# Symbolic algorithm to solve initial value problems for partial differential equations

Srinivasarao Thota, Shiv Datt

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Srinivasarao Thota, Shiv Datt. Symbolic algorithm to solve initial value problems for partial differential equations. Bulletin of Computational Applied Mathematics, 2020, 8 (1), pp.25-48. hal-03536339

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# Symbolic algorithm to solve initial value problems for partial differential equations

Srinivasarao Thota<sup>12</sup>, Shiv Datt Kumar<sup>3</sup>

CompAMa Vol.8, pp.25-48, 2020 - Received November 12, 2019 - Accepted September 30, 2020

## Abstract

In this paper, we present a new symbolic algorithm for finding the Green's function of a given initial value problem for linear partial differential equations of second order with constant coefficients. The proposed algorithm is also applicable for n-th order partial differential equations. We employ the integro-differential algebra to express the initial value problems and the Green's function. Some examples are presented to illustrate the proposed method and compared with other existing method. Link for implemented proposed algorithm in Maple is provided and sample computations are shown.

**Keywords:** Initial value problems; Partial differential equations; Symbolic method; Integro-differential algebra.

## Resumen

En este artículo, presentamos un nuevo algoritmo simbólico para encontrar la función de Green de un problema de valor inicial dado para ecuaciones diferenciales

<footnote n="1"><sup>1</sup>Department of Applied Mathematics, School of Applied Natural Sciences, Adama Science and Technology University, Post Box No. 1888, Adama, Ethiopia. (srinithota@ymail.com, srinivasarao.thota@astu.edu.et). (ORCID iD: https://orcid.org/0000-0002-3265-5656).</footnote>

<footnote n="2"><sup>2</sup>Corresponding author.</footnote>

<footnote n="3"><sup>3</sup>Department of Mathematics, Motilal Nehru National Institute of Technology Allahabad, Prayagraj, Uttar Pradesh, India-211004. (sdt@mnnit.ac.in). (ORCID iD: https://orcid.org/0000-0002-5180-3423).</footnote>
parciales lineales de segundo orden con coeficientes constantes. El algoritmo propuesto también es aplicable para ecuaciones diferenciales parciales de n-ésimo orden. Empleamos el álgebra integro-diferencial para expresar los problemas de valor inicial y la función de Green. Se presentan algunos ejemplos para ilustrar el método propuesto y compararlos con otros métodos existentes. Se proporciona el enlace para el algoritmo propuesto implementado en Maple y se muestran ejemplos de cálculos.

**Palabras claves:** Problemas de valores iniciales; Ecuaciones diferenciales parciales; Método simbólico; Algebra integro-diferencial.

# 1 Introduction

Symbolic computation is an important tool in the scientific field. It is playing central role to solve the mathematical equations, especially problems involving differential equations. In twenty-th century, the science and technology had a very swift progress in various fields, especially in computing, the sub-field of scientific and technological computing. The symbolic computation is a part of scientific computing based on numerical computations. Most of the numerical calculations are carried out with approximate floating point algorithm, whereas symbolic computing emphasizes on the exact solution with symbols representing the mathematical objects. One of the biggest success in the research of symbolic computation is the development of significant software systems. In the recent time, several new symbolic methods have been developed by various researches, for example, see [1-13].

A Green's function, in general, is an integral kernel which represents the analytic solutions of the initial/boundary value problems. Green's function allows us to view the visual interpretation as a result of the actions associated to a source of force or to a charge concentrated at a point [14].

Our focus on this paper is to find the Green's function of a given *initial value problem* (IVP) for Partial differential equations(PDEs) over integro-differential algebra. The key steps to compute the Green's function are the factorization of the given partial differential operator and identifying the quasi-linear PDE of the first order, so-called *Lagrange's equation*. In order to solve IVP in symbolic manner over integro-differential algebra, we first prove certain structures form an integro-differential algebra (see Proposition 1 and Theorem 1). We decompose the solution of the fully inhomogeneous IVP for PDE (inhomogeneous PDE with inhomogeneous initial conditions).
ditions) into the solution of the semi-homogeneous IVP (homogeneous PDE having right hand side zero and inhomogeneous initial conditions) and the solution of the semi-inhomogeneous IVP (PDE having right hand side non-zero (forcing function) and homogeneous initial conditions). We solve the semi-inhomogeneous IVP and semi-homogeneous IVP separately and combine these two solutions to obtain the required Green's function of a given fully inhomogeneous IVP.

Partial differential equations (PDEs) have vast applications in the field of applied sciences and technology, for e.g. in dynamical system with several variables, in the theory of electricity, hydraulics etc. Many physical and engineering problems when formulated in the mathematical language, give rise to PDEs (see, for example, [15–18]).

The general solution of a PDE in a region $\Omega$ contains arbitrary constants or arbitrary functions, the unique solution of a PDE corresponding to a physical problem will satisfy certain initial/boundary conditions.

Mathematically, we solve the general form of the second order PDE of the following form with inhomogeneous initial conditions, suppose $\mathcal{F} = C^\infty(\mathbb{R}^2)$ for simplicity,

$$ 
\begin{aligned}
& a \frac{\partial^2 u(x, y)}{\partial x^2} + b \frac{\partial^2 u(x, y)}{\partial x \partial y} + c \frac{\partial^2 u(x, y)}{\partial y^2} = f(x, y), \\
& u(0, y) = \alpha_1(y), \quad \frac{\partial u(0, y)}{\partial x} = \alpha_2(y),
\end{aligned} 
\quad (1) $$

where $u(x, y) \in \mathcal{F}$ is the solution of the given IVP; $f(x, y) \in \mathcal{F}$ is the forcing function; and $\alpha_1(y), \alpha_2(y) \in \mathcal{F}$ are given initial data. We solve IVP (1) not only for a fixed right hand side functions $f(x, y)$ and $\alpha_1(y), \alpha_2(y)$ but also a generic formulation of the solution on the level of operators over integro-differential algebra.

Some advantages of the proposed symbolic method are: (i) same algorithm is also applicable to solve the IVP for n-th order PDE (ii) one can extend the proposed algorithm to several variables $u(x_1, x_2, \dots, x_n) \in \mathcal{F}$, where $\mathcal{F} = C^\infty(\mathbb{R}^n)$ (iii) this method works directly on operators (iv) this algorithm can help to implement the manual calculations in commercial packages such as Matlab, Mathematica, Singular, Maple, SciLab etc (v) provides a generic formulation of the solution of a given IVP for PDE.

Following is the plan of the paper: In Section 2, we recall the basic definitions and concepts of integro-differential algebra; and Section 2.1 presented
integro-differential algebra for several variables. In Section 3, we describe the proposed symbolic method for IVP and Section 4 provides the pseudo code and link for Maple implementation code of the proposed algorithm. In Section 5 sample computations using the proposed algorithm to demonstrate and verify the symbolic method are provided.

## 2 Integro-Differential Algebra

We recall basic concepts of integro-differential algebras to express the IVP, and the Green's function in operator based notations, see, for example, [1-7] for further details. Throughout this section $K$ denotes the field of characteristic zero.

