TRAVELING WAVE SOLUTIONS OF THE POROUS MEDIUM EQUATION. Laxmi P. Paudel. Dissertation Prepared for the Degree of DOCTOR OF PHILOSOPHY

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1 TRAVELING WAVE SOLUTIONS OF THE POROUS MEDIUM EQUATION Laxi P Paudel Dissertation Prepared for the Degree of DOCTOR OF PHILOSOPHY UNIVERSITY OF NORTH TEXAS May 013 APPROVED: Joseph Iaia, Major Professor Jianguo Liu, Coittee Meber Pieter Allaart, Coittee Meber Su Gao, Chair of the Departent of Matheatics Mark Wardell, Dean of the Toulouse Graduate School

2 Paudel, Laxi P Traveling Wave Solutions of the Porous Mediu Equation Doctor of Philosophy Matheatics, May 013, 5 pp, bibliography, 15 titles We prove the existence of a one-paraeter faily of solutions of the porous ediu equation, a nonlinear heat equation In our work, with space diension 3, the interface is a half line whose end point advances at constant speed We prove, by using axiu principle, that the solutions are stable under a suitable class of perturbations We discuss the relevance of our solutions, when restricted to two diensions, to gravity driven flows of thin fils Here we extend the results of J Iaia and S Betelu in the paper Solutions of the porous ediu equation with degenerate interfaces to a higher diension

3 Copyright 013 by Laxi P Paudel ii

4 ACKNOWLEDGMENTS With highest regard, I would like to thank Dr Joseph Iaia, y ajor professor, for his guidance and unwavering support during this project This work is as a result of the tie, patience and scholarly support he provided to e I would also like to offer y sincere thanks to the coittee ebers Dr Pieter Allaart and Dr Jianguo Liu for their tie and valuable coents while reading the final anuscript I a grateful to y parents and the entire faily ebers for their sacrifices and encourageent for the copletion of this degree My wife Kalpana, daughter Eva, and son Arnov were all a constant source of inspiration Finally, I would like to thank the Departent of Matheatics at the University of North Texas for providing e the opportunity to work and study here iii

5 CONTENTS ACKNOWLEDGMENTS iii CHAPTER 1 INTRODUCTION 1 11 Discussion of the Proble 1 1 Proble in Space Diension n = 13 Proble in Space Diension n = Soe Properties of the Porous Mediu Equation Relevant to Our Work 4 CHAPTER DERIVATION OF THE DIFFERENTIAL EQUATIONS 6 1 Differential Equation in Ters of the Pressure v 6 Differential Equation in Ters of F 8 CHAPTER 3 SOLUTIONS OF THE DIFFERENTIAL EQUATION Existence of the Solution F 15 3 Preliinary Observation 17 CHAPTER 4 THE CASE 0 < c = a 1 CHAPTER 5 THE CASE 0 < c < a 3 CHAPTER 6 THE CASE c > a > 0 34 CHAPTER 7 FINAL COMMENTS 50 BIBLIOGRAPHY 5 iv

6 CHAPTER 1 INTRODUCTION 11 Discussion of the Proble Consider the porous ediu equationpme, 1 u t = u, > 1, R where, u = ux, t > 0, x R n, n is the space diension and t R the tie This parabolic equation has been extensively studied in the literature due to its iportant physical applications The function u = ux, t odels the density of an ideal gas flowing in a hoogeneous porous ediu [1], [3] n 3, the diffusion of strong theral waves [] n 3, and the spreading of viscous gravity currents [6] with n = When n = and = 4 the porous ediu equation odels the spreading of a fil of fluid on a horizontal surface under the action of gravity In particular, if the free surface of the fluid is described by the graph z = hx, y, t, then it has been shown using lubrication theory [6] that h satisfies h t = g 3ν h 3 h = g 1ν h4, where g is the acceleration due to gravity and ν the kineatic viscosity of the fluid After a change of scale Eq becoes Eq 1 with = 4 To derive this equation, we neglect capillary, olecular and inertial forces and we assue the flow has a lateral extension which is uch larger than the typical thickness Several experiental studies see for instance [8, 5] and references therein show that this equation is a good approxiation provided that the lateral extension of the spreading is uch larger Contents in section 11 and section 14 of this chapter have been reproduced, either in part or in full, fro [7] J Iaia and S Betelu, European Journal of Applied Matheatics [doi:101017/s ] with perission fro Cabridge University Press 1

7 than the capillary length γ/ρg, where γ is the surface tension, and provided that the fluid wets the surface ie the contact angle with the substrate at equilibriu is zero In order to get a sipler, quadratic equation easier for analysis, we write the PME in ters of the pressure v where Equation 1 then becoes v = 1 u 1 3 v t = 1v v + v 1 Proble in Space Diension n = In the paper [7], it has been shown that in n = diensions, there are traveling wave solutions of 3 in the for 4 vx, y, t = rx, y, tf θx, y, t, where and r = x ct + y, c > 0, y θ = tan 1, x ct x ct = r cos θ, y = r sin θ Then the function F θ satisfies 5 F cf sin θ + F + cf cos θ + 1F F + F = 0 In that paper it has also been shown that if c > 0 and F π = a > 0, F π = 0, then there is a positive solution of F cf sin θ + F + cf cos θ + 1F F + F = 0 on 0, π

8 such that F 0 = 0, F 0 = 0 if and only if c > a In addition, when c > a then F is π-periodic and there are solutions of 3 in the for 4 13 Proble in Space Diension n = 3 In this work, we attept to prove a siilar theore in n = 3 diensions In particular, we attept to find traveling wave solutions of 3 in n = 3 diensions We assue 6 vx, y, z, t = rx, y, z, tf φx, y, z, t, where 7 r = x + y + z ct, c > 0, 8 θ = tan 1 y x 9 φ = tan 1 x + y, z ct and: We will then show that F φ satisfies, x = r cos θ sin φ, y = r sin θ sin φ, z ct = r cos φ F cf sin φ + F + cf cos φ + 1F F + cos φ sin φ F + F = 0 Proble in section 13 and its solution which is virtually the ost part of y dissertation have been reproduced in elaborated for fro y published article [9] L Paudel and J Iaia, Nonlinear Analysis01, Elsevier, doi:101016/jna that I co-authored with y advisor J Iaia 3

