Solving Poisson’s Equations with fractional order using Haarwavelet

Wavelet is a wave like oscillation with a magnitude that begins at zero, increases, and then decreases back to zero. It can typically be visualized as a brief oscillation like one recorded by a seismograph or heat monitor. Generally, wavelets are purposefully crafted to have specific properties that make them useful for signal processing.

The Fourier transform is a useful tool to analyze the frequency components of the signal. However, if we take the Fourier transform over the whole time axis, we cannot tell at what instant a particular frequency rises. Short time Fourier transform uses a sliding window to find spectrogram, which gives the information of both time and frequency. But still another problem exists: The length of window limits the resolution in frequency. Wavelet transform seems to be a solution to the problem above. Wavelet transforms are based on small wavelets with limited duration. The translated-version wavelets locate where we concern. Whereas the scaled-version wavelets allow us to analyze the signal in different scale, [1] and [2]. In the last few decades many authors pointed out that derivatives and integrals of non-integer order are very suitable for the description of properties of various real material, e.g. polymers. It has been shown that new fractional order models are more adequate than previously used integer models. Fractional derivatives provide an excellent instrument for the description of memory and hereditary properties of various materials and processes. This is the main advantage of fractional derivatives in comparison with classical integer order models [1].

The first literature that relates to the wavelet transform is Haar wavelet. It was proposed by the mathematician AlfrdHaar in 1909. However, the concept of the wavelet did not exist at that time. Until 1981, the concept was proposed by the geophysicist Jean Morlet. Afterward, Morlet and the physicist Alex Grossman invented the term wavelet in 1984. Before 1985, Haar wavelet was the only orthogonal wavelet people know. A lot of researchers even thought that there was no orthogonal wavelet except Haar wavelet. Fortunately, the mathematician Yves Meyer constructed the second orthogonal wavelet called Meyer wavelet in 1985. As more and more scholars joined in this field, the 1^st international conference was held in France in 1987. In 1988, StephaneMallat and Meyer proposed the concept of multi resolution. In the same year, Ingrid Daubechies found a systematical method to construct the compact support orthogonal wavelet. In 1989, Mallat proposed the fast wavelet transform. With the appearance of this fast algorithm, the wavelet transform had numerous applications in the signal processing field, [1]. In 1910, Haar showed that certain square wave functions could be translated and scaled to create a basis set that span the space L². Years later, it was seen that the system of Haar is a particular wavelet system. In comparison with other techniques, which use the same structure of building bases functions and introduce the solution as a linear combination of those base. The Haar wavelet is simple, can implement standard algorithms with high accuracy for a small number of grid points. The simplicity in building the wavelet bases from any function which use only two operations translation and dilation [3], this can be easily seen in Haar wavelet.The simple form of the mother function in Haar wavelet as we see below makes the processes of dilation and translation an easy work and the introduced wavelet family is orthogonal not only linearly independent. Although, the wavelet function appeared in 1910, their use in the solution of differential equations does not appear until recently [4–6], last twenty years.In 2017 Kaoud and El Dewaik, [7] have used Haar wavelet technique to solve Poisson’s equation on a unit square domain with collocation points j/16, j = 1, 3,... , 15. The results obtained here can be seen as a generalization to those we have obtained in [7]. The classical integer case can be seen as limiting process as the order of the fractional derivative appears the integer case.

There are many definitions for fractional order differentiation in fractional calculus e.g: Riemann- Lioville, Caputo fractional and Grünwald-Letnikov fractional. They are given as follows, [8]:

In 1910 Haar showed that certain square wave functions could be translated and scaled to create a basis set that span L²([0, 1]), [9].

The scaling function should have a compact support over 0≤x≤1, therefore

h0(x)={1,0<x≤1,0,otherwise$$\begin{array}{} \displaystyle h_0\mathrm{(}x\mathrm{)=}\left\{ \begin{array}{ll} \mathrm{1,} & 0<x\mathrm{\le }\mathrm{1,} \\ 0, & otherwise \end{array} \right. \end{array}$$(5)

And the mother wavelets function h₁(t) as:

h1(x)={1,0≤x<12,-1,12≤x<1,0,otherwise$$\begin{array}{} \displaystyle h_{\mathrm{1}}\mathrm{(}x\mathrm{)=}\left\{ \begin{array}{ll} \mathrm{1,} & \mathrm{0}\mathrm{\le }x\mathrm{<}\frac{\mathrm{1}}{\mathrm{2}}, \\ \mathrm{-}\mathrm{1,} & \frac{\mathrm{1}}{\mathrm{2}}\mathrm{\le }x\mathrm{<}1, \\ 0, & otherwise \end{array} \right. \end{array}$$(6)

All the other subsequent functions are generated from h₁(x) with two operations: translation and dilation That is

hn(x)=h1(2jx-k);n≥1$$\begin{array}{} \displaystyle h_n\mathrm{(}x\mathrm{)=}h_{\mathrm{1}}\mathrm{(}{\mathrm{2}}^jx\mathrm{-}k\mathrm{)}; n\ \ge 1 \end{array}$$(7)

where n=2^j+k, 0≤j, 0≤k<2^j.

h₀ (t) is also included to make this set complete.

The Haar wavelets are orthogonal in the sense,

∫01hi(t)hl(t)dt=2−jδil={2−ji=l=2j+k0i≠l$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {\mathop \smallint \limits_0^1 {h_i}(t){h_l}(t)dt = \,{2^{ - j}}{\delta _{il}}} \hfill \\ { = \{ \begin{array}{*{20}{c}} {{2^{ - j}}\,\,\,i = l = {2^j} + k} \\ {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i \ne l} \\ \end{array}} \hfill \\ \end{array} \end{array}$$

Therefore, they form a set of basis functions.

