SYMMETRIC POLYNOMIALS IN THE 2D FOURIER EQUATION
Grzegorz Biernat, Sylwia Lara-Dziembek, Edyta Pawlak Institute of Mathematics, Czestochowa University of Technology
Częstochowa, Poland
grzegorz.biernat@im.pcz.pl, sylwia.lara@im.pcz.pl, edyta.pawlak@im.pcz.pl
Abstract. The work is a continuation of the method of calculating the determinant of the block matrix in the two-dimensional case. In this paper we use the Finite Differences Method and the symmetric polynomials.
Keywords: block matrices, symmetric polynomials
Introduction
In this paper we return to the development of the determinant of the block matrix in the 2D case. We express this determinant by the symmetric polynomials and, consequently, by the coefficients of the output matrix (the matrix A1).
The considered problem concerns the effective formulas expressing the sym- metric polynomials of two groups of variables by the symmetric polynomials due to each of these groups.
Solution of the problem
The two-dimensional Fourier equation describing the heat flow is as follows:
( ) ( ) ( )
2 2
2 2
, , , , , ,
T x y t T x y t T x y t
c t
x y
λ ρ
∂ ∂ ∂
+ =
∂ ∂ ∂
(1)
where λ is a thermal conductivity, c is a specific heat, ρ is a mass density and T, x, y, t denote the temperature, geometrical co-ordinates and time, respectively.
Assuming the following shape of difference quotients we get the differential approximation of the second derivatives appearing in the equation (1)
( )
( )
2
1, , , , 1, ,
2 2
2
, 1, , , , 1,
2 2
2 , 1 1
2 , 1 1
i j l i j l i j l
i j l i j l i j l
T T T
T i m
x x
T T T
T j n
y y
− +
− +
− +
∆ = ≤ ≤ −
∆ ∆
− +
∆ = ≤ ≤ −
∆ ∆
(2)
and the approximation of the first derivative of the time
, , , , 1
i j l i j l , 1
T T
T l q
t t
− −
∆ = ≤ ≤
∆ ∆ (3)
Thus, the internal iterations taking the following differential form
2 2
2 2
T T T
c t
x y
λ∆ ∆ ρ ∆
+ =
∆ ∆ ∆
(4)
and the Finite Difference Method leads to the internal system of equations
( ) ( ) ( )
( ) ( ) ( )
1, , , , 1, ,
2 2 2
, 1, , , , 1,
2 2 2
, , , , 1
2
2
i j l i j l i j l
i j l i j l i j l
i j l i j l
T T T
x x x
T T T
y y y
c c
T T
t t
λ λ λ
λ λ λ
ρ ρ
− +
− +
−
− + +
∆ ∆ ∆
+ − + =
∆ ∆ ∆
= −
∆ ∆
(5)
in each time step l.
Moreover, the determinant of the matrix A2 [1]
1 1
1 1 1
2
1 1 1
1 1 n n(block dimension )
A I
I A I
A
I A I
I A
×
=
K K K K K
K K K K
M M M O M M M
K K K K
K K K K K
(6)
is given by the formula
2 4
2 1 1 1
1 2
det det ...
1 2
n n n n n
A A − A − − A −
= − + −
(7)
where
1
1
1 1
1 1
1 m m
a a A
a a
×
=
K K K K K
K K K K
M M M O M M M
K K K K
K K K K K
(8)
and
1
1 0
0 1 0
0 1 0
0 1
m m
I
×
=
K K K K K K K K K
M M M O M M M
K K K K K K K K K
(9)
We consider only the standardized case (b1 =1 in [2]) with the condition detA ≠ . 1 0 We apply the polynomial of degree n
( )
( )( )( ) ( )
2 4
1 2 3
1 2
...
1 2
n n n
n
n n
f x x x x
x p x p x p x p
− −
− −
= − + − =
= − − − ⋅K⋅ −
(10)
and then we obtain
( )( ) ( )
( )
( )
( )
( )
( )
( )
( )
1 2
1 1 1
2 1 1 1 1 2 1 1 1
1 1 1 1 2 1 1 1
1 1 1
1
1 2
2 2
1 2
det det
det det det
... 1 1 ... 1
1 1 ..
