On an ill-posed Problem for the Heat equation Текст научной статьи по специальности «Математика»

УДК 517.956.4

On an Ill-Posed Problem for the Heat Equation

Roman E. Puzyrev* Alexander A. Shlapunov^

Institute of Mathematics, Siberian Federal University, Svobodny, 79, Krasnoyarsk, 660041,

Russia

Received 10.01.2012, received in revised form 10.02.2012, accepted 20.04.2012 A boundary value problem for the heat equation is studied. It consists of recovering a function, satisfying the heat equation in a cylindrical domain, via its values ant the values of its normal derivative on a given part of the lateral surface of the cylinder. We prove that the problem is ill-posed in the natural spaces of smooth functions and in the corresponding Holder spaces; besides, additional initial data do not turn the problem to a well-posed one. Using Integral Representation's Method we obtain Uniqueness Theorem and solvability conditions for the problem.

Keywords: boundary value problems for heat equation, ill-posed problems, integral representation's method.

Introduction

Ill-posed problems are already known in Mathematical Physics for many years. A classical example of problems of this type is the famous Cauchy problem for Laplace Equation (see [1]). Beginning from the middle of XX century they appeared in applications (in Geophysics, Hydrodynamics, Theory of Electronic Signals etc.), see, for example, [2, 3]. One of the effective methods of solving ill-posed problems is the so-called Regularization Method (cf. [4, 5, 6, 7] in the theory of the Cauchy Problem for Elliptic Equations). A combination of Integral Representation's Method and Spectral Theory for self-adjoint operators in Hilbert Spaces was especially effective for ill-posed problems for Elliptic Equations (see [8, 9, 10, 11]).

It is well known that many methods for studying elliptic equations have the corresponding analogues for parabolic ones (see, for instance, [12, 13, 14]). Instead of classic boundary value problems for the Heat Equation we consider the ill-posed problem, consisting in finding a function satisfying the equation in a cylindrical domain via its values and the values of its normal derivative on a given part of the lateral surface of the cylinder. Using the Heat Potentials we prove Uniqueness Theorem and obtain solvability conditions for the problem.

1. The problem

Let Q be a bounded domain (i.e. bounded open connected set) in n-dimensional real space rn with the coordinates x = (xi,..., xn). As usual we denote by Q the closure of Q, and we denote by dQ its boundary. In the sequel we assume that dQ is piece-wise smooth.

Consider a bounded open cylinder QT = {x e Q, 0 < t < T}, having the altitude T > 0 and the base Q, in (n + 1)-dimensional real space rn+i = rn x {—to < t < Let also r c dQ

*effervesce@mail.ru tashlapunov@sfu-kras.ru © Siberian Federal University. All rights reserved

be a non empty connected open (in the topology of dQ) subset of dQ. Set rT = r x (0, T) to obtain rT = r x [0, T].

As usual, for s G z+ (here z+ = n U {0}) and an open subset D C rm we denote Cs(D) the set of all s times continuously differentiable functions in D. The standard topology of this metrisable space induces the uniform convergence on compact subsets in D together with all the partial derivatives up to order s.

For S C dD we denote Cs(D U S) the set of such functions from the space Cs(D) that all their derivatives up to order s can be extended continuously onto D U S. The standard topology of this metrisable space induces the uniform convergence on compact subsets in D U S together with all the partial derivatives up to order s. In particular, for bounded domains, Cs(D U dD) = Cs(D) is a Banach space.

Apart from the standard functional spaces, we need also spaces reflecting the specific properties of parabolic equations in rn+1 = rn x {—to < t < Namely, let C 1,0(QT) be the set of continuous functions u in QT, having in QT the continuous partial derivatives uxi, and let C2,1 (QT) denote the set of continuous functions in QT, having in QT the continuous partial derivatives The standard topology of this metrisable space induces the uniform convergence on compact subsets in D together with all the partial derivatives used in its definition.