**Definition 1.** [2] The structure $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ is called an integro-differential algebra over $K$ if $\mathcal{F}$ is a commutative $K$-algebra with $K$-linear operators $\mathcal{D}$ and $\mathcal{A}$ such that the following conditions are satisfied

$$ \bullet \mathcal{D}(\mathcal{A}f) = f \text{ (section axiom)}, $$

$$ \bullet \mathcal{D}(fg) = (\mathcal{D}f)g + f(\mathcal{D}g) \text{ (Leibniz axiom)}, $$

$$ \bullet (\mathcal{A}Df)(\mathcal{A}Dg) + \mathcal{A}D(fg) = (\mathcal{A}Df)g + f(\mathcal{A}Dg) \text{ (differential Baxter axiom)}. $$

Here $\mathcal{D}: \mathcal{F} \to \mathcal{F}$ and $\mathcal{A}: \mathcal{F} \to \mathcal{F}$ are two maps such that $\mathcal{D}$ is a derivation and $\mathcal{A}$ is a $K$-linear right inverse of $\mathcal{D}$, i.e. $\mathcal{D}\mathcal{A} = 1$ (the identity map).

The operator $\mathcal{E} = 1 - \mathcal{A}\mathcal{D}$ is called the *evaluation* on $\mathcal{F}$ defined $\mathcal{E}f(x) = f(a)$, evaluates at initial point $a$. For an ordinary integro-differential algebra, the evaluation can be treated as a multiplicative linear functional $\mathcal{E}: \mathcal{F} \to K$. We know that the differential algebra $(\mathcal{F}, \mathcal{D})$ induces the algebra of differential operator $\mathcal{F}[\mathcal{D}]$, similarly an integro-differential algebra $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ induces an algebra of integro-differential operators denoted by $\mathcal{F}[\mathcal{D}, \mathcal{A}]$ [2].

**Definition 2.** [1] An integro-differential algebra over $K$ is called ordinary if $\dim(\mathrm{Ker}(\mathcal{D})) = 1$, otherwise partial integro-differential algebra.

### 2.1 Integro-Differential Algebra for Several Variables

Before starting the integro-differential algebra for several variables, we present an example of integro-differential algebra with several variables for simplicity.
**Example 1.** Let $\mathcal{F} = C^\infty(\mathbb{R}^2)$. Consider the IVP for first order PDE

$$ \frac{\partial u(x, y)}{\partial x} + m \frac{\partial u(x, y)}{\partial y} = f(x, y), \qquad (2) $$

$$ u(0, y) = \alpha(y), $$

where $u(x, y) \in \mathcal{F}$ is the solution of the given IVP; $f(x, y) \in \mathcal{F}$ is the forcing function; $\alpha(y) \in \mathcal{F}$ is initial data; and $m \in \mathbb{R}$. The exact solution of the IVP (2) is given by (See Theorem 2)

$$ u(x, y) = \int_{0}^{x} f(t, y - mx + mt) dt + \alpha(y - mx) \in \mathcal{F}. \qquad (3) $$

In operator notations, we have $\mathcal{D} = \partial_x + m\partial_y \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ and the section $\mathcal{A} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ of the operator $\mathcal{D}$ is obtained from the Green's function (3) as follows

$$ \mathcal{A}f = \int_{0}^{x} f(t, y - mx + mt) dt, \qquad (4) $$

such that $\mathcal{D}\mathcal{A} = 1$. Now $\text{Ker}(\mathcal{D})$ is the solution of the homogeneous PDE with $f(x, y) = 0$ and is obtained as

$$ \begin{aligned} \text{Ker}(\mathcal{D}) &= \{v : \mathcal{D}v = 0\} \\ &= \{g(y - mx) : g \in C^\infty(\mathbb{R})\}. \end{aligned} $$

The evaluation $\mathcal{E}: \mathcal{F} \to \mathcal{F}$ is a function defined as

$$ \begin{aligned} \mathcal{E}f(x, y) &= f(x, y) - \mathcal{A}\mathcal{D}f(x, y) \\ &= f(x, y) - (f(x, y) - f(0, y - mx)) \\ &= f(0, y - mx) \end{aligned} $$

The following proposition shows that the structure $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ is an integro-differential algebra, where $\mathcal{D} = \partial_x + m\partial_y$ and $\mathcal{A}$ is defined by equation (4).

**Proposition 1.** Suppose $\mathcal{D} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ and $\mathcal{A} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ are such that $\mathcal{D}f = \partial_x f + m\partial_y f$ and $\mathcal{A}f = \int_0^x f(t, y - mx + mt) dt$, then the algebraic structure $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ form an integro-differential algebra.
Proof. Section axiom:

$$
\begin{align*}
\mathbf{D}(\mathbf{A}f(x, y)) &= (\partial_x + m\partial_y) \left( \int_0^x f(t, y - mx + mt) \, dt \right) \\
&= \left( \partial_x \int_0^x f(t, y - mx + mt) \, dt \right) + m \left( \partial_y \int_0^x f(t, y - mx + mt) \, dt \right) \\
&= \left( -m \int_0^x \partial_x f(t, y - mx + mt) + f(x, y) \right) + m \left( \int_0^x \partial_y f(t, y - mx + mt) \, dt \right) \\
&= f(x, y).
\end{align*}
$$

Leibniz axiom:

$$
\begin{align*}
\mathbf{D}(f(x, y)g(x, y)) &= (\partial_x + m\partial_y)(f(x, y)g(x, y)) \\
&= \partial_x(f(x, y)g(x, y)) + m\partial_y(f(x, y)g(x, y)) \\
&= (\partial_x f(x, y))g(x, y) + f(x, y)(\partial_x g(x, y)) \\
&\quad + m((\partial_y f(x, y))g(x, y) + f(x, y)(\partial_y g(x, y))) \\
&= (\partial_x f(x, y) + m\partial_y f(x, y))g(x, y) + f(x, y)(\partial_x g(x, y) + m\partial_y g(x, y)) \\
&= ((\partial_x + m\partial_y)f(x, y))g(x, y) + f(x, y)((\partial_x + m\partial_y)g(x, y)) \\
&= (\mathbf{D}f(x, y))g(x, y) + f(x, y)(\mathbf{D}g(x, y)).
\end{align*}
$$

Differential Baxter axiom:

$$
\begin{align*}
(\mathbf{A}\mathbf{D}f(x, y))(\mathbf{A}\mathbf{D}g(x, y)) + \mathbf{A}\mathbf{D}(f(x, y)g(x, y)) \\
&= ((1 - \mathbf{E})f(x, y))((1 - \mathbf{E})g(x, y)) + (1 - \mathbf{E})(f(x, y)g(x, y)) \\
&= (f(x, y) - \mathbf{E}f(x, y))(g(x, y) - \mathbf{E}g(x, y)) + f(x, y)g(x, y) - \mathbf{E}f(x, y)g(x, y) \\
&= (f(x, y) - \mathbf{E}f(x, y))(g(x, y) - \mathbf{E}g(x, y)) + f(x, y)g(x, y) - \mathbf{E}f(x, y)\mathbf{D}g(x, y) \\
&= (f(x, y) - (f(x, y) - f(0, y - mx)))(g(x, y) - (g(x, y) - g(0, y - mx))) \\
&\quad + f(x, y)g(x, y) - ((f(x, y) - f(0, y - mx)))(g(x, y) - g(0, y - mx))) \\
&= f(0, y - mx)g(0, y - mx) + f(x, y)g(0, y - mx) \\
&\quad + f(0, y - mx)g(x, y) - f(0, y - mx)g(0, y - mx) \\
&= f(x, y)g(0, y - mx) + f(0, y - mx)g(x, y) \\
&= f(x, y)(g(x, y) - (g(x, y) - g(0, y - mx))) \\
&\quad + (f(x, y) - (f(x, y) - f(0, y - mx)))g(x, y) \\
&= f(x, y)(1 - \mathbf{E})g(x, y) + ((1 - \mathbf{D})f(x, y))g(x, y) \\
&= f(x, y)(\mathbf{A}\mathbf{D}g(x, y)) + (\mathbf{A}\mathbf{D}f(x, y))g(x, y).
\end{align*}
$$
Therefore, the algebraic structure ($\mathcal{F}$, $\mathbf{D}$, $\mathbf{A}$) form an integro-differential algebra, where $\mathbf{D} = \partial_x + m\partial_y \in \mathcal{F}[\mathbf{D}, \mathbf{A}]$ and $\mathbf{A} \in \mathcal{F}[\mathbf{D}, \mathbf{A}]$ is defined by $\mathbf{A}f = \int_0^x f(t, y - mx + mt) dt$. $\square$

