## Fourier Transform Methods for Partial Differential Equations

Department of Mathematics, College of Natural and Computational Science, Wollega University, P. box. 395### Abstract

The purpose of this seminar paper is to introduce the Fourier transform methods for partial differential equations. The introduction contains all the possible efforts to facilitate the understanding of Fourier transform methods for which a qualitative theory is available and also some illustrative examples was given. The resulting Fourier transform maps a function defined on physical space to a function defined on the space of frequencies, whose values quantify the “amount” of each periodic frequency contained in the original function then inverse Fourier transform reconstructs the original function from its transform.

**Keywords:** fourier transform, partial differential equations

*International Journal of Partial Differential Equations and Applications*, 2014 2 (3),
pp 44-57.

DOI: 10.12691/ijpdea-2-3-2

Received May 16, 2014; Revised June 04, 2014; Accepted June 19, 2014

**Copyright**© 2014 Science and Education Publishing. All Rights Reserved.

### Cite this article:

- Negero, Naol Tufa. "Fourier Transform Methods for Partial Differential Equations."
*International Journal of Partial Differential Equations and Applications*2.3 (2014): 44-57.

- Negero, N. T. (2014). Fourier Transform Methods for Partial Differential Equations.
*International Journal of Partial Differential Equations and Applications*,*2*(3), 44-57.

- Negero, Naol Tufa. "Fourier Transform Methods for Partial Differential Equations."
*International Journal of Partial Differential Equations and Applications*2, no. 3 (2014): 44-57.

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### 1. Introduction

The Fourier transform is the natural extension of Fourier series to a function f(x) of infinite period ^{[4]}. This paper develops one of the fundamental topics in analysis and in PDEs, namely orthogonal expansions.

**Definition: **The set of functions {Y_{n}(x):n=0,1,…} each of which is piecewise continuous in an infinite or a finite interval [α,β]*, *is said to be orthogonal* *in [α,β] with respect to the weight function *r*(*x*)>0, if for all m≠n and for all n.

We shall always assume that r(x) has only a finite number of zeros in [α,β] and the integrals exist.

**Definition:** The norm of a function Y_{n}(x) is denoted by ||Y_{n}|| defined as the inner product of a function with itself and written

A real - valued function Y_{n}(x) is called square - integrable on the interval in [α,β] with respect to the weight function r(x) when

The orthogonal set {Y_{n}(x):n=0,1,…} in [α,β] with respect to the weight function *r*(*x*) is said to be orthonormal set if for all *n* ^{[8]}*.*

If {Y_{n}(x)} is an orthonormal set of functions then

(1.1) |

Where is the Kronecker delta ^{[6]}.

Thus, orthonormal functions have the same properties as orthogonal functions, but, in addition, they have been normalized ^{[5]}, i.e., each function Y_{n}(x) of the orthogonal set has been divided by the norm of that function, which is defined as . Hence if an orthogonal set of functions {Y_{n}(x)} ≠0 is defined on the interval [*α,β*], with ||Y_{n}||we can always construct an orthonormal set of functions X_{n}(x) by defining ^{[1]}.

In fact in view of (1.1),

and hence ||X_{n}||=1 for all n.

**Example:** The set of functions is a set of orthogonal functions over the interval 0<X<L With respect to the weight function *r*(*x*)=1 This is shown by calculating the inner products

and The norm squared for each function is given by

For m=1,2,… These results are written in inner product notation as

The seminar paper discusses a periodic function which can be expanded in terms of an infinite sum of sines and cosines in which most functions encountered in engineering are periodic functions.

**Definition****:** Fourier trigonometric series of a function of f(x) defined on is defined on -L≤X≤L*,** *is the infinite trigonometric series

Whose coefficients are given by the inner product formulae

However, if the function f(x) is odd, then since

the Fourier trigonometric series reduces to the Fourier sine series:

Where

Thus, we conclude that if f(x) is odd, or defined only on(0*, π*) and we make its odd extension then the Fourier sine series can be obtained* *Exactly, in the same way if f(x) is even, or defined only on (0*, π*) and we make its even extension then since

the Fourier trigonometric series reduces to the Fourier cosine series:

(1.2) |

Where

**Example**:** **We shall find the Fourier cosine series of the function .

Clearly,

Thus, from (1.2), we have

**Theorem: **(**Fourier’s theorem)** [see ^{[7]} ]Let f(x) and f’(x) be piecewise continuous in the interval [-L,L]. Then, the Fourier trigonometric series of f(x) converges to at each point in the open interval (-L,L) and at x=±L the series converges to .

