Crack Propagation and the Wave Equation on time dependent domains

Transcription

1 Cack Popagation and te Wave Equation on time dependent domains A Majo Qualifying Poject Repot submitted to te Faculty of WPI in patial fulfillment of te equiements fo te Degee of Bacelo of Science in Matematics. Cole Jeznac Poject Adviso: Pofesso Cistope J. Lasen Apil 6, 18

2 1 Abstact Tis poject investigates some poblems associated wit solutions of te wave equation on time dependent domains, a poblem wose solution is elevant to modelling cack popagation in mateials. Mainly, te use of bijective tansfomations fom vaying domains to a fixed domain complicate te PDE, and te elationsip between te beavio of te solution to te oiginal equation and te tansfomed solution becomes difficult to pin down. Despite tese issues, uniqueness of a solution to te poblem can still sometimes be poved. Intoduction Modelling and undestanding gowing cacks emains a matematical callenge. In ode to possibly explain te bancing effect commonly exibited in pysical models, we would like to figue out te diection of te lagest singulaities in tangential deivatives local to te cack fomation. Similaly, to estimate te speed at wic te cack popagates at initiation, we estimate te enegy local to te cack fomation in time inteal fo a ball of adius. In ode to pefom tese calculations, te bounday value poblem wit a time dependent domain can be efomulated as a moe complicated one wit a fixed domain, but only if te speed of te cack, c, is less tan te wave speed [1]. Despite tis, we seek to pove uniqueness of a solution tat does not equie tis efomulation o te estiction of speed c. 3 Calculating enegy flowing into B in [, τ] In ode to estimate speed at initiation, we estimate te enegy in te ball of adius local to te cack fomation. Te integal we wis to calculate is τ EB, τ = u t u B n, 1 B wee B is te ball centeed at te tip consideed to be te oigin, wit adius, and τ. 1

3 At time t between and τ, we ave te following: On te top alf of B wee u t and u ae nonzeo say, fo θ between θ t and π wee θ t is between and π we ave tat u t = 1, u = 1,. We paametize as follows: θ t θ π, xθ = cosθ, yθ = sinθ

4 In wic θ t = accos t = acsin t. Ten note tat we get fo a fixed t between and τ: B u t u B n = wic simplifies to π π θ t dx u t θ uθ B nθ + dθ 1 1, cos θ, sin θ sin θ + cos θ dθ = θ But π dy dθ 3 dθ θ cos θdθ 4 π θ cos θdθ = [sin θ] π θ = t t = t 5 By symmety, ou integal ten becomes: EB, τ = τ Making te substitution t = sinu we obtain t dt 6 EB, τ = acsin τ/ acsin τ/ = cos udu sin u cosudu acsin τ/ = 1 + cosu du 1 τ = acsin + 1 τ 4 sin acsin τ = acsin + 1 τ sin acsin 7 Note toug tat In all, τ sin acsin = sin = τ acsin τ τ cos acsin τ 8 1 3

5 τ EB, τ = acsin + τ τ 1 9 And in paticula, EB,, = π 1 Since tis enegy is at most in fact equal to π, te lengt of te cack at time t is at most t, since te enegy of te cack is its lengt, and total enegy of te system is conseved. 4 Modified PDE fo a solution to te wave equation One metod of dealing wit patial diffeential equations on time dependent domains is tanslating and scaling te domain in suc a way so tat it is fixed. Howeve, suc tansfomations often complicate te patial diffeential equation, making it difficult to elate te two functions. We demonstate by example wit te wave equation in R. Suppose tat Ω R is open, T > and u C Ω [, T ] satisfies te wave equation, i.e., ü = u 11 Fo c > given and α > yet to be detemined, we define te function wx 1, x, t = uαx 1 + ct, x, t 1 We seek to find a value fo α suc tat ẅ w is conveniently expessable in tems of te patial deivatives of u and w. We poceed diectly by calculating, and define fo convenience v := x 1, x, t and z := αx 1 + ct, x, t : w x1 v = αu x1 z, w x1,x 1 x = α u x1,x 1 z w x v = u x z, w x,x v = u x,x z and fo ẅ: w t v = cu x1 z + u t z w t,t v = c [cu x1,x 1 z + u x1,t z] + cu t,x1 z + u t,t z = c u x1,x 1 z + cu x1,t z + cu t,x1 z + u t,t z 4

