Connection to Laplacian in spherical coordinates (Chapter13)

• We might often encounter the Laplace equation and sphericalcoordinates might be the most convenient

∇2u(r , θ, φ) = 0

• We already saw in Chapter 10 how to write the Laplacianoperator in spherical coordinates,

∇2φ =1

r2

∂

∂r

(r2∂u

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+

1

r2 sin2 θ

∂2u

∂φ2

• This is a partial differential equation we will solve by what willbecome a standard approach of separation of variables• We take u(r , θ, φ) = R(r)Θ(θ)Φ(φ), which we substitute in and

then multipy by r2

RΘΦ

Separation of variables for the Laplace equation

• Because of this separation we now wind up with total derivatives

1

R

d

dr

(r2 dR

dr

)+

1

Θ

1

sin θ

d

dθ

(sin θ

dΘ

dθ

)+

1

Φ

1

sin2 θ

d2Φ

dφ2= 0

• Because the first two terms do not depend on φ, we must havefrom the last term

1

Φ

d2Φ

dφ2= −m2

• The functions Φ(φ) must be periodic with period 2π, and thissuggest that m is an integer and Φ = sin mφ or Φ = cos mφ

Separation of variables for the Laplace equation, continued

• Now we can replace 1Φ

d2Φdφ2 with the constant −m2

1

R

d

dr

(r2 dR

dr

)+

1

Θ

1

sin θ

d

dθ

(sin θ

dΘ

dθ

)− m2

sin2 θ= 0

• The first term is a function of only r , and the last two terms arenow functions of only θ, so we can take

1

R

d

dr

(r2 dR

dr

)= k

• Here k is just a constant that we will later take k = l(l + 1)• Then we have the final equation for the θ-dependent terms,

1

sin θ

d

dθ

(sin θ

dΘ

dθ

)+

(k − m2

sin2 θ

)Θ = 0

Separation of variables for the Laplace equation, continued

• We solved for Φ, but we still need to solve for R and Θ• For the moment, let’s focus only on the Θ function that solves

1

sin θ

d

dθ

(sin θ

dΘ

dθ

)+

(k − m2

sin2 θ

)Θ = 0

• Make a change of variables to x = cos θ and then Θ(θ)→ y(x)• Then we have d

dθ = dxdθ

ddx = − sin θ d

dx , and 1sin2 θ

= 11−x2

• We can then obtain the associated Legendre equation

y ′′ − 2xy ′ +

[l(l + 1)− m2

1− x2

]y = 0

• So we have found the associated Legendre equation from Laplaceequation in spherical coordinates!• Hence we know Θ(θ) = Pm

l (cos θ)

Laplace equation in spherical coordinates, continued

• We will see later that R(r) = r l or R(r) = r−l−1

• Recall, we found Φ(φ) = cos(mφ) or Φ(φ) = sin(mφ)• Finally u(r , θ, φ) = R(r)Θ(θ)Φ(φ)• So our solutions are u = r lPm

l (cos θ) sin mφ,u = r lPm

l (cos θ) cos mφ, u = r−l−1Pml (cos θ) sin mφ,

u = r−l−1Pml (cos θ) cos mφ

• Superposition applies! So in general, a solution might be a linearcombination of these solutions for different l and m• Will depend on boundary conditions! For example, maybe we areinterested in solutions near r = 0 where r−l−1 diverges, then

u(r , θ, φ) =l∑

m=−l

∞∑l=0

r lPml (cos θ) [alm cos mφ+ blm sin mφ]

Steady-state temperature in a sphere

• Heat travels by diffusion, so we have

∇2T =1

α2

∂T

∂t

• The constant 1α2 depends on the properties of the medium, in

one can show that 1α2 = C

κ , where C is the specific heat capacityand κ is thermal conductivity• This can be derived from Fourier’s Law ~J = −κ~∇T (For anisotropic medium, actually the conductivity is a rank 2 tensor!)and the continuity equation ∂u

