emmanuel yomba- construction of new solutions to the fully nonlinear generalized camassa-holm...

Construction of new solutions to the fullynonlinear generalized Camassa-Holm equations

by an indirect F function method

Emmanuel Yombaa,b

a Institute for Mathematics and its applications, University of Minnesota,

400 Lind Hall 207 Church Street S.E. Minneapolis, MN 554556-0436 U.S.A.

b Department of Physics, Faculty of Sciences, University of Ngaoundere PO.

BOX 454 Ngaoundere Cameroon.

ABSTRACT

An indirect F function method is introduced to solve the generalized

Camassa-Holm equation with fully nonlinear dispersion and fully nonlinear

convection C(l, n, p). Taking advantage of elliptic equation, this F function is

used to map the solutions of the generalized Camassa-Holm equation to those

of the elliptic equation. As a result, we can successfully obtain in a unified

way and for special values of the parameters of this equation, many new exact

solutions expressed by various single and combined non-degenerative Jacobi

elliptic function solutions and their degenerative solutions (soliton, combined

soliton solutions and triangular solutions) as the modulus m is driven to 1

and 0.

1

1 INTRODUCTION

It is well known that the investigation of the travelling wave solutions of

nonlinear evolution equations (NLEEs), which is an important tool in char-

acterizing many complicated phenomena and dynamical processes in physics,

mechanics, chemistry, biology, etc., plays an essential role in the study of

these physical problems. Hence, seeking for solutions of the NLEEs may

enable help physicists and engineers to better understand the mechanism

that governs these physical problems. Especially, finding explicit solutions

to nonlinear models has become one of the most exciting and extremely ac-

tive areas of research investigation, since they may provide better knowledge

to the physical problems and possible physical applications.

Recently, both mathematicians and physicists have made many attempts

in this direction. A number of works have been done on the construc-

tion of exact solutions of NLEEs and effective and powerful methods have

been developed, such as inverse scattering method [1], Backlund transforma-

tion [2], Darboux transformation [3,4], Hirota bilinear method [5], homoge-

neous balance method [6], Jacobi elliptic function method [7], tanh-function

method [8,9], extended tanh-function method [10-13], improved extended

tanh-function method [14-18], the sine-cosine function method [19].

Very recently, a unified F-expansion method [20-27] has been established

to obtain Jacobi elliptic functions, solitons and periodic functions to a large

variety of NLEEs whose odd-and even-order derivative terms do not coexist.

The main idea of this method is to take full advantage of the elliptic equation

which has more new solutions, to construct exact solutions to the NLEEs.

2

Thus multiple exact solutions can be obtained in a unified way and much

tedious and repeated calculation can be avoided.

In the present paper, a generalized Camassa-Holm equation with fully

nonlinear dispersion and fully nonlinear convection term C(l, n, p) [28] is

considered

Ut + kUx + β1Uxxt + β2(Ul)x + β3Ux(U

n)xx + β4U(Up)xxx = 0. (1)

where k, β1, β2, β3, and β4 are arbitrary real constants.

Eq.(1) is a class of physically important equation. In fact, if one takes

β1 = −1, β2 = 3/2, β3 = −2, β4 = −1, l = 2, n = p = 1, (1) becomes

the new shallow water equation, namely Camassa-Holm equation

Ut + kUx − Uxxt + 3UUx = 2UxUxx + UUxxx, (2)

which has been proposed by Camassa-Holm [29]. They used Hamiltonian

methods to derive the new completely integrable dispersive wave equation for

water by retaining two terms that are usually neglected in small amplitude

shallow water limit. They showed that for all k, Eq.(2) is integrable, and

for k=0, (2) has travelling solutions, which are called peakons. When β1 =

−1, β2 = a, β3 = −2, β4 = −1, l = L + 1, n = p = 1, (1) becomes

another form of generalized Camassa-Holm equation,

Ut + kUx − Uxxt + aULUx = 2UxUxx + UUxxx, (3)