The fractional integrals of the first eight Haar wavelets can be expressed as

q0=J0Rtαh0(t)=1αΓ(α)tα,0≤t<1,q1=J0Rtαh1(t)=1αΓ(α){tα,0≤t<12tα−2(t−12)α,12≤t<1q2=J0Rtαh2(t)=1αΓ(α){tα0≤t<14tα−2(t−14)α,14≤t<12tα−2(t−14)α,12≤t<1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{q_0} = {}_0^RJ_t^\alpha {h_0}(t) = \frac{1}{{\alpha \Gamma (\alpha )}}{t^\alpha },0 \le t < 1,} \\ {{q_1} = {}_0^RJ_t^\alpha {h_1}(t) = \frac{1}{{\alpha \Gamma (\alpha )}}\{ \begin{array}{*{20}{c}} {{t^\alpha },} & {0 \le t < \frac{1}{2}} \\ {{t^\alpha } - 2{{(t - \frac{1}{2})}^\alpha },} & {\frac{1}{2} \le t < 1} \\ \end{array}} \\ {{q_2} = {}_0^RJ_t^\alpha {h_2}(t) = \frac{1}{{\alpha \Gamma (\alpha )}}\{ \begin{array}{*{20}{c}} {{t^\alpha }} \hfill & {0 \le t < \frac{1}{4}} \hfill & {} \hfill \\ {{t^\alpha } - 2{{(t - \frac{1}{4})}^\alpha },} \hfill & {\frac{1}{4} \le t < \frac{1}{2}} \hfill & {} \hfill \\ {{t^\alpha } - 2{{(t - \frac{1}{4})}^\alpha },} \hfill & {\frac{1}{2} \le t < 1} \hfill & {} \hfill \\ \end{array}} \\ \end{array} \end{array}$$

q3=0RJtαh3(t)=1αΓ(α){(t−12)α,12≤t<34(t−12)α−2(t−34)α,34≤t<1q4=0RJtαh4(t)=1αΓ(α){tα,tα−2(t−18)α,tα−2(t−18)α+(t−14)α,0≤t<1818≤t<1414≤t<1q5=0RJtαh5(t)=1αΓ(α){(t−14)α,(t−14)α−2(t−38)α,(t−14)α−2(t−38)α+(t−12)α,14≤t<3838≤t<1212≤t<1q6=0RJtαh6(t)=1αΓ(α){(t−12)α,(t−12)α−2(t−58)α,(t−12)α−2(t−58)α+(t−34)α,12≤t<5858≤t<3434≤t<1q7=0RJtαh7(t)=1αΓ(α){(t−34)α,34≤t<78(t−34)α−2(t−78)α,78≤t<1$$\begin{array}{l} {q_3} = _0^RJ_t^\alpha {h_3}\left( t \right) = \frac{1}{{\alpha \Gamma \left( \alpha \right)}} \left\{ \begin{array}{l} {{{\left( {t - \frac{1}{2}} \right)}^\alpha },}&{\frac{1}{2} \le t < \frac{3}{4}}\\ {{{\left( {t - \frac{1}{2}} \right)}^\alpha } - 2{{\left( {t - \frac{3}{4}} \right)}^\alpha },}&{\frac{3}{4} \le t < 1} \end{array}\right.\\ {{q_4} = _0^RJ_t^\alpha {h_4}\left( t \right) = \frac{1}{{\alpha \Gamma \left( \alpha \right)}} \left\{ \begin{array}{l} {{t^\alpha },}\\ {{t^\alpha } - 2{{\left( {t - \frac{1}{2}} \right)}^\alpha },}\\ {{t^\alpha } - 2{{\left( {t - \frac{1}{8}} \right)}^\alpha } + {{\left( {t - \frac{1}{4}} \right)}^\alpha },} \end{array}\begin{array}{*{20}{c}} {0 \le t < \frac{1}{8}}\\ {\frac{1}{8} \le t < \frac{1}{8}}\\ {\frac{1}{4} \le t < 1} \end{array} \right.}\\ {{q_5} = _0^RJ_t^\alpha {h_5}\left( t \right) = \frac{1}{{\alpha \Gamma \left( \alpha \right)}} \left\{ \begin{array}{l} {{{\left( {t - \frac{1}{4}} \right)}^\alpha },}\\ {{{\left( {t - \frac{1}{4}} \right)}^\alpha } - 2{{\left( {t - \frac{3}{8}} \right)}^\alpha },}\\ {{{\left( {t - \frac{1}{4}} \right)}^\alpha } - 2{{\left( {t - \frac{3}{8}} \right)}^\alpha } + {{\left( {t - \frac{1}{2}} \right)}^\alpha },} \end{array}{\rm{ }}\begin{array}{*{20}{c}} {\frac{1}{4} \le t < \frac{3}{8}}\\ {\frac{3}{8} \le t < \frac{1}{2}}\\ {\frac{1}{2} \le t < 1} \end{array}\right.}\\ {{q_6} = _0^RJ_t^\alpha {h_6}\left( t \right) = \frac{1}{{\alpha \Gamma \left( \alpha \right)}} \left\{ \begin{array}{*{20}{l}} {{{\left( {t - \frac{1}{2}} \right)}^\alpha },}\\ {{{\left( {t - \frac{1}{2}} \right)}^\alpha } - 2{{\left( {t - \frac{5}{8}} \right)}^\alpha },}\\ {{{\left( {t - \frac{1}{2}} \right)}^\alpha } - 2{{\left( {t - \frac{5}{8}} \right)}^\alpha } + {{\left( {t - \frac{3}{4}} \right)}^\alpha },} \end{array}{\rm{ }}\begin{array}{*{20}{c}} {\frac{1}{2} \le t < \frac{5}{8}}\\ {\frac{5}{8} \le t < \frac{3}{4}}\\ {\frac{3}{4} \le t < 1} \end{array}\right.}\\ {{q_7} = _0^RJ_t^\alpha {h_7}\left( t \right) = \frac{1}{{\alpha \Gamma \left( \alpha \right)}} \left\{ \begin{array}{*{20}{l}} {{{\left( {t - \frac{3}{4}} \right)}^\alpha },}&{\frac{3}{4} \le t < \frac{7}{8}}\\ {{{\left( {t - \frac{3}{4}} \right)}^\alpha } - 2{{\left( {t - \frac{7}{8}} \right)}^\alpha },}&{\frac{7}{8} \le t < 1} \end{array}\right.} \end{array}$$