As before, for S C dQT we denote by C1,0 (QT U S) the set of such functions u from the space C1,0 (QT) that their derivatives uxi can be extended continuously onto QT U S. The standard topology of this metrisable space induces the uniform convergence on compact subsets of QT U S of both the functional sequences and the corresponding sequences of the first partial derivatives with respect to Xj. Clearly, C 1,0(QT U dQT) = C 1,0(QT) is a Banach space.

n d2 d

Let now An = Y] be the Laplace operator in rn and Ln+1 = ^— An stand for the heat

i=1 dx2 dt

operator in rn+1. It is well known that the Laplace operator is elliptic and the heat operator is parabolic.

d " d

Besides, let — = Y^ v —— denote the derivative at the direction of the exterior unit normal dv i=1 dxj

vector v = (v1,...,vn) to the surface dQ. As dQ is piece-wise smooth, the normal vector v = (v1,..., vn) is defined almost everywhere on dQ.

Consider two problems for the Heat Equation. Let functions u0(x) G C(Q), u1(x,t) G C 1'0(rT), u2(x,t) G C(FT) and f (x,t) G C(QT) be given.

Problem 1. Find a function u(x, t) G C2'1 (Qt) n C 1'0(Qt U rT) n C(^\ (dQ \F)T) satisfying the Heat Equation

Ln+1u = f in Qt (1)

and boundary conditions

u(x,t)= u1(x, t) on rT, (2)

—u(x,t) = u2(x, t) on rT. (3)

dv

In particular, if n =1 we have QT = (0,1) x (0, T) (i.e. with one spaces variable x = x1 G (0,1) and time variable t G (0, T)). In this case rT = {0} x (0, T) and the following conditions correspond to Problem 1:

u(x, t) — u''(x, t) = f (x, t) b (0, 1) x (0, T), (4)

u(0, t) = u1(t), 0 < t < T, (5)

u'(0,t) = u2, (t), 0 < t < T (6)

. du . du .. d2u (traditionally, here we set u = —, u = , ^ = 2).

Fig. 1

Fig. 1 indicates the sets where the boundary data for Problem 1 are given in the cases of one and two space variables. The process of the heat conduction is described by the function u(x,t) presenting the temperature at the space point x and the time t. Though, of course, it is known that the Heat Equation is not ideal to model the process of the heat conduction.

Problem 2. Find u(x, t) G C2'1(^T) n C 1'0(QT UFT) n C(0t\ \ F)T) satisfying in QT the Heat Equation (1), the boundary conditions (2), (3) and the initial condition

u(x, 0) = uo(x), x G (7)

In the case of one space variable the following initial conditions corresponds to Problem 2:

u(x, 0) = u0(x), x G [0,1]. (8)

Of course one should also take care on the compatibility of the data u0, u1, u2: at least

uo(x) = ui(x, 0) on r, (9)

and, if u0 G C 1(0), even

^U0(x) = U2(x, 0) on r. (10)

Fig. 2

Fig. 2 indicates the sets where the boundary and the initial data for Problem 2 are given in the cases of one and two space variables.

We note that in the classical theory of the (initial and) boundary problems for the Heat

du

Equation equation (1), the initial condition (7) and the boundary condition au + ft— = u3 on the whole lateral surface of the cylinder QT are usually considered. As a rule, such a

problem is well-posed in the proper spaces (Holder spaces, Sobolev spaces etc.), see, for instance, [12].

The motivation of Problems 1 and 2 is transparent. The first one describes the situation where for some reasons at each time t G [0, T] only part r of the boundary of the «body» Q is available for measurements (though the continuity up to the initial time t = 0 is postulated). The second one describes the situation where the whole body Q was available for measurement at the initial time t = 0 but data on dQ \ r were lost as a consequence of extremal temperature conditions for 0 < t < T. Nevertheless we want to preserve the continuity up to the final time t = T in both problems at least on Q U r. It is often important for applications that solutions to Problem 1 and 2 would belong to C1,0 (QT). Later we will indicate the corresponding cases for Problem 1 (see Corollary 2).

