The Wave Equation I. MA 436 Kurt Bryan

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1 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string s it vibrtes. Let s use t for time, x for position long the horizontl xis, u(x, t) for the verticl displcement of the string t position x nd t. We ll lso use T to denote the tension in the string nd λ(x) to denote the liner density of the string t position x. The eqution of motion of the string cn be derived from nothing more thn F = M nd few resonble ssumptions. First, we will ssume tht the motion of the string is of smll mplitude, nd purely verticl. As such, the eqution we derive won t pply to lrge mplitude motion or to string with ny significnt longitudinl motion (e.g., slinky). I m going to use the nottion u x for the first derivtive of u(x, t) with respect to x, u xx for the second derivtive, etc. Consider smll portion of the string stretching from x to x + dx, with ngles θ 1 nd θ 2 s lbelled. 1

2 Our first tsk is to find the totl force on this piece of the string. It s esy to check tht the force on the left side of this smll string element is F(x) = T cos(θ 1 )i T sin(θ 1 )j. Similrly the force on the right end is F(x + dx) = T cos(θ 2 )i + T sin(θ 2 )j (this ll ssumes tension is roughly constnt). Let us ssume tht the string vibrtes in such wy tht the slopes (or ngles θ 1 nd θ 2 ) sty smll; this might be clled smll strin or first order model, depending on your field. In this cse both ngles re close to zero nd we hve the pproximtions (from the Tylor s series) sin(θ) = θ + O(θ 3 ), cos(θ) = 1 O(θ 2 ). In this cse we cn mke the first order pproximtions (dropping the smll qudrtic nd higher terms) F(x) T i T θ 1 j, F(x + dx) T i + T θ 2 j. It s lso esy to combine the bove pproximtions to find tht if θ is smll then tn(θ) θ. From the picture it s cler tht u x (x, t) = tn(θ 1 ) θ 1 nd u x (x + dx, t) = tn(θ 2 ) θ 2, so to first order we hve F(x) T i T u x (x, t)j, F(x + dx) T i + T u x (x + dx, t)j. The totl force on this string element is F(x) + F(x + dx), which is F tot = T (u x (x + dx, t) u x (x, t))j T u xx (x, t) dx j, (1) where I ve used the fct tht u xx (x, t) (u x (x + dx, t) u x (x, t))/dx if dx is smll indeed, tht is the very definition of the derivtive u xx. Note tht the force is (to our level of pproximtion) entirely verticl, in keeping with our originl ssumptions. 2

3 The ccelertion of this piece of the string is just u tt (x, t)j, nd the mss is pproximtely λ(x) dx. Using Newton s second lw, F = m, we cn equte the totl force F tot in eqution (1) with M = λ(x)u tt (x, t) dx to find (I ll switch to Leibnitz nottion for now) λ(x) 2 u t T 2 u = 0, (2) 2 x2 which is one version of the wve eqution. We cn lso write it in the form 2 u t 2 T 2 u λ(x) x = 0. 2 T It s interesting to look t the physicl dimensions of ; tension T hs units λ(x) of mss per length per time squred, while λ hs units of mss per length. T Thus hs units of length squred per time squred, or velocity squred. λ(x) Tht s exctly wht it turns out to be. In the cse tht λ is constnt (tht s wht we ll be most interested in) we frequently write c 2 = T/λ nd write the wve eqution s 2 u t 2 u 2 c2 = 0, (3) x2 where c hs dimensions of velocity. Eqution (3) is clled the wve eqution. It s n exmple of prtil differentil eqution ( PDE for short), i.e., n eqution involving the derivtives of n unknown function of two or more vribles. The wve eqution is one of the big three PDE s from mthemticl physics (the other two re the het eqution nd Lplce s eqution). Our gol is, of course, to find solutions to the wve eqution. There re mny, nd we cn t nil down one without further informtion. It seems obvious tht we ought to need the initil position of the string in order to determine its position t lter times nd positions. Thus, if we tke t = 0 s the initil time, we need the informtion tht u(x, 0) = f(x) for some given function f(x) tht specifies the initil position of the string. Here x will rnge over the length of the string. The initil position might seem like enough informtion to determine the string s motion, but it isn t. Do simple thought experiment, in which two different (but physiclly identicl) strings strt in the sme position, but with different initil velocities; it s cler the strings would hve different 3

4 future motion. So s it turns out, we lso need to know the initil velocity of the string, sy u (x, 0) = g(x) for some specified function g(x). t For now we re going to concentrte on infinite strings, in which x tkes the rnge < x <. So in summry, out gol is to exmine the solvbility of the prtil differentil eqution (3) with initil conditions for some given functions f nd g. 2 Energy u(x, 0) = f(x) u (x, 0) t = g(x) For lter reference, it s convenient to compute the kinetic nd potentil energy of the string s it moves. Agin consider the smll string element from x to x+dx. Its velocity is just u t (x, t) nd its mss is (to first order) λ(x) dx. The kinetic energy of this piece is then 1 2 λ(x)u t(x, t) 2 dx nd the totl kinetic energy of the string is obtined by dding up ll the pieces, i.e., KE = 1 2 λ(x)u t (x, t) 2 dx (4) where nd b re the ends of the string (mybe t plus or minus infinity). The potentil energy of the string in given configurtion is bit more chllenging to find. Suppose tht t some instnt in time the string hs the shpe u(x, t) = ϕ(x); I m thinking of time s frozen, since we re interested only in potentil energy. The potentil energy of the string in this position is the mount of work needed to deform it from the position u(x, t) 0, if we use the u 0 s our reference point. Consider deforming the string from this bse position to u(x, t) = ϕ(x) by tking rϕ(x) nd letting r run from 0 to 1. For given string element stretching from x to x + dx the bove nlysis shows tht the verticl force required to push the element upwrd for ny vlue of r is T rϕ xx dx (this is MINUS the force exerted by the string.) If we chnge r by smll mount dr then we move this string element by distnce ϕ(x) dr. The formul tht work equls force times distnce shows tht we do n mount of work dw = T ϕ(x)ϕ xx (x)r dx dr 4

5 on the string element. The totl work on the element from r = 0 to r = 1 is obtin by dding (integrting) dw from r = 0 to r = 1, nd is just W element = T 2 ϕ(x)ϕ xx(x) dx. The totl work done in moving the entire string from the bse configurtion to the u(x, t) = ϕ(x) configurtion is obtined by dding the work over ech element, nd is thus W totl = T 2 ϕ(x)ϕ xx (x) dx. If we integrte this by prts in x (use u dv = uv v du with u = ϕ, dv = ϕ xx dx) we obtin the potentil energy of the string in the configurtion ϕ(x): PE = T ϕ 2 2 x(x) dx where in doing the integrtion by prts I ve mde the ssumption tht u(, t) = 0 nd u(b, t) = 0 OR u x (, t) = 0 nd u x (b, t) = 0; these boundry conditions model specific physicl situtions tht we ll tlk bout lter. All in ll then, the potentil energy of the string in the position u(x, t) is PE = T 2 u 2 x(x, t) dx. (5) The totl energy of the string, kinetic plus potentil, t time t is Energy = 1 2 (T u 2 x(x, t) + λ(x)u 2 t (x, t)) dx 5

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