apm346h1

apm346h1
APM346H1 Differential Equations
INTRODUCTION
Types of Partial Differential Equations
∂u
∂u
, u y=
, and u  x , y=? .
∂x
∂y
•
Transport equation: u x  x , yu y  x , y=0 , where u x =
•
Shockwave equation: u x  x , yu  x , yu y  x , y=0 .
•
The vibrating string equation: u tt  x , t =c 2 u xx  x , t  , where u t =
•
•
•
∂2 u
∂2 u
u
=
and
.
xx
∂t2
∂ x2
The wave equation: u tt  x , y , z , t =c 2 u xx  x , y , z , t u yy  x , y , z , t u zz  x , y , z , t  .
∂2
∂2
⋯
In general: u tt  x 1 , , x n , t =c 2  u  x 1 , , x n , t  , where =the Laplacian=
and
2
∂ x1
∂ x n2
∂2 u
∂2 u
 u=
⋯
.
2
∂ x1
∂ x n2
Diffusion equation: u t  x , t =c 2 u xx  x , t  .
In general: u t  x 1 , , x n , t =c 2  u  x 1 , , x n , t  .
Steady state: u t =0 .
∂2 u
∂2 u
⋯
=0 .
Laplacian equation:  u=
∂ x 12
∂ x n2
Initial Conditions and Boundary Values for Ordinary Differential Equations
d2 y
dy
d2 y
dy
=F t , y ,  , and think of y t  as the position of the particle,
F t , y ,  as
Consider
2
2 as acceleration, and
dt
dt
dt
dt
dy
force. The state/configuration space is  x 1 t  , x 2 t  , where x 1 t = y t  , x 2 t =
. Then the system of first order
dt
dx 1 dy
= =x 2 t 
dt
dt
equations is
.
dx 2 d 2 y
dy
= 2 =F t , y t  , =F t , x 1 t  , x 2 t 
dt
dt
dt
Theorem: Existence and Uniqueness of Solution
There exists one and only one solution x t = x 1 t  , , x n t  that satisfies x t 0 =x 0 t 0  where x 0 t 0  is the given intial
condition.
Quasi-Linear Partial Differential Equations
Definition: Quasi-Linear Partial Differential Equation
a  x , y , uu x  x , yb x , y , uu y  x , y=c  x , y , u (*) where a, b, c are given functions.
Claim
Let a and b be constant functions, and c=0 , so au x bu y =0 (1) . Then every solution u  x , y of (1) is of the form
u  x , y= f bx−ay for some function of one variable (ex: f =2 ⇒ u  x , y=bx−ay2
).
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APM346H1 Differential Equations
Uniqueness and Initial Conditions
For initial condition, we prescribe u along a given curve  x , so u  x , x=u 0  x is given. Note that when
u  x , y= f bx−ay , u x , y is constant along the line bx−ay=c . So if u 0  x= f bx−a  x , there is a unique f
provided that bx−a  x=c is not constant.
x
 . In conclusion,
Suppose that  x= Ax . Then u 0  x= f bx−aAx⇒ f  x=u 0 
b−aA
b
1) The solution u  x , y is unique for any u 0  x over the line y= Ax provided that A≠ .
a
b
2) When A=
then there are infinitely many solutions provided that u 0  x is constant. If u 0  x is not constant,
a
then there are no solutions.
Method of Characteristic
Define a vector field V  x , y , z=a  x , y , z , b x , y , z , c x , y , z . Normal direction at  x , y , z=u  x , y is
n=u x  x , y , u y  x , y ,−1 , but V⋅n =au x bu y c −1=0 because au x bu y =c . So V lies in the tangent plane.
dx
=a  x t  , y t  , z t 
dt
dy
If  x t  , y t  , z t  is a solution of (1) dt =b x t  , y t  , z t  such that  x 0 , y 0 , z 0 lies in z=u  x , y , ie
dz
dt
=c  x t  , y t  , z t 
u  x 0 , y 0=z 0 , then  x t  , y t  , z t  lies in z t =u  x t  , y t  .
x 0, x 0 = x 0
x t , s= x t , x 0 s
Suppose now that  x t , x 0  , y t , y 0  , z t , z 0  is any solution of (1) such that y 0, y 0 = y 0 where y t , s= y t , y 0  s . In
z 0, z 0 =z 0
z t , s=z t , z 0  s
u