9 Note that this equation has ore ters than 5 and therefore requires a separate analysis than that contained in [7] In this paper we will prove the following theore: Main Theore: Let c > 0 Consider the differential equation 10 F cf sin φ + F + cf cos φ + 1F with F + cos φ sin φ F + F = 0 11 F π = a > 0, F π = 0 There is a positive solution of with F 0 = 0, F 0 = 0 if and only if c > a In addition, when c > a then F is π-periodic and there are solutions of 3 in the for of 6 If c = a then F φ = a cosφ is an explicit solution of 10-11, and so there is not a positive solutions of 3 with F 0 = 0 and F 0 = 0 If c < a and F is a solution of the equations 10-11, then F has a zero, z, with π < z < π and li φ z + F φ φ z 1 > 0 14 Soe Properties of the Porous Mediu Equation Relevant to Our Work Nuerous interesting properties of the solutions of the PME have been studied, and they are nicely suarized in [11] Soe of those properties relevant to our work are suarized as follows: If the initial condition ux, 0 is non-negative, then the solution reains non-negative for all ties 4

10 The boundary of the set S = {x : ux, t = 0} is called the interface In the PME with > 1 the interface can be static or ove forward in the direction where u = 0, but never backwards The function u is sooth in the regions where u > 0 and t > 0 At the interface, where u = 0, u is continuous but its derivative ay be discontinuous The solutions satisfy the axiu principle for parabolic equations: if u 1 x, 0 u x, 0 then u 1 x, t u x, t for t > 0 In this paper, the interface is a half line that terinates abruptly, and since only the end point oves, we call this interface degenerate As in all the applications of the porous ediu equation, in this paper too, the propagating half line can be interpreted as a region epty of fluid 5

11 CHAPTER DERIVATION OF THE DIFFERENTIAL EQUATIONS 1 Differential Equation in Ters of the Pressure v We assue 1 u = c v k, where, c > 0, u > 0, v > 0, k R Then 13 u t = ck v k 1 v t Since differentiating with respect to x, we get u = c v k, 14 u x = c v k x = c k v k 1 v x Again, differentiating with respect to x, we get 15 u xx = c k v k 1 v x x = c k k 1v k v x + v k 1 v xx Proceeding siilary, 16 u yy = c k v k 1 v y y = c k k 1v k v y + v k 1 v yy 17 u zz = c k v k 1 v z z = c k k 1v k v z + v k 1 v zz Adding 15, 16 and 17, we get 18 u = c k k 1v k v + v k 1 v 6

12 Using 13 and 18 in 1, we get Thus Hence ck v k 1 v t = c k k 1v k v + v k 1 v c v k 1 v t = c k 1v k v + v k 1 v v t = c 1 k 1v k k+1 v + v k 1 k+1 v = c 1 k 1v k k 1 v + v k k v Choosing k so that k k 1 = 0 ie k = 1 1 then we get v v t = c v v = c 1 1 1v v + v Choosing c so that then 1 c 1 = 1, 19 v t = 1 v v + v, where ie and c 1 = 1, c =, u = c v k 7

13 1 = 1 1 v 1 1, equivalently, v = 1 u 1 Differential Equation in Ters of F Using spherical coordinates, for x = x, y, z R 3 x = r cos θ sin φ, y = r sin θ sin φ, z ct = r cos φ, we have 0 r = x + y + z ct, tan θ = y x, 1 θ = tan 1 y x, tan φ = x + y, z ct φ = tan 1 x + y z ct For a function ur, θ, φ, fro [10] by Strauss, 3 u = u rr + r u r + 1 u r φφ + cos φ sin φ u φ + 1 sin φ u θθ Hence, 4 φ = cos φ r sin φ, Let 5 vx, y, z, t = rx, y, z, t F φx, y, z, t Then 6 v t = r t F φ + r F φ φ t, 8

14 7 v x = r x F φ + r F φ φ x, 8 v y = r y F φ + r F φ φ y, 9 v z = r z F φ + r F φ φ z, Differentiating 7 with respect to x, we get v xx = r xx F + r x F φ x + r x F φ x + rf φ x + rf φ xx Hence 30 v xx = r xx F + r x F φ x + rf φ x + rf φ xx Siilarly, differentiating 8 with respect to y and 9 with respect to z, we get 31 v yy = r yy F + r x F φ y + rf φ y + rf φ yy 3 v zz = r zz F + r z F φ z + rf φ z + rf φ zz We have 33 r = x + y + z ct Differentiating 33 with respect to t, we get rr t = z ct c, 34 rr t = cz ct, 35 r t = cz ct r = c cos φ Differentiating 33 with respect to x, we get 36 rr x = x, 37 r x = x r 9

15 Differentiating 33 with respect to y, we get 38 rr y = y, 39 r y = y r Differentiating 33 with respect to z, we get 40 rr z = z ct, 41 r z = z ct r Differentiating 36 with respect to x, we get 4 r x + rr xx = 1 Differentiating 38 with respect to y, we get 43 r y + rr yy = 1 Differentiating 40 with respect to z, we get 44 r z + rr zz = 1 Squaring and adding the equations 36, 38 and 40, we get r r x + r y + r z = x + y + z ct = r, So ie r x + r y + r z = 1, 45 r = 1 Adding 4, 43 and 44, we get r + r r = 3 10

16 Using 45, we get r r =, and 46 r = r We have 47 φ = tan 1 x + y z ct Differentiating 47 with respect to t, we get 1 x + y φ t = x +y z ct c z ct Thus = c x + y x + y + z ct = c x + y r 48 φ t = c x + y r = c sin φ r Differentiating 47 with respect to x, we get Hence φ x = = = x +y z ct x x + y z ct xz ct z ct + x + y x + y xz ct r x + y 49 φ x = Siilarly, 50 φ y = xz ct r x + y yz ct r x + y 11