**Example**: Consider the function

Clearly, and has a single jump discontinuity at 0. For this function, the Fourier trigonometric coefficients are a_{0}=1*,* a_{n}=0*,** *. Thus, we have

(1.3) |

From Fourier’s theorem in (1,3) the equality F(x)= f(x) holds at each point in the open intervals (-*π**, 0*) and (0,* **π*) where as at x=0 the right–hand side is 1/2 which is the same as Also, at x=±*π** *the right–hand side is again 1/2 which is the same as .

The Fourier integral is a natural extension of Fourier trigonometric series in the sense that it represents a piecewise smooth function whose domain is semi-infinite or infinite ^{[1]}

A periodic function f(x) defined in a finite interval (-L,L) can be expressed in Fourier series by extending this concept, non periodic functions defined in -∞<x<∞ (for all x) can be expressed as a Fourier integral.

Let f_{p}(x) be a periodic function of period 2p that can be represented by a Fourier series

Where

The problem we shall consider is what happens to the above series when L →∞ for this we insert a_{n} and b_{n}, to obtain

We now set

Then , and we may write the Fourier series in the form

(1.4) |

This representation is valid for any fixed p*, *arbitrarily large, but fixed.

We now let L→∞ and assume that the resulting nonperiodic function is is absolutely integrable on the x-axis, i.e., Then, *, *and the value of the first term on the right side of (1.4) approaches zero. Also, and the infinite series in (1.4) becomes an integral from 0 to ∞*, *which represent f(x), i.e,

(1.5) |

Now if we introduce the notations

(1.6) |

Then (1.5) can be written as

(1.7) |

This representation of f(x) is called** **Fourier integral.

**Theorem: ****(****Fourier Integral Theorem)****: **Let f(x)*,* -∞<x<∞ be piecewise continuous on each finite interval, and (-∞,∞) i.e., f is absolutely integrable on(-∞,∞)*. *Then, f(x) can be represented by a Fourier integral (1.7).

Further, at each x*.*

**Example: **Find the Fourier integral representation of the single pulse function

From (1.6) we have

Thus, (1.7) gives the representation

(1.8) |

Now from this Theorem it is clear that

(1.9) |

This integral is called Dirichlet’s discontinuity factor.

Setting x=0 in (1.9) yields the important integral

(1.10) |

known as the Dirichlet integral.

**Theorem**** ****1: **(**Fourier Cosine Integral Theorem**): If f(x) satisfies the Dirichlet’s conditions on the non negative real line and is absolutely integralble on (0, ∞), then , where

**Theorem**** ****2: **(**Fourier Sine Integral Theorem): **If f(x) satisfies the Dirichlet’s conditions on the non negative real line and is absolutely integralble on (0, ∞), then ; where .

Indeed, if f(x) is an even function, then B(ω)=0 in (3) and and the Fourier integral (1.7) reduces to the Fourier cosine integral,

(1.11) |

Similarly, if f(x) is odd, then in (1.6).we have *A*(*ω*) = 0 and . and the Fourier integral (1.7) reduces to the Fourier sine integral

(1.12) |

**Example: **Express as a Fourier sine integral and hence evaluate

The Fourier sine integral for

At which point of discontinuity of f(x),the value of the above integral

We note that (1.5) is the same as

The integral in bracket is an even function of *ω**, *we denote it by .

Since is an even function of *ω*, the function f does not depend on *ω*, and we integrate with respect to t (not *ω*), the integral of from *ω**=0 *to ∞ is 1/2 times the integral of from -∞ to ∞. Thus,

(1.13) |

From the above argument it is clear that

(1.14) |

A combination of (1.13) and (1.14) gives

(1.15) |

This is called the complex Fourier integral.

