Partial Differential Equations - University of Pittsburghjrm152/ODEs/Presentation7.ppt · PPT file · Web view2016-07-24 · Substitute into the partial differential equation,

Partial Differential Equations

Example A thin iron rod, insulated at the sides, of length , is at at time . At this time, blocks of ice are

held to both ends of the rod. (The temperature of the ice is ). When will the center of the rod reach ?

Solution Temperature depends on two things: position and time.

𝑡

𝑇=𝑇 (𝑥 ,𝑡 )To understand how temperature evolves, we have to zoom in to a short segment of the bar from to

𝑡+Δ𝑡 And consider a short segment of time to Heat transfer from left to center: Heat transfer from center to right:

Change in energy is proportional to heat transfer: Energy is proportional to mass times temperature:𝐸∝𝑚𝑇⟹ Δ 𝐸∝𝑚 Δ𝑇

Putting these together:Δ 𝐸𝑄𝐿−𝑄𝑅=¿𝑘 Δ𝑡 ( (𝑇 𝐿−𝑇 )− (𝑇 −𝑇 𝑅) )=¿ ¿𝑚𝑐 Δ𝑇¿ 𝜌 Δ𝑥 Δ𝑇

𝑘 Δ𝑡 ( (𝑇 (𝑥− Δ 𝑥 ,𝑡 )−𝑇 (𝑥 , 𝑡 ) )− (𝑇 (𝑥 , 𝑡 )−𝑇 (𝑥+Δ𝑥 , 𝑡 ) ) )¿ 𝜌𝑐 Δ 𝑥 (𝑇 (𝑥 , 𝑡+Δ 𝑡 )−𝑇 (𝑥 ,𝑡 ) )

𝑘 Δ𝑡 (− 𝜕𝑇𝜕 𝑥 (𝑥−Δ𝑥 , 𝑡 ) Δ𝑥+𝜕𝑇𝜕 𝑥 (𝑥 , 𝑡 )Δ𝑥 )¿ 𝜌𝑐 Δ 𝑥 (𝑇 (𝑥 , 𝑡+Δ 𝑡 )−𝑇 (𝑥 ,𝑡 ) )

𝑘 𝜕2𝑇𝜕𝑥2

(𝑥− Δ𝑥 , 𝑡 ) Δ𝑥 Δ𝑡¿ 𝜌𝑐 𝜕𝑇𝜕𝑡 (𝑥 , 𝑡 ) Δ𝑥 Δ𝑡

Example A thin iron rod, insulated at the sides, of length , is at at time . At this time, blocks of ice are

held to both ends of the rod. (The temperature of the ice is ). When will the center of the rod reach ?

Solution

𝜕2𝑇𝜕𝑥2

(𝑥− Δ𝑥 ,𝑡 )= 𝜌𝑐𝑘

𝜕𝑇𝜕𝑡

(𝑥 , 𝑡 )Let

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡 )

The heat equation in one dimension

𝑘 𝜕2𝑇𝜕𝑥2

(𝑥− Δ𝑥 , 𝑡 ) Δ𝑥 Δ𝑡¿ 𝜌𝑐 𝜕𝑇𝜕𝑡 (𝑥 , 𝑡 ) Δ𝑥 Δ𝑡

Our analysis shows that the constant depends on density, heat capacity, and conductivity , , respectively, of the material.

is called the thermal diffusivity of the material. For iron, .

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

Setting up a PDE initial value problem

The rod started at : 𝑇 (𝑥 ,0 )=100The ice will cool the ends to zero: 𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

Some differences with partial differential equations:1. An unknown function of two or more independent variables.

Compare to systems of ordinary differential equations: several unknown functions of one independent variable, e.g.

2. Instead of initial values, we have “boundary values”

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

Setting up a PDE initial value problem

The rod started at : 𝑇 (𝑥 ,0 )=100The ice will cool the ends to zero: 𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

Some differences with partial differential equations:1. An unknown function of two or more independent variables.

Compare to systems of ordinary differential equations: several unknown functions of one independent variable, e.g.

