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61
The polynomial
kx
e
b±a
=+¢-¢
¢
1
1
akakkF +++=
...)(
is called the characteristic
n
nn
-
polynomial of the differential equation.
F(k) = 0 – this equation is called the characteristic equation.
Like any algebraic equation of degree n, the characteristic equation
has
-nnn
1
0...
=+++
akak
n roots. Each root of the characteristic
1
equation ki matches to the solution of the differential equation.
Depending on the coefficient k of the characteristic equation may
have either n different real roots, or roots can be multiple roots or they
may be complex – conjugate roots, as various and multiple.
We are not going to consider each case, and will formulate a General
rule for finding solutions of linear homogeneous differential equations
with constant coefficients.
1) Make the characteristic equation and find its roots.
2) Find the particular solutions of a differential equation, with:
· each real root is associated with the solution
;
· each real root of multiplicity m corresponds to m solutions:
1 kxmkxkx
-
....;;
exxee
· each pair of complex – conjugate roots
equation is corresponded to two solutions
xba
cos
of the characteristic
i
and
xe
xba
xe
sin
· each pair of complex – conjugate roots of multiplicity m of the
characteristic equation is corresponded 2m solutions:
1
xmxx
a-aa
1
xmxx
a-aa
,cos...,cos,cos
xexxxexe
bbb
.sin...,sin,sin
xexxxexe
bbb
· Will create a linear combination of the solutions.
This linear combination will be the General solution of the original
linear homogeneous differential equations with constant coefficients.
Example. To solve the equation
IV
.0=- yy
Will develop the characteristic equation:
22
The general solution:
-
Example. To solve the equation
Will develop the characteristic equation:
The general solution:
2
xeCeCy +=
1
2
4
xx
yyy
2
2
xx
.
.01
=-k
.;;1;1;0)1)(1(
ikikkkkk -==-===+-
4321
.sincos
+++=
.044
xCxCeCeCy
4321
.2;044
===+- kkkk
21
62
Example. To solve the equation
=+¢+¢
¢
=¢+¢¢-¢¢¢
=-¢-¢
¢
yyy
.052
yyy
Will develop the characteristic equation:
2
The general solution:
Example. To solve the equation
1
x+=-
;21;16;052
ikDkk +-=-==++
).2sin2cos(
xCxCey
21
yyy
.212ik --=
.067
Will develop the characteristic equation:
223
=+-=+- kkkkkk ;6;1;0
The general solution:
Example. To solve the equation
;0)67(;067
6
xx
;
eCeCCy ++=
321
.02
=== kkk
321
Will develop the characteristic equation:
2
The General solution:
Example. To solve the equation
;2;1;02
=-==-- kkkk
21
-
221xx
.
eCeCy +=
V
- yy
¢¢¢
.09 =
Will develop the characteristic equation:
2335
The General solution:
;0)9(;09
=-=- kkkk
;3;3;0
-===== kkkkk
54321
321
3
2
4
3
xx
-
;
eCeCxCxCCy
++++=
5
Reading questions
1. Do you need to make the characteristic equation for the solution of
differential equations considered in this lecture?
2. What is the final step in the algorithm for solving the considered
type of equations?
3. What is replaced “y” by the character in the drafting of the charac-
teristic equation?
63
Section 7. Elements of field theory
rrr
r
r
r
F
r
¶
¶
¶
F
r
F
r
rrr
r
¶
¶
¶
B
r
¶
¶
¶
¶
¶
¶
rrr
r
Topic 21. Field theory
Introduction. This lecture is devoted to the discussion of the following concepts: vector potential, the gradient of the function f, rotor of vector, the divergence of a vector, Stokes formula.
Let at the space M the surface D is defined. We assume that at each
point P the positive direction of the normal is determined by the unit vec-
tor )(Pnr.
