7 Wave Equation in Higher Dimensions


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1 7 Wave Equaion in Highe Dimensions We now conside he iniialvalue poblem fo he wave equaion in n dimensions, u c u x R n u(x, φ(x u (x, ψ(x whee u n i u x i x i. (7. 7. Mehod of Spheical Means Ref: Evans, Sec..4.; Sauss, Sec. 9. We begin by inoducing a mehod o solve (7. in odd dimensions. Fis, we inoduce some noaion. Fo x R n, le B(x, Ball of adius abou x B(x, Bounday of ball of adius abou x α(n Volume of uni ball in R n nα(n Suface Aea of uni ball in R n. Wih his noaion, he volume of he ball of adius abou x R n, wien as Vol(B(x,, is given by α(n n and he suface aea of he ball of adius abou x R n, wien as S.A.(B(x,, is given by nα(n n. Fo f : R n R, we define he aveage of f ove B(x, as B(x, f(y dy Vol(B(x, B(x, f(y dy We define he aveage of f ove B(x, as f(y ds(y f(y ds(y S.A.(B(x, whee ds(y denoes he suface measue of B(x, in R n. α(n n B(x, f(y dy. f(y ds(y, nα(n n Example. Fo n 3, Vol(B(x, 4 3 π3. fo f : R 3 R, he aveage of f ove B(, is given by f(y dy 3 π π f(ρ, θ, φρ sin φ dρ dθ dφ. 4π 3 B(, B(, Fo n 3, S.A.(B(x, 4π. fo f : R 3 R, he aveage of f ove B(, is given by f(y ds(y π π f(, θ, φ sin φ dθ dφ π π f(, θ, φ sin φ dθ dφ. 4π 4π
2 Ou plan o solve (7. is he following. Fix a poin x R n. Fo >, we define u(x;, u(y, ds(y, he aveage of u(, ove B(x,. Fo, we define u(x;, u(x,. Fo <, we define u(x;, u(x;,. We claim ha fo u smooh, u is a coninuous funcion of, and, heefoe, lim u(x;, u(x,. + In ode o solve (7., we will assume u is a soluion of (7. and look fo an equaion u solves. Noe: We will assume c. Fo c, we can make a change of vaiables o deive he soluion fom he soluion in he case c. Lemma. If u solves u u, x R n, u(x, φ(x u (x, ψ(x, hen u(x;, solves (n u u u, < <, u(x;, φ(x; φ(y ds(y u (x;, ψ(x; ψ(y ds(y fo evey x R n. Poof. u(x;, B(, u(y, ds(y u(x + z, ds(z.
3 u (x;, B(, nα(n n nα(n n nα(n n u(x + z, z ds(z u(y, y x u (y, ds(y ν B(x, B(x, ds(y u (y, ds(y ν u(y, dy u (y, dy by he Divegence Theoem, and using he fac ha u solves he wave equaion, u u. u (x;, u nα(n n (y, dy B(x, which implies which implies n u (x;, u (y, dy. nα(n B(x, ( n u (x;, nα(n n nα(n n n n u (x;,. u (y, ds(y u (y, ds u (y, ds(y ( n u (x;, n u (x;,, (n n u + n u n u. u u (n u 3
4 and Similaly, as claimed. u(x;, u(y, ds φ(y ds φ(x;. u (x;, ψ(x; Soluion fo n 3. We now conside he case of he wave equaion in hee dimensions. soluion of (7. fo n 3. As befoe define he funcion u(x;, such ha u(x;, u(y, ds(y. Assume u is a Nex inoduce a funcion v(x;, such ha v(x;, u(x;, and new funcions g(x; and h(x; such ha g(x; φ(x; h(x; ψ(x; φ( ds(y ψ( ds(y. Lemma 3. Fo each x R n, he funcion v(x;, solves he onedimensional wave equaion on he halfline wih Diichle bounday condiions, v v < <, v(x;, g(x; < < v (x;, h(x; < < v(x;,. Poof. v u [ u + ] u u + u (u + u (u v. 4
