University:
Massachusetts Institute of Technology
Course:
18.152 | Introduction to Partial Differential Equations
Academic year:

2022
Views:

240

Pages:

2
Author:

Vivian Bernard

Lecture Four: The Poincare Inequalities In this lecture we introduce two inequalities relating the integral of a function to the integral of it’s gradient. They are the DirichletPoincare and the NeumannPoincare in equalities. 1 The DirichletPoincare Inequality Theorem 1.1 If u : Br → R is a C 1 function with u = 0 on ∂Br then � � 2 2 u ≤ C(n)r |�u|2 . Br (1) Br We will prove this for the case n = 1. Here the statement becomes � r � r 2 2 f ≤ kr (f � )2 −r (2) −r where f is a C 1 function satisfying f (−r) = f (r) = 0. By the Fundamental Theorem of Calculus � s f (s) = f � (x). (3) −r Therefore s � |f (s)| ≤ |f � (x)|. (4) −r Recall the CauchySchwarz inequality h = 1, g = |f � | �� hg ≤ �� h2 �� r �1/2 �� g2 �1/2 � . Apply this with to get �� s |f (s)| ≤ � 2 (f ) �1/2 1/2 (r + s) ≤ � 2 (f ) �1/2 (2r)1/2 . (5) −r −r Squaring both sides gives 2 � r |f (s)| ≤ 2r −r 1 (f � (s))2 , (6) and ﬁnally we integrate over [−r, r] to give � r � |f (s)|2 ≤ 4r2 −r r |f � (s)|2 (7) −r as required. 2 The NuemanPoincare Inequality � u then Theorem 2.1 If u is C 1 on Br , and we deﬁne A = 1 volBr Br � � (u − A)2 ≤ C(n)r2 |�u|2 . Br (8) Br Again we give the proof in the case n = 1. �r 1 Take a diﬀerentiable function f with A = 2r −r f . Note by the intermediate value theorem that there is a point c in [−r, r] with f (c) = A. We have � s f (s) = f (c) + f � (s). c From this we get |f (s) − A| ≤ �s c |f � (t)| ≤ �r −r |f � (t)|. Apply CauchySchwarz again to give �� r |f (s) − A| ≤ 2r � 2 (f ) �1/2 . (9) −r Squaring and integrating then gives our result. � r � 2 2 (f (s) − A) ≤ (2r) −r r |f � (s)|2 . (10) −r It is not diﬃcult to extend these proofs to higher dimensional cubes. There is another interesting fact related to the NeumannPoincare � � � inequality. If we deﬁne g(x) = Br (u − x)2 we can rearrange to get g(x) = Br u2 − 2x Br u − x2 . We then R u ﬁnd the minimum by taking the derrivative, and see that it occurs at x = Br , so the volBr mean really is the best constant approximation to a function. 2