Remarks on the Maximum Principle for Parabolic-Type PDEs

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1 International Mathematical Forum, Vol. 11, 2016, no. 24, HIKARI Ltd, Remarks on the Maximum Principle for Parabolic-Type PDEs Humberto Serrano Universidad Distrital Francisco Jose de Caldas Bogota, Colombia Copyright c 2016 Humberto Serrano. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract This paper addresses the maximum principle for uniform, parabolic second-order linear differential operators. Specifically, if u C 2 (D) satisfies the parabolic, uniform differential inequality L[u] 0 in the cylinder D = Ω [a, b] R N+1 and there exists (x 0, t 0 ) D, such that u(x 0, t 0 ) 0 and M = u(x 0, t 0 ) u(x, t) for all (x, t) D then u(x, t) u(x 0, t 0 ) = M = a constant for all (x, t) in the region D bounded by Ω [a, t 0 ], where Ω R N bounded domain. Mathematics Subject Classification: 58J10, 58J20, 35Bxx, 35Dxx, 35R30, 35R35, 58J35 Keywords: Operator uniformly parabolic; mximum principle 1 Introduction In elementary calculus courses, it is shown that if a function u C 2 (Ω) satisfies inequality u > 0 in an interval Ω = [a, b], then u attains its maximum at one of the interval s ends. The principle of the maximum represents a generalization of such a fact. In particular, as shown in [2] harmonic functions in a domain Ω R N cannot attain a maximum within Ω unless they are constant, that is, if u 0 in a domain Ω and there exists x 0 Ω such that u(x 0 ) = M = max u. Then u u(x 0 ) = M= a constant in Ω. The PDE that models the Ω heat flow over a thin rod, of length l, made of a homogeneous material is

2 1186 Humberto Serrano L[u] u xx u t = f(x, t), where u = u(x, t) represents the rod s temperature at point x and instant t, and f is the heat dissipation rate of the rod. The maximum principle for the heat equation asserts that if u C 1 ([0, l] [0, T ]), l > 0, T > 0 and there exist partial derivatives u t, u x, u xx, and such derivatives are continuous in the rectangle R = [0, l] [0, T ], then the maximum value of u over clausure R should occur on one of the three sides B, S 1, S 2, where B = {(x, 0) : 0 x l}, S 1 = {(0, t) : 0 t T }, S 2 = {(l, t) : 0 t T } under the condition that L[u] u xx u t 0 in (0, l) (0, T ). Generally, u cannot attain a local maximum if L[u] u u t > 0 because if u has a local maximum at an interior point within the cylinder D then, at such a point u 0 and u t = 0. If L[u] u u t 0 the maximum principle asserts that the maximum of u within the clausure of D D should occur on the boundary of the cylinder D, that is, on Ω {0} or else Ω [0, T ]. In [2] it is shown that if Ω R N, u C 2 (Ω) C(Ω) satisfies L[u](x) N N a ij(x)u xi x j + N b i(x)u xi + c(x)u 0 in a domain Ω where L is uniformly elliptical, the coefficients of L are uniformly bounded continuous functions, c(x, t) 0 and a ij (x) a ji (x) in Ω for all 1 i, j N, and there exists x 0 Ω, such that 0 < M = u(x 0 ) u(x) for all x Ω, then u(x) u(x 0 ) = M= a constant in Ω. Hereinafter, Ω represents a bounded domain of R N, for N being a positive integer, a < b, I = (a, b), D = Ω I. The second order differential operator given by L[u] ( N N a ij (x, t) 2 u + N ) b i (x, t) u + c(x, t)u x i u t is considered for all (x, t) D as well as for all u C 2 (D), where the coefficients of (1) are functions defined in the cylinder D, for 1 i, j N. The operator (1) is called parabolic in (x, t) = (x 1, x 2,..., x N, t) if, for every t fixed, the operator consisting of the first terms of the sum is elliptic in (x, t), this is to say that (1) is parabolic if there exists a constant m > 0 with the following property: for all column vectors ξ = (ξ 1, ξ 2,..., ξ N ) T 0 the following inequality holds: N N a ij (x, t)ξ i ξ j m (1) N ξi 2. (2) The operator (1) is uniformly parabolic in D (see [4]) if (2) holds for the same m > 0 for all (x, t) D = Ω (a, b). Here it is assumed that the operator (1) is uniformly parabolic in D, the coefficients of (1) are continuous functions in D, c(x, t) 0 y a ij (x, t) a ji (x, t) within D for all 1 i, j N. The following section shows the maximum principle for the parabolic case, analogous to the elliptical case [2]. The proof presented herein makes use of a

3 Remarks on the maximum principle for parabolic-type PDEs 1187 slight variation with respect to that presented in [4] and also to the method used by Hopf applied to elliptical-type operators (see [2]) Before listing and proving the maximum principle, an approximation theorem for such a principle is presented. This theorem is also known as the strong maximum principle (see [2] and [4]) Theorem 1 Suppose u C 2,1 (D), L[u] 0 in D, where L is the operator defined in (1) and there exists (x 0, t 0 ) D such that u(x 0, t 0 ) = M u(x, t) for all (x, t) D and u(x 0, t 0 ) > 0 then u u(x 0, t 0 ) = M = a constant in Ω [a, t 0 ] In proving this theorem, four lemmas will be used. The first Lemma asserts that a function u C 2 (D) that satisfies the strict differential inequality L[u] > 0 in a domain D cannot attain a positive maximum in D. Lemma 1 If L[u] > 0 in D then u cannot attain a positive maximum in D Proof. Suppose that u attains a positive maximum in P 0 = (x 0, t 0 ) D. In this case, by using an appropriate change of variable (see [2]), it can be proved that N N a ij (x 0, t 0 ) 2 u (x 0, t 0 ) 0 (3) Furthermore, since u(p 0 ) is a maximum in D, the first partial derivatives of u in (x 0, t 0 ) are zero, that is u x i (x 0, t 0 ) = 0, u(x 0, t 0 ) > 0 and c(x 0, t 0 ) 0 and due to (3) (4) then L[u](x 0, t 0 ) = N N u t (x 0, t 0 ) = 0 (4) a ij (x 0, t 0 ) 2 u (x 0, t 0 ) + c(x 0, t 0 )u(x 0, t 0 ) 0. (5) This contradiction proves the Lemma. The following Lemma asserts that if the maximum of u in D is M > 0 and u < M within an appropriate ellipsoid E D, and u = M at a unique point on the boundary of E then the hyper-plane tangent to E, at such a point, is parallel the the x axis, that is, from a geometrical view-point, the maximum of u should be located at either the north pole or the south pole of E. Lemma 2 If u C 2,1 (D), L[u] 0 in D, u has a positive maximum M in D, (x 1, x 2,..., x N, t ) D, and there exist λ i > 0, i = 0, 1,..., N, R > 0 such that, the solid ellipsoid given by E = {(x 1, x 2,..., x N, t) : N i=0 λ i(x i x i ) 2 + λ 0 (t t ) 2 R 2 } D, u < M within E, and there exists ( x, t) E such that u( x, t) = M, then x = x.

