A new Jacobi wavelets decomposition method for solving partial differential equations

Mohammed Nadjib, Bahri; Bahri, Sidi Mohammed

doi:10.1108/AJMS-05-2024-0081

Purpose

The paper proposes a new spectral method based on Jacobi wavelets for numerically solving partial differential equations (PDEs).

Design/methodology/approach

The authors used Jacobi wavelets as basis functions and employed the operational matrix of derivative to reduce PDEs to systems of ordinary differential equations (ODEs). A dedicated software package was developed to implement the proposed algorithm.

Findings

The authors presented several numerical examples to demonstrate the accuracy and efficiency of the proposed method in solving various types of PDEs.

Originality/value

The method is original in its use of Jacobi wavelets for constructing a wavelet-based spectral method applicable to PDEs.

1. Introduction

Physical and complex phenomena in science and technology, such as fluid flow, fluid motion, chemical reactions, and biological growth, are often modeled using partial differential equations (PDEs). However, analytical solutions to these PDEs are rarely available due to their non-linearity or insufficient solution space. As a result, efficient numerical algorithms have become increasingly important for simulations involving PDEs, as it is typically impossible to solve these equations analytically for most practical applications.

Researchers have developed various numerical methods to obtain approximate solutions for partial differential equations (PDEs). The most commonly used approaches are local methods, such as finite difference, finite element, finite volume, and collocation methods.

In recent years, wavelets have attracted a great deal of interest due to their wide application in various numerical approximations. They are commonly used in numerical methods for solving integral equations and numerical integration [1–3], ordinary differential equations [4–6], partial differential equations [7, 8] and fractional partial differential equations [9, 10]. Various kinds of wavelets were utilized in such applications, for example, Chebyshev wavelets [11, 12], Haar wavelets [13–16], Legendre wavelets [17–19] and Jacobi wavelets [20–22].

The purpose of this article is to propose an effective method for solving partial differential equations (PDEs) in which the unknown function depends on both spatial and temporal variables. The decomposition of this unknown function into a Jacobi wavelet basis will be applied only to the spatial variable. Notably, the coefficients of this decomposition will remain functions of the temporal variable. Consequently, this approach transforms the solution of a PDE into solving a time-dependent ordinary differential equation (ODE). The article concludes by presenting numerical results obtained using the proposed method, demonstrating its validity and practical applicability.

The outline of this article is as follows: Jacobi polynomials and wavelets are introduced in 2. In Section 3, we describe and explain the different steps that lead to the implementation of our approach. Section 4 is devoted to the operational matrix of derivatives. In Section 5, we apply our technique to the partial differential equations. In the last section, the performance of the new method is illustrated by several numerical examples.

2. Decompoition in Jacobi wavelets basis

2.1 Jacobi polynomials and their properties

Jacobi polynomials [17], denoted $J_{m}^{α, β}$ with α > − 1, β > − 1 are a family of orthogonal polynomials defined on the interval [−1, 1] with respect to the weight function

w (x) = {(1 - x)}^{α} {(1 + x)}^{β},

(2.1)

here, m is a positive integer representing the degree of the polynomial. These polynomials belong to the weighted function space $L_{w}^{2} ([- 1,1])$ ⁠. Jacobi polynomials can be expressed in various forms; one of the most notable is the recursive formula, which is given by:

\begin{array}{l} J_{m}^{α, β} (x) & = \frac{(α + β + 2 m - 1) [(α^{2} - β^{2} + x (α + β + 2 m) (α + β + 2 m - 2)]}{2 m (α + β + 2 m - 2) (α + β + m)} J_{m - 1}^{α, β} (x) \\ - \frac{(α + m - 1) (β + m - 1) + (α + β + 2 m)}{m (α + β + 2 m - 2) (α + β + m)} J_{m - 2}^{α, β} (x), \end{array}

where

J_{0}^{α, β} (x) = 1, J_{1}^{α, β} (x) = \frac{α + β + 2}{2} x + \frac{α - β}{2} .

(2.2)

As the Jacobi polynomials are orthogonal with respect to the weight function w, the following orthogonality relation holds:

{< J_{n}^{α, β}, J_{m}^{α, β} >}_{L_{w}^{2}} = h_{n}^{α, β} δ_{m, n}, \forall m, n \in N,

(2.3)

where $h_{n}^{α, β}$ is a normalization constant:

h_{n}^{α, β} = ‖ J_{n}^{α, β} ‖^{2} = \frac{2^{α + β + 1} Γ (α + m + 1) Γ (β + m + 1)}{(2 m + α + β + 1) m! Γ (β + α + m + 1)},

(2.4)

δ_m,n represents the Kronecker function, Γ is the Euler gamma function, and $< ., {. >}_{L_{w}^{2}}$ denotes the inner product of $L_{w}^{2} ([- 1,1])$ ⁠. It has been tablished that the set ${J_{n}^{α, β}}_{n \in N}$ forms an orthogonal basis for $L_{w}^{2} ([- 1,1])$ [20].

Finally, the derivative of Jacobi polynomials is given by

\frac{d}{d x} (J_{n}^{α, β}) (x) = \frac{1}{2} (α + β + n + 1) J_{n - 1}^{α + 1, β + 1} (x) .

(2.5)

2.2 Jacobi wavelets

The Jacobi wavelets are defined by [23].

ψ_{n, m}^{α, β} (x) = \{\begin{cases} \frac{2^{\frac{k + 1}{2}}}{\sqrt{h_{m}^{α, β}}} J_{m}^{α, β} (2^{k + 1} x - 2 n + 1), \frac{n - 1}{2^{k}} \leq x < \frac{n}{2^{k}}, \\ 0 o t h e r w i s e, \end{cases}

(2.6)

where $k \in N$ ⁠, n = 1, …, 2^k represents the number of decomposition levels, m = 0, …, M is the degree of the Jacobi polynomials $(M \in N^{*})$ ⁠. The coefficient $\frac{2^{\frac{k + 1}{2}}}{\sqrt{h_{m}^{α, β}}}$ is for normality. The family ${ψ_{n, m}^{α, β}}_{n = 1, \dots, 2^{k}, m \geq 0}$ forms an orthonormal basis of $L_{w}^{2} ([0,1])$ ⁠.

2.3 Decomposition in Jacobi wavelets basis.

Since the Jacobi wavelet family ${ψ_{n, m}^{α, β}}_{\underset{m \geq 0}{n = 1, \dots, 2^{k}}}$ is an orthonormal basis in $L_{w}^{2} ([0,1])$ ⁠, any function f defined over $L_{w}^{2} ([0,1])$ can be uniquely expressed as:

f (x) = \sum_{n = 1}^{2^{k}} \sum_{m = 0}^{\infty} c_{n, m} ψ_{n, m}^{α, β} (x),

(2.7)

where $c_{n, m} = {〈 f, ψ_{n, m}^{α, β} 〉}_{L_{w}^{2} ([0,1])}$ ⁠. For numerical analysis purposes, the projection of a function f into the linear span of the Jacobi wavelet functions $ψ_{n, m}^{α, β}, m = 0, \dots, M; n = 1, \dots, 2^{k}$ is given by:

f (x) ≃ \sum_{n = 1}^{2^{k}} \sum_{m = 0}^{M} c_{n, m} ψ_{n, m}^{α, β} (x) = C^{T} Ψ^{α, β} (x),

(2.8)

where: C the vector of coefficients, and Ψ^α,β the vector of Jacobi wavelet basis functions are given by

C = (c_{1,0}, \dots, c_{1, M}, \dots, c_{2^{k}, 0}, \dots, c_{2^{k}, M})

and

Ψ^{α, β} (x) = (ψ_{1,0}^{α, β},, \dots, ψ_{1, M}^{α, β}, \dots, ψ_{2^{k}, 0}, \dots, ψ_{2^{k}, M}) .

(2.9)

3. Decompoition in $L_{w}^{2} ([0, T]; L_{w}^{2} ([0,1]))$

Let f be an element in $L_{w}^{2} ([0, T]; L_{w}^{2} ([0,1]))$ ⁠, as defined in [24]:

\begin{matrix} f : [0, T] \to L_{w}^{2} ([0,1]) . \\ t \mapsto f (t, .) \end{matrix}

(3.1)

Since the function f(t, .) belongs to $L_{w}^{2} ([0,1])$ ⁠, then by (2.7) we have

f (t, x) = \sum_{n = 1}^{2^{k}} \sum_{m = 0}^{\infty} c_{n, m}^{α, β} (t) ψ_{n, m}^{α, β} (x),

(3.2)

where the coeficients c_n,m(t) depending on the variable t are defined by

c_{n, m}^{α, β} (t) = {〈 f (t, .), ψ_{n, m}^{α, β} 〉}_{L_{w}^{2} ([0,1])} = \int_{0}^{1} f (t, x) ψ_{n, m}^{α, β} (x) w (x) d x .

