Explicit Solutions to Mixed Type PDEs

Transcription

1 Explicit Solutions to Mixed Type PDEs A. Coons, G. Powell, L. Vo SUMaR 2010 Kansas State University Mentor: Dr. M. Korten July 27, 2010

2 Overview Background One Dimension Two Dimensions The Radial Case Future Work

3 Background The Problem Our problem describes a situation in which one region is governed by a diffusion equation and another region is governed by a conservation law. { (u(x, t) 1)+ u > 1 u t = ( ) div x F (u (x, t)) u < 1 We want to find an explicit solution and suitable boundary data.

4 Background An Application Pipes carrying WCOs through cold climates [1, 3]

5 Background An Application Pipes carrying WCOs through cold climates [1, 3] Wax deposits on the interior walls of the pipes due to cold temperatures

6 Background An Application Pipes carrying WCOs through cold climates [1, 3] Wax deposits on the interior walls of the pipes due to cold temperatures Need to model deposition of wax in order to determine cleaning schedules [2]

7 Our equation becomes { (u(x, t) 1)+xx u > 1 u t = f (u (x, t)) x u < 1 Note: x R 1

8 The Jump Condition A condition which holds at a discontinuity or phase change

9 The Jump Condition A condition which holds at a discontinuity or phase change

10 The Jump Condition Note: our problem can be rewritten as a divergence u t = [ ] x x (u 1) + + x f (u) div x,t [ (u 1)+x + f (u), u ] = 0

11 The Jump Condition Note: our problem can be rewritten as a divergence u t = [ ] x x (u 1) + + x f (u) div x,t [ (u 1)+x + f (u), u ] = 0 Allowing us to use the following standard definition of weak solution Ω + Ω [ φt u φ x [ (u 1)+x + f (u) ]] dxdt = 0 φ C 0 (R R > 0)

12 The Jump Condition We solve separately in the Ω + and Ω regions [ = φ ( u + 1 ) Ω + S + + x φf ( u +), u +] [v x, v t ] ds [ = φ ( u 1 ) Ω S + x φf ( u ), u ] [v x, v t ] ds

13 The Jump Condition We solve separately in the Ω + and Ω regions [ = φ ( u + 1 ) Ω + S + + x φf ( u +), u +] [v x, v t ] ds [ = φ ( u 1 ) Ω S + x φf ( u ), u ] [v x, v t ] ds and then sum them such that [ φ ( u + 1 ) + x + f ( u ), u + u ] [v x, v t ] ds = 0 S

14 The Jump Condition Since φ is arbitrarily chosen, we need [ ( u + 1 ) + x + f ( u ), u + u ] [v x, v t ] = 0

15 The Jump Condition Since φ is arbitrarily chosen, we need [ ( u + 1 ) + x + f ( u ), u + u ] [v x, v t ] = 0 Yay! The Jump Condition has been found!

16 The Jump Condition Since φ is arbitrarily chosen, we need [ ( u + 1 ) + x + f ( u ), u + u ] [v x, v t ] = 0 Yay! The Jump Condition has been found! Now, we must find the individual terms.

17 The Jump Condition Finding (u + 1) +x Use similarity solutions of the form ( ) x u(x, t) = v = v(z) t

18 The Jump Condition Finding (u + 1) +x Use similarity solutions of the form ( ) x u(x, t) = v t to solve the equation when u > 1 = v(z) where u t = α (u (x, t)) xx α (u (x, t)) = (u(x, t) 1) +

19 The Jump Condition Finding (u + 1) +x Use similarity solutions of the form ( ) x u(x, t) = v t to solve the equation when u > 1 = v(z) where u t = α (u (x, t)) xx α (u (x, t)) = (u(x, t) 1) + which gives α (v) (z) zα (v) (z) = 0

20 The Jump Condition Finding (u + 1) +x Now, we use a variant of the complementary error function m (x) = 2 π z1 m (x) = 2 π e kx2 m (x) = 4kx π e kx2 x e kt2 dt

21 The Jump Condition Finding (u + 1) +x Now, we use a variant of the complementary error function m (x) = 2 π z1 m (x) = 2 π e kx2 m (x) = 4kx π e kx2 Plugging into the previous ODE gives x e kt2 dt 0 = ze kz2 (4k 1) So, for m to be a solution, we let k = 1 4

22 The Jump Condition Finding (u + 1) +x Since we find that ( ) x m = α (u (x, t)) = (u (x, t) 1) t + (u (x, t) 1) + = 2 π z1 x t e t 2 4 dt (u (x, t) 1) +x = 2 ( ) 2 ( e 1 x 4 t 1 π t )

23 The Jump Condition Finding the normal vector The curve S is given by t = Ψ (x) = x 2 z 2 1 Normal Tangent so points along it are given by (x, Ψ (x))

