Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Nonlinear Hyperbolic Systems of Conservation Laws and Related Applications I Mapundi K. Banda Applied Mathematics Division - Mathematical Sciences, Stellenbosch University [email protected] Jul 23, 2013 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 1 / 84 All models are wrong, but some models are useful George Box Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 2 / 84 Section 1: Mathematical Models - Evolution Equations Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 3 / 84 Modelling with Partial Differential Equations Evolution equations to be studied are hyperbolic systems of conservation laws: First-order Hyperbolic Partial differential equation. Interesting cases are usually nonlinear. Play a prominent part in modelling flow and transport processes. Discussion of the mathematical aspects. Discussion of the ideas for designing numerical schemes. Presentation of some case studies. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 4 / 84 Why study Hyperbolic Conservation Laws Special Difficulties: shock formation (jump discontinuities). Great deal is not yet known about the mathematical structure. Provide efficient approximations of many complex models. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 5 / 84 Reynold’s Transport Theorem Consider an arbitrary physical quantity ρ (mass, energy, charge, concentration of species of matter, etc). ρ occupies Ω0 ⊂ Rn at time t = 0. Let domain move with time: consider a map Φ : Ω0 × [0, T ] → Rn define Ωt = {x = Φ(y , t) : y ∈ Ω0 } Ω0 - reference configuration; t 7→ Φ(y , t) - trajectory of the particle (or point) y ∈ Ω0 Using x and t is referred to as using the Eulerian description. Using y and t is referred to as the Lagrangian description. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 6 / 84 Reynold’s Transport Theorem The velocity of the particle at point x = Φ(y , t) is given by: u(x, t) = u(Φ(y , t), t) = ∂ Φ(y , t) ∂t The total amount (mass) of the quantity ρ inside the domain Ωt is given by Z Mt = ρ(x, t) dx Ωt How does the total mass change with time - Reynolds Transport theorem Theorem (Reynold’s Transport Theorem) Z Z d ∂ρ ρ(x, t) dx = + ∇ · (ρu) dx dt Ωt Ωt ∂t Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 7 / 84 Reynold’s Transport Theorem d dt Z Z ρ(x, t) dx = Ωt Ωt ∂ρ + ∇ · (ρu) dx ∂t Integral relation between change of total mass, Mt , and local changes of density ρ and the transportation u in an arbitrary domain Ωt ⊂ Rn To derive the partial differential equations from Reynolds Transport Theorem use: Lemma (Variational Lemma) Let f ∈ L2 (Ω) Then f ≡ 0 almost everywhere. Mapundi K. Banda (Maties) Z with fg dx = 0 for all g ∈ L2 (Ω). Ω CIMPA/MPE 2013 Jul 22 - Aug 3 8 / 84 Application: Gas Flow in a pipe Denote gas density by ρ and velocity in x-direction by u(x, t) Domain Ωt = [x1 (t), x2 (t)] of moving gas with d xi (t) = u(xi (t), t) dt Total amount of gas Z x2 (t) Mt = ρ(x, t) dx x1 (t) The Reynolds Transport Theorem yields: Z d ∂ρ ∂(ρu) Mt = + dx dt ∂x Ωt ∂t Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 9 / 84 Application: Gas Flow in a pipe Axiom of Conservation of Mass, Z 0= Ωt d Mt = 0: dt ∂ρ ∂ + (ρu) dx ∂t ∂x for any arbitrary domain Ωt of gas. Apply the Variational Lemma to get the Continuity Equation ∂ρ ∂ + (ρu) = 0 ∂t ∂x (ρu)(x, t) is the mass flux of gas through the cross-section of the pipe at point x and t. To get a closed relation between ρ and u, we need a constitutive relation between ρ and u or additional conservation laws. If u is known, the equation is referred to as an advection equation. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 10 / 84 Mass Conservation Axiom The change of Mass is given by the difference of mass fluxes at the boundaries. Consider the continuity equation on an interval [x1 , x2 ] during time [t1 , t2 ]: ∂ρ ∂ + (ρu) = 0 ∂t ∂x Rewrite the flux as f (ρ) = ρu, assuming u is given Integrating with respect to space: Z x2 ∂ρ dx = f (ρ(x1 , t)) − f (ρ(x2 , t)) x1 ∂t Integrating with respect to time and space: Z x2 Z x2 Z t2 Z ρ(x, t2 ) dx − ρ(x, t1 ) dx = f (x1 , t) dt − x1 Mapundi K. Banda (Maties) x1 t1 CIMPA/MPE 2013 t2 f (x2 , t) dt t1 Jul 22 - Aug 3 11 / 84 Application: Traffic Flow Consider a highway section with car density ρ, cars are a conserved quantity: ∂ ∂ρ + (ρu) = 0 ∂t ∂x To close the