boundary value problems in spherical...

Boundary Value Problems in Spherical

Coordinates

Y. K. Goh

2009

Y. K. Goh

Boundary Value Problems in Spherical Coordinates

Outline

I Laplacian Operator in spherical coordinates

I Legendre Functions

I Spherical Bessel Functions

I Initial-value problem for heat flow in a sphere

I The three-dimensional wave equation

I Laplace Eq. in a sphere and exterior to a sphere

Y. K. Goh


Laplacian Operator in Spherical Coordinates

Laplacian Operator in SphericalCoordinates

Y. K. Goh


Laplacian Operator in Spherical Coordinates

Spherical Coordinates

I (r, θ, φ) : x = r cosφ sin θ, y = r sinφ sin θ, z = r cos θ

I ∇2u =1

r2

∂

∂r

(r2∂u

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+

1

r2 sin2 θ

∂2u

∂φ2

Y. K. Goh


Legendre’s Equations and Legendre’s Functions

Legendre’s Equation arises naturally when solving some PDEin Spherical Coordinate systems. Usually it forms part of theSturm-Liouville problem which requires it to have boundedeigenfunctions over a fixed domain.

Definition (Legendre’s Equation)The Legendre’s Equations is a family of differential equationsdiffer by the parameter λ in the following form

(1− x2)y′′ − 2xy′ + λy = 0, (1)

ord

dx

[(1− x2)

dy

dx

]+ λy = 0. (2)

Y. K. Goh



The Legendre’s equation is a linear 2nd order ODE.

I x = ±1 are two singular points of the ODE.

I A solution near the ordinary point x = 0 is a power series

y =∞∑

n=−∞

anxn, an = 0,∀n < 0.

I The radius of convergence for the power series is thedistance from the centre of the series to the nearestsingular point, i.e. R = 1.

Y. K. Goh



I Substitue the power series into the ODE, we will obtainthe recurrecnce relation

an+2 =n(n+ 1)− λ

(n+ 2)(n+ 1)an, n = 0, 1, 2, . . . .

I The recurrence relation gives two series solutions knownas the Legendre’s functions, where one is an oddfunction and the other one is an even function.

I By using convergence tests, we can show that the twoseries are convergence for |x| < 1.

I However, the series are generally not convergent atx = ±1, except if λ = `(`+ 1).

Y. K. Goh


Legendre’s Polynomials

For the case of λ = `(`+ 1) :

I When n = `, an+2 =n(n+ 1)− `(`+ 1)

(n+ 2)(n+ 1)an = 0,

=⇒ a`+2 = a`+4 = a`+6 = · · · = 0.

I Thus, one of the series solution becomes a polynomial.

I After normalization, we obtain the LegendrePolynomial of degree `,

P`(x) =1

2`

N∑n=0

(−1)n(2`− 2n)!

n!(`− n)!(`− 2n)!x`−2n, (3)

where N = `/2 if ` is even or (`− 1)/2 is ` is odd.

Y. K. Goh


Legendre function of the second kind

For the case of λ = `(`+ 1) :I The Legendre’s equation of order n, for ` = 0, 1, 2, . . .

(1− x2)y′′ − 2xy′ + `(`+ 1)y = 0,−1 < x < 1.

I A solution is the Legendre Polynomial of degree `, P`(x).I The other solution is a series solution known as the

Legendre function of the second kind, Q`(x).I Q`(x) converges on the −1 < x < 1 but unbounded in−1 ≤ x ≤ 1.

I Since P`(x) and Q`(x) are linearly independent, thus thegeneral solution to the ODE is

y = c1P`(x) + c2Q`(x).

Y. K. Goh


Properties of Legendre Polynomials

I Symmetry: Pn(x) is even if n is even, and odd if n is odd.I End points: Pn(1) = 1, and P (−1) = (−1)n for all n.I Boundedness :|Pn(x)| ≤ 1 for all n and x in [−1, 1].I Zeros: Pn(x) has n distinct zeros in [−1, 1].

