Laplace s Equation. Chapter Mean Value Formulas

Chapter 1 Laplace s Equation Let be an open set in R n. A function u C 2 () is called harmonic in if it satisfies Laplace s equation n (1.1) u := D ii u = 0 in. i=1 A function u C 2 () is called subharmonic (superharmonic) in if u 0 ( u 0) in. In this chapter, we discuss some basic properties of harmonic functions, such as mean value theorems, maximum principles, Harnack inequalities, etc. Note that Laplace s equation (1.1) is a special case of Poisson s equation (1.2) u = f in. 1.1. Mean Value Formulas Let u C 2 (). From the divergence theorem it follows (1.3) u dx = div (Du) dx = Du ν ds =: 0 0 0 0 u ν ds for any bounded open subset 0 0 with smooth boundary 0. Here ds is the (n 1)-dimensional surface area element in 0, and ν is the unit outward normal to 0. Fix y and r > 0, such that B r := B r (y) B r. For 0 < ρ r, consider the integral mean value of u over the surface B ρ (y): (1.4) m(ρ) := 1 u(x) ds x = 1 u(y + ρz) ds z, B ρ B ρ(y) σ n B 1 (0) 1

2 1. Laplace s Equation where B ρ = ρ n 1 B 1 = ρ n 1 σ n the (n 1)-dimensional surface area of B ρ. Since u(x) is continuous, m(ρ) u(y) as ρ 0 +. Further, z = (x y)/ρ =: ν x the unit outward normal to B ρ (y) at the point x = y+ρz. Applying the identity (1.3) to the ball B ρ (y), we have m (ρ) = 1 Du(y + ρz) z ds z and (1.5) σ n B 1 (0) = ρ1 n σ n B ρ(y) m(r) u(y) = r 0 Du(x) ν x ds x = ρ1 n σ n m (ρ) dρ = 1 r σ n = 1 0 ρ 1 n σ n x y <ρ<r B ρ(y) B ρ (y) u(x) dx, u(x) dx dρ ρ 1 n u(x) dx dρ. As a byproduct, here we get the following mean value properties. Theorem 1.1 (Mean value formulas). Let u be a function in C 2 () satisfying u = (, ) = 0 in. Than for any ball B r (y) B r (y), we have (1.6) u(y) = (, ) m(r) := 1 u(x) ds x, B r (1.7) u(y) = (, ) 1 B r B r(y) B r (y) u(x) dx. Proof. The relation (1.6) follows immediately from (1.4) (1.5). In order to get (1.7), one can integrate the relation u(y) B ρ = (, ) u(x) ds x with respect to ρ from 0 to r. B ρ (y)

1.2. The Maximum Principle 3 1.2. The Maximum Principle From the mean value formula (1.7), it is easy to derive the following strong maximum principle for subharmonic functions. Theorem 1.2 (Strong maximum principle). Let be an open connected set in R n, and let u C 2 () satisfy u 0 in. Suppose there is a point x 0 such that u(x 0 ) = sup u. Then u = const in. Proof. Denote M := sup u, and := {x : u(x) = M}. The set is not empty, because it contains the point x 0. For an arbitrary point y, one can choose r > 0 such that B r (y) B r (y). By virtue of (1.7), we have M = u(y) 1 u(x) dx M, B r B r (y) hence u M in B r (y), i.e. B r (y). Therefore, the set is both open and closed relative to. Since is connected, we must have =. In turn, from the strong maximum principle it follows immediately the following weak maximum principle, which is usually referred to as the maximum principle. Theorem 1.3 (Weak maximum principle). Let be a bounded open set in R n, and let u C 2 () C() satisfy u 0 in. Then (1.8) sup u = sup u. Remark 1.4. The minimum principles for superharmonic functions are formulated similarly to Theorems 1.2 and 1.3, one only needs to replace the assumption u 0 with u 0, and the sup with the inf. The proof follows from the above theorems applied to the function u. As easy consequences of the maximum principle, we obtain the following two theorems. Theorem 1.5 (Comparison principle). Let be a bounded open set in R n, and let functions u 1, u 2 C 2 () C() satisfy u 1 u 2 in, and u 1 u 2 on. Then u 1 u 2 in. Proof. This assertion follows from Theorem 1.3 applied to the function u := u 1 u 2.

4 1. Laplace s Equation Theorem 1.6 (Uniqueness). Let f C() and f C( ) be given functions, where is a bounded open set in R n. Then there exists at most one classical solution, i.e. solution u C 2 () C() of the boundary value problem (1.9) u = f in, u = g on. Proof. If u 1 and u 2 both satisfy (1.9), then Theorem 1.3 applied to u = ±(u 1 u 2 ) guarantees u 1 u 2 in. 1.3. The Harnack Inequality The following Harnack inequality is another consequence of the mean value formula (1.7) for harmonic functions. Theorem 1.7 (Harnack inequality). Let be a bounded open set in R n, and a constant δ > 0 be such that the set δ := {x : dist(x, ) > δ} is connected. Then for any non-negative harmonic function u in, (1.10) sup u N inf u, δ δ where the constant N depends only on n and δ/diam. Proof. Let r := δ/2. Then for any two points x 0, x 1 in δ, such that x 0 x 1 r, we have B r (x 0 ) B 2r (x 1 ). Since u 0 in, this implies u(x) dx u(x) dx. B r (x 0 ) B 2r (x 1 ) In turn, by the equality (1.7), this yields B r u(x 0 ) B 2r u(x 1 ), so that u(x 0 ) 2 n u(x 1 ). Note that for any x, y δ, one can choose a finite sequence x = x 0, x 1,..., x m = y in δ, such that x j x j 1 r for all j = 1,..., m, and m does not exceed a constant M = M(n, δ/diam ). By the previous argument, u(x) = u(x 0 ) 2 n u(x 1 ) 2 2n u(x 2 )... 2 mn u(x m ) = 2 mn u(y) 2 Mn u(y), and the desired estimate (1.10) follows with N := 2 Mn. Corollary 1.8 (Liouville theorem). Any harmonic function defined over R n and bounded below (or above) is constant

