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Unformatted text preview: ACM 100c  Methods of Applied Mathematics Week 6 May 2, 2009 1 The wave equation In this section we will introduce a new PDE  the wave equation  and analyze the behavior of its solutions. The wave equation in 1D is given by ∂ 2 u ∂t 2 = c ( x ) 2 ∂ 2 u ∂x 2 . The function c ( x ) is a positive function of x . In typical applications, x represents position and t represents time. As a result c will have the units of a velocity. The equation looks similar to the heat equation but the second derivative in time will change drastically the nature of the solutions. In 2D and 3D the equation becomes ∂ 2 u ∂t 2 = c 2 ∇ 2 u. To get a feel for how this equation arises in physical problems, consider a tightly stretched string so that at each point of the string there is a tension force T . We then perturb the string at two points vertically by an amount Δ u 1 and Δ u 2 . The tension T is a force that always points along the tangent to the string and will therefore resolve itself in a horizontal and vertical component when the string is perturbed. This situation is illustrated in Figure 1. If the deformation is small enough, the horizontal component will be of sec ond order in the perturbation Δ u and so the main contribution to the force is the vertical component and this is given by F = bracketleftBigg −  T  Δ u 1 radicalbig (Δ x 1 ) 2 + (Δ u 1 ) 2 +  T  Δ u 2 radicalbig (Δ x 2 ) 2 + (Δ u 2 ) 2 bracketrightBigg j . As long as vextendsingle vextendsingle vextendsingle vextendsingle Δ u Δ x vextendsingle vextendsingle vextendsingle vextendsingle is very small, we can write this force approximately as F ≈  T  ∂ 2 u ∂x 2 Δ x j . 1 Figure 1: A piece of a tensioned string although note this is a linear approximation to the force. The actual force is a nonlinear function of the deformation and so the string is really a nonlinear mechanical system. But to first order it behaves in a linear way. Now that we have the force we can equate it to the mass times an acceleration. We will assume the string has a density ρ and so the mass of the string in the region that we are perturbing is ρ Δ x. The acceleration locally at the point x is ∂ 2 u ∂t 2 . Putting this together, we get ρ Δ x ∂ 2 u ∂t 2 j =  T  ∂ 2 u ∂x 2 Δ x j , and so we have ∂ 2 u ∂t 2 =  T  ρ ∂ 2 u ∂x 2 . 2 The ratio  T  ρ has the dimensions of a velocity squared and we will see that it is related to a characteristic speed for the solutions. We usually write the equation as ∂ 2 u ∂x 2 = c 2 ∂ 2 u ∂x 2 where c 2 =  T  ρ is the square of the characteristic velocity for the wave equation. Note that if the tension and density vary with x then the equation becomes ∂ 2 u ∂t 2 = 1 ρ ( x ) ∂ ∂x parenleftbigg T ( x ) ∂u ∂x parenrightbigg ....
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 Spring '09
 NilesA.Pierce
 Laplace, Boundary value problem, Partial differential equation, Boundary conditions

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