptions about the function $f(x,y,z,p,q)$ These assumptions are imposed by the numerical integration process that we will use.

We will assume that the function $f(x,y,z,p,q)$ admits partial derivatives with respect to $x$ and to $y$ , of the first and second order continuous in $D$ Under these conditions, we can take the numbers $A,B,C$ of the Lipschitz condition (3), upper bounds of $\left|\frac{\partial f}{\partial z}\right|,\left|\frac{\partial f}{\partial p}\right|,\left|\frac{\partial f}{\partial q}\right|$ In $D$ .

It can be demonstrated, with these assumptions, that the functions $z^{(s)}(x,y)$ , given by the partial differential equations (4) with the conditions $z^{(s)}(x,0)=z^{(s)}(0,y)=0$ , have partial derivatives with respect to $x$ and to $y$ , of the first and second order, continuous in $\Delta_{1}$ and that if we ask

f_{s}(x,y)=f\left[x,y,z^{(s-1)},p^{(s-1)},q^{(s-1)}\right]

(6)

Or $s=1,2,\ldots$ , the functions $f_{s}(x,y)$ , Or $s=0,1,2,\ldots$ , with $f_{0}(x,y)==f(x,y,0,0,0)$ , have partial derivatives with respect to $x$ and to $y$ , of the second order, continuous in the rectangle $\Delta_{1}$ We will refer to $N$ an upper bound on the absolute values of all second-order partial derivatives

\frac{\partial^{2}f_{s}}{\partial x^{2}},\frac{\partial^{2}f_{s}}{\partial x\partial y},\frac{\partial^{2}f_{s}}{\partial y^{2}}

(7)

For $s=0,1,2,\ldots$ , v. The number $N$ plays an important role in the numerical integration of the partial differential equation (1).

Given, a positive number $\delta$ sufficiently small, let us designate by $\lambda$ And $\mu$ positive numbers satisfying the following conditions

\lambda=\min\left(a,\frac{\gamma-\delta}{M}\right),\mu=\min\left(b,\frac{\beta-\delta}{M}\right),\quad\lambda\mu\leqq\frac{\alpha-\delta}{M}

(8)

It is obvious that $\lambda\leqq\lambda_{1},\mu\leqq\mu_{1}$ We will refer to it hereafter as $\Delta$ the rectangle formed by the lines $x=0,x=\lambda$ And $y=0,y=\mu$ .

	$\displaystyle\left\|z^{(\nu)}\left(x_{i},y_{k}\right)-z_{ik}^{(\nu)}\right\|<\varepsilon$
	$\displaystyle\left\|p^{(\nu)}\left(x_{i},y_{k}\right)-p_{k}^{(\nu)}\right\|<\varepsilon$		(9)
	$\displaystyle\left\|q^{(\nu)}\left(x_{i},y_{k}\right)-q_{ik}^{(\nu)}\right\|<\varepsilon.$

The search for the algorithm will be based on the cubature formula (21) from § 1 of the first chapter.

§. 2. Application of the cubature formula (21, $1, I)

17.

The first equation (4) of §. 1 and the conditions $z^{(0)}(x,0)==z^{(0)}(0,y)=0$ , domment
(1)

\begin{gathered}z^{(0)}(x,y)=\int_{0}^{x}\int_{0}^{y}f_{0}(s,t)dsdt\\ p^{(0)}(x,y)=\int_{0}^{y}f_{0}(x,t)dt,q^{(0)}(x,y)=\int_{0}^{x}f_{0}(s,y)ds\end{gathered}

To calculate $z^{(0)}(x,y)$ We will use the first cubature formula which leads to the approximate calculation formula (1) (I, § 3). To calculate $p^{(0)}(x,y)$ And $q^{(0)}(x,y)$ we will use the trapezoid quadrature formula

\int_{a}^{b}f(x)dx=\frac{(b-a)}{2}[f(a)+f(b)]+\int_{a}^{b}\frac{(s-a)(s-b)}{2}f^{\prime\prime}(s)ds

which leads to the approximate calculation formula

\int_{a}^{b}f(x)dx=\frac{(b-a)}{2n}\left[f(a)+f(b)+2\sum_{i=1}^{n-1}f\left(x_{i}\right)\right]+R

(2)

where the points $x_{1},x_{2},\ldots,x_{n-1}$ share in $n$ equal parts the interval ( $a,b$ ). As for the rest $R$ we have the following assessment

|R|\leqq\frac{(b-a)^{3}}{12n^{2}}N_{1}

(3)

Or $N_{1}$ is an upper bound of $\left|f^{\prime\prime}(x)\right|$ in the meantime $[a,b]$ .
Either $\varepsilon_{1}$ a positive number which we will specify later and $n,m$ two natural numbers corresponding to $\varepsilon_{1}$ Let's divide the intervals ( $0,\lambda$ ) And $(0,\mu)$ in $n$ And $m$ equal parts by points $x_{1},\ldots,x_{n-1}$ And $y_{1},\ldots,y_{m-1}$ The straight lines $x=x_{i}$ And $y=y_{k}$ Or $i=0,1,\ldots,n$ And $k=0,1,\ldots,m$ , ( $x_{0}=0,x_{n}=\lambda,y_{0}=0,y_{m}=\mu$ form a network $\Gamma$ We will calculate $z^{(0)}(x,y),p^{(0)}(x,y),q^{(0)}(x,y)$ on the nodes of this network. Let us denote by $\Delta_{ik}$ the rectangle formed by the lines $x=0,x=x_{i},y=0,y=y_{k}$ By applying the approximate calculation formulas, we will have
(4)

		$\displaystyle z^{(0)}\left(x_{i},y_{k}\right)=z_{ik}^{(0)}+R_{ik}^{(0)}$
		$\displaystyle p^{(0)}\left(x_{i},y_{k}\right)=p_{ik}^{(0)}+R_{ik}^{(0)}$
		$\displaystyle q^{(0)}\left(x_{i},y_{k}\right)=q_{ik}^{(0)}+R_{ik}^{(0)^{\prime\prime}}$

