Ebook The finite element method (2nd edition) Part 2

(BQ) Part 2 book The finite element method has contents: Further topics in the finite element method, the boundary element method, convergence of the finite element method, computational aspects. (BQ) Part 2 book The finite element method has contents: Further topics in the finite element method, the boundary element method, convergence of the finite element method, computational aspects.

of problems, including parabolic, hyperbolic and non-linear problems It is notintended to be any more than an introduction, and the ideas are presented byway of particular problems The reader with a specific interest in any one subjectarea will find the references useful for further detail.

5.1 The variational approach

We seek a finite element solution of the problem given by eqns (3.30)–(3.32), viz.(5.1) −div(k grad u) = f(x, y) in D,

with the Dirichlet boundary condition

Now, since ˜u e is zero outside element [e], the only non-zero contribution to I[˜ u]

from ˜u ecomes from integration over the element itself Thus

Suppose that element [e] has nodes p, q, , i, , s; see Fig 3.16 Then



ds.

(5.8)

If node i is not associated with element [e], then ∂I e /∂U i= 0; a non-zero

contribution to ∂I e /∂U i will occur only if node i is associated with element [e] This is shown in eqn (5.7) For element [e], as shown in Fig 3.16,

∂x

= 2



∂N e i

∂x

∂N e

∂x .

∂N e i

∂y

∂N p e

∂y .

∂N e i

∂N e

∂x +

∂N e i

∂x

∂N e

∂x +

∂N e i

∂x

∂N e j

∂x +

∂N e i

∂y

∂N e j

in cases where a variational principle does not exist However, the variational

procedure ensures that the resulting stiffness matrix, reduced by enforcingthe essential boundary condition, is positive definite and hence non-singular,provided that the differential operatorL is positive definite.

Example 5.1 Consider the differential operatorL given by

Lu = −div(κ grad u) dx dy.

positive definite provided thatκ is positive definite For a homogeneous Robin

boundary condition of the form

where K is the reduced overall stiffness matrix Now, provided thatκ is positive

definite and σ > 0, L is positive definite; consequently, it follows that

vT Kv > 0,

i.e K is positive definite.

When a variational principle exists, it is always equivalent to a weightedresidual procedure However, the converse is not true, since weighted residualmethods are applied directly to the boundary-value problem under consideration,irrespective of whether a variational principle exists or not

To establish this result, consider the functional

which is stationary when u = u0

Suppose that u = u0+ αv; then the stationary point occurs when (dI/dα) | α=0= 0; see Section 2.6 This yields an equation of the form

and

Equation (5.15) is the so-called Euler equation for the functional (5.13)

If eqns (5.15) and (5.16) are precisely the differential equation and boundaryconditions under consideration, then the variational principle is said to be a nat-ural principle and it follows immediately that eqn (5.14) gives the corresponding

Galerkin method, the weighting function being the trial function v However,

not all differential equations are Euler equations of an appropriate functional;nevertheless, it is always possible to apply a weighted residual method

Thus, if the Euler equations of the variational principle are identical withthe differential equations of the problem, then the Galerkin and Rayleigh–Ritzmethods yield the same system of equations In particular, it follows from Section2.3 that, since the variational principle associated with a linear self-adjointoperator is a natural one, the Galerkin and Rayleigh–Ritz methods yield identicalresults

It is worth concluding this section with a note on the terminology, sincethe method described here is often associated with the name ‘Bubnov–Galerkinmethod’ When piecewise weighting functions other than the nodal functions areused, then the name ‘Petrov–Galerkin’ is associated with the procedure.

5.2 Collocation and least squares methods

Recall the weighted residual method (Section 2.3) for the solution of

where {v i } is a set of linearly independent weighting functions which satisfy

v i ≡ 0 on C1, that part of C on which an essential boundary condition applies.

