Linear Hyperbolic Systems

Transcription

1 Linear Hyperbolic Systems Professor Dr E F Toro Laboratory of Applied Mathematics University of Trento, Italy eleuterio.toro@unitn.it October 8, / 56

2 We study some basic mathematical aspects of linear systems of hyperbolic equations in one space dimension We illustrate the theory through the study, in complete detail, of a simple example with physical meaning, namely a linearised model for blood flow 2 / 56

3 Basic Theory Consider the one-dimensional, time-dependent system of m linear hyperbolic equations with source terms t Q(x, t) + A x Q(x, t) = S(Q(x, t)). (1) Q: unknowns, A: matrix of coefficients, constant, S(Q): source terms: Q = q 1... q i... q m, A = a a 1i... a 1m a i1... a ii... a im a m1... a mi... a mm Special case: the linear advection equation, S(Q) = s 1... s i... s m. (2) t q(x, t) + λ x q(x, t) = 0 (3) 3 / 56

4 Eigenvalues and eigenvectors. Definition: The eigenvalues of system (1) are the roots of the characteristic polynomial P (λ) Det(A λi) = 0. (4) I: m m unit matrix, λ: a parameter, λ i : eigenvalues, in increasing order λ 1 λ 2... λ i... λ m 1 λ m. (5) Definition: A right eigenvector R i of A corresponding to λ i is R i = [r 1i, r 2i,..., r ii,..., r mi ] T, (6) such that AR i = λ i R i. (7) The m right eigenvectors corresponding to the eigenvalues (5) are R 1, R 2,..., R i,..., R m 1, R m. (8) 4 / 56

5 Definition: vector such that A left eigenvector L i of A corresponding to λ i is the row L i = [l i1, l i2,..., l ii,..., l im ], (9) L i A = λ i L i. (10) The m eigenvalues (5) generate corresponding m left eigenvectors L 1, L 2,..., L i,..., L m 1, L m. (11) Definition: Hyperbolic system. A system (1) is said to be hyperbolic if A has m real eigenvalues and a corresponding complete set of m linearly independent eigenvectors. Note: For hyperbolicity, the eigenvalues are not required to be all distinct. What is important is that there is a complete set of linearly independent eigenvectors, corresponding to the real eigenvalues. 5 / 56

6 More definitions Strictly hyperbolic system. A hyperbolic system is said to be strictly hyperbolic if all eigenvalues of the system are distinct. Weakly hyperbolic system. A system may have real but not distinct eigenvalues and still be hyperbolic if a complete set of linearly independent eigenvectors exists. However if all eigenvalues are real but no complete set of linearly independent eigenvectors exists then the system is called weakly hyperbolic. Definition: Orthonormality of eigenvectors. The eigenvectors L i and R j are rthonormal if L i R j = { 1 if i = j, 0 if i j. } (12) 6 / 56

7 Example: an abstract model system Consider the homogeneous linear system with constant coefficients [ ] [ ] [ ] [ ] q1 (x, t) 0 a q1 (x, t) 0 t + q 2 (x, t) b 0 x =, (13) q 2 (x, t) 0 with a and b two real numbers. We first find the eigenvalues ([ ] [ ]) 0 a λ 0 P (λ) Det(A λi) = Det = 0 b 0 0 λ λ 1 = ab, λ 2 = ab. (14) The eigenvalues are real if a and b are real numbers of the same sign. If a = 0 or b = 0 then the eigenvalues are real but not distinct. If a and b are real, distinct from zero and of opposite sign, the eigenvalues are complex and the system is not hyperbolic. 7 / 56

8 Example: an abstract model system Assume a right eigenvector [r 1, r 2 ] T corresponding to an eigenvalue λ [ ] [ ] [ ] 0 a r1 λr1 =, (15) b 0 r 2 λr 2 from which we obtain ar 2 = λr 1, br 1 = λr 2. (16) The equations are not independent. We choose to use the first one r 2 = λ a r 1. (17) For λ = λ 1 = ab and setting r 1 = β 1 we obtain eigenvector R 1 r 1 = β 1, r 2 = ab b a β 1 = a β 1. (18) 8 / 56

