Introduction to Mathematical Modelling:Ordinary Differential Equations and Heat Equations
Knut–Andreas Lie
SINTEF ICT, Dept. Applied Mathematics
INF2340 / Spring 2005 – p. 1
Overview
1. Ordinary differential equations• Example of ODE models
− radioactive decay, Newton’s second law, populationmodels
• Numerical solution of ODEs− forward and backward Euler, Runge-Kutta methods
2. Heat equations• from physical problem to simulator code• steady heat conduction
− finite differences, linear algebra• unsteady heat conduction
− finite differences, boundary conditions
INF2340 / Spring 2005 – p. 2
Ordinary Differential Equations (ODEs)
y(m) = f(
t, y, y, y, . . . , y(n−1))
Simple example: radioactive decay
Given a quantity q of a radioactive matter, which decays ata certain constant rate k. The model reads
dq(t)
dt= r(t) = −k · q.
Solution: q(t) = q(t0)e−k(t−t0).
In general: finding a solution is not so easy, although there areapproaches in certain special cases −→ numerical approach!
INF2340 / Spring 2005 – p. 3
ODEs cont’d
Very simple example from high school physics:
Consider the motion of a body with mass m under constantforce f , which is initially at rest at position x0.
Newton’s second law reads
f = ma = mx(t)
where x(t) is the position of the body at time t. Thecorresponding ODE then reads
f = mx(t), x(0) = x0, x(0) = 0
This equation is easily integrated
x(t) = x0 + f2m t2.
INF2340 / Spring 2005 – p. 4
Population models
Consider the dynamics of a single species (isolated or with nopredators)
• constant birth rate b per time and individual• constant death rate d per time and individual• hence, constant growth rate λ = b − d
The model (Maltus, 1798):
p(t) = λ · p(t)
Solution p(t) = p0eλt predicts exponential growth or decay.
INF2340 / Spring 2005 – p. 5
Population models cont’d
Is there any realism in this?• between 1700 and 1960: growth rate of about 0.02,
population doubles in 34.67 years• generally: limited resources on earth slows down growth
More realism (Verhulst et al., 19th century):• linear rates: b(t) = b0 − b1p(t), d(t) = d0 − d1p(t)
−→ new model:p(t) = −k(p(t) − p∞)
Solution:p(t) = p∞ + (po − p∞)e−kt
INF2340 / Spring 2005 – p. 6
Population models cont’d
What about realism now?• Populations tend to follow a S-shape (logistic model)• New model: p(t) = a · p(t) − b · p2(t)
• New solution: p(t) = a·p0
b·p0+(a−b·p0)e−at
And so on ... adding more than one species (predator–pray) ...
Do we have equilibrium, is it attractive or repellent, ...?
INF2340 / Spring 2005 – p. 7
Population models cont’d
An ODE model from modern research, describing the dynamics ofHIV-1 infection in vivo (Perelson& Nelson, SIAM Review 41/1, 1999) :
The rate of change of uninfected cells T , productively infected cells T ∗,and virus V :
dT
dt= s + pT
`
1 − T/Tmax´
− dT T − kV T
dT ∗
dt= kV T − δT ∗
dV
dt= NδT ∗
− cV.
Here:
dT – death rate of uninfected cells
δ – death rate of infected cells
p – rate of proliferation (continuous development of cells in tissue)
N – virus production per infected cell
c – clearance rate
INF2340 / Spring 2005 – p. 8
Numerical solution of ODEs – Euler’s method
We wish to solve the equation:
y′ = f(y, t), y(a) = α, a ≤ t ≤ b
Obvious solution – use finite differences:• generate a mesh: ti = a + ih, for each i = 0, . . . , N , where
h = (b − a)/N is called stepsize• apply forward differences to the equation
y(ti+1) = y(ti) + hf(y(ti), ti), i = 0, . . . , N − 1
This gives an explicit formula for each y(ti+1) once y0 is known.
