Download Steady State Temperature in a Sphere 4

Lecture 13
Contents
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•
•
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Partial Differential Equations
Sturm-Liuville Problem
Laplace Equation for 3D sphere
Legandre Polynomials
Math for CS
Lecture 13
1
Second order PDE’s
The second order quasi-linear equation is defined by:
 2u
 2u
 2u
u u
A( x, y ) 2  B( x, y )
 C ( x, y ) 2   (u, , , x, y )
x
xy
y
x y
(1)
It is called ‘quasi-linear’ because the left hand side (LHS) is linear in the dependent variable,
but the RHS function may not be. In the short-hand notation this equation looks:
Auxx  Buxy  Cu yy   ( u, u x , u y , x , y )
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Lecture 13
2
Hyperbolic, Parabolic and Elliptic PDE’s
Auxx  Buxy  Cu yy   ( u, u x , u y , x , y )
This PDE can be hyperbolic, parabolic, or elliptic, depending on the sign of the term
B2-4AC (which can vary with x and y, if A, B, and C are not constants):
 0 hyperbolic

B 2  4 AC    0 parabolic
 0
elliptic

(2)
For the motivations of this notation, we consider the simple forms of each of 3 cases. For this
let B=0 and RHS be constant C:
 2
2 
 2  2 u  C
y 
 x
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hyperbolic ;
 2 
 2 u  C
 x 
Lecture 13
parabolic ;
 2
2 
 2  2 u  C elliptic
y 
 x
3
Motivation for the notation
 2
2 
 2  2 u  C
y 
 x
hyperbolic ;
 2 
 2 u  C
 x 
parabolic ;
 2
2 
 2  2 u  C elliptic
y 
 x
parabola ;
u
The simplest solutions for these equations are
u

C 2
x  y2
4

hyperbola ;
u
C 2
x
2


C 2
x  y 2  C ellipse
4
For more complex equation (1) the type change as a function of coordinate, however the
local properties of the solution still depend on the sign of discrimintor B2-4AC.
Math for CS
Lecture 13
4
Examples
Application
Equation
Coefficients
B2-4AC
Class
Wave Equation
utt   2 u xx
A=1,B=0,C=-α2
>0
Hyperbolic
Heat Equation
ut   2 u xx
A=1,B=0,C=-α2
0
Parabolic
Poisson’s Equation
u xx  u yy  f ( x , y )
A=1,B=0,C=1
<0
Elliptic
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Lecture 13
5
Special Coordinate systems
When solving a second order PDE’s in special coordinate systems, the specific
representation of Laplacian arises:
In cylindrical:
 2 f 1 f
1 2 f 2 f
f 

 2
 2
2
2

   
z
And in Spherical:
1   2 f 
1
 
f 
1
2 f
f  2
r
 2
 sin
 2 2
r r  r  r sin  
  r sin   2
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Lecture 13
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Special Coordinate systems
In these cases, the variable separation approach also facilitates the solution. In the Euclidian
case the eigenfunctions were Fourier series. Here, after the substitution
f  P (  )  ( )  Z ( z )
f  P (  )  ( )  ( )
The differential equations arise, which solutions are special functions like Legendre
polinomials or Bessel functions.
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Lecture 13
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Sturm-Liuville Problem
The special functions, which arise in these homogeneous Boundary Value Problems (BVPs) with homogeneous
boundary conditions (BCs) are mostly special cases of Sturm-Liouville Problem, given by:
d 
d

r
(
x
)
y
(
x
)
  q( x )  p( x ) y( x )  0
dx 
dx

On the interval a≤x≤b, with the homogeneous boundary conditions
c1 y(a )  c2 y(a )  0
k1 y(b)  k2 y(b)  0
The values λn, that yield the nontrivial solutions are called eigenvalues, and the corresponding solutions yn(x) are
eigenfunctions.
The set of eigenfunctions, {yn(x)}, form an orthogonal system with respect to the weight function, p(x), over the
interval.
If p(x), q(x), and r(x) are real, the eigenvalues are also real
Math for CS
Lecture 13
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String Equation
d 
d

r
(
x
)
y
(
x
)
  q( x )  p( x ) y( x )  0
dx 
dx

Consider the case r(x)=1, p(x)=1, q(x)=0:
y  y  0
And boundary conditions y(0)=y(π)=0.
Case 1 - Negative Eigenvalues: For this case we try λ=-ν2. With this substitution, the
original ODE becomes:
y   2 y  0
This is just a simple, constant coefficient, second-order ODE with characteristic equation
r1, 2  
and roots
Thus, the general solution for the negative eigenvalue assumption is y( x )  A1e x  A2e  x
The boundary conditions give:
y(0)  A1  A2  0 ;
y( )  A1e  A2e   0 
A1  A2  0
Therefore, the nontrivial solution is only for non-negative eigenvalues, which is a familiar
Fourier series.
Math for CS
Lecture 13
9
Steady State Temperature in a Sphere
Find the steady state temperature of a sphere of radius 1, when the temperature of upper
half is held at T=100 and the lower half at T=0. Inside the sphere, the temperature satisfies
the Laplace equation.
The Laplace equation in spherical coordinates is:
1   2 u 
1
 