I Orthogonality:

∫ 1

−1

Pm(x)Pn(x) dx =2

2n+ 1δmn.

I Rodrigues’ Formula: Pn(x) =1

2nn!

dn

dxn(x2 − 1)n.

I Generating function:

Φ(x, t) = (1− 2xt+ t2)−1/2 =∞∑n=0

Pn(x)tn.

I Pn(x) =1

n!

dn

dtn[Φ(x, t)]t=0.

Y. K. Goh


Graphs of Legendre Polynomials

Figure: Legendre Polynomials, Pn(x).

I P0(x) = 1

I P1(x) = x

I P2(x) = 12(3x2 − 1)

I P3(x) = 12(5x3 − 3x)

I P4(x) =18(35x4 − 30x2 + 3)

I P5(x) =18(63x5 − 70x3 + 15x)

Y. K. Goh


Graphs of Legendre functions of the second kind

Figure: Legendre functions, Qn(x).

I Q0(x) = 12

ln(

1+x1−x

)I Q1(x) = x

2ln(

1+x1−x

)− 1

I Q2(x) = 3x2−14

ln(

1+x1−x

)− 3x

2

I Q3(x) =5x3−3x

2ln(

1+x1−x

)− 5x2

2+ 2

3

I Q4(x) =35x4−30x2+3

16ln(

1+x1−x

)− 35x3

8+

55x24

.

Y. K. Goh


Legendre Series Expansions

I Orthogonality of Legendre Polynomial

〈Pm|Pn〉 =

∫ 1

−1

Pm(x)Pn(x) dx =2

2n+ 1δmn.

I The Legendre series expansions for f(x),−1 ≤ x ≤ 1 is

f(x) =∞∑n=0

cnPn(x)

I Where the generalized Fourier coefficient cn is

cn =〈Pn|f〉‖Pn‖2

Y. K. Goh


Legendre Series Expansions

Examples

I Find the first three terms of the Legendre series expansionfor the function

f(x) =

{0, −1 ≤ x < 0,

1, 0 ≤ x ≤ 1.

I Find the Legendre series expansion for f(x) = x2 + x− 4,defined over the interval −1 ≤ x ≤ 1.

Y. K. Goh


Associated Legendre Functions

DefinitionThe Associated Legendre’s Equation is defined as

(1−x2)y′′−2xy′+

[n(n+ 1)− m2

1− x2

]y = 0, −1 < x < 1,

(4)or

d

dx

[(1− x2)

dy

dx

]+

[n(n+ 1)− m2

1− x2

]y = 0, −1 < x < 1.

(5)

The solution to the associated Legendre’s equation is denotedas Pm

n (x) and Qmn (x) respectively.

Y. K. Goh


Properties of Associated Legendre Functions

I Associated Legendre function of degree n and order m:Pmn (x), n = 0, 1, 2, . . . , |m| ≤ n .

I Legendre polynomial: P 0n(x) = Pn(x).

I Rodrigues’ Formula:

Pmn (x) = (−1)m(1− x2)m/2

dm

dxm[Pn(x)].

I Negative order: Pmn (x) = (−1)m

(n+m)!

(n−m)!P−mn (x).

I Orthogonality:∫ 1

−1

Pmk (x)Pm

l (x) dx =2

2k + 1

(k +m)!

(k −m)!δkl, |m| ≤ k, l.

Y. K. Goh


Associated Legendre Functions

I P 00 (x) = 1

I P 11 (x) = −

√1− x2, P 0

1 (x) = x, P−11 (x) = −1

2P 1

1 (x)

I P 22 (x) = 3(1− x2), P 1

2 (x) = −3x√

1− x2,P 0

2 (x) = 12(3x2 − 1), P−1

2 (x) = −16P 1

2 (x),P−2

2 (x) = 124P 2

2 (x),

I P 33 (x) = −15(1− x2)3/2, P 2

3 (x) = 15x(1− x2),P 1

3 (x) = −32(5x2 − 1)(1− x2)1/2, P 0

3 (x) = 12(5x3 − 3x),

P−13 (x) = − 1

12P 1

3 (x), P−23 (x) = 1

120P 2

3 (x),P−3

3 (x) = − 1720P 3

3 (x).

