APPROXIMATION BY SPLINE FUNCTIONS OF THE SOLUTION OF A BILOCAL LINEAR PROBLEM
COSTICĂ MUSTĂȚA
In the last years the theory of spline functions has become an important tool in the numerical solving of some problems for differential equations (see, for instance, [1], [2], [4]).
In this paper we shall define a space of spline functions of degree 5 which can be used to approximate the solution of a bilocal linear problem.
Let
n
≥
3
n
≥
3
n >= 3 n \geq 3 n ≥ 3 be a natural number and let
Δ
n
:
−
∞
=
t
−
1
<
a
=
t
0
<
t
1
<
…
<
t
n
=
b
<
t
n
+
1
=
+
∞
Δ
n
:
−
∞
=
t
−
1
<
a
=
t
0
<
t
1
<
…
<
t
n
=
b
<
t
n
+
1
=
+
∞
Delta_(n):-oo=t_(-1) < a=t_(0) < t_(1) < dots < t_(n)=b < t_(n+1)=+oo \Delta_{n}:-\infty=t_{-1}<a=t_{0}<t_{1}<\ldots<t_{n}=b<t_{n+1}=+\infty Δ n : − ∞ = t − 1 < a = t 0 < t 1 < … < t n = b < t n + 1 = + ∞ be a division of the real axis.
Denote by
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) the set of all functions s:
R
→
R
R
→
R
RrarrR \mathbf{R} \rightarrow \mathbf{R} R → R having the following properties:
1
0
s
∈
C
4
(
R
)
1
0
s
∈
C
4
(
R
)
1^(0)s inC^(4)(R) 1^{0} s \in C^{4}(\mathrm{R}) 1 0 s ∈ C 4 ( R ) .
2
0
s
|
I
k
∈
P
5
,
I
k
=
[
t
k
−
1
,
t
k
)
,
k
=
1
,
2
,
…
,
n
2
0
s
I
k
∈
P
5
,
I
k
=
t
k
−
1
,
t
k
,
k
=
1
,
2
,
…
,
n
2^(0)s|_(I_(k))inP_(5),I_(k)=[t_(k-1),t_(k)),k=1,2,dots,n \left.2^{0} s\right|_{I_{k}} \in \mathscr{P}_{5}, I_{k}=\left[t_{k-1}, t_{k}\right), k=1,2, \ldots, n 2 0 s | I k ∈ P 5 , I k = [ t k − 1 , t k ) , k = 1 , 2 , … , n .
3
0
s
|
I
0
∈
P
3
,
s
|
I
n
+
1
∈
P
3
,
I
0
=
(
t
−
1
,
t
0
)
,
I
n
+
1
=
[
t
n
,
t
n
+
1
)
3
0
s
I
0
∈
P
3
,
s
I
n
+
1
∈
P
3
,
I
0
=
t
−
1
,
t
0
,
I
n
+
1
=
t
n
,
t
n
+
1
3^(0)s|_(I_(0))inP_(3),s|_(I_(n+1))inP_(3),I_(0)=(t_(-1),t_(0)),I_(n+1)=[t_(n),t_(n+1)) \left.3^{0} s\right|_{I_{0}} \in \mathscr{P}_{3},\left.s\right|_{I_{n+1}} \in \mathscr{P}_{3}, I_{0}=\left(t_{-1}, t_{0}\right), I_{n+1}=\left[t_{n}, t_{n+1}\right) 3 0 s | I 0 ∈ P 3 , s | I n + 1 ∈ P 3 , I 0 = ( t − 1 , t 0 ) , I n + 1 = [ t n , t n + 1 ) .
THEOREM 1. If
s
∈
S
5
(
Δ
n
)
s
∈
S
5
Δ
n
s inS_(5)(Delta_(n)) s \in S_{5}\left(\Delta_{n}\right) s ∈ S 5 ( Δ n ) , then
(1)
s
(
t
)
=
∑
i
=
0
3
A
i
t
i
+
∑
k
=
0
n
a
k
(
t
−
t
k
)
+
5
,
t
∈
R
,
(1)
s
(
t
)
=
∑
i
=
0
3
A
i
t
i
+
∑
k
=
0
n
a
k
t
−
t
k
+
5
,
t
∈
R
,
{:(1)s(t)=sum_(i=0)^(3)A_(i)t^(i)+sum_(k=0)^(n)a_(k)(t-t_(k))_(+)^(5)","quad t inR",":} \begin{equation*}
s(t)=\sum_{i=0}^{3} A_{i} t^{i}+\sum_{k=0}^{n} a_{k}\left(t-t_{k}\right)_{+}^{5}, \quad t \in \mathrm{R}, \tag{1}
\end{equation*} (1) s ( t ) = ∑ i = 0 3 A i t i + ∑ k = 0 n a k ( t − t k ) + 5 , t ∈ R , where
(2)
∑
k
=
0
n
a
k
=
0
and
∑
k
=
0
n
a
k
t
k
=
0
(2)
∑
k
=
0
n
a
k
=
0
and
∑
k
=
0
n
a
k
t
k
=
0
{:(2)sum_(k=0)^(n)a_(k)=0quad" and "quadsum_(k=0)^(n)a_(k)t_(k)=0:} \begin{equation*}
\sum_{k=0}^{n} a_{k}=0 \quad \text { and } \quad \sum_{k=0}^{n} a_{k} t_{k}=0 \tag{2}
\end{equation*} (2) ∑ k = 0 n a k = 0 and ∑ k = 0 n a k t k = 0 Proof. Let
s
∈
S
5
(
Δ
n
)
s
∈
S
5
Δ
n
s inS_(5)(Delta_(n)) s \in S_{5}\left(\Delta_{n}\right) s ∈ S 5 ( Δ n ) . If
t
≥
b
t
≥
b
t >= b t \geq b t ≥ b , then
s
(
4
)
(
t
)
=
0
s
(
4
)
(
t
)
=
0
s^((4))(t)=0 s^{(4)}(t)=0 s ( 4 ) ( t ) = 0 so that
0
=
5
!
∑
k
=
0
n
a
k
(
t
−
t
k
)
+
=
5
!
∑
k
=
0
n
a
k
(
t
−
t
k
)
+
because
0
=
5
!
∑
k
=
0
n
a
k
t
−
t
k
+
=
5
!
∑
k
=
0
n
a
k
t
−
t
k
+
because
0=5!sum_(k=0)^(n)a_(k)(t-t_(k))_(+)=5!sum_(k=0)^(n)a_(k)(t-t_(k))_(+)"because " 0=5!\sum_{k=0}^{n} a_{k}\left(t-t_{k}\right)_{+}=5!\sum_{k=0}^{n} a_{k}\left(t-t_{k}\right)_{+} \text {because } 0 = 5 ! ∑ k = 0 n a k ( t − t k ) + = 5 ! ∑ k = 0 n a k ( t − t k ) + because
(
t
−
t
k
)
+
=
{
0
for
t
<
t
k
t
−
t
k
for
t
≥
t
k
.
t
−
t
k
+
=
0
for
t
<
t
k
t
−
t
k
for
t
≥
t
k
.
(t-t_(k))_(+)={[0," for "t < t_(k)],[t-t_(k)," for "t >= t_(k)].:} \left(t-t_{k}\right)_{+}=\left\{\begin{array}{ll}
0 & \text { for } t<t_{k} \\
t-t_{k} & \text { for } t \geq t_{k}
\end{array} .\right. ( t − t k ) + = { 0 for t < t k t − t k for t ≥ t k . Consequently
∑
k
=
0
n
a
k
=
0
∑
k
=
0
n
a
k
=
0
sum_(k=0)^(n)a_(k)=0 \sum_{k=0}^{n} a_{k}=0 ∑ k = 0 n a k = 0 and
∑
k
=
0
n
a
k
t
k
=
0
∑
k
=
0
n
a
k
t
k
=
0
sum_(k=0)^(n)a_(k)t_(k)=0 \sum_{k=0}^{n} a_{k} t_{k}=0 ∑ k = 0 n a k t k = 0 .
THEOREM 2. a) If
f
:
R
→
R
f
:
R
→
R
f:RrarrR f: \mathbf{R} \rightarrow \mathbf{R} f : R → R verifies the conditions
(3)
f
(
a
)
=
α
1
,
f
(
b
)
=
β
1
,
f
′
′
(
t
k
)
=
λ
k
,
k
=
0
,
1
,
…
,
n
,
f
(
a
)
=
α
1
,
f
(
b
)
=
β
1
,
f
′
′
t
k
=
λ
k
,
k
=
0
,
1
,
…
,
n
,
f(a)=alpha_(1),f(b)=beta_(1),f^('')(t_(k))=lambda_(k),k=0,1,dots,n, f(a)=\alpha_{1}, f(b)=\beta_{1}, f^{\prime \prime}\left(t_{k}\right)=\lambda_{k}, k=0,1, \ldots, n, f ( a ) = α 1 , f ( b ) = β 1 , f ′ ′ ( t k ) = λ k , k = 0 , 1 , … , n , then there exists a unique spline function
s
f
∈
S
5
(
Δ
n
)
s
f
∈
S
5
Δ
n
s_(f)inS_(5)(Delta_(n)) s_{f} \in S_{5}\left(\Delta_{n}\right) s f ∈ S 5 ( Δ n ) such that
(4)
s
f
(
a
)
=
α
1
,
s
f
(
b
)
=
β
1
,
s
f
′
′
(
t
k
)
=
λ
k
,
k
=
0
,
1
,
…
,
n
.
(4)
s
f
(
a
)
=
α
1
,
s
f
(
b
)
=
β
1
,
s
f
′
′
t
k
=
λ
k
,
k
=
0
,
1
,
…
,
n
.
{:(4)s_(f)(a)=alpha_(1)","s_(f)(b)=beta_(1)","s_(f)^('')(t_(k))=lambda_(k)","k=0","1","dots","n.:} \begin{equation*}
s_{f}(a)=\alpha_{1}, s_{f}(b)=\beta_{1}, s_{f}^{\prime \prime}\left(t_{k}\right)=\lambda_{k}, k=0,1, \ldots, n . \tag{4}
\end{equation*} (4) s f ( a ) = α 1 , s f ( b ) = β 1 , s f ′ ′ ( t k ) = λ k , k = 0 , 1 , … , n . b) If
h
:
R
→
R
h
:
R
→
R
h:RrarrR h: \mathbf{R} \rightarrow \mathbf{R} h : R → R verifies the conditions
(5)
h
(
a
)
=
α
2
,
h
′
(
a
)
=
β
2
,
h
′
′
(
t
k
)
=
μ
k
,
k
=
0
,
1
,
…
,
n
,
(5)
h
(
a
)
=
α
2
,
h
′
(
a
)
=
β
2
,
h
′
′
t
k
=
μ
k
,
k
=
0
,
1
,
…
,
n
,
{:(5)h(a)=alpha_(2)","h^(')(a)=beta_(2)","h^('')(t_(k))=mu_(k)","k=0","1","dots","n",":} \begin{equation*}
h(a)=\alpha_{2}, h^{\prime}(a)=\beta_{2}, h^{\prime \prime}\left(t_{k}\right)=\mu_{k}, k=0,1, \ldots, n, \tag{5}
\end{equation*} (5) h ( a ) = α 2 , h ′ ( a ) = β 2 , h ′ ′ ( t k ) = μ k , k = 0 , 1 , … , n , then there exists a unique spline function
s
h
∈
S
5
(
Δ
n
)
s
h
∈
S
5
Δ
n
s_(h)inS_(5)(Delta_(n)) s_{h} \in S_{5}\left(\Delta_{n}\right) s h ∈ S 5 ( Δ n ) such that
(6)
s
h
(
a
)
=
α
2
,
s
h
′
(
a
)
=
β
2
,
s
h
′
′
(
t
k
)
=
μ
k
,
k
=
0
,
1
,
…
,
n
.
(6)
s
h
(
a
)
=
α
2
,
s
h
′
(
a
)
=
β
2
,
s
h
′
′
t
k
=
μ
k
,
k
=
0
,
1
,
…
,
n
.
{:(6)s_(h)(a)=alpha_(2)","s_(h)^(')(a)=beta_(2)","s_(h)^('')(t_(k))=mu_(k)","k=0","1","dots","n.:} \begin{equation*}
s_{h}(a)=\alpha_{2}, s_{h}^{\prime}(a)=\beta_{2}, s_{h}^{\prime \prime}\left(t_{k}\right)=\mu_{k}, k=0,1, \ldots, n . \tag{6}
\end{equation*} (6) s h ( a ) = α 2 , s h ′ ( a ) = β 2 , s h ′ ′ ( t k ) = μ k , k = 0 , 1 , … , n . Proof. a) Using the representation (1) and taking into account conditions (4), we obtain the system
A
0
+
A
1
a
+
A
2
a
2
+
A
3
a
3
=
α
1
A
0
+
A
1
b
+
A
2
b
2
+
A
3
b
3
+
∑
k
=
0
n
−
1
a
k
(
b
−
t
k
)
5
=
β
1
(7)
2
A
2
+
6
A
3
t
j
+
20
∑
k
=
0
n
a
k
(
t
j
−
t
k
)
+
3
=
λ
j
;
j
=
0
,
n
―
∑
k
=
0
n
a
k
=
0
;
∑
k
=
0
n
a
k
t
k
=
0
A
0
+
A
1
a
+
A
2
a
2
+
A
3
a
3
=
α
1
A
0
+
A
1
b
+
A
2
b
2
+
A
3
b
3
+
∑
k
=
0
n
−
1
a
k
b
−
t
k
5
=
β
1
(7)
2
A
2
+
6
A
3
t
j
+
20
∑
k
=
0
n
a
k
t
j
−
t
k
+
3
=
λ
j
;
j
=
0
,
n
¯
∑
k
=
0
n
a
k
=
0
;
∑
k
=
0
n
a
k
t
k
=
0
{:[A_(0)+A_(1)a+A_(2)a^(2)+A_(3)a^(3)=alpha_(1)],[A_(0)+A_(1)b+A_(2)b^(2)+A_(3)b^(3)+sum_(k=0)^(n-1)a_(k)(b-t_(k))^(5)=beta_(1)],[(7)2A_(2)+6A_(3)t_(j)+20sum_(k=0)^(n)a_(k)(t_(j)-t_(k))_(+)^(3)=lambda_(j);j= bar(0,n)],[sum_(k=0)^(n)a_(k)=0;quadsum_(k=0)^(n)a_(k)t_(k)=0]:} \begin{gather*}
A_{0}+A_{1} a+A_{2} a^{2}+A_{3} a^{3}=\alpha_{1} \\
A_{0}+A_{1} b+A_{2} b^{2}+A_{3} b^{3}+\sum_{k=0}^{n-1} a_{k}\left(b-t_{k}\right)^{5}=\beta_{1} \\
2 A_{2}+6 A_{3} t_{j}+20 \sum_{k=0}^{n} a_{k}\left(t_{j}-t_{k}\right)_{+}^{3}=\lambda_{j} ; j=\overline{0, n} \tag{7}\\
\sum_{k=0}^{n} a_{k}=0 ; \quad \sum_{k=0}^{n} a_{k} t_{k}=0
\end{gather*} A 0 + A 1 a + A 2 a 2 + A 3 a 3 = α 1 A 0 + A 1 b + A 2 b 2 + A 3 b 3 + ∑ k = 0 n − 1 a k ( b − t k ) 5 = β 1 (7) 2 A 2 + 6 A 3 t j + 20 ∑ k = 0 n a k ( t j − t k ) + 3 = λ j ; j = 0 , n ― ∑ k = 0 n a k = 0 ; ∑ k = 0 n a k t k = 0 of
n
+
5
n
+
5
n+5 n+5 n + 5 equations with
n
+
5
n
+
5
n+5 n+5 n + 5 unknowns:
A
0
,
A
1
,
A
2
,
A
3
,
a
0
,
a
1
,
…
,
a
n
A
0
,
A
1
,
A
2
,
A
3
,
a
0
,
a
1
,
…
,
a
n
A_(0),A_(1),A_(2),A_(3),a_(0),a_(1),dots,a_(n) A_{0}, A_{1}, A_{2}, A_{3}, a_{0}, a_{1}, \ldots, a_{n} A 0 , A 1 , A 2 , A 3 , a 0 , a 1 , … , a n .