	$\displaystyle\left.z_{ik}^{(0)}=\frac{\lambda\mu}{2nm}\right\rvert\,\sum_{j=0}^{i-1}f_{0}\left(x_{j},0\right)$	$\displaystyle+\sum_{l=1}^{k-1}f_{0}\left(0,y_{e}\right)+\sum_{j=1}^{i}f_{0}\left(x_{j},y_{k}\right)+\sum_{l=1}^{k-1}f_{0}\left(x_{i},y_{l}\right)+$
		$\displaystyle\left.+2\sum_{j=1}^{i-1}\sum_{l=1}^{k-1}f_{0}\left(x_{j},y_{l:}\right)\right]$		(5)

		$\displaystyle p_{ik}^{(0)}=\frac{\mu}{2m}\left[f_{0}\left(x_{i},0\right)+f_{0}\left(x_{i},y_{k}\right)+2\sum_{l=2}^{k-1}f_{0}\left(x_{i},y_{l}\right)\right]$
		$\displaystyle q_{ik}^{(0)}=\frac{\lambda}{2n}\left[f_{0}\left(0,y_{k}\right)+f_{0}\left(x_{i},y_{k}\right)+2\sum_{j=1}^{i-1}f_{0}\left(x_{j},y_{k}\right)\right]$

and mons have

		$\displaystyle\left\|R_{ik}^{(0)}\right\|\leqslant\frac{\lambda\mu}{12}\frac{i}{n}\frac{k}{m}\left(\frac{\lambda^{2}}{n^{2}}+3\frac{\lambda\mu}{nm}+\frac{\mu^{2}}{m^{2}}\right)N.$
		$\displaystyle\left\|R_{ik}^{(0)^{\prime}}\right\|\leqslant\frac{\mu^{3}}{12m^{2}}\frac{k}{m}N,\quad\left\|R_{ik}^{(0)^{\prime\prime}}\right\|\leqslant\frac{\lambda^{3}}{12n^{2}}\frac{i}{n}N$

As $\frac{i}{n}\leqslant 1,\frac{k}{m}\leqslant 1$ we can still write

		$\displaystyle\left\|R_{ik}^{(0)}\right\|\leqslant\frac{\lambda\mu}{12}\left(\frac{\lambda^{2}}{n^{2}}+3\frac{\lambda\mu}{nm}+\frac{\mu^{2}}{m^{2}}\right)N$
		$\displaystyle\left\|R_{ik}^{(1)\prime}\right\|\leqslant\frac{\mu^{3}}{12m^{2}}N,\quad\left\|R_{ik}^{(0)^{\prime\prime}}\right\|\leqslant\frac{\lambda^{3}}{12n^{2}}N$

We will say that $z_{ik}^{(0)},p_{ik}^{(0)},q_{ik}^{(0)}$ given by formulas (5) are the "calculated values" of $z\left(x_{i},y_{k}\right),p\left(x_{i},y_{k}\right),q\left(x_{i},y_{k}\right)$ on the nodes $\left(x_{i},y_{k}\right)$ .

We can choose the smallest natural numbers $n$ And $m$ such as

\frac{\lambda^{3}}{12n^{2}}N<\varepsilon_{1},\quad\frac{\mu^{3}}{12m^{2}}N<\varepsilon_{1},\quad\frac{\lambda\mu}{12}\left(\frac{\lambda^{2}}{n^{2}}+3\frac{\lambda\mu}{mm}+\frac{\mu^{2}}{m^{2}}\right)N<\varepsilon_{1}

(7)

and then we will have in formulas (4)

\left|R_{ik}^{(0)}\right|,\left|R_{ik}^{(n)^{\prime}}\right|,\left|R_{ik}^{(0)^{\prime\prime}}\right|<\varepsilon_{1}

(8)

Natural numbers $n$ And $m$ once chosen by the conditions (7) remain fixed thereafter and consequently the network $\Gamma$ is well determined.
18. Let's move on to the calculation of $z^{(1)}(x,y),p^{(1)}(x,y),q^{(1)}(x,y)$ . We have

		$\displaystyle z^{(1)}(x,y)=\int_{0}^{xy}f\left[s,t,z^{(1)}(s,t),p^{(0)}(s,t),q^{(0)}(s,t)\right]dsdt$
		$\displaystyle p^{(1)}(x,y)=\int_{0}^{y}f\left[x,t,z^{(0)}(x,t),p\quad(x,t),q^{(0)}(x,t)\right]dt$

q^{(1)}(x,y)=\int_{0}^{x}f\left[s,y,z^{(0)}(s,y),p^{(0)}(s,y),q^{(0)}(s,y)\right]ds

and we can apply this to the calculation of $z^{(1}(x,y),p^{(1)}(x,y),q^{(1)}(x,y)$
formulas for approximate calculation on the nodes $\left(x_{i},y_{k}\right)$ formulas for approximate calculation on the nodes $\left(x_{i},y_{k}\right)$

z^{(1)}\left(x_{i},y_{k}\right)=\left[z_{ik}^{(1)}\right]+r_{ik}^{(1)}

	$\displaystyle p^{(1)}\left(x_{i},y_{k}\right)=\left[p_{ik}^{(1)}\right]+r_{ik}^{(1)^{\prime}}$		(10)
	$\displaystyle q^{(1)}\left(x_{i},y_{k}\right)=\left[q_{ik}^{(1)}\right]+r_{ik}^{(1)^{\prime\prime}},$

\left|r_{ik}^{(1)}\right|,\quad\left|r_{ik}^{(1)^{\prime}}\right|,\quad\left|r_{ik}^{(1)^{\prime\prime}}\right|<\varepsilon_{1}

(11)

