Soliton solutions of nonlinear Boussinesq models using the exponential function technique

Recently khalil et al introduced a simplest, most natural and efficient definition of local derivative of order $\alpha \in (0,1]$, known as conformable derivative [33, 34]. For any value of α, we can generalize the definition.

Definition 2.1.1. Let $s:[0,\infty )\to {\mathbb{R}}$ be a function. The conformable derivative having order α for function s is given as [33, 35, 36]:

$$\begin{eqnarray}&&({M}_{\alpha }s)(x)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\displaystyle \frac{s(x+\varepsilon {x}^{1-\alpha })-s(x)}{\varepsilon }.\end{eqnarray} \tag{ 1 }$$

Here $\alpha \in (0,1)$ and x > 0. If in some (0, a) with a > 0, the function s is α-differential and ${\mathrm{lim}}_{x\to {0}^{+}}{s}^{(\alpha )}(x)$ exists, then definition becomes:

$$\begin{eqnarray*}&&{s}^{(\alpha )}(0)=\mathop{\mathrm{lim}}\limits_{x\to {0}^{+}}{s}^{(\alpha )}(x).\end{eqnarray*}$$

Conformable derivative of constant function is zero. The conformable mean value theorem proves s(x) = 0 for $x\in (c,d)$ if we take ${M}_{\alpha }s(x)=0$ [33].

The conformable integral is represented as:

$$\begin{eqnarray}&&{I}_{\alpha }^{c}(r)(x)={\int }_{c}^{x}\displaystyle \frac{r(t)}{{t}^{1-\alpha }}{dt}.\end{eqnarray} \tag{ 2 }$$

Here c ≥ 0 and $\alpha \in (0,1]$. The above integral represents the usual Riemann improper integral.

By using the above definition of derivative, khalil presented the following theorem [33, 36–38].

Theorem. Let $0\lt \alpha \leqslant 1$ and r, s are α-differentiable functions at a point x > 0 then,

• 1.  
  ${M}_{\alpha }({ar}+{bs})={{aM}}_{\alpha }(r)+{{bM}}_{\alpha }(s)$ for all $a,b\in {\mathbb{R}}$.
• 2.  
  ${M}_{\alpha }({t}^{n})={{nt}}^{n-\alpha }$ for all $n\in {\mathbb{R}}$.
• 3.  
  M_α (λ) = 0, for all constant functions r(λ) = 0.
• 4.  
  ${M}_{\alpha }({rs})={{rM}}_{\alpha }(s)+{{sM}}_{\alpha }(r)$.
• 5.  
  ${M}_{\alpha }(\tfrac{r}{s})=\tfrac{{{sM}}_{\alpha }(r)-{{rM}}_{\alpha }(s)}{{s}^{2}}$.
• 6.  
  If r is differentiable, then $({M}_{\alpha }r)(x)={x}^{1-\alpha }\tfrac{{dr}}{{dx}}(x)$.

Theorem. Let us consider two (left) α-differentiable functions r, s such that $r,s:(a,\infty )\to {\mathbb{R}}$, with $\alpha \in (0,1]$. We have h(x) is left α-differentiable when $h(x)=r(s(x))$, ∀x except $s(x)\ne 0$ and $x\ne a$. Hence chain rule will be given as [35]:

$$\begin{eqnarray}&&({M}_{\alpha }^{a}h)(x)=({M}_{\alpha }^{a}r)(s(x)).({M}_{\alpha }^{a}s)(x).s{\left(x\right)}^{\alpha -1},\end{eqnarray} \tag{ 3 }$$

consider x = a, we get

$$\begin{eqnarray}&&({M}_{\alpha }^{a}h)(a)=\mathop{\mathrm{lim}}\limits_{x\to {a}^{+}}({M}_{\alpha }^{a}r)(s(x)).({M}_{\alpha }^{a}s)(x).s{\left(x\right)}^{\alpha -1}.\end{eqnarray} \tag{ 4 }$$

This definition attracted many scientists and they worked more intensively for further investigations [33, 35]. Abdeljawad applied conformable derivative to describe rules for exponential functions, chain rule and integration by parts, Laplace transform and Taylor power series expansion [35]. Researchers worked incredibly on its applications for solving conformable models in mathematical physics [1–3, 39], quantum mechanics [40] and Newton mechanics [41, 42].