The system (7) has a unique solution if and only if the associated homogeneous system (obtained for
α
1
=
β
1
=
0
,
λ
k
=
0
,
k
=
0
,
1
,
…
,
n
α
1
=
β
1
=
0
,
λ
k
=
0
,
k
=
0
,
1
,
…
,
n
alpha_(1)=beta_(1)=0,lambda_(k)=0,k=0,1,dots,n \alpha_{1}=\beta_{1}=0, \lambda_{k}=0, k=0,1, \ldots, n α 1 = β 1 = 0 , λ k = 0 , k = 0 , 1 , … , n ) has only the trivial solution. Suppose that
s
∈
S
5
(
Δ
n
)
s
∈
S
5
Δ
n
s inS_(5)(Delta_(n)) s \in S_{5}\left(\Delta_{n}\right) s ∈ S 5 ( Δ n ) verifies the homogeneous conditions (4) (i.e.,
α
1
=
β
1
=
0
,
λ
k
′
′
=
0
,
k
=
0
,
1
,
…
,
n
)
α
1
=
β
1
=
0
,
λ
k
′
′
=
0
,
k
=
0
,
1
,
…
,
n
{:alpha_(1)=beta_(1)=0,lambda_(k)^('')=0,k=0,1,dots,n) \left.\alpha_{1}=\beta_{1}=0, \lambda_{k}^{\prime \prime}=0, k=0,1, \ldots, n\right) α 1 = β 1 = 0 , λ k ′ ′ = 0 , k = 0 , 1 , … , n ) . Then we have
∫
a
b
[
s
(
4
)
(
t
)
]
2
d
t
=
∫
a
b
s
(
4
)
(
t
)
⋅
(
s
′
′
′
(
t
)
)
′
d
t
=
−
∫
a
b
s
(
5
)
(
t
)
⋅
s
′
′
′
(
t
)
=
=
−
∑
k
=
1
n
∫
t
k
−
1
t
k
s
(
5
)
(
t
)
⋅
s
′
′
′
(
t
)
d
t
=
−
∑
k
=
1
n
c
k
∫
t
k
−
1
t
k
s
′
′
′
(
t
)
d
t
=
=
−
∑
k
=
1
n
c
k
[
s
′
′
(
t
k
)
−
s
′
′
(
t
k
−
1
)
]
=
0
∫
a
b
s
(
4
)
(
t
)
2
d
t
=
∫
a
b
s
(
4
)
(
t
)
⋅
s
′
′
′
(
t
)
′
d
t
=
−
∫
a
b
s
(
5
)
(
t
)
⋅
s
′
′
′
(
t
)
=
=
−
∑
k
=
1
n
∫
t
k
−
1
t
k
s
(
5
)
(
t
)
⋅
s
′
′
′
(
t
)
d
t
=
−
∑
k
=
1
n
c
k
∫
t
k
−
1
t
k
s
′
′
′
(
t
)
d
t
=
=
−
∑
k
=
1
n
c
k
s
′
′
t
k
−
s
′
′
t
k
−
1
=
0
{:[int_(a)^(b)[s^((4))(t)]^(2)dt=int_(a)^(b)s^((4))(t)*(s^(''')(t))^(')dt=-int_(a)^(b)s^((5))(t)*s^(''')(t)=],[=-sum_(k=1)^(n)int_(t_(k-1))^(t_(k))s^((5))(t)*s^(''')(t)dt=-sum_(k=1)^(n)c_(k)int_(t_(k-1))^(t_(k))s^(''')(t)dt=],[=-sum_(k=1)^(n)c_(k)[s^('')(t_(k))-s^('')(t_(k-1))]=0]:} \begin{gathered}
\int_{a}^{b}\left[s^{(4)}(t)\right]^{2} \mathrm{~d} t=\int_{a}^{b} s^{(4)}(t) \cdot\left(s^{\prime \prime \prime}(t)\right)^{\prime} \mathrm{d} t=-\int_{a}^{b} s^{(5)}(t) \cdot s^{\prime \prime \prime}(t)= \\
=-\sum_{k=1}^{n} \int_{t_{k-1}}^{t_{k}} s^{(5)}(t) \cdot s^{\prime \prime \prime}(t) \mathrm{d} t=-\sum_{k=1}^{n} c_{k} \int_{t_{k-1}}^{t_{k}} s^{\prime \prime \prime}(t) \mathrm{d} t= \\
=-\sum_{k=1}^{n} c_{k}\left[s^{\prime \prime}\left(t_{k}\right)-s^{\prime \prime}\left(t_{k-1}\right)\right]=0
\end{gathered} ∫ a b [ s ( 4 ) ( t ) ] 2 d t = ∫ a b s ( 4 ) ( t ) ⋅ ( s ′ ′ ′ ( t ) ) ′ d t = − ∫ a b s ( 5 ) ( t ) ⋅ s ′ ′ ′ ( t ) = = − ∑ k = 1 n ∫ t k − 1 t k s ( 5 ) ( t ) ⋅ s ′ ′ ′ ( t ) d t = − ∑ k = 1 n c k ∫ t k − 1 t k s ′ ′ ′ ( t ) d t = = − ∑ k = 1 n c k [ s ′ ′ ( t k ) − s ′ ′ ( t k − 1 ) ] = 0 where
c
k
=
s
(
5
)
(
t
)
|
I
k
′
k
=
1
,
n
―
c
k
=
s
(
5
)
(
t
)
I
k
′
k
=
1
,
n
¯
c_(k)=s^((5))(t)|_(I_(k)^('))k= bar(1,n) c_{k}=\left.s^{(5)}(t)\right|_{I_{k}^{\prime}} k=\overline{1, n} c k = s ( 5 ) ( t ) | I k ′ k = 1 , n ― .
It follows that
s
(
4
)
=
0
s
(
4
)
=
0
s^((4))=0 s^{(4)}=0 s ( 4 ) = 0 , for all
t
∈
[
a
,
b
]
t
∈
[
a
,
b
]
t in[a,b] t \in[a, b] t ∈ [ a , b ] . Since
s
∈
P
3
s
∈
P
3
s inP_(3) s \in \mathscr{P}_{3} s ∈ P 3 on
I
0
I
0
I_(0) I_{0} I 0 and on
I
n
+
1
I
n
+
1
I_(n+1) I_{n+1} I n + 1 and
s
∈
C
4
(
R
)
s
∈
C
4
(
R
)
s inC^(4)(R) s \in \mathbf{C}^{4}(\mathbf{R}) s ∈ C 4 ( R ) , it follows that
s
(
4
)
(
t
)
=
0
s
(
4
)
(
t
)
=
0
s^((4))(t)=0 s^{(4)}(t)=0 s ( 4 ) ( t ) = 0 for all
t
∈
R
t
∈
R
t inR t \in \mathbf{R} t ∈ R , implying
s
′
′
∈
P
1
s
′
′
∈
P
1
s^('')inP_(1) s^{\prime \prime} \in \mathscr{P}_{1} s ′ ′ ∈ P 1 . As
s
′
′
(
t
k
)
=
0
s
′
′
t
k
=
0
s^('')(t_(k))=0 s^{\prime \prime}\left(t_{k}\right)=0 s ′ ′ ( t k ) = 0 ,
k
=
0
,
1
,
…
,
n
,
(
n
≥
3
)
k
=
0
,
1
,
…
,
n
,
(
n
≥
3
)
k=0,1,dots,n,(n >= 3) k=0,1, \ldots, n,(n \geq 3) k = 0 , 1 , … , n , ( n ≥ 3 ) we conclude that
s
′
′
(
t
)
=
0
s
′
′
(
t
)
=
0
s^('')(t)=0 s^{\prime \prime}(t)=0 s ′ ′ ( t ) = 0 for all
t
∈
R
t
∈
R
t inR t \in \mathbf{R} t ∈ R .
Finally, taking into account the equalities
s
(
a
)
=
s
(
b
)
=
0
s
(
a
)
=
s
(
b
)
=
0
s(a)=s(b)=0 s(a)=s(b)=0 s ( a ) = s ( b ) = 0 , one obtains
s
(
t
)
=
0
s
(
t
)
=
0
s(t)=0 s(t)=0 s ( t ) = 0 for all
t
∈
R
t
∈
R
t inR t \in \mathbf{R} t ∈ R , implying that all the coefficients in representation (1) are null. This shows that the homogeneous system associated to (7) has only the trivial solution.
Assertion b) can be proved similarly, supposing that the function
s
s
s s s given by (1) verifies conditions (6).
COROLLARY 3. There exist the systems of functions
S
=
{
s
0
,
s
1
,
S
0
,
S
1
,
…
,
S
n
}
⊂
S
5
(
Δ
n
)
U
=
(
u
0
,
u
1
,
U
0
,
U
1
,
…
,
U
n
)
⊂
S
5
(
Λ
n
)
S
=
s
0
,
s
1
,
S
0
,
S
1
,
…
,
S
n
⊂
S
5
Δ
n
U
=
u
0
,
u
1
,
U
0
,
U
1
,
…
,
U
n
⊂
S
5
Λ
n
{:[S={s_(0),s_(1),S_(0),S_(1),dots,S_(n)}subS_(5)(Delta_(n))],[U=(u_(0),u_(1),U_(0),U_(1),dots,U_(n))subS_(5)(Lambda_(n))]:} \begin{aligned}
& \mathscr{S}=\left\{s_{0}, s_{1}, S_{0}, S_{1}, \ldots, S_{n}\right\} \subset S_{5}\left(\Delta_{n}\right) \\
& \mathscr{U}=\left(u_{0}, u_{1}, U_{0}, U_{1}, \ldots, U_{n}\right) \subset S_{5}\left(\Lambda_{n}\right)
\end{aligned} S = { s 0 , s 1 , S 0 , S 1 , … , S n } ⊂ S 5 ( Δ n ) U = ( u 0 , u 1 , U 0 , U 1 , … , U n ) ⊂ S 5 ( Λ n ) verifying the conditions
s
0
(
a
)
=
1
,
s
0
(
b
)
=
0
,
s
0
′
′
(
t
k
)
=
0
,
k
=
0
,
n
―
,
s
1
(
a
)
=
0
,
s
1
(
b
)
=
1
,
s
1
′
′
(
t
k
)
=
0
,
k
=
0
,
n
―
,
S
k
(
a
)
=
0
,
S
k
(
b
)
=
0
,
k
=
0
,
n
―
;
S
k
′
′
(
t
j
)
=
δ
k
j
,
k
,
j
=
0
,
n
―
,
u
0
(
a
)
=
1
,
u
0
′
(
a
)
=
0
,
u
0
′
′
(
t
k
)
=
0
,
k
=
0
,
n
―
,
u
1
(
a
)
=
0
,
u
1
′
(
a
)
=
1
,
u
1
′
′
(
t
k
)
=
0
,
k
=
0
,
n
―
,
U
κ
(
a
)
=
0
,
U
k
′
(
a
)
=
0
,
k
=
0
,
n
―
;
U
k
′
′
(
t
j
)
=
δ
k
j
,
k
,
j
=
0
,
n
―
.
s
0
(
a
)
=
1
,
s
0
(
b
)
=
0
,
s
0
′
′
t
k
=
0
,
k
=
0
,
n
¯
,
s
1
(
a
)
=
0
,
s
1
(
b
)
=
1
,
s
1
′
′
t
k
=
0
,
k
=
0
,
n
¯
,
S
k
(
a
)
=
0
,
S
k
(
b
)
=
0
,
k
=
0
,
n
¯
;
S
k
′
′
t
j
=
δ
k
j
,
k
,
j
=
0
,
n
¯
,
u
0
(
a
)
=
1
,
u
0
′
(
a
)
=
0
,
u
0
′
′
t
k
=
0
,
k
=
0
,
n
¯
,
u
1
(
a
)
=
0
,
u
1
′
(
a
)
=
1
,
u
1
′
′
t
k
=
0
,
k
=
0
,
n
¯
,
U
κ
(
a
)
=
0
,
U
k
′
(
a
)
=
0
,
k
=
0
,
n
¯
;
U
k
′
′
t
j
=
δ
k
j
,
k
,
j
=
0
,
n
¯
.
{:[s_(0)(a)=1","s_(0)(b)=0","s_(0)^('')(t_(k))=0","k= bar(0,n)","],[s_(1)(a)=0","s_(1)(b)=1","s_(1)^('')(t_(k))=0","k= bar(0,n)","],[S_(k)(a)=0","S_(k)(b)=0","k= bar(0,n);S_(k)^('')(t_(j))=delta_(kj)","k","j= bar(0,n)","],[u_(0)(a)=1","u_(0)^(')(a)=0","u_(0)^('')(t_(k))=0","k= bar(0,n)","],[u_(1)(a)=0","u_(1)^(')(a)=1","u_(1)^('')(t_(k))=0","k= bar(0,n)","],[U_(kappa)(a)=0","U_(k)^(')(a)=0","k= bar(0,n);U_(k)^('')(t_(j))=delta_(kj)","k","j= bar(0,n).]:} \begin{gathered}
s_{0}(a)=1, s_{0}(b)=0, s_{0}^{\prime \prime}\left(t_{k}\right)=0, k=\overline{0, n}, \\
s_{1}(a)=0, s_{1}(b)=1, s_{1}^{\prime \prime}\left(t_{k}\right)=0, k=\overline{0, n}, \\
S_{k}(a)=0, S_{k}(b)=0, k=\overline{0, n} ; S_{k}^{\prime \prime}\left(t_{j}\right)=\delta_{k j}, k, j=\overline{0, n}, \\
u_{0}(a)=1, u_{0}^{\prime}(a)=0, u_{0}^{\prime \prime}\left(t_{k}\right)=0, k=\overline{0, n}, \\
u_{1}(a)=0, u_{1}^{\prime}(a)=1, u_{1}^{\prime \prime}\left(t_{k}\right)=0, k=\overline{0, n}, \\
U_{\kappa}(a)=0, U_{k}^{\prime}(a)=0, k=\overline{0, n} ; U_{k}^{\prime \prime}\left(t_{j}\right)=\delta_{k j}, k, j=\overline{0, n} .
\end{gathered} s 0 ( a ) = 1 , s 0 ( b ) = 0 , s 0 ′ ′ ( t k ) = 0 , k = 0 , n ― , s 1 ( a ) = 0 , s 1 ( b ) = 1 , s 1 ′ ′ ( t k ) = 0 , k = 0 , n ― , S k ( a ) = 0 , S k ( b ) = 0 , k = 0 , n ― ; S k ′ ′ ( t j ) = δ k j , k , j = 0 , n ― , u 0 ( a ) = 1 , u 0 ′ ( a ) = 0 , u 0 ′ ′ ( t k ) = 0 , k = 0 , n ― , u 1 ( a ) = 0 , u 1 ′ ( a ) = 1 , u 1 ′ ′ ( t k ) = 0 , k = 0 , n ― , U κ ( a ) = 0 , U k ′ ( a ) = 0 , k = 0 , n ― ; U k ′ ′ ( t j ) = δ k j , k , j = 0 , n ― . If
f
,
h
:
R
→
R
f
,
h
:
R
→
R
f,h:RrarrR f, h: \mathbf{R} \rightarrow \mathbf{R} f , h : R → R verify the conditions of Theorem 2, then the functions
s
f
s
f
s_(f) s_{f} s f and
s
h
s
h
s_(h) s_{h} s h admit the representations
(8)
s
f
(
t
)
=
s
0
(
t
)
⋅
f
(
a
)
+
s
1
(
t
)
⋅
f
(
b
)
+
∑
k
=
0
n
S
k
(
t
)
⋅
f
′
′
(
t
k
)
,
t
∈
R
,
(9)
s
h
(
l
)
=
u
0
(
t
)
⋅
h
(
a
)
+
u
1
(
t
)
⋅
h
′
(
a
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
h
′
′
(
t
k
)
,
t
∈
R
.
(8)
s
f
(
t
)
=
s
0
(
t
)
⋅
f
(
a
)
+
s
1
(
t
)
⋅
f
(
b
)
+
∑
k
=
0
n
S
k
(
t
)
⋅
f
′
′
t
k
,
t
∈
R
,
(9)
s
h
(
l
)
=
u
0
(
t
)
⋅
h
(
a
)
+
u
1
(
t
)
⋅
h
′
(
a
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
h
′
′
t
k
,
t
∈
R
.