17 Differentiating 47 with respect to z, we get 1 x + y φ z = x +y z ct z ct Hence = x + y x + y + z ct = x + y r 51 φ z = x + y r Then So, φ = φ x + φ y + φ z = x z ct r 4 x + y + y z ct r 4 x + y + x + y r 4 = z ct + x + y r 4 r 4 = 1 r 5 φ = 1 r Since r φ = r x φ x + r y φ y + r z φ z Using equations 36, 38, 40,49,50 and 51, we have r φ = x r xz ct r x + y + y r = x + y z ct r 3 x + y = z ct x + y r 3 = 0 yz ct z ct r + x + y r z ct x + y r 3 z ct x + y r 3 1 x + y r 1

18 So 53 r φ = 0 v = vx + vy + vz = rx + ry + rzf + rr x φ x + r y φ y + r z φ z F F + φ x + φ y + φ zr F = r F + r r φf F + φ r F = F + F Hence 54 v = F + F Hence v = v xx + v yy + v zz = rf + r φf + φ rf + φrf = r F + 1 r F + cos φ r sin φ F 55 v = r F + 1 r F + cos φ r sin φ F Now, fro the equations 5 and 55, we have v v = rf r F + 1 r F + cos φ r sin φ F Thus = F + F F + cos φ sin φ F F 56 v v = F + F F + cos φ sin φ F F Fro equation 6, we have v t = r t F φ + r F φ φ t 13

19 Using 35 and 48, we get v t = c cos φf + rf c sin φ r = c cos φ F + c sin φ F Hence 57 v t = c cos φ F + c sin φ F Using equations 54, 56 and 57 in v t = 1 v v + v we get c cos φ F + c sin φ F = 1 F + F F + cos φ sin φ F F + F + F Finally, 58 F c sin φ F + F + c cos φ F + 1F F + cos φ sin φ F + F = 0 14

20 CHAPTER 3 SOLUTIONS OF THE DIFFERENTIAL EQUATION We consider the differential equation 59 F c sin φ F + F + c cos φ F + 1F F + cos φ sin φ F + F = 0 By restricting ourselves to syetric solutions about the z axis, we assign the boundary conditions as 60 F π = a, 61 F π = 0 31 Existence of the Solution F Let H = 1 F 1, then the differential equation 59 is reduced to 6 H sin φ + H cos φ c 1 1 c + 1 and satisfies 1 63 Hπ = a 1 64 H π = 0 Rewriting the equation 6, 1 1 H 1 H sin 1 φ + 1 H sin φ 1 1 H cos φ sin φ = 0 1 H sin φ c H 1 sin 1 φ + 1 H sin φ 15

21 c + 1 and integrating over φ to π, we get H sin φ + c H cos φ sin φ = H sin 1 π φ + H sin θ dθ 1 φ [ 1 ] 1 c 1 π + + H 1 sin θ cos θ dθ = φ Dividing by sin φ and integrating once again over φ to π, we get Hπ Hφ = π φ c H 1 sin θ dθ+ π φ 1 π H sin θ dθ dt 1 sin t t [ π 1 1 c 1 ] π + + H 1 sin θ cos θ dθ dt φ 1 sin t 1 t So, 65 Hφ = Hπ π φ P H 1 sin θ dθ π φ Q π π R π H sin θ dθ dt H 1 sin θ cos θ dθ dt sin t t φ sin t t where, P = c 1 1 1, Q = 1, and 1 [ R = c ] 1 1 Fro 65, it can be shown that the ap T defined by T H = right hand side of 65 is a contraction apping, provided φ is sufficiently close to π Hence by the contraction apping principle, T has a fixed point H such that T H = H for φ sufficiently close to π Hence the solution H of equations 6-64 exists in the neighborhood of φ = π This proves the existence of the solution F of in the neighborhood of φ = π 16

22 3 Preliinary Observation Let us suppose 66 W = F c sin φ F + F + c cos φ F Then W = F F c sin φ F c cos φ F + F F + c cos φ F c sin φf = F F + F c sin φf + F = F c sin φf + F Hence 67 W = F c sin φf + F Fro equations 59 and 66, we have 68 W + 1F F + F + cos φ F + sin φ F = 0 sin φ So Then F W + F = 1F cos φ sin φ F F W = F c sin φ W 1F cos φ sin φ F F 69 W + F c sin φ 1F W = F c sin φ cos φ sin φ F F Note that by using 60 and 61 in 66, we get 70 W π = a ac Taking the liit as φ π in equation 59, we get a ac + 1 a F cos φ π + li φ π sin φ F φ + F π = 0 17

23 Using L Hospital s rule to calculate, we get a ac + 1 a F π + F π = 0 And thus a c + 1 F π + F π = 0 71 F π + F π = c a 1 Hence, 7 F π + F π < 0 for a > c, 73 F π + F π = 0 for a = c, and 74 F π + F π > 0 for a < c If we let 75 D = cos φ F + sin φ F, Then 76 D = F + F cos φ Fro 8 W + 1F F + F + cos φ F + sin φ F = 0 sin φ Using 75 and 76, we get D W + 1F cos φ + D = 0 sin φ Therefore 77 W + 1 F cos φ D + 1 F sin φ D = 0 18

24 Fro equation 67 W = F c sin φf + F Using 76, we get 78 W = F c sin φ D cos φ Lea 31 The function F is syetric about the line φ = π, ie F π + φ = F π φ Proof Let F π + φ = G 1 φ and F π φ = G φ, then G 1φ = F π + φ, G φ = F π φ, G 1φ = F π + φ, G φ = F π φ So 79 G 1 0 = a = G 0 and G 10 = 0 = G 0 Replacing φ by π + φ in equation 59, we get G 1 c sinπ + φ G 1 + G 1 + c cosπ + φ G 1 + 1G 1 G 1 + cosπ + φ sinπ + φ G 1 + G 1 = 0 Thus 80 G 1 + c sin φ G 1 + G 1 c cos φ G 1 + 1G 1 G 1 + cos φ sin φ G 1 + G 1 = 0 Replacing φ by π φ in equation 59, we get G + c sinπ φ G + G + c cosπ φ G + 1G G cosπ φ sinπ φ G + G = 0 Thus 81 G + c sin φ G + G c cos φ G + 1G G + cos φ sin φ G + G = 0 Fro equations 80 and 81, we see that G 1 and G satisfy the sae differential equation 19