Y. K. Goh


Graphs of Associated Legendre Functionss

Figure: Associated LegendreFunctions, Pmn (x).

Figure: Associated LegendreFunctions, Qmn (x).

Y. K. Goh


Associated Legendre Series Expansions

I Orthogonality of associated Legendre functions

〈Pmk |Pm

l 〉 =

∫ 1

−1

Pmk (x)Pm

l (x) dx =2

2k + 1

(k +m)!

(k −m)!δkl.

I The associated Legendre series expansions of order m forf(x),−1 ≤ x ≤ 1 is

f(x) =∞∑n=m

cnPmn (x)

I Where the generalized Fourier coefficient cn is

cn =〈Pm

n |f〉‖Pm

n ‖2, n = m,m+ 1,m+ 2, . . . .

Y. K. Goh


Associated Legendre Series Expansions

Example

I The associated Legendre series expansion of order m = 2for

f(x) =

{0, −1 ≤ x < 0,

1, 0 ≤ x ≤ 1,

is given by f(x) =∞∑n=m

cnPmn (x). The first few

coefficients are

Y. K. Goh


Spherical Harmonics Functions

DefinitionThe spherical harmonic function is defined as

Y mn (θ, φ) =

√2n+ 1

4π

(n−m)!

(n+m)!Pmn (cos θ)eimφ. (6)

I Orthogonality:

〈Y mn |Y m′

n′ 〉 =

∫ π

θ=0

∫ 2π

φ=0

Y m∗n Y m′

n′ sin θ dθ dφ = δnn′δmm′

Y. K. Goh


Spherical Harmonics Series Expansions

A function f(θ, φ) defined in 0 ≤ θ < π and 0 ≤ φ < 2π canbe expressed as a spherical harmonics expansion

f(θ, φ) =∞∑n=0

n∑m=−n

cmn Ymn (θ, φ),

where the coefficient cmn is given by

cmn =〈Y m

n |f〉‖Y m

n ‖2= 〈Y m

n |f〉,

since Y mn has be normalized to ‖Y m

n ‖2 = 〈Y mn |Y m

n 〉 = 1.

Y. K. Goh


Spherical Harmonics Functions

Figure: Spherical Bessel’s function, jn(x).

Y. K. Goh


Spherical Bessel Functions

I Spherical Bessel’s equation

x2y′′ + 2xy′ + [x2 − n(n+ 1)]y = 0 (7)

I Two linearly independent solutions are the sphericalBessel’s functions of the first and second kinds.

I jn(x) =

√π

2xJn+1/2(x) = (−x)n

(1

x

d

dx

)nsinx

x.

I yn(x) =

√π

2xYn+1/2(x) = −(−x)n

(1

x

d

dx

)ncosx

x.

I Similar to the Bessel’s functions, jn(x) is bounded nearx = 0, but not bounded at x = 0 for yn(x).

Y. K. Goh


Spherical Bessel’s Functions of the first kind

Figure: Spherical Bessel’s function,jn(x).

I j0(x) =sinx

x

I j1(x) =sinx

x2− cosx

xI j2(x) =(

3

x2− 1

)sinx

x− 3 cosx

x2

I j3(x) =

(15

x3− 6

x

)sinx

x−(

15

x2− 1

)cosx

x

Y. K. Goh


Spherical Bessel’s Functions of the second kind

Figure: Spherical Bessel’s function,yn(x).

I y0(x) = −cosx

x

I y1(x) = −cosx

x2− sinx

xI y2(x) =(

1− 3

x2

)cosx

x− 3 sinx

x2

I y3(x) =(6

x− 15

x3

)cosx

x−(

15

x2− 1

)sinx

x

Y. K. Goh


Laplace Eq. inside a sphere (radial symmetry)

A Dirichlet’s problem inside a sphere with radial symmetry.