1.4. Regularity of harmonic functions 5 Proof. Replacing u by ±u+const, we may assume that inf u = 0. Applying the Harnack inequality with = B 2r (0), δ = r, we obtain 0 sup u N inf u 0 as r. B r (0) B r (0) Since the left side is a non-decreasing function of r, we get u 0. 1.4. Regularity of harmonic functions In this section, we show that the mean value formula implies the interior smoothness, and even analyticity, of harmonic functions. We will use socalled mollifiers η δ (x), δ = const > 0, satisfying the properties (1.11) η δ C (R n ), η δ 0 on R n \ B δ (0), and η δ (x) dx = 1. R n We also assume that η δ are radially symmetric, i.e. η δ (x) η0 δ ( x ), where the functions η0 δ are defined on [0, ). For example, one can take ηδ (x) := δ n η(δ 1 x), where { ( ) c exp 1 if x < 1, x (1.12) η(x) := 2 1 0 if x 1, and the constant c > 0 is chosen so that R n η(x) dx = 1. Lemma 1.9. Let u C 2 () be a harmonic function in an open set R n, and let η δ be a radially symmetric function satisfying the properties (1.11). Then (1.13) u(x) (u η δ )(x) := u(x y)η δ (y) dy B δ (0) on the set δ := {x : dist(x, ) > δ}. Proof. Fix a point x δ. Then B δ (x). Using polar coordinates in B δ (0), we can write δ [ (u η δ )(x) = u(x y)η δ (y) dy = u(x y)η δ (y) ds y ]dr. B δ (0) 0 B r (0) Note that η δ (y) η0 δ(r) on B r(0), and by virtue of (1.6), u(x y) ds y = u(z) ds z = u(x) B r. B r (0) B r (x)

6 1. Laplace s Equation Therefore, δ (u η δ )(x) = u(x) = u(x) 0 δ η0(r) δ B r dr = u(x) η δ (y) dy = u(x) 0 [ B r (0) η δ (y) ds y ]dr on the set δ. B r (0) Theorem 1.10 (Smoothness). Let u C 2 () be a harmonic function in an open set R n. Then u C (). Proof. For x δ, one can rewrite the equality (1.13) in the form (1.14) u(x) = u(y)η δ (x y) dy = u(y)η δ (x y) dy. B δ (x) R n The integrals here are well defined, because B δ (x), and η δ (x y) 0 on R n \ B δ (x). Since η δ C, we have u C ( δ ) for each δ > 0, so that u C (). Theorem 1.11 (Convergence theorem). Let {u m } C 2 () be a sequence of harmonic functions in an open set R n, which converges in L 1 (B r ) to a function u L 1 loc () for each ball B r B r. Then the limit function u (possibly after redefining on a set of zero measure) belongs to C () and is harmonic in. Proof. Fix a ball B r := B r (x 0 ) B r. By virtue of (1.14), we have (1.15) u m (x) = (u m η δ )(x) := u m (y)η δ (x y) dy B δ (x) for all m = 1, 2,... ; x B r/2 (x 0 ), and 0 < δ r/2. Therefore, (u i u j )(x) (u i u j )(y) η r/2 (x y) dy, and B r/2 (x) sup u i u j sup(η r/2 ) B r/2 (x 0 ) B r (x 0 ) u i u j dy 0 as i, j. This means that {u m } is a Cauchy sequence in C(B r/2 (x 0 )). Since this is true for all x 0 and 0 < r < dist(x 0, ), the sequence {u m } converges in C( ) for any bounded open set, and the limit function

1.4. Regularity of harmonic functions 7 u belongs to C(). Then from (1.15) it follows that for all x and 0 < δ < dist(x, ), we have u(x) = lim m u m(x) = lim m (u m η δ )(x) = (u η δ )(x), which implies u C (), and also u(x) = (u η δ )(x) = (u ( η δ ))(x) Theorem is proved. lim (u m ( η δ ))(x) = lim (( u m) η δ )(x) = 0. m m If we take η δ (x) := δ n η(δ 1 x) with η(x) defined in (1.12), then by virtue of (1.14), D l u(x) = δ n l u(y)d l η(δ 1 (x y)) dy, x δ, B δ (0) for any derivative D l u := D l 1 1 Dn ln u(x) of order l := l 1 + +l n. In turn, this implies the estimate (1.16) max sup l =k δ D l u N k δ k sup u for δ > 0, k = 1, 2,..., where N k := max l =k sup Dl η. The following theorem contains an improved version of this estimate. Theorem 1.12 (Estimates of derivatives). Let u C () be a bounded harmonic function in an open set R n. Then the estimate (1.16) holds true with N k := (nk) k. Proof. We first consider the case k = 1. Note that if u is harmonic, then (D i u) = D i ( u) = 0, i.e. D i u is harmonic for each i = 1,..., n. Moreover, we can write D i u = div (ue i ), where e i is the i-th coordinate vector in R n. For x δ, we have B δ (x) B δ (x). The mean value formula (1.7), together with the divergence theorem, yield D i u(x) = 1 D i u(y) dy = 1 u(y)e i ν ds y, B δ B δ B δ (x) B δ (x) where ν is the unit outward vector to B δ (x). Hence D i u(x) B δ B δ sup u n B δ (x) δ sup u for all x δ and i = 1,..., n, i.e. we have (1.16) for k = 1 with N 1 = n.

8 1. Laplace s Equation In the case k 2, we set r := δ/k and apply the above estimate to higher order derivatives of u: max sup l =k δ D l u = max ( n r Theorem is proved. D l u n kr r max l =k 1 ) k 1 ( n max sup D l u l =1 r r l =k sup sup D l u... (k 1)r ) k ( nk sup u = δ ) k sup u. Theorem 1.13 (Analyticity). Let u be a harmonic function in. Then u is analytic in. Proof. Fix an arbitrary point y. We need to show that u(x) = l D l u(y) (x y) l l! in a neighborhood of y, or equivalently, u(x) = l k 1 D l u(y) (x y) l + R k (x), l! where R k (x) 0 as k. Note that by the Taylor formula R k (x) = l =k D l u(z) (x y) l, l! where z belongs to the segment [y, x]. Fix a constant r 0 (0, 1 2dist(y, ). By Theorem 1.12 applied to the ball B 2r0 (y) B 2r0 (y), we have max l =k sup D l u B r0 (y) This estimate implies R k (x) ( nk ) km0, where M 0 := sup u. r 0 B 2r0 (y) ( nk r 0 ) km0 Note that by the multinomial theorem, we also have l =k (a 1 + + a n ) k = l =k l =k 1 l! = nk k!. (x y) l. l! k! l! al,