We represent the general IVP for PDE of second order given in (1) over operator notations as follows: given a forcing function $f \in \mathcal{F}$ and initial data $\alpha_1, \alpha_2 \in \mathcal{F}$, we find a solution $u \in \mathcal{F}$ such that

$$ \begin{aligned} \mathbf{D}u &= f, \\ \tilde{\mathbf{E}}u &= \alpha_1, \tilde{\mathbf{E}}\partial_x u = \alpha_2, \end{aligned} \qquad (5) $$

where $\mathbf{D} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy} \in \mathcal{F}[\mathbf{D}, \mathbf{A}]$; $\tilde{\mathbf{E}}u = u(0, y)$, $\tilde{\mathbf{E}}\partial_x u = (\frac{\partial u}{\partial x})_{x=0} \in \mathcal{F}$; and $a \neq 0, b, c \in \mathbb{R}$.

In general, the IVP for PDE of n-th order is given by

$$ \begin{aligned} \mathbf{D}u &= f, \\ \tilde{\mathbf{E}}u &= \alpha_1, \tilde{\mathbf{E}}\partial_x u = \alpha_2, \tilde{\mathbf{E}}\partial_x^2 u = \alpha_3, \dots = \tilde{\mathbf{E}}\partial_x^{n-1} u = \alpha_n, \end{aligned} \qquad (6) $$

where $\mathbf{D} = a_0\frac{\partial^n}{\partial x^n} + a_1\frac{\partial^n}{\partial x^{n-1}\partial y} + a_2\frac{\partial^n}{\partial x^{n-2}\partial y^2} + \dots + a_n\frac{\partial^n}{\partial y^n} \in \mathcal{F}[\mathbf{D}, \mathbf{A}]$; $\tilde{\mathbf{E}}u = u(0, y)$, $\tilde{\mathbf{E}}\partial_x u = (\frac{\partial u}{\partial x})_{x=0}, \dots, \tilde{\mathbf{E}}\partial_x^{n-1} u = (\frac{\partial^{n-1} u}{\partial x^{n-1}})_{x=0} \in \mathcal{F}$; $\alpha_1, \dots, \alpha_n \in \mathcal{F}$ are initial data; and $a_0 \neq 0, a_i \in \mathbb{R}$, for $i = 0, \dots, n$.

### 3 A Symbolic Algorithm for Initial Value Problem

First we recall the IVP for PDE in operator form

$$ \begin{aligned} \mathbf{D}u &= f, \\ \tilde{\mathbf{E}}u &= \alpha_1, \tilde{\mathbf{E}}\partial_x u = \alpha_2, \end{aligned} \qquad (7) $$

where $\mathbf{D} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy}$; $\tilde{\mathbf{E}}u = u(0, y)$, $\tilde{\mathbf{E}}\partial_x u = (\frac{\partial u}{\partial x})_{x=0}$; and $\alpha_1, \alpha_2$ are initial data. Factorizing the differential operator $\mathbf{D}$ into two linear factors, we get
$$
\begin{aligned}
\mathcal{D} &= a\partial_{xx} + b\partial_{xy} + c\partial_{yy} = a\left(\partial_{xx} + \frac{b}{a}\partial_{xy} + \frac{c}{a}\partial_{yy}\right) \\
&= a\left(\partial_x + \frac{b}{2a}\partial_y + \frac{\sqrt{b^2 - 4ac}}{2a}\partial_y\right)\left(\partial_x + \frac{b}{2a}\partial_y - \frac{\sqrt{b^2 - 4ac}}{2a}\partial_y\right) \\
&= a\left(\partial_x + \left(\frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right)\partial_y\right)\left(\partial_x + \left(\frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right)\partial_y\right).
\end{aligned}
$$

Now, the given IVP (7) can be rewritten in factors as follows

$$
\begin{gathered}
a \left( \partial_x + \left( \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} \right) \partial_y \right) \left( \partial_x + \left( \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a} \right) \partial_y \right) u = f, \\
\tilde{\mathcal{E}}u = \alpha_1, \quad \tilde{\mathcal{E}}\partial_x u = \alpha_2.
\end{gathered}
$$

**Theorem 1.** Suppose $\mathcal{D}$ is linear partial differential operator of second order, i.e. $\mathcal{D}f = a\partial_{xx}f + b\partial_{xy}f + c\partial_{yy}f$ and $\mathcal{A}$ is a section of $\mathcal{D}$, i.e. $\mathcal{DA} = 1$, then the algebraic structure $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ form an integro-differential algebra.

*Proof.* The section axiom is clear from $\mathcal{DA} = 1$, i.e. $\mathcal{DA}f(x, y) = f(x, y)$. For simplicity, suppose $\mathcal{D}_{m_1} = \partial_x + m_1\partial_y$, $\mathcal{D}_{m_2} = \partial_x + m_2\partial_y$, then $\mathcal{D} = \mathcal{D}_{m_1}\mathcal{D}_{m_2}$, where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a} \in \mathbb{R}$ and $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a} \in \mathbb{R}$. The Leibniz axiom can be verified as follows:

$$
\begin{aligned}
\mathcal{D}(f(x, y)g(x, y)) &= (\mathcal{D}_{m_1}\mathcal{D}_{m_2})(f(x, y)g(x, y)) \\
&= \mathcal{D}_{m_1}((\mathcal{D}_{m_2}f(x, y))g(x, y) + f(x, y)(\mathcal{D}_{m_2}g(x, y))) \\
&= \mathcal{D}_{m_1}((\mathcal{D}_{m_2}f(x, y))g(x, y)) + (\mathcal{D}_{m_2}f(x, y))(\mathcal{D}_{m_1}g(x, y)) \\
&\quad + (\mathcal{D}_{m_1}f(x, y))(\mathcal{D}_{m_2}g(x, y)) + f(x, y)(\mathcal{D}_{m_1}(\mathcal{D}_{m_2}g(x, y))) \\
&= (\mathcal{D}f(x, y))g(x, y) + (\mathcal{D}_{m_2}f(x, y))(\mathcal{D}_{m_1}g(x, y)) \\
&\quad + (\mathcal{D}_{m_1}f(x, y))(\mathcal{D}_{m_2}g(x, y)) + f(x, y)(\mathcal{D}g(x, y)).
\end{aligned}
\quad (8)
$$

