Chapter 5 - Partial Differential Equations & Separation of Variables

Question 1

The General Heat/Diffusion Equation

Answer 1

Ut = ΚΔU = Κ (Uxx + Uyy + Uzz), for U(x,y,z,t)
Δ = ∇²
-in the context of heat flow, U represents temperature but it could equally be used to represent the concentration of a chemical in some reaction, smoke in the air or disease in a population
-we need boundary conditions corresponding to the spatial variables and the single time derivative means we need ONE initial condition

Question 2

1D Heat Equation

Answer 2

Ut = Κ Uxx

Question 3

What could the 1D heat equation represent physically?

Answer 3

-with 00 the 1D heat equation can be used to model the evolution of temperature U(x,t) along a finite metal bar

Question 4

Temperature Evolution of a Metal Bar

Two Possible Initial Boundary Value Problems (IVBPs)

Answer 4

1) hold the ends of the bar at fixed temperature:
e.g. U(0,t) = U(L,t) = 0 , ∀t≥0
and an initial temperature profile:
U(x,0) = φ(x)
2) we could insulate the ends meaning no heat can enter or escape from the ends of the bar:
Ux(0,t) = Ux(L,t) = 0 , ∀t≥0
and again an initial temperature profile:
U(x,0) = φ(x)

Question 5

Metal Bar with Ends at Fixed Temperature

Overview

Answer 5

Ut = Κ Uxx for 00
boundary cond. U(0,t) = U(L,t) = 0
initial cond. U(x,0) = φ(x), 0≤x≤1
-steps to solve:
1) Separated Solution
2) Linear Superposition
3) Initial Condition

Question 6

Metal Bar with Ends at Fixed Temperature

Separated Solution

Answer 6

-seek a solution of the form U(x,t) = X(x)T(t) which satisfies the boundary conditions but not the initial condition:
U=XT => Ut = XTt and Uxx = XxxT
XTt = Κ XxxT
Tt/ΚT = Xxx/X
-the LHS depends only on t and the RHS depends only on x so for both sides to be equal they must be equal to a constant, λ :
dT/dt = λΚT and d²X/dx² = λX
-use the boundary conditions U(0,t)=U(L,t)=0 so X(0)=X(L)=0
-solve for Xn(x):
λ = k²>0 => trivial solution
λ = 0 => trivial solution
λ = -ω² < 0 => X(x) = c1cos(ωx) + c2sin(ωx)
λn = - n²π²/L²
Xn(x) = sin(nπx/L)
-solve for Tn(t)
Tn(t) = e^(-n²π²/L² * Κt)
-therefore we have a separated solution:
Un(x,t) = Xn(x)Tn(t) = e^(-n²π²/L² * Κt) * sin(nπx/L)

Question 7

Metal Bar with Ends at Fixed Temperature

Linear Superposition

Answer 7

-since the heat equation is linear, we can take linear combinations of the basic separated solutions:
U(x,t) = Σ bn*Un(x,t)
= Σ bn * e^(-n²π²/L² * Κt) * sin(nπx/L)
-where the sum is taken from n=0 to n=∞
-this is no longer a separated solution
-since each term satisfies boundary conditions so does U(x,t) BUT it is does not generally satisfy the initial condition

Question 8

Metal Bar with Ends at Fixed Temperature

Initial Condition

Answer 8

-imposing the initial conditions on U(x,t) enables us to fix bn:
U(x,0) = Σ bn * e^(-n²π²/L² * Κt) = φ(x)
-i.e. bn are the coefficients of the Fourier sine series of φ(x) calculated by:
bn = 2/L ∫ φ(x) sin(nπx/L) dx
-where the integral is taken between 0 and L

Question 9

Metal Bar with Insulated Ends

Overview

Answer 9

Ut = ΚUxx
boundary cond.s : Ux(0,t) = Ux(L,t) = 0
initial conditions: U(x,0) = φ(x)
-to solve:
1) Separated Solutions
2) Linear Superposition
3) Initial Conditions

