Math 6395: Hyperbolic Conservation Laws and Numerical Methods. Hyperbolic Conservation Laws are a type of PDEs arise from physics: fluid/gas dynamics.

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1 Mth 6395: Hyperbolic Conservtion Lws nd Numericl Methods Course structure Hyperbolic Conservtion Lws re type of PDEs rise from physics: fluid/gs dynmics. Distintive solution structures Other types of PDEs we met before, e.g. the time independent elliptic equtions, or time dependent prbolic equtions. Solutions re smooth. The solutions of hyperbolic problems hve distinctive feture: formtion of discontinuities (shocks). Such feture together with others fctors ttrct mny ttentions mong the scientific nd engineering community, leding to mny interesting reserch work in the PDE level, nd in the design of numericl schemes in the pst yers or so. Structure of the course: mthemticl theory numericl methods methods for liner equtions: locl trunction error, ccurcy, stbility, convergence methods for nonliner sclr equtions low order schemes with good properties corresponding to those in the PDE theory extension to high order schemes in the discontinuous Glerkin frmework. 1 Introduction Hyperbolic conservtion lws: u t + x f( u) = 0, (x, t) R d R + (1) u( x, t = 0) = u 0 ( x) (2) u: conserved quntities, R d R + R m, with d being the dimension of the problem, nd m being the number of components of u. f(u): flux function, R m R m For exmple The cse of m = 1 nd d = 1 corresponds to 1-D sclr conservtion lws. In trffic flow problems, ρ t + f(ρ) x = 0 is of the form of (1). ρ is the density of the trffic, nd f(ρ) = ρv(ρ) is the flux pss loction (mss multiple by the velocity of the trffic). m = 2 + d. In gs/fluid dynmics, the compressible Euler equtions in one dimension reds u t + x f( u) = 0, (3) where u = (ρ, ρv, E) T re the conserved quntities (mss, momentum nd energy) nd f( u) = (ρv, ρv 2 + p, (E + p)v) T re the corresponding flux functions. Here ρ is the density, v is the velocity, ρv is the momentum, E is the energy nd p is the pressure given s function of other stte vribles, known s eqution of stte. Hyperbolic equtions re type of time dependent liner/nonliner PDEs rise from science nd engineering. Most of the time, it is not possible to write down the exct solutions of the equtions. People hevily rely on numerics to ccess the solution structures in future times. 1

2 Definition 1.1. (Hyperbolic for 1-D problem) The 1-D system (1) is hyperbolic if the m m Jcobin mtrix f 1 f u 1 1 u m J f ( u) = f m u 1 of the flux function hs the following property: For ech vlue of u the eigenvlues of J f ( u) re rel, nd the mtrix is digonlizble, i.e. there is complete set of m linerly independent eigenvectors. f m u m Definition 1.2. (Hyperbolic for 2-D problem) The 2-D system in the form of u t + f( u) x + g( u) y = 0 (4) with u: R 2 R + R m, f, g : R m R m is hyperbolic if ny liner combintion of m m mtrix αj f ( u) + βj g ( u) of the flux Jcobins hs rel eigenvlues nd the mtrix is digonlizble, i.e. there is complete set of m linerly independent eigenvectors. Remrk 1.3. If the Jcobin mtrices hve distinct rel eigenvlues, it follows tht it s digonlizble. In this cse the system is clled strictly hyperbolic. If the eigenvlues re rel but not distinct, the Jcobin mtrix my not be digonlizble. Such system is clled wekly hyperbolic. If the Jcobin mtrix is digonlizble but hs complex eigenvlues, the system is not hyperbolic. Typicl fetures of solutions for nonliner hyperbolic systems. Development of discontinuities or shocks in future solutions, even with smooth initil dt. Becuse of this, solutions in clssicl sense could fil to exist. If the solutions re defined in the wek distribution sense, there could be more thn one wek solutions Entropy solution is the uniquely exist physiclly relevnt solution mong wek solutions. In the PDE level, questions to sk/nswer re: How to define entropies solutions? (vnishing viscosity method, entropy inequlities) Wht re pproprite spces (norms) for entropy solutions? (BV norm, L 1 nd L ) Wht re criteri for selecting entropy solutions mong wek solutions? (entropy conditions) Becuse of the bove mentioned difficulties t the PDE level, numericl chllenges in designing schemes for hyperbolic type PDEs re how to design schemes with good ccurcy nd stbility? ( good : high resolution/high order ccurcy, stbility in the sense of totl vritionl diminishing or mximum principle preserving) how to cpture shocks without numericl rtifcts such s oscilltions? how to know (or prove) if the numericl scheme pproximtes wek solutions, or the unique entropy solution mong wek solutions? 2 The Derivtion of Conservtion Lws We cn model the trffic flow s 1-D sclr conservtion lw. Denote density of the trffic s ρ(x, t), then the evolution of the trffic flow density cn be described by the following t ρ(x, t)dx = f(ρ(, t)) f(ρ(b, t)), (5) for ny b. Eqution (5): the rte of chnge of the density over n intervl [, b] is due to the flux flowing into the region from the left boundry x =, nd the flux flowing out of the region from the right boundry x = b. The flux pst given point is give by f(ρ(x, t)) = ρ(x, t)v(x, t), 2