In the space of M define a vector field, putting in line every point of
space vector defined by the coordinates:
.
kzyxRjzyxQizyxPF
),,(),,(),,( ++=
Definition. The surface integral
òò
DdnF
is called the flux of a vector
D
field
through a surface D.
If at the D there exists a function f(x, y, z) has continuous partial de-
k
f
Q
;;; R
=
this func-
z
¶
.
rivatives which satisfy the properties:
f
=
x
¶
f
P
=
y
¶
tion is called a potential function or vector potential
Then the vector
is the gradient of the function f.
f
f
gradfF
f
==
x
¶
j
i
+
+
¶
y
z
¶
The formula of Stokes associates line contour integrals of the second
order with surface integrals of the second order.
æ
R
¶
RdzQdyPdx
ç
=++
ç
òòò
¶
è
SL
ö
Q
¶
÷
-
÷
z
y
¶
ø
dydz
æ
+
ç
è
R
P
¶
¶
¶
ö
-
÷
x
z
¶
ø
dzdx
æ
Q
¶
ç
+
ç
¶
è
ö
P
¶
÷
-
÷
y
x
¶
ø
this formula is called the Stokes formula.
Definition. The vector
R
B
=
y
¶
called the rotor of vector
whose components are respectively equal to
P
R
Q
-
z
¶
P
B
-
=
x
z
¶
¶
and is denoted as: Frotr.
kRjQiPF
++=
Q
B
=
zyx
¶
;;;
-
y
x
¶
dxdy
64
Definition. The vector
rrr
¶
¶
¶
rrr
r
¶
¶
¶
r
(,,)
FxyzMiNjPk
=++
dMdNdP
xyz
divF
¶¶¶
=++
(,,)
fxyz
fff
gradfijk
xyz
¶¶¶
¶¶¶
r
rr
(,,)
FxyzMiNjPk
=++
FdivF
Ñ=
r
rrr
×
×
rrrrrrr
r
++=
×
r
r
rrr
r
ì
¶
=Ñ
í
¶
î
ü
¶
¶
,,
¶
¶
¶
i
=
ý
zyx ¶
x
¶
þ
+
r
rr
j
¶
¶
y
¶
k
+
called the
z
Hamilton operator. Symbol Ñ – “набла”.
kji
rrr
¶
FFrot
=´Ñ=
¶
¶
¶
¶
zyx
¶
RQP
Definition. The expression
vector
kRjQiPF
++=
P
x
¶
and is denoted as:
+
+
¶
is called the divergence of a
z
y
¶
R
Q
P
Fdiv
+
=
y
x
¶
¶
+
z
¶
R
Q
Those look like heavy formulas. They are too much to memorize, unless you use them often. The important point is to connect vector calculus with "scalar calculus," which is not heavy.
May I go back to basic facts about the divergence? First the definition:
has
=ÑF
The divergence is a scalar (not a vector). At each point div F is a
number. In fluid flow, it is the rate at which mass leaves the "flux per
unit volume" or "flux density." The symbol V stands for a vector whose
components are operations not numbers:
r
ì
¶
=Ñ
í
¶
î
ü
¶
¶
,,
¶
¶
¶
i
=
ý
zyx ¶
x
¶
þ
+
r
rr
j
¶
¶
¶
k
+
z
y
This vector is illegal but very useful. First, apply it to an ordinary
function
Second, take the dot product VF with a vector function
= “del dot F”
Example. Find
Find the scalar product: ;zyxar
Find the scalar product:
:
=++
= Vf.
, if
rarrot
)(
.; kjiakzjyixr
++=++=
222
},,{},,{)(
zyzxzyxyyxxzxyxRQPrar ++++++==××
65
kji
rrr
rrr
æ
)(
¶
¶
¶
RQP
¶
¶
¶
rrr
=××
rarrot
¶
R
ç
=
zyx
-
i
ç
¶
y
è
rr
ö
¶
Q
¶
z
)()()( xykxzjyzi -+---=
¶
R
æ
÷
-
÷
ø
-
j
ç
¶
x
è
r
æ
¶
¶
P
¶
z
¶
P
Q
ö
ç
+
÷
ø
-
k
ç
è
¶
¶
y
x
Reading questions
1. Please give a few names of terms which can describe the vector
field.
