Lattice structures
for Laplace transforms
No.2
Takao Saito
1
Thank you for our world
2
Lattice struactures
for Lapalace transforms
3
Preface
This paper has been presented about lattice structure for Laplace
transforms. The lattice form is able to represent as matrix conditions. In
general, lattice structure for Laplace transforms includes Pascal’s triangle
matrix. In this time, I want to explain about the two-dimensional matrix
forms. This matrices forms are satisfied the Cauchy problems. As a
whole, lattice form and Pascal’s matrix are isomorphic. Finally, I have
been extended to two-parameters for Laplace transforms.
Papers
About solvers of differential equations of relatively
for Laplace transforms
Operator algebras for Laplace transforms
Rings and ideal structures for Laplace transforms
Extension and contaction for transrated operators
Operator algebras for group conditions
Some matrices rings for Laplace transforms
Pascal’s triangle matrix for Laplace transforms
Now, let′s consider with me!
Address
695-52 Chibadera-cho
Chuo-ku Chiba-shi
Postcode 260-0844 Japan
URL: http://opab.web.fc2.com/index.html
(Sun) 24.Jul.2011 Takao Saito
4
Contents
Preface
§ Chapter 4
◦ About relations of T (a) and T (0) · · · · · · 7
◦ The matrix condition for lattice structures · · · · · · 15
◦ Some results · · · · · · 17
§ Chapter 5
◦ The solution of lattice operator in two dimensions · · · · · · 18
◦ Lattice and matrix operators for 2-dimensions (first form)· · · 21
◦ Lattice and matrix operators for 2-dimensions (irreducible) 23
◦ Lattice and matrix operators for 2-dimensions (second form) 25
◦ Lattice and matrix operators for 2-dimensions (irreducible) 27
◦ F(a) to L′(a) cobditions · · · · · · 29
◦ Duality of L′(a) operation · · · · · · 30
◦ Some results · · · · · · 32
§ Chapter 6
◦ Formulae for T (a, b) operation (particular case) · · · · · · 33
◦ Formulae for F (a, b) operation · · · · · · 35
◦ Formulae for L(a, b) operation · · · · · · 37
5
F (a, b) (Pascal′s triangle matrix)
◦ Figure of basic operator algebras (Pascal’s triangle) · · · 40
◦ Figure of basic operator algebras (Duality) · · · · · · 40
◦ Basic operator algebras (Pascal’s triangle) · · · · · · 41
◦ Basic operator algebras (Duality) · · · · · · 41
◦ Basic normed operator algebras (Pascal’s triangle) · · · · · · 42
◦ Basic normed operator algebras (Duality) · · · · · · 42
T (a, b) (Laplace transforms)
◦ Figure of basic operator algebras · · · · · · 44
◦ Figure of basic operator algebras (Duality) · · · · · · 44
◦ Basic operator algebras (Laplace transforms) · · · · · · 45
◦ Basic operator algebras (Duality) · · · · · · 45
◦ Basic normed operator algebras (Laplace transforms) · · · 46
◦ Basic normed operator algebras (Duality) · · · · · · 46
◦ T (a, b) and L(a, b) operations · · · · · · 47
◦ Some results · · · · · · 49
◦ Conclusion · · · · · · 50
◦ Reference · · · · · · 51
6
§ Chapter 4
In this chapter, I want to explane the lattice strucrture of T (a) op-
erations. As the first step, T (a) operation could extend to seriese and
satisfied following.
N.B. In this case, T (a) is preserved from T (0).
©About relations of T (a) and T (0).
Now, based system of T (a) operation is represented following.
T (a)f(t) = T (a)∞∑
n=0
f(0)(n)
n!tn =
∞∑
n=0
f(0)(n)
n!T (a)tn
=∞∑
n=0
f(0)(n)
n!{
n−1∑
k=0
(nCk)an−k k!
sk+1+
a0 · n!
sn+1}
N.B. T (a)f(t)def .=
∫ ∞
af(t)e−(t−a)sdt
=∞∑
n=1
n−1∑
k=0
f(0)(n)
n!
n!
(n− k)!k!an−k k!
sk+1+
∞∑
n=1
f(0)(n) · a0
sn+1+
f(0) · a0
s
=∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+
∞∑
n=0
f(0)(n) · a0
sn+1=
∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+
∞∑
n=0
f(0)(n)
n!
a0n!
sn+1
On the other hand, we are able to have following formula.
a0 = e0·s for all a.
So we have
=∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+
∞∑
n=0
f(0)(n)
n!e0·sT (0)tn
7
=∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ R(0)f(t) = −S1(a)f(t) + R(0)f(t)
N.B. We shall attend that the term of a0 is included in R(0) operation.
Becaurse an−k of first term is changed to a0 as n = k. Furthermore this
a0 is changed to 00.
In this time,
R(0) = T (0).
N.B. R(a)f(t)def .=
∫ ∞
0f(t)e−(t−a)sdt = easT (0)f(t)
Therefore
T (a)f(t) = −S1(a)f(t) + R(0)f(t) = −S1(a)f(t) + T (0)f(t).
Hence
T (a)f(t) =∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ T (0)f(t)
↓ ↓ ↓ring ideal ring
(see, p.43, Chapter 2, No.7, N o.2)
N.B. R(0) = T (0)
Of caurse it’s a compact operators.
8
T (a)∫ tν
0· · ·
∫ t2
0
∫ t1
0
f(t)
tdtdt1 · · · dtν−1 =
∞∑
n=1
n+ν−2∑
k=0
f(0)(n)
n(n + ν − 1− k)!
an+ν−1−k
sk+1+
1
sνT (0)
f(t)
t
⇑
T (a)∫ t2
0
∫ t1
0
f(t)
tdtdt1 =
∞∑
n=1
n∑
k=0
f(0)(n)
n(n + 1− k)!
an+1−k
sk+1+
1
s2T (0)
f(t)
t
⇑
T (a)∫ t1
0
f(t)
tdt =
∞∑
n=1
n−1∑
k=0
f(0)(n)
n(n− k)!
an−k
sk+1+
1
sT (0)
f(t)
t
⇑
T (a)f(t)
t=
∞∑
n=1
n−1∑
k=0
f(0)(n+1)
(n + 1)(n− k)!
an−k
sk+1+ T (0)
f(t)
t
⇓
T (a)[f(t)
t]′ =
∞∑
n=1
n−1∑
k=0
f(0)(n+2)
(n + 2)(n− k)!
an−k
sk+1+ sT (0)
f(t)
t− f ′(0)
⇓
T (a)[f(t)
t]′′ =
∞∑
n=1
n−1∑
k=0
f(0)(n+3)
(n + 3)(n− k)!
an−k
sk+1+ s2T (0)
f(t)
t− sf ′(0)− f(0)′′
2
⇓
T (a)[f(t)
t](ν) =
∞∑
n=1
n−1∑
k=0
f(0)(n+ν+1)
(n + ν + 1)(n− k)!
an−k
sk+1+sνT (0)
f(t)
t−
ν∑
k=1
sν−k f(0)(k)
k
N.B. Now T (a) operation is preserved with T (0).
