u( x) = g( y) ds y ( 1 ) U solves u = 0 in U; u = 0 on U. ( 3)

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1 M ath Fall L ecture 4 ( S ep. 6, ) Properties and Estimates of Laplace s and Poisson s Equations In our last lecture we derived the formulas for the solutions of Poisson s equation through Green s function: G( x, y) G( x, y) f( y) dy g( y) ds y ( ) U U n y solves u f in U; u g on U. ( 2) ( When U R n, the boundary term vanishes, and G( x, y) should be replaced by the fundamental solution Φ( x y) ). From this formula we can obtain many regularity estimates for u, see the lecture note Hölder Estimates for the Poisson Equation for Fall 2008 Math527, or the GTM book Partial Differential Equations by J. Jost for more details. Note that, such estimates only apply to u given by the above formula. In other words, before we settle the uniqueness issue, we cannot use the above formula to estimate general solutions of Poisson s equation. Naturally, the uniqueness question leads to the study of Laplace s equation with zero boundary condition: u 0 in U; u 0 on U. ( 3) Thus it is important to study the properties of C 2 functions satisfying u 0. Such functions are called harmonic functions. Remark. We will see soon that, somehow surprisingly, the study of harmonic functions leads to much more than uniqueness of Poisson s equation. It turns out that all estimates can be obtained through several properties of the equation u 0 and the related u ( ) 0, without using the exact formula above. Remark 2. Another reason of studying harmonic functions is that the properties are much more stable under perturbation of the equation itself. While the exact formula only applies to Poisson equation, the properties of harmonic functions are shared by general linear elliptic equation and even nonlinear equations.. Properties of harmonic functions. Recall the definition ( A( x) D u) f ( 4) Definition 3. A C 2 function satisfying u 0 in U is called a harmonic function in U... Mean value formula. Theorem 4. If u C 2 ( U) is harmonic, then for every ball B r ( x) { y y x < r } U. u dx u ds ( 5) Remark 5. It turns out that is easier to prove. Thus we need to first establish u dx for all B r ( x) This is left as an exercise. u ds ( 6) u ds for all B r ( x) U. ( 7)

2 Proof. We prove u ds for all B r ( x) U. Without loss of generality, set x 0 and denote B r ( 0) by B r. We compute [ ] [ d u ds d ] dr B r B r dr u( r w) ds w B B w D u( r w) ds w B B n D u( y) ds y B B u dy 0. ( 8) B B Thus due to the continuity of u. B r Theorem 6. If u C 2 ( U) satisfies or B r u ds lim r 0 B r B r( x ) B r u ds u( 0) ( 9) u ds ( 0) u dx ( ) for all x U and all balls B r ( x) { y y x < r} U, then u is harmonic. Proof. We have already seen that the two conditions are equivalent. Thus we only need to show that u ds for all x U, B r ( x) U u 0 in U. ( 2) This can be fulfilled by simply reverse the argument in the proof of the above theorem. Remark 7. The above Converse to mean-value property is kind of trivial and not very useful. If u is already C 2, we can simply differentiate to see whether at every x u 0 or not, and there is no need to check the mean value condition for every x and every ball. What makes the mean value formula useful is the following theorem, which says we do not need the a priori knowledge that u is C 2. Theorem 8. If u C( U) satisfies u ds ( 3) for all x U and all balls B r ( x) { y y x < r} U, then u is harmonic. Proof. Since we already have shown that the mean value property leads to u harmonic if u C 2, we only need to show u C 2. Take any radially symmetric function φ φ( r) supported in B ε with φ. We will show that B ε ( u φ) ( x) u( y) φ( x y) dy u( x). ( 4) Now recall that R n D α ( u φ) u ( D α φ). ( 5)

3 Obviously we can take φ C 2 and conclude that u C 2. Now we show ( u φ) ( x) u( y) φ( x y) dy u( x). ( 6) R n Without loss of generality, set x 0. Take ε so small such that B ε U. We compute ( u φ) ( 0) u( y) φ( y) dy R n u( y) φ( y) dy B ε ε [ ] u( y) ds y φ( r) dr 0 B r ε B r u( 0) φ( r) dr 0 [ ε ] u( 0) φ( r) ds dr 0 B r u( 0) φ( y) dy B ε u( 0). ( 7) Thus ends the proof. Corollary 9. If u is harmonic, then u C. Proof. As u is harmonic, u satisfies the mean value formula. Therefore u φ u ( 8) for all φ satisfying the condition in the above theorem. Taking φ C gives the conclusion. The same argument can in fact prove the following Weyl s lemma, which relaxes the condition u C( U). Those who are interested in its proof can take a look at the lecture note Harmonic Functions for Fall 2008 s Math 527. Lemma 0. ( Weyl s lemma) Let u: U R be measurable and locally integrable in Ω. Suppose that for all ϕ C 0 ( U), u( x) ϕ( x) dx 0. ( 9) Then u is harmonic and, in particular, smooth. Remark. A question for those who know what a distribution: Let u be a distribution and in the distributional sense. Then can we conclude that u is C? U u 0 ( 20). 2. Local estimates for harmonic functions. Using the mean value formula, we can obtain good estimates for the derivatives of hamonic functions ( recall that harmonic functions are C ). Theorem 2. Assume u is harmonic in U. Then for each x U and B r ( x) U. Proof. See p. 29 of Evans. D α u( x) C k r n+ k B r( x ) u dx ( 2 )