9 For λ = λ 2 = ab, setting r 1 = β 2 we obtain R 2 as r 1 = β 2, r 2 = ab a β 1 = Then the two right eigenvectors are 1 R 1 = β 1 b, R 2 = β 2 a b a β 2. (19) 1 b a. (20) These vectors are linearly independent and thus system (12) is hyperbolic, provided a and b are two non-zero real numbers of equal sign. The system is also strictly hyperbolic, as λ 1 λ 2. Exercise. Find the left eigenvectors corresponding to the two real eigenvalues (14). 9 / 56

10 Diagonalization and characteristic variables Consider R = [R 1,..., R i,..., R m ]: matrix whose columns are the right eigenvectors; Λ: diagonal matrix formed by eigenvalues R = r r 1i... r 1m r i1... r ii... r im r m1... r mi... r mm, Λ = λ λ i λ m (21) Proposition. If A is the coefficient matrix of a hyperbolic system (1) then. A = RΛR 1 or Λ = R 1 AR. (22) In this case A is said to be diagonalisable and consequently system (1) is said to be diagonalisable.11 Proof (omitted). 10 / 56

11 Characteristic variables The existence of R 1 makes it possible to define the characteristic variables C = [c 1, c 2,..., c m ] T via C = R 1 Q Q = RC. (23) Calculating the partial derivatives (constant coefficients) t Q = R t C, x Q = R x C and directly substituting these expressions into equation (1) gives R t C + AR x C = S. Multiplication of this equation from the left by R 1 and use of (22) gives t C + Λ x C = Ŝ, Ŝ = R 1 S. (24) 11 / 56

12 This is called the canonical form or characteristic form of system (1). Assuming Ŝ = 0 and writing the equations in full, we have c 1 λ c t c i λ i x c i... = c m λ m c m 0 Clearly, each equation i-th of this system is of the form. (25) t c i + λ i x c i = 0, i = 1,..., m (26) and involves the single unknown c i (x, t), which is decoupled from the remaining variables. Moreover, this equations is identical to the linear advection equation (3), with characteristic speed is λ i. We have m decoupled equations, each one defining a characteristic curve. 12 / 56

13 Thus, at any chosen point (ˆx, ˆt) in the x-t half-plane there are m characteristic curves x i (t) passing through (ˆx, ˆt) and satisfying the m ODEs dx i dt = λ i, for i = 1,..., m. (27) t dx 1 dt = λ 1 dx i dt = λ i dx m dt = λ m ˆt (ˆx, ˆt) x (0) m ˆx Fig. 1. The solution at a point (ˆx, ˆt) depends on the initial condition at the foot x (0) i of each characteristic x i (t) = x (0) i + λ i t. x (0) i x (0) 1 x 13 / 56

14 Each characteristic curve x i (t) = x (0) i + λ i t intersects the x-axis at the point x (0) i, which is the foot of the characteristic passing through the point (ˆx, ˆt). The point x (0) i is given as x (0) i = ˆx λ iˆt, for i = 1, 2,..., m. (28) Each equation (26) is just a linear advection equation whose solution at (ˆx, ˆt) is given by c i (ˆx, ˆt) = c (0) i (x (0) i ) = c (0) i (ˆx λ iˆt), for i = 1, 2,..., m, (29) where c (0) i (x) is the initial condition, at the initial time. The initial conditions for the characteristic variables are obtained from the transformation (23) applied to the initial condition Q(x, 0). Given the assumed order (5) of the distinct eigenvalues the following inequalities are satisfied. x (0) m < x (0) m 1 <... < x(0) 2 < x (0) 1. (30) 14 / 56

15 Definition: Domain of dependence. The interval [x (0) m, x (0) 1 ] is called the domain of dependence of the point (ˆx, ˆt). See Fig. 1. The solution at (ˆx, ˆt) depends exclusively on initial data at points within the interval [x (0) m, x (0) 1 ]. This is a distinguishing feature of hyperbolic equations. The initial data outside the domain of dependence can be changed in any manner we wish but this will not affect the solution at the point (ˆx, ˆt). 15 / 56

16 The general initial-value problem Proposition: The solution of the general IVP for the linear homogeneous hyperbolic system PDEs: t Q + A x Q = 0, < x <, t > 0, (31) IC: Q(x, 0) = Q (0) (x) is given by Q(x, t) = m c i (x, t)r i. (32) The coefficient c i (x, t) of the right eigenvector R i is a characteristic variable. i=1 16 / 56