2. Instead of initial values, we have “boundary values”

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

Setting up a PDE initial value problem 𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

Solving a PDE initial (boundary) value problemSolution

Idea 1 Try to find solutions of the form Substitute into the partial differential equation, noting and

𝑓 ′ ′ (𝑥 )𝑔 (𝑡 )= 1𝛼 𝑓 (𝑥 )𝑔′ (𝑡 )

Rearrange:𝑓 ′ ′ (𝑥 )𝑓 (𝑥 )

=1𝛼𝑔 ′ (𝑡 )𝑔 (𝑡 )

Brilliant Observation (Fourier)The left side is a function of and the right side is a function of . They must be equal for all and .

Therefore, they are both constant.

− 𝜆=¿

𝑓 ′ ′ (𝑥 )=− 𝜆 𝑓 (𝑥 ) 𝑔 ′ (𝑡 )=− 𝜆𝛼𝑔 (𝑡 )Separation of Variables

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0


Idea 1 Try to find solutions of the form Substitute into the partial differential equation, noting and

𝑓 ′ ′ (𝑥 )𝑔 (𝑡 )= 1𝛼 𝑓 (𝑥 )𝑔′ (𝑡 )

Rearrange:𝑓 ′ ′ (𝑥 )𝑓 (𝑥 )

=1𝛼𝑔 ′ (𝑡 )𝑔 (𝑡 )

Brilliant Observation (Fourier)The left side is a function of and the right side is a function of . They must be equal for all and .

Therefore, they are both constant.

− 𝜆=¿

𝑓 ′ ′ (𝑥 )=− 𝜆 𝑓 (𝑥 ) 𝑔 ′ (𝑡 )=− 𝜆𝛼𝑔 (𝑡 )Separation of Variables

𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

𝑓 (0 )𝑔 (𝑡 )= 𝑓 (2 )𝑔 (𝑡 )=0


Idea 1 Try to find solutions of the form

𝑓 ′ ′ (𝑥 )=− 𝜆 𝑓 (𝑥 ) 𝑔 ′ (𝑡 )=− 𝜆𝛼𝑔 (𝑡 )𝑓 (0 )𝑔 (𝑡 )= 𝑓 (2 )𝑔 (𝑡 )=0

𝑦 (0 )=𝑦 (2)=0Let 𝑦 ′ ′+𝜆 𝑦=0

Solve the characteristic equation

𝜆<0

𝑟2−𝜂 2=0𝜆=−𝜂 2

𝑟=±𝜂𝑦=𝐶1𝑒𝜂 𝑥+𝐶2𝑒−𝜂𝑥

𝜆=0

𝑦=𝐶1+𝐶2𝑥

𝜆>0𝜆=𝜂 2

𝑟2+𝜂 2=0𝑟=±𝜂 𝑖

𝑦=𝐶1 cos (𝜂 𝑥 )+𝐶2sin (𝜂 𝑥 )Initial Conditions

𝐶1=0𝜂𝑒2𝜂𝐶1−𝜂𝑒−2𝜂𝐶2=0

𝐶1=𝐶2=0No Non-Trivial Solutions

𝐶1=0𝐶1+2𝐶2=0𝐶1=𝐶2=0

No Non-Trivial Solutions

𝐶1=0𝐶2 sin (2𝜂 )=0

2𝜂=𝑛𝜋 ,𝑛∈ℕ𝑓 𝑛 (𝑥 )=sin (𝑛𝜋2 𝑥) 𝜆=

𝑛2𝜋 2

4

𝑔 ′ (𝑡 )=− 𝑛2 𝜋 2𝛼4

𝑔 (𝑡 )

Let

𝑑𝑦𝑑𝑡 =− 𝑛

2𝜋 2𝛼4

𝑦

𝑑𝑦𝑦 =− 𝑛

2𝜋 2𝛼4

𝑑𝑡

ln ( 𝑦 )=− 𝑛2𝜋 2𝛼4

𝑡

𝑦=𝑒−𝑛

2𝜋2𝛼4

𝑡

𝑔𝑘 (𝑡 )=𝑒−𝑛

2𝜋2 𝛼4

𝑡

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0


Idea 1 Try to find solutions of the form

𝑓 𝑛 (𝑥 )=sin (𝑛𝜋2 𝑥) 𝑔𝑛 (𝑡 )=𝑒− 𝑛

2 𝜋2𝛼4

𝑡

We didn’t find one solution. We found an infinite series of solutions.

