Week_06 - ACM 100c - Methods of Applied Mathematics Week 6...

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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 1-D 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 2-D and 3-D 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 u 1 and 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 u and so the main contribution to the force is the vertical component and this is given by F = bracketleftBigg | T | u 1 radicalbig ( x 1 ) 2 + ( u 1 ) 2 + | T | u 2 radicalbig ( x 2 ) 2 + ( u 2 ) 2 bracketrightBigg j . As long as vextendsingle vextendsingle vextendsingle vextendsingle u x vextendsingle vextendsingle vextendsingle vextendsingle is very small, we can write this force approximately as F | T | 2 u x 2 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 and so the mass of the string in the region that we are perturbing is x. The acceleration locally at the point x is 2 u t 2 . Putting this together, we get x 2 u t 2 j = | T | 2 u x 2 x j , and so we have 2 u t 2 = | T | 2 u x 2 . 2 The ratio | T | 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 | 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 ( x ) x parenleftbigg T ( x ) u x parenrightbigg ....
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Week_06 - ACM 100c - Methods of Applied Mathematics Week 6...

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