Lect Laplace

Elliptic Equations

Amitabh Bhattacharya

Department of Mechanical Engineering, I.I.T. Bombay

Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 1 / 35

Laplace and Poisson Equation

GivenA∂2u∂x2

+ 2B ∂2u∂x∂y + C ∂2u

∂y2+ Low Order Derivatives = Source Term

I Equation is Elliptic if B2 −AC < 0

We will focus on Poisson Equation for u(x, t):

α2∇2u = F (x, t) forx ∈ D t > 0

BC : a(x)u+ b(x)∂u

∂n= f(x) forx ∈ ∂D

Can be seen as steady state limit of Diffusion and Wave Equations

Physical examples:I Equation for steady state temperature in a heated domain (BC: Heat

Flux or Imposed Temperature)I Equation for loaded membrane stretched across a wireframe (BC:

Height of membrane at wireframe or Slope of membrane at wireframe)


Strategy for Solving Poisson equation

We can write Poisson eqn as :

Lu = F (x, t) whereL = ∇2

BC : a(x)u+ b(x)∂u

∂n= f(x) forx ∈ ∂D

Our general strategy can be as follows:

Separate solution into two parts u(x, t) = u1(x, t) + u2(x, t)

u1 satisfies Laplace equation Lu1 = 0 with inhomogeneous BCa(x)u1 + b(x)∂u1∂n = f(x)

u2 satisfies Poisson equation Lu2 = F (x, t) with homogeneous BCa(x)u2 + b(x)∂u2∂n = 0

Let’s solve for u1 and u2 separately..


Laplace equation with Inhomog BC

Let’s solve for u1, which satisfies Laplace equation Lu1 = 0 withinhomogeneous BC a(x)u1 + b(x)∂u1∂n = f(x)

Unlike before, eigenfunctions of L with homogeneous BC NOT usefulhere: will NOT give a well posed solution

But for Cartesian coordinates we can see that (e.g.)

∇2u = Lyu+ ∂2u∂x2

= Lxu+ ∂2u∂y2

, where Ly = ∂2

∂y2and Lx = ∂2

∂x2

Maybe eigenfunctions/eigenvalues of Lx and Ly are useful, if BC ishomog in x or y directions respectively ?


Example: Laplace equation with Inhomog BC in 2DCartesian coordinates

Problem Statement: ∂2u∂x2

+ ∂2u∂y2

= 0 in D, where

D = {(x, y); 0 ≤ x ≤ a, 0 ≤ y ≤ b}. BC: u(0, y) = u(x, 0) = u(x, b) = 0,u(a, y) = f(y) (Dirichlet BC).Solution:

BC is homogeneous in y direction. We can write down the equationas Lyu+ ∂2u

∂x2= 0.

Find eigenfunctions, eigenvalues (Yn(y), λn) of Ly = ∂2

∂y2, i.e.

∂2Yn∂y2

= −λ2nYn with BC Yn(0) = Yn(b) = 0 (SL problem)

Express solution as u(x, y) =∑

nXn(x)Yn(y) and

f(y) =∑

n fnYn(y)

Solve ODE X ′′n − λ2nXn = 0 with BCs Xn(0) = 0, Xn(a) = fn (BVP,

but NOT SL problem)



Clearly, Yn(y) = sin(nπy/b) and λn = nπ/b

General solution for Xn(x) is Xn(x) = Cn sinh(λnx) +Dn cosh(λnx)

BCs for Xn(x) implies Dn = 0, and

Cn =fn

sinh(λna)=

2

b sinh nπab

∫ b

0f(y) sin

nπy

bdy

Final solution:

u(x, y) =∑n

Cn sinhnπx

asin

nπy

b



Say f(y) = 100 (e.g. constant temperature BC), then

Cn =200

nπ sinh nπab

[1− (−1)n]

All values in the domain are within 0→ 100 (Follows from ”maximumprinciple”)Solution is completely ”controlled” by BC



We can alternately use conventional separation of variables here:u(x, y) = X(x)Y (y)

Substituting into ∇2u = 0...

X ′′

X= −Y

′′

Y= λ2

Y (0) = Y (b) = 0

+ve sign chosen since BC is homog in y direction (SL problem inY (y))

Again, λn = nπ/b and Yn(y) = sin(λny), X ′′n − λ2nXn = 0,

Xn(0) = 0, Xn(a) = fn etc.