3 where v(x, t) is the velocity of the trffic flow, which could be dependent on ρ. For exmple, in simple liner model, v(ρ) = v mx (1 ρ/ρ mx ). At zero density (empty rod), the velocity is v mx ; but decreses to zero s ρ pproches the mximum cpcity of the rod ρ mx (trffic jm). If we ssume f(ρ(x, t)) x exists, then Hence t ρ(x, t)dx = f(ρ(x, t)) x dx (6) ( t ρ(x, t) + f(ρ(x, t)) x)dx = 0 (7) Since the bove eqution is true for ny b, then in terms of the differentil eqution, we hve Remrk 2.1. Note tht ρ t + f(ρ) x = 0. (8) Eqution(5) smoothf(ρ) = Eqution(8); Eqution(8) Eqution(5) The integrl eqution does not require smoothness conditions on the solution. Remrk 2.2. System of conservtion lws such s Euler eqution in gs/fluid dynmics cn be modeled in similr (but more involved) wy concerning the conservtion of mss, momentum nd energy. See Chp. 5 of the book. 3 Sclr Conservtion Lws 3.1 Liner cse u t + u x = 0, u(x, t = 0) = u 0 (x), < x <, (9) chrcteristics in x t plne in deriving the exct solution. Along chrcteristics dx dt =, the solution stys constnt du dt = u t + u x x t = u t + u x = 0. (10) Hence the exct solution is u(x, t) = u 0 (x t). finite speed of propgtion (vs. infinite speed of propgtion for prbolic equtions) If u 0 (x) is smooth function, then u(x, t) is eqully smooth in spce nd in time. If u 0 (x) is not smooth (the concept of clssicl solutions for differentil equtions fils), the solution of u 0 (x t) stisfies the integrl form of the eqution, nd is solution in the wek/distribution sense. 3.2 Nonliner cse e.g. Burgers eqution f(u) = u 2 /2. u t + f(u) x = 0 (11) 3

4 3.2.1 Chrcteristics nd Shock Formtion Along chrcteristics dx dt = f (u) = u, the solution stys constnt du dt = u t + u x x t = u t + f (u)u x = 0. (12) Hence the exct solution u(x, t) = u 0 (x ), where x is the root of the chrcteristic eqution x x t 0 = f (u 0 (x )). (13) Exmple 3.1. Consider the Burgers eqution with initil condition u 0 (x) = sin(x). Plot for crossing of chrcteristics. Shock formtion time. Let d. dt = t + x x t = t + u x be the differentition long chrcteristics. Let v = u x, du dt = 0. dv dt = v t + f (u) v x = 2 u t x + u 2 u x 2. Tking xderivtive on the Burgers eqution we hve, 2 u t x = 2 f(u) x 2, Thus dv dt = f(u) 2 x 2 + u 2 u x 2 = (u x) 2 = v 2. 1 v(t ) 1 v(0) d( 1 v ) = dt v(t ) = = T v(0) T v(0) + 1 (14) v(t ) goes to infinity when T = 1 v(0). Shock forms when T = min x:u 0 (x)<0 { 1 u 0 (x)} Wek Solutions Let φ(x, t) C0(R 1 R + ) be test function, where C0 1 is the spce of function tht re continuously differentible with compct support. If we multiply the eqution (11) by the test function φ nd integrte over spce nd time, we obtin Performing integrtion by prts yields 0 0 (φ t u + φ x f(u))dxdt + φ(u t + f(u) x )dxdt = 0. (15) φ(x, 0)u(x, 0)dx = 0. (16) Definition 3.2. (Wek solution) The function u(x, t) is clled wek solution of the conservtion lws if eqution (16) holds for ll test function φ C 1 0(R R + ). Remrk 3.3. Mthemticlly the definition of wek solution in the integrl form bove is equivlent to the solution for the integrl form of the eqution over ny choices of sptil nd time intervl. The wek form bove could be esier to work with sometimes. 4