(3.3)

Proposition 1.

If $f \in C (] 0, T [; L_{w}^{2} ([0,1]))$ , then the function coefficients $c_{n, m}^{α, β} (t)$ are well defined for all t ∈ [0, T].

Proof. Since the functions f(t, .) and $ψ_{n, m}^{α, β}$ belongs to $L_{w}^{2} ([0,1])$ ⁠, and according to Cauchy-Schwartz inequality, then their product is in $L_{w}^{1} ([0,1])$ ⁠, which allows us to conclude that the coefficients $c_{n, m}^{α, β} (t)$ are well defined for all t ∈ [0, T].□

Lemma 1.

If $f \in C (] 0, T [; L_{w}^{2} ([0,1]))$ , then the function coefficients $c_{n, m}^{α, β} (t)$ are continuous in [0, T].

Proof. Let t₁, t₂ ∈ [0, T], then we have

\begin{matrix} | c_{n, m}^{α, β} (t_{1}) - c_{n, m}^{α, β} (t_{2}) | = \int_{0}^{1} f (t_{1}, x) ψ_{n, m}^{α, β} (x) w (x) d x - \int_{0}^{1} f (t_{2}, x) ψ_{n, m}^{α, β} (x) w (x) d x \\ = | \int_{0}^{1} (f (t_{1}, x) ψ_{n, m}^{α, β} (x) w (x) - f (t_{2}, x) ψ_{n, m}^{α, β} (x) w (x)) d x | \\ \leq \int_{0}^{1} | f (t_{1}, x) - f (t_{2}, x) | \times | ψ_{n, m}^{α, β} (x) w (x) | d x . \end{matrix}

(3.4)

As the function $f \in C (] 0, T [; L_{w}^{2} ([0,1]))$ ⁠, there exist γ(x) > 0 such that

for all t₁, t₂ ∈ [0, T], we have

| f (t_{1}, x) - f (t_{2}, x) | \leq γ (x) | t_{1} - t_{2} | .

(3.5)

Substituting (3.5) in (3.4), we obtain

| c_{n, m}^{α, β} (t_{1}) - c_{n, m}^{α, β} (t_{2}) | \leq \int_{0}^{1} γ (x) | t_{1} - t_{2} | \times | ψ_{n, m}^{α, β} (x) w (x) | d x .

Hence, we have

| c_{n, m}^{α, β} (t_{1}) - c_{n, m}^{α, β} (t_{2}) | \leq κ | t_{1} - t_{2} |,

where

κ = \int_{0}^{1} γ (x) | ψ_{n, m}^{α, β} (x) w (x) | d x .

So, the function coefficients $c_{n, m}^{α, β} (t)$ are continuous. □

Lemma 2.

If $f \in C^{1} (] 0, T [; L_{w}^{2} ([0,1]))$ , then the function coefficients $c_{n, m}^{α, β} (t)$ belong to C¹(]0, T[).

Forthermore, if $\frac{\partial f}{\partial t} \in L_{w}^{2} (] 0, T [; L_{w}^{2} ([0,1]))$ ⁠, then

\frac{\partial c_{n, m}^{α, β} (t)}{\partial t} = \int_{0}^{1} \frac{\partial f (x, t)}{\partial t} ψ_{n, m}^{α, β} w (x) d x .

(3.6)

Proof. Let t ∈ ]0, T[, then

\frac{c_{n, m}^{α, β} (t + Δ t) - c_{n, m}^{α, β} (t)}{Δ t} = \int_{0}^{1} \frac{f (x, t + Δ t) - f (x, t)}{Δ t} ψ_{n, m}^{α, β} (x) w (x) d x .

(3.7)

Since f ∈ C¹(]0, T[; $L_{w}^{2} ([0,1])$ ⁠), we have

\frac{f (x, t + Δ t) - f (x, t)}{Δ t} = \frac{\partial f}{\partial t} (x, t) + θ (x, t, Δ t),

(3.8)

where

\lim_{Δ t \to 0} θ (x, t, Δ t) = 0 .

Substituting (3.8) in (3.7), we get

\frac{c_{n, m}^{α, β} (t + Δ t) - c_{n, m}^{α, β} (t)}{Δ t} = \int_{0}^{1} (\frac{\partial f}{\partial t} (x, t) + θ (x, t, Δ t)) ψ_{n, m}^{α, β} (x) w (x) d x .

As Δt → 0, we obtain

\frac{\partial c_{n, m}^{α, β} (t)}{\partial t} = \int_{0}^{1} \frac{\partial f (x, t)}{\partial t} ψ_{n, m}^{α, β} w (x) d x .

□

Theorem 1.

If $f \in C^{k} (] 0, T [; L_{w}^{2} ([0,1]))$ , then the function coefficients $c_{n, m}^{α, β} (t)$ belong to C^k(]0, T[) for (k ≥ 1). And if there exists $λ (t) \in R^{+}$ such that $| \frac{\partial}{\partial^{2} x} f (x, t) | \leq λ (t)$ , then for all t ∈ [0, T], the series (3.2) is convergent,and the function f can be approximated as

f (t, x) = \sum_{n = 1}^{2^{k}} \sum_{m = 0}^{M} c_{n, m} (t) ψ_{n, m}^{α, β} (x) = C^{T} (t) Ψ^{α, β} (x),

(3.9)

where C and Ψ^α,β are 2^k(M + 1) vectors given by

C (t) = {(c_{1,0} (t), \dots, c_{1, M} (t), \dots, c_{2^{k}, 0} (t), \dots, c_{2^{k}, M} (t))}^{T}

and

Ψ^{α, β} (x) = {(ψ_{1,0}^{α, β} (x),, \dots, ψ_{1, M}^{α, β} (x), \dots, ψ_{2^{k}, 0} (x), \dots, ψ_{2^{k}, M} (x))}^{T} .

(3.10)

Moreover, for m > 1,

| c_{n, m} (t) | \leq \frac{λ {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}}}{n^{\frac{5}{2}} (m - 1) (m + α + β + 1)} .

Proof. For all α > − 1 and β > − 1, we have

c_{n, m} (t) = \int_{0}^{1} w^{α, β} (x) f (t, x) ψ_{n, m}^{α, β} (x) d x .

By using eq. (2.6) we get

c_{n, m} (t) = \frac{2^{\frac{k + 1}{2}}}{\sqrt{h_{m}^{α, β}}} \int_{\frac{n - 1}{2^{k}}}^{\frac{n}{2^{k}}} J_{m}^{α, β} (2^{k + 1} x - 2 n + 1) f (t, x) {(1 - (2^{k + 1} x - 2 n + 1))}^{α} {(1 + (2^{k + 1} x - 2 n + 1))}^{β} d x .

By change of variable s = 2^k+1x − 2n + 1, ds = 2^k+1dx, we obtain

c_{n, m} (t) = \frac{2^{- \frac{k + 1}{2}}}{\sqrt{h_{m}^{α, β}}} \int_{- 1}^{1} w^{α, β} (s) f (2^{- (k + 1)} (s + 2 n - 1)) J_{m}^{α, β} (s) d s .

Applying the integration by parts technique, we have

c_{n, m} (t) = - \frac{2^{- \frac{3 k + 3}{2}}}{2 m \sqrt{h_{m}^{α, β}}} \int_{- 1}^{1} w^{α + 1, β + 1} (s) \frac{\partial f}{\partial x} (2^{- (k + 1)} (s + 2 n - 1)) J_{m - 1}^{α + 1, β + 1} (s) d s .

Integrating by parts again, we have

c_{n, m} (t) = \frac{2^{- \frac{5 k + 5}{2}}}{4 m (m - 1) \sqrt{h_{m}^{α, β}}} \int_{- 1}^{1} w^{α + 2, β + 2} (s) \frac{\partial^{2} f}{\partial x^{2}} (2^{- (k + 1)} (s + 2 n - 1)) J_{m - 2}^{α + 2, β + 2} (s) d s .