I u is radial symmetry, i.e. u(r, θ, φ) = u(r, θ)

I PDE: Laplace Eq. ∇2u(r, θ) = 0, 0 < r < L, 0 ≤ θ < π.

I1r2

∂

∂r

(r2∂u

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)= 0.

I Boundary ConditionsI u(L, θ) = f(θ), 0 ≤ θ < π.

I Regularity Conditions: |u| <∞ in 0 < r < L.

Y. K. Goh



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

I ODEs:{r2R′′ + 2rR′ − λR = 0, 0 < r < L; (Euler Eq.)

Θ′′ + cot θΘ′ + λΘ = 0, 0 ≤ θ < π. (Legendre Eq.)

I B.C.: u(L, θ) = f(θ) acts like initial condition.I The ODE for θ can be converted to Legendre by making

a change of variable s = cos θ.I =⇒ (1− s2)Θ′′ − 2sΘ′ + λΘ = 0, −1 < s < 1,I Bounded solution is only possible for Legendre’s

equation when λ = n(n+ 1), n = 0, 1, 2, . . . , i.e.Θ ∝ Pn(s) = Pn(cos θ).

Y. K. Goh



I Re-write the set of ODEs and corresponding boundedsolutions inside a sphere for n = 0, 1, 2, . . . are{r2R′′ + 2rR′ − n(n+ 1)R = 0, =⇒ R(r) = Cnr

n

Θ′′ + cot θΘ′ + n(n+ 1)Θ = 0, =⇒ Θ(θ) = Pn(cos θ)

I General solution:

u(r, θ) =∞∑n=0

CnrnPn(cos θ), 0 < r < L, 0 <≤ θ < π.

I “Initial” condition gives: CnLn =〈Pn|f〉‖Pn‖2

=

2n+ 1

2

∫ π

0

f(θ)Pn(cos θ) sin θ dθ, n = 0, 1, 2, . . . .

Y. K. Goh



Example

I Solve the Dirichlet problem for the Laplace Equationinside a sphere with f(θ) = L cos2 θ.

Y. K. Goh


Laplace Eq. exterior to a sphere (radial symmetry)

A Dirichlet’s problem outside a sphere with radial symmetry.

I u is radial symmetry, i.e. u(r, θ, φ) = u(r, θ)I PDE: Laplace Eq. ∇2u(r, θ) = 0, r > L, 0 ≤ θ < π.

I1r2

∂

∂r

(r2∂u

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)= 0.

I Boundary ConditionsI u(L, θ) = f(θ), 0 ≤ θ < π.

I Regularity Conditions: |u| <∞ in r > L.

I Separation of variables: u(r, θ) = R(r)Θ(θ) =⇒{r2R′′ + 2rR′ − n(n+ 1)R = 0, =⇒ R(r) = Dnr

−n−1

Θ′′ + cot θΘ′ + n(n+ 1)Θ = 0, =⇒ Θ(θ) = Pn(θ).

Y. K. Goh



I General solution:

u(r, θ) =∞∑n=0

Dnr−n−1Pn(cos θ).

I “Initial” condition gives:

DnL−n−1 =

〈Pn|f〉‖Pn‖2

.

Y. K. Goh



Example

I Find the solution to the Dirichlet’s problem for Laplaceequation exterior to a sphere with the initial conditionu(L, θ) = sin θ.

Y. K. Goh


Laplace Eq. interior to a sphere

A Dirichlet’s problem inside a sphere.

I PDE: Laplace Eq. ∇2u(r, θ) = 0, 0 < r < L, 0 ≤ θ < π.

I 1r2

∂∂r

(r2 ∂u∂r

)+ 1

r2 sin θ∂∂θ

(sin θ ∂u∂θ

)+ 1

r2 sin2 θ∂2u∂φ2 = 0.

I Boundary ConditionsI Periodic B.C.: u(r, θ, φ) = u(r, θ, φ+ 2π).I u(L, θ, φ) = f(θ, φ), 0 ≤ θ < π, 0 ≤ φ < 2π.