The trial functions ˜u are defined in the usual piecewise sense by eqn (5.4) as

˜

u = e

˜

u e ,

with ˜u e interpolated through element [e] in terms of the nodal values The

equations (5.19) then yield a set of algebraic equations for these nodal values.Notice that no restriction is placed on the operatorL; it may be non-linear, in

which case the resulting set of equations is a non-linear algebraic set; see Section5.3 Very often, the equations (5.19) are transformed by the use of an integration-by-parts formula, Green’s theorem, so that the highest-order derivative occurring

in the integrand is reduced, thus reducing the continuity requirement for thechosen trial function

The point collocation method requires that the boundary-value problem be

satisfied exactly at n points in the domain; this is accomplished by choosing

v i (x, y) = δ(x − x i , y − y i), the usual Dirac delta function In practice, the

collo-cation is usually performed at m points in the domain (m  n) and the resulting

overdetermined system is solved by the method of least squares; see Exercise 5.3

In the subdomain collocation method, the region is divided into N mains (elements) D j, and the weighting function is given by

so that the weighting functions are given by ∂r1/∂U i

Example 5.2 Consider Poisson’s equation,

The usual finite element approximation is written in the form

˜

u = e

i dx dy = 0.

Thus element stiffness and force matrices may be obtained, given by

The least squares approach to minimizing the residual errors then leads

to three integral expressions The usual finite element representation for the

unknowns U, ξ, η is then substituted into these expressions to obtain the

neces-sary stiffness and force matrices; see Exercise 5.4

5.3 Use of Galerkin’s method for time-dependent and

or weighted residual method This approach is illustrated in Example 5.4

A Laplace transform approach is also possible, and this is considered inSection 5.5.

Example 5.3 Consider the diffusion equation

∂x

∂N e j

∂x +

∂N e i

∂y

∂N e j

∂y +

1

α N

e i

∂N e j

∂x



dx dy dt,

The essential boundary condition must be enforced in eqn (5.27) together

with the nodal values along the initial plane t = 0 The equations (5.27) then give the solution at time T This solution may then be used to step forward in

time again, and the whole time development of the solution may be obtained

in a stepping manner very similar to that used in the finite difference method(Smith 1985)

To illustrate the procedure, consider the one-dimensional problem

step and the mesh size are each one unit, as shown in Fig 5.1

The element matrices for this problem may be obtained by using the results

and notation of Section 3.8 with y replaced by t, as follows The shape function

1 [1]

[2]

[3]

[4]

5 [7]

[8]

[6]

[5]

x t

Fig 5.1 Eight-element discretization for the problem of Example 5.3.

Enforcing the boundary conditions and the previously calculated value of U5, it

follows that U8= 52

The procedure may be repeated until the required time level is reached Thesolution at points inside an element may be found from the nodal values usingthe usual linear interpolation polynomials

The wave equation may be treated in a similar manner; see Exercise 5.8

Example 5.4 Consider again the initial-boundary-value problem of Example 5.3

but in this case suppose that the approximation in each element is, instead ofeqn (5.26), given by

where K and F are the usual overall stiffness and force matrices and the matrix C

is often referred to as the overall conductivity matrix, since the diffusion equation

models heat conduction Since ce=

[e]Ne TNe

dx dy (cf the mass matrix m e

in Exercise 3.24), it is clear that C is symmetric Some authors, by virtue of the

structural origins of the finite element method, refer to C as the damping matrix.

A finite difference approach to the solution of this system of equations is to

take a sequence of time steps of length Δt from time level j to j + 1.

Using a forward difference scheme given by

Knowing the initial value U0, the time-stepping procedure can start and thesolution be developed by marching forward to the next time level, etc.

To illustrate the method, return to the one-dimensional problem of ple 5.3 In this case, two elements are used in space and these are taken to besimple linear elements of the type discussed in Section 3.5 Thus, using the results

If the three nodes are situated at x = 0, 1, 2 and if Δt = 1, it follows that, since

Fj = 0, eqn (5.33) yields the set of equations

2 + 4U22+ 4− 7 + 8 ×3

2− 7 × 3 = 0, which gives U22= 52

Other difference schemes could of course be used instead of forward

differ-ences in eqn (5.32); this approach actually replaces Uj by a suitably truncatedTaylor series If, instead, a weighted residual approach in time is used, a moregeneral recurrence relation can be set up (Zienkiewicz and Taylor 2000a)

Suppose that between time levels j and j + 1, U i (t) is interpolated by

Ui(t) = [1 − τ τ][U ij Uij+1]T , where t = (j + τ ) Δt with 0 ≤ τ ≤ 1; see Fig 5.2.

from 0 to 1, eqn (5.34) becomes

Equation (5.35) is a recurrence formula for the nodal values at two time

levels By choosing different values for the parameter r, well-known difference formulae are recovered; for example, r = 0,12 and 1 give the forward, Crank–Nicolson and backward difference formulae, respectively Recurrence formulaefor nodal values at three time levels may be obtained by integrating over twoconsecutive time intervals in the weighted residual equation; see Exercise 5.10.Finally, the procedure may also be adopted to solve second-order equations ofthe form

which arise in the finite element solution of propagation problems; see cises 5.11 and 5.12.

