MAT 3724Applied Analysis I

2.1 Part ICauchy Problem for the

Heat Equation

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Chapter 2

PDE on Unbounded Region In one dimension, it is the real line Easier to solve than bounded region

Preview

Initial Value Problem with the Heat equation

Introduce Dimensional Analysis

Cauchy Problem for the Heat Equation

Note the change of notations (u instead of q)

, , 0(2.1)

( ,0) ( )t xxu ku x R t

u x x

x

Set Up

Initially, the temp. distribution is given by ____

, , 0(2.1)

( ,0) ( )t xxu ku x R t

u x x

x

Lateral side insulated

,u x t

Example 1

0

, , 0(2.3)

0 0(2.4) ( ,0)

0

t xxw kw x R t

xw x

u x

Example 1:Thought Experiment

Scenario 1 Scenario 20

, , 0(2.3)

0 0(2.4) ( ,0)

0

t xxw kw x R t

xw x

u x

Example 1:Thought Experiment

0

, , 0(2.3)

0 0(2.4) ( ,0)

0

t xxw kw x R t

xw x

u x

What would happen to w(x,t) as the time moves on?

Example 1: Simplification, , 0

(2.3)0 0

(2.4) ( ,0)1 0

t xxw kw x R t

xw x

x

Example 1: Simplification, , 0

(2.3)0 0

(2.4) ( ,0)1 0

t xxw kw x R t

xw x

x

Interpretations of HW 05 Problem 2

The PDE model and inequality below appear in the last HW

0

, 0 , 0,

0, , 0, 0,

,0 , 0 .

t xxu ku x l l

u t u l t t

u x u x x l

2 2

0

0 0

,l l

u x t dx u x dx

Interpretations of HW 05 Problem 2

Even though the set ups are not exactly the same, some calculations with this example may help us to understand what the inequality means.

2 2

0

0 0

,l l

u x t dx u x dx

Interpretations of HW 05 Problem 2

2 2

0

0 0

,l l

u x t dx u x dx

5 5 5

2 2 2

0

5 5 0

,0 1 5w x dx w x dx dx

Interpretations of HW 05 Problem 2

2 2

0

0 0

,l l

u x t dx u x dx

5 5 5

2 2 2

0

5 5 0

,0 1 5w x dx w x dx dx

52

5

52

5

,1 2.101057720

,10 1.258015936

w x dx

w x dx

Example 1

Solution Method: Dimensional Analysis (units) Guessing

0

, , 0(2.3)

0 0(2.4) ( ,0)

0

t xxw kw x R t

xw x

u x

Inspirations

20

1

2h gt v t 0v

Variables h g t v0 p1 p2

units No No

dimension Dimensionless Dimensionless

Inspirations

20

1

2h gt v t 0v

If we can find___________________, then we can

recover_____________________.

Variables h g t v0 p1 p2

units No No

dimension Dimensionless Dimensionless

Example 1

Variables w x t u0 k wt wxx

units

dimension

If we can find_________________________,

then we can recover_____________________.

0

, , 0(2.3)

0 0(2.4) ( ,0)

0

t xxw kw x R t

xw x

u x

Notations Relaxation for Improper Integrals

Provided that you understand the correct concepts, you are allow to use less rigorous notations. Here is an illustration.

41

2

1

xdx

x