Partial Differential Equations ITim Meagher11/4/2010

Section 2.2

Problem 3: Show that δuδt = k δ

2uδx2 +Q(u, x, t) is linear if Q = α(x, t)u+ β(x, t)

and is homogeneous if beta(x, t) = 0

δuδt = k δ

2uδx2 +Q(u, x, t)

δuδt = k δ

2uδx2 + α(x, t)u+ β(x, t)

−β(x, t) = k δ2uδx2 − δu

δt + α(x, t)uTo test Linearity replace u with c1u1 + c2u2 and see if L(c1u1 + c2u2) =L(c1u1) + (c2u2)

−β(x, t) = k δ2(c1u1+c2u2)

δx2 − δ(c1u1+c2u2)δt + α(x, t)(c1u1 + c2u2)

−β(x, t) = k δ2(c1u1)δx2 − δ(c1u1)

δt +α(x, t)(c1u1) +k δ2(c2u2)δx2 − δ(c2u2)

δt +α(x, t)(c2u2)Therefore linear, and if beta(x, t) = 0

0 = k δ2uδx2 − δu

δt + α(x, t)uThen it is homogenous.

Section 2.3

Problem 2: Consider the differential equationδ2φδx2 + λφ = 0Determine the eigenvalues (and the corresponding eigenfunctions) if φ satisfiesthe following boundary conditions. Analyze three case (λ > 0, λ = 0, λ < 0).You may assume that eigenvalues are real.

b) φ(0) = 0 and φ(1) = 0

δ2φδx2 + λφ = 0 has the characteristics equation k2 − λ = 0

λ = 0⇒ φ(x)c1x+ c2Apply boundary conditions φ(0) = 0 and φ(1) = 0φ(0) = 0⇒ c1(0) + c2 = 0⇒ c2 = 0φ(1) = 0⇒ c1(1) = 0⇒ c1 = 0This means u = 0 therefore can’t be we are looking for nonzero solutions.

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δδt [e(x, t)A∆x] = φ(x, t)A− φ(x+ ∆x)A+Q Since Q=0δδt [e(x, t)A∆x] = φ(x, t)A− φ(x+ ∆x)A therefore the A cancelδδt [e(x, t)∆x] = φ(x, t)− φ(x+ ∆x)δe(x,t)δt = δφ(x,t)

δxTo test Linearity replace u with c1u1 + c2u2 and see if L(c1u1 + c2u2) =

L(c1u1) + (c2u2) δe(x,t)δt = δφ(x,t)

δx

b) Consider the total thermal energy between x=a and x=b

δδt [

∫ bae(x, t)Adx] = φ(a, t)A− φ(b, x)A+Q Since Q=0

δδt [

∫ bae(x, t)Adx] = φ(a, t)A− φ(b, x)A∫ b

aδe(x,t)δt Adx =

∫ baδφ(x,t)δx Adx Since A is a constant∫ b

aδe(x,t)δt dx =

∫ baδφ(x,t)δx dx

Thermal energy e(x, t) = c(x)p(x)u(x, t)

φ(x, t) = −k δu(x,t)δx

cp∫ baδu(x,t)δt dx =

∫ bak δ

2u(x,t)δx2 dx Since and b are arbitrary

cp δu(x,t)δt = k δ2u(x,t)δx2

Problem 3: Derive the heat equation for a rod assuming constant thermal prop-erties with variable cross-section area A(x) assuming no sources by consideringthe total thermal energy between x=a and x=b

From the last problem we have:∫ baδe(x,t)δt A(x)dx =

∫ baδφ(x,t)δx A(x)dx Thermal

energy e(x, t) = c(x)p(x)u(x, t)

φ(x, t) = −k δu(x,t)δx

cp∫ baδu(x,t)δt A(x)dx =

∫ bak δ

2u(x,t)δx2

δA(x)δx dx Since and b are arbitrary

cp δu(x,t)δt = k δ2u(x,t)δx2

δA(x)δx

Problem 8: If u(x, t) is known, give an expression for the total thermal en-ergy contained in a rod (0 < x < L).

Thermal energy e(x, t) = c(x)p(x)u(x, t) then one must sum it over the to-tal length time it area.

Total Thermal Energy=∫ bac(x)p(x)u(x, t)Adx

Section 1.3

Problem 2: Two one-dimensional rods of different materials joined at x = x0are said to be in perfect thermal contact of the temperature is continuous at

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x = x0:u(x0, t) = u(x0+, t)and no heat energy is lost at x = x0 What mathematical equation represents thelatter condition at x = x0? Under what special condition is δu/δx continuousat x = x0?