17 Proof: We find the general solution in terms of the characteristic variables C by solving the IVP PDEs: t C + Λ x C = 0, < x <, t > 0, IC: C(x, 0) = C (0) (x). (33) Here the initial condition is C (0) (x) = [c (0) 1 (x),..., c(0) i (x),..., c (0) m (x)] T = R 1 Q (0) (x) (34) where Q (0) (x) is initial conditions of the original problem, denoted as Q (0) (x) = [q (0) 1 (x),..., q(0) i (x),..., q m (0) (x)] T. The solution of IVP (33) is direct. For each component c i (x, t) we have c i (x, t) = c (0) i (x λ i t), for i = 1,..., m. (35) 17 / 56

18 In terms of the original variables Q the solution is found by transforming back, that is Q = RC, or q 1 = c 1 (x, t)r 11 + c 2 (x, t)r c m (x, t)r 1m, q i = c 1 (x, t)r i1 + c 2 (x, t)r i c m (x, t)r im, q m = c 1 (x, t)r m1 + c 2 (x, t)r m c m (x, t)r mm, or q 1 q 2. q m = c 1(x, t) r 11 r 21. r m1 +c 2(x, t) r 12 r 22. r m c m(x, t) r 1m r 2m. r mm. More succinctly and the sought result follows. Q(x, t) = m c i (x, t)r i (36) i=1 18 / 56

19 Remarks: The function c i (x, t) is the coefficient of R i in an eigenvector expansion of the solution vector Q(x, t) Given a point (x, t) in the x-t plane, the solution Q(x, t) depends only on the initial data at the m points x (i) 0 = x λ i t. See Fig. 1 These points are the intersections of the characteristics of speed λ i with the x axis Solution (32) represents superposition of m waves of unchanged shape c (0) i (x)r i propagated with speed λ i 19 / 56

20 The Riemann problem Proposition: The solution of Riemann problem PDEs: t Q + A x Q = 0, { < x <, t > 0, IC: Q(x, 0) = Q (0) QL if x < 0, (x) = Q R if x > 0, (37) with Q L and Q R two constant vectors, is given by I m Q(x, t) = c ir R i + c il R i, (38) i=1 i=i+1 where m m c il R i = Q L, c ir R i = Q R (39) i=1 i=1 and I = I(x, t) is the maximum value of i for which x λ i t > / 56

21 Remarks on the solution of the Riemann problem The initial data consists of two constant vectors Q L and Q R, separated by a discontinuity at x = 0 This is a special case of IVP (31) The structure of the solution of the Riemann problem (37) is depicted in Fig. 2, in the x-t plane The solution consists of a fan of m waves emanating from the origin, one wave for each eigenvalue λ i. The speed of the wave i is the eigenvalue λ i These m waves divide the x-t half plane into m + 1 constant regions { R i = (x, t)/ < x < ; t 0; λ i < x } t < λ i+1, (40) for i = 0, 1,..., m. 21 / 56

22 Solving the Riemann problem means finding constant values for Q in regions R i for = 1,..., m 1. x t = λ 2 t x t = λ i x t = λ 1 R 1 R i x t = λ m R 0 R m Q L Q R x = 0 Fig. 2. Structure of the solution of the Riemann problem. There are m waves that divide the half x-t plane into m + 1 regions (wedges) R i, with i = 0, 1,..., m. x 22 / 56

23 Proof: First we recall the following notation Q L = [q 1L,..., q il,..., q ml ] T, Q R = [q 1R,..., q ir,..., q mr ] T, C L = [c 1L,..., c il,..., c ml ] T, C R = [c 1R,..., c ir,..., c mr ] T, C L = R 1 Q L, C R = R 1 Q R, where C L, C R are the characteristic variables. (41) The form of the sought solution is the same as that for the general IVP, see (36), for which we only need to find the coefficients c i (x, t). Thus we need to solve the associated Riemann problem for the characteristic variables with initial conditions: C L, C R. To this end one solves the RP for each component c i (x, t) The required data comes from solving the following two linear systems: m m Q L = c il R i = RC L, Q R = c ir R i = RC R. (42) i=1 i=1 23 / 56

24 The two linear systems for the coefficients C L and C R, in full, read r r i1... r 1m r i1... r ii... r im =, r m1... r mi... r mm r r 1i... r 1m r i1... r ii... r im r m1... r mi... r mm c 1L... c il... c ml c 1R... c ir... c mr = q 1L... q il... q ml q 1R... q ir... q mr. (43) These two linear systems for C L, C R are easily solved using standard methods. 24 / 56

