Unformatted text preview: Exam 1: Question 6 Solve the following PDE (20 points) • Solve the boundaryvalue problem:
2 2 ,0 0, ,0 0, , 0 , , 0 0 0, • Find a steadystate solution
2 2 of the boundaryvalue problem: ,0 1, 0, 0 1 0 0 0, ,0 Answer: , , 1, PART A (Solve the boundaryvalue problem) , solution.
2 2 2 2 2 2 , , where V is the homogenous solution and is the particular Because
2 2 is not function of time then 0, then the equation become: The properties of 0, , 0 0, , , 0 function: 0 0, , 0 0 0 0 PARTICULAR PART 0 Characteristic: 0 The solution of
, will be: Finding C1 and C2 using boundary condition: 0, 0,
0 0 0 0 0 , 0 , 2 2 The solution of will be: 2 2 2 HOMOGENOUS PART
2 2 , With the boundary conditions: And Initial condition: ,0 0 0, 0 , 0 Finding initial condition for ,0 ,0 ,0 0 ,0 ,0 Solving homogenous part: , . Solving X 0 Characteristic: 0 , X will be: boundary conditions: 0, 0 0 , 0 0 0 If 0 trivial solution 0 0 0 0 0 0 If sin Finding value 0 Solving T 0 So the solution of T: The solution of V(x,t) will be: , , . sin . , Finding An sin . Using initial condition ,0 sin . ,0 sin From previous we got ,0 Using formula: sin , 2 sin 2 , 0 sin 2 sin 2 . sin 2 . sin 2 1 sin cos 2 1 sin cos 0 2 1 cos 2 cos So the solution of , 2 1 sin cos . , will be: , sin . Putting together for solving 2 1 cos , , , sin PART B Find a steadystate solution
2 2 of the boundaryvalue problem: ,0 1, 0, 0 1 0 0 0, ,0 For the steady state solution
2 2 , , 1, 0, so the equation becomes: 0 , , where V is the homogenous solution and is the particular , solution.
2 2 2 2 0 0 , 1 function: 0 0 0, , 0 0 0 0 The properties of 0, 1, 0 0, 1, PARTICULAR PART 0 We have factor
take
0 0 instead of 0, then we have to solve for will be is the effect of .
0 0 first, after that into account. The solution of 0 and is the solution of to the . Solving for Characteristic: 0 The solution of
, 0 0 will be in the form of: Finding C1 and C2 using boundary condition: 0, 0, 0
0 1, 0
0 0 1, 1 0 1 0 0 0 The solution of 0 will be: 2 sinh 2sinh sinh sinh Solving Because 0 is a constant we assume . 0 0 So So the final solution of sinh sinh sinh 1 sinh is: ...
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 Spring '10
 Ghaleb
 Steady State, Sin, Boundary value problem, Boundary conditions

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