Initial Value Problem of the Whitham Equations for the Camassa-Holm Equation

Initial V alue Problem of the Whitham Equations for the Camassa-Holm Equation T amara Gra v a a , V. U. Pierce b , and F ei-Ran Tian b a SISSA, Via Beirut 2-4, 34014 T rieste, Italy b Dep artment of Mathematics, Ohio State University, 231 W. 18th Avenue, Columbus, OH 43210 Abstract W e study the Whitham equations for the Camassa-Holm equation. The equations are neither strictly h yp erb olic nor genuinely nonlinear. W e are in terested in the initial v alue problem of the Whitham equations. When the initial v alues are giv en b y a step function, the Whitham solution is self-similar. When the initial v alues are giv en by a smo oth function, the Whitham solution exists within a cusp in the x - t plane. On the boundary of the cusp, the Whitham solution matc hes the Burgers solution, whic h exists outside the cusp. Key wor ds: Camassa-Holm equation, Whitham equations, Non-strictly h yp erb olic, Ho dograph transform P A CS: 02.30.Ik, 02.30.Jr 1 In tro duction The Camassa-Holm equation u t + (3 u + 2 ν ) u x −  2 ( u xxt + 2 u x u xx + uu xxx ) = 0 , u ( x, 0;  ) = u 0 ( x ) (1) describ es wa v es in shallo w w ater when surface tension is presen t [2]. Here, ν is a constan t parameter. The solution of the initial v alue problem (1) will dev elop singularities in a finite time if and only if some p ortion of the p ositiv e part of the initial “momen tum” density u 0 ( x ) −  2 u 00 0 ( x ) + ν lies to the left of some p ortion of its negative part [10]. In particular, a unique global solution Email addr esses: grava@sissa.it (T amara Gra v a), virgilpierce@gmail.com (V. U. Pierce), tian@math.ohio-state.edu (F ei-Ran Tian). Preprin t submitted to Elsevier 19 No vem ber 2018 is guaren teed if u 0 ( x ) −  2 u 00 0 ( x ) + ν do es not c hange its sign. These are the non-breaking initial data that w e are in terested in throughout this pap er. Although the zero dispersion limit of equation (1) has not b een established, some of its modulation equations (i.e., Whitham equations) ha v e been deriv ed. The zero phase Whitham equation is u t + (3 u + 2 ν ) u x = 0 , u ( x, 0) = u 0 ( x ) , (2) whic h can b e obtained from (1) by formally setting  = 0. The single phase Whitham equations hav e b een found in [1] and they can b e written in the Riemann in v ariant form u ix + λ i ( u 1 , u 2 , u 3 ) u ix = 0 for − ν < u 3 < u 2 < u 1 , (3) where λ i ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 + 2 ν − I ∂ u i I , (4) and I ( u 1 , u 2 , u 3 ) = Z u 2 u 3 η + ν q ( η + ν )( u 1 − η )( u 2 − η )( η − u 3 ) dη . The constrain t − ν < u 3 < u 2 < u 1 is consisten t with the non-breaking initial data men tioned in the first paragraph of this section. The integral I can be rewritten as a contour integral. Hence, it satisfies the Euler-P oisson-Darb oux equations 2( u i − u j ) ∂ 2 I ∂ u i ∂ u j = ∂ I ∂ u i − ∂ I ∂ u j , i, j = 1 , 2 , 3 (5) since the integrand do es so for each η 6 = u i . The Hamiltonian structure of the single phase Whitham equations for the Camassa-Holm equation w as also obtained in [1] in terms of Abelian in tegrals. The higher phase Whitham equa- tions can also b e derived using this structure. In this pap er w e will study the ev olution of the Whitham solution from the zero phase to the single phase. This problem is similar to that of the zero disp ersion limit of the KdV equation [7,8,17] u t + 6 uu x +  2 u xxx = 0 , u ( x, 0;  ) = u 0 ( x ) . (6) It is kno wn that the zero phase Whitham equation for the KdV equation is u t + 6 uu x = 0 , (7) whic h is equiv alent to (2) for the Camassa-Holm equation. The single phase Whitham equations for the KdV equation are [3,4,18] u ix + µ i ( u 1 , u 2 , u 3 ) u ix = 0 , for u 3 < u 2 < u 1 , (8) 2 where µ i ( u 1 , u 2 , u 3 ) = 2[ u 1 + u 2 + u 3 − ˜ I ∂ u i ˜ I ] , and ˜ I ( u 1 , u 2 , u 3 ) = Z u 2 u 3 1 q ( u 1 − η )( u 2 − η )( η − u 3 ) dη . These equations are also similar to (3) and (4) for the Camassa-Holm equation. In the KdV case, the ev olution from the zero phase to the single phase has b een studied in [14]. There, the Euler-P oisson-Darb oux equations (5) ha v e pla yed an imp ortan t role. The same equations hav e also pla yed a crucial role in the study of the transition from the single phase to the double phase in [5]. Although b oth the Camassa-Holm equation (1) and the KdV equation (6) are disp ersiv e appro ximations to the Burgers equation (2) or (7), there are significan t differences in the limiting dynamics. The biggest difference is that the Whitham equations (3) for the former equation are non-strictly hyperb olic (cf. (32)) while the Whitham equations (8) for the latter equation are strictly h yp erb olic [9]. Non-Strictly h yp erb olic Whitham equations hav e also b een found in the higher order KdV flo ws [11,12,13] and the higher order defocusing NLS flo ws [6]. Self-similar solutions of these Whitham equations hav e b een constructed. They are remark ably different from the self-similar solutions of the KdV-Whitham equations [11,12] or the NLS-Whitham equations [6], b oth of whic h are strictly h yp erb olic. In this pap er, we will mo dify the metho d of pap er [14] so that it can b e used to solv e the non-strictly hyperb olic Whitham equations (3) when the initial function is a smo oth function. W e will then study the ev olution from the zero phase to the single phase for smooth initial data. When the initial function is a step-lik e function, we will use the metho d of pap er [6,11,12] to study the same ev olution. The organization of the pap er is as follows. In Section 2, we will introduce an initial v alue problem which describ es the ev olution of phases. W e will also discuss how to use the ho dograph transform to solv e non-strictly hyperb olic Whitham equations. In Section 3, we will study the properties of the eigen- sp eeds of the single phase Whitham equations. W e will study the initial v alue problem when the initial function u 0 ( x ) is a step-lik e function in Section 4 and when u 0 ( x ) is a smo oth decreasing function in Section 5. 