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Unformatted text preview: Solution of the Wave Equation by Separation of Variables The Problem Let u ( x, t ) denote the vertical displacement of a string from the x axis at position x and time t . The string has length ℓ . Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. For notational convenience, choose a coordinate system so that the left hand end of the string is at x = 0 and the right hand end of the string is at x = ℓ . x u ℓ We assume that the string is undergoing small amplitude transverse vibrations so that u ( x, t ) obeys the wave equation ∂ 2 u ∂ t 2 ( x, t ) = c 2 ∂ 2 u ∂x 2 ( x, t ) for all 0 < x < ℓ and t > (1) The conditions that the left and right hand ends are held at height zero are encoded in the “boundary conditions” u (0 , t ) = 0 for all t > (2) u ( ℓ, t ) = 0 for all t > (3) As we have been told the position and speed of the string at time 0, there are given functions f ( x ) and g ( x ) such that the “initial conditions” u ( x, 0) = f ( x ) for all 0 < x < ℓ (4) u t ( x, 0) = g ( x ) for all 0 < x < ℓ (5) are satisfied. The problem is to determine u ( x, t ) for all x and t . Outline of the Method of Separation of Variables We are going to solve this problem in three steps. Step 1 In the first step, we find all solutions of (1) that are of the special form u ( x, t ) = X ( x ) T ( t ) for some function X ( x ) that depends on x but not t and some function T ( t ) that depends on t but not x . This is where the name “separation of variables” comes from. It is of course too much to expect that all solutions of (1) are of this form. But if we find a bunch of solutions X i ( x ) T i ( t ) of this form, then since (1) is a linear equation, ∑ i a i X i ( x ) T i ( t ) is also a solution for any choice of the constants a i . (Check this yourself!) If we are lucky (and we shall be lucky), we will be able to choose the constants a i so that the other conditions (2–5) are also satisfied. Step 2 We impose the boundary conditions (2) and (3). Step 3 We impose the initial conditions (4) and (5). The First Step – Finding Factorized Solutions The factorized function u ( x, t ) = X ( x ) T ( t ) is a solution to the wave equation (1) if and only if X ( x ) T ′′ ( t ) = c 2 X ′′ ( x ) T ( t ) ⇐⇒ X ′′ ( x ) X ( x ) = 1 c 2 T ′′ ( t ) T ( t ) January 21, 2007 Solution of the Wave Equation by Separation of Variables 1 The left hand side is independent of t . So the right hand side, which is equal to the left hand side, must be independent of t too. The right hand side is independent of x . So the left hand side must be independent of x too. So both sides must be independent of both x and t . So both sides must be constant. Let’s call the constant σ . So we have X ′′ ( x ) X ( x ) = σ 1 c 2 T ′′ ( t ) T ( t ) = σ ⇐⇒ X ′′ ( x ) − σX ( x ) = 0 T ′′ ( t ) − c 2 σT ( t ) = 0 (6) We now have two constant coefficient ordinary differential equations, which we solve in the usual way. WeWe now have two constant coefficient ordinary differential equations, which we solve in the usual way....
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- Spring '10
- Sin, Partial differential equation