Upload
cyma
View
59
Download
0
Embed Size (px)
DESCRIPTION
Zen and the Art of Motorcycle Maintenance Robert Pirsig. The state of “ stuckness ” is to be treasured. It is the moment that precedes enlightenment. The complex exponential function. REVIEW. Hyperbolic functions. Newton’s 2 nd Law for Small Oscillations. =0. ~0. - PowerPoint PPT Presentation
Citation preview
Zen and the Art of Motorcycle Maintenance
Robert Pirsig
• The state of “stuckness” is to be treasured. It is the moment that precedes enlightenment.
The complex exponential functioncos sinixe x i x
2
2
cos
sin
ix ix
ix ixi
e ex
e ex
REVIEW
Hyperbolic functionssinh( ) ; cosh( ) .2 2
sinh( ) 1 2tanh( ) ; sech( ) .cosh( ) cosh( )
x x x x
x xx x x x
e e e ex x
x e ex xe e e ex x
Newton’s 2nd Law for Small Oscillations(3) ( )2
232 1 1 1''(0) (0) (0)2! 3! ! = (0) '(0) n nF x F x F xn
d xd
Fm F xt
=0~0
22 = '(0) '(0) 0 oscilla tion d xm F x
dtF
Differential eigenvalue problems
2
eigenval
''( ) ( ) 0;
(0) 0; ( ) 0
sin( )
0, 1, 2, 3,
sin( ),sin(2 ),sin(3 ), eigenfunctions, or
ues
m odes
f x f x
f f
f A x
f x x x
Partial derivatives
x
y
( , )T x y
0
0
( , ) ( , )( , ) lim
( , ) ( , )( , ) lim .
x
y
T x x y T x yT x y xx
T x y y T x yT x y yy
TT x T yx y
Increment:
x part y part
Total derivatives
x
y
( , , )T x y t
TTT x y tx y tT
yT T xt t tx y t
T T
0limt
dyT dT T dxt dt x dt
T Ty dt t
x part y part t part
Isocontours
x
y
( , )T x y
0
/ isocontour slope/
TT x yx y
Ty xy x
y T xx T y
T
T
Multivariate Calculus 2:
partial integrationseparation of variables
Fourier methods
x
y
( , )T x y
Partial differential equationsAlgebraic equation: involves functions; solutions are numbers.
Ordinary differential equation (ODE): involves total derivatives; solutions are univariate functions.
Partial differential equation (PDE): involves partial derivatives; solutions are multivariate functions.
Notation
2 32 2
subscript notatio
"di" = partial der
n:
; ;
iv
,
ative
x tt xttf f ff f fx t x t
Classification
2
2
2
3 0 linear3 0 nonlinear
3 0 homogeneous3 1 nonhomogeneous
xxt
x t
x t
x t
f f ff f f
f f ff f f
If ( , ) and ( , ) are solutions of a PDE, then any ( ,
Superposition:
linear, homogeneouslinear combinati ) ( , ),
where andon
al are
so con
a sstants,
is .olution
f x t g x taf x t bg x t
a b
Order
2 0 1st order
3 0 3rd orderx t
xxx t
f f gf f f
=order of highest derivative with respect to any variable.
Partial integration
21( ,
. .
) )
,
(
( )
2
x
f x y
e g
f x y x
x c y
Instead of constant,add function of other variable(s)
Partial integration
2
2
boundary condition
(0, ) 1
(0, ) 0 ( ) 1
add :
( , ) ;
1( , ) ( )2
1complete solution: ( , 12
( ) 1
)
x f y y
f y c y y
c
f x y x c y
f x y x
f y x
y
y y
x
Partial integration
( , ) ( )
beca
. .
(
s
)
u e
,x
f x y c y
yd
e g
f x y y
x
dx x
y
x y y
Partial integration
( , ) ( )
because
Now suppose the boundary condition ( ,
.
) 1
(1, ) ( ) 1
.
( , )
( ) 1
( , 1
s
)
1i
x
f y y
f y y c y y
c y
f x y x
e g
f x y y
f x y c y
ydx y dx y x
xy
y
Partial integration. .
( , ) 0
0 means ( , ) does n
(
o
, )
t actually dep
(
end on
)
.
x
x
e g
f x y
f
f x y c
f y
y
x x
Solution by separation of variables
Try to reduce PDE to two (or more) ODEs.
Is it possible for functions of two different variables to be equal?
( ) ( ) ?f x g y
Is it possible for functions of two different variables to be equal?
Assume ( ) ( )
: ( ) ( ).x x
f x g y
f x g yx
Is it possible for functions of two different variables to be equal?
Assume ( ) ( )
: ( ) ( ) 0
: ( ) 0 (0)
( ) ; ( ) .
x x
y y
f x g y
f x g yx
f x gy
f x c g y c
The PlanManipulate PDE into the form ( ) ( ).
Result is 2 ODES: ( ) ; ( ) .
