Optimal Control for 2D Parabolic Problems

Optimal Control for 2D Parabolic Problems
Jonas Holländer
Supervisors: Holger Heumann, Jacques Blum
June 24, 2012
1

Contents
1 Theory of the optimal control problem 2
1.1 Introduction of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Discretization of the constraint . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Discretization of the costfunctional . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Set up of the Lagrangian and first order optimality conditions . . . . . . . 5
1.5 Collecting the optimality conditions . . . . . . . . . . . . . . . . . . . . . 5
2 Implementation of the Optimal Control Problem in Fenics 7
2.1 Motivation to use Fenics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Implementation of the domains and the boundaries . . . . . . . . . . . . . 7
2.3 Implementation of the functionspace . . . . . . . . . . . . . . . . . . . . . 8
2.4 Implementation of the boundary conditions . . . . . . . . . . . . . . . . . 9
2.5 Implementation of the linear system and the initial conditions . . . . . . . 9
2.6 Solving the linear system . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Results 12
3.1 Expected behavior of u and f . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.2 Comparison between expectation and model . . . . . . . . . . . . . . . . . 12
3.3 Numerical limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 Conclusion 17
References 18
1 Theory of the optimal control problem
1.1 Introduction of the problem
In this project we want to solve an optimal control problems like the one given through
equations (1)-(5).
∫ T
min
u(x,t),f(x,t) 0
ω 1
2 〈f(x, t), f(x, t)〉 Ω C
+ ω 2
2 〈u(x, t)−sin(2πt), u(x, t)−sin(2πt)〉 Ω P
dt, (1)
c(x) ∂ t u(x, t) − ∆u(x, t) = f(x, t) in Ω, (2)
{ { ≠ 0 if x ∈ ΩC
1 if x ∈ ΩC
with f(x) =
c(x) =
= 0 else ,
0 else ,
(3)
u(x, 0) = u 0 (x) in Ω, (4)
u(x, t) = 0 on ∂Ω. (5)
2

Figure 1: The definition of the domain of the optimal control problem
Where the domains are defined through:
Ω := [0, 1] × [0, 1],
Ω P := [0.3, 07] × [0.25, 0.75],
⎧
[0.15, 0.25] × [0.1, 0.2]
⎪⎨
[0.15, 0.25] × [0.8, 0.9]
Ω C :=
[0.75, 0.85] × [0.1, 0.2]
⎪⎩
[0.75, 0.85] × [0.8, 0.9],
(6)
which means there are 2 functions searched, which are connected by a partial differential
equation. This function should be furthermore the solution of a minimization problem.
The domain we want to consider in our problem is divided into 5 subdomains:
• The subdomain Ω 5 or Ω P is a rectangular centered in the middle of the domain. It
is of interest since the second part of equation (1) just depends on these domain.
• The subdomains Ω 1 , Ω 2 , Ω 3 and Ω 4 are part of the subdomain Ω C . These subdomains
are small squares surrounding Ω P at the corners. The function f as well
as the constant in front of the time derivative are just different from 0 inside this
domain.
The subdivision of the domain is illustrated in picture (1). We want to solve this problem
with the help of finite element methods. This means we first want to write the discrete
versions of the problem, and derive the Lagrangian of this discrete version. Afterwards we
derivate the Lagrangian to achieve the first order optimality conditions. This conditions
3

