Mathematical Modeling of Engineering Problems
Jaroslav Vlček
Dept. of Mathematics and Descriptive Geometry
VŠB-TU Ostrava
Contents
1 Principles of mathematical modeling
1.1 A problem and its mathematical model . . . . . . . . . . . . . . . . . . . . . . .
1.2 Basic notions and relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 One-dimensional stationary models
2.1 Global balance . . . . . . . . . . .
2.2 Local balance . . . . . . . . . . . .
2.3 Boundary conditions . . . . . . . .
2.4 Boundary problems formulation . .
2.5 Exercises . . . . . . . . . . . . . .
3
3
5
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8
8
11
14
15
23
3 Multidimensional stationary models
3.1 Balance relations . . . . . . . . . . .
3.2 Balance on a boundary . . . . . . . .
3.3 Formulation of boundary problems .
3.4 Fourier method . . . . . . . . . . . .
3.5 Exercises . . . . . . . . . . . . . . .
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24
24
29
31
33
41
4 Non-stationary models
4.1 One-dimensional non-stationary models . . .
4.2 Examples of 1D non-stationary models . . . .
4.3 Partial differential equations of the first order
4.4 Non-stationary multi-dimensional models . .
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43
43
45
54
61
5 Selected problems
5.1 Population and logistic models . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Motion problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
66
69
2
Chapter 1
Principles of mathematical modeling
1.1
A problem and its mathematical model
1.1.1 Example – extremal problem. A bear of rectangular cross-section is manufactured
from a cylindrical trunk of the diameter d to have as much as possible bend stiffness. Determine
the bear dimensions.
Let x, y be the dimensions of bear. It holds well-known for bend modulus of homogeneous
material,
1
w = xy 2 .
6
We aim to find a maximum of this function, especially, a coupled extremum under the condition
x2 + y 2 = d2 . In this simple case, we can put directly y 2 from the last relation into function w.
The solution then runs by following steps:
1
w(x) = x(d2 − x2 ) ,
6
w0 (x) = 0
⇒
d2 − 3x2 = 0
⇒
d
x = ±√ .
3
Evidently, the negative root need not be taken into acount, therefore
y=
Conclusion: y =
√
p
r
d2
−
x2
=d
2
.
3
√
2x, i.e. the ratio of cross-section dimensions is x/y = 1/ 2.
1.1.2 Example – electrical oscilatory circuit. Derive the time dependence of the current
i(t) in serial circuit with C and L by the initial conditions i(0) = i0 , (di(t)/dt)t=0 = 0.
We start with basic balance equation for the voltage:
di(t)
1
L
+
dt
C
Zt
i(τ ) dτ = 0 .
0
After differentiation by t we obtain
d2 i(t)
1
+
i(t) = 0 ,
2
dt
LC
i(t = 0) = i0 ,
3
di(t) =0.
dt t=0
4
CHAPTER 1. PRINCIPLES OF MATHEMATICAL MODELING
We will solve this initial (Cauchy) problem for second-order ordinary linear differential equation
with constant coeffitients using classical method. Characteristic equation
λ2 +
1
=0
LC
has pure imaginary roots
λ1,2 = ± √
i
,
LC
therefore, the general solution consists from goniometric functions:
t
i(t) = C1 cos √
LC
t
+ C2 sin √
LC
.
The constants easy follow from initial conditions, C1 = i0 , C2 = 0, so that
t
i(t) = i0 cos √
LC
.
We obtained stabile periodic solution, i.e. non-damped harmonic oscillations.
1.1.3 Modeling process scheme. Presented examples illustrate simple kinds of model
problems in the two totally different application fields. However, we can discuss in detail their
common features. In the both cases, we needed to choose appropriate balance relation in relevant application branch; and, to supply this one by geometrical and/or physical assumptions.
In this way, a mathematical problem was formulated on certain abstraction level. We specified it (extremal problem, initial problem for differential equation) and selected corresponding
algorithm to solve it. Finally, the results were interpreted regarding the original context.
This strategy should be enforced by the most problems in practice, which we need to mathematize and to solve using effective tools. In other words, we speak about mathematical modeling,
the scheme of which is demonstrated in the following figure.
real problem
exact formulation
formulation ofproblem,
input data
physical description,
simplication
data specification
mathematical model
algorithmization
solution, result
choose of method,
corectness
numerical realization
accuracy, stability
interpretation
Figure 1.1: Modeling process scheme.
1.2. BASIC NOTIONS AND RELATIONS
5
1.1.4 Problem types. It is profitable to be aware of the most frequented types of mathematical problems that we meet by a realization of mathematical model.
Classification by the aim:
• quantitative
−→
values, data, functions,
• qualitative
−→
existence, uniqueness, stability,
• procedures
−→
algorithms, simulations,
•
..
.
By the structure:
• equations,
• extremal problems,
• proofs, verification,
• simulations,
•
1.2
..
.
Basic notions and relations
1.2.1 A classification of real phenomena. Again, the principal categorization can be
carried out by several points of view that are usually applied commonly:
• time dependence:
– stationary phenomena (states),
– non-stationary phenomena (processes)
• the number of spatial co-ordinates:
– one-dimensional,
– multi-dimensional (2D, 3D, ...)
• properties of medium:
– homogenneous vs. non-homogenneous (by the dependence on space variables),
– isotropic vs. anisotropic (directional variability)
1.2.2 The quantities and relations. Obviously, a number of quantities appears in the
mathematical model process with different role, the estimate of which is of key importance. We
distinguish
• local quantity (temperature, velocity etc.) specified in given point and/or time,
• global quantity (mass, electric charge, ...) if considered in a volume or time interval.
Following classification is quite fundamental:
• flux quantity (velocity, heath flux, ...),
• state quantity (potential, temperature, ...).
By a state or process describing the balance relation are principial that are supplied by
especial relations between mentioned ones. We call them constitutive relations. Specific
terminology and notation is known in particular applications; the mostly used examples are
summarized in the table.
6
CHAPTER 1. PRINCIPLES OF MATHEMATICAL MODELING
1.2.3 Density of a quantity. For a 1D case we denote f (x) the density of quantity F on
the interval hx1 , x2 i, d = x2 − x1 . Thus,
Zx2
f (x) dx
F (d) =
x1
represents total ”amount” of the quantity F on this interval.
In multi-dimensional cases, the density f (x1 , x2 , x3 ) is a function of more space variables.
We can define the quantity F
• on a curve k:
F (k) =
R
f ds
(first kind curvilinear integral),
k
• on a surface S:
F (S) =
R
f dS
(first kind surface integral),
f dV
(volume integral).
S
• in a volume V :
F (V ) =
R
V
If f (x1 , x2 , x3 ) is the density of sources distributed in a domain, then the quantity F gives
total production of sources on this one, which becomes oftly a constituent of balance relations.
Balance
relation
energy
conservation rule
balance of forces
(moments)
mass
balance
Gauss
rule
Phenomena,
process
heat
transfer
elastic
deformation
fluid
flow
elektrostatic
field
ϕ
electrostatic potential
p
pressure
u
displacements
T
temperature
State
quantity
D
electric induction
v
velocity
σ
stress
q
heat flux
Flux
quantity
ε = g(u)
D = −ε grad ϕ
v = −g(η) grad p
σ = E.ε,
Hooke’s rule
q = −λ grad T
Constitutive
relation
Atributes of several processes or states
ε
permittivity
η
viscosity
E
Young modulus
λ
heat conductivity
Material
quantity
charges
sources
forces
vnějšı́ch
sources
heat
Source quantity
is the density of . . .
1.2. BASIC NOTIONS AND RELATIONS
7
Chapter 2
One-dimensional stationary models
Notation:
• x ∈ h0, Li
...
space variable,
• f (x)
...
source density,
• u(x)
...
state quantity,
• v(x)
...
flux quantity.
2.1
Global balance
2.1.1 Conservation laws. Let hx1 , x2 i be an element of the interval (0, L) – see Fig. 2.1.
We denote v(x1 ), v(x2 ) the flux in the element end-points, thereby the signs + / − mean out(here) or in- flow with regard to the x axis orientation.
−v(x1 )
v(x2 )
0
-
x1
x2
L
x
-
Figure 2.1: To the flux quantity balance.
A balance of the flux quantity v on the element hx1 , x2 i can be expressed as follows: the
sum of in- and out- flow through the end-points equals to the total production of inner sources
with the density f (x). The symbolic form of the balance relation can be written as
∀ x1 , x2 ∈ (0, L) :
v(x2 ) − v(x1 ) =
Zx2
f (x) dx .
(2.1)
x1
The formula (2.1) represents the global conservation law of the flux quantity called also the
balance relation. In order to be in force, the function f (x) must be integrable. To more
detailled model analysis, a relation between flux and state quantity need be at disposal that will
be discussed bellow.
8
2.1. GLOBAL BALANCE
9
2.1.2 Constitutive relations can be of various form. We present here the most usual variant
that takes place among other by transfer phenomena modeling, relaxation processes etc.
Again, we consider element hx1 , x2 i, for which we denote ∆x = x2 − x1 , and, ∆u(x) =
u(x2 ) − u(x1 ) the change of state. The mean value
∆u(x)
∆x
corresponds to the total change of the flux function
∆v(x) = v(x2 ) − v(x1 ) .
If the two above quantities are globally proportional, we write this property as connstitutive
relation in the form
∆u(x)
,
(2.2)
∆v(x) = − p
∆x
where the coefficient of proportionality p > 0 is a constant or function. The sign ”minus” express
the fact that the flux in direction of decreased state function is positive. In the most cases this
coefficient represents material properties of a medium; in particular
• p = konst.
in homogeneous medium,
• p = p(x)
in non-homogeneous medium,
• p = p(u)
in non-linear medium (dependence on state function).
Combining the flux balance (2.1) with constitutive relation (2.2) we obtain global balance
of state quantity
∆u(x)
=
−p
∆x
Zx2
f (x) dx .
(2.3)
x1
2.1.3 Example – constitutive relation in underground hydraulics. Due to water drawing from a borehole in free surface aquifer, the rotationaly symmetric surface profile get settled
(called a depression cone) – see Fig. 2.2. Find an expression of mean feed velocity to the
borehole.
The instantaneous velocity v(r) in the distance r from the borehole represents the flux
quantity. We express the fact, that its mean value ṽ(r) on the section hr1 , r2 i is proportional to
the change of hydraulic height h(r) what is the state function in discussed problem. Obtained
relation is called Darcy’s law in underground hydraulics:
ṽ(r) = −k
∆h
h(r2 ) − h(r1 )
= −k
.
r2 − r1
∆r
(2.4)
Here the filtration coefficient k depends mainly on material properties of aquifer (porosity). The
negative sign express the circumstance that the hydraulic height fall toward increasing distance
from borehole. Note that obtained formula has identical feature as the general constitutive
relation (2.2).
10
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
Figure 2.2: Depression cone around a borehole.
Figure 2.3: The balance of forces by loading of a string.
2.1.4 Example – balance equation by small deformation of thin elastic string. We
will consider elastic homogeneous string of the length L fixed in the end-points x = 0, x = L,
which is in the whole profile loaded by a force of the density f (x) – Fig. 2.3. Our aim is to
establish global balance equation for the mean bend (displacement) u(x) on the section hx1 , x2 i.
Since the fiber cross-section can be omitted, we solve the one-dimensional problem, where
we have
• f (x)
external forces density producing on the given element total force
F = |F | =
Zx2
f (x) dx ;
x1
• T (x)
• σ(x)
• u(x)
stretching force T = |T | in the point x;
the magnitude of vertical component of T (x) as the flux quantity;
bend (displacement) as the state function.
2.2. LOCAL BALANCE
11
Force balance on the element hx1 , x2 i leads to the equation
−σ(x1 ) + σ(x2 ) = F
⇒
Zx2
σ(x2 ) − σ(x1 ) =
f (x) dx .
(2.5)
x1
Now, we derive constitutive relation with the help of vertical components of the vector T (x) in
the end-points using well-known approximation for very small deformations:
σ(x1 ) = T (x1 ) sin α ≈ T (x1 ) tg α ≈ T (x1 )
∆u ,
∆x x=x1
σ(x2 ) = T (x2 ) sin β ≈ T (x2 ) tg β ≈ T (x2 )
∆u .
∆x x=x2
The mean change in the fiber section is expressed as the difference
∆σ(x) = σ(x2 ) − σ(x1 ) = T (x2 )
∆u ∆u − T (x1 )
,
∆x x=x2
∆x x=x1
(2.6)
or formally for arbitratry difference ∆x,
∆σ(x) = T (x)
∆u
.
∆x
(2.7)
The last two equations represents constitutive relations between the load and displacement.
Note that in more general elasticity problems the stress and deformation are expressed in tensor
form. Mutual relation is called Hooke’s law, in which the other material characteristic take place
– elastic module. In our case, the obtained constitutive relation is quite sufficient. Putting it
into (2.5) leads to global balance equation for displacement
∆u ∆u T (x2 )
− T (x1 )
=
∆x x=x2
∆x x=x1
Zx2
f (x) dx .
(2.8)
x1
2.2
Local balance
2.2.1
Local conservation law of the flux quantity. We turn back to global balance (2.1),
v(x2 ) − v(x1 ) =
Zx2
f (x) dx .
(2.9)
x1
We will suppose that the element hx1 , x2 i ⊂ (0, L) ⊂ R can be arbitrary small in the sense of
continuum hypothesis (obr. 2.4).
0
x1
x
u
x2
L
x
-
Figure 2.4: To the local conservation law.
If the source density f (x) is only integrable, then one holds for arbitrary x ∈ (x1 , x2 )
Zx2
lim
x1 →x−
x2 →x+
f (x) dx = 0.
x1
12
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
Denoting
lim v(x1 ) = v(x−) ,
lim v(x2 ) = v(x+)
x1 →x−
x2 →x+
,
we obtain with regard to (2.9)
∀x :
v(x+) − v(x−) = 0
⇒
v(x−) = v(x+) .
(2.10)
This result is local conservation law of flux quantity that can be simply formulate as the
flux continuity principle: ”the all what flow in certain point must also flow out”.
2.2.2
Local form of constitutive relations. In the introduced global constitutive relation,
v(x2 ) − v(x1 ) = − p(x)
∆u(x)
,
∆x
(2.11)
we realize previous limit process. Therefore, the right-hand difference ratio is substituted by the
derivation:
du(x)
v(x) = − p(x)
.
(2.12)
dx
The flux function v(x) at a point x is proportional to the derivative (local change) of state
function u(x). As the consequence we obtain from (2.10) the transient condition for the
state quantity
du(x+)
du(x−)
= p(x+)
.
(2.13)
p(x−)
dx
dx
2.2.3 Differential form of conservation law. Again we start with global balance of flux
quantity, but by the assumption of continuity of source density function f (x). We divide
the both sides of (2.9) by the difference x2 − x1 ,
v(x2 ) − v(x1 )
1
=
x2 − x1
x2 − x1
Zx2
f (x) dx ,
x1
and, we apply the integral mean-value theorem1 . It results in the quasi-local balance
v(x2 ) − v(x1 )
= f (ξ) .
x2 − x1
The limit transition by ∆x = x2 −x1 → 0 leads to the local differential form of conservation
law for the flux function
dv
= f (x) ,
(2.14)
dx
beacause lim f (ξ) = f (x) for any continuous function f (x).
ξ→x
Applying the local constitutive relation (2.12) in above equation we can write differential
form of the state-function conservation law
du(x)
d
p(x)
dx
dx
−
1
∃ ξ ∈ (x1 , x2 ) :
1
x2 −x1
Rx2
x1
f (x) dx = f (ξ) .
= f (x)
∀ x ∈ (0, L) .
(2.15)
2.2. LOCAL BALANCE
13
2.2.4 Example – differential equation of small deformation of elastic string. We use
the previous approach to the global relation (2.8):
Zx2
∆u ∆u T (x2 )
− T (x1 )
=
∆x x=x2
∆x x=x1
|.
f (x) dx
x1
1
.
x2 − x1
