# Lec6 - Math 124A Viktor Grigoryan 6 Wave equation solution In this lecture we will solve the wave equation on the entire real line x R This

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Math 124A – October 14, 2010 ± Viktor Grigoryan 6 Wave equation: solution In this lecture we will solve the wave equation on the entire real line x R . This corresponds to a string of inﬁnite length. Although physically unrealistic, as we will see later, when considering the dynamics of a portion of the string away from the endpoints, the boundary conditions have no eﬀect for some (nonzero) ﬁnite time. Due to this, one can neglect the endpoints, and make the assumption that the string is inﬁnite. The wave equation, which describes the dynamics of the amplitude u ( x,t ) of the point at position x on the string at time t , has the following form u tt = c 2 u xx , or u tt - c 2 u xx = 0 . (1) As we saw in the last lecture, the wave equation has the second canonical form for hyperbolic equations. One can then rewrite this equation in the ﬁrst canonical form, which is u ξη = 0 . (2) This is achived by passing to the characteristic variables ± ξ = x + ct, η = x - ct. (3) To see that (2) is equivalent to (1), let us compute the partial derivatives of u with respect to x and t in the new variables using the chain rule. u t = cu ξ - cu η , u x = u ξ + u η . We can diﬀerentiate the above ﬁrst order partial derivatives with respect to t , respectively x using the chain rule again, to get u tt = c 2 u ξξ - 2 c 2 u ξη + c 2 u ηη , u xx = u ξξ + 2 u ξη + u ηη . Substituting this expressions into the left hand side of equation (1), we see that u tt - c 2 u xx = c 2 u ξξ - 2 c 2 u ξη + c 2 u ηη - c 2 ( u ξξ + 2 u ξη + u ηη ) = - 4 c 2 u ξη = 0 , which is equivalent to (2). Equation (2) can be treated as a pair two successive ODEs. Integrating ﬁrst with respect to the variable η , and then with respect to ξ , we arrive at the solution u ( ξ,η ) = f ( ξ ) + g ( η ) . Recalling the deﬁnition of the characteristic variables (3), we can switch back to the original variables ( x,t ), and obtain the general solution u ( x,t ) = f ( x + ct ) + g ( x - ct ) . (4) Another way to solve equation (1) is to realize that the second order linear operator of the wave equation factors into two ﬁrst order operators L = 2 t - c 2 2 x = ( t - c∂ x )( t + c∂ x ) . 1

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Hence, the wave equation can be thought of as a pair of transport (advection) equations ( t - c∂ x ) v = 0 , (5) ( t + c∂ x ) u = v. (6) It is no coincidence, of course, that x + ct = constant , (7) and x - ct = constant , (8) are the characteristic lines for the transport equations (5), and (6) respectively, hence our choice of the characteristic coordinates (3). We also see that for each point in the xt plane there are two distinct characteristic lines, each belonging to one of the two families (7) and (8), that pass through the point. This is illustrated in Figure 1 below.
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## This note was uploaded on 01/10/2011 for the course MATH 124A taught by Professor Ponce during the Fall '08 term at UCSB.

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Lec6 - Math 124A Viktor Grigoryan 6 Wave equation solution In this lecture we will solve the wave equation on the entire real line x R This

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