can be represented through a linear system, which was assembled and solved with the
help of the program fenics. 1
1.2 Discretization of the constraint
We want to find a solution of this problem with the help of the finite element method.
Therefor we introduce a discretization of space and time and write:
f(x, t n ) =
N∑
fi n φ i (x) and u(x, t n ) =
i=1
N∑
u n i φ i (x), (7)
i=1
where φ i (x) denotes one of the basisfunctions of the interpolationspace S. This discretization
is applied to the weak formulation of equation (2):
∫
∫
c(x)∂ t u(x, t)ψ(x, t) + ∇u(x, t) · ∇ψ(x, t)dx = f(x, t)ψ(x, t)dx. (8)
Ω
We define 2 finite element matrices now through:
∫
∫
∫
c i,j := φ i (x)φ j (x)dx, m i,j := φ i (x)φ j (x)dx, k i,j :=
Ω C Ω
Ω
Ω
∇φ i (x) · ∇φ j (x)dx, (9)
and use the notation of u n := (u n 1 , un 2 , ..., un I
) to write the discrete version of equation
(8) in the following way:
v t C (un+1 − u n )
δt
+ v t Ku n+1 − v t Cf n = 0 ∀v ∈ R I . (10)
Since this equation holds for any v ∈ R I , it holds especially for the basisvectors with
which it holds for any component and we can write the equation without v t and we get
our constraint:
K n (u, f) := C (un − u n−1 )
+ Ku n − Cf n = 0. (11)
δt
1.3 Discretization of the costfunctional
The next step we have to go for computing the first order optimality condition is to
determine the discrete version of the costfunctional given through equation (1). Here we
use the discretization:
J(u, f) :=
N∑
(f n ) t Mf n δtω 1
2
n=1
+
N∑
(u n − sin(2πt n )) t P(u n − sin(2πt n )) δtω 2
2 , (12)
n=1
where the Matrix P denote the components of 〈φ i , φ j 〉 ΩP .
1 http://fenicsproject.org
4

1.4 Set up of the Lagrangian and first order optimality conditions
For computing the first order optimality conditions we set up the Lagrangian and compute
the derivatives with respect to the components u, f and λ. The Lagrangian can be
written in the form:
L(u, f, λ) = J(u, f) +
N∑
〈λ i , K n (u, f)〉. (13)
From this equation we get the first order optimality conditions as the derivatives of
L(u, f, λ) with respect to the variables u, f and λ. The derivative with respect to λ is
given through equation (11). The optimality conditions are then:
n=1
∂L(u, f, λ)
∂u i = ω 2 δtP(u i − sin(2πt i )) + G t λ i − 1 δt Ct λ i+1 = 0 ∀i ≠ n, (14)
∂L(u, f, λ)
∂u n = ω 2 δtP(u n − sin(2πt n )) + G t λ n = 0, (15)
∂L(u, f, λ)
∂f i = ω 1 δtMf i − C t λ i = 0, (16)
∂L(u, f, λ)
∂λ i = −C 1 δt ui−1 + Gu i − Cf i = 0, (17)
where G was defined through G := 1 δt C + K.
1.5 Collecting the optimality conditions
The next step to take is to form out of this equations a system of equation to be able
to solve it with any linear solver. Therefor we collect all the terms of u, f and λ in a
solution vector and get the following block matix:
⎛ ⎞ ⎛
u A 1 0 A t ⎞ ⎛ ⎞ ⎛ ⎞
2 u 1
A ⎝f⎠ = ⎝ 0 A 3 A t ⎠ ⎝
4 f⎠ = ⎝b
0 ⎠ , (18)
λ A 2 A 4 0 λ b 2
The matrices A 1 , A 3 and A 4 are again blockmatrices with only blocks on the diagonal,
the matrix A 2 has a slightly different shape. All their forms are given by:
⎛
⎞
ω 2 δtP 0 . . . 0
A 1 :=
0 ω 2 δtP . . . .
⎜
⎝
.
. .. . ..
⎟ 0 ⎠ , (19)
0 . . . 0 ω 2 δtP
5

⎛
⎞
G 0 . . . . . . 0
−C 1 δt
G . . .
A 2 :=
0 −C 1 δt
G . . .
. . .. . .. . .. . .. , (20)
.
⎜
.
⎝ .
.. . .. . .. ⎟ 0 ⎠
0 . . . . . . 0 −C 1 δt
G
⎛
⎞
ω 1 δtM 0 . . . 0
A 3 :=
0 ω 1 δtM . . . .
⎜
⎝
.
. .. . ..
⎟ 0 ⎠ , (21)
0 . . . 0 ω 1 δtM
⎛
⎞
−C 0 . . . 0
A 1 :=
0 cC . . . .
⎜
⎝
.
. .. . ..
⎟ 0 ⎠ . (22)
0 . . . 0 −C
It is here nice to notice that A is symmetric which means that A = A t because of the
symmetry of A 1 and A 3 and of the structure and position of the other matrices.
The vector b 1 is determined through equation (14) and (15), here the part −ω 2 δtPsin(2πt n )
is not depending on u, f and λ and is there for moved to the right hand side of the equation.
The other vector b 2 is almost every where 0. The first term depends on u 0 which
is given through the initial conditions of optimal control problem. Since this term also
not depends on any other components of u, f and λ it has to appear on the right side
through 1 δt Cu0 .
We find the desired solution u by solving the linear system (18).
6