Let us show that both Problem 1 and Problem 2 are ill-posed.

Example 1. Take a cube Qn = {0 < xj < 1, 1 ^ j ^ n} as base Q of the cylinder QT. Let r be the face {xn = 0} of the cube Qn. Then rT = Qn-1 x (0, T). Fix N G n and consider the sequence of solutions

efc2(t-T) + fcx„ _

ufc(x, t) = -kw- G C~(Qt)

to problem (1), (2), (3), (7) with the data

fk(x,t) = 0, u0,k(x) =

e-fc(fcT-x„)

ek (t—T ) ek

u1,k(x1,. .. ,xn—1,t) = —-n—, u2,fc(x1, ... ,xn—1,t)

kN ' 1>---> kN —1 '

It is clear, that compatibility conditions (9), (10) hold and

fk —► 0 in CTO(QT), u0,fc —► 0 in CTO(Q), k—' k—

u1,k —► 0 in Cs(rT), u2,k —► 0 in Cs(rT),

if N > 2s + 1. On the other hand, for all xn > 0 and all N G n we have:

ek (T T)+kxn ekxn

uk(x, T) =-kN-= l^k^

Thus there is no continuity with respect to the data and hence Problem 2 is ill-posed. Obviously, Problem 1 is ill-posed, too.

Remark 1. If we replace in settings of Problem 1 and 2 solution's space C2,1(QT) n C1,0(QT U rT) n C(QT \ (dQ \ r)T) with the space C2-1(QT) n C 1-0(QT U rT) n C(QT U rT U (Q x {0})) then Example 1 will be not fit to demonstrate that problems are ill-posed because the uniform convergence on compact sets from QT UTT U(Q x {0}) will be granted for the sequence of solutions. However, in practice this would mean an infinite temperature (and hence a catastrophe) at the final time t = T.

By the way, it is natural to replace solution's space C2,1(QT)nC 1,0(QTUrT)nC(QT\(dQ\r)T) with C2,1 (QT) n C 1,0(QT U rT) in the setting of Problem 1. But then the classical Hadamard example (see [1] or, for instance, [15, Ch. 1, §2]) for the Cauchy problem for the Laplace Equation in rn shows that the problem for Heat Equation in rn+1 will be ill-posed (at least if n ^ 2). Of course, in this case the data for the example do not depend on t.

As both Problems 1 and 2 are ill-posed, we will study Problem 1 only because in addition to (1)-(3) to investigate Problem 2 one needs to know also initial condition (7).

2. Uniqueness Theorem

In this section we will prove that Problem 1 can not have more than one solution. If the surface r and the data of the problem are real analytic then the Cauchy-Kovalevsky Theorem implies that Problem 1 can not have more than one solution in class of (even formal) power series. However the theorem does not imply the existence of solutions to Problem 1 because it grants the solution in a small neighborhood of the surface rT only (but not in a given domain QT!).

To obtain a Uniqueness Theorem for Problem 1 we use an integral representation constructed with the use the fundamental solution

$(x,t) = < (2^)r

|x|2

e if t > 0,

if t < 0,

to Heat Operator Ln+1.

More precisely, consider the cylinder type domain qtl ,t2 = ^t2 \ ^Tl and a closed measurable set S C öa For functions f G C(Ot1,t2), v G C(St), w G C(St), h G C(Ot1,t2) we set

IQ,Ti (h)(x, t) = J $(x - y,t)h(y,Ti)dy,

(11)

t

Gn,Ti (f )(x,t) = y J $(x - y, t - t)f (y,T)dyd

t1 n t

Vs,Ti (v)(x,t)=y J $(x - y,t - t)v(y, t)ds(y)dT

(12)

(13)

Ti s

Ws,Ti (w)(x,t) = - / I d— $(x - y,t - t)w(y,T)ds(y)dT,

Ti s

(14)

where ds is the volume form on S induced from rn. All these functions are called Heat Potentials with densities f, v, w and h respectively. In our situation these are convergent improper integrals depending on vector parameter (x, t) G rn+1 (see, for instance, [12, Ch. 4, §1], [13, Ch. 3, §10], [14, Ch. 1, §3 and Ch. 5, §2]). The potential In,Tl (h) is sometimes called Poisson type integral for the Heat Operator, the functions Gn,Tl (f), Vs,Tl (v), WS,Tl (w) are often referred to as Heat Volume Potential, Heat Single Layer Potential and Heat Double Layer Potential respectively.