5 Specil cses. If u(x, t) is continuously differentible, then u(x, t) is clssicl solution nd wek solution. In the x t plne, if u(x, t) C 1 (R R + ) nd stisfies the strong/differentil form of eqution (11) except on curve x(t), then 0 = u(x, t)dx + f(u(b, t)) f(u(, t)) t ( = x(t) ) b u(x, t)dx + u(x, t)dx + f(u(b, t)) f(u(, t)) t x(t) + x(t) = u t (x, t)dx + u(x(t), t)x (t) + u t (x, t)dx u(x(t) +, t)x (t) + f(u(b, t)) f(u(, t)) x(t) + x(t) = u t (x, t)dx + f(u(x(t), t)) f(u(, t)) + u t (x, t)dx + f(u(b, t)) f(u(x(t) +, t)) x(t) + (f(u(x(t), t)) f(u(x(t) +, t))) + u(x(t), t)x (t) u(x(t) +, t)x (t) = (f(u(x(t), t)) f(u(x(t) +, t))) + u(x(t), t)x (t) u(x(t) +, t)x (t) (17) Let [ f(u) ]. = (f(u(x(t), t)) + f(u(x(t) +, t))) nd [ u ]. = (u(x(t), t) + u(x(t) +, t)) be the jump of f nd u cross discontinuity. From eqution (17), we hve s =. x (t) = [ f(u) ]. (18) [ u ] This is the so clled Rnkine-Hugoniot jump condition. s. = x (t) is the shock speed. Proposition 3.4. A function u(x, t) is piecewise smooth nd stisfies the PDE strong whenever u C 1. If the function stisfies the Rnkine-Hugoniot jump condition long the discontinuity curve, then u(x, t) is wek solution of eqution (11). Remrk 3.5. For curves on which the solution is continuous but not differentible, eqution (17) is stisfied. The solution is wek solution Riemnn Problem nd Not-uniqueness of Wek Solutions Exmple 3.6. Consider Burgers eqution with initil condition consist of two constnt sttes u l nd u r, u 0 (x) = { ul = 1, x < 0 u r = 1, x > 0 (19) Drwing chrcteristics: colliding into ech other (formtion of shocks). By Prop. 3.4, u(x, t) = u 0 (x) is the wek solution of the Burgers eqution. Exmple 3.7. Consider Burgers eqution with initil condition consist of two constnt sttes u l nd u r, u 0 (x) = { ul = 1, x < 0 u r = 1, x > 0 Drwing chrcteristics: spreding out. A wek solution is the rrefction wve 1, x < t x u(x, t) = t, t < x < t 1, x > t (20) (21) Yet by Prop. 3.4, u(x, t) = u 0 (x) is lso wek solution of the Burgers eqution. In fct there re infinitely mny wek solutions for this problem. 5

