Solutions from Chapter 9.1 and 9.2

Transcription

1 Soluions from Chaper 9 and 92 Secion 9 Problem # This basically boils down o an exercise in he chain rule from calculus We are looking for soluions of he form: u( x) = f( k x c) where k x R 3 and k is some fixed vecor Wih his seup we can easily calculae ha: u = cf x u = k x f y u = k y f z u = k z f Here we are wriing k = (k x k y k z ) for he componens of k (which are fixed) By simply repeaing his compuaion we see ha: 2 u = c 2 f 2 xu = k 2 xf 2 yu = k 2 yf 2 z u = k 2 zf Combining all of his we have he relaion: 2 u c 2 u = c 2 ( k 2 )f If we wan he expression on he righ hand side above o vanish we simply need o choose eiher of: Thus we need eiher of: k 2 = f = k = f(ξ) = aξ + b where ξ R is a single real variable and a b R are fixed consans Secion 92 problem # 5 For his problem suppose ha u solves he hree dimensional wave equaion: 2 u c 2 u = u( x) = f( x) u ( x) = g( x) Le us also suppose he iniial daa f and g are such ha hey vanish ouside of a ball of radius ρ Tha is: () From he Kirchhoff formula: u( x ) = 4πc 2 f( x) = g( x) for ρ x [ g( x) ds x + 4πc 2 f( x) ds x ] i is clear ha one always has u( x ) = whenever he poin x is such ha he sphere S c ( x ) cenered a his poin only gives he value zero for g Sc( x ) and f Sc( x ) Tha is whenever he

2 2 iniial daa f g boh vanish on his sphere We are now looking for he se of poins ( x ) such ha boh of he following are rue: g Sc( x ) f Sc( x ) From he condiions () all we need o do is o choose spheres S c ( x ) such ha: S c ( x ) { x < ρ} = I is no hard o check ha his is accomplished whenever: or: x c ρ c + ρ x (You should check his for yourself!) Summing up his means ha he soluion u mus vanish ouside of a spherical shell cenered a he origin of inner radius r = c ρ and ouer radius R = c + ρ Of course for imes where c ρ < he soluion jus vanished ouside of ball of radius R Secion 92 problem # 6 We ll answer hese quesions in order: In his problem he cener poin of he sphere of radius R should probably be called x To do he compuaion we ll se everyhing up in polar coordinaes cenered a x and such ha in hese coordinaes he posiive z axis conains he line segmen beween x and he cener of he ball of radius ρ which was he old origin Is no oo hard o see ha afer a ranslaion and roaion (which does no effec hings like surface areas) we may reduce o his special case regardless of he original posiion of x In his configuraion we see he area we are rying o compue is ha of a spherical cap on he sphere of radius R cenered a he origin where he boundary circle of he cap is precisely he inersecion of he sphere of radius R a ( ) and he ball of radius ρ cenered a he poin ( x ) where x is he original (before ranslaion) cener of he sphere of radius R To compue he area of he spherical cap we now only need o know he wides angle ha his cap makes wih he z-axis If we call his angle φ hen from basic calculus we see ha he area may be compued as: (2) A cap = 2π φ sin(φ)r 2 dφdθ To compue he angle φ we simply use he law of cosines which says ha: ρ 2 = x 2 + R 2 2 x R cos(φ )