And

$\displaystyle{\left[\begin{array}[]{l}(1)\\ lk]=\end{array}=\frac{\lambda\mu}{2nm}\right.}$	$\displaystyle\left\{\sum_{j=0}^{i-1}f\left[x_{j},0,z^{(0)}\left(x_{j},0\right),p^{(0)}\left(x_{j},0\right),q^{(0)}\left(x_{j},0\right)\right]\right.$
	$\displaystyle+\sum_{l=1}^{k-1}f\left[0,y_{l},z^{(0)}\left(0,y_{l}\right),p^{(0)}\left(0,y_{l}\right),q^{(0)}\left(0,y_{l}\right)\right]$
	$\displaystyle+\sum_{j=1}^{i}f\left[x_{j},y_{k},z^{(0)}\left(x_{j},y_{k}\right),p^{(0)}\left(x_{j},y_{k}\right),q^{(0)}\left(x_{j},y_{k}\right)\right]$
	$\displaystyle+\sum_{l=1}^{k-1}f\left[x_{i},y_{l},z^{(0)}\left(x_{i},y_{l}\right),p^{(0)}\left(x_{i},y_{l}\right),q^{(0)}\left(x_{i},y_{l}\right)\right]$
	$\displaystyle+2\sum_{j=1}^{i-1}\sum_{l=1}^{k-1}f\left[x_{j},y_{l},z^{(0)}\left(x_{j},y_{l}\right),p^{(0)}\left(x_{j},y_{l}\right),q^{(0)}\left(x_{j},y_{l}\right)\right]$
$\displaystyle{\left[p_{ik}^{(1)}\right]=}$	$\displaystyle\frac{\mu}{2m}\left\{f\left[x_{i},0,z^{(0)}\left(x_{i},0\right),p^{(0)}\left(x_{i},0\right),q^{(0)}\left(x_{i},0\right)\right]\right.$
	$\displaystyle+f\left[x_{i},y_{k},z^{(0)}\left(x_{i},y_{k}\right),p^{(0)}\left(x_{i},y_{k}\right),q^{(0)}\left(x_{i},y_{k}\right)\right]$	(12)
	$\displaystyle\left.+2\sum_{l=1}^{k-1}f\left[x_{i},y_{l},z^{(0)}\left(x_{i},y_{l}\right),p^{(0)}\left(x_{i},y_{l}\right),q^{(0)}\left(x_{i},y_{l}\right)\right]\right\}$
$\displaystyle{\left[q_{lk}^{(1)}\right]=}$	$\displaystyle\frac{\lambda}{2n}\left\{f\left[0,y^{k},z^{(0)}\left(0,y_{k}\right),p^{(0)}\left(0,y_{k}\right),q^{(0)}\left(0,y_{k}\right)\right]\right.$
	$\displaystyle+f\left[x_{i},y_{k},z^{(0)}\left(x_{i},y^{k}\right),p^{(0)}\left(x_{i},y_{k}\right),q^{(0)}\left(x_{i},y_{k}\right)\right]$
	$\displaystyle\left.+2\sum_{j=1}^{i-1}f\left[x_{j},y_{k},z^{(0)}\left(x_{j},y_{k}\right),p^{(0)}\left(x_{j},y^{k}\right),q^{(0)}\left(x_{j},y_{k}\right)\right]\right\}$

But the values of $z^{(0)}(x,y),p^{(0)}(x,y),q^{(0)}(x,y)$ on the nodes $\left(x_{i},y_{k}\right)$ are given approximately by formulas (4). If $\varepsilon_{1}$ is small enough we can generally replace in formulas (12) the values of $z^{(0)}\left(x_{i},y^{k}\right)p^{(0)}\left(x_{i},y_{k}\right),q^{(0)}\left(x_{i},y_{k}\right)$ by $z_{ik}^{(0)},p_{ik}^{(0)},q_{ik}^{(0)}$ We will demonstrate this in no. 20.

Let us designate by $z_{ik}^{(1)},p_{ik}^{(1)},q_{ik}^{(1)}$ the numbers obtained by substituting into $\left[z_{ik}^{(1)}\right],\left[p_{ik}^{(1)}\right],\quad\left[y_{ik}^{(1)}\right],\quad z^{(0)}\left(x_{j},y_{l}\right),\quad p^{(0)}\left(x_{j},y_{l}\right),\quad q^{(0)}\left(x_{j},y_{l}\right)\quad$ not $z_{jl}^{(0)},\quad p_{jl}^{(0)},q_{jl}^{(0)}$ that's to say

	$\displaystyle z_{ik}^{(1)}$	$\displaystyle=\frac{\lambda\mu}{2nm}\left[\sum_{j=0}^{i-1}f\left(x_{j},0,z_{j0}^{(0)},p_{j0}^{(0)},q_{j0}^{(0)}\right)+\sum_{l=1}^{k-1}f\left(0,y_{l},z_{0l}^{(0)},p_{0l}^{(0)},q_{0l}^{(0)}\right)\right.$
		$\displaystyle+\sum_{j=1}^{i}f\left(x_{j},y_{k},z_{jk}^{(0)},p_{jk}^{(0)},q_{jk}^{(0)}\right)+\sum_{l=1}^{k-1}f\left(x_{i},y_{l},z_{il}^{(0)},p_{il}^{(0)},q_{il}^{(0)}\right)$
		$\displaystyle\left.+2\sum_{j=1}^{i-1}\sum_{l=1}^{k-1}f\left(x_{j},y_{l},z_{jl}^{(0)},p_{jl}^{(0)},q_{jl}^{(0)}\right)\right]$
	$\displaystyle p_{ik}^{(1)}$	$\displaystyle=\frac{\mu}{2m}\left[f\left(x_{i},0,z_{i0}^{(0)},p_{i0}^{(0)},q_{i0}^{(0)}\right)+f\left(x_{i},y_{k},z_{ik}^{(0)},p_{ik}^{(0)},q_{lk}^{(0)}\right)\right.$
		$\displaystyle\left.+2\sum_{l=1}^{k-1}f\left(x_{i},y_{l},z_{ll}^{(0)},p_{il}^{(0)},q_{il}^{(0)}\right)\right]$
	$\displaystyle q_{ik}^{(1)}$	$\displaystyle=\frac{\lambda}{2n}\left[f\left(0,y_{k},z_{0k}^{(0)},p_{0k}^{(1)},q_{0k}^{(0)}\right)+f\left(x_{i},y_{k},z_{ik}^{(0)},p_{ik}^{(0)},q_{ik}^{(0)}\right)\right.$
		$\displaystyle\left.+2\sum_{j=1}^{i-1}f\left(x_{j},y_{k},z_{jk}^{(0)},p_{jk}^{(0)},q_{jk}^{(0)}\right)\right]$