Now

$$
\begin{aligned}
(\mathcal{D}f(x, y))g(x, y) + (\mathcal{D}g(x, y))f(x, y) \\
&= (\mathcal{D}_{m_1}\mathcal{D}_{m_2}f(x, y))g(x, y) + (\mathcal{D}_{m_1}\mathcal{D}_{m_2}g(x, y))f(x, y) \\
&= ((\mathcal{D}_{m_1}\mathcal{D}_{m_2})f(x, y))g(x, y) + (\mathcal{D}_{m_1}f(x, y))(\mathcal{D}_{m_2}g(x, y)) \\
&\quad + f(x, y)((\mathcal{D}_{m_1}\mathcal{D}_{m_2})g(x, y)) + (\mathcal{D}_{m_2}f(x, y))(\mathcal{D}_{m_1}g(x, y)) \\
&= (\mathcal{D}f(x, y))g(x, y) + (\mathcal{D}_{m_2}f(x, y))(\mathcal{D}_{m_1}g(x, y)) \\
&\quad + (\mathcal{D}_{m_1}f(x, y))(\mathcal{D}_{m_2}g(x, y)) + f(x, y)(\mathcal{D}g(x, y)).
\end{aligned}
\quad (9)
$$
Equations (8) and (9) show that the Leibniz axiom is verified. Finally, the differential Baxter axiom is verified as follows:

$$ 
\begin{aligned}
(\mathbf{AD}f(x, y))(\mathbf{AD}g(x, y)) + \mathbf{AD}(f(x, y)g(x, y)) \\
&= ((1 - \mathbf{E})f(x, y))((1 - \mathbf{E})g(x, y)) + (1 - \mathbf{E})(f(x, y)g(x, y)) \\
&= (f(x, y) - \mathbf{E}f(x, y))(g(x, y) - \mathbf{E}g(x, y)) \\
&\quad + f(x, y)g(x, y) - \mathbf{E}f(x, y)g(x, y) \\
&= (f(x, y) - \mathbf{E}f(x, y))(g(x, y) - \mathbf{E}g(x, y)) \\
&\quad + f(x, y)g(x, y) - \mathbf{E}f(x, y)\mathbf{E}g(x, y) \\
&= f(x, y)g(x, y) - f(x, y)\mathbf{E}g(x, y) - g(x, y)\mathbf{E}f(x, y) + f(x, y)g(x, y) \\
&= f(x, y) - g(x, y)\mathbf{E}f(x, y) + f(x, y)g(x, y) - f(x, y)\mathbf{E}g(x, y) \\
&= (f(x, y) - \mathbf{E}f(x, y))g(x, y) + f(x, y)(g(x, y) - \mathbf{E}g(x, y)) \\
&= ((1 - \mathbf{E})f(x, y))g(x, y) + f(x, y)((1 - \mathbf{E})g(x, y)) \\
&= (\mathbf{AD}f(x, y))g(x, y) + f(x, y)(\mathbf{AD}g(x, y)).
\end{aligned}
$$

Therefore, the algebraic structure $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ form an integro-differential algebra, where $\mathcal{D} = \mathcal{D}_{m_1}\mathcal{D}_{m_2} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy}$ and $\mathcal{A}$ is such that $\mathcal{D}\mathcal{A} = 1$. $\square$

Now, we present our main result of this paper in the following theorem to compute the Green's function of a given IVP for PDEs.

**Theorem 2.** Let $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ be an integro-differential algebra. Suppose $\mathcal{D} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ is a linear partial differential operator of second order. Then the IVP

$$ 
\begin{aligned}
\mathcal{D}u &= f, \\
\tilde{E}u &= \alpha_1, \quad \tilde{E}\partial_x u = \alpha_2,
\end{aligned}
\qquad (10) $$

has the unique solution given by

$$ 
\begin{aligned}
u &= \frac{1}{a} \int_0^x \int_0^t f(\xi, y + m_1(t - x) - m_2(t - \xi)) \, d\xi \, dt \\
&\quad + \left(\frac{m_2}{m_2 - m_1}\right) \alpha_1(y - m_1x) + \left(\frac{m_1}{m_1 - m_2}\right) \alpha_1(y - m_2x) \\
&\quad + \left(\frac{1}{m_1 - m_2}\right) \int_{y-m_1x}^{y-m_2x} \alpha_2(s) \, ds \in \mathcal{F}, \quad m_1 \neq m_2.
\end{aligned}
\qquad (11) $$
For $m_1 = m_2$, the unique solution is

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y - m_{1}(x - \xi)) \, d\xi \, dt \\ + \alpha_{1}(y - m_{1}x) + x\alpha_{2}(y - m_{1}x) + m_{1}x\partial_{y}\alpha_{1}(y - m_{1}x) \in \mathcal{F}, \qquad (12) $$

where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}$, $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}$, and $a \neq 0, b, c \in \mathbb{R}$.

*Proof.* We can decompose the Green's function of the given IVP (10) into the solution of semi-inhomogeneous IVP (for $\alpha_1 = \alpha_2 = 0$) and the solution of semi-homogeneous IVP (for $f = 0$). Now, the Green's function of the given fully inhomogeneous IVP (10) is obtained by combining the two solutions of semi-inhomogeneous and semi-homogeneous IVPs.

Consider the given differential equation $Du = f$ satisfying the given initial conditions with $\alpha_1 = \alpha_2 = 0$. Finding the Green's function of the semi-inhomogeneous IVP is nothing but to find the section of the given differential operator. The Green's function of the semi-inhomogeneous IVP is computed as follows.

$$ \begin{aligned} u &= \frac{1}{D} f = \frac{1}{a \left( \partial_x + m_1 \partial_y \right) \left( \partial_x + m_2 \partial_y \right)} f \\ &= \frac{1}{a} \left( \frac{1}{\partial_x + m_1 \partial_y} \right) \left( \frac{1}{\partial_x + m_2 \partial_y} f \right). \end{aligned} \qquad (13) $$

If $u_1$ is the solution of $\frac{1}{(\partial_x + m_2 \partial_y)} f$, then we have

$$ u_1 = \frac{1}{(\partial_x + m_2 \partial_y)} f \quad \text{which gives} \quad \partial_x u_1 + m_2 \partial_y u_1 = f. \qquad (14) $$

Equation (14) is a quasi-linear PDE of the first order and is called *Lagrange's equation* [15]. The subsidiary equations are

$$ \frac{dx}{1} = \frac{dy}{m_2} = \frac{du_1}{f}. \qquad (15) $$

From first two members of (15), we have

$$ \begin{aligned} dy - m_2 dx &= 0 \quad \text{or} \\ y - m_2 x &= c \end{aligned} \qquad (16) $$
From the first and last members of (15), we have

$$ 
\begin{aligned}
du_1 &= f(x, y) \, dx = f(x, c + m_2x) \, dx \quad \text{or} \\
u_1 &= \int_0^x f(t, c + m_2t) \, dt \quad \text{or} \\
u_1 &= \int_0^x f(t, y - m_2x + m_2t) \, dt \in \mathcal{F} \quad (\because c = m_2x - y). \quad (17)
\end{aligned}
$$

From equation (13), we compute the Green's function

$$ 
\begin{aligned}
u &= \frac{1}{a(\partial_x + m_1\partial_y)} u_1 \quad \text{or} \\
\partial_x u + m_1\partial_y u &= \frac{1}{a} u_1.
\end{aligned}
$$

Using the procedure applied for the Lagrange's equation (14), we obtain the Green's function of the semi-inhomogeneous IVP as follows.

$$ 
\begin{aligned}
u &= \frac{1}{a} \int_0^x u_1(t, y - m_1x + m_1t) \, dt \\
&= \frac{1}{a} \int_0^x \int_0^t f(\xi, y - m_1x + m_1t - m_2t + m_2\xi) \, d\xi \, dt \\
&= \frac{1}{a} \int_0^x \int_0^t f(\xi, y + m_1(t - x) - m_2(t - \xi)) \, d\xi \, dt.
\end{aligned}
$$

Therefore, we have

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y + m_{1}(t-x) - m_{2}(t-\xi)) \, d\xi \, dt \in \mathcal{F}, \quad (18) $$

where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}$ and $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}$, $a \neq 0, b, c \in \mathbb{R}$.

Equation (18) represents the Green's function of the semi-inhomogeneous IVP of (10). We compute the Green's function of the semi-homogeneous IVP of (10) as follows.