Fractional Poisson’s equation has the form

∂αu∂xα+∂αu∂yα=F(x,y),1<α≤20≤x≤1,0≤y≤1,$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {\frac{{{\partial ^\alpha }u}}{{\partial {x^\alpha }}} + \frac{{{\partial ^\alpha }u}}{{\partial {y^\alpha }}} = F(x,y),1 < \alpha \le 2} \\ {0 \le x \le 1\,\,,\,0 \le y \le 1,} \\ \end{array} \end{array}$$(8)

With boundary conditions

u(x,0)=f1(x)u(x,1)=f2(x)}0≤x≤1$$\begin{array}{} \displaystyle \left. \begin{array}{c} u\left(x,0\right)=f_1(x) \\ u\left(x,1\right)=f_2(x) \end{array} \right\}\ \ \ \ \ \ \ \ 0\le x\le 1 \end{array}$$(9)

u(0,y)=g1(y)u(1,y)=g2(y)}0≤y≤1.$$\begin{array}{} \displaystyle \left. \begin{array}{c} u\left(0,y\right)=g_1(y) \\ u\left(1,y\right)=g_2(y) \end{array} \right\}\ \ \ \ \ \ \ \ 0\le y\le 1. \end{array}$$(10)

According to the two-dimensional multi-resolution analysis, [11], any function u(x,y) which is square integrable on [0,1]×[0,1] can be expressed in terms of two dimensional Haar series as follows

u(x,y)=∑i=1∞∑j=1∞ai,jhi(x)hj(y)$$\begin{array}{} \displaystyle u\left(x,y\right)=\sum^{\infty }_{i=1}{\sum^{\infty }_{j=1}{a_{i,j}h_i(x)h_j(y)}} \end{array}$$(11)

This series can be taken as an approximation for the solution of Poisson’s equation. Moreover, the expansion of u(x,y) can be terminated.

u(x,y)=∑i=12M1∑j=12M2ai,jhi(x)hj(y)$$\begin{array}{} \displaystyle u\left(x,y\right)=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_i(x)h_j(y)}} \end{array}$$(12)

where the wavelet coefficients a_i,j i=1,2,. . . ,2M₁, j=1,2,. . . ,2M₂ are to be determined.

The approach of Haar wavelet depends on writing the dominant derivative term in the form

uxαyα=∑i=12M1∑j=12M2ai,jhi(x)hj(y)$$\begin{array}{} \displaystyle u_{x^{\alpha }y^{\alpha }}=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_i(x)h_j(y)}} \end{array}$$(13)

Integrating (13) with respect to y in the limits [0,y]

uxαyα−1=∑i=12M1∑j=12M2ai,jhi(x)Pj(y)+C1(x)$$\begin{array}{} \displaystyle u_{x^{\alpha }y^{\alpha -1}}=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_i\left(x\right)P_j\left(y\right)+C_1}(x)} \end{array}$$(14)

Integrating (14) with respect to y (fractional of order α − 1), we get

uxα=∑i=12M1∑j=12M2ai,jhi(x)qj(y)+yα−1C1(x)(α−1)Γ(α−1)+C2(x)$$\begin{array}{} \displaystyle u_{x^{\alpha }}=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_i\left(x\right)q_j\left(y\right)+{y^{\alpha -1}\frac{C_1(x)}{(\alpha -1){\Gamma}(\alpha -1)}+C}_2}(x)} \end{array}$$(15)

Using the boundary and the initial conditions we can get C₁ (x)and C₂ (x). And accordingly one can obtain

uxα(x,y)=∑i=12M1∑j=12M2ai,jhi(x)[qj(y)−yα−1qj(1)]+yα−1∂αf2(x)∂xα+(1−yα−1)∂αf1(x)∂xα$$\begin{array}{} \displaystyle u_{x^{\alpha }}\left(x,y\right)=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_i\left(x\right)\left[q_j\left(y\right)-y^{\alpha -1}q_j\left(1\right)\right]+y^{\alpha -1}\frac{{\partial }^{\alpha }f_2(x)}{\partial x^{\alpha }}+(1-y^{\alpha -1})\frac{{\partial }^{\alpha }f_1(x)}{\partial x^{\alpha }}}} \end{array}$$(16)

Similarly, integrating (13) with respect to x in the limits [0,x]

uxα−1yα=∑i=12M1∑j=12M2ai,jhj(y)Pi(x)+C3(y)$$\begin{array}{} \displaystyle u_{x^{\alpha -1}y^{\alpha }}=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_j\left(y\right)P_i\left(x\right)+C_3}(y)} \end{array}$$(17)

Integrating (17) with respect to x (fractional of order α − 1), we get

uyα=∑i=12M1∑j=12M2ai,jhj(y)qi(x)+xα−1C3(y)(α−1)Γ(α−1)+C4(y)$$\begin{array}{} \displaystyle u_{y^{\alpha }}=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_j\left(y\right)q_i\left(x\right)+{x^{\alpha -1}\frac{C_3(y)}{(\alpha -1){\Gamma}(\alpha -1)}+C}_4}(y)} \end{array}$$(18)

