Introduction to hyperbolic PDEs and Shallow Water Equations

Transcription

1 Introduction to hyperbolic PDEs and Shallow Water Equations Technische Universität München Fundamentals of Wave Simulation - Solving Hyperbolic Systems of PDEs Supervisor Leonhard Rannabauer 8/12/2017

2 Presentation Topics Hyperbolic PDEs Conservation Laws Solution of Advection Equation The Riemann Problem linear and nonlinear characteristics The weak solution Rankine Hugoniot Condition Admissibility Conditions Shock and rarefaction waves Analytical solution of the Dam-Break Problem 2/57

3 Hyperbolic PDEs Definition: An nxn system of equations is strictly hyperbolic if the Jacobian matrix has n real distinct eigenvalues. 3/57

4 Conservation Laws Definition: The overall particular measurable property of an isolated physical system does not change as the system evolves over time. 4/57

5 Conservation Laws formulation [time change of quantity]+[inflow]-[outflow]+[generation]-[destruction]=0 The generation and destruction terms can be omitted for the homogeneous case The mathematical formulation is simplified to ut+f(u)x=0 If f(u)x is smooth ut+fx(u)ux=0 5/57

6 The advection equation ut+ūux=0 u(x,t0)=u0(x,0) (or in conservation form ut+fx(u)ux=0 with fx(u)=ū) ubar=constant Solve with method of characteristics: introduce a new variable s. This is the gradient of the line in the x-t plane along which u remains constant. du ds =0 And by using the chain rule we get dt =1 ds and dx =ū dt Consequently: s=t (t0=0) and x=x0+ūt With these data, solving du =0 ds we get u(x,t)= u0 (x0)=u0(x-ūt) 6/57

7 Characteristics of the advection equation Interpretation of the advection equation solution Image from 1 The lines are parallel as the system is linear 7/57

8 The Riemann problem A conservation system with piecewise initial conditions e.g. ut+ūux=0 With initial conditions q ( x, 0)= { q l for x <0 q r for x > 0 It follows that the solution of the 1d advection equation of the Riemann problem is Image from 1 8/57

9 Generalization for nxn linear conservation system Briefly, since the system is hyperbolic it can be decomposed as n independent equations and the summation of their solutions is the general solution qt+aqx=0 where q is a vector of conserved quantities A is the flux jacobian matrix. In the linear case it is constant Since the system is hyperbolic it has a complete set of eigenvectors and real eigenvalues A R-1ΛR where R ={r1,r2,,rn} and Λ=diag{λ1,λ2,.λn} 9/57

10 System decoupling qt+rλr-1qx=0 R-1qt+ΛR-1qx=0 For ease, a new variable ω is assigned ω=r-1q ωtt+λωx=0 and ω(x,0)= R-1q(x,0) these leads to n scalar advection equations ωtp+λpωxp=0 where ωp is the pth element of the ω vector (scalar) Solution : ωp(x,t)=ωp(x-λpt,0) n q(x,t)=rω(x,t)= ω p ( x λ p t, 0) r p 1 A linear weighted combination of eigenvectors Image from 1 10/57

11 Solution for linear Riemann problem(a) Reminder: q ( x, 0)= { q l for x <0 q r for x > 0 Initial values can be decomposed into a linear combination of the eigenvectors n ql= ω r 1 p l p n and qr= p p ω rr 1 For any point, the solution would be a combination of the eigenvalue decomposition of ql and qr. 11/57

12 Solution for linear Riemann problem(b) E.g. for a 2x2 system there are the following possibilities, depending on the position of x at time t. 1) q(x,t)=ωl1r1+ωl2r2 2)q(x,t)=ωl1r1+ωr2r2 3)q(x,t)=ωr1r1+ωr2r2 Images from [1] Consequently, the overall solution can be divided into three areas 12/57

13 Solution for linear Riemann problem(c) The previous solution can be interpreted as a series of jumps that connect the ql and qr states Image from [1] It can be seen that any two neighboring states differ only on the coefficient of one eigenvector. Also, for the difference between the leftmost and rightmost state the following must hold ql-qr=ωl1r1+ωl2r2 -ωr1r1+ωr2r2 = a1r1+a2r2 where a1=ω1l-ω1r and a2=ω2l-ω2r 13/57

14 Solution for linear Riemann problem(d) Generalizing ql-qr=ωl1r1+ωl2r2 -ωr1r1+ωr2r2 = a1r1+a2r2 ql-qr=ra a can be found initially by decomposing ql-qr Each solution domain can be found by starting from either qr or ql and adding the appropriate number of jumps depending on point coordinates and characteristic speed. q(x,t)=ql+ p a r p p p : λ < x/ t or equivalently q(x,t)=qr+ a pr p p p : λ > x/ t 14/57

