MATH 8.5 COURSE NOTES - CLASS MEETING # 0 8.5 Inroducion o PDEs, Fall 0 Professor: Jared Speck Class Meeing # 0: Inroducion o he Wave Equaion. Wha is he wave equaion? The sandard wave equaion for a funcion u(, x ) (where R, x R n ) is (.0.) u + u = 0. c (.0.) is second order and linear. The consan c > 0 is called he speed (his erminology will be lengh jusified as our course progresses), and i has dimensions of ime. Noe ha heurisically speaking, if we le c, hen (.0.) becomes Laplace s equaion. However, as we will see, in order o have a well-posed problem for (.0.), we will need o specify Cauchy (i.e. iniial daa) for u and also u. The fac ha we need o specify Cauchy daa is in sark conras o Laplace s equaion, bu is analogous o he hea equaion. The fac ha we need o specify wo pieces of Cauchy daa is conneced o he fac ha he wave equaion is second order in ime.. Where does i come from? Equaion (.0.) arises in an incredible variey of physical conexs, especially hose involving disurbances ha propagae a a finie speed. Le s discuss how he wave equaion arises as an approximaion o he equaions of fluid mechanics. For simpliciy, le s only discuss he case of spaial dimension. The equaions of fluid mechanics, which are known as he Euler equaions, ake he following form in + dimensions: (.0.a) ρ + x ( ρv) = 0, (.0.b) ( ρv) + x( ρv ) = x p, where ρ(, x) is he fluid mass densiy, v(, x) is he fluid velociy, and p(, x) is he pressure. Equaion (.0.a) implies he conservaion of mass, and equaion (.0.b) is Newon s second law: he rae of change of fluid momenum is equal o he force, which is creaed by he pressure gradien (i.e., he x p erm). The Euler equaions are highly nonlinear, and we are very far from obaining a full undersanding of how heir soluions behave in general. A fundamenal aspec of fluid mechanics is ha he sysem is no closed because here are no enough equaions. A common mehod of achieving closure is by choosing an equaion of sae, which is a relaionship beween he fluid variables. This relaionship is ofen empirically deermined. A commonly sudied equaion of sae is (.0.3a) p = Kρ γ where γ > and K > 0 are consans. For fuure use, we noe ha under (.0.3a), we have
MATH 8.5 COURSE NOTES - CLASS MEETING # 0 (.0.3b) x p = Kγρ γ x ρ, (.0.3c) Also for fuure use, we differeniae (.0.a) deduce ha xp = Kγρ γ xρ + Kγ( γ ) ρ γ ( x ρ), wih respec o and (.0.b) wih respec o x o (.0.4a) ρ + ρ x v + v x ρ + ρ x v + v x ρ = 0, (.0.4b) ρ xv + v x ρ + ρ x v + v x ρ + x ρ + 4ρv x v + x ρ x v = x p. The heory of acousics is based on linearizing (i.e. hrowing away he nonlinear erms) he equaions (.0.4a) - (.0.4b) around he saic soluions ρ = ρ = cons > 0, v = 0, p = p = cons > 0. These saic soluions describe a fluid a res. Le s assume ha we make a small perurbaion of his soluion, i.e., ha v is small, and ha (.0.5) ρ = ρ + δ, where δ(, x) is a small funcion. Using he expansion (.0.5), we now hrow away (wih he help of (.0.3c)) all of he quadraic and higher-order small erms from (.0.4a) - (.0.4b) o obain he following approximaing sysem (he quaniies ha are assumed o be small are v, δ, and all of heir parial derivaives): (.0.6a) δ + ρ x v = 0, (.0.6b) γ ρ x v = Kγρ xδ. Comparing (.0.6a) and (.0.6b), we see ha δ verifies he following approximaing equaion (.0.7) δ + Kγρ γ xδ = 0. Equaion (.0.7) is a wave equaion for he perurbaion δ(, x )! I models he propagaion of sound waves. This is he linear heory of acousics! Noe ha he speed associaed o he equaion (.0.7) depends on he background densiy ρ (.0.8) c = Kγρ γ. When γ >, higher background densiy faser sound speed propagaion. Remark.0.. For air under normal amospheric condiions, γ =.4 is a prey good model. 3. Some Well-Posed Problems Recall ha well-posed PDEs have hree imporan properies: Given suiable daa, a soluion exiss. The soluion is unique. The soluion depends coninuously on he daa.
