Review Ch8

8/11/2019 Review Ch8

1/22

(1) Method of separation of variables (MSV)

(2) Solution of wave equation (Two versions)

(b) dAlembertsmethod

1

( , ) sin cosnn

n ny x t A x c t

L L

0

2( )sin

L

n

nA f x x dxL L

where

(a)By MSV

1

( , ) ( ) ( )2

y x t f x ct f x ct

(3) Solution of Heat equation

22

1( , ) sin

nc t

L

nn

n

u x t A x eL

0

2( )sin

L

n

nA f x x dx

L L

where

Review Chapter 8


2/22

Example

solve 0x yu xu SolutionLet ( , ) ( ) ( )u x y X x Y y

Then ' '( ) ( ) ( ) ( ) 0X x Y y xX x Y y Hence

' '1 ( ) ( )

( ) ( )

X x Y y

x X x Y y holds for any x and y

Use method of S V to

2


3/22

Therefore

'' '

00

( )1 ( ) ( )

( ) ( ) ( )

Y yX x Y y

x X x Y y Y y for any fixed 0y

Thus' '

1 ( ) ( ) constant( ) ( )

X x Y y kx X x Y y

holds for any x and y

So 'X kxX 'Y kY

for any fixed k

3


4/22

Solve the above two 1storder ODE, get

2 /2( ) kxX x Ae ( ) kyY y Be

Hence

( , ) ( ) ( )u x y X x Y y2 2/2 /2kx ky kx ky

ABe e Ce e

4


5/22

8.2 Wave Equation

Elastic string of length L

tightly stretched

The string is set in motion.

It vibrates in vertical plane.

Animation slide

5

fixed at the end points


6/22

We have2 ( , ) ( , )xx ttc y x t y x t

where 0 , 0x L t

with boundary conditions:

(0, ) 0, ( , ) 0, for all 0y t y L t t

with initial conditions:

( ,0) ( ), ( ,0) 0, where 0ty x f x y x x L

Initial position Initial velocity

6

Proof omitted


7/22

1

( , ) sin cosnn

n ny x t A x c t

L L

0

2( )sin

L

n

nA f x x dx

L L

Solution of wave equation

7


8/22

Example 1

2 ( , ) ( , )xx ttc y x t y x t 0 , 0x t

(0, ) 0, ( , ) 0, for all 0y t y t t

( ,0) , ( ,0) 0, where 0ty x x y x x

Solve the following wave equation



8


9/22

0

2sinnA x nx dx

sin() sin() cos() Use formula

2 sin 2sin() cos()

2cos()

= - (1)= (1)+

9

Solution


10/22

(, ) 21+sin cos

=

is the solution of wave equation


and initial conditions:

2 ( , ) ( , )xx ttc y x t y x t 0 , 0x t

(0, ) 0, ( , ) 0, for all 0y t y t t

( ,0) , ( ,0) 0, where 0ty x x y x x 10


11/22

Example 2

2 ( , ) ( , )xx ttc y x t y x t 0 , 0x t

(0, ) 0, ( , ) 0, for all 0y t y t t

( ,0) sin(10 ), ( ,0) 0, where 0ty x x y x x




Solution

1

( , ) sin cosnn

y x t A nx nct

11


12/22

02 sin(10 )sin

1

sin(10 )sin( )

nA x nx dx

x nx dx

where

Hence 10 1, 0, if 10nA A n

Or we can get the above result

by comparing the coefficients

See next slide

odd odd

12


13/22

1

sin(10 ) sinnn

x A nx

1

( ) sinnn

f x A nx

10 1, 0, if 10nA A n

Hence

( , ) sin 10 cos 10y x t x ct

So

13


14/22

2

( , ) ( , )xx ttc y x t y x t 0 , 0x L t

(0, ) 0, ( , ) 0, for all 0y t y L t t

( ,0) ( ), ( ,0) 0, where 0ty x f x y x x L



dAlembertssolution of the wave equation

Then solution (given by dAlembert) is

1

( , ) ( ) ( )2

y x t f x ct f x ct

14


15/22

Revisit Example 2

2 ( , ) ( , )xx ttc y x t y x t 0 , 0x t

(0, ) 0, ( , ) 0, for all 0y t y t t

( ,0) sin(10 ), ( ,0) 0, where 0ty x x y x x


bydAlembertssolution



15


16/22

16

The solution (given by dAlembert) is

1

( , ) ( ) ( )

2

y x t f x ct f x ct

In this Example 2 ,

( ) sin(10 )f x x

Hence 1( , ) sin 10( ) sin 10( )2

y x t x ct x ct

L

( , ) sin 10 cos 10y x t x ct

Recall by MSV, the solution is

1

sinA cosB sin(A ) sin(A )2

B B

These two solutions are equivalent which can be proved by the following formula


17/22

8.3 Heat Equation

Heat-conducting rod with length L0 L

( , )u x t represents temperature at point x,

at time t

Let2

c be thermal diffusivity

We have2( , ) ( , )t xxu x t c u x t where 0 , 0x L t

17

Proof omitted


18/22


(0, ) 0, ( , ) 0, for any 0u t u L t t

0 0Temperatures are always ZERO at both end points

with initial condition:

( ,0) ( ), where 0u x f x x L

Initial distribution of temperature given by f(x)

18


19/22

0

2

( )sin

L

n

n

A f x x dxL L

22

1

( , ) sin

nc t

L

n

n

nu x t A x e

L

Solution of Heat equation

19


20/22

Suppose ( ) 2 1, 0f x x x L

20

Example 1

0

0 0

2(2 1)sin

4 2sin sin

L

n

L L

nA x x dx

L L

n nx x dx x dx

L L L L

22

1( , ) sin

nc t

L

nn

n

u x t A x eL


21/22

sin() sin() cos() Use the following formulae we can find

sin() 1 cos cos ( 1)

n

n

21


22/22

22

Example 2

Suppose

3,

() sin(10)

0

2( )sin

L

n

nA f x x dx

L L

1

( ) sinnn

nf x A x

L

Now findWe shall compare coeff.

sin(10) sin 3= Note that if

10 3

then 30So

Documents

Review Ch8