I Regularity Conditions: |u| <∞ in 0 < r < L.

Y. K. Goh



I Separation of variables: u(r, θ, φ) = R(r)Θ(θ)Φ(φ) =⇒Φ′′ + λΦ = 0, 0 ≤ φ < π

r2R′′ + 2rR′ − µR = 0, 0 < r < L

Θ′′ + cot θΘ′ + (µ− λ csc2 θ)Θ = 0, 0 ≤ θ < π.

I Periodic B.C. :u(r, θ, φ) = u(r, θ, φ+ 2π), 0 ≤ φ < 2π =⇒

I SL-problem Φ′′ + λΦ = 0, Φ(φ) = Φ(φ+ 2π)I Eigenvalues, λ = λm = m2,m = 0, 1, 2, . . . .I Eigenfunctions, Φm(φ) = Am cosmφ+Bm sinmφ; or

more often for the case of Spherical coordinate problems,the eigenfunctions are writen as Φm(φ) = eimφ.

Y. K. Goh



I ODE for Θ :Θ′′ + cot θΘ′ + (µ−m2 csc2 θ)Θ = 0, 0 ≤ θ < π.

I Could be converted to associated Legendre’s equation bychanging the variable s = cos θ.

I (1− s2)Θ′′ − 2sΘ′ +[µ− m2

1−s2

]Θ = 0

I The associated Legendre’s equation has boundedsolution only when µ = n(n+ 1).

I The bounded solution is the associated Legendrefunction of the first kind, Θmn(θ) = Pmn (cos θ).

I Combining eigenfunctions for φ and θ:I Θmn(θ)Φm(φ) ∝ Y m

n (θ, φ).I The spherical harmonics function is defined as

Y mn (θ, φ) =

√2n+14π

(n−m)!(n+m)!P

mn (cos θ)eimφ.

Y. K. Goh



I ODE for r: r2R′′ + 2rR′ − n(n+ 1)R = 0 is an Eulerequation:

I The bounded solution is R(r) ∝ rn, 0 < r < L.

I General solution:

u(r, θ, φ) =∞∑n=0

n∑m=−n

CnmrnY m

n (θ, φ).

I “Initial” condition: u(L, θ, φ) = f(θ, φ), gives thegeneralized Fourier coefficients

cnmLn =〈Y m

n |f〉‖Y m

n ‖2=

∫ 2π

0

∫ π

0

Y m∗n (θ, φ)f(θ, φ) sin θ dθ dφ

Y. K. Goh


Initial-value problem for heat flow in a sphere

Consider the Dirichlet’s problem for heat equation in a sphere

I PDE:∂u∂t

= c2(

1r2

∂∂r

(r2 ∂u

∂r

)+ 1

r2 sin θ∂∂θ

(sin θ ∂u

∂θ

)+ 1

r2 sin2 θ∂2u∂φ2

),

0 < r < L, 0 ≤ θ < π, 0 ≤ φ < 2π, t > 0.

I Boundary conditions:I periodic boundary: u(r, θ, φ, t) = u(r, θ, φ+ 2π, t).I u(L, θ, φ, t) = 0.

I Regularity condition: |u| <∞ inside the sphere.

I Initial condition: u(r, θ, φ, 0) = f(r, θ, φ).

I Separation of variables : u = T (t)R(r)Θ(θ)Φ(φ).

Y. K. Goh



I ODEsT ′ + c2λ2T = 0 t > 0

Φ′′ +m2Φ = 0 0 ≤ φ < 2π

Θ′′ + cot θΘ′ + (n(n+ 1)−m2 csc2 θ)Θ = 0, 0 ≤ θ < π

r2R′′ + 2rR′ + (λ2r2 − n(n+ 1))R = 0 0 < r < L.

I B.C. :Φ(0) = Φ(2π), =⇒ Φm(φ) = Ame

imφ,m = 0,±1,±2, . . .

|Θ| <∞ =⇒ Θmn(θ) = Pmn (cos θ), n ≥ |m|

R(L) = 0. =⇒ Rk(r) = jn (λnkr) , k = 1, 2, . . .

here, λnk = αnk/L and αnk is the kth root of nth orderspherical Bessel function, jn(x).