25 In terms of the characteristic variables, for i = 1,..., m, we have PDE: t c i + λ i x c i = 0, < x <, t > 0, c il if x < 0, IC: c (0) i (x) = c ir if x > 0, The solutions of these scalar Riemann problems are given by c il if x λ i t < 0 x/t < λ i, c i (x, t) = c (0) i (x λ i t) = c ir if x λ i t > 0 x/t > λ i. For a given (x, t) there is an integer I(x, t) and an associated eigenvalue λ I such that (x, t) belongs to region R I, that is λ I < x/t < λ I+1. See Fig. 3. Then we have } x λ i t > 0 for i = 1, 2,..., I coefficients c Ri, x λ i t < 0 for i = I + 1, I + 2,..., m coefficients c Li. (44) (45) (46) 25 / 56

26 If I = I(x, t) is the maximum value of i for which x λ i t > 0, then Q(x, t) = I m c ir R i + c il R i (47) i=1 i=i+1 and the claimed result is thus proved. Corollary. The solution of the Riemann problem may be expressed as Q(x, t) = Q L + I δ i R i = Q R i=1 m i=i+1 δ i R i, (48) where C = [δ 1,..., δ i,..., δ m ] T, R C = Q = Q R Q L and m δ i R i = Q. (49) i=1 This form is more convenient. We only need to solve one linear system. Proof. Left as exercise. 26 / 56

27 x t = λ I t R i x t = λ I+1 x t = λ 1 ˆt (ˆx, ˆt) x t = λ m R 0 R m Q L Q R x = 0 ˆx x Fig. 3. The solution of the Riemann problem at a point (ˆx, ˆt) depends on the associated index I = I(ˆx, ˆt). 27 / 56

28 Concluding Remarks We have studied the basic mathematical aspects of linear systems of hyperbolic equations in one space dimension. This background on linear hyperbolic equations is useful for studying simplified models for practical problems. The linear theory is also useful background for studying non-linear hyperbolic equations 28 / 56

29 Case study I: Linearised Shallow Water Equations 29 / 56

30 Linearization Consider the one-dimensional non-linear shallow water equations in terms of physical variables: water depth h(x, t) and particle velocity u(x, t) } t h + u x h + h x u = 0, t u + u x u + g x h = gb (50) (x). z Air H(x, t) = b(x) + h(x, t) u(x, t) h(x, t) Water z = 0 Solid Fig. 4. Illustration of the shallow water equations. The source term involves the slope b (x) of the bottom elevation and the acceleration due to gravity g. We are interested in a linearised version of (50), without the source term (homogeneous), b (x) = 0. b(x) x 30 / 56

31 The linear equations Consider small perturbations η(x, t) and v(x, t) in surface elevation and in particle velocity as follows h(x, t) = H + η(x, t), u(x, t) = 0 + v(x, t). (51) H is the, constant, unperturbed water depth and 0 + v(x, t) means a small perturbation v(x, t) to stationary fluid It is also assumed that derivatives of the perturbations are small Then by substituting h and u from equations (51) into (50) and neglecting second order terms we obtain the linearized shallow water equations t η + H x v = 0, (52) t v + g x η = / 56

32 In matrix form the equations read t Q + A x Q = 0, (53) where Q(x, t) and A are respectively [ ] η Q =, A = v [ 0 H g 0 ]. (54) We note in passing that equations (52) reproduce the well-known linear second-order wave equation, for both η and v, namely and (2) t η = gh (2) x η (55) (2) t v = gh (2) x v. (56) The second-order linear wave equation, either (55) for η or (56) for v, is a very popular hyperbolic model for wave propagation phenomena. 32 / 56

33 The eigenstructure The eigenvalues of the matrix A in (53) are obtained from the characteristic polynomial, yielding where λ 1 = a, λ 2 = a, (57) a = gh (58) is the celerity, the speed of propagation of vanishing small-amplitude surface waves. A left eigenvector L = [l 1, l 2 ] of A corresponding to an eigenvalue λ is found from [ ] 0 H [l 1, l 2 ] = [λl g 0 1, λl 2 ], (59) which leads to the algebraic equations l 2 g = λl 1, l 1 H = λl 2. (60) 33 / 56