MATH 8.5 COURSE NOTES - CLASS MEETING # 0 3 Perhaps he mos ofen sudied well-posed problem for he wave equaion is he global Cauchy problem in + n spaceime dimensions: (3.0.9a) n u(, x) + x u(, x) = 0, (, x) R R, (3.0.9b) u( 0, x) = f( x ), x R n, (3.0.9c) u( 0, x) = g( x ), x R n. We now menion some addiional well-posed problems in he case of + dimensions. We assume ha u verifies he wave equaion for (, x) (, ) [0, L] and ha Cauchy daa is given: (3.0.0a) (3.0.0b) (3.0.0c) u(, x) + x u(, x) = 0, (, x) R [ 0, L], u( 0, x) = f( x ), x [ 0, L], u( 0, x) = g( x ), x [ 0, L ]. Unlike in he case of (3.0.9a) - (3.0.9c), because of he finieness of he inerval [0, L], we need o supplemen (3.0.0a) - (3.0.0c) wih addiional condiions in order o generaed a well-posed problem. Here are some well-known ways of generaing a well-posed problem; hey are essenially he same as in he case of he hea equaion. () Dirichle daa: also specifying u(, 0) = a( ), u(, L) = b( ) for > 0 () Neumann daa: also specifying x u(, 0) = a( ), x u(, L) = b( ) for > 0 (3) Robin daa: also specifying x u(, 0) = ku(, 0), x u(, L) = ku(, L) for > 0, where k > 0 is a consan (4) Mixed daa: e.g. one kind of daa a x = 0, and a differen kind a x = L 4. + spaceime dimensions Le s consider he wave equaion wih speed c in + dimensions: (4.0.) c τ u( τ, x) + x u(τ, x) = 0. Le s firs noe he following fac: if f, g are any differeniable funcions, hen u( x, τ) = f( x cτ) and u( x, τ) = g(x + cτ) solve (4.0.). The firs is called a righ-raveling wave, and he second is called a lef-raveling wave. To visualized wave propagaion in + dimensions, you can imagine ha he graph of f( ) and g( ) are ranslaed o he righ/lef a a speed c. This gives a good idea of wha wave moion looks like in + dimensions. In paricular, he ampliudes of he raveling wave soluions are preserved in ime. As we will see, wave propagaion in higher dimensions is quie differen. In higher dimensions, he ampliudes decay in ime due o he spreading ou of he waves. You will sudy he case of + 3 spaial dimension in one of your homework exercises; you will show ha in his case, he ampliudes decay a a rae of order as. Remark 4.0.. No all wave soluions in + dimensions are raveling waves; see Theorem 4.. By making he change of variables = cτ, we can ransform equaion (4.0.) ino a wave equaion wih speed equal o (4.0.) u(, x) + xu (, x) = 0.
4 MATH 8.5 COURSE NOTES - CLASS MEETING # 0 This makes our life a bi easier. Le s now consider he global Cauchy problem by supplemening (4.0.) wih he iniial daa (4.0.3) u(0, x) = f(x), u(0, x) = g(x). As we will see, (4.0.) + (4.0.3) has a unique soluion ha has a nice represenaion. Theorem 4. ( d Alember s formula). Assume ha f C ( R) and g C ( R ). Then he unique soluion u(, x) o (4.0.) + (4.0.3) saisfies u C ([ 0, ) R) and can be represened by d Alember s formula: z=x+ (4.0.4) u(, x) = ( f( x + ) + f( x )) + g z dz. ( ) z = x Remark 4.0.3. For he wave equaion c u + xu = 0 formula (4.0.4) is replaced wih = + (4.0.5) u(, x) = ( f( x + c) + f( x c)) + c z =x c z x c g( z) dz. Remark 4.0.4. Equaion (4.0.4) illusraes he finie speed of propagaion propery associaed o he wave equaion. More precisely, he value of he soluion a (, x) is only influenced by he iniial daa inerval {( 0, y) x y x + }; changes o he iniial daa (4.0.3) ouside of his inerval have no effec on he soluion a (, x ). We will reexamine his propery laer in he course wih he help of energy mehods. Proof. To derive (4.0.4), i is convenien o inroduce a change of variables called null coordinaes: (4.0.6) q = x, (4.0.7) s = + x. The chain rule implies he following relaionships beween parial derivaives: (4.0.8) q = ( x ), s = ( + x ), (4.0.9) = q + s, x = s q. The operaors q and s can be viewed as direcional derivaives in he (, x) Caresian spaceime direcion.5(, ) and.5(, ) respecively. These null direcions, which are someimes called characerisic direcions, are exremely imporan. In he fuure, we will discuss he noion of a characerisic direcion in a general seing. I is now easy o see ha (4.0.) akes he following form in null coordinaes: (4.0.0) s q u = 0. Inegraing (4.0.0) wih respec o s, we have ha (4.0.) q u = H(q ), where H is a funcion of q.
