MATH-M 344 Chapter 10: M344 10.5 Notes (Mar. 29)
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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 𝐿: o 𝑒 π‘₯,𝑑 = temperature at π‘₯ at time 𝑑  π‘₯ = space variable  𝑑 = time variable 2 ο‚· Heat equation- partial differential equation of form𝑑𝑒 = 𝛼 𝑒π‘₯π‘₯ o 𝑒𝑑π‘₯,𝑑 = lim ( 𝑒 π‘₯,𝑑+β„Ž βˆ’π‘’(π‘₯,) = πœ• (𝑒 π‘₯,𝑑 )) β„Žβ†’0 β„Ž πœ•π‘‘ o 𝑒 π‘₯,𝑑 = lim ( 𝑒 π‘₯+π‘˜,𝑑 βˆ’π‘’(π‘₯,) = πœ• (𝑒 π‘₯,𝑑 )) π‘₯ π‘˜β†’0 π‘˜ πœ•π‘₯ o 𝑒 (π‘₯,𝑑 = lim ( 𝑒π‘₯π‘₯+π‘˜,𝑑 βˆ’π‘’ π‘₯,𝑑 ) = πœ•2 (𝑒 π‘₯,𝑑 )) π‘₯π‘₯ π‘˜β†’0 π‘˜ πœ•π‘₯2 o 𝛼 = thermal diffusivity (known constant) π‘™π‘’π‘›π‘”π‘‘β„Ž2  Units: π‘‘π‘–π‘šπ‘’  Depends on materiπœ…l bar made of  Defined by 𝛼 = , where πœ… = thermal conductivity, 𝜌 = density, 𝑠 = specific heat of material πœŒπ‘  o Initially we have a temperature distribution (initial condition) given by 𝑒 π‘₯,0 = 𝑓(π‘₯), where 𝑓(π‘₯) known ( ) ( ) o Boundary conditions given- assume 𝑒 0,𝑑 = 𝑇 an1 𝑒 𝐿,𝑑 = 𝑇 for 𝑑2β‰₯ 0, where 𝑇 ,𝑇 πœ– ℝ1 2  In this section, assume 1 = 𝑇2= 0  Initial and final temperatures o We obtain the following system: 𝑒 𝑑 𝛼𝑒 ,0π‘₯π‘₯ π‘₯ < 𝐿,𝑑 > 0 𝑒 0,𝑑 = 0,𝑑 β‰₯ 0 { ( ) 𝑒 𝐿,𝑑 = 0,𝑑 β‰₯ 0 𝑒 π‘₯,0 = 𝑓 π‘₯ ,0 ≀ π‘₯ ≀ 𝐿  Linear system; 𝑒 only in first power  Heat equation and boundary conditions homogeneous  For physical situations, we want unique solution ο‚· Boundary conditions essential to that- give initial and final temperatures 2  Since heat equation linear, can do linear separation; if 2 functions 𝑒,𝑣 sat{sfy 𝛼 𝑒 π‘₯π‘₯ , 𝑣𝑑= 𝛼 𝑣2 π‘₯π‘₯ then for any π‘Ž,𝑏 πœ– ℝ, we have π‘Žπ‘’ + 𝑏𝑣 ) = 𝛼 π‘Žπ‘’ + 𝑏𝑣 ) , and π‘Žπ‘’ + 𝑏𝑣 is also a solution of 𝑑 π‘₯π‘₯ heat equation ο‚· First, leave out initial condition 𝑒 π‘₯,0 = 𝑓(π‘₯) and find a family of solutions to heat equation using boundary conditions 𝑒𝑑= 𝛼𝑒 ,π‘₯π‘₯< π‘₯ < 𝐿,𝑑 > 0 o Solving {𝑒 0,𝑑 = 0,𝑑 β‰₯ 0 has infinite number of solutions (which makes a family) 𝑒 𝐿,𝑑 = 0,𝑑 β‰₯ 0 o Since we want to find some particular solution, we seek one of form 𝑒 π‘₯,𝑑 = 𝑋 π‘₯ 𝑇(𝑑)  Found by Fourier  Separate variables to get 2 functions that each depend on 1 variable o Substitute 𝑒(π‘₯,𝑑) into heat equation:  (𝑋 π‘₯ 𝑇 𝑑) ) = 𝛼 𝑋(π‘₯ 𝑇 𝑑 ( ) ) β€² 𝑑 2 β€²β€² π‘₯π‘₯  𝑋 π‘₯ 𝑇 𝑑 = 𝛼 𝑋 (π‘₯ 𝑇 𝑑)  If 𝑋 π‘₯ ,𝑇(𝑑) β‰  0, can divide by them 𝑋 π‘₯ 𝑇 𝑑) 2 𝑋′(π‘₯ 𝑇(𝑑)  𝑋 π‘₯ 𝑇 𝑑)= 𝛼 𝑋 π‘₯ 𝑇 𝑑) β€² β€²β€²  𝑇 𝑑)= 𝛼 2 𝑋 (π‘₯) 𝑇 𝑑) 𝑋 π‘₯ ) 𝑋′(π‘₯) 1 𝑇 𝑑 )  𝑋 π‘₯ )= 𝛼 𝑇 𝑑 )  For equation to be valid for 0 < π‘₯ < 𝐿,𝑑 > 0, both sides must be equal to same constant; otherwise one variable being fixed and the other varying would cause inequality 𝑋′(π‘₯) 1 𝑇 𝑑 )  There exists a constant πœ† such that = 2 = βˆ’πœ†; convenient to write negative sign β€²β€² 𝑋 π‘₯) 𝛼 𝑇 𝑑 ) 𝑋 (π‘₯) 𝑋 π‘₯ ) = βˆ’πœ† 𝑋′(π‘₯ = βˆ’πœ†π‘‹(π‘₯) 𝑋 β€²(π‘₯ + πœ†π‘‹ π‘₯ = 0  { 1 𝑇 𝑑) β†’ { β€² 2 β†’ { β€² 2 2 = βˆ’πœ† 𝑇 𝑑 = βˆ’π›Ό πœ†π‘‡(𝑑) 𝑇 𝑑 + 𝛼 πœ†π‘‡ 𝑑 = 0 𝛼 𝑇 𝑑 ) o Now consider boundary conditi
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