6 Energy Methods And The Energy of Waves MATH 22C


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1 6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this to guess the form of the continuum version of this energy for the liner wve eqution. We then verify tht this energy is conserved on solutions of the wve eqution, nd use it to solve the uniqueness problem for the wve eqution. Tht is, we use the energy of wves to prove tht the initil vlue problem for the wve eqution hs unique solution lwys decomposble into the sum of left going nd right going wves. Let s begin by sking the question: How much energy is stored in your cr if it hs mss m kg nd is moving down the freewy t velocity v km per hour? The nswer is Kinetic Energy (KE) = 2 mv2. Tht is, if you hit wll nd come to complete stop t speed v = 00 km hr, nd your cr weighs 2000kg, you will deliver KE = km kg = kg m 2 2 hr s 2 = Newton meter of energy. One Newtonmeter is the energy delivered to mss of bout 0kg flling one meter through Erth s grvity t the surfce. In prticulr, tht s how much energy the wll will deliver to you upon impct. It goes up by the squre of the velocity, so tht s why speeding is so dngerous.
2 2 So how do we know this? There re two min resons. () It comes up in the expression for the work done by force, nd (2) It ppers in the expression for the energy conserved by conservtive force field. When we tret the mss s point mss, the frmework is ODE s, when we view the mss s continuous density, mss per volume insted of point mss, the frmework is PDE s. In this section nd the next, we discuss the point mss cse (ODE s), nd in the lst section we tret the continuum version (PDE s) in the context of the wve eqution. () The rgument for evluting the work done by force F on mss m goes s follows: So T he W ork Done = W = F orce Displcement. W = = vb v F ds = m ds dt dv = m ds = vb v mv dv m dv dt ds = 2 mv2 ] v b v = 2 mv2 2 mv2 b = KE. Thus the work done by force is equl to the chnge in Kinetic Energy, so the work done by the wll s it moves through your cr s you come to rest equls the KE= 2 mv2 you strted with. (2) The second rgument incorportes kinetic energy into the principle of conservtion of energy. We sy F is conservtive force if F = du dy = U (y) for some function U(y) clled the potentil energy. For exmple, let y denote distnce from the Erth s surfce,
3 nd ssume mss m moves verticlly under the influence of grvity ner the surfce. Then Newton s lw F = m gives the ODE, mÿ = mg. To get the energy, we write the force mg s minus the derivtive of the potentil U(y), tht is, so we hve mÿ = d dy {mgy} U(y) = mgy in this cse. This is one instnce to which the following very generl principle pplies. Theorem. If the force is minus the grdient of potentil U(y), so tht the ODE describing the motion is then the energy ÿ = U (y), E = 2 mẏ2 + U(y) is constnt long ech solution of the ODE. This is so importnt tht we give the Proof: We show tht E(y(t)) is constnt long solution y(t) of the eqution ẏ = U (y). For this is suffices to show tht de dt = 0. But by the chin rule we clculte: d dt E(y(t)) = d { } y(t) 2 + U(y(t)) = ẏÿ + U (y)ẏ dt 2 = ẏ { ÿ 2 + U(y) } = 0 becuse y(t) solves the eqution ÿ 2 + U (y) = 0. 3