equation, need a relation between ρ and u: ρ u = u(ρ) = umax 1 − ρmax i.e. the denser the traffic, the slower the cars will move. Define flux as ρ f (ρ) = ρumax 1 − ρmax In general, the equation ∂ ∂ρ + f (ρ) = 0; or ∂t ρ + ∂x f (ρ) = 0 ∂t ∂x is the prototype of a hyperbolic scalar conservation law in one space dimension Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 12 / 84 Application: Euler Equations of Inviscid Gas Dynamics Model the flow of a gas with density ρ, velocity v , energy E and pressure p need additional conservation laws ρt + (ρv )x = 0; 2 conservation of mass, (ρv )t + (ρv + p) = 0; conservation of momentum, Et + (v (E + p))x = 0; conservation of energy. Need constitutive relations, where T is the absolute temperature of the gas: E= ρv 2 2 Mapundi K. Banda (Maties) p = ρT ; ideal gas law, + cv ρT kinetic + thermal (internal) energy. CIMPA/MPE 2013 Jul 22 - Aug 3 13 / 84 Application: Euler Equations of Inviscid Gas Dynamics Model the flow of a gas with density ρ, velocity v , energy E and pressure p need additional conservation laws ρt + (ρv )x = 0; 2 conservation of mass, (ρv )t + (ρv + p) = 0; conservation of momentum, Et + (v (E + p))x = 0; conservation of energy. Obtain 3 equations for 3 variables and p depends on ρ, v , E with ρ ρv u = ρv and F (u) = ρv 2 + p(u) E v (E + p(u)) giving a hyperbolic system of conservation laws in one space dimension: ∂t u + ∂x F (u) = 0 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 14 / 84 Application: Euler Equations of Inviscid Gas Dynamics Assume temperature, T , is constant then p = a2 ρ where a is the speed of sound in the gas. The flow is referred to as isothermal flow: ρt + (ρv )x = 0; 2 2 (ρv )t + (ρv + a ρ) = 0; Mapundi K. Banda (Maties) conservation of mass, conservation of momentum, CIMPA/MPE 2013 Jul 22 - Aug 3 15 / 84 Equations of Internal Energy and Entropy With some manipulation of the viscous heat conducting fluid dynamic equations, one obtains: ρ de = −p(∇ · v ) + εv + ∇ · (k∇T ) + qH dt where: e is the internal energy. p(∇ · v ) is the reversible work of the pressure forces (vanishes in incompressible flows). Other terms are considered as heat additions: εv dissipation term - acting as an irreversible heat source FC = −k∇T is flux due to heat conduction (Fourier’s Law). qH are heat sources. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 16 / 84 Equations of Internal Energy and Entropy Introduce entropy per unit mass s of the fluid through the thermodynamic relation: T ds = de + p d 1 dp = dh − ρ ρ where h is the enthalpy. Thus separation of reversible and irreversible heat additions: T ds = dq + dq 0 where dq is a reversible heat addition. dq 0 is an irreversible heat addition. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 17 / 84 Equations of Internal Energy and Entropy Second Principle of Thermodynamics: dq 0 ≥ 0, i.e. in adiabatic flow (dq = 0) entropy always increases. We obtain: ρT ds = εv + ∇ · (k∇T ) + qH dt where: Last two terms are reversible heat addition by conduction and other sources. For qH = 0 and k = 0 the non-negative dissipation term εv is a non-reversible heat source. Entropy equation of the flow, equation is important but not independent from energy equation Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 18 / 84 Equations of Internal Energy and Entropy Either the entropy equation or the energy equation used with the mass and momentum equation but not both. Entropy is not ”conserved” in the sense of derived conservation equations ρT ds = εv + ∇ · (k∇T ) + qH dt Thus mathematical expression of second principle of thermodynamics for an adiabatic flow without heat conduction or heat sources: ds = εv dt ∂s ρT + v · ∇s = εv ∂t ρT Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 19 / 84 Physical Entropy condition Since εv ≥ 0, viscous dissipation: ρT ∂s + v · ∇s ≥ 0 ∂t Any solution of Euler equations with a physical meaning satisfies the entropy condition above. Hence entropy used as an additional condition, where necessary, to exclude non-physical solutions for uniqueness. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 20 / 84 Some remarks on the Euler Equations Non-viscous, non-heat-conducting fluid, flow in the limit of vanishing viscosity Admits both continuous and discontinuous solutions: Entropy variations: continuous flow variations entropy equation: ρT ∂s + v · ∇s = 0 ∂t i.e. entropy constant along flow path In absence of discontinuities - Euler equation describes isentropic flows. Value of entropy can vary from one flow path to another. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 21 / 84 Application: Cooling of Coffee Consider a pot (domain P ⊂ R3 ) filled with coffee of constant density ρ and with temperature θ. Thermal energy contained in Ω ⊂ P is given by: Z ρcp θ(x, t) dx Et = Ω Assume coffee is at rest u(x, t) ≡ 0. Then Reynolds Transport Theorem yields: Z d ∂θ Et = ρcp dx dt ∂t Ω In general, change of thermal energy is a sum of heat flux through ∂Ω and energy produced in Ω, here assumed to be zero. Also, thermal energy flows from hot to cold i.e. heat flux f : f = −κ∇θ where κ > 0 is heat conductivity. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 22 / 84 Application: Cooling of Coffee Hence Z Z Z ∂θ dx = − fn ds = κn∇θ ds; ∂t Ω ∂Ω ∂Ω where n denotes the outer normal of the domain Ω considering the heat flow out of domain. Z Z ρcp Apply Gauss divergence Theorem Z ∂θ dx = ρcp ∂t Ω ∇ · f dx: fn ds = ∂Ω Ω Z ∇(κ∇θ) dx ∀Ω ⊂ P Ω Assume ρ, cp and κ to be constant, apply variational lemma: ∂θ = α∇2 θ, ∂t κ > 0 is thermal conductivity. ρcp This is the heat equation or the diffusion equation. where α = Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 23 / 84 Application: Cooling of Coffee The heat equation or diffusion equation models molecular diffusion (Fick’s Law)! To get a well-posed problem, specify initial and boundary conditions. For flux due to advection at velocity u, then the equation takes the form: ∂ ∂θ + (θu) = α∇2 θ ∂t ∂x The equation is referred to as the convection-diffusion or advection-diffusion equation. If inside the pot there is an extra source of heat, e.g. a heating element, then a source term could be added: ∂θ ∂ + (θu) = α∇2 θ + S ∂t ∂x Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 24 / 84 Solution of the Advection Equation Consider the continuity equation with u(x, t) ≡ a, known then the problem reduces to: ∂ρ ∂ρ +a = 0. ∂t ∂x Given the initial condition ρ(x, 0) = ρ0 (x) −∞<x <∞ then the solution takes the form: ρ(x, t) = ρ0 (x − at), t>0 Note: initial profile simply moves downstream with velocity a, its shape unchanged. The solution above is true even if ρ0 (x) is not smooth! Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 25 / 84 Example: The Burgers Equation Consider the scalar parabolic equation ∂u ∂2u ∂u +u −ν 2 =0 ∂t ∂x ∂x Introduced by Burgers as simplest differential model for fluid flow (viscous) Burgers equation. Though simple, it is regarded as model for decaying turbulence Burgers studied the limit equation as ν → 0 ∂u ∂ u2 + =0 ∂t ∂x 2 inviscid Burgers Equation Inviscid Burgers Equation (or Burgers Equation) also studied in wave theory to depict distortion of waveform in simple waves Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 26 / 84 Example: Solution of Viscous Burgers Equation Consider the scalar parabolic equation ∂u ∂u ∂2u +u −ν 2 =0 ∂t ∂x ∂x The Cauchy problem of viscous Burgers Equation has an explicit solution: Apply Cole-Hopf transform: u = −2ν φx φ to transform the Burgers Equation to ∂φ ∂2φ =ν 2 ∂t ∂x Use explicit expressions presented above for an explicit expression of the solution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 27 / 84 Solution of the diffusion equation Consider the diffusion equation: ∂θ − ∇2 θ = f ∂t where θ = θ(x, t), t ∈ R+ and x ∈ Ω ⊂ Rn Initial conditions: differential equation first order in time: θ(x, t) = θ0 (x), in Ω Boundary conditions: Dirichlet conditions: θ(x, t) = θb (x) for x ∈ ∂Ω. Neumann conditions: −(n · ∇)θ(x, t) = g (x) for x ∈ ∂Ω - heat flux. Robin conditions: −(n · ∇)θ(x, t) − βθ(x, t) = g (x) for x ∈ ∂Ω and β>0 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 28 / 84 Separation of variables Consider an initial boundary value problem: ∂θ ∂2θ = ∂t ∂x 2 on Q = R+ × (0, 1) with initial condition: θ(x, 0) = θ0 (x), x ∈ (0, 1) and boundary data θ(0, t) = 0 θ(1, t) = 0 for t ∈ R+ Seek a solution of the form: θ(x, t) = T (t) · X (x) Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 29 / 84 Separation of variables Obtain θ(x, t) = ∞ X an e −n 2 π2 t sin(nπx) n=1 and the coefficients are given (using initial conditions) as the Fourier coefficients of θ0 (x): θ0 (x) = ∞ X an sin(nπx) n=1 Note: The solution is infinitely often differentiable - the smoothing property of the diffusion equation. The smoothing property also holds for non-continuous initial data and inconsistent boundary data. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 30 / 84 The Diffusion Kernel Seek the fundamental solution - also called heat kernel: ∂ω − κ∇2 ω = δx δt ∂t in Rn × R+ Using Fourier Transforms: 1 2 ω(x, t) = √ e −x /(4κt) n ( 4πκt) ω(x, t) > 0 for all t > 0, x ∈ Rn Hence for f ≡ 0: 1 θ(x, t) = ω(·, t) ? θ0 = √ ( 4πκt)n Z e −(x−y ) 2 /(4κt) θ0 (y ) dy Rn where ? is a convolution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 31 / 84 Solution of Diffusion Equations 1: Half-plane Consider a 1D problem θt = κθxx , 0 < x < ∞, θ(x, 0) = θ0 (x), θ(0, t) = 0, 0<t<∞ for t = 0 for x = 0 The solution takes the form: θ(x, t) = 1 √ ( 4πκt) Z h i 2 2 × e −(x−y ) /(4κt) − e −(x+y ) /(4κt) θ0 (y ) dy R Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 32 / 84 Solution of Diffusion Equations 1: Half-plane with Neumann Boundary Consider a 1D problem θt = κθxx , 0 < x < ∞, θ(x, 0) = θ0 (x), θx (0, t) = 0, 0<t<∞ for t = 0 for x = 0 The solution takes the form: θ(x, t) = 1 √ ( 4πκt) Z h i 2 2 × e −(x−y ) /(4κt) + e −(x+y ) /(4κt) θ0 (y ) dy R Difference with the previous problem is a sign change. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 33 / 84 The Diffusion Kernel The solution operator is strongly regularizing. The solution of the diffusion equation has infinite speed of propagation Energy estimates: the homogeneous diffusion equation does not conserve energy. The solution of the non-homogeneous equation takes the form: θ(x, t) = ω(·, t) ? θ0 + ω(·, t) ? f Z Z t 1 p ω(x − y , t)θ0 (y ) dy + = n 0 ( 4πκ(t − τ )) Rn Z 2 × e −(x−y ) /(4κ(t−τ )) f (y , τ ) dydτ Rn One can prove that the solution of the non-homogeneous diffusion equation with Dirichlet boundary conditions is unique. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 34 / 84 Cauchy and Riemann Problems: Introduction Burgers possesses features of a scalar convex equation ∂u ∂ + f (u) = 0 ∂t ∂x where f : R → R is a convex smooth function The Cauchy Problem: Equation (1) with (1) u(x, 0) = u0 (x) The Cauchy problem for the Burgers Equation admits discontinuous weak solutions even for a smooth initial function u0 The Riemann Problem: Equation (1) with ( uL for x < 0 u(x, 0) = uR for x ≥ 0 The solution for the Riemann Problem is either a shock propagating or a rarefaction wave Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 35 / 84 Summary Derived the conservation laws based on the Reynolds Transport Theorem. Presented some special cases: linear advection, Burgers equation, Euler equations of gas dynamics Presented solution of the advection equation Presented solutions of the diffusion equations based on the separation of variables as well as the Green’s function. Presented solutions of the advection-diffusion equation Introduced some common concepts in the area of conservation laws Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 36 / 84 Section 2: Mathematical Analysis - Hyperbolic Conservation Laws Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 37 / 84 Preamble: Remarks on Hyperbolic Systems Conservation Laws due to Lax Systems of conservation laws in a single space variable is a well-studied subject Euler equations of compressible flow, are an important example of a hyperbolic system of conservation laws In phenomena governed by nonlinear hyperbolic equations signals propagate with finite speed (recall the linear wave equation) Singularities propagate along characteristics, BUT these arise spontaneously, leading to formation of shocks Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 38 / 84 Preamble: Remarks on Hyperbolic Systems Conservation Laws due to Lax Time is not reversible as for linear equations thus future and past are different, as they are in real life There is loss of information as time moves forward, which can be interpreted as an increase in entropy Basic existence theory of solutions of hyperbolic conservation laws in single space variables is due to Jim Glimm It is a scandal of mathematical physics that, apart from isolated results, no comparable theory exists for more space variables Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 39 / 84 Hyperbolic Conservation Laws In general, seek a weak solution u : RD × [0, T ] → RK to the Cauchy problem: ∂t u + D X ∂xj Fj (u) = 0; u(x, 0) = u0 (x) j=1 with Fj (u) : RK → RK , is a system of K conservation laws in RD with flux functions Fj Consider ∂t u + D X Aj (u)∂xj u = 0; u(x, 0) = u0 (x), j=1 the system is (strictly) hyperbolic iff the matrix D X ξj Aj (u) has only j=1 real (pairwise disjoint) e-values (char. speeds) for all ξ ∈ RD . Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 40 / 84 Examples of Hyperbolic Conservation Laws Linear Equation: Compressible Euler: ∂t u + ∂x u = 0; ∂t ρ + ∂x m = 0; ∂t m + ∂x (ρu 2 + p) = 0; Burgers Equation: ∂t u + ∂x u2 2 ∂t E + ∂x (u(E + p)) = 0; ρ p = (γ − 1)(E − u 2 ). 2 = 0; Incompressible Euler: ∇ · u = 0; ∂t u + ∇ · u ⊗ u + ∇p = 0. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 41 / 84 Hyperbolic Conservation Laws For K = D = 1 i.e. one-dimensional scalar conservation laws, the equation is always strictly hyperbolic For D = 1 one needs to investigate the matrix A(u) = ∂F j ∂uk (u) 1≤j,k≤K F1 F2 with F = . .. FK Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 42 / 84 Example: The p-system Model one-dimensional isentropic gas dynamics in Lagrangian coordinates: ∂u ∂v − = 0; ∂t ∂x ∂u ∂ + p(v ) = 0; ∂t ∂x v is specific volume, u is the velocity, and the pressure p(v ) For a polytropic isentropic ideal gas: p(v ) = Av −γ for some constant A = A(s) > 0 (depending on entropy), γ > 1 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 43 / 84 Example: The p-system ∂v ∂u − = 0; ∂t ∂x ∂u ∂ + p(v ) = 0; ∂t ∂x System is of the form: ∂w ∂ + f(w) = 0; The conservative form ∂t ∂x ∂w ∂w + A(w) = 0; The quasilinear form ∂t ∂x where A(w ) is the Jacobian matrix with two real distinct eigenvalues p p λ1 = − (−p 0 (v )) < λ2 = (−p 0 (v )) i.e. p 0 (v ) < 0. Hence system is strictly hyperbolic provided p 0 (v ) < 0 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 44 / 84 Example: The p-system ∂v ∂u − = 0; ∂t ∂x ∂ ∂u + p(v ) = 0; ∂t ∂x The p-system is the simplest nontrivial example of a nonlinear system of conservation laws. Any nonlinear wave equation ∂2g ∂ ∂g − σ( ) = 0 ∂t 2 ∂x ∂x can be re-written in the form of the p-system: Set u= Mapundi K. Banda (Maties) ∂g , ∂t v= ∂g , ∂x CIMPA/MPE 2013 p(v ) = −σ(v ) Jul 22 - Aug 3 45 / 84 Example: Isothermal Flow For isothermal flow: u= ρ ρv A(u) = F (u) = 0 )2 − (ρv ρ2 + a2 ! ρv (ρv )2 2 ρ +a ρ ! 1 2(ρv ) ρ The matrix has the eigenvalues λ1 = v + a; λ2 = v − a where a is the speed of sound Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 46 / 84 Mathematical Challenges Burgers Equation: 2 ∂t u + ∂x u2 = 0; u(x, 0) = 0.5 + sin(x), x ∈ [0, 2π]. t t t t t t 1.4 1.2 1 = = = = = = 0.0 0.5 1.0 1.5 2.0 2.5 0.8 u 0.6 0.4 0.2 0 −0.2 −0.4 0 1 2 3 4 Dimensionless Distance 5 6 Discontinuous solutions do not satisfy PDE in the classical sense at all points. Need a weaker definition of solution of PDE. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 47 / 84 Example: Euler Equations of Gas Dynamics. ∂t ρ + ∂x m = 0; ∂t m + ∂x (ρu 2 + p) = 0; ∂t E + ∂x (u(E + p)) = 0; ρ p = (γ − 1)(E − u 2 ). 2 Riemann problems. uL , x < 0; u(x, 0) = uR , x > 0. Mapundi K. Banda (Maties) + transparent boundary conditions CIMPA/MPE 2013 Jul 22 - Aug 3 48 / 84 Example: Sod’s Riemann initial data. N = 400,T = 0.1644 N = 400,T = 0.1644 1 1 Exact Upw JX RCWENO 0.9 0.8 0.8 0.7 Pressure Density 0.7 0.6 0.5 0.4 0.6 0.5 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 Exact Upw JX RCWENO 0.9 0.2 0.4 0.6 0.8 1 0 0 x 0.2 0.4 0.6 0.8 1 x Three waves: Rarefaction, Contact Discontinuity, Shock wave Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 49 / 84 Example: Euler Equations of Gas Dynamics in Two dimensions. ∂t ρ + ∂x m + ∂y n = 0; ∂t m + ∂x (ρu 2 + p) + ∂y (ρuv ) = 0; ∂t n + ∂x (ρuv ) + ∂y (ρv 2 + p) = 0; ∂t E + ∂x (u(E + p)) + ∂y (v (E + p)) = 0; ρ p = (γ − 1)(E − (u 2 + v 2 )). 2 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 50 / 84 Double Mach Reflection Domain Ω = [0, 4] × [0, 1]. Reflecting wall in bottom interval: [1/6, 4]. Mach 10 shock at x = 1/6, y = 0, 600 angle with x-axis. Impose exact post-shock condition at bottom boundary [0, 1/6]. Reflective boundary for the rest of the boundary. Top boundary exact motion of a Mach 10 shock is used. Results at t = 0.2. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 51 / 84 Double Mach Reflection ∆x = ∆y = 1/120 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 3 3.5 4 ∆x = ∆y = 1/240 1 0.8 0.6 0.4 0.2 0 0 0.5 Mapundi K. Banda (Maties) 1 1.5 2 CIMPA/MPE 2013 2.5 Jul 22 - Aug 3 52 / 84 Existence of Solution in scalar case Definition (the Cauchy Problem(IVP)) For scalar conservation laws: ut + f (u)x = 0; u : R × R+ → R, u(x, 0) = u0 (x), x ∈R f :R→R Example: Advection Equation: a∈R ut + aux = 0, Burgers Equation: ut + Mapundi K. Banda (Maties) u2 2 x CIMPA/MPE 2013 =0 Jul 22 - Aug 3 53 / 84 The Linear Advection Equation a∈R ut + aux = 0, The solution of the Cauchy problem is: u(x, t) = u0 (x − at) Note that initial data travels unchanged with velocity a. The left side can be interpreted as a directional derivative with: du = 0, dt dx =a dt The equation dx dt = a, x(0) = x0 is the characteristic which travels through x0 in the (x, t)-plane, i.e. t x = x0 + at Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 54 / 84 The Linear Advection Equation ut + aux = 0, a∈R Along the characteristics the solution is constant: Suppose (x, t) is such that x = x0 + at holds Then u(x, t̄) = u0 (x − at̄) = u0 (x0 ) is constant Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 55 / 84 The General Scalar Equation ut + f (u)x = 0 which for smooth f can be written in the form ut + f 0 (u)ux = 0 f 0 (u) is referred to as the characteristic velocity - velocity with which information propagates To find characteristics solve: x 0 (t) = f 0 (u(x(t), t)), x(0) = x0 Along the solution x(t), u(x, t) is constant: d ∂u(x(t), t) ∂ u(x(t), t) = + u(x(t), t)x 0 (t) dt ∂t ∂x = ut + f 0 (u)ux = 0 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 56 / 84 The General Scalar Equation ut + f (u)x = 0 which for smooth f can be written in the form ut + f 0 (u)ux = 0 x 0 (t) is constant since f 0 (u(x(t), t) is constant. Characteristics are straight lines given by: x = x(t) = x0 + f 0 (u(x0 , 0))t = x0 + f 0 (u0 (x0 ))t If the above can be solved for x0 for all (x, t), then the solution of the conservation law takes the form: u(x, t) = u(x0 , 0) = u0 (x0 ) where x0 is implicitly given as: x = x0 + f 0 (u0 (x0 ))t Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 57 / 84 Example of Mathematical challenge ut + f (u)x = 0 which for smooth f can be written in the form ut + f 0 (u)ux = 0 with characteristics: x = x(t) = x0 + f 0 (u(x0 , 0))t = x0 + f 0 (u0 (x0 ))t Equation above cannot always be uniquely solved - the characteristics can intersect after some time. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 58 / 84 Example: Burgers Equation Note: f (u) = u2 2 . Let the initial value be given by u0 (x) = −x. Then the characteristics are x = x(t) = x0 − x0 t and this gives t= x x0 − x =1− x0 x0 and also u(x, t) = u0 (x0 ) = u0 x 1−t . The characteristics intersect at the point (x, t) = (0, 1). Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 59 / 84 Example: Burgers Equation The equations for the characteristics are x(t) = x0 + u0 (x0 ) · t. No unique solution for (x, t) anymore; the characteristics intersect. t x x Figure: Characteristics and shock development Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 60 / 84 Example: Burgers x x Figure: Refraction of waves One attains multivalued solution of the equation. For example, the density of gas cannot be multivalued. In this case the discontinuity propagates as a Shock ! Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 61 / 84 Recall Example Burgers Equation: 2 ∂t u + ∂x u2 = 0; u(x, 0) = 0.5 + sin(x), x ∈ [0, 2π]. t t t t t t 1.4 1.2 1 = = = = = = 0.0 0.5 1.0 1.5 2.0 2.5 0.8 u 0.6 0.4 0.2 0 −0.2 −0.4 0 Mapundi K. Banda (Maties) 1 2 3 4 Dimensionless Distance CIMPA/MPE 2013 5 6 Jul 22 - Aug 3 62 / 84 A case for weak solutions This can now be mathematically accurately treated with the help of weak (not differentiable) solutions. Definition A function u = u(x, t) is called weak solution of the Cauchy–Problem ut + (f (u))x = 0, if Z Z ∞ Z [uφt + f (u)φx ] dt dx + R u(x, 0) = u0 (x), u0 (x) φ(x, 0) dx = 0, ∀φ ∈ C01 (R × R). 0 R Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 63 / 84 A case for weak solutions If u is a smooth solution of the equation, then Z Z ∞ [φut + φf (u)x ] dt dx = 0, ∀φ ∈ C01 (R × R). R 0 Integration by parts gives Z Z ∞ Z ∞ − φt udt + φu dx+ R 0 0 ∞ Z − 0 R ∞ φx f (u)dx + φf (u) = 0. −∞ So smooth solutions are also weak. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 64 / 84 The Riemann Problems ut + f (u)x = 0 ( ul , u(x, 0) = u0 (x) = ur , A weak solution is e.g. the shock wave ( ul , u(x, t) = ur , x <0 x >0 x < st x > st, where s is the shock speed. ul ul ur ur Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 65 / 84 The Riemann Problems A weak solution is e.g. the shock wave ( ul , x < st u(x, t) = ur , x > st, where s is the shock speed. s is given by the Rankine–Hugoniot conditions s= Mapundi K. Banda (Maties) f (ul ) − f (ur ) [f ] =: . ul − ur [u] CIMPA/MPE 2013 Jul 22 - Aug 3 66 / 84 Alternative 1: Integral Form Integrate the differential equation: Z t2 Z x2 n o ∂t u(x, t) + ∂x F (u(x, t)) dx dt = 0; t1 u(x, 0) = u0 (x) x1 Giving the following integral forms: 1 Z t2 Z t2 Z x2 F (u(x2 , t))dt F (u(x1 , t))dt − u(x, t1 )dx + u(x, t2 )dx = t1 x1 t1 x1 Z x2 d u(x, t)dx = F (u(x1 , t)) − F (u(x2 , t)). dt x1 Z 2 x2 More difficult to work with than the differential equation. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 67 / 84 Alternative 2: Weak Formulation Introduce the “weak” form of the PDE: 1 2 Z 0 Allows discontinuous solutions and easier to work with. Fundamental in development and analysis of numerical methods. ∞Z ∞n o φ∂t u(x, t) + φ∂x F (u(x, t)) dx dt = 0; u(x, 0) = u0 (x). −∞ Integration gives: Z ∞Z ∞n Z o φt u(x, t) + φx F (u(x, t)) dxdt = 0 φ∈ −∞ 1 C0 (R ∞ φ(x, 0)u(x, 0)dx −∞ × R) Integration over a bounded domain. Problem: non-uniqueness of solutions with same initial data - seek the physically relevant solution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 68 / 84 Example: Burgers Equation Consider Burgers equation f (u) = u2 , 2 s= 1 ul2 − ur2 1 · = (ul + ur ). 2 ul − ur 2 t t U l U l Ur Ur x x Figure: Characteristics for ul > ur (left) and ul < ur (right) Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 69 / 84 Example: Burgers Equation If the characteristics run into the shock, then it is called stable otherwise unstable The mean value theorem: f (ul ) − f (ur ) s= = f 0 (ξ), ξ ∈ (ul , ur ) or (ur , ul ). ul − ur i.e. one has a stable shock (see figure 5. t U l Ur x Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 70 / 84 Example: Burgers Equation If f is convex, i.e. f 0 is monotone increasing, then ul > ur ⇒ f 0 (ul ) > f 0 (ξ) > f 0 (ur ), If on the other hand ul < ur ⇒ f 0 (ul ) < f 0 (ξ) < f 0 (ur ), then one has an unstable shock. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 71 / 84 Example: Burgers Equation Another weak solution for ul < ur ! Another solution is given by the so called rarefaction wave x 0 ul , t ≤ f (ul ) u(x, t) = v ( xt ), f 0 (ul ) ≤ xt ≤ f 0 (ur ) u , x 0 r t ≥ f (ur ) with f 0 (v (ξ)) = ξ. ur t=0 t=t ul f’l f’r x=0 Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 72 / 84 Example: Burgers Equation One can easily recalculate this, since ut + f (u)x = 0 ξ= x x 1 + f 0 (v )v 0 (ξ) = 0 t2 t x =⇒ f 0 (v ) = = ξ. t =⇒t −v 0 (ξ) u=v (ξ) For the Burger’s equation (f 0 (u) = u, v (ξ) = ξ) then ul , x ≤ ul t u(x, t) = xt , ul ≤ xt ≤ ur , u , x ≥ u t r r i.e. for ul < ur , we have found two weak solutions Not a unique solution. But only one can be physically correct. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 73 / 84 Alternative 3: Rankine-Hugoniot Supplement PDE by additional “jump conditions” that are satisfied across discontinuities. Rankine-Hugoniot conditions give the shock speed, s: s= [F ] F (ul ) − F (ur ) = [u] ul − ur Problem: Consider: Burgers Equation: ∂t u + ∂x s1 = u2 2 Multiply by 2u ∂t (u 2 ) + ∂x = 0; [ 12 u 2 ] 1 = (ul + ur ). [u] 2 s2 = [ 32 u 3 ] [u 2 ] 2 u 3 = 0; 3 2 ur3 − ul3 = 3 ur2 − ul2 Same smooth solution but different discontinuous solutions!!! Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 74 / 84 Why all these Mathematical Challenges? Nature has its own way of solving the same problems, physical intuition necessary: 1 2 Equations are models of reality and some physical effects are ignored. Example: Fluid flow always has viscous effects: strong near discontinuity i.e. discontinuity := thin regions with very steep gradients. Thus, one needs further condition, in order to eliminate the non physical weak solutions. Such a simple additional assumption is the Entropy-condition (Lax): For convex f we get ul > ur , i.e. the shock is stable. For ul < ur the shock solution disappears. In this case the rarefaction wave is the correct physical solution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 75 / 84 Unique Solutions - Vanishing Viscosity Entropy condition I is reasonable for convex f , need a more general solution: 1 Introduce a diffusive term in equation: Consider Linear Advection Equation: ∂t u + a∂x u = 0; 2 u(x, 0) = u0 (x) gives u(x, t) = u0 (x − at). Advection-Diffusion Equation: ∂t u + a∂x u = Duxx ; u(x, 0) = u0 (x) F = F (u, ux ). Parabolic, conservation law, always smooth solution t > 0 even if u is discontinuous, for D 1 good approx. for linear advection equation. 