On a problem of B.A. Karpilovskaya

Costica Mustata
(original), Approximation Theory, ODEs (numerical), paper, splines

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Costica Mustata
“Tiberiu Popoviciu” Institute of Numerical Analysis, Romanian Academy, Romania

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C. Mustăţa, On a problem of B.A. Karpilovskaya, Rev. Anal. Numér. Théor. Approx. 29 (1999) no. 2, pp. 179-189.

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Revue d’Analyse Numer. Theor. Approx.

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https://ictp.acad.ro/jnaat/journal/article/view/1999-vol28-no2-art8

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[7] B.E. Karpilovskaja, The convergence of a method of interpolation for differential equations (russian), U.M.N.t. VIII, 3 (1953), pp.111-118.
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1999-Mustata-On a problem of B.A. Karpilovskaya,-Jnaat

ON A PROBLEM OF B. A. KARPILOVSKAJA

COSTICĂ MUSTĂTA

In [7] one consider the following problem:
(1) y ( 2 p ) ( t ) − φ 1 ( t ) y ( 2 p − 1 ) ( t ) − ⋯ , − φ 2 p ( t ) y ( t ) = f ( t ) , t ∈ [ a , b ] (1) y ( 2 p ) ( t ) − φ 1 ( t ) y ( 2 p − 1 ) ( t ) − ⋯ , − φ 2 p ( t ) y ( t ) = f ( t ) , t ∈ [ a , b ] {:(1)y^((2p))(t)-varphi_(1)(t)y^((2p-1))(t)-cdots","-varphi_(2p)(t)y(t)=f(t)","quad t in[a","b]:}\begin{equation*} y^{(2 p)}(t)-\varphi_{1}(t) y^{(2 p-1)}(t)-\cdots,-\varphi_{2 p}(t) y(t)=f(t), \quad t \in[a, b] \tag{1} \end{equation*}(1)y(2p)(t)−φ1(t)y(2p−1)(t)−⋯,−φ2p(t)y(t)=f(t),t∈[a,b]
(2) y ( q ) ( a ) = y ( q ) ( b ) = 0 , q = 0 , 1 , 2 , … , p − 1 (2) y ( q ) ( a ) = y ( q ) ( b ) = 0 , q = 0 , 1 , 2 , … , p − 1 {:(2)y^((q))(a)=y^((q))(b)=0","quad q=0","1","2","dots","p-1:}\begin{equation*} y^{(q)}(a)=y^{(q)}(b)=0, \quad q=0,1,2, \ldots, p-1 \tag{2} \end{equation*}(2)y(q)(a)=y(q)(b)=0,q=0,1,2,…,p−1
where p ∈ N , p ≥ 1 p ∈ N , p ≥ 1 p inN,p >= 1p \in \mathbb{N}, p \geq 1p∈N,p≥1. In the same paper one determines an approximate solution of the form
(3) y ¯ ( t ) = ( t − a ) ′ ′ ( t − b ) ′ ′ ⋅ ∑ k = 1 n c k t k − 1 , t ∈ [ a , b ] (3) y ¯ ( t ) = ( t − a ) ′ ′ ( t − b ) ′ ′ ⋅ ∑ k = 1 n   c k t k − 1 , t ∈ [ a , b ] {:(3) bar(y)(t)=(t-a)^('')(t-b)^('')*sum_(k=1)^(n)c_(k)t^(k-1)","t in[a","b]:}\begin{equation*} \bar{y}(t)=(t-a)^{\prime \prime}(t-b)^{\prime \prime} \cdot \sum_{k=1}^{n} c_{k} t^{k-1}, t \in[a, b] \tag{3} \end{equation*}(3)y¯(t)=(t−a)′′(t−b)′′⋅∑k=1ncktk−1,t∈[a,b]
where the coefficients c k , k = 1 , 2 , … , n c k , k = 1 , 2 , … , n c_(k),k=1,2,dots,nc_{k}, k=1,2, \ldots, nck,k=1,2,…,n are determined from the system of equations:
(4) y ¯ ( 2 p ) ( t i ) − φ 1 ( t i ) y ¯ ( 2 p − 1 ) ( t i ) − ⋯ − φ 2 p ( t i ) y ¯ ( t i ) = f ( t i ) , i = 1 , 2 , … , n , (4) y ¯ ( 2 p ) t i − φ 1 t i y ¯ ( 2 p − 1 ) t i − ⋯ − φ 2 p t i y ¯ t i = f t i , i = 1 , 2 , … , n , {:(4) bar(y)^((2p))(t_(i))-varphi_(1)(t_(i)) bar(y)^((2p-1))(t_(i))-cdots-varphi_(2p)(t_(i)) bar(y)(t_(i))=f(t_(i))","quad i=1","2","dots","n",":}\begin{equation*} \bar{y}^{(2 p)}\left(t_{i}\right)-\varphi_{1}\left(t_{i}\right) \bar{y}^{(2 p-1)}\left(t_{i}\right)-\cdots-\varphi_{2 p}\left(t_{i}\right) \bar{y}\left(t_{i}\right)=f\left(t_{i}\right), \quad i=1,2, \ldots, n, \tag{4} \end{equation*}(4)y¯(2p)(ti)−φ1(ti)y¯(2p−1)(ti)−⋯−φ2p(ti)y¯(ti)=f(ti),i=1,2,…,n,
where t i , i = l , n ― t i , i = l , n ¯ t_(i),i= bar(l,n)t_{i}, i=\overline{l, n}ti,i=l,n― are the nodes of a partition
(5) Δ n ′ := a < t 1 < t 2 < ⋯ < t n < b (5) Δ n ′ := a < t 1 < t 2 < ⋯ < t n < b {:(5)Delta_(n)^('):=a < t_(1) < t_(2) < cdots < t_(n) < b:}\begin{equation*} \Delta_{n}^{\prime}:=a<t_{1}<t_{2}<\cdots<t_{n}<b \tag{5} \end{equation*}(5)Δn′:=a<t1<t2<⋯<tn<b
of the interval [ a , b ] [ a , b ] [a,b][a, b][a,b].
In the case when the nodes of the partition Δ n ′ Δ n ′ Delta_(n)^(')\Delta_{n}^{\prime}Δn′ are the roots of the Chebyshev polynomial it is given an upper delimitation of the norm ‖ y − y ¯ ‖ ∞ ‖ y − y ¯ ‖ ∞ ||y- bar(y)||_(oo)\|y-\bar{y}\|_{\infty}‖y−y¯‖∞, where y y yyy is the exact solution of the problem (1)-(2). From this delimitation it follows that the order of approximation of the exact solution by the functions y → y → vec(y)\vec{y}y→ given by (3) is O ( ln ⁡ n n ) O ln ⁡ n n O((ln n)/(n))O\left(\frac{\ln n}{n}\right)O(ln⁡nn).
In the following, taking as an approximant of the exact solution of the problem (1)-(2) a spline function belonging to the space S 2 m + 2 p − 1 ( Δ n ) S 2 m + 2 p − 1 Δ n S_(2m+2p-1)(Delta_(n))S_{2 m+2 p-1}\left(\Delta_{n}\right)S2m+2p−1(Δn) of 2 p 2 p 2p2 p2p-derivative-interpolating spline functions, defined in [9], one proves that the
order of approximation is at least O ( 1 n n ) O 1 n n O((1)/(nsqrtn))O\left(\frac{1}{n \sqrt{n}}\right)O(1nn).
DEFINITION 1. Let m , n , p ∈ N , n ≥ 2 , p ≥ 1 , m ≥ 2 , m + p ≤ n + 1 m , n , p ∈ N , n ≥ 2 , p ≥ 1 , m ≥ 2 , m + p ≤ n + 1 m,n,p inN,n >= 2,p >= 1,m >= 2,m+p <= n+1m, n, p \in \mathbb{N}, n \geq 2, p \geq 1, m \geq 2, m+p \leq n+1m,n,p∈N,n≥2,p≥1,m≥2,m+p≤n+1 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) < cdots < t_(n)=b < t_(n+1)=+oo\Delta_{n}:=-\infty=t_{-1}<a=t_{0}<t_{1}<\cdots<t_{n}=b<t_{n+1}=+\inftyΔn:=−∞=t−1<a=t0<t1<⋯<tn=b<tn+1=+∞
be a partition of the interval [ a , b ] [ a , b ] [a,b][a, b][a,b].
A function s : R → R s : R → R s:RrarrRs: \mathbb{R} \rightarrow \mathbb{R}s:R→R satisfying the conditions
1 0 s ∈ C 2 m + p − 2 ( R ) 1 0 s ∈ C 2 m + p − 2 ( R ) 1^(0)s inC^(2m+p-2)(R)1^{0} s \in C^{2 m+p-2}(\mathbb{R})10s∈C2m+p−2(R);