4 4 Conclude: This is mthemticl theorem tht holds for ODE s of form ÿ = U (y) independent of ny interprettion of ÿ s force: The function E = 2 mẏ2 + U(y) is constnt long solutions. The physicl interprettion is tht ll long the motion the KE is the energy of motion, nd it cn be stored nd relesed s potentil energy, but the sum of the two is constnt. Becuse we believe the fundmentl forces of nture re conservtive, fundmentl forces must be conservtive. As first exmple consider the hrmonic oscilltor, ÿ + 2 y = 0. To put it in the form of conservtive system nd find the energy, solve for ÿ nd write the force s grdient: ÿ = 2 y = d { 22 } dy y2. Thus the potentil energy is U(y) = 2 2 y 2, nd the bove theorem tells us tht the totl energy E = 2ẏ y 2, is constnt long solutions of the hrmonic oscilltor. We cn immeditely pply this to see tht the trjectories of solutions in the (y, ẏ) plne must be ellipses. I.e., recll tht ÿ + 2 y = 0 is equivlent to the first order system ( u v ) ( ) ( u v ), () with u = y nd v = ẏ. In (u, v)vribles, constnt energy is E = 2 v u 2, nd this describes n ellipse in the (u, v)plne. Thus the energy immeditely tells us tht solution of the hrmonic oscilltor must evolve on the constnt energy curves which
5 re ellipses. The energy by itself does not, however, tell us nything bout how the solution moves round the ellipse in time. As second nonliner exmple, consider the nonliner pendulum, ÿ + g sin y = 0. L the equtions which describe the swinging motion of mss m t the end of pendulum of length L, swinging bout the equilibrium point y = 0 under the influence of grvity g. (Note the eqution is independent of m!) Here y = θ is the ngle the pendulum mkes with the downwrd verticl rest point of the pendulum, so tht y = θ = 0 is the stble equilibrium t the bottom. (See Strogtz for complete discussion nd derivtion.) Agin, to put this in the form of conservtive system nd find the energy, solve for ÿ nd write the force s grdient: ÿ = g L sin y = d { g } dy L ( cos y). Thus the potentil energy is U(y) = g L ( cos y), nd the bove theorem tells us tht the totl energy 5 E = 2ẏ2 + g ( cos y), (2) L is constnt long solutions of the hrmonic oscilltor. Note tht we hve normlized the ntiderivtive s cos y insted of cos y in order tht the minimum potentil energy occur t y = 0, when the pendulum is t its lowest point. Expnding cos y = y 2 + O()y 4 nd letting g l = 2, we see tht including the lso gives U(y) = g L ( cos y) = g L y2 + O()y 4
6 6 the correct constnt so tht it grees with the potentil energy g L y2 for the hrmonic oscilltor t the leding order. The min difference between the liner hrmonic oscilltor nd the nonliner pendulum is tht the potentil energy 2 y 2 is unbounded in the liner cse, but bounded in the nonliner cse 0 U(y) = g L ( cos y) 2 g L. Note lso tht, to give E the physicl units of energy, sy kg m 2 /s 2, we would need only multiply (2) through by the mss ml 2. (We lost this fctor when we divided through by the mss, nd used the dimensionless ngle θ s n unknown insted of the true distnce long the moving pendulum.) As in the cse of the liner hrmonic oscilltor, we cn gin pply the formul for the energy to get informtion bout the trjectories of solutions in the (y, ẏ) plne. Let s use it to see tht there is criticl vlue of the energy bove which the pendulum swings round with ẏ = θ never chnging sign, but below the criticl energy, there must be point where ẏ = θ chnges sign, the point where it strts swinging the other wy. For this, consider fixed solution of the nonliner pendulum, nd write the energy (2) s 2ẏ2 = E g ( cos y). (3) L where E is constnt long the solution. Now the potentil energy is U(y) = g ( cos y), L which tkes mximum vlue of 2 g L when cos y =. Thus there is criticl energy E = 2 g L, mking for two distinct cses, ccording to whether E > 2 g L or E < 2 g L.