INF2340 / Spring 2005 – p. 9
Example: Euler’s method for u′ = 3u
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
2
4
6
8
10
12
14
16
18
20Exact h=1/4 h=1/16h=1/64
INF2340 / Spring 2005 – p. 10
Another Euler method – backward Euler
Once again we consider:
y′ = f(y, t), y(a) = α, a ≤ t ≤ b
and introduce a mesh: ti = a + ih, for each i = 0, . . . , N
This time we apply backward differences
y(ti+1) = y(ti) + hf(y(ti+1), ti+1), i = 0, . . . , N − 1
This gives an equation for each y(ti+1) once y0 is known.
INF2340 / Spring 2005 – p. 11
Two different methods
• forward Euler: explicit method
y(ti+1) = y(ti) + hf(y(ti), ti)
the new value is given by a formula
• backward Euler: implicit method
y(ti+1) = y(ti) + hf(y(ti+1), ti+1)
the new value is given by an algebraic equation
INF2340 / Spring 2005 – p. 12
Example: u′ = −4π sin(4πx)
u(xi) = u(xi−1) − 4π sin(4πxi−1) u(xi) = u(xi−1) − 4π sin(4πxi)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1 Exact Forward Euler Backward Euler
INF2340 / Spring 2005 – p. 13
Discretisation errors
When using finite differences in forward Euler, we make a localdiscretisation error at each point
y(ti+1) − y(ti)
h= f(y(ti), ti) + error
Consider y = f(y). Expanding y(ti+1) by a Taylor polynomial:
y(ti) + hy(ti) + y(τ) 12h2 − y(ti) = hf(y(ti)) + h · error
for ti ≤ τ ≤ ti+1. Now since y(ti) = f(y(ti)),
error = y(τ) · 12h =
d
dtf(y(τ)) · 1
2h
We say that forward Euler is a first-order method.
INF2340 / Spring 2005 – p. 14
Discretisation errors cont’d
We say that the scheme is consistent if
l(h) = maxt∈[a,b]
∣
∣
∣
y(t + h) − y(t)
h− f(y(t), t)
∣
∣
∣→ 0 as h → 0
Similarly, we define the global discretisation error as
e(h) = maxti∈[a,b]
∣
∣yi − y(ti)∣
∣,
where y(ti) is the exact solution and yi is computed by our scheme.
The scheme is convergent if
e(h) → 0 for t → 0
INF2340 / Spring 2005 – p. 15
Higher order methods – Runge–Kutta
Consider the forward Euler method and try to evaluate f(·) at someother point
yi+1 = yi + hf(yi + β).
Let us repeat the error analysis
y(ti+1) − y(ti) = f(yi)h + f(yi)12h2 + f(τ) 1
6h3
= hf(yi + β) + h · error
Assume now that the error equals 16 f(τ)h2. Now f(yi) = f ′(yi)f(yi)
and we have
f(yi) + f ′(yi)f(yi)12h = f(yi + β)
= f(yi) + βf ′(yi) + β2 + . . .
This means that β = 12h f(yi).
INF2340 / Spring 2005 – p. 16
Runge-Kutta methods cont’d
Thus, we have a new second-order method
yi+1 = yi + hf(
yi + 12h f(yi)
)
There are other alternatives also, e.g.,
yi+1 = yi + 12h
[
f(yi) + f(
yi + hf(yi))]
The Runge-Kutta methods are all on the form:
• Approximate the solution at a point ti ≤ τ ≤ ti+1 by the intermediate stepw = yi + (τ − ti)f(yi)
• Combine yi, w, f(yi), f(w) to get a more accurate solution yi+1.
For higher-order methods we use more intermediate steps.
INF2340 / Spring 2005 – p. 17
From ODEs to PDEs
So far, we have used ODEs:• involve derivatives with respect to only one variable• if the unknown function depends on more variables, these
have been treated as constants
Now, partial differential equations (PDEs):• involve (partial) derivatives with respect to more than one
variable• the unknown function depends on more than one variable• stationary problems: no time-dependence• unsteady problems. time-dependence, but maybe steady
limit
INF2340 / Spring 2005 – p. 18
Mathematical model: ∂tu = ∆u + f
• Models propagation of heat within a given object• Examples
− a heated rod (1D)− a heated plate or smoothing of images (2D)− heat distribution in a furnace (3D)
• Functions of interest− u(x,t), u(x,y), u(x,y,t), u(x,y,z), u(x,y,z,t), ...