{:[(8)s_(f)(t)=s_(0)(t)*f(a)+s_(1)(t)*f(b)+sum_(k=0)^(n)S_(k)(t)*f^('')(t_(k))","quad t inR","],[(9)s_(h)(l)=u_(0)(t)*h(a)+u_(1)(t)*h^(')(a)+sum_(k=0)^(n)U_(k)(t)*h^('')(t_(k))","quad t inR.]:} \begin{align*}
& s_{f}(t)=s_{0}(t) \cdot f(a)+s_{1}(t) \cdot f(b)+\sum_{k=0}^{n} S_{k}(t) \cdot f^{\prime \prime}\left(t_{k}\right), \quad t \in \mathbf{R}, \tag{8}\\
& s_{h}(l)=u_{0}(t) \cdot h(a)+u_{1}(t) \cdot h^{\prime}(a)+\sum_{k=0}^{n} U_{k}(t) \cdot h^{\prime \prime}\left(t_{k}\right), \quad t \in \mathbf{R} . \tag{9}
\end{align*} (8) s f ( t ) = s 0 ( t ) ⋅ f ( a ) + s 1 ( t ) ⋅ f ( b ) + ∑ k = 0 n S k ( t ) ⋅ f ′ ′ ( t k ) , t ∈ R , (9) s h ( l ) = u 0 ( t ) ⋅ h ( a ) + u 1 ( t ) ⋅ h ′ ( a ) + ∑ k = 0 n U k ( t ) ⋅ h ′ ′ ( t k ) , t ∈ R . Remark 1. By Corollary 3 it follows that the set
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) is a (real) linear space of dimension
n
+
3
n
+
3
n+3 n+3 n + 3 and
S
S
S \mathscr{S} S and
U
U
U \mathscr{U} U are two bases in
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) .
Some properties of the space
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) will be presented in what follows.
Let
(10)
W
2
4
(
Δ
n
)
:=
{
g
:
[
a
,
b
]
→
R
,
g
′
′
′
abs.cont.on
I
k
,
k
=
1
,
n
―
and
g
(
4
)
∈
L
2
[
a
,
b
]
}
,
(10)
W
2
4
Δ
n
:=
g
:
[
a
,
b
]
→
R
,
g
′
′
′
abs.cont.on
I
k
,
k
=
1
,
n
¯
and
g
(
4
)
∈
L
2
[
a
,
b
]
,
{:(10)W_(2)^(4)(Delta_(n)):={[g:[a","b]rarrR","g^(''')" abs.cont.on "I_(k)","k= bar(1,n)],[" and "g^((4))inL_(2)[a","b]]}",":} W_{2}^{4}\left(\Delta_{n}\right):=\left\{\begin{array}{c}
g:[a, b] \rightarrow \mathbf{R}, g^{\prime \prime \prime} \text { abs.cont.on } I_{k}, k=\overline{1, n} \tag{10}\\
\text { and } g^{(4)} \in L_{2}[a, b]
\end{array}\right\}, (10) W 2 4 ( Δ n ) := { g : [ a , b ] → R , g ′ ′ ′ abs.cont.on I k , k = 1 , n ― and g ( 4 ) ∈ L 2 [ a , b ] } , (11)
W
2
,
f
4
(
Δ
n
)
:=
{
g
∈
W
2
4
(
Δ
n
)
:
g
′
′
(
t
k
)
=
f
′
′
(
t
k
)
,
k
=
0
,
n
―
}
W
2
,
f
4
Δ
n
:=
g
∈
W
2
4
Δ
n
:
g
′
′
t
k
=
f
′
′
t
k
,
k
=
0
,
n
¯
quadW_(2,f)^(4)(Delta_(n)):={g inW_(2)^(4)(Delta_(n)):g^('')(t_(k))=f^('')(t_(k)),quad k= bar(0,n)} \quad W_{2, f}^{4}\left(\Delta_{n}\right):=\left\{g \in W_{2}^{4}\left(\Delta_{n}\right): g^{\prime \prime}\left(t_{k}\right)=f^{\prime \prime}\left(t_{k}\right), \quad k=\overline{0, n}\right\} W 2 , f 4 ( Δ n ) := { g ∈ W 2 4 ( Δ n ) : g ′ ′ ( t k ) = f ′ ′ ( t k ) , k = 0 , n ― } ,
(12)
W
2
,
f
,
D
4
(
Δ
n
)
:=
{
g
∈
W
2
,
f
4
(
Δ
n
)
:
g
(
t
0
)
=
f
(
t
0
)
,
g
(
t
n
)
=
f
(
t
n
)
}
W
2
,
f
,
D
4
Δ
n
:=
g
∈
W
2
,
f
4
Δ
n
:
g
t
0
=
f
t
0
,
g
t
n
=
f
t
n
quadW_(2,f,D)^(4)(Delta_(n)):={g inW_(2,f)^(4)(Delta_(n)):g(t_(0))=f(t_(0)),g(t_(n))=f(t_(n))} \quad W_{2, f, D}^{4}\left(\Delta_{n}\right):=\left\{g \in W_{2, f}^{4}\left(\Delta_{n}\right): g\left(t_{0}\right)=f\left(t_{0}\right), g\left(t_{n}\right)=f\left(t_{n}\right)\right\} W 2 , f , D 4 ( Δ n ) := { g ∈ W 2 , f 4 ( Δ n ) : g ( t 0 ) = f ( t 0 ) , g ( t n ) = f ( t n ) } ,
(13)
W
2
,
h
,
C
4
(
Δ
n
)
:=
{
g
∈
W
2
,
h
4
(
Δ
n
)
:
g
(
t
0
)
=
h
(
t
0
)
,
g
′
(
t
0
)
=
h
′
(
t
0
)
}
W
2
,
h
,
C
4
Δ
n
:=
g
∈
W
2
,
h
4
Δ
n
:
g
t
0
=
h
t
0
,
g
′
t
0
=
h
′
t
0
W_(2,h,C)^(4)(Delta_(n)):={g inW_(2,h)^(4)(Delta_(n)):g(t_(0))=h(t_(0)),g^(')(t_(0))=h^(')(t_(0))} W_{2, h, C}^{4}\left(\Delta_{n}\right):=\left\{g \in W_{2, h}^{4}\left(\Delta_{n}\right): g\left(t_{0}\right)=h\left(t_{0}\right), g^{\prime}\left(t_{0}\right)=h^{\prime}\left(t_{0}\right)\right\} W 2 , h , C 4 ( Δ n ) := { g ∈ W 2 , h 4 ( Δ n ) : g ( t 0 ) = h ( t 0 ) , g ′ ( t 0 ) = h ′ ( t 0 ) } .
Then we have
THEOREM 4. a) If
s
∈
S
5
(
Δ
n
)
∩
W
2
,
f
,
D
4
(
Δ
n
)
s
∈
S
5
Δ
n
∩
W
2
,
f
,
D
4
Δ
n
s inS_(5)(Delta_(n))nnW_(2,f,D)^(4)(Delta_(n)) s \in S_{5}\left(\Delta_{n}\right) \cap W_{2, f, D}^{4}\left(\Delta_{n}\right) s ∈ S 5 ( Δ n ) ∩ W 2 , f , D 4 ( Δ n ) , then
(14)
‖
s
(
4
)
‖
2
≤
‖
g
(
4
)
‖
2
,
for all
g
∈
W
2
,
f
,
D
4
(
Δ
n
)
.
s
(
4
)
2
≤
g
(
4
)
2
,
for all
g
∈
W
2
,
f
,
D
4
Δ
n
.
||s^((4))||_(2) <= ||g^((4))||_(2)," for all "g inW_(2,f,D)^(4)(Delta_(n))". " \left\|s^{(4)}\right\|_{2} \leq\left\|g^{(4)}\right\|_{2}, \text { for all } g \in W_{2, f, D}^{4}\left(\Delta_{n}\right) \text {. } ‖ s ( 4 ) ‖ 2 ≤ ‖ g ( 4 ) ‖ 2 , for all g ∈ W 2 , f , D 4 ( Δ n ) . b) If
s
∈
S
5
(
Δ
n
)
∩
W
2
,
h
,
c
4
(
Δ
n
)
s
∈
S
5
Δ
n
∩
W
2
,
h
,
c
4
Δ
n
s inS_(5)(Delta_(n))nnW_(2,h,c)^(4)(Delta_(n)) s \in S_{5}\left(\Delta_{n}\right) \cap W_{2, h, c}^{4}\left(\Delta_{n}\right) s ∈ S 5 ( Δ n ) ∩ W 2 , h , c 4 ( Δ n ) , then
(15)
‖
s
(
4
)
‖
2
≤
‖
g
(
4
)
‖
2
,
for all
g
∈
W
2
,
h
,
C
4
(
Δ
n
)
s
(
4
)
2
≤
g
(
4
)
2
,
for all
g
∈
W
2
,
h
,
C
4
Δ
n
||s^((4))||_(2) <= ||g^((4))||_(2)," for all "g inW_(2,h,C)^(4)(Delta_(n)) \left\|s^{(4)}\right\|_{2} \leq\left\|g^{(4)}\right\|_{2}, \text { for all } g \in W_{2, h, C}^{4}\left(\Delta_{n}\right) ‖ s ( 4 ) ‖ 2 ≤ ‖ g ( 4 ) ‖ 2 , for all g ∈ W 2 , h , C 4 ( Δ n ) Proof. We have
But
0
≤
‖
g
(
4
)
−
s
(
4
)
‖
2
2
=
∫
a
b
[
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
]
d
t
=
=
∫
a
b
[
g
(
4
)
(
t
)
]
2
d
t
−
∫
a
b
[
s
(
4
)
(
t
)
]
2
d
t
−
2
∫
a
b
s
(
4
)
(
t
)
[
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
]
d
t
0
≤
g
(
4
)
−
s
(
4
)
2
2
=
∫
a
b
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
d
t
=
=
∫
a
b
g
(
4
)
(
t
)
2
d
t
−
∫
a
b
s
(
4
)
(
t
)
2
d
t
−
2
∫
a
b
s
(
4
)
(
t
)
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
d
t
{:[0 <= ||g^((4))-s^((4))||_(2)^(2)=int_(a)^(b)[g^((4))(t)-s^((4))(t)]dt=],[=int_(a)^(b)[g^((4))(t)]^(2)dt-int_(a)^(b)[s^((4))(t)]^(2)dt-2int_(a)^(b)s^((4))(t)[g^((4))(t)-s^((4))(t)]dt]:} \begin{gathered}
0 \leq\left\|g^{(4)}-s^{(4)}\right\|_{2}^{2}=\int_{a}^{b}\left[g^{(4)}(t)-s^{(4)}(t)\right] \mathrm{d} t= \\
=\int_{a}^{b}\left[g^{(4)}(t)\right]^{2} \mathrm{~d} t-\int_{a}^{b}\left[s^{(4)}(t)\right]^{2} \mathrm{~d} t-2 \int_{a}^{b} s^{(4)}(t)\left[g^{(4)}(t)-s^{(4)}(t)\right] \mathrm{d} t
\end{gathered} 0 ≤ ‖ g ( 4 ) − s ( 4 ) ‖ 2 2 = ∫ a b [ g ( 4 ) ( t ) − s ( 4 ) ( t ) ] d t = = ∫ a b [ g ( 4 ) ( t ) ] 2 d t − ∫ a b [ s ( 4 ) ( t ) ] 2 d t − 2 ∫ a b s ( 4 ) ( t ) [ g ( 4 ) ( t ) − s ( 4 ) ( t ) ] d t
∫
a
b
s
(
4
)
(
t
)
[
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
]
d
t
=
s
(
4
)
(
t
)
[
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
]
|
a
b
−
−
∫
a
b
s
(
5
)
(
t
)
[
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
]
d
t
=
−
∫
a
b
s
(
5
)
(
t
)
[
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
]
d
t
=
=
−
∑
k
=
1
n
∫
t
k
−
1
t
k
s
(
5
)
(
t
)
[
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
]
d
t
=
=
−
∑
k
=
1
n
C
k
[
g
′
′
(
t
k
)
−
s
′
′
(
t
k
)
−
(
g
′
′
(
t
k
−
1
)
−
s
′
′
(
t
k
−
1
)
)
]
=
0
∫
a
b
s
(
4
)
(
t
)
g
(
4
)
(
t
)
−
s
(
4
)
(
t
)
d
t
=
s
(
4
)
(
t
)
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
a
b
−
−
∫
a
b
s
(
5
)
(
t
)
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
d
t
=
−
∫
a
b
s
(
5
)
(
t
)
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
d
t
=
=
−
∑
k
=
1
n
∫
t
k
−
1
t
k
s
(
5
)
(
t
)
g
′
′
′
(
t
)
−
s
′
′
′
(
t
)
d
t
=
=
−
∑
k
=
1
n
C
k
g
′
′
t
k
−
s
′
′
t
k
−
g
′
′
t
k
−
1
−
s
′
′
t
k
−
1
=
0
{:[int_(a)^(b)s^((4))(t)[g^((4))(t)-s^((4))(t)]dt=s^((4))(t)[g^(''')(t)-s^(''')(t)]|_(a)^(b)-],[-int_(a)^(b)s^((5))(t)[g^(''')(t)-s^(''')(t)]dt=-int_(a)^(b)s^((5))(t)[g^(''')(t)-s^(''')(t)]dt=],[=-sum_(k=1)^(n)int_(t_(k-1))^(t_(k))s^((5))(t)[g^(''')(t)-s^(''')(t)]dt=],[=-sum_(k=1)^(n)C_(k)[g^('')(t_(k))-s^('')(t_(k))-(g^('')(t_(k-1))-s^('')(t_(k-1)))]=0]:} \begin{gathered}
\int_{a}^{b} s^{(4)}(t)\left[g^{(4)}(t)-s^{(4)}(t)\right] \mathrm{d} t=\left.s^{(4)}(t)\left[g^{\prime \prime \prime}(t)-s^{\prime \prime \prime}(t)\right]\right|_{a} ^{b}- \\
-\int_{a}^{b} s^{(5)}(t)\left[g^{\prime \prime \prime}(t)-s^{\prime \prime \prime}(t)\right] \mathrm{d} t=-\int_{a}^{b} s^{(5)}(t)\left[g^{\prime \prime \prime}(t)-s^{\prime \prime \prime}(t)\right] \mathrm{d} t= \\
=-\sum_{k=1}^{n} \int_{t_{k-1}}^{t_{k}} s^{(5)}(t)\left[g^{\prime \prime \prime}(t)-s^{\prime \prime \prime}(t)\right] \mathrm{d} t= \\
=-\sum_{k=1}^{n} C_{k}\left[g^{\prime \prime}\left(t_{k}\right)-s^{\prime \prime}\left(t_{k}\right)-\left(g^{\prime \prime}\left(t_{k-1}\right)-s^{\prime \prime}\left(t_{k-1}\right)\right)\right]=0
\end{gathered} ∫ a b s ( 4 ) ( t ) [ g ( 4 ) ( t ) − s ( 4 ) ( t ) ] d t = s ( 4 ) ( t ) [ g ′ ′ ′ ( t ) − s ′ ′ ′ ( t ) ] | a b − − ∫ a b s ( 5 ) ( t ) [ g ′ ′ ′ ( t ) − s ′ ′ ′ ( t ) ] d t = − ∫ a b s ( 5 ) ( t ) [ g ′ ′ ′ ( t ) − s ′ ′ ′ ( t ) ] d t = = − ∑ k = 1 n ∫ t k − 1 t k s ( 5 ) ( t ) [ g ′ ′ ′ ( t ) − s ′ ′ ′ ( t ) ] d t = = − ∑ k = 1 n C k [ g ′ ′ ( t k ) − s ′ ′ ( t k ) − ( g ′ ′ ( t k − 1 ) − s ′ ′ ( t k − 1 ) ) ] = 0 where
C
k
=
s
(
5
)
(
t
)
|
I
k
,
k
=
1
,
2
,
…
,
n
C
k
=
s
(
5
)
(
t
)
I
k
,
k
=
1
,
2
,
…
,
n
C_(k)=s^((5))(t)|_(I_(k)),k=1,2,dots,n C_{k}=\left.s^{(5)}(t)\right|_{I_{k}}, k=1,2, \ldots, n C k = s ( 5 ) ( t ) | I k , k = 1 , 2 , … , n .
