Branch switching to a 3-homoclinic orbit in a
5th-order Korteweg-De Vries model

In ChGr:97 the following water wave model was considered:

$$\frac{2}{15}r''''-b r''+ar+\frac{3}{2}r^2- \frac{1}{2}(r')^2+[rr']' = 0.$$ (27.4)

It represents solitary-wave solutions $r(x+at)$, $r\to 0$ as $x\to \pm\infty$ of the 5th-order PDE

$$r_t+\frac{2}{15}r_{xxxx}-b r_{xxx}+3r r_x+2 r_x r_{xx}+r r_{xxx=0},$$

where $a$ is the wave speed. The ODE corresponds to a Hamiltonian system with Hamiltonian

$$H=-\frac{1}{2}q_1^3-\frac{1}{2}a q_1^2+p_1 q_2-\frac{1}{2}b q_2^2+ \frac{15}{4}p_2^2+\frac{1}{2}q_2^2 q_1$$

and

$$q_1=r, \quad q_2=r', \quad p_1=-\frac{2}{15}r'''+br'-rr', \quad p_2=\frac{2}{15}r''.$$

System (27.4) is also reversible under the transformation

$$t \mapsto -t, (q_1,q_2,p_1,p_2)\mapsto (q_1,-q_2,-p_1,p_2),$$

but we do not exploit the reversible structure (IREV=0), and instead use it as an example of Hamiltonian system. This system exhibits an orbit flip for a reversible Hamiltonian system. In Hamiltonian systems, homoclinic orbits are codimension-zero phenomena, and we have to add an additional parameter $\lambda$ that breaks the Hamiltonian structure in this system, by introducing artificial friction. Thus, the actual system of equations that is used for continuation is

$$\dot x=(\lambda I + J)\nabla H(x),$$

where $x=(q_1,q_2,p_1,p_2)$ and $J$ is the usual skew symmetric matrix in $\mathbb{R}^4$. It is now possible to continue a homoclinic orbit in HomCont in two parameters ($\lambda$ and either $a$ or $b$); see also Be:90a.

An explicit solution exists for $a=3/5(2b+1)(b-2), b\geq -1/2$, and it is given by

$$r(t)=3(b+\frac{1}{2})\mathrm{sech}^2\left([\frac{3}{4}(2b+1)]^{1/2}t\right).$$

It corresponds to a reversible orbit flip for $b>2$ ($a>0$) We start from this explicit solution, using ISTART=2, for $a=3$ and $b=(\sqrt{65}+3)/4$:

demo('kdv')
ld('kdv')
rn()
sv('1')

  BR    PT  TY  LAB    PAR(1)        L2-NORM    ...   PAR(3)     
   1     1  EP    1   3.00000E+00   5.56544E+00 ...  0.00000E+00
   1     2  EP    2   3.04959E+00   5.49141E+00 ...  1.77919E-17

Here PAR(1)=$a$, PAR(2)=$b$, and PAR(3)=$\lambda$. We have only done a very small continuation to give AUTO a chance to create a good mesh and avoid convergence problems later. Next, we set ITWIST=1 and calculate the adjoint:

rn(c='kdv.2',h='kdv.2',s='1')
sv('2')

  BR    PT  TY  LAB    PAR(2)        L2-NORM    ...   PAR(9)
   1     2  EP    3   2.76556E+00   5.49140E+00 ... -7.62944E-08

We now need to move back to the orbit flip at $a=3$:

rn(c='kdv.3',h='kdv.3',s='2')
sv('3')

  BR    PT  TY  LAB    PAR(1)        L2-NORM    ...   PAR(3)     
   1    14  UZ    5   3.00000E+00   5.47612E+00 ...  1.47274E-09

Now all preparations are done to start homoclinic branch switching. This is very similar to the technique used in Sandstede's model in Section 27.1; to find a 3-homoclinic orbit, we open 2 Lin gaps, until $T_1=3.5$, while also varying $\lambda$=PAR(3).

rn(c='kdv.4',h='kdv.4',s='3')
sv('4')

  BR    PT  TY  LAB    PAR(3)     ...   PAR(21)       PAR(22)       PAR(24)
   1    13        8   6.31458E-10 ...  1.65469E+01  -8.57681E-08  -7.30773E-07
   1    23  UZ    9   1.46493E-09 ...  9.92489E+00  -5.84373E-12   1.93098E-07
   1    26       10   4.01320E-09 ...  6.92406E+00   2.59555E-07   7.47534E-07
   1    33  EP   11   2.15487E-06 ...  3.50000E+00   7.92587E-04   3.98390E-04

We then look for an orbit with $a<3$ and close the gap corresponding to $\varepsilon _1$=PAR(22), for decreasing $a$.

rn(c='kdv.5',h='kdv.5',s='4')
sv('5')

  BR    PT  TY  LAB    PAR(2)     ...   PAR(3)        PAR(22)       PAR(24)
   1    10       12   2.58030E+00 ...  2.15869E-06   7.65037E-04   3.82464E-04
   1    13  UZ   13   2.32044E+00 ...  4.02730E-11   1.17522E-10   1.69655E-08
   1    20  EP   14  -1.14985E-01 ... -8.87194E-04  -7.18231E-01  -3.31153E-01

and finally close the gap corresponding to $\varepsilon_2$=PAR(24),

rn(c='kdv.6',h='kdv.6',s='5')
sv('6')

  BR    PT  TY  LAB    PAR(2)     ...   PAR(3)        PAR(23)       PAR(24)
   1    35       15   2.31893E+00 ... -2.16070E-08   7.69046E+00  -1.08126E-05
   1    51  UZ   16   2.34039E+00 ...  2.83533E-07   3.47976E+00   1.42651E-04
   1    58  UZ   17   3.08085E+00 ...  1.84952E-12   3.50004E+00  -1.64827E-10
   1    70  EP   18   3.08870E+00 ... -8.10422E-08   5.87541E+00  -4.82991E-05

so that a three-homoclinic orbit is found. Here the zero at label 16 is the one we are looking for. At label 17, $a$=PAR(1) has changed considerably to the extend that $a>3$ and a second 3-homoclinic orbit is found. Note that for all zeros of PAR(24)= $\varepsilon_2$, the parameter $\lambda$=PAR(3) is also zero (within AUTO accuracy), which it has to be to remain within the original Hamiltonian system. Setting ISTART=1, a normal ``trivial'' continuation (with NMX=1) of the orbit corresponding to label 16 lets HomCont produce a proper concatenated 3-homoclinic orbit:

rn(c='kdv.7',h='kdv.7',s='6')
sv('7')

  BR    PT  TY  LAB    PAR(2)        L2-NORM    ...   PAR(3)     
   1     1  EP   19   2.34039E+00   7.51157E+00 ...  2.83533E-07

This 3-homoclinic orbit is depicted in Figure 27.6.

Figure 27.6: A 3-homoclinic orbit in a 5th-order Hamiltonian Korteweg-De Vries model.