From the above representation of f(x), we have

(1.16) |

**Definition****:** In the Fourier integral of f(x) in the complex form given by

the expression in bracket is a function of *ω*, is denoted by F(*ω*) or F(f) and is called Fourier transform of f. Now writing x for t we get

(1.17) |

And with this (1.16) becomes

(1.18) |

The representation (1.18) is called the inverse Fourier transform of F(*ω*). Finally, as in Theorem-1, if f(x)*, *-∞<x<∞ is piecewise continuous on each finite interval, and Then, the Fourier transform (1.17) of f(x) exists. Further, at each x,

**Example****: **Find the Fourier transform of the square wave function

From (1.17), we have

Further it follows that

The basic technique for solving partial differential equations (PDE) on a bounded spatial domain is the Fourier method ^{[1]}.

The seminar paper deals with the problem of the Fourier transform methods for partial differential equations considering first problems in infinite domains which can be effectively solved by finding the Fourier transform or the Fourier sine or cosine transform of the unknown function. However, for such problems usually the method of separation of variables does not work because the Fourier series are not adequate to yield complete solutions. This is due to the fact that often these problems require a continuous superposition of separated solutions

In this seminar paper we begin by motivating the construction by investigating how Fourier series behave as the length of the interval goes to infinity. Therefore, this paper develops the theory of the Fourier transform methods for partial differential equations in which a qualitative theory exists.

The main objective of this paper is to discuss Fourier transform methods for partial differential equations (PDEs) which often help full in approximation to the true situation and that a more realistic model would include some of the periodic functions can be written in terms of an infinite sum of sines and cosine series by using Fourier transforms which were complicated when we are using Fourier series.

### 2. Transforms of Partial Derivatives

**Definition****:** In the Fourier integral of f(x) in the complex form given by

the expression in bracket is a function of *ω*, is denoted by F(*ω*) or F(f) and is called Fourier transform of f. Now writing x for t we get

(2.1) |

And with this (1.18) becomes

(2.2) |

The representation (2.2) is called the inverse Fourier transform of F(*ω*). Finally, as in Theorem-1, if f(x)*,* -∞<x<∞ is piecewise continuous on each finite interval, and Then, the Fourier transform (2.1) of f(x) exists. Further, at each x,

**Example****: **Find the Fourier transform of the square wave function

From (2.1), we have

Further it follows that

**Theorem (Convolution Theorem**): Suppose that f(x) and g(x) are piecewise continuous, bounded, and absolutely integrable functions on the x-axis. Then

(2.3) |

Where f*g is the convolution of functions f and g defined as

**Proo****f:**** **By the definition and an interchange of the order of integration, we have

Now we make the substitution x-T=v, so that x=T+*ν* and

By taking the inverse Fourier transform on both sides of (2.3) and writing and , and noting that and cancel each other, we obtain

(2.4) |

The solution of a IBVP consisting of a partial differential equation together with boundary and initial conditions can be solved by the Fourier transform method. In one dimensional boundary value problems, the partial differential equation can easily be transformed into an ordinary differential equation by applying a suitable transform. The required solution is then obtained by solving this equation and inverting by means of the complex inversion formula or by any other method. In two dimensional problems, it is sometimes required to apply the transforms twice and the desired solution is obtained by double inversion.

Suppose that u(x,t) is a function of two variables x and t, where -∞<x<∞ and t>0 Because of the presence of two variables, care is needed in identifying the variable with respect to which the Fourier transform is computed. For example, for fixed t, the function u(x,t) becomes a function of the spatial variable x, and as such, we can take its Fourier transform with respect to the x variable. We denote this transform by Thus,

This transform is called Fourier transform in the x variable ^{[2]}. To illustrate the use of this notation we compute some very useful transforms.

**2.1. Fourier Transform and Partial Derivatives**

Given u(x,t) with -∞<x<∞ and t>0, we have

To prove (i) we start with the right side and differentiate under the integral sign with respect to t:

The last expression is the Fourier transform of as a function of x, and (i) follows. Repeated differentiation under the integral sign with respect to t yields (ii).