9
Similarly
T (a)∫ tν
0· · ·
∫ t2
0
∫ t1
0f(t)dtdt1 · · · dtν−1 =
∞∑
n=1
n+ν−2∑
k=0
f(0)(n−1)
n(n + ν − 1− k)!
an+ν−1−k
sk+1+
1
sνT (0)f(t)
⇑
T (a)∫ t2
0
∫ t1
0f(t)dtdt1 =
∞∑
n=1
n∑
k=0
f(0)(n−1)
(n + 1− k)!
an+1−k
sk+1+
1
s2T (0)f(t)
⇑
T (a)∫ t1
0f(t)dt =
∞∑
n=1
n−1∑
k=0
f(0)(n−1)
(n− k)!
an−k
sk+1+
1
sT (0)f(t)
⇑
based system · · · T (a)f(t) =∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ T (0)f(t)
⇓
T (a)f ′(t) =∞∑
n=1
n−1∑
k=0
f(0)(n+1)
(n− k)!
an−k
sk+1+ T (0)f ′(t)
⇓
T (a)f ′′(t) =∞∑
n=1
n−1∑
k=0
f(0)(n+2)
(n− k)!
an−k
sk+1+ T (0)f ′′(t)
⇓
T (a)f (ν)(t) =∞∑
n=1
n−1∑
k=0
f(0)(n+ν)
(n− k)!
an−k
sk+1+ T (0)f (ν)(t)
N.B. Now T (a) operation is preserved with T (0).
10
And
T (a)∫ tν
0· · ·
∫ t2
0
∫ t1
0tf(t)dtdt1 · · · dtν−1 =
∞∑
n=0
n+ν−1∑
k=0
f(0)(n−1)
(n + ν − k)!
an+ν−k
sk+1+
1
sνT (0)tf(t)
⇑
T (a)∫ t2
0
∫ t1
0tf(t)dtdt1 =
∞∑
n=0
n+1∑
k=0
nf(0)(n−1)
(n + 2− k)!
an+2−k
sk+1+
1
s2T (0)tf(t)
⇑
T (a)∫ t1
0tf(t)dt =
∞∑
n=0
n∑
k=0
nf(0)(n−1)
(n + 1− k)!
an+1−k
sk+1+
1
sT (0)tf(t)
⇑
T (a)tf(t) =∞∑
n=1
n−1∑
k=0
nf(0)(n−1)
(n− k)!
an−k
sk+1+ T (0)tf(t)
⇓
T (a)[tf(t)]′ =∞∑
n=1
n−1∑
k=0
(n + 1)f(0)(n)
(n− k)!
an−k
sk+1+ sT (0)tf(t)
⇓
T (a)[tf(t)]′′ =∞∑
n=1
n−1∑
k=0
(n + 2)f(0)(n+1)
(n− k)!
an−k
sk+1+ s2T (0)tf(t)− f(0)
⇓
T (a)[tf(t)](ν) =∞∑
n=1
n−1∑
k=0
(n + ν)f(0)(n+ν−1)
(n− k)!
an−k
sk+1+sνT (0)tf(t)−
ν∑
k=1
sν−k(k−1)f(0)(k−2)
N.B. Now T (a) operation is preserved with T (0).
11
Similarly the −∗ algebras are represented following.
T ∗(a)∫ tν
0· · ·
∫ t2
0
∫ t1
0
f(t)
tdtdt1 · · · dtν−1 =
∞∑
k=1
k+ν−2∑
n=0
f(0)(k)
k(k + ν − 1− n)!
ak+ν−1−n
sn+1+
1
sνT ∗(0)
f(t)
t
⇑
T ∗(a)∫ t2
0
∫ t1
0
f(t)
tdtdt1 =
∞∑
k=1
k∑
n=0
f(0)(k)
n(k + 1− n)!
ak+1−n
sn+1+
1
s2T ∗(0)
f(t)
t
⇑
T ∗(a)∫ t1
0
f(t)
tdt =
∞∑
k=1
k−1∑
n=0
f(0)(k)
k(k − n)!
ak−n
sn+1+
1
sT ∗(0)
f(t)
t
⇑
T ∗(a)f(t)
t=
∞∑
k=1
k−1∑
n=0
f(0)(k+1)
(k + 1)(k − n)!
ak−n
sn+1+ T ∗(0)
f(t)
t
⇓
T ∗(a)[f(t)
t]′ =
∞∑
k=1
k−1∑
n=0
f(0)(k+2)
(k + 2)(k − n)!
ak−n
sn+1+ sT ∗(0)
f(t)
t− f ′(0)
⇓
T ∗(a)[f(t)
t]′′ =
∞∑
k=1
k−1∑
n=0
f(0)(k+3)
(k + 3)(k − n)!
ak−n
sn+1+s2T ∗(0)
f(t)
t−sf ′(0)− f(0)′′
2
⇓
T ∗(a)[f(t)
t](ν) =
∞∑
k=1
k−1∑
n=0
f(0)(k+ν+1)
(k + ν + 1)(k − n)!
ak−n
sn+1+sνT ∗(0)
f(t)
t−
ν∑
n=1
sν−n f(0)(n)
n
N.B. Now T ∗(a) operation is also preserved with T ∗(0).
12
Similarly
T ∗(a)∫ tν
0· · ·
∫ t2
0
∫ t1
0f(t)dtdt1 · · · dtν−1 =
∞∑
k=1
k+ν−2∑
n=0
f(0)(k−1)
k(k + ν − 1− n)!
ak+ν−1−n
sn+1+
1
sνT ∗(0)f(t)
⇑
T ∗(a)∫ t2
0
∫ t1
0f(t)dtdt1 =
∞∑
k=1
k∑
n=0
f(0)(k−1)
(k + 1− n)!
ak+1−n
sn+1+
1
s2T ∗(0)f(t)
⇑
T ∗(a)∫ t1
0f(t)dt =
∞∑
k=1
k−1∑
n=0
f(0)(k−1)
(k − n)!
ak−n
sn+1+
1
sT ∗(0)f(t)
⇑
based system · · · T ∗(a)f(t) =∞∑
k=1
k−1∑
n=0
f(0)(k)
(k − n)!
ak−n
sn+1+ T ∗(0)f(t)
⇓
T ∗(a)f ′(t) =∞∑
k=1
k−1∑
n=0
f(0)(k+1)
(k − n)!
ak−n
sn+1+ T ∗(0)f ′(t)
⇓
T ∗(a)f ′′(t) =∞∑
n=1
k−1∑
n=0
f(0)(k+2)
(k − n)!
ak−n
sn+1+ T ∗(0)f ′′(t)
⇓
T ∗(a)f (ν)(t) =∞∑
k=1
k−1∑
n=0
f(0)(k+ν)
(k − n)!
ak−n
sn+1+ T ∗(0)f (ν)(t)
N.B. Now T ∗(a) operation is also preserved with T ∗(0).