3

studied by Tian and Song [30]. They derived some new exact peaked

solitary wave solutions. When k = 0, β1 = −1, β2 = 3/2, β3 = −2γ, β4 =

−γ, l = 2, n = p = 1, (1) becomes

Ut − Uxxt + 3UUx = γ(2UxUxx + UUxxx), (4)

which has been derived by Dai and Huo [31] when they studied distur-

bances in an initially stretched or compressed rod which is composed of a

compressible Mooney-Rivlin material. By using a non-dimensionalization

process and the reductive perturbation technique, they obtained a new type

of nonlinear dispersive Eq. (4). They also showed that Eq.(4) has a vari-

ety of travelling waves including solitary shock waves, solitary waves, peri-

odic shock waves etc. Liu and Chen [32] showed that Eq.(4) also gener-

ated compactons structures by using bifurcation method of planar dynami-

cal systems and numerical simulation of differential equations. When k = 0,

β1 = −1, β2 = 3/2, β3 = β4 = 0, l = 2, n = p = 1, Eq.(1) becomes the

BBM equation [33] a well known model for surface wave in channel.

More recently, by using four direct ansatze, Tian and Yin [28] obtained

abundant solutions: compactons (solutions with absence of infinite wings),

solitary patterns solutions having infinite slopes or cups, solitary waves and

singular periodic wave solutions and obtained kink compacton solutions and

non-symmetry compacton solutions. In the same paper, they also studied

other form of fully nonlinear generalized Camassa-Holm equation and showed

that their compacton solutions are governed by linear equations.

The present paper is motivated by the desire to extend the Tian and Yin

4

work [28] to make further progress. More importantly, it is the objective

of this work to show that abundant families of Jacobi, combined Jacobi

elliptic functions, solitary wave and combined solitary wave solutions and

triangular functions arise from Eq.(1). To achieve our goal, instead of taking

specific functions as Tian and Yin did in [28], we will use a form of indirect

F function method (not the F-expansion method because the complexity of

Eq.(1) can not allow the use of this method). But our indirect F method

will be very close to the F-expansion method in the sense that, the indirect

F function method will also take advantage of nonlinear ordinary differential

equation (ODE). Thus, one will only need to calculate the function which

is a solution of the ODE, instead of calculating the Jacobi elliptic function

one by one; secondly, the coefficients of the ODE can be selected so that the

corresponding solution is a Jacobi elliptic function, exactly as in F -expansion

method.

The paper is organized as follows: In section 2, a derivation of the pro-

posed method for finding exact solutions is presented. Finally, some discus-

sions and conclusions are given in section 3.

2 General formulas of the solutions

We firstly make the following formal travelling wave transformation

U(x, t) = U(ξ), ξ = λ1x+λ2t, (5)

where λi, (i=1,2) are undetermined constants Substituting (5) into Eq.(1),

5

we have the ODE for U(ξ).

λ2 U′+kλ1U

′+β1λ2λ

21U

′′′+β2λ1(U

l)′+β3λ

31U′(Un)

′′+β4λ

31U(Up)

′′′= 0. (6)

Then a suitable ansatz for the unknown U(ξ) will solve Eq.(6).

Let us assume that Eq.(6) has the solution in the form

U(ξ) = A(F (ξ))B, (7)

where A and B are parameters to be determined later. We also propose that

the function F should be mapped to the solutions of the following elliptic

equation

F′2 = r + aF 2 +

b

2F 4, (8)

r, a and b are constants. Substituting the ansatz (7) along with Eq.(8) into

Eq.(6), collecting coefficients of power of F with the aid of Mathematica, we

can deduce the following polynomial equation

2ABrβ1λ2λ21[2−3B+B2]FB+ 2AB[kλ1+λ2+aB

2β1λ2λ21]F

B+2+ABbβ1λ2λ21[2+

3B+B2]FB+4+[2lBAlβ2λ] F 2+Bl+2A1+nB2nrβ3λ31[nB−1]FB(n+1)+[2aA1+nB3n2β3λ

31]