After limit operations x1 → x−, x2 → x+ we get differential equation
d
du(x)
T (x)
dx
dx
= f (x) .
(2.16)
The positive sign on the left side follows from the orientation of co-ordinate system in the Fig.
2.3.
2.2.5 Consequences of source function properties. Both the relations (2.10) a (2.14)
follow from conservation law of the flux quantity, but by different assumptions. If the function
f (x) is continuous, these relations are equivalent; together, the smoothness of state function is
induced, so that the derivative du
dx exists for any x ∈ (0, L).
On the other hand, the density f (x) need not even be integrable, nevertheless, the problem
is of the good physical nature – let consider an elastic fibre by one-point loading (Fig. 2.5).
f (x0 )
x0 ?
0
L
u
x
-
Figure 2.5: One-point loading of a fibre.
In this case, the source function must be understand in the generalized sense, when we
interpret this one using the Dirac distribution δ(x), for instance. For an illustration, the global
stress balance (2.5) can be written as
Zx2
∆σ(x) =
f (x) δ(x − x0 ) dx
( = f (x0 ) ) .
x1
It is obvious that also state function u(x) will belong to a more general function class. Consequently, we speak about weak formulation of the equation (2.16).
2.2.6 Example – heat transfer in a wall. We will model both the global and local balance
by heat transfer through the wall of thickness s = x2 − x1 with heat conductivity λ(x) by the
surface temperatures T2 < T1 . The heat flux q(x) is the flux quantity, as the state one we take
the temperature T (x). Denoting f (x) a possible density of inner heat sources we can write the
global balance as
q(x2 ) − q(x1 ) =
Zx2
f (x) dx ,
(2.17)
x1
or in local form
dq
= f (x) .
(2.18)
dx
Corresponding local balance of state function arises by application of constitutive relations that
have the global and local form, respectively:
q̃(x) = −λ(x)
T2 − T1
∆T
= −λ(x)
,
x2 − x1
∆x
q(x) = −λ(x)
dT
.
dx
(2.19)
14
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
If we put the last one into local flux balance, then the resulting local conservation law for the
temperature T (x)
d
dT
−
λ(x)
= f (x)
(2.20)
dx
dx
represents well-known stationary Fourier heat-transfer equation in the 1D case.
2.3
Boundary conditions
2.3.1 Balance at end-points. The fluxes v0 , vL or the values of state quantity u0 , uL are
given in end-points of the interval (0, L). Further, f (x) is integrable source density. At first,
with regard to the Fig. 2.6 we can write global flux balance for the both boundary subintervals:
x ∈ (0, x2 )
...
v(x2 ) − v0 =
Zx2
f (x) dx ,
(2.21)
0
x ∈ (x1 , L)
...
v(L) − v(x1 ) =
ZL
f (x) dx .
(2.22)
x1
vL
−v0
-
x1 = 0
x2
x1
x2 = L
x
-
uL
u0
Figure 2.6: To the balance at boundary points.
Local flux balance in the boundary points is derived by limit transient, where we denote
lim v(x2 ) = v(0+) ,
lim v(x1 ) = v(L−) .
x2 →0+
x1 →L−
Similarly as in the paragraph 2.2.1 we obtain
v(0+) − v0 = 0
⇒
v(0+) = v0 ,
vL − v(L−) = 0
⇒
v(L−) = vL .
Local state balance is again a consequence of constitutive relation (2.12), rewritten by the
limits toward to boundaries of interval:
v(0+) = − p(0+)
du(0+)
,
dx
i.e.
− p(0+)
du(0+)
= v0 ,
dx
(2.23)
du(L−)
du(L−)
,
i.e.
− p(L−)
= vL .
(2.24)
dx
dx
In the case, when the values u0 , uL of state function are prescribed at end-points, we usualy
establish corresponding flux on the base of transient conditions
v(L−) = − p(L−)
v0 = α0 (u(0+) − u0 ) ,
(2.25)
vL = αL (u(L−) − uL ) ,
(2.26)
2.4. BOUNDARY PROBLEMS FORMULATION
15
where α0 , αL ∈ R+ are given coefficients. The local balance (2.23) a (2.24) with these right
sides give differential form of boundary conditions for state function:
− p(0+)
− p(L−)
du(0+)
= α0 (u(0+) − u0 ) ,
dx
(2.27)
du(L−)
= αL (u(L−) − uL ) .
dx
(2.28)
Since this conditions follow directly from conservation law, we call this result as natural boundary conditions.
2.3.2 Example – heat exchange at wall surfaces. The temperature profile T (x) in a
homogeneous wall of thickness L is predetermined by boundary conditions on the both surfaces
(Fig. 2.7). We will suppose that the outer temperatures T0 > TL are given as well as the
transient coefficients α0 , αL [W K −1 m−2 ].
Figure 2.7: Temperature field in a wall.
The constant heat conductivity λ [W K −1 m−1 ] is the material characteristics, for which
p(0+) = p(L−) = λ in (2.27) and (2.28):
dT (0+)
= α0 (T (0+) − T0 ) ,
dx
dT (L−)
−λ
= αL (T (L−) − TL )
dx
dT
= α0 (T − T0 ) , x = 0 ,
dx
.
dT
−λ
= αL (T − TL ) , x = L .
dx
λ
λ
or
(2.29)
We changed the sign in the first condition because of T0 ≥ T (x) ≥ TL (the temperature is
decreasing function of the variable x, therefore, it has negative derivative).
The right form is more clear, if we uderstand the terms in the one-side limit sense, therefore,
we prefer this manner in what follows.
2.4
Boundary problems formulation
2.4.1 Fundamental problem. We want determine state function u(x) and flux function v(x)
in the interval (0, L) by following assumptions:
(a) the medium is characterized by the material function p(x);
16
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
(b) u(x), v(x) fulfil constitutive relation (2.12), i.e.
v(x) = − p(x)
du(x)
,
dx
x ∈ (0, L) ;
(c) flux quantity satisfies global balance realation (2.1),
v(x2 ) − v(x1 ) =
Z x2
∀ x1 , x2 ∈ (0, L) ;
f (x) dx
(2.30)
x1
(d) boundary conditions are considered in the form (2.27), (2.28);
(e) f (x), p(x) are continuous.
The last assumption enables to write local form of the boundary problem for state function
u(x) – see also (2.15):
d
du(x)
−
p(x)
= f (x)
dx
dx
du(x)
p(x)
+ α0 u(x) = g0 ,
dx
du(x)
p(x)
+ αL u(x) = gL ,
dx
∀ x ∈ (0, L) ,
(2.31)
x=0,
(2.32)
x=L,
(2.33)
where g0 = α0 u0 , gL = αL uL . In this way, we obtained the most general formulation – the
Newton boundary problem for the second-order differential equation.
2.4.2 Neumann boundary problem for the equation (2.31) folows from Newton one by
α0 = αL = 0 (the equation is unchanged):
du(x)
dx
du(x)
p(x)
dx
p(x)
= h0 ,
x=0,
(2.34)
= hL ,
x=L,
(2.35)
where h0 , hL are given values of the state function derivative in boundary points. In the especial
case, when h0 = hL = 0, we have the boundary problem with homogenneous Neumann
conditions
du(x)
dx
du(x)
p(x)
dx
p(x)
= 0,
x=0,
(2.36)
= 0,
x=L.
(2.37)
2.4.3 Dirichlet boundary problem for the equation (2.31) is formulated by directly prescribed values of the state function u0 , uL :
u(x) = u0 ,
x=0,
(2.38)
u(x) = uL ,
x=L.
(2.39)
It seems that we could obtain these conditions from Newton variant for p(x) = 0, but this idea
is at variance with balance equation. Although the Dirichlet conditions are frequently present
in real situations, we denote them as ”non-natural”. Again, we can work with homogeneous
variant, if u0 = uL = 0.
2.4. BOUNDARY PROBLEMS FORMULATION
17
2.4.4 Boundary problems with non-continuous data. We will study, what consequence
have more soft requirements to the input parameters. We admit a discontinuity, but by bounded
values, i.e. the functions have at least one-side proper limits. The variant with discontinuous
source density f (x) is discussed in the paragraph 2.4.7; here, we show two typical examples of
the non-continuous function p(x) in a point xp ∈ (0, L) – see Fig. 2.8:
• geometrical discontinuity (it appears rather in multidimensional models),
• material discontinuity.
Figure 2.8: Geometrical (at left) and material discontinuity.
The conservation law holds in the both cases (for integrable source density f (x)), but in the
sense of the flux continuity (2.10) as v(xp −) = v(xp +) in the form
p(xp −)
du(xp −)
du(xp +)
= p(xp +)
,
dx
dx
(2.40)
that this is well-known transient condition (2.13). It is essential to appreciate that the
continuity assumption of the state function, u(xp −) = u(xp +), does not ensure the continuity
of its derivative u0 (x) at the point xp , because of
p(xp −) 6= p(xp +)
u0 (xp −) 6= u0 (xp +) .
⇒
Finally, we demonstrate one of many possible variants of general type of 1D boundary
problem with transient condition:
− p(x)u0
0
x 6= xp ,
(2.41)
u(0) = u0 ,
x = 0,
(2.42)
0
x = L,
(2.43)
p(L)u (L) = gL ,
0
x ∈ (0, L) ,
= f (x) ,
0
p(xp −)u (xp −) = p(xp +)u (xp +),
x = xp .
(2.44)
To more clarity we summarize particular terms in this formulation:
(2.41)
...
differential equation (local balance of state function),
(2.42)
...
Dirichlet condition at the point x0 ,
(2.43)
...
Neumann condition at the point xL ,
(2.44)
...
transient condition at the point xp .
Since the boundary conditions are of different kind, we introduce the notion mixed boundary
problem in such case.
18
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
2.4.5 Correctness of model. A correctly formulated problem must fulfil three basic properties:
1. it has a solution (existence),
2. the solution is unique (uniqueness),
3. the solution is well conditioned (stability – continuous dependence on input data).
In practice, a verification of above demands before start of problem solving is not simple
in the most situations, because it requires full theoretical analysis. However, these properties
oftly become clear in the run of solving. Sometimes, a ”compatibility of problem” is taken as
solvability condition that is the extension of the global flux balance on the whole interval h0, Li:
v(x2 ) − v(x1 ) =
Zx2
−→
f (x) dx
ZL
v(L−) − v(0+) =
f (x) dx
x1
(2.45)
0
Generally, the problems with Neumann boundary conditions (2.34), (2.35) can be un-correct
– the solution exist only for several combination of the data f (x), g0 , gL , moreover, it need not
be unique.
2.4.6 Example – heat transfer by Newton boundary conditions. As a demonstration
of typical complete formulation and solution of boundary problem we turn back to temperature
field in the Example 2.3.2 with following assumptions (see Fig. 2.7):
• no heat sources exist inside the wall
→
f (x) = 0,
• heat exchange with outside passes is the same on the both surfaces
• the left surface temperature is more than the on right one
→
→
α0 = αL = α,
T0 > TL .
Our aim is to find general solution and to apply the result to given data: L = 0, 5 m, T0 = 50◦ C,
TL = 20◦ C, λ = 1 WK−1 m−1 (brickwork), α = 2 WK−1 m−2 .
With regard to above assumptions takes the equation (2.41) for the temperature T (x) the
simple form
d2 T
= 0,
x ∈ (0, L).
dx2
Boundary conditions of Newton type
dT
dx
dT
−λ
dx
−λ
= α(T0 − T ),
x=0,
= α(T − TL ),
x=L
were discussed in (2.29). Successive integration in second-order differential equation leads directly to the general solution
T (x) = C1 x + C2 ,
which we put into boundary conditions to obtain the constants C1 , C2 :
x=0
...
−λC1 = α(T0 − C2 ) ,
x=L
...
−λC1 = α(C1 L + C2 − TL ) .
2.4. BOUNDARY PROBLEMS FORMULATION
19
After a short calculation we get
C1 = −
α
(T0 − TL ) ,
αL + 2λ
C2 = T0 −
λ
(T0 − TL ) ;
αL + 2λ
thus, we obtained final solution
T (x) = T0 −
αx + λ
(T0 − TL ) .
αL + 2λ
We verify easy that this result satisfies the equation and boundary equations as well. The linear
function T (x) is continuous on the interval h0, Li that aposteriori implies the correctness of
solved model problem.
An application of given data leads to the straight line T (x) = 40 − 20x, x ∈ h0, 0, 5i; and,
also to the temperature on the surfaces: T (0) = 40◦ C, T (L) = 30◦ C. The graphical output in the
Fig. 2.9 presents the important fact that the Newton conditions do not preserve the continuity
of state function T (x) on the boundary.
Figure 2.9: Temperature field in the wall.
2.4.7 Example – boundary problems by a deformation of elastic fibre. A fibre (string)
is stretched by the force T and fixed at the points x = 0 a x = L. Under the loadeding by a
force of the density f (x) (Fig. 2.10), its deflection u(x) can be established as the solution of the
equation (2.16):
d2 u
f (x)
= q(x) ,
q(x) =
, x ∈ (0, L) .
2
dx
T
f (x)
?
?
?
?
?
?
0
?
L
u(x)
Figure 2.10: Deflection of a string.
x
-
20
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
We will formulate and solve several typical problems that illustrate the notions from previous
text by following assumptions:
• homogenneous string is very long and thin as well, i.e. the length remarkable predominates,
that enables to model this problem as one-dimensional,
• only the transverse stress is considered,
• the ideal elasticity by small deformation is supposed.
Presented analysis is aimed at the right-hand side properties as well as at boundary conditions.
A. Continuous and bounded function f (x), x ∈ (0, L).
A1. f (x) =konst.
The equation u00 (x) = q (= konst.) has general solution
1
u(x) = qx2 + C1 x + C2 ,
2
(2.46)
to which we will apply various boundary conditions.
A1a. Dirichlet problem corresponds to the fixed end-points, i.e. to the homogenneous
conditions
u(0) = u(L) = 0 .
Calculating the constants C1 = − 12 qL, C2 = 0 we obtain the parabola
1
1
1
u(x) = qx2 − qLx = qx(x − L) .
2
2
2
(2.47)
Given problem and its result are presented in the Fig. 2.10; the solution is unique,
the model is correct.
A1b. Neumann boundary problem means that derivative of the function u(x) is prescribed
in the end-points as the tangent of solution (”free ends” – see Fig. 2.11. We write
the boundary conditions in the form
u0 (0) = k0 ,
(k0 = tg α),
u0 (L) = kL ,
(kL = tg β).
f (x)
aa
aa α
aa
0
?
aa
a
?
?
?
?
?
!
!!
a
!
!!β
!
- x
!
!!
?
u(x)
Figure 2.11: Deflection of a string with free ends.
L
2.4. BOUNDARY PROBLEMS FORMULATION
21
Derivative of general solution (2.46) u0 = qx + C1 leads to following relations at the
end-points:
x=0
...
k0 = C1 ,
x=L
...
kL = qL + C1 .
It is obvious that the constant C2 can be arbitrary, therefore, the problem has infinite
number of solutions, but existing for k0 = kL − qL only in the form
1
u(x) = qx2 + k0 x + C2 .
2
It means, that the angles at end-points are invoked by given force, so that these can
not be apriori postulated. Therefore, the formulation with Neumann conditions is
not correct.
A1c. Mixed boundary problem can take many features; one from typical variants is the
bend of one-side fixed thin rod – Fig. 2.12. In this case the boundary conditions have
the form
u(0) = 0
u0 (L) = kL
...
...
on the fixed boundary,
at the free end.
The derivative sign corresponds to the direction of deformation – see the figure. By
kL > 0
0
L
x
-
kL < 0
Figure 2.12: One-side bend of thin rod.
convenient steps we calculate the constants in (2.46): C1 = kL − qL a C2 = 0. The
problem leads correctly to the solution
1
u(x) = qx2 + (kL − qL)x .
2
A2. Let the force density is given as a continuous function f (x) 6= konst.; for simplicity, we set
f (x) = f sin
πx
L
⇒
q(x) =
f
πx
πx
sin
= q sin
,
T
L
L
for instance. Supposing the Dirichlet conditions as in the example A1a, the general solution
is found as
L2 q
πx
u(x) = − 2 sin
+ C1 x + C2 .
π
L
Application of boundary conditions gives C1 = C2 = 0. It means, that resulting deflection
corresponds to the loading force profile:
u(x) = −
L2 q
πx
sin
.
2
π
L
Also in this case the problem is correctly formulated with unique smooth solution.
22
CHAPTER 2. ONE-DIMENSIONAL STATIONARY MODELS
B. Discontinuous (bounded) right-hand side
B1. The function f (x) is piecewise continuous (especially: piecewise constantaneous).
We consider the same boundary problem as in the example A1c, but with constant loading
q = f /T acting only on the part ha, Li, 0 < a < L (Fig. 2.13). We can efficiently approach
? ? ? ? ?
a
0
L
x
-
Figure 2.13: To the example B1.
the model of this problem when consider the particular solution on each string part:
q(x) =




