Classification of partial differential equations

UNIT - III

APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS

Classification of PDE- Method of separation of variables - One Solutions of one dimensional wave equation - dimensional equation of heat conduction - Steady state solution of two dimensional equation of heat conduction (excluding insulated edges).

INTRODUCTION

Partial differential equations arise in connection with several physical and engineering problems in which the functions involved depend on two or more independent variables such as time and co-ordinates in space. In practical problems, the following types of equations are generally used.

1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS

Let a second order p.d.e. in the function u of the two independent variables x, y be of the form

Equation (1) is classified as elliptic, parabolic or hyperbolic depending on

B2 - 4AC < 0 (elliptic equation)

B2 - 4AC = 0 (parabolic equation)

B2 - 4AC > 0 (hyperbolic equation)

Example

Problems on classification of partial differential equations

Example 3.1.1 Find the nature of the partial differential equation

4 uxx + 4uxy + uyy + 2ux − uy = 0

Solution: Given: 4 uxx + 4uxy + uyy + 2ux − uy = 0

Here, A = 4, B = 4, C = 1

B2 - 4 AC = 16 - 4 (4) (1) = 0

⸫ The given equation is parabolic equation.

Example 3.1.2 Classify uxx + uyy = (ux)2 + (uy)2

Solution: Given: uxx + uyy = (ux)2 + (uy)2

Here, A = 1, B = 0, C = 1

B2 - 4AC = 0 - 4 (1) (1) = -4 < 0

⸫ The given equation is elliptic equation.

Example 3.1.3 Classify

Solution : Given: uxy = ux + uy +xy

Here, A = 0, B = 1 C = 0

B2 – 4AC = 1 > 0

⸫ The given equation is hyperbolic equation.

Example 3.1.4 : Classify uxx – y4uyy = 2 y3 uy

Solution: Given: uxx – y4 uyy - 2y3 uy = 0

Here, A = 1, B = 0, C = -y4

B2 – 4AC = 0 − 4(1)(−y4) = 4y4

If y = 0, we get B2 - 4AC = 0 [Parabolic equation]

If y > 0 or y < 0, we get B2 - 4AC > 0 [Hyperbolic equation]

Example 3.1.5 : Classify the p.d.e (1+x)2uxx – 4x uxy + uyy = x

Solution: Given: (1+x)2 uxx - 4x uxy + uyy = x

Example 3.1.6 Classify: x2 uxx + 2xy uxy + (1 + y2) uyy − 2 ux = 0

Solution: Given: x2 uxx + 2xy uxy + (1 + y2) uyy − 2 ux = 0

EXERCISE : 3.1

Classify the following p.d.e