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Introduction to Time Dependent Problems

Problems in One Space Dimension

We first consider a 1D initial-boundary value problem:

∂u

∂t =

∂ 2u

∂x2 + g(x, t), 0 < x

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where f i = I gH idx + H i

∂u∂x

ba

We can write the equations above in matrix form as

M =

I

H 1H 2

( H 1 H 2 ) dx, K =

I

dH 1dxdH 2dx

( dH 1

dxdH 2dx

) dx

and

F =

I

H 1H 2

g(x, t) dx +

−∂u/∂x(a, t)

∂u/∂x(b, t)

and obtain the element equations

M U̇ + KU = F,

where M is called the element mass matrix (in keeping with the structural origins of the fem). Note that theelement load vector contains the partial derivatives of the test function evaluated end points of the interval.These cannot be computed using

ujH j because they would fail to be continuous across the element

boundary – and then the whole method would fail since it is based on the assumption that 10

=

e

xe+1xe

.Fortunately these unknown partial derivatives all cancel out when the element matrices are assembled intothe global matrices – except, that is, for the terms H 1(0)∂u/∂x(0, t) and H 2(1)∂u/∂x(1, t). But since thetest functions w are required to vanish at x = 0, w = H 1 is not admissible in element 1 so this term is tobe omitted, and in the last element nel ∂u/∂x(1, t) = b(t) is given. The fact that all the internal unknowns,∂u/∂x(xe, t), e = 2, . . . , nel − 1, introduced with the element weak form cancel out in the final assemblymakes it fairly reasonable to ignore these unknowns and treat the element load vector as if it consisted of only

F e =

I e

H e1H e2

g(x, t) dx + δ e,ne

0b(t)

,

that is, only in the final element is there a contribution from the flux terms. I think this is probably anacceptable fiction as long as the true state of affairs is kept in mind.

For 1D linear shape functions (see pp. 38-39 of the text) we have:

K ≡ [kij] = 1

h 1 −1−1 1

M ≡ [mij ] = h6 2 1

1 2

It is worthwhile looking at the computation of the mass matrix. In integrating over element of the formI = [a, b] of length h = b − a, it is useful to transform the variable of integration setting r = (x − a)/h, andintegrate over the unit interval [0, 1]. On this interval the basis functions become: H 1(r) = 1− r, H 2(r) = r,so that the integral to be computed becomes

M = h

10

(1− r)2 r(1− r)r(1− r) r2

dr

It is now easy to perform the required integration and verify the results.

The global equations which are assembled, as usual, from the element equations take the form

M U̇ + KU = F.

This is essentially a system of ODEs for the values uj(t) of the solution at the nodes. I say essentially becauseall given boundary value information has to be input in order to create a legitimate system. (In this case,it is required that u1(t) = a(t). This can be done simply by replacing the first equation with u̇1 = ȧ andtacking on the initial condition u1(0) = a(0).)

Generally, the numerical solution of the system of ODEs just derived in accomplished using finite differencemethods not finite element methods. For these methods, we pick a sequence of times tj , j = 1, . . . , wheret1 = 0, and attempt to predict U (t) at these times. Let U

j = U (tj), then by the mean value theorem for

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derivatives asserts that U̇ (tj + θ(tj+1 − tj)) = (U j+1 − U j)/(tj+1 − tj), for some θ, 0 < θ < 1. Letting

∆j = tj+1 − tj we have

M (U j+1 − U j) + ∆jKU (tj + θ∆j) = ∆jF (tj + θ∆j)

So far this equation is exact, but we don’t actually know the value of θ. So we choose a fixed value, andeach such choice creates a new numerical method. (Once θ is picked, the U j will represent a numericalapproximation to the original U , but I will continue to ignore this sort of thing.) The standard ones are:

1 The explicit forward difference method is obtained by choosing θ = 0, and can be written as

M U j+1 = (M − ∆jK )U j + ∆jF

j U 1 = [c(x1), c(x2), . . . , c(1)]T

Note that this determines U j for all j = 1, 2, . . .. It is explicit because it determines U at each new timelevel explicitly in terms of U at the previous level. It is a forward difference method since the derivativeat level j is represented using the difference between U at this level and U at the next (forward) level.

