The Wave Equation. Units. Recall that [z] means the units of z. Suppose [t] = s, [x] = m. Then u = c 2 2 u. [u]

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1 The Wave Equation UBC M57/316 Lecture Notes c 014 by Phiip D. Loewen Many PDE s describe wave motion. The simpest is u tt = c u xx. Here c > 0 is a given constant, and u = u(x, t is some quantity of interest at position x and time t. Depending on the appication of interest, u coud represent sound pressure in air (or water, or rock, microwave ampitude in a waveguide, eectrica potentia (votage in a cabe, or atera string dispacement in a musica instrument. Units. Reca that [z] means the units of z. Suppose [t] = s, [x] = m. Then [ ] [ ] u = c u [u] ( t x s = [u] m [c] m [c] =. s No matter what units u has, c has units of veocity. Does this mean something? A. Derivation Motion of an Eastic String Consider an eastic string whose equiibrium position is on the x-axis. Linear density of the string is ρ; typicay [ρ] = kg/m. The string can move, but ony transversay, i.e., each partice moves at right anges to x-axis. Write u(x, t for the dispacement at time t of the partice whose equiibrium position is x. Bowed String on [0, ] Let τ = τ(x, t denote tension in string at position x, time t: note τ(x, t 0 for a x, t. ( You can t push on a rope. Let θ = θ(x, t denote the ange of eevation for the tangent ine at point x: note π/ < θ(x, t < π/ for each x, and tanθ(x, t = u x (x, t. Assumptions. (i Transverse motion ony (see above. (ii Sma sopes: for a x (0,, sope u x (x, t is so sma that sin θ(x, t cos θ(x, t 1, sin θ(x, t cos θ(x, t = tanθ(x, t = u x(x, t. ] ] [Note: tanθ sin θ = [θ + θ [θ θ3 6 ± = θ3 + O(θ5 extremey sma for θ 0.] Fie 014notes, version of 9 Juy 014, page 1. Typeset at 15:08 August 1, 014.

2 PHILIP D. LOEWEN Transverse Force Baance (u-direction. Study forces on string segment based in interva [x, x + h] at instant t. Assuming θ > 0 here, we have... Right end (x + h: Tension force pus right and up. Vertica component has magnitude τ(x + h, t sinθ(x + h, t τ(x + h, tu x (x + h, t (by (ii. Left end (x: Tension force pus eft and down. Vertica component has magnitude τ(x, t sinθ(x, t τ(x, tu x (x, t (by (ii. Interior points: Externa force per unit ength in u-direction is F ext, some function of u, x, t. Newton II: ma = F tota, so ρhu tt (x CM, t τ(x + h, tu x (x + h, t τ(x, tu x (x, t + hf ext. Taking h 0 + gives 0 = 0 (wow. Dividing by h > 0 and then sending h 0 + gives [ ] τ(x + h, tux (x + h, t τ(x, tu x (x, t im [ρu tt (x CM, t] = im + F ext. h 0 + h 0 + h This eads to the genera wave equation [Variabe inear density is OK.] ρ(xu tt (x, t = x (τ(x, tu x(x, t + F ext. Longitudina Force Baance (x-direction. Net force in x-direction is F = τ(x + h, t cosθ(x + h, t τ(x, t cosθ(x, t + hg ext, where G ext is some externa force per unit ength, possiby depending on u, x, t. To sustain assumption (i [transverse motion ony], this force must be 0: τ(x + h, t cosθ(x + h, t τ(x, t cosθ(x, t = hg ext. Simpest Case. If G ext = 0 (no externa forces, divide by h and send h 0 to get 0 = τ(x, t cosθ(x, t, x whence τ(x, t cosθ(x, t = f(t must be independent of x. We are assuming cos θ 1, so τ(x, t = f(t is independent of x. Without externay changing the tension, its variation wi be due excusivey to stretching of the string, which we are negecting. Thus τ(x, t wi actuay be a constant. When tension τ > 0 and density ρ > 0 are both constant, we get u tt = τ ρ u xx. This fits pattern above, with c = τ/ρ. Check the units: [ ] τ = N ( kg m/s m = ρ kg/m kg/m =. s As expected, c = τ/ρ has units of veocity: Good. ( Fie 014notes, version of 9 Juy 014, page. Typeset at 15:08 August 1, 014.

