Chapter 6 Laplace equation - Home | Department of …siva/ma51515/chapter6.pdfChapter 6 Laplace equation In this chapter we consider Laplace equation in d-dimensions given by ux 1x1

Chapter 6

Laplace equation

In this chapter we consider Laplace equation in d -dimensions given by

ux1 x1+ ux2 x2

+ · · ·+ uxd xd= 0. (6.1)

We study Laplace equation in d = 2 throughout this chapter (excepting Section 6.2), andmost of the ideas can be generalized to general space dimensions d > 2. When d = 2,the independent variables x1, x2 are denoted by x, y, and write x = (x, y). Thus Laplaceequation in two independent variables is

ux x + uyy = 0. (6.2)

The non-homogeneous problem

ux x + uyy = f , (6.3)

where f is a function of the independent variables x, y only is called the Poisson equation.

6.1 Green’s identitiesGreen’s Identities play an important role in the analysis of Laplace equation. They arederived from divergence theorem. Let us recall the divergence theorem now.

Theorem 6.1 (Divergence theorem). Let Ω⊆R2 be a bounded piecewise smooth domain.Let Ψ :Ω→R2 be a function. Let Ψ = (ψ1,ψ2)where ψi ∈C 1(Ω)∩C (Ω) for i = 1,2. Then∫

Ω

∇.Ψ(x, y)d x d y =∫∂ Ω

Ψ.n dσ , (6.4)

where n is the unit outward normal to ∂ Ω.

Theorem 6.2 (Green’s identities). Let Ω ⊆ R2 be a bounded piecewise smooth domain,and let n denote the unit outward normal to ∂ Ω. For u, v ∈C 2(Ω)∩C 1(Ω), we have

(i) (Green’s Identity-I) ∫Ω

∆u(x, y)d x d y =∫∂ Ω

∂n u dσ , (6.5)

147

148 Chapter 6. Laplace equation

where ∂n u :=∇u.n is called the normal derivative of u; as it represents the directionalderivative of u in the dirction of the unit outward normal n.

(ii) (Green’s Identity-II)∫Ω

v∆u − u∆v(x, y)d x d y =∫∂ Ω

v ∂n u − u ∂nv

dσ (6.6)

(iii) (Green’s Identity-III)∫Ω

∇u.∇v d x d y =∫∂ Ω

v ∂n u dσ −∫Ω

v∆u d x d y. (6.7)

Proof. Applying (6.4) with Ψ = ∇u, Ψ = v∇u − u∇v, and Ψ = v∇u yield Green’sidentities I, II, and III respectively.

6.2 Fundamental solutions in Rd

Since Lapalce equation is invariant under translations, and rotations (see Exercise 6.4), welook for solutions to Laplace equation having such symmetric properties.

Let us fix ξ ∈ Rd , and look for solutions to ∆u = 0 having the form vξ (x) = ψ(r ),where

r = ∥x − ξ ∥=√√√√ d∑

i=1

(xi − ξi )2.

Substituting the formula for vξ in the equation, we get

∆vξ (x) =ψ′′(r )+ d − 1

rψ′(r ) = 0.

Solving the last equation, we get

ψ′(r ) =C r 1−d (6.8)

Integrating we get

ψ(r ) =

C ln r if d = 2,C

2−d r 2−d if d ≥ 3.(6.9)

Theorem 6.3. Let K(x ,ξ ) denote the function defined by K(x ,ξ ) :=ψ(∥x − ξ ∥).(i) Let ξ be a point of Ω. For u ∈C 2(Ω) the following identity holds

u(ξ ) =∫Ω

K(x ,ξ )∆u d x −∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ . (6.10)

(ii) The following equality holds in the sense of distributions on Rd :

∆K(x,ξ ) = δξ .

i.e., for every φ ∈C∞0 (Rd ) the following equality holds.

φ(ξ ) =∫Rd

K(x ,ξ )∆φ(x)d x . (6.11)

Sivaji IIT Bombay

6.2. Fundamental solutions in Rd 149

(iii) If u ∈C 2(Ω) and harmonic in Ω (i.e.,∆u = 0 in Ω), then for ξ ∈Ω we get

u(ξ ) =−∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ . (6.12)

Proof. Step 1: Proof of (i):

Let u ∈ C 2(Ω) and ξ be a point of Ω. Note that we cannot apply Green’s identity II(6.6) with v = vξ (x) = ψ(∥x − ξ ∥) since vξ is singular at x = ξ . Thus we cut out a ballB(ξ ,ρ) fromΩ along with its boundary, and then apply Green’s identity II. Note that thedomain Ωρ;=Ω \B[ξ ,ρ] is bounded by ∂ Ω and S(ξ ,ρ). Since∆vξ = 0 in Ωρ, we have∫

Ωρ

vξ∆u d x =∫∂ Ω

vξ ∂n u − u∂nvξ

dσ +∫

S(ξ ,ρ)


dσ . (6.13)

Let us now compute the second term on RHS of the equation (6.13).∫S(ξ ,ρ)


dσ =∫

S(ξ ,ρ)vξ ∂n u dσ −∫

S(ξ ,ρ)u∂nvξ dσ (6.14)

Note that on the circle S(ξ ,ρ), we have vξ (x) = ψ(∥x − ξ ∥) = ψ(ρ). Using thisinformation, and divergence theorem, we get∫

S(ξ ,ρ)vξ ∂n u dσ =ψ(ρ)

∫S(ξ ,ρ)

∂n u dσ =−ψ(ρ)∫

B(ξ ,ρ)∆u d x . (6.15)

Note that the outward unit normal n on the circle S(ξ ,ρ) points towards its center ξ .Also ∂nvξ =−ψ′(ρ) holds at points on the circle S(ξ ,ρ). Thus we get

∫S(ξ ,ρ)

u∂nvξ dσ =−Cρ1−d∫

S(ξ ,ρ)u dσ . (6.16)

Since both u and ∆u are continuous at ξ , we have the following convergences asρ→ 0:

ψ(ρ)∫

B(ξ ,ρ)∆u d x→ 0, ρ1−d

∫S(ξ ,ρ)

u dσ→ωd u(ξ ), (6.17)

whereωd denotes the surface area of the unit sphere in Rd .Now passing to the limit as ρ→ 0 in the equation (6.13), we get∫

Ω

vξ∆u d x =∫∂ Ω


dσ +Cωd u(ξ ) (6.18)

in view of (6.15), (6.16), (6.17).Choosing C = 1

ωdin the formulae (6.9), the equation (6.18) yields (6.10).

Step 2: Proof of (ii): In the equation (6.10) if we take u = φ ∈ C∞0 (Ω), then we get(6.11).

Step 3: Proof of (iii): Equation (6.12) is an immediate consequence of (6.10).

October 4, 2015 Sivaji


6.3 Boundary value problems associated to Laplace equationWe can study Laplace equation or Poisson equation on any open domain Ω⊆R2. Theseproblems will always have an infinite number of solutions. Thus it is interesting to knowif there are solutions to these equations if some constraints are placed on the unknownfunction. WhenΩ is a bounded domain with piecewise smooth boundary, there are atleastthree different kinds of restrictions are placed on the unknown function and its deriva-tives along the boundary ∂ Ω of the domain Ω. Such restrictions are called boundaryconditions.

Let us now describe three boundary value problems (BVP) for Poisson equation. Theydiffer in the nature of boundary conditions. We also define the notions of solutions tothese boundary value problems.

1. For given functions f , g , Dirichlet Problem consists of solving the boundary valueproblem

∆u = f (x) in Ω, (6.19a)u = g on ∂ Ω. (6.19b)

Definition 6.4 (Solution to Dirichlet BVP). Let f ∈ C (Ω), and g ∈ C (∂ Ω). Afunction φ ∈C 2(Ω)∩C (Ω) is said to be a solution to Dirichlet BVP if

(i) the function φ is a solution to Poisson equation (6.19a), i.e., for each x ∈Ω,

∆φ(x) = f (x)

holds, and(ii) for each x ∈ ∂ Ω, the equality φ(x) = g (x) holds.

2. For given functions f , g , Neumann Problem consists of solving the boundaryvalue problem

∆u = f (x) in Ω, (6.20a)∂n u = g on ∂ Ω. (6.20b)

Definition 6.5 (Solution to Neumann BVP). Let f ∈ C (Ω), and g ∈ C (∂ Ω). Afunction φ ∈C 2(Ω)∩C 1(Ω) is said to be a solution to Neumann BVP if


∆φ(x) = f (x)

holds, and(ii) for each x ∈ ∂ Ω, the equality ∂nφ(x) = g (x) holds.

3. For given functions f , g , and α ∈ R, Third Boundary Value Problem a.k.a.Robin Problem consists of solving the boundary value problem

∆u = f (x) in Ω, (6.21a)u +α∂n u = g on ∂ Ω. (6.21b)

Definition 6.6 (Solution to Robin BVP). Let f ∈ C (Ω), and g ∈ C (∂ Ω). A func-tion φ ∈C 2(Ω)∩C 1(Ω) is said to be a solution to Robin BVP if

Sivaji IIT Bombay

6.3. Boundary value problems associated to Laplace equation 151


∆φ(x) = f (x)

holds, and(ii) for each x ∈ ∂ Ω, the equality φ(x)+α∂nφ(x) = g (x) holds.

Remark 6.7.