25 Fro 79, they also satisfy the sae initial conditions Hence by the uniqueness theore [4], G 1 φ = G φ So F π + φ = F π φ This copletes the proof of lea 31 0

26 CHAPTER 4 THE CASE 0 < c = a Lea 41 For c = a, F = a cos φ is a solution for the syste of equations Proof Let Then equation 59 is reduced to W = F c sin φ F + F + c cos φ F W + 1F F + cos φ sin φ F + F = 0 8 W 1F + F + cos φ sin φ F + F = 0 Consider the equation 83 F + cos φ sin φ F + F = 0 Note that for Using in equation 83, we get cos φ + cos φ sin φ F = cos φ, F = sin φ, F = cos φ, sin φ + cos φ = 0 Then, F = cos φ is a solution for equation 83 Besides, equation 83 is a linear hoogeneous differential equation So, F = k cos φ is a solution of equation 83 for any k Also, consider the equation 84 W = F c sin φ F + F + c cos φ F = 0 1

27 Let F = a cos φ, then F = a sin φ, F = a cos φ Since c = a, W = a sin φ c sin φ a sin φ + a cos φ ac cos φ = a ac sin φ + a ac cos φ = a ac = 0 Cobining this with the fact that F = a cos φ satisfies equation 83, we see that F = a cos φ satisfies the equation 8 Hence, whenever c = a, F = a cos φ is a solution for the syste of equations This copletes the proof of lea 41

28 CHAPTER 5 THE CASE 0 < c < a Lea 51 Let 0 < c < a Then F has a zero on π, π Proof By 60 we have F π = a > 0 Contrary to the lea let us assue that F > 0 on π, π Then fro and we have π π Dπ = 0, D = F 0, Using this and 71 we see D π = [F π + F π] cos π = 1[F π + F π] D π > 0 Therefore D is negative to the iediate left of π and D π 0 so it follows that D has a negative local iniu on π, π So there is a p π, π such that D p = 0, D > 0 on p, π] and Dp < 0 Then it follows that 85 Dφ < 0 on [p, π In addition, it follows fro calculus that 86 D p 0 Since D p = 0 and p π, π, then fro 78 it follows that W p = 0 Next, we differentiate 77 and see that we obtain W F F F F D + D + D + D = 0 cos φ cos φ sin φ sin φ Evaluating at φ = p where W p = D p = 0, this reduces to F p F p sin p F p cos p cos p D p + sin Dp = 0 p 3

29 Since p π, π, then cos p < 0 and along with 86, we obtain F p sin p F p cos p sin Dp 0 p Since Dp < 0, we then have F p sin p F p cos p 0 Thus since p π, π we have F p sin p F p cos p < 0 Hence 88 F p < 0 Also fro and 71 we see that F has a local axiu at π Then for soe ɛ > 0 89 F > 0 on π ɛ, π Fro 88 we see that F is decreasing near p and fro 89 we see that F is increasing to the iediate left of π Therefore there exists q p, π where F has a local iniu and thus F q = 0 Since q p, π, using 75 and 85 we see that 0 > Dq = F q cos q + F q sin q = F q sin q which contradicts the assuption F > 0 This copletes the proof of lea 51 Due to the preceding lea, we will now assue that there is a z with π < z < π such that F z = 0 and 90 F > 0 on z, π] Lea 5 Let 0 < c < a Then D > 0 on z, π] Proof We assue by the way of contradiction that there exists w z, π] π, π] such that D w 0 Now recall fro the beginning of the proof of lea 51 that D π > 0 4

30 and Dπ = 0 Hence there exists soe p [w, π such that D p = 0 and D > 0 on p, π] Then 91 D p 0 and Dp < 0 Also, fro 78 we have W p = 0 Using these facts in 87, we obtain F p F p sin p F p cos p 9 cos p D p + sin Dp = 0 p Since p z, π π, π, cos p < 0 and F p > 0, and thus fro 91 we obtain F p sin p F p cos p Dp 0 Since Dp < 0 by 91 we then have F p sin p F p cos p 0 and so F p sin p F p cos p < 0 Thus: F p < 0 So F is decreasing near p and since F has a local axiu at π, F is increasing to the iediate left of π Thus, there exists q p, π such that F q = 0 Then fro 75 we have 93 Dq = F q sin q > 0 Now fro 91 and the coents iediately preceding this lea, Dp < 0, Dπ = 0, and D > 0 on p, π] Thus Dq < 0 which is a contradiction to 93 Thus D > 0 on z, π] This copletes the proof of lea 5 5

31 Note: It follows fro lea 5 that D = F +F cos φ > 0 on z, π Thus F +F < 0 on z, π Since F > 0 on z, π then it follows that F < 0 on z, π] Thus, F is strictly decreasing on z, π Since F π = 0 this iplies F > 0 on z, π Also, since F is strictly decreasing on z, π, we see that li F φ exists but ay be infinite Thus li F φ = A φ z + φ z + where 0 < A Lea 53 Let 0 < c < a Then li φ z + F = Proof We assue by the way of contradiction that li φ z + F φ = A, where 0 < A < Now let E = 1 F + F Fro the note after lea 5 we have F + F < 0 and F > 0 on z, π Thus we see that 94 E = 1 F + F = F F + F < 0 on z, π Therefore E is decreasing on z, π and thus: a A = Eπ li Eφ = φ z + Therefore a A, and since 0 < c < a, we see that 95 0 < c < a A Then using 95 and the fact that sinz 1, we see that 96 li W φ = li φ z + φ z +[F cf sin φ+f +cf cos φ] = A ca sin z = AA c sin z > 0 6

32 Next, dividing 69 by W and taking the liit as φ z +, we see using z π, π and 95 that [ W 97 li φ z + W + F 1 F Next we let c 1 ] sin φ = cos z F sin z 98 G = W F 1 c exp 1 Taking logs and differentiating we obtain Hence fro 97, G 99 li φ z + G = cos z sin z π φ A c sin z > 0 A c sin z sin t F t dt G G = ln G = W W + F 1 F c sin φ 1 F A c sin z B > 0 A c sin z Fro 90 and 96 observe that G > 0 to the iediate right of z So, fro 99, G > 0 near φ = z Thus, G is increasing near φ = z and therefore 100 li φ z + G = L 0 Now fro 99 we have G G B on z, z + ɛ Thus integrating on φ, φ 1 z, z + ɛ we obtain Gφ 1 e Bφ φ 1 Gφ and therefore Thus 0 < Gφ 1 e Bz φ1 li Gφ = L φ z li φ z + G = L > 0 7