u 
1
 2u
u  2
0
r

 sin

r r  r  r 2 sin  
  r 2 sin2   2
Substitute
u  R( r )( )( )
(4)
(3)
r2
and multiply by
:
R
1 d  2 R 
1
d 
 
1
 2
0
r

 sin

R dr  r   sin d 
   sin2   2
Math for CS
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Steady State Temperature in a Sphere 2
1 d  2 R 
1
d 
 
1
 2
0
r

 sin

R dr  r   sin d 
   sin2   2
If we multiply by sin2θ, the last term became a function of φ only, while the first two do not
depend on φ, therefore, the last term is a constant. It must be negative, since the meaningful
solutions must be 2π periodic.
1  2
 m 2 ,
2
 
 sin m
 
cos m
(5)
Now the equation can be rewritten as:
1 d  2 R 
1
d 
 
m2
0
r

 sin

2
R dr  r   sin d 
  sin 
Math for CS
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Steady State Temperature in a Sphere 3
1 d  2 R 
1
d 
 
m2
0
r

 sin

R dr  r   sin d 
  sin2 
The first term is a function of r, while the last two are functions of θ, therefore:
1 d  2 R 
r
k
R dr  r 
(6)
1 d 
 
m2
  k  0
 sin

sin d 
  sin 2 
Making the change x=cosθ, we obtain: dx=sin θdθ, and
d 
m2
2 d 
  k  0
 (1  x )

dx 
dx  1  x 2
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Steady State Temperature in a Sphere 4
d 
m2
2 d 
  k  0
 (1  x )

dx 
dx  1  x 2
This is called the equation for associated Legendre polynomials. It is in fact the specific
case of Sturm-Liuville problem
d 
d

r
(
x
)
y
(
x
)
 q( x )  p( x ) y( x )  0


dx 
dx

When
m2
r ( x )  1  x ; q( x )  
;
p( x )  1;
2
1 x
It has a solutions for k=l(l+1), which is the Legandre’s polynoms:
2
lm
1
2 m/2 d
2
l
Pl ( x )  l (1  x )
(
x

1
)
2 l!
dx l  m
m
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Lecture 13
k
(7)
13
Steady State Temperature in a Sphere 5
The equation (6)
1 d  2 R 
r
  l ( l  1)
R dr  r 
Has the solutions
 rl
R    l 1
r
However, the solution with negative degree is not physical, since it is singular in the center
of the sphere. Combining all together into (4), we obtain:
 sin m 
u  r l Pl m (cos  )
cos m 
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Lecture 13
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Steady State Temperature in a Sphere 6
Now, since the boundary condition does not depend on φ, the solution reduces to m=0:

u   cl r l Pl m (cos  )
l 0
The coefficients cl are determined to satisfy the boundary conditions at r=1:


100
0




2
  r l Pl m (cos  )  

l 0
 0
 
2

0  cos   1

ur 1
 1  cos   0

ur 1   cl r l Pl m ( x )  100 f ( x )
l 0
,where f(x)=0, -1<x<0 and f(x)=1, 0<x<1.
Math for CS
Lecture 13
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Steady State Temperature in a Sphere 7

ur 1   cl Pl m ( x )  100 f ( x )
l 0
,where f(x)=0, -1<x<0 and f(x)=1, 0<x<1. For calculation of cl, we use the Rodriges
formula and normalization of Legendre’s polynomials (given here without proof)
2 lm
1 Pl ( x )Pm ( x )dx  2l  1
1
1 dl
2
l
Pl ( x )  l
(
x

1
)
2 l! dx l
1
1
 f ( x )P ( x )dx  c  P ( x ) dx
2
0
0
1
0
1
1

1
1

1
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0
1
or
0
1
c0  ;
2
2
0
f ( x )P1 ( x )dx  c1   P1 ( x ) dx
2
1
or
0
 xdx  c1 
0
1
f ( x )P2 ( x )dx  c1   P2 ( x ) dx
2
0
 dx  c
2
3
3
c1  ;
4
2
 3 2 1
x

dx

c



2
0  2
2
5
1
or
Lecture 13
c2  0;
16