24 The Jump Condition Finding the normal vector The curve S is given by t = Ψ (x) = x 2 z 2 1 Normal Tangent so points along it are given by (x, Ψ (x)) The tangent vector at that point is [ 1, Ψ (x) ]

25 The Jump Condition Finding the normal vector The curve S is given by t = Ψ (x) = x 2 z 2 1 Normal Tangent so points along it are given by (x, Ψ (x)) The tangent vector at that point is [ 1, Ψ (x) ] and due to orthogonality, the normal vector is n = [ Ψ (x), 1 ] [ ] 2x =, 1 z 2 1

26 The Jump Condition Substitution This leads to 4 e z x f ( u ) + 1 u = 0 z 1 π z 2 1

27 The Jump Condition Substitution This leads to 4 e z x f ( u ) + 1 u = 0 z 1 π Now, we need to find: f (u) and u to solve problem. We chose a linear f z 2 1 f (u) = 3u and f (u) = 3 This makes our conservation law into a transport equation.

28 The Jump Condition Solving for u Substituting in and solving for u gives u = 4 z 1 π e z 2 1 6x z

29 The Jump Condition Solving for u Substituting in and solving for u gives u = 4 z 1 π e z 2 1 6x z Due to the equations of the Characteristic Curves this needs to be the same as the initial u data

30 Finding Initial u Data, u I Finding Initial u Data, u I Take a point on the free boundary (x, x 2 ) z 2 1 Our slope will be 1 f (u) = 1 3

31 Finding Initial u Data, u I Finding Initial u Data, u I Take a point on the free boundary (x, x 2 ) z 2 1 Our slope will be 1 f (u) = 1 3 This gives a line t = 1 3 (y x) + x 2 z 2 1

32 Finding Initial u Data, u I Finding Initial u Data, u I Take a point on the free boundary (x, x 2 ) z 2 1 Our slope will be 1 f (u) = 1 3 This gives a line At t = 0, we can see that y (x) = x + 3x2 z 2 1 t = 1 3 (y x) + x 2 z 2 1

33 Finding Initial u Data, u I Thus, we can say that ( u = u I x + 3x 2 ) z 2 1

34 Finding Initial u Data, u I Thus, we can say that ( u = u I x + 3x 2 ) z 2 1 We rewrite this in terms of a single variable w w = x + 3x 2 z ± x = w z z 2 1

35 Solving for u I Due to Characteristic Curves both expressions for u must be equal. Setting equal and solving gives u I (w) = ± 2 1 z 1 π e z w z 2 1

36 Checking positivity Need 0 < u I < 1 at least in some interval (0, w 0 ) First, examine the numerator: We can see that 4e z2 1 4 > 0 z 1 π 1 4 e z < 1 z 1 π And we graphically find that u I is positive for z 1 > 1.39.

37 Checking Positivity Checking positivity Now we need to check the denominator in each case: Positive Case w z 2 1 3

38 Checking Positivity Checking positivity Now we need to check the denominator in each case: Positive Case Therefore, w z u I (w) = z 1 π e z w z z 1 π e z

39 Checking Positivity Negative Case Need to find an upper bound, so we say we need an h > 0 such that w z 2 1 h (2 h) w z1 2 1 > h 0

40 Checking Positivity Negative Case Need to find an upper bound, so we say we need an h > 0 such that w z 2 1 h (2 h) w z1 2 1 > h 0 We can see that it is possible to choose an h such that 1 4 e z < h and then u I (w) 1 4 z 1 π h 2 1 z 1 π e z 4 < 1

41 Checking Positivity Positive Case Negative Case w 0 = [(2 h)2 1]z

42 The Two Dimensional Case The Two Dimensional Case The Radial Case

43 The Two Dimensional Case The Two Dimensional Case The Radial Case WCOs flowing through a pipeline

44 The Two Dimensional Case The Two Dimensional Case The Radial Case WCOs flowing through a pipeline Can use a lot of the same methods as the one dimensional case if we think of modeling along the radius

45 The Two Dimensional Case Wax Deposition in a Pipeline wax

46 The Two Dimensional Case Rewriting the Problem Recall the original problem in terms of ( x, t) { (u( x, t) 1)+ u > 1 u t = ( ) div x F (u ( x, t)) u < 1

47 The Two Dimensional Case Rewriting the Problem Recall the original problem in terms of ( x, t) { (u( x, t) 1)+ u > 1 u t = ( ) div x F (u ( x, t)) u < 1 We rewrite this in radial coordinates in terms of (r, t) { 1 ( ) u t = r r r (u(x, t) 1)+r u > 1 3u r u < 1

48 The Two Dimensional Case Jump Condition The Jump Condition We use the jump condition found previously [ (u + 1) +r + f (u ) ] n r + (u + u )n t = 0

49 The Two Dimensional Case Jump Condition The Jump Condition We use the jump condition found previously [ (u + 1) +r + f (u ) ] n r + (u + u )n t = 0 Similarly to the one dimensional case, we now find each term