It follows
‖
g
(
4
)
‖
2
2
−
‖
s
(
4
)
‖
2
2
≥
0
g
(
4
)
2
2
−
s
(
4
)
2
2
≥
0
||g^((4))||_(2)^(2)-||s^((4))||_(2)^(2) >= 0 \left\|g^{(4)}\right\|_{2}^{2}-\left\|s^{(4)}\right\|_{2}^{2} \geq 0 ‖ g ( 4 ) ‖ 2 2 − ‖ s ( 4 ) ‖ 2 2 ≥ 0 , which is equivalent to (14).
Inequalities (15) can be proved by a similar argument.
THEOREM 5. a) If
f
∈
W
2
4
(
Δ
n
)
f
∈
W
2
4
Δ
n
f inW_(2)^(4)(Delta_(n)) f \in W_{2}^{4}\left(\Delta_{n}\right) f ∈ W 2 4 ( Δ n ) and
s
f
∈
S
5
(
Δ
n
)
s
f
∈
S
5
Δ
n
s_(f)inS_(5)(Delta_(n)) s_{f} \in S_{5}\left(\Delta_{n}\right) s f ∈ S 5 ( Δ n ) verify conditions (4) from Theorem 2, then
(16)
‖
s
f
(
4
)
−
f
(
4
)
‖
2
≤
‖
s
(
4
)
−
f
(
4
)
‖
2
,
for all
s
∈
S
5
(
Δ
n
)
(16)
s
f
(
4
)
−
f
(
4
)
2
≤
s
(
4
)
−
f
(
4
)
2
,
for all
s
∈
S
5
Δ
n
{:(16)||s_(f)^((4))-f^((4))||_(2) <= ||s^((4))-f^((4))||_(2)","" for all "s inS_(5)(Delta_(n)):} \begin{equation*}
\left\|s_{f}^{(4)}-f^{(4)}\right\|_{2} \leq\left\|s^{(4)}-f^{(4)}\right\|_{2}, \text { for all } s \in S_{5}\left(\Delta_{n}\right) \tag{16}
\end{equation*} (16) ‖ s f ( 4 ) − f ( 4 ) ‖ 2 ≤ ‖ s ( 4 ) − f ( 4 ) ‖ 2 , for all s ∈ S 5 ( Δ n ) b) If
h
∈
W
2
4
(
Δ
n
)
h
∈
W
2
4
Δ
n
h inW_(2)^(4)(Delta_(n)) h \in W_{2}^{4}\left(\Delta_{n}\right) h ∈ W 2 4 ( Δ n ) and
s
h
∈
S
5
(
Δ
n
)
s
h
∈
S
5
Δ
n
s_(h)inS_(5)(Delta_(n)) s_{h} \in S_{5}\left(\Delta_{n}\right) s h ∈ S 5 ( Δ n ) verify conditions (6) from Theorem 2, then
(17)
‖
s
h
(
4
)
−
h
(
4
)
‖
2
≤
‖
s
(
4
)
−
h
(
4
)
‖
2
,
for all
s
∈
S
5
(
Δ
n
)
(17)
s
h
(
4
)
−
h
(
4
)
2
≤
s
(
4
)
−
h
(
4
)
2
,
for all
s
∈
S
5
Δ
n
{:(17)||s_(h)^((4))-h^((4))||_(2) <= ||s^((4))-h^((4))||_(2)","" for all "s inS_(5)(Delta_(n)):} \begin{equation*}
\left\|s_{h}^{(4)}-h^{(4)}\right\|_{2} \leq\left\|s^{(4)}-h^{(4)}\right\|_{2}, \text { for all } s \in S_{5}\left(\Delta_{n}\right) \tag{17}
\end{equation*} (17) ‖ s h ( 4 ) − h ( 4 ) ‖ 2 ≤ ‖ s ( 4 ) − h ( 4 ) ‖ 2 , for all s ∈ S 5 ( Δ n ) Proof. In order to prove (16), we use the identity
‖
s
(
4
)
−
f
(
4
)
‖
2
2
=
∫
a
b
[
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
]
2
d
t
+
∫
a
b
[
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
]
2
d
t
+
+
2
∫
a
b
[
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
]
⋅
[
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
]
d
t
s
(
4
)
−
f
(
4
)
2
2
=
∫
a
b
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
2
d
t
+
∫
a
b
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
2
d
t
+
+
2
∫
a
b
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
⋅
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
d
t
{:[||s^((4))-f^((4))||_(2)^(2)=int_(a)^(b)[s^((4))(t)-s_(f)^((4))(t)]^(2)dt+int_(a)^(b)[s_(f)^((4))(t)-f^((4))(t)]^(2)dt+],[+2int_(a)^(b)[s^((4))(t)-s_(f)^((4))(t)]*[s_(f)^((4))(t)-f^((4))(t)]dt]:} \begin{gathered}
\left\|s^{(4)}-f^{(4)}\right\|_{2}^{2}=\int_{a}^{b}\left[s^{(4)}(t)-s_{f}^{(4)}(t)\right]^{2} \mathrm{~d} t+\int_{a}^{b}\left[s_{f}^{(4)}(t)-f^{(4)}(t)\right]^{2} \mathrm{~d} t+ \\
+2 \int_{a}^{b}\left[s^{(4)}(t)-s_{f}^{(4)}(t)\right] \cdot\left[s_{f}^{(4)}(t)-f^{(4)}(t)\right] \mathrm{d} t
\end{gathered} ‖ s ( 4 ) − f ( 4 ) ‖ 2 2 = ∫ a b [ s ( 4 ) ( t ) − s f ( 4 ) ( t ) ] 2 d t + ∫ a b [ s f ( 4 ) ( t ) − f ( 4 ) ( t ) ] 2 d t + + 2 ∫ a b [ s ( 4 ) ( t ) − s f ( 4 ) ( t ) ] ⋅ [ s f ( 4 ) ( t ) − f ( 4 ) ( t ) ] d t and prove that
T
=
∫
a
b
[
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
]
⋅
[
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
]
d
t
=
0
T
=
∫
a
b
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
⋅
s
f
(
4
)
(
t
)
−
f
(
4
)
(
t
)
d
t
=
0
T=int_(a)^(b)[s^((4))(t)-s_(f)^((4))(t)]*[s_(f)^((4))(t)-f^((4))(t)]dt=0 T=\int_{a}^{b}\left[s^{(4)}(t)-s_{f}^{(4)}(t)\right] \cdot\left[s_{f}^{(4)}(t)-f^{(4)}(t)\right] \mathrm{d} t=0 T = ∫ a b [ s ( 4 ) ( t ) − s f ( 4 ) ( t ) ] ⋅ [ s f ( 4 ) ( t ) − f ( 4 ) ( t ) ] d t = 0 Indeed, integrating by parts, we find
T
=
[
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
]
⋅
[
s
f
′
′
′
(
t
)
−
f
′
′
′
(
t
)
]
]
a
b
−
−
∫
a
b
[
s
(
5
)
(
t
)
−
s
f
(
5
)
(
t
)
]
⋅
[
s
f
′
′
′
(
t
)
−
f
′
′
′
(
t
)
]
d
t
=
=
−
∑
k
=
1
n
c
k
(
s
)
[
s
k
′
′
(
t
k
)
−
f
′
′
(
t
k
)
]
−
[
s
f
′
′
(
t
k
−
1
)
−
f
′
′
(
t
k
−
1
)
]
=
0
,
T
=
s
(
4
)
(
t
)
−
s
f
(
4
)
(
t
)
⋅
s
f
′
′
′
(
t
)
−
f
′
′
′
(
t
)
a
b
−
−
∫
a
b
s
(
5
)
(
t
)
−
s
f
(
5
)
(
t
)
⋅
s
f
′
′
′
(
t
)
−
f
′
′
′
(
t
)
d
t
=
=
−
∑
k
=
1
n
c
k
(
s
)
s
k
′
′
t
k
−
f
′
′
t
k
−
s
f
′
′
t
k
−
1
−
f
′
′
t
k
−
1
=
0
,
{:[T=[s^((4))(t)-s_(f)^((4))(t)]*[s_(f)^(''')(t)-f^(''')(t)]]_(a)^(b)-],[-int_(a)^(b)[s^((5))(t)-s_(f)^((5))(t)]*[s_(f)^(''')(t)-f^(''')(t)]dt=],[=-sum_(k=1)^(n)c_(k)(s)[s_(k)^('')(t_(k))-f^('')(t_(k))]-[s_(f)^('')(t_(k-1))-f^('')(t_(k-1))]=0","]:} \begin{gathered}
\left.T=\left[s^{(4)}(t)-s_{f}^{(4)}(t)\right] \cdot\left[s_{f}^{\prime \prime \prime}(t)-f^{\prime \prime \prime}(t)\right]\right]_{a}^{b}- \\
-\int_{a}^{b}\left[s^{(5)}(t)-s_{f}^{(5)}(t)\right] \cdot\left[s_{f}^{\prime \prime \prime}(t)-f^{\prime \prime \prime}(t)\right] \mathrm{d} t= \\
=-\sum_{k=1}^{n} c_{k}(s)\left[s_{k}^{\prime \prime}\left(t_{k}\right)-f^{\prime \prime}\left(t_{k}\right)\right]-\left[s_{f}^{\prime \prime}\left(t_{k-1}\right)-f^{\prime \prime}\left(t_{k-1}\right)\right]=0,
\end{gathered} T = [ s ( 4 ) ( t ) − s f ( 4 ) ( t ) ] ⋅ [ s f ′ ′ ′ ( t ) − f ′ ′ ′ ( t ) ] ] a b − − ∫ a b [ s ( 5 ) ( t ) − s f ( 5 ) ( t ) ] ⋅ [ s f ′ ′ ′ ( t ) − f ′ ′ ′ ( t ) ] d t = = − ∑ k = 1 n c k ( s ) [ s k ′ ′ ( t k ) − f ′ ′ ( t k ) ] − [ s f ′ ′ ( t k − 1 ) − f ′ ′ ( t k − 1 ) ] = 0 , where
c
k
(
s
)
=
s
(
5
)
(
t
)
−
s
f
(
5
)
(
t
)
,
t
∈
I
k
k
=
1
,
n
―
c
k
(
s
)
=
s
(
5
)
(
t
)
−
s
f
(
5
)
(
t
)
,
t
∈
I
k
k
=
1
,
n
¯
c_(k)(s)=s^((5))(t)-s_(f)^((5))(t),t inI_(k)k= bar(1,n) c_{k}(s)=s^{(5)}(t)-s_{f}^{(5)}(t), t \in I_{k} k=\overline{1, n} c k ( s ) = s ( 5 ) ( t ) − s f ( 5 ) ( t ) , t ∈ I k k = 1 , n ― . (We have used the fact that
(
s
(
4
)
−
s
f
(
4
)
)
(
a
)
==
(
s
(
4
)
−
s
f
(
4
)
)
(
b
)
=
0
s
(
4
)
−
s
f
(
4
)
(
a
)
==
s
(
4
)
−
s
f
(
4
)
(
b
)
=
0
(s^((4))-s_(f)^((4)))(a)==(s^((4))-s_(f)^((4)))(b)=0 \left(s^{(4)}-s_{f}^{(4)}\right)(a)= =\left(s^{(4)}-s_{f}^{(4)}\right)(b)=0 ( s ( 4 ) − s f ( 4 ) ) ( a ) == ( s ( 4 ) − s f ( 4 ) ) ( b ) = 0 .)
Therefore,
(18)
‖
s
(
4
)
−
f
(
4
)
‖
2
2
=
‖
s
(
4
)
−
s
f
(
4
)
‖
2
2
+
‖
s
f
(
4
)
−
f
(
4
)
‖
2
2
,
(18)
s
(
4
)
−
f
(
4
)
2
2
=
s
(
4
)
−
s
f
(
4
)
2
2
+
s
f
(
4
)
−
f
(
4
)
2
2
,
{:(18)||s^((4))-f^((4))||_(2)^(2)=||s^((4))-s_(f)^((4))||_(2)^(2)+||s_(f)^((4))-f^((4))||_(2)^(2)",":} \begin{equation*}
\left\|s^{(4)}-f^{(4)}\right\|_{2}^{2}=\left\|s^{(4)}-s_{f}^{(4)}\right\|_{2}^{2}+\left\|s_{f}^{(4)}-f^{(4)}\right\|_{2}^{2}, \tag{18}
\end{equation*} (18) ‖ s ( 4 ) − f ( 4 ) ‖ 2 2 = ‖ s ( 4 ) − s f ( 4 ) ‖ 2 2 + ‖ s f ( 4 ) − f ( 4 ) ‖ 2 2 , implying that inequality (16) holds.
‖
s
(
4
)
−
h
(
4
)
‖
2
2
=
∫
a
b
[
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
]
2
d
t
+
∫
a
b
[
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
]
2
d
t
+
+
2
∫
a
b
[
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
]
⋅
[
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
]
d
t
s
(
4
)
−
h
(
4
)
2
2
=
∫
a
b
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
2
d
t
+
∫
a
b
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
2
d
t
+
+
2
∫
a
b
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
⋅
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
d
t
{:[||s^((4))-h^((4))||_(2)^(2)=int_(a)^(b)[s^((4))(t)-s_(h)^((4))(t)]^(2)dt+int_(a)^(b)[s_(h)^((4))(t)-h^((4))(t)]^(2)dt+],[+2int_(a)^(b)[s^((4))(t)-s_(h)^((4))(t)]*[s_(h)^((4))(t)-h^((4))(t)]dt]:} \begin{aligned}
\left\|s^{(4)}-h^{(4)}\right\|_{2}^{2} & =\int_{a}^{b}\left[s^{(4)}(t)-s_{h}^{(4)}(t)\right]^{2} \mathrm{~d} t+\int_{a}^{b}\left[s_{h}^{(4)}(t)-h^{(4)}(t)\right]^{2} \mathrm{~d} t+ \\
& +2 \int_{a}^{b}\left[s^{(4)}(t)-s_{h}^{(4)}(t)\right] \cdot\left[s_{h}^{(4)}(t)-h^{(4)}(t)\right] \mathrm{d} t
\end{aligned} ‖ s ( 4 ) − h ( 4 ) ‖ 2 2 = ∫ a b [ s ( 4 ) ( t ) − s h ( 4 ) ( t ) ] 2 d t + ∫ a b [ s h ( 4 ) ( t ) − h ( 4 ) ( t ) ] 2 d t + + 2 ∫ a b [ s ( 4 ) ( t ) − s h ( 4 ) ( t ) ] ⋅ [ s h ( 4 ) ( t ) − h ( 4 ) ( t ) ] d t we have (integrating by parts)
Q
=
∫
a
b
[
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
]
⋅
[
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
]
d
t
=
0
Q
=
∫
a
b
s
(
4
)
(
t
)
−
s
h
(
4
)
(
t
)
⋅
s
h
(
4
)
(
t
)
−
h
(
4
)
(
t
)
d
t
=
0
Q=int_(a)^(b)[s^((4))(t)-s_(h)^((4))(t)]*[s_(h)^((4))(t)-h^((4))(t)]dt=0 Q=\int_{a}^{b}\left[s^{(4)}(t)-s_{h}^{(4)}(t)\right] \cdot\left[s_{h}^{(4)}(t)-h^{(4)}(t)\right] \mathrm{d} t=0 Q = ∫ a b [ s ( 4 ) ( t ) − s h ( 4 ) ( t ) ] ⋅ [ s h ( 4 ) ( t ) − h ( 4 ) ( t ) ] d t = 0 implying that
(
∗
)
‖
s
(
4
)
−
h
(
4
)
‖
2
2
=
‖
s
(
4
)
−
s
h
(
4
)
‖
2
2
+
‖
s
h
(
4
)
−
h
(
4
)
‖
2
2
∗
s
(
4
)
−
h
(
4
)
2
2
=
s
(
4
)
−
s
h
(
4
)
2
2
+
s
h
(
4
)
−
h
(
4
)
2
2
(^(**))quad||s^((4))-h^((4))||_(2)^(2)=||s^((4))-s_(h)^((4))||_(2)^(2)+||s_(h)^((4))-h^((4))||_(2)^(2) \left(^{*}\right) \quad\left\|s^{(4)}-h^{(4)}\right\|_{2}^{2}=\left\|s^{(4)}-s_{h}^{(4)}\right\|_{2}^{2}+\left\|s_{h}^{(4)}-h^{(4)}\right\|_{2}^{2} ( ∗ ) ‖ s ( 4 ) − h ( 4 ) ‖ 2 2 = ‖ s ( 4 ) − s h ( 4 ) ‖ 2 2 + ‖ s h ( 4 ) − h ( 4 ) ‖ 2 2 .