(i) If Dirichlet BVP is posed on a bounded domain Ω, then the corresponding Dirich-let problem is called an interior Dirichlet problem. If Ω is the complement of abounded domain, then corresponding Dirichlet problem is called exterior Dirich-let problem.

(ii) There are other kinds of boundary value problems possible. For example, on a partof the boundary ∂ Ω one of the three boundary value problems described aboveis imposed and on the remaining part another one of the above three. We do notconsider such boundary value problems here.

(iii) Cauchy-Kowalewski theorem guarantees that a solution of an analytic Cauchy prob-lem for an elliptic equation exists and is unique (locally), but is not always well-posed.

The following result says that for Neumann BVP to admit a solution, the data f , gmust be compatible with each other.

Lemma 6.8 (Compatibility of data for Neumann problem). Let f ∈ C (Ω). If u ∈C 2(Ω) is a solution to Neumann BVP on Ω, then∫

Ω

f (x)d x =∫∂ Ω

g (y)dσ(y). (6.22)

Proof. Integrating both sides of the equation∆u = f on Ω yields∫Ω

f (x)d x =∫Ω

∆u(x)d x . (6.23)

Applying Green’s identity-I (6.5), the integral on the right hand side of the equation (6.23)becomes ∫

Ω

∆u(x)d x =∫∂ Ω

∂n u(y)dσ(y). (6.24)

This completes the proof.

In particular, when u is a solution to Neumann BVP with f = 0, we get∫∂ Ω

g (y)dσ(y) = 0.

Theorem 6.9 (Uniqueness theorem). Let Ω be a smooth domain. Then

(a) The Dirichlet problem has at most one solution.



(b) If u solves the Neumann problem, then any other solution is of the form v = u + c forsome real number c.

(c) If α≥ 0, then the Robin problem has at most one solution.

Proof. Proof of (c): Note that (a) is a special case of (c), hence it is enough to prove (c).Let u1, u2 be solutions of the Robin problem. Define w := u1− u2. Observe that w is aharmonic function(why?), and w satisfies the boundary condtion

w +α∂nw = 0. (6.25)

Using the Green’s identity-III (6.7) with u = v = w, we get∫Ω

|∇w|2 d x d y =−α∫∂ Ω

(∂nw)2 d s . (6.26)

Since the left hand side of (6.26) is non-negative while the right hand side is non-positive,we conclude that both sides must be zero. This implies that∇w = 0 inΩ and also ∂nw = 0on ∂ Ω. Since∇w = 0 inΩ, w must be a constant function. Since ∂nw = 0, in view of theequality (6.25), we conclude that w = 0 on the boundary ∂ Ω. Since w ∈C (Ω), it followsthat w ≡ 0 in Ω. This finishes the proof of (c).

Proof of (b): Let u1, u2 be solutions of the Neumann problem. Define w := u1− u2.Observe that w is a harmonic function(why?), and w satisfies the boundary condtion

∂nw = 0. (6.27)

Using the Green’s identity-III (6.7) with u = v = w, we get∫Ω

|∇w|2 d x d y = 0. (6.28)

This implies that w is a constant function. Thus u1 = u2 + c for some constant c ∈ R.

Remark 6.10 (On the formula (6.12)). (i) If Dirichlet problem (6.19) has a solution,then by Theorem 6.9 Dirichlet problem has exactly one solution. Let the solutionto Dirichlet problem be denoted by u. The formula (6.12) gives a representation ofthe solution in terms of the boundary values of u and its normal derivative ∂n u on∂ Ω.

(ii) Note however that in Dirichlet problem, only the values of u are prescribed on ∂ Ω,and ∂n u is an unknown function on ∂ Ω, and thus the formula (6.12) is not usefulfor computing the solution.

(iii) Note that boundary values of u already determine a solution to Dirichlet problem,and thus the quantity ∂n u is already determined.

(iv) A related question concerning Laplace equation is whether Cauchy problem is well-posed for Laplace equation. The answer to this question is negative, and is madeprecise in Theorem 6.31

Sivaji IIT Bombay

6.4. Poisson’s formula and its applications 153

6.4 Poisson’s formula and its applicationsIn this section, we consider Dirichlet boundary value problem for Laplace equation on thedomain Ω= B(0, 1) = (x1, x2) : x2

1 + x22 < 1 in R2. For a given function f ∈C (S(0, 1)),

we consider the boundary value problem

∆u = 0 in B(0, 1), (6.29a)u = f on S(0, 1). (6.29b)

We are going to prove the following result.

Theorem 6.11 (Poisson’s formula). The solution to the BVP (6.29) is given by the Poisson’sformula

u(x) =1−∥x∥2

2π

∫S(0,1)

f (y)∥x − y∥2 d y. (6.30)

Defining v(r,θ) := u(r cosθ, r sinθ), and F (θ) := f (cosθ, sinθ), Poisson’s formula inpolar coordinates is given by

v(r,θ) =1− r 2

2π

∫ 2π0

F (τ)

1− 2r cos (τ−θ)+ r 2

dτ (6.31)

under the condition that F ∈C 1[0,2π] and is a 2π-periodic function.

Proof. Due to the geometry of disk, it is convenient to consider Laplace equation ex-pressed in polar coordinates (see Exercise 6.1). Thus we consider the Dirichlet problemgiven by

∆(r,θ)v ≡ vr r +1r

vr +1r 2

vθθ = 0 for 0< r < 1, 0≤ θ≤ 2π, (6.32a)

v(1,θ) = F (θ) for 0≤ θ≤ 2π, (6.32b)

The proof of theorem is divided into three main steps. In Step 1, we derive a formal so-lution to Dirichlet BVP. In Step 2, we justify that the formal solution derived in the firststep is indded a solution. In Step 3, Poisson’s formula will be obtained.

Step 1: Derivation of a formal solutionWe obtain a solution of the equation (6.32a) using the method of separation of vari-

ables, in which solution is assumed to be of the form

v(r,θ) = h(r )g (θ), (6.33)

where g is assumed to be periodic of period 2π i.e., g (0) = g (2π) and g ′(0) = g ′(2π).Substituting the expression for v into (6.32a), and then dividing the resultant equation

with h(r )g (θ) we get

h ′′(r )h(r )

+1r

h ′(r )h(r )

+1r 2

g ′′(θ)g (θ)

= 0 (6.34)

Re-arranging terms in (6.34) yields

r 2 h ′′(r )h(r )

+ rh ′(r )h(r )

=− g ′′(θ)g (θ)

= λ (6.35)



Note that the LHS of the equation (6.35) is a function of r only, while the RHS is afunction of θ only. Thus both of them must be equal to a constant, and let us denote itby λ. Thus we have

r 2 h ′′(r )h(r )

+ rh ′(r )h(r )

=− g ′′(θ)g (θ)

= λ (6.36)

Thus we get two ODEs from (6.36) which are given by

g ′′(θ)+λg (θ) = 0, (6.37a)r 2h ′′(r )+ r h ′(r ) = λh(r ) (6.37b)

Let us look for solutions of (6.37a) satisfying the periodic boundary conditions

g (0) = g (2π), g ′(0) = g ′(2π) (6.38)

We are interested in finding non-trivial solutions to the boundary value problem (6.32),existence of which depend on the value of λ. Any λ for which the BVP (6.32) admits anon-trivial solution is called an eigenvalue and the corresponding non-trivial solutions arecalled eigenfunctions. Thus we are interested in finding non-trivial solutions to ODEs(6.37a) with boundary conditions (6.38).

(i) (λ = 0) General solution of the ODE (6.37a) is given by g (θ) = aθ+ b . Applyingthe boundary conditions (6.38), we get a = 0, and b arbitrary. Thus λ = 0 is asimple eigenvalue, and g (θ) = 1 is an eigenfunction. We need to solve ODE (6.37b)with µ= 0, which is given by

r 2h ′′(r )+ r h ′(r ) = 0. (6.39)

The general solution of (6.39) is given by

h0(r ) =Alog r +B (6.40)

where A,B are real numbers. Since we are interested in a solution of Laplace equa-tion in the unit disk, we are looking for bounded solutions, and thus A= 0. Thuswe obtain the following solution of Laplace equation:

v0(r,θ) = h0(r )g0(θ) = 1 (6.41)

(ii) (λ > 0) When λ > 0, we may write λ = µ2 where µ > 0. The ODE (6.37a) thenbecomes

g ′′(θ)+µ2 g (θ) = 0,

whose general solution is given by g (θ) = a cos (µθ) + b sin (µθ). Applying theboundary conditions (6.38), we get

a = a cos (2µπ)+ b sin (2µπ)bµ=−aµ sin (2µπ)+ bµcos (2µπ) .