Solution of the given homogeneous PDE using Lagrange's method of multipliers as presented for (14) is given by, for $m_1 \neq m_2$,

$$ u(x, y) = \phi_1(y - m_1x) + \phi_2(y - m_2x), \quad (19) $$
where $\phi_1$ and $\phi_2$ are arbitrary functions to be determined using the given initial conditions. From the given initial conditions, we have

$$ \phi_1(y) + \phi_2(y) = \alpha_1(y), \qquad (20) $$

$$ -m_1\phi'_1(y) - m_2\phi'_2(y) = \alpha_2(y). \qquad (21) $$

From equation (21), we have

$$ m_1\phi'_1(y) = -\alpha_2(y) - m_2\phi'_2(y) \quad \text{or} $$

$$ \phi'_1(y) = -\frac{1}{m_1}\alpha_2(y) - \frac{m_2}{m_1}\phi'_2(y). $$

Integrating both sides, we get

$$ \phi_1(y) = -\frac{1}{m_1} \int \alpha_2(s) \, ds - \frac{m_2}{m_1} \phi_2(y) + C, \qquad (22) $$

where $C$ is an arbitrary constant. Substituting (22) in (20), we get

$$ -\frac{1}{m_1} \int \alpha_2(s) \, ds - \frac{m_2}{m_1} \phi_2(y) + C + \phi_2(y) = \alpha_1(y) \quad \text{or} $$

$$ \left(1 - \frac{m_2}{m_1}\right) \phi_2(y) = \alpha_1(y) + \frac{1}{m_1} \int \alpha_2(s) \, ds - C \quad \text{or} $$

$$ \phi_2(y) = \left(\frac{m_1}{m_1 - m_2}\right) \alpha_1(y) + \left(\frac{1}{m_1 - m_2}\right) \int \alpha_2(s) \, ds - \left(\frac{m_1 C}{m_1 - m_2}\right). \qquad (23) $$

From equations (20) and (23), we have

$$ \phi_1(y) + \left(\frac{m_1}{m_1 - m_2}\right) \alpha_1(y) + \left(\frac{1}{m_1 - m_2}\right) \int \alpha_2(s) \, ds - \left(\frac{m_1 C}{m_1 - m_2}\right) = \alpha_1(y) \quad \text{or} $$

$$ \phi_1(y) = \left(\frac{m_2}{m_2 - m_1}\right) \alpha_1(y) - \left(\frac{1}{m_1 - m_2}\right) \int \alpha_2(s) \, ds + \left(\frac{m_1 C}{m_1 - m_2}\right). \qquad (24) $$

Now, from (19), the Green's function of the semi-homogeneous IVP of (10),
for $m_1 \neq m_2$, is

$$ 
\begin{aligned}
u &= \phi_1(y - m_1x) + \phi_2(y - m_2x) \\
&= \left(\frac{m_2}{m_2 - m_1}\right) \alpha_1(y - m_1x) - \left(\frac{1}{m_1 - m_2}\right) \left(\int \alpha_2(s) \, ds\right)_{s=y-m_1x} \\
&\quad + \left(\frac{m_1C}{m_1 - m_2}\right) + \left(\frac{m_1}{m_1 - m_2}\right) \alpha_1(y - m_2x) \\
&\quad + \left(\frac{1}{m_1 - m_2}\right) \left(\int \alpha_2(s) \, ds\right)_{s=y-m_2x} - \left(\frac{m_1C}{m_1 - m_2}\right) \\
&= \left(\frac{m_2}{m_2 - m_1}\right) \alpha_1(y - m_1x) + \left(\frac{m_1}{m_1 - m_2}\right) \alpha_1(y - m_2x) \\
&\quad + \left(\frac{1}{m_1 - m_2}\right) \int_{y-m_1x}^{y-m_2x} \alpha_2(s) \, ds.
\end{aligned} 
\quad (25) $$

For $m_1 = m_2$, we have

$$ u(x, y) = \phi_1(y - m_1x) + x\phi_2(y - m_1x), \quad (26) $$

where $\phi_1$ and $\phi_2$ are arbitrary functions to be determined using the given initial conditions. From the given initial conditions, we have

$$ 
\begin{aligned}
\phi_1(y) &= \alpha_1(y), \quad \text{and} \\
-m_1\phi'_1(y) + \phi_2(y) &= \alpha_2(y) \quad \text{or}
\end{aligned} 
\quad (27) $$

$$ \phi_2(y) = \alpha_2(y) + m_1\phi'_1(y). \quad (28) $$

Substituting (27) and (28) in (26), we get

$$ 
\begin{aligned}
u &= \phi_1(y - m_1x) + x\phi_2(y - m_1x), \\
&= \alpha_1(y - m_1x) + x(\alpha_2(y - m_1x) + m_1\phi'_1(y - m_1x)).
\end{aligned} 
$$

Therefore, the solution, for $m_1 = m_2$, is given by

$$ u = \alpha_1(y - m_1x) + x\alpha_2(y - m_1x) + m_1x\partial_y\alpha_1(y - m_1x) \in \mathcal{F}. \quad (29) $$

Equation (25) and (29) represent the Green's function of the semi-homogeneous IVP of (10) for $m_1 \neq m_2$ and $m_1 = m_2$ respectively.
Now, the Green's function of the given IVP (10) is computed by summing the solution of the semi-inhomogeneous IVP and the solution of the semi-homogeneous IVP given by, for $m_1 \neq m_2$,

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y + m_{1}(t - x) - m_{2}(t - \xi)) \, d\xi \, dt \\ + \left( \frac{m_{2}}{m_{2} - m_{1}} \right) \alpha_{1}(y + m_{1}x) + \frac{m_{1}}{m_{1} - m_{2}} \alpha_{1}(y + m_{2}x) \quad (30) \\ + \frac{1}{m_{1} - m_{2}} \int_{y - m_{2}x}^{y - m_{1}x} \alpha_{2}(s) \, ds, $$

where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}$, $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}$, and $a \neq 0, b, c \in \mathbb{R}$.

For $m_1 = m_2$, then the Green's function of IVP (10) is given by

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y - m_{1}(x - \xi)) \, d\xi \, dt \quad (31) \\ + \alpha_{1}(y - m_{1}x) + x\alpha_{2}(y - m_{1}x) + m_{1}x\partial_{y}\alpha_{1}(y - m_{1}x) \in \mathcal{F}. $$

In other words, if $b^2 - 4ac = 0$, i.e. $m_1 = m_2 = \frac{b}{2a}$, then the Green's function is

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y - \frac{b}{2a}(x - \xi)) \, d\xi \, dt \\ + \alpha_{1}(y - \frac{b}{2a}x) + x\alpha_{2}(y - \frac{b}{2a}x) + \frac{bx}{2a}\partial_{y}\alpha_{1}(y - \frac{b}{2a}x). $$

It is easy to verify that the Green's function for $m_1 \neq m_2$ and $m_1 = m_2$ given in (30) and (31) respectively satisfy $Du = f$ and $\tilde{E}u = \alpha_1(y)$, $\tilde{E}\partial_x u = \alpha_2(y)$. For uniqueness, suppose there are two solution functions, say $u_1$ and $u_2$ of IVP(10). Then the function $w = u_1 - u_2$ satisfies the IVP

$$ \begin{aligned} \mathcal{D}w &= 0, \\ \tilde{\mathcal{E}}w &= 0, \quad \tilde{\mathcal{E}}\partial_x w = 0. \end{aligned} \quad (32) $$

From the Green's function (30), the IVP (32) has the trivial solution $w = 0$. Therefore $u_1 = u_2$. □

In the following example, we find the Green's function of the wave equation with inhomogeneous initial conditions using the algorithm given in Theorem 2.
**Example 2.** Consider the wave equation with homogeneous initial conditions

$$ 
\begin{aligned}
\partial_{tt}u - c^2\partial_{xx}u &= f(x, t), \\
u(x, 0) &= f(x), \partial_t u(x, 0) = g(x).
\end{aligned} 
\qquad (33) $$

The Green's function of the given wave equation (33) is

$$ u(x,t) = \int_0^t \int_0^s f(\xi, x + c(s-t) + c(s-\xi)) \, d\xi \, ds \\ 
+ \frac{1}{2} (f(x+ct) + f(x-ct)) - \frac{1}{2c} \int_{x+ct}^{x-ct} g(s) \, ds. $$

More details about the calculations are given in Section 5.