3 2 An Initial V alue Problem W e describ e the initial v alue problem for the Burgers equation (2) and Whitham equations (3) as follows (see Figure 1.). Consider a horizon tal motion of the initial curv e u = u 0 ( x ). A t the b eginning, the curv e ev olves according to the Burgers equation (2). The Burgers solution breaks do wn in a finite time. Im- mediately after the breaking, the curv e dev elops three branc hes. Denote these three branc hes by u 1 , u 2 , and u 3 . Their motion is go verned by the Whitham equations (3). As time go es on, the Whitham solution ma y dev elop singular- ities and more branc hes are created. Ho w ever, our fo cus is on the evolution of the solution of the Whitham equations from the one branch regime to the three branc h regime. u u u u 1 x (t) - x u 3 u 2 x (t) + Fig. 1. Profile of the Burgers and Whitham solutions. The Burgers solution u of (2) liv es in the single v alue regions while the Whitham solution u 1 , u 2 and u 3 of (3) reside in the m ultiple v alue region. The Burgers solution u of (2) and the Whitham solution u 1 , u 2 , u 3 of (3) m ust matc h on the trailing edge x = x − ( t ) and leading edge x = x + ( t ). W e see from Figure 1 that      u 1 = u u 2 = u 3 (9) m ust b e imp osed on the trailing edge, and that      u 1 = u 2 u 3 = u (10) m ust b e satisfied on the leading edge. 4 In this paper, w e consider the initial function u 0 ( x ) that is monotone. Since the Burgers solution will not develop an y sho c k if u 0 ( x ) is an increasing function, w e will fo cus on decreasing initial functions. Denoting the inv erse function of u 0 ( x ) by f ( u ), the Burger equation (2) can b e solved using the metho d of c haracteristics; its solution is giv en implicitly by a ho dograph transform, x = (3 u + 2 ν ) t + f ( u ) . (11) The solution metho d (11) has b een generalized to solv e the first order quasi- linear hyperb olic equations whic h can b e written in Riemann inv arian t form and whic h are strictly h yp erb olic ∂ u i ∂ t + s i ( u 1 , · · · , u n ) ∂ u i ∂ x = 0 , i = 1 , 2 , · · · , n. (12) The strict h yp erb olicit y means that the wa v e propagation sp eeds s i ’s do not coincide. W e include Tsarev’s theorem for completeness [14,16]. Theorem 1 If w i ( u 1 , u 2 , · · · , u n ) ’s solve the line ar e quations ∂ w i ∂ u j = A ij ( w i − w j ) (13) with A ij = ∂ s j ∂ u j s i − s j (14) for i, j = 1 , 2 , · · · , n and i 6 = j , then the solution ( u 1 ( x, t ) , · · · , u n ( x, t )) of the ho do gr aph tr ansform, x = s i ( u 1 , u 2 , · · · , u n ) t + w i ( u 1 , u 2 , · · · , u n ) (15) satisfies e quations (12). Conversely, any solution ( u 1 ( x, t ) , · · · , u n ( x, t )) of e quations (12) c an b e obtaine d in this way in the neighb orho o d of ( x 0 , t 0 ) at which u ix ar e not al l vanishing. The strict hyperb olicity , i.e., s i 6 = s j for i 6 = j , of (12) is assumed to ensure that A ij ’s of (14) are not singular. The result is classical when n = 2. The v alidit y of this theorem hinges on tw o factors. First, the linear equations (13) m ust ha ve solutions. Secondly , the hodograph transform (15) must not b e degenerate, i.e., it can b e solved for u i ’s as functions of x and t . One in teresting observ ation is that the Jacobian matrix of (15) is alw a ys diagonal on the solution ( u 1 ( x, t ) , · · · , u n ( x, t )). 5 Corollary 2 A t the solution ( u 1 ( x, t ) , · · · , u n ( x, t )) of x = s i ( u 1 , u 2 , · , u n ) t + w i ( u 1 , u 2 , · · · , u n ) , i = 1 , 2 , · · · , n , the p artial derivatives ∂ ( s i t + w i ) ∂ u j = 0 for i 6 = j . Pr o of. ∂ ( s i t + w i ) ∂ u j = ∂ s i ∂ u j t + ∂ w i ∂ u j = A ij [( s i t + w i ) − ( s j t + w j )] = 0 . 2 Another aspect of Theorem 1 is that it is a lo cal result. Solutions pro duced b y the ho dograph transform are, in general, lo cal in nature. How ever, global solutions can still b e obtained if the conditions of the theorem are satisfied globally [5,14]. The Whitham equations (3) will b e shown to b e non-strictly hyperb olic, i.e., λ i ’s coincide at some p oin ts ( u 1 , u 2 , u 3 ) where 0 < u 3 + ν < u 2 + ν < u 1 + ν . Ho wev er, Theorem 1 can still b e applied to equations (3) since the functions A ij ’s of (14) are still non-singular for the Whitham equations (3), even at the p oin ts of non-strict hyperb olicity . Lemma 3 B ij := ∂ λ i ∂ u j λ i − λ j = 1 2 ( λ i − γ ) − 2( u i − u j ) ( λ j − γ )( u i − u j ) , i 6 = j , wher e γ = u 1 + u 2 + u 3 + 2 ν . Pr o of. By (4), w e calculate λ i − λ j = I ∂ u i I − ∂ u j I ( ∂ u i I )( ∂ u j I ) = 2 I ( u i − u j ) ∂ 2 u i u j I ( ∂ u i I )( ∂ u j I ) , ∂ λ i ∂ u j = 2( u i − u j ) ∂ 2 u i u j I ∂ u i I + I ∂ 2 u i u j I ( ∂ u i I ) 2 , where w e ha ve used (5). Hence, w e get 6 B ij = 2( u i − u j ) + I ∂ u i I 2 I ∂ u j I ( u i − u j ) = 1 2 ( λ i − γ ) − 2( u i − u j ) ( λ j − γ )( u i − u j ) , where w e ha ve used (4) to express I /∂ u i I . 2 3 The Single Phase Whitham Equations In this section, we will summarize some of the prop erties of the sp eeds λ i ’s of (4) for later use. F unction I of (4) is a complete elliptic in tegral; indeed, I ( u 1 , u 2 , u 3 ) = 2( u 3 + ν )Π( ρ, s ) q ( u 1 − u 3 )( u 2 + ν ) , (16) where Π( ρ, s ) is the complete in tegral of third kind, and ρ = u 2 − u 3 u 2 + ν , s = ( u 2 − u 3 )( u 1 + ν ) ( u 1 − u 3 )( u 2 + ν ) . (17) Prop erties of complete elliptic in tegrals of the first, second and third kind are listed in App endix A. Using the w ell kno wn deriv ativ e form ulae (A.3) and (A.4), one is able to rewrite λ i of (4) as [1] λ 1 ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 + 2 ν + 2( u 1 − u 2 ) ( u 3 + ν )Π( ρ, s ) ( u 2 + ν ) E ( s ) , λ 2 ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 + 2 ν + 2( u 3 − u 2 ) (1 − s )Π( ρ, s ) E ( s ) − (1 − s ) K ( s ) , (18) λ 3 ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 + 2 ν + 2( u 2 − u 3 ) ( u 3 + ν )Π( ρ, s ) ( u 2 + ν )[ E ( s ) − K ( s )] . Here K ( s ) and E ( s ) are complete elliptic in tegrals of the first and second kind. Using inequalities (A.9), w e obtain 7 λ 1 − ( u 1 + u 2 + u 3 + 2 ν ) > 0 , (19) λ 2 − ( u 1 + u 2 + u 3 + 2 ν ) < 0 , (20) λ 3 − ( u 1 + u 2 + u 3 + 2 ν ) < 0 , (21) for u 1 > u 2 > u 3 > − ν . In view of (A.5-A.8) and (A.10-A.11), we find that λ 1 , λ 2 and λ 3 ha ve b eha vior (1) A t u 2 = u 3 , λ 1 ( u 1 , u 2 , u 3 ) = 3 u 1 + 2 ν , λ 2 ( u 1 , u 2 , u 3 ) = λ 3 ( u 1 , u 2 , u 3 ) = u 1 + 2 u 3 + 2 ν − 4( u 3 + ν )( u 1 − u 3 ) u 1 + ν . (22) (2) A t u 1 = u 2 , λ 1 ( u 1 , u 2 , u 3 ) = λ 2 ( u 1 , u 2 , u 3 ) = 2 u 1 + u 3 + 2 ν , λ 3 ( u 1 , u 2 , u 3 ) = 3 u 3 + 2 ν . (23) Lemma 4 ∂ λ 3 ∂ u 3 < 3 2 λ 2 − λ 3 u 2 − u 3 < ∂ λ 2 ∂ u 2 for 0 < u 3 + ν < u 2 + ν < u 1 + ν . Pr o of. Comparing form ulae (4) and (18), w e use (16) to obtain ∂ I ∂ u 2 = q ( u 1 − u 3 )( u 2 + ν ) ( u 2 − u 3 )( u 1 − u 2 ) [ E − (1 − s ) K ] . (24) Differen tiating (4) for λ 2 and using (24) yields ∂ λ 2 ∂ u 2 = I ∂ 2 u 2 u 2 I ( ∂ u 2 I ) 2 = ( u 3 + ν )Π ( u 2 + ν )[ E − (1 − s ) K ] 2 × ×  [ s + u 2 − u 3 u 1 − u 3 − 2 u 1 − u 2 u 1 − u 3 ][ E − (1 − s ) K ] + s (1 − s ) K  . (25) Using form ulae (18) for λ 2 and λ 3 , w e obtain λ 2 ( u 1 , u 2 , u 3 ) − λ 3 ( u 1 , u 2 , u 3 ) = 2( u 2 − u 3 )( u 3 + ν )Π M ( u 1 , u 2 , u 3 ) ( u 2 + ν )( K − E )[ E − (1 − s ) K ] , (26) 8 where M ( u 1 , u 2 , u 3 ) = [1 + u 1 − u 2 u 1 − u 3 ] E − [ u 1 − u 2 u 1 − u 3 + (1 − s )] K . (27) W e then obtain from (25) and (26) that ∂ λ 2 ∂ u 2 − 3 2 λ 2 − λ 3 u 2 − u 3 = − ( u 3 + ν )(4 − 3 s )Π E 2 ( u 2 + ν )[ K − E ][ E − (1 − s ) K ] 2  (1 − s )( K E ) 2 − 2( K E ) + 4 + s 4 − 3 s  > 0 , where the inequalit y follows from the negativit y of the function in the brac ket (c.f. (4.18) of [14]). This pro ves part of the lemma. The other part can b e sho wn in the same w ay . 2 W e conclude this section with a few calculations. W e use (A.1) and (A.2) to calculate the deriv ative of (27) ∂ M ( u 1 , u 2 , u 3 ) ∂ u 2 = 1 2( u 1 − u 3 ) { [2 + ( u 1 + ν )( u 3 + ν ) ( u 2 + ν ) 2 ] K − [2 + u 1 + ν u 2 + ν ] E } , (28) ∂ 2 M ( u 1 , u 2 , u 3 ) ∂ u 2 2 = ( u 1 + ν )[4( u 1 + ν ) − 2( u 2 + ν ) + ( u 3 + ν )] E 4( u 1 − u 3 )( u 1 − u 2 )( u 2 + ν ) 2 . (29) Finally , we use the expansions (A.5-A.6) for K and E to obtain M ( u 1 , u 2 , u 3 ) = π 2 ( ( u 2 − u 3 ) s 2( u 1 − u 3 ) + 1 16 (1 − 3( u 1 − u 2 ) u 1 − u 3 ) s 2 + 3 128 (1 − 5( u 1 − u 2 ) u 1 − u 3 ) s 3 + O ( s 4 ) ) = π 2 ( [ u 2 + ν 2( u 1 + ν ) + 1 16 (1 − 3( u 1 − u 2 ) u 1 − u 3 )] s 2 + 3 128 (1 − 5( u 1 − u 2 ) u 1 − u 3 ) s 3 + O ( s 4 ) ) , (30) where w e ha ve used formula (17) for s in the last equalit y . 9 4 Step-lik e Initial Data In this section, w e will consider the step-lik e initial data u 0 ( x ) =      a x < 0 b x > 0 , a 6 = b (31) for equation (1). Since the solution of (2) will nev er develop a sho c k when a ≤ b , w e will b e interested only in the case a > b . W e classify the initial data (31) in to t wo types: • (I) a + ν > 4( b + ν ) , • (II) a + ν ≤ 4( b + ν ) . W e will solv e the initial v alue problem for the Whitham equations for these t wo t yp es of initial data. 4.1 T yp e I: a + ν > 4( b + ν ) 0.0 0.5 1.0 1.5 2.0 0 1 2 3 4 5 6 u x/t x/t = α u u u 1 u 2 u 3 Fig. 2. Self-Similar solution of the Whitham equations for a = 2, b = 0 and ν = 1 / 20 of t yp e I. Theorem 5 (se e Figur e 2.) F or the step-like initial data u 0 ( x ) of (31) with 0 < b + ν < ( a + ν ) / 4 , the solution of the Whitham e quations (3) is given by u 1 = a , x = λ 2 ( a, u 2 , u 3 ) t , x = λ 3 ( a, u 2 , u 3 ) t (32) for (3 a − ν ) / 4 < x/t ≤ α and by u 1 = a , x = λ 2 ( a, u 2 , b ) t , u 3 = b (33) 10 for α t ≤ x < (2 a + b + 2 ν ) t , wher e α = λ 2 ( a, u ∗ , b ) and u ∗ is the unique solution u 2 of λ 2 ( a, u 2 , b ) = λ 3 ( a, u 2 , b ) in the interval b < u 2 < a . Outside the r e gion (3 a − ν ) / 4 < x/t < 2 a + b + 2 ν , the solution of the Bur gers e quation (2) is given by u ≡ a x/t ≤ (3 a − ν ) / 4 (34) and u ≡ b x/t ≥ 2 a + b + 2 ν . (35) The b oundaries x/t = (3 a − ν ) / 4 and x/t = 2 a + b + 2 ν are the trailing and leading edges, resp ectiv ely , of the disp ersiv e sho c k. They separate the solution in to the region go verned by the single phase Whitham equations and the region go v erned by the Burgers equation. The pro of of Theorem 5 is based on a series of lemmas. W e first sho w that the solutions defined by formulae (32) and (33) indeed satisfy the Whitham equations (3) [16]. Lemma 6 (i) The functions u 1 , u 2 and u 3 determine d by e quations (32) give a solution of the Whitham e quations (3) as long as u 2 and u 3 c an b e solve d fr om (32) as functions of x and t . (ii) The functions u 1 , u 2 and u 3 determine d by e quations (33) give a solution of the Whitham e quations (3) as long as u 2 c an b e solve d fr om (33) as a function of x and t . Pr o of. (i) u 1 ob viously satisfies the first equation of (3). T o verify the second and third equations, w e observ e that ∂ λ 2 ∂ u 3 = ∂ λ 3 ∂ u 2 = 0 (36) on the solution of (32). This follows from Lemma 3. W e then calculate the partial deriv ativ es of the second equation of (32) with resp ect to x and t . 1 = ∂ λ 2 ∂ u 2 t ( u 2 ) x , 0 = ∂ λ 2 ∂ u 2 t ( u 2 ) t + λ 2 , whic h giv e the second equation of (3). The third equation of (3) can b e verified in the same wa y . (ii) The second part of Lemma 6 can easily b e prov ed in a similar manner. 2 11 W e now determine the trailing edge. Eliminating x and t from the last tw o equations of (32) yields λ 2 ( a, u 2 , u 3 ) − λ 3 ( a, u 2 , u 3 ) = 0 . (37) In view of form ula (26), w e replace (37) by M ( a, u 2 , u 3 ) s 2 = 0 . (38) Therefore, at the trailing edge where u 2 = u 3 , i.e., s = 0, equation (38), in view of the expansion (30), b ecomes u 3 + ν 2( a + ν ) + 1 16 [1 − 3( a − u 3 ) a − u 3 ] = 0 , whic h giv es u 2 = u 3 = ( a − 3 ν ) / 4. Lemma 7 Equation (38) has a unique solution satisfying u 2 = u 3 . The solu- tion is u 2 = u 3 = ( a − 3 ν ) / 4 . The r est of e quations (32) at the tr ailing e dge ar e u 1 = a and x/t = λ 2 ( a, ( a − 3 ν ) / 4 , ( a − 3 ν ) / 4) = (3 a − ν ) / 4 . Ha ving lo cated the trailing edge, we no w solv e equations (32) in the neigh- b orhoo d of the trailing edge. W e first consider equation (38). W e use (30) to differen tiate M /s 2 at the trailing edge u 1 = a , u 2 = u 3 = ( a − 3 ν ) / 4, to find ∂ ∂ u 2 [ M s 2 ] = ∂ ∂ u 3 [ M s 2 ] = π 8( a + ν ) , whic h sho ws that equation (38) or equiv alently (37) can b e in v erted to giv e u 3 as a decreasing function of u 2 u 3 = A ( u 2 ) (39) in a neigh b orho od of u 2 = u 3 = ( a − 3 ν ) / 4. W e now extend the solution (39) of equation (37) in the region a > u 2 > ( a − 3 ν ) / 4 > u 3 > b as far as p ossible. W e deduce from Lemma 4 that ∂ λ 2 ∂ u 2 > 0 , ∂ λ 3 ∂ u 3 < 0 (40) on the solution of (37). Because of (36) and (40), solution (39) of equation (37) can b e extended as long as a > u 2 > ( a − 3 ν ) / 4 > u 3 > 0. There are t w o p ossibilities: (i) u 2 touc hes a b efore or sim ultaneously as u 3 reac hes b and (ii) u 3 touc hes b b efore u 2 reac hes a . 