midtermF05sol

To calculate k 12 a φ 2 φ 1 dφ 2 dx dφ 1 dx 3 φ

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To calculate K 12 = a ( φ 2 , φ 1 ) = 0 2 dx 1 dx + 3 φ 2 ( x ) φ 1 ( x ) dx , we first focus on 0 2 dx 1 dx dx and see 0 2 dx 1 dx dx = 2 h h 1 h - 1 h dx = - 1 h = - n And we now look at 0 2 φ 2 ( x ) φ 1 ( x ) dx . We see this is equal to 2 2 h h 1 h ( x - h ) - 1 h ( x - 2 h ) dx = - 2 h 2 2 h h ( x - h )[( x - h ) - h ] dx = - 2 h 2 2 h h [( x - h ) 2 - h ( x - h )] dx = - 2 h 2 1 3 ( x - h ) 3 - h ( x - h ) 2 2 2 h h = - 2 h 2 h 3 3 - h 3 2 = h 3 = 3 n and so we see K 12 = - n + 3 n .

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5. (20 points) Fourier Series. Given a thin bar of length = π cm with ρ = 1 g / cm 3 , c = 1 J / gK, and κ = 1 W / cmK, assume that the bar is initially at a uniform temperature of 0 C. Then, beginning at time zero, heat is added at a rate of f ( x, t ) = sin( x ) while the ends of the bar are held constant at 0 C. (a) Set up the corresponding initial-boundary-value problem. (b) Solve this problem using Fourier series methods. Determine the steady-state solution by taking a limit. Draw a rough sketch of how the solution will evolve with time. ( Note : you must show how to derive the solution from first principles and may not make use of a memorized formula.) ( Hint : The solution has a relatively simple form.) Solution: (a) Using the information above with appropriate units (degrees Celsius, Watts per cubic centimeter, etc.), we obtain ∂u ∂t - 2 u ∂x 2 = sin( x ) , 0 < x < π, t > 0 u ( x, 0) = 0 , 0 < x < π u (0 , t ) = u ( π, t ) = 0 , t 0 . (b) Since the x -derivative part of this equation corresponds to the L D operator, we expect the solution to have the form u ( x, t ) = n =1 a n ( t ) sin nπx π = n =1 a n ( t ) sin( nx ) . Substituting this expression into the PDE above, we obtain n =1 da n dt + n 2 a n ( t ) sin( nx ) = sin( x ) . Since sin( x ) is already an eigenfunction, we do not need to expand the right hand side as a Fourier series. By orthogonality of the eigenfunctions, we have da 1 dt + a 1 ( t ) = 1 da n dt + n 2 a n ( t ) = 0 , for n = 2 , 3 , . . . . Since the intial condition is equal to zero, we know that a 1 (0) = 0 and a n (0) = 0 for n = 2 , 3 , . . . . We solve these ODEs using integrating factors. For n = 1, we obtain d dt e t a 1 ( t ) = e t . Integrating from 0 to t yields e t a 1 ( t ) = e t - 1 a t ( t ) = 1 - e - t . Similarly, for n = 2 , 3 , . . . , d dt e n 2 t a n ( t ) = 0 ,
which implies that a n ( t ) = 0 for n = 2 , 3 , . . . . Therefore, the solution is u ( x, t ) = ( 1 - e - t ) sin( x ) . We obtain the steady-state solution u s ( x ) by taking a limit as t → ∞ : u s ( x ) = lim t →∞ u ( x, t ) = lim t →∞ ( 1 - e - t ) sin( x ) = sin( x ) . Of course, this limit can be verified by means of the steady-state differential equation: - d 2 u s dx 2 = sin( x ) , u s (0) = u s ( π ) = 0 . Clearly, u s solves this equation. Therefore, as time increases from zero, the temperature distribution in the bar always main- tains the shape of a sine function with amplitude given by 1 - e - t . Your sketch should show the temperature beginning at zero (a constant function equal to zero), which then grows as a sine curve with amplitude growing from zero to one as time goes to infinity.

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