Exer-This discussion of time-dependent problems has been necessarily brief; for

further details, there are two excellent chapters in the book by Zienkiewicz et al.

(2005)

As well as time-dependent problems, the weighted residual method mayalso be used to tackle non-linear problems as illustrated in Example 5.5 In thesecases, however, the resulting system of algebraic equations is no longer linear,and iterative techniques are usually necessary for their solution

Example 5.5 Consider the equation

In this case suppose that k, f , σ and h all depend on u as well as on position.

The weighted residual method may be applied in the usual manner, and if theweighting functions are the nodal functions, then the element matrices become

(cf eqns (3.45)–(3.48)) However, in this case the unknown u occurs under the

integral sign, so that the matrices themselves are functions of the nodal variables.The overall system is assembled in the usual way to yield a set of equations ofthe form

To illustrate the idea, consider the one-dimensional problem

−(uu ) + 1 = 0, 0 < x < 1, u(0) = 1, u(1) = 0.

Suppose that two linear elements of the type discussed in Section 3.5 are used.Then

ke

(u) =

 1

−1 u



−1/h 1/h

[−1/h 1/h] h

1

1

U2=±1

2.

The non-linear algebraic system has led to two possible solutions, and someknowledge of the behaviour of the solution to the original boundary-value prob-lem is necessary in order that the correct value may be chosen In this case, it may

be shown that the solution U2=−1

2 is not possible, since this would imply that

there is a point x p ∈ (0, 1) with the properties U(x p ) < 0, U (x p ) = 0, U  (x p ) > 0.

These properties are not consistent with the original differential equation Thus

the solution required is U2=12 This situation is typical of non-linear problems,where it is extremely useful to have some intuitive idea of the behaviour of thesolution or the physics of the problem

The problem considered here is of course very simple, leading to a non-linearalgebraic equation whose solution is amenable to hand calculation In practice,this does not occur, and numerical methods of solution are necessary In general,either an iterative or incremental approach is adopted Write eqn (5.37) as

then the Newton–Raphson method gives the following iterative scheme:

F0 of F, eqn (5.38) has the known solution U0; then the solution proceeds by

adding small increments to F and finding the corresponding increments in U.

Write eqn (5.37) as

H(U) − λF ∗ = 0,

where

H(U) = K(U)U and λF ∗ = F(U).

Then, differentiating with respect to λ,

Thus the incremental scheme is

(5.42) Un+1= Un+ J−1 n ΔFn ,

where ΔFn = Δλ nF ∗ is the nth increment in F Simple problems illustrating

the use of these methods may be found in Exercises 5.13–5.16

It has been the intention of this section to illustrate the way in which theGalerkin finite element approach may be used to solve time-dependent and non-linear problems Only simple examples have been introduced, the general area ofsuch problems being well beyond the scope of this text Much research has beenconducted in these areas, and the interested reader is recommended to consultthe books by Zienkiewicz and Taylor (2000a,b) for references to specific subjectareas such as plasticity, electromagnetic theory and viscous flow

5.4 Time-dependent problems using variational principles which are not extremal

In Section 2.9, variational principles for time-dependent problems were duced In this section, they are used to develop finite element solutions tothe wave and heat equations Some authors feel that a direct approach to theproblem at hand is better than using such variational problems, either becausethe solutions do not yield extrema for the functionals involved (Finlayson andScriven 1967) or because a weighted residual approach is more general, sincethere is always a Galerkin method equivalent to a given variational principle(Zienkiewicz and Taylor 2000a) However, it is a possible approach when suchprinciples exist and a suitable functional is known, and it gives users anotherweapon in their finite element armoury

intro-Example 5.6 (Noble 1973) Consider the wave equation

From Section 2.9, it follows that the functional (2.75) is stationary at the solution

of eqn (5.43) subject to eqn (5.44) and (5.45) and a condition at time Δt of

the form

u(x, Δt) = h(x),

which replaces eqn (5.46)