5 Nex, Similaly, Now, v(x;, u(x;, u(y, ds(y φ(y ds(y φ(x, g(x; v (x;, h(x;. v(x;, u(x;,. v(x;, solves he onedimensional wave equaion on a halfline wih Diichle bounday condiions, as claimed. Now we use his fac o consuc he soluion of (7.. By d Alembe s fomula, we know ha fo, he soluion v(x;, is given by Now and Now implies Similaly, v(x;, [g(x; + g(x; ] + v(x;, u(x, lim + lim + u(x, lim u(x;, + v(x;, u(x;,. + { [g(x; + g(x; ] + d g(x; + h(x;. d g(x; φ(x; + + h(x; y dy. + g(x; φ(x; φ(y ds(y. h(x; ψ(x; ψ(y ds(y. } h(x; y dy 5
6 he soluion of he wave equaion in R 3 (wih c is given by u(x, [ ] φ(y ds(y + ψ(y ds(y. If φ is smooh, he soluion can be simplified fuhe. In paicula, fo φ smooh, we have d d g(x; d ( φ(y ds(y d d ( φ(x + z ds(z d B(, φ(x + z ds(z + φ(x + z z ds(z B(, B(, ( y x φ(y ds(y + φ(y ds(y φ(y ds(y + φ(y (y x ds(y. And, h(x; ψ(x; ψ(y ds(y. we have u(x, [φ(y + φ(y (y x + ψ(y] ds(y. We noe ha in R 3,. nα(n n 4π he soluion of he IVP fo he wave equaion in R 3 (wih c and φ smooh is given by u(x, [φ(y + φ(y (y x + ψ(y] ds(y. (7. 4π This is known as Kichoff s fomula fo he soluion of he iniial value poblem fo he wave equaion in R 3. Remak. Above we found he soluion fo he wave equaion in R 3 in he case when c. If c, we can simply use he above fomula making a change of vaiables. In paicula, conside he iniialvalue poblem v c v x R n v(x, φ(x v (x, ψ(x. 6 (7.3
7 Suppose v is a soluion of (7.3. Then define u(x, v(x,. Then c implies u is a soluion of u u c v v u u xx x R n u(x, φ(x u (x, c ψ(x. u is given by Kichoff s fomula above. Now by making he change of vaiables, we see ha c v(x, u(x, c, and we aive a he soluion fo (7.3, v(x, [φ(y + φ(y (y x + ψ(y] ds(y. 4πc 7. Mehod of Descen B(x,c In his secion, we use Kichoff s fomula fo he soluion of he wave equaion in hee dimensions o deive he soluion of he wave equaion in wo dimensions. This echnique is known as he mehod of descen. This echnique can be used in geneal o find he soluion of he wave equaion in even dimensions, using he soluion of he wave equaion in odd dimensions. Soluion fo n. Suppose u is a soluion of he iniial value poblem fo he wave equaion in wo dimensions, u u, x R, u(x, φ(x u (x, ψ(x. We will find a soluion in he D case, by using he soluion o he 3D poblem. u(x, x, be he soluion o he D poblem. Define ũ(x, x, x 3, u(x, x,. ũ(x, x, x 3, u(x, x, φ(x, x ũ (x, x, x 3, u(x, x, ψ(x, x. Clealy, ũ(x, x, x 3, is a soluion of he 3D wave equaion wih iniial daa φ(x, x and ψ(x, x, ũ ũ x x ũ x x ũ x3 x 3 ũ(x, x, x 3, φ(x, x, x 3 φ(x, x ũ (x, x, x 3, ψ(x, x, x 3 ψ(x, x. 7 Le