34 These two equations are not independent (verify). Using the second equation we may write l 2 = H λ l 1. (61) Now, setting l 1 = α 1 and λ = λ 1 = a in (61) gives the two components of the left eigenvector L 1 corresponding to the eigenvalue λ 1 = a, as l 1 = α 1, l 2 = H a α 1, (62) where α 1 is a scaling parameter open to choice. To find the left eigenvector L 2 corresponding to λ = λ 2 = a we set l 1 = α 2 and λ = λ 2 = a in (61) to obtain l 1 = α 2, l 2 = H a α 2, (63) where α 2 is again a scaling parameter open to choice. Then the two left eigenvectors corresponding to the eigenvalues λ 1 = a and λ 2 = a are respectively given by L 1 = α 1 [1, H a ], L 2 = α 2 [1, H a ]. (64) 34 / 56

35 Analogously, a right eigenvector [r 1, r 2 ] T corresponding to an eigenvalue λ satisfies [ ] [ ] [ ] 0 H r1 λr1 =, (65) g 0 r 2 λr 2 from which we obtain the relation Setting r 1 = β 1 and λ = λ 1 = a in (66) gives Setting r 1 = β 2 and λ = λ 2 = a in (66) gives r 2 = λ H r 1. (66) r 1 = β 1, r 2 = a H β 1. (67) r 1 = β 2, r 2 = a H β 2. (68) Then the right eigenvectors corresponding to the eigenvalues α 1 = a and α 2 = a are 1 1 R 1 = β 1, R 2 = β 2. (69) a/h a/h 35 / 56

36 If we want the left and right eigenvectors to be orthonormal, then the scaling parameters must be chosen to satisfy α 1 β 1 = 1/2, α 2 β 2 = 1/2. (70) Choosing α 1 = α 2 = 1 gives β 1 = β 2 = 1 2. Then the matrices L and R formed by the left and right eigenvectors are given by 1 H/a L =, R = (71) 2 1 H/a a/h a/h It is easy to show that R 1 = L, (72) where the matrix R 1 denotes the inverse matrix of R (verify). 36 / 56

37 Equations in characteristic variables First we define the characteristic variables C = [c 1, c 2 ] T as C = LQ. (73) Note that we could also use C = R 1 Q, but (73) is more direct. For our system, written in full, we have c 1 1 H/a η =, (74) 1 H/a v c 2 which gives the characteristic variables as c 1 = η H a v, c 2 = η + H a v. (75) 37 / 56

38 In terms of the characteristic variables C we have t Q = L 1 t C, x Q = L 1 x C. (76) Then (53) becomes L 1 t C + AL 1 x C = 0. (77) Multiplying (77) from the left by L gives t C + (LAL 1 ) x C = 0. (78) It is easily verified that [ LAL 1 λ1 0 = Λ = 0 λ 2 ] = [ a 0 0 a ] (79) and thus the equations in characteristic variables become t C + Λ x C = 0. (80) 38 / 56

39 The equations have become completely decoupled, namely t c 1 + λ 1 x c 1 = 0, or The general initial-value problem t c 2 + λ 2 x c 2 = 0, t c 1 a x c 1 = 0, t c 2 + a x c 2 = 0. PDEs: t Q + A x Q = 0, ICs: Q(x, 0) = Q (0) (x). (81) (82) (83) Here, the initial condition Q (0) (x) at time t = 0 is an arbitrary function of x alone. 39 / 56

40 We may now replace the IVP (83) by the equivalent IVP PDEs: t C + Λ x C = 0, ICs: C(x, 0) = C (0) (x) = LQ (0) (x) (84) in terms of characteristic variables C. From (75) the initial conditions are c (0) 1 (x) = η(0) (x) H a v(0) (x), (85) c (0) 2 (x) = η(0) (x) + H a v(0) (x). Hence the solution of (84) is c 1 (x, t) = c (0) 1 (x λ 1t) = η (0) (x + at) H a v(0) (x + at), c 2 (x, t) = c (0) 2 (x λ 2t) = η (0) (x at) + H a v(0) (x at). (86) 40 / 56

41 In terms of the original variables Q =RC the complete solution to the original IVP (83) is η(x, t) = 1 { η (0) (x + at) + η (0) (x at) + H } 2 a [ v(0) (x + at) + v (0) (x at)], v(x, t) = 1 2 (Verify). { v (0) (x + at) + v (0) (x at) + a } H [ η(0) (x + at) + η (0) (x at)]. (87) 41 / 56