12 It follo ws from (23) that λ 2 ( a, a, u 3 ) > λ 3 ( a, a, u 3 ) for b ≤ u 3 < a . This shows that (i) is imp ossible. Hence, u 3 will touch b b efore u 2 reac hes a . When this happ ens, equation (37) b ecomes λ 2 ( a, u 2 , b ) − λ 3 ( a, u 2 , b ) = 0 . (41) Lemma 8 Equation (41) has a simple zer o in the r e gion b < u 2 < a , c ounting multiplicities. Denoting the zer o by u ∗ , then λ 2 ( a, u 2 , b ) − λ 3 ( a, u 2 , b ) is p ositive for u 2 > u ∗ and ne gative for u 2 < u ∗ . Pr o of. W e use (26) and (29) to pro ve the lemma. In equation (26), K − E and E − (1 − s ) K are all p ositiv e for 0 < s < 1 in view of (A.9). W e claim that M ( a, u 2 , b ) = 0 for u 2 = b , M ( a, u 2 , b ) < 0 for u 2 near b , and M ( a, u 2 , b ) > 0 for u 2 = a . The equality and the first inequality follo w from expansion (30) and a + ν > 4( b + ν ). The second inequality is obtained by applying (A.7) and (A.8) to (27). W e conclude from the t wo inequalities that M ( a, u 2 , b ) has a zero in b < u 2 < a . This zero is unique b ecause M ( a, u 2 , b ), in view of (29), is a conv ex function of u 2 . This zero is exactly u ∗ and the rest of the theorem is prov en easily . 2 Ha ving solv ed equation (37) for u 3 as a decreasing function of u 2 for ( a − 3 ν ) / 4 ≤ u 2 ≤ u ∗ , we turn to equations (32). Because of (36) and (40), the second equation of (32) gives u 2 as an increasing function of x/t , for (3 a − ν ) / 4 ≤ x/t ≤ α , where α = λ 2 ( a, u ∗ , b ) . Consequen tly , u 3 is a decreasing function of x/t in the same interv al. Lemma 9 The last two e quations of (32) c an b e inverte d to give u 2 and u 3 as incr e asing and de cr e asing functions, r esp e ctively, of the self-similarity variable x/t in the interval (3 a − ν ) ≤ x/t ≤ α , wher e α = λ 2 ( a, u ∗ , b ) and u ∗ is given in L emma 8. W e no w turn to equations (33). W e wan t to solve the second equation when x/t > α or equiv alently when u 2 > u ∗ . According to Lemma 8, λ 2 ( a, u 2 , b ) − λ 3 ( a, u 2 , b ) > 0 for u ∗ < u 2 < a , whic h, by Lemma 4, sho ws that ∂ λ 2 ( a, u 2 , b ) ∂ u 2 > 0 . 13 Hence, the second equation of (33) can b e solv ed for u 2 as an increasing func- tion of x/t as long as u ∗ < u 2 < a . When u 2 reac hes a , w e hav e x/t = λ 2 ( a, a, b ) = 2 a + b + 2 ν , where we ha v e used (23) in the last equality . W e ha ve therefore pro v ed the follo wing result. Lemma 10 The se c ond e quation of (33) c an b e inverte d to give u 2 as an incr e asing function of x/t in the interval α ≤ x/t ≤ 2 a + b + 2 ν . W e are ready to conclude the pro of of Theorem 5. The Burgers solutions (34) and (35) are trivial. According to Lemma 9, the last tw o equations of (32) determine u 2 and u 3 as functions of x/t in the region (3 a − v ) / 4 ≤ x/t ≤ α . By the first part of Lemma 6, the resulting u 1 , u 2 and u 3 satisfy the Whitham equations (3). F urthermore, the b oundary condition (9) is satisfied at the trailing edge x/t = (3 a − v ) / 4. Similarly , b y Lemma 10, the second equation of (33) determines u 2 as a func- tion of x/t in the region α ≤ x/t ≤ 2 a + b + 2 ν . It then follo ws from the second part of Lemma 6 that u 1 , u 2 and u 3 of (33) satisfy the Whitham equa- tions (3). They also satisfy the b oundary condition (10) at the leading edge x = (2 a + b + 2 ν ) t . W e hav e therefore completed the pro of of Theorem 5. A graph of the Whitham solution for the initial data (31) of type I is giv en in Figure 2. It is obtained b y plotting the exact solutions of (32) and (33). 4.2 T yp e II: a + ν ≤ 4( b + ν ) Theorem 11 (se e Figur e 3.) F or the step-like initial data (31) with 0 < ( a + ν ) / 4 ≤ b + ν < a + ν , the solution of the Whitham e quations (3) is given by u 1 = a , x = λ 2 ( a, u 2 , b ) t , u 3 = b (42) for λ 2 ( a, b, b ) < x/t < λ 2 ( a, a, b ) , wher e λ 2 ( a, b, b ) = a + 2 b + 2 ν − 4( a − b )( b + ν ) / ( a + ν ) and λ 2 ( a, a, b ) = 2 a + b + 2 ν . Outside this interval, the solution of (2) is given by u ≡ a x/t ≤ λ 2 ( a, b, b ) and u ≡ b x/t ≥ λ 2 ( a, a, b ) . 14 0 1 2 3 4 5 6 0.0 0.5 1.0 1.5 2.0 u u u 3 u 2 u 1 u x/t Fig. 3. Self-Similar solution of the Whitham equations for a = 2, b = 0 . 8 and ν = 1 / 20 of type I I. Pr o of. W e will giv e a brief pro of, since the argumen ts are, more or less, similar to those in the pro of of Theorem 5. It suffices to show that λ 2 ( a, u 2 , b ) is an increasing function of u 2 for b < u 2 < a . Using the inequalit y (A.9) to estimate the righ t hand side of (28), we obtain dM ( a, u 2 , b ) du 2 > ( a + ν )( u 2 − b ) E ( s ) 2(2 − s )( a − b ) 2 ( u 2 + ν ) 2 { 2( u 2 + ν ) + 2( b + ν ) − − ( a + ν ) } > 0 for b < u 2 < a , where we ha ve used ( a + ν ) / 4 ≤ b + ν in the second inequalit y . Since M ( a, u 2 , b ) = 0 at u 2 = b in view of (30), this implies that M ( a, u 2 , b ) > 0 for b < u 2 < a . It then follo ws from (26) that λ 2 ( a, u 2 , b ) − λ 3 ( a, u 2 , b ) > 0. By Lemma 4, w e conclude that dλ 2 ( a, u 2 , b ) du 2 > 0 for b < u 2 < a . 2 A graph of the Whitham solution for initial data (31) of t yp e II is giv en in Figure 3. It is obtained b y plotting the exact solution of (42). 15 5 Smo oth Initial Data In this section, we will study the initial v alue problem of the Whitham equa- tions when the initial v alues are giv en b y a smo oth monotone function u 0 ( x ). Since the Burgers solution of (2) will nev er dev elop a sho ck when u 0 ( x ) is an increasing function, we will b e in terested only in the case that u 0 ( x ) is a decreasing function. W e consider the initial function u 0 ( x ) whic h is a decreasing function and is b ounded at x = ±∞ lim x →−∞ u 0 ( x ) = a , lim x → + ∞ u 0 ( x ) = b . (43) By Theorem 1 and Lemma 3, we can use the ho dograph transform, x = λ i ( u 1 , u 2 , u 3 ) t + w i ( u 1 , u 2 , u 3 ) , i = 1 , 2 , 3 , (44) to solv e the Whitham equations (3). Here, w i ’s satisfy a linear o ver-determined system of t yp e (13) ∂ w i ∂ u j = B ij ( w i − w j ) , (45) where B ij ’s are giv en in Lemma 3. The b oundary conditions on w i ’s are obtained by observing that the ho dograph solution (44) of the Whitham equations (3) must matc h the characteristic solution (11) of the Burgers equation (2) at the trailing and leading edges in the fashion of (9-10). By (22-23), w i ’s m ust satisfy the b oundary conditions,      w 1 ( u 1 , u 1 , u 3 ) = w 2 ( u 1 , u 1 , u 3 ) , w 3 ( u 1 , u 1 , u 3 ) = f ( u 3 ) , (46)      w 1 ( u 1 , u 3 , u 3 ) = f ( u 1 ) , w 