M257-316Notes_Lecture20

M257-316Notes_Lecture20 - Chapter 16 Lecture 20 -...

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Chapter 16 Lecture 20 - Inhomogeneous Derivative BC Example 16.1 Inhomogeneous Derivative BC: u t = α 2 u xx 0 <x<L, t> 0 (16.1) BC: u x (0 ,t )= Au x ( L, t B (16.2) IC: u ( x, 0) = g ( x ) . (16.3) Try for a steady solution: u 0 ( x )=0 , u ( x αx + β , u x = α but then we cannot match both BC unless A = B = α . This means that if we are pumping and removing heat from the rod at diFerent rates then the temperature does not reach a steady state. Instead of subtracting oF a steady solution we subtract a particular solution which depends on x and t of the form: w ( x, t ax 2 + bx + ct (16.4) w t = c = α 2 w xx =2 α 2 a c α 2 a. (16.5) 101
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Lecture 20 - Inhomogeneous Derivative BC Then w ( x, t )= ax 2 + bx +2 α 2 at (16.6) solves the heat equation. Now we determine the constants a and b so that w ( x, t ) satisfes the inhomogeneous BC: w x =2 ax + b : w x (0 ,t b = A, w x ( L, t )=2 aL + A = B. (16.7) ThereFore a =( B A ) / 2 L . ThereFore w ( x, t ( B A ) 2 L x 2 + Ax + α 2 µ B A L t. (16.8) Now let u ( x, t w ( x, t )+ v ( x, t ) . (16.9) u t = w t + v t = α 2 ( w xx + v xx ) v t = α 2 v xx u x (0 A = w x (0 v x (0 A + v x (0 ) v x (0 )=0 u x ( L, t B = w x
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M257-316Notes_Lecture20 - Chapter 16 Lecture 20 -...

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