15 Non linear system Shallow water equations. Conserved quantities: h (hρ:mass) and hu (huρ:momentum) They are derived from the mass and momentum equilibrium accounting for hydrostatic pressure due to wave height. q q= h = 1 uh q2 f (q)= hu 1 hu2 + gh2 2 for smooth solutions the normal conservation expression qt+f(q)x=0 can be re written as qt+f (q)qx=0 With f ' (q)= u + gh 2u 15/57

16 Solution of a non linear hyperbolic system Question: are there still characteristic curves on which the solution remains constant? d q( x (t ), t )=q t + X ' (t )q x =0 dt For smooth solutions the conservation law takes the form qt+f (q)qx=0 Combining the 2 it follows that f (q)qx=x (t)qx This is valid if X (t) is an eigenvalue of the flux Jacobian The slope of the characteristic curve is the eigenvalue BUT the eigenvalue is now q dependent! 16/57

17 Eigenvalues of the shallow water equation The eigenvectors and eigenvalues pairs of the flux jacobian can found 1 λ =u gh 2 λ =u+ gh 1 r= 2 r = 1 u- gh 1 u+ gh Which are q dependent. Therefore, lines exist on the x-t plane along which q is constant, BUT they are not parallel! 17/57

18 Linear and Non Linear Characteristics(b) In this case seeing the gradient we can deduce that λ1(ql)<λ1(qr) This leads to a rarefaction A continuous solution to the Riemann problem In this case seeing the gradient we can deduce that λ2(ql)>λ2(qr) This leads to a shock A discontinuous solution to the Riemann problem Image from [1] 18/57

19 Linear and Non Linear Characteristics(a) Images from [1] 19/57

20 The weak solution Discontinuities cant be handled by using the smooth conservation law Analytical solution is not always feasible (especially for larger systems) Solution: weak solution Idea :Instead of solving the equation exactly and require the solution to be continuously differentiable, loosen the constrain by solving for u(t,x) for which it holds that ( uφ t + f ( u) φ x dxdt )= 0 For every continuously differentiable function with compact support φ Cc1 20/57

21 Rankine Hugoniot Jump condition(a) As seen, the Riemann problem can result in discontinuities which must be resolved. { ul for x < λt ur for λt > 0 Definition: φ=φ(t,x) any continuously differentiable function with compact support Let Ω be the domain. Compact support means φ(t)=0 on Ω boundaries U ( x, 0)= ( Uφ t + f (U ) φ x dxdt )= 0 ( + ) ( Uφ t + f (U ) φ x dxdt )=0 Apply divergence theorem Ω - Ω + ( + ) (div(uφ,f(u)φ))dxdt =0 Ω - Ω + 21/57

22 Rankine Hugoniot Jump condition(b) Remark. Due to compact support, φ(x,0) 0 only on the interface The interface is the line x=λt ds the differential of the interface Image from [4] (along the line x=λt) n+ds=(λ,-1)dt and n-ds=(-λ,1)dt 22/57

23 Rankine Hugoniot Jump condition(c) Splitting the integrals along the interface normal n vds + n vds=0 + Ω - + Ω - Substituting [ λu u f (u )+f (u )]φ(t, λt )dt =0 For the above to hold for every φ it follows that λ (u u )=f (u ) f (u ) Rankine Hugoniot Condition 23/57

24 Admissibility conditions Weak solution doesn't provide unique solutions, therefore the following constrains can be adopted Vanishing viscosity ute+f(ue)=euexx which converges to u as ε 0 i.e. the second order term acts as a smoother Entropy inequality Call entropy a continuously differentiable function η R n R And entropy flux a q Rn R So that it holds Dη(u)Df(u)=Dq(u) It follows that if u=u(t,x) us a C1 solution of the system of conservation laws η(u)t+q(u)x=0 i.e. we require an additional conservation law to hold thus decreasing the weak solution s possible breach of conservation. Stability condition As seen before, a shock is split into two minor shocks connected by an intermediate state. In order for the shock to be preserved and not dissipate the hint shock must be faster than the front one f ( u* ) f (u- ) * u u - f (u + ) f (u * ) u+ u * 24/57

25 Addressing the strange formations The solution can t be decomposed as the summation of weighted eigenvectors (due to nonlinearity) Idea: Find an intermediate state. Image from [1] Transition:depending on the speed difference of the characteristics choose either Shock transition Rarefaction transition 25/57