3 The Wave Equation 3 Interesting Aternative. If the x-axis points straight up, then the externa force acting on [x, x + h] is its weight: this gives In this case we woud have hg ext = mg = ρhg. τ(x + h, t cosθ(x + h, t τ(x, t cosθ(x, t = ρhg. Dividing by h > 0 and sending h 0 + gives (τ(x, t cosθ(x, t = ρg. x Hence τ(x, t cosθ(x, t = ρgx + C for some C. Again using cos θ 1 and assuming τ has no time-dependence, we get τ(x = τ(0 + ρgx. Finay, if the string is hanging freey with its oose end at eve x = 0, then τ(0 = 0, so τ(x = ρgx. In this case our wave equation (assuming constant density ρ becomes ρu tt = x (ρgxu x, i.e., u tt = g x (xu x = gu x + gxu xx. I have seen this outside my office window! We study this in detai ater, if time permits. B. Boundary Conditions and Boundary Vaue Probems Boundary Conditions. If string segment [0, h] has eft end attached to a spring with force constant k (Hooke s Law, arguments above give this modification of ( : ρhu tt (x CM, t τ(h, tu x (h, t ku(0, t + hf ext. Taking h 0 + here gives 0 = τ(0, tu x (0, t ku(0, t; in constant-tension case one gets τu x (0, t ku(0, t = 0. Three cases arise: (i k = 0 [Free vertica motion at x = 0]: u x (0, t = 0. (ii k [Camped string (no motion at x = 0]: u(0, t = 0. (iii 0 < k < [Some springy resistance at x = 0]: u x (0, t αu(0, t = 0. (α = k/τ. Other Interpretations. For pipes of air, u(x, t measures the excess pressure above ambient at ocation x and time t. Therefore u(0, t = 0 if the end of the pipe at x = 0 is open (e.g., a fute has both ends open, whereas u x (0, t = 0 if the end of the pipe at x = 0 is cosed (e.g., bowing across the open top of a soda botte. Reference: Iain G. Main, Vibrations and Waves in Physics (3/e, pages Fie 014notes, version of 9 Juy 014, page 3. Typeset at 15:08 August 1, 014.

4 4 PHILIP D. LOEWEN Boundary Vaue Probem. A we-formed probem for the 1D wave equation in 0 < x < has 3 ingredients. Thinking about the atera vibrations of a tight string makes each one seem reasonabe. (PDE expresses Newton s nd aw: concavity drives acceeration. [More genera forms possibe here.] u tt = c u xx 0 < x <, t > 0. (BC force-baance reation at each end of string. Choose between (i u(0, t = 0 = u(, t... fixed ends, (ii u x (0, t = 0 = u x (, t... free ends, (iii u x (0, t au(0, t = 0 = u x (, t bu(, t... springy ends, (iv u(0, t = φ(t, u(, t = ψ(t... end positions decreed by prescribed fcns of time [e.g., skipping rope] (v τu x (0, t = φ(t, τu x (, t = ψ(t... given time-varying vertica forces act on ends of rope. Cdx (i (iii are homogeneous; (iv (v are not [uness φ = 0 = ψ]. Aow different cdx at different ends. (IC initia conditions. Need two: (i u(x, 0 = f(x, 0 < x <... initia position, (ii u t (x, 0 = g(x, 0 < x <... initia veocity. Both f and g must be given in probem statement. C. Separation of Variabes and Modes Consider the simpest wave equation for 0 < x <, ignoring initia conditions for now. (PDE u tt = c u xx, 0 < x <, t > 0, (BC x = 0: Choose fixed (u = 0 or free (u x = 0, t > 0, x = : Choose fixed (u = 0 or free (u x = 0, t > 0, Trying u(x, t = X(xT(t in (PDE eads to T (t c T(t = X (x X(x = 444 tan ( πα, ( for some α ( 1, 1. This form of the separation constant is deiberatey chosen to be ugy, in order to make a point. (But notice that any rea constant can be expressed this way, for some α ( 1, 1. Spit ( : ( πα T (t 444c tan T(t = 0, (1 X (x 444 tan ( πα X(x = 0. ( Fie 014notes, version of 5 Juy 014, page 4. Typeset at 15:08 August 1, 014.