2 Implementation of the Optimal Control Problem in Fenics
2.1 Motivation to use Fenics
The biggest problem of solving equation (18) is the implementation of all the finite element
matrices needed. This matrices make it hard to implement this problem in 1
dimension and even more complicated to handle this problem in 2 or more dimensions.
However, the form of equation (18) will not change, just the submatrices inside change.
To solve this problem we there for take the software fenics, since it makes the implementation
incredibly much easier. This software has the idea to assemble in a easy way
all the finite element matrices, apply the boundary conditions and solve the constructed
linear system. Furthermore it allows us to define all the necessary subdomains and the
right functionspaces.
The last important reason for using fenics is the possibility to adapted the code fast to
other optimal control problem, which would take without it incredible much time since
it needs all the finite element matrices to be changed.
2.2 Implementation of the domains and the boundaries
The first necessary step to take in the implementation is the definition of the domains.
We need to define a unit-square with the 5 subdomains Ω 1 to Ω 5 , to define the variational
formulation of our problem. To be able to solve our problem we need the appropriate
boundary conditions, in our case Dirichlet boundary conditions, for which we need to
define the boundary. All this is shown in the following lines of code exemplary for the
left boundary and the domain Ω 1 :
1
# D e f i n i t i o n o f the domain and the f u n c t i o n s p a c e
3 mesh = UnitSquare ( 6 4 , 64)
5 c l a s s L e f t ( SubDomain ) :
d e f i n s i d e ( s e l f , x , on boundary ) :
7 r e t u r n near ( x [ 0 ] , 0 . 0 )
9 c l a s s Areaone ( SubDomain ) :
d e f i n s i d e ( s e l f , x , on boundary ) :
11 r e t u r n ( between ( x [ 0 ] , ( 0 . 1 5 , 0 . 2 5 ) ) and between ( x [ 1 ] , ( 0 . 1 , 0 . 2 ) ) )
Line 2 defines the mesh as a unit-square with 64 × 64 elements. In line 4 the initialization
of the left boundary starts. It is define as all nodes of the mesh which have
|x| < ɛ. It is not take the exact value x = 0 to prevent the loss of boundary points
through rounding errors. We define the subdomain Ω 1 in a similar way in the area
(0, 15 ≤ x ≤ 0, 25 × 0, 1 ≤ x ≤ 0, 2)
7

After the initialization of the mesh function for the interior domains and boundaries,
which is done in line 1-11 of the following code block. It is necessary to define symbolic
measures for the domains and boundaries. This makes the process of creating the
variational formulation easier.
1
# I n i t i a l i z e mesh f u n c t i o n f o r i n t e r i o r domains
3 domains = CellFunction ( ” u i n t ” , mesh )
domains . s e t a l l ( 0 )
5 areaone . mark ( domains , 1)
7 # I n i t i a l i z e mesh f u n c t i o n f o r boundary domains
boundaries = FacetFunction ( ” u i n t ” , mesh )
9 boundaries . s e t a l l ( 0 )
l e f t . mark ( boundaries , 1 )
11
# Define new measures a s s o c i a t e d with the i n t e r i o r domains and
13 # e x t e r i o r boundaries
dx = Measure ( ”dx” ) [ domains ]
15 ds = Measure ( ” ds ” ) [ boundaries ]
2.3 Implementation of the functionspace
Our aim is to solve the optimal control problem in the space of continuous function.
Because of this we want to chose Lagrange elements to interpolate our solution. Every
time step we have to solve a linear system for the 3 unknowns u, f and λ, the problems
for different time steps depend on each which makes it hard to solve them separately.
To avoid this problem we convert it to a finite element problem of a vectorial function,
where the components are one of the unknown functions u, f and λ at a certain timestep.
The creation of this functionspaces is done in the programm through the following lines
of code:
1 #d e f i n i t i o n o f the f u n c t i o n s p a c e s
V = FunctionSpace ( mesh , ’ Lagrange ’ , 1)
3 S = MixedFunctionSpace ( [V] ∗ 3 ∗ number )
u = T r i a l F u n c t i o n s ( S )
5 v = TestFunctions ( S )
The second line of codes defines the functionspace for one of the unknowns u, f or λ, and
states that the space should be the space of first order Lagrange elements on the previous
defined mesh. The next line defines the space of functions which we actually use, as the
productspace of V 3N , with N as the number of time steps. In the last lines some trial and
test functions are initialized, which are later necessary to define the variational problem.
8

