THE WAVE EQUATION 43. (S) Le u(x, ) be a soluion of he wave equaion u u xx = 0. Show ha Q43(a) (c) is a. Any ranslaion v(x, ) = u(x + x 0, + 0 ) of u(x, ) is also a soluion (where x 0, 0 are consans). par hand-in for week 9 b. Any dilaion v(x, ) = u(λx, λ) of u(x, ) is also a soluion (where λ is consan). c. Any derivaive of u(x, ) is also a soluion (assume u is as smooh as you like). d. Define he energy densiy e(x, ) by e = 2 u2 + 2 u2 x and he momenum densiy p(x, ) by p = u u x. Soluion: i. Show ha e = p x and e x = p. ii. Show ha boh e and p solve he wave equaion, i.e. e e xx = 0 and p p xx = 0. a. Clearly v xx (x, ) = u xx (x + x 0, + 0 ) and v (x, ) = u (x + x 0, + 0 ), herefore ( v)(x, ) = ( u)(x + x 0, + 0 ) = 0. (Noaion: u(x, ) = u (x, ) u xx (x, ) for -dimensional problems, and in general u(x, ) = u (x, ) 2 u(x, ) in higher dimensions.) b. v xx (x, ) = λ 2 u xx (λx, λ) and v (x, ) = λ 2 u (λx, λ), herefore ( v)(x, ) = λ 2 ( u)(λx, λ) = 0. c. To see for example ha u x is a soluion differeniae u u xx = 0 wih respec o x o ge u x u xxx = 0 and hen change he order of differeniaion (we can do his if our funcion is sufficienly smooh) o ge (u x ) (u x ) xx = 0. d. i. Assume as usual ha u is a wice coninuously differeniable soluion of he wave equaion u u xx = 0. Then ( e = 2 u2 + ) 2 u2 x = u u + u x u x ( e x = 2 u2 + ) 2 u2 x = u u x + u x u xx p = (u u x ) = u u x + u u x x and p x = (u u x ) x = u x u x + u u xx e p x = u (u u xx ) = 0 e x p = u x (u u xx ) = 0 ii. Assume now ha u is hree imes coninuously differeniable so ha we can ake second derivaives of e and p. We have e e xx = (e ) (e x ) x = p x p x = 0 and p p xx = (p ) (p x ) x = e x e x = 0
44. (S) Le u(x, ) be a real soluion of he damped wave equaion Q44 is u u xx + u = 0. (9) Show ha he energy E() = ( 2 u (x, ) 2 + 2 u x(x, ) 2) dx decreases (make any convenenien assumpions of regulariy and decay a infiniy). Soluion: Clearly we have o assume ha u is C 2 o make sense of he equaion. Moreover, we need o make sure ha all he generalized Riemann inegrals ha come in he proof exis. This will be rue if all inegrands below decay sufficienly fas a infiniy. We work exacly as in he case of he wave equaion: muliply boh sides of (9) by u o ge u u u u xx + u 2 = 0. Now u u = ( ) 2 u2 and u u xx = (u u x ) x u x u x = (u u x ) x ( ) 2 u2 x. ( ) ( ) 2 u2 (u u x ) x + 2 u2 x + u 2 = 0. Inegrae wih respec o x over R o ge (( ) 2 u2 ( ) ) (u u x ) x + 2 u2 x + u 2 dx = 0. Working as in Problem 85 we can show ha if lim u (x, )u x (x, ) = 0 for all (add his o our lis of decay x ± a infiniy assumpions, if i s no already here), hen (u u x ) x dx = 0. We are hus lef wih This can be wrien as In oher words, he energy is decreasing. (( ) ( ) 2 u2 + 2 u2 x + u 2 ) dx = 0. d ( d 2 u2 + ) 2 u2 x dx + u 2 dx = 0. de d () = u 2 dx 0. 