The general form of nonlinear conformable PDEs is described as follows:

$$\begin{eqnarray}&&F\left(u,\displaystyle \frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }},\displaystyle \frac{\partial u}{\partial x},\displaystyle \frac{\partial u}{\partial y},\displaystyle \frac{{\partial }^{2\alpha }u}{\partial {t}^{2\alpha }},\displaystyle \frac{{\partial }^{2}u}{\partial {x}^{2}},\displaystyle \frac{{\partial }^{2}u}{\partial {y}^{2}}\right)=0.\end{eqnarray} \tag{ 5 }$$

Now, consider a following transformation:

$$\begin{eqnarray}&&u(x,y,t)=u(\omega ),\,\,\omega ={px}+{qy}-r\displaystyle \frac{{t}^{\alpha }}{\alpha }.\end{eqnarray} \tag{ 6 }$$

Here $q\ne 0$, $p\ne 0$ and $r\ne 0$ are constants, these constants are calculated later in process.

Utilizing equation (3), we have

$$\begin{eqnarray}&&\displaystyle \frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }}=-r\displaystyle \frac{{du}}{d\omega },\,\,\displaystyle \frac{\partial u}{\partial x}=p\displaystyle \frac{{du}}{d\omega },\,\,\displaystyle \frac{\partial u}{\partial y}=q\displaystyle \frac{{du}}{d\omega },\,\ldots .\end{eqnarray} \tag{ 7 }$$

Using equations (7) into (5), we get a nonlinear ODE as

$$\begin{eqnarray}&&Q\left(u,{u}_{\omega },{u}_{\omega \omega },{u}_{\omega \omega \omega },{u}_{\omega \omega \omega \omega },\,\ldots \right)=0.\end{eqnarray} \tag{ 8 }$$

Assume the above ODE has solution of the following form

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{{\sum }_{i=-c}^{d}{a}_{i}{e}^{i\omega }}{{\sum }_{j=-m}^{n}{b}_{j}{e}^{j\omega }}.\end{eqnarray} \tag{ 9 }$$

Here c, d, n and m are positive integers determined by homogeneous balancing principle, a_i and b_j are constants which are unknown. The next step is to balance the lowest order linear term in equation (8) with the lowest order non-linear term, in order to evaluate c and m. Similarly balance the highest order linear term in equation (8) with highest order nonlinear term, in order to calculate the value of d and n. As a result, we get the analytical solution of nonlinear PDEs.

The nonlinear time conformable Boussinesq equation has the following form [22, 49, 50]:

$$\begin{eqnarray}&&\displaystyle \frac{{\partial }^{2\alpha }u(x,t)}{\partial {t}^{2\alpha }}-\displaystyle \frac{{\partial }^{2}u(x,t)}{\partial {x}^{2}}-\displaystyle \frac{{\partial }^{2}{u}^{2}(x,t)}{\partial {x}^{2}}+\displaystyle \frac{{\partial }^{4}u(x,t)}{\partial {x}^{4}}=0,\end{eqnarray} \tag{ 10 }$$

where $\alpha \in (0,1)$, $u(x,t)$ is a field function and $\tfrac{{\partial }^{2\alpha }u(x,t)}{\partial {t}^{2\alpha }}$ represents the conformable derivative of u(x, t).