The above system of linear equations may be written as1− cos (2µπ) − sin (2µπ)

sin (2µπ) 1− cos (2µπ)

ab

=

00

. (6.42)

Sivaji IIT Bombay


Note that the system (6.42) possesses non-trivial solutions if and only if cos (2µπ) =1. This means thatµ ∈N. Thus we get a sequence of eigenvalues λn = n2 indexed byn ∈N, and a basis for the corresponding eigenspace is given by cos (nθ) , sin (nθ).For each n ∈N, we need to solve ODE (6.37b) with µ= n2, which is given by

r 2h ′′(r )+ r h ′(r )− n2h(r ) = 0. (6.43)

Note that this is an ODE of Cauchy-Euler type, which reduces to a constant coeffi-cient ODE by a change of variable s = log r , and hence can be easily solved, whichis left as an exercise to the reader. The solutions are: For each n ∈ N, the generalsolution of (6.43) is given by

hn(r ) =Ar n +B r−n (6.44)

where A,B are real numbers. Since we are interested in a solution of Laplace equa-tion in the unit disk, we are looking for solutions that remain bounded as r → 0, andthus B = 0. For each n ∈N, we obtain the following solution of Laplace equation:

vn(r,θ) = hn(r )gn(θ) = r n (an cos (nθ)+ bn sin (nθ)) (6.45)

(iii) (λ < 0) When λ < 0, we may write λ = −µ2 where µ > 0. The ODE (6.37a) thenbecomes

g ′′(θ)−µ2 g (θ) = 0,

whose general solution is given by g (θ) = aeµθ + b e−µθ. Applying the boundaryconditions (6.38), we get

a+ b = ae2µπ+ b e−2µπ

aµ− bµ= aµe2µπ− bµe−2µπ.

The above system of linear equations may be written as1− e2µπ 1− e−2µπ

1− e2µπ −1+ e−2µπ

ab

=

00

. (6.46)

Note that the determinant of the coefficient matrix in (6.46) is zero if and only ife2µπ + e−2µπ = 2. Note that this relation is of the form α+ 1

α = 2 for which α = 1is the only solutoin. Since µ > 0, the equation e2µπ + e−2µπ = 2 has no positivesolution for µ. Thus any λ with λ < 0 is not an eigenvalue.

Consider the function v defined by superposition of v0 and vn for n ∈N (which wereobtained above) as follows:

v(r,θ) =a0

2+∞∑

n=1

r n (an cos (nθ)+ bn sin (nθ)) , (6.47)

where an , bn are real numbers. We choose these constants so that v satisfies the boundarycondition v(1,θ) = F (θ). Thus we are led to the relation

v(1,θ) = F (θ) =a0

2+∞∑

n=1

(an cos (nθ)+ bn sin (nθ)) . (6.48)



Let the constants an , bn be chosen as the Fourier coefficients of F (θ), i.e.,

an =1π

∫ 2π0

F (θ)cos (nθ) dθ, bn =1π

∫ 2π0

F (θ) sin (nθ) dθ. (6.49)

This finishes the construction of a formal solution of the Dirichlet boundary valueproblem. The solution is given by (6.47), where the coefficients an , bn are given by (6.49).

Step 2: Formal solution is indeed a solution

We are going to show that v defined by (6.47) is indeed a solution of the Dirichletboundary value problem (6.32).

Substituting the values of an , bn from (6.49) into (6.47), we get

v(r,θ) =1

2π

∫ 2π0

F (τ)dτ+1π

∞∑n=1

r n∫ 2π

0F (τ)cos (n(τ−θ)) dτ

(6.50)

As in the proof of Theorem 4.21, from the definition of an , bn it follows that thereexists an M > 0 such that for all n ∈N we have

|an | ≤M , n|an | ≤M ; and |bn | ≤M , n|bn | ≤M ,

which follow from the formulae (6.49) and using integration by parts once in them.

As a consequence of the above estimates on an , bn , the Fourier series for F (θ) con-verges uniformly, and hence the formal series in (6.47) for all r ≤ R< 1 where 0< R< 1is an arbitrary but fixed number. It also follows that series in (6.47) can be differentiatedterm-by-term and the resultant series also converges uniformly for all r ≤ R < 1. Sinceeach term in the series is a solution to Laplace equation, so will be v(r,θ). This provesthat the formal solution is indeed a solution.

Step 3: Poisson’s formula

Since the series in (6.47) is uniformly convergent for r ≤ R < 1, the summation andintegral in (6.50) can be interchanged. Thus we get

v(r,θ) =1π

∫ 2π0

F (τ)

12+∞∑

n=1

r n cos (n(τ−θ))

dτ

=1π

∫ 2π0

F (τ)

12+∞∑

n=1

r n

e i n(τ−θ)+ e−i n(τ−θ)2

dτ

=1

2π

∫ 2π0

F (τ)1+

r e i (τ−θ)1− r e i (τ−θ) +

r e−i (τ−θ)1− r e−i (τ−θ)

dτ

Simplifying the last equation, we get the following Poisson’s summation formula forthe solution of Laplace equation for r < 1:

v(r,θ) =1− r 2

2π

∫ 2π0

F (τ)

1− 2r cos (τ−θ)+ r 2

dτ. (6.51)

Sivaji IIT Bombay


In cartesian coordinates, the formula (6.51) becomes

u(x) =1−∥x∥2

2π

∫S(0,1)

f (y)∥x − y∥2 d y. (6.52)

Checking that u(x) given by (6.52) satisfies the boundary condition (6.29b) is left asan exercise (see Exercise 6.13).

6.4.1 Harnack’s inequality

Theorem 6.12 (Harnack’s inequality). Let B(0, R) ⊂ Rd be the open ball with center at0 and having radius R. Let u : B(0, R) → [0,∞) be a harmonic function. Then for anyx ∈ B(0, R), the following inequalities hold:

Rd−2(R−∥x∥)(R+ ∥x∥)d−1

u(0)≤ u(x)≤ Rd−2(R+ ∥x∥)(R−∥x∥)d−1

u(0). (6.53)

Proof. Let us prove this for d = 2. Recall from Poisson’s formula (6.52)

u(x) =1−∥x∥2

2π

∫S(0,1)

u(y)∥x − y∥2 d y. (6.54)

As a consequence of triangle inequality, for y ∈ S(0, 1) the following inequalities hold:

1−∥x∥ ≤ ∥x − y∥ ≤ 1+ ∥x∥.Using these inequalities in the Poisson’s formula (6.54) yields

u(x)≤ 1+ ∥x∥1−∥x∥

12π

∫S(0,1)

u(y)d y =1+ ∥x∥1−∥x∥ u(0),

and

u(x)≥ 1−∥x∥1+ ∥x∥

12π

∫S(0,1)

u(y)d y =1−∥x∥1+ ∥x∥ u(0).

This completes the proof of Harnack’s inequalities for d = 2.

Theorem 6.13 (Liouville’s theorem). If u : Rd → R is a harmonic function that isbounded below, then u must be a constant function.

Proof. Let m ∈ R be such that u(x) ≥ m for all x ∈ Rd . Then the function v(x) :=u(x)− m is a non-negative harmonic function. Applying Harnack’s inequality to thefunction v yields for every R> 0, and for every x ∈ B(0, R)

Rd−2(R−∥x∥)(R+ ∥x∥)d−1

v(0)≤ v(x)≤ Rd−2(R+ ∥x∥)(R−∥x∥)d−1

v(0). (6.55)



Fix an x ∈Rd , and R> 0 such that R> ∥x∥. Passing to the limit in the inequalities (6.55),we get

v(0)≤ v(x)≤ v(0).

Thus we get v(x) = v(0). Since x is arbitrary, we conclude that v is a constant function.As a consequence, u is a constant function.

6.5 Dirichlet problem on a rectangleIn this section, Dirichlet problem for Laplace equation is considered on a rectangle. Asolution is obtained using the method of separation of variables.

ux x + uyy = 0 for 0< x <π, 0< y <A, (6.56a)

u(0, y) = u(π, y) = 0 for 0≤ y ≤A, (6.56b)u(x,A) = 0 for 0≤ x ≤π, (6.56c)

u(x, 0) = f (x) for 0≤ x ≤π. (6.56d)

We use the method of separation of variables to solve the boundary value problem (6.56).Substituting

u(x, y) =X (x)Y (y)

in the equation (6.56a), and re-arranging the terms, we get

X ′′(x)X (x)

=−Y ′′(y)Y (y)

(6.57)

Since the LHS and RHS of the equation (6.57) are functions of the variables x and yrespectively, each of them must be a constant function. Thus we have, for some constantλ,

X ′′(x)X (x)

=−Y ′′(y)Y (y)

=−λ (6.58)

From the equation (6.58), we get the following two ODEs

X ′′(x)+λX (x) = 0, (6.59a)Y ′′(y)−λY (y) = 0. (6.59b)

Since we are interested in finding a non-trivial solution, the boundary conditions (6.56b)-(6.56c) give rise to the following conditions on the functions X and Y :

X (0) = 0, X (π) = 0, (6.60a)Y (A) = 0. (6.60b)

The ODE (6.59a) has non-trivial solutions satisfying the boundary conditions (6.60a) ifand only if λ= n2 for some n ∈N. Thus λn = n2 are eigenvalues and the correspondingeigenfunctions are given by Xn(x) = sin nx.

Sivaji IIT Bombay

6.6. Weak maximum principle 159

For each n ∈N, the solution of ODE (6.59b) withλ= λn = n2 satisfying the condition(6.60b) (upto a constant multiple) is given by Yn(y) = sinh n(A− y).

