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Chapter 10

# MATH-M 344 Chapter 10: M344 10.5 Notes (Mar. 29) Premium

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Department
Mathematics
Course
MATH-M 344
Professor
Benjamin Melinand
Semester
Spring

Description
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 πΏ: 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’ = πΌ π’π₯π₯ o π’π‘π₯,π‘ = lim ( π’ π₯,π‘+β βπ’(π₯,) = π (π’ π₯,π‘ )) ββ0 β ππ‘ o π’ π₯,π‘ = lim ( π’ π₯+π,π‘ βπ’(π₯,) = π (π’ π₯,π‘ )) π₯ πβ0 π ππ₯ o π’ (π₯,π‘ = lim ( π’π₯π₯+π,π‘ βπ’ π₯,π‘ ) = π2 (π’ π₯,π‘ )) π₯π₯ πβ0 π ππ₯2 o πΌ = thermal diffusivity (known constant) πππππ‘β2 ο§ Units: π‘πππ ο§ Depends on materiπl bar made of ο§ Defined by πΌ = , where π = thermal conductivity, π = density, π  = specific heat of material ππ  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’ πΏ,π‘ = π for π‘2β₯ 0, where π ,π π β1 2 ο§ In this section, assume 1 = π2= 0 ο§ Initial and final temperatures o We obtain the following system: π’ π‘ πΌπ’ ,0π₯π₯ π₯ < πΏ,π‘ > 0 π’ 0,π‘ = 0,π‘ β₯ 0 { ( ) π’ πΏ,π‘ = 0,π‘ β₯ 0 π’ π₯,0 = π π₯ ,0 β€ π₯ β€ πΏ 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 πΌ π’ π₯π₯ , π£π‘= πΌ π£2 π₯π₯ then for any π,π π β, we have ππ’ + ππ£ ) = πΌ ππ’ + ππ£ ) , and ππ’ + ππ£ 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‘= πΌπ’ ,π₯π₯< π₯ < πΏ,π‘ > 0 o Solving {π’ 0,π‘ = 0,π‘ β₯ 0 has infinite number of solutions (which makes a family) π’ πΏ,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‘) ) = πΌ π(π₯ π π‘ ( ) ) β² π‘ 2 β²β² π₯π₯ ο§ π π₯ π π‘ = πΌ π (π₯ π π‘) ο§ If π π₯ ,π(π‘) β  0, can divide by them π π₯ π π‘) 2 πβ²(π₯ π(π‘) ο§ π π₯ π π‘)= πΌ π π₯ π π‘) β² β²β² ο§ π π‘)= πΌ 2 π (π₯) π π‘) π π₯ ) πβ²(π₯) 1 π π‘ ) ο§ π π₯ )= πΌ π π‘ ) ο§ For equation to be valid for 0 < π₯ < πΏ,π‘ > 0, both sides must be equal to same constant; otherwise one variable being fixed and the other varying would cause inequality πβ²(π₯) 1 π π‘ ) ο§ There exists a constant π such that = 2 = βπ; convenient to write negative sign β²β² π π₯) πΌ π π‘ ) π (π₯) π π₯ ) = βπ πβ²(π₯ = βππ(π₯) π β²(π₯ + ππ π₯ = 0 ο§ { 1 π π‘) β { β² 2 β { β² 2 2 = βπ π π‘ = βπΌ ππ(π‘) π π‘ + πΌ ππ π‘ = 0 πΌ π π‘ ) o Now consider boundary conditi
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