∂t = −~∇ · ~J = κ∇2T and ∂u∂t = C ∂T

∂t

• In steady state ∂T∂t = 0, so ~∇T = 0

Steady-state temperature in a sphere

• Consider a sphere of radius r = 1, with the temperature T = 100on the top half (z > 0 or 0 < θ < π/2) and T = 0 on the bottomhalf (z < 0 or π/2 < θ < π)• We know that our solution is a solution to Laplace equation∇2T = 0 most conveniently in spherical coordinates

T (r , θ, φ) =l∑

m=−l

∞∑l=0

r lPml (cos θ) [alm cos mφ+ blm sin mφ]

• Based on the problem, we see there is no φ dependence, so weonly require m = 0 and cos mφ = 1, so we simplify

T (r , θ) =∞∑l=0

cl rlPm=0

l (cos θ)

Steady-state temperature in a sphere

• Solution is a series in the Legendre polynomials! At r = 1 weknow T (r = 1, θ)

T (r = 1, θ) =∞∑l=0

clPl(cos θ)

• Find the coefficents cl !• With T (r = 1, x = cos θ) = 100f (x), we have f (x) = 0 for−1 < x < 0 and f (x) = 1 for 0 < x < 1, and then

cl = 100

(2l + 1

2

)∫ 1

−1f (x)Pl(x)dx

c0 = 100

(1

2

)∫ 1

0dx = 100

(1

2

)c1 = 100

(3

2

)∫ 1

0xdx = 100

(3

4

)

Steady-state temperature in a sphere, continued

c2 = 100

(5

2

)∫ 1

0

(3

2x2 − 1

2

)dx = 0

c3 = 100

(7

2

)∫ 1

0

(5

2x3 − 3

2x

)dx = 100

(− 7

16

)• The integral for c4 = 0... we can even see this by symmetry... sowe can go to c5

c5 = 100

(11

2

)∫ 1

0

(1

8

)(63x5 − 70x3 + 15x

)dx = 100

(11

32

)

Steady-state temperature in a sphere, continued

• Now that we have our Legendre coefficients, c0 = 100(

12

),

c1 = 100(

34

), c2 = 0, c3 = 100

(− 7

16

),c4 = 0, c5 = 100

(1132

), etc.,

we can write a series solution

T (r , θ) =∞∑l=0

cl rlPl(cos θ)

T (r , θ) = 100

[1

2+

3

4r cos θ − 7

16r3

(5

2cos3 θ − 3

2cos θ

)+ ...

]• Notice that we write T (r , θ) since we determined from thebeginning that the solution is independent of φ• It is crucial to remember that we solved this only for r < 1! Oursolution does not apply outside of this region

Laplace equation in Cartesian coordinates

• The Laplace equation is written

∇2φ = 0

• For example, let us work in two dimensions so we have to findφ(x , y) from,

∂2φ

∂x2+∂2φ

∂y2= 0

• We use the method of separation of variables and writeφ(x , y) = X (x)Y (y)

Laplace equation in Cartesian coordinates

• The Laplace equation is written

∇2φ = 0

• For example, let us work in two dimensions so we have to findφ(x , y) from,

∂2φ

∂x2+∂2φ

∂y2= 0

• We use the method of separation of variables and writeφ(x , y) = X (x)Y (y)

X ′′

X+

Y ′′

Y= 0

Laplace equation in Cartesian coordinates, continued

• Again we have two terms that only depend on one independentvariable, so

Y ′′

Y= −k2

• This is called a Helmholtz equation (we’ve seen in before), andwe can write it

Y ′′ + k2Y = 0

• The we have another equation to solve,

X ′′ − k2X = 0

• Here k is real and k ≥ 0• Could we have done this a different way? Yes!

Laplace equation in Cartesian coordiates, continued

• We could have a different sign for the constant, and then

Y ′′ − k2Y = 0

• The we have another equation to solve,

X ′′ + k2X = 0

• We will see that the choice will determine the nature of thesolutions, which in turn will depend on the boundary conditions

Steady-state temperature in a semi-infinite plate

• Imagine a metal plate bounded at y = 0 but extending to infinityin the +y direction, and from x = 0 to x = 10• Hold the y = 0 surface at T = 100, and as y → +∞, T = 0• Surface at x = 0 and x = 10 are both held at T = 0• Find the scalar temperature field T (x , y) inside the plate

• Again we solve ∇2T = 0, or ∂2T∂x2 + ∂2T

∂y2 = 0

• We apply separation of variables T (x , y) = X (x)Y (y), andchoose the sign of the constant carefully!