26 Resolving the shock(a) Idea: use the Rankine-Hugoniot condition f(ql)-f(qr)=s(ql-qr) Consequently, the speed of the shock is s and can be found solving the above equation 26/57

27 Resolving the shock(b) Image from [1] The shock is resolved by separating the crashing regions by a line with slope=s as calculated by the Rankine Hugoniot condition 27/57

28 Resolving the shock(c) The family of possible q* that provide a shock can be calculated as follows: First, determine that at the interface between q and q* there is a shock. This can be done by taking the λp(qr) and comparing it to λp(ql) if at any characteristic it holds that λp(qr)<λp(ql), or vice versa, there is a shock formation. The Rankine Hugoniot condition must hold, so: s(q-q*)=f(q*)-f(q) where q* is the intermediate state and q the extreme state Specifically for the shallow water equation we have s(h-h*)=hu-h*u* s(hu-h*u*)=hu2-h*u*u*+g(h2-h*h*)/2 Solving for s and substituting we get a quadratic equation that has the following solution 28/57

29 Resolving the shock(d) g (h) h* u (h)=u± (h h *) 2 h* h Each sign choice corresponds to one family of characteristics. In our case the sign corresponds to the λ1 family and the + to the λ2 (the - family approaches q* tangent to the vector r1 and the + tangent to the vector r2) It can be easily determined how many, if any, are physically correct. (entropy condition) 29/57

30 The entropy condition (Lax) For the shock to be physically correct, it must hold that The characteristics of the shock must satisfy λp(ql)>s>λp(qr)>0 or λp(ql)<s<λp(qr)<0 λj(ql)<s λj(qr)<s for j<p λj(ql)>s λj(qr)>s for j>p with the eigenvalues being ordered as λ1<λ2<.λn 30/57

31 Resolving the rarefaction(a) An integral curve is a parametric curve that represents a specific solution to an ordinary differential equation or system of equations. If the differential equation is represented as a vector field or slope field, then the corresponding integral curves are tangent to the field at each point. 31/57

32 Resolving the rarefaction(b) It follows that if q(ξ) is an integral curve of the vector field rp, it holds that q (ξ)=a(ξ)rp(q(ξ)) Take the shallow water equation as example q1(ξ) =α(ξ) which is satisfied with q1(ξ)=α(ξ)ξ We will examine the parameter value a=1 for ease and readability of the formulas 32/57

33 Resolving the rarefaction(c) q2(ξ) =q2(ξ)/q1(ξ)- (gq1(ξ) Fixing a point on the curve (any point) and solving for q2(ξ) and r1 q2(ξ)=ξul+2ξ( (gq1*- (gq1(ξ)) 33/57

34 Resolving the rarefaction(d) Or in terms of actual physical quantities hu=hlul+2h( (gq1*- (gq1(ξ)) similar procedure is followed for r2 Plotting the curves for the SWE gives (with fixed point being ql or qr depending on the problem configuration Image from [2] 34/57

35 Resolving the rarefaction(e) hu=h*u*+2h( (gh*- (gh) this equation can be rewritten as hu+2h (gh=h*u*+2h (gh*) Remark: since the points can be anywhere on the curve w1(q)=u+2h (gh) is constant along the curve equally w2(q)=u-2h (gh) Constant functions on the integral curve of rp are called p-riemann invariants 35/57

36 Resolving the rarefaction(f) Since wp is constant along the rp integral curve, d p it follows that w (q (ξ ))=0 dξ Expanding and substituting q (ξ)=q(ξ)rp wprp=0 wp is the gradient with respect to q 36/57

37 Resolving the rarefaction(g) The rarefaction is a continuous solution, therefore a strong one. The conservation law must be fulfilled qt=q (ξ(x,t))ξt and qx=q (ξ(x,t))ξx Substituting into conservation law ξt+λp(q(ξ))ξx=0 d p p λ = λ q ' (ξ )=0 dξ 37/57

38 Resolving the rarefaction(h) The rarefaction is a continuous solution, therefore a strong one. The conservation law must be fulfilled qt=q (ξ(x,t))ξt and qx=q (ξ(x,t))ξx Substituting into conservation law ξt+λp(q(ξ))ξx=0 The variation of λ along the curve is given by d p p λ = λ q ' (ξ )=0 dξ 38/57

39 Resolving the rarefaction(i) A centered rarefaction wave is a special wave, were the solution is constant along rays through the origin. The solution for this wave is self similar i.e. q(x,t)=q(x/t) in this case, x/t=ξ (ξ speed of propagation) consequently ξ=λp(q(ξ)) 39/57