Since for each n ∈ N, the function un(x, y) = Xn(x)Yn(y) is a solution (6.56a), wepropose a formal solution of (6.56a) by superposition of the sequence of solutions (un) asfollows:

u(x, y) =∞∑

n=1

bn sin nx sinh n(A− y). (6.61)

The coefficients bn in the formula (6.61) will be determined using the condition (6.56d).Thus we get

f (x) =∞∑

n=1

bn sin nx sinh nA. (6.62)

Choosing cn to be the Fourier sine coefficients of f , we get bn =cn

sinh nA . That is, if f isgiven by

f (x) =∞∑

n=1

cn sin nx, (6.63)

then cn are given by

cn =2π

∫ π0

f (s) sin ns d s . (6.64)

Thus the formal solution of (6.56) is given by

u(x, y)≈∞∑

n=1

2

π sinh nA

∫ π0

f (s) sin ns d s

sin nx sinh n(A− y). (6.65)

The formal solution (6.65) is indeed a solution to (6.56) under some conditions, andis proved in Theorem 6.21.

6.6 Weak maximum principleLet Ω ⊆ R2 be a bounded domain. Let u : Ω −→ R be a continuous function such thatu can be extended to the closure of Ω as a continuous function. Such class of functions isdenoted by C (Ω). When Ω is bounded every function in C (Ω) attains both its maximumand minimum values, somewhere in Ω. Weak Maximum principle says that if u happensto be a harmonic function, then the maximum and minimum values of u are definitelyattained on the boundary of the domain Ω whether or not they are attained in Ω.

Theorem 6.14 (Weak maximum principle). Let Ω ⊆ R2 be a bounded domain. Letu ∈ C 2(Ω) ∩ C (Ω) be a harmonic function in Ω. Then the maximum value of u in Ω isachieved on the boundary ∂ Ω.

Proof. Step 1: Recall from differential calculus of two variables that at a point of interiormaximum, ∂ 2 u

∂ x2 ≤ 0 and ∂ 2 u∂ y2 ≤ 0. As a consequence, ∆u ≤ 0 at an interior maximum

point. Thus if v is a function such that∆v > 0 in Ω, then the maximum value of v on Ω



cannot be attained in Ω. Hence v attains its maximum value only on the boundary ∂ Ω.The idea to prove Weak maximum principle is to find such a function v starting from thegiven harmonic function u.

Step 2: Define the function vε by

vε(x, y) := u(x, y)+ ε(x2+ y2). (6.66)

Then vε ∈C 2(Ω)∩C (Ω). Note that∆vε > 0 in Ω and thus vε attains its maximum onlyon the boundary ∂ Ω. Denoting

M :=max∂ Ω

u, L := max∂ Ω(x2+ y2), (6.67)

we havevε(x, y)≤M + εL, ∀(x, y) ∈Ω. (6.68)

Since u(x, y)≤ vε(x, y) for all (x, y) ∈Ω, we have

u(x, y)≤M + εL, ∀(x, y) ∈Ω. (6.69)

Note that last inequality holds for every ε > 0. Thus taking limits as ε→ 0, we get

u(x, y)≤M , ∀(x, y) ∈Ω. (6.70)

This finishes the proof.

Corollary 6.15 (Weak minimum principle). Let Ω ⊆ R2 be a bounded domain. Letu ∈ C 2(Ω) ∩ C (Ω) be a harmonic function in Ω. Then the minimum value of u in Ω isachieved on the boundary ∂ Ω.

Proof. Define by v(x, y) = −u(x, y) on Ω. Then v is a harmonic function. Now applythe weak maximum principle to the harmonic function v and conclude.

Remark 6.16. We have to understand clearly what the Weak maximum principle saysand what it does not talk about. The weak maximum principle says that on boundeddomains, any harmonic function takes its maximum value on the boundary surely. Theweak maximum principle is silent on whether the harmonic function will take or will nottake the maximumum value in the domain.

For example, consider Ω to be the unit disk centered at the origin. Then the constantfunction u ≡ 1 is definitely a harmonic function in Ω. It attains its maximum in Ω also,apart from attaining on ∂ Ω.

There is another maximum principle, called Strong maximum priniciple, which saysthat on bounded domains non-constant harmonic functions will never attain maxima inΩ and maxima are attained only on the boundary ∂ Ω.

Sivaji IIT Bombay


6.6.1 Consequences of weak maximum principle

We already discussed uniqueness of solutions to all the three boundary value problems forLaplace equation posed on bounded domains using Green’s identities. For Wave and Heatequations, we discussed uniqueness via Energy method and same analysis can be doen forLaplace equation as well.

We are now going to discuss uniqueness questions concerning the boundary value prob-lems using another tool, namely, the weak maximum principle for harmonic functionson a bounded domain.

Theorem 6.17 (Uniqueness of solutions to Dirichlet problem). Let Ω ⊆ R2 be abounded domain and consider the Dirichlet problem on Ω:

∆u = f on Ω,u = g on ∂ Ω,

where g is a continuous function on ∂ Ω. Then the Dirichlet problem has at most one solutionin the class C 2(Ω)∩C (Ω).

Proof. Assume that u and v belonging to the class C 2(Ω) ∩ C (Ω) solve the Dirichletproblem. Define w := u − v. Then w belongs to the class C 2(Ω)∩C (Ω) and solves thehomogeneous Dirichlet problem:

∆w = 0 on Ω,w = 0 on ∂ Ω.

By weak maximum principle w ≤ 0 and by weak minimum principle w ≥ 0. Thus w ≡ 0.This finishes the proof of the theorem.

Remark 6.18. (i) Note that uniqueness of solutions to Dirichlet problem was alreadyproved in Theorem 6.9. Recall that its proof required that ∂n u is defined on theboundary, and thus the uniqueness of solutions to Dirichlet problem could be provedonly for u ∈ C 1(Ω). Thanks to maximum principle, uniqueness result is valid forany harmonic function which belongs to the space C (Ω).

(ii) In the above theorem, it is essential that Ω is a bounded domain. For unboundeddomains the theorem may not hold good. For example, consider the followingDirichlet problem on the upper half plane:

∆u(x, y) = 0 for x ∈R, 0< y <∞,u(x, 0) = 1 for all x ∈R.

This problem has at least two solutions, namely u1(x, y) = xy, and, u2(x, y) = 0.

Theorem 6.19. Let Ω⊆R be a bounded domain. For i = 1,2, let ui ∈C 2(Ω)∩C (Ω) solvethe following Dirichlet problems on Ω:

∆ui = f on Ω,u = gi on ∂ Ω,



where gi is a continuous function on ∂ Ω for each i = 1,2. Then u1 and u2 satisfy

maxΩ|u1(x, y)− u2(x, y)| ≤max

∂ Ω|g1(x, y)− g2(x, y)|. (6.71)

Proof. Define w := u1− u2. Then w solves the following Dirichlet problem

∆w = 0 on Ω,w = g1− g2 on ∂ Ω.

Applying the weak maximum principle and the weak minimum principle, we get

min∂ Ω(g1− g2)≤ w(x, y)≤max

∂ Ω(g1− g2) ∀(x, y) ∈Ω, (6.72)

which finishes the proof of the theorem.

Theorem 6.20 (Harnack). Let Ω be a bounded domain in R2. Let (un) be a sequence offunctions in C 2(Ω)∩C (Ω) that are harmonic on Ω, such that un = fn on ∂ Ω. Let fn → fin C (∂ Ω) i.e., fn→ f uniformly on ∂ Ω. Then

(i) The sequence (un) converges uniformly on ∂ Ω.(ii) Let u denote the uniform limit of the sequence (un). Then u is harmonic on Ω, and

u ∈C (∂ Ω).(iii) u = f on ∂ Ω.

Proof. For each (m, n) ∈N×N, the function un − um is harmonic on Ω, and belongs toC (Ω). On the boundary ∂ Ω,

un − um = fn − fm

Since ( fn) converges uniformly on ∂ Ω, it is a uniformly Cauchy sequence. Thus for agiven ε > 0, there exists an N ∈ N such that for all (x, y) ∈ ∂ Ω, and for all m ≥ N andn ≥N , the following inequality holds.

| fm(x, y)− fn(x, y)|< ε. (6.73)

Applying Theorem 6.19, we get

maxΩ|um(x, y)− un(x, y)| ≤max

∂ Ω| fm(x, y)− fn(x, y)|. (6.74)

From the inequalities (6.73) and (6.74), we get for all m ≥ N and n ≥ N the followinginequality

maxΩ|um(x, y)− un(x, y)| ≤max

∂ Ω| fm(x, y)− fn(x, y)|< ε. (6.75)

Thus the sequence (un) is a Cauchy sequence in C (Ω), and hence converges uniformly, tou ∈ C (Ω). Since each member of the sequence un satisfies mean value property, u alsosatisfies mean value property being the uniform limit of the sequence (un). Since u is acontinuous function and satisfies mean value property, it follows from Theorem 6.28 thatu is harmonic in Ω. It is clear that u = f on ∂ Ω.

Sivaji IIT Bombay


Theorem 6.21. The formal solution (6.65) for the boundary value problem (6.56) is indeeda solution if the function f : [0,π] → R is continuous, and satisfies f (0) = f (π) = 0,∫π

0 ( f′)2(s)d s <∞.

Proof.Recall the formal solution of (6.56) given by

u(x, y) =∞∑

n=1

bn sin nxsinh n(A− y)

sinh nA, (6.76)

where

bn =2π

∫ π0

f (s) sin ns d s

In order to establish that the formal solution given by (6.76) is indeed a solution to theBVP (6.56), we need to show that

(i) The series in (6.76) is twice continuously differentiable w.r.t. both the variables xand y, thus making sure that ux x and uyy are meaningful. Then we need to showthat equation (6.56a) is satisfied.