(1) 2 0 s | I k ∈ R 2 m + p − 1 , I k = ( t k − 1 , t k ) , k = 1 , 2 , … , n ; (1) 2 0 s I k ∈ R 2 m + p − 1 , I k = t k − 1 , t k , k = 1 , 2 , … , n ; {:(1)2^(0)s|_(I_(k))inR_(2m+p-1)","I_(k)=(t_(k-1),t_(k))","k=1","2","dots","n;:}\begin{equation*} \left.2^{0} s\right|_{I_{k}} \in \mathscr{R}_{2 m+p-1}, I_{k}=\left(t_{k-1}, t_{k}\right), k=1,2, \ldots, n ; \tag{1} \end{equation*}(1)20s|Ik∈R2m+p−1,Ik=(tk−1,tk),k=1,2,…,n;
3 0 s | I n ∈ B m + p − 1 , I 0 = [ t − 1 , t i ) , I n + 1 = [ t n , t n + 1 ) 3 0 s I n ∈ B m + p − 1 , I 0 = t − 1 , t i , I n + 1 = t n , t n + 1 3^(0)s|_(I_(n))inB_(m+p-1),I_(0)=[t_(-1),t_(i)),I_(n+1)=[t_(n),t_(n+1))\left.3^{0} s\right|_{I_{n}} \in \mathscr{B}_{m+p-1}, I_{0}=\left[t_{-1}, t_{i}\right), I_{n+1}=\left[t_{n}, t_{n+1}\right)30s|In∈Bm+p−1,I0=[t−1,ti),In+1=[tn,tn+1),
is called a natural spline function of degree 2 m + p − 1 2 m + p − 1 2m+p-12 m+p-12m+p−1.
Here P r ( r ∈ N ) P r ( r ∈ N ) P_(r)(r inN)\mathscr{P}_{r}(r \in \mathbb{N})Pr(r∈N) stands for the set of polynomials of degree at most r r rrr.
Denoting by S 2 m + p − 1 ( Δ n ) S 2 m + p − 1 Δ n S_(2m+p-1)(Delta_(n))S_{2 m+p-1}\left(\Delta_{n}\right)S2m+p−1(Δn) the set of all functions verifying the conditions 1 0 − 3 0 1 0 − 3 0 1^(0)-3^(0)1^{0}-3^{0}10−30 from Definition 1, one sees that each s ∈ S 2 m + p − 1 ( Δ n ) s ∈ S 2 m + p − 1 Δ n s inS_(2m+p-1)(Delta_(n))s \in S_{2 m+p-1}\left(\Delta_{n}\right)s∈S2m+p−1(Δn) admits a representation of the form.
(97202 bry in
(6) s ( t ) = ∑ i = 0 m + p − 1 A i t i + ∑ k = 0 n a k ( t − t k ) + 2 m + p − 1 , t ∈ R (6) s ( t ) = ∑ i = 0 m + p − 1   A i t i + ∑ k = 0 n   a k t − t k + 2 m + p − 1 , t ∈ R {:(6)s(t)=sum_(i=0)^(m+p-1)A_(i)t^(i)+sum_(k=0)^(n)a_(k)(t-t_(k))_(+)^(2m+p-1)","t inR:}\begin{equation*} s(t)=\sum_{i=0}^{m+p-1} A_{i} t^{i}+\sum_{k=0}^{n} a_{k}\left(t-t_{k}\right)_{+}^{2 m+p-1}, t \in \mathbb{R} \tag{6} \end{equation*}(6)s(t)=∑i=0m+p−1Aiti+∑k=0nak(t−tk)+2m+p−1,t∈R
where
(7) ∑ k = 0 n a k t k j = 0 , j = 0 , 1 , 2 , … , m − 1 (7) ∑ k = 0 n   a k t k j = 0 , j = 0 , 1 , 2 , … , m − 1 {:(7)sum_(k=0)^(n)a_(k)t_(k)^(j)=0","quad j=0","1","2","dots","m-1:}\begin{equation*} \sum_{k=0}^{n} a_{k} t_{k}^{j}=0, \quad j=0,1,2, \ldots, m-1 \tag{7} \end{equation*}(7)∑k=0naktkj=0,j=0,1,2,…,m−1
and
(8) ( t − t k ) + = { 0 , if t ≤ t k , t − t k , if t > t k , t ∈ [ a , b ] . (8) t − t k + = 0 ,  if  t ≤ t k , t − t k ,  if  t > t k , t ∈ [ a , b ] . {:(8)(t-t_(k))_(+)={[0","," if "t <= t_(k)","],[t-t_(k)","," if "t > t_(k)","t in[a","b].]:}:}\left(t-t_{k}\right)_{+}=\left\{\begin{array}{cc} 0, & \text { if } t \leq t_{k}, \tag{8}\\ t-t_{k}, & \text { if } t>t_{k}, t \in[a, b] . \end{array}\right.(8)(t−tk)+={0, if t≤tk,t−tk, if t>tk,t∈[a,b].
(see Theorem 2 from [9]).
