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MATH-M 344
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Benjamin Melinand
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Chapter 10

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Mathematics

MATH-M 344

Benjamin Melinand

Spring

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M344 Section 10.5 Notes- Separation of Variables; Heat Conduction in a Rod (The Heat Equation) 3-29-17
ο· Want to study heat conduction problem for a rod of length πΏ:
o π’ π₯,π‘ = temperature at π₯ at time π‘
ο§ π₯ = space variable
ο§ π‘ = time variable
2
ο· Heat equation- partial differential equation of formπ‘π’ = πΌ π’π₯π₯
o π’π‘π₯,π‘ = lim ( π’ π₯,π‘+β βπ’(π₯,) = π (π’ π₯,π‘ ))
ββ0 β ππ‘
o π’ π₯,π‘ = lim ( π’ π₯+π,π‘ βπ’(π₯,) = π (π’ π₯,π‘ ))
π₯ πβ0 π ππ₯
o π’ (π₯,π‘ = lim ( π’π₯π₯+π,π‘ βπ’ π₯,π‘ ) = π2 (π’ π₯,π‘ ))
π₯π₯ πβ0 π ππ₯2
o πΌ = thermal diffusivity (known constant)
πππππ‘β2
ο§ Units:
π‘πππ
ο§ Depends on materiπ
l bar made of
ο§ Defined by πΌ = , where π
= thermal conductivity, π = density, π = specific heat of material
ππ
o Initially we have a temperature distribution (initial condition) given by π’ π₯,0 = π(π₯), where π(π₯)
known
( ) ( )
o Boundary conditions given- assume π’ 0,π‘ = π an1 π’ πΏ,π‘ = π for π‘2β₯ 0, where π ,π π β1 2
ο§ In this section, assume 1 = π2= 0
ο§ Initial and final temperatures
o We obtain the following system:
π’ π‘ πΌπ’ ,0π₯π₯ π₯ < πΏ,π‘ > 0
π’ 0,π‘ = 0,π‘ β₯ 0
{ ( )
π’ πΏ,π‘ = 0,π‘ β₯ 0
π’ π₯,0 = π π₯ ,0 β€ π₯ β€ πΏ
ο§ Linear system; π’ only in first power
ο§ Heat equation and boundary conditions homogeneous
ο§ For physical situations, we want unique solution
ο· Boundary conditions essential to that- give initial and final temperatures
2
ο§ Since heat equation linear, can do linear separation; if 2 functions π’,π£ sat{sfy πΌ π’ π₯π₯ ,
π£π‘= πΌ π£2 π₯π₯
then for any π,π π β, we have ππ’ + ππ£ ) = πΌ ππ’ + ππ£ ) , and ππ’ + ππ£ is also a solution of
π‘ π₯π₯
heat equation
ο· First, leave out initial condition π’ π₯,0 = π(π₯) and find a family of solutions to heat equation using
boundary conditions
π’π‘= πΌπ’ ,π₯π₯< π₯ < πΏ,π‘ > 0
o Solving {π’ 0,π‘ = 0,π‘ β₯ 0 has infinite number of solutions (which makes a family)
π’ πΏ,π‘ = 0,π‘ β₯ 0
o Since we want to find some particular solution, we seek one of form π’ π₯,π‘ = π π₯ π(π‘)
ο§ Found by Fourier
ο§ Separate variables to get 2 functions that each depend on 1 variable
o Substitute π’(π₯,π‘) into heat equation:
ο§ (π π₯ π π‘) ) = πΌ π(π₯ π π‘ ( ) )
β² π‘ 2 β²β² π₯π₯
ο§ π π₯ π π‘ = πΌ π (π₯ π π‘)
ο§ If π π₯ ,π(π‘) β 0, can divide by them
π π₯ π π‘) 2 πβ²(π₯ π(π‘)
ο§ π π₯ π π‘)= πΌ π π₯ π π‘) β² β²β²
ο§ π π‘)= πΌ 2 π (π₯)
π π‘) π π₯ )
πβ²(π₯) 1 π π‘ )
ο§ π π₯ )= πΌ π π‘ )
ο§ For equation to be valid for 0 < π₯ < πΏ,π‘ > 0, both sides must be equal to same constant;
otherwise one variable being fixed and the other varying would cause inequality
πβ²(π₯) 1 π π‘ )
ο§ There exists a constant π such that = 2 = βπ; convenient to write negative sign
β²β² π π₯) πΌ π π‘ )
π (π₯)
π π₯ ) = βπ πβ²(π₯ = βππ(π₯) π β²(π₯ + ππ π₯ = 0
ο§ { 1 π π‘) β { β² 2 β { β² 2
2 = βπ π π‘ = βπΌ ππ(π‘) π π‘ + πΌ ππ π‘ = 0
πΌ π π‘ )
o Now consider boundary conditi

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