φ(x0−, t) = φ(x0+, t) and for this will be continuous if−k− δu(x0−,t)δx = −k+ δu(x0+,t)

δx

Problem 3: consider a bath containing a fluid of specific heat cf and massdensity ρf that surrounds the end x=L of a one-dimensional rod. Suppose thatthe bath is rapidly stirred in a manner such that the bath temperature is approx-imately uniform throughout, equaling the temperature at x=L, u(L, t). Assumethe rod is thermally insulated except at it perfect thermal contact with the rod,where bath may be heated or cooled by the rod. Determine an equation for thetemperature in the bath.

Let V be the volume of the bath. Therefore the Thermal energy of the bath=V cfpfu(L, t) since the bath has the same temperature as the rod. The changeof this in time is equal to the flux acrossed the boundary.

V cfpfδu(L,t)δt = φ(L, t)A

V cfpfδu(L,t)δt = −K0

δu(x,t)δx A

Section 1.4

Problem 1: Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sourcesand boundary conditions:

a) Q = 0, u(0) = 0, u(L) = T

We have:ODE:δ2u(x,t)δx2 = 0

BC:u(0) = 0u(0) = TThe Solution to the ODE is a u(x) = c1x+ c2 then apply boundary conditions:u(0) = c1(0) + c2 = 0⇒ c2 = 0u(L) = c1L+ c2 = T ⇒ c2 = T

L

therefore u(x) = TxL

d) Q = 0, u(0) = T, δu/δx(L) = α

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We have:ODE:δ2u(x,t)δx2 = 0

BC:u(0) = Tδu/δx(L) = αThe Solution to the ODE is a u(x) = c1x+ c2 then apply boundary conditions:u(0) = c1x+ c2 = T ⇒ c2 = Tδu/δx(L) = α = c1 ⇒ c1 = αtherefore u(x) = T + αx

h) Q = 0, δu/δx(0)− u(0) + T = 0, δu/δx(L) = α

We have:ODE:δ2u(x,t)δx2 = 0

BC:δu/δx(L) = αδu/δx(0)− u(0) + T = 0The Solution to the ODE is a u(x) = c1x+ c2 then apply boundary conditions:δu/δx(L) = α = c1 ⇒ c1 = αδu/δx(0) = u(0)− T ⇒ α = α(0) + c2− T ⇒ c2 = T + αtherefore u(x) = T + α(x+ 1)

Problem 2: Consider the equilibrium temperature distribution for a uniformone-dimensional rod with Q/K0 = x of thermal energy, subject to the bound-ary conditions u(0)=0 and u(L)=0.a) Determine the heat energy generated per unit time inside the entire rod.

Sum the energy generated over the whole rod.

Energy produce per unit time=∫ L0Qdx

Energy produce per unit time=∫ L0K0xdx

Energy produce per unit time=K0L2

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b) Determine the heat energy flowing out of the rod per unit time at x=0and at x=L.

φ(x, t) = −K0δu(x,t)δx and

We have:ODE:δ2u(x,t)δx2 = K0(x)

BC:

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u(0) = 0u(0) = 0The Solution to the ODE is a u(x) = K0

6 x3 + c1x + c2 then apply boundary

conditions:u(0) = c0(0) + c2 = 0⇒ c2 = 0u(L) = c1L+ c2 = 0⇒ c2 = 0therefore u(x) = K0

6 x3

So φ(x, t) = −K0δu(x,t)δx = K0x

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2c) What relationships should exist between the answers in part (a) and part (b).

These quantities should be equal, which is the case here, for the heat in therod should flow out of the rod at unless the rod heats up to infinity, which isimpossible.

Problem 4: If both ends of a rod are insulated, derive from the partial dif-ferential equation that the total thermal energy in the rod is constant.

δδt [

∫ L0e(x, t)Adx] = φ(0, t)A− φ(L, x)A+Q Since Q=0

δδt [

∫ L0e(x, t)Adx] = φ(0, t)A− φ(L, x)A+Q Since φ(0, t) = 0 and φ(0, t) = 0

δδt [

∫ L0e(x, t)Adx] = 0

δe(x,t)δt = 0

e(x, t) = K for some constant K.

Problem 11: Suppose δuδt = δ2u

δx2 + x, u(x, 0) = f(x), δuδx (0, t) = β, δuδx (L, t) = 7.a) calculate the total thermal energy in the one-dimensional rod (as a functionof time).