5 The Wave Equation 5 Different (BC above have different consequences for X. E.g., suppose both ends fixed: u(0, t = 0 = u(, t gives X(0T(t = 0, X(T(t = 0, t > 0. This forces X(0 = 0 = X(, so we have an eigenvaue probem for X: ( πα X (x 444 tan X(x = 0, 0 < x < ; X(0 = 0 = X(. ( nπx The eigenfunctions are known (FSS: X n (x = sin for n = 1,, 3,.... The corresponding vaues of α are too horribe to contempate. But the eigenfunctions are a we need to know to express ( nπx u(x, t = T n (t sin for some T n s. Pug this into PDE: 0 = u tt c u xx ( [ nπx = T n (t sin ( ] ( nπ nπx c T n (t sin [ ( ( nπc nπx = T n (t + T n (t] sin. By FSS theory, this gives (for each n [ ( nπc 0 = T n(t + T n (t] ( nπc = T n (t = A n cos t + δ n for some ampitude constant A n and phase shift constant δ n determined by the initia conditions. So a typica soution [basic wave eq n, fixed-end case] wi ook ike u(x, t = ( nπc A n cos t + δ n sin ( nπx Notice: We never had to know the horribe α n s in ( and use them in (1. The anaogue of (1 came up naturay when pugging our series form in the PDE, and the appropriate constants were provided automaticay. Modes. Simpest nontrivia soutions have exacty one nonzero term of product form, ike ( ( 5πc 5πx cos [choose δ 5 = 0, A 5 = 1, a other A n = 0], t sin ( ( 15πc 15πx cos t sin [choose δ 5 = π/, A 15 = 1, a other A n = 0].. Fie 014notes, version of 5 Juy 014, page 5. Typeset at 15:08 August 1, 014.

6 6 PHILIP D. LOEWEN ( nπx For fixed n, the basic eigenfunction shape X n (x = sin is preserved but with a time-varying ampitude that varies harmonicay. The time-variation (what you hear has anguar frequency of nπc ; the mode shape variation (what you see has anguar frequency nπ. Note that when discussing osciatory signas in time, ike sound waves or radio waves, we usuay don t use anguar frequency (radians per second; instead we use Hertz (cyces per second. Since one cyce equas π radians, the time-variations associated with motion in mode n have frequency nπc rad sec = nπc rad sec 1 cyce π rad = nc cyce sec = nc Hz. D. Fourier s Ring and Traveing Waves In cassic wave-motion probems driven by u tt = c u xx, 0 < x <, t > 0, we are we acquainted with boundary conditions that wi ead to eigenfunction series from one of the Big Four famiies. Separating u(x, t = X(xT(t works out as foows: ( nπx [FSS] u(0, t = 0 = u(, t = X n (x = sin, n = 1,, 3,.... ( nπx [FCS] u(0, t = 0 = u(, t = X n (x = cos ( (n 1 [HPSS] u(0, t = 0 = u x (, t = X n (x = sin ( (n 1 [HPCS] u(0, t = 0 = u x (, t = X n (x = cos, n = 0, 1,, 3,.... πx, n = 1,, 3,.... πx, n = 1,, 3,.... In probems with FSS or FCS eigenfunctions, the choice = π is convenient because then the eigenfunction famiies reduce to sin(nx and cos(nx. For HPSS and HPCS probems, the convenient choice is = π/, because then the eigenfunction famiies become sin((n 1x and cos((n 1x. With the right vaue of, the appropriate postuate for each of these probems becomes some specia case of the foowing genera pattern: u(x, t = 1 A 0(t + [A n (t cos(nx + B n (t sin(nx]. ( In detai, choosing a A n (t = 0 gives the FSS postuate when = π; choosing a B n (t = 0 gives the FCS postuate for = π; choosing a A n (t = 0 and a B k (t = 0 when k is even gives the HPSS postuate for = π/; choosing a B n (t = 0 and a A k (t = 0 when k is even gives the HPCS postuate for = π/. But ( has another, more direct, interpretation: it s aso the natura FFS postuate for the probem on < x < with periodic boundary conditions. So that setup, informay known as Fourier s Ring, captures a the others. Fie 014notes, version of 01 August 014, page 6. Typeset at 15:08 August 1, 014.

