soln3 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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Unformatted text preview: MATH 300 Fall 2007 Advanced Boundary Value Problems I Solutions to Problem Set 3 To Be Completed by: Friday October 26, 2007 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 148, #4.4.7] Small vertical vibrations u of a uniform vibrating string which is initially at rest are governed by the initial value – boundary value problem ∂ 2 u ∂t 2 = c 2 ∂ 2 u ∂x 2 u (0 , t ) = 0 , u ( L, t ) = 0 for < x < L u ( x, 0) = f ( x ) , ∂u ∂t ( x, 0) = 0 for t > . Show that u ( x, t ) = 1 2 F ( x- ct ) + F ( x + ct ) , where F ( x ) is the odd periodic extension of f ( x ) . Hint : Use separation of variables to solve the problem and then use the addition formula for the sine function: sin a cos b = 1 2 sin( a + b ) + sin( a- b ) to write the solution in the form shown above. Solution: We assume a solution of the form u ( x, t ) = X ( t ) · T ( t ) and separating variables we have two ordinary differential equations X 00 ( x ) + λX ( x ) = 0 , < x < L T 00 ( t ) + λc 2 T ( t ) = 0 , t > X (0) = 0 T (0) = 0 . X ( L ) = 0 The eigenvalues and eigenfunctions for the X-equation are λ n = n 2 π 2 L 2 and X n ( x ) = sin nπ L x for n ≥ 1 , and the corresponding solutions of the T-equation are T n ( t ) = a n cos nπ L ct + b n sin nπ L ct for n ≥ 1 . Using the superposition principle, we write the solution as u ( x, t ) = ∞ X n =1 ( a n cos nπ L ct + b n sin nπ L ct ) sin nπ L x, and ∂u ∂t ( x, t ) = ∞ X n =1 (- a n nπc L sin nπ L ct + b n nπc L cos nπ L ct ) sin nπ L x. We determine the coefficients using the initial conditions and the orthogonality of the eigenfunctions on the interval [0 , L ] . From the first initial condition we have f ( x ) = u ( x, 0) = ∞ X n =1 a n sin nπ L x, ( * ) so that a n = 2 L Z L f ( x ) sin nπ L x dx for n ≥ 1 . From the second initial condition we have 0 = ∂u ∂t ( x, 0) = ∞ X n =1 b n nπc L sin nπ L x, so that b n = 0 for n ≥ 1 . The solution is u ( x, t ) = ∞ X n =1 a n cos nπ L ct sin nπ L x = ∞ X n =1 a n 1 2 sin nπ L ( x- ct ) + 1 2 sin nπ L ( x + ct ) for 0 < x < L and t > . Note that if f ∈ P W S [0 , L ] , that is, f is peicewise smooth on the interval [0 , L ] , then ( * ) is the Fourier sine series for f, and converges for all real numbers x, and, except for at most countably many values of x, it converges to the odd periodic extension F of f. Therefore, assuming the odd periodic extension of F is continuous, the solution is u ( x, t ) = 1 2 ∞ X n =1 a n sin nπ L ( x- ct ) + 1 2 ∞ X n =1 a n sin nπ L ( x + ct ) = 1 2 F ( x- ct ) + 1 2 F ( x + ct ) for 0 < x < L and t > . Question 2. [p 147, #4.4.4] Small vertical vibrations u of a uniform vibrating string which is initially unperturbed are governed by the initial value – boundary value problem ∂ 2 u ∂t 2 = c 2 ∂ 2 u ∂x 2 u (0 , t ) = 0 , u ( L, t ) = 0 for < x < L u ( x, 0) = 0 , ∂u ∂t ( x, 0) = g ( x ) for t > ....
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This homework help was uploaded on 04/15/2008 for the course APMA 3102 taught by Professor Courdurier during the Spring '00 term at Columbia.

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soln3 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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