Outline This lecture Diffusion and advection-diffusion Riemann - PowerPoint PPT Presentation

Outline This lecture • Diffusion and advection-diffusion • Riemann problem for advection • Diagonalization of hyperbolic system, reduction to advection equations • Characteristics and Riemann problem for acoustics Reading: Chapter 3 Recall: Some slides have section numbers on footer. $CLAW/book Examples from the book. www.clawpack.org/doc/apps.html Gallery of applications. R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011

Diffusive flux q ( x, t ) = concentration β = diffusion coefficient ( β > 0 ) diffusive flux = − βq x ( x, t ) q t + f x = 0 = ⇒ diffusion equation: q t = ( βq x ) x = βq xx (if β = const) . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.2]

Diffusive flux q ( x, t ) = concentration β = diffusion coefficient ( β > 0 ) diffusive flux = − βq x ( x, t ) q t + f x = 0 = ⇒ diffusion equation: q t = ( βq x ) x = βq xx (if β = const) . Heat equation: Same form, where q ( x, t ) = density of thermal energy = κT ( x, t ) , T ( x, t ) = temperature, κ = heat capacity, flux = − βT ( x, t ) = − ( β/κ ) q ( x, t ) = ⇒ q t ( x, t ) = ( β/κ ) q xx ( x, t ) . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.2]

Advection-diffusion q ( x, t ) = concentration that advects with velocity u and diffuses with coefficient β : flux = uq − βq x . Advection-diffusion equation: q t + uq x = βq xx . If β > 0 then this is a parabolic equation. Advection dominated if u/β (the Péclet number) is large. Fluid dynamics: “parabolic terms” arise from • thermal diffusion and • diffusion of momentum, where the diffusion parameter is the viscosity. R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.2]

The Riemann problem The Riemann problem consists of the hyperbolic equation under study together with initial data of the form � q l if x < 0 q ( x, 0) = q r if x ≥ 0 Piecewise constant with a single jump discontinuity from q l to q r . The Riemann problem is fundamental to understanding • The mathematical theory of hyperbolic problems, • Godunov-type finite volume methods Why? Even for nonlinear systems of conservation laws, the Riemann problem can often be solved for general q l and q r , and consists of a set of waves propagating at constant speeds. R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.8]

The Riemann problem for advection The Riemann problem for the advection equation q t + uq x = 0 with � q l if x < 0 q ( x, 0) = q r if x ≥ 0 has solution � q l if x < ut q ( x, t ) = q ( x − ut, 0) = q r if x ≥ ut consisting of a single wave of strength W 1 = q r − q l propagating with speed s 1 = u . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.8]

Riemann solution for advection q ( x, T ) x – t plane q ( x, 0) R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.8]

Discontinuous solutions Note: The Riemann solution is not a classical solution of the PDE q t + uq x = 0 , since q t and q x blow up at the discontinuity. Integral form: � x 2 d q ( x, t ) dx = uq ( x 1 , t ) − uq ( x 2 , t ) dt x 1 Integrate in time from t 1 to t 2 to obtain � x 2 � x 2 q ( x, t 2 ) dx − q ( x, t 1 ) dx x 1 x 1 � t 2 � t 2 = uq ( x 1 , t ) dt − uq ( x 2 , t ) dt. t 1 t 1 The Riemann solution satisfies the given initial conditions and this integral form for all x 2 > x 1 and t 2 > t 1 ≥ 0 . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.7]

Discontinuous solutions Vanishing Viscosity solution: The Riemann solution q ( x, t ) is the limit as ǫ → 0 of the solution q ǫ ( x, t ) of the parabolic advection-diffusion equation q t + uq x = ǫq xx . For any ǫ > 0 this has a classical smooth solution: R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 11.6]

Discontinuous solutions Vanishing Viscosity solution: The Riemann solution q ( x, t ) is the limit as ǫ → 0 of the solution q ǫ ( x, t ) of the parabolic advection-diffusion equation q t + uq x = ǫq xx . For any ǫ > 0 this has a classical smooth solution: R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 11.6]