4 1188 Humberto Serrano Proof. Suppose that x x. It can be assumed that P = ( x, t) is the unique point in E such that u( P ) = M. Let us construct 0 < r < x x such that the closed ball B r ( P ) D. Let C = B r ( P ), C 1 = C E, C 2 = C C 1. Since C 1 is a compact set, there exist δ > 0( such[ that u < M δ in C 1. Let h(x, t) be N ]) the following function h(x, t) = exp α i=0 λ i(x i x i ) 2 + λ 0 (t t ) 2 exp { αr 2 }, where α > 0 is a constant that will be determined as convenient. By definition h > 0 within E, h = 0 in E, h < 0 outside E. Using a simple calculation it is known that: L[h](x, t) = ( N N a ij (x, t) 2 h + ( [ N ]) N = exp α λ i (x i x i ) 2 + λ 0 (t t ) 2 {4α 2 i=0 [ N 2α a ii λ i + N ) b i (x, t) h + c(x, t)h h x i t N a ij λ i λ j (x i x i )(x j x j) ] N b i λ i (x i x i ) λ 0 (t t ) + c} c exp( αr 2 ) (6) Since L is uniformly parabolic in D, by (2), there exists m > 0 such that N N a ij λ i λ j (x i x i )(x j x j) m N λ 2 i (x i x i ) 2 (7) Also, for (x, t) B r ( P ), it can be stated that x x x x + x x, then x x x x x x > x x r > 0, so for sufficiently large α, L[h] > 0 for all (x, t) B r ( P ). Let us define function v(x, t) = u(x, t)+ɛh(x, t) in B r ( P ), where ɛ > 0. Since u < M δ for (x, t) C 1, ɛ > 0 can be chosen sufficiently small ɛ < δ/h such that v < M in C 1. Also, since h < 0 and u M in C 2, then v < M in C 2 and therefore v < M over the entire bound C = B r ( P ); furthermore, v( P ) = u( P ) + ɛh( P ) = M. Due to the continuity of v and the compactness of B then v has a positive maximum within B r ( P ); additionally, L[v] = L[u] + ɛl[h] > 0 in B r ( P ). This fact contradicts the Lemma 1 The following shows that if L[u] 0 in D and u has a positive maximum M at point (x 0, t 0 ) D then u M on the copy of Ω on the hyper-plane t = t 0 denoted as Ω {t 0 }, where a t 0 b. Lemma 3 If L[u] 0 in D and u has a positive maximum at point P 0 = (x 0, t 0 ) D, then u u(p 0 ) = M = a constant in Ω {t 0 } Proof. Suppose u(p 0 ) = M > 0 and there exists a point P 1 = (x 1, t 0 ) Ω {t 0 } such that u(p 1 ) < u(p 0 ) = M. Since Ω is a connected set, there exists

5 Remarks on the maximum principle for parabolic-type PDEs 1189 a continuous function γ : [0, 1] Ω {t 0 } such that γ(0) = P 1, γ(1) = P 0. Let G = {s [0, 1] : u(γ(s)) < M}; since G is bounded, there exists s (0, 1) such that s = sup G, P = γ(s ), that is P = (x, t 0 ) γ([0, 1]) such that u(p ) = u(p 0 ) = M, u(γ(s)) < u(p 0 ) for all 0 s < s. Over γ([0, 1]) let us take a point P = ( x, t 0 ) between P 1 and P such that dist( P, P ) < d/2, where 0 < d < dist(γ([0, 1]), Ω t 0 ). Since u( P ) < u(p ), there exists ɛ > 0 such that u(p ) < u(p ) for all P within the segment σ = { x} [t 0 ɛ, t 0 + ɛ]. Now consider the family of ellipsoids E λ : x x 2 + λ(t t 0 ) 2 λɛ 2. It can be easily observed that the ends of σ are located over E λ and that E λ approximates σ as λ tends to zero. As a result, there exists λ = λ > 0 with the following properties: u(p ) < u(p 0 ) for all P within E λ and there exists a point Q = (y, t) E λ such that u(q) = u(p 0 ). Since u(p ) < u(p ) for all P σ, Q is not in σ, that is y x. The following Lemma asserts that if L[u] 0 in D and there exists (x 0, t 0 ) such that u(x 0, t 0 ) = M u(x, t) for all (x, t) in a closed rectangle S D, (x 0, t 0 ) S and M > 0, then u u(x 0, t 0 ) = M = a constant in S Lemma 4 Suppose (x 0, t 0 ) = (x 01, x 02,..., x 0N, t 0 ) D, a 0, a 1,..., a N are positive, real numbers and