7 If E > 2 g L, then the potentil energy is never lrge enough to mke the right hnd side of (3) zero, in which cse the left hnd side, /2ẏ 2 cn never be zero long the solution, so the pendulum keeps swinging in the sme ngulr direction, round nd round. On the other hnd, if E < 2 g L, then there must be n ngle y(t) in the solution where the potentil energy is equl to the energy E, i.e., E = g L ( cos y(t)). At this point, the KE must be zero, so the velocity ẏ(t) must be zero t tht point. Tht is, there must be point where the pendulum comes to rest. But t this point we clim tht the velocity must ctully chnge sign, tht is, the pendulum must chnge its direction of rottion. Indeed, if it did not, then the cosine would continue chnging in the sme direction t tht point, mking the potentil energy lrger thn the energy E, the right hnd side of (3) would go negtive, mking the KE=ẏ2 2 < 0 on the left hnd side of (3) s well. But the KE is lwys positive becuse ẏ 2 0, so to void the contrdiction, the velocity ẏ must chnge sign t the point where ẏ = 0, s climed. Now generl principle of differentil equtions is tht the qulittive property of solutions cn only chnge t criticl vlues of dimensionless constnts. To find the dimensionless constnt, note tht ẏ chnges sign nd the pendulum chnges direction when E = g ( cos y(t)), L nd it does not chnge direction when E exceeds the lrgest vlue on the right hnd side. Dividing by g L, our condition is tht ẏ chnges sign when the dimensionless constnt γ meets the condition γ = L E = ( cos y(t)). g 7
8 8 The prmeter γ is dimensionless becuse ll terms in physicl eqution must hve the sme dimensions, nd cos y on the right hnd side is dimensionless. Thus the qulittive property of solutions chnges t vlue of the energy which mkes γ equl to the lrgest vlue of cos y bove this vlue the pendulum hs enough energy to swing over the top, but below this it swings bck nd forth. Conclude: The dimensionless constnt is γ = L g E, nd its criticl vlue γ = γ is γ = Energy Conservtion for Point Msses (ODE s) in Three Dimensions The physicl principle of conservtion s expressed in () nd (2) bove cn be generlized to more relistic motion in three dimensions by introducing line integrls. This is topic of Vector Clculus. In this section we describe the mthemticl theory nd its connection to the physics of motion for multiple intercting point msses described by systems of ODE s. This is very beutiful mthemtics, nd completes the ODE picture of conservtion of energy for point msses. In the next section we extend the notion of conservtion to PDE s in the context of the wve eqution. To generlize () to three dimensions, we first need to define the work done by force field, nd for this we must recll the line integrl. So let F(x, y, z) denote vector field in R 3 which we interpret s force field, F(x, y, z) = F (xy, z)i + F 2 (x, y, z)j + F 3 (x, y, z)k, where (F, F 2, F 3 ) re the components of the force t (x, y, z). Then given curve C : r(t) = x(t)i + y(t)j + z(t)k,
9 t b, the work done by F on point mss m s it moves long C is defined s F Tds, (4) C which is clled the line integrl of F long C. Recll tht the line integrl is defined s the limit of the Riemnn sum N F Tds = lim F i T i s, (5) N where is the force, C i= F i = F(r(t i )) T i = r (t i ) r (t i ) is the unit tngent vector on C t time t i, ds is rc length long C becuse ds dt = v(t), v(t) = r (t), v(t) is the velocity vector tngent to C t r(t), nd s usul, t i denote the mesh points obtined by dividing [, b] into N equl intervls of size b t = N, t 0 =, t i = + i t, t N = b. Thus the line integrl is the nturl generliztion of work becuse it is the limit of the sum of force times displcement long the curve. Recll tht by the substitution principle, it hs the nice property tht it cn be evluted in terms of ny prmeteriztion r(t) of the curve C by F Tds = F(r(t)) r (t)dt, (6) C thereby reducing the line integrl for work to n elementry integrl of first qurter clculus. 9