Using the boundary and the initial conditions we can get C₃ (y)and C₄ (y). And accordingly one can obtain

uxα(x,y)=∑i=12M1∑j=12M2ai,jhj(y)[qi(x)−xα−1qi(1)]+xα−1∂αg2(x)∂yα+(1−xα−1)∂αg1(x)∂yα$$\begin{array}{} \displaystyle u_{x^{\alpha }}\left(x,y\right)=\sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}h_j\left(y\right)\left[q_i\left(x\right)-x^{\alpha -1}q_i\left(1\right)\right]+x^{\alpha -1}\frac{{\partial }^{\alpha }g_2(x)}{\partial y^{\alpha }}+(1-x^{\alpha -1})\frac{{\partial }^{\alpha }g_1(x)}{\partial y^{\alpha }}}} \end{array}$$(19)

Then we Integrate equation (16) two times with respect to x and using equation (10), we obtain

u(x,y)=∑i=12M1∑j=12M2αi,j[qi(x)−xα−1qi(1)][qj(y)−yα−1qj(1)]+xα−1g2(y)+(1−xα−1)g1(y)+yα−1f2(x)+(1−yα−1)f1(x)−xα−1yα−1f2(1)+xα−1(1−yα−1)f1(1)−(1−xα−1)yα−1f2(0)−(1−xα−1)(1−yα−1)f1(0)$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u(x,y) = \mathop \sum \limits_{i = 1}^{2{M_1}} \mathop \sum \limits_{j = 1}^{2{M_2}} {\alpha _{i,j}}[{q_i}(x) - {x^{\alpha - 1}}{q_i}(1)][{q_j}(y) - {y^{\alpha - 1}}{q_j}(1)]} \hfill \\ { + {x^{\alpha - 1}}{g_2}(y) + (1 - {x^{\alpha - 1}}){g_1}(y) + {y^{\alpha - 1}}{f_2}(x)} \hfill \\ { + (1 - {y^{\alpha - 1}}){f_1}(x) - {x^{\alpha - 1}}{y^{\alpha - 1}}{f_2}(1) + {x^{\alpha - 1}}(1 - {y^{\alpha - 1}}){f_1}(1)} \hfill \\ { - (1 - {x^{\alpha - 1}}){y^{\alpha - 1}}{f_2}(0) - (1 - {x^{\alpha - 1}})(1 - {y^{\alpha - 1}}){f_1}(0)} \hfill \\ \end{array} \end{array}$$(20)

The wavelet collocation points are defined by

xl=l−0.52M1,l=1,2,…,2M1$$\begin{array}{} \displaystyle x_l=\frac{l-0.5}{2M_1},\ \ \ \ \ \ \ \ l=1,2,\dots ,\ 2M_1 \end{array}$$(21)

yn=n−0.52M2,n=1,2,…,2M2$$\begin{array}{} \displaystyle y_n=\frac{n-0.5}{2M_2},\ \ \ \ \ \ \ \ n=1,2,\dots ,\ 2M_2 \end{array}$$(22)

Substituting equations (17) and (18) in equation (8), and replacing x by x_l and y by y_n in the obtained equations and equation (16), we arrive at

∑i=12M1∑j=12M2ai,jA(i,j,l,n)=∅(xl,yn)$$\begin{array}{} \displaystyle \sum^{2M_1}_{i=1}{\sum^{2M_2}_{j=1}{a_{i,j}\ A\left(i,j,l,n\right)=\emptyset (x_l,y_n)}} \end{array}$$(23)

Where

A(i,j,l,n)=hi(xl)[qj(yn)−ynα−1qj(1)]+[qi(xl)−xlα−1qi(1)]hj(yn)$$\begin{array}{} \displaystyle A\left(i,j,l,n\right)=h_i\left(x_l\right)\left[q_j\left(y_n\right)-y^{\alpha -1}_nq_j\left(1\right)\right]+\left[q_i\left(x_l\right)-x^{\alpha -1}_lq_i(1)\right]h_j(y_n) \end{array}$$(24)

∅(xl,yn)=(ynα−1−1)f1"1(xl)−ynα−1∂αf2(xl)∂xα+(xlα−1−1)∂αg1(yn)∂yα−xlα−1∂αg2(yn)∂yα+F(xl,yn)$$\begin{array}{} \displaystyle \emptyset \left( {{x_l},{y_n}} \right) = \left( {y_n^{\alpha - 1} - 1} \right)f_1^1\left( {{x_l}} \right) - y_n^{\alpha - 1}\frac{{{\partial ^\alpha }{f_2}({x_l})}}{{\partial {x^\alpha }}} + \left( {x_l^{\alpha - 1} - 1} \right)\frac{{{\partial ^\alpha }{g_1}({y_n})}}{{\partial {y^\alpha }}} - x_l^{\alpha - 1}\frac{{{\partial ^\alpha }{g_2}({y_n})}}{{\partial {y^\alpha }}} + F({x_l},{y_n}) \end{array}$$(25)

