FALLSEM2015-16_CP0667_12-Oct-2015_RM01_51-Applications-of-pde---One-Dimensional-Wave-Equation.pdf - Applications of Partial Differential Equations Wave

# FALLSEM2015-16_CP0667_12-Oct-2015_RM01_51-Applications-of-pde---One-Dimensional-Wave-Equation.pdf

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Applications of Partial Differential Equations Wave Equation Dr. T. Phaneendra Professor of Mathematics, School of Advanced Sciences, VIT University, Vellore-632014,Tamil Nadu, India, e-mail: [email protected]  1 Vibrating Strings Consider a thin copper string of length l cm, stretched between the fixed ends x = 0 and x = l . Suppose that it is released from the initial position f ( x ) by giving a velocity g ( x ) to it. We wish to determine the subsequent vertical displacement u ( x , t ) at any point x of the string. The transverse displacement u ( x , t ) in vibrating strings is governed by the homoge- neous partial differential equation of second order: 2 u t 2 = a 2 2 u x 2 , (1.1) where a 2 is a physical constant, called the velocity constant of the string. Since both the ends x = 0 and x = l are fixed , there will not be any vertical displacement at the ends at any time t . Thus the boundary conditions are: u ( 0 , t ) = 0 for all t > 0 , (1.2) and u ( l , t ) = 0 for all t > 0 . (1.3) We assume a trial solution of the form u ( x , t ) XT = X ( x ) T ( t ) . Substituting this expression in the wave equation (1.1), XT 00 = a 2 X 00 T . Now separating the variables, we get T 00 a 2 T = X 00 X · Since x and t are independent variables, each of these fractions reduces to a constant, say T 00 a 2 T = X 00 X = - λ . 1 This results in two ordinary differential equations: X 00 + λ X = 0 , (1.4) T 00 + λ a 2 T = 0 . (1.5) We consider three cases: Case (a): λ < 0 , say λ = - μ 2 , where μ 6 = 0 . The general solution of (1.4) will then be given by X ( x ) = Ae μ x + Be - μ x . This satisfies the boundary conditions (1.2) and (1.3) only if A + B = Ae μ l + Be - μ l = 0 , that is only if A = 0 = B . This leads to the trivial solution : X 0 for all x [ 0 , l ] of (1.4). Case (b): λ = 0 Then the general solution of (1.4) is X ( x ) = Ax + B , which satisfies the boundary con- ditions (1.2) and (1.3) only if A = 0 = B . This again is the trivial solution X 0 of (1.4). Case (c): λ > 0 , say λ = μ 2 , where μ 6 = 0 . Vibrations in strings are in general periodic in nature and the general solution of (1.4) is given by X ( x ) = A cos μ x + B sin μ x . (1.6) Applying the first boundary condition (1.2), we see that X ( 0 ) = A = 0. Thus (1.6) reduces to X ( x ) = B sin μ x . But then using the second boundary condition (1.3), this implies that X ( l ) = B sin μ l = 0 . In order to get a nontrivial solution, we must have B 6 = 0 so that sin μ l = 0 μ l = n π or μ = n π l , n = ± 1 , ± 2 , .... (1.7) Thus all the nontrivial solutions of (1.4) are of the form: X n ( x ) = B n sin n π x l n = ± 1 , ± 2 ,..., where B 0 n s are arbitrary constants depending on n . Since sin - k π x l = - sin k π x l for all k = 1 , 2 , 3 ,..., 2 we see that X - k is a constant multiple of X k for each k . Therefore, we shall discard the indices n = - 1 , - 2 , - 3 ,... so that the linearly indepen- dent solutions are given by X n ( x ) = B n sin n π x l , n = 1 , 2 , 3 , .... (1.8) Now the general solution of (1.5) is T ( t ) = C cos μ at + D sin μ at .  #### You've reached the end of your free preview.

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