33 We already know fro 96 that li W φ = AA c sin z > 0 So, fro 98 and 101 φ z + we see [ li F φ z + Taking natural logs iplies 1 c φ exp 1 π φ ] sin t F t dt [ π ] sin t li ln F + c φ z + φ F t dt is positive and finite exists and is finite Rewriting, we see that 10 li φ z + ln F [1 + c π φ sin t dt F t ln F ] exists and is finite Since li φ z + F φ = A it follows fro L Hopital s rule that li φ z + F φ φ z = li φ z + F φ = A Using this and the fact that π < z < π we see that the integral in 10 becoes unbounded as φ z + Hence applying L Hopital s rule we see that c 103 li φ z + π φ sin t dt F t ln F = li φ z + c sin φ F F F = c sin z A Now, fro 95 we have A a > c > c sin z and so π sin t c φ li dt F t = 1 c sin z φ z + ln F A > 0 However, since 105 li ln F = φ z + we see that 104 and 105 contradict 10 Therefore it ust be the case that A = This copletes the proof of lea 53 Lea 54 Let 0 < c < a Then F F li = 1 φ z + F 1 8

34 Proof Dividing 59 by F gives c sin φ 1F cf cos φ 1F F 1F cos φ F F F F F sin φ Since F z = li F φ = 0 and li F φ = by lea 53, it follows then that φ z + φ z + = li φ z + c sin φ F = li φ z + 1F F = li φ z + cf cos φ F = 1F cos φ F sin φ = 0 Taking the liit as φ z + in 106 and using 107 we see that 1 + li φ z + 1F F F = 0 Thus, This copletes the proof of lea 54 F F li = 1 φ z + F 1 Lea 55 Let 0 < c < a Then φ zf li φ z + F = 1 Proof Fro lea 54, it follows that given ɛ > 0 there exists soe δ > 0 such that 1 1 ɛ < F F < 1 + ɛ for z < φ < z + δ F 1 Since F F F = 1 F, F we ay rewrite the above as 1 ɛ < F F < + ɛ for z < φ < z + δ 1 Integrating on φ 1, φ gives 1 ɛ φ φ 1 < F F F φ 1 F φ 1 < 1 + ɛ φ φ 1 for z < φ 1 < φ < z + δ Using the facts li F φ 1 = F z = 0 and li F φ = by lea 53, we see when φ 1 z + φ z + taking the liit as φ 1 z + that 1 ɛ φ z F F 1 + ɛ φ z for z < φ < z + δ 9

35 Thus Therefore Thus 1 ɛ F + ɛ for z < φ < z + δ φ zf 1 li φ z + F φ zf = 1 φ zf li φ z + F = 1 This copletes the proof of lea 55 Lea 56 Let 0 < c < a Then li F F 1 1 φ z + exists and is positive Proof Dividing 69 by W and then dividing the nuerator and denoinator of the right hand side of this expression by F and rewriting we see that W W + F 1 F c sin φ 1 F Then using lea 5, we have G li φ z + G = li φ z +[W W + F 1 F = c 1 c sin φ F 1 c sinφ F + F cos φ sin φ + F F F sin φ F ] = cosz sinz Therefore G is decreasing near z and so for soe L with 0 < L cf cosφ F C < 0 li G = L φ z + Then integrating on φ, φ 1 z, z + ɛ we have Thus ln Gφ1 φ1 G φ1 = Gφ φ G dt > C dt = Cφ 1 φ φ L = li φ z + Gφ Gφ 1e Cφ 1 z 30

36 Thus the liit of G is a positive, finite nuber Next, choosing 0 < ɛ < 1 then by lea 55 we have for soe δ > 0 that: φ zf F < ɛ for z < φ < z + δ Dividing by φ z, and integrating on φ, z + δ where z < φ < z + δ gives F z + δ ln 1 1 δ F φ + ɛ ln φ z Therefore Thus, we have F z + δ F 1 F where Cδ = δ1 1 +ɛ F z+δ Since 0 < ɛ < 1 δ 1 1 +ɛ for z < φ < z + δ φ z 1 1 +ɛ Cδ for z < φ < z + δ, φ z 1 1 +ɛ we see that Hence 1 F φ z 1 F t dt φ z Cδ Cδ 1 dt = t z 1 1 +ɛ 1 φ z ɛ ɛ < on z, z + δ is integrable on z, z + δ for soe δ > 0 Also, on [z + δ, π] we know that F is bounded fro below by a strictly positive constant and therefore we see that 1 F on z, π] Also, since π φ sin t F t dt = sin φ F φ c π 108 li exp φ z + 1 φ sin t F t dt < 0 on z, π we see that exists and is finite and positive is integrable Since a > c > 0 we see fro 108 and 98 that the liit of G is positive and finite so that li W φf 1 φ exists and is positive φ z + Rewriting this using 66 we see therefore that there exists T > 0 such that [ 109 li F F 1 1 c sin φ + F ] cf cos φ + = T > 0 φ z + F F F Again, since F z = li F = 0 and li F = by lea 53 it follows fro 107 and φ z + φ z that 110 li F F 1 = T > 0 φ z + 31

37 Fro the note between lea 5 and lea 53 we know that F > 0 on z, π and therefore it follows fro 110 that 111 li F F 1 1 = T > 0 φ z + This copletes the proof of the lea 56 Lea 57 Let 0 < c < a Then li φ z + F φ z 1 = 1 T 1 Proof Letting, then we see that and so by 111 K = Thus fro L Hopital s rule we see that 1 F li φ z + φ z = li K = F 1, 1 F 1 1 F, li K = T φ z + 1 φ z + K φ z = li K = T φ z + 1 > 0 This copletes the proof of lea 57 and the proof of the ain theore in the case 0 < c < a Note: We have assued throughout chapter 5 that F is defined and positive on z, π However, if F is defined and positive only on x 0, π where z < x 0 < π, it is easy to show that D does not have a negative local iniu on x 0, π A siilar proof as in Lea 5 shows D > 0 on x 0, π If x 0 π, we have shown earlier in the chapter that F has a zero on π, π which contradicts that F > 0 on x 0, π If x 0 > π, then by lea 5 F + F < 0 on x 0, π Also F > 0 on x 0, π So, F < 0 on x 0, π Hence F is decreasing on x 0, π So, li φ x + 0 F φ exists possibly Since 3