2.4 Implementation of the boundary conditions
This subsection is on the implementation of the boundary conditions of our optimal
control problem. The easiest boundary conditions applicable are Dirichlet boundary
conditions and these are the ones we want to use. The problem here is to find the right
boundary values since we are now in the space V 3N , which means we have to define
boundary values for all components.
The boundary values of all components associated to u are 0, because of equation (5).
The values associated to f are 0, because of (3) and the fact that there is no intersection
of Ω C and the boundary. That all components of λ are 0 comes from equation (14) and
the fact that u = 0 on the boundary. Since it is now clear that all components vanish
on the boundary, we can have a look at the implementation:
1 #v e c t o r f o r the boundary c o n d i t i o n s
vec = [ 0 . 0 ] ∗ 3 ∗ number
3 # Define D i r i c h l e t boundary c o n d i t i o n s at top and bottom boundaries
bcs = [ DirichletBC (S , vec , boundaries , 1) ,
5 DirichletBC (S , vec , boundaries , 2) ,
DirichletBC (S , vec , boundaries , 3) ,
7 DirichletBC (S , vec , boundaries , 4) ]
Here we do two things, first we create a vector with all components 0 and the appropriate
size for our boundary conditions, second we create a list of boundary conditions. The
components of the list are 0 boundary conditions, of the functionspace S applied to the
boundaries 1-4. Where the numbers 1-4 correspond to the markings previously done.
2.5 Implementation of the linear system and the initial conditions
In this section I want to present the way equation (18) was inserted into Fenics. To
illustrate this I will use the first line of equation (18):
⎛ ⎞
⎛ ⎞
u
B ⎝f⎠ = ( A 1 0 A t )
u
⎝
2 f⎠ = b 1 . (23)
λ
λ
The Matrix we want to consider has N ×3N components where each component contains
a finite element matrix. The Matrix A 1 is a block diagonal Matrix with the Matrix ω 2 δtP
on the diagonal. It is created through:
1 #d e f i n i t i o n o f the matrix
aa = omega2 ∗ dt ∗ dot ( u [ 0 ] , v [ 0 ] ) ∗dx ( 5 )
3 #e n t e r i n g Matrix A 1
f o r i i n range ( 1 , number ) :
5 aa = aa + omega2 ∗ dt ∗ i n n e r ( u [ i ] , v [ i ] ) ∗dx ( 5 )
9

The definition of aa contains the test and the trial function from which it is clear that
we want to create a matrix. The part dot(u[0],v[0]) stands for the finite element matrix
associated to mass matrix m in equation (9), and the two [0] indicate that it’s the 0/0
block-matrix of the matrix B. The multiplication with dx(5) stands for integrating this
part of the equation just over the domain marked with number 5, in our case P. In
the loop afterwards all further matrices of A 1 are added. The implementation of the
matrices A 2 is done similar by:
1 #e n t e r i n g Matrix A 2t
f o r i i n range ( number ) :
3 f o r j i n range ( 6 ) :
aa = aa + dot ( grad ( u [ i +2∗number ] ) , grad ( v [ i ] ) ) ∗dx ( j )
5 f o r i i n range ( number ) :
f o r j i n range ( 1 , 5 ) :
7 aa = aa + (1/ dt ) ∗ dot ( u [ i +2∗number ] , v [ i ] ) ∗dx ( j )
f o r i i n range ( number−1) :
9 f o r j i n range ( 1 , 5 ) :
aa = aa −(1/ dt ) ∗ dot ( u [ i +2∗number +1] , v [ i ] ) ∗dx ( j )
The matrix A t 2 consists of 3 parts, 2 matrices on the diagonal blocks and one up the
diagonal. The matrices on the diagonal blocks are added again in the same way, the 2
differences are first the shifting of u by a factor of 2*number and second the different
term in the first matrix. The command dot(grad(u),grad(v)) stands for another finite
difference matrix, the last one given through equation (9). The last matrix up of the
diagonal is implemented similar to the previous one, just with an additional factor of 1
added to indicate that the different position. All the summations over j indicate over
which domains the integration takes part, e.g. for the second and third matrices over
Ω 1 to Ω 4 .
The initial conditions we want to consider are given through:
⎛ ⎞
⎛ ⎞
⎛
1
sin(2πt
⎝b n ⎞
1
)
δt Cu0
0 ⎠ ⎜ ⎟
0
with b 1 := ω 2 δtP ⎝ . ⎠ and b 2 := ⎜ ⎟
b 2 sin(2πt n ⎝ . ⎠ , (24)
)
0
They are implemented via the following lines:
#d e f i n i t i o n o f the i n i t i a l c o n d i t i o n s
2 u0 = Expression ( ”− 10∗ exp(− pow( x [ 1 ] − 0 . 5 , 2) ) ” )
f = (1/ dt ) ∗u0∗v [ 2 ∗ number ] ∗ dx ( 1 )+ (1/ dt ) ∗u0∗v [ 2 ∗ number ] ∗ dx ( 2 ) +(1/ dt ) ∗u0∗v
[ 2 ∗ number ] ∗ dx ( 3 ) +(1/ dt ) ∗u0∗v [ 2 ∗ number ] ∗ dx ( 4 )
4 f o r i i n range ( number ) :
f = f +s i n (2∗ i ∗ dt ∗ 3 . 1 4 1 5 ) ∗omega2∗ dt ∗v [ i ] ∗ dx ( 5 )
10

