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 + eμx ) = 0, so, since e μx + eμx 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 n1) π 2 for n = 1, 2, 3, . . . So we get the eigenvalues λ n = ± (2 n1) π 2 L ² 2 with associated eigenfunctions X n ( x ) = sin (2 n1) πx 2 L . The diﬀerential equations for t become T n + (2 n1) 2 π 2 α 2 4 L 2 T n = 0, or T n ( t ) = e(2 n1) 2 π 2 α 2 t/ 4 L 2 . So we get u n ( x,t ) = e(2 n1) 2 π 2 α 2 t/ 4 L 2 sin (2 n1) πx 2 L . 1...
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 Spring '07
 COSTIN
 Math, Differential Equations, Equations, Partial Differential Equations, Tn, Boundary value problem, boundary condition

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