By asking

		$\displaystyle{\left[z_{ik}^{(1)}\right]=z_{ik}^{(1)}+\varrho_{ik}^{(1)}}$
		$\displaystyle{\left[p_{ik}^{(1)}\right]=p_{ik}^{(1)}+\varrho_{ik}^{(1)^{\prime}}}$
		$\displaystyle{\left[q_{ik}^{(1)}\right]=q_{ik}^{(1)}+\varrho_{ik}^{(1)^{\prime\prime}},}$

we will have

		$\displaystyle\begin{array}[]{l}\varrho_{ik}^{(1)}=\frac{\lambda\mu}{2nm}\left\{\ldots+\left[f\left(x_{j},y_{l},z^{(0)}\left(x_{j},y_{l}\right),p^{(0)}\left(x_{j},y_{l}\right),q^{(0)}\left(x_{j},y_{l}\right)\right.\right.\right.\\ \left.\left.\quad-f\left(x_{j},y_{l},z_{jl}^{(0)},p_{jl}^{(0)},q_{jl}^{(0)}\right)\right]+\cdots\right\}\end{array}$
		$\displaystyle\varrho_{ik}^{(1)\prime}=\ldots\ldots$
		$\displaystyle\varrho_{ik}^{(1)^{\prime\prime}}=\ldots\ldots$

18 - Mathematica

From these formulas, it follows that

	$\displaystyle\left\|\varrho_{ik}^{(1)}\right\|\leqslant\lambda\mu\frac{i}{n}\frac{k}{m}(A+B+C)\varepsilon_{1}\leqslant\lambda\mu(A+B+C)\varepsilon_{1}$
	$\displaystyle\left\|\varrho_{ik}^{(1)^{\prime}}\right\|\leqslant\mu\frac{k}{m}(A+B+C)\varepsilon_{1}\leqslant\mu(A+B+C)\varepsilon_{1}$		(14)
	$\displaystyle\left\|\varrho_{ik}^{(1)^{\prime\prime}}\right\|\leqslant\lambda\frac{i}{n}(A+B+C)\varepsilon_{1}\leqslant\lambda(A+B+C)\varepsilon_{1}$

By designating

Q=\max[\lambda\mu(A+B+C),\mu(A+B+C),\lambda(A+B+C)]

(15)

we will have
(14')

\left|\varrho_{ik}^{(1)}\right|,\quad\left|\varrho_{ik}^{(1)^{\prime}}\right|,\quad\left|\varrho_{ik}^{(1)^{\prime\prime}}\right|<Q\varepsilon_{1}.

So we have

$\displaystyle z^{(1)}\left(x_{i},y_{k}\right)$	$\displaystyle=z_{ik}^{(1)}+R_{ik}^{(1)}$
$\displaystyle p^{(1)}\left(x_{i},y_{k}\right)$	$\displaystyle=p_{ik}^{(1)}+R_{ik}^{(1)^{\prime}}$	(16)
$\displaystyle q^{(1)}\left(x_{i},y_{k}\right)$	$\displaystyle=q_{ik}^{(1)}+R_{ik}^{(1)^{\prime\prime}}$

R_{ik}^{(1)}=r_{ik}^{(1)}+\varrho_{ik}^{(1)},\quad R_{ik}^{(1)^{\prime}}=r_{ik}^{(1)^{\prime}}+\varrho_{ik}^{(1)^{\prime}},\quad R_{ik}^{(1)^{\prime\prime}}=r_{ik}^{(1)^{\prime\prime}}+\varrho_{ik}^{(1)^{\prime\prime}}.

According to inequalities (11) and (14) we have in formulas (16)

\left|R_{ik}^{(1)}\right|,\quad\left|R_{ik}^{(1)^{\prime}}\right|,\quad\left|R_{ik}^{(1)^{\prime\prime}}\right|<(1+Q)\varepsilon_{1}.

(17)

19.

Let's move on to the general case. Suppose we have proven the formulas

	$\displaystyle z^{(s-1)}\left(x_{i},y_{k}\right)=z_{ik}^{(s-1)}+R_{ik}^{(s-1)}$
	$\displaystyle p^{(s-1)}\left(x_{i},y_{k}\right)=p_{ik}^{(s-1)}+R_{ik}^{(s-1)^{\prime}}$		(18)
	$\displaystyle q^{(s-1)}\left(x_{i},y_{k}\right)=q_{ik}^{(s-1)}+R_{ik}^{(s-1)^{\prime\prime}}$

analogous to formulas (16) and inequalities

\left|R_{lk}^{(s-1)}\right|,\quad\left|R_{lk}^{(s-1)}\right|,\quad\left|R_{lk}^{(s-1)^{\prime\prime}}\right|<\left(1+Q+\ldots+Q^{s-1}\right)\varepsilon_{1}

(19)

analogous to inequalities (17).
For $s=2$ formulas (18) and inequalities (19) reduce to formulas (16) and inequalities (17).

The function $z^{(s)}(x,y)$ and its partial derivatives $p^{(s)}(x,y),q^{(s)}\left(x,y^{\prime}\right)$ are given by the formulas

	$\displaystyle z^{(s)}(x,y)$	$\displaystyle=\int_{0}^{xy}f\left[\xi,\eta,z^{(s-1)}(\xi,\eta),p^{(s-1)}(\xi,\eta),q^{(s-1)}(\xi,\eta)\right]d\xi d\eta$
	$\displaystyle p^{(s)}(x,y)$	$\displaystyle=\int_{0}^{y}f\left[x,\eta,z^{(s-1)}(x,\eta),p^{(s-1)}(x,\eta),q^{(s-1)}(x,\eta)\right]d\eta$
	$\displaystyle q^{(s)}(x,y)$	$\displaystyle=\int_{0}^{x}f\left[\xi,y,z^{(s-1)}(\xi,y),p^{(s-1)}(\xi,y),q^{(s-1)}(\xi,y)\right]d\xi$

and we will have on the network nodes $\Gamma$

	$\displaystyle z^{(s)}\left(x_{i},y_{k}\right)$	$\displaystyle=\int_{0}^{x_{i}y_{k}}\int_{0}f\left[\xi,\eta_{1},z^{(s-1)}(\xi,\eta),p^{(s-1)}(\xi,\eta),q^{(s-1)}(\xi,\eta)\right]d\xi d\eta$
	$\displaystyle p^{(s)}\left(x_{i},y_{k}\right)$	$\displaystyle=\int_{0}^{y_{k}}f\left[x_{i},\eta,z^{(s-1)}\left(x_{i},\eta\right),p^{(s-1)}\left(x_{i},\eta\right),q^{(s-1)}\left(x_{i},\eta\right)\right]d\eta$
	$\displaystyle q^{(s)}\left(x_{i},y_{k}\right)$	$\displaystyle=\int_{0}^{x_{i}}f\left[\xi,\eta_{k},z^{(s-1)}\left(\xi,\eta_{k}\right),p^{(s-1)}\left(\xi,\eta_{k}\right),q^{(s-1)}\left(\xi,\eta_{k}\right)\right]d\xi$