u(xl,yn)=∑i=12M1∑j=12M2ai,j[qi(xl)−xlα−1qi(1)][qj(yn)−ynα−1qj(1)]+xlα−1g2(yn)+(1−xlα−1)g1(yn)+ynα−1f2(xl)+(1−ynα−1)f1(xl)−ynα−1xlα−1f2(1)−xlα−1(1−ynα−1)f1(1)−(1−xlα−1)ynα−1f2(0)−(1−xlα−1)(1−ynα−1)f1(0)$$\begin{array}{} \displaystyle \begin{array}{*{20}{l}} {u\left( {{x_l},{y_n}} \right) = \sum\limits_{i = 1}^{2{M_1}} {\sum\limits_{j = 1}^{2{M_2}} {{a_{i,j}}} } \left[ {{q_i}\left( {{x_l}} \right) - x_l^{\alpha - 1}{q_i}\left( 1 \right)} \right]\left[ {{q_j}\left( {{y_n}} \right) - y_n^{\alpha - 1}{q_j}\left( 1 \right)} \right]}\\ { + x_l^{\alpha - 1}{g_2}\left( {{y_n}} \right) + \left( {1 - x_l^{\alpha - 1}} \right){g_1}\left( {{y_n}} \right) + y_n^{\alpha - 1}{f_2}\left( {{x_l}} \right)}\\ { + \left( {1 - y_n^{\alpha - 1}} \right){f_1}\left( {{x_l}} \right) - y_n^{\alpha - 1}x_l^{\alpha - 1}{f_2}\left( 1 \right) - x_l^{\alpha - 1}\left( {1 - y_n^{\alpha - 1}} \right){f_1}\left( 1 \right)}\\ { - \left( {1 - x_l^{\alpha - 1}} \right)y_n^{\alpha - 1}{f_2}\left( 0 \right) - (1 - x_l^{\alpha - 1})(1 - y_n^{\alpha - 1}){f_1}(0)} \end{array} \end{array}$$(26)

The coefficients a_i,j, i=1,2,. . . ,2M₁, j=1,2,. . . ,2M₂ are found from equation (19). Then we substitute in equation (22) to obtain the Haar solution at the collocation points x_l,l = 1,2,...,2M₁, j = 1,2,... ,2M₂.

This section is a generalization of a previous work done for the integer case [7]. The properties of the resulting linear system using Haar wavelet method are investigated.

Where D1α,D2α,D3α,D4α,A1α,A2α,A3α,B1α,B2α,$\begin{array}{} \displaystyle D_1^\alpha \;,D_2^\alpha \;,D_3^\alpha \;,D_4^\alpha \;,A_1^\alpha \;,A_2^\alpha \;,A_3^\alpha \;,B_1^\alpha \;,B_2^\alpha \;, \end{array}$ and C1α$\begin{array}{} \displaystyle C_1^\alpha \end{array}$ are illustrated in the appendix. For the integer case (at α = 2) we have

D12=164[−7−11−11−7−11−15−15−11−11−15−15−11−7−11−11−7],D22=164[−3−333−3−33333−3−333−3−3],D32=1128[04314−8−3−13−1001−100],D42=1128[00−1100−33−1−3−841340],A12=164[−5−555−9−999−9−999−5−555],A22=1128[−71131−151931−151931−71131],B22=1128[−1−3−117−1−3−1915−1−3−1915−1−3−117],B12=1128[−3731−37313−7−3−13−7−3−1],B22=1128[−1−3−73−1−3−73137−3137−3],C12=1128[−1−3−40138−4003−3001−1].$$\begin{array}{*{20}{c}} {D_1^2 = \frac{1}{{64}}{\rm{ }}\left[ {\begin{array}{*{20}{c}} { - 7}&{ - 11}&{ - 11}&{ - 7}\\ { - 11}&{ - 15}&{ - 15}&{ - 11}\\ { - 11}&{ - 15}&{ - 15}&{ - 11}\\ { - 7}&{ - 11}&{ - 11}&{ - 7} \end{array}} \right],D_2^2 = \frac{1}{{64}}\left[ {\begin{array}{*{20}{c}} { - 3}&{ - 3}&3&3\\ { - 3}&{ - 3}&3&3\\ 3&3&{ - 3}&{ - 3}\\ 3&3&{ - 3}&{ - 3} \end{array}} \right],D_3^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{c}} 0&4&3&1\\ 4&{ - 8}&{ - 3}&{ - 1}\\ 3&{ - 1}&0&0\\ 1&{ - 1}&0&0 \end{array}} \right],}\\ {D_4^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{c}} 0&0&{ - 1}&1\\ 0&0&{ - 3}&3\\ { - 1}&{ - 3}&{ - 8}&4\\ 1&3&4&0 \end{array}} \right],A_1^2 = \frac{1}{{64}}\left[ {\begin{array}{*{20}{c}} { - 5}&{ - 5}&5&5\\ { - 9}&{ - 9}&9&9\\ { - 9}&{ - 9}&9&9\\ { - 5}&{ - 5}&5&5 \end{array}} \right],}\\ {A_2^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{c}} { - 7}&{11}&3&1\\ { - 15}&{19}&3&1\\ { - 15}&{19}&3&1\\ { - 7}&{11}&3&1 \end{array}} \right],B_2^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 3}&{ - 11}&7\\ { - 1}&{ - 3}&{ - 19}&{15}\\ { - 1}&{ - 3}&{ - 19}&{15}\\ { - 1}&{ - 3}&{ - 11}&7 \end{array}} \right],}\\ {B_1^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{r}} { - 3}&7&3&1\\ { - 3}&7&3&1\\ 3&{ - 7}&{ - 3}&{ - 1}\\ 3&{ - 7}&{ - 3}&{ - 1} \end{array}} \right],B_2^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{r}} { - 1}&{ - 3}&{ - 7}&3\\ { - 1}&{ - 3}&{ - 7}&3\\ 1&3&7&{ - 3}\\ 1&3&7&{ - 3} \end{array}} \right],}\\ {C_1^2 = \frac{1}{{128}}\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 3}&{ - 4}&0\\ 1&3&8&{ - 4}\\ 0&0&3&{ - 3}\\ 0&0&1&{ - 1} \end{array}} \right].} \end{array}$$

Finite difference approximations of fractional derivatives

In (2014) I.K.Youssef and A.M.Shoukr, [12], represented the structure of the coefficient matrix of fractional Poisson’s equation using finite difference method.