The integral formula, that we need, is similar to the famous Green Formula for the Laplace Operator.

Lemma 1. For all 0 < T1 < T2 and all u G C2'1(QTljT2) n C 1'0(QTljT2) with Ln+1u G C(QTl,T2) the following formula holds:

u(x,t), (x,t) G Q.t1,t2 0, (x, t) G ^T

—j = ^In,Ti (u) + Gn,Ti (Ln+1 u) + Vdn,Ti + Wdn,Ti(u)^ (x,t).

Proof. See, for instance, [16, Ch. 6, §12].

(15) □

Theorem 1 (Uniqueness Theorem). If r has at least one interior point (on dQ), and function u G C2'1(Qt) n C 1'0(QT U rT) satisfies (1), (2), (3) with f = 0, u1 = u2 = 0 then u = 0 in QT.

1

0

Proof. Under the hypothesis of the theorem there is an interior point x0 on r. Then there is such a number r > 0 that B(x0, r) n dQ c r where B(x0, r) is ball in rn with center at x0 and radius r. Fix an arbitrary point (x',t') G QT. It is clear that there is a domain Q' 9 x' satisfying Q' C Q and Q' n dQ c r n B(x0, r). Then (x',t') G Qt1jt2 with some 0 < T < T2 < T.

But u G C2,1 (QT1 T2) n C 1'0(QT1 T2) and Ln+1w = 0 in QT1 T2 under the hypothesis of the theorem. Hence formula (15) implies:

u(x,t), (x,t) G

QT1 T2

o, (x,t) G QT1iT2

I = /n',Ti(u)(x,t) + Van'\r,T^lu) (x,t) + WW\r,Ti (u)(x,t), (16)

du

because u = — = 0 on rT. dv

Taking into account the character of the singularity of the kernel $(x — y,t — t) we conclude that the following properties are fulfilled for the integrals, depending on parameter, from the right hand side of identity (16):

in'.Ti (u) G C2-1({x G rn,T1 < t < T2}), WSn'\r,T1 (u),Van'\r,T1<t<T^ d^) G C2,1({x G rn \ (dQ' \ r),T1 < t < T2})

(see, for instance, [12, Ch. 4, §1], [13, Fn. 3, §10] or [14, Ch . 1, §3 and Ch. 5, §2]). Moreover, as $ is a fundamental solution to Heat Operator then Ln+1(x,t)$(x — y,t —t) =0 for (x,t) = (y, t), and therefore, using Leibniz rule for differentiation of integrals depending on parameter we obtain:

¿n+1ln',T1 (u) = 0 in the domain {x G rn, T1 < t < T2}, Ln+1Van'\r,T^du) = Ln+1Wafi'\r,Ti (u) =0 in qT1,t2 = {x G rn \ (dQ' \ r), T1 < t < T2}.

Hence the function

P(x,t) = /n',T1 (u)(x,t)+ Vdn'\r,T^Tj^j (x,t)+ Wan'\r,T1 (u)(x,t),

satisfies the heat equation

(Ln+1P)(x,t) = 0 in qT1,t2.

This implies that the function P(x,t) is real analytic with respect to the space variable x G rn \ (dQ' \ r) for any T1 < t < T2 (see, for instance, [15, Ch. VI, §1, Theorem 1]). In particular, by the construction the function P(x,t) is real analytic with respect to x in the ball B(x0, r) and it equals to zero for x G B(x0,R) \ Q for all T1 < t < T2. Therefore, the Uniqueness Theorem for real analytic functions yields P(x,t) = 0 in QT1 T2, and in the cylinder QT1 T2, the containing point (x',t'). Now it follows from (16) that u(x',t') = P(x',t') = 0 and then, since the point (x',t') G QT is arbitrary we conclude that u = 0 in QT. The proof is complete. □

Corollary 1. Problem 1 has no more that one solution.