4 Remark 3. From the above estimates, it is easy to show that u is not only C, but in fact analytic. See pp of Evans. Remark 4. The mean value formulas cease to be true for Poisson s equation or the more general elliptic equations. As a consequence, one can not obtain local estimates for these equations using the above method. A more robust way is to estimate through the following maximum principles.. 3. Harnack inequality. It turns out that, for nonnegative harmonic functions, its value at two different points are always comparable. Theorem 5. ( Harnack s inequality) For each connected open set V U, there exists a positive constant C, depending only on V, such that for all nonnegative harmonic functions u in U. Proof. First consider two points x, y ( denote r formula we have u B 2 r B 2 r The conclusion easily follows. B 2 r( x ). 4. Uniqueness for Poisson equation. It suffices to establish the following maximum principle: sup u C inf u ( 22) V V dist( x, y) ), such that B 2 r ( x) U. Using mean value B r( y) u u( y). ( 23) 2 n Theorem 6. ( Weak maximum principle) Suppose u C 2 ( U) C ( ) is harmonic in U. Then max u max u. ( 24) U Proof. Assume the contrary, that is max u > max U u. Then there must be a x 0 such that u( x 0 ) max u ( 25) U but u u( x 0 ) in some neighborhood of x 0. This contradicts the mean value formula. In fact, one can establish the stronger Theorem 7. ( Strong maximum principle) Suppose u C 2 ( U) C ( ). If U is connected and there exists a point x 0 U such that then u u( x 0 ) in U. u( x 0 ) max u, ( 26) Proof. See p. 27 of Evans. Remark 8. It is clear that the strong maximum principle ceases to be true when U is not connected. Theorem 9. The solution to Poisson s equation is unique. Proof. It follows from applying the weak maximum principle to the equation u 0 ( 27) with 0 boundary condition. 2. Maximum principles.

5 2.. S ubharmonic and superharmonic functions. We consider, instead of u 0, the inequalities u ( ) 0. ( 28) A simple adaptation of the proof for the Laplace equation then gives u ( ) 0 u( x) ( ) u dx, u ds ( 29) for all x U, B r ( x) U. This naturally leads to the following definition. Definition 20. Let u be continuous. It is called subharmonic (superharmonic) if for every B r ( x) U, we have u( x) ( ) u dx or u ds. ( 30) Remark 2. It is easy to see that subharmonic/ superharmonic functions are not necessarily differentiable, as the D example u x shows. Remark 22. One can show that, v is subharmonic( superharmonic) if and only if for every V U, and every harmonic function u on V such that u ( ) v on V, we have This further justifies the terminology subharmonic ( superharmonic ). u ( ) v in V. ( 3 ) Remark 23. Question: Do we still have or not? u( x) ( ) B r( x ) u dx u( x) ( ) u ds ( 32) It is easy to show that u subharmonic, then max and if u( x 0 ) max u for some x 0 U, then u u( x 0 ) ; u superharmonic, then inf and if u( x 0 ) inf u for some x 0 U, then u u( x 0 ). u max u, ( 33) U u inf u, ( 34) U This can be applied to obtain various estimates for Laplace s and Poisson s equations. For example, we can prove the estimate sup D u C sup u ( 35) B / 2 B for harmonic function u. To see this, we take a cut-off function η C 0 ( B ) such that η in B / 2. Then we compute ( ) η 2 2 D u 2 η η D u D η 2 D u η ( Dη) D 2 u D u + 2 η 2 D 2 u 2 [ 2 η η D u D η 2 D u 2 ( ) η η 6 D η D u 8 D η 2 D u η 2 D 2 u 2 ] + 2 η 2 D 2 u 2 C D u 2. ( 36)

6 Next we notice that, if u 0, then As a consequence, we have for some constant α. Thus η 2 D u 2 + α u 2 is subharmonic, and we have ( u ) 2 2 D u 2. ( 37) ( ) η 2 D u 2 + α u 2 0 ( 38) max D u 2 max η 2 D u 2 + α u 2 max η 2 D u 2 + α u 2 α B / 2 B B ( max B ) 2 u. ( 39) Remark 24. Note that the above argument does not involve mean value formula. Thus this method is more robust than estimating through mean value formula.

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