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32
Section 5. Mathematics
deleted numbers. Total «11» numbers are deleted. Thus number m = 6 deleted 2 times, as 1 + 5 -1 = 6 + 7 • 0 = 6. Therefore, in general the number of deletions is «10», as there is in (2).
To find the double, triple or more deletions necessary to solve Diophantine equations of the AS, included in (l). In this example, there are the equations:
1 + 5k = 1 + 7t gives m =1 + 35k l; 1 + 5k = 6 + 7t gives m = 6 + 35k; 4 + 5k = 1 + 7t gives m = 29 + 35k; 4 + 5k = 6 + 7t gives m = 34 + 35k .
They give bad m, which are deleted twice by the first four equations. When taking the next m, there are triple, quadruple and so on deletions. For example, 1 + 5k = 6 + 7t = 2 + 13s give m = 41 + 455k that only these AS always deleted three times. Thus, we note that if the number in is deleted n times, his contribution in Nab (k) is Ckn = n!/ k!(n - k) times.
As a final example, we calculate the number of PT in the buffer zone L3. We use twelve AS on mi = 1, i2 = 2, i3 = 3. The first not deleted number is i4 = 5. It gives the PT {29; 31}. We are looking for the right border area:
M3 =(((6• 5-1)2±l)/6)-1 = ((841 ±l)/6)-1 . We have L3 = [5,139] .These AS delete N3 (l) = 166 numbers one time, deleted N3 (2) = 79 numbers twice, deleted N3 (3) = 21 numbers three times, N 3 (4 ) = 2. numbers four times. These numbers are m = 41 and m = 54. In result the number of PT coincides with the fact L3 = [5,139] (5;139) = 139 - 5 +1 -166 + 79 - 21 + 2 = 29.
This calculation scheme tested by us up to m = 10000000 and gives the exact number of PT. This scheme of calculating the number of PT also points to their distribution on the whole real axis, up to infinity. In sum
r (d,i) = (- n +1) - Nne (t) + Nne (2) - Nah (3) +. ..^(-l)+t Nab (k)
(6)
we have an alternating structure in which each subsequent summand modulo less than the previous one, but with the growth of b these modules increase. Therefore, obtaining a finite number of PT on the entire axis is impossible, i. e. it is infinite.
References:
1. Druzhinin V V, NTVP, № 1, 22-25, 2014.
Mustapokulov Khamdam Yangiboevich, senior teacher of Tashkent state technical university named after Abu Raikhon Beruni E-mail: mhy1506@yandex.ru Minarova Nigora Xudayberganovna, senior teacher of Tashkent state technical university named after Abu Raikhon Beruni
About invariance in problem heat of exchange with border management
Abstract: In given work is considered the question about strong and weak invariance of constant ambiguous image for equations heat of exchange with border management. Sufficient conditions are received for strong or weak invariance given ambiguous image.
Keywords: the invariant ensemble, management, ambiguous image, system management with portioned parameters.
1. Introduction
/
Let =S dx"
,(x)-dz- , Pz = — + h(x)z, x efi,
'ôx, dn K '
1 y
where atj (x) = ajt (x) e C1 (Q), i, j = 1,...,n, Q- limited area in Rn with the piecewise-smooth boundary, A - elliptical differential operator, i. e. exists positive constant y such that
^ aij (xfor any x eQ and real number
i, j=l
li,...,|„, * 0, h(x)- given positive continuous function,
i=1
— - derivative with respect to the outer normal to the
dn
boundary of n at point x edQ.
Consider the problem ofthermal management [1, 30-35]
^^ = Az(t, x), 0 < t < T, x eQ (l)
with boundary and initial conditions
Pz(t,x) = u(t,x),0 < t < T, x edQ, (2)
z (0, x ) = z0 (x), x eQ, (3)
here z = z (t, x)- unknown function, T - arbitrary positive constant, z0 ( ) e L2 (Q) - the initial function. Governance is a measurable function u (•, ■) e L2 (ST), where ST ={(t,x) t e [0,T], x edQ}.
In [1, 23-39] that in any u (•,•) e L2 and z0 (•) e L2 (Q) the problem (l) - (3) has a unique solution z = z(t,^ Hilbert space fV2'-0 (QT), where QT ={(t, x)| t e(0, T), x e^, consisting of elements of the space L2 (QT), with square-
About invariance in problem heat of exchange with border management
integrable on QT generalized derivatives z , i = 1,...,n. It is known that an elliptic operator A with the boundary condition Pz (t, x) = 0,0 < t < T, x e dQ has a discrete spectrum, ie, eigenvalues \, such that 0 <Xl < ^ <... <lk ^ +<», and the corresponding eigenfunctions qk (x), x efi, constitute a complete orthonormal system L2 (Q).
