Journal of Applied Mathematics and Stochastic Analysis, 13:2 (2000), 181-195. ON A THIRD ORDER PARABOLIC EQUATION WITH A NONLOCAL BOUNDARY CONDITION

In this paper we demonstrate the existence, uniqueness and continuous dependence of a strong solution upon the data, for a mixed problem which combine classical boundary conditions and an integral condition, such as the total mass, flux or energy, for a third order parabolic equation. We present a functional analysis method based on an a priori estimate and on the density of the range of the operator generated by the studied problem.


Introduction
In the rectangle Q (0, l)x (0, T), with < oe and T < oe, we consider the one-dimen- sional third order parabolic equation Ot Ox 2 a(x, t) f(x,t). (1.1) Assumption A: We shall assume that Oa(x,t) c o < a(x, t) <_ Cl, Ot <--c2' where c > O, (i O, 1, 2).We pose the following problem for equation (1.1)" to determine its solution v in Q where O(x), x(t), O(t), re(t), a(x, t) and f(x, t)are known functions.
The data satisfies the following compatibility conditions: The first investigation of problems of this type goes back to Cannon [12] and Batten [2] independently in 1963.The author of [12] proved, with the aid of an integral equation, the existence and uniqueness of the solution for a mixed problem which combine Dirichlet and integral conditions for the homogeneous heat equation.Kamynin [21] extended the result of [12] to the general linear second order parabolic equation in 1964, by using a system of integral equations.
Mixed problems with only integral conditions for a 2m-parabolic equation was studied in Bouziani [6], and for second order parabolic and hyperbolic equations in Bouziani-Benouar [8, 9].
In this paper, we demonstrate that problem (1.1)-(1.5)possesses a unique strong solution that depends continuously upon the data.We present a functional analysis method which is an elaboration of that in Bouziani [4,5] and nouziani-nenouar [10].
where zfxu: f u(, t)d, and let L2(O, T; BI(o, 1)) be the space of all functions which x are square integrable on (0, T) in the Bochner sense, i.e., Bochner integrable and (1.8)-(1.10); the operator L is on B into F; B is the Banach space obtained by the completion of D(L) in the form iI0 11 +sup ii0 , , 11 and F is the Hilbert space of the vector-valued functions ff (f, ) with the norm Let L be the closure of the operator L with the domain D(L ).
Definition: A solution of the operator equation Luis called a strong solution of the problem (1.6)-(1.10)., We now introduce the family of operators p 10 and (p-1)0 defined by the formulas p 10 l _ ( i el-d(r t)O(x' v)dr, e > O, 0 T which we use as smoothing operators with respect to t.These operators provide the solutions of the problems and (gfle--10 r Oq -t-Pc-10 0, ( fl-10(X, 0)--0 (,o<-o(,, T) 0 (2.4) respectively.They have the following properties.

PJ +- O(x
For the proof of the above lemma, it suffices to integrate by parts the expression -100 P 0--g" Lemma 4: For all 0 G L2(0, T), we have (i) Proof of Lemma 4 is similar to the proof of the lemma of Section 2.18 in [1].
We easily get the following lemma.
Lemma 5: where A'(t) is the operator of form (2.5) whose coefficient is the first derivative with respect to t of the corresponding coefficient of A(t).
3. A Priori Estimate and Its Consequences Theorem 1" u, such that Under Assumption A, there exists a positive constant c, independent of II-I I B < I I L I I F.