Discontinuous solutions Vanishing Viscosity solution: The Riemann solution q ( x, t ) is the limit as ǫ → 0 of the solution q ǫ ( x, t ) of the parabolic advection-diffusion equation q t + uq x = ǫq xx . For any ǫ > 0 this has a classical smooth solution: R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 11.6]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Suppose hyperbolic: • Real eigenvalues λ 1 ≤ λ 2 ≤ · · · ≤ λ m , • Linearly independent eigenvalues r 1 , r 2 , . . . , r m . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Suppose hyperbolic: • Real eigenvalues λ 1 ≤ λ 2 ≤ · · · ≤ λ m , • Linearly independent eigenvalues r 1 , r 2 , . . . , r m . Let R = [ r 1 | r 2 | · · · | r m ] m × m matrix of eigenvectors. Then Ar p = λ p r p means that AR = R Λ where λ 1   λ 2    ≡ diag ( λ 1 , λ 2 , . . . , λ m ) . Λ =   ...    λ m R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Suppose hyperbolic: • Real eigenvalues λ 1 ≤ λ 2 ≤ · · · ≤ λ m , • Linearly independent eigenvalues r 1 , r 2 , . . . , r m . Let R = [ r 1 | r 2 | · · · | r m ] m × m matrix of eigenvectors. Then Ar p = λ p r p means that AR = R Λ where λ 1   λ 2    ≡ diag ( λ 1 , λ 2 , . . . , λ m ) . Λ =   ...    λ m ⇒ A = R Λ R − 1 R − 1 AR = Λ . AR = R Λ = and Similarity transformation with R diagonalizes A . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Multiply system by R − 1 : R − 1 q t ( x, t ) + R − 1 Aq x ( x, t ) = 0 . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9, 3.1]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Multiply system by R − 1 : R − 1 q t ( x, t ) + R − 1 Aq x ( x, t ) = 0 . Introduce RR − 1 = I : R − 1 q t ( x, t ) + R − 1 ARR − 1 q x ( x, t ) = 0 . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9, 3.1]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Multiply system by R − 1 : R − 1 q t ( x, t ) + R − 1 Aq x ( x, t ) = 0 . Introduce RR − 1 = I : R − 1 q t ( x, t ) + R − 1 ARR − 1 q x ( x, t ) = 0 . Use R − 1 AR = Λ and define w ( x, t ) = R − 1 q ( x, t ) : w t ( x, t ) + Λ w x ( x, t ) = 0 . Since R is constant! R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9, 3.1]

Diagonalization of linear system Consider constant coefficient linear system q t + Aq x = 0 . Multiply system by R − 1 : R − 1 q t ( x, t ) + R − 1 Aq x ( x, t ) = 0 . Introduce RR − 1 = I : R − 1 q t ( x, t ) + R − 1 ARR − 1 q x ( x, t ) = 0 . Use R − 1 AR = Λ and define w ( x, t ) = R − 1 q ( x, t ) : w t ( x, t ) + Λ w x ( x, t ) = 0 . Since R is constant! This decouples to m independent scalar advection equations: w p t ( x, t ) + λ p w p x ( x, t ) = 0 . p = 1 , 2 , . . . , m. R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 2.9, 3.1]

Solution to Cauchy problem ◦ ( x ) Suppose q ( x, 0) = q for − ∞ < x < ∞ . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.1]

Solution to Cauchy problem ◦ ( x ) Suppose q ( x, 0) = q for − ∞ < x < ∞ . From this initial data we can compute data ◦ ( x ) ≡ R − 1 q ◦ ( x ) w R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.1]

Solution to Cauchy problem ◦ ( x ) Suppose q ( x, 0) = q for − ∞ < x < ∞ . From this initial data we can compute data ◦ ( x ) ≡ R − 1 q ◦ ( x ) w The solution to the decoupled equation w p t + λ p w p x = 0 is ◦ p ( x − λ p t ) . w p ( x, t ) = w p ( x − λ p t, 0) = w R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.1]

Solution to Cauchy problem ◦ ( x ) Suppose q ( x, 0) = q for − ∞ < x < ∞ . From this initial data we can compute data ◦ ( x ) ≡ R − 1 q ◦ ( x ) w The solution to the decoupled equation w p t + λ p w p x = 0 is ◦ p ( x − λ p t ) . w p ( x, t ) = w p ( x − λ p t, 0) = w Putting these together in vector gives w ( x, t ) and finally q ( x, t ) = Rw ( x, t ) . R.J. LeVeque, University of Washington IPDE 2011, June 22, 2011 [FVMHP Sec. 3.1]