F 2+B(n+1) + A1+nB2nbβ3λ31[nB + 1]F 4+B(n+1) + 2A1+pBprβ4λ

31[2 − 3Bp +

B2p2]FB(1+p) +[2aA1+pB3p3β4λ31] F

2+B(p+1)+

A1+pBpbβ4λ31[2 + 3Bp+B2p2]F 4+B(1+p) = 0 (9)

One may easily see that in the specific cases, where r = 1, a = −1 and

b = 0 which lead to F = cos ξ, when r = 1, a = −1 and b = 0 which lead

to F = sin ξ, when r = −1, a = 1 and b = 0 which lead to F = cosh ξ and

6

when r = 1, a = 1 and b = 0 which lead to F = sinh ξ, then we are in the

situation of ansatz 1, ansatz 2, Ansatz 3 and ansatz 4 in [28] respectively.

We are interested on the general case where r × a× b 6= 0.

In view of the study of Eq.(9), we may observe carefully the different

powers of F intervening in this equation. Then, it appears that it is power

2 + Bl, 2 + B(n + 1) and 2 + B(1 + p) whose coefficients are singles which

may determine our discussion.

Thus, from Eq.(9), we get the follows possible cases to be discussed.

Case 1 2 +Bl = 2 +B(n+ 1) = 2 +B(1 + p), we have l = n+ 1, p = n,

Case 2: 2+Bl = 2+B, 2+B(n+1) = 2+B(1+p), we have l = 1, p = n,

Case 3:

2 +Bl = 2 +B(1 + p), 2 +B(n+ 1) = 2 +B, we have l = 1 + p, n = 0,

Case 4:

2 +Bl = 2 +B(1 + n), 2 +B(p+ 1) = 2 +B, we have l = 1 + n, p = 0.

But the cases 3.and 4 produce no solution.

From the case 1, choosing l = 1+n and p = n, substituting in Eq.(9) and

collecting all terms with same power of F , we obtain the following equation

2ABrβ1λ2λ21[2−3B+B2]FB+2AB[kλ1+λ2+aB

2β1λ2λ21]F

B+2+ABbβ1λ2λ21[2+

3B+B2]FB+4 + 2A1+nBnrλ31[β3B(nB− 1) + β4(2− 3Bn+B2n2)]FB(n+1) +

2A1+nB[(n+1)β2λ1 +aB2n2λ31(β3 +nβ4)] F

2+B(n+1) +A1+nBnbλ31[Bβ3(Bn+

1) + β4(2 + 3Bn+B2n2)]F 4+B(1+n) = 0 (10)

7

From (10) it appears that when setting to zero the coefficient of FB i.e.

2ABrβ1λ2λ21[2 − 3B + B2] = 0 which leads to [2 − 3B + B2] = 0 (B = 1

or B = 2) then the coefficient of power FB+4 should be different of zero

i.e. ABbβ1λ2λ21[2 + 3B + B2] 6= 0. Thus the power of FB+4 should be

shifted to another power, this may help the coefficient of this power to enter

another relationship with other terms. Thus the following two subcases may

be satisfied

First subcase B+4 = B(n+1) ↗when B = 1, =⇒ n = 4; l = 5, p = 4

↘when B = 2, =⇒ n = 2; l = 3, p = 2

Substituting the first type of relation of the first subcase in Eq.(10) and

solving the set of given system of equations we obtain the following results

λ1 =

√52

4

√β2

aβ4, λ2 = − kλ1

1+aβ1λ21, A = ±1

2

(bβ1λ2

rβ4λ1

) 14, β3 = −6β4. (11)

a, b and r are arbitrary constants.