x
,
y=z
t

x
,
y
,
s

x
,
y
most situations, we can solve for t and s in terms of x and y. Then
.
Note: When the Jacobian J =det
[ ]
∂x
∂t
∂y
∂t
∂x
∂s
∂y
∂s
≠0
, then we can solve for t and s in terms of x and y locally.
Note: If J =0 , then if u  x , y =z that contains u  x 0  s  , y 0  s  =z 0  s  satisfies
dz 0  s 
= c  x 0  s  , y 0  s  , z 0  s   , there are
ds
infinitely many solutions; if not, then there is no solution.
Second Order Equations
a  x , y ⋅u xx 2 b  x , y ⋅u xy c  x , y ⋅u yy d  x , y ⋅u x e  x , y ⋅u y  f  x , y ⋅u=0
functions.
(1), where a, b, c, d, e, f are given
Canonical Types
1. Hyperbolic type: b 2−ac0 .
2. Parabolic type: b 2−ac=0 .
3. Elliptic type: b 2 −ac0 .
Fact
[
]
=  x , y 
x  y
If we make a (one-to-one) change in variables =  x , y  and require that det   ≠0⇔ x eta y −xi y  x ≠0
x
y
there is a transformation such that (1) is transformed into:
1. u  lower order terms=0 in the hyperbolic type;
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, then
APM346H1 Differential Equations
2. u  lower order terms=0 in the parabolic type;
3. u u  lower order terms=0 in the elliptic type;
Special Case: a, b, c constants
Linear change of coordinates  x , y     , 
det
[
] [
x
x
xi y

=det x
y
x
= x y
given by = x y
]
xi y
= − ≠0
y
such that
.
Then (1) becomes Au 2 Bu Cu  lower order terms
A=a  22 b  c 2
, where B=a  2b    2c 
C =a  2 b  c 
.
1. In the hyperbolic case, choose =−b  b 2−ac  , =−b−  b 2−ac  , ==a
, =−b−   b 2 −ac  , ==a .
Then A=C =0, B≠0 .
2. In the parabolic case, choose ==−b , ==a
. Then B=C =0, A≠0 or A=B=0 ,C ≠0 .
c
−c
, =
, =0, =1
3. In the elliptic case, choose =
, then A=C ≠0, B=0 .
  ac−b2 
  ac−b2 
THE WAVE EQUATION
u tt  x , t =c 2 u xx  x , t  ,−∞x∞ with initial conditions u  x , 0 =  x  , u t  x , 0 =  x 
1
1
The solution is u  x , t =    xct   x−ct  
2
2c
∫

x−ct
.
  z  dz .
xct
DIFFUSION EQUATION
u t  x , y , z , t =k  u=k  u xx u yy u zz  .
In one dimension, u t  x , t =k u xx  x , t  is a parabolic type.
In One Dimension
u t  x , t =k u xx  x , t  ,−∞x∞ with given initial conditions u  x , 0 =  x  where   x  is a given function.
∞
1
The solution is u  x , t =
∫  ye
 4  k t −∞
− x− y 
4 kt
2
dy
−x 2
. If S  x , t =
Properties of the Kernel
The heat kernel/Gaussian/diffusion kernel S  x , t  has the following properties:
1. Symmetric: S  x , t =S −x , t  .
∞ , x=0
2. lim S  x , t ={
.
0, x≠0
t 0
∞
3.
∫ S  x , t  dx=1, ∀ t0
.
−∞
∞
4. lim ∫   x  S  x , t  dx=  0  , ∀  .
t  0 −∞
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∞
1
e 4 kt , t0 , then u  x , t = ∫   y  S  x− y , t  dy .
4