Easier, but not clear why it works



BCs can be more complex, e.g. u(a, y) = f(y) and u(x, b) = g(x)

in this case, we use superposition: u(x, y) = u1(x, y) + u2(x, y),where

u1(x, 0) = u1(0, y) = u1(x, b) = 0, u1(a, y) = f(y)

u2(x, 0) = u2(0, y) = u2(a, y) = 0, u2(x, b) = g(x)

Final solution is:

u(x, y) =∑n

[An sinh

nπy

bsin

nπx

a+Bn sinh

nπx

asin

nπy

b

]An =

2

a sinh nπba

∫ a

0g(x) sin

nπx

adx

Bn =2

b sinh nπab

∫ b

0f(y) sin

nπy

bdy


Poisson Equation with Homog BC in 2D Cartesiancoordinates

Problem Statement: Find solution u(x, y), which satisfies∂2u∂x2

+ ∂2u∂y2

= F (x, y), along with BCs

u(0, x) = u(x, 0) = u(a, y) = u(x, b) = 0Solution: Again, we solve Eigenvalue problem for L = ∇2. We alreadyknow the eigenvalues and eigenfunctions via separation of variables:

Φmn(x, y) = sinmπx

asin

nπy

b

λ2mn =

(mπa

)2+(nπb

)2


Poisson Equation with Homog BC in 2D CartesiancoordinatesNow represent u(x, y) and F (x, y) in terms of eigenfunctions Φmn(x, y)

u(x, y) =∑m,n

umnΦmn(x, y)

F (x, y) =∑m,n

FmnΦmn(x, y)

Substituting into Poisson eqn..

−λ2mnumn = Fmn (Algebraic Equation !)

⇒ u(x, y) = −∑m,n

Fmnλ2mn

Φmn(x, y)

= −∑m,n

[(mπa

)2+(nπb

)2]−1

Fmn sinmπx

asin

nπy

b


Poisson Eqn With Inhomogeneous BC

Problem Statement: Find solution u(x, y), which satisfies∂2u∂x2

+ ∂2u∂y2

= F (x, y), along with BCs u(0, x) = u(x, 0) = u(x, b) = 0 and

u(a, y) = f(y)Solution: We can simply add the solutions u1 and u2

u(x, y) =∑n

Cn sinhnπx

asin

nπy

b−

∑m,n

[(mπa

)2+(nπb

)2]−1

Fmn sinmπx

asin

nπy

b

where

Cn =fn

sinh(λna)=

2

b sinh nπab

∫ b

0f(y) sin

nπy

bdy


Properties of Laplace Eqn

Theorem I: Laplace Eqn has the following properties:

If ∇2u = 0 on D and u = 0 on ∂D then u = 0 inside DIf ∇2u = 0 on D and ∂u

∂n = 0 on ∂D then u = constant inside DProof: Note that ∫

D∇ ·wdv =

∮∂D

w · ndS

Now say w = Φ∇Φ(x), where ∇2Φ = 0 in D∫D∇ · (Φ∇Φ)dv =

∮∂D

Φ∇Φ · ndS

⇒∫D

(∇Φ · ∇Φ + Φ∇2Φ)dv =

∮∂D

Φ∇Φ · ndS


Properties of Laplace EqnProof (Contd):

⇒∫D∇Φ · ∇Φdv =

∮∂D

Φ∇Φ · ndS

Neumann BC for Φ on ∂D implies:

∂Φ

∂n(x) = 0 forx ∈ ∂D ⇒ ∇Φ = 0 forx ∈ D

⇒ Φ = constant inD

Dirichlet BC for Φ on ∂D implies:

Φ(x) = 0 forx ∈ ∂D ⇒ ∇Φ = 0 forx ∈ D⇒ Φ = constant inD

But Φ = 0 on ∂D, hence the constant is 0. Therefore Φ = 0 in D for thiscase. Hence Proved.



Theorem II: If u is prescribed on the domain boundary ∂D and ∇2u = 0then solution is uniqueProof: Let there be 2 solutions, ψ(x) and ψ(x). Now ψ(x) = Φ(x) forx ∈ ∂D. If γ(x) = ψ(x)− Φ(x), then ∇2γ = 0 in D, and γ = 0 forx ∈ ∂D. Hence, from Theorem I, γ = 0 is the only solution possible.Hence ψ(x) = ψ(x). Hence Proved.Theorem III: If ∂u

∂n is prescribed on the domain boundary ∂D and∇2u = 0 then solution is unique to within an additive constant.Proof: Let there be 2 solutions, ψ(x) and ψ(x). Now ∂ψ

∂n (x) = ∂Φ∂n (x) for

x ∈ ∂D. If γ(x) = ψ(x)− Φ(x), then ∇2γ = 0 in D, and ∂γ∂n = 0 for

x ∈ ∂D. Hence, from Theorem I, γ = constant is the only solutionpossible. Hence ψ(x) = ψ(x) + constant. Hence Proved.



