CC BY
31
УДК 517.929.7
DOI: 10.20310/1810-0198-2016-21-1-76-81
POISSON PROBLEM FOR A LINEAR FUNCTIONAL DIFFERENTIAL EQUATION
© S. Labovskiy, M. F. Getimane
The solvability, existence and positiveness of the Green function of the Poisson problem
-Au -J (u(y) - u(x)) r(x, dy) = pf, u|r(n) = 0 Q
are showed. The spectral properties of corresponding eigenvalue problem are considered. Here is an open set in 1" , and r(l) is the boundary of the . For almost all x G , r(x, ■) is a measure satisfying certain symmetry condition. The function p is a positive weight. This problem has a clear mechanical interpretation.
Key words: poisson problem; discreteness of spectrum; Green function positivity; functional differential equation.
1. The Poisson problem 1.1. The problem
Let l be an open set in 1n, and r(l) be the boundary of the l. We study the Poisson problem
-Au - I(u(y) - U(X)) 'dy) =
П
ulr(n) =0 (2)
where x = (x\xn), Au = u'X1x1 + ■ ■ ■ + u'£nXn . For almost all x G l, the function r(x, ■) is a measure satisfying certain symmetry condition. The function p is a positive weight.
The boundary value problem (1),(2) may describe behavior of an mechanical object. Due to this mechanical interpretation, we can predict some of the properties of the boundary value problem. Problems with a similar approach were considered in [1, 2, 3].
The scheme of method is as following. Represent the problem (1),(2) in the form
Lu = f,
where
Lu = - - ( Au + J (u(y) — u(x)) r(x, dy) 1 . (3)
Using variational method we show unique solvability of the problem. Multiplying by v, and integrating, we obtain an equation of the form
[u, v] = / fv dx, (4)
Jo,
where [u, v] is a bilinear form. Therefore, we start with the bilinear form4
[u,v] d= J u'xv'x dx + 1 J (u(y) - u(x))(v(y) - v(x)) £(dx x dy), (5)
QxQ
dx = dxi ■ ■ ■ dxn . Then we show that the equation (4) with relation to u has a unique solution in a space. The operator L will be constructed automatically.
1.2. Assumptions and notation
Let Q C Rn be a nonempty bounded open set, r(Q) be the boundary of the Q , and X = Q be the closure of Q . For a real function u = u(x) defined on Q and having derivative of first order,
ux = (uxi, • • • , uxn) ,
where x = (xi, xn). For two such functions u and v ,
uxvx = ux1 vx1 + ■ ■ ■ + uxn vxn ■
1.2.1. The form (5) we use under following assumptions
Let M be the set of all Lebesgue measurable subsets in X = Q . Let the function r: X xM ^ R satisfy two conditions: for almost all x € X , the function r(x, ■) is a measure on (X, M) , for any e €M , r(■, e) is measurable on X .
The set functiom £ defined by the equality
£(E) = f r(x, Ex) dx, Ex = {y: (x, y) € E} (6)
JX
is a measure. Assume that £ is symmetric, i.e.
£(ei x e2) = £(e2 x ei), Vei,e2 € M. (7)
The form [u,v] defined by (5) we will use in the Sobolev space W^'2(Q). Definition 1. Let W be the vector subspace of all elements from W0' (Q) satisfying [u, u] < TO .
Define the operator T: W ^ L2(Q) by the equality Tu(x) = u(x), x € Q . The operator T is continuous.
1.2.2. L2(X,/J.)
Let p(x), x € X , be a positive measurable weight (density). The measure j(S) = fS p(x)dx we may imagine as the mass of a measurable part S C X . Let
(f,g) = j f (x)g(x)p(x) dx. (8)
n
Let L2(X,j) (or simply L2(X), or L2(Q)) be the set of all j-measurable functions on X with finite integral f f (x)2p(x) dx . The L2(X,j) is Hilbert space.
n
4the notation £(dx x dy) is equivalent to
1.3. Representation of the operator L
So, our bilinear form [u, v] is inner product in the space W. The image T(W) is dense in L2(X). Note first that the equation [u,v] = (f,Tv) has the form
J u'xv'x dx + 1 J (u(y) - u(x))(v(y) — v(x))£(dx x dy)
Q QxQ
= / f (x)v(x)p(x) dx, Vv € W. (9)
JQ
For any f € L2(Q) it has a unique solution u = T*f € W . Now we confirm the representation (3) of operator L .