From this equality it follows (17).
COROLLARY 6. Iff,
h
∈
W
2
4
(
Δ
n
)
h
∈
W
2
4
Δ
n
h inW_(2)^(4)(Delta_(n)) h \in W_{2}^{4}\left(\Delta_{n}\right) h ∈ W 2 4 ( Δ n ) and
s
p
,
s
h
∈
S
5
(
Δ
n
)
s
p
,
s
h
∈
S
5
Δ
n
s_(p),s_(h)inS_(5)(Delta_(n)) s_{p}, s_{h} \in S_{5}\left(\Delta_{n}\right) s p , s h ∈ S 5 ( Δ n ) verify conditions (4) and (6) from Theorem 2, then
(19)
‖
f
(
4
)
‖
2
2
=
‖
s
f
(
4
)
‖
2
2
+
‖
f
(
4
)
−
s
f
(
4
)
‖
2
2
,
(20)
‖
h
(
4
)
‖
2
2
=
‖
s
h
(
4
)
‖
2
2
+
‖
h
(
4
)
−
s
h
(
4
)
‖
2
2
,
(19)
f
(
4
)
2
2
=
s
f
(
4
)
2
2
+
f
(
4
)
−
s
f
(
4
)
2
2
,
(20)
h
(
4
)
2
2
=
s
h
(
4
)
2
2
+
h
(
4
)
−
s
h
(
4
)
2
2
,
{:[(19)||f^((4))||_(2)^(2)=||s_(f)^((4))||_(2)^(2)+||f^((4))-s_(f)^((4))||_(2)^(2)","],[(20)||h^((4))||_(2)^(2)=||s_(h)^((4))||_(2)^(2)+||h^((4))-s_(h)^((4))||_(2)^(2)","]:} \begin{align*}
& \left\|f^{(4)}\right\|_{2}^{2}=\left\|s_{f}^{(4)}\right\|_{2}^{2}+\left\|f^{(4)}-s_{f}^{(4)}\right\|_{2}^{2}, \tag{19}\\
& \left\|h^{(4)}\right\|_{2}^{2}=\left\|s_{h}^{(4)}\right\|_{2}^{2}+\left\|h^{(4)}-s_{h}^{(4)}\right\|_{2}^{2}, \tag{20}
\end{align*} (19) ‖ f ( 4 ) ‖ 2 2 = ‖ s f ( 4 ) ‖ 2 2 + ‖ f ( 4 ) − s f ( 4 ) ‖ 2 2 , (20) ‖ h ( 4 ) ‖ 2 2 = ‖ s h ( 4 ) ‖ 2 2 + ‖ h ( 4 ) − s h ( 4 ) ‖ 2 2 ,
(29:मे ,760ig brom)
‖
s
f
(
4
)
‖
2
≤
‖
f
(
4
)
‖
2
,
(29:मे ,760ig brom)
s
f
(
4
)
2
≤
f
(
4
)
2
,
{:(29:मे ,760ig brom)||s_(f)^((4))||_(2) <= ||f^((4))||_(2)",":} \begin{equation*}
\left\|s_{f}^{(4)}\right\|_{2} \leq\left\|f^{(4)}\right\|_{2}, \tag{29:मे ,760ig brom}
\end{equation*} म े (29:मे ,760ig brom) ‖ s f ( 4 ) ‖ 2 ≤ ‖ f ( 4 ) ‖ 2 ,
(21)
‖
s
h
(
4
)
‖
2
≤
‖
h
(
4
)
‖
2
,
(21)
s
h
(
4
)
2
≤
h
(
4
)
2
,
{:(21)||s_(h)^((4))||_(2) <= ||h^((4))||_(2)",":} \begin{equation*}
\left\|s_{h}^{(4)}\right\|_{2} \leq\left\|h^{(4)}\right\|_{2}, \tag{21}
\end{equation*} (21) ‖ s h ( 4 ) ‖ 2 ≤ ‖ h ( 4 ) ‖ 2 ,
(22)
‖
f
(
4
)
−
s
f
(
4
)
‖
≤
‖
f
(
4
)
‖
2
,
(22)
f
(
4
)
−
s
f
(
4
)
≤
f
(
4
)
2
,
{:(22)||f^((4))-s_(f)^((4))|| <= ||f^((4))||_(2)",":} \begin{equation*}
\left\|f^{(4)}-s_{f}^{(4)}\right\| \leq\left\|f^{(4)}\right\|_{2}, \tag{22}
\end{equation*} (22) ‖ f ( 4 ) − s f ( 4 ) ‖ ≤ ‖ f ( 4 ) ‖ 2 , (24)
(23)
‖
h
(
4
)
−
s
h
(
4
)
‖
2
≤
‖
h
(
4
)
‖
2
.
(23)
h
(
4
)
−
s
h
(
4
)
2
≤
h
(
4
)
2
.
{:(23)||h^((4))-s_(h)^((4))||_(2) <= ||h^((4))||_(2).:} \begin{equation*}
\left\|h^{(4)}-s_{h}^{(4)}\right\|_{2} \leq\left\|h^{(4)}\right\|_{2} . \tag{23}
\end{equation*} (23) ‖ h ( 4 ) − s h ( 4 ) ‖ 2 ≤ ‖ h ( 4 ) ‖ 2 . Proof. Equalities (19) and (20) follow from (18) and (*) for
s
≡
0
s
≡
0
s-=0 s \equiv 0 s ≡ 0 . The remaining inequalities follow from (19) and (20).
Application. Consider the bilocal linear problem
(D)
y
′
′
=
p
(
t
)
⋅
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
y
′
′
=
p
(
t
)
⋅
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
y^('')=p(t)*y+q(t),t in[a,b] y^{\prime \prime}=p(t) \cdot y+q(t), t \in[a, b] y ′ ′ = p ( t ) ⋅ y + q ( t ) , t ∈ [ a , b ] ,
y
(
a
)
=
α
,
y
(
b
)
=
β
y
(
a
)
=
α
,
y
(
b
)
=
β
y(a)=alpha,y(b)=beta y(a)=\alpha, y(b)=\beta y ( a ) = α , y ( b ) = β .
If
p
,
q
p
,
q
p,q p, q p , q are continuous functions on
[
a
,
b
]
[
a
,
b
]
[a,b] [a, b] [ a , b ] and
p
(
t
)
>
0
,
t
∈
[
a
,
b
]
p
(
t
)
>
0
,
t
∈
[
a
,
b
]
p(t) > 0,t in[a,b] p(t)>0, t \in[a, b] p ( t ) > 0 , t ∈ [ a , b ] , then the problem
(
D
)
(
D
)
(D) (D) ( D ) has a unique solution
y
y
y y y (see [3], Theorem 10.1, p. 519).
Consider the Cauchy problems
(
C
1
)
y
′
′
=
p
(
t
)
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
C
1
y
′
′
=
p
(
t
)
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
(C_(1))y^('')=p(t)y+q(t),t in[a,b] \left(C_{1}\right) y^{\prime \prime}=p(t) y+q(t), t \in[a, b] ( C 1 ) y ′ ′ = p ( t ) y + q ( t ) , t ∈ [ a , b ] ,
y
(
a
)
=
α
,
y
′
(
a
)
=
0
y
(
a
)
=
α
,
y
′
(
a
)
=
0
y(a)=alpha,y^(')(a)=0 y(a)=\alpha, y^{\prime}(a)=0 y ( a ) = α , y ′ ( a ) = 0 ,
(
C
2
)
y
′
′
=
p
(
t
)
y
,
t
∈
[
a
,
b
]
C
2
y
′
′
=
p
(
t
)
y
,
t
∈
[
a
,
b
]
(C_(2))y^('')=p(t)y,t in[a,b] \left(C_{2}\right) y^{\prime \prime}=p(t) y, t \in[a, b] ( C 2 ) y ′ ′ = p ( t ) y , t ∈ [ a , b ] ,
y
(
a
)
=
0
,
y
′
(
a
)
=
1
y
(
a
)
=
0
,
y
′
(
a
)
=
1
y(a)=0,y^(')(a)=1 y(a)=0, y^{\prime}(a)=1 y ( a ) = 0 , y ′ ( a ) = 1 .
The Cauchy problems have unique solutions
y
1
,
y
2
y
1
,
y
2
y_(1),y_(2) y_{1}, y_{2} y 1 , y 2 , respectively (see [3], Theorem 5.15, p. 263), and the function
(25)
y
(
t
)
=
y
1
(
t
)
+
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
with
y
2
(
b
)
≠
0
,
t
∈
[
a
,
b
]
,
(25)
y
(
t
)
=
y
1
(
t
)
+
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
with
y
2
(
b
)
≠
0
,
t
∈
[
a
,
b
]
,
{:(25)y(t)=y_(1)(t)+(beta-y_(1)(b))/(y_(2)(b))y_(2)(t)" with "y_(2)(b)!=0","quad t in[a","b]",":} \begin{equation*}
y(t)=y_{1}(t)+\frac{\beta-y_{1}(b)}{y_{2}(b)} y_{2}(t) \text { with } y_{2}(b) \neq 0, \quad t \in[a, b], \tag{25}
\end{equation*} (25) y ( t ) = y 1 ( t ) + β − y 1 ( b ) y 2 ( b ) y 2 ( t ) with y 2 ( b ) ≠ 0 , t ∈ [ a , b ] , is the solution of the problem (
D
D
D D D ) (see [3]).
Applying Theorem 2b) to the solutions
y
1
,
y
2
y
1
,
y
2
y_(1),y_(2) y_{1}, y_{2} y 1 , y 2 of the problems (
C
1
C
1
C_(1) C_{1} C 1 ), (
C
2
C
2
C_(2) C_{2} C 2 ), it follows that there exist the functions
s
y
1
,
s
y
2
∈
S
5
(
Δ
n
)
s
y
1
,
s
y
2
∈
S
5
Δ
n
s_(y_(1)),s_(y_(2))inS_(5)(Delta_(n)) s_{y_{1}}, s_{y_{2}} \in S_{5}\left(\Delta_{n}\right) s y 1 , s y 2 ∈ S 5 ( Δ n ) such that
s
y
1
(
a
)
=
α
,
s
y
1
′
(
a
)
=
0
,
s
y
1
′
′
(
t
k
)
=
y
1
′
′
(
t
k
)
,
k
=
0
,
n
―
,
(**)
s
y
2
(
a
)
=
0
,
s
y
2
′
(
a
)
=
1
,
s
y
2
′
′
(
t
k
)
=
y
2
′
′
(
t
k
)
,
k
=
0
,
n
―
.
s
y
1
(
a
)
=
α
,
s
y
1
′
(
a
)
=
0
,
s
y
1
′
′
t
k
=
y
1
′
′
t
k
,
k
=
0
,
n
¯
,
(**)
s
y
2
(
a
)
=
0
,
s
y
2
′
(
a
)
=
1
,
s
y
2
′
′
t
k
=
y
2
′
′
t
k
,
k
=
0
,
n
¯
.
{:[s_(y_(1))(a)=alpha","quads_(y_(1))^(')(a)=0","quads_(y_(1))^('')(t_(k))=y_(1)^('')(t_(k))","quad k= bar(0,n)","],[(**)s_(y_(2))(a)=0","quads_(y_(2))^(')(a)=1","quads_(y_(2))^('')(t_(k))=y_(2)^('')(t_(k))","quad k= bar(0,n).]:} \begin{align*}
& s_{y_{1}}(a)=\alpha, \quad s_{y_{1}}^{\prime}(a)=0, \quad s_{y_{1}}^{\prime \prime}\left(t_{k}\right)=y_{1}^{\prime \prime}\left(t_{k}\right), \quad k=\overline{0, n}, \\
& s_{y_{2}}(a)=0, \quad s_{y_{2}}^{\prime}(a)=1, \quad s_{y_{2}}^{\prime \prime}\left(t_{k}\right)=y_{2}^{\prime \prime}\left(t_{k}\right), \quad k=\overline{0, n} . \tag{**}
\end{align*} s y 1 ( a ) = α , s y 1 ′ ( a ) = 0 , s y 1 ′ ′ ( t k ) = y 1 ′ ′ ( t k ) , k = 0 , n ― , (**) s y 2 ( a ) = 0 , s y 2 ′ ( a ) = 1 , s y 2 ′ ′ ( t k ) = y 2 ′ ′ ( t k ) , k = 0 , n ― . We call the functions
s
y
1
,
s
y
2
s
y
1
,
s
y
2
s_(y_(1)),s_(y_(2)) s_{y_{1}}, s_{y_{2}} s y 1 , s y 2 spline solutions in
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) of the problems
(
C
1
)
C
1
(C_(1)) \left(C_{1}\right) ( C 1 ) ,
(
C
2
)
C
2
(C_(2)) \left(C_{2}\right) ( C 2 ) , and the function
(26)
s
y
(
t
)
=
s
y
1
(
t
)
+
β
−
s
y
1
(
b
)
s
y
2
(
b
)
s
y
2
(
t
)
,
s
y
2
(
b
)
≠
0
,
t
∈
[
a
,
b
]
,
(26)
s
y
(
t
)
=
s
y
1
(
t
)
+
β
−
s
y
1
(
b
)
s
y
2
(
b
)
s
y
2
(
t
)
,
s
y
2
(
b
)
≠
0
,
t
∈
[
a
,
b
]
,
{:(26)s_(y)(t)=s_(y_(1))(t)+(beta-s_(y_(1))(b))/(s_(y_(2))(b))s_(y_(2))(t)","s_(y_(2))(b)!=0","t in[a","b]",":} \begin{equation*}
s_{y}(t)=s_{y_{1}}(t)+\frac{\beta-s_{y_{1}}(b)}{s_{y_{2}}(b)} s_{y_{2}}(t), s_{y_{2}}(b) \neq 0, t \in[a, b], \tag{26}
\end{equation*} (26) s y ( t ) = s y 1 ( t ) + β − s y 1 ( b ) s y 2 ( b ) s y 2 ( t ) , s y 2 ( b ) ≠ 0 , t ∈ [ a , b ] , is called a spline solution in
S
5
(
Δ
n
)
S
5
Δ
n
S_(5)(Delta_(n)) S_{5}\left(\Delta_{n}\right) S 5 ( Δ n ) of the problem
(
D
)
(
D
)
(D) (D) ( D ) .
THEOREM 7. Consider the problem
(C)
y
′
′
=
p
(
t
)
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
y
(
a
)
=
α
,
y
′
(
a
)
=
γ
,
(C)
y
′
′
=
p
(
t
)
y
+
q
(
t
)
,
t
∈
[
a
,
b
]
y
(
a
)
=
α
,
y
′
(
a
)
=
γ
,
{:[(C)y^('')=p(t)y+q(t)","quad t in[a","b]],[y(a)=alpha","y^(')(a)=gamma","]:} \begin{gather*}
y^{\prime \prime}=p(t) y+q(t), \quad t \in[a, b] \tag{C}\\
y(a)=\alpha, y^{\prime}(a)=\gamma,
\end{gather*} (C) y ′ ′ = p ( t ) y + q ( t ) , t ∈ [ a , b ] y ( a ) = α , y ′ ( a ) = γ , where
p
(
t
)
>
0
,
t
∈
[
a
,
b
]
p
(
t
)
>
0
,
t
∈
[
a
,
b
]
p(t) > 0,t in[a,b] p(t)>0, t \in[a, b] p ( t ) > 0 , t ∈ [ a , b ] and
p
,
q
p
,
q
p,q p, q p , q are continuous on
[
a
,
b
]
[
a
,
b
]
[a,b] [a, b] [ a , b ] .
