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Transcript of hello...............this is lesson 6
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7/29/2019 hello...............this is lesson 6
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Lesson 5
Method of Weighted Residuals
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Classical Solution Technique
The fundamental problem in calculus of variations is toobtain a function f(x) such that small variations in the
function f(x) will not change the original function
The variational function can be written in general form for asecond-order governing equation (no first derivatives) as
Where ,, and are prescribed values
2
2
V
1 dfJ(f ) f 2 f dV
2 dx
= +
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Classical solution (continued)
An equation containing first-order derivatives may
not have a corresponding variational function. In
some cases, a pseudovariational function can beused
where C=C(x)
2
2
r
V
1 dC dCJ(C) D Cu K C 2mC dV
2 dx dx
= + +
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Classical solution (continued)Example:
Consider a rod of length L. The equation defining heat transfer in the rod is
with boundary conditions
Integrating twice, one obtains
Applying boundary conditions, the final result is
2
2
d T Q
dx k=
T(0) T(L) 0= =
2
1 2
QxT C x C
2k= + +
2Q(Lx x )T
2k
=
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Rayleigh-Ritz Method
FEM variational approach attributed to Lord Rayleigh
(1842-1919) & Walter Ritz (1878-1909)
Let
Assume a quadratic function
with boundary conditions
ni 1
i i
i 1
T(x) C x (where C are the unknowns)
=
=
2
1 2 3T C C x C x= + +
1
2 3
T(0) 0 C 0
T(L) 0 C C L
= == =
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R-R Method (continued)
Hence,
Now integrate
Thus
2
3T C (x Lx)=
3
dT
C (2x L)dx =
2
2
V
1 dfJ f 2 f dV (let f T, k, 0, Q)
2 dx
= +
2 2 2
3 3
V
1J kC (2x L) 2QC (x Lx) Adx
2 =
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R-R Method (continued)
which becomes
To find the value of C3 that makes J a minimum,
Therefore,
2 3 3
3 3AkC L AC QLJ6 6
= +
3 3
3
3
2AkC LJ AQL0
C 6 6
= + =
2
3
Q Q(Lx x )C or T
2k 2k
= =
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Variation Methods
Given a function u(x), the following constraints must
be met
(1) satisfies the constraints u(x1)=u1 andu(x2)=u2
(2) is twice differentiable in x1
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Variational methods (continued)
Then it can be shown that u(x) is also the solution of
the Euler-Lagrange equation
where
(1)
d F F0
dx u u
=
i(i )
i
d uu
dx=
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Variational methods (continued)
For higher derivatives of u,
Hence,( )
2
1
x
(1) (2) (n)
xJ F x, u, u , u ,..., u dx=
2 n
n(1) 2 (2) n (n)
F d F d F d F( 1) 0
u dx u dx u dx u + = L
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Variational methods (continued)
In 2-D, the constraints are(1) satisfies the constraint u = u0 on
(2) is twice differentiable in domain A(x,y)
(3) minimizes the functional
and
u uJ F x, y,u, , d
x y
=
F F F0
u x ( u / x) y ( u / y)
=
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Variational methods (continued)
Example:
Find the functional statement for the 2-D heat
diffusion equation
Applying the Euler-Lagrange relation
x yT Tk k Q 0
x x y y + + =
x y
F d F d F0
u dx u dy u =
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Variational methods (continued)
we find that
which yields the final functional form
x y
x y
22
yx1 2 3
22
yx
F T F T Fk , k , Q
u x u y u
kk dT dTF C , F C , F QT C
2 dx 2 dy
kk dT dTJ(T) QT d2 dx 2 dy
= + = + = +
= +
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A Rayleigh-Ritz Example
Begin with the equation
with boundary conditions
First find the variational statement (J)
2
2
d uu x 0 0 x 1
dx+ + =
u(0) u(1) 0= =
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R-R example (continued)
The variational statement is
Assume a quadratic approximation
with boundary conditions
12
2
0
du duJ x, u, u 2xu dx
dx dx
=
2
0 1 2u(x) a a x a x= + +
0
1 2
u(0) a 0
u(1) a a 0
= == + =
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R-R example (continued)Thus,
Now,
To be a minimum,
Finally
1 1 1
1
u(x) a [x(1 x)] a (x)
dua (1 2x)
dx
=
=
1 2 2 2 2 2 2
1 1 1 10
J(a ) a (1 2x) a x (1 x) 2a x (1 x) dx =
1
11
J 3a 1 5
0 aa 10 6 18
= = =
5u(x) x(1 x)
18=
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The Weak Statement
Method of Weighted Residuals - one does not
need a strong mathematical background to use
FEM. However, one must be able to integrate.
To illustrate the MWR, let us begin with a simpleexample - Conduction of heat in a rod of length L
with source term Q.
2
2
L
d TK Q 0 x Ldx
dTK q for x 0
dxT T for x L
= < < = =
= =
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weak statement (continued)
Integrating,
or
This analytical solution serves as a useful benchmark
for verifying the numerical approach.
( ) ( ) ( )L y
L x 0
q 1T x T L x Q z dz dyK K = + +
( ) ( )
( )
2 2L
q QT x T L x L x
K 2K
= + +
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weak statement (continued)
There are basically two ways to numerically solve
this equation using the FEM: Rayleigh Ritz
Method and the Galerkin Method ( which
produces a weak statement)
Consider
T2 0
2
0
A f 0 in Bu=g on
A B= , ,x x x x
=
+ +
L L
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weak statement (continued)
Hence,
Example:
The inner product becomes
( )i i j jd A c f d 0
+ =
( )2 2
2 2
u uf x, y
x y
=
( ) ( )2 2
i i i 2 2
u u, dxdy f x, y dxdy 0
x y
= = + + =
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weak statement (continued)
Integrating by parts (Green-Gauss Theorem)
Problem: Use Galerkins Method to solve
( )i ii iu u u u
dy dx f x, y dxdy 0x y x x y y
+ + + =
2
2
d uf 0 0
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weak statement (continued)
The inner product is
or
The weak statement becomes
( )1
i i0
, dx 0 = =
( )2
L2
i 1 1 120
d uf dx 0 where u=c c x 2xL
dx
+ = =
L LL 1 1 1
1 0 10 0
du d dcdx fdx 0
dx dx dx
+ =
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weak statement (continued)
Since
we obtain
This is the same as the Rayleigh-Ritz Method
but no variational principle is required.
du0 at x=L
dx=
( )
1
2
fc
2
fu x 2xL
2
=
=
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Weighting Function Choices
Galerkin
Least Squares
Method of Moments
( )i i
i
W
, 0
=
=
( )i
, which is the square of the errorc
( )i
i
x , 0 i=0,1,2,
W any set of linearly independent functions
=
=
L
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Weighting Function Choices (continued)
Collocation
Sub domain
( )
( )( )i
i i
i x
W x x
x x , 0
=
=
i
1 for x in subinterval iW
0 for x outside i
=
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Least Squares Example
Let
For the previous example problem
( ) 2
i i i
, dxdy 2 dxdy 0c c c
=
( )
( )
2
1 1 1
1
2
12
u c c x 2xL
uc 2x 2L
x
u2c
x
= =
=
=
( )i i
1
, 2 dx 0c c
2c f
= =
= +
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