Deriving Holder’s inequality for this identity, we obtain

\begin{matrix} | c_{n, m} (t) | & \leq & \frac{λ (t) 2^{- \frac{5 k + 5}{2}}}{4 m (m - 1) \sqrt{h_{m}^{α, β}}} {(\int_{- 1}^{1} {(w^{α + 2, β + 2} (s) J_{m - 2}^{α + 2, β + 2} (s))}^{2} d s)}^{\frac{1}{2}} \\ \leq & \frac{λ (t) 2^{- \frac{5 k + 5}{2}}}{4 m (m - 1) \sqrt{h_{m}^{α, β}}} {(\max_{x \in [0,1]} w^{α + 2, β + 2} (x))}^{\frac{1}{2}} {(\int_{- 1}^{1} w^{α + 2, β + 2} (s) {(J_{m - 2}^{α + 2, β + 2} (s))}^{2} d s)}^{\frac{1}{2}} . \end{matrix}

Since

\max_{x \in [0,1]} w^{α + 2, β + 2} (x) = {(\frac{2 α + 4}{α + β + 4})}^{α + 2} {(\frac{2 β + 4}{α + β + 4})}^{β + 2},

and according to equation

\int_{- 1}^{1} w^{α + 2, β + 2} (s) (J_{m - 2}^{α + 2, β + 2} (s)) d s = h_{m - 2}^{α + 2, β + 2} .

We have

\begin{matrix} | c_{n, m} (t) | & \leq & \frac{λ (t) 2^{- \frac{5 k + 5}{2}} \sqrt{h_{m - 1}^{α + 2, β + 2}}}{4 m (m - 1) \sqrt{h_{m}^{α, β}}} {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}} \\ \leq & \frac{λ (t) 2^{- \frac{5 k + 9}{2}}}{m (m - 1)} {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}} \sqrt{\frac{h_{m - 1}^{α + 2, β + 2}}{h_{m}^{α, β}}} \\ \leq & \frac{λ (t) 2^{- \frac{5 k + 9}{2}}}{m (m - 1)} {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}} \sqrt{\frac{16 m (m - 1)}{(m + α + β + 2) (m + α + β + 1)}} \\ \leq & \frac{λ (t) {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}}}{2^{\frac{5 k + 5}{2}} \sqrt{m (m - 1) (m + α + β + 2) (m + α + β + 1)}} . \end{matrix}

Since n ≤ 2^k−1, we get

\begin{matrix} | c_{n, m} (t) | & \leq & \frac{λ (t) {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}}}{{(2 n)}^{\frac{5}{2}} \sqrt{m (m - 1) (m + α + β + 2) (m + α + β + 1)}} \\ \leq & \frac{λ (t) {(\frac{2 α + 4}{α + β + 4})}^{\frac{α + 2}{2}} {(\frac{2 β + 4}{α + β + 4})}^{\frac{β + 2}{2}}}{n^{\frac{5}{2}} (m - 1) (m + α + β + 1)} . \end{matrix}

□

4. Operational matrix of derivative of the Jacobi wavelets

The derivative of the Jacobi wavelets vector Ψ^α,β, as given in (2.9), can be expressed as follows:

\frac{d Ψ^{α, β} (x)}{d x} = D_{x}^{α, β} Ψ^{α, β} (x),

(4.1)

where D^α,β denotes the 2^k(M + 1) × 2^k(M + 1) operational matrix of derivative given by

D_{x}^{α, β} = (\begin{matrix} F^{α, β} & 0 & . & . & . & 0 \\ 0 & . & . \\ . & . & F^{α, β} & . & . \\ . & . & . & . \\ . & . & 0 \\ 0 & . & . & . & 0 & F^{α, β} \end{matrix}),

(4.2)

F^α,β is (M + 1) × (M + 1) matrix, where its (i,j)^th element is given by

F_{i, j}^{α, β} = \{\begin{cases} 2^{k + 1} \frac{\sqrt{h_{j - 1}^{α, β}}}{\sqrt{h_{i - 1}^{α, β}}} γ_{i - 1, j - 1}^{α, β} i f i > j, \\ 0 o t h e r w i s e, \end{cases}

(4.3)

in which $h_{i - 1}^{α, β}$ and $h_{j - 1}^{α, β}$ are defined from (2.4), and $γ_{i - 1, j - 1}^{α, β}$ are given by

\begin{matrix} γ_{i - 1, j - 1}^{α, β} & = & \frac{Γ (i + β)}{2 Γ (i + α + β)} \frac{(2 (j - 1) + α + β + 1) Γ (α + β + j)}{Γ (α + j)} \\ \times & [\sum_{d = j - 1}^{i - 1} {(- 1)}^{d - j - 1} \frac{(2 (d + 1) + α + β) Γ (α + d + 1)}{Γ (β + d + 2)}] \end{matrix}

Corollary 1.

The operational matrix for n^th derivative can be obtained using (4.1) as

\frac{d^{n} Ψ^{α, β} (x)}{d x^{n}} = {(D_{x}^{α, β})}^{n} Ψ^{α, β} (x),

(4.5)

where ${(D_{x}^{α, β})}^{n}$ is the n^th power of matrix D^α,β.

5. Method of solution

5.1 Application to telegraph equations

The telegraph equation is typically expressed in standard form as follows:

\frac{\partial^{2} u (t, x)}{\partial x^{2}} = a \frac{\partial^{2} u (t, x)}{\partial t^{2}} + b \frac{\partial u (t, x)}{\partial t} + c u (t, x),

(5.1)

with boundary conditions

u (t, 0) = g_{1} (t) \frac{\partial u (t, 0)}{\partial x} = g_{2} (t),

(5.2)

and initial conditions

u (0, x) = g_{3} (x) \frac{\partial u (0, x)}{\partial t} = g_{3} (x),

(5.3)

where a, b and c are constants,g₁(t), g₂(t) are continuous functions in [0, T], and g₃(x), g₄(x) are continuous in [0, 1].

Let us assume that the function u(x, t) can be expressed as

u (x, t) = C^{T} (t) Ψ^{α, β} (x) .

(5.4)

Then, from (4.1), the derivatives of u(x, t) on x and t respectively are given as

\frac{\partial u (x, t)}{\partial x} = C^{T} (t) D_{x}^{α, β} Ψ^{α, β} (x),

(5.5)

and

\frac{\partial u (x, t) (x, t)}{\partial t} = \frac{\partial C (t)}{\partial t} Ψ^{α, β} (x) .

(5.6)

Substituting (5.4), (5.5) and (5.6) in (5.1), we obtain

C^{T} (t) {(D_{x}^{α, β})}^{2} Ψ^{α, β} (x) - a \frac{\partial^{2} C (t)}{\partial t^{2}} Ψ^{α, β} (x) - b \frac{\partial C (t)}{\partial t} Ψ^{α, β} (x) + c C^{T} (t) Ψ^{α, β} (x) = 0

(5.7)

Using boundary conditions in (5.2) we have

C^{T} (t) Ψ^{α, β} (0) = g_{1} (t) C^{T} (t) D_{x}^{α, β} Ψ^{α, β} (0) = g_{2} (t) .

(5.8)

Collocating the obtained equation in (5.7) with the collocation points x_ii = 1, 2, …2^k(M + 1) − 2, we get a system of 2^k(M + 1) − 2 linear differential equations, which can be combined with the equations in (5.8), we obtain the following system of 2^k(M + 1) linear differential equations

\{\begin{cases} - a \frac{\partial^{2} C^{T} (t)}{\partial t^{2}} Ψ^{α, β} (x_{i}) - b \frac{\partial C^{T} (t)}{\partial t} Ψ^{α, β} (x_{i}) + c . C^{T} (t) Ψ^{α, β} (x_{i}) + C^{T} (t) {(D_{x}^{α, β})}^{2} Ψ^{α, β} (x_{i}) = 0, i = 1, \dots 2^{k} (M + 1) - 2, \\ C^{T} (t) Ψ^{α, β} (0) = g_{1} (t), \\ C^{T} (t) D_{x}^{α, β} Ψ^{α, β} (0) = g_{2} (t) . \end{cases}

Hence, we have

\{\begin{cases} - a C^{″} (t) - b C^{'} (t) + (c + {({(D_{x}^{α, β})}^{T})}^{2}) C (t) = 0, \\ C^{T} (t) Ψ^{α, β} (0) = g_{1} (t), \\ C^{T} (t) D_{x}^{α, β} Ψ^{α, β} (0) = g_{2} (t); \end{cases}

with initial conditions

C^{T} (0) Ψ^{α, β} (x) = g_{3} (x), C^{'} (0) Ψ^{α, β} {(x)}^{T} = g_{4} (x) .