Equation (10) includes two competing effects i.e. (i) the linear dispersive term represented by $\tfrac{{\partial }^{4}u(x,t)}{\partial {x}^{4}}$ that explains the out spreads and (ii) the nonlinear term $\tfrac{{\partial }^{2}{u}^{2}(x,t)}{\partial {x}^{2}}$ which describes the translation of wave. If both of these effects are presents then solitary wave can be determined. Hence Boussinesq equation gives rise to solitary waves because of the balance between the dissipative effects of $\tfrac{{\partial }^{4}u(x,t)}{\partial {x}^{4}}$ and nonlinearity effects of $\tfrac{{\partial }^{2}{u}^{2}(x,t)}{\partial {x}^{2}}$. These waves are confined as they travel with constant shapes and speeds. For large distances these waves are asymptotically zero. Solitary waves have special kind called solitons. Solitons possess a property which keeps its identity after interacting with other solitons. Now, using the wave transformation and chain rule, we get

$$\begin{eqnarray}&&\displaystyle \frac{{\partial }^{2\alpha }u(x,t)}{\partial {t}^{2\alpha }}={r}^{2}\displaystyle \frac{{d}^{2}u(x,t)}{d{\omega }^{2}},\,\displaystyle \frac{{\partial }^{2}u(x,t)}{\partial {x}^{2}}=\displaystyle \frac{{d}^{2}u(x,t)}{d{\omega }^{2}},\,\displaystyle \frac{{\partial }^{2}{u}^{2}(x,t)}{\partial {x}^{2}}=\displaystyle \frac{{d}^{2}{u}^{2}(x,t)}{d{\omega }^{2}}.\end{eqnarray} \tag{ 11 }$$

Using equations (11) into (10), we have

$$\begin{eqnarray}&&{u}_{\omega \omega }(\omega )+({r}^{2}-1)u(\omega )-{u}^{2}(\omega )=0.\end{eqnarray} \tag{ 12 }$$

Assume a solution of the above ODE as follows:

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{{a}_{d}{e}^{d\omega }\,+\,...\,+\,{a}_{-c}{e}^{-c\omega }}{{b}_{n}{e}^{n\omega }\,+\,...\,+\,{b}_{-m}{e}^{-m\omega }}.\end{eqnarray} \tag{ 13 }$$

In order to evaluate the constants c, d, n and m, here we apply homogeneous balancing principle. According to this principle, we have to compare ${u}_{\omega \omega }$ and ${u}^{2}(\omega )$ of equation (12). Now consider

$$\begin{eqnarray}&&{u}_{\omega \omega }=\displaystyle \frac{...\,+\,{k}_{1}{e}^{(d+3n)\omega }+...}{...\,+\,{k}_{2}{e}^{4n\omega }+\,...},\end{eqnarray} \tag{ 14 }$$

and

$$\begin{eqnarray}&&{u}^{2}(\omega )=\displaystyle \frac{...\,+\,{k}_{3}{e}^{2d\omega }+...}{...\,+\,{k}_{4}{e}^{2n\omega }+...}=\displaystyle \frac{...\,+\,{k}_{3}{e}^{(2d+2n)\omega }+...}{...\,+\,{k}_{4}{e}^{4n\omega }+...}.\end{eqnarray} \tag{ 15 }$$

For unknowns d and n compare the highest order of exp-function in equations (14) and (15) which results

$$\begin{eqnarray}&&d+3n=2d+2n,\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&d=n.\end{eqnarray} \tag{ 17 }$$

Similarly,

$$\begin{eqnarray}&&{u}_{\omega \omega }=...\displaystyle \frac{...\,+\,{l}_{1}{e}^{-(c+3m)\omega }+...}{...\,+\,{l}_{2}{e}^{-4m\omega }+...},\end{eqnarray} \tag{ 18 }$$

and

$$\begin{eqnarray}&&{u}^{2}(\omega )=...\displaystyle \frac{...\,+\,{l}_{3}{e}^{-2c\omega }+...}{...\,+\,{l}_{4}{e}^{-2m\omega }+...}=...\displaystyle \frac{...\,+\,{l}_{3}{e}^{-(2c+2m)\omega }+...}{...\,+\,{l}_{4}{e}^{-4m\omega }+...},\end{eqnarray} \tag{ 19 }$$

where l_i are coefficients. Comparing the lowest order of exp-function in equations (18) and (19) we get

$$\begin{eqnarray}&&c+3m=2c+2m,\end{eqnarray} \tag{ 20 }$$

$$\begin{eqnarray}&&c=m.\end{eqnarray} \tag{ 21 }$$

Setting d = c = 1, equation (13) becomes

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{{a}_{1}{e}^{\omega }+{a}_{0}+{a}_{-1}{e}^{-\omega }}{{b}_{1}{e}^{\omega }+{b}_{0}+{b}_{-1}{e}^{-\omega }}.\end{eqnarray} \tag{ 22 }$$