6 Combining te above, we obtain tat ẅ w v = u t,t z+c u x1,x 1 z+c [u x1,t z + u t,x1 z] α u x1,x 1 z u x,x z If we take α = c + 1, ten c α = 1 and tus ẅ w v = ü uc [u x1,t z + u t,x1 z] = c [u x1,t z + u t,x1 z] Since u C Ω [, T ] we ave tat ẅ w v = cu x1,t z 13 Altoug te PDE in 13 as few tems, it also complicates te pocess of undestanding te beavio of u wen we only know te beavio of w wit fo example, finding tangential deivatives. 4.1 Finding te tangential deivative of a tansfomed solution of te wave equation On domains canging in time, like Ω t := Ω \ Γt, fomal calculation suggests tat if u solves te wave equation on Ω and n u = wee n is te unit nomal to Γt on Γt, a cack gowing at speed c, ten u admits a decomposition as follows []: If u H 1 Ω wit u L Ω ten tee is a w H Ω and a c R suc tat wee ψ given by u = w + cψ 14 ψ : R R, ψ, θ := sinθ/ 15 Using tis function, we demonstate te complexity in undestanding te beavio of a simple tansfomation tat is, tanslation and scaling in te x 1 diection. Define by ψ c te function ψ as given in catesian coodinates, i.e., ψ c x 1, x = x 1 + x sin actanx1 /x and define te function f c x 1, x := ψ c x 1 ct, x β 5

7 Fo < c < 1 and β = 1 c. Ou goal in tis section is to calcuate f θ fo time t = wee f is te function f c witten in pola coodinates. To wite f as a composite of functions, we define te map φ : R R cos θ φ, θ = + sin θ, actan sin θ β cos θ β cos = θ β + sin θ, actan β tan θ So tat in paticula, and tat f = ψ φ, θ 16 f θ, θ = ψ φ, θφ θ, θ + ψ θ φ, θφ θ θ, θ 17 wee φ and φ θ ae te fist and second θ components of te function φ espectively. Calculating diectly, we obtain [ φ cos θ sin θ θ = cos θ β + sin θ β [ cos θ sin θ = 1 1 ] cos θ β + sin θ β ] + cos θ sin θ and φ θ 1 θ, θ = β tan θ + 1 β sec θ β = β sin θ + cos θ ψ, θ = 1 sinθ/ ψ θ, θ = cosθ/ 6

8 and tus 1 actan β tan θ ψ φ, θ = 1/4 sin cos θ β + sin θ ψ θ φ, θ = cos θ β + sin θ 1/4 actan β tan θ cos and tus we may calculate f θ, θ diectly fo θ =, π/4, π/. 4. f θ, φ θ, = φ θ θ, = β + 1 = β and Altogete ten, 1 actan ψ φ, = 1/4 sin 1 β = actan ψ θ φ, = β cos = β f θ, = ψ φ, φ θ, + ψ θ φ, φ θ θ, = + β β β = 4.3 f θ, π/ Note fo θ = π/ we may extend by continuity te definition of cos actanβ tanπ/ actanβ tan θ to include π/ so tat = cosπ/4 = cos π/4 = /. Altoug te same cannot be applied to sin actanβ tan θ, tis will not matte 7

9 since φ θ, π/ = and since bot lim θ π/ + sin actanβ tan θ and lim θ π/ sin exist in R. actanβ tan θ φ θ, π/ = φ θ θ, π/ = β β = 1 β and 1 ψ φ, π/ = sin±π/4 1/4 1 = ± 1 cannot be extended by continuity, but is bounded wit espect to left and igt limits 1/4 1 actanπ/ ψ θ φ, π/ = cos = Altogete ten, f θ, π/ = ψ φ, π/φ θ, π/ + ψ θ φ, π/φ θ θ, π/ = + 1 β = β 4.4 Comments Note tat as c 1, we ave tat β and tus β f θ, = f θ, π/ = β + In te pysical model of cack popagation, tis lage deivative fo θ fo lage enoug c coesponds to angles in wic cacks ae moe likely to continue pogessing. Tis may explain te penomena of bancing in cack popagation. 8