38 F π = 0 and F < 0 on x 0, π then F φ 0 on x 0, π and thus li φ x + F φ exists 0 Next we show that if li φ x + 0 F φ 0 then li φ x + 0 F φ Suppose li φ x + F φ = L > 0 and li 0 φ x + F φ = We have fro 10, F F + F = F c sin φ c cos φ F 1F Since c sin φ as φ approaches to x + 0 Then, 1 cos φ F sin φ 1 cos φ F is positive, the right hand side of equation 11 converges to sin φ F F 1 1 = F F F 1 1 F = 1 F F F + F which converges to by equation11 This contradicts the lea 56 that li φ x + 0 F F 1 1 exists and is finite Thus if li φ x + 0 F φ > 0 then li φ x + 0 F φ is finite Then we can find an interval x 0 ɛ, π, for soe ɛ > 0, where F is defined contradicting the axiality of the interval x 0, π So x 0 = z and F is defined on all of z, π 33

39 CHAPTER 6 THE CASE c > a > 0 We first observe that if a = c then F = c 1 cos φ is a positive solution on 0, π of 58 such that and F π = a = c, F π = 0, F 0 = 0, F 0 = 0 We will now show that a siilar result is true for all other values of a where 0 < a < c Lea 61 Let c > a > 0 Then D < 0 on π, π and F + F > 0 on π, π Proof We have fro that Dπ = 0 and D π < 0 Thus D is decreasing near π and if the lea were not true then there would exist a q π, π with 113 D q = 0, D q 0, D < 0 on q, π, and D > 0 on [q, π Thus fro 75 F cos φ + F sin φ > 0 on [q, π, and therefore we see that Integrating on φ, π gives and thus F > 0 on [q, π cos φ a cos π F cos φ > 0 on [q, π, 114 F > a cos φ > 0 on [q, π 34

40 Next, fro 77 we have and therefore fro 113 and 114 W + 1 F cos φ D + 1 F sin φ D = 0, 115 W q = 1 F q Dq < 0 sin q Fro 113 we have D q = 0 and so by 78 W q = 0 Also, fro and the fact that q π, π, we see 116 F q cos q D q 0 Using these facts in 87, we then see F q sinq F q cosq 117 sin Dq 0 q Since Dq > 0 then 117 iplies 118 F q sin q F q cos q 0 In addition, fro 115 Then, using 118, we obtain F q cf q sin q + F q + cf q cos q = W q < F q + F q < cf q sin q cf q cos q 0 This forces F q = F q = 0 which iplies Dq = 0 but this contradicts 113 This copletes the proof of the first part of the lea The second part of the lea follows fro 76 and the fact that φ π, π This copletes the proof of lea 61 Lea 6 Let c > a > 0 Then F > 0 on π, π] Proof By the previous lea, D < 0 on π, π] Also since Dπ = 0 we then have that D > 0 on π, π Thus fro 75 F cos φ + F sin φ > 0 on π, π, 35

41 and therefore Integrating over φ, π, we get F > 0 on cos φ a π, π F π cos φ for φ, π, and thus F a cos φ > 0 for φ π, π As F π = a > 0, we see that F > 0 on π, π ] This copletes the proof of lea 6 Lea 63 Let c > a > 0 Then W < 0 on π, π] Proof Fro leas 61 and 6 we have F > 0 and D < 0 on π, π ] Thus 10 F cos φ D > 0 on π, π ] Also since: Dπ = 0 and D < 0 on π, π ], then D 0 on π, π ] Thus 11 F π ] sin φ D 0 on, π Hence using 10 and 11 in 77, we have W < 0 on π, π ] This copletes the proof of the lea 36

42 Figure 61 The function F φ for a = 01, = 4 and for diensionless speed values c = 40, 0, 10, 050, 05, 015, 01 and 011 in lexicographical order Note: Let c > a > 0 Since Dπ = 0 and D < 0 on π, π, it follows fro 75 that F π = D π > 0 Lea 64 Let c > a > 0 Then F π > 0 37

43 Proof Fro lea 6, we have F > 0 on π, π] Fro lea 63, we have W < 0 on π, π] Then by continuity, π W 0, and so fro 66 we have π π π 1 F + F c F By the note before this lea, F π > 0 Then fro 1 we see that π F > 0 This copletes the proof of the lea Lea 65 Let c > a > 0 Then F π + F π > 0 Proof Fro lea 61, we have F + F > 0 on π, π ] And so by continuity F + F 0 on [ π, π ] Now let us assue by way of contradiction that π 13 F + F = 0 Since F + F > 0 on π, π] it follows fro 13 that π 14 F + F 0 Fro equations 76-77, we have 15 W + 1F F + F + 1 F sin φ D = 0 38

44 Differentiating 15 we obtain 16 W 1 + F F + F + F F + F + Also note fro 67, 76, and 13 that W π = D π = 0 Therefore using this and 14 in 16, we obtain 17 F π Dπ 0 Fro the note before lea 64, we have π 18 F and thus by 17 π = D > 0 π 19 F 0 F F D + D = 0 sin φ sin φ Evaluating 66 and 15 at φ = π and using 13, 18, and 19 gives π π π 0 < F cf + F = W π π = 1F < 0, which is ipossible Thus the lea ust hold This copletes the proof of lea 65 Lea 66 Let c > a > 0 Then D has a local axiu at π Proof Fro 76 we have D π = F + F π cosπ = 0 Differentiating 76 we have D = F + F cos φ F + F sin φ Thus by lea 65 we have D π π = F + F < 0 Hence D has a local axiu at π 39