Proof. Let v(x, t) and w(x, t) be two solutions to Problem 1. Then function u = (v — w) G C2,1(qt) n C 1,0(qt U rT) n C(QT \ (dQ \ r)T) is a solution to the corresponding problem with f = 0, u1 =0, u2 = 0. Using 1 we conclude that u is identically zero in QT. □

Thus, the Uniqueness Theorem implies that the data of Problem 1 are suitable in order to uniquely define its solution. Moreover, the theorem clarify why the problem is ill-posed. The reason is the redundant data. Indeed, if r has at least one interior point (on dQ), then taking a smaller set r' C r we again obtain a problem with no more than one solution.

3. Solvability Conditions

From now on we will study Problem 1 under the assumption that its data belong to Holder spaces (cf., [14, Ch. 1, §1] for other boundary problems for parabolic equations). We recall that a function u(x), defined on a set M e rm, is called Holder continuous with a power 0 < A < 1 on M, if there is such a constant C > 0 that

|u(x) - u(y)| < C|x - y|A for all x,y e M (17)

(|x — y| = j J2 (xj — yj)2 being Euclidean distance between points x and y in rm). Let CA(QT) V j=i

stand for the set of Holder continuous functions with a power A over QT. Besides, let C1+A'A(QT) be the set of Holder continuous functions with a power A over QT, having Holder continuous derivatives uxi, 1 < i < n, with the same power in QT.

We choose a set Q+ in such a way that the set D = Q U r U Q+ would be a bounded domain with piece-wise smooth boundary. It is possible since r is an open connected set. It is convenient to set Q- = Q. For a function v on DT we denote by v+ its restriction to Q+ and, similarly, we denote by v- its restriction to Q. It is natural to denote limit values of v± on rT, when they are defined, by v±T.

Theorem 2 (Solvability criterion). Let r G C 1+\ f G CA(QT), u1 G C 1+A'A(rT), u2 G CA(rT). Problem (1), (2), (3) is solvable in the space C2,1(QT) n C 1,o(QT U rT) if and only if there is a function F G C2,1(Dt) satisfying the following conditions:

1) L„+1F = 0 in DT,

2) F = Gn,o(f)+ Vr,o(u2) + Wr_o(u1) in Q+.

Proof. Necessity. Let a function u(x,t) G C2'1(QT) n C 1'0(QT U rT) satisfies (1), (2), (2). Consider the function

F = Gn,o(f) + Vrj0(u2) + Wro(u1) — u.

in the domain DT, where xm is a characteristic function of the set M c rn+1. By the very construction condition 2) is fulfilled for it.

Clearly, the function u(x, t) belongs to the space C1,2 (QT) for each cylindrical domain QT with such a base Q' that Q' c Q and Q' n dQ c r. Besides, Ln+1u = f G CA(QT). Without loss of the generality we may assume that the interior part r' of the set Q' n dQ is non-empty.

We note that xnTu = x^Tu in D^, where D' = Q'Ur'UQ+. Then using Lemma 2 we obtain:

F = Gq\q7o(f) + VF\r,,o(u2) + Wp\r, o(u1) — /n',o(u) in DT. (18)

Arguing as in the proof of Theorem 1 we conclude that each of the integrals in the right hand side of (18) satisfies homogeneous Heat Equation outside the corresponding integration set. In particular, we see that Ln+1F = 0 in D^. Obviously, for any point (x,t) G DT there is a domain D^ containing (x, t). That is why Ln+1 F = 0 in DT, and hence F belongs to the space C2,1(Dt). Thus this function satisfies condition 1), too.

