INTRODUCTION

Partial differential equations (PDEs), like ordinary differential equations (ODEs), are classified as linear or nonlinear. Analogous to a linear ODE (see (6) of Section 1.1), the dependent variable and its partial derivatives appear only to the first power in a linear PDE. In this and the chapters that follow, we are concerned only with linear partial differential equations.

Linear Partial Differential Equation

If we let u denote the dependent variable and x and y the independent variables, then the general form of a linear second-order partial differential equation is given by

(1)

where the coefficients A, B, C, … , G are constants or functions of x and y. When G(x, y) = 0, equation (1) is said to be homogeneous; otherwise, it is nonhomogeneous.

EXAMPLE 1 Linear Second-Order PDEs

The equations

are examples of linear second-order PDEs. The first equation is homogeneous and the second is nonhomogeneous.≡

Solution of a PDE

A solution of a linear partial differential equation (1) is a function u(x, y) of two independent variables that possesses all partial derivatives occurring in the equation and that satisfies the equation in some region of the xy-plane.

It is not our intention to examine procedures for finding general solutions of linear partial differential equations. Not only is it often difficult to obtain a general solution of a linear second-order PDE, but a general solution is usually not all that useful in applications. Thus our focus throughout will be on finding particular solutions of some of the important linear PDEs, that is, equations that appear in many applications.

Separation of Variables

Although there are several methods that can be tried to find particular solutions of a linear PDE, the one we are interested in at the moment is called the method of separation of variables. In this method if we are seeking a particular solution of, say, a linear second-order PDE in which the independent variables are x and y, then we seek to find a particular solution in the form of product of a function x and a function of y:

u(x, y) = X(x)Y(y).

With this assumption, it is sometimes possible to reduce a linear PDE in two variables to two ODEs. To this end we observe that

,

where the primes denote ordinary differentiation.

EXAMPLE 2 Using Separation of Variables

Find product solutions of .

SOLUTION

Substituting u(x, y) = X(x)Y(y) into the partial differential equation yields

X″Y = 4XY′.

After dividing both sides by 4XY, we have separated the variables:

Since the left-hand side of the last equation is independent of y and is equal to the right-hand side, which is independent of x, we conclude that both sides of the equation are independent of x and y. In other words, each side of the equation must be a constant. As a practical matter it is convenient to write this real separation constant as –λ. From the two equalities,

,

we obtain the two linear ordinary differential equations

X″ + 4λX = 0 and Y′ + λY = 0.(2)

See Example 2, Section 3.9 and Example 1, Section 12.5.

For the three cases for λ: zero, negative, or positive; that is, λ = 0, λ = –α² < 0, and λ = α² > 0, where α > 0, the ODEs in (2) are, in turn,

X″ = 0and Y′ = 0, (3)

X″ – 4α²X = 0 and Y′ – α²Y = 0, (4)

X″ + 4α²X = 0 and Y′ + α²Y = 0.(5)

Case I (λ = 0): The DEs in (3) can be solved by integration. The solutions are X = c₁ + c₂x and Y = c₃. Thus a particular product solution of the given PDE is

u = XY = (c₁ + c₂x)c₃ = A₁ + B₁x, (6)

where we have replaced c₁c₃ and c₂c₃ by A₁ and B₁, respectively.

Case II (λ = –α²): The general solutions of the DEs in (4) are

X = c₄ cosh 2αx + c₅ sinh 2αx and Y = c₆,

respectively. Thus, another particular product solution of the PDE is

or (7)

where A₂ = c₄c₆ and B₂ = c₅c₆.

Case III (λ = α²): Finally, the general solutions of the DEs in (5) are

respectively. These results give yet another particular solution

(8)

where A₃ = c₇c₉ and B₃ = c₈c₉.≡

It is left as an exercise to verify that (6), (7), and (8) satisfy the given partial differential equation u_xx = 4u_y. See Problem 29 in Exercises 13.1.

Separation of variables is not a general method for finding particular solutions; some linear partial differential equations are simply not separable. You should verify that the assumption u = XY does not lead to a solution for ∂²u/∂x² – ∂u/∂y = x.

Superposition Principle

The following theorem is analogous to Theorem 3.1.2 and is known as the superposition principle.

Throughout the remainder of the chapter we shall assume that whenever we have an infinite set u₁, u₂, u₃, … of solutions of a homogeneous linear equation, we can construct yet another solution u by forming the infinite series

,

where the c_k , k = 1, 2, … , are constants.

Classification of Equations

A linear second-order partial differential equation in two independent variables with constant coefficients can be classified as one of three types. This classification depends only on the coefficients of the second-order derivatives. Of course, we assume that at least one of the coefficients A, B, and C is not zero.

The linear second-order partial differential equation

where A, B, C, D, E, F, and G are real constants, is said to be

hyperbolic if  B² – 4AC > 0,

parabolic if    B² – 4AC = 0,

elliptic if        B² – 4AC < 0.

EXAMPLE 3 Classifying Linear Second-Order PDEs

Classify the following equations:

(a)

(b)

(c) .

SOLUTION

(a) By rewriting the given equation as

we can make the identifications A = 3, B = 0, and C = 0. Since B² – 4AC = 0, the equation is parabolic.

(b) By rewriting the equation as

we see that A = 1, B = 0, C = –1, and B² – 4AC = –4(1)(–1) > 0. The equation is hyperbolic.

(c) With A = 1, B = 0, C = 1, and B² – 4AC = –4(1)(1) < 0, the equation is elliptic.≡

13.1 ExercisesAnswers to selected odd-numbered problems begin on page ANS-33.

In Problems 1–16, use separation of variables to find, if possible, product solutions for the given partial differential equation.

1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. a constant

In Problems 17–26, classify the given partial differential equation as hyperbolic, parabolic, or elliptic.

In Problems 27 and 28, show that the given partial differential equation possesses the indicated product solution.

27. 
    
28. 
    
29. Verify that each of the products u = X(x)Y(y) in (6), (7), and (8) satisfies the second-order PDE in Example 2.
30. Definition 13.1.1 generalizes to linear PDEs with coefficients that are functions of x and y. Determine the regions in the xy-plane for which the equation

    

    is hyperbolic, parabolic, or elliptic.

Discussion Problems

In Problems 31 and 32, discuss whether product solutions u = X(x)Y(y) can be found for the given partial differential equation. [Hint: Use the superposition principle.]