• More generally, heat propagation depends on theconductivity of the material
ut = ∇(K(x, u)∇u) + f(x, t, u)
INF2340 / Spring 2005 – p. 19
Heat conduction in the continental crust
x
sT
x=b
x=0
-Q
• Knowing the temperature at the earth’s surface and the heat flowfrom the mantle, what is the temperature distribution through thecontinental crust?
• Interesting question in geology and geophysics – and for thosenations exploring oil resources...
INF2340 / Spring 2005 – p. 20
Physical and mathematical model• Our prototype differential equation (an ODE in 1D):
−∂2u
∂x2= f(x)
Here u is the temperature.
• Needs to be equipped with boundary conditions
• Very simple equation, but it has applications to− fluid flow in channels− deflection of electric cables− strength analysis of beams− ...
• In multidimensions −∆u = f is the elliptic Poission equation,which is fequently occuring in mathematical models
INF2340 / Spring 2005 – p. 21
Heat conduction in the continental crust cont’d
x
sT
x=b
x=0
-Q
Physical assumptions:
• Crust of infinite area
• Steady state heat flow
• Heat generated by radioactivedecay
Physical quantities:
• u(x) : temperature
• q(x) : heat flux (velocity of heat)
• s(x) : heat release per unit timeand mass
INF2340 / Spring 2005 – p. 22
Derivation of the model
x
x=0
x=b
outflow
inflow
s(x)=R exp(-x/L)
Physical principles:
• First law of thermodynamics:
net outflow of heat = total generated heat
• Fourier’s law: heat flows from hot to cold regions (i.e. heat velocity is poportionalto changes in temperature)
q(x) = −λu′(x)
• Heat generation due to radioactive decay:
s(x) = R exp(−x/L)
INF2340 / Spring 2005 – p. 23
Derivation of the model cont’d
x
s(x) h
q(x-h/2)
q(x+h/2)
x=0
x=b
From the first law of thermodynamics:
q(x + h/2) − q(x − h/2)
h= s(x)
Using a Taylor expansion:
q(x + h/2) − q(x − h/2)
h= q′(x) +
1
24q′′′(x)h2 + . . .
Hence, as h → 0 we have
q′(x) = s(x)
INF2340 / Spring 2005 – p. 24
Derivation of the model cont’d
x
sT
x=b
x=0
-Q
Combining the 1st law of thermodynamics (q′ = s) with Fourier’s law (q = −λu), we get
−d
dx
„
λdu
dx
«
= s(x)
Boundary conditions:
• u(0) = Ts (at the surface of the earth)
• q(b) = −Q (at the bottom of the crust)
INF2340 / Spring 2005 – p. 25
Mathematical model
−d
dx
(
λdu
dx
)
= Re−x/L, u(0) = Ts, λ(b)u′(b) = −Q
Observe that u depends upon seven parameters:u = u(x; λ, R, L, b, Ts, Q)!
Suppose that we want to investigate the influence of the differentparameters. Assume (modestly) three values of each parameter:−→ Number of possible combinations: 36 = 729.
Using scaling we can reduce the six physical parametersλ, R, L, b, Ts, Q to only two!
INF2340 / Spring 2005 – p. 26
Scaling
We introduce dimensionless quantities (and assume that λ isconstant):
x = xb, u = Ts + Qbu/λ, s(bx) = Rs(x)
This gives
−d2u
dx2= γe−x/β, u = 0,
du
dx(1) = 1
where we have two dimensionless quantities
β = b/L, γ = bR/Q
Dropping the bars, we get an equation on the form
−u′′(x) = f(x), x ∈ (0, 1), u(0) = 0, u′(1) = 1
INF2340 / Spring 2005 – p. 27
Discretisation
• Introduce a grid xi = (i − 1)h
and compute the unknown atgrid points ui = u(xi)
u
u u
u u3
4 51
2
x=1x=0x
• Differential equation fulfilled at each node
−u′′(xi) = f(xi)
• Approximate by standard finite differences
ui+1 − 2ui + ui−1 = −h2fi, i = 1, . . . , n − 1
As opposed to the ODEs we have seen earlier, this is a linearsystem of unknowns
INF2340 / Spring 2005 – p. 28
Discretising boundary conditions• u(0) = 0 simply becomes u1 = 0
• u′(1) = 1 can be approximated as
un+1 − un−1
2h= 1
• Problem: un+1 is not in the mesh!