Fourier method define the solution of the problem (l) -(3). If after fk ( ) It marked the Fourier coefficients f ( ) relative to the system {(pk}, the solution of the problem (l) -(3) has the form
(t, x ) = £ zleh' + J J a (т, s )(pk (s )ds
-hk ('-
dx
Vt (x),
(4)
0 < t < T, X £ñ.
Further, through U the set of controls that are specified below by some positive number p .
Definition 1. The multi-valued mapping D: [0,T] — 2R, where R = (-<»,<») called strongly invariant with respect to the problem (l) - (3), if any (z0 (•)) ( ) e D (0) u (■, -)eU the inclusion (z(t,-)g D (t) for all 0 < t < T, where ( ) — corresponding norm, z (■, ■)— an appropriate solution of the problem (l)-(3) [2, 266-276; 3, 232-233].
Definition 2. The multi-valued mapping D: [0,T] — 2s, where R = (-<», <») It is weakly invariant with respect to the problem (l) - (3), if any {z0 O))^) e D (0) there is management u(•,-)eU such that (z(t,-)g D(t) for all 0< t < T, where {} — the relevant rules, z (■, ■)— an appropriate solution of the problem (l) - (3).
2. Statement of the problem
In this paper we investigate the weak and strong invariance continuous multivalued mapping type
D(t) = [0,b], 0 < t < T, where b - the positive constant.
Our, next goal is to find the connection between the parameters T, b, p so as to provide strong or weak invariance multi-valued mapping D (t), t e [0,T] with respect to the problem (1) - (3) [4, 27-29].
3. Main results.
A) Let (z (v)) = ||z (y)|| a
U = I M(•'•) J1 u(t,s(s)dx | < p, t e [0,r]
Here \\z(v)|| = J¡||z(t,dt = í¿¡z2k (t)dt.
k=1 0
We introduce the following function
g (í ) = be+p
1 - e
, 0 < í < T.
(5)
Proposition: For any positive parameters b, p the following equalities
b, if p<
(6)
sup g (t ) =
P + [ b, if p>\b.
Proof of Proposition: We investigate the extremum function g (t ) a segment [0,T]. We compute the derivative
g '(t) = e (p-Xfi) .Hence it is easy to show that if p^\b then g'(t) # 0, if p<\b the function g (t) kills, but p>\b the function g (t) increases. Consequently,
b, if p< ,
sup g (t ) =
P + h -f\e~pT, if P>Pb.
Theorem 1. If either p<\b, T < 1, or 1 <p/(\b)<(l-y/fe-^T)((f(l-, the multi-valued mapping D ( t), t e [0, T], strongly invariant with respect to the problem (l) - (3).
Theorem 2. If 2X1 > 1, the multi-valued mapping D (t), t e [0, T], weakly invariant with respect to the problem (l) - (3).
Proof ofTheorem 1. We show that for any ||z0 b and \\u(•,•)< p the inclusion |[z (•,-)|e D (t),0 < t < T.
Ilz (r ' 2 =ilz (t 'x )2 dx = (t ) =
jr zle* u (t, s )<pk (s)dsdr
=n
j e
-\(t -t)
II-
j u (t, s)<pk (s~)ds
dr +
j e (< j u (t, s )(Pk (s )dsdr
k=i v o
Now, applying double inequality Koshi-Bunyakovskiy
have
:(t, -)|2 < b2e+ 2bppe* j (t~z)dx -
1 - e
A
-
p2-
(7)
Consequently, |\z(t, < be + p From (5), (6) and (7) the following relations
T T ✓ N.2
Ilz (v)||2 = f|\z (t,-)|2 dt < f |z(t)|2 dt <| sup g(t)) T.
J0" J0' ^ o<, <T J
Let p ^ ^b, T < 1. Then we have ||z (•,•)< sup g(t)JT = g(0)Vt = bsT < b.
0<t <T
Let 1 < P(\b ) < (l-,jTe )/(( ( - e .Then
>/f < b.
beT +P(-eT )
||z(•,•)< supg(t= g(T)JT =
0<t <T
Consequently, D (t), 0 < t < T much invariant. Theorem 1 is proved.
The proof of Theorem 2, the proof of Theorem 1.
B) Let (z(t,-)H|z(tand
U = u(t,s)<t>k (s)dxj dt <p
Theorem 3. If p> 0, the multi-valued mapping D (t), t g [0, T], not much is invariant with respect to the problem (l) - (3) in the time interval [0,T], where T - any positive number.
Proof of Theorem 3. Let p > 0 . In the proof of the theorem essentially it takes advantage of the fact that the value
2
k=1
k=1
2
2
0