(ii) We need to show that u(x, y) given by (6.76) is continuous upto the boundary of(0,π)× (0,A) and that the boundary conditions (6.56b) - (6.56d) are satisfied.

Step 1: Proof of (i)Note that

sinh n(A− y)sinh nA

=en(A−y)− e−n(A−y)

enA− e−nA(6.77)

= e−ny 1− e−2n(A−y)

1− e−2nA(6.78)

≤ e−ny

1− e−2A. (6.79)

In view of the following inequalitiesbn sin nxsinh n(A− y)

sinh nA

≤ 11− e−2A

e−ny ,

and the fact that the series∑∞

n=1 e−ny converges uniformly for y > y0, it follows thatthe series in (6.76) converges uniformly y > y0. Since y0 > 0 is arbitrary, it follows thatu(x, y) is a continuous function in y > 0. Thus boundary values at x = 0, x = π, y = Aare satisfied.

By a similar reasoning, it follows that the series in (6.76) can be differentiated term-by-term w.r.t. x and y twice and the resultant series define continuous functions.

Step 2: Proof of (ii) In view of the assumptions on f , the fourier sine series of fconverges to f uniformly. Denoting the k th partial sum of the series (6.76) by sk , we have

sk (x, y) =k∑

n=1

bn sin nxsinh n(A− y)

sinh nA. (6.80)

We wish to show that sk (x, 0)→ f uniformly on [0,π], by showing that the sequence (sk )is a Cauchy sequence in C [0,π]. Note that for q > p, the function sp (x, y)−sq (x, y) solves



the IBVP (6.56) with sp (x, 0)− sq (x, 0) =∑q

n=p bn sin nx. Since the fourier sine series off converges uniformly to f , the corresponding sequence of partial sums

∑qn=1 bk sin nx

is Cauchy in C [0,π]. The required conclusion follows on applying maximum principle.Theorem 6.19.

6.7 Mean value property and its consequences

Definition 6.22. LetΩ⊆R2 be an open domain. A continuous function u :Ω−→R is saidto possess the Mean Value Property-I on Ω if for every point P = (x, y) ∈ Ω and for everyr > 0 such that the open disk Br (P )$Ω, the function u satisfies the relation

u(x, y) =1πr 2

∫|ξ−x|2+|η−y|2≤r 2

u(ξ ,η)dξ dη. (6.81)

Definition 6.23. Let Ω ⊆ R2 be a domain. A continuous function u : Ω −→ R is said topossess the Mean Value Property-II on Ω if for every point P = (x, y) ∈ Ω and for everyr > 0 such that the open disk Br (P )$Ω, the function u satisfies the relation

u(x, y) =1

2πr

∮Cr

u(s)d s (6.82)

where Cr denotes the circle centered at P having radius r .

On expanding the line integral in the equation (6.82), we get the following form of theequation (6.82):

u(x, y) =1

2π

∫ 2π0

u(x + r cosθ, y + r sinθ)dθ. (6.83)

Lemma 6.24. Let u : Ω−→R be a continuous function. Then the following statements areequivalent.

(i) u has the mean value property-I on Ω.(ii) u has the mean value property-II on Ω.

Proof. We will show (ii) implies (i) and all the steps in this proof can be reversed therebyproving that (i) implies (ii). Let P = (x, y) ∈Ω be an arbitrary point. Further let R> 0 besuch that the open disk BR(P ) $ Ω. Then if u satisfies mean value property-II, then forevery 0< τ ≤ R we have

u(x, y) =1

2π

∫ 2π0

u(x +τ cosθ, y +τ sinθ)dθ. (6.84)

Multiply the equation (6.84) with τ and then integrate w.r.t. τ over the interval [0, r ].This yields∫ r

0τ u(x, y)dτ =

12π

∫ r0

∫ 2π0

u(x +τ cosθ, y +τ sinθ)dθτ dτ. (6.85)

Sivaji IIT Bombay

6.7. Mean value property and its consequences 165

The last equation simplifies to

r 2

2u(x, y) =

12π

∫ r0

∫ 2π0

u(x +τ cosθ, y +τ sinθ)τ dθ dτ

=1

2π

∫|ξ−x|2+|η−y|2≤r 2

u(ξ ,η)dξ dη.

Thus we get

u(x, y) =1πr 2

∫|ξ−x|2+|η−y|2≤r 2

u(ξ ,η)dξ dη, (6.86)

which proves that u has mean value property-I as well.

Remark 6.25. In view of Lemma 6.24, we say that a continuous function u has the meanvalue property on a domain Ω if either of the mean value properties holds on Ω and as aconsequence both the mean value properties hold on Ω. The mean value property-I mayalso be called the Solid mean value property (as the average is taken over the entire disk)and the mean value property-II may also be called Surface/Circle mean value property.This nomenclature is not standard.

Theorem 6.26 (Mean value principle). Let Ω ⊆ R2 be a domain and u be a harmonicfunction in Ω. Then u has the mean value property on Ω.

Proof. Let P0 = (x0, y0) ∈Ω and let R> 0 be such that B(P0, R)⊂ B[P0, R]⊂Ω.

We would like to prove that the function u has the surface mean value property,namely, for 0< r < R

u(x0, y0) =1

2π

∫ 2π0

u(x + r cosθ, y + r sinθ)dθ. (6.87)

Note that the left hand side of (6.87) is a real number, whereas the right hand side of (6.87)depends on r . Thus to prove (6.87), the strategy is to prove that the right hand side is aconstant function of the variable r . To this end, define a function V : (0, R)→R by

V (r ) =1

2π

∫ 2π0

u(x0+ r cosθ, y0+ r sinθ)dθ. (6.88)

Let us compute dVd r for 0 < r < R and prove that it is zero. As a consequence, we will

then haveV (r ) = lim

ρ→0V (ρ) = u(x0, y0), (6.89)

which completes the proof of the mean value principle. It remain to show that dVd r = 0

on (0, R).

dVd r(r ) =

12π

∫ 2π0

∂

∂ ru(x0+ r cosθ, y0+ r sinθ)dθ=

∫S(0,r )

∂νu dσ

=∫

B(0,r )∆u d x = 0. (6.90)



In proving the last equality, we have used the equation (6.24) and the fact that u is a har-monic function on Ω.

Theorem 6.27. If u is continuous and has the mean value property on a domain Ω, then ithas continuous derivatives of all orders and all of them have the mean value property on Ω.

Proof. Step 1: It is sufficient to show that the first order partial derivatives of u exist andare continuous. Once it is known that first order partial derivatives of u exist, it followsby computing partial derivatives from the equation

u(x, y) =1

2π

∫ 2π0

u(x + r cosθ, y + r sinθ)dθ (6.91)

which holds as u satisfies mean value property on Ω to obtain

ux (x, y) =1

2π

∫ 2π0

ux (x + r cosθ, y + r sinθ)dθ,

uy (x, y) =1

2π

∫ 2π0

uy (x + r cosθ, y + r sinθ)dθ.

By repeating the arguments by replacing u with its partial derivatives, we conclude thatu has partial derivatives of all orders and all of them satisfy the mean value relation.

Step 2: Let us now prove that first order partial derivatives of u exist and are continuous.Since u has mean value property -I (6.81), we have

u(x, y) =1πr 2

∫|ξ−x|2+|η−y|2≤r 2

u(ξ ,η)dξ dη. (6.92)

Writing the double integral on the right hand side as iterated integrals, the equation (6.92)takes the form

u(x, y) =1πr 2

∫|ξ−x|2+|η−y|2≤r 2

u(ξ ,η)dξ dη=1πr 2

∫ y+r

y−rdη∫ x+pr 2−(η−y)2

x−pr 2−(η−y)2u(ξ ,η)dξ

The last equation tells us that ux (x, y) exists, in view of fundamental theorem of integralcalculus. In fact ux (x, y) is given by

ux (x, y) =1πr 2

∫ y+r

y−r

hu(x +Æ

r 2− (η− y)2,η)− u(x −Ær 2− (η− y)2,η)i

dη. (6.93)

Since the right hand side of the equation (6.93) is a continuous function of (x, y), weconclude that ux is a continuous function on Ω. Similarly one can deduce the existenceof uy at each point and continuity of uy on Ω.

Theorem 6.28. If u : Ω−→R is continuous and u has the mean value property on Ω, thenu is harmonic in Ω.

Sivaji IIT Bombay


Proof. If u has the mean value property, then by Theorem 6.27 we know that all partialderivatives of u exist and all of them are continuous onΩ. We want to show that∆u = 0.On the contrary, suppose that there exists a point P = (x0, y0) in Ω such that∆u(P ) = 0;and without loss of generality assume that∆u(P )> 0. By continuity of the second orderpartial derivatives, there exists a disk of radius ε > 0 centered at P (denote it by B(P,ε) onwhich∆u > 0.

We will be using the following identity in the computation that follows.

dd r

u(x0+ r cosθ, y0+ r sinθ) =∇x u(x0+ r cosθ, y0+ r sinθ).(cosθ, sinθ)

Also note that the outward unit normal to the circle S(P, r ) is given by 1r (x−x0,−y0).

