Q8B_230

# Q8B_230 - x = 0 tells us that C 1 =-C 2 and the boundary...

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Math 415 Quiz 8B Consider the equation au xx - bu t + cu = 0, where a , b , and c are constants. If we let u ( x,t ) = e δt w ( x,t ), where δ is a constant, what diﬀerential equation does w satisfy. Plugging in our formula for u ( x,t ) into the diﬀerential equation gives us ae δt w xx - bδe δt w - be δt w t + ce δt w = 0 . Or, simplifying aw xx - bw t + ( c - ) w = 0 . Consider a uniform bar of length L . Assume that the temperature at the end x = 0 is kept at 0 C, while the other end x = L is insulated. Find the functions u n ( x,t ) = X n ( x ) T n ( t ) that satisfy the diﬀerential equation and its boundary condition. We have the boundary value problem α 2 u xx = u t , u (0 ,t ) = 0, u x ( L,t ) = 0. As usual, taking u ( x,t ) = X ( x ) T ( t ), this leads to the boundary value problem in X : X 00 + λX = 0, X (0) = 0, X 0 ( L ) = 0. So we consider the alternatives: λ = 0: The general solution is X ( x ) = C 1 x + C 2 . The boundary condition at x = 0 tells us that C 2 = 0 and the other tells us that C 1 = 0. So there are no interesting solutions with λ = 0. λ < 0, λ = - μ 2 : The general solution is X ( x ) = C 1 e μx + C 2 e - μx . The boundary condition at
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Unformatted text preview: x = 0 tells us that C 1 =-C 2 , and the boundary condition at x = L then tells us that C 1 μ ( e μx + e-μx ) = 0, so, since e μx + e-μx 6 = 0 for real x and μ , and since by assumption μ 6 = 0, C 1 = C 2 = 0. λ > 0, λ = μ 2 : The general solution is X ( x ) = C 1 sin( μx ) + C 2 cos( μx ). The ﬁrst boundary condition tells us that C 2 = 0, and then the second tells us that C 1 μ cos( μL ) = 0. This can happen when μL = (2 n-1) π 2 for n = 1, 2, 3, . . . So we get the eigenvalues λ n = ± (2 n-1) π 2 L ² 2 with associated eigenfunctions X n ( x ) = sin (2 n-1) πx 2 L . The diﬀerential equations for t become T n + (2 n-1) 2 π 2 α 2 4 L 2 T n = 0, or T n ( t ) = e-(2 n-1) 2 π 2 α 2 t/ 4 L 2 . So we get u n ( x,t ) = e-(2 n-1) 2 π 2 α 2 t/ 4 L 2 sin (2 n-1) πx 2 L . 1...
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