A A A n
n
n
W p W p W p
n m
m
A A p I A p I A p I
A p I A p I A p I
p p p
p p
λ λ
λ λ λ
λ λ
= − − ⋅ ⋅ − =
= − ⋅ − ⋅ ⋅ − =
= ⋅ ⋅ − − ⋅ ⋅ −
⋅ − − ⋅
K K
1442443 1442443 1442443
( )
2
1
1 2
. 1 ...
1 ... 1
1 ... 1
m
n n
n m n
mn
p
p p
S S S
λ
λ λ
⋅
⋅ −
⋅ − ⋅ ⋅ − =
= − + − + −
(11)
where S1 (the first symmetric polynomial of the variables indicated above) is equal to (τj =τj
(
λ1,λ2, ...,λm)
- fundamental symmetric polynomials 1≤ ≤j m)( )
1
1 1 2
1 1 1 2 2 2
1 1
1 2 1 2 1 2
1
1 2 1
1 2
1 1 1
, ,...,
, ,..., ; , ,..., ; ...; , ,...,
1 1 1
... ...
m
n n n
m m m
m n
m m
p p p p p p p p p
S S
p p p
ω
τ
λ λ λ
λ λ λ λ λ λ λ λ λ
ω τ
λ λ λ τ
−
= =
= + + + + + + = ⋅
144424443
144424443
(12)
while Smn (the last symmetric polynomial of the indicated above variables) is of the form
( )
1 2
1 1 1 2 2 2
1 2 1 2 1 2
1 2
1 2
1 1 1
, ,...,
, ,..., ; , ,..., ; ...; , ,...,
1 1 1 1
... ...
n
m
m
n n n
mn mn
m m m
n
m m
n n n
m m
p p p p p p p p p
S S
p p p
ω
τ
λ λ λ
λ λ λ λ λ λ λ λ λ
ω
λ λ λ τ
= =
= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅
1442443
144424443
(13)
The other symmetric polynomials (S2, …, Smn–1) can be calculated from the Newton formulas [3]
1
2 1
1 2 3 1
1 2 2 1
1 0 0 0
2 0 0
1det , 2 1
! 1
l
l l l
l l l
S l mn
l
l δ
δ δ
δ δ δ δ
δ δ δ δ δ
− − −
− −
= ≤ ≤ −
−
K K
M M M O M M
K K
(14)
determining the value of the power symmetric polynomialsδj, 1≤ ≤ . j l We have
1 1 1 2 2 2
1 2 1 2 1 2
1 1 1 2 2 2
1 2 1 2
1 2
, ,..., ; , ,..., ; ...; , ,...,
... ... ...
n n n
j j
m m m
j j j j j j
m m
j j
n n
p p p p p p p p p
p p p p p p
p p
δ δ
λ λ λ λ λ λ λ λ λ
λ λ λ λ λ λ
λ λ
= =
= + + + + + + + + +
+ + +
(
1 2)
1 2
1 1 1
... ... ...
j
j j
n j j j
n j j j
m m
p
p p p
ω
τ δ
δ
λ λ λ λ
+ = + + + + + +
144424443
144424443 (15)
where 1≤ ≤j mn− (1 δ1=S1).
The polynomials j , j
ω τ
δ δ are calculated again by using the Newton formulas
( )
( )
1
2 1
1 2 3 1
1 2 2 1
1 0 0 0
2 1 0 0
det , min ,
1 1
j
j j l
j j j
j m n
j j
ω
ω
ω ω
δ
ω ω ω ω
ω ω ω ω ω
− − −
− −
= ≤
−
K K
M M M O M M
K K
(16)
(ωk =ωk
(
p1, p2, ..., pn)
- fundamental symmetric polynomials,1 k≤ ≤ ) j and so( )
( )
1
2 1
1 2 3 1
1 2 2 1
1 0 0 0
2 1 0 0
det , min ,
1 1
j
j j l
j j j
j m n
j j
τ
τ
τ τ
δ
τ τ τ τ
τ τ τ τ τ
− − −
− −
= ≤
−
$ K
$ $ K
M M M O M M
$ $ $ K $
$ $ $ K $ $
(17)
where
1 2
1 1 1
, ,..., m k , 1
k k
m m
k j τ τ τ
λ λ λ τ
−
= = ≤ ≤
$ (18)
Formulas (16) and (17) are true also for min
(
m n,)
< ≤j max(
m n,)
and( )
max ,
j> m n . We need only the missing number of variables are assumed to be zero.