Thus for 0< r < ε, we have

0<∫

B(P,r )∆u(x, y)d x d y =

∫S(P,r )

∂νu dσ

=∫ 2π

0

∂

∂ ru(x0+ r cosθ, y0+ r sinθ) r dθ

= r∂

∂ r

∫ 2π0

u(x0+ r cosθ, y0+ r sinθ)dθ

= r∂

∂ r

2πu(x0, y0)= 0.

This contradiction means that our assumption that∆u at some point is non-zero is wrong.Thus u is a harmonic function.

Theorem 6.29 (Derivative estimate). Let u : Ω → R be a harmonic function, whereΩ ⊆ R2 is an open set. Let k be a non-negative integer. There exists a Ck > 0 such that foreach multi-index α with |α|= k, and for each ball B(x0, r )⊆Ω,

|Dαu(x0)| ≤ (16k)k

πr k+2

∫B(x 0,r )

|u(x, y)|d x d y. (6.94)

Proof. Since the inequality (6.94) involves a non-negative integer k, we prove it by in-duction. For k = 0, the inequality (6.94) follows from mean value property of harmonicfunctions. We will prove the case k = 1 as the arguments used to prove that the requiredinequalities hold for k = l + 1, assuming that they hold for k = l are similar, and whenk = 1 the proof can be presented cleanly without extra clutter of notations.

Step 1: Case of k = 1: We prove the estimate for ux , as the proof is similar for uy . ByTheorem 6.27, the partial derivative ux possesses mean value property. As a consequence,we have

|ux (x0)|= 1π 4

r 2

∫B(x 0, r

2 )ux (x, y)d x d y

=

1π 4r 2

∫S(x 0, r

2 )u.nx dσ

≤ 4

rmax

S(x 0, r2 )|u|. (6.95)



Let us estimate the quantity maxS(x 0, r2 )|u|. Let x ∈ S(x0, r

2 ). Then

B

x ,r2

⊂ B(x0, r )⊂Ω.

Using (solid) mean value property, we get the estimate

|u(x)| ≤ 4πr 2

∫B(x 0,r )

|u(x, y)|d x d y. (6.96)

Thus we get

maxS(x 0, r

2 )|u| ≤ 4

πr 2

∫B(x 0,r )

|u(x, y)|d x d y. (6.97)

The desired estimate (6.94) for k = 1 follows from (6.95) and (6.97).

Step 2: Induction step: Let us assume that the result is true for k = l . Let us provethat the result is true for k = l + 1. Let α be a multi-index with |α| = l + 1. Consider aball B(x0, r )⊂Ω. Note that Dαu will be a first order partial derivative of Dβu for somemulti-index β such that |β| = l . Proceeding as in Step 1 after writing down the (solid)mean value property for the function Dαu on the ball B(x0, r

l+1 ), we get

|Dαu(x0)| ≤ 2(l + 1)r

maxS(x 0, r

l+1 )|Dβu|. (6.98)

Note that for x ∈ S(x0, rl+1 ), we have

B

x ,l

l + 1r⊂ B(x0, r )⊂Ω.

Applying induction hypothesis, we have following estimate for Dβu(x) for x ∈ S(x0, rl+1 ).

|Dβu(x)| ≤ (16l )l

πr l+2

∫B(x 0,r )

|u(x, y)|d x d y. (6.99)

The inequalities (6.98) and (6.99) together imply the estimate (6.94) for k = l +1.

Recall that harmonic functions are infinitely differentiable (see Theorem 6.27). In factharmonic functions are nicer than being infinitely differentiable, they have convergentTaylor series expansions at every point ofΩ, and such functions are known as real-analyticfunctions.

Theorem 6.30 (Harmonic functions are analytic). Let u : Ω → R be a harmonicfunction, where Ω⊆R2 is an open set. Then u is analytic in Ω.

Proof. Let x0 ∈ Ω be fixed. Since u is an infinitely differentiable function, we may writedown the Taylor series for u around the point x0 which is given by

u(x)≈ ∑α∈(N∪0)2

Dαu(x0)α!

(x − x0)α (6.100)

Sivaji IIT Bombay


In order to show that u is analytic, we need to show that the series in (6.100) converges tou. This is shown by expressing u(x) = TN (x)+RN (x), where TN is the Taylor polynomialof degree N associated with u and RN is the remainder. We then show that RN → 0 asN →∞. Note that TN (x is given by

TN (x) =∑

α∈(N∪0)2, |α|≤N

Dαu(x0)α!

(x − x0)α. (6.101)

By mean value theorem, there exists z on the line joining x and x0 such that RN (x) hasthe following expression:

RN (x) = u(x)−TN (x) =∑

α∈(N∪0)2, |α|=N+1

Dαu(z )α!

(x − x0)α. (6.102)

Let r = 14 d (x0,∂ Ω), and M = 1

πr 2

∫B(x 0,2r ) |u(x, y)|d x d y. For each z ∈ B(x0, r ), we

haveB(z , r )⊂ B(x0, 2r )⊂Ω.

Applying Theorem 6.29 with B(x , r ), and |α|=N + 1, we get

|Dαu(z )| ≤ (16(N + 1))N+1

πr N+1+2

∫B(z ,r )|u(x, y)|d x d y

≤ (16(N + 1))N+1

πr N+1+2

∫B(x 0,2r )

|u(x, y)|d x d y

≤M(16(N + 1))N+1

r N+1

From the last string of inequalities, we conclude that

supz∈B(x 0,r )

|Dαu(z )| ≤M

16r

N+1(N + 1)N+1 (6.103)

We now state two general inequalities. They are

(i) From Stirling’s formula [3]which is concerned with growth rate of factorials, thereexists a C > 0 such that for all n ∈N the following inequality holds:

nn ≤C en n! (6.104)

(ii) This is an estimate on n!. By multinomial theorem, we have

d k = (1+ 1+ · · ·+ 1)k =∑|α|=k

|α|!α!

.

Hence we get

|α|!≤ d |α|α! (6.105)

In view of (6.104) and (6.105), from the estimate (6.103) we get

supz∈B(x 0,r )

|Dαu(z )| ≤C M

32er

N+1α! (6.106)



Let us now estimate the remainder term RN given by (6.102).

|RN (x)| ≤∑

α∈(N∪0)2, |α|=N+1

|Dαu(z )|α!

∥x − x0∥|α|

≤ ∑α∈(N∪0)2, |α|=N+1

C M

32er

N+1 ∥x − x0∥N+1 (6.107)

Let s > 0 (to be chosen later) be such that ∥x − x0∥ ≤ s . Then the estimate (6.107) yields

|RN (x)| ≤∑

α∈(N∪0)2, |α|=N+1

C M

32e sr

N+1(6.108)

Choose an s such that 32e sr =

14 . In other words, choose s = r

128e . With this choice of s ,the estimate (6.108) takes the form

|RN (x)| ≤∑

α∈(N∪0)2, |α|=N+1

C M

14

N+1 ≤C M 2N+2

14

N+1 ≤C M1

2N. (6.109)

Thus the remainder term RN → 0 as N → ∞. Thus u is an analytic function on Ω.

As a consequence of the last result, it follows that Cauchy problem for Laplace equa-tion need not have solutions. This is the content of the next result.

Theorem 6.31. Let Γ ⊂ R2 denote a segment of x-axis. Let g be a function defined on Γ .Then the Cauchy problem for∆u = 0 where Cauchy data is given by

u(x, 0) = 0,∂ u∂ y(x, 0) = g (x) for (x, 0) ∈ Γ . (6.110)

has no solutions unless x 7→ g (x) is a real-analytic function.

Proof. Let P (ξ , 0) ∈ Γ . Let B be an open disk with center at P such that B ∩ x-axis iscontained in Γ . Let B+ denote the part of B which lies above x-axis, i.e.,

B+ = (x, y) ∈ B : y ≥ 0.Let u ∈ C 2(B+) be a solution of Laplace equation corresponding to the Cauchy data(6.110). Let u be extended to whole of B by reflection by setting for (x, y) ∈ B ,

u(x, y) :=−u(x,−y) for y < 0.

Verifying that the extended function u : B→R has the following properties

(i) u ∈C 2(B), and(ii) ∆u = 0 on B

is left as an exercise to the reader. By Theorem (6.30), the harmonic function u is analyticin B . In particular, u is analytic at P , and so is its derivative ∂ u

∂ y . Thus g , which is the

Sivaji IIT Bombay


restriction of an analytic function to x-axis is itself real-analytic. This completes the proofof the theorem.

Theorem 6.32 (Harnack’s inequality). Let O be a connected open subset of Ω such thatO ⊆ O ⊂Ω. There exists a constant C > 0 such that

supO

u ≤C infO u (6.111)

holds for all non-negative harmonic functions u :Ω→ [0,∞).Proof. We will first prove the result on each of the balls contained in the domain Ω, andthen deduce the result for O using connectedness of O and compactness of O .

