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Introduction to Galerkinand Variational Methods
We have seen that the idea of using the weak form of a
differential equation together with the idea
of looking for the solution of the weak form in a nite
dimensional subspace lead to an effectivemeans for generating
approximate solutions to our model boundary value problem. This
method of generating numerical approximations is called the
Galerkin method.
Before looking further at further examples of the Galerkin
method methods, I will list a few commentsintended to enhance your
understanding of the method, and how it relates to the concept of
bestapproximation in a subspace.
1. Energy inner product and norm . It is easy to see that the
bilinear form a(u, v ), u ,v V satises all the requirements for an
inner product on the linear space V. In particular, it isclear that
a(u, u ) = 0 AE (u )2 + cu2 dx ≥ 0, and if a(u, u ) = 0 it must be
true that u ≡ 0.This is obvious when c > 0, if c ≡ 0 then u = 0
so u = b =constant. Then since u V entailsu(0) = 0, it follows that
b = 0 . Thus, the pair V, a(·, ·) forms an inner product space
(wecall a the energy inner product), and we can dene the energy
norm u E =
a(u, u ) so that
V, · E is a normed space.
2. Best approximation . We now see that the problem of nding u V
such that a(u, v) =(f, v ) v V takes place in a normed space, and
we can now show that the Galerkin methodleads to the best
approximation un to the solution from the nite dimensional subspace
V n .That is un = min w n V n wn − u E . This follows directly from
the observations: a(un , vn ) =(f, v n ) vn V n V , and a(u, vn ) =
( f, v n ) for u the actual solution. Subtracting we nd thata(u −
un , vn ) = 0 vn V n so that u − un is perpendicular to V n . This
property, as we’vepreviously seen, implies that un is the best
approximation from V n .
3. Need for sparse stiffness matrices. An objection to the use
of polynomials as the basiselements {φi } of our approximating
subspace is that the resulting stiffness matrix, [ a(φi , φj )],may
have many nonzero entries, and this fact can pose a heavy
computational burden. For
example, with approximating subspace of, say, dimension 100, a
fully populated stiffness matrixwould have 10,000 elements to be
computed and would, in addition, be difficult to invert. Theideal
basis functions should be easy to work with and lead to sparse
stiffness matrices (i.e.,matrices with many 0 elements).
4. Finite element basis functions . The nite element approach to
choosing basis elements forthe Galerkin method is (in 1D) to divide
the interval over which the equation is to be satisedinto a nite
number of parts, and in each of these parts to use a polynomial
approximation tothe solution. E.g., choose points x0 = 0 < x 1 ,
· · · < x n < x n +1 = in the interval [0 , ], andtake V n as
the set of continuous functions that are linear in each subinterval
[ x i , x i +1 ], i =0, . . . , n and vanish at 0 , (i.e.,
continuous, piecewise linear functions). V n is a subspace of C
([0, ]) of dimension n – corresponding to values of v V n at the
points x i , i = 1 , . . . , n . Thischoice of basis functions
leads to sparse stiffness matricies as we will discover
shortly.
5. A technical difficulty There is problem that has to be
mentioned with the nite element basisfunctions just described.
Consider the continuous piecewise linear functions with, say, n =
2for simplicity. A typical v V 2 looks like
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x 0 x1 x2 x 3
It is clear that the function v is continuous but has only one
derivative, and this derivative isonly piecewise continuous
(actually piecewise constant). Thus, V 2 is not a subspace of V
whoseelements must have two continuous derivatives. The way around
this dilemma is to extend thenotion of a solution, i.e., enlarge
the space V so that it contains subspaces of nite elementpiecewise
polynomials. We won’t worry about how to do this, and will verify
by examples thatwe haven’t made the approximating subspaces too
large. Thus, at least within the context of this course, the
difficulties just discovered may be forgotten, and we may assume
whenevernecessary that the space V contains the nite element
subspaces to be introduced.
To present the basic ideas of the nite element method and at the
same time generalize our sampleproblem, we’ll consider a boundary
value problem (BVP) for the ordinary differential equation of the
type we’ve been studying, but with some extra features. Let p = 0 R
be given and setV p = {w C 2 ([0, 1]) : w(0) = p}, i.e. functions
having two continuous derivatives and taking aspecied value at the
left end point. We formulate our BVP as follows:
Problem P 1 : nd u V p satsifying L(u) = − u + u = f (x), 0 <
x < 1, u (1) = q,
where q R and f C ([0, 1]) are given.