45 Lea 67 Let c > a > 0 If b 0 and F > 0 on b, π, then F 0 on b, π Proof Contrary to the lea, let us assue By lea 64, F p < 0 for soe p π F > 0 b, π Thus F is decreasing near p, increasing near π, and so there exists a local iniu q p, π such that 130 F q = 0 and F q 0 Fro equations 59 and 66 we have 131 W + 1F At φ = q, we have Using this in 131 we obtain F + cos φ sin φ F + F = 0 F q = 0, F qf q 0, and F q > 0 W q < 0 On the other hand, fro 66 and 130 we have W q = F q + cf q cos q > 0 which is a contradiction This copletes the proof of the lea Lea 68 Let c > a > 0 If b 0 and F > 0 on b, π, then D does not have a positive local iniu on b, π Proof On the contrary, let us assue that D has a positive local iniu on b, π Then there is a p b, π with the following properties 13 Dp > 0, D p = 0, D φ > 0 on p, π, and D p 0 40

46 Fro 78, we see that W p = 0 Using 13 and 87 at φ = p, we see F p sinp F p cosp sin Dp 0 p Since Dp > 0 we then have 133 F p sin p F p cos p 0 Also note fro 77 and 13 that Using this in 66 we see W p = 1 F p Dp < 0 sin p W p = F p cf p sin p + F p + cf p cos p < 0 Thus using 133 and rewriting this iplies < F p + F p c [F p sin p F p cos p] 0 which is ipossible Hence the lea ust be true This copletes the proof of the lea Lea 69 Let c > a > 0 If there exists z [ 0, π such that F z = 0 and F > 0 on z, π then D 0 and F + F 0 on z, π Proof By lea 68, we see that D does not have a positive local iniu on z, π Since F > 0 on z, π, by lea 65 we have F 0 on z, π This iplies 135 D = F cos φ + F sin φ > 0 on z, π By lea 66, D has a local axiu at π Therefore if there were a q z, π with D q < 0 then these facts along with 135 iply that D has a positive local iniu in 41

47 z, π contradicting lea 68 Hence, we see that D 0 on z, π This proves the first part of the lea The second part of the lea follows fro 76 and the fact that cos φ > 0 on z, π 0, π This copletes the proof of the lea Lea 610 Let c > a > 0 Then F has a zero on [ 0, π Proof Contrary to the lea, let us assue F > 0 on [ 0, π By lea 67, F 0 on 0, π, and by lea 69, F + F 0 on 0, π Therefore 136 F F + cos φ sin φ F + F > 0 on 0, π Now, fro 59 and 66 we have W + 1F F + cos φ sin φ F + F = 0, and therefore by 136 W = 1F F + cos φ sin φ F + F < 0 on 0, π By continuity [ W = F cf sin φ + F + cf cos φ 0 on 0, π Evaluating at φ = 0, we obtain F 0 + F 0 + cf 0 0 4

48 which is ipossible since we are assuing F 0 > 0 Thus the lea ust be true This copletes the proof of the lea So now we assue that there exists z [0, π such that F z = 0 and F > 0 on z, π Lea 611 Let c > a > 0 If z [ 0, π such that F z = 0 and F > 0 on z, π then W < 0 on z, π Proof By assuption F > 0 on z, π, and by lea 67 we see that F 0 on z, π, and by lea 69 we see that F + F 0 on z, π So F F + cos φ sin φ F + F > 0 on z, π Using this inequality along with 59 and 66, we see W = 1F F + cos φ sin φ F + F < 0 on z, π This copletes the proof of the lea Lea 61 Let c > a > 0 If z is a zero of F on [ 0, π and F > 0 on z, π then 0 < F < c sin φ on z, π Proof We have F > 0 on z, π By lea 67 F 0 on z, π, and by lea 611 W = F cf sin φ + F + cf cos φ < 0 on z, π 43

49 Thus F cf sin φ < F cf cos φ < 0 on z, π This yields 137 F F c sin φ < 0 on z, π Since F 0 then 137 iplies F > 0 and F c sin φ < 0 Thus we conclude that 0 < F < c sin φ on z, π This copletes the proof of the lea Lea 613 Let c > a > 0 Then there exists an A with 0 A c sin z such that li F φ = A φ z + Proof Let E = 1 [ F + F ] Then E = F [F + F ] By lea 67, we know that F 0 on z, π Also, by lea 69, we know that F + F 0 on z, π and thus E 0 on z, π 44

50 Thus li E exists As li F = 0, we see that li F exists Again fro lea 67 we φ z + φ z + φ z + have that F 0 on z, π and so we see that there exists an A such that We also know by lea 611 that li F = A 0 φ z + W = F cf sin φ + F + cf cos φ < 0 on z, π Taking the liit as φ z +, we obtain A ca sin z = AA c sin z 0 Hence 0 A c sin z This copletes the proof of the lea Notes: By lea 63, lea 611, and continuity we have W = F cf sin φ + F + cf cos φ 0 on z, π] Therefore copleting the square, we have and therefore F c sin φ + F + c cos φ c, 138 F F + c cos φ + c cos φ c + c = c and 139 F F c sin φ + c sin φ c + c = c Hence F and F are uniforly bounded on z, π] Lea 614 li F = 0 φ z + 45

51 Proof Fro lea 613 we know that li φ z + F = A where 0 A c sin z Note that if z = 0 then A = 0 and the lea is proved So now let us suppose z > 0 and also that 0 < A < c sin z Multiplying 59 by F F and rewriting gives: F F [F cf sin φ] + 1F F + cf cos φ + 1F F + Integrating this on φ, φ where z < φ < z + ɛ = φ and ɛ > 0 gives φ F 140 φ F [F cf sin t] dt = 1 1 cos φ F = 0 sin φ [F φ F ] + 1 [F φ F ] φ φ cos t +c[f φ cos φ F cos φ] + c F sin t dt + 1 φ φ sin t F dt Fro 138 and 139 we have that F and F are bounded and thus we see that the right-hand side of 140 is bounded Thus we see there exists an M > 0 such that 141 z F F [F cf sin t] dt M φ Also, by assuption we have 0 < A < c sinz and therefore it follows fro 66 that 14 [ li W φ = li φ z + F cf sin φ + F + cf cos φ ] = li F [F c sin φ] = A[A c sin z] φ z + φ z + Thus if φ is sufficiently close to z and φ > z then we see that [F cf sin φ] A A c sin z > 0 However using 141 we see that for ɛ sufficiently sall we have 143 M φ φ F F [F cf sin t] dt A A c sin z φ φ F F = A A c sin z ln [ ] F φ F φ Since we have assued 0 < A < c sin z, we see that the right-hand side of 143 goes to + as φ z + since li F φ = F z = 0 whereas the left-hand side of 143 is bounded This φ z + is a contradiction and thus we see that either A = 0 or A = c sin z 46