This system can be solved for unknown coefficients of the vector C(t). Consequently, the solution u(t, x) given in (5.4) can be calculated.

5.2 Application to linear hyperbolic partial differential equations

Consider the general form of the first-order linear hyperbolic equation

\frac{\partial u (t, x)}{\partial x} + \frac{\partial u (t, x)}{\partial t} + u (t, x) = f (t, x),

(5.9)

subject to the conditions:

u (t, 0) = p (t) u (t, 1) = q (t), u (0, x) = r (x)

(5.10)

where p(t), q(t), r(x) and f(t, x) are continuous real-valued functions. Assume that,

u (x, t) = C^{T} (t) Ψ^{α, β} (x) .

(5.11)

The partial derivatives of u(x, t) with respect to x and t are given

\frac{\partial u (x, t)}{\partial x} = C^{T} (t) D_{x}^{α, β} Ψ^{α, β} (x),

(5.12)

and

\frac{\partial u (x, t) (x, t)}{\partial t} = \frac{\partial C (t)}{\partial t} Ψ^{α, β} (x) .

(5.13)

Substituting (5.11), (5.12) and (5.13) in (5.9), we obtain

C^{T} (t) (D_{x}^{α, β}) Ψ^{α, β} (x) + \frac{\partial C^{T} (t)}{\partial t} Ψ^{α, β} (x) + C^{T} (t) Ψ^{α, β} (x) = f (t, x)

(5.14)

Using the conditions in (5.10) we have

C^{T} (t) Ψ^{α, β} (0) = p (t) C^{T} (t) Ψ^{α, β} (1) = q (t) .

(5.15)

Now, by collocating the obtained equation in (5.14) with the collocation points x_i where i = 1, 2, …2^k(M + 1) − 2, we obtain a system of 2^k(M + 1) − 2 linear differential equations, which can be combined with the equations in (5.15), we obtain system of 2^k(M + 1) linear differential equations, with initial conditions

C^{T} (0) Ψ^{α, β} (x) = r (x) .

To extract the values of unknown coefficients, we can either apply the Jacobi wavelet method again or use finite difference schemes. Finally, by substituting the obtained values of the unknown coefficients into equation (5.9), the numerical solution of the given PDE is obtained.

6. Numerical illustrations

To illustrate the approach outlined above and to evaluate the developed method, we have applied it to several numerical examples and solved them using the proposed technique. For the error analysis, we consider the following types of errors:

L_{error}^{2} = \sqrt{\sum_{i}^{n} e_{i}^{2}}, L_{error}^{\infty} = \max e_{i}, R = \log_{2} [\frac{u_{app} (k - 1) - u_{e x}}{u_{app} (k) - u_{e x}}]

where $e_{i} = | {(u_{e x})}_{i} - {(u_{app})}_{i} |$ ⁠, and R is the rate convergence [16].

6.1 Example 1

Consider the linear hyperbolic equation offirst-order in [0, 1]

u_{t} + u_{x} + u = {(x - t)}^{2},

(6.1)

with

u (x, 0) = x^{2} a n d u (0, t) = t^{2} .

(6.2)

The exact solution for this example is given by

u_{e x} = {(x - t)}^{2} .

We solve the problem (6.1) and (6.2) using the presented method with different values of (α, β) for k = 1 and M = 6, The obtained results are as follows: Figure 1 represents the time–space graph of the solution computed by the present method alongside the exact solution. Figure 2 shows that the error of the present method is smaller than those of the Legendre wavelet collocation method, and the Chebyshev wavelet collocation method. Table 1 compares the absolute errors obtained from the present method, the Legendre wavelet collocation method [25], and the Chebyshev wavelet collocation method [25]. Figures 3 and 4 illustrate the graphical comparison of the present method solution with the exact solution for different values of x, α and β. Table 2 compares the error norms obtained from the present method for level k = 1 and other methods in the literature [25]. Table 3 displays the maximum absolute errors (L^∞) and the convergence rates for the present method, evaluated for various choices of levels k, α and β.

Figure 1

Two 3-D surface plots compare the exact solution and numerical approach.

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The illustration contains two side-by-side 3-D surface plots, (a) and (b), that compare the exact solution with a numerical solution approach. The plot on the left is titled “exact solution,” and the plot on the right is titled “solution approach.” Both plots depict a surface in a 3-D coordinate system. The x- and y-axes for both plots range from 0 to 1 with increments of 0.5, while the z-axis (representing the solution value) also ranges from 0 to 1 with increments of 0.5. The surfaces in both plots have a bowl-like or concave shape, starting at a value of 1 on the edges of the x-y plane and dipping to a minimum value near the center. The surfaces are colored with a gradient, with blue representing the lowest values in the center and green and yellow representing higher values as they move toward the edges. The grid lines on the surface provide a visual sense of its curvature and shape.

Graphical illustration of the present method and the exact solution for (α, β) = (0, 0). Source: Authors’ own creation

Figure 2

Four plots compare the numerical approximation with the exact solution for different alpha and beta values.

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The line graphs are arranged in a 2 by 2 grid, each comparing an approximate solution, u (a p p), with an exact solution, u (e x), for different parameter values of alpha and beta. In all four subplots, the horizontal axis ranges from 0 to 1 with increments of 0.5, and the vertical axis ranges from 0 to 0.3 with increments of 0.1. The approximate solution, “u (a p p),” is represented by blue asterisks, while the exact solution, “u (e x),” is shown as a solid red line. The general shape of all four curves is a parabola-like function that starts at (0, 0.25), dips to a minimum near 0.5 on the horizontal axis, rises again, and ends at (1, 0.25). The top-left subplot is titled “u (a p p) for alpha equals negative 0.5, beta equals negative 0.5.” The top-right subplot is titled “u (a p p) for alpha equals 0, beta equals 2.” The bottom-left subplot is titled “u (a p p) for alpha equals 0, beta equals 0.” The bottom-right subplot is titled “u (a p p) for alpha equals 0.5, beta equals 1.5.” In all four cases, the blue asterisks representing the approximate solution fall very close to the red line of the exact solution, indicating a high degree of accuracy for the numerical method across various parameter combinations. Each subplot has a legend in the top-right corner, clarifying the symbols for “u (a p p)” and “u (e x).” Note: All the numerical data values are estimated.

Graphical evaluation of the present method solution and the exact solution for different value of (α, β) at t = 0.5. Source: Authors’ own creation

Table 1

Comparison of absolute errors for different values of (α, β), and various wavelet methods for example 1

(t, x)	An absolute error by Legendre wavelet [25]	An absolute error by Chebychev wavelet [25]	An absolute error by our method α = β = 0	An absolute error by our method α = β = −0.5
(0.1,0.1)	8.20 × 10⁻⁵	4.16 × 10⁻⁴	1.52 × 10⁻⁵	1.28 × 10⁻⁵
(0.2,0.2)	7.10 × 10⁻⁶	5.37 × 10⁻⁴	1.18 × 10⁻⁶	2.67 × 10⁻⁵
(0.3,0.3)	4.65 × 10⁻⁵	3.55 × 10⁻⁵	2.32 × 10⁻⁵	1.94 × 10⁻⁵
(0.4,0.4)	7.86 × 10⁻⁵	2.65 × 10⁻⁴	1.64 × 10⁻⁵	1.43 × 10⁻⁵
(0.5,0.5)	8.93 × 10⁻⁵	1.27 × 10⁻⁵	1.28 × 10⁻⁵	1.72 × 10⁻⁵
(0.6,0.6)	7.86 × 10⁻⁵	1.93 × 10⁻⁵	3.15 × 10⁻⁵	1.39 × 10⁻⁵
(0.7,0.7)	4.65 × 10⁻⁵	2.11 × 10⁻⁵	2.67 × 10⁻⁵	2.91 × 10⁻⁵
(0.8,0.8)	7.10 × 10⁻⁶	3.22 × 10⁻⁵	1.86 × 10⁻⁵	1.52 × 10⁻⁵
(0.9,0.9)	8.22 × 10⁻⁵	1.20 × 10⁻⁴	1.53 × 10⁻⁵	1.79 × 10⁻⁵