 0 , x ∈ h0, a),



 q,
=⇒
u(x) =




 u1 (x) , x ∈ h0, a),



 u2 (x) , x ∈ ha, Li.
x ∈ ha, Li,
Both the functions are connected at the point x = a by the contact conditions following
from the flux continuity principle: the string remains non-disturbed after a deformation.
Therefore, the flux function T is continuous (constant) by the assumption. With regard
to the conclusions in the paragraph 2.4.4 we get the transient conditions
u1 (a) = u2 (a) ,
u01 (a) = u02 (a)
that we add to above boundary problems:
u001 = 0 ,
x ∈ (0, a) ,
u1 (0) = 0 ,
u002 = q ,
x ∈ (a, L) ,
u02 (L) = kL .
The four constants in general solutions
u1 (x) = C1 x + C2 ,
1
u2 (x) = qx2 + D1 x + D2
2
follow from boundary conditions. Thus,
u(x) =




 u1 (x) = (kL − qL + aq)x ,
x ∈ h0, a),



 u2 (x) = 12 qx2 + (kL − qL)x + 12 qa2 , x ∈ ha, Li.
It is easy to ascertain that this function fulfil both differential equation and boundary
conditions.
B2. If a loading f (x) acts in one point x0 ∈ (0, L) of a string only, the situation becomes
very specific. Since the right-hand side is not a function in classical sense, this physically
reasonable problem leads to the question, in what function class the solution should be
2.5. EXERCISES
23
expected. Among other, the ”function” f (x) can be written with the help of Dirac impulse
q(x) = qδ(x − x0 ). In the model, the bellowed relation can be used2 :
Z
δ(x − x0 ) dx = h(x) =




 C1 ,
x < x0 ,



 C2 ,
x > x0 ,
Z
h(x) dx =




 C1 x + C3 ,
x < x0 ,



 C2 x + C4 ,
x > x0 .
For the problem solved by the same way as in B1 we can not suppose the continuity of
state function derivative at the point x0 , because of u01 (x0 ) 6= u02 (x0 ). Therefore, we apply
the moments balance as the fourth condition,
T
L
Z L
Z L
u(x) dx =
0
f (x) δ(x − x0 ) dx ,
0
which is formulated globally on the interval h0, Li. Note, that the right-hand side equals
to f (x0 ) at arbitrary point.
2.5
Exercises
2.5.1 Formulate mixed boundary problem for the one-dimensional heating of a wall. On the
outside surface the constant temperature is given, T0 = 32 ◦ C. At the inside one, the heat transfer
proceeds with the coefficient α into the medium of the temperature TL = 18 ◦ C. Solve by the
same assumptions as in the Example 2.4.6 with following input data: L = 0, 3 m, λ = 0, 2
WK−1 m−1 , α = 4 WK−1 m−2 . Establish the temperature at inner surface; demonstrate the
resulting temperature profile graphically.
2.5.2 Solve the Example 2.4.7A1a, if the string stretched by the force T is loaded by the
constant force density f on the section hx0 − a, x0 + ai. Determine the string deflection in the
point x0 by these values:
• L = 3 m, x0 = L/3, a = L/6,
• f = 12 Nm−1 , T = 325 N.
2.5.3 A string of the length L stretched by the force T = 60 is fixed at the both ends. It is
loaded at the point x0 dividing its length by the ratio 1:2. For the force density f = 10 Nm−1
solve this problem to establish the resulting form of string and the highest deflection (use the
conclusions of the Example 2.4.7B2).
2
these relation are not purely exact because of their rather formal character; indeed, an use of them in presented
context leads to required result
Chapter 3
Multidimensional stationary models
3.1
Balance relations
3.1.1
Fundamental notions.
Ω . . . a domain in R2 or R3 ,
∂Ω . . . its boundary – (piecewise) smooth curve (in R2 ) or surface (in R3 ),
Ω̄ = Ω ∪ ∂Ω . . . closed domain,
Ω0 ⊂ Ω . . . reference sub-domain,
n . . . unit normal vector of ∂Ω0 .
Figure 3.1: To fundamental notions.
Further notation (see Fig. 3.1)
M ∈ Ω0 . . . inner point,
P ∈ ∂Ω0 . . . boundary point,
u = u(M ) . . . state function (usualy scalar quantity),
v = v(M ) . . . flux function (usualy vector field),
f = f (M ) . . . source function, source density (integrable).
The normal and tangential flux belong to basic quantities on a boundary (Fig. 3.2).
Denote τ the tangential vector that lies (in R3 ) in the plane determined by the vectors n and v.
24
3.1. BALANCE RELATIONS
25
Similarly, n0 , τ 0 are the unit items. Thus, we can express projections of the flux v into tangent
and normal direction:
normal flux v n = (v · n0 ) n0 ,
vn = v · n0 ,
tangential flux v τ = (v · τ0 ) τ 0 ,
vτ0 = v · τ .
(3.1)
(3.2)
Note that v n + v τ = v.
Figure 3.2: Normal and tangential flux v.
3.1.2 Global and local field characteristics. The group of global characteristics is defined
for the functions u, v ∈ C 1 (Ω0 ) by integral operators:
• total production of sources in Ω0
Z
F =
f (M ) dV ,
(3.3)
Ω0
• circulation of the field v along closed curve K ⊂ Ω0
I
C=
v(M ) dl ,
(3.4)
K
• flux of the field v through the surface S
Z
T =
Z
v(P ) dS =
S
v(P ) · n dS .
S
On the other hand, we can estimate locally at arbitrary point of the domain Ω0 :
(3.5)
26
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
1. a change of the scalar field u in given direction s with the help of direction derivative
∂u
= s0 · grad u,
∂s
where s0 is unit direction vector; the field gradient
grad u =
∂u ∂u ∂u
,
,
∂x ∂y ∂z
represents the direction of the largest decrease (or increase) of the function u;
2. source character with regard to the divergence, because
• div v 6= 0 for source field,
• div v = 0 for non-source field,
3. vorticity regarding field rotation established using curl operator, because
• curl v 6= o in vortex field,
• curl v = o in non-vortex (potential) field.
3.1.3 From global to local characteristics. This step is realized by limit process, when
the volume (surface, curve, ...) is reduced into single point. Following explanation is slightly
simplified without loss of generality. At first we recall two important theorems of vector analysis.
The Gauss-Ostrogradsky theorem express the equality of volume integral of divergence on
a domain and the total flux through its boundary:
Z
Z
div v dV =
Ω0
v dS .
(3.6)
∂Ω0
The Stokes formula allows to express flux circulation along the curve K by surface integral,
Z
I
rot v dS ,
v dl =
(3.7)
S
K
where S is arbitrary smooth surface passed through the curve K and being accordantly oriented
to this one. Similar statements hold also for volume integral of gradient or divergence:
Z
Z
grad u dV =
Ω0
u dS ,
(3.8)
v × dS .
(3.9)
∂Ω0
Z
Z
rot v dV =
Ω0
∂Ω0
Let µ(Ω0 ) denotes a measure of the domain Ω0 (volume, surface, diameter etc.), for instance.
Supposing continuity of the function f on the reference domain, i.e. f ∈ C(Ω̄0 ), the process of
continuous reduction of Ω0 into the point M can be expressed in symbolic form as
Ω0 −→ M
⇐⇒
µ(Ω0 ) −→ 0 .
If we apply this principle to volume integral representing total source production in Ω0 , then we
obtain generalized average integral value of the density:
1
lim
µ→0 µ(Ω0 )
Z
f dV = f (M ) .
Ω0
(3.10)
3.1. BALANCE RELATIONS
27
Above relation enables defining the divergence of the field v by (3.6) as local characteristics at
the point M :
Z
1
lim
v dS = div v .
(3.11)
µ→0 µ(Ω0 )
∂Ω0
In the same way, the remaining differential operators can be locally defined using (3.8) and (3.9):
1
µ→0 µ(Ω0 )
Z
lim
1
µ→0 µ(Ω0 )
(3.12)
v × dS = rot v .
(3.13)
∂Ω0
Z
lim
3.1.4
tions.
u dS = grad u ,
∂Ω0
Balance equations. As in 1D models, we will proceed from global to local formula-
(A) Conservation laws
Global conservation law of the flux quantity
∀Ω0 ⊂ Ω :
Z
Z
v dS =
f (M ) dV .
(3.14)
Ω0
∂Ω0
By (3.6), it means that the total flux through the domain boundary equals to accumulate production of the sources inside.
Local form of conservation law for the flux quantity
Executing the previous limit process we obtain by (3.10) and (3.6) local conservation law:
div v(M ) = f (M ) .
(3.15)
(B) Constitutive relations are introduced in local vectorial form that is typical for transfer
phenomena as
v(M ) = − p(M ) grad u(M ) ,
M ∈Ω.
(3.16)
More general feature is obtained, if the material quantity p depends also on state function u:
v(M ) = − p(M, u) grad u(M ) .
(3.17)
(C) Local balance of state function
Finally, we put constitutive formula (3.16) into conservation law (3.15) supposing that the
density f as well as constitutive parameter p are continuous in Ω:
− div (p. grad u) = f .
(3.18)
Especially, if p = konst., then we have
− ∆u =
f
,
p
(3.19)
that is the well-known Poisson equation because of div grad = ∇ · ∇ = ∆. Having medium free
of sources (f = 0) we obtain directly the Laplace equation
∆u = 0 .
(3.20)
28
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
3.1.5 Example – laminar flow of ideal liquid. Our aim is to create mathematical model
of laminar flow in given region (Fig. 3.3). The global balance of mass flux m [kg.m2 s−1 ] in
the volume Ω is the starting relation, where the source production is described by the density
f [kg.m3 s−1 ]. In the case of ideal (non-viscous) liquid we can write
Z
Z
m dS =
f dV .
Ω
∂Ω
Figure 3.3: Reference volume in the flow model.
According the procedure introduced previously we obtain local balance of the flux quantity,
div m = f
on
Ω.
In practice, the use of the flow velocity v instead of mass flux is more advantegeous, therefore,
we put m = ρv, where ρ is mass density of the liquid. In the resulting relation
div (ρv) = f
can be transcribed even the divergence operator:
ρ div v + v · grad ρ = f .
Here both left-hand parts are called as source and advective term, respetively.
In what follows, we will assume an uncompressible liquid, for which ρ = konst. The resulting
equation
ρ div v = f
(3.21)
passes into well-known continuity equation div v = 0 in a volume free of sources.
Finally, we apply the requirement of laminar flow. It means, that velocity field circulation
along arbitrary closed curve in given region is equal to zero:
I
C=
Z
v dl =
K
rot v dS = 0
∀ K ⊂ Ω.
S
It follows from the Stokes formula that velocity field is non-vortex in Ω,
rot v = o .
3.2. BALANCE ON A BOUNDARY
29
Moreover, the velocity can be written as the gradient of certain scalar function that we call
velocity potential ψ:
v = grad ψ .
Putting it into (3.21) we obtain
ρ div ( grad ψ) = f ;
(3.22)
and, the classical Laplace equation ∆ψ = 0 for the potential ψ, if f = 0.
There exists the other way to the continuity equation directly from global balance without
sources. It suffices to rewrite the flux term as the sum
Z
S3
S2
S1
∂Ω
ρv dS = 0 .
ρv dS +
ρv dS +
ρv dS =
Z
Z
Z
Since the surfaces S1 , S2 have reversely oriented normals thereby the flux through the mantle
S3 is zero-valued, one holds for ρ = konst.
S1 v1 = S2 v2 .
3.2
Balance on a boundary
3.2.1 Interfacial condition. We will analyse situation on inner boundary Γ dividing the
region Ω into two parts (Fig. 3.4). Again, we consider reference sub-domain Ω0 , where
Γ0 = Ω0 ∩ Γ,
If no sources appears in Ω0 , then
∂Ω0 = Γ1 ∪ Γ2 ∪ Γ3 ∪ Γ4 .
Z
v dS = 0 .
(3.23)
∂Ω0
n(P )
Γ3
'
$
Γ
Ω
&
%
v
Γ2
Ω0
P
Γ4
v
Γ0
n
Γ1
n
Γ
Figure 3.4: Interfacial flux balance.
The limiting steps µ(Ω0 ) → 0 along the normal n from inside of Ω0 and in opposite direction
from outside lead to µ(Γ1 ) → 0, µ(Γ3 ) → 0, but the boundary Γ0 = Γ ∩ Ω0 retains unchanged.
Thus, we can write
1
lim
µ→0 µ(Ω0 )
Z
Z
v dS =
∂Ω0
Γ0
[v(P +) − v(P −)] · n dS
∀ P ∈ Γ0
∀ Γ0 ⊂ Γ .
(3.24)
30
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
As the local consequence of (3.23) for arbitrary boundary point P ∈ Γ we obtain [v(P +) −
v(P −)] · n = 0, therefore, by (3.1)
vn (P +) = vn (P −) .
(3.25)
In this way we formulated local condition of normal flux continuity or interfacial condition.
3.2.2 Condition on outer boundary. Now, we attend to the boundary ∂Ω, where the
outside flux v 0 (P ) is prescribed (Fig. 3.5). The limit step occurs at the same principle as in
previous case with following result:
[v(P +) − v 0 (P )] · n = 0
⇒
vn (P ) = v0n ,
v0n = v 0 (P ) · n .
(3.26)
n(P )
v 0
P
Γ0
Ω0
∂Ω
v
Ω
Figure 3.5: Flux balance on outer boundary.
3.2.3
Local consequences of constitutive relations.
(A) Interfacial condition at P ∈ Γ:
(3.25)
⇒
p(P +). grad u(P +) · n = p(P −). grad u(P −) · n ,
or
p(P +)
∂u(P +)
∂u(P −)
= p(P −)
.
∂n
∂n
(3.27)
(B) The flux v 0 (P ) given at P ∈ ∂Ω:
∂u(P +)
= v0n ,
v0n = v 0 (P ) · n .
∂n
Usualy in the transfer phenomena models, we suppose that
(3.26)
⇒
−p(P +)
v0n = α( u(P +) − u0 (P ) ) ,
(3.28)
(3.29)
where u0 (P ) is given function and α denotes a transfer coefficient. Thus, the above condition takes the form
∂u(P +)
p(P +)
+ α u(P +) = αu0 (P ) ,
(3.30)
∂n
or, if the right side is directly precribed as g(P ),
p(P +)
∂u(P +)
+ α u(P +) = g(P ) .
∂n
(3.31)
3.3. FORMULATION OF BOUNDARY PROBLEMS
3.3
31
Formulation of boundary problems
3.3.1 Boundary conditions typology. We will consider derived local balance of state
function u in the form of second-order partial differential equation (PDE)
− div (p. grad u) = f
on
Ω.
(3.32)
that constitutes boundary problem if supplied by several conditions below. The classification of
boundary problems is analogous as by 1D models.
1. Newton condition (3.31) in quite general form:
p
∂u
+ αu = g
∂n
na ∂Ω
(3.33)
by given inputs p on Ω̄, q, f on Ω and α, g on ∂Ω.
2. Neumann condition. If the gradient of state function is prescribed on the boundary, we
obtain required condition formally from previous relation (3.33) for α = 0:
p
∂u
=g
∂n
na ∂Ω .
(3.34)
In the case of α = 0 in (3.30), i.e. g = 0 in (3.34), we speak about homogeneous problem.
3. Dirichlet condition. In this simplest variant, directly the state function is given on the
boundary:
u = u0
na ∂Ω .
(3.35)
If various conditions are prescribed for different parts of the boundary ∂Ω, the boundary
problem is called as mixed.
3.3.2 Corectness of multidimensional models. Principially, we set the same requirements
as in one-dimensional problems, but the interpretation is naturally generalized.
1. Existence of solution (solvability); the properties of solution depend not only on the
input data properties (f, p, . . .) as for 1D models, but also on quality of the domain
Ω and its boundary.
2. Uniqueness of solution is based on completing of conditions, for which the homogeneous
problem
− div (p grad u) = 0
∂u
p
+ αu = 0
∂n
on Ω ,
(3.36)
on ∂Ω
has only trivial solution u = 0 by sufficiently smooth functions f and p.
3. Stability – see 1D models.
32
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
3.3.3 Example – temperature field in rectangular plate. We want formulate boundary problem describing stationarized down-cooling of rectangular plate (Fig. 3.6) by following
assumptions:
(1) material is homogeneous (heat conductivity λ = konst.),
(2) no heat source are inside the plate,
(3) the side no. 3 keeps constant temperature T0 ,
(4) the sides no. 1 and 4 are termally isolated,
(5) heat losses take the place along the side no. 2 expressed by the continuous heat flux q(y) ≥ 0
Wm−2 .
Figure 3.6: To the model of temperature field in a plate.
Equation
Let the rectangle sides have the size a, b, respectively. Thus, the plate inward is the domain
Ω = (0, a)×(0, b), where we will estimate the temeperature field T (x, y). For this state quantity,
the local conservation law has previously derived form
− div (λ grad T ) = f (x, y)
on Ω .
(3.37)
Regarding first two assumptions the equation takes more simple Laplace PDR form:
(1)
(2)
⇒