Substitute equations (22) into (12), then equating the coefficients of ${e}^{i\omega }$, we obtain

$$\begin{eqnarray}&&{h}_{3}=0,\,{h}_{2}=0,\,{h}_{1}=0,\,{h}_{0}=0,\,{h}_{-1}=0,\,{h}_{-2}=0,\,{h}_{-3}=0,\end{eqnarray} \tag{ 23 }$$

where ${h}_{i}(i=-3,\,\ldots ,\,0,\,\ldots ,\,3)$ are algebraic equations evaluated by the help of Maple 13.

By inspection, we get

$$\begin{eqnarray}&&r=\sqrt{2},\,{a}_{0}=-2{b}_{0},\,{a}_{1}=\displaystyle \frac{{b}_{0}^{2}}{4{b}_{-1}},\,{a}_{-1}={b}_{-1},\,{b}_{0}={b}_{0},\,{b}_{1}=\displaystyle \frac{{b}_{0}^{2}}{4{b}_{-1}},\,{b}_{-1}={b}_{-1}.\end{eqnarray} \tag{ 24 }$$

Using equations (24) into (22), it becomes

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{\tfrac{{b}_{0}^{2}}{4{b}_{-1}}{e}^{\omega }-2{b}_{0}+{b}_{-1}{e}^{-\omega }}{\tfrac{{b}_{0}^{2}}{4{b}_{-1}}{e}^{\omega }+{b}_{0}+{b}_{-1}{e}^{-\omega }},\end{eqnarray} \tag{ 25 }$$

where $\omega =x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha }$, after simplifying equation (25) we obtain

$$\begin{eqnarray}&&u(x,t)=\displaystyle \frac{{b}_{0}^{2}{e}^{\left(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha }\right)}-8{b}_{-1}{b}_{0}+4{b}_{-1}^{2}{e}^{-(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha })}}{{b}_{0}^{2}{e}^{\left(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha }\right)}+4{b}_{-1}{b}_{0}+4{b}_{-1}^{2}{e}^{-(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha })}}.\end{eqnarray} \tag{ 26 }$$

Setting ${b}_{0}={b}_{-1}=1$, equation (26) becomes

$$\begin{eqnarray}&&u(x,t)=\displaystyle \frac{{e}^{\left(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha }\right)}-8+4{e}^{-(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha })}}{{e}^{\left(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha }\right)}+4+4{e}^{-(x-\sqrt{2}\tfrac{{t}^{\alpha }}{\alpha })}}.\end{eqnarray} \tag{ 27 }$$

Equation (27) reveals that Boussinesq equation can exhibits explosive wave form. Figures 2–4 show the solution of time conformable Boussinesq equation for α = 0.25, α = 0.5, α = 0.75 and α = 1, respectively. Figure 5 represents the solution of time conformable Boussinesq equation by allocating various values to α. This solution describes an explosive solitons. For different values of α these solitons are moving with constant speed and shape. The solutions presented in figures 1–4 are the singular type of solitons. The similar type of solutions for α = 1 (c.f. figure 4) can be observed in the literature [50, 51].

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The nonlinear space-time conformable Boussinesq equation is described as follows [25, 29, 52, 53]:

$$\begin{eqnarray}&&{D}_{t}^{2\alpha }u+{{AD}}_{x}^{2\alpha }u+{{BD}}_{x}^{2\alpha }({u}^{2})+{{CD}}_{x}^{4\alpha }u=0,\end{eqnarray} \tag{ 28 }$$

where t > 0, $\alpha \in (0,1)$ and the diffusion parameter C = constant where C > 0 depends on the hardness and compression properties of the substance, u(x, t) is the vertical diversion, B is the nonlinearity coefficient and the curvature of the bending beam is described by the quadratic nonlinearity ${({u}^{2})}_{{xx}}$, α is a parameter which explains the conformable order of the space and time derivatives.

