Научная статья на тему 'On one proble for higher-order equation'

On one proble for higher-order equation Текст научной статьи по специальности «Математика»

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Ключевые слова
partial differential equations of higher order / not well posed problem / method of separation of variables / simple continued fractions

Аннотация научной статьи по математике, автор научной работы — Yuldasheva Asal Victorovna

In this paper not well posed problem for the even-order equation is studied. The stability of the problem is restored by additional conditions and conditions to domain.

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Текст научной работы на тему «On one proble for higher-order equation»

Bulletin KRASEC. Phys. & Math. Sci, 2014, vol. , no. 2, pp. 18-22. ISSN 2313-0156

MSC 35C05

ON ONE PROBLE FOR HIGHER-ORDER EQUATION

A.V. Yuldasheva

National University of Uzbekistan by Mirzo Ulugbeka, 100174, Uzbekistan, Tashkent c., VUZ gorodok st. E-mail: [email protected]

In this paper not well posed problem for the even-order equation is studied. The stability of the problem is restored by additional conditions and conditions to domain.

Key words: partial differential equations of higher order, not well posed problem, method of separation of variables, simple continued fractions.

Problem definition

The present paper considers for the equation

d2ku d2u

= 0, k = 2n + 1, n G N, (1)

dx2k dt2

in the domain D = {(x,t) : 0 < x < n, 0 < t < 2n} a problem with the following conditions: d 2m u d 2m u

dx^(0,t) = dX2u(n,t) = 0 m = 0, l,...,k-l, 0 < t < 2n (2)

u(an,t)= f (t), 0 < t < 2n, (3)

where a is some constant from (0,1) and f(t)— is the given quite smooth function.

We shall show that if a is an irrational number, then the theorem of solution uniqueness

2k 2

of the problem (1), (2), (3) is valid in the class u e Cxt' (D).

Note that this problem is ill-posed, since a small change in the function f(t) under the norm Cs(s e N) may cause arbitrary large change of the solution u under the norm L2.

This problem may be regularized by a side condition, for example, by a priori estimate

n2n ' dku \2

dtdx < E2, 0 < t < 2n, (4)

d xk

0 0

where E is the defined constant.

Yuldasheva Asal Victorovna - Ph.D. (Phys. & Math.), Lecturer of the Dep. Differential Equations and Mathematical Physics, of the National University of Uzbekistan, Tashkent. ©Yuldasheva A.V., 2014.

The problem well-posedness

-,2k,2 ,

(2), (3), then u may be presented in the form of a series

^ i \ u(x,t) = ^ sinnx(ancosnkt + bnsinnkt\, (5)

n=1

and it follows from this representation that the function f (t) should have the form

^ / \ f (t) = ^ sin nan (an cos nkt + bn sin nkt\. (6)

n=1

Theorem 1. If a is a irrational number, the problem (1), (2), (3) does not have more

2k 2

than one solution of u e Cxy (D).

Proof. Indeed, if in (6) f = 0 , than an = bn = 0. Consequently, h u = 0. □ Remark. If a is a rational number, there is no uniqueness.

For example, let q be some natural number, then the function u (x,t) = sinqxcosqkt satisfies (1), (2) and

n

u(-, t)= 0, 0 < t < 2n. q

Defination. We shall indicate that the irrational number a have the order Q, if Q is the upper boundary of the numbers a, satisfying the inequality

P

a--

1

q

for any f g Q. It is known that almost all the numbers a have the order Q = 1 [3, ?].

H

The next statement is associated with the question on the solution stability depending on a and f. Here is an example.

Theorem 2. Let a be an irrational number. Then there is a sequence 2n of periodic

2k 2

functions fn g C™ (R), uniformly vanishing, and it is such un g Cxt' (D), satisfying (1), (2) and

un(an, t ) = fn(t ), (7)

the following relation holds

Proof. Let

\\m\\Un\\L2{D)=+c

(8)

fn(t ) =

Vnk

k

sin n t,

then

Unix, t ) =

nk sin nan)

i

k

sin n t sin nx.

It is known that [2] there is a sequence of such integers pn, qn that

\im qn = +<*>,

n—> œ

Pn

a - —

qn

1

< ~2 qn

q

and then the theorem statement appears from the following estimation

|sinqnan| = |sin - n| < —.

qn

Note, that for any given integer s there is such an irrational number a (for example, of the order s + 2) that the solution

un(x, t) = n-1-s(sin nan )-1 sin nkt ■ sin nx

of the problem (1), (2) and un(an,t) = n-1-ssinnkt satisfies the following estimation

lim \\un\\L2(D) = + ~ lim Ifn\\as(D) = 0,

n—y^ 2\ ) n—y^ ^ '

from which it is clear that the problem is ill-posed. □

Further we shall show that the problem is also unstable relative to a.