13
And
T ∗(a)∫ tν
0· · ·
∫ t2
0
∫ t1
0tf(t)dtdt1 · · · dtν−1 =
∞∑
k=0
k+ν−1∑
n=0
f(0)(k−1)
(k + ν − n)!
ak+ν−n
sn+1+
1
sνT ∗(0)tf(t)
⇑
T ∗(a)∫ t2
0
∫ t1
0tf(t)dtdt1 =
∞∑
k=0
k+1∑
n=0
kf(0)(k−1)
(k + 2− n)!
ak+2−n
sn+1+
1
s2T ∗(0)tf(t)
⇑
T ∗(a)∫ t1
0tf(t)dt =
∞∑
n=0
n∑
n=0
kf(0)(k−1)
(k + 1− n)!
ak+1−n
sn+1+
1
sT ∗(0)tf(t)
⇑
T ∗(a)tf(t) =∞∑
k=1
k−1∑
n=0
nf(0)(k−1)
(k − n)!
ak−n
sn+1+ T ∗(0)tf(t)
⇓
T ∗(a)[tf(t)]′ =∞∑
k=1
k−1∑
n=0
(k + 1)f(0)(k)
(k − n)!
ak−n
sn+1+ sT ∗(0)tf(t)
⇓
T ∗(a)[tf(t)]′′ =∞∑
k=1
k−1∑
n=0
(k + 2)f(0)(k+1)
(k − n)!
ak−n
sn+1+ s2T ∗(0)tf(t)− f(0)
⇓
T ∗(a)[tf(t)](ν) =∞∑
k=1
k−1∑
n=0
(k + ν)f(0)(k+ν−1)
(k − n)!
ak−n
sn+1+sνT ∗(0)tf(t)−
ν∑
n=1
sν−n(n−1)f(0)(n−2)
N.B. Now T ∗(a) operation is also preserved with T ∗(0).
14
©The matrix condition for lattice structures
Now, since based system for lattice structure, we have following.
T (a)f(t) =∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ T (0)f(t)
This operator algebra is following.
T (a) = −S1(a) + T (0)
As a whole, operator T (a), R(a) are satisfied ring condition and the
term of∑
is ideal. I could change to following matrix representation.
In general,
Lattice =
α1a0
β1a α2a0
γ1a2 β2a α3a
0
. . . . . . . . .
γn−2a2 βn−1a αna
0
=
α1
β1a α2
γ1a2 β2a α3
. . . . . . . . .
γn−2a2 βn−1a αn
(1)
Now, we define L(a) operation.
L(a) =
α1
β1a α2
γ1a2 β2a α3
. . . . . . . . .
γn−2a2 βn−1a αn
(2)
15
ideal ring
L(a) =
β1a
γ1a2 β2a
. . . . . .
γn−2a2 βn−1a
+
α1
α2
α3
. . .
αn
(3)
decomposed
βn =f(0)(n−1)
snγn =
f(0)(n−2)
2
1
sn−1
and
αn = f(0)(n) a0
sn+1etc
Since αn → 0 as n →∞, the matrix representation is satisfied compact
operator. Therefore the operator T (a) is compact. If a 6= 0 clearly a0 = 1.
However “Now Laplace transforms T (0)” is defined as a = 0.
N.B. αn = Asn+1
A is arbitrary constant.
Similarly, F (a) is also finite rank. So it has a property of compact-
ness.
16
Some results
◦ The polynomial form for Laplace transforms is represented as following.
T (a)f(t) =∞∑
n=0
n∑
k=0︸ ︷︷ ︸lattice form
f(0)(n)
(n− k)!
an−k
sk+1
This form is able to be represented by lattice condition.
◦ T (a) operation is able to extend centre of T (0) operation. This T (a)
opreration is preserved with T (0).
◦ The process of extension of T (0) is represented as ideal structures
(S1(a)).
◦ The operator algebras for T (a) operation is following.
T (a)f(t) =∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ T (0)f(t)
T (a) = {−S1(a)} + T (0)
◦ The lattice structure for Laplace transforms is seemed to same condition
with the Pascal’s triangle matrix operation (F (a)).
◦ Also the lattice form is able to represent as triangle matrices operations.
17
§ Chapter 5
In this chapter, I want to explain about Pascal’s matrix F (a) operation
and lattice structure for Laplace transforms. At the first time, the Cauchy
probrems for two-dimensional operations. The next is relation of Pascal’s
matrix and Lattice conditions.
©The solution of lattice operator in two dimensions.
(Cauchy problem)
Now, T (a) operations for polynomial form is following.
T (a)f(t) =∞∑
n=0
n∑
k=0
f (n)(0)
(n− k)!
an−k
sk+1
In general, the lattice form is represented following
L(a) =
f(0)s
a0
f ′(0)s
a f ′(0)s2 a0
f ′′(0)2!
1sa2 f ′′(0)
s2 a f ′′(0)s3 a0
f (3)(0)3!
1sa3 f (3)(0)
2!1s2 a
2 f (3)(0)s3 a f (3)(0)
s4 a0
......
. . .
(4)
For example, let the following
f ′(0)
s2a0 =
f ′′(0)
s3a0
So we have
sf ′(0) = f ′′(0)
In general, this diagonal has following differential equations form.
f (n+1)(t) = sf (n)(t) , s is generator. (Cauchy problems)
condition f (n)(0) = sn.
18
Solution 1
This solution is represented following matrix condition. Especially, if s is
real number then the matrix has Hermitian form.
ps est est est . . .
est ps est est . . .
est est ps est . . .
est est est ps. . .
. . . . . . . . . . . . . . .
(5)
N.B. ps is the solution of power serirs for Cauchy problems.
Now, I want to consider the ring condition for Cauchy problems.
This solution represented for ideal structures. Of caurse, the diagonal
condition has a property of Cauchy problems.
Solution 2
This solution is able to decompose following matrix condition.
ps
est ps
est est ps...
......
. . .
=
ps
0 ps
0 0 ps...
......
. . .
+
est
est est
......
. . .
(6)
ring ideal
N.B. ps is the solution of power serirs for Cauchy problems.
Now, I want to consider the ring condition for Cauchy problems. This
solution represented for ideal structures. Of caurse, the diagonal condi-
tion has a property of Cauchy problems. L(a) operations are based from
Laplace transforms. Essentially this operations are same for F (a) opera-
tion and it’s compactness. In this case, I treated two dimensional forms.