Substituting the second type of relation of the first subcase in Eq.(10) and


λ1 =√

34

√β2

aβ4, λ2 = − kλ1

1+4aβ1λ21, A2 = −

(kbβ1

r(3β1β2+4β4)

),

β3 = −3β4. (12)

8


second subcase

B + 4 = 2 +B(n+ 1)↗when B = 1, =⇒ n = 2; l = 3, p = 2

↘when B = 2, =⇒ n = 1; l = 2, p = 1

Substituting the first type of relation of the second subcase in Eq.(10) and


λ2 = −A2(3β2−8aβ4λ21)

3bβ1λ1, A2 = λ2

1

(3bkβ1

3β2+λ21[a(3β1β2−8β4)+8β1β4λ

21(3br−a2)]

),

β3 = −4β4 (13)

λ1, a, b and r are arbitrary constants.

Substituting the second type of relation of the second subcase in Eq.(10) and


λ2 = −λ1(4arβ4λ21−k)

1+4aβ1λ21

, A = 3bkβ1λ21

β2+λ21[2a(2β1β2−β4)+4β1β2λ

21(3br−2a2)]

,

β3 = −2β4 (14)

λ1, a, b and r are arbitrary constants

Now, if we put the coefficient of FB+4 to zero i.e. [2 + 3B + B2] = 0

(B = −1 or B = −2) then the coefficient 2ABbβ1λ2λ21[2 − 3B + B2] of

9

power FB+4 should be different of zero, in order to allow this coefficient to

enter new relationship with other coefficients of the other power of F , we

have the following two subcases

third subcase

B = 2 +B(n+ 1)↗when B = −1, =⇒ n = 2; l = 3, p = 2

↘when B = −2, =⇒ n = 1; l = 2, p = 1

Substituting the first type of relation of the third subcase in Eq.(10) and

solving the set of given system of equation we obtain the following results

λ2 = − kλ1(8aβ4λ21−3β2)

3β2+aλ21(3β1β2−8β4)+8β1β4(3br−a2)

, A2 = 6krβ1λ21

3β2+aλ21(3β1β2−8β4)+8β1β4(3br−a2)

,

β3 = −4β4. (15)


Substituting the second type of relation of the third subcase in Eq.(10) and


λ2 = − kλ1(2aβ4λ21−β2)

12brβ1β4λ41−(1+4aβ1λ

21)(2aβ4λ

21−β2)

, A =(

6krβ1λ21

12brβ1β4λ41−(1+4aβ1λ

21)(2aβ4λ

21−β2)

),

β3 = −2β4. (16)


fourth subcase

10

B = 4 +B(n+ 1)↗when B = −1, =⇒ n = 4; l = 5, p = 4

↘when B = −2, =⇒ n = 2; l = 3, p = 2

Substituting the first type of relation of the fourth subcase in Eq.(10) and


λ2 = − kλ1

1+aβ1λ21, A = ±

(−rkβ1

4bβ4(1+aβ1λ21)

) 14,

β3 = −6β4 (17)

where λ1, satisfies the following relation

β2 + a(5β1β2 − 32β4)λ21 − 32a2β1β4λ

41 = 0, (18)


Substituting the second type of relation of the fourth subcase in Eq.(10) and


λ2 = − kλ1

1+4aβ1λ21, A2 = − rkβ1

bβ4(1+aβ1λ21),

β3 = −3β4, (19)

where λ1, should satisfy the following relation

3β2 + 4a(3β1β2 − 4β4)λ21 − 64a2β1β4λ

41 = 0, (20)

11


From the case 2, taking l = 1 and p = n, substituting these relations in

Eq.(9) and collecting all terms with same power of F , we obtain the following

equation

2ABrβ1λ2λ21[2−3B+B2]FB+2AB [β2λ1 +kλ1 +λ2 +aB2β1λ2λ

21]F

B+2 +

ABbβ1λ2λ21[2 + 3B +B2]FB+4 + 2A1+nBnrλ3

1[β3B(nB − 1) + β4(2− 3Bn+

B2n2)]FB(n+1) +2aA1+nB3n2λ31[β3 +nβ4] F

2+B(n+1) +A1+nBnbλ31[Bβ3(Bn+

1) +β4(2 + 3Bn+B2n2)]F 4+B(1+n) = 0 (21)