The following corollaries present the Green's function semi-inhomogeneous and semi-homogeneous IVPs.

**Corollary 1.** Let $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ be an integro-differential algebra. Suppose $\mathcal{D} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ is a linear partial differential operator of second order. Then the semi-inhomogeneous IVP

$$ 
\begin{aligned}
\mathcal{D}u &= f, \\
\tilde{E}u &= 0, \tilde{E}\partial_x u = 0,
\end{aligned} 
$$

has the unique solution given by

$$ u = \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(\xi, y + m_{1}(t - x) - m_{2}(t - \xi)) \, d\xi \, dt, $$

where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}$, $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}$, and $a \neq 0, b, c \in \mathbb{R}$.

**Corollary 2.** Let $(\mathcal{F}, \mathcal{D}, \mathcal{A})$ be an integro-differential algebra. Suppose $\mathcal{D} = a\partial_{xx} + b\partial_{xy} + c\partial_{yy} \in \mathcal{F}[\mathcal{D}, \mathcal{A}]$ is a linear partial differential operator of second order. Then the semi-homogeneous IVP

$$ 
\begin{aligned}
\mathcal{D}u &= 0, \\
\tilde{E}u &= \alpha_1, \tilde{E}\partial_x u = \alpha_2,
\end{aligned} 
$$

has the unique solution given by

$$ u = \left( \frac{m_2}{m_2 - m_1} \right) \alpha_1(y - m_1 x) + \left( \frac{m_1}{m_1 - m_2} \right) \alpha_1(y - m_2 x) \\ 
+ \left( \frac{1}{m_1 - m_2} \right) \int_{y-m_1 x}^{y-m_2 x} \alpha_2(s) \, ds \in \mathcal{F}, \quad m_1 \neq m_2. $$
For $m_1 = m_2$, the unique solution is

$$u = \alpha_1(y - m_1x) + x\alpha_2(y - m_1x) + m_1x\partial_y\alpha_1(y - m_1x) \in \mathcal{F},$$

where $m_1 = \frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a}$, $m_2 = \frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a}$, and $a \neq 0, b, c \in \mathbb{R}$.

# 4 Proposed Algorithm in Maple

In this section, we discuss the Maple package IVPforPDE of the proposed algorithm by creating different data types. We have ParialDiffOperator(a,b,c), where a,b,c are the coefficients as in (1); GreensFunction(PDO), where PDO is the partial differential operator of the given PDE; ExactSolution(PDO, ff, ic1, ic2), where ff is the forcing function f(x, y) of the given IVP and ic1, ic2 are the initial data of the given problem; and finally, we have ApplyPartialDiffOp(PDO,u) to verify the solution u of the given IVP obtained from u:= ExactSolution(PDO, ff, ic1, ic2). The Maple package IVP for PDE is available at http://www.sinivasaraothota.webs.com/research with example worksheet.

## 4.1 Pseudo-Code of Proposed Algorithm

<table><caption>**Algorithm 1. Exact Solution**</caption><tr><td>1.</td><td>PDO ← Partial Differential Operator</td></tr><tr><td>2.</td><td>ff ← Forcing Function f(x, y)</td></tr><tr><td>3.</td><td>α1 ← Initial Data1</td></tr><tr><td>4.</td><td>α2 ← Initial Data2</td></tr><tr><td>5.</td><td>coeff1 ← a; coeff2 ← b; coeff3 ← c</td></tr><tr><td>6.</td><td>m1 ← Root1 of Auxiliary Equation of PDO</td></tr><tr><td>7.</td><td>m2 ← Root2 of Auxiliary Equation of PDO</td></tr><tr><td>8.</td><td>f ← ff(s, y + m1 * (t - x) - m2 * (t - s))</td></tr><tr><td>9.</td><td>if m1 = m2 then</td></tr><tr><td>10.</td><td>a1 ← α1(y - m1 * x)</td></tr><tr><td>11.</td><td>a2 ← α2(y - m1 * x)</td></tr><tr><td>12.</td><td>u ← <sup>1</sup>&frasl;<sub>a</sub> &int;<sub>0</sub><sup>x</sup> &int;<sub>0</sub><sup>t</sup> f ds dt + a1 + x * a2 + m1 * x * <sup>∂</sup>&frasl;<sub>∂y</sub>(a1);</td></tr><tr><td>13.</td><td>else</td></tr><tr><td>14.</td><td>a1 ← α1(y - m1 * x)</td></tr><tr><td>15.</td><td>a12 ← α1(y - m2 * x)</td></tr><tr><td>16.</td><td>a2 ← α2(s)</td></tr><tr><td>17.</td><td>u ← <sup>1</sup>&frasl;<sub>a</sub> &int;<sub>0</sub><sup>x</sup> &int;<sub>0</sub><sup>t</sup> f ds dt + (<sup>m2</sup>&frasl;<sub>m2-m1</sub>) * a1 * (<sup>m1</sup>&frasl;<sub>m1-m2</sub>) * a12 + (<sup>1</sup>&frasl;<sub>m1-m2</sub>) &int;<sub>y-m1*x</sub><sup>y-m2*x</sup> a2ds</td></tr><tr><td>18.</td><td>end if</td></tr></table>
**Algorithm 2.** Greens Function

1. PDO ← Partial Differential Operator

2. coeff1 ← a; coeff2 ← b; coeff3 ← c

3. m1 ← Root1 of PDO

4. m2 ← Root2 of PDO

5. **if** m1 = m2 **then**

$$
6. \quad \left|
\begin{array}{l}
u \leftarrow \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(s, y - m1(x - s)) \, ds \, dt + \alpha1(y - m1 * x) + x * \alpha2(y - m1 * x) + \\
m1 * x + \frac{\partial}{\partial y} \alpha1(y - m1 * x)
\end{array}
\right|
$$

7. **else**

$$
8. \quad \left|
\begin{array}{l}
u \leftarrow \frac{1}{a} \int_{0}^{x} \int_{0}^{t} f(s, y + m1 * (t - x) - m2 * (t - s)) \, ds \, dt + \left(\frac{m2}{m2-m1}\right) * \alpha1(y - \\
m1 * x) + \left(\frac{m1}{m1-m2}\right) * \alpha1(y - m2 * x) + \left(\frac{1}{m1-m2}\right) \int_{y-m1*x}^{y-m2*x} \alpha2(s) \, ds
\end{array}
\right|
$$

9. end if

**Example 3.** *Consider the IVP for wave equation*

$$
\begin{align*}
& \partial_{xx} u - c^2 \partial_{yy} u = f(x, y), \\
& u(0, y) = f(y), \partial_y u(0, y) = g(y).
\end{align*}
$$

If $c = 1$ and $f(x, y) = xy$, $f(y) = y$, $g(y) = y^2 + 1$ for simplicity, then we have