−λ div grad T = f 
⇒




f (x, y) = 0
∆T (x, y) = 0
on Ω .
(3.38)
Boundary conditions
Denote Γ1 , Γ2 , Γ3 , Γ4 the rectangle sides in agreement with enumeration in the figure; thus,
∂Ω = ∪Γi , i = 1, . . . , 4. We exploit the fact, that normal vectors of boundaries (rectangle sides)
are parallel to coordinate axes, therefore, we can write directly partial derivatives regarding the
3.4. FOURIER METHOD
33
variables x, y instead the gradients. It is obvious, that the orientation of normal vector n must
be taken into account on each boundary with regard to the identity
∂u
= n0 · grad u ,
∂n
where n0 is unit vector of outer normal.
• The third condition means the ideal temperature contact with surrounding medium on Γ3
that we write as Dirichlet condition
T = T0
for y = b .
(3.39)
• The boundaries Γ1 and Γ4 are isolated by the assumption (4); zero-valued heat flux implies
homogenneous Neumann conditions
∂T
∂y
∂T
∂x
= 0
for y = 0 ,
(3.40)
= 0
for x = 0 .
(3.41)
• The last condition on the boundary Γ2 is of Newton type; we write this one in the wellknown form
∂T
−λ
= q(y)
for x = a .
(3.42)
∂x
Obtained results represents the mixed boundary problem for Laplace equation on a rectangle.
Remarks
(a) The equation ∆T = 0 by itself is less telling. The essential properties (domain form,
characteristics of boundaries etc.) are contained in boundary conditions.
(b) Verification of correctness requirements follows from the fact that input data are constants
or continuous functions. However, the situation at rectangle corners should be discussed
(what is the conclusion?).
(c) The solution of presented problem can be found e.g. by the Fourier separation method –
see the next section.
3.4
Fourier method
3.4.1 Othogonal expansion. We define scalar product of real functions f (x) and g(x) on
the interval ha, bi as
Zb
f (x) g(x) dx .
(3.43)
a
This product induces norm of function f (x),
2
kf (x)k =
Zb
| f (x) |2 dx .
a
In what follows, we will consider the square-integrable functions, for which kf (x)k < ∞.
(3.44)
34
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
A functional sequence {gn (x), x ∈ ha, bi}∞
n=1 is called as orthogonal system on interval
ha, bi, if
Zb
if m 6= n .
gm (x) gn (x) dx = 0
(3.45)
a
Moreover, if kgn (x)k = 1, we have orthonormal system. Any orthogonal system can be
normed with the elements {gn (x)/kgn (x)k}.
Let {gn (x)} be an orthogonal system on ha, bi, and, function f (x) is square-integrable therein.
The series
∞
X
an gn (x)
(3.46)
n=1
express expansion of the function f (x) with respect to the system {gn (x)}, if one holds
for the coefficients an ,
1
an =
kf (x)k2
Zb
f (x) g(x) dx .
(3.47)
an gn (x) .
(3.48)
a
Generally, this fact is written as
f (x) ∼
∞
X
n=1
Multiplying both sides of this relation by gm (x) and integrating over interval ha, bi, we obtain
by (3.45) only single non-zero term on the right side for m = n that equals to square of the
norm. However, it is obvious, that we need to find answer to two basic questions (related to the
given interval):
1. Does converge the series?
2. If yes, it is f (x) the sum of this one?
The answers are contained in following theorems.
(1) The series
∞
P
an gn (x) converges (in the norm k · k), if and only if
n=1
∞
X
| an |2 < ∞
(Bessel inequality).
(3.49)
n=1
(2) The series
∞
P
an gn (x) converges on the interval ha, bi to the function f (x), if and only if
n=1
2
kf (x)k =
∞
X
| an |2
(Parseval equality).
(3.50)
n=1
Therefore,
f (x) =
∞
X
n=1
only in this case.
an gn (x)
(3.51)
3.4. FOURIER METHOD
3.4.2
35
Example. Consider the system
1
, cos x, sin x, cos 2x, sin 2x, . . . , cos nx, sin nx, . . .
2
(3.52)
on the interval h0, 2πi. We will prove its orthogonality and derive normed form.
To this puprose, we use following integrals:
Z2π
cos mx cos nx dx =
0
Z2π
sin mx sin nx dx =
0




0





π







 2π




0





π







 0
, m 6= n
, m = n 6= 0
, m = n = 0,
, m 6= n
, m = n 6= 0
, m = n = 0,
Z2π
cos mx sin nx dx = 0 .
0
The orthogonality is obvious; further, we calculate the norms of the functions contained in given
system:
Z2π 2
1
2
Z2π
π
dx = ,
2
Z2π
2
cos mx dx = π,
0
0
0
sin2 mx dx = π .
Orthonormal system that corresponds to (3.52) results as
1 cos x sin x cos 2x sin 2x
√ , √ , √ , √ , √ ,...
π
π
π
π
2π
.
(3.53)
3.4.3 Fourier series. If we seek for expansion of the function f (x) with respect to the system
(3.52), then we obtain using (3.47) and (3.48)
f (x) ∼
where
1
an =
π
∞
a0 X
(an cos nx + bn sin nx) ,
+
2
n=1
(3.54)
Z2π
f (x) cos nx dx ,
n = 0, 1, 2, . . . ,
(3.55)
n = 1, 2, . . . .
(3.56)
0
1
bn =
π
Z2π
f (x) sin nx dx ,
0
The series (3.54) is called as Fourier expansion of the function f (x) on interval h0, 2πi,
the coefficients of which are given by (3.55) and (3.56).
36
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
This result can be easily generalized for interval h0, Li by the transform x → 2πx/L:
∞
a0 X
2πnx
2πnx
+
an cos
+ bn sin
2
L
L
n=1
f (x) ∼
,
(3.57)
n = 0, 1, 2, . . . ,
(3.58)
n = 1, 2, . . . .
(3.59)
where
2
an =
L
ZL
f (x) cos
2πnx
dx ,
L
0
2
bn =
L
ZL
f (x) sin
2πnx
dx ,
L
0
The converegence properties of Fourier series is summarized in Dirichlet theorem.
Dirichlet theorem. Let the function f (x)
(D1) is integrable on the interval h0, Li,
(D2) has only finite number of extrems therein.
Thus there exists uniquely determined Fourier expansion on h0, Li of the form
∞
a0 X
2πnx
2πnx
+
an cos
+ bn sin
2
L
L
n=1
s(x) =
with coefficients defined by above derived relations. This series converges
• to the function f (x) at each point of interval (0, L), where this function is continuous,
• to the mean of one-side limits at inner points of discontinuity,
• to the periodic enlargement f˜(x) outside interval (0, L) or to the mean of one-side limits,
where f˜(x) is not continuous.
Thus, it holds at arbitrary x ∈ h0, Li:
s(x) =




f (x),





in continuous case,
(3.60)







 f (x+) + f (x−)
2
otherwise.
3.4.4 Example. The function f (t) = (t − 2)2 is given for t ∈ h0, 2i. We will create the graph
of its periodic enlargement and find Fourier expansion.
The part for x ∈ h0, 4i of the parabola f (t) = (t − 2)2 is in Fig. 3.7 on the left. Function f (t)
is continuous, therefore, the discontinuities arrise after periodization only at bounds of periods,
where the expansion will converge to mean of one-side limits (it is equal to unity) – see the right
part of the figure. The Fourier coefficients are calculated by (3.58), (3.59) on the interval h0, 2i:
3.4. FOURIER METHOD
37
f˜(x)
f (x)
c
c
c
c 4
c
c
c
s
s
s
s
s
s
s
c
c
c
c
c
c
c
−6
−4
−2
2
4
6
4
x
0
2
4
0
x
Figure 3.7: Periodic enlargement.
a0 =
an =
R2
0
R2
(x − 2)2 dx = 83 ,
(x − 2)2 . cos nπx dx =
0
=
bn =
h
(x−2)2
nπ
R2
sin nπx −
2
nπ
x2
− nπ
cos nπx +
i2
1
n2 π 2
sin nπx
=
0
4
,
n2 π 2
(x − 2)2 . sin nπx dx =
0
=
h
2
− (x−2)
nπ cos nπx +
2
nπ
x−2
nπ
sin nπx +
1
n2 π 2
i2
cos nπx
0
=
4
nπ
.
Now, we can represent given function by its Fourier expansion as
∞
X
4
cos nπx sin nπx
f (x) ∼ + 4
+
3
n2 π 2
nπ
n=1
,
x ∈ h0, 2i.
3.4.5 Fourier method for Laplace equation. We show the basic principle for simple
Dirichlet problem on a rectangle:
∂2u ∂2u
+ 2 =0
∂x2
∂y
Ω = {0 < x < a, 0 < y < b},
on
(3.61)
u(x, 0) = 0,
y = 0,
x ∈ h0, ai,
(3.62)
u(a, y) = 0,
x = a,
y ∈ h0, bi,
(3.63)
u(x, b) = ϕ(x),
u(0, y) = 0,
y = b,
x = 0,
x ∈ h0, ai,
y ∈ h0, bi.
(3.64)
(3.65)
In the first step we will work with the assumption that the problem has a solution of the form
u(x, y) = X(x).Y (y). Putting this expression into Laplace equation we obtain after simple
rearrangement
1 d2 X
1 d2 Y
=
−
on Ω.
X dx2
Y dy 2
The left side depends only on the variable x, but the right side only on y. Since the relation must
be fulfilled at arbitrary point (x, y) ∈ Ω, the both sides must be equal to the same constant that
we denote −µ2 (the choose of sign will be explained later). In this way we obtain the double of
ordinary second-order differential equations
X 00 (x) + µ2 X(x) = 0 ,
Y 00 (y) − µ2 Y (y) = 0
38
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
with general solutions
X(x) = A cos µx + B sin µx ,
Y (y) = C cosh µy + D sinh µy .
(3.66)
Thus, we can write
u(x, y) = (A cos µx + B sin µx)(C cosh µy + D sinh µy) .
(3.67)
Described procedure is usually referred as Fourier separation method.
Boundary conditions.
We derive the constants A, B, C, D and unknown value µ from boundary conditions. It is very
advantageous to start with homogeneous conditions, therefore, we consider at first the relation
(3.62). In this case,
0 = (A cos µx + B sin µx).C
for arbitrary x ∈ h0, ai. Thus, C = 0, and
u(x, y) = (α cos µx + β sin µx) sinh µy ,
where
α = AD, β = BD.
Now, we take into account the condition (3.65):
0 = α sinh µy,
what implies α = 0 ,
and,
u(x, y) = β sin µx sinh µy .
Next condition (3.63) holds at x = a:
0 = β sin µa sinh µy .
Since sinh µy 6= 0 almost everywhere for y ∈ h0, bi,
sin µa = 0,
i.e.
µa = nπ, n = 0, 1, 2, . . . .
This approach leads to infinite sequence of eigen-values
µn =
nπ
;
a
(3.68)
and, to corresponding eigen-functions
un (x, y) = βn sin µn x sinh µn y .
(3.69)
We must write the solution of problem as the superposition of all items as the series
u(x, y) =
∞
X
un (x, y) =
n=1
∞
X
βn sin µn x sinh µn y
n=1
(if n = 0, then u0 (x, y) = 0). This expansion must satisfy the conditiion (3.64) at y = b:
ϕ(x) =
∞
X
n=1
βn sin µn x sinh µn b .
(3.70)
3.4. FOURIER METHOD
39
It means, that this series is the Fourier expansion of the function ϕ(x) on the interval h0, ai,
because of
∞
ϕ(x) =
X
β̃n sin µn x ,
where
β̃n = βn sinh µn b.
(3.71)
n=1
It can be shown that the system gn (x) = sin µn x is orthogonal on this interval,
Za
sin
0