Step 1: Let r < d (O ,∂ Ω). Let x ∈ O and y ∈ O be such that ∥x − y∥< r . By (solid)mean value property, we have

u(x) =1

4πr 2

∫B(x ,2r )

u(z )d z . (6.112)

Since B(y, r )⊆ B(x , 2r ), from the equation (6.112) we get

u(x) =1

4πr 2

∫B(x ,2r )

u(z )d z ≥ 14πr 2

∫B(y ,r )

u(z )d z =14

u(y). (6.113)

Thus we have 14 u(y) ≤ u(x). Interchanging the roles of x and y we get 1

4 u(x) ≤ u(y).Combining these two estimates, we get the following estimates, valid for all x ∈ O andy ∈ O satisfying ∥x − y∥< r :

14

u(y)≤ u(x)≤ 4u(y). (6.114)

Step 2: Since O is connected and its closure is compact, there exists finitely manyballs B1,B2, · · · ,BN having radius r , and Bi ∩ Bi+1 = ; for i = 1,2, · · · ,N − 1. Let x ∈ Oand y ∈ O . Then there exists an i and a j such that x ∈ B j and y ∈ Bk . Without loss ofgenerality, we may assume that j ≤ k. For i = 1,2, · · · ,N − 1, let z i ∈ Bi ∩Bi+1 be fixed.

u(x)≤ 4u(z j )≤ 42u(z j+1)≤ · · · ≤ 4k− j u(z k−1)≤ 4k+1− j u(y)≤ 4N u(y) (6.115)

Using the first inequality in (6.114), and following a similar argument, we get

u(x)≥ 14N

u(y) (6.116)

Combining the inequalities (6.115) and (6.116), we get

14N

u(y)≤ u(x)≤ 4N u(y), (6.117)

which is valid for every x ∈ O and y ∈ O . The required conclusion (6.111) now follows.



6.7.1 Strong maximum principle

Let Ω ⊆ R2 be a bounded domain and u : Ω −→ R be a harmonic function. The Weakmaximum principle asserted that maximum value of u on the closure of Ω is attained onthe boundary of Ω, and was silent about attaining the said maximum value in Ω. Strongmaximum principle that we are going to prove says that if such a maximum is also attainedinside Ω, then necessarily the harmonic function u is a constant function.

Theorem 6.33 (Strong maximum principle). LetΩ⊆R2 be a domain (domain need notbe bounded) and u : Ω −→ R be a harmonic function. If u attains its maximum in Ω, thenu is constant.

Proof. Step 1: Let u assume its maximum at P0 ∈ Ω and let this maximum value bedenoted by M . We want to prove that u is the constant function that takes the value Meverywhere in Ω. Let P be an arbitrary point in Ω. We wil show that u(P ) =M .

Step 2: Join P to P0 by a smooth curve γ 12. Since γ is a compact set, it maintains a positivedistance from ∂ Ω (in case ∂ Ω is non-empty), let us denote this distance by dγ . Take a disk

of radiusdγ2 with center at P0 denoted by B dγ

2

(P0). Since u is a harmonic function in Ω, it

has the mean value property in Ω. Applying the mean value property13 on this disk, weconclude that u equals M on the entire disk.

Step 3:(Continuation argument) We now take a point P1 on γ which lies in the disk

B dγ2

(P0) which is at least at a distancedγ4 from P0. Note that u(P1) = M . Now repeat the

arguments of Step 2 with the disk B dγ2

(P1) and then find a P2 in a similar way. Continuing

this process till we get a k ∈ N with the property that P ∈ B dγ2

(Pk )14. We will then have

u(P ) =M . This finishes the proof of the theorem.

Remark 6.34. This is a continuation of Remark 6.16. It is important to note that Strongmaximum principle says that if a harmonic function attains its maximum in a domain Ω(bounded domain or otherwise), then it is necessarily a constant function. In particularStrong maximum principle does not talk about where the maximum will be attained;sometimes a harmonic function may not have a maximum value also. This happens whenthe domainΩ is not bounded, because in that case though u is continuous onΩwe cannotassert the existence of a maximum value, since Ω is not compact. See the next example inthis context.

Example 6.35. Let Ω be the domain exterior to the unit disk. The function u(x, y) =log(x2+y2) is a harmonic function onΩ. It has neither a maximum value nor a minimumvalue in Ω.

12This is possible since Ω is open and connected and hence it is path-connected. Prove this.13Mean Value Property-I or Mean Value Property-II ?14Why is it possible to find such a k?

Sivaji IIT Bombay

6.8. Dirichlet principle 173

6.8 Dirichlet principleDirichlet principle characterises the solution of Dirichlet BVP in terms of minimizer of afunctional. An analogous result in the context of systems of linear equations says that thesolutoin of the linear system Ax = b (where A is a symmetric positive definite matrix) isthe minimizer of the functional 1

2 xT Ax − xT b.

Theorem 6.36 (Dirichlet principle). LetΩ⊂R2 be a bounded domain with smooth bound-ary. Let f ∈C (Ω), g ∈C (∂ Ω), and u ∈C 2(Ω). The following statements are equivalent.

(i) The function u is a solution to Dirichlet BVP

−∆u = f (x) in Ω, (6.118a)u = g on ∂ Ω. (6.118b)

(ii) u is a minimizer of the functional J : S→R where S := v ∈C 2(Ω) : v = g on∂ Ω,and

J (v) :=12

∫Ω

∥∇v∥2 d x −∫Ω

f (x)v(x)d x (6.119)

Proof. Proof of (i)=⇒(ii): Let u be a solution to Dirichlet BVP (6.118). Let v ∈ S.Multiplying the equation (6.118a) with u − v and then integrating on Ω yields

−∫Ω

∆u(x) (u(x)− v(x))d x =∫Ω

f (x) (u(x)− v(x))d x

Integrating by parts on the LHS of the last equation gives∫Ω

∇u(x). (∇u −∇v) (x)d x =∫Ω

f (x) (u(x)− v(x))d x (6.120)

Note that the boundary terms resulting from integration by parts do no appear as u = von ∂ Ω. Re-arranging terms in the equation (6.120), we get

∫Ω

∥∇u(x)∥2 d x −∫Ω

f (x) u(x)d x =∫Ω

∇u(x).∇v(x)d x −∫Ω

f (x)v(x)d x

(6.121)

Note that ∫Ω

∇u.∇v d x ≤∫Ω

∥∇u∥∥∇v∥d x

≤∫Ω

∥∇u∥2 d x 1

2∫Ω

∥∇v∥2 d x 1

2

≤ 12

∫Ω

∥∇u∥2 d x +∫Ω

∥∇v∥2 d x

(6.122)

Using the inequality (6.122) in the equation (6.121), we get



∫Ω

∥∇u(x)∥2 d x −∫Ω

f (x) u(x)d x ≤ 12

∫Ω

∥∇u∥2 d x +∫Ω

∥∇v∥2 d x−∫Ω

f (x)v(x)d x

(6.123)

Re-arranging the terms in (6.123), we get

12

∫Ω

∥∇u(x)∥2 d x −∫Ω

f (x) u(x)d x ≤ 12

∫Ω

∥∇v∥2 d x −∫Ω

f (x)v(x)d x

(6.124)

Note that the equation (6.123) is nothing but

J (u) ≤ J (v)

Thus u is the minimizer of the functional J over the set S.

Proof of (ii)=⇒(i): Let v ∈C∞0 (Ω), and t ∈R. We have

J (u + t v) = J (u)+ t∫Ω

(∇u.∇v − f v) d x +t 2

2

∫Ω

∥∇v∥2 d x

Rewriting the last equality, we get

J (u + t v)− J (u) = t∫Ω

(∇u.∇v − f v) d x +t 2

2

∫Ω

∥∇v∥2 d x . (6.125)

Since the functional achieves its minimum at u, the real valued function h :R→R definedby

h(t ) = J (u + t v)− J (u)

achieves its minimum at t = 0. Thus h ′(0) = 0, which yields∫Ω

∇u(x).∇v(x)d x =∫Ω

f (x)v(x)d x (6.126)

Integrating by parts on the LHS of the equation (6.126) gives

−∫Ω

∆u(x)v(x)d x =∫Ω

f (x)v(x)d x (6.127)

Since the equality (6.127) holds for every v ∈C∞0 (Ω), we conclude that for each x ∈Ω,

−∆u(x) = f (x).

Note that u satisfies the boundary condition (6.118b) as u ∈ S. This proves that u is asolution to the BVP (6.118).

Sivaji IIT Bombay

Exercises 175

ExercisesGeneral

6.1. [16] Show that the laplace operator in two space dimensions takes the form

∆u =1r∂

∂ r

r∂ u∂ r

+

1r 2

∂ 2u∂ θ2

in the polar coordinates x1 = r cosθ, x2 = r sinθ.6.2. [16] Show that the laplace operator in three space dimensions takes the form

∆u =1r∂

∂ r

r∂ u∂ r

+

1r 2

∂ 2u∂ θ2

+∂ 2u∂ z2

in the cylindrical coordinates x1 = r cosθ, x2 = r sinθ, x3 = z.6.3. [16] Show that the laplace operator in three space dimensions takes the form

∆u =∂ 2u∂ r 2

+2r∂ u∂ r+

1

r 2 sin2ϕ

∂ 2u∂ θ2

+1r 2

∂ 2u∂ ϕ2

++cotϕ

r 2

∂ u∂ ϕ

in the spherical coordinates x1 = r cosθ sinϕ, x2 = r sinθ sinϕ, x3 = r cosϕ.6.4. Let Ω⊆R2 be an open set and let u :Ω−→R.

(a) Let (a, b ) ∈ R2 and let T : R2 → R2 be given by T (x, y) = (x + a, y + b ).Define Ω be given by

Ω := (x − a, y − b ) : (x, y) ∈Ω.Let T denote the restriction of T to Ω. Show that u ∈ C 2(Ω) if and only ifuT ∈C 2(Ω) and that∆(uT ) = (∆u)T . In particular,

∆u = 0⇐⇒∆(uT ) = 0.