Question 10

Metal Bar with Insulated Ends

Separated Solution

Answer 10

-seek a solution of the form U(x,t)=X(x)T(t) which satisfies the boundary conditions but not the initial condition
U=XT => Ut=XTt and Uxx=XxxT
Tt/ΚT = Xxx/X = λ
=> dT/dt = λΚT and d²X/dx² = λX
-boundary conditions:
Ux(0,t) = Xx(0)T(t) = Ux(L,0) = Xx(L)T(t) = 0
so, Xx(0) = Xx(L) = 0
-solve for X(x):
λ = k²>0 => trivial solution
λ = 0 => X(x) = c1 + c2*x => Xo = 1
λ = -ω² < 0 => X(x) = c1*cos(ωx) + c2*sin(ωx)
boundary conditions => λn= -n²π²/L²
Xn(x) = cos(nπx/L)
-solve for Tn(t):
Tn(t) = e^(-n²π²Κt/L²)
-separated solution:
Un(x,t) = e^(-n²π²Κt/L²) * cos(nπx/L)
-for n=0, To(t) = 1, Xo(x) = 1
so Uo(x,t) = To(t)Xo(x) = 1

Question 11

Metal Bar with Insulated Ends

Linear Superposition

Answer 11

-take linear combinations of the basic separated solutions:
U(x,t) = ao/2 + Σ an*Un(x,t)
= ao/2 + Σ an * e^(-n²π²Κt/L²) * cos(nπx/L)
-where the sum is taken from n=1 to n=∞, this is no longer a separated solution
-since each term satisfies the boundary conditions, so does function U(x,t) BUT the initial condition is not generally satisfied

Question 12

Metal Bar with Insulated Ends

Initial Condition

Answer 12

-imposing the initial condition enables us to fix the coefficients an:
U(x,0) = ao/2 + Σ an * cos(nπx/L) = φ(x)
-which means that an are the coefficients in the Fourier cosine series of φ(x) given by:
an = 2/L ∫ φ(x) cos(nπx/L)
-where the integral is taken from 0 to L

Question 13

1D Wave Equation

Answer 13

Utt = c²Uxx

• as for the metal bar we require two boundary conditions
• also as there is a second time derivative we need two initial conditions

Question 14

String Pegged at Both Ends

Overview

Answer 14

Utt = c²Uxx
boundary cond.s : U(0,t) = U(L,t) = 0
initial cond.s : U(x,0) = φ(x) & Ut(x,0) = ψ(x)
-to solve:
1) Separated Solution
2) Linear Superposition
3) Initial Conditions

Question 15

String Pegged at Both Ends

Separated Solution

Answer 15

Utt = c²Uxx
-suppose U(x,t)=X(x)T(t) satisfying the boundary conditions but not the initial conditions:
Utt = XTtt and Uxx = XxxT
so:
Ttt/c²T = Xxx/X = λ
-boundary conditions:
U(0,t) = U(L,t) = 0 => X(0) = X(L) = 0
-solve for Xn(x):
d²X/dx² = λX
λ = k² > 0 => trivial solution
λ = 0 => trivial solution
λ = -ω² < 0 => X(x) = c1*sin(ωx) + c2*cos(ωx)
so:
λn = -n²π²/L²
Xn(x) = sin(nπx/L)
-solve for Tn(t):
Tntt = -n²π²c²/L² * Tn
=> Tn(t) = an*cos(nπct/L) + bn*sin(nπct/L)
-separated solution:
Un(x,t) = sin(nπx/L) [ an*cos(nπct/L) + bn*sin(nπct/L) ]

Question 16

String Pegged at Both Ends

Linear Superposition

Answer 16

-take linear combinations of the separated solutions:

U(x,t) = Σ sin(nπx/L) [ ancos(nπct/L) + bnsin(nπct/L) ]

Question 17

String Pegged at Both Ends

Initial Conditions

Answer 17

-for U(x,0) = φ(x):
Σ an * sin(nπx/L) = φ(x)
-where an is given by the half-range Fourier series:
an = 2/L ∫ φ(x)*sin(nπx/L)