a



= k sin µn xk2
nπx
mπx
2
sin
dx =

a
a


 0
, m = n,
.
, m 6= n
There are two ways to estimate of coefficients β̃n : a modification of general Fourier formulae
to obtain sine expansion or direct use of orthogonality. We apply the second one so, that we
multiply equation (3.71) by the function gm (x) = sin µm x and integrate over period L = a:
Za
ϕ(x) sin µm x dx =
∞
X
Za
β̃n
n=1
0
sin µm x sin µn x dx .
0
The right-side integrals are non-zero only if m = n that yields
Za
ϕ(x) sin µm x dx = β̃n
a
.
2
0
Thus
2
β̃n =
a
Za
ϕ(x) sin µn x dx
and
βn =
0
β̃n
.
sinh µn b
(3.72)
Finally, we create resulting solution of given problem:
u(x, y) =
∞
X
β̃n
sin µn x sinh µn y .
sinh µn b
n=1
(3.73)
Now, we turn back to the choose of negative sign of the constant µ2 . It follows from the fact,
that we have function of variable x in non-homogeneous boundary condition (3.64), therefore, we
require the goniometric eigen-functions also in this variable. In the opposite case (the condition
as a function of y), we should take the positive sign to have othogonal system in this variable.
In the case of non-homogeneous boundary conditions in the both variables, one from conditions
can be replaced in the equation as a source term by appropriate substitution.
3.4.6 Example – 3.3.3 continued. Derive the temperature field in the rectangular plate
(see Fig. 3.6) by following input data: a = 1 m, b = 2 m, T0 = 60 ◦ C, λ = 2 WK−1 m−1 ; and, by
heat flux
q0
q(y) = 2 (by − y 2 )
with q0 = 80 Wm−2 .
b
Demonstrate graphically temperature profile on the side, where the heat losses are acting.
We start with boundary problem (3.38)-(3.42), where we introduce substitution T − T0 = ϑ
to obtain homogeneous Dirichlet condition. This step results in mixed boundary problem for
Laplace equation on the rectangle:
∆ϑ(x, y) = 0
on Ω ,
(3.74)
40
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
∂ϑ(x, y) = 0,
∂y y=0
(3.75)
∂ϑ(x, y) = 0,
∂x x=0
(3.76)
ϑ(x, b) = 0 ,
(3.77)
∂ϑ(x, y) = q(y) .
λ
∂x x=a
(3.78)
Separation of variables – Fourier method
We seek for solution of the form ϑ(x, y) = X(x)Y (y). We must choose positive constant µ2 in
the relation
1 d2 X
1 d2 Y
=
= µ2
X dx2
Y dy 2
to obtain goniometric functions of y in the solution due to the condition (3.78). Therefore, we
write
X(x) = A cosh µx + B sinh µx ,
Y (y) = C cos µy + D sin µy .
(3.79)
Application of boundary conditions
(3.75) Y 0 (y) = µ(−C sin µy + D cos µy)
(3.76) X 0 (x) = µ(A sinh µx + B cosh µx)
(3.77) 0 = C cos µb
⇒
µk = (2k − 1)
ϑ(x, y) =
∞
X
⇒
⇒
π
,
2b
0 = µD
⇒
for y = 0
0 = µB
for x = 0
D = 0;
⇒
B = 0;
k = 1, 2, 3 . . . ,
αk cosh µk x cos µk y ,
αk = Ak Ck .
k=1
(3.78) λ
∂ϑ(x, y) = q(y)
∂x x=a
⇒
∞
X
αk µk sinh µk a cos µk y =
k=1
q(y)
.
λ
(3.80)
The system {cos µk y}∞
1 is orthogonal on h0, bi because of
Zb
cos µk y cos µm y dy =
0




0,





k 6= m ,







 b = k cos µk yk2 , k = m .
2
The scalar product with function cos µm y on the both sides of (3.80) on h0, bi leads to the
relation
Zb
b
1
αk µk sinh µk a =
q(y) cos µk y dy ,
2
λ
0
3.5. EXERCISES
41
Figure 3.8: Temperature profile on the plate side.
from which
2qk
αk =
,
λbµk sinh µk a
Zb
q(y) cos µk y dy .
qk =
(3.81)
0
Final solution
Using these results we can create following expansion of the temperature field that satisfies to
Laplace equation as well as given boundary conditions:
T (x, y) = T0 +
∞
2 X
qk
cosh µk x cos µk y .
λb k=1 µk sinh µk a
(3.82)
Numerical results
Calculation of qk :
Zb
q(k) =
(by − y 2 ) cos µk y dy = . . . =
0
µk b − 2 sin µk b
.
µ3k
Temperature field:
T (x, y) = T0 +
∞
2q0 X
µk b − 2 sin µk b
cosh µk x cos µk y .
λb3 k=1 µ4k sinh µk a
(3.83)
Temperature profile at x = a
The graphical output is presented in the Fig. 3.8. Computation has been realized in Matlab
with the first ten terms of series.
3.5
Exercises
3.5.1 Stationary temperature field. The half-strip by the Fig. 3.9 is considered. On
the finite border, parabolic temperature profile is given symmetrically along the x axis with
the maximum 2T0 at the centre of co-ordinate system. The horizontal sides of half-strip are
42
CHAPTER 3. MULTIDIMENSIONAL STATIONARY MODELS
maintained at the temperature T0 . Create mathematical model of stationary temperature field
in the half-strip. Calculate and graphically demonstrate temperature profile along the half-stripe
axis for appropriate chosen data (use symmetry of problem).
Figure 3.9: Temperature profile on the half-strip edge.
3.5.2 Electrostatic field. Our aim is to estimate distribution of electrostatic potential in
homogeneous rectangle with the sides a, b, where no free charges occur. No potential gradient
exists on the side of size b; the potential profile on neighboring side is described by the function
ϕ0
r2
,
a2
where the variable r denotes distance from common vertex. The two opposite sides have the
same constant potential ϕ0 .
Create mathematical model and solve obtained problem. Calculate the potential at the
rectangle center. Demonstrate graphically both the lines of force and equipotential lines by
appropriate scale for ϕ0 = 24, a = 10, b = 5.
3.5.3 Diffusive contamination. Find the distribution of contamining matter in rectangular
horizontally placed water reservoir of the dimensions L a d by the conditions presented below.
As the local balance use the example (a) in the subsection 4.3.3.
• steady state is considered,
• two neighboring sides are isolated (watertight),
• the opposite side to the side of the length L is ”clean”, i.e. the zero concentration of
pollutant is keeping here,
• the side opposite to the side of the length d acts as contamination source that is given
by gradient of concentration: this one is zero-valued at the clean side; and, it increases
linearly toward to neighboring vertex, where it reaches the value c0 /d,
• influence of gravitation force is neglected.
Represent graphically the result by following data: L = 20 m, d = 10 m, c0 = 5 g/m2 and
k = 0, 1 m−1 (diffusion coefficient).
Chapter 4
Non-stationary models
4.1
One-dimensional non-stationary models
4.1.1 Global balance. Again as in Sect. 2 we consider flux function v, state function u and
source function f on the interval h0, Li. Besides the space variable x these functions depend
also on the time t ∈ h0, ∞):
f = f (x, t),
v = v(x, t),
u = u(x, t) .
Further, we introduce several important expressions to describe global balance of flux quantity.
Total source production on the element hx1 , x2 i ⊂ (0, L) and in time interval ht1 , t2 i ⊂
(0, ∞):
Zx2 Zt2
F =
f (x, t) dt dx .
(4.1)
x1 t1
The flux at a point x ∈ hx1 , x2 i from t1 to t2 :
Zt2
v(x, t) dt .
t1
The change of flux balance as a response on time variability of state function:
Zx2
[w(x, t2 ) − w(x, t1 )] dx .
x1
Thus, the global balance in space-time domain hx1 , x2 i × ht1 , t2 i can be written in the form
Zt2
v(x2 , t) dt −
Zx2 Zt2
v(x1 , t) dt =
{z
(a)
}
|
f (x, t) dt dx −
x1 t1
t1
t1
|
Zt2
{z
(b)
}
|
Zx2
[w(x, t2 ) − w(x, t1 )] dx .
x1
{z
(c)
Meaning of single terms:
(a) flux at point x2 ,
(b) flux at point x1 ,
43
}
|
{z
(d)
}
(4.2)
44
CHAPTER 4. NON-STATIONARY MODELS
(c) source production in the element hx1 , x2 i,
(d) this function plays different role regarding modeled process: energy dissipation by transfer
phenomena due to friction, viscosity etc.; inertial term in movement modeling; eventually,
this one need not be contained in balance equation.
4.1.2 Local balance at internal point. The possibility of limit transition at arbitrary point
[x, t] ∈ (x1 , x2 ) × (t1 , t2 ) is supposed:
x1 → x− ,
x2 → x+ ,
t1 , t2 → t .
As the requirements to the functions in (4.2) are enhanced, the limiting leads to different balance
relations.
(A) If f (x, t), w(x, t) are only integrable, the equation (4.2) reduces in pure flux continuity:
v(x−, t) = v(x+, t) .
(4.3)
(B) If f (x, t), w(x, t) are continuous in t and together w(x, t) will differentiable in this variable,
then we obtain after limiting in t
Zx2
v(x2 , t) − v(x1 , t) =
f (x, t) dx −
x1
Zx2
x1
∂w(x, t)
dx .
∂t
(4.4)
(C) Moreover, if f (x, t), w(x, t) are continuous also in x, the analogous process results in local
balance
∂v(x, t)
∂w(x, t)
= f (x, t) −
.
(4.5)
∂x
∂t
It can be prove, that differentiability of the function v(x, t) by x follows from used assumptions.
(D) Consider the previously introduced local constitutive relation modified for time dependence:
∂u(x, t)
.
(4.6)
∂x
Its application in flux balance equation leads to local differential conservation law for state
quantity,
∂
∂u(x, t)
∂w(x, t)
−
p(x)
= f (x, t) −
,
(4.7)
∂x
∂x
∂t
that has the form of second-order PDR specified by nature of the function w(x, t).
v(x, t) = −p(x)
4.1.3 Boundary conditions. We will constitute boundary conditions directly at end-points
x = 0 a x = L of considered interval. Again, we start with the situation demonstrated in the
Fig. 2.6. Even now we can anticipate that balance relations have the same form as in chapter
2.3 with time-dependent governing functions.
(A) If the fluxes v0 (t), vL (t) are prescribed at boundaries, then we obtain the flux balance
v(0+, t) = v0 (t) ,
v(L−, t) = vL (t) .
(4.8)
Thus, we can rewrite above relations with the help of constitutive relation (4.6) as the
balance of state quantity u(x, t) in the form of Neumann conditions for state function:
p(0+)
∂u(0+, t)
= v0 (t) ,
∂x
p(L−)
∂u(L−, t)
= vL (t) .
∂x
(4.9)
4.2. EXAMPLES OF 1D NON-STATIONARY MODELS
45
(B) If the transfer conditions are released with the coefficients α0 , αL and given functions
u0 (t), uL (t), we linearize the flux balance as
v0 (t) = α0 [u(0+, t) − u0 (t)] ,
(4.10)
vL (t) = αL [u(L−, t) − uL (t)] .
Local balance of state function leads by application of constitutive relations to the Newton
conditions
∂u(0+, t)
p(0+)
= α0 [u(0+, t) − u0 (t)] ,
∂x
(4.11)
∂u(L−, t)
p(L−)
= αL [u(L−, t) − uL (t)] .
∂x
(C) In the case of prescribed values of state function we have simply the Dirichlet conditions
u(0+, t) = u0 (t),
u(L−, t) = uL (t) .
(4.12)
Remark. When boundary condition at some point can not be prescribed, we can write (if no
variance with physical reality takes place)
u(x, t) −→ u0 (t)
for
x −→ ∞ .
Oftly it suffices to assume only the boundness of solution for x → ∞ (see the example 4.2.3).
4.1.4 Initial conditions. If time development of the function u(x, t) is modeled for t > 0,
we need to know initial state as a function of space variable:
u(x, 0) = ϕ(x) .
A feature of initial conditions can be extended, if also initial time derivative is required (e.g. as
the initial velocity in kinematic models):
u(x, 0) = ϕ(x) ,
∂u(x, 0)
= u0t (x, 0) = ψ(x),
∂t
... .
(4.13)
These conditions can be also of the other meaning, when no time dependence is studied. If
we seek for solution of a differential equation at given point or on given line, the same problem
is called as Cauchy problem.
Summary. The full formulation of 1D non-stationary problems consist of the equation (4.7), initial conditions (4.13) and boundary conditions of the type (4.9), (4.11), (4.12),
eventually, of mixed ones.
4.2
Examples of 1D non-stationary models
4.2.1 Heat conduction in a rod. Let q(x, t) denotes the heat flux. Global balance of 1D
temperature field (state function) on the domain hx1 , x2 i × ht1 , t2 i without internal heat sources
(f = 0) has the form
Zt2
Zt2
q(x1 , t) dt =
t1
Zx2
q(x2 , t) dt +
t1
x1
cρ(x) [T (x, t2 ) − T (x, t1 )] dx ,
(4.14)
46
CHAPTER 4. NON-STATIONARY MODELS
where c [J kg−1 K−1 ] is heat capacity and ρ [kg m−1 ] the density (related to unit length). Rightside dissipation function w(x, t) = cρ(x)T (x, t) express heat accumulation in considered medium
like as ∆Q = cρ∆T . In the constitutive relation
q(x, t) = −λ
∂T (x, t)
∂x
the symbol λ denotes heat conductivity of the rod. Putting it into local balance (4.7) we obtain
∂T (x, t)
∂
λ(x)
∂x
∂x
= cρ(x)
∂T (x, t)
.
∂t
(4.15)
Supposing homogeneous material properties, the above relation leads to well-known Fourier
equation
∂2T
∂T
a 2 =
,
(4.16)
∂x
∂t
where we write a = λ/(cρ) for the temperature conductivity. The three following examples
illustrate boundary problems with this equation.
4.2.2 Example – heat conduction in a thin finite rod I. We shall analyze temperature
field in a finite rod of the length `, which keeps constant temperature T0 at the left end-point;
and, retains isolated at the right end-point. Initial temperature distribution is described by the
function T0 (1 + sin πx/`).
Obviously, we need to solve Fourier equation (4.16) on the domain Ωx,t = (0, `) × (0, ∞) with
the initial condition
πx
T (x, 0) = T0 1 + sin
.
(4.17)
`
Boundary conditions are given as
T (0, t) = T0 ,
(4.18)
∂T
= 0, x = ` .
∂x
Moreover, the asymptotic stability of solution can be ensured by the condition
T (x, t) < ∞
for t → ∞ .
(4.19)
(4.20)
Before problem solving we introduce an efficient notation that simplifies the calculation:
u(x, t) =
T − T0
,
T0
ξ=
x
,
`
κ=
a
.
`2
(4.21)
After easy rearrangement we obtain following formulation of mixed non-stationary boundary
problem for second order partial differential equation (PDE):
κu00ξξ = u0t
1. u(ξ, 0) = sin πξ ,
2. u(0, t) = 0 ,
3. u0ξ (1, t) = 0 ,
4. u < ∞ .
na
(0, 1) × (0, ∞) ,
(4.22)
4.2. EXAMPLES OF 1D NON-STATIONARY MODELS
47
We will modify formerly introduced Fourier method for this type of PDE. Setting u(ξ, t) =
w(ξ).v(t) into the equation leads to the identity
w00
1 v0
=
= −α2
w
κv
that must hold on the whole region Ωx,t . The double of ordinary differential equations results
as the consequence, where the negative sign follows from the condition 4 (solution boundness):
w00 + α2 w = 0
→
v 0 = κv
w(ξ) = A cos αξ + B sin αξ ,
v(t) = e−α
→
2 κt
.
(4.23)
(4.24)
The condition 2 implies A = 0, the No. 3 leads to the relation
⇒
B cos αξ = 0 ,
π
αn = (2n − 1) , n = 1, 2, . . .
2
(4.25)
In this way we derived the solution in the form of infinite expansion
u(ξ, t) =
∞
X
wn (ξ).vn (t) =
n=1
∞
X
2
Bn sin αn ξe−αn κt .
(4.26)
n=1
Now, we apply the last condition No. 1 to determine the coefficients Bn by solving the equation
sin πξ =
∞
X
Bn sin αn ξ .
(4.27)
n=1
To this purpose we use the orthogonality of eigen-function system {sin αn ξ}∞
1 on the interval
[0, 1] that yields