In this method the integral in Caputo’s formula is replaced by a finite sum of integrals at the discretization points, and approximate the second order derivative by using the standard finite difference formula, then the finite difference formula of fractional Poisson’s equation takes the form:

∑k=0i−1bk(Ui−k+1,j−2Ui−k,j+Ui−k−1,j+bs*(Ui,j−s+1−2Ui,j−s+Ui,j−s−1)=fi,j$$\begin{array}{} \displaystyle \sum^{i-1}_{k=0}{b_k(U_{i-k+1,j}-2U_{i-k,j}}+U_{i-k-1,j\ \ \ \ }+b^*_s\left(U_{i,j-s+1}-2U_{i,j-s}+U_{i,j-s-1}\right)=f_{i,j} \end{array}$$

The structure of coefficient matrix

A=[A″1(b0*)*I00(b0*−2b1*)*IA1(b0*)*I0(b1*−2b2*)*I(b0*−2b1*−2b2*)*IA1(b0*)*I(b2*−2b3*)*I(b1*−2b2*−2b3*)*I(b0*−2b1*−2b2*)*IA1],A1=[(b1−2b0−2b0*+b1*)b000b0−2b1(b1−2b0−2b0*+b1*)b00b1−2b2(b0*−2b1*−2b2*)*I(b1−2b0−2b0*+b1*)b0b2−2b3(b1*−2b2*−2b3*)*I(b0*−2b1*−2b2*)*I(b1−2b0−2b0*+b1*)](A"1)ij={−2b0−2b0•ifi=jb1−2b0−2b0*+b1•ifi=j=2,3,...,N−1b0ifi=i+1,i=1,2...,N−2bi−2−2bi−1ifi=2,3,...,N−1,j=1bi−j−1−2bi−j−2bi−j+1ifi=2,3,...M−1,j=10othereise$$\begin{array}{*{20}{l}} {{\rm{ }}A = \left[ {\begin{array}{*{20}{c}} {A''1}&{(b_0^*)*I}&0&0\\ {(b_0^* - 2b_1^*)*I}&{A1}&{(b_0^*)*I}&0\\ {(b_1^* - 2b_2^*)*I}&{(b_0^* - 2b_1^* - 2b_2^*)*I}&{A1}&{(b_0^*)*I}\\ {(b_2^* - 2b_3^*)*I}&{(b_1^* - 2b_2^* - 2b_3^*)*I}&{(b_0^* - 2b_1^* - 2b_2^*)*I}&{A1} \end{array}} \right],}\\ {A1 = \left[ {\begin{array}{*{20}{c}} {({b_1} - 2{b_0} - 2b_0^* + b_1^*)}&{{b_0}}&0&0\\ {{b_0} - 2{b_1}}&{({b_1} - 2{b_0} - 2b_0^* + b_1^*)}&{{b_0}}&0\\ {{b_1} - 2{b_2}}&{(b_0^* - 2b_1^* - 2b_2^*)*I}&{({b_1} - 2{b_0} - 2b_0^* + b_1^*)}&{{b_0}}\\ {{b_2} - 2{b_3}}&{(b_1^* - 2b_2^* - 2b_3^*)*I}&{(b_0^* - 2b_1^* - 2b_2^*)*I}&{({b_1} - 2{b_0} - 2b_0^* + b_1^*)} \end{array}} \right]}\\ {{\rm{ }}{{\left( {A1} \right)}_{ij}} = \left\{ \begin{array}{*{20}{c}} { - 2{b_0} - 2b_0^ \bullet }&{ifi = j}\\ {{b_1} - 2{b_0} - 2b_0^* + b_1^ \bullet }&{ifi = j = 2,3,...,N - 1}\\ {{b_0}}&{ifi = i + 1,i = 1,2...,N - 2}\\ {{b_{i - 2}} - 2{b_{i - 1}}}&{ifi = 2,3,...,N - 1,j = 1}\\ {{b_{i - j - 1}} - 2{b_{i - j}} - 2{b_{i - j + 1}}}&{ifi = 2,3,...M - 1,j = 1}\\ 0&{othereise} \end{array}\right.} \end{array}$$

While in case of the finite difference method the resulting coefficient matrix is block tri diagonal matrix with the natural ordering is considered [13], [14].

The following fractional Poisson’s equation:

∂αU(x,y)∂xα+∂αU(x,y)∂yα=f(x,y)$$\begin{array}{} \displaystyle \frac{{\partial }^{\alpha }U(x,y)}{\partial \ x^{\alpha }}+\frac{{\partial }^{\alpha }U(x,y)}{\partial \ y^{\alpha }}=f(x,y) \end{array}$$

Was considered on a finite domain 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 with the non-homogeneous function f (x,y) = Γ(α + 1)(x^α + y^α) and the boundary conditions:

U(x,0)=U(0,y)=0,U(x,1)=xα,U(1,y)=yα.$$\begin{array}{} \displaystyle U\left(x,0\right)=U\left(0,y\right)=0, U\left(x,1\right)=\ x^{\alpha },\ U\left(1,\ y\right)=\ y^{\alpha }. \end{array}$$

This fractional Poisson’s equation has the exact solution U (x, y) = (xy)^α. The fractional Poisson absolute error is defined by:

Error=1(m−1)2∑i,j=1m−1(Ui,j−ui,j)2$\begin{array}{} \displaystyle {\rm{Error}} = \frac{1}{{{{(m - 1)}^2}}}\sqrt {{{\sum\nolimits_{i,j = 1}^{m - 1} {\left( {{U_{i,j}} - {u_{i,j}}} \right)} }^2}} \end{array}$, in which U_i,j and u_i,j are the exact and numerical solutions respectively, [15].

This problem is solved using Haar wavelet method. The results show higher accuracy compared with the finite difference method, [15].