The second line gives the possibility to implement a function as initial conditions. In
the next line a right hand side vector is created which contains b 2 . Afterwards the initial
conditions for b 1 are set in a loop.
It is to mention that we just have to change this part if we want to use the program to
solve another optimal control problem in the same domain.
2.6 Solving the linear system
The last step left now is to solve the linear system. Here for we use the default solver of
fenics UMFpack. The solution command does 3 things it, first it assembles out of the
variational formulation the corresponding matrices and in a second step the boundary
conditions are applied. The system is solved in the last step. The code takes here for
just two lines.
1 #s o l v i n g the system
U = Function ( S )
3 s o l v e ( aa==f ,U, bcs )
We specify the solution variable U, and give it as an input for the solution procedure.
Then the linear system aaU = f with the boundary conditions is solved.
11

3 Results
3.1 Expected behavior of u and f
I want to represent the results I gained in this section. I want to begin with some words
on the what results I would expect. This depend on the factors ω 1 and ω 2 of the costfunctional,
more precisely on their ratio.
With a large factor ω 1 compared to ω 2 the first term of the costfunctional becomes dominating
which means that the function f tends to 0 in Ω C . The relation between u and
f given through equation (2) together with the boundary conditions force u to be 0 in
the whole domain as well.
If the ratio between ω 1 and ω 2 is converse, the second term of the costfunctional is dominant
which makes the function u behave sinusoidal in the subdomain Ω P . The influence
of the first term of the costfunctional on the function f will become negligible. The
function f will be totally determined by the maximization of u.
3.2 Comparison between expectation and model
That the computed results fit with this properties is visible from the graphs on the next
pages. The values of u and f were plotted for both ratios once (ω 1 = 1, ω 2 = 10 13 and
ω 1 = 1, ω 2 = 1). It is here to notice that we are in the first case for the same value
of ω 1 and ω 2 , the value of ω 2 has to be on the other hand very large to see the results
associated to the second case.
To be able to see the oscillations in the subdomain Ω P only the values of u in there
are shown, since u attains large values in Ω C which distract the view. A full picture,
without this restrictions is for example shown on the first page.
It is visible that the expected results correspond with the model. For small values of ω 2
the values in both graphs u and f tend to 0. It is further visible that they satisfy the
boundary conditions, and that f is just different from 0 in the domain Ω C .
The graphs for large values ω 2 show the expected oscillation in the domain Ω P , furthermore
they satisfy the boundary conditions and the fact that f should be different from
0 just in Ω C .
For the visualizations the following parameters were taken. For the time stepping differences
of δt = 0.1 were taken. There are no graphs of the initial conditions included,
so the time starts at 0.1 and finished at 0.6. The values for ω 1 and ω 2 are as above
mentioned: ω 1 = 1, ω 2 = 10 13 and ω 1 = 1, ω 2 = 1. And the number of basisfunctions
taken on the unitsquare are 50 × 50.
12