(t, x)	An absolute error by Legendre wavelet [25]	An absolute error by Chebychev wavelet [25]	An absolute error by our method α = β = 0	An absolute error by our method α = β = −0.5
(0.1,0.1)	8.20 × 10⁻⁵	4.16 × 10⁻⁴	1.52 × 10⁻⁵	1.28 × 10⁻⁵
(0.2,0.2)	7.10 × 10⁻⁶	5.37 × 10⁻⁴	1.18 × 10⁻⁶	2.67 × 10⁻⁵
(0.3,0.3)	4.65 × 10⁻⁵	3.55 × 10⁻⁵	2.32 × 10⁻⁵	1.94 × 10⁻⁵
(0.4,0.4)	7.86 × 10⁻⁵	2.65 × 10⁻⁴	1.64 × 10⁻⁵	1.43 × 10⁻⁵
(0.5,0.5)	8.93 × 10⁻⁵	1.27 × 10⁻⁵	1.28 × 10⁻⁵	1.72 × 10⁻⁵
(0.6,0.6)	7.86 × 10⁻⁵	1.93 × 10⁻⁵	3.15 × 10⁻⁵	1.39 × 10⁻⁵
(0.7,0.7)	4.65 × 10⁻⁵	2.11 × 10⁻⁵	2.67 × 10⁻⁵	2.91 × 10⁻⁵
(0.8,0.8)	7.10 × 10⁻⁶	3.22 × 10⁻⁵	1.86 × 10⁻⁵	1.52 × 10⁻⁵
(0.9,0.9)	8.22 × 10⁻⁵	1.20 × 10⁻⁴	1.53 × 10⁻⁵	1.79 × 10⁻⁵

Source(s): Authors’ own creation

Figure 3

Four plots comparing numerical approximations with exact solutions for various alpha and beta values.

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The four line graphs are arranged in a 2 by 2 grid, each comparing an approximate solution, u (a p p), with an exact solution, u (e x), for different parameter values of alpha and beta. In all four subplots, the horizontal axis ranges from 0 to 1 with increments of 0.5, and the vertical axis ranges from 0 to 1 with increments of 0.5. The approximate solution, “u (a p p),” is represented by blue asterisks, while the exact solution, “u (e x),” is shown as a solid red line. The general shape of all four curves is a concave-up ascending that starts at (0, 0), rises, and ends at (1, 0.75). The top-left subplot is titled “u (a p p) for alpha equals negative 0.5, beta equals negative 0.5.” The top-right subplot is titled “u (a p p) for alpha equals 0, beta equals 2.” The bottom-left subplot is titled “u (a p p) for alpha equals 0, beta equals 0.” The bottom-right subplot is titled “u (a p p) for alpha equals 0.5, beta equals 1.5.” In all four cases, the blue asterisks are slightly above the red line and follow the same trend as the red line. Each subplot has a legend in the top-right corner, clarifying the symbols for “u (a p p)” and “u (e x).” Note: All the numerical data values are estimated.

Graphical evaluation of the present method solution and the exact solution for different value of (α, β) at t = 0.1. Source: Authors’ own creation

Figure 4

A line graph comparing the absolute error of a numerical method with the Legendre multiwavelet and TOM method.

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The vertical axis ranges from 0 to 6 in increments of 1, while the horizontal axis ranges from 0 to 1 in increments of 0.2. The legend at the top right defines the four lines plotted on the graph. A red dashed line with circles represents “error for alpha, beta equals (0,2).” A blue dashed line with circles represents “error for alpha, beta equals (0,0).” A green dashed line with circles represents “Legendre multiwavelet.” A purple dashed line with circles represents the “TOM method.” The red and blue lines generally show very low error values, staying close to the horizontal axis and mostly below 0.5. The purple line also shows relatively low error, mostly staying below 1. The green line, however, exhibits much more erratic behavior with significant spikes in error, reaching a peak value of over 5 at 0.2 on the horizontal axis.

Absolute error graphical comparison of the present method solution, Legendre wavelet collocation method, and Chebyshev wavelet collocation method, for example 1. Source: Authors’ own creation

Table 2

Comparison of error norms between the present method for level k = 1 and other methods in the literature [25], for example 1

M	L² via LW [25]	L² via CW [25]	L² via our method (α, β) = (0, 0)	L² via our method (α, β) = (−0.5, − 0.5)
2	4.52 × 10⁻¹	5.40 × 10⁻¹	1.29 × 10⁻²	1.68 × 10⁻²
4	3.48 × 10⁻⁴	1.20 × 10⁻³	2.26 × 10⁻⁶	1.64 × 10⁻⁴

M	L² via LW [25]	L² via CW [25]	L² via our method (α, β) = (0, 0)	L² via our method (α, β) = (−0.5, − 0.5)
2	4.52 × 10⁻¹	5.40 × 10⁻¹	1.29 × 10⁻²	1.68 × 10⁻²
4	3.48 × 10⁻⁴	1.20 × 10⁻³	2.26 × 10⁻⁶	1.64 × 10⁻⁴

Source(s): Authors’ own creation

Table 3

Covergence rate and L^∞ errors for M = 6 for example, 1

k	N = 2^k	L^∞ for (α, β) = (0, 0)	R for (α, β) = (0, 0)	L^∞ for α = β = −0.5	R for α = β = −0.5
1	2	1.49 × 10⁻¹	–	1.04 × 10⁻¹	–
2	4	2.61 × 10⁻²	2.51	2.10 × 10⁻²	2.30
3	8	4.76 × 10⁻³	2.57	3.51 × 10⁻³	2.58
4	16	5.06 × 10⁻⁴	3.24	2.37 × 10⁻⁴	3.90
5	32	2.19 × 10⁻⁵	4.54	1.42 × 10⁻⁵	4.07
6	64	4.64 × 10⁻⁷	5.94	1.20 × 10⁻⁶	4.86

k	N = 2^k	L^∞ for (α, β) = (0, 0)	R for (α, β) = (0, 0)	L^∞ for α = β = −0.5	R for α = β = −0.5
1	2	1.49 × 10⁻¹	–	1.04 × 10⁻¹	–
2	4	2.61 × 10⁻²	2.51	2.10 × 10⁻²	2.30
3	8	4.76 × 10⁻³	2.57	3.51 × 10⁻³	2.58
4	16	5.06 × 10⁻⁴	3.24	2.37 × 10⁻⁴	3.90
5	32	2.19 × 10⁻⁵	4.54	1.42 × 10⁻⁵	4.07
6	64	4.64 × 10⁻⁷	5.94	1.20 × 10⁻⁶	4.86

Source(s): Authors’ own creation

We calculated the absolute error for different values of k, and the results are as follows:

Table 3 shows the effect of the number of levels on the solution. For k = 1, 2, …, 6 (N = 2^k-levels), the absolute errors with respect to analytic solutions are presented. As k increase from 1 to 6 for M = 6, the errors decrease to 10^–6

6.2 Example 2

Let us consider the telegraph equation

u_{t t} + 4 u_{t} + 2 u = u_{x x}, 0 < x < 1,

(6.3)

under the initial conditions

u (x, 0) = \sin x, u_{t} (x, 0) = - \sin x,

(6.4)

with the exact solution

u_{e x} (x, t) = e^{- t} \sin x .

We applied the present method to solve equation (6.3) for different values of (α, β), with k = 2 and M = 5. Table 4 presents a comparison of the absolute errors obtained using the proposed method with those reported in the literature [26, 27]. Figures 5 and 6 provide a graphical representation of the solution obtained using the current method for k = 2 and M = 5 with (α, β) = (0, 2) and (α, β) = (0, 0), respectively. These figures compare the method’s solution to the exact solution and display the corresponding absolute error. Figure 7 shows that the error of the present method is smaller than the errors of both the Legendre Multiwavelet and Taylor polynomial methods. Table 5 details the performance of our method by presenting its maximum absolute errors (L^∞) and L² errors, which were calculated for different values ofk, α and β.