the rectangle S = {(x 1, x 2,..., x N, t) R N+1 : x 0i a i x i x 0i +a i, t 0 a 0 t t 0 +a 0, i = 1, 2,, N} D. Si L[u] 0 in D and u(x 0, t 0 ) = M u(p ) for all P S, then u(x, t) u(x 0, t 0 ) = M = a constant for all (x, t) S where t 0 a 0 t t 0 + a 0, under the condition that u(x 0, t 0 ) = M > 0. Proof. Suppose there is Q = (x, t ) S such that u(q) < u(p 0 ), then it is possible to assume that t < t 0. Over the segment γ that joins points Q and P 0 there exists a point P 1 = (x 1, t 1 ) such that u(p 1 ) = u(p 0 ) and u(p ) < u(p 0 ) for every point P over the segment γ between Q and P 1, x 1 = (x 11, x 21,..., x N1 ). Le us suppose that P 1 = P 0, t = t 1 ã 1, for some real number ã 1 > 0. Since P 1 lies within S, there exist N real numbers b 1 > 0, b 2 > 0,..., b N, such that the rectangle S 1 = {(x 1, x 2,... x N, t) R N+1 : x i1 b i x i x i1 +b i, t 1 ã 1 t t 1, i = 1, 2,, N} S. By Lemma 3, if P = (x, t) S 1 y t < t 1, then u < u(p 1 ) in Ω {t}. Consider the following function: h(x, t) = (t 1 t) k x x 1 2, k > 0. Then it can be assumed that the paraboloid given by M = {(x, t) R N+1 : t 1 t = k x x 1 2 } = {(x 1, x 2,... x N, t) : N k[(x i x i1 ) 2 +(t t 1 )] = 0} intersects the set (Ω {t }) S 1. By the definition of function h, h 0 in M, h < 0 in the set above the paraboloid M and h > 0 in the subset bellow M. Furthermore, L[h](x, t) = 1 2k N a ii 2k N b i (x i x i1 ) + c(x, t)[(t 1 t) k x x 1 2 ] > 0 in S 1 if the dimensions of S 1 are allowed to be sufficiently small, and k > 0 is such that 4k N a ii < 1 in S 1. Let R be the set bounded by M and

6 1190 Humberto Serrano Ω {t } and with B = R M, B = R B. In B, u < u(p 0 ) δ for δ > 0 sufficiently small, then there exists ɛ > 0 such that v u + ɛh < 0 in B, also v u in B {P 1 } and v(p 1 ) = u(p 1 ). Since L[h] > 0 in S 1, L[v] = L[u] + ɛl[h] > 0 in S 1. From this last inequality and by Lemma 1 then: max R v = max R v = v(p 1 ) = u(p 1 ) = u(p 0 ), it can be concluded that 0 v t (P 1) = u t (P 1)+ɛ h t (P 1) = ɛ+ u t (P 1), u t (P 1) ɛ > 0. Since u(p 1 ) is maximum in S, it is true that u t (P 1) = 0. This contradiction proves the Lemma. Proof Theorem. Suppose there exists P = ( x, t) D such that u( P ) < u(p 0 ), t t 0, by Lemma 3, t < t 0. Since D is a connected set, there exists a continuous function γ : [0, 1] D such that γ(0) = P and γ(1) = P 0, and exists P 1 = (x 11, x 21,..., x N1 ) = γ(s ) γ([0, 1]) such that u(p 1 ) = u(p 0 ) and u(γ(s)) < u(p 0 ) for all 0 s < s. Since D is an open set, there exists a > 0 such that the rectangle S = {(x 1, x 2,... x N, t) R N+1 : x i1 a x i x i1 + a, t 1 a t t 1 + a, i = 1, 2,..., N} D. By Lemma 4 u = u(p 0 ) in S = {(x, t) S : t t 1 } and S γ([0, s ]) is non-empty. This fact leads to a contradiction. Theorem 2 Suppose u C 2,1 (D), L[u] 0 in D. If u attains a non-negative maximum at point P 0 = (x 0, t 0 ) D, then u is constant in Ω [a, t 0 ] The proof is exactly the same to that of Theorem 1 References [1] David Gilbarg and Neil S. Trudinger, Elliptic Partial Differential Equations of Second Order, Springer, [2] M. Protter and H. Weinberger, Maximum Principles in Differential Equations, Prentice-Hall, Inc., [3] K. Gustafson, Partial Differential Equations, John Wiley and Sons, [4] Avner Friedman, Partial Differential Equations of Parabolic Type, Robert, Krieger Publishing Company Malabar, Florida, [5] Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer, Received: October 4, 2016; Published: December 16, 2016