45. (E) For a (coninuously) differeniable funcion u(x, ) define D + u and D u by compleion of hand-in for Week 9 D + u = u + u x, D u = u u x Show ha if u is wice coninuously differeniable hen D + D u = D D + u = u u xx Soluion: A sraighforward calculaion. 46. (S) Le u(x, ) be a C 2 soluion of he wave equaion in one space dimension. Le ABCD be a parallelogram on he (x, ) plane whose sides are characerisics. Using he general soluion of he wave equaion, or oherwise, show ha u(a) + u(c) = u(b) + u(d) (0) Soluion: In characerisic coordinaes ξ := x +, η := x he soluion is of he form u(ξ, η) = f(ξ) + g(η). The characerisics become ξ = cons and η = cons so he parallelogram looks like his:
y C ξ = ξ2 η = η2 B D ξ = ξ η = η A Then O u(a) = u(ξ, η ) = f(ξ ) + g(η ), u(b) = u(ξ 2, η ) = f(ξ 2 ) + g(η ) u(c) = u(ξ 2, η 2 ) = f(ξ 2 ) + g(η 2 ), u(d) = u(ξ, η 2 ) = f(ξ ) + g(η 2 ) x so boh sides of (0) are equal o f(ξ ) + f(ξ 2 ) + g(η ) + g(η 2 ). 47. (E) Le f : R 2 R and a, b : R R be coninuously differeniable funcions. Show ha d d b() f(, s)ds = b() f (, s)ds + f(, b())b () f(, )a () (Hin: Consider he funcion of hree variables F(x, y, ) = y x f(, s)ds. Then b() f(, s)ds = F(, b(), ). Use he chain rule from mulivariable calculus.) Soluion: We have Bu d d b() b() f(, s)ds = d d( F(, b(), )) f(, s)ds = F(, b(), ) = F x (, b(), )a () + F y (, b(), )b () + F (, b(), ) d d b() F x (x, y, ) = f(, x) F y (x, y, ) = f(, y) F z (x, y, ) = y x f (, s)ds b() f(, s)ds = f(, )a () + f(, b())b () + f (, s)ds
48. Le u, v : R R R be wo soluions of he wave equaion in one space dimension wih iniial daa u(x, 0) = f (x), u (x, 0) = 0, v(x, 0) = f 2 (x), v (x, 0) = 0 where f, f 2 : R R are C 2 funcions vanishing ouside some bounded inerval [ M, M]. Define Q 0, Q : R R R by Q 0 = u v u x v x, Q = u v x u x v a. (S) Express Q 0 and Q in erms of he iniial daa f and f 2. (Hin: Use d Alember s formula). b. (P) Show ha Q 0 L 2 (R R) C f L 2 (R) f 2 L 2 (R) Q L2 (R R) C f L 2 (R) f 2 L 2 (R) where C is some posiive consan independen of he iniial daa. (Hin: Use he expression you found in (a) and change o characerisic coordinaes). We have used he following noaion: For a funcion g : R R we define ( + /2 g L2 (R) = g(x) dx) 2, and similarly for a funcion G : R R R we define G L 2 (R) = ( provided ha all hese generalised Riemann inegrals exis. ) /2 G(x, ) 2 dxd [Ans (a) 2 [f (x + )f 2(x ) + f (x )f 2(x + )], 2 [f (x + )f 2(x ) f (x )f 2(x + )],] Soluion: D Alember s formula gives Using hese we ge u(x, ) = 2 (f (x + ) + f (x )), v(x, ) = 2 (f 2(x + ) + f 2 (x )) u x (x, ) = 2 (f (x + ) + f (x )) u (x, ) = 2 (f (x + ) f (x )) v x (x, ) = 2 (f 2(x + ) + f 2(x )) v (x, ) = 2 (f 2(x + ) f 2(x )) Q 0 (x, ) = u (x, )v (x, ) u x (x, )v x (x, ) = 4 [(f (x + ) f (x ))(f 2(x + ) f 2(x )) (f (x + ) + f (x )) (f 2(x + ) + f 2(x ))] = 2 [f (x + )f 2(x ) + f (x )f 2(x + )] Q 0 (x, ) 2 dxd = R 4 (x + )f 2(x ) + f (x )f 2(x + ) 2 dxd 2 R 2 2 (x + )f 2(x ) 2 dxd () R 2 + 2 (x )f 2(x + ) 2 dxd (2) R 2