> PDOp:=PartialDiffOperator(1, 0, -1);

$$
\mathrm{PDOp} := \partial(xx) - \partial(yy)
$$

> g:=GreensFunction(PDOp);

$$
g := \int_{0}^{x} \int_{0}^{t} f(s, y - 2t + x + s) \, ds \, dt + \frac{1}{2} a1(y + x) + \frac{1}{2} a1(y - x) - \frac{1}{2} \int_{y+x}^{y-x} a2(s) \, ds
$$

> u:=ExactSolution(PDOp,xy,y,y²+1);

$$
u := -\frac{1}{6}x^4 + \frac{1}{3}\left(\frac{1}{2}y + \frac{1}{2}x\right)x^3 + y + x - \frac{1}{6}(y-x)^3 + \frac{1}{6}(y+x)^3
$$

> ApplyPartialDiffOp(PDOp,u);

$xy, y, y^2 + 1$

> ApplyPartialDiffOp(PDOp,g);

$f(x, y), a1(y), a2(y)$
## 5 Sample Computations

In this section, we provide some examples to illustrate the proposed algorithm presented in Theorem 2.

**Example 4.** Consider an IVP of the following form to demonstrate the proposed algorithm.

$$ \frac{\partial^2 u(x, y)}{\partial x^2} + \frac{\partial^2 u(x, y)}{\partial x \partial y} - 6 \frac{\partial^2 u(x, y)}{\partial y^2} = y \cos x, \quad (34) $$
$$ u(0, y) = y - 1, \quad \frac{\partial u(0, y)}{\partial x} = y^2. $$

In symbolic representation of the given IVP (34) is

$$ \begin{aligned} & Du = f, \\ & \tilde{E}u = \alpha_1(y), \tilde{E}\partial_x u = \alpha_2(y), \end{aligned} \quad (35) $$

where

$textttD = \partial_{xx} + \partial_{xy} - 6\partial_{yy}$; $f(x, y) = y \cos x$; $\tilde{E}u = u(0, y)$, $\tilde{E}\partial_x u = \left(\frac{\partial u}{\partial x}\right)_{x=0}$ and $\alpha_1(y) = y - 1, \alpha_2(y) = y^2$.

Factorizing the given differential operator, we get

$$ \begin{aligned} u &= \frac{1}{D} f = \frac{1}{(\partial_x - 2\partial_y)(\partial_x + 3\partial_y)} f \\ &= \left( \frac{1}{(\partial_x - 2\partial_y)} \right) \left( \frac{1}{(\partial_x + 3\partial_y)} f \right). \end{aligned} $$

We have $m_1 = -2$ and $m_2 = 3$. Now, the Green's function of the given IVP (35) is computed as in Theorem 2 given by

$$ \begin{aligned} u(x, y) = & \int_0^x \int_0^t f(\xi, y - 2(t - x) + 3(\xi - t)) \, d\xi \, dt \\ & + \frac{3}{5}\alpha_1(y + 2x) + \frac{2}{5}\alpha_1(y - 3x) - \frac{1}{5}\int_{y+2x}^{y-3x} \alpha_2(s) \, ds. \end{aligned} \quad (36) $$

In [15], authors solved the equation (34) without initial conditions for particular forcing function $f(x, y) = y \cos x$. However, we have found the generic
formulation for the solution of the IVP (34) as given in equation (36). Therefore, if $f(x, y) = y \cos x$, and $\alpha_1(y) = y - 1$, $\alpha_2(y) = y^2$, then the exact solution of the given IVP (34) is obtained from the Green's function (36) as follows

$$ 
\begin{aligned}
u(x, y) &= \int_0^x \int_0^t (y - 2(t - x) + 3(\xi - t)) \cos \xi \, d\xi \, dt \\
&\quad + \frac{3}{5}(y + 2x - 1) + \frac{2}{5}(y - 3x - 1) - \frac{1}{5} \int_{y+2x}^{y-3x} s^2 \, ds \\
&= (y - x - y \cos x + \sin x) + \left( y - 1 + xy^2 - yx^2 + \frac{7}{3}x^3 \right) \\
&= x(y^2 - 1) + y(2 - x^2 - \cos x) + \sin x + \frac{7}{3}x^3 - 1.
\end{aligned}
$$

One can verify that $Du = y \cos x$ and $\tilde{E}u = y - 1$, $\tilde{E}\partial_x u = y^2$, i.e.,

$$ \frac{\partial^2 u(x, y)}{\partial x^2} + \frac{\partial^2 u(x, y)}{\partial x \partial y} - 6 \frac{\partial^2 u(x, y)}{\partial y^2} = y \cos x \quad \text{and} \\ u(0, y) = y - 1, \frac{\partial u(0, y)}{\partial x} = y^2. $$

**Example 5.** Consider an IVP

$$ \frac{\partial^2 u(x, y)}{\partial x^2} + \frac{\partial^2 u(x, y)}{\partial x \partial y} - 2 \frac{\partial^2 u(x, y)}{\partial y^2} = (y - 1)e^x, \qquad (37) \\ u(0, y) = \frac{1}{y}, \frac{\partial u(0, y)}{\partial x} = y^2 + 3y. $$

In symbolic representation, the given IVP (37) is

$$ \begin{aligned} Du &= f, \\ \tilde{E}u &= \alpha_1(y), \tilde{E}\partial_x u = \alpha_2(y), \end{aligned} \qquad (38) $$

where $D = \partial_{xx} + \partial_{xy} - 2\partial_{yy}$; $f(x, y) = (y-1)e^x$; $\tilde{E}u = u(0, y)$, $\tilde{E}\partial_x u = (\frac{\partial u}{\partial x})_{x=0}$; and $\alpha_1(y) = \frac{1}{y}$, $\alpha_2(y) = y^2 + 3y$.

Following the proposed algorithm in Section 3 similar to Example 4, we get

$$ u = \left( \frac{1}{(\partial_x - \partial_y)} \right) \left( \frac{1}{(\partial_x + 2\partial_y)} f \right). $$
We have $m_1 = -1$ and $m_2 = 2$, and the Green's function of the given IVP (38) is

$$ u(x, y) = \int_{0}^{x} \int_{0}^{t} f(\xi, y - (t-x) + 2(\xi-t)) \, d\xi \, dt \\ + \frac{2}{3}\alpha_{1}(y+x) + \frac{1}{3}\alpha_{1}(y-2x) - \frac{1}{3}\int_{y+x}^{y-2x} \alpha_{2}(s) \, ds. \qquad (39) $$

If $f(x, y) = (y-1)e^x$, and $\alpha_1(y) = \frac{1}{y}$, $\alpha_2(y) = y^2+3y$, then the exact solution of (37) from the Green's function (39) is

$$ \begin{aligned} u(x, y) &= \int_{0}^{x} \int_{0}^{t} (y - (t - x) + 2(\xi - t) - 1)e^{\xi} \, d\xi \, dt \\ & \quad + \frac{2}{3} \left( \frac{1}{y+x} \right) + \frac{1}{3} \left( \frac{1}{y-2x} \right) - \frac{1}{3} \int_{y+x}^{y-2x} (s^2 + 3s) \, ds \\ &= (1 + x - e^x)(2 - y) + \frac{1}{2}x^2 + \frac{2}{3} \left( \frac{1}{y+x} \right) + \frac{1}{3} \left( \frac{1}{y-2x} \right) \\ & \quad - \frac{1}{3} \left( \frac{1}{3}(y-2x)^3 + \frac{3}{2}(y-2x) - \frac{1}{3}(y+x)^3 - \frac{3}{2}(y+x)^2 \right) \\ &= 2 - y + 2x - yx + \frac{1}{2}x^2 + e^x y - 2e^x \\ & \quad + \frac{1}{2(y+x)(2x-y)}(-2y + 2x - 2xy^4 + 4y^3x^2 - 2yx^4 + 4x^5 - 6xy^3 + 9y^2x^2 + 9yx^3 - 6x^4) \end{aligned} $$

One can verify that $Du = (y-1)e^x$ and $\tilde{E}u = \frac{1}{y}$, $\tilde{E}\partial_x u = y^2 + 3y$.