1.4. Regularity of harmonic functions 9 Therefore, for x B r (y), 0 < r r 0, we get ( n 2 kr ) k M0 R k (x) M k := k!, k 1. We see that M k+1 M k r 0 = n2 r (1 + 1 ) k en 2 r θ :=. r 0 k r 0 Finally, we choose 0 < r < r 0 en 2. Then 0 < θ < 1, and by iteration, we obtain M k M 0 θ k. This means that R k (x) 0 uniformly on the ball B r (y) as k, i.e. u is analytic at the point y. Since y is an arbitrary point in, the harmonic function u is analytic in. As a standard property of harmonic functions, the following statement hold true. Corollary 1.14 (Uniqueness of continuation). Let u 1 and u 2 be harmonic functions in an open connected set R n, such that u 1 u 2 in an ball B r (y). Then u 1 u 2 in. At the beginning of Section 1.1, we already used the classical divergence theorem in a special case of the vector field F = Du. For our reference, we formulate this theorem under general assumptions on F and. Theorem 1.15. Let be a bounded domain in R n with the boundary of class C 1, and let ν be the unit outward normal to. Then for any vector field F = (F 1,..., F n ) C 1 () C(), such that div F := D 1 F 1 + + D n F n L 1 (), we have (1.17) div F dx = F ν ds. Note that C 1 if there exists a function Ψ C 1 () such that (1.18) = {x R n : Ψ(x) > 0}, and DΨ 0 on. Since = {Ψ = 0} - a level set of the function Ψ, the unit outward normal ν = DΨ/ DΨ C( ), so that F ν C( ). Lemma 1.16 (Green s formula). Let be as in the previous theorem, and let functions u, v C 2 () C 1 () be such that u v v u L 1 (). Then ( (1.19) (u v v u) dx = u v ν v u ) ds. ν Proof. It suffices to apply the divergence theorem to the vector field F = udv vdu.

10 1. Laplace s Equation Corollary 1.17. Let u be a harmonic function in an arbitrary domain R n. Then (1.20) u v dx = 0 v C0 (). Proof. By definition, each function v C0 () belongs to C (), and v 0 on \ for some bounded subdomain. Since dist(, ) > 0, one can always replace by another subdomain with smooth boundary, such that. Then u, v C 2 ( ), v = Dv = 0 on, and the desired equality (1.20) follows from Green s formula applied to the domain. By our previous considerations, from u C 2 () and u = 0 in it follows that u C () and u satisfies (1.20). Note that the integral in (1.20) is well-defined for functions u L 1 loc (). It turns out that if u L 1 loc () satisfies (1.20), then u C () and u = 0 in. This is the statement of Weyl s lemma below. For its proof, we need some of the following properties of the operation of convolution. Proposition 1.18. Let u be a function in L 1 loc (), and let ηδ, δ > 0 be non-negative functions satisfying (1.11). Then the functions (1.21) u (δ) (x) := (u η δ )(x) := u(x y)η δ (y) dy B δ (y) belong to C ( δ, where δ := {x : dist(x, ) > δ}, and satisfies the following properties: (i) For each x δ, u (δ) (x) u(x) ω x (δ) := sup u(x + h) u(x). h δ In particular, if u C(), then for any bounded subdomain, u (δ) u in C( ) as δ 0 + : sup u (δ) u sup ω x (δ) 0 as δ 0 +. x (ii) The operation of convolution does not increase the L 1 -norm: u (δ) 1, δ u 1, := u(x) dx. (iii) For any bounded subdomain, u (δ) u in L 1 ( ) as δ 0 + : u (δ) u 1, 0 as δ 0 +.

1.5. Growth lemma and its application 11 Lemma 1.19 (Weyl s lemma). Let u L 1 loc () satisfy the property (1.20). Then u C (), and u is harmonic in. Proof. Fix y and r > 0 such that the ball B := B r (y) B 2r (y). For 0 < δ < r, the functions u (δ) C (B), and u (δ) (x) = (u η (δ) )(x) = u(y) x η (δ) (x y) dy. B δ (0) Here x denotes the Laplacian with respect to the variables x = (x 1,..., x n ). It is easy to see that x η (δ) (x y) y η (δ) (x y). For fixed x B and 0 < δ < r, the function v(y) := η (δ) (x y) C0 (), so that by virtue of (1.20) we have u (δ) (x) = 0. Hence u (δ), 0 < δ < r, are harmonic functions on B. By Proposition 1.18 (iii), u (δ) u in L 1 (B) as δ 0 +. Now it remains to apply the convergence Theorem 1.11. 1.5. Growth lemma and its application Lemma 1.20 (Growth lemma). Let be an open set in R n, and let x 0 and r > 0 be such that the measure (1.22) B r (x 0 ) µ B r, µ = const (0, 1). Let v be a function in C 2 () C(), such that (1.23) v > 0, v 0 in B r (x 0 ); v = 0 on ( ) B r (x 0 ). Then (1.24) v(x 0 ) µ M, where M := sup v, B r(x 0 ) and µ is the constant in (1.22). Proof. Fix ε > 0 and choose a function Φ ε C (R 1 ) (depending on ε) such that Φ ε, Φ ε, Φ ε 0 on R 1, Φ ε 0 on (, ε], and Φ ε 1 on [2ε, ). It is easy to check that the convolution Φ ε (x) := (Φ 0 η ε )(x 2ε), where Φ 0 (x) := x + := max(x, 0), satisfies all these properties, if we take η ε (x) := ε n η(ε 1 x), where the function η is defined in (1.12). Further, define (1.25) v ε := Φ ε (v) in B r (x 0 ), v ε 0 on B r (x 0 ) \. From the above properties of the function Φ ε it follows (1.26) (v 2ε) + v ε (v ε) + < M in B r (x 0 ).

12 1. Laplace s Equation Since v = 0 on the set ( ) B r (x 0 ), the function v ε vanishes near this set. Hence we have v ε C 2 (B r (x 0 ))) and v ε 0 in B r (x 0 ). Moreover, Dv ε = Φ ε(v) Dv, and (1.27) v ε = Φ ε(v) v + Φ ε(v) Dv 2 0 in B r (x 0 ). Now we can use the mean value formula (1.7), which gives us v ε (x 0 ) 1 v ε (x) dx B r (1.28) 1 B r B r (x 0 ) B r(x 0 ) M dx B r(x 0 ) B r M µ M. Since v ε (x 0 ) v(x 0 ) as ε 0 +, the desired estimate (1.24) follows. Definition 1.21. An open set R n satisfies the condition (A) with a constant µ 0 (0, 1) if the measure (1.29) B r (y) µ 0 B r for all y and r > 0. Theorem 1.22. Let be a bounded open set in R n satisfying the condition (A) with a constant µ 0 > 0. Let an open subset and a function u C 2 ( ) C( ) be such that u > 0 in and u = 0 on. Then there are constants γ > 0 and N > 0, which depend only on n and µ 0, such that (1.30) M := sup d γ u N F, where F := sup d 2 γ ( u), d = d(x) := dist(x, ), ( u) := max( u, 0). Proof. Without loss of generality, we may assume that dist(, ) > 0. Indeed, for δ > 0, the function u δ and the domain δ := {u > δ} satisfy these assumptions. Therefore, if we first prove the estimate (1.30) for u δ in δ, then this estimate for the original function u in follows by taking δ 0 +. This additional assumption guarantees that d γ is bounded in for γ > 0 (the constant γ (0, 1) will be specified later). Hence the function d γ u attains its maximum on at some point x 0, i.e. (1.31) M = r γ u(x 0 ), where r := d(x 0 ) := dist(x 0, ) > 0. We will take into consideration the auxiliary function which satisfies w(x) := 1 2n (4r2 x x 0 2 ) (1.32) w > 0, w = 1 in B 2r (x 0 ).

