NUMERICAL ANALYSIS PROBLEM DIFERENTIAL EQUATIONS II

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NUMERICAL ANALYSIS PROBLEM DIFFERENTIAL EQUATION II

6200


1.
Answer :





From the figure above, we have :
T T T T
T
T
T
min
T
min min
T
h
T
1 1

c s (c s) cot cot ,c h , : the radius of the inscribed
circle
2 2 2 2
Let the three int erior angles of T be a b c
h
1 1
cot cot
2 2 2 2
We have :
by a cons tant for all h
h
T then for som
α β
= + − = ρ + ρ = ρ
θ = α ≤ β ≤ γ ⇒ ≤ ≤
α β
⇒ = +
ρ
θ ≤ θ
⇔ θ ≤ α ≤ β ≤ γ θ
∀ ∈ Τ ≤ σ
ρ
T
T
min
T
T
min min
e
h

1 1
We have : cot cot
2 2 2 2
Because and cot is a decrea sing function so
h
1 1
cot cot
2 2 2 2
cot cot where cot
2 2 2
σ
α β
= +
ρ
θ ≤ α ≤ β
α α
≤ +
ρ
α θ θ
≤ ≤ σ =

2.
Answer :


a) Difference between regular and uniform grid

T
T T h
T

T
T
h
Regular : , such that h h h,where h maxh ,T
h
Uniform grid : , h max h , K 0
K
The regular grid is stronger than the uniform grid
b)The mesh in the figure includes both uniform and regular grid
≤ σ ∃α α ≤ ≤ = ∈ Τ
ρ
≤ σ = >


3.

Answer :

j
j j
1
j
1
j
1 1
j z
j
1
z z
j

h
a)Show that tan
x 2
h
tan
x 2
h h
tan dx tan .x
2 2
We know that :
1
h h
tan .x 1 x tan
2 2
h
This is true when we have known that z tan , 0
2
−
−
− −
−
∂
 
φ = θ
 
∂
 
∂
 
φ = θ

 
∂
 
   
⇒ φ = θ = θ
   
   
φ =
 
⇔ θ = ⇒ = θ
 
 
 
= θ
 
 
∫

( )
2
2
j
I
b)Consider the function u(x,y) y .
Its linear int erpo tant vanishes at z and equals
h / 4 at the two other vertices
Show that
h
u u
x 2tan

=
∂
− =
∂ θ


( )
( )
j
j
j
j j
I
2
2
z
I z
2
z
2
z z
j
Answer :
h
u u
x 2tan
h
With u(x,y) y (u)
4
h h

u u dx .x
2tan 2tan
We know that :
h
(u)
4
h h h
.x x tan
2tan 4 2
h
This is true when we have known that z tan , 0
2
From the parts above, we can see that
Genera
∂
− =
∂ θ
= ⇒ φ =
⇒ − = =
θ θ
φ =
⇔ = ⇒ = θ
θ
 
= θ
 
 
∫
lly, then the are of traingle T,which have creat
ed

by (0,h / 2);(0, h / 2) and (h / 2tan ,0) is− θ


j
j
2
z
2
0
z
1 h h 1 h h h
(u) . . tan . . tan tan
2 2 2 2 2 2 4
h
Hence, if (u) then tan 1 45
4
But, we have the shape regularity condition :
φ = θ + θ = θ
φ = θ = ⇒ θ =

T min
T
min
h
cot cot
2 2
1
tan
2
θ

θ
= ≤
ρ
≤
θ


min T
T
0
min
min
tan tan
2 h 2
45
tan 1
Thus,the tan may be cause large error
θ ρ
θ
⇔ ≤ =
θ ≤ θ =
θ ≤
θ




4.

