Chapter 5 - Partial Differential Equations & Separation of Variables Flashcards
The General Heat/Diffusion Equation
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
1D Heat Equation
Ut = Κ Uxx
What could the 1D heat equation represent physically?
-with 00 the 1D heat equation can be used to model the evolution of temperature U(x,t) along a finite metal bar
Temperature Evolution of a Metal Bar
Two Possible Initial Boundary Value Problems (IVBPs)
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)
Metal Bar with Ends at Fixed Temperature
Overview
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
Metal Bar with Ends at Fixed Temperature
Separated Solution
-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)
Metal Bar with Ends at Fixed Temperature
Linear Superposition
-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
Metal Bar with Ends at Fixed Temperature
Initial Condition
-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
Metal Bar with Insulated Ends
Overview
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
Metal Bar with Insulated Ends
Separated Solution
-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
Metal Bar with Insulated Ends
Linear Superposition
-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
Metal Bar with Insulated Ends
Initial Condition
-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
1D Wave Equation
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
String Pegged at Both Ends
Overview
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
String Pegged at Both Ends
Separated Solution
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) ]
String Pegged at Both Ends
Linear Superposition
-take linear combinations of the separated solutions:
U(x,t) = Σ sin(nπx/L) [ ancos(nπct/L) + bnsin(nπct/L) ]
String Pegged at Both Ends
Initial Conditions
-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
Normal Modes of Vibration
-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
The Plucked String
-the string has some initial triangular shape, φ(x)
piecewise linear with two sections
-released from rest so ψ(x)=0
Laplace’s Equation
ΔU = Uxx + Uyy + Uzz = 0
-or in 2D:
Uxx + Uyy = 0
-only involves spatial coordinates so is related to boundary value problems
Dirichlet Boundary Value Problem
Description
- 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
Dirichlet Boundary Value Problem
Rectangular Domain
-use Cartesian coordinates
AB: y=0, BC: x=a, DC: y=b, AD: x=0
-we require that Uxx + Uyy = 0 for 0
Dirichlet Boundary Value
Circular Domain - Overview
-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
Dirichlet Boundary Value
Circular Domain - Separated Solution
-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
Dirichlet Boundary Value Problem
Circular Domain - Linear Superposition
U(r,θ) = ao/2 + Σ r^n * (ancos(nθ) + nbsin(nθ))
Dirichlet Boundary Value Problem
Circular Domain - Boundary Condition
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
Separation of Variables in Spherical Polars
-separate variables twice:
->first separation of variables:
= R(r) Y(θ,φ)
->second separation:
Y(θ,φ) = T(θ)P(φ)