Z1
sin αn ξ sin αm ξ dξ =
0



 12
for m = n,



 0
for m 6= n.
(4.28)
Multiplicating the both sides of (4.27) by sin αm ξ, the integration from 0 to 1 gives
Bn =
(−1)n
8
.
π (2n − 3)(2n + 1)
(4.29)
The figure 4.1 demonstrates initial condition (full line) and the solution at three other time
points.
4.2.3 Example – heat conduction in a thin finite rod II. Unlike previous problem the
situation is described by rather modified assumptions.
1. Start of the heat exchange process: thin rod of the length L has constantaneous temperature T0 that is the same as outside one.
2. The first boundary point: fixed temperature T1 > T0 .
3. The second border: isolated.
4. Temperature interchange with surrounding medium of temperature T0 : along the whole
rod length with heat transfer coefficient α = const. [WK−1 m−2 ].
48
CHAPTER 4. NON-STATIONARY MODELS
40
teplota [ oC ]
35
30
t
t
t
t
25
20
0
0.2
0.4
ξ
= 0
= 6
= 18
= 60
0.6
0.8
1
Figure 4.1: Time development of temperature profile (the legend data in seconds).
We aim to find local formulation to estimate the temperature T (x, t) at arbitrary point in
x ∈ (0, L) for the time t > 0. At first we emphasize several important atributes of this problem:
• as the rod is thin, the instantaneous temperature is the same in the entire cross-section →
one-dimensional model can be considered;
• the assumption 4 will be realized by introduction of source term depending on state function T (x, t):
α
f (x, t, T ) = (T − T0 ) ,
x ∈ (0, L), t > 0 ;
(4.30)
L
• λ = konst. [WK−1 m−1 ] because of homogenity of the rod.
We use differential equation (4.15) supplied by source function (4.30):
∂T
∂2T
α
− λ 2 = (T − T0 ) ,
x ∈ (0, L), t > 0 .
(4.31)
∂t
∂x
L
Thus, the heat capacity c [Jkg−1 K−1 ] and density ρ [kgm−3 ] need be added to input data of our
model. Note, that all terms in obtained non-homogeneous PDR are of the same physical nature
– heat flux density [Wm−3 ].
Initial condition is prescribed by the assumption 1, i.e.
cρ
T (x, 0) = T0 ,
x ∈ h0, Li .
(4.32)
Without loss of generality we place the assumption No. 2 into the point x = 0; and, we write
the No. 3 at the point x = L. The first one implies directly the Dirichlet condition
T (0, t) = T1 ,
t>0,
(4.33)
the second one is represented by homogenneous Neumann condition
∂T (x, t)
=0,
x = L, t > 0 .
(4.34)
∂x
Performed model consists from the equation (4.31) with conditions (4.32)–(4.34). Again, it is
possible to solveit using Fourier method.
4.2. EXAMPLES OF 1D NON-STATIONARY MODELS
49
4.2.4 Example – heat conduction in a thin long rod. Very long thin rod has the initial
temperature T0 . At certain time (t = 0), it will at one end-point heated up to temperature
T1 > T0 that will be conserved. We shall determine the temperature at arbitrary point and
time.
Unlike previous problem no source term arises here that means simple equation
cρ
∂T
∂2T
−λ 2 =0 ,
∂t
∂x
x ∈ (0, ∞),
t ∈ (0, ∞) .
(4.35)
Suppose that the Dirichlet condition is located at the point x = 0,
T (0, t) = T1 ,
t>0;
(4.36)
we see, that second end-point is not specified. Since the rod is very long, the straight semi-line
x > 0 can be considered as the interval, where we seek for solution. Therefore, in agreement
with physical principles, the boundness of solution toward infinity can be considered, especially
T (x, t) = T1 ,
x → ∞, t → ∞
(4.37)
x ∈ h0, ∞) .
(4.38)
with initial condition
T (x, 0) = T0 ,
The formulated problem can be solved advantageously using Laplace transform. Here, we
apply quite specific approach based on appropriate substitution. At first, we simplify our formulation to have at least one homogeneous boundary condition. If we set u(x, t) = T −T1 , a = λ/cρ,
then
1 ∂u ∂ 2 u
− 2 =0,
x ∈ (0, ∞), t ∈ (0, ∞) ,
(4.39)
a ∂t
∂x
u(x, 0) = −(T1 − T0 ) = −ϑ ,
(4.40)
u(0, t) = 0 ,
u(x, t) = 0 ,
x → ∞, t → ∞ .
Further, we exploit an observation that
x
τ=√
at
is dimensionless quantity; therefore, we use the above relation as the substitution that yields
∂u
∂u τ
=−
,
∂t
∂τ 2t
∂2u
∂2u 1
.
=
∂x2
∂τ 2 at
We get ordinary second-order differential equation
d2 u τ du
+
=0
dτ 2
2 dτ
with the conditions
u(0, t) = u(τ = 0) = 0 ,
u(x, 0) = u(τ = ∞) = −ϑ .
50
CHAPTER 4. NON-STATIONARY MODELS
Putting du/dτ = y(τ ) we obtain
dy
τ
= − dτ
y
2
with general solution
Zτ
u(τ ) = C.
2
e
− z4
√
π
dz = C.
erf
2
τ
2
.
0
The ,,error function” erf(x) is defined as (see Fig. 4.2)
2
erf(x) = √
π
Zx
2
e−z dz .
0
Figure 4.2: The function erf(x).
√
The condition at infinity implies C = −2ϑ/ π, because of erf(∞) = 1. Inverse substitution
leads to resulting expression
T (x, t) = T1 − (T1 − T0 ) erf
x
√
2 at
.
(4.41)
Following results were calculated with the temperatures T0 = 20◦ , T1 = 120◦ by a = 0.02 m2 s−1 .
Particularly, the temperature profile after 30 seconds is illustrated; and, time dependence of the
temperature during first 30 seconds at x = 0.5 m from heated end-point (Fig. 4.3).
4.2.5 Vibration of a string. Let we derive non-stationary boundary problem that describes
vibration of elastic fibre (string). Remember local form of stationary state balance of instantaneous deviation u(x) by stress force T (x) and external forces density f (x) in the paragraph
2.2.4:
d
du
T (x)
= f (x) ,
(4.42)
dx
dx
4.2. EXAMPLES OF 1D NON-STATIONARY MODELS
51
Figure 4.3: Temperature profile after 30 s (left), the temperature T (x = 0.5, t) .
In particular we model non-damped oscillation invoked by initial conditions without an
influence of external forces. Thus, besides the time dependence of each quantity we must supply
the inertial term into global balance that represents the momentum,
w(x, t) = ρ(x)
∂u
.
∂t
Here, we see the other role of previously discussed function w. Supposing time continuity of the
all functions we can write in the interval hx1 , x2 i,
Zx2
ρ(x) a(x, t) dx ,
a(x, t) =
x1
∂ 2 u(x, t)
,
∂t2
where ρ(x) and a(x, t) are the mass density and the acceleration, respectively. Obviously, the
result corresponds to the second Newton law F = ma. Finally, the local equation of free
oscillations of string follows as
∂
∂u(x, t)
T (x)
∂x
∂x
= ρ(x)
∂ 2 u(x, t)
.
∂t2
(4.43)
The classical wave equation results by constant stress T ,
∂2u
∂2u
c2
=
,
∂x2
∂t2
s
c=
T
,
ρ
(4.44)
where the parameter c denotes stiffness of the string.
Boundary conditions of Dirichlet type express fixation of string in its end-points 0 and L:
u(0, t) = 0,
u(L, t) = 0 ,
t≥0.
(4.45)
In this model, two initial conditions are required reagarding second-order time derivation of
state function. The first one characteries initial deviation,
u(x, 0) = ϕ(x) ,
x ∈ (0, L) ,
(4.46)
52
CHAPTER 4. NON-STATIONARY MODELS
the second one describes initial velocity as time derivative,
∂u(x, 0)
= ψ(x) ,
∂t
x ∈ (0, L) .
(4.47)
The wave equation (4.44) with boundary conditions (4.45) and initial conditions (4.46), (4.47)
constitute full formulation of non-stationary problem for free non-damped string vibrations. To
solve this one, the Fourier method can be used.
4.2.6 Example – oscillation of a string. Solve the problem for non-damped vibration of
the string of the length L = 1 m and stiffness c = 0.3 m/s by initial deviation
1
x2
ϕ(x) =
x−
50
L
!
and initial velocity ψ(x) = 0. Demonstrate graphically instantaneous deviation at the time 1,
3, 5 and 7 seconds.
4.2.7 Typology of second-order PDE’s. For simplicity, we restrict the theory in this
subsection to linear problems with no more than two variables that we denote x and y without
physical interpretation. Partial differential equation of second order for the function u(x, y) can
be written as
∂2u
∂2u
∂2u
a11 (x, y) 2 + 2a12 (x, y)
+ a22 (x, y) 2 = f
∂x
∂x∂y
∂y
∂u ∂u
,
, u, x, y
∂x ∂y
.
(4.48)
For the solving on given domain, the continuity of coefficients aij (x, y) as well as continuity of
the function f (x, y) on the right-side are assumed. The classification of PDE’s is based on the
properties of the coefficients belonging to second-derivative terms. We write them into matrix


