NUMERICAL ANALYSIS PROBLEM DIFERENTIAL EQUATIONS II

WEBSITE : http://maths-minhthe.violet.vn NUMERICAL ANALYSIS PROBLEM DIFFERENTIAL EQUATION II 6200 1... It is not the most general form.. coercive: there is a unique solution to the equ

WEBSITE : http://maths-minhthe.violet.vn

NUMERICAL ANALYSIS PROBLEM DIFFERENTIAL EQUATION II

6200

1

Answer :

From the figure above, we have :

T T T T

T

T

T

T h

T

c s (c s) cot cot ,c h , : the radius of the inscribed circle

Let the three int erior angles of T be a b c

cot cot

We have :

by a cons tan t for all h h

T then for som

θ = α ≤ β ≤ γ ⇒ ≤ ≤

ρ

θ ≤ θ

ρ

T T

min

T

T

e

We have : cot cot

Because and cot is a decrea sin g function so

cot cot

cot cot where cot

σ

ρ

θ ≤ α ≤ β

ρ

2

Answer :

a) Difference between regular and uniform grid

T

T

T

T

h

Re gular : , such that h h h,where h max h ,T

h Uniform grid : , h max h , K 0

K The regular grid is stronger than the uniform grid

ρ

3

Answer :

j

1 j

1 j

j

1

j

h a)Show that tan

h

tan

tan dx tan x

We know that :

1

tan x 1 x tan

h This is true when we have known that z tan , 0

2

−

−

−

φ = θ

φ = θ

⇒ φ =  θ = θ

φ =

= θ 

∫

2

2 j

I

b)Consider the function u(x, y) y

Its linear int erpo tan t vanishes at z and equals h / 4 at the two other vertices Show that

h

u u

=

∂

− =

( )

j

j

j

I

2 2

z

2 z

2

j

Answer :

h

u u

h With u(x, y) y (u)

4

2 tan 2 tan

We know that :

h

(u)

4

h This is true when we have known that z tan , 0

2 From the parts above, we can see that

Genera

∂

− =

θ

= θ 

∫

lly, then the are of traingle T, which have created

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

j

j

2 z

2

0 z

(u) tan tan tan

h Hence, if (u) then tan 1 45

4 But, we have the shape regularity condition :

T

min

h

cot cot

1 tan 2

θ θ

ρ

≤

θ

T 0 min

min

45

Thus,the tan may be cause large error

θ ≤ θ =

θ ≤

θ

4

"

Consider the BVP problem : find a periodic function u such that

u (x) f(x) on ( 1,1 )

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

Answer :

First We multiply the equation by test function v(x)

[ ]

"

'

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

−

−

−

∫

' '

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

2 coercive:

there is a unique solution to the equation

a (u,v) = l(v)

Answer :

+ We will prove the property 1 : bounded:

a(u,v) u v dx , (v) fv dx ,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

(u) fv dx f dx v dx

H

f v ,C f

C v

≤

+ We continue to prove the property 2 : coercive:

( ) ( ) ( )

2

2

1

2

2

2

L

L

We will prove the Poincare inequality :

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

− < <

( )

( )

2

1

2

2

1

1

L 2

H

2

2 H

u H with u( 1) u(1) 0 u 0 , is boundary

Poincare inequality means :

Where a(u,u) u u dx u

1 Thus, a u,u u

3

! u H such that a u,u (u), v H

Ω

Ω

Γ

Ω

Ω

Ω

≥

∫

c)

i n n

i 1

i n

Partition as 1 x x x x 1

a(u ,v ) (v ) , v H

The stiffness matrix is symmetric

if A A a( , ) a( , )

≠

≠

≠

Ω − = < < < < =

= ⇔ φ φ = φ φ

= ϕ = ϕ + ϕ + ϕ

= ϕ + ϕ + ϕ

∑

{ , ϕn 1−}

ij ij ij

Thus,

2 ij

h

36

This stiffness matrix is symmetric

d)

Set up a set equations based on finite differences

1

0 1

ij

From the figure above , we have the equations :

Usin g the Intergrating to compute

∫



0 1

11

0 0

∂ϕ ∂ϕ ∂ϕ ∂ϕ

∫

∫ ∫





1

0 1

22

0 0

∫

∫ ∫



1

0 1

33

1 1

0 0

∫

∫ ∫



1

0 1

0 0

∂ϕ ∂ϕ ∂ϕ ∂ϕ

∫

∫ ∫





1

0 1

1 1

0 0

∫

∫ ∫



1

0 1

0 0

∫

∫ ∫



The element stiffness matrix

2 1 1

−

We have the matrix four the aparts so :

( )

1

2 1

3

1

3

∫

e) We have

1

1 1

1

ău,v) u (x)v (x)dx

(v) f(x)v(x)dx ặ,.) is binear form

−

−

=

=

∫

∫

1 x x x − x 1

− = < < < < <

{ }

h

h

h

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 have n 1 dim ensions Let u is solution of the subspace

u

Basic f , , , ,

We h

−

= α ϕ = α ϕ + α ϕ + α ϕ

2

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

−

∃ >

2

i i 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 lines to devive error estimates for finite element and u sin g higher order polynomials of the subspace V

+

−

Bonus Problem 2 :

( )

1

ij

0

1

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 :

K

∫

∫

( )

( )

1

2 31

0 1

22

0

1

2

0 1

0 1

33

0

1 2

0 1

44

0

∫

∫

∫

∫

∫

∫

∫

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

T hus, the local stencil for u u sin g the bilinears is :

1

3

∆

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

Thus,

h

36