Strong, Weak and Finite Element Formulations of 1-D Scalar ...

Strong, Weak and Finite Element Formulations of 1-D Scalar Problems

ME 964; Krishnan Suresh

1. From Strong to Weak

Strong statement:

d dx

-a

du dx

=

f

u(0) = 0;-au,x (1) = -Q

To convert into weak form: 1. Multiply ODE by a virtual function uv(x) satisfying uv(0) = 0 :

( ) -au,x

uv

,x

= fuv

2. Shift Derivatives: Recall that

( ) ( ) ( ) -au,xuv

,x

-au,x

uv +

,x

-au,x

u,vx

Thus, replacing LHS

( ) ( ) -au,xuv ,x - -au,x u,vx = fuv

3. Integrate:

1

1

1

( ) ( ) -au,xuv

dx -

,x

-au,x u,vxdx =

fu v (x )dx

0

0

0

i.e.,

1

1

-au,xuv 10 + au,xu,vxdx = fuv(x)dx

0

0

4. Apply boundary conditions:

1

1

-Quv (1) + au,xu,vxdx = fuv(x)dx

0

0

i.e.,

1

1

au,xu,vxdx = fuv (x)dx + Quv (1)

0

0

5. Arrive at weak statement:

Find u(x),where,u(0) = 0,s.t.

1

1

au,xu,vxdx = fuv (x)dx + Quv (1)

0

0

uv(x)where,uv(0) = 0

(1.1) (1.2) (1.3) (1.4) (1.5) (1.6) (1.7) (1.8)

(1.9)

2. Graphical Interpretation of Strong & Weak Forms

Strong form:

Strong Form

u(x)

Kinetic Law

f (x)

Balance Law

g(x)

q(x)

Material Law

Kinetic Law : g(x) = - du dx

Fourier Law : q(x) = -a du dx

Balance Law : Heat balance in x

q(x + dx) - q(x) = f (x)dx

dq = f dx

d dx

-a

du dx

=

f

Differential Equation :

d dx

-a

du dx

=

f

4

u(x)

Kinetic Law

Weak Forms

f (x)

uv (x)

g(x)

q(x)

Material Law

gv (x)

Kinetic Law : g = - du & g v = - duv

dx

dx

Fourier Law : q = -a du dx

1

1

Virtual Work Law: qg vdx = fuvdx

0

0

1

0

-a

du dx

-

du v dx

dx

=

1 0

fuvdx

1

0 a

du dx

du v dx

dx

=

1 0

fuvdx

7

3. Equivalence

Equation (1.1) is 'equivalent' to Equation (1.9). Consider the example:

d dx

-

du dx

=

1

u(0) = 0;u,x (1) = -1

(1.10)

i.e., f = 1 & Q = -1 The exact solution is u(x) = -x 2 / 2 as one can easily verify. Let us

find the same solution via Equation (1.9). Search for a solution of the form u(x) = Ax + Bx 2 (satisfies the essential boundary condition). Similarly, let

uv(x) = Avx + Bvx 2 . Substituting in Equation (1.9):

Find A & B,s.t.

1

1

(A + 2Bx)(Av + 2Bvx)dx = (Avx + Bvx 2)dx - (Av + Bv )

0

0

Av & Bv

i.e.,

1

1

(Av + 2Bvx)(A + 2Bx)dx = (Avx + Bvx 2)dx - (Av + Bv )

0

i.e.,

{1

Av

0

0

} { } { Bv

21x 1

2x

BAdx =

1 0

Av

} { Bv xx2dx - Av

}Bv 11

i.e.,

{Av

} { Bv

1 0

21x

1

} { 2x dx BA = Av

} { Bv

1 0

xx2 dx

-

Av

}Bv 11

i.e.,

{Av

} Bv

1 0

21x

{ 2x

4x 2

dx

BA

=

Av

} { Bv

1 0

xx2dx -

Av

}Bv 11

i.e.,

{Av

}Bv 11

{ 4

1 /

3

BA

=

Av

} { Bv 11// 23 - Av

}Bv 11

Find A & B,s.t.