2. What mathematical concepts are connected by a Stokes formula?
3. What can you say about the partial derivatives of the potential
function?
ö
÷
=
÷
ø
66
Section 8. Series
1...
24
+++=
1
1
x
-
1
2
110
==¥
1()...
22
+++
2
2
=
...
++
n
aaa
+++
Topic 22. Number series
Introduction. Infinite series can be a pleasure (sometimes). They
throw a beautiful light on sinx and cosx. They give famous numbers like
π and e. Usually they produce total unknown functions which might be
good. But on the painful side is the fact than infinite series has infinitely
many terms. It is not easy to know the sum of those terms. More than
that, it is not certain that there is a sum. We need tests, to decide if the
series converges. We also need ideas, to discover what the series converges to.
Here are examples of convergence, divergence, and oscillation:
11
The first series converges. The second series (the sum of 1's) obviously diverges to infinity. All those and more are special cases of one infinite series which is absolutely the most important of all:
The geometric series is
1...
++++=
This is a series of functions. It is a "power series." When we substitute numbers for x, the series on the left may converge to the sum on the
2, 1+1+1+…= ∞
23
xxx
.
right. We need to know when it doesn't. Choose x = 1 and x =
1+1+1+… is divergent. Its sum is
11
The partial sum Sn , of the series
Sn = sum of the fist n terms=
Thus Sn is part of the total sum. When we speak about convergence or
divergence of a series, we are really speaking about convergence or divergence of its "partial sums."
2
is convergent series. Its sum is
aa
12
12
11
-
...
,
1
1
1
.
-
stops at an,:
:
67
Definition. The sum of the members of the infinite numerical se-
111
248
+++
1...
xx
+++
0nx
®
1
³
nn
ab
££
n
b
å
n
a
å
n
212
2
quence called numerical series.
¥
å
n
uuuu
nn
=
1
The numbers
=++++
21
will be called members of a number series and
,...,21uu
......
un -the term of the series.
n
=+++=
Definition. The sum
...
21
å
k
uuuuS
n = 1, 2, …
knn
=
1
are called the partial sums of the series.
The series
Definition. Numerical series
converges to 1.
¥
=++++
21
......
å
n
uuuu
=
1
nn
is called
convergent, if the sequence of its partial sums is converges
¥
.,lim
uSSS
å
nn
1
===n
If a series converges (sn → s) then its terms must approach zero
(un → 0).
Example. For the geometric series
the same as
. The test is failed if
2
., the test un → 0 is
x
, because the powers of x
don't go to zero. Automatically the series diverges. The test is passed if 1 < x< 1. But to prove convergence, we cannot rely on un → 0. It is the
partial sums that must converge.
We may never know the exact limit s. But it is still possible to decide
convergence – whether there is a sum by comparison with another series
that is known to converge.
(Comparison test) Suppose
0
and
, converges. Then
converges.
¥
1
Example. To explore the convergence of a number series
Because
1
progression), then
, and
<
nn
¥
1
å
n
2
n
=1
n
1
converges (as the decreasing geometric
å
n
also converges.
å
n
.
n
2
n
=1
68
(Ratio test) If
n
a
1
()
n
å
å
å
å
n
a
å
n
a
å
n
a
å
n
a
å
å
å
4..
357
=-+-+
...
-+-
,0
£®
...
-+-
a
1n
+
approaches a limit L < 1, the series converges.
(Root test) If the nthroot
approaches L < 1, the series converges.
a
n
They give no decision when L = 1.