**2.2. Fourier Sine and Cosine Transform**

For an even function the Fourier cosine integral (1.15) where *A*(*ω*) is given by (1.16). We set , where c indicates cosine. Then replacing t by x

We get

(2.5) |

and

(2.6) |

Formula (2.5) gives from f(x) a new function called the Fourier cosine transform of f(x) whereas (2.6) gives back f(x) from , and we call it the inverse Fourier cosine transform of *. *Relations (2.5) and (2.6) together form a Fourier cosine transform pair.

Similarly, for an odd function f(x) the Fourier sine transform is and the inverse Fourier sine transform is

(2.7) |

and the inverse Fourier sine transform is

(2.8) |

**Example**** ****1:** Find the Fourier cosine transform of

Solution

**Example 2:** Consider the function

Express f using an inverse Fourier cosine and then an inverse sine transform.

**Solution****:** We first start by computing the Fourier cosine transform. From (2.8),

Using (2.6), we obtain the inverse cosine transform representation

We compute the sine transform similarly by using (2.6).

and thus the inverse sine transform representation

**2.2.1. Cosine and Sine Transforms of Derivatives of Functions.**

If f(x) is absolutely integrable on the positive x-axis and piecewise continuous on every finite interval, then the Fourier cosine and sine transforms of f exist. Furthermore, it is clear that F_{c} and F_{s} are linear operators, i.e.,

**Theorem: **Let f(x) be continuous and absolutely integrable on the x-axis, let f’(x) be piecewise continuous on each finite interval, and let f(x)→0 as x→∞. Then,

(2.9) |

(2.10) |

**Proof:**** **To show (i), we integrate by parts, to obtain

Also we integrate by parts, to obtain

Similarly, (a)

By formula (ii) with f’ instead of f gives

(b)

By formula (2.10) with f’ instead of f gives

hence by (2.9)

We have by the Fourier transform,

(2.11) |

By similar procedure we can find a relation between the sine and cosine Fourier transforms of the derivatives of a function, such as

(2.12) |

Under the assumptions, as and as x→0

Similarly, integrating,

(2.13) |

Equations, (2.12) and (2.13) yield, repeating the procedure may be expressed as the sum of a’s and either or or . will occur when x is odd and in that case we can write in place of .

We thus have

And

Similar procedure with help of (2.12) and (2.13) will yield

and

Similarly the following results are easily deducible,

(i)

When =0,

(ii)

(iii)

when x=0,

(iv)

**Example:** We found that

Applying (ii) with , we obtain

Hence

**2.2.2. Convolution theorems for Fourier sine and cosine transform**

**Theorem**:** **Let and be the Fourier cosine transform of f(x) and g(x), respectively, and let and be the Fourier sine transform of f(x) and g(x) respectively

Then

We have

**Example:**** (Convolution w****ith Cosine)**

Suppose that f is integrable and even f(-x)= f(x) for all x and let Show that, for all real numbers a; .

Solution From the definition and the fact that ; f*g=g*f, we have

Since f is even, the product is odd, hence and so

**2.3. The Fourier Transform method.**

We summarize the Fourier transform method as follows:

**Step 1:** Fourier transform the given boundary value problem in u(x,t) and get ordinary differential equation in in the variable t.

**Step 2:** solve the ordinary differential equation and find .

**Step 3:** inverse Fourier transform to get u(x,t).

This method is successful in treating a variety of partial differential equations, but it has its limitations, since we have to assume that the functions in the problem and its solution have Fourier transforms. Nevertheless, the method offers us opportunities beyond these limitations, as we now illustrate.

The method of solution is best explained through the following example.

**Example 1:** We will show how the Fourier transform applies to the heat equation. We consider the heat flow problem of an infinitely long thin bar insulated on its lateral surface, which is modeled by the following initial-value problem

(2.14) |

where the function f is piecewise smooth and absolutely integrable in (-∞,∞).