𝑇 𝑛 (𝑥 ,𝑡 )= 𝑓 𝑛 (𝑥 )𝑔𝑛 (𝑡 )

𝑇 𝑛 (𝑥 ,𝑡 )=sin(𝑛𝜋2 𝑥 )𝑒−𝑛 2𝜋 2𝛼4

𝑡,𝑛∈ℕ

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

They all satisfy (check it!)But they do not satisfy .

Idea 2 Superposition: any linear combination will also satisfy our differential equation. Let’s find a linear combination that satisfies the boundary condition

𝑇 (𝑥 ,𝑡 )=𝑏1𝑇 1+𝑏2𝑇2+𝑏3𝑇 3+…

𝑇 (𝑥 , 𝑡)=𝑏1sin ( 𝜋2 𝑥 )𝑒−𝜋2𝛼4

𝑡+𝑏2 sin(2𝜋2 𝑥)𝑒−

4 𝜋2𝛼4 +𝑏3sin ( 3𝜋2 𝑥 )𝑒−

9𝜋2 𝛼4

𝑡+…



𝑇 (𝑥 , 𝑡)=𝑏1sin ( 𝜋2 𝑥 )𝑒−𝜋2𝛼4

𝑡+𝑏2 sin(2𝜋2 𝑥)𝑒−

4 𝜋2𝛼4 +𝑏3sin ( 3𝜋2 𝑥 )𝑒−

9𝜋2 𝛼4

𝑡+…

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0

¿∑𝑛=1

∞

𝑏𝑛 sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4

𝑡

We want this to satisfy the boundary condition . Let’s write it.

𝑇 (𝑥 ,0 )=100¿∑𝑛=1

∞

𝑏𝑛sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4

( 0)

100=∑𝑛=1

∞

𝑏𝑛sin(𝑛𝜋2 𝑥)

𝑇 (𝑥 ,𝑡 ) ¿∑𝑛=1

∞


𝑡

– periodic Fourier Series Formula

𝑓 (𝑥 )=𝑎02

+∑𝑛=1

∞

𝑎𝑛cos ( 2𝜋𝑛𝑥𝐿 )+𝑏𝑛 sin( 2𝜋𝑛𝑥𝐿 )𝑎𝑛=

2𝐿∫0

𝐿

𝑓 (𝑥 ) cos ( 2𝜋𝑛𝑥𝐿 )𝑑𝑥 𝑏𝑛=2𝐿∫0

𝐿

𝑓 (𝑥 ) sin( 2𝜋𝑛𝑥𝐿 )𝑑𝑥

𝑇 (𝑥 ,𝑡 )¿∑𝑛=1

∞


𝑡

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0



100=∑𝑛=1

∞

𝑏𝑛sin(𝑛𝜋2 𝑥)Our sine series has to add up to on the interval in order to satisfy the boundary condition . But our sine series will add up to an odd -periodic function. The only such function is the square wave :

And the coefficients must be the Fourier coefficients of .

𝑏𝑛=24∫0

4

𝑓 (𝑥 )sin ( 2𝜋 𝑛𝑥4 )𝑑𝑥𝑇 (𝑥 ,𝑡 )

𝜕2𝑇𝜕𝑥2

(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)

𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0


And the coefficients must be the Fourier coefficients of .

¿∑𝑛=1

∞


𝑡

{𝑏𝑛}=400𝜋 ,0 , 4003𝜋 ,0 ,4005𝜋 ,0 ,…

𝛼=2.3×10−5

Define the partial sum:

This is our solution. Let’s plot the partial sums until it looks “right”.

See lab for explanation of instead of . (important)

When will the rod be safe to touch at the center?