2 ( u 1 , u 3 , u 3 ) = w 3 ( u 1 , u 3 , u 3 ) , (47) where f ( u ) is the inv erse of the initial function u 0 ( x ). Analogous to the KdV case [14,15], equations (45) sub ject to boundary con- ditions (46-47) are related to a b oundary v alue problem of a linear o ver- determined system of Euler-P oisson-Darb oux type (cf. (5)) 16 2( u i − u j ) ∂ 2 q ∂ u i ∂ u j = ∂ q ∂ u i − ∂ q ∂ u j , i, j = 1 , 2 , 3 , (48) q ( u, u, u ) = f ( u ) , (49) for i, j = 1 , 2 , 3. The solution is unique and symmetric in u 1 , u 2 , u 3 . It is giv en explicitly b y [14] q ( u 1 , u 2 , u 3 ) = 1 2 √ 2 π Z 1 − 1 Z 1 − 1 f ( 1+ u 2 1+ v 2 u 1 + 1+ u 2 1 − v 2 u 2 + 1 − u 2 u 3 ) q (1 − u )(1 − v 2 ) dudv . (50) Theorem 12 If q ( u 1 , u 2 , u 3 ) is a solution of (48) and (49), then w i ( u 1 , u 2 , u 3 ) ’s given by w i = [ λ i ( u 1 , u 2 , u 3 ) − γ ] ∂ q ( u 1 , u 2 , u 3 ) ∂ u i + q ( u 1 , u 2 , u 3 ) , (51) wher e γ = u 1 + u 2 + u 3 + 2 ν , solve e quations (45) and satisfy b oundary c on- ditions (46-47). Pr o of. By Lemma 3 and (48), w e obtain [ λ j − γ ] " ∂ q ∂ u i − ∂ q ∂ u j # B ij = ∂ q ∂ u j − ∂ q ∂ u i ! + [ λ i − γ ] ∂ 2 q ∂ u i ∂ u j . (52) Using (51), w e calculate ∂ w i ∂ u j = ∂ λ i ∂ u j ∂ q ∂ u i + [ λ i − γ ] ∂ 2 q ∂ u i ∂ u j + ∂ q ∂ u j − ∂ q ∂ u i , w i − w j = [ λ i − γ ] [ ∂ q ∂ u i − ∂ q ∂ u j ] + [ λ i − λ j ] ∂ q ∂ u i . Substituting these in to (52), w e find that w i ’s satisfy (45). Finally , we shall c heck the b oundary conditions (46-47). W e only consider the leading edge, and the trailing edge can b e handled in the same wa y . Since q ( u 1 , u 2 , u 3 ) is symmetric in u 1 , u 2 and u 3 , the first condition of (46) follo ws from (51) and (23). 17 F or the second condition, it follows from (23) and (51) again that w 3 ( u 1 , u 1 , u 3 ) = 2( u 3 − u 1 ) ∂ q ∂ u 3 + q . (53) Differen tiating this with resp ect to u 1 yields, ∂ w 3 ∂ u 1 = − 2 ∂ q ∂ u 3 + 2( u 3 − u 1 )[ ∂ 2 q ∂ u 1 ∂ u 3 + ∂ 2 q ∂ u 2 ∂ u 3 ] + ∂ q ∂ u 1 + ∂ q ∂ u 2 = 0 , where we ha v e used (48) in the last equalit y . Since w 3 ( u 1 , u 1 , u 3 ) is independent of u 1 , w e replace u 1 b y u 3 in (53) and use (49) to obtain the second condition of (46). 2 Theorem 12 has b een rep orted in [1] in the case of ν = 0. In the rest of this section, we study the hodograph transform (44) with w i ’s giv en b y (50) and (51). W e shall sho w that the transform can b e solv ed for u 1 , u 2 and u 3 as functions of ( x, t ) within a cusp in the x - t plane. Since t b = − [3 min ( u 0 0 ( x ))] − 1 is the breaking time of the Burgers solution of (2), the breaking is caused b y an inflection point in the initial data. If x 0 is this inflection p oint, then ( x b , t b ) is the breaking p oin t on the ev olving curv e where x b = x 0 + [3 u 0 ( x 0 ) + 2 ν ] t b , and t b is the breaking time. Without loss of generalit y , we ma y assume x b = 0, t b = 0 and denote u 0 (0) by ˆ u . The effect of these choices is that we are starting at the breaking time, and the evolving curv e is ab out to turn ov er at the p oin t (0 , ˆ u ) in the x - u plane. It immediately follo ws that f ( ˆ u ) = f 0 ( ˆ u ) = f 00 ( ˆ u ) = 0 , (54) where x = f ( u ) is the in v erse function of the decreasing initial data u = u 0 ( x ). On the assumption that x = f ( u ) has only one inflection p oin t, it follo ws from the monotonicit y of the function f ( u ) that f 00 ( u ) =              < 0 u > ˆ u = 0 u = ˆ u > 0 u < ˆ u . (55) Under a little bit stronger condition than (55), w e will b e able to sho w that ho dograph transform (44) can b e inv erted to give u 1 , u 2 and u 3 as functions of ( x, t ) in some domain of the x - t plane. Theorem 13 Supp ose u 0 ( x ) is a de cr e asing function satisfying (43) with a + ν > b + ν > 0 . If, in addition to (54), the inverse function f ( u ) satisfies f 000 ( u ) 18 < 0 for b < u < a , then tr ansform (44) with w i ’s given by (50) and (51) c an b e solve d for u 1 , u 2 and u 3 as functions of ( x, t ) within a cusp in the x - t plane for al l t > 0 . F urthermor e, these u 1 , u 2 and u 3 satisfy b oundary c onditions (46-47) on the b oundary of the cusp. The pro of is based on a series of lemmas. The organization is as follo ws: w e eliminate x from transform (44) to obtain t w o equations inv olving u 1 , u 2 , u 3 , and t . These t wo equations can b e sho wn, for eac h fixed time after the breaking, to determine u 1 and u 3 as decreasing functions of u 2 within an in terv al whose end p oin ts dep end on t . Substituting u 1 and u 3 as functions of u 2 in to the ho dograph transform, w e find that, within a cusp in the x - t plane, u 2 is a function of ( x, t ), and so, therefore, are u 1 and u 3 . First, w e conclude from form ula (50) Lemma 14 If f ( u ) satisfies the c onditions of The or em 13, then q ( u 1 , u 2 , u 3 ) given by (50) satisfies ∂ 3 q ∂ u i ∂ u j ∂ u k < 0 , i, j, k = 1 , 2 , 3 . Eliminating x from (44) yields ( λ 1 t + w 1 ) − ( λ 2 t + w 2 ) = 0 , (56) ( λ 2 t + w 2 ) − ( λ 3 t + w 3 ) = 0 . (57) Using (51) for w 1 and w 2 , and (18) for λ 1 and λ 2 , w e write ( λ 1 t + w 1 ) − ( λ 2 t + w 2 ) = ( λ 1 − γ )[ t + ∂ q ∂ u 1 ] − ( λ 2 − γ )[ t + ∂ q ∂ u 2 ] = 2( u 1 − u 2 )( u 3 + ν )Π ( u 2 + ν ) E F ( u 1 , u 2 , u 3 ) , where F = ( t + ∂ q ∂ u 1 ) + sE E − (1 − s ) K u 2 + ν u 1 + ν [ t + ∂ q ∂ u 2 ] . (58) Similarly , we use (51) for w 2 and w 3 to write ( λ 2 t + w 2 ) − ( λ 3 t + w 3 ) = ( λ 2 − λ 3 )( t + ∂ q ∂ u 2 ) + ( λ 3 − γ )( ∂ q ∂ u 2 − ∂ q ∂ u 3 ) = 2( u 2 − u 3 )( u 3 + ν )Π ( u 2 + ν )[ K − E ][ E − (1 − s ) K ] G ( u 1 , u 2 , u 3 ) , 19 where G = M ( u 1 , u 2 , u 3 )( t + ∂ q ∂ u 2 ) − 2( u 2 − u 3 )[ E − (1 − s ) K ] ∂ 2 q ∂ u 2 ∂ u 3 . (59) In the deriv ation, we ha ve used formula (26) for λ 2 − λ 3 , formula (18) for λ 3 and equation (48). Since (A.9) implies that K ( s ) − E ( s ) > 0 and E ( s ) − (1 − s ) K ( s ) > 0, equations (56) and (57) are equiv alent to F ( u 1 , u 2 , u 3 ) = 0 , G ( u 1 , u 2 , u 3 ) = 0 (60) for 0 < s < 1. 