10 5 Uniqueness of Solution to Wave Equation wit Time-Dependent Domain Wile te calculation above depends on te assumption tat c be less tan te speed of te wave, we seek in tis section to pove uniqueness of a solution independent of te speed c fo te simple case tat te cack is a line gowing in te positive x 1 diection in time. Let Ω R be an open set, and let Γ : [, T ] PΩ be an inceasing set function s t Γs Γt wit H 1 Γt bounded. Define te set Ω t := Ω \ Γt, and te time-dependent bilinea fom fo t [, T ] and f, g H 1 Ω t B[f, g; t] := f g Ω t Moeove, we wok in te function spaces wic ae useful fo solving weak fomulations of te wave equation [1] V t := {v H 1 Ω t v = on Ω} Conside te following initial/bounday value poblem: üt ut = on Ω t 18 ut = on Ω 19 u = g, u = f, g H 1 Ω, f L Ω A weak solution of te wave equation wit a time-dependent domain [1] is a function ux 1, x, t suc tat and u satisfies u H 1, T ; V T W 1,, T ; L Ω t [, T ] ut V t s [, T u W, s, T ; V sup ü L s,t ;Vs < + s [,T s, T te functions s t 1 ut ut L Ω,, s ae equiintegable on s, T üt, v H 1 Ω t + B[ut, v; t] = v H 1 Ω t u = g, u = f 9

11 Poposition 1. Fo any t [, T ], B[f, g; t] is a continuous bilinea fom on H 1 Ω t. Poof. Fo f, g H 1 Ω t we ave tat B[f, g; t] = f x1 g x1 + f x g x Ω t Ω t f x1 g x1 + f x g x Ω t Ω t f x1 L Ω t g x1 L Ω t + f x L Ω t g x L Ω t f x1 L Ω g t x 1 L Ω + f t x L Ω g t x L Ω t f L Ω + f t x 1 L Ω + f t x L Ω t g L Ω + g t x 1 L Ω + g t x L Ω t = f H 1 Ω t g H 1 Ω t Hence, B is continuous. Poposition. If v H 1, T ; H 1 Ω wee spt v Ω ten d dtb[vt, vt; ] exists and equals B[vt, vt, ] fo t a.e. in [, T ]. Poof. Fo, we ave tat 1 [B[vt +, vt + ; ] B[vt, vt; ]] = 1 [B[vt + vt, vt + ; ] B[vt, vt + vt] [ ] vt + vt vt + vt = B[, vt + ; ] B[vt, Ten since v H 1, T ; H 1 Ω, we ave tat v is absolutely continuous and ence continuous, and diffeentiable almost eveywee [3]. Hence, as, we obtain fo t a.e. in [, T ], vt + vt vt in H 1 Ω vt + vt in H 1 Ω We ten see fom Poposition 1 tat vt + vt B[, vt + ; ] Ḃ[ vt, vt; ] vt + vt B[vt, ; ] Ḃ[vt, vt; ] Fo t a.e. in [, T ]. Since B is symmetic, te claim is poven. 1