𝑏𝑛=24∫0

4

𝑓 (𝑥 )sin ( 2𝜋 𝑛𝑥4 )𝑑𝑥𝑇 (𝑥 ,𝑡 )¿∑

𝑛=1

∞

𝑏𝑛 sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4

𝑡

𝛼=2.3×10−5

4𝑎𝑛𝑑 𝑎h𝑎𝑙𝑓 h𝑜𝑢𝑟𝑠 .

Example An insulated golden rod has temperature distribution as pictured at time .

𝑥=0 𝑥=6𝑇=0 𝑇=100 𝑇=0

Initial temperature

At this time, ice blocks are held to both ends.Find the temperature . Use for gold.

SolutionThe first step is to write the problem as PDE/Boundary Value ProblemWe still have the heat equation

𝑇 𝑥𝑥=1𝛼 𝑇 𝑡

𝑇 ( 0 ,𝑡 )=𝑇 (6 , 𝑡 )=0𝑇 (𝑥 ,0 )= 𝑓 (𝑥 )


𝑥=0 𝑥=6𝑇=0 𝑇=100 𝑇=0

Initial temperature


SolutionThe first step is to write the problem as PDE/Boundary Value ProblemWe still have the heat equation


𝑇 ( 0 ,𝑡 )=𝑇 (6 , 𝑡 )=0𝑇 (𝑥 ,0 )= 𝑓 (𝑥 )

Initial temperature

Solution


𝑇 (0 ,𝑡 )=𝑇 (6 , 𝑡 )=0𝑇 (𝑥 ,0 )= 𝑓 (𝑥 )

We attempt to find solutions of the form becomes:

𝑓 ′ ′ (𝑥 )𝑔 (𝑡 )= 1𝛼 𝑓 (𝑥 )𝑔′ (𝑡 )

𝑓 ′ ′ (𝑥 )𝑓 (𝑥 )

=1𝛼𝑔 ′ (𝑡 )𝑔 (𝑡 )

The LHS is a function of ; the right is a function of . So both sides must be constant.

− 𝜆=¿

𝑓 ′ ′ (𝑥 )+𝜆 𝑓 (𝑥 )=0 𝑔 ′ (𝑡 )+𝜆𝛼𝑔 (𝑡 )=0

𝑓 ′ ′ (𝑥 )+𝜆 𝑓 (𝑥 )=0 𝑇 (0 ,𝑡 )=𝑇 (6 , 𝑡 )=0

If ,

𝑓 ′ ′−𝜂 2 𝑓 =0

𝑓 (0 )𝑔 (𝑡 )= 𝑓 (6 )𝑔 (𝑡 )=0𝑓 (0 )= 𝑓 (6 )=0

𝑟2−𝜂 2=0𝑟=±𝜂

𝑓 (𝑥 )=𝐶1𝑒𝜂 𝑥+𝐶2𝑒−𝜂 𝑥

𝐶1=𝐶2=0

𝜆=0

𝑓 ′ ′=0

𝑟=0,0𝑓 (𝑥 )=𝐶1𝑥+𝐶2

𝑟2=0

𝐶1=𝐶2=0No non-trivial solutions No non-trivial solutions

If ,

𝑓 ′ ′+𝜂2 𝑓 =0𝑟2+𝜂 2=0

𝑟=±𝜂 𝑖𝑓 (𝑥 )=𝐶1 cos (𝜂𝑥 )+𝐶2 sin (𝜂 𝑥 )0=𝐶1

0=𝐶2sin (6𝜂 )

No non-trivial solutions unless

Non-trivial solutions only occur if , in which case

Initial temperature

Solution


𝑇 (0 ,𝑡 )=𝑇 (6 , 𝑡 )=0𝑇 (𝑥 ,0 )= 𝑓 (𝑥 )

We attempt to find solutions of the form becomes:

𝑓 ′ ′ (𝑥 )𝑔 (𝑡 )= 1𝛼 𝑓 (𝑥 )𝑔′ (𝑡 )

𝑓 ′ ′ (𝑥 )𝑓 (𝑥 )

=1𝛼𝑔 ′ (𝑡 )𝑔 (𝑡 )

The LHS is a function of ; the right is a function of . So both sides must be constant.