If
y
∈
W
2
4
(
△
n
)
y
∈
W
2
4
△
n
y inW_(2)^(4)(/_\_(n)) y \in W_{2}^{4}\left(\triangle_{n}\right) y ∈ W 2 4 ( △ n ) is the exact solution of
(
C
)
(
C
)
(C) (C) ( C ) and
s
y
∈
S
5
(
Δ
n
)
s
y
∈
S
5
Δ
n
s_(y)inS_(5)(Delta_(n)) s_{y} \in S_{5}\left(\Delta_{n}\right) s y ∈ S 5 ( Δ n ) is its spline solution (cf. Theorem 2b)), then we have
(27)
‖
y
′
′
−
s
′
′
‖
∞
≤
2
‖
Δ
n
‖
3
/
2
⋅
‖
y
(
4
)
‖
2
,
(27)
y
′
′
−
s
′
′
∞
≤
2
Δ
n
3
/
2
⋅
y
(
4
)
2
,
{:(27)||y^('')-s^('')||_(oo) <= sqrt2||Delta_(n)||^(3//2)*||y^((4))||_(2)",":} \begin{equation*}
\left\|y^{\prime \prime}-s^{\prime \prime}\right\|_{\infty} \leq \sqrt{2}\left\|\Delta_{n}\right\|^{3 / 2} \cdot\left\|y^{(4)}\right\|_{2}, \tag{27}
\end{equation*} (27) ‖ y ′ ′ − s ′ ′ ‖ ∞ ≤ 2 ‖ Δ n ‖ 3 / 2 ⋅ ‖ y ( 4 ) ‖ 2 , where
‖
Δ
n
‖
=
max
{
t
k
−
t
k
−
1
,
k
=
1
,
n
―
}
Δ
n
=
max
t
k
−
t
k
−
1
,
k
=
1
,
n
¯
||Delta_(n)||=max{t_(k)-t_(k-1),k= bar(1,n)} \left\|\Delta_{n}\right\|=\max \left\{t_{k}-t_{k-1}, k=\overline{1, n}\right\} ‖ Δ n ‖ = max { t k − t k − 1 , k = 1 , n ― }
Proof. We have
y
′
′
(
t
i
)
−
s
′
′
y
(
t
i
)
=
0
,
i
=
0
,
n
―
y
′
′
t
i
−
s
′
′
y
t
i
=
0
,
i
=
0
,
n
¯
y^('')(t_(i))-s^('')y(t_(i))=0,quad i= bar(0,n) y^{\prime \prime}\left(t_{i}\right)-s^{\prime \prime} y\left(t_{i}\right)=0, \quad i=\overline{0, n} y ′ ′ ( t i ) − s ′ ′ y ( t i ) = 0 , i = 0 , n ― so that, by Rolle's theorem, there exist
t
i
(
1
)
∈
(
t
i
,
t
i
+
1
)
,
i
=
0
,
n
−
1
―
t
i
(
1
)
∈
t
i
,
t
i
+
1
,
i
=
0
,
n
−
1
¯
t_(i)^((1))in(t_(i),t_(i+1)),i= bar(0,n-1) t_{i}^{(1)} \in\left(t_{i}, t_{i+1}\right), i=\overline{0, n-1} t i ( 1 ) ∈ ( t i , t i + 1 ) , i = 0 , n − 1 ― such that
y
′
′
′
(
t
i
(
1
)
)
−
s
y
′
′
′
(
t
i
(
1
)
)
=
0
,
i
=
0
,
n
−
1
y
′
′
′
t
i
(
1
)
−
s
y
′
′
′
t
i
(
1
)
=
0
,
i
=
0
,
n
−
1
y^(''')(t_(i)^((1)))-s_(y)^(''')(t_(i)^((1)))=0,quad i=0,n-1 y^{\prime \prime \prime}\left(t_{i}^{(1)}\right)-s_{y}^{\prime \prime \prime}\left(t_{i}^{(1)}\right)=0, \quad i=0, n-1 y ′ ′ ′ ( t i ( 1 ) ) − s y ′ ′ ′ ( t i ( 1 ) ) = 0 , i = 0 , n − 1 Applying again Rolle's theorem, it follows the existence of
t
i
(
2
)
∈
(
t
i
(
1
)
,
t
i
+
1
(
1
)
)
t
i
(
2
)
∈
t
i
(
1
)
,
t
i
+
1
(
1
)
t_(i)^((2))in(t_(i)^((1)),t_(i+1)^((1))) t_{i}^{(2)} \in\left(t_{i}^{(1)}, t_{i+1}^{(1)}\right) t i ( 2 ) ∈ ( t i ( 1 ) , t i + 1 ( 1 ) ) ,
i
=
0
,
n
−
2
―
i
=
0
,
n
−
2
¯
i= bar(0,n-2) i=\overline{0, n-2} i = 0 , n − 2 ― such that
y
(
4
)
(
t
i
(
2
)
)
−
s
y
(
4
)
(
t
i
(
2
)
)
=
0
,
i
=
0
,
n
−
2
―
.
y
(
4
)
t
i
(
2
)
−
s
y
(
4
)
t
i
(
2
)
=
0
,
i
=
0
,
n
−
2
¯
.
y^((4))(t_(i)^((2)))-s_(y)^((4))(t_(i)^((2)))=0,i= bar(0,n-2). y^{(4)}\left(t_{i}^{(2)}\right)-s_{y}^{(4)}\left(t_{i}^{(2)}\right)=0, i=\overline{0, n-2} . y ( 4 ) ( t i ( 2 ) ) − s y ( 4 ) ( t i ( 2 ) ) = 0 , i = 0 , n − 2 ― . The inequalities
|
t
i
+
1
(
1
)
−
t
i
(
1
)
|
≤
2
‖
Δ
n
‖
and
|
t
i
+
1
(
2
)
−
t
i
(
2
)
|
≤
3
‖
Δ
n
‖
t
i
+
1
(
1
)
−
t
i
(
1
)
≤
2
Δ
n
and
t
i
+
1
(
2
)
−
t
i
(
2
)
≤
3
Δ
n
|t_(i+1)^((1))-t_(i)^((1))| <= 2||Delta_(n)||" and "|t_(i+1)^((2))-t_(i)^((2))| <= 3||Delta_(n)|| \left|t_{i+1}^{(1)}-t_{i}^{(1)}\right| \leq 2\left\|\Delta_{n}\right\| \text { and }\left|t_{i+1}^{(2)}-t_{i}^{(2)}\right| \leq 3\left\|\Delta_{n}\right\| | t i + 1 ( 1 ) − t i ( 1 ) | ≤ 2 ‖ Δ n ‖ and | t i + 1 ( 2 ) − t i ( 2 ) | ≤ 3 ‖ Δ n ‖ hold for
i
=
0
,
n
−
2
i
=
0
,
n
−
2
i=0,n-2 i=0, n-2 i = 0 , n − 2 and
i
=
0
,
n
−
3
―
i
=
0
,
n
−
3
¯
i= bar(0,n-3) i=\overline{0, n-3} i = 0 , n − 3 ― , respectively.
For every
t
∈
[
a
,
b
]
t
∈
[
a
,
b
]
t in[a,b] t \in[a, b] t ∈ [ a , b ] there is an index
i
0
∈
{
0
,
1
,
…
,
n
−
1
}
i
0
∈
{
0
,
1
,
…
,
n
−
1
}
i_(0)in{0,1,dots,n-1} i_{0} \in\{0,1, \ldots, n-1\} i 0 ∈ { 0 , 1 , … , n − 1 } such that
|
t
−
t
i
0
(
1
)
|
≤
2
‖
Δ
n
‖
t
−
t
i
0
(
1
)
≤
2
Δ
n
|t-t_(i_(0))^((1))| <= 2||Delta_(n)|| \left|t-t_{i_{0}}^{(1)}\right| \leq 2\left\|\Delta_{n}\right\| | t − t i 0 ( 1 ) | ≤ 2 ‖ Δ n ‖ so that, taking into account (24), we have
|
y
′
′
′
(
t
)
−
s
′
′
′
y
(
t
)
|
=
|
∫
t
i
0
(
1
)
t
(
y
(
4
)
(
u
)
−
s
y
(
4
)
(
u
)
)
d
u
|
≤
≤
|
∫
t
i
0
(
1
)
t
d
u
|
1
/
2
⋅
|
∫
t
i
0
(
1
)
(
1
)
t
[
y
(
4
)
(
u
)
−
s
y
(
4
)
]
2
d
u
|
1
/
2
≤
≤
2
‖
Δ
n
‖
⋅
|
∫
a
b
[
y
(
4
)
(
u
)
−
s
y
(
4
)
(
u
)
]
2
d
u
|
1
/
2
,
≤
2
⋅
‖
Δ
n
‖
1
/
2
⋅
‖
y
(
4
)
‖
2
.
y
′
′
′
(
t
)
−
s
′
′
′
y
(
t
)
=
∫
t
i
0
(
1
)
t
y
(
4
)
(
u
)
−
s
y
(
4
)
(
u
)
d
u
≤
≤
∫
t
i
0
(
1
)
t
d
u
1
/
2
⋅
∫
t
i
0
(
1
)
(
1
)
t
y
(
4
)
(
u
)
−
s
y
(
4
)
2
d
u
1
/
2
≤
≤
2
Δ
n
⋅
∫
a
b
y
(
4
)
(
u
)
−
s
y
(
4
)
(
u
)
2
d
u
1
/
2
,
≤
2
⋅
Δ
n
1
/
2
⋅
y
(
4
)
2
.
{:[|y^(''')(t)-s^(''')y(t)|=|int_(t_(i_(0))^((1)))^(t)(y^((4))(u)-s_(y)^((4))(u))du| <= ],[ <= |int_(t_(i_(0))^((1)))^(t)(d)u|^(1//2)*|int_(t_(i_(0)^((1)))^((1)))^(t)[y^((4))(u)-s_(y)^((4))]^(2)(d)u|^(1//2) <= ],[ <= sqrt(2||Delta_(n)||)*|int_(a)^(b)[y^((4))(u)-s_(y)^((4))(u)]^(2)(d)u|^(1//2)","],[ <= sqrt2*||Delta_(n)||^(1//2)*||y^((4))||_(2).]:} \begin{gathered}
\left|y^{\prime \prime \prime}(t)-s^{\prime \prime \prime} y(t)\right|=\left|\int_{t_{i_{0}}^{(1)}}^{t}\left(y^{(4)}(u)-s_{y}^{(4)}(u)\right) \mathrm{d} u\right| \leq \\
\leq\left|\int_{t_{i_{0}}^{(1)}}^{t} \mathrm{~d} u\right|^{1 / 2} \cdot\left|\int_{t_{i_{0}^{(1)}}^{(1)}}^{t}\left[y^{(4)}(u)-s_{y}^{(4)}\right]^{2} \mathrm{~d} u\right|^{1 / 2} \leq \\
\leq \sqrt{2\left\|\Delta_{n}\right\|} \cdot\left|\int_{a}^{b}\left[y^{(4)}(u)-s_{y}^{(4)}(u)\right]^{2} \mathrm{~d} u\right|^{1 / 2}, \\
\leq \sqrt{2} \cdot\left\|\Delta_{n}\right\|^{1 / 2} \cdot\left\|y^{(4)}\right\|_{2} .
\end{gathered} | y ′ ′ ′ ( t ) − s ′ ′ ′ y ( t ) | = | ∫ t i 0 ( 1 ) t ( y ( 4 ) ( u ) − s y ( 4 ) ( u ) ) d u | ≤ ≤ | ∫ t i 0 ( 1 ) t d u | 1 / 2 ⋅ | ∫ t i 0 ( 1 ) ( 1 ) t [ y ( 4 ) ( u ) − s y ( 4 ) ] 2 d u | 1 / 2 ≤ ≤ 2 ‖ Δ n ‖ ⋅ | ∫ a b [ y ( 4 ) ( u ) − s y ( 4 ) ( u ) ] 2 d u | 1 / 2 , ≤ 2 ⋅ ‖ Δ n ‖ 1 / 2 ⋅ ‖ y ( 4 ) ‖ 2 . Similarly, for every
t
∈
[
a
,
b
]
t
∈
[
a
,
b
]
t in[a,b] t \in[a, b] t ∈ [ a , b ] there exist
j
0
∈
{
0
,
1
,
…
,
n
−
1
}
j
0
∈
{
0
,
1
,
…
,
n
−
1
}
j_(0)in{0,1,dots,n-1} j_{0} \in\{0,1, \ldots, n-1\} j 0 ∈ { 0 , 1 , … , n − 1 } such that
|
t
−
t
j
0
(
1
)
|
≤
‖
Δ
n
‖
t
−
t
j
0
(
1
)
≤
Δ
n
|t-t_(j_(0))^((1))| <= ||Delta_(n)|| \left|t-t_{j_{0}}^{(1)}\right| \leq\left\|\Delta_{n}\right\| | t − t j 0 ( 1 ) | ≤ ‖ Δ n ‖ , implying
|
y
′
′
(
t
)
−
s
y
′
′
(
t
)
|
=
|
∫
t
s
0
(
1
)
t
[
y
′
′
′
(
u
)
−
s
y
′
′
′
(
u
)
d
u
]
|
≤
≤
‖
y
′
′
′
−
s
y
′
′
′
‖
∞
⋅
‖
Δ
n
‖
y
′
′
(
t
)
−
s
y
′
′
(
t
)
=
∫
t
s
0
(
1
)
t
y
′
′
′
(
u
)
−
s
y
′
′
′
(
u
)
d
u
≤
≤
y
′
′
′
−
s
y
′
′
′
∞
⋅
Δ
n
{:[|y^('')(t)-s_(y)^('')(t)|=|int_(t_(s_(0))^((1)))^(t)[y^(''')(u)-s_(y)^(''')(u)du]| <= ],[ <= ||y^(''')-s_(y)^(''')||_(oo)*||Delta_(n)||]:} \begin{gathered}
\left|y^{\prime \prime}(t)-s_{y}^{\prime \prime}(t)\right|=\left|\int_{t_{s_{0}}^{(1)}}^{t}\left[y^{\prime \prime \prime}(u)-s_{y}^{\prime \prime \prime}(u) \mathrm{d} u\right]\right| \leq \\
\leq\left\|y^{\prime \prime \prime}-s_{y}^{\prime \prime \prime}\right\|_{\infty} \cdot\left\|\Delta_{n}\right\|
\end{gathered} | y ′ ′ ( t ) − s y ′ ′ ( t ) | = | ∫ t s 0 ( 1 ) t [ y ′ ′ ′ ( u ) − s y ′ ′ ′ ( u ) d u ] | ≤ ≤ ‖ y ′ ′ ′ − s y ′ ′ ′ ‖ ∞ ⋅ ‖ Δ n ‖ It follows that inequality (27) holds.
COROLLARY 8. If
y
∈
W
2
4
(
Δ
n
)
y
∈
W
2
4
Δ
n
y inW_(2)^(4)(Delta_(n)) y \in W_{2}^{4}\left(\Delta_{n}\right) y ∈ W 2 4 ( Δ n ) is the exact solution of the problem (
C
C
C C C ), then
(28)
‖
y
−
s
y
‖
∞
≤
2
(
b
−
a
)
2
‖
Δ
n
‖
3
/
2
⋅
‖
y
(
4
)
‖
2
.
(28)
y
−
s
y
∞
≤
2
(
b
−
a
)
2
Δ
n
3
/
2
⋅
y
(
4
)
2
.