10 0 Things strt to get very interesting, s we sw in Vector Clculus, when we pply these principles to the cse when the force is conservtive. We sy F is conservtive if F (y) = f(y), for some sclr function f(y) clled the potentil. The following mthemticl theory generlizes the fundmentl theorem of clculus to line integrls: Theorem 2. Assume F is conservtive, so F (y) = f(y) for sclr function f. Then F Tds = f(r(b)) f(r()). (7) C Proof: This follows directly from the definition of line integrl s follows: F Tds = F(r(t)) r (t)dt (8) C d = dt F(r(t))dt = F(r(b)) F(r()), where we hve pplied the chin rule nd then the regulr Fundmentl Theorem of Clculus. Comment: The definition of the line inertil in (6) is purely mthemticl, but its mening s work comes to light when we impose the vector form of Newton s second lw F = m. Two problems then immeditely come to mind, one of them esy nd the other one hrd. The esy problem is this: Given curve r(t), find the the totl force tht is creting the motion. This is esy if we hve formuls for the components of r(t) = (x(t), y(t), z(t)) becuse we cn then just clculte = r (t) = (x (t), y (t), z (t)) nd solve for the force F = m. The hrd problem is: Given the force F, find the trjectory of the prticle r(t)
11 tht gives the motion creted by tht force. Then F = m gives you n eqution to solve for, nmely, r (t) = mf. In prticulr, if the force depends on the position, this cn be complicted ODE to solve for the position r(t), nd initil conditions will be required. To keep the following rguments cler, it is importnt to keep in mind tht mny forces my be cting on mss m to crete the motion r(t), nd for ech seprte force we cn define the line integrl of the work done by tht force, but it is only the sum of ll the forces, the totl force, tht cretes the totl ccelertion through F = m. The finl justifiction of (6) s the correct generliztion of work to three spce dimensions comes from the following generliztion of () nd (2). In () we prove tht the work done is equl to the chnge in kinetic energy. In (2) we prove tht Kinetic Energy+Potentil Energy is conserved when the totl force creting the motion is conservtive. For this we must ssume tht F is the TOTAL force which is ccelerting the mss long the curve C. () Assume r(t) is the trjectory of point mss m moving ccording to Newton s lw F = m, mening implicitly tht F is the sum of ll the forces creting the ccelertion, nd therefore (t) = r (t) is the vector ccelertion of the prticle determined by its trjectory r(t). Then W = = = F Tds = m v dt = F(r(t)) r (t)dt m r (t) r (t) dt 2 m d dt {r (t) r (t)} dt = = 2 m v(b) 2 2 m v() 2 = KE. 2 m d { } v(t) 2 dt dt
12 2 Thus the work done by force is equl to the chnge in Kinetic Energy for generl motions in three dimensions! (2) The generl principle of conservtion of energy now holds when the totl force F cting on m is conservtive. I.e., ssume F = U for some sclr function U(y) clled the potentil energy. Applying (7) with f = U we conclude the work done by F long C is not just the chnge in kinetic energy KE (s shown in ()), but lso, but Theorem 2, is equl to minus the chnge in potentil energy P E s well, nmely, we hve: W = F Tds (9) = U(r(b)) + U(r()) = P E. Since by () W equls the chnge in KE, nd by (9) it equls minus the chnge in P E, so KE + P E = 0 ll long the curve, we must hve E = KE + P E = 2 m v(t) 2 + U(r(t)) = const ll long the curve. Conclude: Energy is conserved in conservtive force field. Another wy to express conservtion of energy is in terms of ODE s, nd this is the expression used in differentil equtions. For this, notice tht Newton s lw F = m for conservtive force field cn be viewed s system of equtions for the curve y(t), (we use y in plce of r for ODE s), nmely mÿ (t) = U(y), (0) where y = (x, y, z). Conservtion of energy cn then be reexpressed s the following theorem bout equtions of form (0).
13 3 Theorem 3. The quntity E = 2 m ẏ 2 + U(y), which we cll the sum of the kinetic nd potentil energy, is lwys constnt long solutions of (0). We hve lredy proven this through the notion of the work done W which ws expressed two wys, s chnge in KE nd chnge in PE. But we cn prove this directly without ppeling to the notion of work t ll: Proof: Assume y = r(t) is solution of (0), so ẏ = v(t). Then for this solution, the energy is E(t) = 2 m v(t) 2 + U(r(t)). To prove E(t) is constnt long the solution r(t), we prove Ė = 0 for every t. Writing v 2 = v v nd tking the derivtive using the product rule for the dot product together with the chin rule, we obtin Ė = d { } dt 2 m (v(t) v(t))2 + U(r(t)) () = mv (t) v(t) + U r (t) = m(t) v(t) + U v(t) = U v(t) + U v(t) = 0. We used the product rule d dt (v(t) v(t))2 = v (t) v(t) + v(t) v (t) = 2(t) v(t), nd the chin rule d dt U(r(t)) = = U d U(x(t), y(t), z(t)) dt x x (t) + U y y (t) + U z z (t) = U r (t).