Introducing the transformation $\omega =p\tfrac{{x}^{\alpha }}{\alpha }-r\tfrac{{t}^{\alpha }}{\alpha }$, equation (28) is converted into an ODE of the form:

$$\begin{eqnarray}&&{{Cp}}^{4}{u}_{\omega \omega }+({r}^{2}+{{Ap}}^{2})u+{{Bp}}^{2}{u}^{2}=0\end{eqnarray} \tag{ 29 }$$

Assume a solution of the form,

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{{a}_{d}{e}^{d\omega }\,+\,...\,+\,{a}_{-c}{e}^{-c\omega }}{{b}_{n}{e}^{n\omega }\,+\,...\,+\,{b}_{-m}{e}^{-m\omega }}.\end{eqnarray} \tag{ 30 }$$

Using the same approach in example 1, equating the highest order of exp-function in ${u}_{\omega \omega }$ and ${u}^{2}(\omega )$, we get the result

$$\begin{eqnarray*}&&d=n.\end{eqnarray*}$$

Similarly, equating lower order of exp-function in ${u}_{\omega \omega }$ and ${u}^{2}(\omega )$, we get

$$\begin{eqnarray*}&&c=m.\end{eqnarray*}$$

Taking d = c = 1, equation (30) becomes

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{{a}_{1}{e}^{\omega }+{a}_{0}+{a}_{-1}{e}^{-\omega }}{{b}_{1}{e}^{\omega }+{b}_{0}+{b}_{-1}{e}^{-\omega }}.\end{eqnarray} \tag{ 31 }$$

Putting equation (31) into equation (29), then equating coefficients of ${e}^{i\omega }$ we obtain a system of algebraic equations. Then obtained algebraic system is solved by using Maple and we get the following results:

$$\begin{eqnarray}&&p=p,\,r=\sqrt{-{{Cp}}^{2}-A}p,\,{a}_{0}=\displaystyle \frac{3{{Cp}}^{2}{b}_{0}}{B},\,{a}_{1}=0,\,{a}_{-1}=0,\,{b}_{0}={b}_{0},\,{b}_{1}=\displaystyle \frac{{b}_{0}^{2}}{4{b}_{{}_{1}}},\,{b}_{-1}={b}_{-1},\end{eqnarray} \tag{ 32 }$$

using equations (32) into (31), we have

$$\begin{eqnarray}&&u(\omega )=\displaystyle \frac{12{{Cp}}^{2}}{B({e}^{\omega }+4+{e}^{-\omega })}.\end{eqnarray} \tag{ 33 }$$

As $\omega =p\tfrac{{x}^{\alpha }}{\alpha }-\sqrt{-{{Cp}}^{2}-A}p\tfrac{{t}^{\alpha }}{\alpha }$ therefore, equation (33) becomes

$$\begin{eqnarray}&&u(x,t)=\displaystyle \frac{12{{Cp}}^{2}}{B({e}^{\left(p\tfrac{{x}^{\alpha }}{\alpha }-\sqrt{-{{Cp}}^{2}-A}p\tfrac{{t}^{\alpha }}{\alpha }\right)}+4+{e}^{-(p\tfrac{{x}^{\alpha }}{\alpha }-\sqrt{-{{Cp}}^{2}-A}p\tfrac{{t}^{\alpha }}{\alpha })})}.\end{eqnarray} \tag{ 34 }$$

The real valued domain of u(x, t) is positive values of x. Figures 6–9 show the solution of space-time conformable Boussinesq equation for α = 0.25, α = 0.5, α = 0.75 and α = 1, respectively. Figure 10 represents the solution u(x, t) of space-time conformable Boussinesq equation by using different values of α. Figure 10 reveals that the flatten curves are observed for less values of α, and sharp curves for values of α near to 1. Figure 9 represents the bell-shaped solitary wave solution for α = 1 which is already presented in literature [50].

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