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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 2 / 35
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..
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 3 / 35
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 ?
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 4 / 35
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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 5 / 35
Example: Laplace equation with Inhomog BC in 2DCartesian coordinates
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 6 / 35
Example: Laplace equation with Inhomog BC in 2DCartesian coordinates
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 7 / 35
Example: Laplace equation with Inhomog BC in 2DCartesian coordinates
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 8 / 35
Example: Laplace equation with Inhomog BC in 2DCartesian coordinates
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 9 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 10 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 11 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 12 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 13 / 35
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 14 / 35
Properties of Laplace Eqn
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 15 / 35
Properties of Laplace Eqn
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).
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 16 / 35
Properties of Laplace Eqn
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 17 / 35
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 18 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 20 / 35
Curved Geometry (2D Polar)
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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 21 / 35
Curved Geometry (2D Polar)
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]
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 22 / 35
Curved Geometry (2D Polar)
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 23 / 35
Curved Geometry (2D Polar)
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, ..
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 24 / 35
Curved Geometry (2D Polar)
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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 25 / 35
Curved Geometry (2D Polar)
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 26 / 35
Curved Geometry (2D Polar)
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 27 / 35
Curved Geometry (2D Polar)
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.
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 28 / 35
Curved Geometry (2D Polar)
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 29 / 35
Curved Geometry (2D Polar)
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]
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 30 / 35
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
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 31 / 35
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
)...
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 32 / 35
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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 33 / 35
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)
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 34 / 35
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 <∞
Amitabh Bhattacharya (Department of Mechanical Engineering, I.I.T. Bombay)Elliptic Equations 35 / 35