5.1 The tr ailing e dge W e first study the trailing edge. W e use (A.5), (A.6), and (30) to expand F = t + ∂ q ∂ u 1 + 2( u 2 + ν ) ( u 1 + ν ) ( t + ∂ q ∂ u 2 ) − 3( u 2 + ν ) 4( u 1 + ν ) ( t + ∂ q ∂ u 2 ) s − 3( u 2 + ν ) 32( u 1 + ν ) ( t + ∂ q ∂ u 2 ) s 2 + O ( s 3 ) , (61) and G = π 2 ( [ u 2 + ν 2( u 1 + ν ) + 1 16 (1 − 3( u 1 − u 2 ) u 1 − u 3 )]( t + ∂ q ∂ u 2 ) − − ( u 1 − u 3 )( u 2 + ν ) u 1 + ν ∂ 2 q ∂ u 2 ∂ u 3 ) s 2 + π 2 ( 3 128 (1 − 5 u 1 − u 2 u 1 − u 3 )( t + ∂ q ∂ u 2 ) − − ( u 1 − u 3 )( u 2 + ν ) 8( u 1 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 ) s 3 + O ( s 4 ) . (62) T aking the limits of F = 0 and G/s 2 = 0 as s → 0 and simplifying the results a bit, w e obtain the equations go verning the trailing edge U ( u 1 , u 3 ) := ( u 1 + ν )( t + ∂ q ( u 1 , u 3 , u 3 ) ∂ u 1 )+2( u 3 + ν )( t + ∂ q ( u 1 , u 3 , u 3 ) ∂ u 2 ) = 0 , (63) 20 and V ( u 1 , u 3 ) := [( u 3 + ν ) − 1 4 ( u 1 + ν )]( t + ∂ q ∂ u 2 ) − 2( u 1 − u 3 )( u 3 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 = 0 . (64) Solving for t from (63) and substituting it in to (64), w e use (48) to simplify the result and get W ( u 1 , u 3 ) := [( u 3 + ν ) − 1 4 ( u 1 + ν )]( u 1 + ν ) ∂ 2 q ∂ u 1 ∂ u 3 + [( u 1 + ν ) + 2( u 3 + ν )]( u 3 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 = 0 . (65) Ob viously , equations (63) and (64) are equiv alent to equations (63) and (65). W e now solve equation (65) for u 3 as a function of u 1 in the neigh b orho o d of u 1 = u 3 = ˆ u . W e use formula (50) and the symmetry of q to write ∂ 2 q ( u 1 , u 3 , u 3 ) ∂ u 1 ∂ u 3 = 1 16 √ 2 π Z 1 − 1 f 00 ( 1 + µ 2 u 3 + 1 − µ 2 u 1 ) (1 − µ 2 ) √ 1 − µ dµ , (66) ∂ 2 q ( u 1 , u 3 , u 3 ) ∂ u 2 ∂ u 3 = 1 64 √ 2 π Z 1 − 1 f 00 ( 1 + µ 2 u 3 + 1 − µ 2 u 1 ) (1 + µ ) 2 √ 1 − µ dµ . (67) F or u 1 = ˆ u , it follo ws from (55), (66) and (67) that equation (65) has only the solution u 3 = ˆ u in the neigh b orho o d of u 1 = u 3 = ˆ u . F or u 1 whic h is a little bigger than ˆ u , we will show that there is a unique u 3 suc h that equation (65) holds. By (55), (66) and (67), w e hav e W ( ˆ u, ˜ u 3 ) > 0 for some ˜ u 3 < ˆ u . Hence, W ( u 1 , ˜ u 3 ) > 0 for u 1 a bit larger than ˆ u . F or each of such u 1 ’s, w e deduct from (55) and (65) again that W ( u 1 , ˆ u ) < 0 . By the mean v alue theorem, we show that, for eac h u 1 that is sligh tly larger than ˆ u , there exists a u 3 < ˆ u suc h that (65) holds. It is easy to chec k the uniqueness of u 3 . Therefore, (65) determines u 3 as a function of u 1 , u 3 = A ( u 1 ), for small non- p ositiv e u 1 with A ( ˆ u ) = ˆ u . The smo othness of the function f ( u ) and Lemma 14 imply that A ( u 1 ) is a smo oth decreasing function of u 1 . 21 Next, substituting u 3 = A ( u 1 ) in to (63), it is not hard to show that (63) determines u 1 as a function of t . W e hav e therefore pro ved the short time v ersion of the follo wing lemma. Lemma 15 Under the c onditions of The or em 13, e quations (63) and (64) have a unique solution ( u − 1 ( t ) , u − 2 ( t ) , u − 3 ( t ) ) with u − 2 ( t ) = u − 3 ( t ) for al l t ≥ 0. The solution has the pr op erty that u − 1 ( t ) > u − 2 ( t ) = u − 3 ( t ) for t > 0 and that u − 1 (0) = u − 2 (0) = u − 3 (0) = ˆ u . Pr o of. W e will no w extend the solution ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )) of equations (63) and (64) for all t > 0. Before doing this, w e need a lemma. Lemma 16 Under c onditions of The or em 13, the fol lowing hold: ∂ 2 q ∂ u 1 ∂ u 2 = ∂ 2 q ∂ u 1 ∂ u 3 < 0 , ∂ 2 q ∂ u 2 1 < 0 , t + ∂ q ∂ u 1 < 0 , and t + ∂ q ∂ u 2 = t + ∂ q ∂ u 3 > 0 , at the solution ( u 1 , u 3 , u 3 ) of (65) wher e u 1 > u 3 . Pr o of. By (48) and Lemma 14, ∂ 2 q ∂ u 1 ∂ u 3 − ∂ 2 q ∂ u 2 ∂ u 3 = ∂ ∂ u 3 [ ∂ q ∂ u 1 − ∂ q ∂ u 2 ] = 2( u 1 − u 2 ) ∂ 3 q ∂ u 1 ∂ u 2 ∂ u 3 < 0 , (68) whic h when com bined with (65) gives ∂ 2 q ∂ u 1 ∂ u 3 < 0 as long as 4( u 3 + ν ) − ( u 1 + ν ) > 0. This inequalit y holds even when 4( u 3 + ν ) − ( u 1 + ν ) ≤ 0. T o see this, supp ose the inequality fails at some p oin t, at whic h ∂ 2 u 2 u 3 q must v anish b ecause of (65). This would violate (68). The other inequalities of Lemma 16 are shown in the same wa y . 2 W e now calculate the partial deriv ativ es of U and V at the solution ( u 1 , u 3 , u 3 ) of (63) and (64), 22 ∂ U ∂ u 1 = t + ∂ q ∂ u 1 + ( u 1 + ν ) ∂ 2 q ∂ u 2 1 + 2( u 3 + ν ) ∂ 2 q ∂ u 1 ∂ u 3 < 0 , ∂ U ∂ u 3 = 2( t + ∂ q ∂ u 3 ) + 2( u 1 + ν ) ∂ 2 q ∂ u 1 ∂ u 3 + 8( u 3 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 = 1 u 1 − u 3 " ( u 1 + ν )( t + ∂ q ∂ u 1 ) + 2( u 3 + ν )( t + ∂ q ∂ u 3 ) # = 0 , ∂ V ∂ u 3 = t + ∂ q ∂ u 2 + [(8( u 3 + ν ) − 3( u 1 + ν )] ∂ 2 q ∂ u 2 ∂ u 3 − 2( u 1 − u 3 )( u 3 + ν )( ∂ 3 q ∂ u 2 2 ∂ u 3 + ∂ 3 q ∂ u 2 ∂ u 2 3 ) = 3( u 1 + ν ) 2 − 12( u 1 + ν )( u 3 + ν ) + 24( u 3 + ν ) 2 8( u 1 − u 3 )( u 3 + ν ) ( t + ∂ q ∂ u 2 ) − 2( u 1 − u 3 )( u 3 + ν )( ∂ 3 q ∂ u 2 2 ∂ u 3 + ∂ 3 q ∂ u 2 ∂ u 2 3 ) > 0 , where we hav e used Lemmas 14 and 16 to determine the signs of the deriv a- tiv es. These sho w that the Jacobian ∂ ( U, V ) ∂ ( u 1 , u 3 ) 6 = 0 on the solution ( u 1 , u 3 , u 3 ) of equations (63) and (64) where u 1 > u 3 . Therefore, by the Implicit F unction Theorem, equations (63) and (64) can b e solv ed for u − 1 ( t ), u − 2 ( t ) = u − 3 ( t ) for all t ≥ 0. F urthermore, it is easy to chec k that u − 1 ( t ) is an increasing function of t . 2 5.2 The le ading e dge A t the leading edge, u 1 = u 2 , i.e., s = 1, it follo ws from (A.7), (A.8), (58), and (59) that equations (60) turn out to b e t + ∂ q ∂ u 1 ( u 1 , u 1 , u 3 ) = 0 , t + ∂ q ∂ u 3 ( u 1 , u 1 , u 3 ) = 0 . In the same wa y as w e handle Lemma 15, w e can solv e the ab o v e equations for u 1 and u 3 as functions of t , leading to the lemma. Lemma 17 Under the c onditions of The or em 13, system (60) has a unique solution ( u + 1 ( t ) , u + 2 ( t ) , u + 3 ( t ) ) with u + 1 ( t ) = u + 2 ( t ) for al l t ≥ 0 . The solution has the pr op erty that u + 1 ( t ) = u + 2 ( t ) > u + 3 ( t ) for t > 0 and that u + 1 (0) = u + 2 (0) = u + 3 (0) = ˆ u . 