1.5. Growth lemma and its application 13 For fixed constants γ, ε (0, 1), consider the function (1.33) v(x) := u(x) (εr) γ M (εr) γ 2 F w(x). First we assume that v(x 0 ) > 0. Then x 0 belongs to the set (1.34) V := B 2r (x 0 ) {x : d(x) > εr} {x : v(x) > 0}, so that V is not empty. We are going to apply the Growth lemma to the function v in the domain V, with the ball B 2r (x 0 ) in place of B r (x 0 ). Note that r := dist(x 0, ) = x 0 y 0 for some point y 0. Then B r (y 0 ) B 2r (x 0 ), and also V. Since satisfies the condition (A), i.e. inequality (1.29), at the point y = y 0, we get B 2r (x 0 ) \ V B r (x 0 ) \ = B r B r (y 0 ) (1 µ 0 ) B r = (1 µ 0 )2 n B 2r, V B 2r (x 0 ) = B 2r B 2r (x 0 ) \ V µ B 2r, where µ := 1 (1 µ 0 )2 n (0, 1). This means the condition (1.22) is satisfied with this constant µ = µ(n, µ 0 ) (0, 1). Further, we claim that (1.35) v = 0 on ( V ) B 2r (x 0 ). Suppose otherwise, i.e. v(y) > 0 at some point y ( V ) B 2r (x 0 ). Since y is an interior point of B 2r (x 0 ), we have w(y) > 0. Moreover, since 0 < v(y) < u(y) and u = 0 on, the point y must be an interior point of. By definition of M, we have u d γ M, hence 0 < v(y) < u(y) (εr) γ M [d γ (y) (εr) γ ] M, which implies d(y) > εr, i.e. y is an interior point of the set {x : d(x) > εr}. Summarizing these arguments, we see that y is an interior point of V, in contradiction with our assumption y V. This contradiction proves the property (1.35). Finally, since d > εr in V and γ 2 < 0, we have which in turn implies u ( u) d γ 2 F (εr) γ 2 F in V, v = u (εr) γ 2 F w = u + (εr) γ 2 F 0 in V. Therefore, we can apply the Growth lemma (Lemma 1.20) to the function v < u in the domain V B 2r (x 0 ), which yields (1.36) v(x 0 ) µ sup V v µ sup B 2r (x 0 ) u.

14 1. Laplace s Equation Since d(x 0 ) = r, we also have d = d(x) x x 0 + d(x 0 ) < 3r, u d γ M (3r) γ M on the set B 2r (x 0 ). From the previous estimate we now obtain (1.37) v(x 0 ) µ (3r) γ M. In the proof of this estimate, we assumed v(x 0 ) > 0, which guarantees that x 0 belongs to the set V in (1.34). In the case v(x 0 ) 0, the estimate (1.37) is obvious, so that it is true without any additional assumptions (for fixed γ, ε (0, 1)). Having in mind that u(x 0 ) = r γ M and w(x 0 ) = 2r 2 /n, by virtue of (1.33) and (1.37), we conclude v(x 0 ) = r γ M (3r) γ M (3r) γ 2 F 2 n r2 µ (3r) γ M or equivalently, (1.38) (1 ε γ 3 γ µ) M 2 n εγ 2 F. Since µ = µ(n, µ 0 ) < 1, now it remains to choose a small constant γ = γ(n, µ 0 ) > 0, and then a small constant ε = ε(n, µ 0 ) > 0 in such a way that the constant c 0 = c 0 (n, µ 0 ) := 1 ε γ 3 γ µ is strictly positive. Then the desired estimate (1.30) follows immediately with the constant N = N(n, µ 0 ) := 2 nc 0 ε γ 2. 1.6. Model problems for the Laplace operator In this section, we consider three boundary value problems for Laplace s equation. They are called model problems here because they serve as a foundation for more general theory of elliptic equations with Hölder coefficients. The corresponding Theorems 1.24 1.26 assert the existence of harmonic functions, satisfying specific boundary conditions for each problem, and having certain regularity properties. First we consider the Dirichlet problem in a ball: (1.39) u = f in B r := B r (x 0 ), u = g on B r, with polynomials f and g. For k = 0, 1, 2,..., denote P k the collection of all polynomials p(x) = p(x 1,..., x n ) of degree at most k: P k := {p = p(x) = } c l x l, c l constants. l k Lemma 1.23. Let polynomials f P k and g P k+2 be given. Then for arbitrary x 0 R n and r > 0, the Dirichlet problem (1.39) has a unique solution in C 2 (B r ) C(B r ). This solution belongs to P k+2 and is represented in the form u = g + (r 2 x x 0 2 )h, where h P k.