2 The backward difference or implicit method is obtained using θ = 1. This gives

(M + ∆jK )U j+1 = M U j + ∆jF

j+1 U 0 = [c(x1), c(x2), . . . , c(xn+1)]T

Thus, U at the new level us obtained only by solving a system of equations, i.e., the difference formulaonly defines U at the new level implicitly. The backward difference designation obviously refers to thefact the derivative is approximated at the new level using U at the previous (backward) level. Since,our problem is linear we can actually solve for U j+1 to obtain

U j+1 = (M + ∆jK )−1(MU j + ∆jF

j+1)

3 The Crank-Nicholson method is obtained using θ = 1/2, and using the additional approximations

U (1

2(tj + tj+1)) =

1

2(U j+1 + U j), F (

1

2(tj + tj+1)) =

1

2(F j+1 + F j).

From these we obtain

(2M + ∆jK )U j+1 = (2M − ∆jK )U

j + ∆j(F j+1 + F j)

.

These methods have different properties. To keep the discussion simple, assume that tj+1 − tj = ∆, j =0, 1, . . . It can be shown that the explicit method is only works (technically, is stable) for small ∆. Theimplicit method is stable for all ∆, and its accuracy increases linearly as delta decreases. The Crank-Nicholson method is also stable for all ∆, its accuracy increases quadratically as ∆ decreases, but it involvesmore work per step than the implicit method.

I have included a small m-file which uses the backward difference method to solve an initial value problem.

function transient2(n, nstep, dt)

% 1D transient heat conduction problem:

% u_t=u_xx, 0

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% dt = time increment

%

sdof=n+1;

xvec=linspace(0,1,n+1); % define the node points

h=1/n;

nodes=[[1:n]’ [2:n+1]’]; % configuration matrix

mm=zeros(sdof,sdof); kk=zeros(sdof,sdof); ff=zeros(sdof,1);ke=elstif(h);

me=elmass(h);

for iel=1:n % assembly loop all elements have same mass and stiffness

for ii=1:2 % assemble mass and stiffness

gn1=nodes(iel,ii);

for jj=1:2

gn2=nodes(iel,jj);

mm(gn1,gn2)=mm(gn1,gn2)+me(ii,jj);

kk(gn1,gn2)=kk(gn1,gn2)+ke(ii,jj);

end

end

end

uold=init_val(xvec’); % set inital conditionseffrc=zeros(sdof,1);

t=0;

esol=exac(xvec,0);

close all

hold on

plot(xvec, uold, xvec,esol,’o’); % plot initial condition

for k=1:nstep % use implicit method

ff(sdof)=1.0; % forcing at time t

effrc=mm*uold+dt*ff; % effective forcing

aa=mm+dt*kk;

aa(1,:)=0; aa(1,1)=1; effrc(1)=0; % impose bc at x=0 and time t

fsol=aa\effrc;

esol=exac(xvec,t+dt);

plot(xvec’, fsol, xvec, esol, ’o’); % plot solution at new time, t=t+dt

uold=fsol; t=t+dt;

end

legend(’fem soln’, ’exact soln’,4)

hold off

%============================================

% end of main program

%============================================

function me=elmass(h)

% element mass matrix

me=(h/6)*[2 1; 1 2];

%============================================function ke=elstif(h)

% element stiffness matrix

ke=(1/h)*[1 -1; -1 1];

%============================================

function h=init_val(x)

h=sin(pi*x./2)+x;

%============================================

function u=exac(x,t)

u=x+exp(-pi^2*t/4)*sin(pi*x/2)

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Problems in Two Space Dimensions

We now shift our attention to problems with 2 space dimensions. As with the 1D case we consider only theheat equation in the form

∂u

∂t =

∂ 2u

∂x2 +

∂ 2u

∂y2 + g(x, y, t), (x, y) ∈ D, t > 0.

On the boundary ∂ D of D, u takes the values

u = a(x, y, t) on ∂ Du, ∂u

∂n = b(x, y, t) on ∂ Dq,

with ∂ D = ∂Du ∪ ∂Dq, and a, b given functions. At time t = 0, u satisfies the initial condition

u(x,y, 0) = c(x, y), (x, y) ∈ D,

where, again, c is given. (Again, the usual consistency requirements need to be imposed, e.g., c(x, y) →a(x0, y0, 0), as (x, y) → (x0, y0) ∈ ∂Du.) The weak form is derived just as in the 1D case. We define theresidual

r = ∂u

∂t

− ∂ 2u

∂x2 −

∂ 2u

∂y2 − g(x, y, t),

multiply r by a test function w(x, y) which must vanish on ∂ Du (note that w is independent of time t), andthen integrate over D. Using the divergence theorem to integrate by parts just as in the stationary case, wefind

D

wrdA =

D

(w∂u

∂t +

∂ w

∂x

∂u

∂x +

∂ w

∂y

∂u

∂y −wg) dA −

∂Dq

w∂u

∂n ds.