kt

−∞
APM346H1 Differential Equations
Evaluation Techniques
∞
Useful formula:
1
−∞
If =1, ' =0 , then
•
[
∞
∞
−∞
−∞
∫ '  y  S  x− y , t  dy= 2 kt ∫ y   y  S  x− y , t  dy−x ∫   y  S  x− y , t  dy
∞
∫ y S  x− y , t  dy= x . So if
−∞
∞
]
.
  x =x , then
∞
u  x , t = ∫   y  S  x− y , t  dy= ∫ y S  x− y , t  dy=x and u  x , 0 =x=  x  .
−∞
If = y , then
•
−∞
∞
∫y
2
S  x− y , t  dy=x 22 kt . So if   x =x 2 , then u  x , t =x 22 kt and u  x , 0 =x 2=  x  .
−∞
∞
2
•
If = y , then
∫ y 3 S  x− y , t  dy=x 36 ktx . So if
  x =x 3 , then u  x , t =x 36 ktx and u  x , 0 =x 3=  x  .
−∞
Theorem
2
  x  e−x =∞
Suppose that   x  is such that ∣lim
∣
x ∞
∞
∫   y  S  x− y , t  dy
∞
, then lim ∫   y  S  x− y , t  dy=  x  , ∀ x
t  0 −∞
. In that sense
is a solution with u  x , 0 =  x  .
−∞
The Maximum Principle
Let u  x , t  be a solution of u t =k u xx on a rectangle 0≤ x≤l , 0≤t≤T . The maximum of u  x , t  occurs only on the
part of the boundary { x , 0  : 0≤ x≤l }∪ { 0, t  : 0≤t≤T }∪ { l , t  : 0≤t≤T }
.
Theorem: Uniqueness of Solution
Suppose that we seek a solution u  x , t  that satisfies u  x , 0 =  x  , 0≤x≤l . Suppose further that u  x , t  satisfies
u  0, t =  t  and u  l , t =  t  , where   t  and   t  are prescribed functions. Then the solution is unique, i.e. there is at
most one solution.
T
0
l
DIFFUSION EQUATION ON HALF LINE
Equation: u t  x , t =k u xx  x , t  , 0 x∞ .
Initial data: u  x , 0 =  x  , x0 .
Boundary conditions:
• Dirichlet Condition: prescribe u  0, t =  t  (usually   t =0 ).
• Neumann Condition: prescribe u x  0, t =  t  (usually   t =0 ).
• Robin Condition: prescribe u  0, t a u x  0, t =0 .
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APM346H1 Differential Equations
Method of Solution: Dirichlet Boundary Condition
Take the case with u t  x , t =k u xx  x , t  , x0 , u  x , 0 =  x  , x0 , u  0, t =0, ∀ t≥0 .
We want to extend  to the entire line −∞ x∞ such that the solution u  x , t  induced by this extension satisfies
u  0, t =0 .
  x =  x  , x0 . Now, u  0, t =0 for all t0 iff 
 is an odd function ( 
 −x =−
  x  ).
Note that 
∞
∞
−∞
0
Then u  x , t = ∫ 
  y  S  x− y , t  dy=∫   y   S  x− y , t −S  x y , t   dy .
Method of Solution: Neumann Boundary Condition
Solve u t =k u xx , x0 , with initial data u  x , 0 =  x  , x0 and Neumann condition u x  0, t =0 .
If u  x , t  is even (i.e. u −x , t =−u  x , t  ), then u x  x , t  is odd (i.e. u x −x , t =−u x  x , t  ).
∞
∞
−∞
0
The solution is u  x , t = ∫ 
  y  S  x− y , t  dy=∫   y   S  x− y , t S  x y , t   dy .
WAVE EQUATION ON HALF LINE
Solve u tt =c 2 u xx , x0 .
Initial data u  x , 0 =  x  , and u t  x , 0 =0 for simplicity.
Dirichlet Boundary Condition
Dirichlet condition u  0, t =0 .
1
 . Then the solution is u  x , t =  
  xct 
  x−ct  
Extend  to odd function 
2
Note: u  x , t =−u  x , t  ⇒ u  0, t =0 .
.
WAVE EQUATION ON FINITE INTERVAL
Solve: u tt =c 2 u xx , 0x L .
Initial data:   x =u  x , 0  , 0 xL
and   x =u t  x , 0  , 0 xL .
Dirichlet Boundary Condition
Dirichlet condition: u  0, t =u  L , t =0 .
 so that u  x , t  is odd about x=0 (i.e. u −x , t =−u  x , t  ) and odd about x=L (i.e.
 and  to 
Extend  to 
x−ct
1
1
u  xL , t =−u  L−x , t  ). Then the solution is u  x , t =  
  xct 
  x−ct  
∫   z  dz .
2
2 c xct