1.6. Model problems for the Laplace operator 15 Proof. Replacing x by r 1 (x x 0 ), we may assume without loss of generality x 0 = 0, r = 1. The uniqueness of solution is contained in Theorem 1.6. In order to prove the existence, we set v := u g, so that (1.39) is reduced to the equivalent problem (1.40) v = f 0 in B 1, v = 0 on B 1, with f 0 = f g P k. Next, consider the linear mapping T p := ((1 x 2 )p) : P k P k. If T p = 0, then v := (1 x 2 )p is a solution of the problem (1.40) with f 0 0. By uniqueness, we have v 0 and p 0. Hence T has the trivial null space: T 1 (0) = {0}. Since the dimension of P k is finite, we must have T (P k ) = P k. Then T h = f 0 for some h P k, and v = (1 x 2 )h is a solution of (1.40). Finally, u = g + v = g + (1 x 2 )h is a solution of the problem (1.39). Theorem 1.24. For any continuous function g on B r := B r (x 0 ), the Dirichlet problem (1.41) u = 0 in B r := B r (x 0 ), u = g on B r, has a unique classical solution (i.e. solution in C 2 (B r ) C(B r )). This solution belongs to C (B r ) and satisfies the estimate (1.42) max l =k sup D l u N k δ k sup u for 0 < δ < r, k = 1, 2,..., B r δ B r with the constant N k := (nk) k. Proof. Based on the Weierstrass approximation theorem, we can choose a sequence of polynomials {g m } such that g g m 1/m on B r for m = 1, 2,.... By Lemma 1.23, there exists a sequence of polynomials {u m } satisfying u m = 0 in B r, u m = g m on B r. From the comparison principle (Theorem 1.5) it follows sup u i u j = sup g i g j sup g i g + sup g j g B r B r B r B r 1 i + 1 j 0 as i, j. Therefore, {u m } is a Cauchy sequence in C(B r ), so it converges uniformly to a function u C(B r ), which implies the convergence in L 1 (B r ). By Convergence theorem (Theorem 1.11), the limit function u C (B r ) C(B r ) and satisfies (1.41). The uniqueness of solution is guaranteed by Theorem 1.6.

16 1. Laplace s Equation In the remaining part of this section, we consider two problems in a half ball B + r := B + r (x 0 ) := {x = (x 1,..., x n ) B r (x 0 ) : x n > 0}, where x 0 R n 0 := {x Rn : x n = 0}. Note that B + r = Γ S, where (1.43) Γ := B r {x n = 0}, S := ( B + r ) \ Γ = ( B r ) {x n 0}. In other words, Γ and S are correspondingly the flat and the spherical portions of B + r, and Γ is open relative to B + r. Theorem 1.25. For any continuous function g on B + r, such that g 0 on Γ, the Dirichlet problem (1.44) u = 0 in B + r, u = g on B + r, has a unique classical solution. This solution belongs to C (B r Γ) and satisfies the estimate (1.45) max l =k sup B + r δ with N k := (nk) k. D l u N k δ k sup u for 0 < δ < r, k = 1, 2,..., B r + Proof. One can extend the function g from S to B r as an odd function of x n. Then g(x) = g(x 1,..., x n 1, x n ) = g(x 1,..., x n 1, x n ) on B r. The extended function g is continuous on B r, so that by the previous theorem, the problem (1.41) has a unique solution u C (B r ) C(B r ). Then the function v(x) = v(x 1,..., x n 1, x n ) := u(x 1,..., x n 1, x n ) satisfies both the equation v = u = 0 in B r and the boundary condition v = g on B r. By uniqueness, we must have u v in B r, which implies u 0 on Γ. This means that u is also a solution of the problem (1.44), and the estimate (1.45) follows from (1.42). The next theorem deals with different kinds of conditions on different parts on B + r. These are the mixed boundary conditions (1.46) D n u = 0 on Γ, u = g on S, where Γ and S are defined in (1.43). Theorem 1.26. For any continuous function g on S, the equation u = 0 in B + r, with the boundary conditions (1.46), has a unique solution u C (B r Γ) C(B + r ). This solution satisfies the estimate (1.45).

1.7. Removable singularities for harmonic functions 17 Proof. Now we extend the function g from S to B r as an even function of x n : g(x) = g(x 1,..., x n 1, x n ) = g(x 1,..., x n 1, x n ) on B r. As in the previous proof, the problem (1.41) has a unique solution u C (B r ) C(B r ). By uniqueness, From this equality it follows u(x) v(x) := u(x 1,..., x n 1, x n ) in B r. D n u = D n v = D n u on Γ := B r {x n = 0}, so that D n u = 0 on Γ. Therefore, the function u satisfies all the desired properties in this theorem. Now it remains to show that the solution is unique. If u 1 and u 2 are two harmonic functions in B r +, both satisfying (1.46), then u := u 1 u 2 satisfies (1.46) with g = 0. Hence we need to show that from D n u = 0 on Γ and u = 0 on S it follows u = 0 in B r +. Note that for any ε > 0, the function u ε := u + εx n is harmonic in B r + and belongs to C (B r ) C(B r ). By the maximum principle (Theorem 1.3) it attains its maximum on B r + at some point y B r + = Γ S. However, since D n u ε = D n u + ε = ε > 0 on Γ, the point y cannot belong to Γ. Then y S, and sup B + r u ε = sup S u ε = sup(εx n ) = εr. S This implies that u = lim ε 0 uε 0 in B + + r. Quite similarly (or replacing u by u), we also get u 0 in B r +. Therefore, u 0, and the uniqueness follows. 1.7. Removable singularities for harmonic functions For radially symmetric functions v(x) := v 0 (r), r := x, the Laplace s operator has the form (1.47) v(x) = v 0(r) + n 1 v r 0(r) = r 1 n(. r n 1 v 0(r)) Solving a simple ODE, one can describe all harmonic functions h(x) := h 0 (r) on R n \ {0} as follows { c 1 + c 2 ln x, if n = 2; (1.48) h(x) = c 1 + c 2 x 2 n, if n 2, where c 1 and c 2 are arbitrary constants.

18 1. Laplace s Equation Theorem 1.27 (Removable singularity). Let be an open set in R n, and let u be a harmonic function in \ {x 0 } for some point x 0, such that { ln x, if n = 2; (1.49) lim x x 0 u(x) = 0, where h(x) := h(x x 0 ) x 2 n, if n 2, Then u can be extended as a harmonic function to the whole set. Proof. Without loss of generality, we can assume x 0 = 0. Choose r (0, 1) small enough, so that B r := B r (0) B r. By Theorem 1.24, there exists a solution ũ C 2 (B r ) C(B r ) of the problem ũ = 0 in B r, ũ = u on B r. Next, for 0 < δ < r, we compare u(x) with the functions where w δ 1,2(x) := ũ(x) ± β(δ) h(x) on δ := B r \ B δ, u(x) ũ(x) 0 β(δ) := sup 0 as δ 0 + B δ h(x) by the assumption (1.49). All three functions u, w1 δ, wδ 2 are harmonic in δ, and w1 δ u wδ 2 on δ. By the comparison principle, we have w1 δ u wδ 2 in δ. Then for arbitrary x B r \ {0}, u(x) ũ(x) β(δ) h(x) 0 as δ 0 +. Therefore, u ũ on B r \ {0}, and the function u, extended as u(0) := ũ(0), is harmonic in. Remark 1.28. It follows from this theorem that the Dirichlet problem (1.50) u = 0 in := B 1 \ {0}; u = 0 on B 1, u(0) = 1; does not have solutions in C 2 () C(). One can use similar arguments in order to fix boundary singularities. Namely, the maximum principle (Theorem 1.3) and the comparison principle (Theorem 1.5) are applied to functions in C 2 () C(), where is a bounded open set in R n. These statements mat fail if we allow discontinuity even at a single point x 0. Example 1.29. Note that any derivative of a harmonic function is harmonic. Differentiating the function h(x) in (1.48), we get: (1.51) h 0 (x) := x n x n = const D nh(x) is harmonic in R n \ {0}.