19
This differential equation is represented as Cauchy problems. In general,
it’s produced all same forms in L(a) conditions. Moreover f(0) = e0·s = 1
iff s is finite condition. (field condition) If s is infinite then we should
treat as ring conditions. And this diagonal is treated as 00 = e0·s. Now,
this condition is not defined.
In general,
a0 = eas on T (0) operation.
Solution 3
This solution is represented following matrix condition.
est 0 · · · 0
0 est 0...
... 0 est 0 0
0 0 est 0...
... 0 est 0
0 · · · 0 · · · 0 est
n− th. (7)
Now, I want to consider the ring condition for Cauchy problems. This
solution represented on diagonal. Of caurse, the diagonal condition has a
property of Cauchy problems. Ideal structures has pure null conditions.
This operation is refered to D(a) operation as t = a. Ideal is O(a)
operations. As a whole, the matrix condition has a property of Hermitian
iff s is real number.
20
©Lattice and matrix operators for 2− dimensions
(first form)
Now, T (a) operations for polynomial form is following.
T (a)f(t) =∞∑
n=0
n∑
k=0
f (n)(0)
(n− k)!
an−k
sk+1
If it’s 2-dimensional form then we have following.
L(a) =
f ′(0)s2 a0
f ′′(0)s2 a f ′′(0)
2!2!s3 a
0
(8)
Now, we are considering the only cofficient of function s. s is related
with function t. Therefore, in this case, I omitted the part of function s.
Therefore we obtain following.
L′(a) =
(f ′(0)a0
f ′′(0)a f ′′(0)2!
a0
)(9)
Now, I want to generate that diagonal is identity for F(a) matrix
condition.
Therefore let f ′(t) = f ′′(t)2
and f ′(0) = 1.
2Lf ′(t) = Lf ′′(t)
2sLf = s2Lf + C1s + C2
Lf =c1s + C2
s(s− 2)= C3
1
s+ C4
1
s− 2
Therefore
f(t) = C3 + C4e2t and f ′(t) = 2C4e
2t
Moreover since f ′(0) = 1 then f ′(0) = 2C4 = 1. So C4 = 12. And
f ′′(t) = 4C4e2t.
f ′′(0) = 4C4 = 4 · 1
2= 2
21
So we havef ′′(0)
2= 1.
Therefore L′(a) is represented following.
L′(a) =
(a0
2a a0
)ring. −
(
2a
)ideal (10)
Now, F(a) operation is represented following.
F(a) =
(a0
2a a0
)ring. −
(
2a
)ideal (11)
Therefore ideal structure is same with L′(a) operator. (2-dimensionals)
So we have L′(a) is isomorphic with F(a). The proof was completed.
Hence
L′(a)iso←→ F(a)
L′ = F
22
©Lattice and matrix operators for 2− dimensions
(first form - irreducible)
Now, T (a) operations for polynomial form is following.
T (a)f(t) =∞∑
n=0
n∑
k=0
f (n)(0)
(n− k)!
an−k
sk+1
If it’s 2-dimensional form then we have following.
L(a) =
f ′(0)s2 a0
f ′′(0)s2 a f ′′(0)
2!2!s3 a
0
(12)
Now, we are considering the cofficient of function s. s is related with
function t. Therefore, in this case, I omitted the part of function s.
Therefore we obtain following.
L′(a) =
(f ′(0)a0
f ′′(0)a f ′′(0)2!
a0
)(13)
Now, I want to generate that diagonal is identity for F(a) matrix
condition.
Therefore let f ′(t) = f ′′(t)2
and f ′(0) = eas.
2Lf ′(t) = Lf ′′(t)
2sLf = s2Lf + C1s + C2
Lf =c1s + C2
s(s− 2)= C3
1
s+ C4
1
s− 2
Therefore
f(t) = C3 + C4e2t and f ′(t) = 2C4e
2t
Moreover since f ′(0) = eas then f ′(0) = 2C4 = eas. So C4 = 12· eas.
And f ′′(t) = 4C4e2t.
f ′′(0) = 4C4 = 4 · 1
2· eas = 2 · eas
23
So we havef ′′(0)
2= eas.
Therefore L′(a) is represented following.
L′(a) = eas
(a0
2a a0
)ring. eas
(
2a
)ideal (14)
Now, F(a) operation is represented following.
F (a) = eas
(a0
2a a0
)ring. eas
(
2a
)ideal (15)
Therefore ideal structure is same with L′(a) operator. (2-dimensionals).
So we have L′(a) is isomorphic with F(a). The proof was completed.
Hence
L′(a)iso←→ F (a)
L′ = F
Of caurse, the norm is generated as eas.
24
©Lattice and matrix operators for 2− dimensions
(second form)
Now, T (a) operations for polynomial form is following.
T (a)f(t) =∞∑
n=0
n∑
k=0
f (n)(0)
(n− k)!
an−k
sk+1
If it’s 2-dimensional form then we have following.
L(a) =
( f(0)s
a0
f ′(0)s
a f ′(0)s2 a0
)(16)
Now, we are considering the cofficient of function s. s is related with
function t. Therefore, in this case, I omitted the part of function s.
Therefore we obtain following.
L′(a) =
(f(0)a0
f ′(0)a f ′(0)a0
)(17)
Now, I want to generate that diagonal is identity for F(a) matrix
condition.
Therefore let f(t) = f ′(t) and f(0) = 1.
Lf(t) = Lf ′(t)
Lf = sLf − 1
(s− 1)Lf = 1 Lf =1
s− 1Therefore
f(t) = et and f ′(t) = et
So we have f ′(0) = 1.
Therefore L′(a) is represented following.
L′(a) =
(a0
a a0
)ring.
(
a
)ideal (18)
25
Now, F(a) operation is represented following.
F(a) =
(a0
a a0
)ring.
(
a
)ideal (19)
Therefore ideal is same with L′(a) operator. (2-dimensionals) So we
have L′(a) is equal to F(a). The proof was completed.
Hence
In this case,
L′(a) = F(a).
L′ = F
26
©Lattice and matrix operators for 2− dimensions
(second form - irreducible)
Now, T (a) operations for polynomial form is following.
T (a)f(t) =∞∑
n=0
n∑
k=0
f (n)(0)
(n− k)!
an−k
sk+1
If it’s 2-dimensional form then we have following.
L(a) =
( f(0)s
ao
f ′(0)s
a f ′(0)s2 a0
)(20)
Now, we are considering the cofficient of function s. s is related with
function t. Therefore, in this case, I omitted the part of function s.
Therefore we obtain following.
L′(a) =
(f(0)a0
f ′(0)a f ′(0)a0
)(21)
Now, I want to generate that diagonal is identity for F(a) matrix
condition.
Therefore let f(t) = f ′(t) and f(0) = eas.