X ′′

X+

Y ′′

Y= 0

Steady-state temperature in semi-infinite plate, continued

• We choose the sign of the constant to give us reasonablebehavior based on the boundary conditions

X ′′ + k2X = 0

Y ′′ − k2Y = 0

• From the first ordinary differential equation, we getX (x) = sin kx and X (x) = cos kx• Since T = 0 at x = 0, the cos kx solution does not work• Boundary condition T = 0 at x = 10 means we have solutionsXn(x) = sin nπx

10 , with n = 1, 2, 3, ...

Steady-state temperature in semi-infinite plate, continued

• Now the equation for Y (y), taking care that we now havek2n = (nπ

10 )2

Y ′′ − k2nY = 0

• We get solutions Yn(y) = ekny and Yn(y) = e−kny

• Since T = 0 as y →∞, we have Yn(y) = e−kny = e−nπy/10

• We can then write the general solutions that satisfy theboundary conditions as y →∞, and at x = 0 and x = 10

T (x , y) =∞∑

n=1

bne−nπy/10 sin

nπx

10

Steady-state temperature in semi-infinite plate, continued

• To determine bn coefficients, use that T = 100 at y = 0

T (x , y = 0) =∞∑

n=1

bn sinnπx

10= 100

• A Fourier series! We find the bn from

bn =2

10

∫ 10

0100 sin

nπx

10dx = (20)

(10

nπ

)(− cos

nπx

10

)|100

• We find bn = 400nπ for odd n, and bn = 0 for even n

T =400

π

(e−πy/10 sin

πx

10+

1

3e−3πy/10 sin

3πx

10+

1

5e−πy/2 sin

πx

2+ ...

)

Time-dependent diffusion or heat flow

• If ∇2T 6= 0, then the temperature field becomes time-dependent• Imagine a strip of metal, extending to ±∞ in the y direction, butbounded at x = 0 and x = l• At t = 0, the system is in steady-state with u = 0 at x = 0, andu = 100 at x = l• The temperature profile at t = 0 satisfies ∇2u = 0, or d2u

dx2 = 0• We find initially a temperature profile u = 100 x

l• Now abruptly set u = 0 at x = 0 and x = l• The temperature profile is now time-dependent

d2u

dx2=

1

α2

du

dt

Time-dependent diffusion or heat flow, continued

• We apply separation of variables u(x , t) = T (t)X (x)

X ′′

X=

1

α2

T ′

T= −k2

• We find T = e−k2α2t

• From the boundary conditions u = 0 at x = 0 and x = l , we findX = sin nπx

l with n = 1, 2, 3, ...

u(x , t) =∞∑

n=1

bn sinnπx

le−( nπx

l)2t

• We find the bn from the initial condition u(x , t = 0) = 100 xl

Time-dependent diffusion or heat flow, initial conditions

• We can solve for the initial conditions (t = 0)

u(x , t = 0) =100

lx =

∞∑n=1

bn sinnπx

l

• Notice compared to before, when we expanded periodicfunctions, the y0(x) and v0(x) are not periodic with period l• However, we can show that the sin nπx

l make a complete,orthogonal set over the interval 0 < x < l• We find the bn then, using that∫ l

0sin

mπx

lsin

nπx

ldx =

l

2δm,n

• We find that bn = 200π

(−1)n−1

n

Time-dependent diffusion or heat flow, final answer

• Then we can finally write the series solution

u =200

π

[e−(πα/l)2t sin

πx

l− e−(2πα/l)2t sin

2πx

l+ e−(3πα/l)2t sin

3πx

l− ...

]• Notice that as t →∞, we reach equilibrium T = 0 everywhere• Another example, slightly different... final conditions involve atemperature gradient