3 3 (3) Noice ha we are compuing hings wih respec o he riangle wih verices ( ) ( x ) and he poin of inersecion of our wo spheres which lies in he firs quadran of he z-x plane (you may wan o draw a picure of all of his o ge a feeling for i) Therefore we see ha he angle φ is defined implicily via he relaion: cos(φ ) = ρ2 x R 2 2 x R This easily allows us o compue he inegral (2) and he answer is: ( ) A cap = 2πR 2 cos(φ) φ ( ρ = 2πR 2 2 x R 2 ) + 2 x R = πr( ρ 2 ( x R) 2) x Noice ha his las formula is no valid when here is some nonrivial inersecion beween hese wo spheres In he case where ρ+r < x here is no inersecion and herefore no area On he oher hand in he case where x + R < ρ here is also no inersecion bu his ime he enire sphere of radius R cenered a x is conained in he ball of radius ρ cenered a he origin Therefore in his second case he area is jus 4πR 2 To compue he soluion for his specific iniial value problem we use he Kirchhoff formula which in his case says ha: u( x ) = 4πc 2 A χ ρ ( x) ds x where χ ρ is he cuoff funcion: (4) χ( x) = { where x ρ oherwise Now his inegral is jus A imes he area of he inersecion of he sphere of radius c cenered a x and he solid ball of radius ρ cenered a he origin which is wha we jus compued There are hree cases depending on he geomery (as explained a he end of he las paragraph) and we see from using formula (3) ha he answer is: A ) if x < ρ c (ρ (5) u( x) = A 2 ( x c) 2 4c x if ρ c x ρ + c if ρ + c < x For problems c) and d) is no so easy o ype up a picure so I ll jus skip i The above formula can be used o obain any informaion one needs o draw a picure

4 4 (6) If we compue he soluion u( x + v) for large and x fixed wih v = c and x < ρ hen we will be in he middle case of formula (5) above Tha is we are rying o compue he limi: ( ρ 2 ( x + v c) 2) lim u( x + v) A 4c x + v ( ρ 2 ( x + v c) 2) A 4c x + v To complee he compuaion of he limi i suffices o be o compue: lim ( x + v c) = lim ( x + v c) x + v 2 (c) 2 x + v + c x x v x + v + c x x v x + v + c = x v c Plugging his compuaion ino he line (6) above we have: lim u( x + v) = ( x + v + c) x + v + c) A ( ρ 2 4c 2 ( x v) 2 ) c 2 ) Secion 92 Problem # 9 We ll solve each of hese separaely: This problem is almos idenical o par of Problem 5 above In paricular if he iniial daa f g boh vanish ouside of he sphere of radius ρ (cenered a he origin) hen he corresponding soluion u mus vanish inside he sphere of radius R() = ρ c for laer imes R() such ha < R() To obain his uniform behavior les jus look a he siuaion where he iniial displacemen f( x) is idenically zero In he more general case where i is no zero one can also prove decay on he order of by essenially he same argumen bu one needs o use a change of variables in he Kirchhoff inegral formula as we had discussed in reciaion We are now rying o show ha if u( x ) is given by he formula: u( x ) = 4πc 2 g( x) ds x

5 5 hen one has he uniform esimae: u( x ) C (7) (8) (9) To prove his i clearly suffices o show ha one simply has a uniform bound on he inegral in he Kirchhoff formula Tha is we only need o show ha: g( x) ds x C for some consan C which will depend on g bu no on he poin ( x ) To prove his bound we ll need o use he following basic esimae from analysis: g( x) ds x g( x) ds x where he bound holds for any funcion where he inegrals converge This las bound reduces our ask o esimaing some basic inegrals because by he condiions we are placing on he iniial daa g is no oo difficul o see ha: g( x) M χ ρ ( x) where M = max g and χ ρ is he radial cuoff funcion defined on line (4) above Here ρ is chosen so ha g( x) whenever ρ x Using his noaion he bound (8) implies ha: g( x) ds x M χ ρ ( x) ds x Therefore (by changing he definiion of he consan C) we see ha o esablish (7) we only have o prove he bound: χ ρ ( x) ds x C where C is a uniform consan ha only depends on ρ (and herefore on he properies of g) bu no on he poin ( x ) The proof of (9) is quie easy if one does no insis on being oo precise The poin is ha is more or less geomerically obvious ha he sphere of radius c cenered a he poin x only inersecs he ball of radius ρ in a surface pach of area uniformly bounded by a consan depending only on ρ I ll leave he deails of his o he reader (ie see if you can come up wih a precise esimae)