8 Now we can solve he 3D wave equaion using Kichoff s fomula. In paicula, ou soluion is given by ũ(x, x,, [ φ(y + φ(y (y x + ψ(y] ds(y whee B(x, is he ball of adius in R 3 abou he poin x (x, x,. Now we noe ha φ(y ds(y φ(y ds(y 4π φ(y( + γ(y / dy π B(x, whee B(x, is he ball in R of adius abou he poin x (x, x and γ(y ( y x /. y x γ(y ( y x / which implies Similaly, and ( ( + γ(y / /. y x φ(y ds(y φ(y dy. π B(x, ( y x / ψ(y ds(y ψ(y dy π B(x, ( y x / φ(y (y x ds(y φ(y (y x dy. π B(x, ( y x / he soluion of he iniialvalue poblem fo he wave equaion in R (wih c is given by u(x, φ(y + ψ(y + φ(y (y x dy. (7.4 π B(x, ( y x / Again, by making a change of vaiables, we see ha he soluion of he wave equaion in wo dimensions is given by u(x, cφ(y + c ψ(y + c φ(y (y x dy. πc (c y x / B(x,c 8
9 7.3 Huygen s Pinciple Noe ha fo he iniialvalue poblem fo he wave equaion in hee dimensions, he value of he soluion a any poin (x, R 3 (, depends only on he values of he iniial daa on he suface of he ball of adius c abou he poin x R 3 ; ha is, on B(x, c. Tha is o say, disubances all avel a exacly speed c. This is known as Huygens s pinciple. In conas, in wo dimensions, he value of he soluion u a he poin (x, depends on he iniial daa wihin he ball of adius c abou he poin x R. Signals don all avel a speed c. In fac, as we will see, fo n 3 and odd, Huygens s pinciple holds. Tha is, all signals avel a exacly speed c. In even dimensions, howeve, ha is no he case. 7.4 Wave Equaion in R n, n > 3 Ref: Evans, Sec..4. Noe: In his secion, we assume c. Fo c, we can make a change of vaiables o find he soluion. Odd dimensions. Fo he case of odd dimensions, we use he mehod of spheical means as we did fo he case of n 3. Le n k +. Le x R n. Define ( k v(x;, ( k u(x;, ( k g(x; ( k φ(x; h(x; k ( k ψ(x;. ( Noice ha fo k, hese definiions educe o hose funcions inoduced in he case n 3. Fis, we will show ha v(x;, solves he wave equaion on he halfline wih Diichle bounday condiions. Lemma 4. Fo each inege k, fo each x R n, he funcion v(x;, defined above solves v v > v(x;, g(x; v (x;, h(x; v(x;,. The poof elies on he following lemma. Lemma 5. Le φ : R R be C k+. Then fo k,,.... ( d d ( k d ( k φ( 9 ( d d k ( k dφ d (
10 . ( k d ( k φ( whee each β k j is independen of φ. k j β k j j+ dj φ d j ( 3. β k 3 5 (k. Poof. Use inducion. Poof of Lemma 4. [ ( k v d ( k u(x;, ] ( k d ( k u (x;, by Lemma 5 ( k ( d d ( k u (x;, ( k ( d [kk u + k u ] ( k [ ] d k ( k u + u ( k ( [ ] d n k u + u ( k d ( k u ( k d ( k u v Clealy, v(x;, g(x;, v (x;, h(x; and v(x;,. he lemma is poved. Now v(x;, is a soluion of he onedimensional wave equaion on he halfline wih Diichle bounday condiion implies fo, he soluion is given by v(x;, [g(x; + g(x; ] + + h(x; y dy. Recall: u(x, lim u(x;,.
11 Now v(x;, ( k d ( k u(x;, k βj k j j+ j u(x;, j β k u(x;, + β k u (x;, βk k k k u(x;,. k which implies β k u(x;, v(x;, β k u (x;,... βk k k k u(x;,, k u(x;, v(x;, β k βk β k u (x;,... βk k β k k k u(x;,. k [ v(x;, u(x, lim β k v(x;, lim β k lim β k β k βk β k u (x;,... βk k β k [ g(x; + g(x; + [ g(x; + h(x; ] whee β k 3 5 (k. Recall g(x; + ( k ( k φ(x;. Now n k + implies k (n /, and, heefoe, g(x; ( n 3 ( n ] u(x;, k k k ] h(x; y dy φ(y ds(y. And, h(x; h(x; ( ( k ( k ψ(x;. n 3 ( n ψ(y ds(y.