Theorem 1. Wq'2 (Q) C D(L) and in Wq'2 (Q) operator L has representation (3). Proof. For any u € C™(Q) or u € W^'2(Q) and v € W integrating by parts one can easily obtain the identity
I u'xv'x dx = — I Au ■ v dx.
Hence, if Au = g , then
J u'xv'x dx = — J g ■ v dx, Vv € W. (10)
Q Q
Note that C^ C D(L). We take equation (10) as definition of operator A on space D(L) in a weak sense.
For any u, v € W
1 J (u(y) — u(x))(v(y) — v(x)) d£ = —j dxv(x)J(u(y) — u(x)) r(x,dy).
QxQ Q Q
From (9) and by definition of the A
Au = —j (u(y) — u(x)) r(x,dy) — pf. (11)
Q
Thus, the operator L has representation
Lu = - - ( Au + j (u(y) — u(x)) r(x, dy) I .
1.4. Eigenvalue problem and spectrum of L
Theorem 2. Let Q satisfy the cone condition [4, Paragraph 4.6]. The eigenvalue problem
—Au — J(u(y) — u(x)) r(x, dy) = \pu, (12)
n
ulr(n) =0 (13)
has in W a system of nontrivial solutions un(x) corresponding to positive eigenvalues Xn. This
1 2
system forms an orthogonal basis in the space W .
Remark. Expression Au is defined in the weak sense (10).
1 2 1 Remark. Each element u € W0 ' satisfies the boundary condition u|r(n) =0 in the following
sense. Since limit value of any u € Cq°(Q) on r(Q) is zero, it may be taken as value of u € Wq'2 on the r(Q).
2. Positivity of solutions
Lemma Suppose f > 0 and the solution u(x) of the problem (1),(2) is in C0X(Q). Then u(x) > 0 in Q .
Proof. Suppose u(x0) < 0 . Let u(x0) <h< 0 and E = {x € Q: u(x) < h} . Since
dx (u(y) — u(x))r(x,dy) = (u(y) — u(x))d£ = 0 Je Je Jexe
dx (u(y) — u(x))r(x,dy) = dx I (u(y) — u(x))r(x,dy) > 0. Je Jx Je Jx\e
we have
Now
Audx = I ds,
Je Jr(E)
where in the right side5 is integral on boundary ) of E of the derivative of u along on outer normal v of u. This integral is positive for some h € (u(x0), 0). So
J ^Au + J (u(y) — u(x))r(x,dy) + f^ dx > 0.
This contradiction proves the lemma.
L e m m a 1. Suppose f > 0 and u(x) is the solution of the problem (1),(2). Then u(x) > 0 in Q .
Proof. Since W is closure of Cq°(Q) the u can be represented as limit u = lim un , where un € CI. Let fn = Lun , fn = f+ — f- where f+ = f + \fn\)/2 . Let Dn = {x : f+(x) = 0} . Since
f(f — fn)2 dx >i (f — fn)2 dx =f (f + f-)2 dx >i (f-)2 dx Jn J Dn J Dn J Dn
the sequence f- tends to zero in L2(Q). Since un = u+ — u- , where u± = T*f± ,
u = lim un = lim u+.
This limit is a nonnegative function because by lemma 1 u+ > 0 .
Theorem 3. Suppose f >^ 0 and u(x) is the solution of the problem (1),(2). Then u(x) > 0 in Q.
Proof. From (1)
—Au + p(x)u = i u(s)r(x,ds) + p(x)f (x), Jn
where p(x) = r(x, Q). By lemma 1 the right side is nonnegative. Since p > 0 the solution of this equation with boundary condition (2) is positive in Q .
Corollary. The minimal eigenvalue Ao of the problem (12),(13) is positive and simple. It associated with a positive in Q eigenfunction.