-for Ut(x,0) = ψ(x)
Σ nπc/L* bn * sin(nπx/L)
-where bn is given by:
nπc/L* bn = 2/L ∫ ψ(x)*sin(nπx/L)

-integrals between 0 and L

Question 18

Normal Modes of Vibration

Answer 18

-when ψ(x)=0, bn=0 for all n
-the general solution is built from basic functions:
Un(x,t) = sin(nπx/L)*cos(nπct/L)
n=1 -> fundamental mode
n=2 -> second normal mode
n=3 -> third normal mode
-for n≥2, the modes are the harmonics

Question 19

The Plucked String

Answer 19

-the string has some initial triangular shape, φ(x)
piecewise linear with two sections
-released from rest so ψ(x)=0

Question 20

Laplace’s Equation

Answer 20

ΔU = Uxx + Uyy + Uzz = 0
-or in 2D:
Uxx + Uyy = 0
-only involves spatial coordinates so is related to boundary value problems

Question 21

Dirichlet Boundary Value Problem

Description

Answer 21

• specifies a region Ω in the plane enclosed by the boundary Γ
• function U should satisfy ΔU=0 inside Ω with U specified on the boundary Γ
• coordinate system used depends on the shape of the region

Question 22

Dirichlet Boundary Value Problem

Rectangular Domain

Answer 22

-use Cartesian coordinates
AB: y=0, BC: x=a, DC: y=b, AD: x=0
-we require that Uxx + Uyy = 0 for 0

Question 23

Dirichlet Boundary Value

Circular Domain - Overview

Answer 23

-use cylindrical polar coordinates:
ΔU = Urr + 1/r Ur + 1/r² Uθθ = 0
-for a circle or radius 1 centred on the origin:
Urr + 1/r Ur + 1/r² Uθθ = 0 for 0

Question 24

Dirichlet Boundary Value

Circular Domain - Separated Solution

Answer 24

-let U(r,θ) = R(r)T(θ), then Laplace’s Equation is given by:
(r²Rrr + rRr)/R = - Tθθ/T = λ²

->for λ≠0:
Tθθ + λ²T = 0 => T(θ) = acos(λθ) + bsin(λθ)
-periodicity implies λ=n (a positive constant)
-for λ=n>0 we have Tn(θ) = ancos(nθ) + bnsin(nθ)
-and R(r) satisfying: r²Rrr + rR - n²R = 0:
Rn(r) = cnr^n + dnr^(-n)
-requiring our function to be bounded for 0 dn=0

-separated solution:
Un(r,θ) = r^n * (ancos(nθ) + nbsin(nθ)) for n≠0

-for λ=0:
Tθθ = 0 => T(θ) = ao + bo*θ
-periodicity => bo=0 so To = 1
-and R(r) satisfies: rRrr + Rr = 0
=> Ro = co + do*log(r)
-requiring our function to be bounded for 0 do=0
-so Uo(r,θ) = ao/2

Question 25

Dirichlet Boundary Value Problem

Circular Domain - Linear Superposition

Answer 25

U(r,θ) = ao/2 + Σ r^n * (ancos(nθ) + nbsin(nθ))

Question 26

Dirichlet Boundary Value Problem

Circular Domain - Boundary Condition

Answer 26

U(r,θ) = ao/2 + Σ r^n * (an*cos(nθ) + nb*sin(nθ))
-on the boundary r=1:
 ao/2 + Σ an*cos(nθ) + nb*sin(nθ) = f(θ)
-where:
an = 1/π ∫ f(θ)*cos(nθ) dθ
bn = 1/π ∫ f(θ)*sin(nθ) dθ

Question 27

Separation of Variables in Spherical Polars

Answer 27

-separate variables twice:
->first separation of variables:
= R(r) Y(θ,φ)
->second separation:
Y(θ,φ) = T(θ)P(φ)