• Solution: Use the discrete differential equation for i = n:
un−1 − 2un + ui+1 = −h2fn
and the discrete boundary condition to eliminate un+1
• The result is2un−1 − 2un = −2h − h2fn
INF2340 / Spring 2005 – p. 29
Linear system of equations
u1 = 0
u1 −2u2 +u3 = −h2f2
u2 −2u3 +u4 = −h2f3
. . . . . . . . ....
. . . . . . . . ....
. . . . . . . . ....
. . . . . . . . ....
un−2 −2un−1 +un = −h2fn−1
2un−1 −2un = −2h − h2fn
INF2340 / Spring 2005 – p. 30
Using linear algebraWe write the system as Au = b:
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 · · · · · · · · · · · · 0
1 −2 1 0. . .
. . .. . . 0
0 1 −2 1. . .
. . .. . . 0
.... . .
. . .. . .
. . .. . .
......
. . .. . .
. . ....
.... . .
. . .. . .
. . ....
.... . .
. . .. . .
. . .. . .
......
. . .. . .
. . . 1 −2 1
0 · · · · · · · · · · · · · · · 0 2 −2
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
u1
u2
u3
...
...
...
...
un−1
un
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0
−h2f2
−h2f3
...
...
...
...
−h2fn−1
−2h − h2fn
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
This system can be solved by Gaussian elimination.
INF2340 / Spring 2005 – p. 31
ImplementationAssembly of matrix in pseudocode:
Matrix( real ) A(n,n)Vector(real ) b(n ), u(n)
// Assemble matrix and left−hand sidefor i=1:n
if i==1 // Left boundaryA(1,1) = 1;b(1) = 0
else if i==n // Right boundaryA(i , i−1) = 2; A(i , i ) = −2;b( i ) = −2∗h − h∗h∗f(x(i ));
else // InteriorA(i , i−1) = 1; A(i , i ) = −2; A(i , i +1) = 1;b( i ) = −h∗h∗f(x(i ));
end
INF2340 / Spring 2005 – p. 32
In C++int main(int argc, char ∗∗argv){
:double ∗b, ∗u;double ∗∗A;
:b = new double[n];u = new double[n];A = new double∗[n];A [0] = new double[n∗n];for ( i =1; i<n; i++)
A[i ] = A[i−1]+n;
for ( i =0; i<n; i ++) {b[ i ] = u[ i ] = 0.0;for ( j =0; j<n; j++)A[i ][ j ] = 0.0;
}
h = 1.0/( n−1);for ( i =0; i<n; i ++) {
x = i∗h;if ( i==0) {
A[ i ][ i ] = 1; b[ i ] = 0; }else if ( i>0 && i<n−1) {
A[ i ][ i−1] = 1; A[i ][ i ] = −2; A[i ][ i +1] = 1;b[ i ] = h∗h∗(alpha+1)∗pow(x,alpha);
} else {A[ i ][ i−1] = 2; A[i ][ i ] = −2;b[ i ] = −2∗h + h∗h∗(alpha+1)∗pow(x,alpha);
}}solveSys(A, u, b , n);
// output solution ...return 0;
}
INF2340 / Spring 2005 – p. 33
Solution of linear systemAbstract formulation of Gaussian elimination:
• Compute the LU factorization: A = LU, i.e., LUx = b
• Solve Ly = b (forward elimination)
• Solve Ux = y (backward substitution)
LU factorization is feasible since A is tridiagonal.