Theorem IV: If ∇2Φ = 0, then∮∂D

∂Φ∂ndS = 0 Proof: We can write

∇2Φ = ∇ · ∇Φ. Integrating over the volume:∫D∇2Φdv =

∫D∇ · ∇Φdv =

∮∂D∇Φ · ndS = 0

Hence proved.Means that if pure Neumann condition ∂Φ

∂n is applied all over ∂D, thenabove constraint has to hold. Otherwise problem will be ill-posed andsolution Φ will not exist (e.g. To keep temperature in a body constantheat flux in has to be equal to heat flux out at the boundaries).



Theorem V (Mean Value Theorem): Let φ(x, y) be satisfying∇2φ = 0, and let R ∈ D be a circle with radius r centered at (ζ, η) ∈ D.Boundary ∂R. φ(ζ, η) is always the average of its value on thecircumference of the circle.Proof: From Theorem IV, since ∇2φ = 0 inside R, therefore∮∂R

∂φ∂nds = 0. Choosing origin such that ζ = η = 0, and using polar

coordinates, so that Φ(r, θ) = φ(x, y) and Φ(0, θ) = φ(ζ, η) we get∮∂R

∂φ

∂nds =

∫ 2π

0

∂Φ

∂r(r, θ)rdθ = 0

⇒ d

dr

∫ 2π

0Φ(r, θ)rdθ = 0

⇒∫ 2π

0Φ(r, θ)rdθ = Constant


Properties of Laplace EqnProof(contd):Now let ε < r, then∫ 2π

0Φ(ε, θ)rdθ =

∫ 2π

0Φ(r, θ)rdθ

Observe that limε→0

∫ 2π0 Φ(ε, θ)rdθ = 2πΦ(0, θ) = 2πφ(ζ, η). Therefore

taking this limit for LHS term above..

φ(ζ, η) =1

2π

∫ 2π

0Φ(r, θ)rdθ (Hence Proved)

Theorem VI (Maximum Theorem): Maximum of non-constantHarmonic function φ(x) (i.e. ∇2Φ = 0) is always at the boundary.Proof: For any point (ζ, η) inside D, we can always draw a circle aroundthe point. For some points on the circle, φ will always have a higher valuethan φ(ζ, η). At a boundary point x ∈ ∂D, Mean Value Theorem cannothold, since we cannot draw such a circle. Hence, extremum of φ is possibleonly at boundary ∂D.


Curved Geometry (2D Polar)Let’s solve Laplace equation in a curved geometry in 2D:

€

R1

€

R2

€

φ

Equation for ∇2u = 0 in Polar coordinates is:

1

r

∂

∂

[r∂u

∂r

]+

1

r2

∂2u

∂θ2= 0

BC’s: Homogeneous in θ and inhomogeneous in r,or

Inhomogeneneous in θ and homogeneous in rAmitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 19 / 35

Curved Geometry (2D Polar)

Separation of variables: u(r, θ) = R(r)Θ(θ) leads to:

r

R

d

dr

[r

dR

dr

]= − 1

Θ

d2Θ

dθ2= ±λ2

r2R′′ + rR′ − (±λ2)R = 0, Θ′′ ± λ2Θ = 0

Case 1: Say u(r, 0) = u(r, φ) = 0, u(R1, θ) = 0, u(R2, θ) = f(θ)BC is homog in θ direction, so choose +λ2 on RHS

Θ(θ) = A sin(λθ) +B cos(λθ)

R(r) =

[C +D ln r ifλ = 0Crλ +Dr−λ ifλ > 0



Case1(contd) Applying Θ(0) = Θ(φ) = 0..

B = 0, λnφ = nπ

⇒ Θn(θ) = sin

(nπθ

φ

), Rn(r) = Cnr

λn +Dr−λn

⇒ u(r, θ) =

∞∑n=1

sin

(nπθ

φ

)[Cnr

λn +Dnr−λn]

Applying BCs at r = R1 and r = R2..

u(R1, θ) =

∞∑n=1

sin

(nπθ

φ

)[CnR

λn1 +DnR

−λn1

]= 0

u(R2, θ) =

∞∑n=1

sin

(nπθ

φ

)[CnR

λn2 +DnR

−λn2

]= f(θ)



Case1(contd) We can expand f(θ) as:

f(θ) =

∞∑n=1

fn sin

(nπθ

φ

), fn =

2

φ

∫ φ

0f(θ) sin

(nπθ

φ

)dθ

From BC’s in r direction..