δδt [

∫ L0e(x, t)Adx] = φ(0, t)A− φ(L, x)A+Q Since in this case Q=x we have

δδt [

∫ L0e(x, t)Adx] = φ(0, t)A− φ(L, x)A+ x A is constant

δδt [

∫ L0e(x, t)dx] = φ(0, t)− φ(L, x) + x Plug in boundary conditions.

δδt [

∫ L0e(x, t)dx] = β − 7 + x∫ L

0e(x, t)dx = (β − 7 + x)t+ C

use other boundary condition.∫ L0e(x, t)dx = (β − 7 + x)t+ f(x)

b) From part (a), determine a value of β for which an equilibrium exists. Forthis value of β, determine limt→∞u(x, t).

Let β = 7 − x therefore solution don’t depend on time, there fore it will hita equilibrium.Section 1.5

Problem 2: For conduction of thermal energy, the heat flux vector isφ =

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−K0∇u. If in addition the molecules move at an average velocity V, a pro-cess called convection, then briefly explain why φ = −K0∇u+ cpuV. Derive thecorresponding equation for heat flow, including both conduction and convectionof thermal energy.

cp δu(x,t)δt = ∇ · φ+Q since no Q

cp δu(x,t)δt = ∇ · φ φ = −K0∇u+ cpuV.

cp δu(x,t)δt = ∇ · (−K0∇u+ cpuV )

cp δu(x,t)δt = K0∇2u− cpu · ∇V − cp∇u · V )δu(x,t)δt = k∇2u− u · ∇V −∇u · V )

δu(x,t)δt +−u · ∇V = k∇2u−∇u · V )

Problem 5: Assume that the temperature is circularly symmetric: u=u(r,t),where r2 = x2 + y2. We will derive the heat equation for this problem. Con-sider any circular annulus a ≤ r ≤ b.a) Show that the total heat energy is 2π

∫ bacpurdr.

heatenergy =∫ ba

∫ π−π cpurdθdr

heatenergy =∫ π−π dθ

∫ bacpurdr

heatenergy = 2π∫ bacpurdr

b) Show that the flow of heat energy per unit time out of the annulus at r=b is−2πK0δu/δr|r=b. A similar results holds at r=a.

φ = −K0circumferenceδu(x,t)δt

circumference = 2πr

φ = −K02πr δu(x,t)δt

c) Use parts (a) and (b) to derive the circularly symmetric heat equation with-out sources:

δδt2π

∫ bacpurdr = − δφδr

δδt2π

∫ bacpurdr = − δ

δr (−K02πr δu(x,t)δr )δδt

∫ baurdr = k δ

δr ( δu(x,t)δr )

r δδtu = k δδr ( δu(x,t)δr )

δuδt = k

rδδr (r δφδr )

Problem 9: Determine the equilibrium temperature distribution inside a cir-cular annulus (r1 ≤ r ≤ r2) :a) If the outer radius is at temperature T2 and the inner at T1

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We have:ODE:krδδr (r δφδr )=0

BC:u(r1) = T1u(r2) = T2The Solution to the ODE is a u = c1ln(r) + c2 then apply boundary conditions:u(r1) = T1 ⇒ c1ln(r1) + c2 = T1 ⇒ c2 = c1ln(r1)− T1u(r2) = T2 ⇒ c1ln(r2) + c2 = T2 ⇒ c1ln(r2) + c1ln(r1) − T1 = T2 ⇒ c1 =

T2+T1

ln(r2)−ln(r1)c2 = T2+T1

ln(r2)−ln(r1) ln(r1)− T1u(x) = T2+T1

ln(r2)−ln(r1) ln(r) + T2+T1

ln(r2)−ln(r1) ln(r1)− T1

b) If the outer radius is insulated and the inner radius is at temperature T1.

We have:ODE:krδδr (r δφδr ) = 0

BC:δu(r1)δr = 0

u(r2) = T1The Solution to the ODE is a u = c1ln(r) + c2 then apply boundary conditions:δu(r1)δr = 0⇒ c1 = 0

u(r2) = T1 ⇒ c2 = T2u(r) = T1

Problem 15: Derive the heat equation in three dimension assuming constantthermal properties and no sources.

heatenergy =∫ ∫ ∫

cpudVδδt

∫ ∫ ∫cpudV = −

∫ ∫φ · nds apply the divergence theorem

δδt

∫ ∫ ∫cpudV = −

∫ ∫ ∫∇ · φdV apply the Fourier’s law of heat conduction.

φ = −K0∇uδδt

∫ ∫ ∫cpudV = −

∫ ∫ ∫∇ · −K0∇udV

δδtcpu = K0∇2udVδδtu = k∇2udV

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