Proceeding in the same manner as above, for [2− 3B + B2] = 0 (B = 1

or B = 2) then the coefficient of power FB+4 which should be different of

zero, may enter into another relation if only we have the following relations

First subcase B+ 4 = B(n+ 1) ↗when B = 1, =⇒ n = 4; l = 1, p = 4

↘when B = 2, =⇒ n = 2; l = 1, p = 2

Substituting the first type of relation of the first subcase in Eq.(21) and

solving the set of given system of equations we have found that there is no

solution for this first subcase.

second subcase B+4 = 2+B(n+1) ↗when B = 1, =⇒ n = 2; l = 1, p = 2

↘when B = 2, =⇒ n = 1; l = 1, p = 1

12

Substituting the first type of relation of the second subcase in Eq.(21) and


λ2 = 8aA2β4λ1

3bβ1, A2 = −

(3bβ1(k+β2)

8β2[a−(3br−a2)β1λ21]

),

β3 = −4β4 (22)


Substituting the second type of relation of the second subcase in Eq.(21) and


λ2 = 2aAβ4λ1

3bβ1, A = −

(3bβ1(k+β2)

2β2[a−2(3br−2a2)β1λ21]

),

β3 = −2β4 (23)


If now, it’s [2+3B+B2] = 0 (B = −1 or B = −2) then it’s the coefficient of

power FB which should be different of zero, may enter into another relation

if only we have the following relations

third subcase B=2+B(n+1) ↗when B = −1, =⇒ n = 2; l = 1, p = 2

↘when B = −2, =⇒ n = 1; l = 1, p = 1

Substituting the first type of relation of the third subcase in Eq.(21) and

solving the set of given system of equations we have found the following

13

results

λ2 = − aλ1(k+β2)

a−(3br−a2)β1λ21, A2 =

(3rβ1λ2

4aβ4λ1

),

β3 = −4β4 (24)


Substituting the second type of relation of the third subcase in Eq.(21) and


λ2 = −(

aλ1(k+β2)

a−2(3br−2a2)β1λ21

), A = 3rβ4λ2

aβ4λ1,

β3 = −2β4 (25)


fourth subcase

B = 4 +B(n+ 1)↗when B = −1, =⇒ n = 4; l = 1, p = 4

↘when B = −2, =⇒ n = 2; l = 1, p = 2

Substituting the first type of relation of the second subcase in Eq.(21)

and solving the set of given system of equations we have found that there is

no solution for this subcase.

Since in all this cases, r, a and b are arbitrary constants, we may choose

them properly such that the corresponding solution F of the ODE (8) is one

14

of the Jacobi elliptic, combined Jacobi elliptic functions.

If r = 1, a = −(1 +m2), b = 2m2, then the solution is

U1 = A[sn(ξ | m)]B, (26.a)

or

U2 = A[cd(ξ | m)]B ≡ A

[cn(ξ | m)

dn(ξ | m)

]B, (26.b)

where 0 ≤ m ≤ 1, is called modulus of Jacobi elliptic sine functions, and

sn(ξ | m) is Jacobi elliptic functions see [34,35].

If r = 1−m2, a = 2m2 − 1, b = −2m2, then the solution is

U3 = A[cn(ξ | m)]B, (27)

where cn(ξ | m) is Jacobi elliptic cosine functions see [34,35].

If r = m2 − 1, a = 2−m2, b = −2, then the solution is

U4 = A[dn(ξ | m)]B, (28)

where dn(ξ | m) is Jacobi elliptic function of third kind see [34,35].