f x g y
f x c g y c
Example 1
Try ( , ) ( ) ( )
0x yy
f x y X x Y y
f f
Example 1
Try ( , ) ( ) ( )
Substitute: ( ) ( ) ( ) ( ) 0
( ) ( ) ( ) ( )Divide by : 0( ) ( )
( )( )Rearrange: 0( ) ( )
0
x yy
x yy
yyx
x yy
f x y X x Y y
X x Y y X x Y y
X x Y y X x Y yXY X x Y y
Y yX xX x Y y
f f
Example 1
Try ( , ) ( ) ( )
Substitute: ( ) ( ) ( ) ( ) 0
( ) ( ) ( ) ( )Divide by : 0( ) ( )
( )( )Rearrange: 0( ) ( )
( )( ) Viola!( ) ( )
0
x yy
x yy
yyx
yyx
x yy
f x y X x Y y
X x Y y X x Y y
X x Y y X x Y yXY X x Y y
Y yX xX x Y y
Y yX x cX x Y y
f f
Example 1
( )( )( ) ( )
; , 2 ODEs
; sin( ), cos( )
( , ) ( ) ( ) sin( ) (or something like that)
yyx
x yy
Cx
Cx
Y yX x cX x Y y
X cX Y cY
X e Y cy cy
f x y X x Y y e cy
Thermal diffusion in a 1D bar
0( ,0) ( )T x T x
0x x L( , ) ; xxtTT T x t T
(0, ) 0 ; ( , ) 0T t T L t Boundary conditions:
Initial condition:
Applications
Diffusion of:
• Heat• Salt• Chemicals, e.g. O2, CO2, pollutants• Critters, e.g. phytoplankton• Diseases• Galactic civilizations• Money (negative diffusion)
Thermal diffusion in a 1D bar
0( ,0) ( )T x T x
0x x L( , ) ; xxtTT T x t T
(0, ) 0 ; ( , ) 0T t T L t Boundary conditions:
Initial condition:
Solution by separation of variables
Try ( , ) ( ) ( )
Substitute: ( ) ( ) ( ) ( )
( ) ( )Divide by : c( ) ( )
( ) ( ) ; ( ) ( )
xxt
xxt
t xx
xxtt
T T
T x t X x t
X x t X x t
t X xX t X x
ct c t X x X x
First try
2
22 2
;
Assume 0, or ;
boundary co
;
( )
( ) ) 0?
NO GOO
nditions: (0) 0
) 2 sinh(
2 sinh(
0, or D! Try
x
a ax xa t
a ax x
xtt
xxtt
cc X X
ac c a a X X
X A B B A
e X Ae Be
aX A e e x
aX L
c
L
A
A
c
2a
sinh( ) 2x xe ex
Second try
2
22 2
; sin( ) cos( )
sin( )
( ) sin( ) 0?
NO PROBLEM! , 0,
;
Assume 0, or ;
boundary conditions:
1, 2,
(0) 0
xxt
x
t
x
a
t
a a
cc X X
ac c a a X
e X A x B x
aX A
X
X B
x
aX L A L
a L n n
2
, 2
sin( )
, or
sin ; 1,2,3,
and
( , ) ( ) ( ) sin
a t
n tL
aX A x
a a nL n L
xX A n nL
ne a L
xT x t X x t A n eL
2
( , ) ( ) ( ) sinn tLxT x t X x t A n eL
Characteristics of time dependence:
•T 0 as t ∞ , i.e. temperature equalizes to the temperature of the endpoints.
•Higher (diffusivity) leads to faster diffusion.
•Higher n (faster spatial variation) leads to faster diffusion.
In fact:
22
2So diffusion time
"Wavelength"
.
n
ntn
L
n
t
Ln
e e
Time scale is proportional to (length scale)2.
In fact:
22
2So diffusion time
"Wavelength"
.
n
ntn
L
n
t
Ln
e e
E.G. Water 1/2 as deep takes 1/4 as long to boil!
Sharp gradients diffuse rapidly.
This observation is surprising.
Now what about the initial condition?
0( ,0) ( )T x T x
0x x LxxtT T
(0, ) 0 ; ( , ) 0T t T L t Boundary conditions:
Initial condition: ?
2
( , ) sinn tLxT x t A n eL
0( ,0) ( )T x T xInitial condition: ?
2
.
20Suppose
( , ) sin
( ,0) sin
( ) sin
Choose , 1 so ( , ) sin
n tL
tL
xT x t A n eL
xT x A
xT
n L
xT x t eL
x L
A n
0x x L
0T
T
2
.
23
0Suppose ( ) sin 3
Choose , 3
( , ) sin
so ( , ) si 3 n
n tL
tL
xT x L
A n
xT x t A n eL
xT x t eL
0x x L
0T
T
01Suppose ( ) sin sin 3 ?2
x xT x L L
0x x L
22
2
0
3
1Suppose ( ) sin sin 3 ?2
!
and are solusin sin 3
Asin
If tions,
h
T en
tt LL
L
x xe eL L
x
x xT x L L
SUPERPOSITION
eL
2
22
3
3
sin 3
1(
is a sol
, ) sin s
ution.
1Choose 1, in 3 22
tt L
tt LL
xB eL
x xT x t e eA B L L
0x x L
Fourier’s Theorem
0x x L
0 0
1
0
0
ANY FUNCTION ( ) that obeys the boundary conditions (0) 0 and ( ) 0 can be represented as a Fo
( ) s
urier series:
The corresponding solution for the diff
in
u
nn
xT x A n
x T L
L
T T
2
1 (
sion probl
, ) s
em :
n
is
in tL
nn
xT x t A n eL
To find the constants:
002 ( )sin
Ln nL
xA T x dxL
Problem solved
0( ,0) ( )T x T x
0x x LxxtT T
(0, ) 0 ; ( , ) 0T t T L t Boundary conditions:
Initial condition:
2
001
2 ( , ) sin ; ( )si n n t LL
n nn
nLx xT x t A n e A T x dxL L
Homework clarification3.2.1 partial integration3.2.2 separation of variables3.2.3 Fourier series solution
1. Solve for a single Fourier mode• Separate• Choose sign of separation constant • Satisfy boundary conditions and initial condition ht=0 • Write down the general solution satisfying h(x,0)=h0(x), but don’t derive the Fourier series.
2. Interpret• Time dependence: exponential or …?• Describe in physical terms.• How might you create the modes n=1, n=2, …?
3. Time scale• Is the time scale for a mode proportional to the length scale squared? • If not, what?
0x x L
0( ,0) ( )h x h xh
( ,0) 0th x