− 𝜆=¿

𝑓 ′ ′ (𝑥 )+𝜆 𝑓 (𝑥 )=0 𝑔 ′ (𝑡 )+𝜆𝛼𝑔 (𝑡 )=0

𝜆𝑛=𝑛2 𝜋 2

36, 𝑓 𝑛 (𝑥 )=sin (𝑛𝜋6 𝑥 ) 𝑔𝑛

′ (𝑡 )+𝛼𝑛2 𝜋236

𝑔𝑛 (𝑡 )=0

𝑔𝑛 (𝑡 )=exp(−𝛼𝑛2𝜋 2

36𝑡)

𝑇 𝑛 (𝑥 ,𝑡 )= 𝑓 𝑛 (𝑥 )𝑔𝑛 (𝑡 )=sin (𝑛𝜋6 𝑥)exp (−𝛼𝑛2 𝜋 2

36𝑡 )

As before, we now find a linear combination

𝑇 (𝑥 ,𝑡 )=𝑏1𝑇 1+𝑏2𝑇2+𝑏3𝑇 3+…¿∑𝑛=1

∞

𝑏𝑛sin(𝑛𝜋6 𝑥)exp (− 𝛼𝑛2 𝜋236

𝑡)

𝑇 (𝑥 ,𝑡 )

Solution


𝑇 (0 ,𝑡 )=𝑇 (6 , 𝑡 )=0𝑇 (𝑥 ,0 )= 𝑓 (𝑥 )

¿∑𝑛=1

∞

𝑏𝑛 sin(𝑛𝜋6 𝑥)exp (− 𝛼𝑛2 𝜋236

𝑡)𝑇 (𝑥 ,𝑡 )

This solution will satisfy the PDE and .

We need it to also satisfy

𝑓 (𝑥 )=𝑇 (𝑥 ,0 )=∑𝑛=1

∞

𝑏𝑛sin(𝑛𝜋6 𝑥) For

However, this is the Fourier Series for an odd -perioid function.

Desired:

𝑓 (𝑥 )=¿𝑏𝑛=

212∫0

12

𝑓 (𝑥 ) sin(𝑛𝜋6 𝑥)𝑑𝑥

Solution

¿∑𝑛=1

∞


𝑡)𝑇 (𝑥 ,𝑡 ) 𝑏𝑛=212∫0

12

𝑓 (𝑥 ) sin(𝑛𝜋6 𝑥)𝑑𝑥

𝑇 (𝑥 ,𝑡 )≈ 800𝜋 2 sin( 𝜋6 𝑥)exp (−𝛼 𝜋

2

36𝑡)

− 8009𝜋 2 sin( 𝜋 𝑥2 )exp (− 9𝜋

2𝛼36

𝑡)+…


𝑥=0 𝑥=6𝑇=0 𝑇=100 𝑇=0


Solution

¿∑𝑛=1

∞


𝑡)𝑇 (𝑥 ,𝑡 ) 𝑏𝑛=212∫0

12

𝑓 (𝑥 ) sin(𝑛𝜋6 𝑥)𝑑𝑥For example, we can ask the question, when is the center of the rod ?

𝑇 (3 ,𝑡 )=50 , solve for 𝑡Almost four hours.

We only used 15 terms. Since this is only an approximation of an infinite sum, it’s a good idea to check that our answer is reasonably accurate.

Since using 15 more terms gave us the same answer to within a tenth of a second, we conclude the answer we have is accurate—we’ve used plenty of terms.

Interestingly, if we use just one term in the series we get a decent approximation of the correct answer, off by less than two minutes out of nearly four hours. Fourier series tend to converge fast—this is immensely useful.

Documents

Partial Differential Equations - University of Pittsburghjrm152/ODEs/Presentation7.ppt · PPT file · Web view2016-07-24 · Substitute into the partial differential equation,