6 Remrk 3.8. (from wiki) A Riemnn problem, nmed fter Bernhrd Riemnn, consists of conservtion lw together with piecewise constnt dt hving single discontinuity. The Riemnn problem is very useful for the understnding of hyperbolic prtil differentil equtions like the Euler equtions becuse ll properties, such s shocks nd rrefction wves, pper s chrcteristics in the solution. It lso gives n exct solution to some complex nonliner equtions, such s the Euler equtions. In numericl nlysis, Riemnn problems pper in nturl wy in finite volume methods for the solution of eqution of conservtion lws due to the discreteness of the grid. For tht it is widely used in computtionl fluid dynmics nd in MHD simultions. In these fields Riemnn problems re clculted using Riemnn solvers Entropy Solution We hve seen tht the clssicl/strong solution my fil to exist for hyperbolic equtions; but wek solutions my not be unique. Additionl criteri re needed to pick out the unique physiclly relevnt solution, i.e. the entropy solution, mong wek solutions. In physicl models, the blnce lws only come with some physicl viscosity. For exmple, in the trffic flow modeling, the viscosity tkes the form of slow response of drivers nd utomobiles; in the fluid dynmics, the viscosity corresponds to the informl notion of thickness. For exmple, honey hs higher viscosity thn wter. Conservtion lws with viscous terms provide more physiclly relevnt models. Definition 3.9. (Vnishing viscosity method for the entropy solution) Consider the viscous eqution u ε t + f(u ε ) x = εu ε xx (22) An entropy solution of (11) is the limit (.e.) of u ε of eqution (22) when ε 0. There re severl issues to resolve to mke sense of the bove definition. For exmple, the solution for the viscous eqution (22) hs to exist nd is unique for ech ε; there hs to exist limit of u ε s ε 0; such limit should be one of the wek solutions (in fct the physiclly relevnt entropy solution) for eqution (11). These questions hve been nswered in the PDE theory, but they re out of the scope of this course. Proposition (Equivlent definition for entropy solution) A wek solution of (11) is n entropy solution if for ll convex entropy function U(u) with U (u) 0 nd the ssocited entropy flux function F (u) with F (u) = U (u)f (u), we hve U(u) t + F (u) x 0, (23) in the distribution sense. Tht is for ll φ 0, φ C 1 0(R R + ), we hve 0 Proof: see Section of the textbook. (φ t U(u) + φ x F (u))dxdt φ(x, 0)U(u(x, 0))dx 0. (24) Remrk If solution is continuously differentible (clssicl solution), then it is n entropy solution. The entropy inequlity becomes n equlity. If solution u(x, t) C 1 (R R + ) nd stisfies the strong/differentil form of eqution (11) except on curve x(t), then by using similr procedure in deriving the Rnkine-Hugoniot condition, we hve the following inequlity for the entropy solution s[ U ] + [ F (U) ] 0, for ny entropy-entropy flux pirs (25) The bove definitions for entropy solutions re mthemticlly rigorous; however, it will be difficult to directly pply these definitions to select the entropy solution mong wek solutions (e.g. for the exmple of Riemnn problem bove). Below, we look for simpler nd more prcticl entropy conditions tht one cn directly pply in exmples. Proposition (Oleinik entropy condition) A discontinuity propgting with speed s = [ f(u) ] [ u ] given by the Rnkine-Hugoniot jump condition stisfies the Oleinik entropy condition if for ll u between u l nd u r, f(u) f(u l ) u u l s f(u) f(u r) u u r (26) where u l nd u r re left nd right stte long the discontinuity respectively. 6

7 Proposition (Lx entropy condition) A discontinuity propgting with speed s given by the Rnkine- Hugoniot jump condition stisfies the Lx entropy condition if f (u l ) > s > f (u r ), (27) where u l nd u r re left nd right stte long the discontinuity respectively. Remrk One cn see tht Oleinik entropy condition implies Lx entropy condition; the converse does not hold. Lx entropy condition is necessry but not sufficient condition to single out the entropy condition. However, if f(u) is strictly convex with f (u) > 0 (or strictly concve with f (u) < 0), then the Lx entropy condition is equivlent to the Oleinik entropy condition, nd is sufficient to single out the entropy condition. In fct, under the convexity (or concvity) ssumption on f(u), the Lx entropy condition is reduced to f (u l ) > f (u r ), i.e., the chrcteristics propgting into the shock, rther thn diverging from the shock. Entropy solutions of the sclr hyperbolic eqution (11) re proved to exist nd re unique in the clss of BV L 1 L. Here BV stnds for bounded vrition. A function is of bounded vrition on given intervl [, b], if V b (f). = sup i f(x i ) f(x i+1 ) <, {x i } i is ny prtition of[, b]. If f(u) is continuously differentible, then V b (f). = f (x) dx. Proposition (BV) If u 0 (x) is function of loclly bounded vrition on (, ), then for ech t > 0, u(, t) is lso function of loclly bounded vrition on (, ), nd where s = mx x f (u). V R Ru(, t) T V R+st Rst u 0( ), Proposition (L 1 contrction property) If u(x, t) nd v(x, t) re solutions of the sclr hyperbolic eqution (11) with initil dt u 0 (x) nd v 0 (x) respectively, then u(, t) v(, t) L 1 u 0 ( ) v 0 ( ) L 1. Specificlly, consider v 0, then u(, t) L 1 u 0 ( ) L 1. Proposition (L mximum principle) If u(x, t) is solution of the sclr hyperbolic eqution (11) with initil condition u 0 (x), then mx x u(x, t) mx u 0(x), min x x u(x, t) min u 0 (x), x 7