1.7. Removable singularities for harmonic functions 19 If we take := B + 1 (0) := {x Rn : x < 1, x n > 0}, and x 0 := 0, then h 0 is harmonic in, belongs to C 2 () C(\{0}), and does not satisfy the maximin principle: = sup h 0 > sup h 0 = 1. ( )\{0} In this example, the harmonic function h 0 has singularity of order x 1 n at the point x 0 = 0. The following theorem states that the maximum principle, and hence the comparison principle, are applicable for weaker singularities. Theorem 1.30. Let be a bounded open set in R n, n 2, such that B r (0) = B + r := {x R n : x < r, x n > 0} for some r > 0. Let u C 2 () C( \ {0}) be a harmonic function in satisfying the property (1.52) u(x) x n 1 0 as x, x 0. Then (1.53) M := sup u = M 0 := sup u. ( )\{0} Proof. Denote (1.54) m r := sup S r u, where S r := {x R n : x = r, x n > 0} B + r. First we reduce the proof to the case = B + r, i.e. (1.55) sup u = sup u. B r + ( B r + )\{0} Indeed, the set ( B + r ) \ {0} is contained in the union of two sets: ( ) \ {0} and S r, so that from (1.55) it follows sup u max(m 0, m r ). B r + Moreover, the boundary of \ B r + is also contained in the union of these two sets, hence by the maximum principle, sup \B + r u = ( sup ) u max(m 0, m r ). \B + r Since we always have M 0 M and m r M, these inequalities imply M := sup u = max(m 0, m r ). It is easy to see that the case M > M 0 is impossible: in this case we must have M = m r, and the maximum of u would be attained at some point

20 1. Laplace s Equation x 0 S r, which contradicts to the strong maximum principle. We have proved that from (1.55) it follows (1.53). Now it remains to prove (1.55), i.e. (1.53) in the case := B + r. Replacing u by u const is necessary, we may assume M 0 = 0. Since the function h 0 in (1.51) is harmonic in R n \ {0}, the functions h δ (x) = h δ (x, x n ) := h 0 (x, x n + δ) = x n + δ [ x 2 + (x n + δ) 2] n/2, 0 < δ r, belong to C (B + r ) and satisfy h δ > 0, h δ = 0 in B + r ; 2 n δ 1 n h δ 2 δ 1 n in B + δ. By virtue of (1.52), β(δ) := sup S δ For each δ (0, r], the function u(x) 0 as δ 0+ h δ (x) u δ (x) := u(x) β(δ) h δ (x) is harmonic in δ := B + r \ B + ρ. By definition of β(δ), u δ 0 on S δ. On the remaining part of δ, which is a subset of ( B + r ) \ {0}, we have u δ u M 0 = 0. By the maximum principle, For arbitrary x B + r, we obtain u δ sup δ u 0 in δ = B + r \ B + ρ. u(x) = lim δ 0 + u δ(x) 0. This means M 0 = M 0. Since the opposite inequality is always true, the estimate (1.53) is proved. Corollary 1.31. In addition to the assumptions of the previous theorem, suppose that u = g on ( )\{0}, where g C( ). Then u can be extended as a continuous function from \ {0} to by setting u(0) := g(0). Proof. By defining g := u on S r := { x = r, x n > 0} = ( B + r ) \ ( ), we extend g as a continuous function on B + r. Consider the Dirichlet problem (1.56) ũ = 0 in B + r, ũ = g on B + r.

1.8. Dirichlet problem for Laplace s equation in general domain 21 In the next section, we prove that this problem has a classical solution ũ C 2 (B r + ) C(B r + ). By the previous theorem, the functions v := ±(ũ u) satisfy sup v = sup v = 0. B r + ( B r + )\{0} Therefore, u ũ on (B r + ) \ {0}, and the extended function u coincides with ũ C(B r + ) on B r +. Remark 1.32. A bit longer argument helps to bypass the existence result for the problem (1.56) (which is not proved yet), when g is an arbitrary continuous function on B + r. We do have the existence theorem (Theorem 1.25) in the case g 0 on Γ r := { x r, x n = 0} - the flat portion of B + r. Replacing u, g by u c, g c, we easily extend this theorem to the case g c = const on Γ r. For 0 < ρ r, we can now define ũ ρ C 2 (B + ρ ) C(B + ρ ) as a solution to the problem where ũ ρ = 0 in B + ρ, ũ ρ = g ρ on B + r, r ρ := c ρ := sup g on Γ ρ, g ρ := max(u, c ρ ) on S ρ. Γ ρ By Theorem 1.30, sup B + ρ ( u ũρ ) = sup ( B + ρ )\{0} ( u ũρ ) 0, 0 < ρ r. Then lim sup u(x) lim ũ ρ = c ρ, 0 < ρ r. x 0 x 0 Here c ρ g(0) as ρ 0 +. Therefore, lim sup u(x) g(0). x 0 Similarly (or replacing u by u and g by g), we also get lim inf x 0 u(x) g(0). From here the desired property follows: u(x) g(0) as x 0. 1.8. Dirichlet problem for Laplace s equation in general domain Let be a bounded domain in R n satisfying the condition (A) in Definition 1.21 with a constant µ 0 (0, 1). In this section, we prove the existence of a classical solution, i.e. solution in the class C 2 () C(), to the Dirichlet problem (1.57) u = 0 in, u = g on,

22 1. Laplace s Equation where g is a continuous function on. By results in Section A.1, we can extend u as a continuous function to the whole space R n, so that we can assume g C(R n ). 1.8.1. Existence of a weak solution. First we assume g C 1 (R n ). Consider the energy functional (1.58) E[v] := 1 Dv 2 dx 2 on the class of admissible functions A := {v C 1 () : v = g on }. This class is not empty, because it contains the function g. Denote (1.59) m := inf E[v] 0, v A and choose a sequence u 1, u 2,..., such that m k := E[u k ] m as k. For each pair i, j, the function 1 2 (u i + u j ) belongs to A, which implies 1 ] m E[ 2 (u i + u j ) = 1 D(u i + u j ) 2 dx. 8 Using the parallelogram identity D(u i + u j ) 2 + D(u i u j ) 2 = 2 Du i 2 + 2 Du j 2, we can write (1.60) 1 ] D(u i u j ) 2 dx = 4 E[u i ] + 4 E[u j ] 8 E[ 2 (u i + u j ) 4m i + 4m j 8m 0 as i, j. Since u i u j = 0 on, by Poincaré inequality, u i u j 2 dx 0 as i, j. Therefore, there is a function u L 2 () L 1 () such that u k u in L 2 () as k 0, which implies the convergence in L 1 (): (1.61) u k u 1 := u k u dx 0 as k.