Lf(t) = Lf ′(t)
Lf = sLf − C1
(s− 1)Lf = eas Lf =eas
s− 1Therefore
f(t) = easet and f ′(t) = easet
So we have f ′(0) = eas.
Therefore L′(a) is represented following.
L′(a) = eas
(a0
a a0
)ring. eas
(
a
)ideal (22)
27
Now, F(a) operation is represented following.
F (a) = eas
(a0
a a0
)ring. eas
(
a
)ideal (23)
Therefore ideal is same with L′(a) operator. (2-dimensionals) So we
have L′(a) is equal to F (a). The proof was completed.
Hence
In this case,
L′(a) = F (a).
L′ = F
Of caurse, the norm is generated as eas.
Therefore, let choice the part of 2-dimensions matrix of both matrix
conditions L′(a) and F(a). In this case, L′(a) is compact operator for
finite rank. So F(a) is also compact operators. In general, L′(a) and
F(a) operators are not equivalent. It’s permitted iff 2-dimensions. Other
structure is isomorphic each other. Moreover if this dimensions are not
less than 3 dimensions then there operations are not compact for it has
not convergence. The possibility to be compact, I want to be included
in L′(a) operation. Then we will be able to extend to infinite conditions.
Of caurse, if F(a) operator has finite rank then we have compactness.
Now, we have
L′(a)iso←→ F(a)
iso←→ T (a)
Therefore
T (a) is compact operator.
28
©F(a) to L′(a) conditions
In general, Pascal’s matrix is satisfied following.
a00
a10 a11
a20 a21 a22
a30 a31 a32 a33
...... aij
.... . .
an0 an1 an2 · · · · · · ann
(24)
Now, if the coodinate of n for lattice condition takes to down then we
have following conditions.
(0, 0) −→ (k, 0)
↓ ↘ T ∗(a) (n ≤ k)
(0, n) T (a) n = k
(n ≥ k)
Therefore we are able to have following relation.
aij ←→ (j, i)
N.B. F(a)iso←→ T (a)
iso←→ L′(a)
This form is coffiecient of matrix representation for L′(a) operation.
Therefore the lattice structure for T (a) operation is able to be represented
by L′(a) operation. (see p.11, Chapter 4, No.2, N o.2) On this time F(a)
is commuted with power function. So this F(a) operation is able to
include in L′(a).
Especially, if these matrices are 2-dimensional operation and f(t) = tn
then we have following.
F(a) = L(a)
Furthermore if there operations are satisfied finite rank then we have
compactness.
29
Since this matrix form I suggest new form as L′(a) conditions. This
concept is generated from relations coordinate and matrix conditions.
In general, if f(t) = tn then we have
T (a)tn =n∑
k=0
n!
(n− k)!· an−k
sk+1=
n∑
k=0
n!
(n− k)!k!· an−k · k!
sk+1
=n∑
k=0
nCk · an−k · T (0)tk =n∑
k=0
nCk · ak · T (0)tn−k = L(a)tn.
This polynomial condition is applied to F (a) operations.
(see, p.22, Chapter 2, No.7, N o.2)
©Duality of L′(a) operation
Essentially F∗(a) and L′(a)∗ are same relations. So we could obtain
following.
k − th
−→
n− th ↓ L′(a)∗ =
a00 a01 a02 · · · a0n
a11 a12 · · · a1n
a22 · · · .... . . an−1n
ann
(25)
N.B. ann is all real number.
Similarly, we will have following conditions.
T ∗(a)iso←→ L′∗(a)
iso←→ F∗(a.)
For examlpe, it’s able to apply to the lattice form for T ∗(a) operation.
T ∗(a)f(t) =∞∑
k=1
k∑
n=1
f (k)(0)
(k − n)!
ak−n
sn
30
=∞∑
k=2
k−1∑
n=1
f (k)(0)
(k − n)!
ak−n
sn+ R∗(a)f(t)
= −S∗(a)f(t) + R∗(a)f(t)
In general, T ∗(a)f(t) has following polynomial forms.
T ∗(a)f(t) =∞∑
k=0
k∑
n=0
f (k)(0)
(k − n)!· ak−n
sn+1
in real spaces.
31
Some results
◦ The coefficient of lattice matrix form generates Cauchy problems.
f (n+1)(t) = sf (n)(t)
Especially, in this case, I defined the condition as
f (n)(0) = sn.
◦ Since this solusion has est, I could have D(a) operation.
◦ In two dimensionnal form on L′(a) operation, this matrix condition is
same with Pascal’s matrix operations (F (a)).
L′(a) = F (a) iff two− dimensions.
◦ In general, L′(a) operation is isomorphic with Pascal’s triangle matrix
operation.
◦ As a whole, these operations have following relation.
T (a) = L(a)iso←→ L′(a)
iso←→ F (a)
◦ L(a) operation is sum of all elements in this matrix. On the contrary,
F (a) operation is sum of row in this matrix.
32
§ Chapter 6
In this chapter, I want to explain about 2-variables for Laplace trans-
forms. This condition is extended from one-parameter. In general, it’s
able to represent as Freadfolm’s equation.
©Formulae for T (a, b) operation (particular case)
Now, we have
(T (a) cos ωt
T (a) sin ωt
)=
(cos aω − sin aω
sinaω cos aω
) (T (0) cos ωt
T (0) sin ωt
)(26)
This form generats following formula.
T (a) = easT (0)
(see p.47, Chapter 2, No.1, N o.1)
Now,
T (a, b) cos ωt =∫ ∞
acos ωte−(t−b)sdt = ebs
∫ ∞
acos ωte−stdt
= e(b−a)s∫ ∞
acos ωte−(t−a)sdt = e(b−a)sT (a) cos ωt
So
T (a, b) cos ωt = e(b−a)sT (a) cos ωt
Similarly
T (a, b) sin ωt = e(b−a)sT (a) sin ωt
Therefore
(T (a, b) cos ωt
T (a, b) sin ωt
)= e(b−a)s
(cos aω − sin aω
sinaω cos aω
) (T (0, 0) cos ωt
T (0, 0) sin ωt
)(27)
33
On the other hand, since T (a) = easT (0), we have
T (a, b) cos ωt = e(b−a)s · easT (0) cos ωt = ebsT (0) cos ωt
T (a, b) sin ωt = e(b−a)s · easT (0) sin ωt = ebsT (0) sin ωt
So
T (a, b) cos ωt = ebsT (0) cos ωt
(coincide)
T (a, b) sin ωt = ebsT (0) sin ωt
(28)
T (a, b)
(cos ωt
sin ωt
)= ebsT (0)
(cos ωt
sin ωt
)s = iω. (29)
Moreover, in general
T (0)f(t) =∫ ∞
0f(t)e−stdt =
∫ ∞
0f(t)e−(t−0)sdt = T (0, 0)f(t)
So
T (0)f(t) = T (0, 0)f(t)
Since this formura, then we have
T (a, b) cos ωt = ebsT (0, 0) cos ωt
(coincide)
T (a, b) sin ωt = ebsT (0, 0) sin ωt
(30)
T (a, b)
(cos ωt
sin ωt
)= ebsT (0, 0)
(cos ωt
sin ωt
)s = iω. (31)
Hence
34
T (a, b) = e(b−a)sT (a) = e(b−a)sT (a, a)
T (0, b) = ebsT (0) = ebsT (0, 0)
or
T (a, b) = ebsT (0) = ebsT (0, 0)
N.B.
