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Numerical Solutions with Linearization Techniques of the Fractional Harry Dym Equation

Asıf Yokuş
Asıf Yokuş
 oraz 
Sema Gülbahar
Sema Gülbahar
   | 27 mar 2019
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Data publikacji: 27 mar 2019
Zakres stron: 35 - 42
Otrzymano: 14 sty 2019
Przyjęty: 21 lut 2019
DOI: https://doi.org/10.2478/AMNS.2019.1.00004
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© 2019 Asıf Yokuş and Sema Gülbahar, published by Sciendo
This work is licensed under the Creative Commons Attribution 4.0 International License.
Introduction

Fractional analysis; is the generalization of the classic analysis of integration and differentiation of process (noninteger) order. This issue is an old issue as much as differential calculus. G. W. Leibniz and Marquis de L' Hospital correspondence in 1695, is known as the first exit point of fractional calculus. Leibniz expressed fractional order derivatives of noninteger order f (x) = emx, m ∈ R, as follows:

dnemxdxn=mnemx, $$\frac{{{d}^{n}}{{e}^{mx}}}{d{{x}^{n}}}={{m}^{n}}{{e}^{mx}},$$

here, n is a value, where i is the noninteger.

Later, many scientists, such as Liouville, Riemann,Weyl, Lacroix, Leibniz, Grunward and Letnikov (cf. [4]), expanded the range of this derivative.

Since the beginning of the definition of fractional order derivatives first handled by Leibniz, fractional partial differential equations have attracted the attention of many scientists and have also shown a progressive development.(cf. [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 15, 16, 17, 18, 19]).

Beside these facts, the third order fractional Harry Dym partial differential equation is studied in mathematics and especially in the theory of solitons. This equation is given as

∂γu∂tγ=u3∂3u∂x3. $$\frac{{{\partial }^{\gamma }}_{u}}{\partial {{t}^{\gamma }}}={{u}^{3}}\frac{{{\partial }^{3}}u}{\partial {{x}^{3}}}.$$

Here, 0 < γ ≤ 1 is the order of fractional derivative and u(x, t) is a function of x and t.

The Harry Dym equation first appeared in a study by Kruskal [10]. Harry Dym equation represents a system in which dispersion and nonlinearity were coupled. Furthermore, the Harry Dym equation is a completely integrable nonlinear evolution that may be solved by means of the inverse scattering transform. It does not possess the Painlevé property.

This paper is organized as follows:

In the second section, some basic facts dealing with the finite difference method are mentioned and three Linearization techniques are presented. In the third section, stability analysis of the proposed method is investigated and it is shown that the Harry Dym equation is stable under which conditions. Also, numerical examples are given. In the fourth section, conclusions obtained throughout the paper are discussed.
Finite Difference Methods

In this section, we first need to define a set of grid points in a domain D to obtain a numerical solution to Eq. (1:1) using finite difference methods as follows:

Let Δx(h)=b−aN $\Delta x\left( h \right)=\frac{b-a}{N}$(N is an integer) denotes a state step size and Δt denotes a time step size. Draw a set of horizontal and vertical line across D, and get all intersection points (xj, tn) or simply ( j,n) where x j = a+ jΔx, j = 0,1,2, · · · ,N, and tn = nΔt, n = 0,1, . . . ,M. If we write D = [a,b]×[0,T] then we may choose Δt=TM( M $\Delta t=\frac{T}{M}\left( M \right.$is an integer) and tn = nΔt, n = 0,1, . . . ,M.

Then, an appropriate finite difference approximation is given in Eq. (1.1) instead of derivatives and its variable. In this case, the solution problem of Eq. (1.1) is reduced to solution problem of algebraic differential systems of linear and nonlinear equations consisting of finite difference equation. But when applied to non-linear problems, it normally leads to nonlinear system of equations and they cannot be solved directly. Therefore, we use three linearization techniques for a nonlinear term as given in Eq. (1.1).