Separation of Variables and Boundary Value Problems
Method of Separation of Variables
The method of separation of variables assumes that any solution u  x , t  can be written as u  x , t = X  x  T  t  .
Solutions
With the diffusion or wave equation, we need to solve X ' '  X =0 , where  is an unknown constant:
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APM346H1 Differential Equations
For 0 , X  x = A cos    x B sin    x  .
• For 0 , X  x = A cosh  − x B sinh  − x  .
• For =0 , X  x = AxB .
n
n 2 2
x
In the Dirichlet case ( u  0, t =u  L , t =0 ), 0 and n= 2 . So X n  x =sin
L
L
•
 
In the Neumann case ( u x  0, t =u x  L , t =0 ), =0 so X  x =constant ; or 0 and n=
X n  x =cos
 nL x .
Dirichlet Boundary Condition
For the wave equation u tt  x , t =c 2 u xx  x , t  , we have T n  t =a n cos
[
u n  x , t = X n  x  T n  t = a n cos
−k
n
n
For the diffusion equation u t  x , t =k u xx  x , t  , we have T  t =c e−k
n
n
u n  x , t = X n  x  T n  t =c n e
n 2 2
, so
L2
 cnL t b sin  cnL t  . So
 cnL t b sin  cnL t ]sin  nL x .
2
.
2
n 
t
2
L
sin
2
2
n 
t
L2
. So
 nL x .
Neumann Boundary Condition
cn 
cn 
Wave equation: u n  x , t = a n cos
t b n sin
t
L
L
[

−k
Diffusion equation: u n  x , t =c n e
n 2 2
t
L2

 ]cos nL x .
n
cos 
x .
L 
Mixed Boundary Condition
Mixed boundary condition u  0, t =u x  L , t =0 , then X  0 = X '  L =0 .
  2 n1 
We have  n=
.
2L
Robin Condition
Take u x  0, t −h u  0, t =0
•
''
and u x  L , t =0 . We have X  X =0 .
Assume 0 . Then X  x = A cos    x B sin    x 
, and we get tan    L =
h
. Setting y=   L0 , we get

Lh c
= a transcendental equation. On y0 , we get infinitely many solutions y 1 y 2⋯⇒ 12 ⋯
y y
with the difference approaching  .
c
• Assume 0 . Then X  x = A cosh  − x B sinh  − x  , and setting y= − L0 we get tanh y=− a
y
transcendental equation. We get no solution.
So there are infinitely many eigenvalues 12⋯ with corresponding eigenfunctions X 1  x  , X 2  x  , .
tan y=
VECTOR SPACES: INTRODUCTION TO FOURIER SERIES
6 of 15
,
APM346H1 Differential Equations
Let V n be the space of all linear combinations of f =b1 sin
Define L : V n  V n L  f =−
 
 L xb sin  2L x⋯b sin  nL x
2
n
k 2 2
k
d2 f
=
b
sin
∑
k
2
2
L
d x k =1
L
2
[
. Then the matrix of L relative to this basis
n
2
0
2
⋮
2
2 / L
2
⋯
0
.
.
{  L x , v =sin  2L x , , v =sin  nL x }
 L  0 ⋯
0
 2L  0
Choose basis: v 1= sin
n
⋱
0
⋮
0
n
L
 
2
]
 
2
is a diagonal matrix since L  v k =
k
vk .
L
∞
Let n  ∞ and consider the space of functions f on 0≤ xL which can be written as f  x =∑ b k sin
k =1
L
Fourier coefficients b k =
 
k
2
∫ f  x  sin L x dx k =1, 2, 3,
L 0
of f relative to X n .
FULL FOURIER SERIES
Definition
Let −LxL . The full Fourier series of f  x  is
 