 a11


A=
a12 

a12 a22


(4.49)
and calculate its determinant
det(A) = δ = a11 a22 − a212 .
(4.50)
The symmetric matrix A represents quadratic form related to a conic by value of δ. In this
sense we denote corresponding PDE in agreement with following definition.
If δ has stable sign on the domain Ω, then partial differential equation of second order is called
on this domain as
• elliptical for δ > 0,
• parabolical for δ = 0,
• hyperbolical for δ < 0.
Examples of classification:
(a) ∆u = 0, i.e. u00xx + u00yy = 0
...
elliptical (δ = 1 > 0),
4.2. EXAMPLES OF 1D NON-STATIONARY MODELS
53
(b) the same holds in the case of Poisson’s equation ∆u = f ,
(c) a∆u = u0t
...
(d) a∆u = u00tt
(e)
∂2u
∂x∂y
2
...
= h(x, y)
2
(f ) y ∂∂xu2 + x ∂∂yu2
quadrant.
parabolical (δ = 0),
hyperbolical (δ < 0),
...
...
hyperbolical (δ = −1/4 < 0),
(δ = xy) elliptical in 1st and 3rd quadrant, hyperbolical in 2nd and 4th
4.2.8 Free advection. Suppose a very long stream, for which the flow can be considered
as one-dimensional with the velocity v = v(x, t). Concentration of a pollutant is described by
continuous function c = c(x, 0) = c0 (x) at the time t = 0. Due to water flow, the transfer of
contaminant is in progress by free advection. The other pollution sources of the local density
q(x, t) can also take the place. Our aim is to constitute balance equation to estimate of contamination c = c(x, t) at arbitrary point.
At first we write global concentration balance of pollutant in the time interval ht1 , t2 i on a
flow section hx1 , x2 i:
Zt2
Zt2 Zx2
t1
t1 x1
[c(x2 , t)v(x2 , t) − c(x1 , t)v(x1 , t)] dt =
q(x, t) dx dt −
Zx2
[c(x, t2 ) − c(x, t1 )] dx .
(4.51)
x1
The left side express total change of the flow function c.v, the right side can be considered as
a production term, where w(x, t) = c(x, t). The analogy to the general global balance (4.2) is
obvious. Since both the concentration and flow velocity are continuous in the time, we divide
the equation by ∆t and apply limit process ∆t → 0 that yields
c(x2 , t)v(x2 , t) − c(x1 , t)v(x1 , t) =
Zx2
q(x, t) dx −
x1
Zx2
x1
∂c
dx .
∂t
(4.52)
Now, we reply this step for the variable x that results in the first-order PDE
∂
∂c
(c.v) +
= q(x, t)
∂x
∂t
∂c
∂v
∂c
v+c
+
= q(x, t) .
∂x
∂x ∂t
i.e.
(4.53)
4.2.9 Free advection – model problems. Solve the equation (4.53) without internal
sources (q = 0) for initial state
c0 (x) =
cmax
,
cosh kx
k>0.
(a) By constant flow velocity v = v0 = const.
(b) By time-dependent velocity
v(t) =
v0
.
1 + α2 t2
(c) By the velocity being the function of space variable x as
v(x) = v0 e−kx .
In all these cases the method of characteristics can be used – see the next subsection.
54
CHAPTER 4. NON-STATIONARY MODELS
4.3
Partial differential equations of the first order
4.3.1 Basic notions. We will seek for a function of two variables u(x, y) satisfying the
equation
∂u ∂u
F
,
, u, x, y = 0.
(4.54)
∂x ∂y
If an equation does not contain another terms with derivatives of the function u, we call it as
linear or quasi-linear by following examples:
∂u
(a) y ∂u
∂x − x ∂y = 0
...
∂u
2
(b) y ∂u
∂x − x ∂y = xyu
(c) y
∂u
∂x
2
linear equation,
...
2
2
− x ∂u
∂y = x + y
quasi-linear equation,
...
non-linear equation.
The equation (a) is homogeneous because of zero-valued right side. In the opposite case (a
function g(x, y) 6= 0 as r.-h. term) it is non-homogeneous.
As the solution of the 1st-order PDE on the domain Ω we call any function u(x, y) fulfilling
the equation and having continuous derivatives. The last property can be written as u ∈ C1 (Ω).
4.3.2 Method of characteristics for homogeneous linear equation. This equation type
has the general form
∂u
∂u
P (x, y)
+ Q(x, y)
=0,
(4.55)
∂x
∂y
where the coefficients P, Q are continuous on a domain Ω, where we seek for solution, i.e.
P, Q ∈ C(Ω). Any solution u(x, y) represents an integral surface, equiscalar lines of which are
called as characteristics of the form
ϕ(x, y) = C .
(4.56)
Thus, ϕ(x, y(x)) = C are their projections into the z = 0 plane, which we will differenciate:
ϕ0x + ϕ0y .y 0 = 0 ,
ϕ0x
dy
= −y 0 = −
.
0
ϕy
dx
⇒
(4.57)
The characteristics also satisfy given PDE, therefore,
P ϕ0x + Qϕ0y = 0 ,
tj.
ϕ0x
Q
=− .
ϕ0y
P
(4.58)
Combining the last two relations we obtain characteristic equation
dx
dy
=
.
P (x, y)
Q(x, y)
(4.59)
General solution of the equation (4.55) is of the form
u(x, y) = G (ϕ(x, y)) ,
(4.60)
where G(ϕ) is arbitrary differentiable function. This fact can be verified by putting it into
original equation:
P G0x + QG0y = P G0ϕ u0x + QG0ϕ u0y = G0ϕ (P u0x + Qu0y ) .
4.3. PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
55
The expression in the last brackets is equal to zero in agreement of problem formulation that
implies correctness of the statement.
It remains to show, how to derive particular function G with regard to given conditions. We
can demonstrate them by the following way: the graph of solution u(x, y) is passed through
prescribed curve of parametrization x = x(τ ), y = y(τ ), z = z(τ ), τ ∈ hτ1 , τ2 i. Substituting x
and y in (4.56) the relation ϕ(τ ) = C is obtained, from which we express τ as a function of ϕ.
Finally, u = z(τ ) is the particular solution – see the following problem as an illustration.
4.3.3 Homogenneous equation – an example. We will seek for the solution of the equation (a) in 4.3.1 that is passed through
1) the parabola x = τ, y = τ, z = τ 2 , τ ∈ h−1, 1i,
2) the circle x = 1, y = cos τ, z = sin τ, τ ∈ h0, 2πi.
In the given equation yu0x − xu0y = 0 we have P (x, y) = y, Q(x, y) = −x. Characteristic equation
follows as
dx
dy
=−
,
i.e.
ydy = −xdx .
y
x
General solution x2 + y 2 = C represents the characteristics, therefore, the general solution can
be written as
u(x, y) = G(x2 + y 2 ) .
The calculation for given conditions runs as follows:
1) Putting of the parametrization into characteristics,
ϕ(τ ) = 2τ 2 ;
u = z = τ2 ,
i.e.
u = ϕ/2 = G(ϕ).
The resulting function
1
u(x, y) = (x2 + y 2 )
2
represents the rotationally symmetric parabolic surface.
2) In this case, the relation
ϕ(τ ) = 1 + cos2 τ = 2 − sin2 τ
follows from the equation of characteristics; as we have z = sin τ = u, then ϕ = 2 − u2 , therefore,
the spherical surface
x2 + y 2 + u2 = 2
√
has been obtained. We can supply that G(ϕ) = 2 − ϕ in this case.
4.3.4 Existence and uniqueness of solution. For the problems of above type, following
general formulation is used, named as Cauchy problem or initial problem for a first-order
PDE:
∂u
∂u
+ Q(x, y)
=0,
u(x(τ ), y(τ )) = z(τ ), τ ∈ hτ1 , τ2 i .
(4.61)
P (x, y)
∂x
∂y
The Cauchy uniqueness theorem (without a proof):
Let P (x, y), Q(x, y) ∈ C1 (Ω), Ω ⊂ R2 . The curve K parametrization (x(τ ), y(τ ), z(τ )), τ ∈
56
CHAPTER 4. NON-STATIONARY MODELS
hτ1 , τ2 i is given so, that the plane curve k = {x(τ ), y(τ )} lies in Ω. If one holds for any
τ ∈ hτ1 , τ2 i in the Cauchy problem (4.61) that
Q(x(τ ), y(τ )) 6 0,
=
ẏ(τ )
P (x(τ ), y(τ ))
ẋ(τ )
then a subdomain Ω∗ ⊂ Ω exists containing the curve k, where the solution is unique.
We show an application of this theorem to the example 2) in 4.3.3. The related determinant
is of the form
τ
1
−τ = 2τ .
1 It means that obtained solution is unique in arbitrary domain, which does not contain the point
τ = x = y = 0, i.e. the vertex of parabolic surface.
4.3.5 Application – free advection. We use method of characteristics to solve of the
problem 4.2.9(b) formulated as the PDE for the concentration c(x, t),
v(t)
∂c
∂c
+
=0,
∂x ∂t
v(t) =
v0
,
1 + α2 t2
(4.62)
where v0 , α are given constants. Initial condition is introduced by the function
cmax
cosh kx
c(x, 0) = c0 (x) =
with the input data cmax and k > 0.
At first the characteristic equation need be derived, where P (x, t) = v(t) , Q(x, t) = 1:
dx
= dt,
v(t)
i.e.
from which we obtain
ϕ(x, t) = x −
dx = v(t) dt ,
Z
v(t) dt = K .
Parametrization of initial condition is the next step:
x = τ, t = 0, c = c0 (τ ) .
If we put t = 0 into the characteristics, then ϕ = x = τ ; and, by the third parametrization
component,
Z
c(x, t) = c0 (ϕ(x, t)) = c0 x − v(t) dt .
(4.63)
In our particular case, we supply the velocity integral
Z
v(t) dt =
v0
arctan αt
α
4.3. PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
57
into initial condition, that implies
c(x, t) =
cmax
.
cosh k x − vα0 arctan αt
(4.64)
A short discussion can be added. The maximum of concentration at fixed time tk will located
at the point, where the denominator is minimized, it means at
v0
x=
arctan αtk .
α
Since v(t) → 0 as t → ∞, this process ceases by time increase. Finally, the concentration will
stationarized having the maximum at the point
v0
v0 π
xs = lim
arctan αt =
.
t→∞ α
α 2
4.3.6 Method of characteristics for quasilinear equation. We seek for a function that
satisfy the equation
∂u
∂u
P (x, y, u)
+ Q(x, y, u)
= R(x, y, u)
(4.65)
∂x
∂y
with P, Q, R ∈ C(Ω) ⊂ R3 , the characteristics of which is supposed in the form
ϕ(x, y, u(x, y)) = C .
We express partial derivatives by the both variables x, y:
ϕ0x + ϕ0u u0x = 0 ,
ϕ0y + ϕ0u u0y = 0 .
Now, we add the first equation multiplied by the function P (x, y, u) to the second one multiplied
by the function Q(x, y, u):
P ϕ0x + Qϕ0y + ϕ0u (P u0x + Qu0y ) = 0 .
Since the bracketted term is equal to R(x, y, u) by (4.65), we can write
P ϕ0x + Qϕ0y + Rϕ0u = 0 .
Further, we differenciate equation of characteristics ϕ(x, y, u) = C:
ϕ0x dx + ϕ0y dy + ϕ0u du = 0 .
The last two relation hold together only if one holds
dx
dy
du
=
=
.
P (x, y, u)
Q(x, y, u)
R(x, y, u)
(4.66)
We obtained three characteristic differential equations, from which it suffices to solve arbitrary
couple that yields two different characteristics
ϕ(x, y, u) = C1 ,
ψ(x, y, u) = C2 .
(4.67)
Putting them into quasilinear equation (4.65) we can prove, that general solution is of the form
G(ϕ, ψ) = 0 ,
(4.68)
where G is a function of two variables differentiable on Ω.
A Cauchy problem is formulated in the same way as for homogeneous equation. The solving
process should be as follows:
58
CHAPTER 4. NON-STATIONARY MODELS
1. derive the characterics of the form (4.67) by the above presented procedure;
2. put into them the parametrization (x(τ ), y(τ ), z(τ ) to obtain
ϕ(τ ) = C1 ,
ψ(τ ) = C2 ;
3. eliminate the parameter τ that leads to
G(C1 , C2 ) = 0 ;
4. substitute C1 a C2 by the characteristics ϕ and ψ.
In this case, the range 2 of the matrix


 P (τ )



ẋ(τ )
Q(τ ) R(τ ) 

ẏ(τ )
ż(τ )


is required to unique solution on the prescribed interval for the parameter τ .
4.3.7
Quasilinear equation – an example. Our problem is to solve the equation
xu0x + yu0y = u
(4.69)
by the condition given as the cut of the sphere x2 + y 2 + z 2 = 5 by the plane y = 1.
1. The continuous coefficients P = x, Q = y, R = u lead to system of characteristic equations
dx
dy
du
=
=
.
x
y
u
We easily calculate that y = C1 x, u = C2 x, for instance, therefore,
ϕ(x, y, u) =
y
= C1 ,
x
ψ(x, y, u) =
u
= C2 .
x
(4.70)
2. The circle of radius 2 in the y = 1 plane is given curve (bold dashed in the Fig. 4.4) with
parametrization
x = 2 cos τ, y = 1, z = 2 sin τ, τ ∈ h0, 2πi .
The matrix


 P (τ ) Q(τ ) R(τ )



  2 cos τ
 
=
 
ẋ(τ )
ẏ(τ )
ż(τ )