We can apply the approximate calculation formulas we used to the integrals on the right-hand side, and we will have
where

	$\displaystyle z^{(s)}\left(x_{i},y_{k}\right)=\left[z_{lk}^{(s)}\right]+r_{ik}^{(s)}$		(20)
	$\displaystyle p^{(s)}\left(x_{i},y_{k}\right)=\left[p_{ik}^{(s)}\right]+r_{ik}^{(s)^{\prime}}$
	$\displaystyle q^{(s)}\left(x_{i},y_{k}\right)=\left[q_{ik}^{(s)}\right]+r_{ik}^{(s)^{\prime\prime}}$
	$\displaystyle{\left[z_{ik}^{(s)}\right]=\frac{\lambda_{i}}{2nm}\left\{\sum_{j=0}^{i-1}f\left[x_{j},0,z^{(s-1)}(xj,0),p^{(s-1)}\left(x_{i},0\right),q^{(s-1)}\left(x_{j},0\right)\right]+\right.}$
	$\displaystyle+\sum_{l=1}^{k-1}f\left[0,y_{l},z^{(s-1)}\left(0,y_{l}\right),p^{(s-1)}\left(0,y_{l}\right),q^{(s-1)}\left(0,y_{l}\right)\right]+$
	$\displaystyle+\sum_{j=1}^{i}f\left[x_{j},y_{k},z^{(s-1)}\left(x_{j},y_{k}\right),p^{(s-1)}\left(x_{j},y_{k}\right),q^{(s-1)}\left(x_{j},y_{k}\right)\right]+$
	$\displaystyle+\sum_{l=1}^{k-1}f\left[x_{i},y_{l},z^{(s-1)}\left(x_{i},y_{l}\right),p^{(s-1)}\left(x_{i},y_{l}\right),q^{(s-1)}\left(x_{i},y_{l}\right)\right]+$

+2\sum_{j=1}^{l-1}\sum_{l=1}^{k-1}f\left[x_{j},y_{l},z^{(s-1)}\left(x_{j},y_{l}\right),p^{(s-1)}\left(x_{j},y_{l}\right),q^{(s-1)}\left(x_{j},y_{l}\right)\right]

$\displaystyle{\left[p_{ik}^{(s)}\right]=}$	$\displaystyle\frac{\mu}{2m}\left\{f\left[x_{i},0,z^{(s-1)}\left(x_{i},0\right),p^{(s-1)}\left(x_{i},0\right),q^{(s-1)}\left(x_{i},0\right)\right]+\right.$	(21)
	$\displaystyle+f\left[x_{i},y_{k},z^{(s-1)}\left(x_{i},y_{k}\right),p^{(s-1)}\left(x_{i},y_{k}\right),q^{(s-1)}\left(x_{i},y_{k}\right)\right]+$	(24)
	$\displaystyle\left.+2\sum_{l=1}^{k-1}f\left[x_{i},y_{l},z^{(s-1)}\left(x_{i},y_{l}\right),p^{(s-1)}\left(x_{i},y_{l}\right),q^{(s-1)}\left(x_{i},y_{l}\right)\right]\right\}$
$\displaystyle{\left[q_{ik}^{(s)}\right]=}$	$\displaystyle\frac{\lambda}{2n}\left\{f\left[0,y_{k},z^{(s-1)}\left(0,y_{k}\right),p^{(s-1)}\left(0,y_{k}\right),q^{(s-1)}\left(0,y_{k}\right)\right]+\right.$
	$\displaystyle+f\left[x_{i},y_{k},z^{(s-1)}\left(x_{i},y_{k}\right),p^{(s-1)}\left(x_{i},y_{k}\right),q^{(s-1)}\left(x_{i},y_{k}\right)\right]+$
	$\displaystyle\left.+2\sum_{j=1}^{i-1}f\left[x_{j},y_{k},z^{(s-1)}\left(x_{j},y_{k}\right),p^{(s-1)}\left(x_{j},y_{k}\right),q^{(s-1)}\left(x,y_{k}\right)\right]\right\}$

\left.+2\sum_{j=1}^{i-1}f\left(x_{j},y_{k},z_{ik}^{(s-1)},p_{jk}^{(s-1)},q_{jk}^{(s-1)}\right)\right].

By asking

		$\displaystyle{\left[\sum_{ik}^{(s)}\right]=z_{ik}^{(s)}+\varrho_{ik}^{(s)}}$
		$\displaystyle{\left[p_{ik}^{(s)}\right]=p_{ik}^{(s)}+\varrho_{ik}^{(s)^{\prime}}}$
		$\displaystyle{\left[q_{ik}^{(s)}\right]=q_{ik}^{(s)}+\varrho_{ik}^{(s)^{\prime\prime}}}$

we will have

		$\displaystyle\varrho_{ik}^{(s)}=\frac{\lambda_{il}}{2nm}\left\{\cdots+\left[f\left(x_{j},y_{l},z^{(s-1)}\left(x_{j},y_{l}\right),p^{(s-1)}\left(x_{j},y_{l}\right),q^{(s-1)}\left(x_{j},y_{l}\right)\right)-\right.\right.$
		$\displaystyle\left.\left.\quad-f\left(x_{j},y_{l},z_{ll}^{(s-1)},p_{jl}^{(s-1)},q_{ll}^{(s-1)}\right)\right]+\ldots\right\}$
		$\displaystyle\quad\varrho_{ik}^{(s)\prime}=\ldots$
		$\displaystyle\varrho_{ik}^{(s)\prime\prime}=\ldots$

and where

\left|r_{ik}^{(s)}\right|,\quad\left|r_{ik}^{(s)^{\prime}}\right|,\quad\left|r_{ik}^{(s)^{\prime\prime}}\right|<\varepsilon_{1}.