void solveSys(double ∗∗A, double ∗x, double ∗b, int n) {int i , j , k ; double m;
for ( i =0; i<n−1; i++)for ( j=i+1; j<n; j ++) {
m = A[j ][ i ]/ A[ i ][ i ];for ( k=i ; k<n; k++)
A[j ][ k] −= m∗A[i][k ];b[ j ] −= m∗b[i];A[ j ][ i ] = −m;
}
/∗ Backward substitution ∗/for ( i=n−1; i>=0; i−−) {
x[ i ] = b[ i ];for ( k=i+1; k<n; k++)
x[ i ] −= A[i ][ k]∗x[k ];x[ i ] /= A[i ][ i ];
}
return;}
INF2340 / Spring 2005 – p. 34
Evaluation of algorithm
Observation:
A is tridiagonal, i.e., the only nonzero entries are ai,i−1,ai,i, and ai,i+1
Gaussian elimination:
• is designed for a general dense matrix
• storage requirement: n2 real numbers
• number of operations: O(n3)
Save memory and CPU-time by utilizing the tridiagonal structure.
INF2340 / Spring 2005 – p. 35
Tridiagonal matrices
Tridiagonal matrix:
• nonzero elements: 3n − 2
• in Gaussian elimination:− forward elimination: only need to eliminate lower diagonal− backward substitution: only need to substitute values along
upper diagonal
−→ number of operations is O(n)!
Linear systems arising from the discretization of differential equationstypically contain a lot of zeros (as we saw above). Gaussianelimination therefore performs a lot of unnecessary computions.Generally, one therefore prefers methods that utilize the specialstructure of the linear system.
INF2340 / Spring 2005 – p. 36
Implementation in C++int main(int argc, char ∗∗argv){
:double ∗b, ∗u, ∗Am, ∗Ac, ∗Ap;
:Am = new double[n];Ac = new double[n];Ap = new double[n];
for ( i =0; i<n; i++)b[ i ] = u[ i ] = Am[i] = Ac[i ] = Ap[i ] = 0.0;
h = 1.0/( n−1);
.
for ( i =0; i<n; i ++) {x = i∗h;if ( i==0){
Am[i ] = 0; Ac[i ] = 1; Ap[i ] = 0; b[ i ] = 0; }else if ( i>0 && i<n−1) {
Am[i ] = 1; Ac[i ] = −2; Ap[i ] = 1;b[ i ] = h∗h∗(alpha+1)∗pow(x,alpha);
} else {Am[i] = 2; Ac[i ] = −2; Ap[i ] = 0;b[ i ] = −2∗h + h∗h∗(alpha+1)∗pow(x,alpha);
}}
solveSys(Am, Ac, Ap, u, b, n);:
INF2340 / Spring 2005 – p. 37
Gaussian elimination
void solveSys(double ∗Am, double ∗Ac, double ∗Ap, double ∗x, double ∗b, int n){
int i , j , k ; double m;
/∗ Forward elimination ∗/for ( i =1; i<n; i++) {
m = Am[i]/Ac[i−1];Am[i] = −m; Ac[i] −= m∗Ap[i−1];b[ i ] −= m∗b[i−1];
}
/∗ Backward substitution ∗/i = n−1;x[ i ] = b[ i ]/ Ac[i ];for ( i−−; i>=0; i−−)
x[ i ] = ( b[ i ] − Ap[i]∗x[ i +1])/Ac[i ];return;
}
INF2340 / Spring 2005 – p. 38
Evaluation of program
Advantages:
• Reduced memory requirementn2 double + n double∗ + 1 double∗∗ −→ 3n double + 3 double∗
• Reduced CPU requirements:
Table: CPU time in seconds for grid with n unknowns
125 250 500 1000 2000
heat 0.07 0.6 5.9 48.9 391.5
heatTri 0.01 0.01 0.01 0.03 0.05
Disadvantages:
• We had to rewrite major portions of the code
• Storage scheme for tridiagonal matrix is shown explicit
We will come back to this later in the lecturesINF2340 / Spring 2005 – p. 39
Heat conduction in 2D
• Matrix for the discretization of−∇2 = f :
• Only 5n out of n2 entries arenonzero.
• Storage: store only nonzeroentries 0 5 10 15 20 25 30 35
0
5
10
15
20
25
30
35
nz = 166
You will get to know the discretisation of the 2D operatorintimately in the first assignment...