23 5.3 Ne ar the tr ailing e dge By Lemma 15, ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )) satisfies equations (63) and (64). F or eac h fixed t > 0, we need to solve equations (60) for u 1 and u 3 as functions of u 2 in the neigh b orho od of u − 2 ( t ). This is carried out in Lemma 18 F or e ach t > 0, e quations (60) c an b e solve d for u 1 and u 3 in terms of u 2 in the neighb orho o d of ( u − 1 ( t ) , u − 2 ( t ) , u − 3 ( t ) )      u 1 = M ( u 2 ) u 3 = N ( u 2 ) (69) such that u − 1 ( t ) = M ( u − 2 ( t )) and u − 3 ( t ) = N ( u − 2 ( t )) , Mor e over, for u 2 > u − 2 ( t ) , N ( u 2 ) < u 2 < M ( u 2 ) . (70) Pr o of. Calculating the first partial deriv atives of F and G/s 2 of (61) and (62) at ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )), where u − 2 ( t ) = u − 3 ( t ), and using (48), w e find ∂ F ∂ u 1 = ∂ 2 q ∂ u 2 1 − 2( u 2 + ν ) ( u 1 + ν ) 2 ( t + ∂ q ∂ u 2 ) + 2( u 2 + ν ) u 1 + ν ∂ 2 q ∂ u 1 ∂ u 2 < 0 , ∂ F ∂ u 2 = ∂ 2 q ∂ u 1 ∂ u 2 + [ 2 u 1 + ν − 3 4 1 u 1 − u 3 ]( t + ∂ q ∂ u 2 ) + + 6( u 2 + ν ) u 1 + ν ∂ 2 q ∂ u 2 ∂ u 3 = 0 , ∂ F ∂ u 3 = ∂ 2 q ∂ u 1 ∂ u 3 + 3 4 1 u 1 − u 3 ( t + ∂ q ∂ u 2 ) + 2( u 2 + ν ) u 1 + ν ∂ 2 q ∂ u 2 ∂ u 3 = 0 , ∂ ( G/s 2 ) ∂ u 2 = π 2 ( 19( u 1 + ν ) − 16( u 3 + ν ) 32( u 1 − u 3 )( u 1 + ν ) ( t + ∂ q ∂ u 2 ) − u 1 − u 2 2( u 1 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 ) = 2( u 1 + ν ) 2 + 9( u 1 + ν )( u 3 + ν ) − 8( u 3 + ν ) 2 64( u 1 − u 3 )( u 1 + ν )( u 2 + ν ) π ( t + ∂ q ∂ u 2 ) > 0 , ∂ ( G/s 2 ) ∂ u 3 = π 2 ( [ − 3 16( u 1 − u 3 ) + 3( u 1 + ν ) 32( u 1 − u 3 )( u 2 + ν ) ]( t + ∂ q ∂ u 2 )+ + 3( u 2 + ν ) 2( u 1 + ν ) ∂ 2 q ∂ u 2 ∂ u 3 − ( u 1 − u 3 )( u 2 + ν ) u 1 + ν ∂ 3 q ∂ u 2 ∂ u 2 3 ) = π 2 ( 3[( u 1 + ν ) 2 − 4( u 1 + ν )( u 3 + ν ) + 8( u 3 + ν ) 2 ] 32( u 1 − u 3 )( u 1 + ν )( u 2 + ν ) ( t + ∂ q ∂ u 2 ) − ( u 1 − u 3 )( u 2 + ν ) u 1 + ν ∂ 3 q ∂ u 2 ∂ u 2 3 ) > 0 , 24 where we hav e used (63) and (64) to simplify the results, and Lemmas 14 and 16 to determine the signs of the deriv ativ es. These pro v e the non-v anishing of the Jacobian ∂ ( F , G/s 2 ) ∂ ( u 1 , u 3 ) at ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )). Hence, equations (60) can b e solved for      u 1 = M ( u 2 ) u 3 = N ( u 2 ) in a neigh b orhoo d of u − 2 ( t ) suc h that u − 1 ( t ) = M ( u − 2 ( t )) and u − 3 ( t ) = N ( u − 2 ( t )). F urthermore, N ( u 2 ) is a decreasing function of u 2 and so (70) holds. 2 5.4 The p assage fr om the tr ailing e dge to the le ading e dge W e shall sho w that, for each fixed t > 0, solutions (69) of equations (60) can b e further extended as long as N ( u 2 ) < u 2 < M ( u 2 ). The Jacobian of system (56) and (57) with resp ect to ( u 1 , u 3 ) has to b e estimated along the extension. Lemma 19 Under the c onditions of The or em 13, the fol lowing ine qualities hold for e ach t > 0 . ∂ ( λ 1 t + w 1 ) ∂ u 1 < 0 , ∂ ( λ 2 t + w 2 ) ∂ u 2 > 0 , ∂ ( λ 3 t + w 3 ) ∂ u 3 < 0 (71) on the solution ( u 1 , u 2 , u 3 ) of (56) and (57) (or e quivalently (60)) in the r e gion 0 < u 3 + ν < u 2 + ν < u 1 + ν . Pr o of. Using formulae (51) for w 1 , w 2 , and w 3 , we see that (56) and (57) are equiv alent to [ λ 1 − ( u 1 + u 2 + u 3 + 2 ν )]( t + ∂ q ∂ u 1 ) = [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )]( t + ∂ q ∂ u 2 ) , (72) [ λ 2 − 2( u 1 + u 2 + u 3 + 2 ν )]( t + ∂ q ∂ u 2 ) = [ λ 3 − ( u 1 + u 2 + u 3 + 2 ν )]( t + ∂ q ∂ u 3 ) . (73) By Lemma 16, ∂ 2 q ( u 1 , u 2 , u 3 ) ∂ u 1 ∂ u 2 < 0 , ∂ 2 q ( u 1 , u 2 , u 3 ) ∂ u 1 ∂ u 3 < 0 (74) 25 at the trailing edge. W e claim that inequalities (74) hold on the solution ( u 1 , u 2 , u 3 ) of (56) and (57) as long as 0 < u 3 + ν < u 2 + ν < u 1 + ν . W e justify the claim b y contradiction. Suppose otherwise, for instance at some p oin t ( ¯ u 1 , ¯ u 2 , ¯ u 3 ) on the solution of (56) and (57), with 0 < ¯ u 3 + ν < ¯ u 2 + ν < ¯ u 1 + ν , ∂ 2 q ∂ u 1 ∂ u 2 = 0 . In view of (48), this giv es ∂ q ∂ u 1 = ∂ q ∂ u 2 at ( ¯ u 1 , ¯ u 2 , ¯ u 3 ) , whic h together with (19), (20), and (72) imply t + 1 2 ∂ q ∂ u 1 = t + 1 2 ∂ q ∂ u 2 = 0 . (75) By (20), (21), (73), and (75), we obtain t + ∂ q ∂ u 3 = 0 , whic h together with (48) giv es ∂ 2 q ∂ u 1 ∂ u 2 = ∂ 2 q ∂ u 1 ∂ u 3 = 0 (76) at ( ¯ u 1 , ¯ u 2 , ¯ u 3 ). On the other hand, b y (48) and Lemma 14, ∂ 2 q ∂ u 1 ∂ u 2 − ∂ 2 q ∂ u 1 ∂ u 3 = 2( u 2 − u 3 ) ∂ 3 q ∂ u 1 ∂ u 2 ∂ u 3 < 0 at ( ¯ u 1 , ¯ u 2 , ¯ u 3 ). This con tradicts (76) and the claim has b een justified. By (48), w e ha ve 2( u 1 − u 3 ) ∂ 2 q ∂ u 1 ∂ u 3 = ∂ q ∂ u 1 − ∂ q ∂ u 3 . Differen tiating this with resp ect to u 1 yields ∂ 2 q ∂ u 2 1 = 3 ∂ 2 q ∂ u 1 ∂ u 3 + 2( u 1 − u 3 ) ∂ 3 q ∂ u 2 1 ∂ u 3 < 0 , (77) where w e ha ve used (74) and Lemma 14 in the last step. 26 It follo ws from (74) and (48) that ∂ q ∂ u 1 < ∂ q ∂ u 2 , ∂ q ∂ u 1 < ∂ q ∂ u 3 , whic h when com bined with (19), (20), (21), (72) and (73) gives t + 1 2 ∂ q ∂ u 1 < 0 , t + 1 2 ∂ q ∂ u 2 > 0 , t + 1 2 ∂ q ∂ u 3 > 0 (78) on the solution ( u 1 , u 2 , u 3 ) of (56) and (57) in the region 0 < u 3 + ν < u 2 + ν < u 1 + ν . Therefore, b y (51), ∂ ( λ 1 t + w 1 ) ∂ u 1 = ∂ λ 1 ∂ u 1 ( t + ∂ q ∂ u 1 ) + [ λ 1 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 1 < 0 , where in the last inequalit y w e hav e used (19), (77), (78), and ∂ λ 1 ∂ u 1 = I ∂ 2 u i u i I ( ∂ ui I ) 2 > 0 . This pro v es the first inequality of (71). Next w e shall pro ve the rest of Lemma 19. By (48), w e hav e 2( u 2 − u 3 ) ∂ 2 q ∂ u 2 ∂ u 3 = ∂ q ∂ u 2 − ∂ q ∂ u 3 . Differen tiating this with resp ect to u 2 yields ∂ 2 q ∂ u 2 2 = 3 ∂ 2 q ∂ u 2 ∂ u 3 + 2( u 2 − u 3 ) ∂ 3 q ∂ u 2 2 ∂ u 3 . (79) Using (48) to rewrite (73), w e obtain ( λ 2 − λ 3 )[ t + ∂ q ∂ u 3 ] + 2[ λ 3 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 ∂ u 3 ( u 2 − u 3 ) = 0 (80) whic h together with (79) giv es 3 λ 2 − λ 3 u 2 − u 3 ( t + ∂ q ∂ u 3 ) + 2[ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 2 = 4[ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )]( u 2 − u 3 ) ∂ 3 q ∂ u 2 2 ∂ u 3 > 0 , (81) 27 where w e ha ve used (21) and Lemma 14 in the last inequalit y . It follo ws from (51) that ∂ ( λ 2 t + w 2 ) ∂ u 2 = ∂ λ 2 ∂ u 2 ( t + ∂ q ∂ u 2 ) + [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 2 > 3 2 λ 2 − λ 3 u 2 − u 3 ( t + ∂ q ∂ u 2 ) + [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 2 = 3 2 λ 2 − λ 3 u 2 − u 3 ( t + ∂ q ∂ u 3 ) + [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 2 + 3( λ 2 − λ 3 ) ∂ 2 q ∂ u 2 ∂ u 3 > 3( λ 2 − λ 3 ) ∂ 2 q ∂ u 2 ∂ u 3 ≥ 0 , where we hav e used Lemma 4, (78) in the first inequalit y , and (81) in the second one. The last inequality is due to the fact that λ 2 − λ 3 and ∂ 2 u 2 u 3 q ha ve the same sign b ecause of (80). This prov es the second inequality of (71). In the same wa y , we can prov e the last one. 2 W e are ready to conclude the pro of of Theorem 13. Pro of of Theorem 13: By Lemma 18, equations (60) can b e solv ed for      u 1 = M ( u 2 ) u 3 = N ( u 2 ) in the neigh b orho od of ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )). F urthermore, (70) holds if u 2 > u − 2 ( t ). W e shall extend the solution in the p ositive u 2 direction as far as p ossible. It follo ws from Corollary 2 and Lemma 19 that, along the extension of (69) in the region u 1 + ν > u 2 + ν > u 3 + ν > 0, the Jacobian matrix of (56) and (57) is diagonal and therefore is nonsingular. F urthermore, equations (56) and (57) determines (69) as tw o decreasing functions of u 2 . This immediately guarantees that (69) can b e extended as far as necessary in the region u 1 + ν > u 2 + ν > u 3 + ν > 0. Since M ( u 2 ) is decreasing, (69) stops at some p oin t u + 2 ( t ) where, ob viously , M ( u + 2 ( t )) = u + 2 ( t ). Therefore, w e hav e sho wn that (56) and (57) determine u 1 and u 3 as decreasing functions of u 2 o ver the in terv al [ u − 2 ( t ) , u + 2 ( t )]. 