Theorem The Green operator G = T* has the integral representation
T f (x)= i G(x,s)f (s)p(s) ds. (14)
J n
5this follows from the formula f vAudx = — f vxux dx + f v^^ ds
Je JE Jt(E) dv
Proof. Let u = T*f and xo = (xO,... ,x%) € Q is a fixed point. Then u ^ u(x0) is a linear
functional defined on . It is bounded as
fs du u(xo) = — I - dX1 dxl
" x0
„2 ™n\
where x0 = (s,xQ,..., xn) is the nearest point of r(Q) on the line x% = xQ , i = 2,... ,n , x1 > xQ and
\u(x0)\2 < J^dx^ dx1 • J^ dx1 < C[u, u].
у 1 V д x J f /у 1
x0x0
Let yxo be the extension of this functional on all space W . This functional has the representation yx0 (u) = [u, gx0] , gx0 € W ,
Vx0(u) = [T*f,gX0} = (f,TgX0) = j gxo(s)f (s)p(s) ds.
n
Let G(x, s) = gx(s). For u € CQt°(Q) the formula (14) gives exact value for any x € Q .If u € W , it is lim un , un € CQt°(Q), and the (14) is fulfilled almost everywhere on Q .
REFERENCES
1. Labovskii S.M. On the Sturm-Liouville problem for a linear singular functional-differential equation // Russ. Math., 1996. V. 40, № 11. P. 50-56. (English. Russian original. Translation from Izv. Vyssh. Uchebn. Zaved., Mat. 1996, № 11 (414), 48-53).
2. Labovskii S. Little vibrations of an abstract mechanical system and corresponding eigenvalue problem // Functional Differential Equations, 1999. V. 6, № 1-2. P. 155-167.
3. Labovskiy S. On spectral problem and positive solutions of a linear singular functional differential equation // Functional Differential Equations, 2013. V. 20, № 3-4. P. 179-200.
4. Adams R.A. and Fournier J. Sobolev Spaces // Elsevier, 2003.
Received 12 January 2016.
Labovskiy Sergei Mikhailovich, Plekhanov Russian University of Economics, Moscow, the Russian Federation, Candidate of Physics and Mathematics, Associate Professor of the Higher Mathematics Department, e-mail: [email protected]
Getimane Mario Frengue, Instituto Superior de Transportes e Comunicacoes, Maputo, Mozambique, Associate Professor, Mathematics Department, e-mail: [email protected]
UDC 517.929.7
DOI: 10.20310/1810-0198-2016-21-1-76-81
ЗАДАЧА ПУАССОНА ДЛЯ ЛИНЕЙНОГО ФУНКЦИОНАЛЬНО-ДИФФЕРЕНЦИАЛЬНОГО УРАВНЕНИЯ
© С. М. Лабовский, М. Ф. Жетимане
Показаны разрешимость, существование и положительность функции Грина задачи Пуассона
-AU - I(U(y) - U(X)) Г(Х, dy) = u|r(n) = 0
Q
Рассмотрены спектральные свойства соответствующей задачи на собственные значения. Здесь И - открытое множество в RN и Г(О) является границей И . Для почти всех х £ Q , r(x, ■) является мерой, удовлетворяющей определенному условию симметрии. Функция р - положительный вес. Задача имеет ясную механическую интерпретацию.
Key words: задача Пуассона; дискретность спектра; положительность функции Грина; функционально-дифференциальное уравнение.
СПИСОК ЛИТЕРАТУРЫ
1. Labovskii S.M. On the Sturm-Liouville problem for a linear singular functional-differential equation // Russ. Math., 1996. V. 40, № 11. P. 50-56. (English. Russian original. Translation from Izv. Vyssh. Uchebn. Zaved., Mat. 1996, № 11 (414), 48-53).
2. Labovskii S. Little vibrations of an abstract mechanical system and corresponding eigenvalue problem // Functional Differential Equations, 1999. V. 6, № 1-2. P. 155-167.
3. Labovskiy S. On spectral problem and positive solutions of a linear singular functional differential equation // Functional Differential Equations, 2013. V. 20, № 3-4. P. 179-200.
4. Adams R.A. and Fournier J. Sobolev Spaces // Elsevier, 2003.
Поступила в редакцию 12 января 2016 г.
Лабовский Сергей Михайлович, Российский экономический университет им. Г. В. Плеханова, г. Москва, Российская Федерация, кандидат физико-математических наук, доцент кафедры высшей математики, e-mail: [email protected]
Жетимане Марио Френге, Институт транспорта и сообщений, Мапуто, Мозамбик, доцент кафедры математики, e-mail: [email protected]