(b) Laplace equation is invariant under any real orthogonal transformation. LetU :R2→R2 be any real orthogonal transformation i.e.,

U (x, y) =MU

xy

∀(x, y) ∈R2,

where MU is a real 2×2 invertible matrix whose inverse is equal to its trans-pose. Define Ω := U−1(Ω) and let U be the restriction of U to Ω. Showthat u ∈C 2(Ω) if and only if uU ∈C 2(Ω) and that∆(uU ) = (∆u)U . Inparticular,

∆u = 0⇐⇒∆(uU ) = 0.

6.5. Let u :R2→R be a polynomial function, which is also radial and harmonic. Showthat u is a constant function.



Solutions on subdomains

6.6. [33] Let Ω be the domain Ω= (x, y) ∈R2 : x2+ y2 < 4. Consider the Neumannproblem

∆u = 0 on Ω,∂n u = αx2+βy + γ for (x, y) ∈ ∂ Ω,

where α,β,γ are real numbers.

(i) Find the values of α,β,γ for which the problem is not solvable.(ii) Solve the problem for those values of α,β,γ for which a solution exists.

6.7. Let a, b ∈ R be such that a < b . Let A,B ∈ R. Solve ux x + uyy = 0 in the annular

region 0 < a <p

x2+ y2 < b with the boundary conditions u = A on r = a andu = B on r = b . (Hint: Look for a solution depending only on r =

px2+ y2.)

6.8. Solve ux x + uyy = 1 inp

x2+ y2 < a with u vanishing on x2 + y2 = a2. (Ans:

u(x, y) = 14 (x

2+ y2− a2).)6.9. Solve∆u = 0 in the disk D = (x, y) : x2+ y2 < a2 with the boundary condition

u = 1+ 3sinθ on the circle r = a. (Ans: u(r,θ) = 1+ 3ra sinθ.)

6.10. Solve the equation ∆u = 0 in the domain D = (x, y) : x2 + y2 > a2 with theboundary condition u = 1+3sinθ on the circle r = a and with the condition thatthe solution is bounded for r →∞. (Ans: u(r,θ) = 1+ 3a

r sinθ.)6.11. [33] Solve the problem

∆u = 0 for 0< x <π, 0< y <π,u(x, 0) = u(x,π) = 0 for 0≤ x ≤π,

u(0, y) = 0 for 0≤ y ≤π,u(π, y) = sin y for 0≤ y ≤π.

Is there a point (x, y) in the square (0,π)× (0,π) such that u(x, y) = 0 ?6.12. [33] Solve the Laplace equation∆u = 0 in the domainΩ= (x, y) ∈R2 : x2+y2 >

4, subject to the boundary condition u(x, y) = y on ∂ Ω, and the decay condition

lim|x|+|y|→∞ u(x, y) = 0.

6.13. [38] This exercise is concerned with Poisson’s formula, to show that the functiondefined by Poisson’s formula satisfies the boundary conditions (see Theorem 6.11).Let ξ ∈ S(0, 1).

(a) Show that

1−∥x∥22π

∫S(0,1)

1∥x − y∥2 d y = 1,

and then deduce

u(x)− u(ξ ) =1−∥x∥2

2π

∫S(0,1)

f (y)− f (ξ )∥x − y∥2 d y.

Sivaji IIT Bombay

Exercises 177

(b) For δ > 0, write

u(x)− u(ξ ) =1−∥x∥2

2π

∫S(0,1)∩B(ξ ,δ)

f (y)− f (ξ )∥x − y∥2 d y +

∫S(0,1)\B[ξ ,δ]

f (y)− f (ξ )∥x − y∥2 d y

= I + I I .

Fix ε > 0. Using the continuity of f show that there exists a δ0 such that forall δ <δ0, |I |< ε holds.

(c) Show that if ∥x − ξ ∥ < δ2 and ∥y − ξ ∥ > δ, then ∥x − y∥ > δ

2 . Furtherdeduce that

limx→ξ I I = 0.

Maxmimum principles

6.14. Formulate the strong minimum principle and prove it by following the method ofproof of the strong maximum principle.

6.15. [23] Let Ω⊆R2 be given by

Ω := (x, y) ∈R2 : x2+ y2 > 1. (6.130)

Let u ∈C 2(Ω) be such that∆u = 0 in Ω, and lim|(x,y)|→∞u(x, y) = 0. Show that

maxΩ|u|=max

∂ Ω|u|. (6.131)

6.16. [23] Let Ω⊆R2 be a bounded domain. Let u ∈C 2(Ω)∩C (Ω) be a solution of

∆u + a(x, y)ux + b (x, y)uy + c(x, y)u = 0,

where a, b , c ∈ C (Ω) and c(x, y) < 0 in Ω. Show that u = 0 on ∂ Ω implies u = 0in Ω. (Hint: Show that max u ≤ 0 and min u ≥ 0.)

6.17. [23] Let Ω⊆R2 be given by

Ω := (x, y) ∈R2 : y > 0. (6.132)

Let u ∈ C 2(Ω) ∩ C (Ω) be a harmonic function in Ω. Further assume that u isbounded above in Ω. Prove that

supΩ

u = sup∂ Ω

u. (6.133)

(Hint: Take for ε > 0 the harmonic function

u(x, y)− ε log(p

x2+(y + 1)2),

Apply the maximum principle to a region (x, y) ∈R2 : x2+(y+1)2 < a2, y ≥ 0with large a. Let ε→ 0.)

6.18. Check the validity of the maximum principle for the harmonic function 1−x2−y2

1−2x+x2+y2

in the disk D = (x, y) : x2+ y2 ≤ 1. Explain.



6.19. Let Ω ⊆ R2 be a bounded domain. u ∈ C 2(Ω)∩C (Ω) be such that ∆u ≥ 0 in Ω.Prove that its maximum value is attained on the boundary ∂ Ω. What can you sayabout the minimum value?

6.20. Let u be a harmonic function on the whole plane such that u = 3sin(2θ) + 1 onthe circle x2+ y2 = 2. Without finding the concrete form of the solution, find thevalue of u at the origin.

6.21. [33] Let u be a non-constant harmonic function in the disk (x, y) : x2+y2 < R2.Define a function M : (0, R)→R by

M (r ) = maxx2+y2=r 2

u(x, y).

Show that M is a strictly increasing function on the interval (0, R).6.22. [33] Let Ω= (x, y) : 0< x <π, 0< y <π.

(i) Assume that vx x + vyy + xvx + yvy > 0 in Ω. Prove that v has no localmaximum in Ω.

(ii) Consider the problem

ux x + uyy + x ux + y uy = 0 on Ω,

u(x, y) = f (x, y) for (x, y) ∈ ∂ Ω,

where f is a given continuous function. Show that if u is a solution, then themaximum of u is achieved on the boundary ∂ Ω. (Hint: Use the auxiliaryfunction vε(x, y) = u(x, y)+ εx2)

(iii) Show that the problem in (ii) above has at most one solution.

6.23. [33] Let v(x, y) be a smooth vector field defined on a domainΩ. Let u be a smoothsolution of the Dirichlet problem

∆u + v.∇u = F on Ω,u(x, y) = g (x, y) for (x, y) ∈ ∂ Ω,

where F > 0 in Ω, and g < 0 on ∂ Ω. Show that u(x, y)< 0 for (x, y) ∈Ω.6.24. [33] Let Ω be the domain Ω = (x, y) ∈ R2 : x2+ y2 < 36. Let u be a harmonic

function on Ω which satisfies the Dirichlet boundary condition on ∂ Ω

u(x, y) =§

x if x < 0,0 otherwise.

(i) Prove that u(x, y) < minx, 0 in Ω. (Hint: Prove that u(x, y) < x andu(x, y)< 0 in Ω.)

(ii) Evaluate u(0,0) using the mean value principle.(iii) Using Poisson’s formula, evaluate u(0, y) for 0≤ y < 6.(iv) Using separation of variables method, find the solution u.(v) Is the solution u obtained in (iv) above classical?

6.25. Deduce the following version of Liouville’s theoerm from Theorem 6.29: If u :R2→ R is a harmonic function that is bounded, then u must be a constant func-tion.

6.26. Derive mean value property-II starting from the formula (6.10) as follows:

Sivaji IIT Bombay

Exercises 179

(i) For every w ∈C 2(Ω) satisfying∆w = 0, the function G(x ,ξ ) defined by

G(x ,ξ ) =K(x ,ξ )+w(x)

is a fundamental solution of Lapalce equation.(ii) Letψ be as defined by (6.9). Now apply (6.10) when the domain Ω is the ball

B(ξ ,ρ), and G(x ,ξ ) =ψ(∥x − ξ ∥)−ψ(ρ), to get

u(ξ ) =∫

B(ξ ,ρ)(ψ(∥x − ξ ∥)−ψ(ρ))∆u d x +

1ωdρd−1

∫S(ξ ,ρ)

u dσ .

for ξ ∈ B(ξ ,ρ), and u ∈C 2(B[ξ ,ρ]).(iii) Deduce mean value property-II when the function u is harmonic.


Documents

Chapter 6 Laplace equation - Home | Department of …siva/ma51515/chapter6.pdfChapter 6 Laplace equation In this chapter we consider Laplace equation in d-dimensions given by ux 1x1