Table 4

Comparison of absolute errors by the present method for different values for (α, β), and various wavelet methods, for example 2

(x, t)	(α, β) = (0, 2)	(α, β) = (0, 0)	Legendre multiwavelet [27]	TPM [26]
(0,0)	0	0	0	0
(0.1,0.1)	1.05 × 10⁻¹⁰	2.10 × 10⁻¹⁰	8.31 × 10⁻¹⁰	8.05 × 10⁻¹³
(0.2,0.2)	1.18 × 10⁻⁹	3.81 × 10⁻¹⁰	2.00 × 10⁻¹⁰	4.62 × 10⁻¹⁰
(0.3,0.3)	5.25 × 10⁻⁹	2.32 × 10⁻¹⁰	1.00 × 10⁻¹⁰	2.04 × 10⁻⁸
(0.4,0.4)	7.15 × 10⁻¹⁰	7.60 × 10⁻¹⁰	4.00 × 10⁻⁹	3.07 × 10⁻⁷
(0.5,0.5)	8.02 × 10⁻¹⁰	6.59 × 10⁻⁹	1.11 × 10⁻⁹	2.57 × 10⁻⁶
(0.6,0.6)	1.43 × 10⁻⁹	3.07 × 10⁻⁹	5.01 × 10⁻¹⁰	1.49 × 10⁻⁵
(0.7,0.7)	2.19 × 10⁻⁹	2.00 × 10⁻¹⁰	3.36 × 10⁻⁷	6.68 × 10⁻⁵
(0.8,0.8)	8.50 × 10⁻¹⁰	1.06 × 10⁻¹⁰	8.32 × 10⁻⁸	2.48 × 10⁻⁴
(0.9,0.9)	3.10 × 10⁻¹⁰	5.42 × 10⁻¹⁰	2.80 × 10⁻⁶	7.96 × 10⁻⁴

(x, t)	(α, β) = (0, 2)	(α, β) = (0, 0)	Legendre multiwavelet [27]	TPM [26]
(0,0)	0	0	0	0
(0.1,0.1)	1.05 × 10⁻¹⁰	2.10 × 10⁻¹⁰	8.31 × 10⁻¹⁰	8.05 × 10⁻¹³
(0.2,0.2)	1.18 × 10⁻⁹	3.81 × 10⁻¹⁰	2.00 × 10⁻¹⁰	4.62 × 10⁻¹⁰
(0.3,0.3)	5.25 × 10⁻⁹	2.32 × 10⁻¹⁰	1.00 × 10⁻¹⁰	2.04 × 10⁻⁸
(0.4,0.4)	7.15 × 10⁻¹⁰	7.60 × 10⁻¹⁰	4.00 × 10⁻⁹	3.07 × 10⁻⁷
(0.5,0.5)	8.02 × 10⁻¹⁰	6.59 × 10⁻⁹	1.11 × 10⁻⁹	2.57 × 10⁻⁶
(0.6,0.6)	1.43 × 10⁻⁹	3.07 × 10⁻⁹	5.01 × 10⁻¹⁰	1.49 × 10⁻⁵
(0.7,0.7)	2.19 × 10⁻⁹	2.00 × 10⁻¹⁰	3.36 × 10⁻⁷	6.68 × 10⁻⁵
(0.8,0.8)	8.50 × 10⁻¹⁰	1.06 × 10⁻¹⁰	8.32 × 10⁻⁸	2.48 × 10⁻⁴
(0.9,0.9)	3.10 × 10⁻¹⁰	5.42 × 10⁻¹⁰	2.80 × 10⁻⁶	7.96 × 10⁻⁴

Source(s): Authors’ own creation

Figure 5

A 3-D surface plot showing the exact solution of a function u of (t, x) as it changes over time and space.

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The 3-D surface plot is titled “The exact solution.” The vertical axis is labeled “u of (t, x),” ranging from 0 to 0.8 in increments of 0.2. The horizontal plane contains two axes: the bottom left is labeled “t,” ranging from 0 to 1 in increments of 0.5, and the bottom right axis is labeled “x,” also ranging from 0 to 1 in increments of 0.5. The surface, which has grid lines on its surface, is colored with a gradient, transitioning from blue at low values (near the x equals 0 and t equals 0 planes) to yellow at high values (near t equals 1, x equals 0). The surface has a sloping or ramp-like shape. The values of the plane are as follows: bottom left corner (t equals 0, x equals 0, u of (t, x) equals 0); bottom right corner (t equals 0, x equals 0.85, u of (t, x) equals 0); top right corner (t equals 1, x equals 0.95, u of (t, x) equals 0.3); and top left corner (t equals 1, x equals 0.1, u of (t, x) equals 0.75). Note: All the numerical data values are estimated.

The graph of the exact solution for example 2. Source: Authors’ own creation

Figure 6

Four 3-D plots show approximate solutions and their absolute errors for u (t, x) with different alpha and beta values.

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The four 3-D surface plots are arranged in a 2 by 2 grid format. The plots on the left column show the approximate solution for two different parameter sets, while the plots on the right column show the corresponding errors. The vertical axis of the plots on the left column is labeled “u of (t, x)” and ranges from 0 to 0.8 in increments of 0.2, while the vertical axis of the plots on the right column is also labeled “u of (x, t),” ranging from 0 to 8 times 10 to the negative 9 power in increments of 2 times 10 to the negative 9 power. In all graphs, the horizontal plane contains two axes: the bottom left axis is labeled “t,” and the bottom right axis is labeled “x,” with a scale from 0 to 1 in increments of 0.5. The grid lines on each surface provide a visual sense of its curvature and shape. The top-left plot is titled “The approximate solution for (alpha, beta) equals (0, 2).” The surface is colored with a gradient from blue to yellow and has a sloping shape. The top-right plot is titled “er for (alpha, beta) equals (0, 2).” The error surface is mostly flat and blue, with a small spike near x equals 0.5 for any fixed t values. After this peak, the surface drops (yellow region) and again becomes flat. The bottom-left plot is titled “The approximate solution for (alpha, beta) equals (0, 0).” This surface is also sloping, as in the top-left graph. The bottom-right plot is titled “er for (alpha, beta) equals (0, 0).” This error surface has two spikes of varying heights: one occurs at (x equals 0.25, u of (t, x) equals 5 times 10 to the negative 9 power) and the other occurs at (x equals 0.75, u of (t, x) equals 1 times 10 to the negative 9 power) for fixed values of t. The plane features a small yellow region after the first peak; otherwise, the plane is in blue shades. Note: All the numerical values are estimated.

Graphical illustration of the present method along with its absolute error, for example 2. Source: Authors’ own creation

Figure 7

A multiple-line graph comparing the absolute error of the present method solution.

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The horizontal axis ranges from 0 to 1 with increments of 0.2, while the vertical axis ranges from 0 to 4 in increments of 0.5. Four dashed lines with circular markers are plotted. The legend in the top-left corner identifies each line: a red line for “error for alpha, beta equals (0, 2),” a blue line for “error for alpha, beta equals (0, 0),” a green line for “Legendre multiwavelet,” and a purple line for “TOM method.” The red and blue lines consistently show very low error values, staying close to the horizontal axis and rarely exceeding 0.6. The purple line and the green line show much higher and more erratic error values. The purple line has a large spike, reaching a peak error of approximately 4 at 0.9 on the horizontal axis, while the green line has a peak error of approximately 3.3 at 0.75 on the horizontal axis. Note: All the numerical data values are estimated.

Absolute error graphical comparison of the present method solution, Legendre multiwavelet and Taylor polynomial method, for example 2. Source: Authors’ own creation

Table 5

The errors L² and L^∞ from example 2

k	L^∞ via our method α = β = 0	L² via our method α = β = 0	L^∞ via our method (α, β) = (0, 2)	L² via our method (α, β) = (0, 2)
1	2.56 × 10⁻³	1.95 × 10⁻³	2.49 × 10⁻²	3.32 × 10⁻²
2	2.61 × 10⁻⁵	3.06 × 10⁻⁵	1.57 × 10⁻⁴	6.51 × 10⁻⁴
3	1.85 × 10⁻⁶	4.20 × 10⁻⁷	3.15 × 10⁻⁶	2.84 × 10⁻⁶
4	6.92 × 10⁻⁹	1.82 × 10⁻¹⁰	2.01 × 10⁻¹⁰	1.24 × 10⁻¹⁰
5	1.52 × 10⁻¹²	2.76 × 10⁻¹²	1.01 × 10⁻¹²	3.21 × 10⁻¹¹

k	L^∞ via our method α = β = 0	L² via our method α = β = 0	L^∞ via our method (α, β) = (0, 2)	L² via our method (α, β) = (0, 2)
1	2.56 × 10⁻³	1.95 × 10⁻³	2.49 × 10⁻²	3.32 × 10⁻²
2	2.61 × 10⁻⁵	3.06 × 10⁻⁵	1.57 × 10⁻⁴	6.51 × 10⁻⁴
3	1.85 × 10⁻⁶	4.20 × 10⁻⁷	3.15 × 10⁻⁶	2.84 × 10⁻⁶
4	6.92 × 10⁻⁹	1.82 × 10⁻¹⁰	2.01 × 10⁻¹⁰	1.24 × 10⁻¹⁰
5	1.52 × 10⁻¹²	2.76 × 10⁻¹²	1.01 × 10⁻¹²	3.21 × 10⁻¹¹

Source(s): Authors’ own creation

6.3 Example 3

Consider the hyperbolic equation of first-order of the form

u_{t} - u - u_{x} = - \sqrt{2} e^{- \sqrt{2} t - x} 0 < x < 1,0 < t < 1

(6.5)

subject to initial and boundary conditions

u(x, 0) = e^−x, 0 < x < 1 and $u (0, t) = e^{- \sqrt{2} t}, 0 < t < 1$ ⁠.