(cos ωt
sin ωt
)is eigenvector. (32)
©Formula for F (a, b) operation
N.B. T (a, b)iso←→ F (a, b)
F (a, b) operation is given by
F(a)
F (a, b) = ebsF(a) = ebs
a0
a a0
a2 2a a0
a3 3a2 3a a0
......
......
. . .
an−1n−1C1a
n−2n−1C3a
n−3n−1Cka
n−1−k a0
(33)
N.B. F (a, b) = ebsF(a)
and F(a) =
a0
a a0
a2 2a a0
a3 3a2 3a a0
......
......
. . .
an−1n−1C1a
n−2n−1C3a
n−3n−1Cka
n−1−k a0
(34)
35
= ebse−as · easF(a) = e(b−a)seas
a0
a a0
a2 2a a0
a3 3a2 3a a0
......
......
. . .
an−1n−1C1a
n−2n−1C3a
n−3n−1Cka
n−1−k a0
(35)
= e(b−a)sF (a) = e(b−a)sF (a, a)
Therefore
F (a, b) = e(b−a)sF (a) = e(b−a)sF (a, a)
If we could consider the eigenvalues for F (a, b) operation then we have
λ = ebsa0
N.B. λ is eigenvalue for F (a, b) operation.
So we have
F (a, b) = ebsa0I = ebsD(a) = ebsPF(a)
= ebsF(0) = ebsF(0, 0) = ebsF (0) = ebsF (0, 0)
Hence
F (a, b) = e(b−a)sF (a) = e(b−a)sF (a, a)
F (0, b) = ebsF (0) = ebsF (0, 0)
or
F (a, b) = ebsF (0) = ebsF (0, 0)
36
if it’s eigenspace
©Formula for L(a, b) operation
Now, we consider the lattice form for T (a, b) operation
T (a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt = e(b−a)s
∫ ∞
af(t)e−(t−a)sdt
= e(b−a)sT (a)f(t) = e(b−a)s∞∑
n=0
n∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1
= e(b−a)sL(a)f(t) = e(b−a)sL(a, a)f(t) = L(a, b)f(t)
= ebs∞∑
n=0
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
eas · sk+1+ ebs
∞∑
n=0
f(0)(n)
sn+1· a0
eas
If a = 0 then we have
T (0, b)f(t) = ebs∞∑
n=0
f(0)(n)
sn+1· 00
e0s= ebs
∞∑
n=0
f(0)(n)
sn+1
= ebsL(0)f(t) = L(0, b)f(t) = ebsL(0, 0)f(t)
N.B. e0s def.= 00
In general, the lattice form is represented following
L(a, b)f(t) = ebs
f(0)s
a0
f ′(0)s
a f ′(0)s2 a0
f ′′(0)2!
1sa2 f ′′(0)
s2 a f ′′(0)s3 a0
f (3)(0)3!
1sa3 f (3)(0)
2!1s2 a
2 f (3)(0)s3 a f (3)(0)
s4 a0
......
. . .
(36)
N.B. L(a, b) is finited rank.
37
This condtion is compared with F (a, b) operation.
Hence
L(a, b) = e(b−a)sL(a) = e(b−a)sL(a, a)
L(0, b) = ebsL(0) = ebsL(0, 0)
or
L(a, b) = ebsL(0) = ebsL(0, 0)
N.B.
(cos ωt
sin ωt
)is eigenvector. (37)
38
On the next time, since p.20, Chapter 2, No.7, N o.2, let extend to
F (a, b), G(a, b) and H(a, b). It’s generated n+1-th matrices. F (a, b), G(a, b)
are definded following.
Definitions
F (a, b) = ebsF(a) = ebs
a0
a a0
a2 2a a0
......
. . .
annC1a
n−1 na a0
↓ n + 1− th(38)
G(a, b) = −ebsG(a) = −ebs
a
a2 2a...
. . .
annCka
n−k na
(ideal). (39)
Similarly involution conditions are defined following.
F ∗(a, b) = ebsF∗(a) = ebs
a0 a a2 · · · an
a0 2a · · · ...
a0 · · · nCkan−k
. . ....
a0
↓ n + 1− th(40)
G∗(a, b) = −ebsG∗(a) = −ebs
a a2 · · · an
2a · · · .... . . nCka
n−k
na
(ideal). (41)
39
©Figure of basic operator algebras. (Pascal′s triangle)
F (a, b) + G(a, b) = ebs · F (0, 0) = H(a, b)
↖ extended ↗F (0, 0) is identity.
↙ decomposition ↘F(a, b) + G(a, b) = F(0, 0) = F (0, 0)
{ F(a, b) + G(a, b) = F(0, 0) = F (0, 0)
ebsF(a) = F (a, b)(42)
↘ contracted ↙F (a,−b) + G(a,−b) = e−bs · F (0, 0) = H(a,−b)
©Figure of basic operator algebras. (Duality of Pascal′s triangle)
F ∗(a, b) + G∗(a, b) = ebs · F ∗(0, 0) = H∗(a, b)
↖ extended ↗F ∗(0, 0) is identity.
↙ decomposition ↘F∗(a, b) + G∗(a, b) = F∗(0, 0) = F ∗(0, 0)
{ F∗(a, b) + G∗(a, b) = F∗(0, 0) = F ∗(0, 0)
ebsF∗(a) = F ∗(a, b)(43)
↘ contracted ↙F ∗(a,−b) + G∗(a,−b) = e−bs · F ∗(0, 0) = H∗(a,−b)
40
©Basic operator algebras. (Pascal′s triangle)
F (a, b) + G(a, b) = ebsF (0, 0) = H(a, b)
F (0, 0) = I
F(a, b) + G(a, b) = F (0, 0) = I
ebsF(a) = F (a, b)
F (a,−b) + G(a,−b) = e−bsF (0, 0) = H(a,−b)
(44)
and
F (a, b)G(a, b) = G′(a, b)
N.B. S(a, b) and G(a, b) are ideals.