Linearization 1

First, we use the Caputo fractional derivative approximation for ∂γu∂tγ $\frac{{{\partial }^{\gamma }}u}{\partial {{t}^{\gamma }}}$defined by

∂ γ u ∂ t γ ≃ 1 h 2 Δ t − γ Γ 2 − γ ∑ k = 0 n u m n + 1 − k − u m n − k k + 1 1 − γ − k 1 − γ n ≥ 1 Δ t − γ Γ 2 − γ u m 1 − u m 0 n = 0 , $$\frac{{{\partial }^{\gamma }}u}{\partial {{t}^{\gamma }}}\simeq \frac{1}{{{h}^{2}}}\left\{ \begin{matrix}\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n}{\left[ u_{m}^{n+1-k}-u_{m}^{n-k} \right]\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]} & n\ge 1 \\\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\left( u_{m}^{1}-u_{m}^{0} \right) & n=0, \\\end{matrix} \right.$$

and Crank–Nicolson derivative approximation given by

u=un+1+un2, $$u=\frac{{{u}^{n+1}}+{{u}^{n}}}{2},$$

in Eq. (1.1) at the nodal point (m,n+1) [12]. Then, if we discretize time derivative of the fractional Harry Dym equation by using Caputo fractional derivative formula between two successive time levels n and n +1, Crank Nicolson derivative formula and usual finite difference formula between two successive time levels n and n +1,respectively, we obtain

  (Δt)−γΓ(2−γ)∑k=0n[ umn+1−k−umn−k ][ (k+1)1−γ−k1−γ ]=[ (u3)mn+1+(u3)mn2 ][ um+2n−2um+1n+2um−1n−um−2n2h3 ]. $$\begin{align}& \,\,\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n}{\left[ u_{m}^{n+1-k}-u_{m}^{n-k} \right]\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]} \\ & =\left[ \frac{\left( {{u}^{3}} \right)_{m}^{n+1}+\left( {{u}^{3}} \right)_{m}^{n}}{2} \right]\left[ \frac{u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n}}{2{{h}^{3}}} \right]. \\ \end{align}$$

The nonlinear term in the above mentioned equation is linearized by using the following equation:

(u3)mn+1=3(u2)nun+1−2(u3)n. $$\left( {{u}^{3}} \right)_{m}^{n+1}=3{{\left( {{u}^{2}} \right)}^{n}}{{u}^{n+1}}-2{{\left( {{u}^{3}} \right)}^{n}}.$$

If the necessary arrangements are made in Eq. (2.1), we have the following equation:

  (Δt)−γΓ(2−γ)∑k=0n[ umn+1−k−umn−k ][ (k+1)1−γ−k1−γ ]=14h3[ 3(u2)mnumn+1−(u3)n ][ um+2n−2um+1n+2um−1n−um−2n ]. $$\begin{align}& \,\,\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n}{\left[ u_{m}^{n+1-k}-u_{m}^{n-k} \right]\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]} \\ & =\frac{1}{4{{h}^{3}}}\left[ 3\left( {{u}^{2}} \right)_{m}^{n}u_{m}^{n+1}-{{\left( {{u}^{3}} \right)}^{n}} \right]\left[ u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n} \right]. \\ \end{align}$$
Linearization 2

Let us use the Caputo fractional derivative approximation for ∂γu∂tγ., $\frac{{{\partial }^{\gamma }}u}{\partial {{t}^{\gamma }}}.,$Crank–Nicolson derivative approximation and usual finite difference approximation for Uxxx given by

Uxxx≃12h3(um+2n−2um+1n+2um−1n−um−2n), $${{U}_{xxx}}\simeq \frac{1}{2{{h}^{3}}}\left( u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n} \right),$$

in Eq. (1.1) at the nodal point (m,n+1), respectively [12]:

If we use the following linearization technique for the non-linear term U 3Uxxx:

U3Uxxx≃(Umn)312h3(um+2n−2um+1n+2um−1n−um−2n), $${{U}^{3}}{{U}_{xxx}}\simeq {{\left( U_{m}^{n} \right)}^{3}}\frac{1}{2{{h}^{3}}}\left( u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n} \right),$$

then we have the following system of algebraic equation:

  (Δt)−γΓ(2−γ)∑k=0n[ umn+1−k−umn−k ][ (k+1)1−γ−k1−γ ]=(Umn)312h3(um+2n+1−2um+1n+1+2um−1n+1−um−2n+1). $$\begin{align}& \,\,\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n}{\left[ u_{m}^{n+1-k}-u_{m}^{n-k} \right]}\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right] \\ & ={{\left( U_{m}^{n} \right)}^{3}}\frac{1}{2{{h}^{3}}}\left( u_{m+2}^{n+1}-2u_{m+1}^{n+1}+2u_{m-1}^{n+1}-u_{m-2}^{n+1} \right). \\ \end{align}$$

Eq. (2.3) can be solved using an approximate algorithm.
Linearization 3

Let us use the Caputo fractional derivative approximation for ∂γu∂tγ $\frac{{{\partial }^{\gamma }}u}{\partial {{t}^{\gamma }}}$and usual finite difference approximation for Uxxx in Eq. (1.1) at the nodal point (m,n+1) respectively [12]: If we use the following linearization technique for the nonlinear term U 3Uxxx, then we have

U 3 U x x x ≃ U m n + U m + 1 n 2 3 1 2 h 3 u m + 2 n − 2 u m + 1 n + 2 u m − 1 n − u m − 2 n . $${{U}^{3}}{{U}_{xxx}}\simeq \left( \frac{U_{m}^{n}+U_{m+1}^{n}}{2} \right)^{3}\frac{1}{2{{h}^{3}}}\left( u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n} \right).$$

Thus, we get the following system of algebraic equation:.

  (Δt)−γΓ(2−γ)∑k=0n−1[ umn−k−umn−k−1 ][ (k+1)1−γ−k1−γ ]=((Umn+Um+1n)2)312h3(um+2n−2um+1n+2um−1n−um−2n). $$\begin{align}& \,\,\frac{{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n-1}{\left[ u_{m}^{n-k}-u_{m}^{n-k-1} \right]\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]} \\ & ={{\left( \frac{\left( U_{m}^{n}+U_{m+1}^{n} \right)}{2} \right)}^{3}}\frac{1}{2{{h}^{3}}}\left( u_{m+2}^{n}-2u_{m+1}^{n}+2u_{m-1}^{n}-u_{m-2}^{n} \right). \\ \end{align}$$

Eq. (2.4) can be solved using an approximate algorithm.
Stability analysis

In this section, we investigate whether this method is stable based on von-Neumann analysis. If the Fourier method analyzes the stability, then the growth factor of a typical Fourier mode is defined as:

umn=ξneℓmφ,     ℓ=−1, $$u_{m}^{n}={{\xi }^{n}}{{e}^{\ell m\varphi }},\,\,\,\,\,\ell =\sqrt{-1},$$

where ξ n is considered as the amplification factor. First, by substituting the Fourier mode (3.1) into the recurrence relationship (2.3), one can obtain

                          4h3(Δt)−γΓ(2−γ)∑k=0n−1[ (k+1)1−γ−k1−γ ][ ξn−k+1−ξn−k ]eℓmφ=(ξneℓmφ)3(ξn+1eℓ(m+2)φ−2ξn+1eℓ(m+1)φ+2ξn+1eℓ(m−1)φ−ξn+1eℓ(m−2)φ). $$\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4{{h}^{3}}{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n-1}{\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]\left[ {{\xi }^{n-k+1}}-{{\xi }^{n-k}} \right]{{e}^{\ell m\varphi }}} \\ & ={{\left( {{\xi }^{n}}{{e}^{\ell m\varphi }} \right)}^{3}}\left( {{\xi }^{n+1}}{{e}^{\ell \left( m+2 \right)\varphi }}-2{{\xi }^{n+1}}{{e}^{\ell \left( m+1 \right)\varphi }}+2{{\xi }^{n+1}}{{e}^{\ell \left( m-1 \right)\varphi }}-{{\xi }^{n+1}}{{e}^{\ell \left( m-2 \right)\varphi }} \right). \\ \end{align}$$