If r = m2, a = −(1 +m2), b = 2, then the solution is

15

If r = 1/4, a = (1− 2m2)/2, b = 1/2, then the solution is

U13 = A[ns(ξ | m)± cs(ξ | m)]B. (36)

If r = (1−m2)/4, a = (1 +m2)/2, b = (1−m2)/2, then the solution is

U14 = A[nc(ξ | m)± sc(ξ | m)]B. (37)

If r = m4/4, a = (m2 − 2)/2, b = 1/2, then the solution is

U15 = A[ns(ξ | m) + ds(ξ | m)]B. (38)

If r = m2/4, a = (m2 − 2)/2, b = m2/2, i2 = −1 then the solution is

U16 = A[sn(ξ | m)± icn(ξ | m)]B. (39)

In addition, we see that other solutions are obtained in case of degeneracies:

when m −→ 0, the Jacobi elliptic and combined Jacobi elliptic functions

degenerate to the trigonometric functions of the given NLPDE (1) i.e.

sn(ξ | m) −→ sin(ξ), cn(ξ | m) −→ cos(ξ), dn(ξ | m) −→ 1, ns(ξ |

m) −→ csc(ξ | m), cs(ξ | m) −→ cot(ξ), ds(ξ | m) −→ sec(ξ),

when m −→ 1, the Jacobi elliptic and combined Jacobi elliptic functions

degenerate to the soliton and combined soliton wave solutions of the given

NLPDE (1) i.e.

sn(ξ | m) −→ tanh(ξ), cn(ξ | m) −→ sech(ξ), dn(ξ | m) −→ sech(ξ), ns(ξ |

m) −→ coth(ξ), cs(ξ | m) −→ csch(ξ), ds(ξ | m) −→ csch(ξ).

17

So we can derive solutions expressed in terms of trigonometric functions

and hyperbolic functions. We omit them for simplicity.

It is worth noticing that the Jacobi transformation dn(ξ | m) = cn(√mξ |

m−1) [36] implies that any solution found by the dn-function may be trans-

formed into an equivalent one that can be obtained by cn-function. Moreover,

since other Jacobi elliptic function and combined Jacobi elliptic function so-

lutions obtained here for the point of view of mathematics as solutions of

Eq.(8) have singularities, and it’s well known for the point of view of physics

that the singular solutions cannot have any meaning in the applications, we

may focus only on non singular solutions among the listed mathematical so-

lutions. When B > 0 (i.e. B = 1 or B = 2) non singular solutions are

given by (26.a), (27) and (28). When B < 0 (i.e. B = −1 or B = −2) non

singular solutions are selected to be (29.a), (30) and (31). Other interesting

fact here is that by selecting B = −1 (or B = −2), the solutions (29.a), (30)

and (31) are reduced to the same family of solutions as (26.a), (27) and (28)

respectively when B = 1 (or B = 2).

Taking into account the above mentioned remarks, we have plotted for B = 1

and B = 2, some figures to illustrate our study or to raise the physics value

of our study.

18

Figure 1: The plot of sn−, tanh−, cn− and sech− solutions respectively.

These non singular solutions are the structure graphes of Eq.(11) that is

Ut+kUx+β1Uxxt+β2(U5)x = β4(6Ux(U

4)xx−U(U4)xxx), the two first graphes

are obtained for the values β1 = −1.00, β2 = 1.50, β4 = −0.75, k = −2.00,

meanwhile the last two graphes have the same parameter values except that

β4 = 0.75.

Figure 2: The plot of sn2−, tanh2−, cn2− and sech2− solutions respectively.


Ut+kUx+β1Uxxt+β2(U3)x = β4(3Ux(U

2)xx−U(U2)xxx), the two first graphes

are obtained for the values β1 = −1.00, β2 = 1.50, β4 = −0.75, k = −2.00,

meanwhile the last two graphes have the same parameter values except that

β4 = 0.75, k = 2.00.