©Basic operator algebras. (Dual Pascal′s triangle)
F ∗(a, b) + G∗(a, b) = ebsF ∗(0, 0) = H∗(a, b)
F ∗(0, 0) = I
F∗(a, b) + G∗(a, b) = F ∗(0, 0) = I
ebsF∗(a) = F ∗(a, b)
F ∗(a,−b) + G∗(a,−b) = e−bsF ∗(0, 0) = H∗(a,−b)
(45)
and
F ∗(a, b)G∗(a, b) = G′∗(a, b)
N.B. S∗(a, b) and G∗(a, b) are ideals.
41
©Basic normed operator algebras. (Pascal′s triangle)
‖H(a, b)‖ = ‖F (a, b) + G(a, b)‖ = |ebs|‖F (0, 0)‖ = |ebs|
‖F (0, 0)‖ = 1
‖F(a, b) + G(a, b)‖ = ‖F (0, 0)‖ = 1
|ebs|‖F(a)‖ = ‖F (a, b)‖
‖H(a,−b)‖ = ‖F (a,−b) + G(a,−b)‖ = |e−bs|‖F (0, 0)‖ = |e−bs|
‖G(a, b)‖ = 0
(46)
N.B. ‖T (a, b)‖ = ‖detF (a, b)‖ = |ebs|©Basic normed operator algebras. (Dual Pascal′s triangle)
‖H∗(a, b)‖ = ‖F ∗(a, b) + G∗(a, b)‖ = |ebs|‖F ∗(0, 0)‖ = |ebs|
‖F ∗(0, 0)‖ = 1
‖F∗(a, b) + G∗(a, b)‖ = ‖F ∗(0, 0)‖ = 1
|ebs|‖F∗(a)‖ = ‖F ∗(a, b)‖
‖H∗(a,−b)‖ = ‖F ∗(a,−b) + G∗(a,−b)‖ = |e−bs|‖F ∗(0, 0)‖ = |e−bs|
‖G∗(a, b)‖ = 0
(47)
N.B. ‖T ∗(a, b)‖ = ‖detF ∗(a, b)‖ = |ebs|
42
Now, we could be extended to dual spaces of two variables.
Background is following.
Definitions
T (a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt
S(a, b)f(t) =∫ a
0f(t)e−(t−b)sdt
T (a, b)f(t) =∫ ∞
af(t)e−stdt
S(a, b)f(t) =∫ a
0f(t)e−stdt
and
R(a, b)f(t) = T (a, b)f(t) + S(a, b)f(t) = e(b−a)s · T (0, 0)f(t)
∗-algebra is following.
T ∗(a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt
S∗(a, b)f(t) =∫ a
0f(t)e−(t−b)sdt
T ∗(a, b)f(t) =∫ ∞
af(t)e−stdt
S∗(a, b)f(t) =∫ a
0f(t)e−stdt
and
R∗(a, b)f(t) = T ∗(a, b)f(t) + S∗(a, b)f(t) = e(b−a)s · T ∗(0, 0)f(t)
43
©Figure of basic operator algebras.
T (a, b) + S(a, b) = ebs · T (0, 0) = R(a, b)
↖ extended ↗T (0, 0) is unitary.
↙ decomposition ↘T (a, b) + S(a, b) = T (0, 0) = T (0, 0)
{ T (a, b) + S(a, b) = T (0, 0) = T (0, 0)
ebsT (a) = T (a, b)(48)
↘ contracted ↙T (a,−b) + S(a,−b) = e−bs · T (0, 0) = R(a,−b)
©Figure of basic operator algebras. (Duality)
T ∗(a, b) + S∗(a, b) = ebs · T ∗(0, 0) = R∗(a, b)
↖ extended ↗T ∗(0, 0) is unitary.
↙ decomposition ↘T ∗(a, b) + S∗(a, b) = T ∗(0, 0) = T ∗(0, 0)
{ T ∗(a, b) + S∗(a, b) = T ∗(0, 0) = T ∗(0, 0)
ebsT ∗(a) = T ∗(a, b)(49)
↘ contracted ↙T ∗(a,−b) + S∗(a,−b) = e−bs · T ∗(0, 0) = R∗(a,−b)
44
I defined operator algebras and it’s following.
©Basic operator algebras.
T (a, b) + S(a, b) = ebsT (0, 0) = R(a, b)
T (0, 0) = 1
T (a, b) + S(a, b) = T (0.0) = 1
ebsT (a) = T (a, b)
T (a,−b) + S(a,−b) = e−bsT (0, 0) = R(a,−b)
(50)
and
T (a, b)S(a, b) = S ′(a, b)
©Basic operator algebras. (Duality)
T ∗(a, b) + S∗(a, b) = ebsT ∗(0, 0) = R∗(a, b)
T ∗(0, 0) = 1
T ∗(a, b) + S∗(a, b) = T ∗(0, 0) = 1
ebsT ∗(a, b) = T ∗(a, b)
T ∗(a,−b) + S∗(a,−b) = e−bsT ∗(0, 0) = R∗(a,−b)
(51)
and
T ∗(a,−b)S∗(a,−b) = S ′∗(a,−b)
45
Normed condition is following.
©Basic normed operator algebras.
‖R(a, b)‖ = ‖T (a, b) + S(a, b)‖ = |ebs|‖T (0, 0)‖ = |ebs|
‖T (0, 0)‖ = 1
‖T (a, b) + S(a, b)‖ = ‖T (0, 0)‖ = 1
|ebs|‖T (a)‖ = ‖T (a, b)‖
‖R(a,−b)‖ = ‖T (a,−b) + S(a,−b)‖ = |e−bs|‖T (0, 0)‖ = |e−bs|
‖S(a, b)‖ = 0
(52)
and
‖T (a, b)‖‖S(a, b)‖ = ‖S ′(a, b)‖ = 0
©Basic normed operator algebras.
‖R∗(a, b)‖ = ‖T ∗(a, b) + S∗(a, b)‖ = |ebs|‖T ∗(0, 0)‖ = |ebs|
‖T ∗(0, 0)‖ = 1
‖T ∗(a, b) + S∗(a, b)‖ = ‖T ∗(0, 0)‖ = 1
|ebs|‖T ∗(a)‖ = ‖T ∗(a, b)‖
‖R∗(a,−b)‖ = ‖T ∗(a,−b) + S∗(a,−b)‖ = |e−bs|‖T ∗(0, 0)‖ = |e−bs|
‖S∗(a, b)‖ = 0
(53)
and
‖T ∗(a, b)‖‖S∗(a, b)‖ = ‖S ′∗(a, b)‖ = 0
Essentially, T (a, b) = L(a, b).
46
©T (a, b) and L(a, b) operations.
Now, we defined the T (a, b)f(t).
T (a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt
On the other hand , T (a, b) operator is represented by lattice form.
