And the solution to the PDE problem must be of the form Π 1

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1 5. Self-Similar Soluions b Dimensional Analsis Consider he diffusion problem from las secion, wih poinwise release (Ref: Bluman & Cole, 2.3): c = D 2 c x + Q 0δ(x)δ() 2 c(x,0) = 0, c(±,) = 0 Iniial release wihin infiniel narrow neighborhood of x = 0, such ha Π (x)/d = δ (x) and L / d. Noe Q 0 has differen dimension as he previous Q because of he cross-secional area S and ime conained in δ (). 1. Dimensional analsis: {c}=ml -3, {D}=L 2 T -1, {Q 0 }=ML -2 (Mass releases per uni cross-secional area) {x}=l, {}=T Thus, we expec 2 Pi groups: = D Q 0 c, Π 2 = x D And he soluion o he PDE problem mus be of he form = f (Π 2 ) or c = Q 0 D f x D Normall we expec dimensional analsis o reduce he number of variables and parameers. Bu here we reduced he number of independen variables from 2 o 1! 2. Transformaion of PDE o ODE: Now we can plug his form back ino he PDE. Firs, he parial derivaives: c = Q 0 2 D f Q x 0 2D 2 f ', c x = Q 0 D f ', 2 c x 2 = Q 0 (D) 3/2 f " For > 0, here is no more injecion: δ () = 0. Afer insering he above ino he PDE: f 2 x 2 D f ' = f " or f "+ ξ 2 f '+ f = 0, (1) 2 where ξ = x D is our new independen variable. We have successfull ransformed he PDE ino an ODE. How abou he iniial and boundar condiions? Noe ha = 0 and x= boh correspond o ξ =, so ha he iniial and boundar condiions can be rolled ino one: 1

2 f( ± ) = 0. (2) Bu we need anoher condiion on f, one ha reflecs he amoun of iniial injecion. This is obained b inegraing he PDE over he following inervals: d [PDE]dx, where = 0 - means jus before = 0. 0 Now he lef-hand-side is c d dx c = dx 0 d = [c(x,) c(x, 0)]dx 0 = c(x,)dx and he wo erms on he righ-hand-side are: 2 c d x dx = d Now we have c 0 x d = 0 because c = 0 a ± ; Q 0 δ()δ(x)dx = Q 0 b virue of he definiion of he dela funcion. c(x,)dx = Q 0, which can be ransformed, using he variable ξ, ino f (ξ)dξ = 1 (3) ODE (1), along wih condiions (2) and (3) will uniquel deermine f(ξ), from which we ge c(x,). We are no concerned wih he acual soluion of he new ODE problem. Raher, he ineresing quesion is: How did we manage o urn a PDE o an ODE? 3. Discussion a. The problem admis a self-similar soluion: if x is scaled b he diffusion lengh (D) 1/2, hen he c(x,) profiles a differen imes can be collapsed ono each oher if c is scaled b Q 0 /(D) 1/2. b. This means ha x and are no reall 2 independen variables; as far as c is concerned, he can be rolled ino one independen variable ξ. c. Similari soluions are happ coincidences in phsical processes. Can we alwas find hem for an PDE s? No. This problem is special in ha here is no inheren lengh scale. Thus, we are no able o form dimensionless groups for each of he variables x,, and c; insead, we have o combine hem and end up wih onl 2 Pi groups. Tha s how we ended up wih he ODE. If we had he release lengh ds or he domain lengh L, he self-similari will be ruined. d. Can we alwas find similari soluions b dimensional analsis? No. Bu we will sud anoher example nex, and hen inroduce he general sreching ransformaion idea for deecing similari soluions. 2