Y. K. Goh



I The remaining ODE for t has the solutionTkn(t) ∝ e−(cλn

k )2t

I General solution:

u(r, θ, φ, t) =∞∑k=1

∞∑n=0

n∑m=−n

Akmne−(cλn

k )2tjn(λnkr)Pmn (cos θ)eimφ

or

u(r, θ, φ, t) =∞∑k=1

∞∑n=0

n∑m=−n

akmne−(cλn

k )2tjn(λnkr)Ymn (θ, φ)

Y. K. Goh



I The generalized Fourier coefficient

akmn =

∫ L0

∫ 2π

0

∫ π0f(r, θ, φ)jn(λnkr)Y

mn (θ, φ) r2 sin θ dθ dφ dr∫ L

0

∫ 2π

0

∫ π0|jn(λnkr)Y

mn (θ, φ)|2 r2 sin θ dθ dφ dr

,

k = 1, 2, . . . , n = 0, 1, 2, . . . ,m = 0,±1,±2, . . . ,±n.

Y. K. Goh


The three-dimensional wave equation

Consider the Dirichlet’s problem for wave equation in a sphere

I PDE: ∂2u∂t2

=

c2(

1r2

∂∂r

(r2 ∂u

∂r

)+ 1

r2 sin θ∂∂θ

(sin θ ∂u

∂θ

)+ 1

r2 sin2 θ∂2u∂φ2

),

0 < r < L, 0 ≤ θ < π, 0 ≤ φ < 2π, t > 0.I Boundary conditions:

I periodic boundary: u(r, θ, φ, t) = u(r, θ, φ+ 2π, t).I u(L, θ, φ, t) = 0.

I Regularity condition: |u| <∞ inside the sphere.I Initial condition:

I u(r, θ, φ, 0) = f(r, θ, φ).I ∂u

∂t (r, θ, φ, 0) = g(r, θ, φ).I Separation of variables : u = T (t)R(r)Θ(θ)Φ(φ).

Y. K. Goh



I ODEsT ′′ + c2λ2T = 0 t > 0

Φ′′ +m2Φ = 0 0 ≤ φ < 2π

Θ′′ + cot θΘ′ + (n(n+ 1)−m2 csc2 θ)Θ = 0, 0 ≤ θ < π

r2R′′ + 2rR′ + (λ2r2 − n(n+ 1))R = 0 0 < r < L.

I B.C. :Φ(0) = Φ(2π), =⇒ Φm(φ) = Ame

imφ,m = 0,±1,±2, . . .

|Θ| <∞ =⇒ Θmn(θ) = Pmn (cos θ), n ≥ |m|

R(L) = 0. =⇒ Rk(r) = jn (λnkr) , k = 1, 2, . . .

I ODE for t has the solution T (t) = A cos cλnkt+B sin cλnkt

Y. K. Goh



I General solution: u(r, θ, φ, t) =∞∑k=1

∞∑n=0

n∑m=−n

(akmn cos cλnkt+ bkmn sin cλnkt)jn(λnkr)Ymn (θ, φ).

I Fourier coefficients:

akmn =

∫ L0

∫ 2π

0

∫ π0f(r, θ, φ)jn(λnkr)Y

mn (θ, φ) r2 sin θ dθ dφ dr∫ L

0

∫ 2π

0

∫ π0|jn(λnkr)Y


,

bkmn =

∫ L0

∫ 2π

0

∫ π0g(r, θ, φ)jn(λnkr)Y

mn (θ, φ) r2 sin θ dθ dφ dr

cλnk∫ L

0

∫ 2π

0

∫ π0|jn(λnkr)Y


,

k = 1, 2, . . . , n = 0, 1, 2, . . . ,m = 0,±1,±2, . . . ,±n.Y. K. Goh


THE END

Y. K. Goh


boundary value problems in spherical...

Documents