The new features of this BVP are: 1.) A nonzero boundary
condition on the unknown functionhas been added at the left end
point. 2.) The right hand boundary condition has been changed
torequire that the rst derivative take a given value. (In the
elastic rod case, this would correspondto a prescribed value of
stress at x = 1 – recall σ = Eu . Note that the left end condition
hasbeen incorporated into the denition of V p – conditions on the
unknown function are called essentialconditions – whereas the right
hand condition is given explicitly.
Our approach to numerial approximation requires that we start
with the weak form. How do we getit for this new problem? Recall
that we showed the equvalence of the weak form and the BVP simplyby
using integration by parts. We can do this with our new problem.
Dene the linear subspaceof C 2 ([0, 1]) by V 0 = {w C 2 ([0, 1]) :
w(0) = 0 } (note only the value at the left hand end point
isrequired to be 0). For u V p satisfying P 1 and any v V 0 we
have
0 = 1
0 v(Lu − f ) dx = 1
0 (− u v + uv − fv ) dx,
and integrating the rst term by parts
1
0− u v dx =
1
0(− (u v) + u v ) dx = − u v|10 +
1
0u v dx = − qv(1) +
1
0u v dx
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where we’ve used the facts that v(0) = 0 and u (1) = q.
Substuting we have
− qv(1) + 1
0(u v + uv − fv ) dx = − qv(1) + a(u, v) − (f, v ) = 0 ,
where as before
a(u, v) = 1
0(u v + uv) dx, (f, v ) =
1
0fvdx, u V p , v V 0
In summary, u V p satises P 1 implies
a(u, v) = ( f, v ) + qv(1) for all v V 0 .
But now we can reverse the process just completed and nd the P 1
is equivalent to
Problem P 2 : nd u V p such that a(u, v) = ( f, v ) + qv(1) for
all v V 0
Note that the only difference between V p and V 0 is that
functions in V p are required to take the value p = 0 at x = 0,
where as those in V 0 vanish there. V 0 is a linear space (since v,
w V 0 v + w andαv V 0 ), but V p is not – e.g. if v(0) = p then αv
/ V p unless α = 1. In fact, if v̂ V p is any xedelement of V p ,
then all other elements v are of the form v = v̂ + w, w V 0 (since
the difference of any two elements of V p belongs to V 0 ). Thus, V
p is just a translated version of the linear space V 0– the
dimension of V p is dened as the dimension of V 0 (it is called an
affine subspace of V ratherthan a linear subspace).
Ignoring the fact P 1 and hence P 2 can be solved exactly for
any f , we will try to nd an approximatenumerical solution. Using a
small generalization of the previous Galerkin approach, we seek
theapproximate solution u in a nite dimensional subspace of V p,n V
p , and we require that P 2 besatised for all v V 0 ,n where V p,n
and V 0 ,n are n dimensional subspaces of V (any u V p,n ison the
form û + v with û a xed element of V p,n and v V 0 ,n ). (u will
usually be called the trialfunction and any v will be called a test
function.)
Of course, we could work with polynomials again, but now we
introduce the nite element ideas.We dene a grid of points in the
interval [0 , 1], and call these points xi , i = 1 , . . . , n + 1
where0 = x1 < · · · < x n +1 = 1, and xi +1 − x i ≡ hi , i =
1 , . . . , n is the grid spacing. Usually, these pointswill be
evenly spaced, hi = 1 /n , but at this point we won’t enforce this
restriction. We take V p,nto be the space of continuous, piecewise
linear functions u on [0, 1] such that u(0) = p, u is linearin each
subinterval [ x i , x i +1 ], with u(x i ) = u i , i = 1 , . . . ,
n . We take V 0 ,n to be the same space butwith the value 0 specied
at x = 0 . Any function in either space is determined entirely by
its valuesat the n points x i , i = 2, . . . , n + 1, and this
implies that the spaces are n dimensional. In fact we’llnow dene a
useful system of basis functions with n elements.
Consider three grid points xe − 1 , xe , xe +1 , and let v V 0
,n (or V p,n ) have the values ve − 1 , ve , ve +1at these points
as shown in the sketch below.