3 6. Similari Soluions b Sreching Transformaion I is rare ha similari soluions can be obained from dimensional analsis. In his secion we inroduce he idea of sreching ransformaion which is a more general procedure for seeking ou similari in PDE problems. The maerials are based on Barenbla ( 5.2) and Bluman & Cole ( 2.5). As a concree example, we will ake Prandl s boundar laer equaion for flow over a fla semi-plane. Afer he boundar laer approximaion (ha viscosi acs onl wihin a hin laer, ha he gradien in he flow direcion (x) is much smaller han in he ransverse direcion (), and ha he pressure is consan in he direcion), he governing equaions are u u x + v u = ν 2 u 2 u x + v = 0 u(x,0) = 0, v(x,0) = 0 u(x, ) = U, u(0, ) = U where U is he free-sream veloci, and ν is he kinemaic viscosi. If ou recall our fluid mechanics, his problem does have a similari soluion (Blasius soluion), and he PDE can be reduced o ODE. (Tr o disinguish he veloci v from he viscosi ν. We could use differen smbols bu hese are he convenional ones.) 1. Would dimensional analsis work? Le s wrie ou he dimensions of all he variables and parameers: {u}={v}={u }=L/T, {ν}=l 2 /T, {x}={}=l There are 2 independen dimensions involved (L and T), and we can consruc 4 dimensionless groups ou of hese. For insance: = u U, Π 2 = v, Π 3 = U x U ν and we expec soluions such as = f (Π 3,Π 4 ), Π 2 = g(π 3,Π 4 )., Π 4 = U ν Plugging hese back ino he equaions, and we will see ha we have NOT achieved a reducion of he number of independen variable. Dimensional analsis has failed o give us he similari soluion. Wh? Even hough he problem has no inrinsic ime or lengh scales, similar o he diffusion problem in he las secion, here are onl 2 independen dimensions (L and T) insead of 3. Thus, i is possible for x and o form heir own Pi groups; he don have o be forced ino a single one. I urns ou ha in his paricular example, a rivial manipulaion can cure he above problem. This is no a general echnique, bu neverheless is fun o illusrae here. We will 3

4 ake his lile deour before marching ino he general echnique ha is he focus of his secion. Based on he phsical insigh ha hings happen a differen scales along he x and direcions, which is he fundamenal idea behind he boundar laer approximaion, we assign wo differen dimensions o x and, L and H, and for he momen preend ha he are differen dimensions. Now he lis of variables and unknowns are scaled as such: {u}={u }= L/T, {v}= H/T, {ν}=h 2 /T, {x}= L, {}= H. There are now 3 independen dimensions involved (L, H and T), and we can consruc onl 3 dimensionless groups ou of hese: = u U, Π 2 = v νu / x, Π 3 = νx /U = ζ. Now we expec a similari soluion in his form: u = U f (ζ ), v = νu x g(ζ). Plugging his ino he original PDE will show ha indeed, we have reduced he PDE problem o a couple of ODEs, whose soluion is deailed in Fluid Mechanics exbooks. For anoher example of such ingenious dimensional analsis, see he Raleigh problem analzed in he nex secion (see also Bluman & Cole, p. 195). We picall seek o increase he number of independen dimensions (as done above) or decrease he number of dimensional parameers (as done in Bluman & Cole s example). 2. Sreching ransformaion The ingenious dimensional analsis mehod is specific o he problems. There is, however, a general scheme for seeking ou possible similari soluions. The scheme someimes goes b he name of renormalizaion groups or invarian ransformaion groups, and is based on raher formalisic mahemaical manipulaions. We will skip he proofs and focus on he echnique iself. Since he essence of similari is ha he soluion is invarian afer cerain scaling of he independen and dependen variables, we consider he following sreching ransformaion, and see if such ransformaions will leave he PDE and he boundar condiions invarian. Consider: U = α a u, V = α b v, X = α c x, Y = α d where α is a posiive number. Under his ransformaion, we have u x = α c a U X, u = α d a U Y, v = α d b V Y, 2 u = α 2d a 2 U 2 Y. 2 Plugging hese ino he original PDE s and boundar condiions, we ll see wha choices of a, b, c, d ma mainain he invariance of he problem. The coninui equaion ields: c a = d b. 4

