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Unformatted text preview: Solutions to Final Exam Final Exam Introduction to PDE MATH 336325820 (Fall 2009) This exam has 4 questions, for a total of 40 points. Please answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. Upon finishing PLEASE write and sign your pledge below: On my honor I have neither given nor received any aid on this exam. 1 Rules Be sure to show a few key intermediate steps and make statements in words when deriving results  answers only will not get full marks. You are free to use any of the information given in Section 2, without proof, on any question in the exam. 2 Given You may use the following without proof: The eigensolution to X 00 + λX = 0 , X (0) = 0 = X ( L ) is X n ( x ) = sin nπx L , λ n = nπ L 2 , n = 1 , 2 ,... Orthogonality condition for sines: for any L > 0, Z L sin mπx L sin nπx L dx = L/ 2 , m = n m 6 = n A useful result derived from the Divergence Theorem, ZZ D v Δ vdV = ZZ D ∇ v  2 dV + Z ∂D v ∇ v · ndS for and 2D or 3D region D with closed boundary ∂D . Page 1 of 16 Please go to the next page. . . Solutions to Final Exam 3 Questions 1. 10 points A bar with initial temperature profile f ( x ) > 0, with ends held at 0 o C, will cool as t → ∞ ; and approach a steadystate temperature 0 o C. However, whether or not all parts of the bar start cooling initially depends on the shape of the initial temperature profile. The following heat problem may enable you to discover the relationship: (a) Solve the heat problem on the interval 0 ≤ x ≤ 1, u t = u xx , u (0 ,t ) = 0 = u (1 ,t ) , u ( x, 0) = f ( x ) , where f ( x ) = 1 2 sin 3 πx + 3 2 sin πx. Solutoin: Using separation of variables, we let u ( x,t ) = X ( x ) T ( t ) and substitute this into the PDE to obtain X 00 X = T 00 T = λ where λ is a constant because the left hand side depends only on x and the middle only depends on t . The SturmLiouville problem for X ( x ) is X 00 + λX = 0; X (0) = 0 = X (1) whose solution is (given), X n ( x ) = sin( nπx ) , λ n = n 2 π 2 , n = 1 , 2 ,... The equation T 00 = λT gives the solution T n ( t ) = B n e n 2 π 2 t , n = 1 , 2 ,..., Therefore, the solution u n ( x,t ), for n = 1 , 2 ,... , to the PDE is u n ( x,t ) = X n ( x ) T n ( t ) = B n sin( nπx ) e n 2 π 2 t for constants B n . Summing all u n ( x,t ) together gives u ( x,t ) = ∞ X n =1 u n ( x,t ) = ∞ X n =1 B n sin( nπx ) e n 2 π 2 t Page 2 of 16 Please go to the next page. . . Solutions to Final Exam Imposing the IC gives f ( x ) = u ( x, 0) = ∞ X n =1 B n sin( nπx ) Multiplying by sin( mπx ), for m = 1 , 2 ,... , and integrating from x = 0 to x = 1 gives Z 1 f ( x ) sin( mπx ) dx = ∞ X n =1 B n Z 1 sin( nπx ) sin( mπx ) dx Using the given orthogonality condition gives B m = 2 Z 1 f ( x ) sin( mπx ) dx, m = 1 , 2 ,......
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 Spring '08
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 Math, Sin, Boundary value problem, Solutoin

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