Of course, h(x) is not known; however, it is treated as if it is known and

then, at the end of the analysis, eqn (5.46) is used to eliminate this unknownfunction

Suppose that l = 1 and that [0, 1] is divided into E elements Consider the

usual finite element approximation

˜

u = e

For the linear element of length h discussed in Section 3.5, the x integrations

are easily performed to yield the two matrices

To satisfy the essential conditions at node i at times t = 0 and t = Δt, set

U (0) = f i and U (Δt) = h i Now U (t) may be interpolated between t = 0 and

At this stage a i is treated as unknown, so that substitution of eqn (5.48) into

eqn (5.47) gives I = I(a1, , a n)

Stationary values of I occur when ∂I/∂a i = 0, i = 1, , n If I =,

e I e,then

A similiar result may be obtained when element [e] contains nodes i and

i + 1 Assembling the complete equations,

e ∂I e /∂a i= 0 yields



h i+

1

This system of equations allows h i to be found, since f i and a i are known

To illustrate the technique, suppose that c = 1 and

Take four equal elements along the x-axis and suppose that Δt = 0.1, so that

μ = 0.4 Then the difference equation for the unknown function h is

h2= h4= 0.254, h3= 0.431

(5.51) U (0.1) = 20(h˙ i − f i)− a i = b i , say.

However, h i − f i is the difference between two nearly equal numbers and, to

avoid inaccuracies, h i is eliminated between eqns (5.50) and (5.51) to give

0.00529b i −1 + 0.02382b i + 0.00529b i+1= 0.0426(f i −1 + f i+1)− 0.1066f i

+ 0.00689(a i −1 + a i+1) + 0.01849a i

Using the initial conditions and boundary conditions previously stated,

together with b1= b5= 0, three equations may be obtained, to yield the solutions

b = b4= 0.082, b =−1.380.

The procedure may now be repeated using the calculated values at t = Δt to step

forward in time once again Thus a step-by-step method has been developed, and

the solution at any time t may be obtained by taking a sufficiently large number

5.5 The Laplace transform

Until recently, this approach has not been particularly popular However, theuse of Stehfest’s numerical inversion method (Stehfest 1970a,b) has developed

a new interest in the Laplace transform associated with finite element methods(Moridis and Reddell 1991) A very good introduction to the use of the Laplacetransform for diffusion problems has been given, in the context of boundaryintegral methods, by Zhu (1999)

Example 5.7 Consider the diffusion equation

is the Laplace transform in time of u(x, t), then, using the Laplace transform,

eqn (5.52) becomes the ordinary differential equation

1

λ2

+λ6

1

Inverting then gives

(5.53) U2(t) =12+ t −1

8e −3t .

A drawback in this approach is the fact that the transform variable λ is

carried through the analysis This causes no difficulty in the simple hand lation given, but it is not at all convenient for a numerical procedure In practice,

calcu-the equations could be solved for discrete set of values λ1, λ2, , λ M and theresulting transforms inverted numerically Before we consider this approach, weshall illustrate the use of the convolution theorem

Example 5.8 Consider the wave equation with boundary and initial conditions

as in Example 5.6 Then, taking the Laplace transform,

will be used here, is to assume a time variation for u in the interval 0 ≤ t ≤ Δt,

evaluate the convolution integral and use a finite element analysis to solve theresulting two-point boundary-value problem

Suppose that the approximation is

˜

u = e

space variable x (see Fig 3.7), it follows that the element matrices are



2 1

1 2

Thus the overall system is assembled in the usual way to give the following

difference equation at node i:

Then, just as for eqn (5.49), this set of equations can be solved to find h i,

since f i and a i are known The value of U i (Δt) is then found using

Δt (h i − f i)− a i and then eliminating h i between eqns (5.55) and (5.56) as before

In Examples 5.7 and 5.8, the inversion of the Laplace transform has beeneffected in closed form In practice, this is usually not possible, and numericalinversion is required There are many possibilities (Davies and Martin 1979),and for diffusion problems it has been shown that Stehfest’s method provides asuitable approach One advantage of the method is that the solution is found

at a specific time τ without the necessity for solutions at preceding times, as

is the case with the finite difference approach (see Example 5.4) Consequently,the method does not exhibit the stability problems associated with the finitedifference method; see Section 7.5

In Stehfest’s method we choose a specific time value, τ , and a set of

Example 5.9 Consider again the problem of Example 5.7 with two linear

ele-ments, leading to the transformed value

We shall evaluate the solution at t = 1 with M = 2:

This compares rather poorly with the exact inversion, U2(1)≈ 1.994, given by eqn (5.53) This is due to the fact that we have used M = 2 Much better results are obtained with higher values of M ; see Exercise 5.20.

