Maximum Principles for Parabolic Equations

Transcription

1 Maximum Principles for Parabolic Equations Kamyar Malakpoor 24 November 2004 Textbooks: Friedman, A. Partial Differential Equations of Parabolic Type; Protter, M. H, Weinberger, H. F, Maximum Principles in Differential Equations; 1/22

2 Outline Review of MP for the elliptic equations; MP for the heat equation L(u) = 2 u x u 2 t Weak MP for the parabolic equations; Applications; Comparison Principle; Uniqueness Results; Strong MP for the parabolic equations; 2/22

3 Review of MP for the elliptic equations Consider the operator n 2 u Lu a ij (x) + x i x j i,j=1 n i=1 b i (x) u x i + c(x)u (1) in an n-dimensional domain Ω (open and bounded). (A) We say that L is elliptic in Ω, if there exists λ > 0 such that for every x Ω and for any real vector ξ 0, n a ij (x)ξ i ξ j > λ ξ 2 i,j=1 (B) We assume that the coefficients in L are bounded and continuous functions in D 3/22

4 Lu n 2 u a ij (x) + x i x j i,j=1 n i=1 b i (x) u x i + c(x)u A,B c = 0 u C 2 (Ω) C(Ω) and Lu 0 in Ω = supu = max u = max u Ω Ω Ω Weak A,B c 0 u C 2 (Ω) C(Ω) and Lu 0 in Ω = supu = max u max u + Ω Ω Ω u + = max(u,0) MP-elliptic A,B c = 0 Ω Open, bounded and connected, if Lu 0 and u attains maximum at an interior point, = u constant in Ω Strong A,B c 0 Ω Open, bounded and connected, if Lu 0 and u attains a non-negative maximum, = u constant in Ω 4/22

5 MP for the Heat Equation L(u) = 2 u x 2 u t Suppose u(x, t) satisfies the inequality L(u) > 0 in the rectangular region = (0, l) (0, T ] then u cannot have a (local) maximum at any interior point. For at such a point 2 u x t 0 and u t = 0, thereby violating Lu > 0 (0,T) S 4 S 1 S 3 S 2 (l,0) x 5/22

6 Suppose u(x, t) satisfies in L(u) 0 in. Then max u = max S 1 S 2 S 3 u Define M := max u. S 1 S 2 S 3 Let (x 0, t 0 ), such that M 1 =: u(x 0, t 0 ) > M. Define v(x) := u(x) + M 1 M (x x 2l 2 0 ) 2, then v(x) < M 1 on S 1 S 2 S 3 and v(x 0, t 0 ) = M 1, Furthermore L(v) = L(u) + M 1 M > 0 on v cannot have an interior maximum. At a maximum on S 4, 2 v/ x 2 0 and therefore v/ t < 0 and this contradicts with u(x 0, t 0 ) = v(x 0, t 0 ) < M. l 2 6/22

7 Weak MP for the Parabolic Equations Consider the operator n 2 u Lu a ij (x, t) + x i x j n b i (x, t) u + c(x, t)u x i,j=1 i=1 i }{{} Au u t in = Ω (0, T ], with T > 0, and Ω domain in R n, (open and bounded). (A) We say that L is parabolic in, if there exists λ > 0 such that for every (x, t) and for any real vector ξ 0, n a ij (x, t)ξ i ξ j > λ ξ 2 i,j=1 (2) (B) We assume that the coefficients in L are bounded functions in 7/22

8 Weak MP for the Parabolic Equations(1) Lu Notation: n 2 u n a ij (x, t) + x i,j=1 i x j i=1 }{{} Au b i (x, t) u x i + c(x, t)u C (2,1) ( ) = {u : R; u, u t, u x i, Define = Ω {T }. u t 2 u x i x j C( )} Theorem: Let (A),(B) hold and c = 0. If u C (2,1) ( ) C( ) satisfies L(u) = A(u) u t 0, then sup u = max u = max u 8/22

9 Proof. Suppose L(u) > 0 and max is attained at (x 0, t 0 ). Therefore u/ x i = u/ t = 0 at (x 0, t 0 ) and D 2 u := ( 2 u x i x j (x 0, t 0 )) i,j is negative semi-definite, therefore 0 <L(u) = (a ij ) : D 2 (u) 0, contradiction!! If the max is attained at (x 0, T ), then u/ t(x 0, T ) 0 0 <L(u) = (a ij ) : D 2 (u) u t 0, contradiction!! If L(u) 0, then take u ɛ = u ɛt L(u ɛ ) = (A t )(u ɛt) = L(u) + ɛ > 0 This implies that max u ɛ = max u ɛ for every ɛ > 0. The assertion follows as ɛ 0. 9/22

10 Weak MP for the Parabolic Equations(2) Lu n 2 u n a ij (x, t) + x i,j=1 i x j i=1 }{{} Au b i (x, t) u x i + c(x, t)u u t Theorem: Let (A),(B) hold and c 0 implies that, if u C (2,1) ( ) C( ) satisfies L(u) = A(u) u t 0, then sup u = max u max u + where u = u + u, u + = max(u, 0). 10/22

11 Proof. Suppose L(u) > 0, and that u has a nonnegative maximum at (x 0, t 0 ), then 0<L(u) = ((a ij ) : D 2 (u) + c(x }{{} 0, t 0 )) }{{}}{{} u , contradiction!! If the max is attained at (x 0, T ), then u/ t(x 0, T ) 0 0 <L(u) = (a ij ) : D 2 (u) }{{} 0 u t }{{} 0 + c(x 0, T )u }{{} 0 0, contradiction!! 11/22