14 4 Thus the energy is constnt becuse when the force is conservtive, the derivtive of the kinetic energy is minus the derivtive of the potentil energy. In fct, this mthemticl expression of conservtion extends immeditely to equtions with ny number of unknowns y = y,..., y n ). For exmple, it extends to the motion of n = 0 23 molecules in continer. The generliztion goes s follows. We sy F is conservtive if F = U for some sclr function U. Then the equtions re (we tke m = ) ÿ = U(y), nd the theorem sys tht energy is conserved in the sense tht the energy E = 2 ẏ 2 + U(y), is constnt long solutions. The proofs of () nd (2) re identicl for generl n s in the cse n = 3. Just retin the vector form of the equtions in ech step, nd interpret y s being in R n insted of R 3. This is very generl mthemticl principle. 3. The Energy of Wves Recll tht the initil vlue problem for the wve eqution is u tt c 2 u xx = 0, u(x, 0) = h(x), (2) u t (x, 0) = k(x), problem tht sks for solution u(x, t) given initil dt functions h nd k. Tht is, the wve eqution is second order in time, so by nlogy with the second order hrmonic oscilltor, we should be ble to ssign the initil vlue u(x, 0) nd the initil velocity u t (x, 0). Now in Section 4 we found solution to the initil vlue problem (2)
15 in terms of left going wve nd right going wve. The result ws tht one prticulr solution of (2) is given by where u(x, t) = f(x + ct) + g(x ct), (3) f(x) = 2 g(x) = 2 with the requirement x 0 x 0 [h (ξ) + c k(ξ) ] dξ + f(0), [h (ξ) c k(ξ) ] dξ + g(0), f(0) + g(0) = h(0). 5 (4) This then defines the functions f nd g in terms of the given h nd k so tht (3) solves (2). But lthough (4) is one solution, how do we know there ren t some other solutions hnging round? Mybe solutions tht cnnot be decomposed into left nd right going wves? The nswer is settled once we prove tht for given h nd k, the initil vlue problem (2) hs one nd only one unique solution. This is wht we cn prove by using the energy. So to this end, we sk, wht is the energy contined in solution u(x, t) of the wve eqution? The nswer is tht t ech point, the kinetic energy is KE = 2 u2 t, (not too hrd to guess), nd the potentil energy is P E = 2 c2 u 2 x, (not so esy to guess, but not unresonble). But given the energy is KE+PE t ech point, the totl energy t given
16 6 time t should be the sum of ll the kinetic plus potentil energies, summed over ll the points t time t. But this is clculus, nd this sum only mkes sense s n integrl. Here is the energy of solution u(x, t) t given time t: E(t) = 2 u2 t + 2 c2 u 2 x dx. But the only rel proof tht this is the correct energy of the wve u(x, t) is the following theorem, which sttes tht the energy so defined, is constnt in time long ny solution of the wve eqution. Theorem 4. If u(x, t) solves the wve eqution, u tt c 2 u xx = 0, then E(t) is constnt in time. Tht is, d dt E(t) = d dt 2 u2 t (x, t) + 2 c2 u 2 x(x, t) dx = 0. (5) Before proving the Theorem, we first demonstrte its utility by using it to prove the uniqueness of solutions of the initil vlue problem (2). Uniqueness: Assuming Theorem 4, we cn prove: Theorem 5. Assume u(x, t) nd v(x, t) re smooth functions of finite energy which both solve the initil vlue problem (2) for the sme given functions h(x) nd k(x). Then for ll x R, t 0. u(x, t) = v(x, t) Proof: Assume then tht both u(x, t) nd v(x, t) solve (2). We prove tht w(x, t) = u(x, t) v(x, t) = 0 for ll (x, t). Since the wve eqution u tt c 2 u xx = 0 is liner nd
17 homogeneous, superposition holds, nd so we know tht w = u v lso solves the wve eqution. Tht is (u v) tt c 2 (u v) xx = u tt c 2 u xx + v tt c 2 v xx = 0. But u nd v stisfy the sme initil conditions, u(x, 0) v(x, 0) = h(x) h(x) = 0, u t (x, 0) v t (x, 0) = k(x) k(x) = 0. Tht is, the function w(x, t) = u(x, t) v(x, t), solves the wve eqution with zero initil dt, w(x, 0) = 0 = w t (x, 0). Thus by Theorem 4, the energy is constnt in time long the solution w(x, t) nd the energy strts out zero, so E(t) = E(0) = 2 w2 t (x, 0) + 2 c2 w 2 x(x, 0) dx = 0, becuse w nd w t re both zero t t = 0. Therefore, t ech time t 0 we hve E(t) = 7 2 w2 t (x, t) + 2 c2 w 2 x(x, t) dx = 0. (6) But the integrnd in the integrl (6) is lwys nonnegtive, 2 w2 t (x, t) + 2 c2 w 2 x(x, t) 0, nd the only wy n integrl over nonegtive function cn be zero is if the integrnd is zero, 2 w2 t (x, t) + 2 c2 w 2 x(x, t) 0. Tht is, the only wy the re under the grph of nonnegtive function cn be zero is if the function is zero. It follows then tht both w t = 0, w x = 0