"

Consider the BVP problem : find a periodic funct
ion u such that
u (x) f(x) on ( 1,1 )− = Ω = −


a) State the weak formulation a(u,v) (v)
=



Answer :

First. We multiply the equation by test function v(x)
[ ]
" 1
1 1
"
1 1
'
" '
1 1
" ' ' '
1 1
1
' ' ' '
1
u v fv on ( 1,1 ), v H ( 1,1 )
We compute the weak form by int egrating over 1,1
u v dx fv dx
Set p v dp v dx

dq u dx q u
1
So : u v dx u .v u v dx
1
u (1).v(1) u ( 1).v( 1) u v dx
Where we have used t
− −
− −
−
− = Ω = − ∈ −
−
− =
=
⇒
=
=
⇒
=
− = − + =
−
= − + − − +
∫ ∫
∫ ∫
∫
' '
1 1
' '
1 1
1
2

1
hat u( 1) u(1) 0 u ( 1) u (1) 0
Hence, a(u,v) u v dx , (v) fv dx
u( 1) u(1) 0 f L and f dx 0
so u c is a solution
− −
−
− = =
⇒
− = =
= =
− = = ∈ =
+
∫ ∫
∫


b) Use the Lax-Milgram theorem to prove the existence and stability of the solution
Recall
: The Lax–Milgram theorem
This is a formulation of the
Lax–Milgram theorem
which relies on properties of the
symmetric part of the bilinear form. It is not the most general form.
Let
V
be a Hilbert space and a bilinear form on
V
, which is
1. bounded: and

2. coercive:
there is a unique solution to the equation
a
(
u
,
v
) =
l
(
v
)


Answer :

+ We will prove the property 1 : bounded:
We have
( )
1 1
' ' 1
1 1
a(u,v) u v dx , (v) fv dx ,H H
− −
= = = Ω
∫ ∫

1 1
' ' ' '
1 1

1/2 1/2
1 1
2 2
' '
1 1
H H
a(u,v) u v dx u v dx
u dx v dx (applying to Cauchy Schazt inequality)
C. u . v ,C 1
− −
− −
= ≤
   
≤ −
   
   
≤ =
∫ ∫
∫ ∫

1/2 1/2
1 1 1
2 2
1 1 1
(u) fv dx f dx v dx
− − −
   
= ≤
   
   

∫ ∫ ∫

2 2
L . H L .
H
f v ,C f
C. v
≤ =
≤


+ We continue to prove the property 2 : coercive:


( ) ( )
( )
( )
( )
( )
( )
(
)
( )
2 2
2
2
1
L L
2
x x

2
2 ' 2 '
1 1
2
2 2
2
2
2 '
L
L
We will prove the Poincare inequality :
u C. u , u H
We see that
u (x) u( 1) u (t)dt 2 u ( 1) u (t) dt
for 1 x 1
( We have used Cauchy 's inequality : a b 2 a b )
We have : u 2. u ( 1) u
For
Ω Ω
− −
Ω
Ω
≤ ∇ ∀ ∈ Ω
   
= − + ≤ − +
   
   
− < <
+ ≤ +
≤ − +

∫ ∫
( )
( )
( )
( )
( )
( )
( )
( ) ( ) ( )
(
)
( )
( )
( )
( ) ( )
2
2
2
1
2
2 2 2 2
L
L L L L
2
L
1
1
L
2
2

'
L
L
2 2 2 2
2 2
' ' ' '
H
2
' ' '
2
H
1 1
u H with u( 1) u(1) 0 u 0 , is boundary
u 2 u
Poincare inequality means :
u u u u 2 u 3 u 3a u,u
Where a(u,u) uu dx u
1
Thus, a u,u u
3
! u H such that a u,u (u), v H
Ω
Ω Ω Ω Ω
Ω
Γ
Ω
Ω
Ω
Ω
Ω

∈ Ω − = =
⇒
= Γ
⇒
≤
≤ + ≤ + = =
= =
≥
⇒
∃ ∈ Ω = ∀ ∈
∫

( )
Ω

c)
0o 1 2 n
n h
n n n n n
ij ji j i i j
n n
h i i i i n n i i
i 1 i 1
i n
n
n 1 n i i
i 1
i n
h 1 n 2
Partition as 1 x x x . . . x 1

u H H
a(u ,v ) (v ) , v H
The stiffness matrix is symmetric
if A A a( , ) a( , )
We have u u u u u
u ( ) u
So H span , , .
= ≠
≠
≠
≠
Ω − = < < < < =
∈ ⊂
= ∀ ∈
= ⇔ φ φ = φ φ
= ϕ = ϕ + ϕ + ϕ
= ϕ + ϕ + ϕ
= ϕ + ϕ ϕ
∑ ∑
∑