The exact solution is $u (x, t) = e^{- \sqrt{2} t - x}$ ⁠. This problem was solved using the present method. Figures 8 and 9 provide a graphical comparison of the solution obtained by the present method with the exact solution. Figures 10 and 11 illustrate the graphical evaluation of the present method solution for (α, β) = (0.5, 0.5), at k = 12 and M = 9 compared with the exact solution at different values of x and t, respectively. Tables 6 and 7 compare the absolute errors at different values of t obtained using the present method with those from the methods in [28, 29]. Table 8 displays the maximum absolute errors (L^∞) and the convergence rates for the present method, evaluated for various choices of levels k, α and β.

Figure 8

A 3-D surface plot showing the exact solution of a function u of (t, x) with respect to time and space.

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The 3-D surface plot is titled “The exact solution.” The vertical axis is labeled “u of (t, x),” ranging from 0 to 1 in increments of 0.2. The horizontal plane contains two axes: the bottom left is labeled “t,” ranging from 0 to 1 in increments of 0.5, and the bottom right axis is labeled “x,” also ranging from 0 to 1 in increments of 0.5. The surface is colored with a gradient, transitioning from blue to yellow. The surface has a sloping or ramp-like shape. The values of the plane are as follows: bottom left corner (t equals 0.25, x equals 0.25, u of (t, x) equals 0); bottom right corner (t equals 0.75, x equals 0.98, u of (t, x) equals 0); top right corner (t equals 0, x equals 0.85, u of (t, x) equals 0.3); and top left corner (t equals 0, x equals 0, u of (t, x) equals 0.1). The top left region of the surface is in a yellow shade, while the remaining area is in blue shades. Note: All the numerical data values are estimated.

The graph of the exact solution for example 3. Source: Authors’ own creation

Figure 9

Two 3-D surface plots show the approximate solution of u of (x, t) for two different values of alpha and beta.

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Two side-by-side 3-D surface plots, each showing the approximate solution of a function u of (x, t) as it varies with space “x” and time “t.” The plot on the left is titled “The approximate solution for alpha equals beta equals 0” and shows a surface over an x-axis ranging from 0 to 1 with increments of 0.5 and a t-axis ranging from 0 to 1 with increments of 0.2. The z-axis, labeled “u of (x, t),” represents the solution value and ranges from 0 to 1 with increments of 0.2. The surface has a downward slope from the “t equals 0” plane toward the “t equals 1” plane. The plot on the right is titled “The approximate solution for alpha equals beta equals negative 0.5” and shows a similar surface. The x and z axes and their ranges are identical to the first plot, while the t-axis scales from 0 to 1 in increments of 0.5. The shape of the surface is also similar, showing a decrease in value as time “t” increases. Both plots are colored with a gradient, transitioning from yellow and orange at the highest values (near t equals 0) to green and blue at the lowest values (near t equals 1). Note: All the numerical data values are estimated.

Graphical illustration of the present method at different values of (α, β), for example 3. Source: Authors’ own creation

Figure 10

A line graph comparing the approximate and exact solutions of a function u of (t) for different values of x.

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The line graph is titled “Solution for alpha equals beta equals 0.5 and different value of x.” The horizontal axis is labeled “t” and ranges from 0 to 1 with increments of 0.2, while the vertical axis is labeled “u of (t)” and ranges from 0.2 to 0.8 in increments of 0.1. The plot shows six different lines, representing three pairs of data: approximate solutions and their corresponding exact solutions for three different x-values. A legend in the top-right corner identifies each line. The approximate solutions are shown as dashed lines with circular or star-shaped markers, while the exact solutions are shown as solid lines. The red dashed line with asterisks represents “Approximate solution at x equals 0.1,” and the solid brown line represents “Exact solution at x equals 0.1.” The blue dashed line with circles represents “Approximate solution at x equals 0.2,” and the solid purple line represents “Exact solution at x equals 0.2.” The blue dashed line with asterisks represents “Approximate solution at x equals 0.3,” and the solid cyan line represents “Exact solution at x equals 0.3.” All lines show a similar decreasing trend as “t” increases. The approximate solutions, with their corresponding solid exact solution lines, are perfectly coinciding, indicating a high degree of accuracy. The lines with x equals 0.1 have the highest u of (t) values, followed by lines with x equals 0.2, and then lines with x equals 0.3.

Graphical evaluation of the present method’s solution and the exact solution at different values of x, with α = β = 0.5, for example 3. Source: Authors’ own creation

Figure 11

A line graph comparing approximate and exact solutions of function u(t) across different t values.

View large Download slide

The line graph is titled “Solution for alpha equals beta equals 0.5 and different value of t.” The horizontal axis is labeled “x” and ranges from 0 to 1 with increments of 0.2, while the vertical axis is labeled “u of (t)” and ranges from 0.2 to 0.9 in increments of 0.1. The plot shows six different lines, representing three pairs of data: approximate solutions and their corresponding exact solutions for three different t-values. A legend in the top-right corner identifies each line. The approximate solutions are shown as dashed lines with circular or star-shaped markers, while the exact solutions are shown as solid lines. The red dashed line with asterisks represents “Approximate solution at t equals 0.1,” and the solid brown line represents “Exact solution at t equals 0.1.” The blue dashed line with circles represents “Approximate solution at t equals 0.2,” and the solid purple line represents “Exact solution at t equals 0.2.” The blue dashed line with asterisks represents “Approximate solution at t equals 0.3,” and the solid cyan line represents “Exact solution at t equals 0.3.” All lines show a similar decreasing trend as “x” increases. The approximate solutions, with their corresponding solid exact solution lines, are perfectly coinciding, indicating a high degree of accuracy. The lines with t equals 0.1 have the highest u of (t) values, followed by lines with t equals 0.2, and then lines with t equals 0.3.

Graphical evaluation of the present method’s solution and the exact solution at different values of t, with α = β = 0.5, for example 3. Source: Authors’ own creation

Table 6

Comparison of absolute errors of projected method and different collocation methods, for example 3, at t = 0.1

x	An absolute error in [28]	An absolute error in [29]	An absolute error by our method for α = β = 0.5	An absolute error by our method for α = β = 0
0.1	3.52 × 10⁻¹⁴	2.84 × 10⁻⁷	5.31 × 10⁻¹⁴	8.05 × 10⁻¹⁵
0.2	9.87 × 10⁻¹⁵	8.79 × 10⁻⁶	1.86 × 10⁻¹⁵	9.22 × 10⁻¹⁵
0.3	5.89 × 10⁻¹⁶	1.20 × 10⁻⁵	4.59 × 10⁻¹⁶	1.16 × 10⁻¹⁵
0.4	2.24 × 10⁻¹⁴	1.12 × 10⁻⁵	2.72 × 10⁻¹⁴	1.07 × 10⁻¹⁴
0.5	1.70 × 10⁻¹⁴	7.95 × 10⁻⁶	1.43 × 10⁻¹⁴	2.52 × 10⁻¹⁴
0.6	2.85 × 10⁻¹⁴	3.29 × 10⁻⁶	1.69 × 10⁻¹⁴	7.63 × 10⁻¹⁵
0.7	1.23 × 10⁻¹⁵	1.78 × 10⁻⁶	5.36 × 10⁻¹⁵	1.13 × 10⁻¹⁵
0.8	6.85 × 10⁻¹⁴	6.53 × 10⁻⁶	7.82 × 10⁻¹⁴	5.78 × 10⁻¹⁵
0.9	2.23 × 10⁻¹²	1.04 × 10⁻⁵	1.89 × 10⁻¹⁴	8.96 × 10⁻¹⁴