3 Let D → 0 - “vanishing viscosity” method: useful for analysis, in general, not optimal! Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 76 / 84 Unique Solutions - Entropy Let η and ψ be two convex functions, for which we have for all smooth solutions u of the conservation law: η(u)t + ψ(u)x = 0, i.e. η 0 (u)ut + ψ 0 (u)ux = 0. For smooth solutions ut + f 0 (u)ux = 0 or η 0 (u)ut + η 0 (u)f 0 (u)ux = 0. Giving η 0 (u)f 0 (u) = ψ 0 (u) and such functions (η, ψ) are called entropy–entropy flux pairs Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 77 / 84 Unique Solutions - Entropy Sufficient for existence and uniqueness of weak solutions. The function u(x, t) is the entropy solution if, for all convex entropy functions, η(u), and corresponding entropy fluxes, ψ(u), the inequality η(u)t + ψ(u)x ≤ 0, is satisfied in the weak sense: i.e. Z Z Z∞ (φt η(u) + φx ψ(u)) dt dx + φ(x, 0)η(u)(x, 0) dx ≤ 0 R 0 (2) R ∀φ ∈ C01 (R × R), φ ≥ 0. Definition Let u be a weak solution of the conservation law. Moreover,suppose u fulfills (2) for any entropy–entropy flux pair (η, ψ), then the function u is called an entropy solution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 78 / 84 Unique Solutions - Entropy For scalar conservation laws every convex function η leads to an entropy: i.e. there exist infinite many entropies!? Theorem Let f : R → R be a strict convex function and u is a piecewise smooth solution of the conservation law ut + f (u)x = 0. If u satisfies the entropy condition for one convex entropy η, then it will be fulfilled for every entropy. Therefore u is an entropy solution. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 79 / 84 Unique Solutions - Entropy Appropriate function spaces. Definition Let u ∈ L∞ (Ω), Ω ⊂ Rn be open. Then the total variation of u is defined by Z 1 TV (u) = lim sup |u(x + ε) − u(x)| dx. ε→0 ε Ω The space of bounded variation is BV (Ω) := {u ∈ L∞ (Ω) : TV (u) < ∞}. If u 0 ∈ L1 (Ω) holds, then Z TV (u) = |u 0 | dx. Ω Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 80 / 84 Unique Solutions - Entropy Theorem (Kruskov) The scalar Cauchy-Problem ut + (f (u))x = 0, f ∈ C 1 (R) u(x, 0) = u0 (x), u0 ∈ L∞ (R) has a unique entropy solution u ∈ L∞ (R × R+ ), having the following properties: (i) ku(·, t)kL∞ ≤ ku0 (·)kL∞ , t ∈ R+ (ii) u0 ≥ v0 ⇒ u(·, t) ≥ v (·, t), t ∈ R+ (iii) u0 ∈ BV (R) ⇒ u(·, t) ∈ BV (R) and TV (u(·, t)) ≤ TV (u0 ) Z Z (iv) u0 ∈ L1 (R) ⇒ u(x, t) dx = u0 (x) dx, t ∈ R+ (i)-(iv) are called R ∞ L -stability, Mapundi K. Banda (Maties) R monotonicity, TV-stability, conservativity. CIMPA/MPE 2013 Jul 22 - Aug 3 81 / 84 Take Note! The theorem can be extended to several dimensions x ∈ Rd , d > 1. The theorem cannot be extended to the case of systems (n > 1). Till now there is no general proposition proved for the system case. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 82 / 84 References E. Godlewski, P.-A. Raviart, Numerical Approximation of Hyperbolic systems of Conservation Laws, Springer, Berlin, 1996 A. Klar, Numerical methods of partial differential equations II: Hyperbolic conservation laws, Lecture Notes, Kaiserslautern. C. Hirsch, Numerical Computation of Internal and External Flows I, Wiley, New York, 2001. C. Hirsch, Numerical Computation of Internal and External Flows II, Wiley, New York, 2000. B. Perthame, Transport Equations in Biology, Birkhäuser, Berlin, 2000. P. D. Lax, Hyperbolic Partial Differential Equations, AMS, New York, 2006. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 83 / 84 Acknowledgements CIMPA, IMU, LMS-AMMSI, ICTP, AIMS Prof. J. Banasiak, College of Mathematics, Computer Science and Statistics, University of KwaZulu-Natal, Durban. National Research Foundation, South Africa. University of KwaZulu-Natal, Witwatersrand and Stellenbosch. Mapundi K. Banda (Maties) CIMPA/MPE 2013 Jul 22 - Aug 3 84 / 84