Coefficients
The coefficients are uniquely determined from orthogonality of functions cos
.
 nL sin  mL dx={ 0L n≠m
n=m
n
m
∫ sin  L  cos  L  dx=0 .
n
m
.
∫ cos  L  cos  L  dx={ 0L n≠m
n=m
L
•
∫ sin
−L
L
•
−L
L
•
−L
These relations imply that:
L
n
1
a
=
•
∫ f  x  cos L dx n=0, 1, 2,
n
L −L
 
n
1
b = ∫ f  x  sin 
dx n=0, 1, 2, .
L
L 
.
L
•
 
∞
n
n
1
a 0 ∑ a n cos
b n sin
2
L
L
n=1
n
−L
7 of 15
 nL 
..
and sin
 nL  :
 kL x 
for
APM346H1 Differential Equations
Relation To Differential Equations
Take 0≤x≤L .
L
2
Dirichlet Condition: Take f  x  an odd extension of   x  . Then a n=0 and b n= ∫ f  x  sin
L 0
L
Neumann Condition: Take f  x  an even extension of   x  . Then a n=
 
 nL dx n=1, 2, .
n
2
f  x  cos
dx n=0, 1, 2, and b n=0 .
∫
L 0
L
GENERAL EIGENVALUES AND EIGENFUNCTIONS
X ' '  X =0
on 0≤x≤L .
  and X  x =sin  nL x .
n
n
2. If X  0 = X  L =0 , then  =
and X  x =cos 
x .
L 
L 
1. If X  0 = X  L =0 , then n=
n
L
2
n
2
'
'
n
n
3. If X  0 −hX  0 =0 and X  L =0 , then 12⋯ (eigenvalues) and X 1, X 2,  (eigenfunctions).
2
2n
4. If X  0 = X  L =0 and X '  0 = X '  L =0 , then =0 and X  x =constant or n=
and
L
2n
2 n
X n  x = An sin
x B n cos
x where An and B n are arbitrary constants.
L
L
'
'



 

General Boundary Conditions
Solve X ' '  X =0 , a≤ x≤b subject to the boundary conditions
constants 1, , n , 1,  , n
1 X  a  2 X  b 3 X '  a  4 X '  b =0
1 X  a 2 X  b 3 X '  a 4 X '  b =0
for some
.
Definition: Symmetric Boundary Conditions
Let f and g be any functions that satisfies the above boundary condition. Then conditions are called symmetric if
'
'
'
'
f '  x  g  x − f  x  g '  x ∣x=b
.
x=a =0⇔ f  b  g  b − f  b  g  b − f  a  g  a  − f  a  g  a  =0
Fact
Conditions 1 to 4 are symmetric.
Theorem
Suppose that X n and X m are eigenfunctions on [ a , b ] that corresponds to distinct eigenvalues n and m ( n ≠m ),
and suppose that the boundary conditions are symmetric. Then X n and X m are orthogonal in the sense that
b
∫ X n  x  X m  x  dx=0
.
a
HILBERT SPACE
Basic Space
{
b
L 2 [ a , b ]= f : [ a , b ]  ℝ |∫ f 2  x  dx∞
a
}
.
8 of 15
APM346H1 Differential Equations
Fact
L 2 [ a , b ] is a vector space.
Inner Product
b
Take f and g in L . Then define the inner product to be  f , g =∫ f  x  g  x  dx .
2
a
Norm