This leads to the weak form. That is, we require that u take the given values on ∂ Du, satisfy D

(w∂u

∂t +

∂ w

∂x

∂u

∂x +

∂ w

∂y

∂u

∂y) dA =

D

wg dA +

∂Dq

wbds

for all test functions w vanishing on ∂Du, and take the initial condition u = c at time 0. It is easy to seethat for smooth functions the weak form is equivalent to the original problem.

Now we introduce a division of D into finite elements – linear triangles, bilinear rectangles, etc. – I’ll onlyconsider linear triangles here but other elements can be handled in a similar manner. We specify that thetrial function be linear on each triangular element and continuous throughout the domain. We can thenwrite it in the form u =

n+1j=1 φj(x, y)uj(t) where the global shape functions, φj satisfy the conditions

φi(xj , yj) = δ ij . We require that the trial solution u satisfy the Dirichlet boundary condition on ∂Du sothat uj(t) = a(xj , yj , t) for nodes on this boundary. We demand that the weak form be satisfied for anytest function w from the set {φk, k = 1, . . . , n + 1} ( φj corresponding to boundary nodes j on ∂ Du shouldnot be included since w must vanish there, but as usual we ignore this restriction and impose the properconditions at the end). Substituting for u and w gives a system of linear equations:

n+1

j=1mij u̇j +

n+1

j=1kijuj = f i, i = 1, . . . , n + 1

Rather than go into the details of these global matrices here, I will focus immediately on a single elementsince the global matrices can be assembled from the lower dimensional element matrices. We assume theelement number e is fixed, and therefore to simplify notation do not explicitly indicate it in our formulas.

Letting the element shape functions be denoted by H j , j = 1, 3 as usual, we can write for an element

M =

I

H 1H 2

H 3

( H 1 H 2 H 3 ) dA,

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K =

I

∂H 1∂x∂H 2∂x∂H 3∂x

( ∂H 1

∂x∂H 2∂x

∂H 3∂x

) +

∂H 1∂y∂H 2∂y∂H 3∂y

∂H 1∂y ∂H 2∂y ∂H 3∂y

dA,

and

F = I

H 1H 2H 3 g dA +

∂I ∩∂Dq

H 1H 2H 3 b ds

Using these matrices we obtain the element equations

M U̇ + KU = F.

For linear triangles the element stiffness matrix is the same as in the static case. The element mass matrixM is most easily computed by using a transformation of variables to covert the integral over the element toone over a simpler triangle T = {(r, s) : 0 < r < 1, 0 < s < 1 − r}. If (xj , yj), j = 1, 2, 3 are the verticesof the element triangle, a suitable transformation is x =

j H jxj , y =

j H jyj , where H j are the basis

functions for the triangle T :H 1(r, s) = 1 − r − sH 2(r, s) = r

H 3(r, s) = sThe indicated functions map T linearly onto the element triangle, and the Jacobian of the transformation(equal to the elemental area ratio) is ∂ (x, y)/∂ (r, s) = 2A, where A is the area of the element triangle.Changing the integration variables then gives

M = 2A

10

dr

1−r0

ds

1− r − sr

s

( 1− r − s r s ) = A

12

2 1 11 2 1

1 1 2

.

The corresponding global equations are of the form

M U̇ + KU = F.

This is essentially a system of ODEs for the values uj(t) of the solution at the nodes. As in the 1D case,all given boundary value information has to be input in order to create a legitimate system. (In this case,it is required that uj(t) = a(xj , yj , t) whenever node j is on ∂Du. This can be imposed by substitutingu̇j = ȧ(xj , yj , t) for the j th equation, and adding the boundary condition uj(0) = a(xj , yj , 0) = c(xj , yj).)

The remaining considerations involving integrating these equations forward in time are essentially the sameas in the 1D case. Let’s look at a slightly altered version of the file ex5111.m from the text which illustratesthe 2D case. The explicit method of time integration is chosen in this file. The numerical solution could becompared to the exact as computed using the separation of variables technique, but the authors have notdone this. It is easy to show that the exact solution at the center node is given by

u = 1001 − 2∞

n=1

(−1)n−1

λnexp(−

λ2n2

t) ,where λn = (2n−1)π/2, = 2.5. Thus I have included a computation of this value in the file below. Clearlythere are some inaccuracies at the early stages of the computation.