Ut + kUx + β1Uxxt + β2(U3)x = β4(4Ux(U

2)xx − U(U2)xxx), the two first

graphes are obtained for the parameter values β1 = −1.00, β2 = 1.5, β4 =

−0.75, k = −2.00, λ1 = 0.65, meanwhile the last two graphes have the same

parameter values but for β4 = 0.75, k = 2.00

19



Ut+kUx+β1Uxxt+β2(U2)x = β4(2Ux(U)xx−U(U)xxx), the two first graphes

are obtained for the values β1 = −1.00, β2 = 1.50, β4 = −0.75, k =

−2.00, λ1 = 0.65, meanwhile the last two graphes have the same parameter

values except that β4 = 0.75, k = 2.00.



Ut+kUx+β1Uxxt+β2(U)x = β4(4Ux(U2)xx−U(U2)xxx), the two first graphes



values but for β4 = 0.75, k = 2.00.



Ut +kUx +β1Uxxt +β2(U)x = β4(2Ux(U)xx−U(U)xxx), the two first graphes



values except that β4 = 0.75, k = 2.00.

20

3 Conclusion

In this work, we have applied an indirect F function method very close

to the F -expansion method to solve the generalized Camassa-Holm equation

with fully nonlinear dispersion and fully nonlinear convection term C(l, n, p).

By using this F function method, we have been able to obtain in a unified

way simultaneously many periodic wave solutions expressed by various single

and combined non degenerative Jacobi elliptic function solutions and their

degenerative solutions (When the modulus m is driven to 1 and 0). This

method gives elliptic solutions for specific values of the parameters n, p, l

and for arbitrary values of β1 and β2, but β3 and β4 must be proportional

with the proportionality constants for various cases given explicitly. In the

various Camassa-Holm classes listed below, β3 is replaced by the appropriate

expression of β4. The following Camassa-Holm Family of equations has been

solved by this indirect F function method


4)xx − U(U4)xxx),


2)xx − U(U2)xxx),


2)xx − U(U2)xxx),

Ut + kUx + β1Uxxt + β2(U2)x = β4(2Ux(U)xx − U(U)xxx),

Ut + kUx + β1Uxxt + β2(U)x = β4(4Ux(U2)xx − U(U2)xxx),

Ut + kUx + β1Uxxt + β2(U)x = β4(2Ux(U)xx − U(U)xxx).

21

Acknowledgements

The author would like to thank the referee for his/her valuable sugges-

tion. He is grateful to Professor George Sell of the School of Mathematics

for valuable discussions. He will like to thank Profs. Douglas Arnold, Debra

Lewis and Pamela Cook of the Institute for Mathematics and its Applications

of Minneapolis for their warmhearted help. This research was supported in

part by the Institute for Mathematics and its Applications with funds pro-

vided by National Science Foundation.

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Hi Janet, Thank you for the draft. is my reaction.

- Problem 23: b) Since there are 3 transformations, the question can count

only 3 points instead of 5. The additional points of the question b) may be

put in question c (now named second b) and in question d (now named c).

Please also update the questions (repetition of question b in the problem).

- Problem 25: c) ... that the first one is red and the other two are white

should be replaced by ” that one red marble and two white marbles are

drawn.

- For Multiple choice problems

Problem 17: there is a problem here. P(show profit)+P(absorb a loss)=1,

but in the problem that sum is greater than 1. Please check.

Here are my solutions for multiple choice Problems. 1) d 2) b 3) c 4) e 5)

a 6) c 7) e 8) e 9) d 10) a 11) d 12) c (please can you include the parentheses

in the function log) 13) b 14) e 15) a 16) e 17) 18) d 19) d 20) a.

Here are the statistics (without problem 17) 4 answers with a 2 answers

with b 3 answers with c 5 answers with d 5 answers with e.

Millions Thanks for your job.

best, Emmanuel.

24

emmanuel yomba- construction of new solutions to the fully nonlinear generalized camassa-holm...

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