18 8 for ll x nd t 0. But w t nd w x re the two components of the grdient of w, so we hve w (x, t) = 0. Now the only wy the grdient of function cn vnish is if the function is constnt, so we must hve w(x, t) = const. And since w(x, 0) = 0, it follows tht the constnt must be zero, w(x, t) = u(x, t) v(x, t) = 0. We conclude tht the two solutions u nd v tht solve the sme initil vlue problem must be equl, nd therefore there cn only be one unique solution to the initil vlue problem, s climed. To finish the section, we now give the proof of Theorem 4. The min technique we will need is integrtion by prts for prtil derivtives. It s relly no different thn regulr integrtion by prts so long s you ignore the other vribles hnging round in the expression. For integrtion by prts, ssume you hve two functions u(x) nd v(x), nd use the PDE nottion v x = v (x). Then by the product rule, (uv) x = u x v + uv x, so by the Fundmentl Theorem of Clculus uv x dx = (uv) x u x v dx = uv] + u x vdx. In prticulr, if we ssume the functions u nd v vnish t x = ±, then uv x dx = (uv) x u x v dx = u x vdx.
19 Note tht the formul would be no different if u nd v depended on x nd t, so long s nmely lim u(x, t) = 0 = lim v(x, t), x ± x ± u(x, t)v x (x, t) dx = u x (x, t)v(x, t) dx. All the ction is going on in the vrible x tht is being integrted, nd the t just goes long for the ride. The finl thing we need in order to see tht the energy is constnt long solutions of the wve eqution, is formul for differentiting through n integrl sign. Tht is, consider the generl problem of differentiting G(t) with respect to t when We wnt to see tht G (t) = d dt G(t) = f(x, t)dx = f(x, t)dx. f(x, t)dx. t We cn see this directly by writing the derivtive s limit, nd using properties of limits to pss the limits through the integrl sign. Tht is, d dt f(x, t)dx = lim t 0 = lim { f(x, t + t)dx t { f(x, t + t)dx t t 0 = lim = t 0 f(x, t + t) f(x, t)dx t f(x, t)dx. t 9 } f(x, t)dx } f(x, t)dx The point is tht to justify ll the steps, you only need the differentibility of f nd tht lim x ± f(x, t) = 0 rpidly
20 20 enough tht ll the integrls nd sums re uniformly finite. (Getting the precise conditions stright is topic in Advnced Clculus.) So, to prove the energy E(t) is constnt long solutions, differentite (5), d dt E(t) = = = = = d dt 2 u2 t (x, t) + 2 c2 u 2 x(x, t) dx 2 t u2 t (x, t) + 2 c2 t u2 x(x, t) dx u t u tt (x, t) + c 2 u x (x, t)u xt (x, t) dx u t u tt (x, t) c 2 u xx (x, t)u t (x, t) dx u t { utt (x, t) c 2 u xx (x, t) } dx = 0, where we hve first differentited through the integrl sign, then pplied the chin rule, nd finlly integrtion by prts on the lst term. The finl expression hs the wve eqution s fctor in the integrnd, which therefore vnishes becuse u is ssumed to be solution of the wve eqution t the strt. We conclude tht if the energy is finite, then it is constnt on solutions of the wve eqution, so this completes the proof of Theorem 4. Conclude: We hve shown tht the energy E = 2 u2 t (x, t) + 2 c2 u 2 x(x, t) dx is constnt long solutions of the wve eqution. Using this, together with the linerity of solutions, we prove tht the difference between two solutions stys zero if it strts out zero, thereby proving uniqueness of the initil vlue
21 problem. It follows tht the solution of the initil vlue problem we constructed s the superposition of left going wve nd right going wve, is the only solution, (t lest mong those of finite energy!) 2
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