(1.11) (1.12) (1.13) (1.14) (1.15) (1.16)

{ } Av

Bv 11

Av & Bv

{ 4

1 /

3

BA

=

Av

}Bv --21// 23

(1.17)

This is true if and only if:

11

4

1 /

3

BA

=

--21

/ /

23

i.e., A = 0, B = -1/ 2 , i.e., u(x) = -x 2 / 2 . We obtain the same solution.

(1.18)

4. Why Weak Formulation?

Consider the example:

d dx

-

du dx

=

1

-1 + x2

u(0) = 0;u,x (1) = 0

(1.19)

It is difficult to integrate Equation (1.19) to obtain the exact solution. So, let us try an approximate solution: u(x) = Ax + Bx 2 where A and B are constants (satisfies the

essential boundary condition). Substituting in Equation (1.19), we have:

-B

=

1

-1 + x2

A + 2B = 0

(1.20)

The above suggests that B is not a constant (that's about it!!). Now consider the weak formulation. As before, let u(x) = Ax + Bx 2 and

uv(x) = Avx + Bvx 2 . Substituting:

Find A & B,s.t.

{ } Av

Bv 11

4

1 /

3

BA

=

1

(Avx

+

B

vx

2

)1

-1 +x

2

dx

0

Av & Bv

Evaluate right hand side numerically, we have:

11

4

1 /

3

BA

=

--00..231446399

i.e., u(x) -0.75x + 0.4x 2

(1.21) (1.22)

Weak formulations naturally promote computing approximate solutions to challenging problems, and are 'equivalent' to strong forms.

5. Finite Element Solutions of Weak Formulation

Consider the model problem: Find u(x),where,u(0) = 0,s.t.

1

1

au,xu,vxdx = fuv (x)dx + Quv (1)

0

0

uv(x)where,uv(0) = 0

Break up the domain into finite elements; thus:

xi+1 au,xu,vxdx =

xi +1

fuvdx + Quv (1)

e xi

e xi

Define linear shape functions:

L1()

=

1

- 2

; L2 ()

=

1

+ 2

,-1

1

Now, let (over a single element):

(1.23) (1.24) (1.25)

x = L1()xi + L2()xi+1

u = L1()ui + L2()ui+1

uv = L1()uiv + L2()uiv+1

Note that:

dx d

= L1,xi

+ L2,xi+1

= xi+1 - xi 2

=h 2

( ) { } u,x

=

2 h

L1,ui + L2,ui+1

=2 h

L1,

L2, uui+i 1

( ) { u,vx

=

2 h

L1,uiv

+

L uv 2, ,i+1

=2 h

uiv

} uv ,i +1

LL21,,

Considering integration over a single element:L

{ xi+1 au,xu,vxdx

xi

=

1

a

-1

2 h

uiv

} { uv ,i +1

LL12,,

2 h

L1,

} L2,

uui+i 1

h 2

d

{ = 2 h

uiv

} uv ,i +1

1 -1

a

LL12,,LL11,,

L1,L2, L2,L2,

d

uui+i 1

and

{ xi+1

1

fuvdx = f uiv

xi

-1

} uv ,i +1

LL21

h 2

d

Thus, from Equation (1.24) and Equation (1.28), we have:

{ e

2 h

uiv

} uv ,i +1

1 -1

a

LL21,,LL11,,

L1,L2, L2,L2,

d

uui+i 1

{ } 1

=

f uiv

e -1

uv ,i +1

LL21

h 2

d

+

Quv

(1)

The element stiffness matrix and element forcing vector are :

Ke

=

h 2

1 -1

a

LL12,,xx

L1,x L1,x

L1,x L2,x

L2,x L2,x

d

fe

=

1 -1

f

LL21

h 2

d

After assembly, and eliminating the virtual variables, we have: Ku = f

where

K = Ke e

f = fe + fQ e

(1.26) (1.27)

(1.28)

(1.29) (1.30) (1.31) (1.32)

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