12 Proof. If L(u) 0. Suppose Ω { x 1 < d}. Consider u ɛ = u + ɛe αx 1 L(u ɛ ) = (A t )(u + ɛe αx 1 ) = L(u) + ɛ(α 2 a 11 (x, t) + αb 1 (x, t) + c(x, t))e αx 1 ɛ(α 2 λ α b 1 c )e αx 1. By choosing α large enough, L(u ɛ ) > 0, therefore sup u sup u ɛ max u + ɛ = max u + ɛ max u + + ɛe αd Ω T for every ɛ > 0. The assertion follows as ɛ 0. /22

13 Lu n 2 u a ij (x, t) + x i x j } {{ } Au i,j=1 i=1 Weak A,B c 0 MP-parabolic A,B c = 0 Strong A,B c 0 n b i (x, t) u x i + c(x, t)u u t A,B c = 0 u C (2,1) ( ) C( ) and Lu 0 in = supu = max u = max u u C 2 ( ) C( ) and Lu 0 in = supu = max u max u + u + = max(u,0) 13/22

14 Applications In this section we derive bounds on solution u of the equation L(u) = f in. (1). Let (A) and (B) hold and c(x, t) 0. If L(u) = 0 in, then max u max u (apply the weak MP to u and to u). (2). Let (A) and (B) hold and c(x, t) η. If L(u) = 0 in, then max u e ηt max u (apply (1) to v := ue ηt. Indeed, (A t )(ue ηt ) = e ηt (A(u) t u+ηu). 14/22

15 Applications (Continue) (3). Let (A) and (B) hold and c(x, t) 0. Also assume that Ω { x 1 < d} and a 11 λ 2 +b 1 λ 1 in, for some positive constant λ. If L(u) = f in, then max u max u + (e λd 1)max f define w := ±u max u (1 e λx 1 )eλd max f, then L(w) 0 in Ω, therefore w 0 on, and this results the above inequality. 15/22

16 Applications (Continue) (4). If in (3) the assumption c(x, t) 0 replaced by c(x, t) η, then [ ] max u e ηt max u This follows by applying (3) to v := ue ηt. + (e λd 1)max f 16/22

17 Comparison Principle Theorem. Let (A) and (B) hold. Let c 0 and suppose that f(x, t, u) is a continuous function of variables x, t and u and satisfies the one-sided uniform Lipschitz condition in u f(x, t, v) f(x, t, u) k(v u), x, t, u, v, v > u, If u, v C (2,1) ( ) C( ) satisfy Lu + f(x, t, u) 0 and Lv + f(x, t, v) 0 in, and u v in, then u v, in. Proof. 0 L(u v)+f(x, t, u) f(x, t, v) (L+k)(u v), therefore max(u v) e (k+ c T ) max(u v) 0 17/22

18 Uniqueness Results The First initial boundary value problem consists of solving the differential equation Lu(x, t) = f(x, t), in ; u(x, 0) = ϕ(x), on Ω {0}; u(x, t) = g(x, t), on Ω (0, T ]. Theorem. Let (A) and (B) hold. Then there exists at most one solution to the above problem. Proof. The assumption (B) implies that c(x, t) is bounded, c(x, t) η. Define v := ue ηt. This transformation carries Lu = 0 into Lv := Lv ηv = 0. Now the assertion of the theorem follows from the weak MP for v and v. 18/22

19 Nonlinear Parabolic Equations Consider the nonlinear differential operator Lu F (x, t, u, u x i, 2 u ) u x i x j t, where F is a nonlinear function of its arguments. We say that F is parabolic at a point (x 0, t 0 ) if for any p, p 1,, p n, p 11,, p nn, the matrix ( ) F (x0, t 0, p, p i, p ii ) is positive definite. p hk If Lu 1 = Lu 2 in the domain then, by the mean value theorem, 19/22

20 (u 1 u 2 ) t = F (x, t, u 1, u1 2 u 1, ) F (x, t, u 2, u2 2 u 2, ) x i x i x j x i x i x j = 2 (u 1 u 2 ) a hk + (u 1 u 2 ) b h + c(u 1 u 2 ), x h x k x h where a hk, b h, c are continuous functions provided F/ p, F/ p h, F/ p h k are continuous functions. (a hk ) is positive definite matrix. Applying the previous theorem, we conclude that there exists at most one solution to Lu = 0. 20/22

21 Strong MP for the Parabolic Equations Theorem. Let Ω be open, bounded, and connected in R n. Let (A) and (B) hold. Let u C (1,2) ( ) C( ) with Lu = Au t u 0, then If c 0, then u cannot have a global maximum in, unless u is constant. If c 0, then u cannot have a global nonnegative maximum in, unless u is constant. 21/22

22 Lu n 2 u a ij (x, t) + x i,j=1 i x j i=1 }{{} Au n b i (x, t) u x i + c(x, t)u u t A,B c = 0 u C (2,1) ( ) C( ) and Lu 0 in = supu = max u = max u Weak A,B c 0 u C 2 ( ) C( ) and Lu 0 in = supu = max u max u + u + = max(u,0) MP-parabolic A,B c = 0 ΩT Open, bounded and connected, if Lu 0 and u attains maximum at an interior point, = u constant in Strong A,B c 0 Open, bounded and connected, if Lu 0 and u attains a non-negative maximum, = u constant in 22/22