Let be the Fourier transform of u(x,t). Thus, from the Fourier transform pair, we have

Assuming that the derivatives can be taken under the integral, we get

In order for u(x,t) to satisfy the heat equation, we must have

Thus, must be a solution of the ordinary differential equation

The initial condition is determined by

Therefore, we have

and hence

(2.15) |

Now since

if in (2.4) we denote and then from (2.5) it follows that

(2.16) |

This formula is due to Gauss.

This formula is due to Gauss and Weierstrass.

For each μ the function is a solution of the heat equation and is called the fundamental solution. Thus, (2.16) gives a representation of the solution as a continuous superposition of the fundamental solution.

The standard normal distribution function* *Ф is defined as

This is a continuous increasing function with If a<b, then we can write

(2.17) |

From (2.16) and (2.17) it is clear that the solution of the problem

Can be written as

Now using the properties we can verify that

**Example 2:** Consider the problem

(2.18) |

u and u_{x} finite as

which appears in heat flow in a semi–infinite region. In (2.18) the function f is piecewise smooth and absolutely integrable in [0, ∞)

We define the odd function

Then from (2.16) we have

In the first integral we change μ to -μ and use the oddness of *, *to obtain

Thus, the solution of the problem (2.18) can be written as

The above procedure to find the solution of (1.18) is called the method of images.

In an analogous way it can be shown that the solution of the problem

(2.19) |

u and u_{x} finite as

can be written as

Here, of course, we need to extend f(x) to an even function

In (2.19) the physical significance of the condition is that there is a perfect insulation, i.e., there is no heat flux across the surface.

**Example 3:**** **Consider the initial-value problem for the wave equation

u and u_{x} finite as

(2.20) |

where the functions f_{1} and f_{2} are piecewise smooth and absolutely integrable in (-∞,∞).

To find the solution of this problem, we introduce the Fourier transforms

and its inversion formulas

We also need the Fourier representation of the solution u(x,t),

Where is an unknown function, which we will now determine. For this, we substitute this into the differential equation (2.20), to obtain

Thus, must be a solution of the ordinary differential equation

whose solution can be written as

To find and *, *we note that

and hence and .

Therefore, it follows that

and hence the Fourier representation of the solution is

Now since we have

Similarly,

Putting these together yields d’ Alembert’s formula

**Example 4:** Consider the following problem involving the Laplace equation in a half-plane:

where the function f is piecewise smooth and absolutely integrable in (-∞,∞)

If , then we also have the implied boundary conditions

For this, we let

and

We find that

Thus, must satisfy the ordinary differential equation

and the initial condition for each ω.

The general solution of the ordinary differential equation is *. *If we impose the initial condition and the boundedness condition, the solution becomes

Thus, the desired Fourier representation of the solution is

To obtain an explicit representation, we insert the formula for F(ω) and formally interchange the order of integration, to obtain

Now the inner integral is

Therefore, the solution u(x,y) can be explicitly written as

(2.21) |

This representation is known as Poisson’s integral formula.

In particular, for

Thus, (2.21) become

Using the substitution we have *, *so that

**2.4. The Fourier Sine and Cosine Transform Methods**

We will motivate the introduction of Fourier sine and cosine transforms by considering a simple physical problem.** **In order to use the Fourier sine and cosine transform to solve a partial differential equation:

If the boundary conditions are of the Dirichlet type: where the function value is prescribed on the boundary, then the Fourier sine transform is used.

If the boundary conditions are of the Neumann type: where the derivatives of function is prescribed on boundary, then Fourier cosine transform is applied.

In either case, the PDE reduces to an ODE in Fourier transform which is solved. Then the inverse Fourier sine (or cosine) transforms will give the solution to the problem.

**2.4.1. Infinite Fourier cosine and sine Transform Method**

To solve a partial differential equation (containing a second derivative) defined on a semi-infinite interval x≥0, using Fourier cosine transform, must be known. In case, is given then we employ cosine transform to remove .

**Definition:** The infinite Fourier cosine transform of a function f(x) for 0<x<∞,is defined as n being a positive integer.

Here f(x) is called as the inverse Fourier cosine transform of F_{c}(n) and is defined as

Similarly, the Fourier sine transform may be used for semi-infinite problems if f(0) is given.