where we have used (a + b) 2 2a 2 + 2b 2. We change o characerisic coordinaes ξ = x +, η = x. For he inegral in () we have: (x + )f 2(x ) 2 dxd = R 2 (ξ) 2 2(η) 2 dξdη 2 R R = ( ) ( ) 2 (ξ) 2 dξ 2(η) 2 dη R R Similarly he inegral in (2) is: herefore = 2 f L 2 (R) 2 f 2 L 2 (R) 2 R 2 (x )f 2 (x + ) 2 dxd = 2 f L2 (R) 2 f 2 L2 (R) 2 R 2 Q 0 (x, ) 2 dxd 2 f L2 (R) 2 f 2 L2 (R) 2 The proof for Q is similar. 49. (S) Le g : R 3 R be given by g(x) = g 0 (r) = r 2. We wish o solve he following IVP for he wave equaion in hree space dimensions wih radial daa: We sar by guessing ha u u = 0, u(x, 0) = 0, u (x, 0) = g(x) = g 0 (r) u(x, ) = u 0 (r, ) where u 0 : R R R is even in r i.e. u 0 ( r, ) = u 0 (r, ) (why is his a good guess?). a. Show ha he wave equaion becomes r(u 0 ) r(u 0 ) rr 2(u 0 ) r = 0 b. Show ha his means ha w(r, ) = ru 0 (r, ) solves he wave equaion in one space dimension wih iniial daa w(r, 0) = 0 and w (r, 0) = rg 0 (r). c. Use d Alember s formula o find w. w w rr = 0 d. Conclude ha u 0 (r, ) = 2r r+ r ρg 0 (ρ)dρ e. Find u. [Ans: (c) r 3 + r 3, (e) x 2 + 3 ] Soluion: Since he iniial daa are radial funcions (hey depend only on r = x ) we guess ha he soluion will also be radial (his can acually be proved), say u(x, ) = u 0 ( x, ) = u 0 (r, ), for some funcion u 0 (r, ) defined for r 0, R. Since he iniial daa are smooh even funcions of r we guess ha u 0 could be exended o all r s as a smooh even funcion of r, i.e. u 0 (r, ) = u 0 ( r, ). Using problem 32(a) wih n = 3 we have u u = (u 0 ) [(u 0 ) rr + 2r ] (u 0) r herefore, for all r 0 and R we have r(u 0 ) r(u 0 ) rr 2(u 0 ) r = 0 This is saisfied for r < 0 oo. Indeed, using u 0 ( r, ) = u 0 (r, ), we have (u 0 ) r (r, ) = (u 0 ) r ( r, ) (u 0 ) rr (r, ) = (u 0 ) rr ( r, ) (u 0 ) (r, ) = (u 0 ) ( r, )
for r < 0, R we have r(u 0 ) (r, ) r(u 0 ) rr (r, ) 2(u 0 ) r (r, ) = = r(u 0 ) ( r, ) r(u 0 ) rr ( r, ) + 2(u 0 ) r ( r, ) = [( r)(u 0 ) ( r, ) ( r)(u 0 ) rr ( r, ) (u 0 ) r ( r, )] = 0 Define w : R R R by w(r, ) = ru 0 (r, ). Then w w rr = r(u 0 ) r(u 0 ) rr 2(u 0 ) r = 0 i.e. w solves he wave equaion in one space dimension wih iniial daa w 0 (r, 0) = ru 0 (r, 0) = 0, (w 0 ) (r, 0) = r(u 0 ) (r, 0) = r 3 By d Alember s formula hence w(r, ) = 2 u 0 (r, ) = r+ r w(r, ) r ρ 3 dρ = r 3 + r 3 = r 2 + 3 u(x, ) = x 2 + 3 50. (S) Le u(x, ) be a soluion of he hree-dimensional wave equaion, and le v(, x) = u(, x + x 0 ). Verify ha v also saisfies he wave equaion. If v(x, 0) = 0 and v (x, 0) = g(x), x R 3 i is shown in class ha v(0, ) = 4π y = g(y)ds. Deduce ha u(x 0, ) = u (y, 0)dS. 4π y x 0 = Soluion: Wih v i (x, ) = x i v(x, ), ec, we have Hence resul. Nex, v (x, 0) = u (x + x 0, 0) = g(x). Hence v i (x, ) = u i (x + x 0, ) v ii (x, ) = u ii (x + x 0, ) v(x, ) = u(x + x 0, ) +v (x, ) = u (x, ). u(x 0, ) = v(0, ) = 4π = 4π = 4π y = y = g(y)ds y x 0 = u (y + x 0, 0)dS u (y, 0)dS where y = y + x 0.