Next, if we assume that ξ n+1 = ζ ξ n and ζ = ζ (φ) are independent of time, we can easily obtain the following expression:

4 h 3 Δ t − γ Γ 2 − γ ∑ k = 0 n − 1 k + 1 1 − γ − k 1 − γ ζ − k − 3 n + 1 − ξ − k − 3 n = ξ e 3 ℓ m φ e 2 ℓ φ − 2 e ℓ φ + 2 e − ℓ φ − e − 2 ℓ φ . $$\begin{align}& \,\,\,\frac{4{{h}^{3}}{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n-1}{\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]\left[ {{\zeta }^{-k-3n+1}}-{{\xi }^{-k-3n}} \right]} \\ =& \xi {{e}^{3\ell m\varphi }}\left( {{e}^{2\ell \varphi }}-2{{e}^{\ell \varphi }}+2{{e}^{-\ell \varphi }}-{{e}^{-2\ell \varphi }} \right). \\ \end{align}$$

Hence, we get

ξ=X1+iX2Y1+iY2, $$\xi =\frac{{{X}_{1}}+i{{X}_{2}}}{{{Y}_{1}}+i{{Y}_{2}}},$$ X1=2(Δt)−γΓ(2−γ)∑k=0n−1[ (k+1)1−γ−k1−γ ][ ζ1−3n−k−ζ−3n−k ],X2=0,Y1=−24cos2[ mγ ]sin[ mφ ]sin2[ φ2 ]sin[ φ ]+8sin3[ mφ ]sin2[ φ2 ]sin[ φ ],Y2=−24cos[ mγ ]sin2[ mφ ]sin2[ φ2 ]sin[ φ ]+8cos3[ mφ ]sin2[ φ2 ]sin[ φ ], $$\begin{align}& {{X}_{1}}=\frac{2{{\left( \Delta t \right)}^{-\gamma }}}{\Gamma \left( 2-\gamma \right)}\sum\limits_{k=0}^{n-1}{\left[ {{\left( k+1 \right)}^{1-\gamma }}-{{k}^{1-\gamma }} \right]\left[ {{\zeta }^{1-3n-k}}-{{\zeta }^{-3n-k}} \right]}, \\ & {{X}_{2}}=0, \\ & {{Y}_{1}}=-24{{\cos }^{2}}\left[ m\gamma \right]\sin \left[ m\varphi \right]{{\sin }^{2}}\left[ \frac{\varphi }{2} \right]\sin \left[ \varphi \right]+8{{\sin }^{3}}\left[ m\varphi \right]{{\sin }^{2}}\left[ \frac{\varphi }{2} \right]\sin \left[ \varphi \right], \\ & {{Y}_{2}}=-24\cos \left[ m\gamma \right]{{\sin }^{2}}\left[ m\varphi \right]{{\sin }^{2}}\left[ \frac{\varphi }{2} \right]\sin \left[ \varphi \right]+8{{\cos }^{3}}\left[ m\varphi \right]{{\sin }^{2}}\left[ \frac{\varphi }{2} \right]\sin \left[ \varphi \right], \\ \end{align}$$

which shows that

| ζ |=| X1+X2Y1+Y2 |. $$\left| \zeta \right|=\left| \frac{{{X}_{1}}+{{X}_{2}}}{{{Y}_{1}}+{{Y}_{2}}} \right|.$$

For the Fourier stability definition and for the examined scheme to be stable, the condition |ξ | ≤ 1 must be gratified. Therefore, if the following inequality is provided, the schema is unconditionally stable.

X12Y12+Y22≤1. $$\frac{X_{1}^{2}}{Y_{1}^{2}+Y_{2}^{2}}\le 1.$$

Other schemes can be studied by following a similar way.