Now consider the two-dimensional problem

Using the results of Exercise 3.24, the overall system of equations for theLaplace transform ¯U = [ ¯ U2 U¯3]T is

Solving these equations yields

5.6 Exercises and solutions

Exercise 5.1 A beam on an elastic foundation has a functional given by

Exercise 5.2 Repeat Exercise 3.7 using the variational approach.

Exercise 5.3 Consider the boundary-value problem

Illustrate the procedure for the case

L ≡ − d2

dx2, f (x) = x, 0 < x < 1, with u(0) = u(1) = 0.

Exercise 5.4 Poisson’s equation (5.22) is to be solved using the least squares

finite element method Introduce auxiliary variables ξ and η so that the system

of equations to be solved is given by eqns (5.23) and (5.24)

By taking linear variations in ξ and η and quadratic variations in u, set up

the element approximation in the form ve= NeUefor a triangular element withsix nodes; see Fig 4.5 Hence obtain expressions for the element stiffness and

force matrices The results of Exercise 4.6 allow an explicit form for ke to beobtained in this case.

Exercise 5.5 Show how Galerkin’s method may be used to solve a system of

equations of the form

where L is a matrix of differential operators given by

L = [L ij]and

u = [u1 u p]T

By considering the one-dimensional form of Poisson’s equation, −d2u/dx2= f ,

which may be written as a system of equations by introducing the variable q =

−du/dx, show that the Galerkin approach is not unique and that the symmetry

of the resulting system of equations depends on the ordering of eqn (5.61)

Exercise 5.6 The diffusion equation is to be solved using triangular space–time

elements Obtain the stiffness matrices for the two triangles in the xt plane shown

in Fig 5.4 Hence obtain the difference equation for node i at time step j.

Exercise 5.7 Find the solution at time t = 1 to the problem in Example 5.4

using six space–time elements of the type used in Exercise 5.6

Exercise 5.8 Obtain an expression for the stiffness matrix for the wave equation

using elements in which the shape functions are functions of position and time

Exercise 5.9 Rework Exercise 5.7 using three equal space elements and a forward

difference scheme in time given by eqn (5.33)

Exercise 5.10 Write down the difference equation equivalent to eqn (5.34),

valid for (j − 1)Δt ≤ t ≤ j Δt Using Galerkin’s method in time, show that the

1

3 3

following three-level difference equation may be obtained:

Exercise 5.12 For the overall system of equations (5.62), suppose that U (t) is

interpolated between the three time levels j − 1, j, j + 1 by

U i (t) =1

2(τ2− τ) 1 − τ2 1

2(τ2+ τ ) [U ij −1 U ij U ij+1]T ,

where t = (j + τ )Δt with −1 ≤ τ ≤ 1 Suppose also that F i is interpolated in

the same way By using a weighting function W , set up a recurrence relation between U j −1 , U j and U j+1 in terms of the parameters

using two linear elements

Exercise 5.15 Repeat Exercise 5.13 making use of the Newton–Raphson scheme

given by eqn (5.41); perform one iteration

Exercise 5.16 Repeat Exercise 5.13 using an incremental method Choose

suit-able starting vectors and use two equal increments for the force vector

Exercise 5.17 The magnetic scalar potential φ in a magnetic material with

zero current density satisfies the equation div{μ(H) grad φ} = 0, where the

permeability μ is a function of the magnetic field strength; φ is related to the

magnetic field by H =−grad φ (see Appendix A) Show that the matrix J of

eqn (5.40) can be assembled from the element matrices given by

Exercise 5.18 Using the functional (2.76) and the method of Example 5.6, show

that the difference equation for the heat equation

4(1 + μ)h i+ (1− 2μ)(h i −1 + h i+1) = 4(1− 2μ)f i + (1 + 4μ)(f i −1 + f i+1),

where f i = U (0), h i = U (Δt) and μ = k Δt/h2

Exercise 5.19 Using the Laplace transform and two linear space elements, find

the solution at x = 12 to the wave equation

Exercise 5.20 Solve the problem of Example 5.9 using two linear elements

together with the Stehfest Laplace transform inversion with M = 8.