Define ∥ f ∥=
b
∫ f 2  x  dx
a

1
2
1
= f , f  2
to be the norm of f .
Cauchy-Schwartz Inequality
∣ f , g ∣≤∥ f ∥∥g∥
Note: This implies
or
∣
∣∣
b
b
∣∣
∣
1 b
1
∫ f  x  g  x  dx ≤ ∫ f 2  x  dx 2 ∫ g 2  x  dx 2 .
a
a
a
∣∥ ff ∥ , ∥gg∥∣≤1 , so define cos =∣∥ ff ∥ , ∥gg∥∣ .
Definition: Convergence
2
∥ f n− f ∥=0 .
{ f n }∈ L is said to converge to f if nlim
∞
Definition: Cauchy Sequence
2
{ f n }∈ L is called a Cauchy Sequence if ∥ f n− f m∥ 0 as n , m  ∞ .
Basic Properties of Inner Product and Norm
1. Symmetric:  f , g = g , f  .
2. Bilinear:  f ,  g h =  f , g   f , h  for  , ∈ℝ and f , g , h∈ L 2 .
b
3.  f , f ≥0 ∀ f ∈L 2 ; if  f , f =∫ f 2  x  dx=0 then f =0 “almost everywhere”.
a
4.
L 2 is complete in the sense that any Cauchy sequence in L 2 converges to an element in L2 .
Definition: Hilbert Space
Any vector space H with an inner product  ,  that satisfies properties 1 to 4 is called a Hilbert space.
Theorem
If X 1, X 2,  , X n , are the eigenfunctions corresponding to symmetric boundary problem, then the Fourier series of any
function f converges to f in L 2 norm.
LEAST SQUARE APPROXIMATION
Let V n denote the linear span of X 1, X 2,  , X n (i.e. f ∈V n ⇔ f =1 X 1 2 X 2⋯ n X n
Problem
9 of 15
.
APM346H1 Differential Equations
Question: Let f ∈ L 2 . For which values of 1,  2,  ,  n
 f , X i
i=1,, n .
Answer: i =
∥ X i∥
is the distance
∥ f −∑ i X i∥
minimum?
CONVERGENCE OF FOURIER SERIES
Theorem
N
The Fourier series relative to X 1, X 2,  of any element f ∈ L 2 converges to f . That is, if S N =∑  f i , X i  X i , then
i=1
lim ∥S N − f ∥=0 .
N ∞
Definition: Piecewise Continuous
A function is piecewise continuous if it is continuous at all but a finite number of points. At a point of discontinuity f has
both a right and a left limit (ie f has a jump discontinuity).
f  c + =lim f  x 
f  c - =lim f  x 
So if c is a point of discontinuity of f , then both
and
exist.
xc
xc
xc
xc
Theorem: Point-wise Convergence of Fourier Series
Assume f is such that:
•
f is periodic of period 2  .
•
f and its derivative f ' are “piecewise continuous”.
1
+
Then lim S N  x =  f  x  f  x  .
2
n∞
Note: If f is continuous at x , then f  x + = f  x - = f  x  , so lim S N  x = f  x  .
n∞
Auxiliary Results
2
g , X k
1. Bessel's Inequality: ∥g∥ ≥∑
where X 1, X 2,  are eigenfunctions on [ a , b ] with symmetric boundary
2
k =1 ∥ X k∥
2
g , X k
2
lim
=0 .
values and g ∈L [ a , b ] . This implies that
2
k ∞ ∥X ∥
k
2
∞
N
2. Let K N   =12 ∑ cos k 
3.

k =1

. Then
pi
pi
−
−
1
∫ K N    d =2 ⇔ 2
∫ K N    d =1 .
1
sin N  
2
K N   =
.

sin
2

Definition: Uniform Convergence
max ∣ f n  x − f  x ∣=0
f n converges to f uniformly if nlim
 ∞ a xb
.
10 of 15
APM346H1 Differential Equations
Harmonic Functions and Laplace's Equation
LAPLACE'S EQUATION
In n dimensions,
 u=div  grad u =
∂2 u
∂2 u
⋯ 2 =0 .
2
∂ x1
∂ xn
Notes:
•
•
 is the Laplacian, and it is an operator that acts on functions of n variables.
∂u
∂u
, ,
The gradient of a scalar function u  x 1,  , x n  is grad u=∇ u=
∂ x1
∂ xn


.
•
The divergence of a vector field V  x 1,  , x n = V 1  x 1,  , x n  , , V n  x 1,  , x n   is div  V =
•
∂u
∂u
, ,
When V =∇ u=
∂ x1
∂ xn