{:(28)||y-s_(y)||_(oo) <= sqrt2(b-a)^(2)||Delta_(n)||^(3//2)*||y^((4))||_(2).:} \begin{equation*}
\left\|y-s_{y}\right\|_{\infty} \leq \sqrt{2}(b-a)^{2}\left\|\Delta_{n}\right\|^{3 / 2} \cdot\left\|y^{(4)}\right\|_{2} . \tag{28}
\end{equation*} (28) ‖ y − s y ‖ ∞ ≤ 2 ( b − a ) 2 ‖ Δ n ‖ 3 / 2 ⋅ ‖ y ( 4 ) ‖ 2 . Proof. For every
t
∈
[
a
,
b
]
t
∈
[
a
,
b
]
t in[a,b] t \in[a, b] t ∈ [ a , b ] we have
|
y
(
t
)
−
s
y
(
t
)
|
=
|
∫
a
t
(
y
′
(
u
)
−
s
y
′
(
u
)
)
d
u
|
≤
(
b
−
a
)
⋅
‖
y
′
−
s
y
′
‖
∞
y
(
t
)
−
s
y
(
t
)
=
∫
a
t
y
′
(
u
)
−
s
y
′
(
u
)
d
u
≤
(
b
−
a
)
⋅
y
′
−
s
y
′
∞
|y(t)-s_(y)(t)|=|int_(a)^(t)(y^(')(u)-s_(y)^(')(u))du| <= (b-a)*||y^(')-s_(y)^(')||_(oo) \left|y(t)-s_{y}(t)\right|=\left|\int_{a}^{t}\left(y^{\prime}(u)-s_{y}^{\prime}(u)\right) \mathrm{d} u\right| \leq(b-a) \cdot\left\|y^{\prime}-s_{y}^{\prime}\right\|_{\infty} | y ( t ) − s y ( t ) | = | ∫ a t ( y ′ ( u ) − s y ′ ( u ) ) d u | ≤ ( b − a ) ⋅ ‖ y ′ − s y ′ ‖ ∞ and
|
y
′
(
t
)
−
s
y
′
(
t
)
|
=
|
∫
a
t
(
y
′
′
(
u
)
−
s
y
′
′
(
u
)
)
d
u
|
≤
(
b
−
a
)
⋅
‖
y
′
′
−
s
y
′
′
‖
∞
.
y
′
(
t
)
−
s
y
′
(
t
)
=
∫
a
t
y
′
′
(
u
)
−
s
y
′
′
(
u
)
d
u
≤
(
b
−
a
)
⋅
y
′
′
−
s
y
′
′
∞
.
|y^(')(t)-s_(y)^(')(t)|=|int_(a)^(t)(y^('')(u)-s_(y)^('')(u))du| <= (b-a)*||y^('')-s_(y)^('')||_(oo). \left|y^{\prime}(t)-s_{y}^{\prime}(t)\right|=\left|\int_{a}^{t}\left(y^{\prime \prime}(u)-s_{y}^{\prime \prime}(u)\right) \mathrm{d} u\right| \leq(b-a) \cdot\left\|y^{\prime \prime}-s_{y}^{\prime \prime}\right\|_{\infty} . | y ′ ( t ) − s y ′ ( t ) | = | ∫ a t ( y ′ ′ ( u ) − s y ′ ′ ( u ) ) d u | ≤ ( b − a ) ⋅ ‖ y ′ ′ − s y ′ ′ ‖ ∞ . From these inequalities and from (20) we obtain (28).
holds, too.
The approximative determination of the values of the spline solution
s
y
s
y
s_(y) s_{y} s y of the problem (
D
D
D D D ) on the nodes of the division
Δ
n
Δ
n
Delta_(n) \Delta_{n} Δ n First observe that the exact solution
y
∈
W
2
4
(
Δ
n
)
y
∈
W
2
4
Δ
n
y inW_(2)^(4)(Delta_(n)) y \in W_{2}^{4}\left(\Delta_{n}\right) y ∈ W 2 4 ( Δ n ) of the problem (
D
D
D D D ) and its spline solution
s
y
∈
S
5
(
Δ
n
)
s
y
∈
S
5
Δ
n
s_(y)inS_(5)(Delta_(n)) s_{y} \in S_{5}\left(\Delta_{n}\right) s y ∈ S 5 ( Δ n ) given by (26) verify
|
y
(
t
)
−
s
y
(
t
)
|
=
|
y
1
(
t
)
+
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
−
s
y
1
(
t
)
−
β
−
s
y
1
(
b
)
s
y
2
(
b
)
⋅
s
y
2
(
t
)
|
≤
≤
|
y
1
(
t
)
−
s
y
1
(
t
)
|
+
|
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
−
β
−
s
y
1
(
b
)
s
y
2
(
b
)
⋅
s
y
2
(
t
)
|
y
(
t
)
−
s
y
(
t
)
=
y
1
(
t
)
+
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
−
s
y
1
(
t
)
−
β
−
s
y
1
(
b
)
s
y
2
(
b
)
⋅
s
y
2
(
t
)
≤
≤
y
1
(
t
)
−
s
y
1
(
t
)
+
β
−
y
1
(
b
)
y
2
(
b
)
y
2
(
t
)
−
β
−
s
y
1
(
b
)
s
y
2
(
b
)
⋅
s
y
2
(
t
)
{:[|y(t)-s_(y)(t)|=|y_(1)(t)+(beta-y_(1)(b))/(y_(2)(b))y_(2)(t)-s_(y_(1))(t)-(beta-s_(y_(1))(b))/(s_(y_(2))(b))*s_(y_(2))(t)| <= ],[ <= |y_(1)(t)-s_(y_(1))(t)|+|(beta-y_(1)(b))/(y_(2)(b))y_(2)(t)-(beta-s_(y_(1))(b))/(s_(y_(2))(b))*s_(y_(2))(t)|]:} \begin{aligned}
\left|y(t)-s_{y}(t)\right|= & \left|y_{1}(t)+\frac{\beta-y_{1}(b)}{y_{2}(b)} y_{2}(t)-s_{y_{1}}(t)-\frac{\beta-s_{y_{1}}(b)}{s_{y_{2}}(b)} \cdot s_{y_{2}}(t)\right| \leq \\
& \leq\left|y_{1}(t)-s_{y_{1}}(t)\right|+\left|\frac{\beta-y_{1}(b)}{y_{2}(b)} y_{2}(t)-\frac{\beta-s_{y_{1}}(b)}{s_{y_{2}}(b)} \cdot s_{y_{2}}(t)\right|
\end{aligned} | y ( t ) − s y ( t ) | = | y 1 ( t ) + β − y 1 ( b ) y 2 ( b ) y 2 ( t ) − s y 1 ( t ) − β − s y 1 ( b ) s y 2 ( b ) ⋅ s y 2 ( t ) | ≤ ≤ | y 1 ( t ) − s y 1 ( t ) | + | β − y 1 ( b ) y 2 ( b ) y 2 ( t ) − β − s y 1 ( b ) s y 2 ( b ) ⋅ s y 2 ( t ) | for every
t
∈
[
a
,
b
]
t
∈
[
a
,
b
]
t in[a,b] t \in[a, b] t ∈ [ a , b ] , where
s
y
1
s
y
1
s_(y_(1)) s_{y_{1}} s y 1 and
s
y
2
s
y
2
s_(y_(2)) s_{y_{2}} s y 2 are determined by the conditions (**).
Using (28), we obtain
β
−
y
1
(
b
)
y
2
(
b
)
=
β
−
s
y
1
(
b
)
s
y
2
(
b
)
+
O
(
‖
Δ
n
‖
3
/
2
)
,
β
−
y
1
(
b
)
y
2
(
b
)
=
β
−
s
y
1
(
b
)
s
y
2
(
b
)
+
O
Δ
n
3
/
2
,
(beta-y_(1)(b))/(y_(2)(b))=(beta-s_(y_(1))(b))/(s_(y_(2))(b))+O(||Delta_(n)||^(3//2)), \frac{\beta-y_{1}(b)}{y_{2}(b)}=\frac{\beta-s_{y_{1}}(b)}{s_{y_{2}}(b)}+O\left(\left\|\Delta_{n}\right\|^{3 / 2}\right), β − y 1 ( b ) y 2 ( b ) = β − s y 1 ( b ) s y 2 ( b ) + O ( ‖ Δ n ‖ 3 / 2 ) , showing that
‖
y
(
t
)
−
s
y
(
t
)
‖
=
O
(
‖
Δ
n
‖
3
/
2
)
.
y
(
t
)
−
s
y
(
t
)
=
O
Δ
n
3
/
2
.
||y(t)-s_(y)(t)||=O(||Delta_(n)||^(3//2)). \left\|y(t)-s_{y}(t)\right\|=O\left(\left\|\Delta_{n}\right\|^{3 / 2}\right) . ‖ y ( t ) − s y ( t ) ‖ = O ( ‖ Δ n ‖ 3 / 2 ) . a) The approximative determination of the solution
s
y
1
s
y
1
s_(y_(1)) s_{y_{1}} s y 1 on the nodes of the division
Δ
n
Δ
n
Delta_(n) \Delta_{n} Δ n Representation (9) yields
s
y
1
(
t
)
=
u
0
(
t
)
⋅
α
+
∑
k
=
0
n
U
k
(
t
)
⋅
y
1
′
′
(
t
k
)
=
=
u
0
(
t
)
⋅
α
+
∑
k
=
0
n
U
k
(
t
)
[
p
(
t
k
)
⋅
y
1
(
t
k
)
+
q
(
t
k
)
]
.
s
y
1
(
t
)
=
u
0
(
t
)
⋅
α
+
∑
k
=
0
n
U
k
(
t
)
⋅
y
1
′
′
t
k
=
=
u
0
(
t
)
⋅
α
+
∑
k
=
0
n
U
k
(
t
)
p
t
k
⋅
y
1
t
k
+
q
t
k
.
{:[s_(y_(1))(t)=u_(0)(t)*alpha+sum_(k=0)^(n)U_(k)(t)*y_(1)^('')(t_(k))=],[=u_(0)(t)*alpha+sum_(k=0)^(n)U_(k)(t)[p(t_(k))*y_(1)(t_(k))+q(t_(k))].]:} \begin{aligned}
& s_{y_{1}}(t)=u_{0}(t) \cdot \alpha+\sum_{k=0}^{n} U_{k}(t) \cdot y_{1}^{\prime \prime}\left(t_{k}\right)= \\
= & u_{0}(t) \cdot \alpha+\sum_{k=0}^{n} U_{k}(t)\left[p\left(t_{k}\right) \cdot y_{1}\left(t_{k}\right)+q\left(t_{k}\right)\right] .
\end{aligned} s y 1 ( t ) = u 0 ( t ) ⋅ α + ∑ k = 0 n U k ( t ) ⋅ y 1 ′ ′ ( t k ) = = u 0 ( t ) ⋅ α + ∑ k = 0 n U k ( t ) [ p ( t k ) ⋅ y 1 ( t k ) + q ( t k ) ] . Letting
v
i
:=
s
y
i
(
t
i
)
,
i
=
0
,
n
―
e
i
:=
y
1
(
t
i
)
−
s
y
1
(
t
i
)
,
i
=
0
,
n
―
v
i
:=
s
y
i
t
i
,
i
=
0
,
n
¯
e
i
:=
y
1
t
i
−
s
y
1
t
i
,
i
=
0
,
n
¯
{:[v_(i):=s_(y_(i))(t_(i))","quad i= bar(0,n)],[e_(i):=y_(1)(t_(i))-s_(y_(1))(t_(i))","quad i= bar(0,n)]:} \begin{gathered}
v_{i}:=s_{y_{i}}\left(t_{i}\right), \quad i=\overline{0, n} \\
e_{i}:=y_{1}\left(t_{i}\right)-s_{y_{1}}\left(t_{i}\right), \quad i=\overline{0, n}
\end{gathered} v i := s y i ( t i ) , i = 0 , n ― e i := y 1 ( t i ) − s y 1 ( t i ) , i = 0 , n ― one obtains the system
s
y
1
(
t
i
)
=
u
0
(
t
i
)
α
+
∑
k
=
0
n
U
k
(
t
i
)
[
p
(
t
k
)
(
e
k
+
v
k
)
+
q
(
t
k
)
]
=
=
u
0
(
t
i
)
α
+
∑
k
=
0
n
U
k
(
t
i
)
[
p
(
t
k
)
v
k
+
q
(
t
k
)
]
+
o
(
‖
Δ
n
‖
3
/
2
)
,
i
=
0
,
n
―
s
y
1
t
i
=
u
0
t
i
α
+
∑
k
=
0
n
U
k
t
i
p
t
k
e
k
+
v
k
+
q
t
k
=
=
u
0
t
i
α
+
∑
k
=
0
n
U
k
t
i
p
t
k
v
k
+
q
t
k
+
o
Δ
n
3
/
2
,
i
=
0
,
n
¯
{:[s_(y_(1))(t_(i))=u_(0)(t_(i))alpha+sum_(k=0)^(n)U_(k)(t_(i))[p(t_(k))(e_(k)+v_(k))+q(t_(k))]=],[=u_(0)(t_(i))alpha+sum_(k=0)^(n)U_(k)(t_(i))[p(t_(k))v_(k)+q(t_(k))]+o(||Delta_(n)||^(3//2))","],[i= bar(0,n)]:} \begin{aligned}
s_{y_{1}}\left(t_{i}\right)= & u_{0}\left(t_{i}\right) \alpha+\sum_{k=0}^{n} U_{k}\left(t_{i}\right)\left[p\left(t_{k}\right)\left(e_{k}+v_{k}\right)+q\left(t_{k}\right)\right]= \\
& =u_{0}\left(t_{i}\right) \alpha+\sum_{k=0}^{n} U_{k}\left(t_{i}\right)\left[p\left(t_{k}\right) v_{k}+q\left(t_{k}\right)\right]+o\left(\left\|\Delta_{n}\right\|^{3 / 2}\right), \\
i & =\overline{0, n}
\end{aligned} s y 1 ( t i ) = u 0 ( t i ) α + ∑ k = 0 n U k ( t i ) [ p ( t k ) ( e k + v k ) + q ( t k ) ] = = u 0 ( t i ) α + ∑ k = 0 n U k ( t i ) [ p ( t k ) v k + q ( t k ) ] + o ( ‖ Δ n ‖ 3 / 2 ) , i = 0 , n ― The approximative values of the spline solution
s
y
1
s
y
1
s_(y_(1)) s_{y_{1}} s y 1 on the nodes of
Δ
n
Δ
n
Delta_(n) \Delta_{n} Δ n are the solutions
ν
κ
ν
κ
nu_(kappa) \nu_{\kappa} ν κ of the linear system
v
i
=
u
0
(
t
i
)
α
+
∑
k
=
0
n
U
k
(
t
i
)
[
p
(
t
k
)
v
k
+
q
(
t
k
)
]
,
i
=
0
,
n
―
v
i
=
u
0
t
i
α
+
∑
k
=
0
n
U
k
t
i
p
t
k
v
k
+
q
t
k
,
i
=
0
,
n
¯
v_(i)=u_(0)(t_(i))alpha+sum_(k=0)^(n)U_(k)(t_(i))[p(t_(k))v_(k)+q(t_(k))],quad i= bar(0,n) v_{i}=u_{0}\left(t_{i}\right) \alpha+\sum_{k=0}^{n} U_{k}\left(t_{i}\right)\left[p\left(t_{k}\right) v_{k}+q\left(t_{k}\right)\right], \quad i=\overline{0, n} v i = u 0 ( t i ) α + ∑ k = 0 n U k ( t i ) [ p ( t k ) v k + q ( t k ) ] , i = 0 , n ― b) The approximative determination of the solution
s
y
2
s
y
2
s_(y_(2)) s_{y_{2}} s y 2 on the nodes of
Δ
n
Δ
n
Delta_(n) \Delta_{n} Δ n Using again representation (9), one obtains
s
y
2
(
t
)
=
u
1
(
t
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
y
2
′
′
(
t
k
)
=
=
u
1
(
t
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
p
(
t
k
)
⋅
y
2
(
t
k
)
.
s
y
2
(
t
)
=
u
1
(
t
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
y
2
′
′
t
k
=
=
u
1
(
t
)
+
∑
k
=
0
n
U
k
(
t
)
⋅
p
t
k
⋅
y
2
t
k
.
{:[s_(y_(2))(t)=u_(1)(t)+sum_(k=0)^(n)U_(k)(t)*y_(2)^('')(t_(k))=],[=u_(1)(t)+sum_(k=0)^(n)U_(k)(t)*p(t_(k))*y_(2)(t_(k)).]:} \begin{aligned}
& s_{y_{2}}(t)=u_{1}(t)+\sum_{k=0}^{n} U_{k}(t) \cdot y_{2}^{\prime \prime}\left(t_{k}\right)= \\
& =u_{1}(t)+\sum_{k=0}^{n} U_{k}(t) \cdot p\left(t_{k}\right) \cdot y_{2}\left(t_{k}\right) .