**Example 6.** Consider an IVP for PDE of the following form

$$ \begin{aligned} & 4\frac{\partial^2 u(x, y)}{\partial x^2} - 4\frac{\partial^2 u(x, y)}{\partial x \partial y} + \frac{\partial^2 u(x, y)}{\partial y^2} = 16 \log(x + 2y), \\ & u(0, y) = 0, \frac{\partial u(0, y)}{\partial x} = 0. \end{aligned} \qquad (40) $$

Operator notation of the given IVP (40) is

$$ \begin{aligned} Du &= f, \\ \tilde{E}u &= 0, \tilde{E}\partial_x u = 0, \end{aligned} \qquad (41) $$
where $D = 4\partial_{xx} - 4\partial_{xy} + \partial_{yy}$, $f(x, y) = 16\log(x+2y)$ and $\tilde{E}u = u(0, y)$, $\tilde{E}\partial_x u = (\frac{\partial u}{\partial x})_{x=0}$. Following the algorithm in Theorem 2, we have the exact solution of IVP (41) as

$$u(x, y) = 2x^2 \log(x + 2y).$$

It is easy to verify that $Du = 16\log(x + 2y)$ and $\tilde{E}u = 0$, $\tilde{E}\partial_x u = 0$.

The Maple implementation is as follows:

> PDO:=PartialDiffOperator(4,-4,1);

$$PDO := 4\partial(xx) - 4\partial(xy) + \partial(yy)$$

> g:=GreensFunction(PDO);

$$g := \frac{1}{4} \int_{0}^{x} \int_{0}^{t} f(s, y + \frac{1}{2}x - \frac{1}{2}s) \, ds \, dt \\ + a1(y + \frac{1}{2}x) + xa2(y + \frac{1}{2}x) - \frac{1}{2}xD(a1)(y + \frac{1}{2}x)$$

> u:=ExactSolution(PDO,16\*log(x+2\*y),0,0);

$$u := 2\ln(x + 2y)x^2$$

> ApplyPartialDiffOp(PDO,u);

$$16\ln(x + 2y), 0, 0$$

> ApplyPartialDiffOp(PDOp,g);

$$f(x, y), a1(y), a2(y)$$

# 6 Conclusion

We have discussed a new symbolic algorithm to solve a given IVP for second order linear PDEs for Green's function, which can also be extended to n-th order PDEs. Several numerical examples are presented to illustrate the proposed method. Implemented Maple package IVPforPDE is provided at http://www.sinivasaraothota.webs.com/research with example worksheet.
References

[1] M. Rosenkranz and G. Regensburger. Solving and factoring boundary problems for linear ordinary differential equations in differential algebras. *Journal of Symbolic Computation*, 43(8):515-544, 2008. doi: 10.1016/j.jsc.2007.11.007.

[2] M. Rosenkranz, G. Regensburger, L. Tec, and B. Buchberger. Symbolic analysis for boundary problems: From rewriting to parametrized gröbner bases. In Peter Paule Ulrich Langer, editor, *Numerical and Symbolic Scientific Computing: Progress and Prospects, Texts & Monographs in Symbolic Computation*, pages 273-331. Springer Science & Business Media, Vienna, 2011. ISBN 9783709107942, http://gregensburger.com/papers/sfbbook.pdf.

[3] S. Thota and S.D. Kumar. Symbolic algorithm for a system of differential-algebraic equations. *Kyungpook Mathematical Journal*, 56(4):1141-1160, 2016. doi: 10.5666/KMJ.2016.56.4.1141.

[4] S. Thota and S.D. Kumar. Solving system of higher-order linear differential equations on the level of operators. *International Journal of Pure and Applied Mathematics*, 106(4):11-21, 2016. doi: 10.12732/ij-pam.v106i4.1.

[5] S. Thota and S.D. Kumar. Symbolic method for polynomial interpolation with stieltjes conditions. In *International Conference on Frontiers in Mathematics*, 1:225-228, 2015. https://wwwsemanticscholar.org/paper/Symbolic-method-for-polynomial-interpolation-with-Thota-Kumar/23e0679d8284268d6e6f90a9345eaae353b3cdb0.

[6] S. Thota. On a symbolic method for error estimation of a mixed interpolation. *Kyungpook Mathematical Journal*, 58(3):453-462, 2018. doi: 10.5666/KMJ.2019.59.1.13.

[7] S. Thota. On a new symbolic method for initial value problems for systems of higher-order linear differential equations. *International Journal of Mathematical Models and Methods in Applied Sciences*, 12:194-202, 2018. https://www.naun.org/main/NAUN/mcs/2019/a502002-abh.pdf.
[8] S. Thota. A symbolic algorithm for polynomial interpolation with integral conditions. *Applied Mathematics & Information Sciences*, 12(5):995-1000, 2018. doi: 10.18576/amis/120512.

[9] S. Thota. Initial value problems for system of differential-algebraic equations in maple. *BMC Research Notes*, 11, 2018. doi: 10.1186/s13104-018-3748-0.

[10] S. Thota. On a symbolic method for fully inhomogeneous boundary value problems. *Kyungpook Mathematical Journal*, 59(1):13-22, 2019. doi: 10.5666/KMJ.2019.59.1.13.

[11] S. Thota. A symbolic algorithm for polynomial interpolation with stieltjes conditions in maple. *Proceedings of the Institute of Applied mathematics*, 8(2):112-120, 2019. https://www.academia.edu/41892366/A_Symbolic_Algorithm_for_Photometric Interpolation with Stieltjes Conditions in Maple.

[12] S. Thota. On a new symbolic method for solving two-point boundary value problems with variable coefficients. *International Journal of Mathematics and Computers in Simulation*, 13:160-164, 2019. https://www.naun.org/main/NAUN/mcs/2019/a502002-abh.pdf.

[13] S. Thota and P. Shanmugasundaram. On a symbolic method for neutral functional-differential equations with proportional delays. *Coherent Mathematics & Statistics*, 7(1), 2020. Article 1813961, doi: 10.1080/25742558.2020.1813961.

[14] Q.H. Qin. Green's functions of magneto-electro-elastic plate under thermal loading. In Richard B. Hetnarski, editor, *Encyclopedia of Thermal Stresses*, volume 1, pages 2096-2103. Springer Netherlands, Netherlands, 2014. ISBN 978-94-007-2738-0, doi: 10.1007/978-94-007-2739-7.

[15] R.K. Jain and S.R.K. Iyengar. *Advanced Engineering Mathematics*. Narosa Publishing House, Third edition, India, 2008. ISBN 9781842651858.

[16] V. Grigoryan. *Partial differential equations*. Department of Mathematics, University of California, Saanta Barbara, USA, 2010. https://web.math.ucsb.edu/grigoryan/124A.pdf.
[17] S. Pal and S.C. Bhunia. *Engineering Mathematics*. Oxford University Press, 2015. ISBN 9780198070894.

[18] Y. Pinchover and J. Rubinstein. *An Introduction to Partial Differential Equations*. Cambridge University Press, 2015. ISBN 978-0521613231.