12 implies u(x, γ n ( u(x, γ n [ g(x; + h(x; ] ( + γ n ( n 3 n 3 ( n ( n φ(y ds(y ψ(y ds(y whee γ n 3 5 (n. Even dimensions. As in he case of n dimensions, we use he mehod of descen. In paicula, suppose u(x,..., x n, is a soluion of he wave equaion in R n wih iniial daa u(x,..., x n, φ(x,..., x n and u (x,..., x n, ψ(x,..., x n. Then define ũ(x,..., x n+, u(x,..., x n, φ(x,..., x n+ φ(x,..., x n ψ(x,..., x n+ ψ(x,..., x n. ũ is a soluion of he wave equaion in R n+, whee now n + is odd. fom he fomula above fo he case when he dimension is odd, ou soluion a he poin (x, (x,..., x n,, is given by ũ(x, ( γ n+ + γ n+ ( ( n n ( n ( n φ(y ds(y ψ(y ds(y whee γ n+ 3 5 (n, and whee B(x, is he ball in R n+ of adius abou he poin x (x,..., x n,. Now, φ(y ds(y φ(y ds(y. (n + α(n + n Bu, noice B(x, {y n+ } is he gaph of he funcion γ(y ( y x /. And, similaly, B(x, {y n+ } is he gaph of γ. φ(y ds(y φ(y( + γ(y / dy (n + α(n + n (n + α(n + n B(x, Now ( + γ(y / ( y x /.
13 φ(y ds(y (n + α(n + n B(x, α(n (n + α(n + α(n n α(n (n + α(n + ou soluion fomula is given by u(x, ( γ n+ + γ n+ γ n+ ( ( n n ( n ( n [ ( ( α(n (n + α(n + ( n ( + n Now γ n+ 3 5 (n and B(x, φ(y dy ( y x / B(x, φ(y dy ( y x / φ(y dy. ( y x / φ(y ds(y ψ(y ds(y n ( n ψ(y dy ( y x / φ(y dy ( y x / ]. whee Γ(n is he gamma funcion, α(n πn/ Γ ( n+, Γ(n e x x n dx. γ n+ α(n (n + α(n (n Using popeies of he gamma funcion, namely ha πn/ Γ( n+ (n + π(n+/ Γ( n+3 n (n + Γ( π/ Γ( n+. Γ(m + mγ(m and Γ(/ π /, 3
14 we can conclude ha ( ( ( n + 3 n + n Γ and And, heefoe, ( n + ( n ( n Γ ( Γ (. α(n γ n+ (n + α(n + 4 (n n ( he soluion of he wave equaion in even dimensions is given by u(x, γ n [ ( whee γ n 4 (n n. ( + n ( ( n n B(x, ( n φ(y dy ( y x / B(x, 7.5 Wave Equaion in R n wih a souce. ψ(y dy ( y x / In his secion, we conside he inhomogeneous wave equaion in R n. Fis, ecall Duhamel s Pinciple. If S( is he soluion opeao fo he fisode iniialvalue poblem { U + AU U( Φ, hen he soluion of he inhomogeneous poblem { U + AU F U( Φ ] should be given by U( S(Φ + S( sf (s ds. Now conside he iniialvalue poblem fo he wave equaion in R n, u u f(x, x R n u(x, φ(x u (x, ψ(x. (7.5 4
15 Inoducing a new funcion v u, we can ewie his equaion as [ ] [ ] [ ] [ ] u u + x R n v v f [ ] [ ] u(x, φ(x. v(x, ψ(x o { U + AU F (7.6 whee U [ ] u v F U(x, Φ(x [ ] A [ ] [ ] φ Φ. f ψ Now in ode o solve (7.5, we look fo he soluion opeao S( associaed wih he fisode sysem (7.6. Fis, conside he case n 3. In hee dimensions, we can find he soluion opeao S( by using Kichoff s fomula. Recall ha he soluion of he iniialvalue poblem fo he homogeneous wave equaion in hee dimensions (wih c is given by u(x, 4π [φ(y + φ(y (y x + ψ(y] ds(y, which implies