(22)

Let us designate by $z_{ik}^{(s)},p_{ik}^{(s)},q_{ik}^{(s)}$ the numbers obtained by substituting into the second members of formulas (21) in general $z^{(s-1)}\left(x_{j},y_{l}\right),p^{(s-1)}\left(x_{j},y_{l}\right)$ , $q^{(s-1)}\left(x_{j},y_{l}\right)$ by $z_{jl}^{(s-1)},p_{jl}^{(s-1)},q_{jl}^{(s-1)}$ We will demonstrate in section 11r. 20 that this is possible. We will therefore have
$z_{ik}^{(s)}=\frac{\lambda\mu}{2nm}\left[\sum_{j=0}^{l-1}f\left(x_{j},0,z_{j0}^{(s-1)},p_{j0}^{(s-1)},q_{j0}^{(s-1)}\right)+\sum_{l=1}^{k-1}f\left(0,y_{l},z_{0l}^{(s-1)},p_{0l}^{(s-1)},q_{0l}^{(s-1)}\right)+\right.$

\begin{array}[]{r}+\sum_{j=1}^{l}f\left(x_{j},y_{k},z_{jk}^{(s-1)},p_{jk}^{(s-1)},q_{jk}^{(s-1)}\right)+\sum_{l=1}^{k-1}f\left(x_{i},y_{l},z_{ll}^{(s-1)},p_{il}^{(s-1)}\right.\\ \left.+2\sum_{j=1}^{i-1}\sum_{l=1}^{k-1}f\left(x_{j},y_{l},z_{jl}^{(s-1)},p_{jl}^{(s-1)},q_{jl}^{(k-1)}\right)\right]\end{array}

(23) $p_{ik}^{(s)}=\frac{\mu}{2m}\left[f\left(x_{i},0,z_{i0}^{(s-1)},p_{i0}^{(s-1)},q_{i0}^{(s-1)}\right)+f\left(x_{i},y_{k},z_{ik}^{(s-1)},p_{ik}^{(s-1)},q_{ik}^{(s-1)}\right)+\right.$

	$\displaystyle\left.+2\sum_{l=1}^{k-1}f\left(x_{i},y_{l},z_{il}^{(s-1)},p_{il}^{(s-1)},q_{il}^{(s-1)}\right)\right]$
	$\displaystyle q_{ik}^{(s)}=\frac{\lambda}{2n}\left[f\left(0,y_{k},z_{ok}^{(s-1)},p_{ok}^{(s-1)},q_{ok}^{(k-1)}\right)+f\left(x_{i},y_{k},z_{ik}^{(s-1)},p_{ik}^{(s-1)},q_{ik}^{(s-1)}\right)+\right.$		(27)

and according to formulas (18), (19) we can write

		$\displaystyle\left\|\varrho_{ik}^{(s)}\right\|\leqslant\lambda\mu(A+B+C)\left(1+Q+\ldots+Q^{s-1}\right)\varepsilon_{1}$
		$\displaystyle\left\|\varrho_{ik}^{(s)^{\prime}}\right\|\leqslant\mu(A+B+C)\left(1+Q+\ldots+Q^{s-1}\right)\varepsilon_{1}$
		$\displaystyle\left\|\varrho_{ik}^{(s)^{\prime\prime}}\right\|\leqslant\lambda(A+B+C)\left(1+Q+\ldots+Q^{s-1}\right)\varepsilon_{1},$

or, according to the meaning of the number $Q$ ,

\left|\varrho_{ik}^{(s)}\right|,\left|\varrho_{ik}^{(s)^{\prime}}\right|,\left|\varrho_{ik}^{(s)^{\prime\prime}}\right|\leqq\left(Q+Q^{2}+\cdots+Q^{s}\right)\varepsilon_{1}

(25)

Returning to formulas (20), (24) we will have

		$\displaystyle z^{(s)}\left(x_{i},y_{k}\right)=z_{ik}^{(s)}+R_{ik}^{(s)}$
		$\displaystyle p^{(s)}\left(x_{i},y_{k}\right)=p_{ik}^{(s)}+R_{ik}^{(s)^{\prime}}$
		$\displaystyle q^{(s)}\left(x_{i},y_{k}\right)=q_{ik}^{(s)}+R_{ik}^{(s)\prime^{\prime\prime}}$

		$\displaystyle R_{ik}^{(s)}=r_{ik}^{(s)}+\varrho_{ik}^{(s)}$
		$\displaystyle R_{ik}^{(s)^{\prime}}=r_{ik}^{(s)^{\prime}}+\varrho_{ik}^{(s)^{\prime}}$
		$\displaystyle R_{ik}^{(s)^{\prime\prime}}=r_{ik}^{(s)^{\prime\prime}}+\varrho_{ik}^{(s)^{\prime\prime}}$

and taking into account inequalities (22) and (25) we will have

\left|R_{ik}^{(s)}\right|,\quad\left|R_{ik}^{(s)^{\prime}}\right|,\quad\left|R_{ik}^{(s)^{\prime\prime}}\right|\leqslant\left(1+Q+\ldots+Q^{s}\right)\varepsilon_{1}.

Comparing formulas (26), (27) with formulas (18), (19) shows that formulas (26), (27) are valid for $s=12,\ldots$ , v.
20. We have therefore established an algorithm for calculating numbers $z_{ik}^{(s)},p_{ik}^{(s)},q_{ik}^{(s)}$ The numbers $z_{ik}^{(0)},q_{ik}^{(0)},q_{ik}^{(0)}$ are given by formulas (5), and the numbers $z_{ik}^{(s)},p_{ik}^{(s)},q_{ik}^{s)}$ For $s=1,2,\ldots,v$ , are given by the recurrence formulas (23).

In order to apply the recurrence formulas (23), it is necessary to show that one can choose $\varepsilon_{1}$ so that the coordinate points $\left(x_{j},y_{l}\right.$ , $\left.z_{jl}^{(s)},p_{jl}^{(s)},q_{jl}^{(s)}\right)$ Or $s=0,1,2,\ldots,\nu$ are located in the domain $D$ function definition $f(x,y,z,p,q)$ . For $s=v$ We have

		$\displaystyle z_{jl}^{(v)}=z^{(v)}\left(x_{j},y_{l}\right)-R_{jl}^{(v)}$
		$\displaystyle p_{jl}^{(v)}=p^{(v)}\left(x_{j},y_{l}\right)-R_{jl}^{(v)^{\prime}}$
		$\displaystyle q_{jl}^{(v)}=q^{(v)}\left(x_{j},y_{l}\right)-R_{jl}^{(v)^{\prime\prime}}.$

Since the nodes $\left(x_{j},y_{l}\right)$ are taken from the rectangle $\Delta$ , we have according to formulas (8)

