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8/11/2019 ChewMA1506-14 Ch8
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Chapter 8
Partial Differential equationsPDEs
MA1506
Mathematics II
Chew T S MA1506-14 Chapter 8 1
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Chew T S MA1506-14 Chapter 8 2
In this Chapter, we will study
(1) Method of separation of variables (MSV)
(2) Solution of wave equation (Two versions)
(b) dAlemberts method
1
( , ) sin cosnn
n n y x t A x c t
L L
=
=
0
2( )sin
L
n
n A f x x dx L 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
Ln
n
nu x t A x e L
=
=
0
2( )sin
L
n
n A f x x dx
L L = where
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Notation
( , )u x y x xudenoted by
( , )u x y y
yudenoted by2
2
( , )u x y
x
xxudenoted by
2 ( , )u x y y x
xyudenoted by
Chew T S MA1506-14 Chapter 8 3
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Partial Differential Equations
0 xxu u = 0 x yu xu+ =
y xxu u= yy xxu u=
0 xx yyu u+ =Equations above involving partial derivativesare called partial differential equations
Chew T S MA1506-14 Chapter 8 4
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By method of separation of variables, weassume that solution ( , )u x y can be written as
( , ) ( ) ( )u x y X x Y y=Hence' ( ) ( ) xu X x Y y=
'( ) ( ) yu X x Y y=
' '( ) ( ) xyu X x Y y=
'' ( ) ( ) xxu X x Y y= ''( ) ( ) yyu X x Y y=
8.1 SV for PDE
8.1 Separation of variables for PDE N L pp 7-12
In this section, , we study one simple case
by method of separation of variables .
Chew T S MA1506-14 Chapter 8 5
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Example
solve 0 x yu xu+ =SolutionLet ( , ) ( ) ( )u x y X x Y y=
Then ' '( ) ( ) ( ) ( ) 0 X 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 to8.1 SV for PDE
Chew T S MA1506-14 Chapter 8 6
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Therefore'' '
00
( )1 ( ) ( )( ) ( ) ( )
Y y X x Y y x X x Y y Y y= = for any fixed 0 yThus
' '
1 ( ) ( ) constant( ) ( )
X x Y y k x X x Y y
= = =
holds for any x and y
So ' X kxX = 'Y kY =
8.1 SV for PDE
for any fixed k
Chew T S MA1506-14 Chapter 8 7
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Solve the above two 1 st order ODE, get
2 /2( ) kx X x Ae= ( ) kyY y Be=Hence
( , ) ( ) ( )u x y X x Y y=2 2/2 /2kx ky kx ky
ABe e Ce e
= =
8.1 SV for PDE
Chew T S MA1506-14 Chapter 8 8
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USEFUL FACTS for studying wave and heat equationsUSEFUL FACTS
Sturm-Liouville equationconsider ( ) + ( ) = 0 with initial conditions (0) = 0, (0) = 0 Then we know that zero is a solution.Zero is the only solution .
Now, if we replace the above initial conditions byboundary conditions
Then we have Sturm-Liouville equation
( ) + ( ) = 0 (0) = 0,
( ) = 0
(0) = 0, ( ) = 0 Chew T S MA1506-14 Chapter 8 9
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We shall write down solutions ofSturm-Liouville equation without proofs.
Case 1 < 0 Zero is the only solution
Case 2 = 0
Zero is the only solution
Case 3 > 0
There are two subcasesSubcase 1 When 2, = 1,2, . . .
Zero is the only solution
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 10
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Subcase 2 = 2, = 1,2, . . .
= 2, = 1,2, . . . called eigenvalueFor eachthere exists nonzero solution
called eigenfunction
USEFUL FACTS
( ) = sin
Chew T S MA1506-14 Chapter 8 11
This case is very important
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1. if
n m
n m=1
sin sin L
L
n m x x dx
L L L
=
0
1
if
where m, n are positive integers
USEFUL FACTSSome special integrals
1cos cos
L
L
n m x x dx
L L L
= 2.0
1
if
if
n mn m=
1sin cos 0
L
L
n m x x dx
L L L
= 3.for any positive
integers m, n
where m, n are positive integers
Chew T S MA1506-14 Chapter 8 12
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You dont have to spend time on working out
Examples 1, 2,3.Just need to know the results.
Use 1, 2 and 3,
21 (sin 4sin 2 23cos3 ) x x x dx
+ +
2 2 21 4 23= + +
we get
expand
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 13
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1. Odd function f(x) : symmetric w.r.t. origin
2. Even function f(x): symmetric w.r.t. y-axis
USEFUL FACTSOdd and even functions
( ) = ( )
( ) = ( ) Chew T S MA1506-14 Chapter 8 14
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3. sinx, sin x are odd, where is any real number
4. cosx, cos x are even, where is any real number
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 15
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5. oddeven =odd functionoddodd =even functioneveneven =even function
6.