L2 and L∞ error norms

The equation of the numerical results was obtained for the test problem used in this study and all computations have been run on using double precision arithmetic. To show how accurate the results, both the error norm L2

L2=| | Uexact−UN | |2=h∑J=0N| Ujexact−(UN)j |2, $${{L}_{2}}={{\left| \left| {{U}^{exact}}-{{U}_{N}} \right| \right|}_{2}}=\sqrt{h\sum\limits_{J=0}^{N}{{{\left| U_{j}^{exact}-{{\left( {{U}_{N}} \right)}_{j}} \right|}^{2}}}},$$

and L∞

L∞=| | Uexact−UN | |∞=maxj| Ujexact−(UN)j |, $${{L}_{\infty }}={{\left| \left| {{U}^{exact}}-{{U}_{N}} \right| \right|}_{\infty }}=\underset{j}{\mathop{\max }}\,\left| U_{j}^{exact}-{{\left( {{U}_{N}} \right)}_{j}} \right|,$$

are going to be computed and presented.
Test Problem

The analytical solution of the fractional Harry Dym equation is given as follows [11].

u(x,t)=(4−32(x+t))23,                t≥t0,                   0≤x≤1. $$u\left( x,t \right)={{\left( 4-\frac{3}{2}\left( x+t \right) \right)}^{\frac{2}{3}}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t\ge {{t}_{0}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\le x\le 1.$$

In our computations, for the numerical solution of the test problem three different linearization techniques have been applied. The values of the error norms L2 and L∞ have been computed at t = 1 for different values of Δt, h. The comparison of the error norms L2 and L∞ obtained by the linearization techniques is summarized in Table 1. As summarized in Table 1, it is obvious that the obtained results using linearization 1 are better than the obtained results using other linearizations. As shown in Figure 1, one can compare the exact and approximate solution of the collocation method using four radial basis functions for h = 0.01. Similarly, as shown in Figures 2 and 3 the exact and approximate solutions can be compared for linearization 2 and linearization 3 successively.

Fig. 1

Comparison of the exact and approximate solution for linearization 2 at h = 0.01

Fig. 2

Comparison of the exact and approximate solution for linearization 3 at h=0.01

Fig. 3

Comparison of the exact and approximate solution for linearization 1, linearization 2, and linearization 3 at h = 0.01

Comparison of the error norms L2 and L∞ that are obtained using the linearization techniques at t = 1,α = 0.9 for different values of Δt, h.

 Lin. I
 Lin. II
 Lin. III
L2 ×102 L∞ ×102 L2 ×102 L∞ ×102 L2 ×102 L∞ ×102
Δt = h = 0.001 0.00002 0.00002 0.00566 0.05664 0.00565 0.05659
Δt = h = 0.05 0.05254 0.08344 0.64556 0.94172 0.58516 0.84499
Δt = h = 0.04 0.02957 0.05128 0.50690 0.82106 0.47224 0.76100
Δt = h = 0.03 0.01418 0.02773 0.36800 0.68391 0.35113 0.65086
Δt = h = 0.02 0.00506 0.01186 0.23124 0.52313 0.22502 0.50855
Δt = h = 0.01 0.00088 0.00285 0.10161 0.32318 0.10046 0.31945
Δt = h = 0.2 2.44568 0.00285 3.12756 3.78842 0.80756 1.29370
Δt = h = 0.1 3.27620 0.41645 1.31211 1.41287 0.94027 0.94563
Conclusions

Finite difference methods based on using three different linearization techniques have been proposed for the numerical solutions of the fractional Harry Dym equation. In addition, numerical results were obtained by using three different linearization techniques. The proposed method has been tested on a problem and demonstrated how effective it is. The error norms L2 and L∞ have been calculated and given. The third linearization technique, as shown in Figure 3, yielded better results. Because the third linearization technique gives better results, this technique can be suggested in the next problems and the situation of the problem should be considered. The obtained results show that the error norms are sufficiently small during all computer runs. It has been observed that the considered method is a power numerical scheme to solve the fractional Harry Dym equation.
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