\left|z^{(\nu)}\left(x_{j},y_{l}\right)\right|\leqq\alpha-\delta,\quad\left|p^{(\nu)}\left(x_{j},y_{l}\right)\right|\leqq\beta-\delta,\quad\left|q^{(\nu)}\left(x_{j},y_{l}\right)\right|\leqq\gamma-\delta

And so the previous formulas show that

		$\displaystyle\left\|z_{jl}^{(v)}\right\|\leqq\alpha-\delta+\left\|R_{jl}^{(v)}\right\|\leqq\alpha-\delta+\left(1+Q+\ldots+Q^{\nu}\right)\varepsilon_{1}$
		$\displaystyle\left\|p_{jl}^{(v)}\right\|\leqq\beta-\delta+\left\|R_{jl}^{(v)}\right\|\leqq\beta-\delta+\left(1+Q+\ldots+Q^{\nu}\right)\varepsilon_{1}$
		$\displaystyle\left\|q_{jl}^{(v)}\right\|\leqq\gamma-\delta+\left\|R_{jl}^{(v)^{\prime\prime}}\right\|\leqq\gamma-\delta+\left(1+Q+\ldots+Q^{\nu}\right)\varepsilon_{1}.$

If we take
we will have

\varepsilon_{1}<\frac{\delta}{1+Q+\ldots+Q^{v}}

(28)

\left|z_{ll}^{(v)}\right|\leqq\alpha,\quad\left|p_{ll}^{(v)}\right|\leqq\beta,\quad\left|q_{jl}^{(v)}\right|\leqq\gamma

and consequently the point with coordinates $\left(x_{j},y_{l},z_{jl}^{(v)},p_{jl}^{(v)},q_{jl}^{(v)}\right)$ is located in the domain $D$ .

For any index $s<v$ , we have, through similar calculations
$\left|z_{jl}^{(s)}\right|\leqq\alpha-\delta+\left(1+Q+\ldots+Q^{s}\right)\varepsilon_{1}<\alpha-\delta+\left(1+Q+\ldots+Q^{\nu}\right)\varepsilon_{1}<\alpha\left|p_{jl}^{(s)}\right|\leqq\beta-\delta+\left(1+Q+\ldots+Q^{s}\right)\varepsilon_{1}<\beta-\delta+\left(1+Q+\ldots+Q^{v}\right)\varepsilon_{1}<\beta\left|q_{ji}^{(s)}\right|\leqq\gamma-\delta+\left(1+Q+\ldots+Q^{s}\right)\varepsilon_{1}<\gamma-\delta+\left(1+Q+\ldots+Q^{\nu}\right)\varepsilon_{1}<\gamma$ , which proves that the points of coordinates $\left(x_{j},y_{l},z_{jl}^{(s)},p_{jl}^{(s)},q_{ji}^{(s)}\right)$ For $s=0,1,2,\ldots,v-1$ are located in the domain $D$ .

So, assuming that $\varepsilon_{1}$ verifying inequality (28), formulas (13), and in general formulas (23) are valid for $s=1,2,\ldots,\nu$ 21.
We can now specify the number $\varepsilon_{1}$ By working with formulas (20) and inequalities (27) $s=v$ and taking into account inequality (28), let

\varepsilon_{1}=\min\left(\frac{\delta}{1+Q+\ldots+Q^{\nu}},\frac{\varepsilon}{1+Q+\ldots+Q^{\nu}}\right).

(29)

We will then have in formulas (26) for $s=v$ ,

\left|R_{ik}^{(v)}\right|,\quad\left|R_{ik}^{(v)^{\prime}}\right|,\quad\left|R_{ik}^{(v)^{\prime\prime}}\right|<\varepsilon

(30)

and for $s<v$ we will have in formulas (26)

\left|R_{ik}^{(s)}\right|,\quad\left|R_{ik}^{(s)^{\prime}}\right|,\quad\left|R_{ik}^{(s)^{\prime\prime}}\right|\leqslant\left(1+Q+\cdots+Q^{s}\right)\varepsilon_{1}<\left(1+Q+\ldots+Q^{v}\right)\varepsilon_{1}

that's to say

\left|R_{ik}^{(s)}\right|,\quad\left|R_{ik}^{(s)^{\prime}}\right|,\quad\left|R_{ik}^{(s)^{\prime\prime}}\right|<\varepsilon.

(31)

22.

In summary, the method for numerically calculating the values of the integral $z(x,y)$ and its partial derivatives $p(x,y),q(x,y)$ on the network nodes $I$ is as follows: first we calculate the numbers $z_{ik}^{(s)},p_{ik}^{(s)},q_{ik}^{(s)}$ by formulas (5) and (23). Then we have formula (26) and, according to inequalities (30) and (31), we have

\left|z^{(s)}\left(x_{i},y_{k}\right)-z_{ik}^{(s)}\right|,\quad\left|p^{(s)}\left(x_{i},y_{k}\right)-p_{ik}^{(s)}\right|,\quad\left|q^{(s)}\left(x_{i},y_{k}\right)-q_{ik}^{(s)}\right|<\varepsilon

(32)

For $s=0,1,\ldots$ Let 's
consider inequalities

		$\displaystyle z\left(x_{i},y_{k}\right)-z_{ik}^{(v)}=\left[z\left(x_{i},y_{k}\right)-z^{(v)}\left(x_{i},y_{k}\right)\right]+\left[z^{(v)}\left(x_{i},y_{k}\right)-z_{ik}^{(v)}\right]$
		$\displaystyle p\left(x_{i},y_{k}\right)-p_{ik}^{(v)}=\ldots$
		$\displaystyle q\left(x_{i},y_{k}\right)-q_{ik}^{(v)}=\ldots$

and taking into account inequalities (5) of § 1 and (32). We will have

	$\displaystyle\left\|z\left(x_{i},y_{k}\right)-z_{ik}^{(\nu)}\right\|<2\varepsilon$
	$\displaystyle\left\|p\left(x_{i},y_{k}\right)-p_{ik}^{(\nu)}\right\|<2\varepsilon$		(33)
	$\displaystyle\left\|q\left(x_{i},y_{k}\right)-q_{ik}^{(\nu)}\right\|<2\varepsilon$

Thus we have attached to a positive number $\varepsilon$ given a network $I$ formed by the nodes ( $x_{i},y_{k}$ and an algorithm for calculating numbers $z_{ik}^{(s)},p_{ik}^{(s)},q_{ik}^{(s)}$ with respect to this network, for $s=0,1,\ldots,v$ ,
these numbers being for $s=v$ , the approximate values of the integral $z(x,y)$ and its partial derivatives $p(x,y),q(x,y)$ on the network nodes, the absolute values of the differences $z\left(x_{i},y_{k}\right)-z_{ik}^{(v)},p\left(x_{i},y_{k}\right)-p_{ik}^{(v)},q\left(x_{i},y_{k}\right)-q_{ik}^{(i)}$ being smaller than $2\varepsilon$ .