∂V 1
∂V n
⋯
.
∂ x1
∂ xn
, then  u=div  grad u  .
Definition: Harmonic
Any solution of Laplace's equation is called harmonic.
Properties
•
•
d2u
=0 , so u  x = AxB .
d x2
n=2 : Connection to complex functions f  z =u  x , y iv  x , y  . Then f analytic (can be expressed as a Taylor
series, i.e. differentiable) implies u and v are harmonic.
n=1 :
MAXIMUM/MINIMUM PRINCIPLE
Definitions
n
1. D is an open subset of ℝ if for all x ∈D , there exists r0 such that for all y∈D , ∥x −y∥r .
2. ∂ D is the boundary of D . A point b is a boundary point if for all 0 , B= {∥x −b∥ }
has non empty
intersection with both D and the complement of D in ℝ n .
3. D is connected if there exists a polynomial curve joining any two points in D and is lying in D .
4. D is bounded if it is contained in some ball B={x |∥x∥R } 0R∞ .
Maximum/Minimum Principle
Assume D to be an open, connected subset of ℝ n such that D∪∂ D is bounded. Let u be any solution of (the Laplace
∂2 u
∂2 u
⋯
=0 equation) in D such that u is defined and continuous in D∪∂ D . Then:
∂ x 12
∂ x 2n
• Maximum Principle: u  x ≤max u  x  ∀ x ∈D .
∂D
• Minimum Principle: u  x ≥min u  x  ∀ x ∈D .
∂D
That is, u attains its maximum/minimum on ∂ D .
BOUNDARY VALUE PROBLEMS
11 of 15
APM346H1 Differential Equations
Dirichlet Problem
∂2 u
∂2 u
⋯
=0 subject to u |∂ D=  x  (given).
∂ x 12
∂ x 2n
By the Maximum/Minimum Principle, the solution of the Dirichlet problem is unique.
Neumann Problem
∂2 u
∂2 u
∂u
⋯
=0 subject to
| =  x  (given). Here, n is the external normal.
∂ n ∂ D
∂ x 12
∂ x 2n
Robin/Mixed Problem
∂2 u
∂2 u
∂u
⋯
=0 subject to
a  x  u  x =k  x  (given).
2
2
∂ n
∂ x1
∂ xn
BASIC PROPERTY
Solutions to the Laplace equation are invariant under rigid motions   x =T  x R  x 
• T  x =
a x is a translation,
•
R  x = A x ( AT = A−1 det A=±1 ) is a rotation.
RECTANGULAR HARMONICS
{
u  x , 0 = f  x 
u  a , y =g  x 
with ∂ D=
u  x , b =h  x 
u  0, y =i  x 
∂2 u ∂2 u
 2 =0 on D= { x , y  : 0≤x≤a , 0≤ y≤b }
2
∂x ∂y
functions.
}
, where
where f , g , h , i are given
Separation of Variables
We assume u  x , y = X  x  Y  y  .
u  x , 0 = f  x 
n
n
n
yb n sinh
y sin
x .
u  x , b =h  x 
When ∂ D=
, then u n  x , y = a n cosh
a
a
a




u 0, y =u a , y =0
u  0, y =g  x 
n
n
n
u  a , y =i  x 
When ∂ D=
, then u n  x , y = a n cosh
xb n sinh
x sin
y.
b
b
b
u  x , 0 =u  x , b =0
u  x , 0 = f  x 
u  x , 0 = f  x 
u  0, y =g  x 
u  a , y =g  x  =

u  x , b =h  x 
u  a , y =i  x 
Note: ∂ D=
.
u  x , b =h  x 
u  0, y =u  a , y =0
u  x , 0 =u  x , b =0
u  0, y =i  x 
{
{
{
∞
}{
}
}




}{
}
Note: u  x , y =∑ u n  x , y  .
n=1
Case 1
Suppose that we want the solution with ∂ D=
u  x , 0 = f  x 
.
{u  x , b =u
 0, y =u  a , y =0 }
12 of 15
APM346H1 Differential Equations
 
a
The sine Fourier coefficients of f are a n=
∞

Since u  x , b =0 ⇔ ∑ a n cosh
n=1
n
2
∫ f  z  sin a z dz .
a 0

n
n
n
n
n
bb n sinh
b sin
x=0  a n cosh
bb n sinh
b=0
a
a
a
a
a
, so
n
b
a
bn=−
.
n
sinh
b
a
a n cosh
Then
∞


n
n
n
yb n sinh
y sin
x
a
a
a
n=1
n
cosh
b
∞
n
n
n
a
=∑ a n cosh
y−
sinh
y sin
x
a
a
a .
n