he soluion opeao S( associaed wih (7.6 is given by [ ] [ ] φ [φ(y + φ(y (y x + ψ(y] ds(y 4π S(Φ S( ( ψ [φ(y + φ(y (y x + ψ(y] ds(y. 4π S( sf (s S( s [ ] [ ( f(s ] f(y, s ds(y B(x, s f(y, s ds(y. 4π( s B(x, s 4π( s Now using he fac ha he soluion of (7.6 is given by [ ] u(x, U(x, S(Φ(x + S( sf (x, s ds, v(x, we see ha he soluion of (7.5 is given by he fis componen of U. he soluion of he iniialvalue poblem fo he inhomogeneous wave equaion in hee dimensions (wih c (7.5 is given by u(x, [φ(y + φ(y (y x + ψ(y] ds(y 4π + 4π( s B(x, s 5 f(y, s ds(y ds.
16 Similaly, in wo dimensions, he fis componen of he soluion opeao is given by [ ] φ S (Φ S ( φ(y + ψ(y + φ(y (y x dy. ψ π ( y x / B(x, he soluion of he iniialvalue poblem fo he inhomogeneous wave equaion in wo dimensions (wih c (7.5 is given by u(x, π + B(x, φ(y + ψ(y + φ(y (y x dy ( y x / π( s B(x, s ( s f(y, s dy ds. (( s y x / 7.6 Wave Equaion on a Bounded Domain in R n. In his secion, we conside he iniialvalue poblem fo he wave equaion on a bounded domain Ω R n, u c u, x Ω u(x, φ(x u (x, ψ(x u saisfies ceain bounday condiions on Ω, As befoe, we look fo a soluion using sepaaion of vaiables. In paicula, we look fo a soluion of he fom u(x, X(xT (. Subsiuing a funcion of his fom ino ou PDE, we aive a he equaion T X c T X. This equaion implies he funcions T and X saisfy he following equaion fo some scala λ, T c T X X λ. Consequenly, we ae lead o he following eigenvalue poblem { X λx, x Ω X saisfies ceain bounday condiions on Ω. Suppose we find eigenvalues λ n wih coesponding eigenfuncions X n (x. Then fo each n, we jus need o solve T n ( + c λ n T n (. If λ n is posiive, his means T n ( A n cos( λ n c + B n sin( λ n c. If λ n, his means T n ( A n + B n. 6
17 If λ n is negaive, his means Then defining he funcion T n ( A n cosh( λ n c + B n sinh( λ n c. u(x, n T n (X n (x, fo X n, T n as defined above fo any choice of consans A n, B n, we have found a soluion of he wave equaion on he bounded domain Ω R n, which saisfies ou bounday condiions. Now in ode fo ou iniial condiions o be saisfied, ha is, u(x, φ(x and u (x, ψ(x, we need o choose consans A n, B n such ha u(x, n A n X n (x φ(x and u (x, λ n c λ n B n X n (x + λ n B n X n (x ψ(x. If ou eigenfuncions ae ohogonal, hen we can find coefficiens A n, B n saisfying he above equaions, by muliplying hese equaions by X m fo a fixed m and inegaing ove Ω. Doing so, we see ha ou coefficiens A n ae given by A n X n, φ X n, X n Ω X n(xφ(x dx Ω X n(x dx, and c λ n B n X n, ψ X n, X n B n X n, ψ X n, X n X Ω n(xψ(x dx Ω X n(x dx X Ω n(xψ(x dx Ω X n(x dx fo λ n fo λ n. 7
7 Wave Equation in Higher Dimensions
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