42 The Riemann problem PDEs: t Q + A x Q = 0, Q L if x < 0, ICs: Q(x, 0) = Q (0) (x) = Q R if x > 0. The problem is to find Q, the solution in region R 1 : Star Region. (88) Star region t x λ 1t = 0 x λ 2t = 0 R 1 R 0 R 2 Q L Q R Fig. 5. Structure of the solution of the Riemann problem for the linearized shallow water equations. 0 x 42 / 56

43 From (48) the solution at any point (x, t) can be written as Q(x, t) = Q L + I δ i R i, (89) i=1 where the positive integer I(x, t) is such that for λ I < x t < λ I+1 (90) The coefficients δ i are the solution of the m m linear system m δ i R i = Q R Q L = Q. (91) i=1 Here m = 2 and the linear algebraic system is η η R η L δ 1 + δ 2 = =. (92) 1 2 a/h 1 2 a/h v v R v L 43 / 56

44 This gives The solution is δ 1 + δ 2 = 2 η, δ 1 + δ 2 = 2 H a v. (93) δ 1 = η H a v, δ 2 = η + H v. (94) a Here we are interested in the solution Q = [η, v ] T in the star region R 1, which is given by the points (x, t) such that λ 1 = a < x t < λ 2 = a, (95) with I = 1 in (90). See also (46). Then the solution is given by Q = Q L + δ 1 R 1. (96) 44 / 56

45 This gives η = 1 2 (η L + η R ) 1 H 2 a (v R v L ), v = 1 2 (v L + v R ) 1 a 2 H (η R η L ). A numerical example. Initial conditions η L = 1.0m, η R = 0.1m, v L = 0.0m/s, v R = 0.0m/s. Initial discontinuity at time t = 0s is placed at x 0 = 500m. As parameters, choose H = 10m and g = 9.8m/s 2, so that the celerity is a = 9.9m/s. The exact solution in the star region between the two waves is η = 0.55m; v = m/s. Solution profiles are shown in Fig. 6 at time t = 25s. (97) 45 / 56

46 Distance along channel Free surface perturbation Velocity perturbation Distance along channel Fig. 6. Solution of Riemann for the linearised shallow water equations. Free-surface elevation and velocity at time t = 25s. 46 / 56

47 Case study II: Linearised Blood Flow Equations 47 / 56

48 Equations: conservation of mass and momentum Blood flow in medium-size to large arteries and veins can be represented by the non-linear system of hyperbolic equations } t A + x (ua) = 0 t (ua) + x (Au 2 ) + A ρ (98) xp = Ru. Assumed axially symmetric vessel configuration in 3D at time t. Cross-sectional area A(x, t) and wall thickness h 0 (x) are illustrated 48 / 56

49 Unknowns and a closure condition: tube law A(x, t): cross-sectional area of the vessel at position x and time t u(x, t): averaged velocity of blood at a cross section p(x, t) is pressure Blood density ρ is constant and R > 0, the viscous resistance, prescribed There are two PDEs (98) and three unknowns: A(x, t), u(x, t) and p(x, t) An extra relation is required to close the system: the tube law. This relates p(x, t) to wall displacement via A(x, t), thus coupling elastic properties of the vessel to the fluid dynamics inside the vessel where p = p e (x, t) + ψ(a; K) (99) ψ(a; K) = p p e p trans (100) is the transmural pressure, the difference between the pressure in the vessel, the internal pressure, and the external pressure. 49 / 56

50 More on the tube law Here we adopt ( ) ψ(a; K) = K(x) A A0, (101) with K(x) = π E(x)h 0 (x) (1 ν 2. (102) ) A0 (x) A 0 (x) is the cross-sectional area of the vessel at equilibrium, that is when u = 0; h 0 (x) is the vessel wall thickness; E(x) is the Young s modulus of elasticity of the vessel and ν is the Poisson ratio, taken to be ν = 1/2. The external pressure is assumed to be a known function of space and time and may be decomposed as follows p e (x, t) = p atm + p musc (x, t), (103) where p atm is the atmospheric pressure, assumed constant here, and p musc (x, t) is the pressure exerted by the surrounding tissue. 50 / 56

51 Simplified model in conservative form Assume h 0 = constant; A 0 = constant; E = constant. Therefore K in (102) is constant. We also assume p ext = constant and R = 0. Then and thus A ρ xp in (98) is x p = K 2 A xa (104) A ρ xp = K 3ρ xa 3/2. (105) Then (98) can be written in conservation-law form as where t Q + x F(Q) = 0, (106) [ ] [ ] q1 A Q = q 2 Au [ ] f1 Au F(Q) = Au 2 + K 3ρ A3/2 f 2 (107) 51 / 56

