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Transcript of Partial Differential Equations - University of Pittsburghjrm152/ODEs/Presentation7.ppt · PPT...
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
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
𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0
𝑓 (0 )𝑔 (𝑡 )= 𝑓 (2 )𝑔 (𝑡 )=0
Solving a PDE initial (boundary) value problemSolution
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
Solving a PDE initial (boundary) value problemSolution
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+…
Solving a PDE initial (boundary) value problemSolution
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)=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
∞
𝑏𝑛sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4
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
∞
𝑏𝑛sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4
𝑡
𝜕2𝑇𝜕𝑥2
(𝑥 ,𝑡 )= 1𝛼𝜕𝑇𝜕𝑡 (𝑥 ,𝑡)
𝑇 (𝑥 ,0 )=100𝑇 (0 ,𝑡 )=𝑇 (2 ,𝑡 )=0
Solving a PDE initial (boundary) value problemSolution
Idea 2 Superposition: any linear combination will also satisfy our differential equation. Let’s find a linear combination that satisfies the boundary condition
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
Solving a PDE initial (boundary) value problemSolution
And the coefficients must be the Fourier coefficients of .
¿∑𝑛=1
∞
𝑏𝑛sin(𝑛𝜋2 𝑥)𝑒−𝑛2 𝜋2𝛼4
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 )
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 )
Initial temperature
Solution
𝑇 𝑥𝑥=1𝛼 𝑇 𝑡
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 𝑥𝑥=1𝛼 𝑇 𝑡
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 𝑥𝑥=1𝛼 𝑇 𝑡
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
∞
𝑏𝑛 sin(𝑛𝜋6 𝑥)exp (− 𝛼𝑛2 𝜋236
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)+…
Example An insulated golden rod has temperature distribution as pictured at time .
𝑥=0 𝑥=6𝑇=0 𝑇=100 𝑇=0
At this time, ice blocks are held to both ends.Find the temperature . Use for gold.
Solution
¿∑𝑛=1
∞
𝑏𝑛 sin(𝑛𝜋6 𝑥)exp (− 𝛼𝑛2 𝜋236
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.