n=1
sinh
b
a
∞
an
n
n
 b− y  sin
=∑
sinh
x
a
a
n

n=1
sinh
y
a
u  x , y =∑ a n cosh





LAPLACE'S EQUATION ON CIRCULAR REGIONS
1. Annulus: D= { r ,   ,−≤ , a≤r≤b }
, ∂ D={ a ,   ,−≤ }∪{ b ,   ,−≤ } .
2. Disk: D= { r ,  ,−≤ , 0≤r≤b } , ∂ D={ b ,   ,−≤ } .
3. Wedge: D= { r ,   , 0≤≤ , 0≤r≤b } .
Polar Coordinates and Separation of Variables
x=r cos 
1
1
Using polar coordinates y=r sin  , u xx u yy =0 becomes u rr  u r  2 u  =0 .
r
r
Assuming u  r , =R  r     
2
, we get the two equations r R ' ' r R ' − R=0
and ' ' =0
.
Annulus
Eigenfunctions:
n
−n
• When n≠0 , n   = An cos n B n sin n  and R n  r =C n r D n r .
• When n=0 , 0   = An and R0  r =c 0c 1 ln r .
∞
∞
n=0
n=1
n
−n
So the solution is u  r , =∑ R n  r  n   =c 0c 1 ln r∑  C n r  D n r   An cos n B n sin n  
are the Fourier coefficients of the boundary conditions u  a , = f   
. The coefficients
and u  b ,  =g    .
Disk
The usual assumption is that u  0,   bounded. This forces c 1=D n=0 . So the solution is
∞
∞
1
u  r ,  =c 0 ∑ C n r n  An cos n B n sin n  = a 0 ∑ r n  a n cos n b n sin n   , where a n and b n are the Fourier
2
n=1
n=1
coefficients determined by the boundary condition u  b ,  = f    .
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APM346H1 Differential Equations
Wedge
Consider the special case that u  r , 0 =u  r ,  =0
∞
The solution is u  r ,  =∑ a n r
n

n=1
and u  b ,  = f    .

n
2
n
sin
 , where a n = b n /  ∫ f    sin   d 
0

is the Fourier coefficient
determined by u  b ,  = f    .
POISSON FORMULA AND POISSON KERNEL
Poisson Formula
On the disk D= { r ,   ,−≤ , 0≤r≤b } with u  b ,  = f    ,
u  r ,  =
1
2

[

∫ f 
−

∞
12 ∑
n=1
  cos n −d ]
r
a
n
= ∫ f    P  −  d 
−

 b 2−r 2  f   
d
2
2
− b −2 b r cos  − r
=∫

∞

1
r
where P  − = 2  12 ∑ a
n=1
n

cos n  −  =
1
b 2−r 2
2
2  b −2 b r cos  − r 2
.
Poisson Kernel
1
a 2−r 2
P a  r ,  =
.
2  a 2−2 a r cos r 2
Basic Properties
2
1.
1
2
a 2−r 2
∫ P a  r ,−  d =1⇔ 2  ∫ a 2−2 a r cos  − r 2 d =1 . In this case
0
u  r ,  =1 ∀ r ,  also.
P a  r , ={ 0 ≠0
2. lim
r a
∞ =0 .
u  a , = f   =1 , but
0
2
3. lim ∫ f    P a  r , −  d = f    whenever f is a continuous function of  .
r a 0
4. Averaging Property of Harmonic Functions: x= y=0⇔ r=0 , so
2
2
 a 2−02  f   
1
1
u  0, 0  =u
d =
 0,   = ∫ 2
∫ f  d 


2  0 a −2 a 0 cos  − 02
2 0
cartesian
polar
is the average value of f .
Consequences of Poisson Representation
1. A harmonic function u defined on some domain D cannot attain a maximum (nor minimum) in the interior of D .
Here the interior of D are the points p in D such that there exists a disk centered at p that is entirely contained in
D.
2. u  r ,   has partial derivatives of all orders, even when f is only continuous.
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APM346H1 Differential Equations
2
u  r ,  =∫ f    P a  r ,−  d ⇒
0
2
∂u
∂
=∫ f   
P  r ,−  d  .
∂r 0
∂r a
15 of 15
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