Sufficiency. Let there be a function F G C2,1(DT), satisfying conditions 1) and 2) of the theorem. Consider on the set DT the function

U = Gn,o(f) + Vr,o(u2) + Wr,o (u1) — F. (19)

As f G CA(QT) then the results of [14, Ch. 1, §3] imply

Gn,o(f) G C2,1(Q±) n C1^(Dt) n C(Dt) (20)

and, moreover,

Ln+1G—,0(f) = f in Qt, ¿n+1G+, 0(f) = 0 in Q+. (21)

Since u2 G CA(fT) then the results of [14, Ch. 5, §2] yield

/T,0

Vr0(u2) G C2,1(Q±) n C 1-0((Q± U r)T) n C(Dt \ (dr)T), (22)

Ln+1Vr0(u2) = 0 in QT U Q+. (23)

On the other hand, the behavior of the Double Layer Potential Wp 0(u1) is similar to the behavior of the normal derivative of Single Layer Potential Vp 0(u1). Hence

Wr,0(u1) G C2 -1(Q±) n C(Q± \ (dQ± \ r)T), (24)

Ln+1Wr 0(u1) = 0 in QT U Q+. (25)

Lemma 2. Let S C r G C1+A. If u1 G C 1+A,A(rT), then the potential W^(u1) belongs to the space C 1-0(qt U St) if and only if W^0(u2) G C 1-0(Q+ U St). '

Proof. It is similar to the proof of the analogous lemma for Newton Double Layer Potential (see, for instance, [9, lemma 1.1]). Actually, one needs to use Lemma 2 instead of the standard Green formula for the Laplace operator. □

Since F G C 1,0(DT) then it follows from the discussion above that W^0(u2) G C 1,0((Q+ U r)T). Thus, formulas (19)-(25) and Lemma 2 imply that

U G c2-1(q±) n C 1-0((Q± U r)T) n C(Q± \ (dQ \ r)T),

Ln+1 U = xdtf in qt U Q+.

In particular, (1) is fulfilled for U—.

Let us show that the function U— satisfies (2) and (3).

Since F G C 1i0(DT) we see that daF— = daF + on rT for a G z+ with |a| < 1 and

daF+T = (5 aG+,0(f) + daV+0(u2) + 5 aW+0(u1))

It follows from formulas (20) and (22) that Heat Volume Potential and Single Layer Potential are continuous if the point (x, t) passes over the surface rT. Then

U—T = Wr—0(u1)|rT — Wi0(u1VT = u1.

because of the theorem on jump behavior of the Heat Double Layer Potential (see, for instance, [14, Ch. 5, §2, theorem 1]), i.e. equality (2) is valid for U— .

Formula (20) means that that the normal derivative of the Heat Volume Potential is continuous if the point (x,t) passes over the surface rT. Therefore

dv irT va>i>2>) — (¿Vi>2>) + (£"¥■,(«.>) ^^Wic(«10,rr. (26)

By theorem on jump behavior of the normal derivative of the Heat Single Layer Potential (see, for instance, [13, Ch. 3, §10, theorem 10.1])

(l^lrr — (^V+c(«20| ^ = »2. <27>

Finally, we need the following lemma which is an analogue of the famous Theorem on jump behavior of the normal derivative of the Newton's Double Layer Potential.

i'

T

Lemma 3. Let r G C1+A and u G CA(rT). If Wr~0(ui) G C1 -0((O U r)T) or W+(ui) G C1 -0((O+ U r)T) then

(1^0(ui) - ^WT+o(ui^|rT =0. (28)

Proof. Really, let, for instance, Wr"0(u1) G C1 '0((Q U r)T). Then using Lemma 2 we obtain

W+0(ui) G C1 -0((^+ U r)T) and (^W]±0(w1^ G C(rT).

Let ^ G ) be a function with compact support in . Then Gauss-Ostrogradskii

formula yields:

I ^(¿W^ui) - ¿^0(ui^ ds(x)dt = (29)

^AWp 0(u1)dxdt + (V^)'VWp 0(u1)dxdt =

i ^^Wp 0(u1)dxdt + i (V^)'VWr 0(u1)dxdt ./nT un+ ot ' Jqt uq+

because Ln+1W±0(u1) =0 in

according to (25).