Exercise 5.21 Solve the problem of Example 5.10 using four equal elements, and

compare the results with the exact solution

Solution 5.1 The procedure for obtaining the finite element equations follows

in exactly the same way as in Chapter 3; in this case

Thus it follows that

gives the residual

r = e

This residual is now minimized by taking m e collocation points inside element

[e] If there are M e degrees of freedom in each element, then m e  M e Suppose

that at the collocation point P i,L(N e) = Le

i and f = f i ; then the residual at P iis

r e i = Le iUe − f i ,

and this residual is chosen to be zero

Thus, in each element, there is a set of equations of the form

LeUe

= fe

Since Le is a rectangular m e × M e matrix, this constitutes an overdeterminedset, which may be solved by the least squares method, setting

∂

∂U i j

The overall system is then assembled in the usual way

For the problem −u  = x, u(0) = u(1) = 0, it is necessary that the shape

functions have a continuous first derivative sinceL is a second-order differential

operator Thus suppose that the Hermite elements of the type shown in Fig 4.6are used The shape functions are given by eqn (4.8) and (4.9), so that

L(N e

) = 1

h2[−6ξ h(1 − 3ξ) 6ξ − h(1 + 3ξ)] Consider the one-element solution with collocation at x = 0,14,12,34, 1; then

Enforcing the essential boundary conditions U1= U2= 0 and solving the

resulting equations yields U1 = 16, U2 =−1

where the L i are the usual triangular coordinates and the N i are the shapefunctions for the quadratic triangle (see eqns (4.4) and (4.5)), i.e.

ve= NeUe, where

ve

= [ξ η u] T , Ue

= [ξ1, , η1, , U1, , U6]Tand

Using the results and notation of Exercise 4.6, it follows that

Solution 5.5 The finite element approximation, in terms of nodal functions, is

easily generalized to problems involving more than one dependent variable togive

w qi Q i =

e i

N qi e Q i ,

u = i

;thus it follows that the element stiffness matrix is of the form

If, however, the equations are written in the reverse order so that

h

2 dξ, s = q, u,and a different form is obtained in which the symmetry is lost

Solution 5.6 For element 1,

The difference equation for node i at time step j is thus

(−r − r)U ij −1+ (0− r)U i +1j−1+ (−3)U i −1j + (3 + r + r + 3)U ij

+ (−3 + r)Ui +1j = 0,

i.e

(6 + 2r)U ij − (2rU ij −1 + rU i +1j−1 + 3U i −1j+ (3− r)U i +1j ) = 0.

Solution 5.7 In this case, r = 4

9 and the only unknowns are U22 and U32; thus

the difference equations are

62

9U i2−8

9U i1+49U i+11+ 3U i2+239U i+12

= 0, i = 2, 3.

Enforcing the initial and boundary conditions

U21= 29, U31=89, U41= 2, U12= 1, U42= 3and solving the resulting pair of equations yields

the only difference occurring in the Galerkin formulation from that for Poisson’sequation in Sections 5.1 and 5.2 is that the time derivative term becomes

∂t

∂N e j

∂t dt

dx dy

to the element stiffness matrix

Solution 5.9 In this case h = 2

⎤

⎥

⎦ ,

which yields U22=119, U32=179

Solution 5.10 Equation (5.34) gives an approximation to the system (5.32) for

values of t between j Δt and (j + 1) Δt Similarly, for values of t between (j − 1) Δt and j Δt,

Collecting the terms together gives the required result

Solution 5.11 The finite element approximation is

˜

u = e

9 and the only unknowns are U22 and U 32< /sup>; thus

the difference equations are

62< /small>

9U... U 42< /sub>= 3and solving the resulting pair of equations yields

the only difference occurring in the Galerkin formulation from that for Poisson’sequation in Sections 5.1 and 5 .2 is that the. .. data-page="39">

Enforcing the initial and boundary conditions

U21 = 2< /sup>9, U31=89, U41= 2, U 12< /sub>=