Review Ch8
Transcript of Review Ch8
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(1) Method of separation of variables (MSV)
(2) Solution of wave equation (Two versions)
(b) dAlembertsmethod
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( , ) 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
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( , ) ( ) ( )2
y x t f x ct f x ct
(3) Solution of Heat equation
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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
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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
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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
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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
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8.2 Wave Equation
Elastic string of length L
tightly stretched
The string is set in motion.
It vibrates in vertical plane.
Animation slide
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fixed at the end points
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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
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Proof omitted
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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
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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
with boundary conditions:
with initial conditions:
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0
2sinnA x nx dx
sin() sin() cos() Use formula
2 sin 2sin() cos()
2cos()
= - (1)= (1)+
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Solution
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(, ) 21+sin cos
=
is the solution of wave equation
with boundary conditions:
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
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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
Solve the following wave equation
with boundary conditions:
with initial conditions:
Solution
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( , ) sin cosnn
y x t A nx nct
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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
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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
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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
with boundary conditions:
with initial conditions:
dAlembertssolution of the wave equation
Then solution (given by dAlembert) is
1
( , ) ( ) ( )2
y x t f x ct f x ct
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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
Solve the following wave equation
bydAlembertssolution
with boundary conditions:
with initial conditions:
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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
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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
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Proof omitted
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with boundary conditions:
(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)
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0
2
( )sin
L
n
n
A f x x dxL L
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1
( , ) sin
nc t
L
n
n
nu x t A x e
L
Solution of Heat equation
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Suppose ( ) 2 1, 0f x x x L
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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
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1( , ) sin
nc t
L
nn
n
u x t A x eL
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sin() sin() cos() Use the following formulae we can find
sin() 1 cos cos ( 1)
n
n
21
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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