52 A linear model A linearised version of (98) is obtained as follows: Consider a small perturbation a(x, t) of the equilibrium cross-sectional area A 0 and a small velocity perturbation v(x, t) of a stationary flow A(x, t) = A 0 + a(x, t), u(x, t) = 0 + v(x, t). (108) Assume a(x, t) is small compared to A 0, that v is small and that derivatives of the perturbations are also small and thus products of these small quantities can be neglected. Then by substituting A(x, t) and u(x, t) from equations (108) into equations (98) and neglecting second order terms we obtain a system of linearised blood flow equations where t a + A 0 x v = 0, t v + c2 0 A 0 x a = 0, (109) c 0 = K A 0 2ρ : wave velocity, constant. (110) 52 / 56

53 Tasks P0: Verify that the equations (109) in matrix form read t Q + M x Q = 0, (111) where the vector of unknowns Q(x, t) and the coefficient matrix M are respectively given by [ ] a Q =, M = 0 A 0 c v 2 0. (112) 0 A 0 P1: Justify the derivation of the linearised equations explaining each step. P2: Find the eigenvalues of the linear system. P3: Find the corresponding left eigenvectors with general scaling coefficients α 1, α / 56

54 P4: Find the corresponding right eigenvectors with general scaling coefficients β 1, β 2. P5: Find relations for the coefficients α i, β j so that the left and right eigenvectors are orthonormal. P6: Verify that R 1 = L, where L is the matrix of left eigenvectors and R is the matrix of right eigenvectors, suitably normalized. P7: Find the characteristic variables. P8: Solve analytically the general initial value problem PDEs: t Q + M x Q = 0, ICs: Q(x, 0) = Q (0) (x). Here, the initial condition Q (0) (x) at time t = 0 is an arbitrary function of x alone. (113) 54 / 56

55 P9: Consider the Riemann problem PDEs: t Q + M x Q = 0, ICs: Q(x, 0) = Q (0) (x) = { QL if x < 0, Q R if x > 0, (114) where Q L and Q R are any two constant vectors. Solve the Riemann problem exactly, verifying that the solution is actually given by a = 1 2 (a L + a R ) 1 A 0 (v R v L ), 2 c 0 v = 1 2 (v L + v R ) 1 c (115) 0 (a R a L ). 2 A 0 P10: Invent an example for the Riemann problem (114) and plot the exact solution at a given time ˆt of your choice. 55 / 56

56 Exercises for the Linear Hyperbolic Systems 56 / 56

57 Problem 1: linear system with source terms. Consider the abstract inhomogeneous linear system t Q + A x Q = S(Q), (116) where the vector of unknowns Q(x, t), the constant coefficient matrix A and the source term vector S(Q) are Q = [ q1 q 2 ], A = [ 0 a a 0 ], S = [ s1 1 Show that the eigenvalues of the system are s 2 ], a > 0. (117) λ 1 = a, λ 2 = a. (118) 2 Show that the corresponding right eigenvectors are 1 1 R 1 = α 1, R 2 = α 2. (119) / 56

58 3 Show that the corresponding left eigenvectors are L 1 = β 1 [1, 1], L 2 = β 2 [1, 1]. (120) 4 Show that imposing orthonormality of left and right eigenvectors leads to the condition α 1 β 1 = 1/2, α 2 β 2 = 1/2. (121) 5 Choose α 1 = α 2 = 1 and β 1 = β 2 = 1/2. Show that the inverse matrix of R formed by the columns of right eigenvectors is R 1 = (122) Note that L = R 1, where L is the matrix whose rows are the left eigenvectors of A. Problem 2: characteristic variables. 56 / 56

59 1 Show that the vector of characteristic variables is 1 c 1 2 (q 1 q 2 ) C = =. (123) 1 c 2 2 (q 1 + q 2 ) 2 Show that decoupled equations with source terms in characteristic variables are t c 1 a x c 1 = 1 2 (s 1 s 2 ), (124) t c 2 + a x c 2 = 1 2 (s 1 + s 2 ). See equation (24). 3 Show that if the original source term in (116) is chosen as S = 1 (γ 1 + γ 2 )q 1 + (γ 2 γ 1 )q 2, (125) 2 (γ 2 γ 1 )q 1 + (γ 1 + γ 2 )q 2 where γ 1 and γ 2 are any two arbitrary real numbers, then the 56 / 56