Theorem 3. Let p,q G N,p < q and {an} be a sequence of irrational numbers

converging to P. And let un g C^'2 (D) be the solution of the problem (1), (2) and

q '

un (an,t) = sinqkt, then

Hm\\un\\L2(D) =+<*>.

k

The solution is written in the form un(x,t) = S1"q *anqx , from which the theorem

^ 1 ' sin qann

statement is obvious.

Thus, a side condition is required.

d k \ ^

—r | dtdx < E2. dxk

0 0

Problem with a bounded solution

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Let a,£,E be some positive constants, and a g (0,1).

2k 2

Let f g L2 (0,2n). The function class u g Cx t' (D) satisfying (1), (2) and

\\U (an, ■) - f 11L2(0,2n) < £,

(9)

d ku

d xk

< E.

(10)

¿2(D)

is indicated by r (e, E). The condition (9) substitutes the condition (3), and the priori estimate (10) is required for the problem to be well-posed. We introduce the following notations

2n

|u|| = sup / u2(x,t)dt,

(11)

xG[0,n ]

0

A(£, E )= sup ||v — w|

v,wGr(£ ,E )

Theorem 4. Let a be a rational number, a = P, (p, q) = 1 and

q

E

Then

If q =

q2 < 2- . £

E

A (£, E) < 3 e .

2E\k £

, then

A (e,E) <

2k 2

Proof. Let v,w e r(e,E), then u = v — w e Cx,t, (D) satisfies (1), (2) and

d ku

\u (an, 011^(0,2*) < 2£,

d xk

< 2E.

L2(D)

Since u is represented as (5), the conditions (15) are rearranged as

p 4e2

£ (al + b2) sin2 n-% < ,

n=1 q n

C 17 2

J2kf „2 , u2\ - 8E

£ n2k(a2n + b2n) <

n=1

n2

whence it follows

From (5) we have

p £2 8£2 £ (sin2 nqn + n2^ ) (an + bü < —.

n=l

= n max £ (a2 + b^) sin2 nx < % £ (al + ^.

xG[0,n ] n=1

n=1

It follows from (12)

A2 (e, E) < n supj £ (a2n + bl) : an, bn\ ,

\n=1

satisfying (18).

According to the Lagrange multiplier role we find

2 2 p 2k e2 1

A2 (£, E) < 8£2 min (sin2 n + r2k)-1, r G N

q

E2

From (13) we obtain

p £2 £2 sin2 r-n + r2k—> q2k, 1 < r < q.

q E2 E2

(13)

(14)

(15)

(16)

(17)

(18)

(19)

Substituting this estimation into (19) we obtain (14). The theorem has been proved. □

2

u

Assume that a is an irrational number expanded into a continued fraction

1

a = —

ai + 1

«2 + ...

Theorem 5. Let a g (0,1) be an irrational number and at < Ka, then

i

A (£, E) < 3 . (20)

Ka + 2 X2 2

Proof. To make sure that this estimation is valid, note that it follows from the theorem 4 that the estimation (19) is true for A (e,E) then from the condition at < Ka [3] we obtain

2. - 27

Then

sin ran >-^—, r> 1.

" 4 (Ka + 2)2r2 "

• , 27 2k £2 1 £ V27

min <-^--+ r2k—> >

rGNv| 4 (Ka + 2)2 r2 E2 J -E (Ka + 2)' Thus, the required estimation follows from the above

A2 (£,E) < 9ee[K0+2

References

1. Frosali G. Papi. On the stability of the Dirichlet problem for the vibrating string equation. Ann. Scuola Norm. Sup. Pisa, 1979, vol. 6, pp. 719-728.

2. Yuldasheva A.V. Ob odnoj zadache dlya uravneniya vysokogo poryadka [A problem for the equation of high order]. Doklady Akademii nauk Respubliki Uzbekistan - Reports of the Academy of Sciences of the Republic of Uzbekistan, 2012, no. 5, pp. 11-14.

3. Hinchin A.Ya. Cepnye drobi [Continued fractions]. Leningrad, 1961. 112 p.

Original article submitted: 23.10.2014

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