\end{aligned} s y 2 ( t ) = u 1 ( t ) + ∑ k = 0 n U k ( t ) ⋅ y 2 ′ ′ ( t k ) = = u 1 ( t ) + ∑ k = 0 n U k ( t ) ⋅ p ( t k ) ⋅ y 2 ( t k ) . Letting
w
i
:=
s
y
2
(
t
i
)
,
i
=
0
,
n
―
e
¯
i
:=
y
2
(
t
i
)
−
s
y
2
(
t
i
)
,
i
=
0
,
n
―
w
i
:=
s
y
2
t
i
,
i
=
0
,
n
¯
e
¯
i
:=
y
2
t
i
−
s
y
2
t
i
,
i
=
0
,
n
¯
{:[w_(i):=s_(y_(2))(t_(i))","quad i= bar(0,n)],[ bar(e)_(i):=y_(2)(t_(i))-s_(y_(2))(t_(i))","quad i= bar(0,n)]:} \begin{gathered}
w_{i}:=s_{y_{2}}\left(t_{i}\right), \quad i=\overline{0, n} \\
\bar{e}_{i}:=y_{2}\left(t_{i}\right)-s_{y_{2}}\left(t_{i}\right), \quad i=\overline{0, n}
\end{gathered} w i := s y 2 ( t i ) , i = 0 , n ― e ¯ i := y 2 ( t i ) − s y 2 ( t i ) , i = 0 , n ― it follows that
w
i
w
i
w_(i) w_{i} w i are the solutions of the system
w
i
=
u
1
(
t
i
)
+
∑
k
=
0
n
U
k
(
t
i
)
p
(
t
k
)
w
k
+
O
(
‖
Δ
n
‖
3
/
2
)
w
i
=
u
1
t
i
+
∑
k
=
0
n
U
k
t
i
p
t
k
w
k
+
O
Δ
n
3
/
2
w_(i)=u_(1)(t_(i))+sum_(k=0)^(n)U_(k)(t_(i))p(t_(k))w_(k)+O(||Delta_(n)||^(3//2)) w_{i}=u_{1}\left(t_{i}\right)+\sum_{k=0}^{n} U_{k}\left(t_{i}\right) p\left(t_{k}\right) w_{k}+O\left(\left\|\Delta_{n}\right\|^{3 / 2}\right) w i = u 1 ( t i ) + ∑ k = 0 n U k ( t i ) p ( t k ) w k + O ( ‖ Δ n ‖ 3 / 2 ) Therefore, the approximative values of
s
y
2
(
t
i
)
s
y
2
t
i
s_(y_(2))(t_(i)) s_{y_{2}}\left(t_{i}\right) s y 2 ( t i ) can be obtained from the linear system
(30)
w
i
=
u
1
(
t
i
)
+
∑
k
=
0
n
U
k
(
t
i
)
p
(
t
k
)
⋅
w
k
,
i
=
0
,
n
―
(30)
w
i
=
u
1
t
i
+
∑
k
=
0
n
U
k
t
i
p
t
k
⋅
w
k
,
i
=
0
,
n
¯
{:(30)w_(i)=u_(1)(t_(i))+sum_(k=0)^(n)U_(k)(t_(i))p(t_(k))*w_(k)","i= bar(0,n):} \begin{equation*}
w_{i}=u_{1}\left(t_{i}\right)+\sum_{k=0}^{n} U_{k}\left(t_{i}\right) p\left(t_{k}\right) \cdot w_{k}, i=\overline{0, n} \tag{30}
\end{equation*} (30) w i = u 1 ( t i ) + ∑ k = 0 n U k ( t i ) p ( t k ) ⋅ w k , i = 0 , n ― The approximative values of the spline solution
s
y
∈
S
5
(
Λ
n
)
s
y
∈
S
5
Λ
n
s_(y)inS_(5)(Lambda_(n)) s_{y} \in S_{5}\left(\Lambda_{n}\right) s y ∈ S 5 ( Λ n ) on the nodes of the division
Δ
n
Δ
n
Delta_(n) \Delta_{n} Δ n are given by
(31)
s
y
(
t
i
)
=
v
i
+
β
−
v
n
w
n
w
i
,
i
=
0
,
n
―
s
y
t
i
=
v
i
+
β
−
v
n
w
n
w
i
,
i
=
0
,
n
¯
s_(y)(t_(i))=v_(i)+(beta-v_(n))/(w_(n))w_(i),quad i= bar(0,n) s_{y}\left(t_{i}\right)=v_{i}+\frac{\beta-v_{n}}{w_{n}} w_{i}, \quad i=\overline{0, n} s y ( t i ) = v i + β − v n w n w i , i = 0 , n ― A numerical example. The problem
(D)
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
y^('')=4y,t in[0,1] y^{\prime \prime}=4 y, t \in[0,1] y ′ ′ = 4 y , t ∈ [ 0 , 1 ]
y
(
0
)
=
1
,
y
(
1
)
=
e
−
2
y
(
0
)
=
1
,
y
(
1
)
=
e
−
2
y(0)=1,y(1)=e^(-2) y(0)=1, y(1)=e^{-2} y ( 0 ) = 1 , y ( 1 ) = e − 2
has the exact solution
y
=
e
−
2
t
y
=
e
−
2
t
y=e^(-2t) y=e^{-2 t} y = e − 2 t .
The associated Cauchy problems are
(
C
1
)
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
C
1
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
(C_(1))y^('')=4y,t in[0,1] \left(C_{1}\right) y^{\prime \prime}=4 y, t \in[0,1] ( C 1 ) y ′ ′ = 4 y , t ∈ [ 0 , 1 ]
y
(
0
)
=
1
,
y
′
(
0
)
=
0
y
(
0
)
=
1
,
y
′
(
0
)
=
0
y(0)=1,y^(')(0)=0 y(0)=1, y^{\prime}(0)=0 y ( 0 ) = 1 , y ′ ( 0 ) = 0
(
C
2
C
2
C_(2) C_{2} C 2 )
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
y
′
′
=
4
y
,
t
∈
[
0
,
1
]
y^('')=4y,t in[0,1] y^{\prime \prime}=4 y, t \in[0,1] y ′ ′ = 4 y , t ∈ [ 0 , 1 ]
y
(
0
)
=
0
,
y
′
(
0
)
=
1
y
(
0
)
=
0
,
y
′
(
0
)
=
1
y(0)=0,y^(')(0)=1 y(0)=0, y^{\prime}(0)=1 y ( 0 ) = 0 , y ′ ( 0 ) = 1
and have the exact solutions
y
1
(
t
)
=
1
2
[
e
2
t
+
e
−
2
t
]
y
2
(
t
)
=
1
4
[
e
2
t
−
e
−
2
t
]
.
y
1
(
t
)
=
1
2
e
2
t
+
e
−
2
t
y
2
(
t
)
=
1
4
e
2
t
−
e
−
2
t
.
{:[y_(1)(t)=(1)/(2)[e^(2t)+e^(-2t)]],[y_(2)(t)=(1)/(4)[e^(2t)-e^(-2t)].]:} \begin{aligned}
& y_{1}(t)=\frac{1}{2}\left[e^{2 t}+e^{-2 t}\right] \\
& y_{2}(t)=\frac{1}{4}\left[e^{2 t}-e^{-2 t}\right] .
\end{aligned} y 1 ( t ) = 1 2 [ e 2 t + e − 2 t ] y 2 ( t ) = 1 4 [ e 2 t − e − 2 t ] . For
n
=
5
n
=
5
n=5 n=5 n = 5 , let
Δ
5
:=
{
t
0
=
0
,
t
1
=
0.2
,
t
2
=
0.4
,
t
3
=
0.6
,
t
4
=
0.8
,
t
5
=
1
}
Δ
5
:=
t
0
=
0
,
t
1
=
0.2
,
t
2
=
0.4
,
t
3
=
0.6
,
t
4
=
0.8
,
t
5
=
1
Delta_(5):={t_(0)=0,t_(1)=0.2,t_(2)=0.4,t_(3)=0.6,t_(4)=0.8,t_(5)=1} \Delta_{5}:=\left\{t_{0}=0, t_{1}=0.2, t_{2}=0.4, t_{3}=0.6, t_{4}=0.8, t_{5}=1\right\} Δ 5 := { t 0 = 0 , t 1 = 0.2 , t 2 = 0.4 , t 3 = 0.6 , t 4 = 0.8 , t 5 = 1 } .
Using representation (1), one obtains Table 1 for the coefficients of
s
y
1
s
y
1
s_(y_(1)) s_{y_{1}} s y 1 and
s
y
2
s
y
2
s_(y_(2)) s_{y_{2}} s y 2 .
Table 1
n
=
5
n
=
5
n=5 n=5 n = 5
s
y
1
s
y
1
s_(y_(1)) s_{y_{1}} s y 1
s
y
2
s
y
2
s_(y_(2)) s_{y_{2}} s y 2
A
0
A
0
A_(0) A_{0} A 0 1
0
A
1
A
1
A_(1) A_{1} A 1 0
1
A
2
A
2
A_(2) A_{2} A 2 2
0
A
3
A
3
A_(3) A_{3} A 3 0.1576268148
0.6678202118
a
0
a
0
a_(0) a_{0} a 0 0.8446081893
0.1257524863
a
1
a
1
a_(1) a_{1} a 1 -0.6853940844
0.07800170745
a
2
a
2
a_(2) a_{2} a 2 -0.6014825811
-0.1770186129
a
3
a
3
a_(3) a_{3} a 3 3.020199906
0.7804651905
a
4
a
4
a_(4) a_{4} a 4 -5.717416684
-1.970643804
a
5
a
5
a_(5) a_{5} a 5 3.139485253
1.163443033
n=5 s_(y_(1)) s_(y_(2))
A_(0) 1 0
A_(1) 0 1
A_(2) 2 0
A_(3) 0.1576268148 0.6678202118
a_(0) 0.8446081893 0.1257524863
a_(1) -0.6853940844 0.07800170745
a_(2) -0.6014825811 -0.1770186129
a_(3) 3.020199906 0.7804651905
a_(4) -5.717416684 -1.970643804
a_(5) 3.139485253 1.163443033 | $n=5$ | $s_{y_{1}}$ | $s_{y_{2}}$ |
| :--- | :--- | :--- |
| $A_{0}$ | 1 | 0 |
| $A_{1}$ | 0 | 1 |
| $A_{2}$ | 2 | 0 |
| $A_{3}$ | 0.1576268148 | 0.6678202118 |
| $a_{0}$ | 0.8446081893 | 0.1257524863 |
| $a_{1}$ | -0.6853940844 | 0.07800170745 |
| $a_{2}$ | -0.6014825811 | -0.1770186129 |
| $a_{3}$ | 3.020199906 | 0.7804651905 |
| $a_{4}$ | -5.717416684 | -1.970643804 |
| $a_{5}$ | 3.139485253 | 1.163443033 |
For the values of
s
y
s
y
s_(y) s_{y} s y on the nodes of
Δ
5
Δ
5
Delta_(5) \Delta_{5} Δ 5 , one uses
s
y
(
t
i
)
=
s
y
1
(
t
i
)
+
e
−
2
−
s
y
1
(
1
)
s
y
2
(
1
)
s
y
2
(
t
i
)
,
i
=
0
,
5
―
s
y
t
i
=
s
y
1
t
i
+
e
−
2
−
s
y
1
(
1
)
s
y
2
(
1
)
s
y
2
t
i
,
i
=
0
,
5
¯
s_(y)(t_(i))=s_(y_(1))(t_(i))+(e^(-2)-s_(y_(1))(1))/(s_(y_(2))(1))s_(y_(2))(t_(i)),i= bar(0,5) s_{y}\left(t_{i}\right)=s_{y_{1}}\left(t_{i}\right)+\frac{e^{-2}-s_{y_{1}}(1)}{s_{y_{2}}(1)} s_{y_{2}}\left(t_{i}\right), i=\overline{0,5} s y ( t i ) = s y 1 ( t i ) + e − 2 − s y 1 ( 1 ) s y 2 ( 1 ) s y 2 ( t i ) , i = 0 , 5 ― Table 2 contains the values
s
y
(
t
i
)
,
i
=
0
,
5
―
s
y
t
i
,
i
=
0
,
5
¯
s_(y)(t_(i)),i= bar(0,5) s_{y}\left(t_{i}\right), i=\overline{0,5} s y ( t i ) , i = 0 , 5 ― , and the errors
E
i
=
|
y
(
t
i
)
−
s
y
(
t
i
)
|
,
i
=
0
,
5
―
E
i
=
y
t
i
−
s
y
t
i
,
i
=
0
,
5
¯
E_(i)=|y(t_(i))-s_(y)(t_(i))|,quad i= bar(0,5) E_{i}=\left|y\left(t_{i}\right)-s_{y}\left(t_{i}\right)\right|, \quad i=\overline{0,5} E i = | y ( t i ) − s y ( t i ) | , i = 0 , 5 ―
Table 2
t
i
t
i
t_(i) \boldsymbol{t}_{i} t i
s
y
(
t
i
)
s
y
t
i
s_(y)(t_(i)) \boldsymbol{s}_{y}\left(\boldsymbol{t}_{i}\right) s y ( t i )
E
i
E
i
E_(i) \boldsymbol{E}_{i} E i
0
1
0
0.2
0.6708587727
0.5387267
⋅
10
−
3
0.5387267
⋅
10
−
3
0.5387267*10^(-3) 0.5387267 \cdot 10^{-3} 0.5387267 ⋅ 10 − 3
0.4
0.4506125215
0.12835574
⋅
10
−
2
0.12835574
⋅
10
−
2
0.12835574*10^(-2) 0.12835574 \cdot 10^{-2} 0.12835574 ⋅ 10 − 2
0.6
0.303315766
0.21215541
⋅
10
−
2
0.21215541
⋅
10
−
2
0.21215541*10^(-2) 0.21215541 \cdot 10^{-2} 0.21215541 ⋅ 10 − 2
0.8
0.204249434
0.2352916
⋅
10
−
2
0.2352916
⋅
10
−
2
0.2352916*10^(-2) 0.2352916 \cdot 10^{-2} 0.2352916 ⋅ 10 − 2
1
0.135335284
0.8
⋅
10
−
9
0.8
⋅
10
−
9
0.8*10^(-9) 0.8 \cdot 10^{-9} 0.8 ⋅ 10 − 9
t_(i) s_(y)(t_(i)) E_(i)
0 1 0
0.2 0.6708587727 0.5387267*10^(-3)
0.4 0.4506125215 0.12835574*10^(-2)
0.6 0.303315766 0.21215541*10^(-2)
0.8 0.204249434 0.2352916*10^(-2)
1 0.135335284 0.8*10^(-9) | $\boldsymbol{t}_{i}$ | $\boldsymbol{s}_{y}\left(\boldsymbol{t}_{i}\right)$ | $\boldsymbol{E}_{i}$ |
| :---: | :---: | :---: |
| 0 | 1 | 0 |
| 0.2 | 0.6708587727 | $0.5387267 \cdot 10^{-3}$ |
| 0.4 | 0.4506125215 | $0.12835574 \cdot 10^{-2}$ |
| 0.6 | 0.303315766 | $0.21215541 \cdot 10^{-2}$ |
| 0.8 | 0.204249434 | $0.2352916 \cdot 10^{-2}$ |
| 1 | 0.135335284 | $0.8 \cdot 10^{-9}$ |
REFERENCES
P. Blaga and G. Micula, Polynomial natural spline of even degree, Studia Univ. "BabesBolyai", Mathematica 38, 2 (1993), 31-40.
P. Blaga, R. Gorenflo and G. Micula, Even degree spline technique for numerical solution of delay differential equations, Freic Universität Berlin, Preprint No. A-15 (1996), Serie A-Mathematik.
R. L. Burden and T. Douglas Faires, Numerical Analysis, Third Edition, PWS-KENT Publishing Company, Boston, 1985.
G. Micula, P. Blaga and M. Micula, On even degree polynomial spline functions with applications to numerical solution of differential equations with retarded argument, Technische Hochschule Darmstadt, Preprint No. 1771, Fachbereich Mathematik (1995).
Received May 15, 1996
3400 Cluj-Napoca, 1