60 decoupled inhomogeneous system reads t c 1 a x c 1 = γ 1 c 1, t c 2 + a x c 2 = γ 2 c 2. (126) Problem 3: solutions with source terms. 1 Show that the exact solution of the IVP above for the inhomogeneous system (126) is c 1 (x, t) = c (0) 1 (x + at)eγ 1t, c 2 (x, t) = c (0) 2 (x at)eγ 2t, where c (0) 1 (x) and c(0) 2 (x) are the initial conditions for the characteristic variables. (127) 2 Find the final expression for the corresponding solution in terms of the original variables. 56 / 56

61 Problem 4: a computational problem. Choose a = 1, a spatial domain [ 5, 5] and initial conditions q 1 (x, 0) = αe βx2, q 2 (x, 0) = 0, with α = 1 and β = 8. 1 For γ 1 = γ 2 = 1 plot the exact solutions q 1 (x, T ) and q 2 (x, T ) at the output times T = 1, T = 2 and T = 3. 2 For γ 1 = 1 and γ 2 = 5 exact plot the solutions q 1 (x, T ) and q 2 (x, T ) at the output times T = 1, T = 2 and T = 3. 3 Solve the homogeneus problem numerically and plot the numerical (in symbols) and the exact (in full line) solutions q 1 (x, T ) and q 2 (x, T ) at the output times T = 1, T = 2 and T = 3. Problem 5: electrical transmission line. Consider an electrical transmission line. The problem is to determine the current I(x, t) and the potential E(x, t) as functions of space and time. These quantities are solutions of the following system of linear hyperbolic equations with source terms t I(x, t) + 1 L xe(x, t) = R L I, (128) t E(x, t) + 1 C xi(x, t) = G C E, 56 / 56

62 where C is the capacitance to ground per unit length, G is the conductance to ground per unit length, R is the resistance per unit length and L is the inductance per unit length. Further details on these equations are found in [?]. 1 Find eigenvalues, left and right eigenvectors. 2 Apply the orthonormality condition to determine the choice of the scaling parameters in the left and right eigenvectors above. 3 Find the characteristic variables. 4 Write the equations in characteristic variables, keeping the appropriate source terms. Problem 6:. Consider initial-value problem for a distorsionless line, which is required to satisfy the condition RC = LG. 1 Verify that the governing equations are t I(x, t) + 1 L xe(x, t) = βi, t E(x, t) + 1 C xi(x, t) = βe, (129) 56 / 56

63 with β = R L. (130) 2 Write system (129) in terms of characteristic variables. 3 Suppose that the initial distribution of the current I(x, 0) = I (0) (x) and the potential E(x, 0) = E (0) (x) are known functions of distance x. Using the canonical form of the equations (or equations in characteristic variables) show that the exact solution of the general initial-value problem is and I(x, t) = [ I (0) (x λ 1 t) + I (0) (x λ 2 t) ] e βt C L [E(0) (x λ 1 t) E (0) (x λ 2 t)]e βt E(x, t) = 1 2 [E(0) (x λ 1 t) + E (0) (x λ 2 t)]e βt 1 2 L C [I(0) (x λ 1 t) I (0) (x λ 2 t)]e βt. (131) (132) 56 / 56

64 Problem 7: the general Riemann problem. Consider the general Riemann problem for the homogeneous version (no source terms) of equations (128) in which the initial conditions are { I(x, 0) = I (0) IL if x < 0, (x) = I R if x > 0, E(x, 0) = E (0) (x) = { EL if x < 0, E R if x > 0. (133) 1 Display the structure of the solution of the Riemann problem through a figure in the x-t plane identifying the waves present and the unknown region of space-time. 2 Show that the solution in the star region is given as follows I = 1 2 (I L + I R ) 1 2 (E R E L )/B, E = 1 2 (E L + E R ) 1 2 (I R I L )B, (134) where B = L/C. 56 / 56

65 3 Consider a spatial domain [ 100, 100], with the following specific initial conditions { IL = 1 if x < 0, I(x, 0) = I R = 3/4 if x < 0, { (135) EL = 1/4 if x < 0, E(x, 0) = E R = 5/4 if x > 0. Assume C = G = R = L = 1, for simplicity. Show that the solution in the star region is I = and I = Plot the solution profiles for I(x, T ) and E(x, T ) at the output times T = 12.5 and T = / 56