5 The hree erms of he momenum equaion requires: c 2a = d a b = 2d a. Noe ha he firs equaion above is idenical o he preceding equaion, and hus he momenum equaion adds onl 1 addiional consrain on he power indices. Finall he boundar condiions require a = 0 because for he problem in he new variables o be invarian, he non-homogeneous BC should remain as U (X, ) = U. Now we have 3 equaions consrain he 4 indices, and we rewrie he ransformaion as: U = u, V = v ε, where ε = α d. X = ε 2 x, Y = ε This ransformaion will leave he problem he same as before, in he new sreched and scaled variables. The fac ha his one-parameer famil of ransformaions will mainain he invariance of he PDE problem reveals he inrinsic self-similari of he problem. In oher words, if we srech he coordinae b a facor ε, hen we mus srech x b ε 2 and he veloci componen v b 1/ ε in order o collapse he veloci profiles. From his argumen, we recognize ha u, v x, x shall remain he same no maer how we srech he coordinaes. These are known as he invarians of he ransformaion, and immediael sugges he following similari soluion: u = f (ζ ) v = 1 x g(ζ), wih he similari variable ζ = x. This is he same form as obained from he ingenious dimensional analsis, aside from a few consan facors. Noe ha we reached he conclusion here no hrough dimensional consideraions, bu hrough he idea of invariance under general sreching ransformaions. Now i s a simple maer o plug hese forms ino he original PDE problem, and ransform i ino he following ODE problem: ζ ν f "+ f ' 2 f g = 0, ζ f ' 2g' = 0,, f ( ) = U, f (0) = 0, g(0) = 0 he soluion of which will no be of immediae ineres o us here. Noe ha he wo BC s a x = 0 and = boh projec ono ζ =. 5

6 3. Similari Soluion for he Raleigh Problem The Raleigh problem is anoher classical example wih a self-similar soluion. Consider he ransien moion in a viscous fluid induced b a fla plae moving in is own plane. Iniiall boh he plae and he fluid are a res. Saring a = 0, he plae moves wih a consan veloci U 0. The Navier-Sokes equaions, simplified for his problem, along wih he iniial and boundar condiions, can be wrien as: u = ν 2 u 2 u(,0) = 0, u(0,) = U 0, u(,) = 0 u(,) U 0 (a) Dimensional analsis. From he following dimensions: {u}={u 0 }= L/T, {ν}=l 2 /T, {}= T, {}= L, we can make 3 Pi groups, sa u/u 0, U 0 /ν, U 02 / ν, and here is no reducion o ODE. Again, we can pla ricks here, b eiher increasing he number of independen dimensions, or decreasing he number of parameers, so as o reduce he number of Pi groups. Using he phsical observaion ha he viscous diffusion happens along he direcion, while he primar flow is in he x direcion, we can inroduce differen lengh scales: {u}={u 0 }= L/T, {ν}=h 2 /T, {}= T, {}= H. Now here are onl 2 Pi groups: = u U 0, Π 2 = ν and we can r a similari soluion of he form u(,) = U 0 f ν. Alernaivel, we can reduce he number of parameers b scaling u b U 0, and calling u(,) = u(,) /U 0 he new dependen variable. Now he problem has one less parameer, and again onl admis 2 Pi groups. In he following, however, le us carr ou he formal procedure of sreching ransformaion as an exercise. (b) Sreching ransformaion. Consider: U = α a u, Y = α b, T = α c, where α is a posiive number. Under his ransformaion, we have 6

7 u = α c a U T, 2 u = α 2b a 2 U 2 Y. 2 To mainain invariance of he PDE, we require c a = 2b a, or c = 2 b. The boundar condiion u ( 0, ) = U 0 requires a = 0. Thus, we have he following ransformaion ha renders he problem invarian: U = u, Y = ε, T = ε 2, which ε = α b. This ransformaion dicaes ha and be ransformed in a coordinaed wa. Thus u and ζ = / shall be our new variables ha remain unchanged for an sreching α or ε: u = f = f (ζ ). This reduces he original PDE ino he following ODE problem: 2ν f "+ ζ f ' = 0, f (0) = U, f ( ) = 0 which can be inegraed analicall o give: ζ f = c 1 exp z2 0 4ν dz + c 2. Noing ha exp z2 0 4ν dz = 2 ν exp( ξ 2 ) dξ = πν, he wo consans of 0 inegraion are deermined: c 1 = U 0 / πν and c 2 = U 0. Finall he soluion can be wrien in erms of he complemenar error funcion: ζ f = U 0 erfc 2 ν = U erfc 0 wih erfc(x)=1-2 exp z 2 π dz. 0 x ( ) 4ν, 7

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