7 The Wave Equation 7 Infinite Extent. At each fixed instant t, the function u(x, t in ine ( is certain to be π-periodic with respect to x. (This is true for any choice of time-varying coefficients A n (t and B n (t. Therefore u(x, t is actuay defined for a rea x, not just the x-vaues in the physica interva of interest. And, as we wi soon show, the behaviour of soutions in the physica segment gets easier to predict once we understanding the behaviour of u(x, t on the whoe rea ine. Periodicity and Visuaization. For any π-periodic function f, the identity f(x + π = f(x, x R, (0 means that knowing the vaues of f on any interva of ength π is equivaent to knowing the vaues of f everywhere. To iustrate this, it s convenient to change the etter x to θ, and to imagine that θ is the centra ange that abes a particuar point on the unit circe. Instead of stretching out a representative interva ike π < θ π onto a horizonta axis, we imagine wrapping it around the circe and joining the ends. Now (assuming f is continuous the graph of f has the parametric representation (cos θ, sinθ, f(θ, θ R. On the graph, the periodicity identity abeed (0 above (where now x = θ corresponds to the fact that the anges θ and θ + π seect the same point on the unit circe. Here are two views of the same function to iustrate the idea: y=f(θ A π periodic function f(θ \pi 0 \pi θ, 3π<θ<3π Signa f(θ graphed above unit circe z x y Fie 014notes, version of 01 August 014, page 7. Typeset at 15:08 August 1, 014.

8 8 PHILIP D. LOEWEN With these observations in mind, we now change the etter x to θ and imagine u = u(θ, t having the unit circe for its domain. This exacty captures the periodicity property of u in the FFS series form (. As noted above, each of the Big Four eigenfunction series soutions can then be obtained by insisting on 0-vaues for certain coections of coefficient functions, and by restricting the physica interva of interest to a suitabe sub-arc of the whoe circe. The physica domains for each case are shown here: [FFS] the whoe ring, π < θ < π; [FSS] haf the ring, 0 < θ < π (reca = π; [FCS] haf the ring, 0 < θ < π (reca = π; [HPSS] the quarter-ring 0 < θ < π/ (reca = π/; [HPCS] the quarter-ring 0 < θ < π/ (reca = π/. Modes. In (, the modes are the functions cos(nx and sin(nx. Pugging ( into the wave equation shows how each mode moves: Ä n (t + n c A n (t = 0, Bn (t + n c B n (t = 0. For probems with zero initia veocity, both ODE famiies have cosine-type soutions, and we have u(θ, t = 1 a 0 + [a n cos(nct cos(nθ + b n cos(nct sin(nθ]. ( This is an infinite superposition of simpe separated pieces, and in each one the index n infuences both the mode shape and the frequency with which it osciates in time. Traveing Waves. Reca that in ine (, u t (θ, 0 = 0 for a θ. Define Then reca the identities f(θ = u(θ, 0 = 1 a 0 + [a n cos(nθ + b n sin(nθ]. sin(a ± B = sin(a cos(b ± cos(a sin(b, ( cos(a ± B = cos(a cos(b sin(a sin(b. ( Adding the two equations condensed into the singe ine in ( gives sin(a cos(b = 1 [ sin(a B + sin(a + B]. For the two equations packed into ine (, adding and subtracting both produce usefu resuts: sin(a sin(b = 1 [cos(a B cos(a + B], cos(a cos(b = 1 [cos(a B + cos(a + B]. Fie 014notes, version of 01 August 014, page 8. Typeset at 15:08 August 1, 014.