12 Poposition 3. If u H 1, T ; H 1 Ω T, ut H 1 Ω t t [, T ], and Γs as measue zeo fo all s [, T ], ten d dt ut L Ω t exists a.e. fo t [, T ] and d dt ut L Ω t = ut, ut L Ω t Poof. As befoe, fo we calculate 1 ut + L Ω t+ ut L Ω t = 1 ut +, ut + L Ω t+ ut, ut L Ω t = 1 ut +, ut + L Ω t ut, ut L Ω t Since Ω t+ \ Ω t and Ω t \ Ω t+ ave measue zeo = ut + ut, ut + L Ω t + ut, ut + ut L Ω t Again, since u H 1, T ; H 1 Ω T, u is continuous in t and diffeentiable almost eveywee fo t [, T ] [3]. Hence as we get ut + ut ut in H 1 Ω t ut + vt in H 1 Ω t Recall tat H 1 Ω t convegence implies L Ω t convegence, so by Caucy Scwaz, we see tat ut + ut, ut + L Ω t ut, ut L Ω t ut + ut ut, L Ω t ut, ut L Ω t fo t a.e. in [, T ], poving te claim. Poposition 4. Assume tat ux 1, x, t is a weak solution to te above wit g = f =, wee Γt is a line extending in te positive x 1 diection wose position at time t is given by φt wee φ is an inceasing continuous function fom R to R. In paticula, Γl as measue zeo fo eac l [, T ]. Assume also tat ut is compactly suppoted in Ω t fo eac t [, T ] and tat sptut φt Ω fo eac t [, T ]. Ten u =. Poof. We modify a poof fom Evans [4]. Define fist te function wx 1, x, t := ux 1 + φt, x, t 1 11

13 and emak tat since ut H 1 Ω t and by te estictions of sptut we ave tat wt H 1 Ω fo eac t [, T ]. Fix some s, T ] and define te following function: vt := χ [,s] t Fo te inteval [, s], note tat s t wτdτ vt = s t = vs + wτdτ t s wτdτ 3 4 Ten since wτ H 1 Ω we get tat v H 1, s; H 1 Ω see poposition.4 - iii in [3]. Since vs = and v is extended to fo t s, ten v H 1, T ; H 1 Ω. Fom Popositions and 3 we ave tat d B[vt; vt; ] = B[vt, vt; ] dt d dt ut L Ω = ut, ut t L Ω t Moeove, note tat fo t [, s], vt = wt = ux 1 +φt, x, t. Hence Similaly, B[vt, vt; ] = vt ux 1 + φt, x, t Ω = vx 1 φt, x, t ut Ω t = B[ut, vx 1 φt, x, t; t] s ut, ut L Ω t = ut, vx 1 φt, x, t L Ω t Combining tese two, we get tat d 1 dt ut L Ω 1 t B[vt; vt; ] = s = üt, vx 1 φt, x, t H 1 Ω t = s ut, ut L Ω t B[vt, vt; ]dt üt, vx 1 φt, x, t H 1 Ω t + B[ut, vx 1 φt, x, t; t]dt 1

14 Now vx 1 φt, x, t H 1 Ω t so tat we may apply te definition of weak solution to obtain s d 1 dt ut L Ω 1 t B[vt; vt; ] = 5 = Howeve, we also note tat s d 1 dt ut L Ω 1 1 t B[vt; vt; ] = ut L Ω 1 t B[vt; vt; ] t=s t= = 1 us L Ω + 1 s B[v, v; ] since u L Ω = B[vs, vs; ] =. Finally, tis implies us L Ω s =, sowing tat u =. 6 Conclusion Studying PDEs on time-dependent domains is difficult since many metods of tansfoming domains do not wok. Howeve, suc PDEs ae impotant to pysical modeling poblems, suc as cack popagation. Wile tansfoming te domain of te PDE can alleviate tis issue in some sense, it poses its own poblems: tese tansfomed solutions solve new and sometimes moe complicated PDEs, and tei elationsip between te beavio of te oiginal solutions and te tansfomed solutions is not always eadily seen. In paticula, tese tansfomation metods equie te cack speed to be below te wave speed, yet existence can be sown independently of te cack speed [1]. Hee we ave sown tat fo a staigt cack, uniqueness also olds independent of cack speed. Refeences [1] C.J. Lasen G. Dal Maso. Existence fo wave equations on domains wit abitay gowing cacks. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natu. Rend. Lincei 9 Mat. Appl., :387 38, 11. [] Anna-Magaete Sandig Sege Nicaise. Dynamic cack popagation in a d elastic body: te out-of-plane case. Jounal of Matematical Analysis and Applications, 39:1 3, 7. [3] Macel Keute Wolfgang Aendt. Mapping teoems fo sobolev-spaces of vecto-valued functions. Studia Mat, 43:75 99, 16. [4] Lawence C. Evans. Patial Diffeential Equations. Ameican Matematical Society, pg