50 The Two Dimensional Case Jump Condition Finding (u + 1) +r Use similarity solutions of the form ( ) r u(x, t) = v t

51 The Two Dimensional Case Jump Condition Finding (u + 1) +r Use similarity solutions of the form ( ) r u(x, t) = v t Following the same process as before, we obtain ( ) r (u + e 1) + = v = c t r t s 2 4 s ds (u + 1) +r = v r = ce r 2 4t r

52 The Two Dimensional Case Jump Condition Finding the normal vector Our free boundary is given by (R r)2 t = z0 2 This can be written as wax φ(r, t) = (R r)2 z 2 0 t = 0 normal

53 The Two Dimensional Case Jump Condition Finding the normal vector Our free boundary is given by (R r)2 t = z0 2 This can be written as wax φ(r, t) = (R r)2 z 2 0 t = 0 normal And we find [ ] 2(R r) 2 n = φ(r, t) =, 1 z 2 0

54 The Two Dimensional Case Jump Condition Substituting In r 2 ce 4t r ( 3u 2 z 2 0 ) (R r) + [ 1 u ] ( 1) = 0

55 The Two Dimensional Case Jump Condition Substituting In r 2 ce 4t r ( 3u 2 z 2 0 ) (R r) + [ 1 u ] ( 1) = 0 We rewrite all in terms of t, knowing from our free boundary that r = R z 0 t

56 The Two Dimensional Case Jump Condition Substituting In r 2 ce 4t r ( 3u 2 z 2 0 ) (R r) + [ 1 u ] ( 1) = 0 We rewrite all in terms of t, knowing from our free boundary that r = R z 0 t and solving for u u = 1 2 cz 0 texp ( ) (R z 0 t) 2 4t z 2 0 (R z 0 t) z 0 t

57 The Two Dimensional Case Finding Initial u Data Finding initial u data Points on the free boundary can be given as ) (r, t) = (R z 0 t, t wax

58 The Two Dimensional Case Finding Initial u Data Finding initial u data Points on the free boundary can be given as ) (r, t) = (R z 0 t, t wax We write our transport equation as u r u t = 0 so g(u) = 1 3 u and we know that the slope is 1 g (u) = 3

59 The Two Dimensional Case Finding Initial u Data Finding initial u data Points on the free boundary can be given as ) (r, t) = (R z 0 t, t wax We write our transport equation as u r u t = 0 so g(u) = 1 3 u and we know that the slope is 1 g (u) = 3 yielding a line T (t) = t + z 0 3 t

60 The Two Dimensional Case Finding Initial u Data We say ( u = U R t + z 0 ) t 3 and rewrite in terms of a single variable w w = t + z 0 t 3 t = z 0 3 ± z w 2

61 The Two Dimensional Case Finding Initial u Data We say ( u = U R t + z 0 ) t 3 and rewrite in terms of a single variable w w = t + z 0 t 3 t = z 0 3 ± z w Note, we will only use the (+) in order to have t > 0 2

62 The Two Dimensional Case Setting both expressions for u equal and writing in terms of w

63 The Two Dimensional Case Setting both expressions for u equal and writing in terms of w U R (w) = 1 ( ) z cz z 0 2 R z 0 2 z z w 9 +4w exp z z w ( )] z0 [R 2 z 02 z z w z 0 ( z z w )

64 Future Work Future Work A more realistic model

65 Future Work Future Work A more realistic model Nonuniform wax deposition

66 Future Work Future Work A more realistic model Nonuniform wax deposition Gravity Convection

67 Future Work Future Work A more realistic model Nonuniform wax deposition Gravity Convection Different free boundaries

68 Future Work Future Work A more realistic model Nonuniform wax deposition Gravity Convection Different free boundaries Nonlinear f (u)

69 Future Work Thank You Dr. Marianne Korten Kansas State University Mathematics Department Bryan Bischof SUROP NSF grant DMS

70 Future Work References [1] Correra, S.; Fasano, A.; Fusi, L.; Merino-Garcia,D. Calculating deposit formation in the pipelining of waxy crude oils. (English summary) Meccanica 42 (2007), no. 2, [2] Correra, S.; Fasano, A.; Fusi, L.; Primicerio, M.; Rosso, F. Wax diffusivity under given thermal gradient: a mathematical model. (English summary) ZAMMZ. Angew. Math. Mech. 87 (2007), no. 1, [3] Fasano, A.; Primicerio, M. Heat and mass transport in non-isothermal partially saturated oil-wax solutions. (English summary) New trends in mathematical physics, 34 44, World Sci. Publ., Hackensack, NJ, [4] Bouillet, J.E. Signed solutions to diffusion-heat conduction equations. Free boundary problems: theory and applications, Vol. II (Irsee, 1987), , Longman Sci. Tech., Harlow, [5] Smoller, J. Shock waves and reaction-diffusion equations. Second edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 258. Springer-Verlag, New York, 1994.