28 Let      u + 1 ( t ) = M ( u + 2 ( t )) u + 3 ( t ) = N ( u + 2 ( t )) . Clearly , ( u + 1 ( t ) , u + 2 ( t ) , u + 3 ( t )) solves system (60) at the leading edge u 1 = u 2 . Hence, these u + 1 ( t ), u + 2 ( t ) and u + 3 ( t ) are exactly the ones app earing in Lemma 17. Substituting (69) in to (44), w e obtain x = λ 2 ( M ( u 2 ) , u 2 , N ( u 2 )) t + w 2 ( M ( u 2 ) , u 2 , N ( u 2 )) whic h by Corollary 2 and Lemma 19 clearly determines x as an increasing function of u 2 o ver interv al [ u − 2 ( t ) , u + 2 ( t )]. It follo ws that, for each fixed t > 0, u 2 is a function of x ov er the in terv al [ x − ( t ), x + ( t )], and that therefore so are u 1 and u 3 , where x ± ( t ) = λ 2 ( u ± 1 ( t ) , u ± 2 ( t ) , u ± 3 ( t )) t + w 2 ( u ± 1 ( t ) , u ± 2 ( t ) , u ± 2 ( t )) . (82) Th us, (44) can b e solved for u 1 = u 1 ( x, t ) , u 2 = u 2 ( x, t ) , u 3 = u 3 ( x, t ) within a w edge x − ( t ) < x < x + ( t ) f or t > 0 , x − (0) = x + (0) = 0 , (83) where w e ha ve used (82), Lemma 15, and Lemma 17 in the last equations. Boundary conditions (9) and (10) can b e c hec k ed easily . The pro of of Theorem 13 would b e completed if we can v erify that the wedge is indeed a cusp. First w e need a lemma. Lemma 20 A t ( u − 1 ( t ) , u − 2 ( t ) , u − 3 ( t )) , ∂ ( λ 2 t + w 2 ) ∂ u 2 = ∂ ( λ 3 t + w 3 ) ∂ u 3 = 0 , (84) while at ( u + 1 ( t ) , u + 2 ( t ) , u + 3 ( t )) , ∂ ( λ 1 t + w 1 ) ∂ u 1 = ∂ ( λ 2 t + w 2 ) ∂ u 2 = 0 . (85) Pr o of. 29 Using expansions (A.5) and (A.6), w e obtain from (25) that ∂ λ 2 ∂ u 2 = 6[( u 3 + ν ) − 1 4 ( u 1 + ν )] u 1 + ν at ( u − 1 ( t ), u − 2 ( t ), u − 3 ( t )). By (51), w e ha ve ∂ ( λ 2 t + w 2 ) ∂ u 2 = 6[( u 3 + ν ) − 1 4 ( u 1 + ν )] ( u 1 + ν ) ( t + ∂ q ∂ u 2 ) − 4( u 1 − u 3 )( u 3 + ν ) u 1 + ν ∂ 2 q ∂ u 2 2 = 0 , where we ha v e used (22) in the first equalit y , and (64) and (79) in the last equalit y . The pro of also applies to the other equation of (84). T o prov e (85), we pro ceed as follo ws. By (18), (A.7), and (A.8), we find ∂ λ 2 ∂ u 1 = O ( K ( s )) as s → 1 . (86) By Corollary 2 and formula (51), w e calculate the deriv ativ e on the solution of (56) and (57) 0 = ∂ ( λ 2 t + w 2 ) ∂ u 1 = ∂ λ 2 ∂ u 1 ( t + ∂ q ∂ u 2 ) + 1 2 [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 1 ∂ u 2 + ∂ q ∂ u 1 − ∂ q ∂ u 2 , whic h when com bined with (23) and (86) gives lim s → 1 [ t + ∂ q ∂ u 2 ] K ( s ) = 0 . (87) On the other hand, as in (86), we hav e ∂ λ 2 ∂ u 2 = O ( K ( s )) as s → 1 . (88) Therefore, b y (51) w e see that at ( u + 1 , u + 2 , u + 3 ) ∂ ( λ 2 t + w 2 ) ∂ u 2 = ∂ λ 2 ∂ u 2 ( t + 1 2 ∂ q ∂ u 2 ) + [ λ 2 − ( u 1 + u 2 + u 3 + 2 ν )] ∂ 2 q ∂ u 2 2 = 0 , 30 where, by (87) and (88), the first term v anishes, while the second term v anishes b ecause of (23). This prov es the second equality of (85). In the same w a y , we can c hec k the other equality of (85). 2 No w we contin ue to finish the pro of of Theorem 13. Differentiating (82) with resp ect to t , by Corollary 2 and Lemma 20 we obtain dx + ( t ) dt = ∂ ( λ 2 t + w 2 ) ∂ u 2 du ± 2 dt + λ 2 ( u ± 1 , u ± 2 , u ± 3 ) = λ 2 ( u ± 1 , u ± 2 , u ± 3 ) , whic h when com bined with (22), (23), Lemma 15, and Lemma 17 gives dx ± ( t ) dt = 3 ˆ u + 2 ν at t = 0 . Therefore, w edge (83) is a cusp. This completes the pro of of Theorem 13. W e immediately conclude from Theorem 1 and Theorem 13 the following result on the initial v alue problem of the Whitham equations. Theorem 21 F or a de cr e asing initial function u 0 ( x ) whose inverse function f ( u ) satisfying the c onditions of The or em 13, the Whitham e quations (3) have a solution ( u 1 ( x, t ) , u 2 ( x, t ) , u 3 ( x, t )) within a cusp for al l p ositive time. The Bur gers solution of (2) exists outside the cusp. The Whitham solution matches the Bur gers solution on the b oundary of the cusp in the fashion of (9) and (10). W e close this pap er with t wo observ ations. First, it is ob vious from the pro of of Theorem 21 that one should expect lo cal (in time) results if local conditions are assumed. Namely , if the global condition f 000 ( u ) < 0 for all b < u < a in Theorem 21 is replaced b y a local condition f 000 ( u ) < 0 in the neighborho o d of the breaking p oin t ˆ u , the results of Theorem 21 are only true for a short time after the breaking time. Second, a hump-lik e initial function can be decomp osed in to a decreasing and an increasing parts. It is kno wn that the decreasing part causes the Burgers solution of (2) to develop finite time singularities while the increasing part do es not. These t w o pieces of data w ould not in teract with eac h other for a short time after the breaking of the Burgers solution. As a consequence, a short time result also holds for a hump-lik e initial function. 31 A Complete Elliptic In tegrals In this Appendix, w e list some of the w ell-known prop erties of the complete elliptic in tegrals of the first, second and third kind. These are the deriv ative form ulae dK ( s ) ds = E ( s ) − (1 − s ) K ( s ) 2 s (1 − s ) , (A.1) dE ( s ) ds = E ( s ) − K ( s ) 2 s , (A.2) d Π( ρ, s ) dρ = ρE ( s ) + ( s − ρ ) K ( s ) + ( ρ 2 − s )Π( ρ, s ) 2 ρ (1 − ρ )( ρ − s ) , (A.3) d Π( ρ, s ) ds = E ( s ) − (1 − s )Π( ρ, s ) 2(1 − s )( s − ρ ) . (A.4) K ( s ) and E ( s ) ha ve the expansions K ( s ) = π 2 [1 + s 4 + 9 64 s 2 + · · · + ( 1 · 3 · · · (2 n − 1) 2 · 4 · · · 2 n ) 2 s n + · · · ] , (A.5) E ( s ) = π 2 [1 − s 4 − 3 64 s 2 − · · · − 1 2 n − 1 ( 1 · 3 · · · (2 n − 1) 2 · 4 · · · 2 n ) 2 s n − · · · ] (A.6) for | s | < 1. They also ha ve the asymptotics K ( s ) ≈ 1 2 log 16 1 − s , (A.7) E ( s ) ≈ 1 + 1 4 (1 − s )[log 16 1 − s − 1] (A.8) when s is close to 1. 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