1.8. Dirichlet problem for Laplace s equation in general domain 23 Further, for arbitrary function v C0 () and τ R1, the functions u k + τv belong to A, hence m E[u k + τv] = m k + 2τ Du k Dv dx + τ 2 E[v] = m k 2τ u k v dx + τ 2 E[v]. Taking the limit as k, we obtain 0 2τ u v dx + τ 2 E[v] for all τ R 1, which is only possible if u v dx = 0, v C 0 (). By Weyl s lemma (Lemma 1.19) we conclude that u C (), and u = 0 in. 1.8.2. Comparison principle. We still need to show that under appropriate assumptions on, the function u belongs to C(), and u = g on. Suppose this is the case. Then by the comparison principle, from (1.62) w C 2 (), w < 0 on, and w g on, it follows u w in. It turns out that the function u in (1.61) satisfies this property, without any additional assumptions on. Lemma 1.33. Let w be a function satisfying (1.62). Then u w in. Proof. Fix ε > 0, and let Φ ε C (R 1 ) be a function in the proof of Growth lemma (Lemma 1.20). Since u k C 1 () and u k = g w on, the functions v k := Φ ε (u k w) belong to C0 1 (). By the chain rule, Dv k = Φ ε(u k w) D(u k w). Since 0 Φ ε 1, we have 0 Dv k 2 dx = (Φ ε) 2 D(u k w) 2 dx Φ ε D(u k w) 2 dx = = A k + B k, Dv k D(u k w) dx

24 1. Laplace s Equation where A k := Dv k D(u k u) dx, B k := Dv k D(u w) dx. From the convergence in (1.60) it follows that Du L 2 (), and D(u k u) 2 dx 0 as k. By the Cauchy-Schwartz inequality, we obtain A 2 k Dv k 2 dx D(u k u) 2 dx const D(u k u) 2 dx 0 as k. Further, since v k C0 1 (), we have B k = v k (u w) dx = From (1.61) it follows This implies v k w dx. v k v 1 0 as k, where v := Φ ε (u w) 0. 0 lim(a k + B k ) = k 0 v w dx. Since v 0, w < 0 in, and v C(), we must have v = Φ ε (u w) 0 in. This inequality is true for arbitrary ε > 0. Hence i.e. u w in. (u w) + = lim ε 0 + Φ ε(u w) 0 in, 1.8.3. Existence in the case C 2, g C 2. Now we assume that the domain has boundary of class C 2. Then it satisfies the following exterior sphere condition with a constant r 0 > 0: (S) For each point x 0, there is a ball B := B r0 (y 0 ) such that B =, x 0 B.

1.8. Dirichlet problem for Laplace s equation in general domain 25 Lemma 1.34. Let satisfy the above condition (S), and let g C 2 (R n ). Then the weak solution u satisfies (1.63) lim x x x 0 u(x) = g(x) for each x 0. Therefore, the function u, extended as u := g on, belongs to C () C(), and solves the problem (1.57) in a classical sense. Proof. Fix constants γ > n 2 0 and R > diam () + r 0, where r 0 > 0 is a constant in the condition (S). By virtue of this condition, for each x 0, there exists y 0 R n such that Note that {x R n : r 0 < x y 0 < R}, and x 0 B r0 (y 0 ). For fixed x 0, set ( x γ) = γ(γ n + 2) x γ 2 > 0 for x 0. w(x) := g(x) + N We have w C (), w g on, and [r γ 0 x y 0 γ], where N = const > 0. w(x) = g(x) Nγ(γ n + 2) x y 0 γ 2 sup( g) Nγ(γ n + 2) R γ 2 < 0 on, provided the constant N > 0 is large enough. Then from Lemma 1.33 it follows u w in, therefore, lim sup x x x 0 u(x) lim x x x 0 w(x) = g(x 0 ). Replacing u by u and g by g, one can also get a similar estimate lim inf u(x) g(x 0 ) as x x 0. From these estimates, the equality (1.63) follows. 1.8.4. Existence in the general case. Theorem 1.35. Let be a bounded domain in R n satisfying the condition (A). Then for any function g C(R n ), the Dirichlet problem (1.57) has a unique solution u C () C(). Proof. We first assume g C 2 (R n ). For small ε > 0, choose open sets ε satisfying the properties (1.64) ε C 2, and ( {d > ε} ) ε ε, where d = d(x) := dist(x, ). By Lemma 1.34, the problems (1.65) u ε = 0 in ε, u ε = g on ε,

26 1. Laplace s Equation have classical solutions u ε C 2 ( ε ) C( ε ). Since we assume g C 2 (R n ), the functions v ε := u ε g satisfy (1.66) v ε = f := g in ε, u ε = 0 on ε. From Theorem 1.22, applied to the functions ±v ε on := ε, it follows the estimate sup d γ v ε N sup d 2 γ g =: N 0, ε where the constants γ > 0 and N > 0 depend only on n and µ 0 in (1.29). Set v ε := 0 on \ ε. Then v ε C(), and (1.67) v ε (x) N 0 d γ (x) in, uniformly with respect to ε > 0. Fix ε 0 > 0. Then for 0 < ε 1, ε 2 < ε 0, we have (v ε1 v ε2 ) = (u ε1 u ε2 ) in {d > ε 0 }. The maximum principle together with (1.67) imply sup v ε1 v ε2 = sup v ε1 v ε2 2N 0 d γ (x) for 0 < ε 1, ε 2 < ε 0. {d ε 0 } Therefore, lim sup sup v ε1 v ε2 2N 0 d γ (x). ε 1,ε 2 0 + Here the left side does not depend on ε 0 > 0, which can be made arbitrary small. This means v ε converge in C() to a function v C() as ε 0 +. Equivalently, u ε extended as u ε := g on \ ε, converge in C() to u := v + g C(), By the convergence theorem (Theorem 1.11), from u ε = 0 in {d ε 0 }, 0 < ε < ε 0, it follows that the limit function u is harmonic in {d ε 0 }. Since ε 0 > 0 is arbitrary, the function u is harmonic in and belongs to C () C(). This completes the proof in the case g C 2 (R n ). In the general case, when the function g belongs to C(R n ), we can assume u C 0 (R n ), and then approximate it uniformly on R n by standard convolutions g (δ) C (R n ): sup g g (δ) 0 as δ 0 +. R n For each δ > 0, there exists a solution u δ C () C() to the problem u (δ) = 0 in, u = g (δ) on. By the comparison principle, u δ converges in C() to a function u C() as δ 0 +. By Theorem 1.11, the limit function u C C() is harmonic in.

