PARTIAL DIFFERENTIAL PARTIAL DIFFERENTIAL EQUATIONSEQUATIONS

IntroductionIntroduction

Given a function u that depends on both x and y, the partial derivatives of u w.r.t. x and y are:

y

yxuyyxu

y

u

x

yxuyxxu

x

u

y

x

),(),(lim

),(),(lim

0

0

An equation involving partial derivatives of an unknown function of two or more independent variables is called Partial Differential EquationPartial Differential Equation (PDE). Examples:

xyx

u

x

u

yuy

ux

yx

u

uy

uxy

x

u

2

33

2

2

2

2

2

2

2

2

2

2

6

58

12 The order of a PDE is that of the highest-order partial derivative appearing in the equation.

A PDE is linear if it is linear in the unknown function and all its derivatives, with coefficients depending only on the independent variables

e.g.

x’’ + ax’ + bx + c = 0 – linear

x’ = t2x – linear

x’’ = 1/x – nonlinear

For linear, two independent variables second order equations can be expressed as:

02

22

2

2

Dy

uC

yx

uB

x

uA

where A, B and C are functions of x and y and D is a function of x, y, u/x and u/y.

Above equation can be classified into categories in the next slide based on values of A, B, and C.

B2 – 4AC Category Example

< 0 EllipticElliptic Laplace equation (Steady state with two spatial dimension)

= 0 ParabolicParabolic Heat conduction equation (time variable with one spatial dimension)

> 0 HyperbolicHyperbolic Wave equation (time variable with one spatial dimension)

02

2

2

2

y

T

x

T

2

2'

x

Tk

t

T

2

2

22

2 1

t

y

cx

T

Elliptic EquationsElliptic Equations

Typically used to characterize steady-state distribution of an unknown in two spatial dimensions.

Laplace EquationLaplace Equation

0

y

q

x

q

The PDE as an expression of the conservation of energy

Need to reformulate the equation in terms of temperature. Use Fourier’s Law:

and

substituting back results in

i

TCkqi

CV

HT

02

2

2

2

y

T

x

T (Laplace equation)

Parabolic EquationsParabolic Equations

Heat conduction

Hot Cool

Heat balance (the amount of heat stored in the element) over a unit time, t

TCzyxtzyxxqtzyxq )()(

t

TC

x

xxqxq

)()(

t

TC

x

q

Input – Output = Storage

Dividing by volume of the element (xyz) and t

Taking the limit yields:

i

TCkqi

Substituting Fourier’s Law:

t

T

x

Tk

2

2

Gives:

SolutionSolution

Finite Difference

A grid used for the finite difference solution of elliptic PDEs in two independent variables.

Numerical Differentiation using Centred-Numerical Differentiation using Centred-Finite Divided DifferenceFinite Divided Difference

First Derivative

Second Derivative

Third Derivative

h

xfxfxf ii

2

)()()(' 11

211 )()(2)(

)("h

xfxfxfxf iii

32112

2

)()(2)(2)()("'

h

xfxfxfxfxf iiii

SolutionSolution

Finite Element

Finite Element AnalysisFinite Element Analysis

Two interpretations1.Physical Interpretation:

The continous physical model is divided into finite pieces called elements and laws of nature are applied on the generic element. The results are then recombined to represent the continuum.

2.Mathematical Interpretation:The differentional equation representing the system is converted into a variational form, which is approximated by the linear combination of a finite set of trial functions.

Group AssignmentGroup Assignment

Group Task

Group A Problem 1

Group B Problem 2

Group C Problem 3

Group D Problem 4