{ }
n 1
. . ,
−
ϕ

ij ij ij
i 1,j ij i 1,j i,j 1 ij i,j 1
u u f
u 4u u u 4u u
6 6
+ + − +
+ =
+ + + +
⇔ +


Thus,
2
ij
1 4 1
h
A 4 16 4
36
1 4 1
 
 
=
 
 
 
. This stiffness matrix is symmetric

d)
Set up a set equations based on finite differenc

es










1
0 1
1 2 3 4
j j
i i
ij
From the figure above ,we have the equations :
(1 y)(1 x) ; x(1 y) ; xy ; (1 x)y
Using the Intergrating to compute
K h ( )dx dy ; [0,h] [0,h]
x x y y
ϕ = − − ϕ = − ϕ = ϕ = −
∂ϕ ∂ϕ
∂ϕ ∂ϕ
= + ×
∂ ∂ ∂ ∂
∫


1
0 1
1 1 1 1
11
3 3
1 1
2 2
0 0
K ( )dxdy ; [0,1] [0,1]
x x y y
1 1
(1 y) (1 x) 2
h (1 y) (1 x) dxdy h h
3 0 3 0 3
∂ϕ ∂ϕ ∂ϕ ∂ϕ
= + = ×
∂ ∂ ∂ ∂
 
− −
 
= − + − = + =
 
 
 
∫
∫ ∫



1
0 1
2 2 2 2
22
3 3
1 1
2 2
0 0
K h ( )dxdy
x x y y
1 1
(1 y) x 2
h (1 y) x dxdy h h
3 0 3 0 3
∂ϕ ∂ϕ ∂ϕ ∂ϕ
= +
∂ ∂ ∂ ∂
 
−
 
= − + = + + =
 
 
 
∫
∫ ∫


1
0 1

3 3 3 3
33
3 3
1 1
2 2
0 0
K h ( )dxdy
x x y y
1 1
(1 y) x 2
h (1 y) x dx dy h h
3 0 3 0 3
∧ ∧
∂ϕ ∂ϕ ∂ϕ ∂ϕ
= +
∂ ∂ ∂ ∂
 
−
 
= − + = + + =
 
 
 
∫
∫ ∫


1
0 1
1 2 1 2

12 21
3 2 3
1 1
2
0 0
K K ( )dxdy ; [0,1] [0,1]
x x y y
1 1
(1 y) x x 1
h (1 y) (1 x)x dxdy h h
3 0 2 3 0 2
∂ϕ ∂ϕ ∂ϕ ∂ϕ
= = + = ×
∂ ∂ ∂ ∂
 
 
−
 
= − − + − = − + − =
 
 
 
 
 
∫
∫ ∫



[ ]

1
0 1
3 3
1 1
13 31
2 3 2 3
1 1
0 0
K K h ( )dxdy
x x y y
1 1 1 1
y y x x 1
h (1 y)y (1 x)x dx dy h h
2 0 3 0 2 0 3 0 3
∂ϕ ∂ϕ
∂ϕ ∂ϕ
= = +
∂ ∂ ∂ ∂
 
= − − − − = − + − + = −
 
 
∫
∫ ∫


1
0 1
3 3
2 2

23 32
2 3 3
1 1
2
0 0
K K h ( )dxdy
x x y y
1 1
y y x 1
h (1 y)y x dxdy h h
2 3 0 3 0 2
∂ϕ ∂ϕ
∂ϕ ∂ϕ
= = +
∂ ∂ ∂ ∂
 
 
 
= − − = − + + =
 
 
 
 
 
∫
∫ ∫


The element stiffness matrix

2 1 1
3 2 3
1 2 1
K h
2 3 2
1 1 2
3 2 3
 
−
 
 
 
⇒ =
 
 
 
−
 
 


We have the matrix four the aparts so :
(
)
( )
( ) ( )

