Parabolic Partial Differential Eq’n

Parabolic Partial Differential Eq’n

Obillo, Alvin

Gonzales, Patricia Shaine

INTRODUCTION

GENERAL FORM OF SECOND ORDER LINEAR PDE WITH TWO INDEPENDENT VARIABLES AND ONE DEPENDENT VARIABLE:

Where u = u(x, y) and a, b, c, d, e, f, and g are functions of x and y ONLY. If g is zero, then the equation is

homogeneous. The PDE is said to be elliptic if The PDE is said to be hyperbolic if The PDE is said to be parabolic if

HEAT-CONDUCTION EQUATION

Where a = k, b = 0, c = 0, and d = -1

Then:

THEREFORE, HEAT-CONDUCTION EQUATION IS A PARABOLIC DIFFERENTIAL EQUATION

THE EXPLICIT METHOD

where:

T is the temperature

x is the location

j is the time step

i is the interval of location

THE EXPLICIT METHODSubstituting the approximations from the last slide to the general equation…

Solving for the temperature at time interval j + 1 would give….

Where:

If we know the temperature at the first time step and the boundary temperatures, it is possible to get the temperature of each location at the next time step.

THE EXPLICIT METHODEXAMPLE 1: Consider a steel rod that is subjected to a temperature of 100 degrees Celsius on the left end and 25 degrees Celsius on the right end. If the rod is of length 0.05 m, use the explicit method to find the temperature distribution in the rod from t = 0 and t = 9 seconds. Use ∆x = 0.01 m, and t = 3 seconds. = 1.4129 x / s. The initial temperature of the rod is 20 degrees Celsius.

0 1 2 3 4 5 6

25 deg C

100 deg C

EXAMPLE 1SOLUTION:

When t = 0 seconds, j = 0:

= 100 degrees C

= 20 degrees C

= 20 degrees C

= 20 degrees C

= 20 degrees C

= 25 degrees C

When t = 3 seconds, j = 1

= 100 degrees C

= 53.912 degrees C

= 20 degrees C

= 20 degrees C

= 22.120 degrees C

= 25 degrees C

EXAMPLE 1SOLUTION:

When t = 6 seconds, j = 2:

= 100 degrees C

= 59.073 degrees C

= 34.375 degrees C

= 20.889 degrees C

= 22.442 degrees C

= 25 degrees C

When t = 9 seconds, j = 3

= 100 degrees C

= 65.953 degrees C

= 39.132 degrees C

= 27.266 degrees C

= 22.872 degrees C

= 25 degrees C

EXAMPLE 1If the temperature vs the location is to be plotted in a graph…

THE EXPLICIT METHOD Based on the graph, as the time goes on, the graph

seems to become linear. This means that as time approaches infinity, the temperature becomes linear – which corresponds to a steady-state system.

CRANK – NICHOLSON METHOD

where:

T is the temperature

x is the location

j is the time step

i is the interval of location

CRANK – NICHOLSON METHODSubstituting the approximations from the last slide to the general equation…

Solving for the temperature at time interval j + 1 would give….

Where:

In order to solve for the temperature of every location at any time, this method makes use of simultaneous linear equations.

CRANK – NICHOLSON METHOD

EXAMPLE 1: Consider a steel rod that is subjected to a temperature of 100 degrees Celsius on the left end and 25 degrees Celsius on the right end. If the rod is of length 0.05 m, use the explicit method to find the temperature distribution in the rod from t = 0 and t = 9 seconds. Use ∆x = 0.01 m, and t = 3 seconds. = 1.4129 x / s. The initial temperature of the rod is 20 degrees Celsius.

0 1 2 3 4 5 6

25 deg C

100 deg C

EXAMPLE 1SOLUTION:

For time = 3 seconds;

i = 1, j = 0:

(-0.4239)(100) + (2)(1+0.4239) - (0.4239) (0.4329)(100) + (2)(1-0.4329)(20) + (0.4329)(20)

Which would then give:

2.8478 - 0.4239 = 116.30

EXAMPLE 1SOLUTION:

By further solving i = 2, 3, and 4, this matrix can be created:

EXAMPLE 1SOLUTION:

At time = 3 seconds;

EXAMPLE 1SOLUTION:

At time = 6 seconds;

EXAMPLE 1SOLUTION:

At time = 6 seconds;

EXAMPLE 1SOLUTION:

At time = 9 seconds;

EXAMPLE 1SOLUTION:

At time = 9 seconds;

COMPARISON OF RESULTSLocation Explicit Implicit Crank-

Nicholson

Analytical

65.953 59.043 62.604 62.510

39.132 36.292 37.613 37.084

27.266 26.809 26.562 25.844

22.872 24.243 24.042 23.610