Furthermore, problems are more readily solved if the boundary conditions are homogeneous. Thus, if f(0)=0, separation of variable motivates the use of sines only. Similarly, implies the use of cosine.

**Definition:** The infinite Fourier sine transform of a function f(x) of x such that 0<x<∞ is denoted by F_{s}(n), n being a positive integer and is defined as

Here f(x) is called as the inverse Fourier sines transform of F_{s}(n) and defined as

**Example: **We shall employ the Fourier sine transform to find the solution of the following problem involving the Laplace equation in a semi-infinite strip:

where the function f is piecewise smooth and absolutely integrable in [0, ∞). We shall also need the boundary conditions and

For this, we let

and

We find that

Thus, * *must satisfy the ordinary differential equation

and hence

Now the boundary condition yields

Thus, we have

Now since , we find , and therefore

This gives the solution

**2.4.2. Finite Fourier cosine and sine Transform Method**

When the domain of the physical problem is finite, it is generally not convenient to use the transforms with an infinite range of integration. In many cases, finite Fourier transform can be used with advantage.

**Definition: **The finite Fourier sine transforms* *of is defined as

Where n≥0 is an integer. The function f(x) is then called the inverse finite Fourier sine transform* *of F_{s}(n)* *and is given by

(2.22) |

**Definition: **The finite Fourier cosine transforms* *of is defined as

Where n≥0 is an integer. The function f(x) is then called the inverse finite Fourier cosine transform* *of F_{c}(n) and is given by

Finite Fourier transforms are useful in solving partial differential equations. For this, we note that

And hence

and similarly,

(2.23) |

**Example:**** **Find the solution of the problem

Taking the finite Fourier sine transform with L* *= 4 of both sides of the partial differential equation gives

Writing for F_{s}(n) and using (2.23) with leads to

which can be solved to obtain .

Now taking the finite Fourier sine transform of the condition u(x=0)=2x* *we have

Since it follow that

Thus, from (2.22) we get

### 3. Conclusion

However, Physical problems never In this work, when modeling problems over regions that extended very far in at least one direction, we often idealized the situation to that of a problem having infinite extent in one or more directions, where any boundary conditions that would have applied on the far-away boundaries are discarded in favor of simple boundedness conditions on the solution as the appropriate variable is sent to infinity. Such problems were mathematically modeled by differential equations defined on infinite regions. For one-dimensional problems we distinguish two types of infinite regions: infinite intervals extending from -∞ to ∞ and semi-infinite intervals extending from one point (usually the origin) to infinite (usually +∞) are infinite, but by introducing a mathematical model with infinite extent, we are able to determine behavior of problems in the situations in which the influence of actual boundaries are expected to be negligible. Thus the seminar paper developed the Fourier transform method and applied it to solve: heat flow problem of an infinitely long thin bar insulated on its lateral surface, heat flow in a semi–infinite region, wave equation, Laplace equation in a half-plane and in a semi-infinite strip, and some partial differential equation on the entire real line. Even though a survey of this seminar paper shows that what is actually studied Fourier transform method to PDE is that, we taken the Fourier transform of PDE and its initial and boundary conditions to reduce it into an ODE. We then solved this ODE for the transformed function. We inverted this function to determine the solution to our PDE. This is not just a method that is specific to the Fourier transform because this method also works for the Laplace transform and in general for many integral transforms. The integrals defining the Fourier transform and its inverse are remarkably alike, and this symmetry was often exploited, for example when assembling appendix given for Fourier transforms. One condition on this is that the variable you taken to the integral transform its domain must match the range of integration of the integral transform. The type of boundary and initial conditions that are given should also played a role in which transform should be used. In case, the Fourier transform is used to analyze boundary value problems on the entire line.

The extension of Fourier methods to the entire real line leads naturally to the Fourier transform, an extremely powerful mathematical tool for the analysis of non-periodic functions.

It is reasonable to expect a Fourier transform method apply to solve different forms of partial differential equations such as

Telegraph equation: for the case

The Fourier transform is of fundamental importance in a broad range of applications, including both ordinary and partial differential equations, quantum mechanics, signal and image processing, control theory, and probability, to name but a few.

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