9 The Wave Equation 9 Adapt these for use in ( : with A = nθ and B = nct, we have cos(nct cos(nθ = 1 [ cos(n[θ ct] + cos(n[θ + ct]], cos(nct sin(nθ = 1 [ sin(n[θ ct] + sin(n[θ + ct]]. Spitting a 0 = 1 a a 0 then gives ( u(θ, t = 1 1 a 0 + [a n cos(n[θ ct] + b n sin(n[θ ct]] ( 1 1 a 0 + [a n cos(n[θ + ct] + b n sin(n[θ + ct]] = 1 f(θ ct + 1 f(θ + ct. Shifts. For any number p, the curve y = g(x p is an exact copy of y = g(x, shifted to the right by p units. Thus y = g(x ct is an exact copy of y = g(x, shifted right by p = ct units. That is, the shift increases with time at the constant rate c: the whoe curve moves to the right, with speed c, and no change in shape. Likewise y = g(x+ct represents a curve moving to the eft, with speed c, no change in shape. In our wave-equation probem, the profie f(θ = u(θ, 0 spits into two equa parts f = 1 f + 1 f, and then these parts move to the eft and the right at speed c. The movies in cass ceary show this! Exercise. Show that for genera interva 0 < x <, and for each Big-Four eigenfunction famiy, a wave BVP with zero initia veocity is soved by the appropriate version of u(x, t = 1 f(x ct + 1 f(x + ct, where f(x = u(x, 0, 0 < x <. Unwinding. After understanding how u(θ, t moves in graphs based on the unit circe, it is time to undo the change of variabes x = θ and think again about stretching the variabe x out aong the straight infinite rea axis. Now u(x, 0 admits a πperiodic extension caed f(x, defined for a rea x, and this function defines the shapes that side sideways aong the rea ine. Graphica Soutions. To sove a standard string probem with any combination of fixed and free ends, and zero initia veocity,... (1 Buid the extension f appropriate to the eigenfunctions of the probem. This provides the initia condition for the whoe rea axis. ( Haf the extended initia condition f wi trave to the right with speed c, the other haf to the eft; neither shape wi change. At each instant t, the sum of these two contributions gives the soution for u. This mathematica motion Fie 014notes, version of 01 August 014, page 9. Typeset at 15:08 August 1, 014.

10 10 PHILIP D. LOEWEN happens on the entire x-axis, and the segment in the interva 0 < x < correcty captures the actua physica motion on the interva of interest. In particuar, the BC s wi aways be satisfied. (3 Refections [good exam question]: Puses hitting a pinned end bounce back upright; puses hitting a free end bounce back inverted. (Optics appications: Newton s Rings, 3D gasses in mirror. Verification. For u(x, t = 1 f(x + ct + 1 f(x ct, we have u(x, 0 = f(x = f(x at every point of the basic interva, whie the Chain Rue gives u t (x, t = c f (x + ct c 1 f (x ct = u t (x, 0 = c f (x c 1 f (x = 0. Thus the initia conditions check out. The Chain rue aso shows that this function satisfies the PDE. Ony the BC s present a rea chaenge. We show two cases here, and encourage readers to practice on the others. Fixed/Fixed. In probems where the BC s are u(0, t = 0 = u(, t, the FSS is appropriate. So in the soution form u(x, t = 1 f(x + ct + 1 f(x ct, the function f is odd and -periodic. Consequenty u(0, t = 1 f(ct + 1 f( ct = 0 (since f is odd, u(, t = 1 f( + ct + 1 f( ct = 1 f( + ct + 1 f( ct = 0 (since f is -periodic and odd. Free/Free. In probems where the BC s are u x (0, t = 0 = u x (, t, the FCS is appropriate. So in the soution form u(x, t = 1 f(x + ct + 1 f(x ct, the function f is even and -periodic. This impies that f is odd and -periodic. The Chain Rue gives u x (x, t = 1 f (x + ct + 1 f (x ct. Consequenty u x (0, t = 1 f (ct + 1 f ( ct = 0 (since f is odd, u x (, t = 1 f ( + ct + 1 f ( ct = 1 f ( + ct + 1 f ( ct = 0 E. D Aembert s Soution (since f is odd. For any two smooth functions φ and ψ of one variabe, possiby unreated to each other, the function u(x, t = φ(x + ct + ψ(x ct ( wi satisfy the wave equation u tt = c u xx. This is easy to check using the Chain Rue. But there s more: Every smooth soution of the wave equation has the form ( Fie 014notes, version of 31 Juy 014, page 10. Typeset at 15:08 August 1, 014.