Cn =fn

Rλn2 −Rλn1

, Dn = − fnR2λn1

Rλn2 −Rλn1

Final solution:

u(r, θ) =

∞∑n=1

sin

(nπθ

φ

)fnR

λn1

Rλn2 −Rλn1

[(r

R1

)λn−(r

R1

)−λn]



Case 2: Say u(r, 0) = 0, u(r, φ) = g(r), u(R1, θ) = u(R2, θ) = 0.Again, using u(r, θ) = R(r)Θ(θ)...:

r

R

d

dr

[r

dR

dr

]= − 1

Θ

d2Θ

dθ2= ±λ2

r2R′′ + rR′ − (±λ2)R = 0, Θ′′ ± λ2Θ = 0

BC homogeneous in r direction, so choose −λ2 on RHS

Θ(θ) = A sinh(λθ) +B cosh(λθ)

R(r) =

[C +D ln r ifλ = 0

C ′riλ +D′r−iλ = C cos(λ ln r) +D sin(λ ln r) ifλ > 0

cos(λ ln r) and sin(λ ln r) are highly oscillatory..can satisfy homog BC in rif R1 > 0.



Case 2(contd): Applying homog BCs in r direction:

C ′Riλ1 +D′R−iλ1 = 0

C ′Riλ2 +D′R−iλ2 = 0

Characteristic Eqn of SL problem:(R1

R2

)iλ−(R2

R1

)iλ= 0

⇒(R2

R1

)2iλ

= 1

⇒ cos(2λ lnR2

R1) + i sin(2λ ln

R2

R1) = 0

⇒ λn =nπ

ln(R2/R1), n = 1, 2, 3, ..



Case 2(contd): General solution is then:

u(r, θ) =

∞∑n=1

[Cn cos(λn ln r) +Dn sin(λn ln r)]×

[An sinh(λnθ) +Bn cosh(λnθ)]

λn =nπ

ln(R2/R1)

Applying BC at θ = 0, i.e. u(r, 0) = 0 implies Bn = 0

u(r, φ) = g(r)

⇒∞∑n=1

[Cn cos(λn ln r) +Dn sin(λn ln r)] sinh(λnφ) = g(r)



Case 2(contd): Orthogonality condition for sin(λn ln r), cos(λn ln r) canbe seen by comparing eqn for R(r) with SL equation: SL form:

[p(x)y′]′ + q(x)y + λw(x)y = 0 a ≤ x ≤ b

Inner product is∫ ba w(x)ym(x)yn(x)dx. In our problem:

r2R′′ + rR′ + λ2R = 0

or [rR′]′ +λ2

rR = 0⇒ w(r) = 1/r

Relevant inner product between any ψ(r) and φ(r) in our case is∫ R2

R1φ(r)ψ(r)1

rdr =∫ R2

R1φ(r)ψ(r)d ln r



Case 2(contd): So if we representg(r) =

∑∞n=1[En cos(λn ln r) + Fn sin(λn ln r)], then

En =

∫ R2

R1g(r) cos(λn ln r)d ln r∫ R2

R1cos2(λn ln r)d ln r

=2

ln(R2R1

) ∫ R2

R1

g(r) cos(λn ln r)d ln r

Fn =

∫ R2

R1g(r) sin(λn ln r)d ln r∫ R2

R1sin2(λn ln r)d ln r

=2

ln(R2R1

) ∫ R2

R1

g(r) sin(λn ln r)d ln r

From BC at θ = 0,

Cn =En

sinh(λnφ), Dn =

Fnsinh(λnφ)

Case 1 and case 2 can be superposed to solve any inhomogeneous BC onthis domain, for R1 > 0.



Case 3: Now suppose R1 = 0, and BC’s are homogeneous in r direction.u(r, 0) = 0, u(r, φ) = g(r), u(R2, θ) = 0, u(0, θ) is finite.Technique used in case 2 no longer works, since sin(λ ln r) and cos(λ ln r)are both singular at r = 0. We instead break the problem into two parts:

u(r, θ) = v(r, θ) + w(r, θ)

∂2w

∂θ2= 0, w(r, 0) = 0, w(r, φ) = g(r)

1

r

∂

∂

[r∂v

∂r

]+

1

r2

∂2v

∂θ2= 0 v(r, 0) = v(r, φ) = 0, v(R2, θ) = −w(R2, θ)

Solution for w(r, θ) is w(r, θ) = g(r) θφProblem in v is inhomogeneous in r, homogeneous in θ; can be easilysolved using method in case 1.



