We study the first model of the (3+1)-dimensional modified KdV equations [2,3] that reads
$${u}_{t}+6{u}^{2}{u}_{x}+{u}_{xyz}=0.$$(11)
We introduce the wave variable
$$\xi =kx+ry+sz-\omega t,$$(12)
where *k*, *r* and *s* are constants, and *ω* is the dispersion relation. A variety of ansatze will be applied to derive a set of solutions characterized by distinct physical structures.

(i) Using the sech / csch method:

The sech method admits the use of the solution in the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sech}}^{M}(kx+ry+sz-\omega t),$$(13)
where *R* and M are are parameters that will be determined. Using first the balance method to determine *M*, we find that solutions exist only if *M* = 1. Substituting (13) with *M* = 1, into (11), collecting the coefficients of cosh^{i}(*ξ*), *i* = 0, 2, and equating each coefficient to zero we find
$$\begin{array}{rl}& \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\omega +krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0,\\ & 6k{R}^{2}-6krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0.\end{array}$$(14)
This will give
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pm \sqrt{rs}.\end{array}$$(15)
The bell shaped soliton solution is thus given by
$$u(x,y,z,t)=\pm \sqrt{rs}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sech}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-krs\phantom{\rule{thinmathspace}{0ex}}t).$$(16)

We also assume the solution takes the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\mathrm{csch}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t).$$(17)
Proceeding as presented earlier we obtain
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pm \sqrt{-rs},rs<0.\end{array}$$(18)
This gives the singular solution
$$u(x,y,z,t)=\pm \sqrt{-rs}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{csch}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-krs\phantom{\rule{thinmathspace}{0ex}}t).$$(19)

(ii) Using the tanh / coth method:

The tanh method admits the use of the solution in the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{tanh}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t),$$(20)
where *R* is a parameter that will be determined. Substituting (20) into (11), collecting the coefficients of tanh^{i}(*ξ*), *i* = 0, 2, and equating each coefficient to zero we find
$$\begin{array}{rl}& \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\omega -2krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0,\\ & 6k{R}^{2}+6krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0.\end{array}$$(21)
This will give
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-2krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pm \sqrt{-rs}.\end{array}$$(22)
The kink solution is thus given by
$$u(x,y,z,t)=\pm \sqrt{-rs}\phantom{\rule{thinmathspace}{0ex}}\mathrm{tanh}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+2krs\phantom{\rule{thinmathspace}{0ex}}t).$$(23)

We may also assume the solution takes the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\mathrm{coth}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t).$$(24)
Proceeding as presented earlier we obtain
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-2krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\pm \sqrt{-rs},\phantom{\rule{thinmathspace}{0ex}}rs<0.\end{array}$$(25)
This gives the singular solution
$$u(x,y,z,t)=\pm \sqrt{-rs}\phantom{\rule{thinmathspace}{0ex}}\mathrm{coth}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+2krs\phantom{\rule{thinmathspace}{0ex}}t).$$(26)

(iii) Using the sec / csc method:

The sec ansatz sets the solution in the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\mathrm{sec}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t),$$(27)
where *R* is a parameter that will be determined. Substituting (27) into (11), collecting the coefficients of cos^{i}(*ξ*), *i* = 0, 2, and equating each coefficient to zero we find
$$\begin{array}{rl}& \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\omega +krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0,\\ & -6k{R}^{2}-6krs\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0.\end{array}$$(28)
This will give
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\pm \sqrt{-rs},rs<0.\end{array}$$(29)
The exact solution is thus given by
$$u(x,y,z,t)=\pm \sqrt{-rs}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sec}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+krs\phantom{\rule{thinmathspace}{0ex}}t).$$(30)

We also assume the solution takes the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\mathrm{csc}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t).$$(31)
Proceeding as presented earlier we obtain
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pm \sqrt{-rs},rs<0.\end{array}$$(32)
This gives the singular solution
$$u(x,y,z,t)=\pm \sqrt{-rs}\phantom{\rule{thinmathspace}{0ex}}\mathrm{csc}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+krs\phantom{\rule{thinmathspace}{0ex}}t).$$(33)

(iv) Using the tan / cot method:

The tan ansatz assumes the solution in the form
$$u(x,y,z,t)=R\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-\omega t),$$(34)
where *R* is a parameter that will be determined. Proceeding as before, collecting the coefficients of tan^{i}(*ξ*), *i* = 0, 2, and equating each coefficient to zero we find
$$\begin{array}{rl}& \omega \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2krs,\\ & R\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\pm \sqrt{-rs},rs<0.\end{array}$$(35)
The periodic solution is thus given by
$$u(x,y,z,t)=\pm \sqrt{-rs}\mathrm{tan}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-2krs\phantom{\rule{thinmathspace}{0ex}}t).$$(36)

In a like manner, we can derive the exact solution
$$u(x,y,z,t)=\pm \sqrt{-rs}\mathrm{cot}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-2krs\phantom{\rule{thinmathspace}{0ex}}t).$$(37)

Concerning the other two forms of the (3+1)-dimensional modified KdV equations
$${u}_{t}+6{u}^{2}{u}_{y}+{u}_{xyz}=0,$$(38)
and
$${u}_{t}+6{u}^{2}{u}_{z}+{u}_{xyz}=0,$$(39)
we can follow the same approach as presented for the first model. Based on this, we just list two solitary waves for each model. For Eq. (38) one obtains the soliton solution
$$u(x,y,z,t)=\pm \sqrt{ks}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sech}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-krs\phantom{\rule{thinmathspace}{0ex}}t),$$(40)
and the kink solution
$$u(x,y,z,t)=\pm \sqrt{-ks}\phantom{\rule{thinmathspace}{0ex}}\mathrm{tanh}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+2krs\phantom{\rule{thinmathspace}{0ex}}t),ks<0.$$(41)

However, for Eq. (39) one obtains the soliton solution
$$u(x,y,z,t)=\pm \sqrt{kr}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sech}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz-krs\phantom{\rule{thinmathspace}{0ex}}t),$$(42)
and the kink solution
$$u(x,y,z,t)=\pm \sqrt{-kr}\phantom{\rule{thinmathspace}{0ex}}\mathrm{tanh}\phantom{\rule{thinmathspace}{0ex}}(kx+ry+sz+2krs\phantom{\rule{thinmathspace}{0ex}}t),kr<0.$$(43)

Some main conclusions can be made here. Hereman [2,3] introduced one (3+1)-dimensional modified KdV equation with one solitary wave solution. However, in this work we introduced two more equations, and we obtained a variety of solitary wave solutions and periodic solutions for each model. Moreover, we showed that each model gives rise to soliton and kink solutions as well.

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