INF2340 / Spring 2005 – p. 40
The heat equation ut = ∆u
Two types of boundary conditions
Dirichlet boundary conditions• fix u on part of the boundary
u(x, y, z) = g(x, y, z)
Neumann boundary conditions• fix the normal derivative of u on part of the boundary
∂u
∂n(x, y, z) = f(x, y, z)
⇒ the solution is unique if the boundary data are Dirichlet or amixture of Dirichlet and Neumann
⇒ the solution is determined up to an additive constant for pureNeumann conditions
INF2340 / Spring 2005 – p. 41
Finite-difference methods
ut = uxx, u(0, t) = u(1, t) = 0, u(x, 0) = u0(x)
Introduce a grid in time and space (ih, nk) and define uni = u(ih, nk).
Then we use finite-differences
• Spatial discretisation: central difference as above
∂2u
∂x2≈
uni+1 − 2un
i + uni−1
h2
• Temporal discretisation: forward or backward Euler
∂u
∂t≈
uni − un−1
i
kor
∂u
∂t≈
un+1i − un
i
k
INF2340 / Spring 2005 – p. 42
Finite-difference methods
Forward Euler — explicit marching algorithm (r = k/h2)
un+1i = un
i + r(
uni+1 − 2un
i + uni−1
)
, u0i = u0(ih)
The scheme is stable for r < 1/2, meaning that |uni | may grow
uncontrolled if k is choosen too large −→ possibly very small timesteps.
Backward Euler — solution of linear system
−run+1i−1 + (1 + 2r)un+1
i − run+1i+1 = un
i
The system is written (I − rA)un+1 = un, where A was introduced
above. The scheme is unconditionally stable, meaning that k can bechoosen indenpendent of h. However, the scheme is less accurate forlarge k.
INF2340 / Spring 2005 – p. 43
Boundary conditions (explicit scheme)
Two types of boundaries:
• For Dirichlet conditions we can simply set the value
un0 = g(0, nk), un
m = g(mh, nk)
and not discretise the PDE at the boundary points
• For Neumann conditions, we can either introduce extra cells(ghost cells) outside the domain, for which
∂u
∂n≈
1
2h(un
1 − un−1) = f(0, nk)
−→ un−1 = un
1 − 2hf(0, nk),
or we can modify the stencil
un+10 = un
0 + 2r(un1 − un
0 − hf(0, nk))
INF2340 / Spring 2005 – p. 44
Two spatial dimensions
ut = uxx + uyy
Explicit discretisation
un+1ij = un
ij + r(
uni+1,j + un
i,j+1 − 4uni,j + un
i−1,j + uni,j−1
)
This scheme is stable provided r = k/h2 < 1/4.
The implicit scheme is defined analogously and gives apentadiagonal matrix as seen above.
INF2340 / Spring 2005 – p. 45
Boundary conditions in 2D
Consider Neumann boundary conditions:
∂u
∂n≡ ∇u · n = 0
Assume a rectangular domain. At the vertical (x =constant)boundaries the condition reads:
0 =∂u
∂n= ∇u · (±1, 0) = ±
∂u
∂x
Similarly at the horizontal boundaries (y =constant)
0 =∂u
∂n= ∇u · (0,±1) = ±
∂u
∂y
INF2340 / Spring 2005 – p. 46
Implementing boundary conditions
Consider the left boundary (i = 1, j = 1, . . . , ny). Now, let us apply thefinite difference stencil:
−1 0 1 2 3 4 5 6 7 8 9 10−1
0
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10Ghost cells
The computations involve cells outside our domain. This is a problem.The obvious answer is to use the boundary condition, e.g.,
u2,j − u0,j
2∆x= 0 ⇒ u0,j = u2,j
But how do we include this into the scheme..?
INF2340 / Spring 2005 – p. 47
Implementing boundary conditions cont’d
The two approaches are as before: ghost cells or modifiedstencils
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10Ghost cells
−1 0 1 2 3 4 5 6 7 8 9 10−1
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10Modified stencile
Which approach to choose depends upon the application andthe complexity of the boundary.
INF2340 / Spring 2005 – p. 48