Case 4: Now say φ = 2π, so that u(r, θ) = u(r, θ + 2π). Also,u(R1, θ) = 0, u(R2, θ) = f(θ)Similar to case 1, +λ2 is used on RHSIn this case, λn = n and eigenfunctions in θ direction are 1, sinnθ, cosnθwhere n = 0, 1, 2, 3, ..General solution is:

u(r, θ) = A+B ln r +

∞∑n=1

[Cn sinnθ +Dn cosnθ][Enrn + Fnr

−n]

From u(R1, θ) = 0..

A+B lnR1 = 0 EnRn1 + FnR

−n1 = 0

⇒ A = −B lnR1, Fn = −EnR2n1



Case 4(contd) So..

u(r, θ) = B lnr

R1+

∞∑n=1

[Cn sinnθ +Dn cosnθ]

[(r

R1

)n−(r

R1

)−n]

Now if f(θ) = a0 +∑∞

n=1[an cosnθ + bn sinnθ], then

B =a0

ln R2R1

, Cn =bn[(

R2R1

)n−(R2R1

)−n] ,Dn =

an[(R2R1

)n−(R2R1

)−n]


Spherical Geometry

Spherical to Cartesian transformation:

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ

0 < r <∞, 0 ≤ φ < 2π, 0 ≤ θ ≤ π

Let’s consider Laplace Eqn for u(r, θ, φ) assuming axi-symmetry in φ:

∂

∂r

[r2∂u

∂r

]+

1

sin θ

∂

∂θ

(sin θ

∂u

∂θ

)= 0


Spherical Geometry

Use separation of variables: u(r, θ) = R(r)Θ(θ)

1

R

∂

∂r

[r2∂R

∂r

]= − 1

Θ sin θ

∂

∂θ

[sin θ

∂

∂θΘ

]= µ

Let’s focus on eqn for Θ and use transformation ζ = cos θ

d

dζ

[(1− ζ2)

dΘ

dζ

]+ µΘ = 0 (SL Form!)

This is the Legendre equation, with solutions y1(ζ) and y2(ζ), which aresingular at ζ = ±1, or θ = 0, π.If µ = n(n+ 1), where n is an integer, then one of the solutions, Pn(ζ) isbounded at ζ = ±1Pn(ζ) are Legendre polynomials of degree n. e.g. P0(ζ) = 1, P1(ζ) = ζ,P2(ζ) = 1

2

(3ζ2 − 1

)...


Spherical Geometry

Let’s use µ = n(n+ 1) in the Eqn for R(r)..

r2 d2R

dr2+ 2r

dR

dr− n(n+ 1)R = 0 (Cauchy Euler Eqn)

⇒ R(r) = Anrn +Bnr

−(n+1)

General solution is:

u(r, θ) =

∞∑n=0

(Anr

n +Bnrn+1

)Pn(cos θ)


Spherical Geometry

Example: Consider a spherical surface at radius r = R maintained at atemperature u(R, θ) = E cos2 θ. Obtain temperature u(r, θ) for0 < r <∞ and 0 < θ < π.For 0 < r < R, boundedness implies that Bn = 0, so that :

u(r, θ) =

∞∑n=0

AnrnPn(cos θ)

Now note that cos2 θ = 23P2(cosθ) + 1

3P0(cos θ). Thus

u(R, θ) =

∞∑n=0

AnRnPn(cos θ) = E cos2 θ

⇒ A0 =E

3, A2 =

2E

3R2

An = 0 for n = 1, 3, 4, .. (Using orthogonality of Pn(cos θ) for different n)


Spherical Geometry

Example(Contd): For r ≥ R, boundedness implies that An = 0, so that :

u(r, θ) =

∞∑n=0

Bn

r(n+1)Pn(cos θ)

⇒ u(R, θ) =

∞∑n=0

BnR−(n+1)Pn(cos θ) = E cos2 θ

⇒ B0 =ER

3; B2 =

2

3ER3

Bn = 0 for n = 1, 3, 4, ... Final solution is:

u(r, θ) =

{E3 + 2

3E(rR

)2P2(cos θ) for 0 ≤ r ≤ R

E3Rr + 2

3E(Rr

)3P2(cos θ) forR ≤ r <∞


Lect Laplace

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