Again integrating by parts and using Theorem on jump behavior of Heat Double Layer Potential we see that

d

l/1/p 0(ui)uxut + l (v^) V Wp 0(

i ^^Wp 0(u1)dxdt + / (V^)'VWp 0(u1)dxdt = (30)

J ^Wr, 0(ui)dxdt - J (A^)'Wr- 0(ui)dxdt + J ^(WT0(ui) - W+0(ui))ds(x)dt =

f d^uids(x)dt - f (Ln+i^)Wp 0(ui)dxdt. ./rT ./nT un+

But the kernel $(x — y,t — t) is a fundamental solution of the parabolic operator with

respect to variables (y, t). Hence

/ ¿n+1^(x,t)$(x - y,t - t)dxdt = ^(y, t), (y,T) G . j dt

Then the type of the singularity of the fundamental solution allows us to apply Fubini Theorem and to conclude that

f (L+i^)Wp 0(ui)dxdt = f ui — f ¿n+i^(x,t)$(x - y,t - t)dxdtds(y)dT =

iflTun+ ./ r T jdt (31)

= / d^uids(y)dT.

T

Finally, formulas (29)-(31) imply that

/p T - ¿^>0) Us = 0

for all ^ G ). As such functions are dense in the Lebesgue space L1(K) for any compact

K c rT then formula (28) holds true. □

Now using lemma 3 and formulas (26), (27), we conclude that |r T = u2, i.e. (3) is fulfilled for U—.

Thus, function u(x, t) = U—(x,t) satisfies conditions (1)-(3). The proof is complete. □

It follows from formula (19) that properties of a solution to Problem 1 depend on properties of the extension F of the sum of heat potentials, described in Theorem 2.

Corollary 2. Let S C dQ \ r. Under the hypotheses of Theorem 2, Problem 1 is solvable in the space

c 21(qt ) n c 10(QT u rT) n C(QT \ st )

if and only if there exists a function

F G C2'1 (DT) n C 10(QT U rT) n C(QT \ ST), satisfying conditions 1) and 2) of Theorem 2.

In particular, if S = 0 then corollary 2 gives criterion for the existence of solution to Problem 1 in the space C(QT).

We note that Theorem 2 is an analogue of Theorem by Aizenberg and Kytmanov [8]) describing solvability conditions of the Cauchy problem for the Cauchy-Riemann system (cf. also [9] in the Cauchy Problem for Laplace Equation or [7] in the Cauchy problem for general elliptic systems). Formula (19), obtained in the proof of Theorem 2, gives the unique solution to Problem 1. Clearly, if we will be able to write the extension F of the sum of potentials Gfi'0(f) + Vr 0(u2) + Wr 0(u1) from Q+ onto DT as a series with respect to special functions or a limit of parameter depending integrals then we will get a Carleman type formula for solutions to Problem 1 (cf. [8]). However this is a topic for another paper. In the present article we will discuss polynomial and formal solutions only.

4. Polynomial Solutions and Dense Solvability

It is not difficult to prove dense solvability of Problem 1 in the case where r is an open connected set of the hyperplane {xn = 0}.

Lemma 4. If r is an open connected set if the hyperplane {xn = 0} the Problem 1 is densely solvable.

Proof. First let us prove that if in this case the data of Problem 1 are polynomials then the problem is solvable and its solution is a polynomial.