x	An absolute error in [28]	An absolute error in [29]	An absolute error by our method for α = β = 0.5	An absolute error by our method for α = β = 0
0.1	3.52 × 10⁻¹⁴	2.84 × 10⁻⁷	5.31 × 10⁻¹⁴	8.05 × 10⁻¹⁵
0.2	9.87 × 10⁻¹⁵	8.79 × 10⁻⁶	1.86 × 10⁻¹⁵	9.22 × 10⁻¹⁵
0.3	5.89 × 10⁻¹⁶	1.20 × 10⁻⁵	4.59 × 10⁻¹⁶	1.16 × 10⁻¹⁵
0.4	2.24 × 10⁻¹⁴	1.12 × 10⁻⁵	2.72 × 10⁻¹⁴	1.07 × 10⁻¹⁴
0.5	1.70 × 10⁻¹⁴	7.95 × 10⁻⁶	1.43 × 10⁻¹⁴	2.52 × 10⁻¹⁴
0.6	2.85 × 10⁻¹⁴	3.29 × 10⁻⁶	1.69 × 10⁻¹⁴	7.63 × 10⁻¹⁵
0.7	1.23 × 10⁻¹⁵	1.78 × 10⁻⁶	5.36 × 10⁻¹⁵	1.13 × 10⁻¹⁵
0.8	6.85 × 10⁻¹⁴	6.53 × 10⁻⁶	7.82 × 10⁻¹⁴	5.78 × 10⁻¹⁵
0.9	2.23 × 10⁻¹²	1.04 × 10⁻⁵	1.89 × 10⁻¹⁴	8.96 × 10⁻¹⁴

Source(s): Authors’ own creation

Table 7

Comparison of the absolute errors between the projected method and various collocation methods, for example 3 at t = 0.5

x	An absolute error in [28]	An absolute error in [29]	An absolute error by our method for (α, β) = (0.5, 0.5)	An absolute error by our method for (α, β) = (0, 0)
0.1	2.59 × 10⁻¹⁴	8.95 × 10⁻⁶	7.29 × 10⁻¹⁴	9.25 × 10⁻¹⁵
0.2	4.25 × 10⁻¹⁴	4.10 × 10⁻⁶	6.93 × 10⁻¹⁵	8.53 × 10⁻¹⁵
0.3	1.55 × 10⁻¹⁵	1.39 × 10⁻⁵	7.89 × 10⁻¹⁵	6.19 × 10⁻¹⁵
0.4	9.83 × 10⁻¹⁴	2.07 × 10⁻⁵	5.38 × 10⁻¹⁴	1.70 × 10⁻¹⁴
0.5	3.33 × 10⁻¹²	2.47 × 10⁻⁵	3.53 × 10⁻¹⁴	3.93 × 10⁻¹⁴
0.6	5.37 × 10⁻¹¹	2.62 × 10⁻⁵	2.42 × 10⁻¹⁴	8.79 × 10⁻¹⁵
0.7	3.71 × 10⁻¹⁰	2.55 × 10⁻⁵	8.26 × 10⁻¹⁵	5.31 × 10⁻¹⁵
0.8	1.72 × 10⁻⁹	2.31 × 10⁻⁵	1.86 × 10⁻¹⁴	7.82 × 10⁻¹⁵
0.9	6.26 × 10⁻⁹	1.93 × 10⁻⁵	2.59 × 10⁻¹⁴	2.86 × 10⁻¹⁴

x	An absolute error in [28]	An absolute error in [29]	An absolute error by our method for (α, β) = (0.5, 0.5)	An absolute error by our method for (α, β) = (0, 0)
0.1	2.59 × 10⁻¹⁴	8.95 × 10⁻⁶	7.29 × 10⁻¹⁴	9.25 × 10⁻¹⁵
0.2	4.25 × 10⁻¹⁴	4.10 × 10⁻⁶	6.93 × 10⁻¹⁵	8.53 × 10⁻¹⁵
0.3	1.55 × 10⁻¹⁵	1.39 × 10⁻⁵	7.89 × 10⁻¹⁵	6.19 × 10⁻¹⁵
0.4	9.83 × 10⁻¹⁴	2.07 × 10⁻⁵	5.38 × 10⁻¹⁴	1.70 × 10⁻¹⁴
0.5	3.33 × 10⁻¹²	2.47 × 10⁻⁵	3.53 × 10⁻¹⁴	3.93 × 10⁻¹⁴
0.6	5.37 × 10⁻¹¹	2.62 × 10⁻⁵	2.42 × 10⁻¹⁴	8.79 × 10⁻¹⁵
0.7	3.71 × 10⁻¹⁰	2.55 × 10⁻⁵	8.26 × 10⁻¹⁵	5.31 × 10⁻¹⁵
0.8	1.72 × 10⁻⁹	2.31 × 10⁻⁵	1.86 × 10⁻¹⁴	7.82 × 10⁻¹⁵
0.9	6.26 × 10⁻⁹	1.93 × 10⁻⁵	2.59 × 10⁻¹⁴	2.86 × 10⁻¹⁴

Source(s): Authors’ own creation

Table 8

Covergence rate and L^∞ errors for M = 2 for example 3

k	N = 2^k	L^∞ for (α, β) = (0, 0)	R for (α, β) = (0, 0)	L^∞ for α = β = −0.5	R for α = β = −0.5
1	2	1.18 × 10⁻¹	–	1.23 × 10⁻¹	–
2	4	1.76 × 10⁻²	2.75	1.86 × 10⁻²	2.73
3	8	2.06 × 10⁻³	3.10	2.19 × 10⁻³	3.09
4	16	3.60 × 10⁻⁴	2.78	3.56 × 10⁻⁴	2.62
5	32	2.61 × 10⁻⁶	2.80	5.82 × 10⁻⁵	2.64
6	64	1.92 × 10⁻⁸	3.77	9.56 × 10⁻⁷	5.9

k	N = 2^k	L^∞ for (α, β) = (0, 0)	R for (α, β) = (0, 0)	L^∞ for α = β = −0.5	R for α = β = −0.5
1	2	1.18 × 10⁻¹	–	1.23 × 10⁻¹	–
2	4	1.76 × 10⁻²	2.75	1.86 × 10⁻²	2.73
3	8	2.06 × 10⁻³	3.10	2.19 × 10⁻³	3.09
4	16	3.60 × 10⁻⁴	2.78	3.56 × 10⁻⁴	2.62
5	32	2.61 × 10⁻⁶	2.80	5.82 × 10⁻⁵	2.64
6	64	1.92 × 10⁻⁸	3.77	9.56 × 10⁻⁷	5.9

Source(s): Authors’ own creation

7. Conclusion

The aim of this paper is to develop a novel Jacobi wavelet decomposition method for approximating the solutions to partial differential equations (PDEs). Our work demonstrates that the Jacobi wavelet method, combining the Jacobi wavelet method with an operational matrix of derivatives provides an efficient technique for transforming partial differential equations into a system of ordinary differential equations. We have tested this numerical method on various examples, and the comparison of the results obtained with those from other methods underscores its effectiveness in computing numerical solutions to such equations.

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Bahri Mohammed Nadjib and Sidi Mohammed Bahri

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at Link to the terms of the CC BY 4.0 licence.

A new Jacobi wavelets decomposition method for solving partial differential equations

1. Introduction

2. Decompoition in Jacobi wavelets basis

2.1 Jacobi polynomials and their properties

2.2 Jacobi wavelets

2.3 Decomposition in Jacobi wavelets basis.

3. Decompoition in $L_{w}^{2} ([0, T]; L_{w}^{2} ([0,1]))$

4. Operational matrix of derivative of the Jacobi wavelets

5. Method of solution

5.1 Application to telegraph equations

5.2 Application to linear hyperbolic partial differential equations

6. Numerical illustrations

6.1 Example 1

6.2 Example 2

6.3 Example 3

7. Conclusion

References

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A new Jacobi wavelets decomposition method for solving partial differential equations Open Access

1. Introduction

2. Decompoition in Jacobi wavelets basis

2.1 Jacobi polynomials and their properties

2.2 Jacobi wavelets

2.3 Decomposition in Jacobi wavelets basis.

3. Decompoition in Lw2([0,T];Lw2([0,1]))

4. Operational matrix of derivative of the Jacobi wavelets

5. Method of solution

5.1 Application to telegraph equations

5.2 Application to linear hyperbolic partial differential equations

6. Numerical illustrations

6.1 Example 1

6.2 Example 2

6.3 Example 3

7. Conclusion

References

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A new Jacobi wavelets decomposition method for solving partial differential equations

3. Decompoition in $L_{w}^{2} ([0, T]; L_{w}^{2} ([0,1]))$