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The trial or test functions are linear on each interval, or
element, I e = [xe , x e +1 ] of the grid.Thus, on element e − 1 we
have v = ve − 1 (xe − x)/h e − 1 + ve (x − xe − 1 )/h e − 1 and on
element e,v = ve (xe +1 − x)/h e + ve +1 (x − xe )/h e . That is,
on each element I e , a test or trial function v canbe expressed as
a linear combination of two element basis functions H e1 (x) = ( xe
+1 − x)/h e andH e2 (x) = ( x − xe )/h e , i.e., we can write v| I
e =
2j =1 ve + j − 1 H
ej , where the notation v| I e indicates the
restriction of v to element e. Combining the expressions for v
on the separate elements, we arrive ata formula which holds for all
of I
v(x) =n +1
i =1
φi (x)vi
with
φ1 (x) = H 11 (x), x1 = 0 ≤ x < x 2 ,
0, x2 ≤ x < 1 φn +1 (x) =
0, 0 ≤ x ≤ xn ,H n2 (x), xn < x ≤ 1 = xn +1
and
φj (x) =
0, 0 ≤ x < x j − 1H j − 12 (x), xj − 1 ≤ x < x j
H j
1 (x), xj ≤ x < x j+1
0, xj +1 ≤ x ≤ 1
, j = 2, . . . , n ,
where v1 = 0 for a test function and v1 = p for the trial
function. The φi ’s, sketched below, areoften called tent functions
for obvious reasons.
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and
φj (x) =
0, x ≤ xj − 11/h j − 1 , xj − 1 < x < x j− 1/h j , xj <
x < x j +10, x ≥ xj +1
, j = 2, . . . , n
Thus, if c = 0 and the grid spacing is constant grid spacing we
have
K |c=0 = 1h
1 − 1 0 0 · · · 0− 1 2 − 1 0 · · · 0
0 . . . . . . . . .
......
0 0 · · · − 1 2 − 10 0 · · · 0 − 1 1
If c = 0, we must add the elements kc,ij = c 1
0 φi φj dx. We nd for the diagonal elements
kc,ee =
I e − 1
H e − 12 (x)2 dx +
I e
H e1 (x)2 dx =
1
3(he − 1 + he ), e = 2 , . . . , n ,
and kc, 11 = h1 / 3, kc,n +1 ,n +1 = hn / 3. Of the off diagonal
elements, only those on the rst subdiagonaland superdiagonal are
nonzero, and we nd
kc,e − 1 ,e = I e − 1 H e − 12 (x)H e1 (x) dx = he − 1 / 3,and
kc,e,e +1 = he / 3.
We now have all the elements necessary for obtaining the
solution of the differential equation.
function fem_ex00(n, c, f, p, q)% finite element solution of
-u’’+cu=f, u(0)=p,u’(1)=q % using n intervals in [0,1]. Known value
of u at x=0 is% included as a additional equation so we start with
n+1 unknowns.x=linspace(0,1,n+1)’; h=1/n; % define grid points and
spacing% construct the load vectorff=zeros(n+1,1);ff(2:n)=h*f;
ff(n+1)=h*f/2+q;kk=kstif(n,c);% enforce boundary condition
u(0)=u1=p at x=0kk(1,:)=0; kk(1,1)=1; ff(1)=p;uu=kk\ff; % solve for
uuex=uexac(x,c,f, p, q);fprintf(1,’%10s%10s%10s%10s\n’,’x’,
’uapprox’, ’uexact’, ’error’);disp([x, uu, uex,
abs(uex-uu)]);plot(x,uu,’k-’, x,uex,
’ko’);%===================================function kmat=kstif(n,c)%
stiffness matrix for c=0% kk(1,1)=kk(n+1,n+1)=1/h, all other
diagonal elements kk(i,i)=2/h% first sub and super diagonal
elements kk=-1/h
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A more quantative comparison is given in the output data
>> fem_ex00(10,4,1,1,2)x uapprox uexact error0 1.0000
1.0000 0.0000
0.1000 0.9226 0.9230 0.00040.2000 0.8723 0.8730 0.00070.3000
0.8470 0.8480 0.00100.4000 0.8458 0.8470 0.00120.5000 0.8685 0.8700
0.00140.6000 0.9162 0.9178 0.00160.7000 0.9907 0.9925 0.00180.8000
1.0951 1.0969 0.00190.9000 1.2334 1.2354 0.00201.0000 1.4114 1.4134
0.0020
The computation above is not arranged in the usual FEM fashion.
We have not taken advantage
of the fact that all computations can be performed locally on
elements . We will in our futurework arrange things so that the
element stiffness matrix and load vectors are computed and
thenassembled into the global stiffness matrix and load vector. For
the problem just completed, thiselement approach would not achieve
any gains in simplicity, but in more complex problems theassembly
process becomes necessary for computational efficiency.
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