−2 sin τ
1 2 sin τ 

0 2 cos τ


is of the range 2 on the whole interval h0, 2πi, the solution will be unique.
3. We use the parametrization relations in characteristics and eliminate τ :
C1 =
y
1
=
,
x
2 cos τ
C2 =
u
sin τ
=
x
cos τ
⇒
C2 =
q
4C12 − 1 .
4.3. PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
59
4. This implies that
q
4C12 − 1 = 0 .
G(C1 , C2 ) = C2 −
Setting C1 = ϕ, C2 = ψ leads to the result:
u(x, y) =
q
4y 2 − x2 .
(4.71)
In particular, we found the conic surface
1
y 2 = (x2 + u2 )
4
with the axis y and the vertex in the origin of coordinate system. In the Fig. 4.4 we see
its part, where τ ≥ 0).
Figure 4.4: To the Example 4.3.7.
4.3.8
Quasilinear equation – the variant 4.2.9c. Rewriting (4.53) as
v(x)
∂c
∂c
+
= −v 0 (x)c ,
∂x ∂t
we find out that it is quasilinear. From the system
dx
dt
dc
=
=−
v(x)
1
v(x)c
we chose a couple, from which the deriving of characteristic ϕ, ψ will be advantageous:
dx
v(x)
= dt
→
dx
v(x)
dc
= − v(x)c
→ ψ(x, t, c) = v(x)c(x, t) = K2 .
ϕ(x, t, c) = t −
R dx
v(x) = K1 ,
60
CHAPTER 4. NON-STATIONARY MODELS
In partiular,
1 kx
e ,
v0 k
cmax
2v0 cmax
ψ = v0 e−kx
= 2kx
.
cosh kx
e
+1
The parametrization of initial condition can be introduced as x = ξ, t = 0, c = c0 (ξ). Thus
ϕ=t−
1
v0
Z
ekx dx = t −
ekξ = −v0 kK1 ,
2v0 cmax
= K2 .
e2kξ + 1
Substituting ekξ from the first relation into second one yields
2v0 cmax
K2 − 2 2 2
=0.
v0 k K1 + 1
Finally, we put ϕ, ψ instead K1 , K2 ; a short rearrangement leads to the solution
c(x, t) =
2cmax ekx
.
(v0 kt − ekx )2 + 1
Validity of result can be easily verified.
There is of interess to review obtained solution for particular numerical data. Suppose that
emission of the concentration c(x, t) is abducted by wind of the velocity 40 km/h that ceases
with increasing distance from the point of its origin with the parameter k = 0.05 km−1 . At first
we estimate, when maximum of contamination shows itself at the distance xm = 10 km. This
extreme comes on by v0 kt − ekx = 0, therefore,
1 kxm
tm =
= · · · ≈ 49, 5 min.
e
kv0
Corresponding concentration of pollution results as
c(xm , tm ) = cmax ekxm = 1, 65 cmax > cmax .
Obviously, the maximum of contamination increases at a place with lower advection velocity by
comparison with initial maximum. Theoretically, this process is in progress by growing spacing
from the point of origin. However, this is only a consequence of quite simplified model.
4.3.9 Heat advection model. A liquid flows through very long narrow pipeline by following
conditions:
(a) flow velocity is invariable;
(b) the process is adiabatical;
(c) heat transfer proceeds in the surrounding of the constant temperature T0 on the surface of
pipeline;
(d) a temperature change of running media takes effect instantaneously in the entire pipeline
cross-section; therefore, the heat transfer coefficient [α] = WK−1 m−3 relates to the volume
unit;
(e) input medium temperature T1 > T0 is constant as well as its density ρ and specific heat c;
(f) initial temperature in the entire pipeline is equal to the surrounding temperature.
Find the solution using method of characteristics by following data:
α = 700 WK−1 m−3 , c = 4186 Jkg−1 K−1 , ρ = 1000 kgm−3 , v = 0, 2 ms−1 , T0 = 5◦ C, T1 = 85◦ C.
Demonstrate the graphs of
• temperature profile in the pipeline from the input point to the distance 250 m at the time
tk = 15 minutes,
• time course of temperature at the point xp = 200 within the first twenty minutes.
4.4. NON-STATIONARY MULTI-DIMENSIONAL MODELS
4.3.10
61
Excersises.
1. Find the solution and postulate the uniqueness condition:
∂u
∂u
+ y2
=0,
x = −1, y = τ, z = τ + 1, τ ∈ R .
∂x
∂y
u(x, y) =
xy+2y+1
xy+y+1 ,
y 6= 0
2. yu0x − xu0y = 2xy, u(x, x) = 2x2 − 1,
3. 2yu0x − xu0y = 4xyu2 , u(x, x) = 1/x2 ,
u(x, y) = 32 x2 + 21 y 2 − 1
u(x, y) =
3
4y 2 −x2
4. Show that solving of the examples 4.2.9a,b can be generalized as an use of the substitution
ξ = x − vt
ξ =x−
or
Zt
v(τ ) dτ .
0
5. A water contamination in neighbourhood of the point x = 0 is described by pollution
concentration as
cm
,
c0 (x) =
1 + (bx)2
where x ≥ 0 is the distance in meters, b [m−1 ] is given parameter. The water stream of
the velocity
v0
v(t) =
1 + e−at
is characterized by deceleration factor a > 0; hydraulic properties of stream are omitted.
Establish the point of contamination maxima past 15 minutes. Take following data in
account: b = 1, a = 10−3 s−1 , v0 = 0, 1 ms−1 .
4.4
Non-stationary multi-dimensional models
4.4.1
Balance relations. The stationary flux conservation law (3.14)
∀M ∈ Ω, P ∈ ∂Ω :
Z
Z
f (M ) dV
v(P ) dS =
(4.72)
Ω
∂Ω
needs to be supplied – similarly as in one-dimensional case – by a term representing time
variability of state function. Thus, by continuity of the functions f and w the requested equation
is analogous as its 1D variant (4.4):
∀M ∈ Ω, P ∈ ∂Ω :
Z
Z
f (M, t) dV −
v(P, t) dS =
∂Ω
Ω
Z
∂w(M, t)
.
∂t
(4.73)
Ω
Supposing also continuity related to space variables we can derive local formulation applying
the Gauss-Ostrogradsky theorem on the left side and by the use of limit transient µ(Ω) → 0:
div v(M, t) = f (M, t) −
∂w(M, t)
.
∂t
(4.74)
An introducing of non-stationary constitutive relation
v = −p(M ) grad u(M, t)
(4.75)
leads to the local balance of state quantity u(M, t) in the form of the second-order PDE
− div (p. grad u) = f −
∂w
∂t
on Ω .
(4.76)
62
CHAPTER 4. NON-STATIONARY MODELS
4.4.2 Boundary conditions. As in previous paragraphs, we suppose data continuity relating
to space variables including time variability. The formerly applied steps in the subsection 3.1
for stationary models needs not be repeated, therefore, we present directly the local variant of
basic relations:
(A) Boundary conditions (P ∈ ∂Ω, t > 0) by given functions p(P, t) and g(P, t):
• Newton conditions – see (3.33)
p(P, t)
∂u(P, t)
+ α u(P, t) = g(P, t) .
∂n
(4.77)
• Neumann conditions – see (3.34)
p(P, t)
∂u(P, t)
= g(P, t) ;
∂n
(4.78)
• Dirichlet conditions – see (3.35)
u(P, t) = g(P, t) .
(4.79)
(B) Initial conditions (M ∈ Ω, t = 0) describe initial distribution of state function, eventually,
its time derivative:
u(M, 0) = u0 (M ) ,
(4.80)
or (if needed)
∂u(M, t) = u1 (M ) .
∂t
t=0
(4.81)
Formulation as well as typology of non-stationary multi-variable boundary problems corresponds to stationary variants presented in the subsection 3.3, although include the time dependence of governing functions; and, supplied by initial conditions. We illustrate such situation
in the following example, where a local balance is well-known (Fourier heat transfer equation),
but the attention must be oriented to boundary conditions.
4.4.3 Non-stationary 2D temperature field.
Constitute non-stationary boundary problem for temperature field T (x, y, t) in cooled rectangular plate composed from two square parts of the size a (see Fig. 4.6). Use following input
data:
1. Temperature of surrounding: T0 = const.
2. Initial temperature: Tp > T0 (the same in the both parts).
3. Thermal conductivity: λ1 inside the segment ABEF, λ2 in the neighboring one; corresponding temperature conductivity a1 , a2 given as well.
4. The sides AC, AF and EF isolated.
5. Heat transfer into outside medium with the coefficient α on the rest of of boundary.
6. Perfect thermal contact on the interface BE.
4.4. NON-STATIONARY MULTI-DIMENSIONAL MODELS
63
E
F
D
C
B
A
Figure 4.5: The cooled plate in 4.3.3.
Let Ω1 , Ω2 denote open squares ABEF and BCDE, respectively. Coordinate system with
the origin A is introduced in this way that the rectangle is located in the first quadrant and the
vertex F lies on the y axis. We will model given situation as a couple of boundary problems for
temperature fields T1 (x, y, t), T2 (x, y, t) related to the domains Ω1 , Ω2 . A coupling on common
interface will be expressed by transient conditions.
Using formerly obtained knowledge of local balance equations (4.16), (3.38) and (4.76) we
can write regarding assumptions 2 and 3
aj ∆Tj (x, y, t) =
∂Tj (x, y, t)
∂t
on Ωj ,
j = 1, 2 .
(4.82)
Also the initial condition is the same for the both regions (see 2):
Tj (x, y, t = 0) = Tp ,
j = 1, 2 .
(4.83)
Since any boundary segment is an abscissa parallel to a coordinate axis, the normal derivatives
in (4.77), (4.78) simplify as derivatives by single space variables (see the paragraph 3.3.3, for
instance). Homogeneous Neumann conditions take place on each isolated boundary part; the
Newton conditions hold on the sides CD and DE. The transient conditions on the boundary BE
contains besides usual flux continuity also equality of temperatures by the assumption 6:
T1 = T2 ,
λ1
∂T1
∂T2
= λ2
,
∂x
∂x
x = a, y ∈ h0, ai .
(4.84)
4.4.4 Diffusion with advection. This combined phenomena represents typical situation,
where variability of model approaches can be showed for non-stationary multi-dimensional problem. Our aim is to derive global and local balance by atmosphere contamination on the region
Ω – see Fig. 4.6. The contamination level is expressed by dimensionless concentration of a
pollutant, c(M, t), M ∈ Ω, t > 0. The soil transfer is realized partly by advection (air flow)
of the velocity v(M, t), partly as a diffusion with the coefficient k(M ). Thus, the flux function
q(M, t) is considered in the form
q = vc − k grad c .
(4.85)
Further, let f (M, t) is the density of inner pollution sources (as local hearths etc.).
The mass conservation law will be the basic balance relation, where we take the injurant
concentration in volume unit c(M, t) as the state quantity. We express the fact that production
of internal sources is equal to the sum of pollution exchange with surrounding through the
64
CHAPTER 4. NON-STATIONARY MODELS
Figure 4.6: To the problem 4.3.3.
boundary ∂Ω employed by flux q; and, by the time-dependent change of contaminant amount
in the volume Ω – cf. (4.53):
Z
∂Ω
q · dS =
Z
f (M, t) dV −
Z
∂c(M, t)
dV .
∂t
(4.86)
Ω
Ω
Having the global balance, we continue toward local one. At first, we apply the GaussOstrogradsky theorem to the left-hand side:
Z
div (vc − k grad c) dV =
Z
f (M, t) dV −
Ω
Ω
Z
∂c(M, t)
dV .
∂t
(4.87)
Ω
Thus, the limit transient µ(Ω) → 0 yields
∂c
+f .
(4.88)
∂t
We obtained two terms on the left-hand side, the advection and the diffusion one, respectively.
Following special cases show that this equation acquires various forms with regard to the weight
of single phenomena. We set f (M, t) = 0 and k = const. on the domain Ω, for simplicity.
div (c.v) − div (k grad c) = −
(a) If kvk << 1, the diffusion character of contamination process predominates; the parabolic
second-order PDE results from (4.88) called also as Fick’s law:
∂c
= k∆c ,
∂t
(4.89)
(b) Advection process prevails for k → 0 and/or kvk >> 1 that is described by 1st-order PDE
∂c
= div (vc) .
∂t
Rewriting the divergence operator we have
−
−
∂c
= c div v + v · grad c .
∂t
It is obvious that this relation implies the equation (4.53) in one-dimensional case.
(4.90)
(4.91)
4.4. NON-STATIONARY MULTI-DIMENSIONAL MODELS
65
4.4.5 Example - non-stationary diffusion on a rectangle. A shallow rectangular reservoir of the sides a, b (a > b) is isolated on two neighbouring sides. Infiltration of a contaminant
runs through the side opposite to the shorter wall thereby the highest concentration is at the
middle (ten times greater than the normal value c0 ); and, this one decreases toward the borders,
where it achieves normal value. The filtration (cleaning) on the remaining wall keeps here also
the standardized value c0 . The same is supposed as the initial state in the whole reservoir.
Constitute mathematical model of pollution distribution in the reservoir by prevailng diffusion transfer (i.e. by negligible advection).
Chapter 5
Selected problems
5.1
Population and logistic models
The models based on differential equations can be used also in the problems with discretized
data as there are number of persons in economic models, number of individuals in a population
etc. The important assumption need be accepted that the increment of a quantity is very small
compared by frequency of modeled ensemble.
5.1.1 Reproduction with migration. Constitute the problem describing time variability
of the number of residents in a closed region as a consequence of reproduction and/or migration
processes. Take into account move in, move away, nativity and deaths. Characterize the problem
type and suggest method to solve.
Let x0 denotes initial number of residents and x(t) their number at a time t. Futher, let
∆x(t) be the change corresponding to time interval ∆t. A population progress caused by move
in and away will be incorporated into common function f (t). In order to describe reproduction
processes depending on time as well as on instantaneous number of residents, we introduce
functions α(x, t) and β(x, t) for the number of born or dead residents in the time section ∆t.
In- or decrease of inhabitants ∆x in the time section ∆t can be balanced with the help of
difference equation
∆x(t)
= α(x, t) − β(x, t) + f (t),
∆t
where α, β are non-negative functions. Supposing |∆x| << x, |∆t| << t we can set time
variance of resident number as continuous that leads by well-known steps to the differential
equation
dx(t)
= α(x, t) − β(x, t) + f (t).
dt
This one constitutes first-order Cauchy problem with initial condition x(0) = x0 . Generally,
it is correctly formulated problem, but solvable only using numerical methods. An analytical
approach can be used only if the functions α, β are of favourable form.
As the simplest example we present linearized model, where
α(x, t) = k1 x,
β(x, t) = k2 x
(k1 , k2 ≥ 0).
At first, we consider isolated system (f (t) = 0), for which we obtain separable linear equation
that can be easy resolved:
dx
= (k1 − k2 )x ,
dt
x(0) = x0
66
⇒
x(t) = x0 e(k1 −k2 )t .
5.1. POPULATION AND LOGISTIC MODELS
67
This result represents three possible variants:
1. die-out of the population, if k1 < k2 (asymptotically stabile solution),
2. over-reproduction, if k1 > k2 (un-stabile solution),
3. steady state, if k1 = k2 (stabile solution).
5.1.2 Reproduction with migration - open linear system. Solve the previous problem
with non-zero migration term
f (t) = xm sin ωt,
xm << x0 .
Find the solution by following conditions: time unit 1 day, x0 = 6000, xm = 5, ω = 0, 1 1/day.
Discuss the result regarding three above variants. Calculate and demonstrate graphically the
solution for the first three years by k1 = 1, 18.10−3 [1/day], k2 = 1, 15.10−3 [1/day]. Estimate
the number of inhabitants at the end of given time interval.
5.1.3 The model of prey–predator system. A closed natural system contains (in a reduced sense) the population of predators and prey population that serves as a food. Constitute
linear model of (continuous) time progression of the both populations with regard to following
activities:
• reproduction abibility of any population,
• prey decrease due to predator activity,
• increase of predator population as a consequence of food sufficiency.
Consider velocity of individuals number change as modeled quantity. Thus, resulting time
courses result as integrals of these one with respect to time variable. Solve the problem with
constant coefficients given as follows:
a = 0.05, d = −0.05 for reproduction ability of prey and predator population, respectively,
b = −0, 02 for prey decrease due to predator activity,
c = 0, 17 for increase of predator population.
Initial population amounts are given as 600 (prey) and 10 (predators); initial velocities of state
changes related to one day are prescribed as 0.2 by prey and 0.02 by predator population.
Solve the problem in the one-year time section. Illustrate graphically the course of population
progress and characterize the system from stability point of view.
5.1.4 Advertising effectivity. A product has N potential consumers, but only N0 of them
is informed about it. At this time, advertising campaigne is started thereby the propagation
velocity of trade references (i.e. the number of informed consumers in a time unit) is direct
proportional to geometric mean of actually informed and non-informed consumers. Formulate
mathematical model to estimate effectivity of advertising (take one day as the time unit), if N7
of consumers is acquainted with the product at the first week end. Solve by the following input
data: N = 5000, N0 = 50, N7 = 737.
Let n(t) denotes number of informed consumers at the time t. The basic balance relation is
obvious:
q
∆n(t)
= k n(N − n) ,
∆t
where k is a constant that we determine later. As ∆n << N for small increment ∆t, we can
introduce initial problem for differential equation in the form
q
dn(t)
= k n(N − n) ,
dt
n(0) = N0 .
(5.1)
68
CHAPTER 5. SELECTED PROBLEMS
This separable equation has generall solution
arcsin
2n − N
= kt + C ,
N
where the constant C folows from initial condition by t = 0,
C = arcsin
2N0 − N
.
N
Thus, the function n(t) results as
n(t) =
N
[1 + sin(kt + C)] .
2
(5.2)
Since we know that n = N7 for t = 7, the constant k can be obtained putting this data into
above solution:
N
N7 =
[1 + sin(7k + C)] ,
2
from whence
1
2N7 − N
2N0 − N
k=
arcsin
.
− arcsin
7
N
N
Finally, we set n = N in (5.2). After some rearranging we obtain
1
tp =
k
π
−C
2
,
(5.3)
therefore, the needed period five weeks results for input data in statement – see also Fig. 5.1.
Figure 5.1: To the problem 5.1.4.
5.1.5
Advertising effectivity 2. Modify previous model by following assumptions:
1. propagation velocity of trade references is direct proportional to the number of informed as
well as non-informed clients (i.e. to the product of items);
2. The percent amount of informed consuments need be estimated at the end of tenth day, if
N = 3000, N0 = 60 a k = 0.0002.
5.2. MOTION PROBLEMS
69
5.1.6 Combat against epidemic. An infection mushrooms in a population of 106 persons.
Eight percent until now healthful peoples becomes ill every day. Once three percent of residents
sicken, vaccination action is launched that dailly safes 600 healthful persons. Determine, when
epidemic threat vanish (i.e. number of threatened persons will equal to zero).
5.2
Motion problems
5.2.1 Movement in gravitation field. Mass particle is throwed up crossways with velocity
v0 under the angle α. It moves in gravitation field without aerodynamic drag. State and solve
the problem to determine the particle trajectory, maximum output level and fall distance.
Generalize problem to describe movement with medium resistance. Consider the relation
F r = −r.v for the drag force, where r is resistance coefficient.
5.2.2 Movement of charged particle in magnetic field. A particle of the mass m with
electric charge Q is emitted into homogeneous magnetic field under the angle α to flux lines.
Constitute mathematical model to calculate the particle trajectory. Discuss obtained solution
with regard to the angle α.
5.2.3 Water outflow a vase. Consider a vase of the form arrising by rotation of the curve
y = S(x) around the x axis. The water flows freely away through a small outlet close to the
bottom with the velocity
p
v = k 2gh
(value of the constant k is 0.6 for the water).
1. Determine the period of total vase emptying.
2. Derive the vase form, for which is the level decrease uniform (a ”water clock” principle).