Now we return to the formula (14) to calculate the missing symmetric polynomials.
Moreover, by the equation (14) it is enough to restrict considerations to the case 1
2 l Emn−
≤
. Indeed, the following equality takes place:
1 1 1 2 2 2
1 2 1 2 1 2
1 2 1 2 1 2
1 1 1 2 2 2
, ,..., ; , ,..., ; ...; , ,...,
, ,..., ; , ,..., ; ...; , ,...,
n n n
mn j
m m m
m m m
j mn
n n n
p p p p p p p p p
S
S S
p p p p p p p p p
λ λ λ λ λ λ λ λ λ
λ λ λ λ λ λ λ λ λ
−
=
= ⋅
(19)
For example
( )
1
1
1 1 1 2 2 2
1
1 2 1 2 1 2
1 2 1 2 1 2
1
1 1 1 2 2 2
1 2
1 2
1
, ,..., ; , ,..., ; ...; , ,...,
, ,..., ; , ,..., ; ...; , ,...,
1 1 1
... ...
n n n
mn
m m m
m m m
mn
n n n
m
n
p p p p p p p p p
S
S S
p p p p p p p p p
p p p
τ
ω
λ λ λ λ λ λ λ λ λ
λ λ λ λ λ λ λ λ λ
λ λ λ
−
=
= ⋅ =
= + + + + + +
144424443
( )
1 2
1 2
1 1
, ,...,
1 1
1 1 1 2
1 2
1 1 1
, ,...,
1 1
1
1 1 1
... ...
n
n
m
m mn
p p p
n
n n m
mn n
n n m
m
n n
n m
S
S p p p
ω
τ
λ λ λ
ω ω
τ τ
ω ω λ λ λ
τ
ω ω
τ
− −
−
−
⋅ =
= ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅
144424443
144424443
144424443
(20)
(we assume that 0 ωn ≠ ).
Now, it remains to express the fundamental polynomials and
j k
τ ω by the terms of the matrix A1 (by the element a).
From the development of the characteristic polynomial of the matrix A1
( ) ( )
1 1 1
1
1 1
det det
1 1
1
A
m m
a a
W A I
a a λ
λ
λ λ
λ λ
×
−
−
= − = =
−
−
K K K K K
K K K K
M M M O M M M
K K K K
K K K K K
( ) ( ) ( )
( ) ( ) ( )
2 4
2 4
1 2
1 2 ...
1 2
( 1) ...
1 2
n n n
n n n
n
n n
a a a
n n
a a a
λ λ λ
λ λ λ
− −
− −
− −
= − − − + − − =
− −
= − − − − + − −
(21)
we obtain the polynomialsτj
2
4 6
1 2
1 2
2 4 3 6
2 4 3 6 ...
j j
j
j j
m m m
a a
j j
m m m m
a a
j j
τ −
− −
− −
= − − +
− − − −
+ − − − +
(22)
By the expression (10) we get the polynomialsωk
( )
20 if is even
1 2 if is odd
2
k k
k n k
k k
ω
−
= −
(23)
So
2
4
1 1
2 2 n
n ω
ω
−
= −
−
=
KKKKK
(24)
Remark
The procedure given above constitutes an introduction to the general procedure for calculating the determinants of the matrix block in the n-dimensional case.
This will be the subject of our subsequent paper.
References
[1] Biernat G., Lara-Dziembek S., Pawlak E., The determinants of the three-band block matrices, Scientific Research of the Institute of Mathematics and Computer Science 2012, 3(11), 5-8.
[2] Biernat G., Lara-Dziembek S., Pawlak E., The determinants of the block band matrices based on the n-dimensional Fourier equation. Part 1, Journal of Applied Mathematics and Computational Mechanics 2013, 3(12), 5-13.
[3] Mostowski A., Stark M., Elementy algebry wyższej, PWN, Warszawa 1975.