The approximate solution at α = 2 error of order 10⁻¹⁹

The approximate solution at α = 1.75 error of order 10⁻¹⁸

Fig. 1

Fig. 2

Fig. 3

The wavelet solution gives reliable results for the fractional order Poisson’s equation as in the integer case. The numerical results obtained generalize the results of the classical integer case. Moreover, the matrix structure

Fig. 4

of the linear system (27) has the symmetric block structure. Comparison with the corresponding matrix appears in the finite difference treatment help in building the block structure [12]. The memory and hereditary behaviors of the fractional order derivatives appears with the coefficients through the Gamma function factors.

D1α=[d1ij]d111=−14+21−3α,d122=−34+21−3α(3)α,d133=−54+21−3α(5)α,d144=−74+21−3α(7)αd112=d121=−12+(38)α+8−α,d113=d131=−34+(58)α+8−αd114=d41=−1+(78)α+8−α,d123=d132=−1+(38)α+(58)αd124=d42=−54+(38)α+(58)α,d134=d143=−32+(58)α+(78)α.$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {D_1^\alpha = \;\left[ {d{1_{ij}}} \right]} \\ {d{1_{11}} = \frac{{ - 1}}{4} + {2^{1 - 3\alpha }},\,d{1_{22}} = \frac{{ - 3}}{4} + {2^{1 - 3\alpha }}{{(3)}^\alpha },d{1_{33}} = \frac{{ - 5}}{4} + {2^{1 - 3\alpha }}{{(5)}^\alpha },d{1_{44}} = \frac{{ - 7}}{4} + {2^{1 - 3\alpha }}{{(7)}^\alpha }} \\ {d{1_{12}} = d{1_{21}} = \frac{{ - 1}}{2} + {{(\frac{3}{8})}^\alpha } + {8^{ - \alpha }},d{1_{13}} = d{1_{31}} = \frac{{ - 3}}{4} + {{(\frac{5}{8})}^\alpha } + {8^{ - \alpha }}} \\ {d{1_{14}} = {d_{41}} = - 1 + {{(\frac{7}{8})}^\alpha } + {8^{ - \alpha }},\;d{1_{23}} = d{1_{32}} = - 1 + {{(\frac{3}{8})}^\alpha } + {{(\frac{5}{8})}^\alpha }} \\ {d{1_{24}} = {d_{42}} = \frac{{ - 5}}{4} + {{(\frac{3}{8})}^\alpha } + {{(\frac{5}{8})}^\alpha },d{1_{34}} = d{1_{43}} = \frac{{ - 3}}{2} + {{(\frac{5}{8})}^\alpha } + {{(\frac{7}{8})}^\alpha }.} \\ {} \\ \end{array} \end{array}$$

D2α=[d2ij]d211=−14+21−3α+2−1−α,d222=−34+21−3α(3)α+3(2)−1−α,d233=2−2−3α(16−5(2)1+2α−8(5)α+5(8)α,d244=2−1−3α(4−3(2)1+2α+4(3)α−2(5)α−2(7)α+3(8)α),d212=d221=−12+(38)α+3(2)−2−α+8−α,d213=d231=−12+(58)α+(2)−α+3(8)−αd214=d221=2−2−3α(−4+3(2)1+2α−8(3)α+4(7)α−3(8)α),d223=d232=2−2−3α(−8+(2)1+2α−4(3)α+4(5)α−(8)α),d224=d242=−12+(78)α+2−α+31+α8−α,d234=d243=2−1−3α(4−3(2)1+2α+4(3)α−2(5)α−2(7)α+3(8)α),$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {D_2^\alpha = \;\left[ {d{2_{ij}}} \right]} \\ {d{2_{11}} = \frac{{ - 1}}{4} + {2^{1 - 3\alpha }} + {2^{ - 1 - \alpha }},d{2_{22}} = \frac{{ - 3}}{4} + {2^{1 - 3\alpha }}{{(3)}^\alpha } + 3{{(2)}^{ - 1 - \alpha }},} \\ {d{2_{33}} = {2^{ - 2 - 3\alpha }}(16 - 5{{\left( 2 \right)}^{1 + 2\alpha }} - 8\;{{\left( 5 \right)}^\alpha } + 5{{(8)}^\alpha },} \\ {d{2_{44}} = {2^{ - 1 - 3\alpha }}(4 - 3{{\left( 2 \right)}^{1 + 2\alpha }} + 4\;{{\left( 3 \right)}^\alpha } - 2\left( {5{)^\alpha } - 2{{\left( 7 \right)}^\alpha } + 3{{\left( 8 \right)}^\alpha }} \right),} \\ {d{2_{12}} = d{2_{21}} = \frac{{ - 1}}{2} + {{(\frac{3}{8})}^\alpha } + 3{{\left( 2 \right)}^{ - 2 - \alpha }} + {8^{ - \alpha }},d{2_{13}} = d{2_{31}} = \frac{{ - 1}}{2} + {{(\frac{5}{8})}^\alpha } + {{\left( 2 \right)}^{ - \alpha }} + 3{{(8)}^{ - \alpha }}} \\ {d{2_{14}} = d{2_{21}} = {2^{ - 2 - 3\alpha }}( - 4 + 3{{\left( 2 \right)}^{1 + 2\alpha }} - 8\;{{\left( 3 \right)}^\alpha } + 4\left( {7{)^\alpha } - 3{{\left( 8 \right)}^\alpha }} \right),} \\ {d{2_{23}} = d{2_{32}} = {2^{ - 2 - 3\alpha }}( - 8 + {{\left( 2 \right)}^{1 + 2\alpha }} - 4\;{{\left( 3 \right)}^\alpha } + 4\left( {5{)^\alpha } - {{\left( 8 \right)}^\alpha }} \right),\;} \\ {d{2_{24}} = d{2_{42}} = \frac{{ - 1}}{2} + {{(\frac{7}{8})}^\alpha } + {2^{ - \alpha }} + {3^{1 + \alpha }}\;{8^{ - \alpha }}\;\;,} \\ {d{2_{34}} = d{2_{43}} = {2^{ - 1 - 3\alpha }}(4 - 3{{\left( 2 \right)}^{1 + 2\alpha }} + 4\;{{\left( 3 \right)}^\alpha } - 2\left( {5{)^\alpha } - 2{{\left( 7 \right)}^\alpha } + 3{{(8)}^\alpha }} \right),} \\ \end{array} \end{array}$$