MATH 8.5 COURSE NOTES - CLASS MEETING # 0 5 Noe ha he value of q is he same for he pair of Caresian spaceime poins ( τ, y) and ( 0, y τ ). Thus, using he iniial condiions (4.0.3), we have ha (4.0.) q u(τ, y) = q u(0, y τ) = ( ( x )u)(0, y τ) = (g(y τ) f ( y τ )). Similarly, inerchanging he parial derivaives in (4.0.0) o deduce s qu = 0, we conclude ha (4.0.3) su( τ, y) = (g(y + τ) + f ( y + τ )). Adding (4.0.) and (4.0.3), and using (4.0.8), we have ha (4.0.4) u(, x) = (f (x + ) f (x ) + g(x + ) + g(x )). Inegraing (4.0.4) in ime wih respec o from 0 o, and again using he iniial condiions (4.0.3), we have ha f( ) (4.0.5) u(, x) = u(0, x x ) + (f(x + ) f(x) + f(x ) f(x)) + τ g ( x + ) + g ( x τ ) dτ τ=0 z= x+ = ( f( x + ) + f( x )) + g z dz, ( ) z = x where o derive he las equaliy, we made he inegraion change of variables z = x + τ for he g( x + τ) erm, and he change of variables z = x τ for he g( x τ) erm. We have hus derived (4.0.4). Wihou a lo of addiional effor, we can exend Theorem 4. o apply o he following iniial + boundary value PDE in + dimensions; he resul is saed and proved in he nex corollary. This PDE would arise in he sudy of e.g. he following idealized problem: a descripion of he propagaion of waves on an infiniely long vibraing sring wih one end fixed. Furhermore, he corollary will laer play a role in our exension of Theorem 4. o he case of + 3 dimensions. Corollary 4.0.. Le f C ([ 0, )), g C ([0, )), and assume ha f(0) = g(0) = 0. Then he unique soluion o he following + dimensional iniial + boundary value problem (4.0.6a) u(, x) + xu (, x) = 0, (, x) [ 0, ) ( 0, ), (4.0.6b) u(, 0) = 0, [ 0, ), (4.0.6c) u( 0, x) = f( x ), x (0, ), (4.0.6d) u( 0, x) = g( x ), x ( 0, ) saisfies u C([0, ) [0, )). Furhermore, i can be represened as z x (4.0.7) u(, x) = (f = + (x + ) + f(x )) + z= x g( z) dz, if 0 x, z x f x f x = + ( ( + ) ( )) + z x g z dz, if 0 x = ( ).
6 MATH 8.5 COURSE NOTES - CLASS MEETING # 0 Proof. The idea is ha if we exend u o be odd in x, hen we can reduce he problem o he case of Theorem 4.. Moivaed by his, we ine u(, x), if 0, x 0, (4.0.8) ũ(, x) = { u (, x ), if 0, x 0, (4.0.9) (4.0.30) f(x), f( x), if x 0, if x 0, f ( x ) = {, ( ) g(x) = { g x, if x 0, g( x ), if x 0. Since u(, x) solves (4.0.6a), i follows ha (, x) R R wih iniial daa u ( 0, x) = f ( u (, x) is a soluion o he wave equaion (4.0.) for x), u (, x) = g (x). Thus, by (4.0.4), we have ha (4.0.3) ũ(, x) = z=x+ (f ( x + ) + f (x )) + g(z) dz. z= x The expression (4.0.7) now easily follows from considering (4.0.3) separaely in he spaceime regions {(, x) 0 x} and {(, x) 0 x }, and from he iniions (4.0.8) - (4.0.30); noe ha in he case {(, x) 0 x}, since g is odd, he par of he inegral from x o x cancels and hus he only ne conribuion comes from he inegraion inerval [ x, x + ].,
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