This operation defines for L(a, b). Essentially the lattice form L(a, b) and
T (a, b) are equivalent. Therefore
T (a, b)f(t) = e(b−a)s∞∑
n=0
n∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1= L(a, b)f(t)
L(a, b)f(t) = ebs∞∑
n=0
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
eassk+1+ ebs
∞∑
n=0
f(0)(n)
sn+1
a0
eas
= ebs∞∑
n=0
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
eassk+1+ ebsL(a, 0)f(t)
= ebs∞∑
n=0
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
eassk+1+ X(a, b)f(t)
The is ideal terms is next condition..
−M(a, b) = ebs∞∑
n=0
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
eassk+1
As a whole, this form is represented following.
L(a, b) = −M(a, b) + X(a, b)
ring ideal ring
Especially, if a = 0 then we have
L(0, b) = X(0, b)
47
This adjunction form is represented following.
L(a, b) = ebsX(a, 0)︸ ︷︷ ︸X(a,b)
+ {−M(a, b)}
Or
X(a, b) = ebsL(a, 0)︸ ︷︷ ︸L(a,b)
+ {M(a, b)}
Now, we will need the concept of under form. The upper form should
X(a, 0) → X(0, 0). However, this condition loses the value of a. In
fact, the element a is necesarry for separating the ring conditions L(a, b)
and the ideal structures M(a, b). In this case, bare ideal have two kinds
of additive forms. ‖M(a, b)‖ = 0 because the diagonal of this matrix
conditions are zero. About under form, this form does not concreat the
part of L(a, 0). If this operation has a property of irreducibility then
we could satisfied L(0, 0). However, in general, it’s not satisfied. This
key process is that M(a, 0) is preserved null condition. In this time, we
are able to consider L(a, 0) as L(0, 0). This representation of L(a, 0) has
a property of group structures. However it is not generated subgroups.
(see A.E.Zalesski) And this matrix L(a, 0) has a property of semi-groups.
Moerover it commutes following integral forms.
(a0
a a0
)iso←→
∫ ∞
ae−(t−a)sdt (54)
Therefore we are able to replace integral condition (Laplace trans-
forms) to the lattice condition (the kind of matrix operators). This op-
erations are able to treat as same. Furthermore, we have that F (a, b)
operation is kind of this lattice forms.
In this time, this irreducible condition is represented as D(a, b) opera-
tions. In general, if b = 0 then we have unitary operators. T (a, 0),L(a, 0)
and F (a, 0) are all unitary operators. In general, ‖T (a, b)‖ = ‖L(a, b)‖=‖F (a, b)‖=|ebs|. Therefore this operations are satisfied the property of all
semi-groups.
48
Some results
◦ T (a, b) operation is defined as following.
T (a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt
(Fredholm’s equation)
◦ T (a) = T (a, a) operations to T (a, b) operation is translated by following
coefficient.
T (a, b) = e(b−a)sT (a) = e(b−a)sT (a, a)
◦ The irreducible form is represented following.
T (a, b) = ebsT (0) = ebsT (0, 0)
◦ T (a, b) operation is L(a, b) operation. Of caurse, it’s Laplace transforms
both T (a, b) and L(a, b) operations.
Two variable integral︸ ︷︷ ︸T (a,b) operation
= Polynomial condition︸ ︷︷ ︸L(a,b) operation
◦ Similarly with one-dimensional operations, we have
L(a, b) = T (a, b)iso←→ F (a, b).
◦ As a whole, we have following important forms.
Polynomial condition︸ ︷︷ ︸L(a,b) operation
iso←→ Pascal′s triangle matrix︸ ︷︷ ︸F (a,b) operation
49
Conclusion
Now, T (a) operation is able to represent the polynomial form for
Laplace transforms. This polynomial condition constitutes the lattice
structures. As a whole, I have following form.
T (a)f(t) =∞∑
n=0
n∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1
Moreover, if I attention to the ring conditions and ideal structure then
I have
T (a)f(t) =∞∑
n=1
n−1∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1+ T (0)f(t)
Ring = Ideal + Ring
In this time, T (0) operation generates 00 condition on ring boundary.
This T (0) operation calls “Now Laplace transforms”.
Since T (a) operation is isomorphic with F (a) operation, also the lat-
tice forms for Laplace transform have something relation. To prove the
relation, I treated the two dimensional matrix form as examples. These
matrices have a property of Cauchy problems. In this time, I defined
f(0)(n) = sn as condition to fix the coefficients with Pascal’s triangle
matrix (F (a)) operation. Therefore we have following relations.
L′(a) = F (a) iff two− dimensions
In general, this relation is able to extend to isomorphic conditions.
Lattice formiso←→ Pascal′s matrix
Finally, if the kernel of Laplace transform is independent for a then
we have following form.
T (a, b)f(t) =∫ ∞
af(t)e−(t−b)sdt = e(b−a)s
∞∑
n=0
n∑
k=0
f(0)(n)
(n− k)!
an−k
sk+1= L(a, b)f(t)
This formura is kind of Fredholm’s equation. Also this case, we are
able to extend to following.
T (a, b) = L(a, b)iso←→ F (a, b)
(Sun)24.Jul.2011
Now let′s go to next papers with me!
50
References
[1] Bryan P.Rynne and Martin A.Youngson, Linear functional anal-
ysis, Springer, SUMS, 2001.
[2] P.M.Cohn, Springer, SUMS, An Introduction to Ring Theory.
SUMS, 2002.
[3] M.A.Naimark, Normed Rings, P.Noordhoff,Ltd, 1959.
[4] Iain T.Adamson, Introduction to Field Theory, Cambridge Uni-
versity Press, 1982.
[5] Charles E.Rickart, General theory of Banach algebras, D.Van
Nostrand Company,Inc,1960.
[6] Shoichiro Sakai, C∗−algebras and W ∗−algebras, Springer, 1971.
[7] Hille and Philips, Functional analysis and semi-Groups, AMS,
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[8] J.L.Kelley · Isaac Namioka Linear topological spaces, D.Van
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(Mon)1.Dec.2010 (the latest).
[10] Takao Saito, Rings and ideal structures for Laplace transforms,
(reprint) (Mon)10.Dec.2007.
Furthermore
[11] Micheal O Searcoid, Elements of Abstract Analysis, Springer,
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[12] Israel Gohberg and Seymour Goldberg, Basic operator theory,
Birkhauser, 1980.
[13] Harry Hochtadt, Integral Equations, John & Sons,Inc, 1973.
[14] Emil G.Milewski, Rea’s Problem Solvers Topology, 1998.
[15] Irina V.Melnikova, Abstract Cauchy problems, Chapman, 2001.
[16] Paul L.Butzer Hubert Berens, Semi-groups of Operators and
Approximation, Springer, 1967
[17] Kosaku Yoshida, Functional Analysis, Springer, 1980.
51
[18] Richard V. Kadison John R.Ringrose Fundamentals of the the-
ory of Operator Algebras, AMS
[19] Dunford & Schwartz Linear operators I,II,III, Wiley.
52
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