11 The Wave Equation 11 for we-chosen φ and ψ. We have aready seen how the choices φ = ψ = 1 f come up in separation-of-variabes soutions associated with a of our favourite eigenfunction famiies. Now we wi show that the more genera form in ( covers every conceivabe soution. Suppose u = u(x, t is a soution of the wave equation... smooth enough that u xt = u tx everywhere. Introduce new variabes r = x ct, s = x + ct; observe x = 1 (s + r, t = 1 (s r. Let c ( s + r U(r, s = u(x, t = u, s r. c Then use the Chain Rue to cacuate ( s + r U r (r, s = u x, s r [ ] ( 1 s + r + u t c, s r [ 1 ] c c and simiary [ 1 U rs (r, s = [ + 1 c = 1 ( s + r 4 u xx, s r c ]{ u xx ( s + r ] { u tx ( s + r, s r [ ] ( 1 s + r + u xt c, s r [ ] } 1 c c, s r [ ] ( 1 s + r + u tt c, s r [ ]} 1 c c 1 ( s + r 4c u tt, s r = 0. c This shows s U r = 0, so U r is independent of s, i.e., U r (r, s = Φ(r. This forces U(r, s = φ(r + ψ(s for some smooth φ and ψ. Concusion: Every smooth soution of u tt = c u xx can be expressed as u(x, t = φ(x ct + ψ(x + ct ( for suitabe choices of φ and ψ. This expression has a egitimate caim to be caed the genera soution of u tt = c u xx. Matching Given IC s. Now suppose IC s u(x, 0 = f(x and u t (x, 0 = g(x are given. What φ and ψ shoud be used in (? First, Simiary, f(x = u(x, 0 = φ(x + ψ(x for a x. (1 g(x = u t (x, 0 = [ cφ (x ct + cψ (x + ct] t=0, so ψ (x φ (x = 1 c g(x. Fie 014notes, version of 31 Juy 014, page 11. Typeset at 15:08 August 1, 014.

12 1 PHILIP D. LOEWEN This is required to hod for a rea x. So we write it again with the dummy variabe p repacing x and then integrate: x p=0 [ψ (p φ (p] dp = 1 c x p=0 g(p dp ψ(x φ(x = ψ(0 φ(0 + 1 c x p=0 g(p dp. ( Now add equations (1 and ( to get ψ(x = 1 f(x + 1 [ψ(0 φ(0] + 1 c Subtract (it s simiar and the resut is φ(x = 1 f(x 1 [ψ(0 φ(0] 1 c x 0 x 0 g(p dp. g(p dp. Once again, these formuas are vaid for a rea x, so we can repace x throughout by any rea-vaued expression. Thus when we use these forms in the soution form (, we get u(x, t = φ(x ct + ψ(x + ct = 1 f(x ct + 1 f(x + ct + 1 c x+ct x ct g(r dr. This is D Aembert s Soution for u tt = c u xx on the whoe x-axis with initia dispacement f and initia veocity g. Reconciiation. How does D Aembert s soution reate to the probem on a finite string? Before answering, we must note that the f in D Aembert s soution must have a of R for its domain in order to use that formua for arge t-vaues. Let s consider the case g = 0. Then to start from the eigenfunction series and and express that soution in form ( requires doing an appropriate eigenfunction extension and appying D Aembert s formua to the function f, as outined above. To work backwards, supposing that a soution of form ( obeys the given BC s and respects the condition of zero initia veocity, one can deduce that u must have the form written above, and f must have the expected periodicity and symmetry properties. (Suppementary homework and practice probems. Domain of Infuence. An observer at the point P with position x istening at time t wi observe an ampitude u(x, t determined by the the initia dispacement f at the points ct units to the eft and right of x, together with a contribution combining initia veocity information between the two. This can be iustrated by drawing a two-dimensiona cone: information in the shaded region infuences the vaue u(x, t, whie information outside it gets ignored. (As time advances, the cone moves upward, and incudes more and more of the (x, t pane. So information that was previousy ignored eventuay reaches the observer. Information traves at speed c. Fie 014notes, version of 31 Juy 014, page 1. Typeset at 15:08 August 1, 014.

13 The Wave Equation 13 t (x,t x ct x+ct x Fie 014notes, version of 31 Juy 014, page 13. Typeset at 15:08 August 1, 014.