( ) ( )
ij i j
1 1
1 2 1 1 2 3
1 3 3 1
A a ,
2 8
a , 4
3 3
1
a , . d x d y 2 1 a ,
2
1
a , a ,
3
1 3 1
1
A 3 8 3
3
1 3 1
= ϕ ϕ
⇒ ϕ ϕ = × =
⇒ ϕ ϕ = ∇ ϕ ∇ ϕ = × = = ϕ ϕ
ϕ ϕ = ϕ ϕ = −
− −
 
 
⇒ =
 
 

− −
 
∫

e)
We have
1
' '
1
1
1
a(u,v) u (x)v (x)dx
(v) f(x)v(x)dx
a(.,.) is binear form
−
−
=
=
∫
∫

0 1 n 1 n
1 x x . . . x x 1
−
− = < < < < <

{ }
h
h
h
n n
h j j 0 0 n n j j
j 0 j 0
n 1 2 n 1 0 n
V is the space of all continuous functions and be subspace of V
V have the value 0 at the endpoint s of [ 1,1] and ha
ve n 1 dimensions
Let u is solution of the subspace
u
Basic f , , . . . , ,
We h
= ≠
−
− −
= α ϕ = α ϕ + α ϕ + α ϕ
= ϕ ϕ ϕ ϕ + ϕ
∑ ∑
2
h h
h
'
h h
h h
' ' "
L ( 1,1)
ave a(u ,v) (v) , v V

u is a linear approximation of the exact solution u
By Ce a's lemma , C 0 such that
u u C u v , v V
Choose u u in V , K depend on the endpoints 1 and 1
such that u (x) ( x) x Kh u ; x [ 1,1 ]
h is the lar
−
= ∀ ∈
∃ >
− ≤ − ∀ ∈
= π ∃ −
− π ≤ ∈ −

2
i i 1
"
h
L ( 1,1)
'
h
gest length of the subint ervals [x ,x ] in the partition
u u Ch u
Thus,Ce a's lemma can be applied along the same l
ines to devive error
estimates for finite element and using higher o
rder polynomials of the
subspace V
+
−
⇒ − ≤

Bonus Problem 2 :





( )
1 2 3 4
1
j j
i i
ij
0
1
2 2
11
0
13
From the figure above ,we have the equations :
N (x,y) (1 y)(1 x) ; N (x,y) x(1 y) ; N (x,y) xy ;N (x,
y) (1 x)y
The stiffness matrix form :
N N
N N
K dxdy i, j 1,2 ,3 ,4
x x y y
K (1 y) (1 x) dxdy 2 / 3
K

= − − = − = = −
∂ ∂
 
∂ ∂
= + =
 
∂ ∂ ∂ ∂
 
= − + − =
∫
∫
( )
( )
( )
( )
( )
( )
( )
1
2
31
0
1
2 2
22
0
1
2
23 32
0

1
24 42
0
1
2 2
33
0
1
2
34 43
0
1
2 2
44
0
K (1 y) (1 x)x dxdy 1/ 6
K (1 y) x dxdy 2 / 3
K K (1 y)y x dxdy 1/ 6
K K (1 y)y x(1 x) dxdy 1/ 3
K y x dxdy 2 / 3
K K ( y (1 x)x dxdy 1/ 6
K y x dxdy 2 / 3
= = − − − − = −
= − + =
= = − − = −
= = − − − − = −
= + =
= = − + − = −
= + =
∫

∫
∫
∫
∫
∫
∫





We can write out the element stiffness matrix completely


2 / 3 1/ 6 1/ 3 1/ 6
1/ 6 2 / 3 1/ 6 1/ 3
1/ 3 1/ 6 2 / 3 1/ 6
1/ 6 1/ 3 1/ 6 2 / 3
− − −
 
 
− − −
 
 
− − −
 
− − −
 

We can add four numbers to get the local stencil












2 8
At A : 4
3 3
1 1
At B ,C,D,E : 2
6 3
Thus, the local stencil for u u sin g the bilinear
s is :
1 1 1
1
K 1 8 1
3
1 1 1
× =
− −
× =

∆
− − −
 
 
= − −
 
 
− − −
 


And computing as the problem 4c) we have the local stencil for the mass matrix is


Thus
,
2
1 4 1
h
M 4 16 4
36
1 4 1
 
 
=
 
 
 
.