Figure 2: This picture shows the
graph of u at time t=0.1 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
Figure 3: This picture shows the
graph of f at time t=0.1 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
Figure 4: This picture shows the
graph of u at time t=0.2 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
Figure 5: This picture shows the
graph of f at time t=0.2 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
Figure 6: This picture shows the
graph of u at time t=0.3 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
Figure 7: This picture shows the
graph of f at time t=0.3 restricted
to the domain Ω P with ω 1 = ω 2 = 1.
To produce this graph 50 × 50 basisfunctions
were used.
13

Figure 8: This picture shows the
graph of u at time t=0.1 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 9: This picture shows the
graph of u at time t=0.2 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 10: This picture shows the
graph of u at time t=0.3 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 11: This picture shows the
graph of u at time t=0.4 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 12: This picture shows the
graph of u at time t=0.5 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 13: This picture shows the
graph of f at time t=0.6 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
14

Figure 14: This picture shows the
graph of f at time t=0.1 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 15: This picture shows the
graph of f at time t=0.2 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 16: This picture shows the
graph of f at time t=0.3 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 17: This picture shows the
graph of f at time t=0.4 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 18: This picture shows the
graph of f at time t=0.5 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
Figure 19: This picture shows the
graph of f at time t=0.6 restricted
to the domain Ω P with ω 1 = ω 2 =
10 13 . To produce this graph 50 × 50
basisfunctions were used.
15

3.3 Numerical limits
The main problem in computing the solution of the optimal control problem in the
amount of memory needed. For every time step we need to store 3 finite element matrices.
This exceeds the available amount of memory fast, especially in higher dimensions. It
is also necessary to pick the more finite elements for the space dimensions, to achieve a
resolution of the different domains.
I computed the results in two dimensions with up to 6 time steps and a mesh of 50 × 50
elements. To determine the maximum number of elements possible to use, i gradually
increased the number until the program is no more able to invert the matrix. This was
achieved by a mesh with 86 × 86 elements. The problem is caused by UMFpack which
is not able to handle a matrix of this size.
Another problem of computation appeared here. By increasing the number of timesteps,
or the resolution, the computational time increases also significant.
A way to solve the problem with the total memory needed could be use another way of
solving the matrix. The structure of the linear system has many repeated part which
are saved several times. To optimize this part would be a way to increase the number of
possible timesteps. This would be on the other hand no solution for the computational
time needed. A possibility to solve this issue could be the splitting of the program on
several CPU’s.
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4 Conclusion
This semester I listen courses on finite differences and finite volumes, finite elements, optimization
and algorithms. It was for me therefor a good opportunity to see how these
lectures are connected. In this project we want to solve an optimal control problem
which combines topics from all lectures. The general setting of minimizing a function
with a constraint and the way it is solved, by means of a Lagrangian is typically for optimization.
The constrain has the shape of a partial differential equation and introduces
therefor finite differences and finite element aspects, from which we for example take
the implicit scheme to avoid CFL conditions. The last part, the implementation needs
knowledge about algorithms to be able to solve the problem.
The work with fenics was especially interesting for me. The part which interested me
the most was the way finite element methods are implemented. The program uses a
syntax which is very similar to the abstract formalism used by the variational formulation
of finite element methods. This makes it easy to understand the code without
much background in python, the language of the fenics interface. It leaves on the other
hand flexibility in the problem itself as well as in the computation of the results. It is
possible to use arbitrary functionspaces, domain, or partial differential equations. The
choice of the linear solver used to compute the solution is one freedom connected to the
computation.
Another big part of the project was finding the appropriate notation and calculating
carefully everything before starting. Mistakes in the beginning have later big consequences,
and lead to wrong results or make it impossible to find a solution. It takes
usually more time to erase this mistakes then being carefully from the beginning. I did
such mistakes and they cost lots of time and nerves.
And in the end it is always satisfying to have after several days a working program.
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References
[1] Automated Solution of Differential Equations by the Finite Element
Method,A. Logg and K.-A. Mardal and G. N. Wells
[2] DOLFIN User Manual, Logg, Wells, et al.
[3] The Mathematical Theory of Finite Element Methods,Brenner, Susanne
C., Scott, Ridgway
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