40 Resolving the rarefaction(almost done) ξt+λp(q(ξ))ξx=0 in this case becomes ξ=λp(q(ξ)) Differentiate w.r.t. ξ 1= λ q ' (ξ )=0 Substitute q (ξ)=a(ξ)rp(q(ξ)) p p 1= λ ar p Solving for a and substituting we get rp q ' (ξ )= p p λ r 40/57

41 Resolving the rarefaction(last one I swear) rp q ' (ξ )= p p λ r Can be solved. As boundary condition we use the known state which the rarefaction connects to the intermediate state Image from [1] 41/57

42 Finding the intermediate state As discussed above, we can find a series of intermediate points that connect the left and right state with shocks and rarefactions. To find the correct intermediate point 2 conditions must hold It must the states in a physically correct way It must connect both states Solution: Find the common point that fulfills both those conditions (analytically or numerically (or graphically)) 42/57

43 Hugoniot loci The intermediate state must satisfy both the conditions coming from the left and from the right. All possible states that connect ql q* and qr are plotted. Admissibility conditions must be satisfied or the state discarded (this includes expected transition form or unphysical behavior, eg negative height) q* is chosen as as the only state satisfying all the above conditions Image from [1] 43/57

44 Structure of the rarefaction The rarefaction can be seen as a set of intermediate points from the intermediate point to the state it connects to. Therefore, structure of the rarefaction is given by the the intermediate states between q* and the side it connects to. Images from [1] 44/57

45 General strategy for solving the non linear 2x2 Riemann problem Determine the nature of the wave (shock or rarefaction) either by direct eigenvalue comparison or by an appropriate entropy condition Determine the intermediate state between the two waves Determine the structure of any rarefaction wave 45/57

46 Remark It might happen that a shock and rarefaction collide Images from [5] It is treated precisely as if two characteristic lines collided, i.e. with a shock or a rarefaction (in this example shock) that fulfills all the necessary conditions (Rankine Hugoniot, entropy) 46/57

47 Image from [5] 47/57

48 Dam break problem (a) Following the above procedure we can approach the dam breaking problem A system of shallow water equations with initial conditions { q ( x, 0)= h=hl u=0 for x <0 h=h r u=0 for x > 0 Assuming hl=6 hr=4 As discussed previously Consequently λ2(ql)>λ2(qr)>0>λ1(qr)>λ1(ql) 1 λ =u gh 2 and λ =u+ gh λ1(ql)<λ1(qr)<0 the left state is connected to the intermediate via a rarefaction 0<λ2(ql)<λ2(qr) the right state is connected to the intermediate via a shock 48/57

49 Dam break problem (b) The solution has the form As discussed above, we can use Image from [1] the shock and rarefaction equations to create families of intermediate states. From the resulting curves we only need to take into account the 1 rarefaction that connects the intermediate with ql and the 2 shock that connects qr with the intermediate 49/57

50 Dam break problem (c) Rarefaction connects the 1 characteristics and ql with q* therefore we use the function q ' (ξ )= with 1 r= 1 u- gh 1 λ =u gh 1 r 1 1 λ r 1 λ = u/h- 0.5 g/h 1/hu 50/57

51 Dam break problem (d) 1 r q ' (ξ )= 1 1 λ r Substituting into h=(a-ξ)2/9g with A the integration constant The left side of the rarefaction is known (q=ql) at ξ=ul- ghl A=ul+ ghl Equally, all the intermediate points must satisfy the same conditions This way using h=(ul+ ghl-ξ)2/9g we can find the intermediate points Alternatively, we can use the Riemann invariants and determine the points from A=ul+ ghl=u*+ gh* 51/57

52 Dam break problem (e) Shock connects the 2 characteristics and q* with qr therefore we use this function g (h) h * u (h)=u + (h h* ) 2 h* h the entropy condition excludes all the unphysical points Finding the common intermediate we get Images from [1] 52/57

53 Dam break problem (f) And the physical behavior Images from [1] 53/57

54 Different initial conditions Depending on the initial conditions the possible solutions are qualitatively Two shocks Two rarefactions Shock-rarefaction Rarefaction-shock Determined with the procedure outlined in the previous slides 54/57

55 Literature [1] Lectures of subject tsunami simulation at [2] LeVeque, R. (2002). Finite Volume Methods for Hyperbolic Problems (Cambridge Texts in Applied Mathematics) [3] Hyperbolic Conservation Laws An Illustrated Tutorial, Lectures Notes [4] Modelling and Optimization of Flows on Networks Cetraro School 2007 [5] Riemann Problems of the Shallow Water Equations at 55/57

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