test1_solns

# 1 4 points derive the steady state equilibrium

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1. 4 points Derive the steady-state (equilibrium) solution u E ( x ) = b 12 (1 - x 4 ) + 1 It is insufficient to simply verify that the solution works. Solutoin: The steady-state solution u E ( x ) satisfies the PDE and BCs, yielding 0 = u E + bx 2 , 0 < x < 1 u E (0) = 0 , u E (1) = 1 Integrating (twice) the ODE, i.e., u E = - bx 2 , for u E gives u E ( x ) = - b 12 x 4 + C 1 x + C 2 Imposing the BCs: u E (0) = 0 C 1 = 0 and u E (1) = 1 C 2 = b 12 + 1, yielding u E ( x ) = b 12 (1 - x 4 ) + 1 2. 3 points (a) Using u E ( x ), transform the given heat problem for u ( x, t ) into the following problem for a function v ( x, t ): v t = v xx , 0 < x < 1 , t > 0 v x (0 , t ) = 0 , v (1 , t ) = 0 , t > 0 v ( x, 0) = f ( x ) , 0 < x < 1 , where f ( x ) will be determined by the transformation. Show your work, which involves writing v = u - u E and using the information from u Page 2 of 5 Please go to the next page. . . [1 point] [1 point] [1 point] [1 point]

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Solutions to Test 1 and u E to derive the problem for v . (b) State f ( x ) in terms of u 0 , b and x . Solutoin: Writing v ( x, t ) = u ( x, t ) - u E ( x ) we have v t = u t v xx = u xx - u E = u xx + bx 2 and hence the PDE becomes v t = v xx The BCs for v are v x (0 , t ) = u x (0 , t ) - u E (0) = 0 - 0 = 0 v (1 , t ) = u (1 , t ) - u E (1) = 1 - 1 = 0 The IC is v ( x, 0) = u ( x, 0) - u E ( x ) = u 0 - b 12 (1 - x 4 ) - 1 thus f ( x ) = u 0 - 1 - b 12 (1 - x 4 ) . 3. 10 points Derive the solution v ( x, t ) = n =1 v n ( x, t ) = n =1 A n cos ( ( n - 1 2 ) πx ) e - ( n - 1 2 ) 2 π 2 t and derive equations for A n in terms of f ( x ). Be sure to give the intermediate steps:
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