52 So now we suppose z > 0 and A = c sin z > 0 Then we see fro 14 that li W φ = 0 φ z + Also 144 li φ z +[F c sinφ] = A c sinz = c sinz c sinz = c sinz > 0 And fro 67 W = F c sin φf + F By lea 69 and 144, we see that W 0 on z, z + ɛ and since li W φ = 0, we see φ z + that W is nonnegative to the iediate right of z contradicting lea 611 Thus A = 0 and this copletes the proof of the lea Lea 615 z = 0 Proof By lea 610, we have z [ 0, π So, z 0 Let us assue that z > 0 By lea 611, W < 0 on z, π Then 145 cf sin φ + cf cos φ F cf sin φ + F + cf cos φ < 0 on z, π Since F > 0 then by lea 61, F > 0 on z, π, and so by < F sin φ z, F cos φ on π, and so F F is bounded near z Then there exists soe finite positive nuber M such that 146 where ɛ > 0 and z + ɛ < π F F M on z, z + ɛ, Since F > 0 on z, z + ɛ we ay divide 59 by F and after rewriting we obtain F + 1 F F F + c F F cos φ + 1 F F F + 1 cos φ F = c sin φ sin φ Taking the liit as φ z + and using 146 and lea 614 we obtain li c φ z + FF F cos φ + 1F = c sin z > 0 47 F

53 Thus there is an ɛ 1 > 0 with 0 < ɛ 1 < ɛ such that c F F cos φ + 1F > c F F sin z on z, z + ɛ 1 Thus Therefore 1 F F F > c sin z c F F cos φ on z, z + ɛ F > c sin z F 1 F c 1 cos φ for z < φ < z + ɛ 1 Now we let δ > 0 and integrate 147 on z + δ, φ where z + δ < φ < z + ɛ 1 and we obtain 148 F F z + δ c sin z [ ] 1 ln F c [sin φ sinz + δ] F z + δ 1 Now as δ 0 the left hand side of 148 goes to F φ by lea 614 while the right hand side goes to + since li F φ = F z = 0 yielding a contradiction Thus we see that φ z + z = 0 This copletes the proof of the lea It then follows fro lea 614 that li F φ = 0 φ 0 + We also saw earlier in lea 31 that F π φ = F π + φ and therefore we see that F is defined on [0, π] And also we know fro lea 615 that F 0 = F π = 0 Thus we can extend F to be π-periodic on all of R and so F satisfies 59 on all of R Thus it follows that: Lea 616 F is π periodic on R This copletes the proof of the ain theore in the case c > a > 0 Note: Throughout chapter 6 we assued that F is defined on 0, π However, if we suppose F is defined only on x 0, π for soe x 0 > 0 Then by the siilar arguent as in the note after lea 613, we can show that F and F are uniforly bounded on x 0, π by c Then fro equation 10, it follows that F M on x 0, π ɛ for soe ɛ > 0 and soe finite M > 0 So, F, F and F are bounded on x 0, π ɛ So li φ x F φ and

54 li φ x + 0 F φ exist Then we can extend F to a larger interval, contradicting axiality of x 0, π This gives x 0 0 Hence F exists on all of 0, π 49

55 CHAPTER 7 FINAL COMMENTS One can also look for solutions of 3 in spherical coordinates in R n where n 3 Suppose vx 1, x,, x n, t = rx 1, x,, x n, tf φx 1, x,, x n, t where and rx 1, x,, x n, t = One can then show that F will satisfy x x n 1 + x n ct, φx 1, x,, x n, t = tan 1 x x n 1 x n ct F cf sin φ + F + cf cos φ + 1F F + n cos φ sin φ F + n 1F = 0 Note that this reduces in the n = and n = 3 cases to the equations obtained in the introduction We conjecture that a theore siilar to the Main Theore is true in this case as well In [7], the authors considered the behavior of solutions when a and also when a 0 + Preliinary investigations indicate that a result siilar to the result in [7] is also true In fact, we conjecture that if we denote the solution of as F a then Fa li = H a a where H satisfies H + H + 1H H + cos φ sin φ H + H = 0, 50

56 and for soe C > 0 Fa li = C sinφ a 0 + ax F a 51

57 BIBLIOGRAPHY [1] DG Aronson, The porous ediu equation, Nonlinear Diffusion Probles, Lecture Notes in Matheatics , 1 [] G Barenblatt, Scaling, self-siilarity, and interediate asyptotics, Cabridge University Press 1996 [3] GI Barenblatt, On soe unsteady fluid and gas otions in a porous ediu, Prikl Mat Mekh , 67 [4] G Birkhoff and G C Rota, Ordinary differential equations, Ginn and Copany 196 [5] J Diez, L Thoas, S Betelú, R Gratton, B Marino, J Gratton, D Aronson, and S Angenent, Noncircular converging flows in viscous gravity currents, Phys Rev E , 618 [6] H Huppert, The propagation of two-diensional and axisyetric viscoucs gravity currents over a rigid horizontal surface, J Fluid Mech , 43 [7] J Iaia and S Betelú, Solutions of the porous ediu equation with degenerate interfaces, Euro J App Math 5 01, 1 [8] B Marino, L Thoas, R Gratton, J Diez, and S Betelú, Waiting-tie solutions of a nonlnear diffusion euqation: Experiental study of a creeping flow near a waiting front, Phys Rev E , 68 [9] L Paudel and J Iaia, Traveling wave solutions of the porous ediu equation with degenerate interfaces, Nonlinear Analysis 01 [10] Walter A Strauss, Partial differential equations, an introduction, John Wiley & Sons, Inc 199 [11] J L Vazquez, The porous ediu equation, Oxford Matheatical Monographs 007 5