Indeed, Problem 1 is easily can be reduced to the following one:

Ln+1 v = g in (32)

v(xi,..., xn-1,0, t) = 0 on rT, (33)

dv „ „ — , N

-—(xi,...,xn_ 1,0,t)=0 Ha Ft. (34)

dx„

with g(x,t) = f (x,t) - (L„ui)(xi,... ,x„_i,t) - x„(L„M2)(xi,... ,x„_i,t). Besides, u(x,t) = v(x,t) + ui(xi,... ,xn_i, v) + xnU2(xi,... ,xn_i,t). Now consider data g(j'a)(x,t) = tjxa with a multi-index a £ z+. If 0 < ai + ... an-i < 1, we easily obtain (unique) polynomial solutions

v(ja)(x,t) = x?1 ••• x^n-1 w(j'a")(x„,t), an j £ z+, (35)

yk+2 k!

to problem (32)-(34) where w(0'fc)(y,t) = --,

(k + 2)!

z+, y G r and, by the induction with respect to j G z+

w(1-fc)(y,i) = -

tyfc+2k! yk+4k!

(k + 2)! (k + 4)!

k G

(y,t) = -E

= 0

tj-Myk+2M+2k!j! (k + 2M + 2)!(j - M)!,

k G z+, y G r.

(36)

+ . Namely, let for s > 2 and all a' with |a'| = s the solutions

ai

To finish the arguments we use the induction with respect to |a'| G z+ where a' (ai,...,an-i) G zn-1 to the problem are polynomial. If |a'|

tjxa — w(j'an)(xn,t)An-1 (x?1 ••• x?--1) . Clearly, the degree of the polynomial pjja(x,t) = w(j'a")(i„,i)A„_1 (x?1 • • • x^-1) with respect to x' G rn-1 equals to s -1. Then, by the induc-

s +1 then Ln+1 (x?

x

n-1

-11 w(j'an)(xn,t))

tion, problem (1)-(3) with data pjj0,(x,t) admits a polynomial solution, say, rjia(x,t). Therefore the solution v(j'a)(x,t) to problem (1)-(3) with data g(j'a)(x,t) = tjxa, |a'| = s + 1, is given as follows: v(j'a)(x,t) = x^1 • • • x^--1 w(j,a")(x„,t) + rjja(x,t), i.e. it is a polynomial, too.

Now Problem 1 with zero boundary data in the case r c {xn = 0} is densely solvable because any continuous function g on the compact set QT can be approximated by polynomials. But the reducing to zero boundary data was organized in such a way that one easily sees, in this case Problem 1 is densely solvable for non-zero boundary data, too. □

The dense solvabilty of Problem 1 in general setting is natural to expect if the set dQ \ r has at least one interior point in dQ (cf. [10] in the Cauchy Problem for elliptic equations).

Finally, we note that polynomial solutions indicated in the proof of Lemma 4 can be used in order to construct formal solutions to Problem 1. For example, if n =1 and the data for problem

(4)-(6) are written as (formal) power series f (x,t) = ckjxktj, wi(t) =5^ «jtj, u2(t) =

fc,j=0

j=0

y bjthen (35), (36) imply that its formal solution is given by the power series

j=0

dfcj = <

u(x,t) = J2 dfcjxktj with fc,j=0

((j + 1)aj+1 - c0,j) /2, ((j + 1)bj+1 - C1,j) /6,

k-2 V c • z^p=0 ¿-i cp,i

¿^j,2i+p=fc+2j

p!i! kj!,

k = 0, k = 1, k = 2, k = 3,

k > 4.

a

j

b

The work was supported by RFBR grant 11-01-00852a.

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Об одной некорректной задаче для уравнения теплопроводности

Роман Е. Пузырев Александр А. Шлапунов

В работе исследована одна краевая .задача для уравнения теплопроводности. Она состоит в восстановлении функции, удовлетворяющей уравнению теплопроводности в цилиндрической области, по заданным ее значениям и значениям ее нормальной производной на части боковой поверхности цилиндра. Доказано, что задача является некорректной в естественных для нее пространствах гладких функций и соответствующих пространствах Гельдера, а добавление к условиям начальных данных не превращает задачу в корректную. С помощью метода интегральных представлений получены теорема единственности, условия 'разрешимости и формулы для решения задачи.

Ключевые слова: краевые задачи для уравнения теплопроводности, некорректные задачи, метод интегральных представлений.