D3α=[d3ij]d311=2−2−3α(8+(3)α(2)1+α−(4)α−(8)α),d322=2−2−3α(16−8(3)α+3(4)α−(6)1+α+3(8)α),d333=d344=0],d312=d321=2−2−3α(−12+4(3)α+(3)α(2)1+α−(4)α−(8)α),d313=d331=2−3(1+α)(8−16(3)α+5(3)α(2)1+α−5(4)α+8(5)α−5(8)α),d314=d341=2−3(1+α)(8(3)α+7(3)α(2)1+α−7(4)α−16(5)α+8(7)α−7(8)α)],[d323=d332=2−3(1+α)(−8+16(3)α−5(3)α(2)1+α+5(4)α−8(5)α+5(8)α),d324=d342=2−3(1+α)(−8(3)α−7(3)α(2)1+α+7(4)α+16(5)α−8(7)α+7(8)α),d334=d343=0$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {D_3^\alpha = \;\left[ {d{3_{ij}}} \right]}\\ {d{3_{11}} = {2^{ - 2 - 3\alpha }}(8 + {{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} - \;{{\left( 4 \right)}^\alpha } - \left( {8{)^\alpha }} \right),d{3_{22}} = {2^{ - 2 - 3\alpha }}(16 - 8{{\left( 3 \right)}^\alpha } + 3\;{{\left( 4 \right)}^\alpha } - \left( {6{)^{1 + \alpha }} + 3{{(8)}^\alpha }} \right),d{3_{33}} = d{3_{44}} = 0],}\\ {d{3_{12}} = d{3_{21}} = {2^{ - 2 - 3\alpha }}( - 12 + 4\;{{\left( 3 \right)}^\alpha } + {{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} - \left( {4{)^\alpha } - {{\left( 8 \right)}^\alpha }} \right),}\\ {d{3_{13}} = d{3_{31}} = {2^{ - 3\left( {1 + \alpha } \right)}}(8 - 16{{\left( 3 \right)}^\alpha } + 5{{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} - 5\left( {4{)^\alpha } + 8{{\left( 5 \right)}^\alpha } - 5{{\left( 8 \right)}^\alpha }} \right),}\\ {d{3_{14}} = d{3_{41}} = {2^{ - 3\left( {1 + \alpha } \right)}}(8{{\left( 3 \right)}^\alpha } + 7{{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} - 7\left( {4{)^\alpha } - 16{{\left( 5 \right)}^\alpha } + 8{{(7)}^\alpha } - 7{{\left( 8 \right)}^\alpha }} \right)],[d{3_{23}} = d{3_{32}} = {2^{ - 3\left( {1 + \alpha } \right)}}( - 8 + 16{{\left( 3 \right)}^\alpha } - 5{{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} + 5\left( {4{)^\alpha } - 8{{\left( 5 \right)}^\alpha } + 5{{(8)}^\alpha }} \right),}\\ {d{3_{24}} = d{3_{42}} = {2^{ - 3\left( {1 + \alpha } \right)}}( - 8{{\left( 3 \right)}^\alpha } - 7{{\left( 3 \right)}^\alpha }{{\left( 2 \right)}^{1 + \alpha }} + 7\left( {4{)^\alpha } + 16{{\left( 5 \right)}^\alpha } - 8{{(7)}^\alpha } + 7{{\left( 8 \right)}^\alpha }} \right),}\\ {d{3_{34}} = d{3_{43}} = 0}\\ {} \end{array} \end{array}$$

D4α=[d4ij]d411=d422=0,d433=2−2−3α(8+5(2)1+α−5(4)α),d444=2−2−3α(16−7(2)1+α−8(3)α+7(4)αd412=d421=0,d413=d431=d414=d441=−2−3−2α(−2+(2)α)d423=d432=−3(2)−3−2α(−2+2α)d424=d442=3(2)−3−2α(−2+2α),d434=d443=(2)−3−2α(−12+2α+1+4(3)α−4α$$\begin{array}{} \displaystyle \begin{array}{l} {D_4^\alpha = [d{4_{ij}}]} \\ {d{4_{11}} = d{4_{22}} = 0,d{4_{33}} = {2^{ - 2 - 3\alpha }}(8 + 5{{(2)}^{1 + \alpha }} - 5\,{{(4)}^\alpha }),d{4_{44}} = {2^{ - 2 - 3\alpha }}(16 - 7{{(2)}^{1 + \alpha }} - 8\,{{(3)}^\alpha } + 7{{(4)}^\alpha }d{4_{12}} = d{4_{21}} = 0,} \\ {d{4_{13}} = d{4_{31}} = d{4_{14}} = d{4_{41}} = - {2^{ - 3 - 2\alpha }}( - 2 + {{(2)}^\alpha })d{4_{23}} = d{4_{32}} = - 3{{(2)}^{ - 3 - 2\alpha }}( - 2 + \,{2^\alpha })} \\ {d{4_{24}} = d{4_{42}} = 3{{(2)}^{ - 3 - 2\alpha }}( - 2 + \,{2^\alpha }),d{4_{34}} = d{4_{43}} = {{(2)}^{ - 3 - 2\alpha }}( - 12 + \,{2^{\alpha + 1}} + 4{{(3)}^\alpha } - {4^\alpha }} \\ \end{array} \end{array}$$