( ) 0
L
L odd dx =sin( ) 0
L
L x dx
=
sin cos 0 L L
n m x x dx L L
= for any integers n, m
odd even
means
multiplication
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 16
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7.0
( ) 2 ( ) L L
L
even dx even dx
=
8. Let f be odd.
( )sin = 2 ( )sin 0 odd odd
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 17
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sin( ) =sin( )
2 cos( )
2sin( ) = 2 sin( )2 + 2 22)cos(3
9.
11.
sin( ) = 1 cos 10.
USEFUL FACTS
Chew T S MA1506-14 Chapter 8 18
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2 1
cos (1 cos 2 )2 x x= +
2 1sin (1 cos 2 )
2 x x=
1sin cos sin 2
2
x x x=
1sin sin [cos( ) cos( )]
2 x y x y x y= +
1cos cos [cos( ) cos( )]
2 x y x y x y= + +
1sin cos [sin( ) sin( )]2
x y x y x y= + +
12.
Chew T S MA1506-14 Chapter 8 19
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14.
1sin(2 1) ( 1)2
nn + =
sin 0n =1,2,3,...n =
1,2,3,...n =
USEFUL FACTS
sin(2
) =
sin(
2
) = Chew T S MA1506-14 Chapter 8 21
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8.2 Wave Equation8.2 wave equation
Elastic string of length L
tightly stretched
The string is set in motion.It vibrates in vertical plane.
Animation slide
Chew T S MA1506-14 Chapter 8 22
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Let be vertical displacement at pt , at time( , ) y x t x t Let c be constant velocity of propagation
of wave along string
8.2 wave equation
Animation slide
x
Chew T S MA1506-14 Chapter 8 23
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Now we shall use the method of S Vto solve the above wave equation
Solution
By MSV, let ( , ) ( ) ( ) y x t u x v t =
Then ( , ) ( ) ( ) xx y x t u x v t =( , ) ( ) ( )tt y x t u x v t =
Hence2
( , ) ( , ) xx tt c y x t y x t =implies 2 ( ) ( ) ( ) ( )c u x v t u x v t =
8.2 wave equation
Chew T S MA1506-14 Chapter 8 25
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Therefore2
( ) 1 ( )( ) ( )
u x v t u x c v t
= holds for
ALL x, and t
So 2( ) 1 ( )( ) ( )
u x v t u x c v t
= = for any fixed
Hence ( ) ( ) 0u x u x + =2( ) ( ) 0v t c v t + =
For convenience, we use -
8.2 wave equation
Chew T S MA1506-14 Chapter 8 26
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Now we shall solve
( ) ( ) 0u x u x + =with boundary conditions:
(0, ) 0, ( , ) 0, for all 0 y t y L t t = = >i.e., (0) ( ) 0, ( ) ( ) 0, for all 0u v t u L v t t = = >
i.e., (0) 0, ( ) 0u u L= =
8.2 wave equation
Chew T S MA1506-14 Chapter 8 27
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So now we solve
( ) ( ) 0u x u x + =with boundary conditions: (0) 0, ( ) 0u u L= =Recall the above equation is
Sturm-Liouville equation.
= 2, = 1,2, . . . Whenthere exists nonzero solution
( ) = sin
8.2 wave equation
Chew T S MA1506-14 Chapter 8 28
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now look at corresponding
For
2( ) ( ) 0v t c v t + =
The general solution is
( ) cos sinn nn n
v t A c t B c t L L = +
However, from one of the initial conditions( ,0) 0t y x = i.e., (0) 0v = we have 0n B =
= 2, = 1,2, . . . 8.2 wave equation
Chew T S MA1506-14 Chapter 8 29
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So for each n,
sin cosnn n
A x c t L L
is a solution of the given PDE
with boundary conditions and initial condition( ,0) 0t y x =
1sin cos
m
nn
n n A x c t L L
=
Hence
is again a solution
8.2 wave equation
Chew T S MA1506-14 Chapter 8 30
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1
( , ) sin cosnn
n n y x t A x c t
L L
=
=
In fact
is the general solution of the given PDE
Now we shall use the 2nd
initial condition
to find the value ofn
A( ,0) ( ) y x f x=
8.2 wave equation
Chew T S MA1506-14 Chapter 8 31
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1 1
( ) ( ,0) sin cos 0 sinn nn n
n n n f x y x A x c A x
L L L
= =
= = =
1
( ) sinnn
n f x A x
L
=
=
How to find n A
where 0 x L<
( ,0) , ( ,0) 0, where 0t y x x y x x = = <
(0, ) 0, ( , ) 0, for all 0 y t y t t = = >( ,0) , ( ,0) 0, where 0t y x x y x x = = <
(0, ) 0, ( , ) 0, for all 0 y t y t t = = >
( ,0) sin(10 ), ( ,0) 0, where 0t y x x y x x = = <
(0, ) 0, ( , ) 0, for all 0 y t y L t t = = >
( ,0) ( ), ( ,0) 0, where 0t y x f x y x x L= = <
Chew T S MA1506-14 Chapter 8 48
8 3 Heat Equation
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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=