%----------------------------------------------------------------------------

%----------------------------------------------------------------------------

function ex5111_alt(nx,ny)

% to solve the transient two-dimensional Laplace’s equation

% u,t = u,xx + u,yy , 0 < x < 5, 0 < y < 2

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% boundary conditions:

% u(0,y,t) = 100, u(5,y,t) = 100,

% u,y(x,0,t) = 0, u,y(x,2,t) = 0

% initial condition:

% u(x,y,0) = 0 over the domain

% using linear triangular elements and forward difference method

%% Variable descriptions

% k = element matrix for time-independent term (u,xx + u,yy)

% m = element matrix for time-dependent term (u,t)

% f = element vector

% kk = system matrix of k

% mm = system matrix of m

% ff = system vector

% gcoord = coordinate values of each node

% nodes = nodal connectivity of each element

% index = a vector containing system dofs associated with each element

% bcdof = a vector containing dofs associated with boundary conditions

% bcval = a vector containing boundary condition values associated with

% the dofs in ’bcdof’%----------------------------------------------------------------------------

nx1=nx+1; ny1=ny+1;

nel=2*nx*ny; % number of elements

nnel=3; % number of nodes per element

ndof=1; % number of dofs per node

nnode=nx1*ny1; % total number of nodes in system

sdof=nnode*ndof; % total system dofs

deltt=0.1; % time step size for transient analysis

stime=0.0; % initial time

ftime=10; % termination time

ntime=fix((ftime-stime)/deltt); % number of time increment

%---------------------------------------------

% input data for nodal coordinate values

% gcoord(i,j) where i->node no. and j->x or y

%---------------------------------------------

[gcoord, nodes]=gridgen([0,0],[5,2],nx,ny,2);

gridplot(nx,ny,gcoord,nodes,2)

input(’Strike any key to continue solution’);

bcdof=[1:nx1:nnode-nx, nx1:nx1:nnode ]

lbc=length(bcdof);

bcval=100*ones(1,lbc);

%-----------------------------------------

% initialization of matrices and vectors

%-----------------------------------------

ff=zeros(sdof,1); % initialization of system vector

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fn=zeros(sdof,1); % initialization of effective system vector

fsol=zeros(sdof,1); % solution vector

sol=zeros(2,ntime+1); % vector containing time history solution

kk=zeros(sdof,sdof); % initialization of system matrix

mm=zeros(sdof,sdof); % initialization of system matrix

index=zeros(nnel*ndof,1); % initialization of index vector

%-----------------------------------------------------------------

% computation of element matrices and vectors and their assembly

%-----------------------------------------------------------------

for iel=1:nel % loop for the total number of elements

nd(1)=nodes(iel,1); % 1st connected node for (iel)-th element

nd(2)=nodes(iel,2); % 2nd connected node for (iel)-th element

nd(3)=nodes(iel,3); % 3rd connected node for (iel)-th element

x1=gcoord(nd(1),1); y1=gcoord(nd(1),2);% coord values of 1st node

x2=gcoord(nd(2),1); y2=gcoord(nd(2),2);% coord values of 2nd node

x3=gcoord(nd(3),1); y3=gcoord(nd(3),2);% coord values of 3rd node

index=feeldof(nd,nnel,ndof);% extract system dofs associated with element

k=felp2dt3(x1,y1,x2,y2,x3,y3); % compute element matrix

m=felpt2t3(x1,y1,x2,y2,x3,y3); % compute element matrix

kk=feasmbl1(kk,k,index); % assemble element matrices

mm=feasmbl1(mm,m,index); % assemble element matrices

end

%-----------------------------

% loop for time integration

%-----------------------------

for in=1:sdof

fsol(in)=0.0; % initial condition

end

sol(1,1)=fsol(8); % store time history solution for node no. 8

sol(2,1)=fsol(9); % store time history solution for node no. 9

for it=1:ntime % start loop for time integration

fn=deltt*ff+(mm-deltt*kk)*fsol; % compute effective column vector

[mm,fn]=feaplyc2(mm,fn,bcdof,bcval); % apply boundary condition

fsol=mm\fn; % solve the matrix equation

sol(1,it+1)=fsol(8); % store time history solution for node no. 8

sol(2,it+1)=fsol(9); % store time history solution for node no. 9

end

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%------------------------------------

% plot the solution at nodes 8 and 9

%------------------------------------

close all

time=0:deltt:ntime*deltt;

for k=1:ntime+1esol(k)=cntr((k-1)*deltt, 20, 2.5);

end

plot(time,sol(1,:),’*’,time,sol(2,:),’-’, time, esol,’o’);

xlabel(’Time’)

ylabel(’Solution at specified nodes’)

legend(’node 8’, ’node 9’, ’node 8 exact’,0)

%---------------------------------------------------------------

function [index]=feeldof(nd,nnel,ndof)

%----------------------------------------------------------

% Purpose:% Compute system dofs associated with each element

%

% Synopsis:

% [index]=feeldof(nd,nnel,ndof)

%

% Variable Description:

% index - system dof vector associated with element "iel"

% iel - element number whose system dofs are to be determined

% nnel - number of nodes per element

% ndof - number of dofs per node

%-----------------------------------------------------------

edof = nnel*ndof;

k=0;

for i=1:nnel

start = (nd(i)-1)*ndof;

for j=1:ndof

k=k+1;

index(k)=start+j;

end

end

function [k]=felp2dt3(x1,y1,x2,y2,x3,y3)

%-------------------------------------------------------------------

% Purpose:

% element matrix for two-dimensional Laplace’s equation

% using three-node linear triangular element

%

% Synopsis:

% [k]=felp2dt3(x1,y1,x2,y2,x3,y3)

%

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% k - element stiffness matrix (size of 3x3)

% x1, y1 - x and y coordinate values of the first node of element

% x2, y2 - x and y coordinate values of the second node of element

% x3, y3 - x and y coordinate values of the third node of element

%-------------------------------------------------------------------

% element matrix

A=0.5*(x2*y3+x1*y2+x3*y1-x2*y1-x1*y3-x3*y2); % area of the triangule

k(1,1)=((x3-x2)^2+(y2-y3)^2)/(4*A);

k(1,2)=((x3-x2)*(x1-x3)+(y2-y3)*(y3-y1))/(4*A);

k(1,3)=((x3-x2)*(x2-x1)+(y2-y3)*(y1-y2))/(4*A);

k(2,1)=k(1,2);

k(2,2)=((x1-x3)^2+(y3-y1)^2)/(4*A);

k(2,3)=((x1-x3)*(x2-x1)+(y3-y1)*(y1-y2))/(4*A);

k(3,1)=k(1,3);

k(3,2)=k(2,3);

k(3,3)=((x2-x1)^2+(y1-y2)^2)/(4*A);

function [m]=felpt2t3(x1,y1,x2,y2,x3,y3)

%-------------------------------------------------------------------

% Purpose:

% element matrix for transient term of two-dimensional

% Laplace’s equation using linear triangular element

%

% Synopsis:

% [m]=felpt2t3(x1,y1,x2,y2,x3,y3)

%


% m - element stiffness matrix (size of 3x3)

% x1, y1 - x and y coordinate values of the first node of element

% x2, y2 - x and y coordinate values of the second node of element

% x3, y3 - x and y coordinate values of the third node of element

%-------------------------------------------------------------------

% element matrix

A=0.5*(x2*y3+x1*y2+x3*y1-x2*y1-x1*y3-x3*y2); % area of the triangle

m = (A/12)* [ 2 1 1;

1 2 1;

1 1 2 ];

function [kk]=feasmbl1(kk,k,index)

%----------------------------------------------------------

% Purpose:

% Assembly of element matrices into the system matrix

%

% Synopsis:

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% [kk]=feasmbl1(kk,k,index)

%


% kk - system matrix

% k - element matri

% index - d.o.f. vector associated with an element

%-----------------------------------------------------------

edof = length(index);

for i=1:edof

ii=index(i);

for j=1:edof

jj=index(j);

kk(ii,jj)=kk(ii,jj)+k(i,j);

end

end

function [kk,ff]=feaplyc2(kk,ff,bcdof,bcval)

%----------------------------------------------------------

% Purpose:

% Apply constraints to matrix equation [kk]{x}={ff}

%

% Synopsis:

% [kk,ff]=feaplybc(kk,ff,bcdof,bcval)

%


% kk - system matrix before applying constraints

% ff - system vector before applying constraints

% bcdof - a vector containging constrained d.o.f

% bcval - a vector containing contained value

%

% For example, there are constraints at d.o.f=2 and 10

% and their constrained values are 0.0 and 2.5,

% respectively. Then, bcdof(1)=2 and bcdof(2)=10; and

% bcval(1)=1.0 and bcval(2)=2.5.

%-----------------------------------------------------------

n=length(bcdof);

sdof=size(kk);

for i=1:n

c=bcdof(i);

for j=1:sdof

kk(c,j)=0;

end

kk(c,c)=1;

ff(c)=bcval(i);

end

%-----------------------------------------------------------

function u=cntr(t,n,lx)

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% exact solution at the center node

% t=time, n=number of terms in sum, lx=half-length of region

v=1:2:2*n+1; %n+1 terms

w=cos((0:n)*pi);

lam=pi*v/(2*lx);

series=w.*(v.^-1).*exp(-(lam.^2)*t);

u=100-(400/pi)*sum(series);

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me471s03_lec09