1.9. Fundamental solution 27 1.9. Fundamental solution 1.9.1. Properties of fundamental solution. Changing the order of integration in (1.5) yields m(r) u(y) = 1 [ r ] ρ 1 n dρ u(x) dx (1.68) where = σ n B r (y) B r (y) (1.69) Γ 0 (r) = x y [Γ 0 (r) Γ 0 ( x y )] u(x) dx, { 1 (2 n)σ n r 2 n, n 2; ln r, n = 2. 1 2π Definition 1.36. The function Γ(x) := Γ 0 ( x ), x 0, is called the fundamental solution of Poisson s equation u = f. (1.70) (1.71) Note that for x 0, D i Γ(x) = Γ(x) = x i σ n x n, D ijγ(x) = 1 n D ii u(x) = 0. i=1 σ n x n ( δ ij nx ) ix j x 2 ; The choice of constants in (1.69) is motivated by the following property. Theorem 1.37. If u C0 2(Rn ), then (1.72) u(y) = (Γ u)(y) := Γ(y x) u(x) dx for all y R n. R n Proof. For fixed y R n, choose r > 0 large enough, such that u 0 on R n \ B r (y). Then in (1.68) we have m(r) = 0, and also by virtue of (1.3), u Γ 0 (r) u(x) dx = Γ 0 (r) ds = 0. ν B r(y) B r(y) Therefore, the equality (1.68) is reduced to (1.72). Corollary 1.38. Assume f C 2 0 (Rn )). Then the function u := Γ f belongs to C 2 (R n ), and u = f in R n.

28 1. Laplace s Equation Proof. Since f(x) has compact support and Γ(x) is locally integrable in R n, one can differentiate the equality u(y) := Γ f(y) := Γ(x)f(y x) dy twice, which implies D ij u(y) = Γ(x)D ij f(y x) dy =: Γ D ij f(y) C(R n ), R n i.e. u C 2 (R n ). By the previous theorem, u = Γ f = f in R n. R n 1.9.2. Green s function. Let be a bounded domain in R n with boundary of class C 1. Then the unit outward normal ν = ν y is well defined at each point y. In this section, we derive a general formula for solutions to the Dirichlet problem (1.73) u = f in, u = g on, where f and g are given functions. For arbitrary u, v C 2 (), by virtue of the divergence theorem, div (udv vdu) dx = (udv vdu) ν ds. We can rewrite this equality in the following form, which is known as the Green s identity: ( (1.74) (u v v u) dx = u v ν v u ) ds. ν In particular, if u = v = 0 on, then (1.75) u v dx = v u dx. Definition 1.39. Suppose that for each x, there is a function h x = h x (y) C 2 () satisfying y h x (y) = 0 in, h x (y) = Γ(y x) for all y. Then the function (1.76) G(x, y) := Γ(y x) h x (y), x, y. x y, is called the Green s function for the domain. Note that the Green s function G(x, y) satisfies (1.77) y G(x, y) = 0 in \{x}; G(x, y) = 0 for all x, y.

1.9. Fundamental solution 29 Theorem 1.40. If u C 2 (), then (1.78) u(x) = G(x, y) u(y) dy + for all x. G(x, y) u(y) ds y ν y Proof. Fix x and choose a function ζ C0 () such that ζ 1 on B ε (x) for some ε > 0. Then the function u is represented in the form u = u 1 + u 2, where u 1 := ζu C 2 0(), u 1 u on B ε (x); u 2 := (1 ζ)u u near, u 2 0 on B ε (x). By Theorem 1.37, u(x) = u 1 (x) = Γ u 1 (x) := Γ(y x) u 1 (y) dy. Moreover, since u 1 vanishes near, and the function h x is harmonic in, from the Green s identity (1.74) it follows ( h x u 1 dy = u 1 h x dy + h x u 1 ν u h x ) 1 ds = 0. ν This gives us where I 1 := u(x) = G(x, y) u(y), ε G(x, y) u 1 (y) dy = I 1 + I 2, I 2 := G(x, y) u 2 (y). Since u 2 0 on B ε (x) and y G(x, y) = 0 on \ B ε (x), we have I 2 = [u 2 (y) y G(x, y) G(x, y) u 2 (y)] dy ε [ G(x, y) = u 2 (y) G(x, y) u ] 2(y) ds y. ν y ν y The boundary ε consists of the exterior part and the interior part B ε (x). In the last surface integral, the integral function vanishes on B ε (x) together with u 2, and G = 0 on. Therefore, I 2 = G(x, y) u 2 (y) ds y = ν y G(x, y) u(y) ds y, ν y and the representation (1.78) for u(x) = I 1 + I 2 follows.

30 1. Laplace s Equation The following theorem states that the Green s function G(x, y) is symmetric in x and y. In particular, this implies that one can interchange x and y in the relations (1.77). Theorem 1.41. The Green s function G(x, y) satisfies (1.79) G(x, y) = G(y, x) for all x, y, x y. Proof. Fix x, y, x y, and a constant ε (0, x y /2). Let u ε, v ε be functions in C 2 (), such that u ε = G(x, z) on \ B ε (x), v ε = G(y, z) on \ B ε (y). By virtue of (1.78), G(x, y) = u ε (y) = G(y, z) u ε (z) dz. Note that u ε (z) = z G(x, z) = 0 on the set \ B ε (x), which contains the ball B ε (y). Therefore, G(x, y) = v ε (z) u ε (z) dz, and quite similarly, G(y, x) = v ε (x) = G(x, z) v ε (z) dz = u ε (z) v ε (z) dz. Since u ε = v ε = 0 on, the desired equality G(x, y) = G(y, x) now follows from (1.75).