This section finally allows the numbers a, to be negative
Definition. The series
ries
converges
u
n
Definition. The series
converges, and the series
If
converges then
might converge, even if
D'Alembert test. If
lim
n
is called absolutely convergent if the se-
u
n
u
is called conditionally convergent if it
n
u
diverges.
n
converges (absolutely). But
,
diverges to infinity.
u
+
n
1
r=
¥®
u
n
then if r<1 the series
,
con-
u
n
verges absolutely, if r>1 the series is divergent. If r=1 then the test does
not answer the convergence of the series.
Cauchy test. If
n
¥®
n
r=
ulim
, then if r<1 the series
n
u
converges
n
absolutely, if r>1 the series is divergent. If r=1 then the test does not
answer the convergence of the series.
Alternating series.
p
444
That alternating series is really remarkable. It is typical of this lecture,
because its pattern is clear.
aaa
Leibniz test. An alternating series
conditionally, may be not absolutely) if every
123
aaa
1
nnn
+
converges (at least
.
Reading questions.
1. Can we rely on un → 0 to prove convergence?
2. Can we get some decision using the Root test when L = 1?
3. Can we get some decision using the Ratio test when L = 1?
4. There is the series
aaa
123
The name of this series.?
69
Topic 23. Exponential and functional series
0
õõ
=
Introduction. This lecture is devoted to the discussion of the following concepts: the convergence of functional series partial sums of
functional series, the convergence of functional series convergence of
functional series, the power series, the idea of matching derivatives
by powers.
Definition. If members are not numbers, as a function of x, then the
series is called functional.
A study on the convergence of functional series are harder to study
numerical series. The same functional series may, at some values of the
variable x to converge, while others diverge. Therefore, the question of
convergence of functional series is reduced to the determination of those
values of the variable x, in which the series converges.
Definition. Partial sums of functional series
n
==
nxuxS
tions
å
k
kn
=
1
Definition. Functional series
point (
), if at this point converges to the sequence of its partial
,...2,1),()(
¥
xu
å
=1)(n
is called convergent at the
n
sums.
Theorem. (A Weierstrass test of convergence)
¥
The series
å
xu
n
=1)(n
converges uniformly and, moreover, absolutely
on the interval [a,b], if the modules of its members on the same interval
does not exceed the relevant members of convergent numerical series
with positive members:
so the inequality:
Mxu £)(
nn
21
MMM
Example. To study the convergence of the series
Because
, it is obvious that
1cos £nx
¥
xu
å
=1)(n
......
++++
n
£
is called func-
n
¥
cos
nx
å
3
n
=1
n
1cosnnnx
.
33
.
70
¥
3
2
n
41312
312
n
It is known that the harmonic series å
converges if a=3>1 , in
n
=a11n
accordance with the Weierstrass test these analyzed series converges uniformly and then in any interval.
n
¥
x
Example. To study the convergence of the series
n
x
On the interval [-1,1] is the inequality
å
1
£
that is, according to
33
nn
.
n
=13n
the Weierstrass test in this interval analyzed series converges, and at intervals (-µ, -1) È (1, µ) diverges.
Definition. The power series called the series of type
2
210
n
n
¥
=+++++
......
å
n
n
xaxaxaxaa
n
=
0
.
To investigate the convergence of the power series it is convenient to
use the test of D'alembert.
To study the convergence of the series
32
x
n
xxx
......
+++++
Use the test of D'alembert:
n
+
1
x
u
n
+
1
n
¥®
u
n
n
+
=
limlim
n
1
n
¥®
x
xn
=
lim
n
n
We find that this series converges when
=
+
1
x
x
lim
nn
¥®¥®
=
1
+
1
n
and diverges if
1<x
.
1>x
We now study the convergence of the points on the boundary of the
interval 1 and –1.
When х = -1:
1
1 -+-+-
...
the series converges according to
.
the Leibniz test.
When х = 1:
1 +++++
1
...
1
...
the series diverges (harmonic se-
ries).
The radius of convergence can be found by the formula:
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