BIBLIOGRAPHY

[1] Radon J., Restausdrücke bei Interpolations und Quadratur Formeln durch bestimmte Integralen. Monatshefte für Mathematik und Physik 42, 389 (1935).
[2] Ionescu DV, Cuadraturi numerice. București, Editura Tehnică 1957. Received on 25. XI. 1959.

MATHEMATICA VOL. 1 (24), 2, 1959. pp. 281-286

APPLICATIONS OF GENERALIZED CONVEX FUNCTIONS

1.

In this work we will give applications of the notion of convex function with respect to a set of interpolating functions, which we introduced in our work [3].

Either $E$ a set of points on the real axis and $f_{n}$ a set of real functions and a real variable. In [3] we gave the

Definition 1. The set $\mathscr{F}_{n}$ is called order interpolation $n$ on $E$ or simply a set of the type $I_{n}\{E\}$ if:
(A) The elements of $\mathscr{F}_{n}$ are continuous functions on $E$ (
B) Whatever the $n$ distinct points of $E$
(1)

x_{1},x_{2},\ldots,x_{n}

and whatever the $n$ numbers
(2)

y_{1},y_{2},\ldots,y_{n}

there is one function and only one $\varphi(x)\in\mathscr{F}_{n}$ such as one might have

\varphi\left(x_{i}\right)=y_{i},\quad i=1,2,\ldots,n.

(3)

To highlight conditions (3) and the set $\mathscr{F}_{n}$ , we use for the function $\varphi(x)$ which satisfies conditions (3) the notation
and also

L\left(\mathcal{F}_{n};x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}\mid x\right)

L\left(\mathscr{F}_{n};x_{1},x_{2},\ldots,x_{n};f\mid x\right)

when $f(x)$ is a function that takes the values (2) at the corresponding points (1).

Let us consider a system of $n+1$ distinct points of the set $E$

	$\displaystyle\left\|z(x,y)-z^{(\nu)}(x,y)\right\|\leqslant\varepsilon$
	$\displaystyle\left\|p(x,y)-p^{(\nu)}(x,y)\right\|\leqslant\varepsilon$		(5)
	$\displaystyle\left\|q(x,y)-q^{(\nu)}(x,y)\right\|\leqslant\varepsilon$

	$\displaystyle\left\|z^{(\nu)}\left(x_{i},y_{k}\right)-z_{ik}^{(\nu)}\right\|<\varepsilon$
	$\displaystyle\left\|p^{(\nu)}\left(x_{i},y_{k}\right)-p_{k}^{(\nu)}\right\|<\varepsilon$		(9)
	$\displaystyle\left\|q^{(\nu)}\left(x_{i},y_{k}\right)-q_{ik}^{(\nu)}\right\|<\varepsilon.$

	$\displaystyle\left\|\varrho_{ik}^{(1)}\right\|\leqslant\lambda\mu\frac{i}{n}\frac{k}{m}(A+B+C)\varepsilon_{1}\leqslant\lambda\mu(A+B+C)\varepsilon_{1}$
	$\displaystyle\left\|\varrho_{ik}^{(1)^{\prime}}\right\|\leqslant\mu\frac{k}{m}(A+B+C)\varepsilon_{1}\leqslant\mu(A+B+C)\varepsilon_{1}$		(14)
	$\displaystyle\left\|\varrho_{ik}^{(1)^{\prime\prime}}\right\|\leqslant\lambda\frac{i}{n}(A+B+C)\varepsilon_{1}\leqslant\lambda(A+B+C)\varepsilon_{1}$

	$\displaystyle\left\|z\left(x_{i},y_{k}\right)-z_{ik}^{(\nu)}\right\|<2\varepsilon$
	$\displaystyle\left\|p\left(x_{i},y_{k}\right)-p_{ik}^{(\nu)}\right\|<2\varepsilon$		(33)
	$\displaystyle\left\|q\left(x_{i},y_{k}\right)-q_{ik}^{(\nu)}\right\|<2\varepsilon$

Cubature formulas. Application to the numerical integration of second-order partial differential equations of hyperbolic type

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CUBATURE FORMULAS. APPLICATION TO THE NUMERICAL INTEGRATION OF SECOND-ORDER HYPERBOLIC PARTIAL DIFFERENTIAL EQUATIONS

I. CUBATURAL FORMULAS

§. First cubature formula

§ 2. Second cubature formula

$. 3. Approximate calculation of a double integral in a rectangle $D$

II. NUMERICAL INTEGIATION OF THE SECOND-ORDER HYPERBOLIC PARTHELIC DIFFERENTIAL EQUATION

s. 1. General considerations and assumptions

§. 2. Application of the cubature formula (21, $1, I)

BIBLIOGRAPHY

MATHEMATICA VOL. 1 (24), 2, 1959. pp. 281-286

APPLICATIONS OF GENERALIZED CONVEX FUNCTIONS

Related Posts

Cubature formulas. Application to the numerical integration of second-order partial differential equations of hyperbolic type

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

CUBATURE FORMULAS. APPLICATION TO THE NUMERICAL INTEGRATION OF SECOND-ORDER HYPERBOLIC PARTIAL DIFFERENTIAL EQUATIONS

I. CUBATURAL FORMULAS

§. First cubature formula

§ 2. Second cubature formula

$. 3. Approximate calculation of a double integral in a rectangleDD

II. NUMERICAL INTEGIATION OF THE SECOND-ORDER HYPERBOLIC PARTHELIC DIFFERENTIAL EQUATION

s. 1. General considerations and assumptions

§. 2. Application of the cubature formula (21, $1, I)

BIBLIOGRAPHY

MATHEMATICA VOL. 1 (24), 2, 1959. pp. 281-286

APPLICATIONS OF GENERALIZED CONVEX FUNCTIONS

Related Posts

$. 3. Approximate calculation of a double integral in a rectangle $D$