l:i
Returning to the proof of Theorem 1, we denote the first term on the left-hand side of (3.9) by fl(r), the remaining term on the same side on (3.9) by f2(v), and the sum of two first terms on the right-hand side of (3.9) by f3(r).Consequently, Lemma 6 implies the inequality -0-{ L2(O,7";BI(o,t)) + Oz L2(o,t) Since the right-hand side of the above inequality does not depend on r, in the left- hand side we take the upper bound with respect tofrom 0 to T. Therefore, we obtain inequality (3 1) where c-c 1/2 Proposition 1" The operalor L from B into F is closable.The proof of this proposition is analogous to the proof of the proposition in [7].
Since the points of the graph of L are limits of the sequences of points of the graph of L, we can extend (3.1) to apply to strong solutions by taking the limits.Corollary 1" Under Assumption A, there is a constant c > O, independent of u, such that I] u I] B -c I I L u I I F, Vu G D(L). (3.11) Let R(L) and R(L) denote the set of values taken by L and L, respectively.Inequality (3.11) implies the following corollary.
Corollary 2: The range R() is closed in F, R(L)-R() and ,-1_ L-1 where L-1 is the extension of L-1 by continuity from R(L) to R(L).
4. Solvability of the Problem Theorem 2: Let Assumption A be satisfied and let Oa and be bounded.Then o L2a(O 1), roblem (1 6)-(1 10) admits a unique strong for arbitrary f E L2(Q) and-x p solution uay L-13.
Proof: Corollary 1 asserts that, if a strong solution exists, it is unique and depends continuously on 3. (If u is considered in the topology of B and is considered in the topology of F.) Corollary 2 states that, to prove that (1.6)-(1.10)has a strong solu- tion for an arbitrary Y-(f,o) F it is sufficient to show the equality R(L)-F.
To this end, we need the following proposition.Proposition 2: Let the assumptions of Theorem 2 hold and let Do(L be the set of all u D(L) vanishing in a neighborhood of t O. If, for h L2(Q) and for all u Do(L), we have (u,h)L2(Q)-0, (4.1) then h vanishes almost everywhere in Q.
o o Note that for the determination of A( and A2, the corresponding calculations are not difficult, but they are long.Therefore, we only give the final results of the compu- tations" (a(x, "r)) The lea-hand side of (4.7)shows that the mapping f fAu.K(p21 dd is a continuous linear %notional of u, where 1).
K(pC -(I + cA:)(p . (4.1a) Consequently, this assertion holds true, if the function K( has the following properties and satisfies the following conditions: O, OKe 0 0 and 0. (4.14) From (4.12), we deduce that the operator A is bounded on L2(Q).Hence, the norm of cA on L2(Q) is smaller than 1 for sufficiently small c.So, the operator K has the continuous inverse operator in L2(Q).
From (4.12)and (4.14), we deduce that I+a(x,t) (4.18) For each fixed x E [0, l] and sufficiently small , the operator e 1 l)*0a(;c, 7") 1)* has the continuous inverse operator on L2(0, T).Hence, (4.15)-(4.18)imply that 0 p-I h In (4.2), we replace h by its representation (4.20).Consequently, (4.23) Substituting (2.3) in (4.23) (with 0-z) and integrating by parts (with respect to x), by taking into account (4.22), we obtain 1 )*z 02(p dxdt i i ((.-x)xz-2J'z)dxdt --i i a(x't)(l-x) grating by parts with respect to t on each term of the right-hand side of the obtained equality, we obtain, by taking into account (2.4) and due to u E Do(L that ( -OxOt .dxdti/ -(-)'a(, ) dd tOu Ou dxd (l x)e CTta,(x .Ox OxOt Hence, for sufficiently small _< 1, we have l ' (4.30) Passing to the limit in the above inequality and integrating by parts with respect to x, we obtain, by Lemma 4, that ecT'(xz)2dxdt <_ 0 and thus z-0.Hence, h-0, which completes the proof.Now, we return to the proof of Theorem 2. Since F is a Hilbert space, we have that R(L)-F is equivalent to the orthogonality of vector (h, ho)E F to the set R(L), i.e., if and only if, the relation (u,h)o,Q +( Ou Oh) =0, (4.31) OX OX L2(0 l) 10) is equivalent to the operator equation where =(f,p),L=(,e) with the domain D(L) consisting of all functions