Taking into account the representation (7) and the conditions (8), it follows that each s ∈ S 2 m + p − 1 ( Δ n ) s ∈ S 2 m + p − 1 Δ n s inS_(2m+p-1)(Delta_(n))s \in S_{2 m+p-1}\left(\Delta_{n}\right)s∈S2m+p−1(Δn) depends on n + p + 1 n + p + 1 n+p+1n+p+1n+p+1 free parameteres, so that S 2 m + p − 1 ( Δ n ) S 2 m + p − 1 Δ n S_(2m+p-1)(Delta_(n))S_{2 m+p-1}\left(\Delta_{n}\right)S2m+p−1(Δn) is a vector space of dimension n + p + 1 n + p + 1 n+p+1n+p+1n+p+1 with respect to the usual (pointwise) of addition and multiplication by scalar of real functions.
The following theorem will allow us to use a spline function from S 2 m + 2 p − 1 ( Δ n ) S 2 m + 2 p − 1 Δ n S_(2m+2p-1)(Delta_(n))S_{2 m+2 p-1}\left(\Delta_{n}\right)S2m+2p−1(Δn) as an approximant for the solution of the problem (1)-(2).
where t k , k = 0 , n t k , k = 0 , n t_(k),k=0,nt_{k}, k=0, ntk,k=0,n are the nodes of the partition Δ n Δ n Delta_(n)\Delta_{n}Δn and α ( q ) , β ( q ) , q = 0 , p − 1 α ( q ) , β ( q ) , q = 0 , p − 1 alpha^((q)),beta^((q)),q=0,p-1\alpha^{(q)}, \beta^{(q)}, q=0, p-1α(q),β(q),q=0,p−1 and γ k , k = 0 , n ― γ k , k = 0 , n ¯ gamma_(k),k= bar(0,n)\gamma_{k}, k=\overline{0, n}γk,k=0,n―, are given numbers.
Then there exists a unique spline function s ∈ S 2 m + 2 p − 1 ( Δ n ) s ∈ S 2 m + 2 p − 1 Δ n s inS_(2m+2p-1)(Delta_(n))s \in S_{2 m+2 p-1}\left(\Delta_{n}\right)s∈S2m+2p−1(Δn) such that
(10) s f ( q ) ( a ) = α ( q ) , q = 0 , p − 1 ― , s f ( q ) ( b ) = β ( q ) , q = 0 , p − 1 ― , s f ( 2 p ) ( t k ) = γ k , k = 0 , n ― . (10) s f ( q ) ( a ) = α ( q ) , q = 0 , p − 1 ¯ , s f ( q ) ( b ) = β ( q ) , q = 0 , p − 1 ¯ , s f ( 2 p ) t k = γ k , k = 0 , n ¯ . {:(10){:[s_(f)^((q))(a)=alpha^((q))",",q= bar(0,p-1)","],[s_(f)^((q))(b)=beta^((q))",",q= bar(0,p-1)","],[s_(f)^((2p))(t_(k))=gamma_(k)",",k= bar(0,n).]:}:}\begin{array}{ll} s_{f}^{(q)}(a)=\alpha^{(q)}, & q=\overline{0, p-1}, \\ s_{f}^{(q)}(b)=\beta^{(q)}, & q=\overline{0, p-1}, \tag{10}\\ s_{f}^{(2 p)}\left(t_{k}\right)=\gamma_{k}, & k=\overline{0, n} . \end{array}(10)sf(q)(a)=α(q),q=0,p−1―,sf(q)(b)=β(q),q=0,p−1―,sf(2p)(tk)=γk,k=0,n―.
Proof. If s f s f s_(f)s_{f}sf is of the form (6), fulfilling the conditions (7), then, imposing the conditions (10), we find the system:
∑ i = 0 m + 2 p − q − 1 ( q + 1 ) ! i ! A q + i t 0 i = α ( q ) , q = 0 , p − 1 ― ∑ i = 0 m + 2 p − q − 1 ( q + 1 ) ! i ! A q + i t n + ∑ k = 0 n ( 2 m + 2 p − 1 ) ! ( 2 m + 2 p − q − 1 ) ! a k ( t n − t k ) 2 m + 2 p − q − 1 = β ( q ) (11) (11) ∑ i = 0 m − 1 ( 2 p + 1 ) ! i ! A 2 p + i t j i = ∑ k = 0 n ( 2 m + 2 p − 1 ) ! ( 2 m − 1 ) ! a k ( t j − t k ) 2 m − 1 = γ j , j = 0 , n ― ∑ k = 0 n a k t k i = 0 , i = 0 , m − 1 ― ∑ i = 0 m + 2 p − q − 1   ( q + 1 ) ! i ! A q + i t 0 i = α ( q ) , q = 0 , p − 1 ¯ ∑ i = 0 m + 2 p − q − 1   ( q + 1 ) ! i ! A q + i t n + ∑ k = 0 n   ( 2 m + 2 p − 1 ) ! ( 2 m + 2 p − q − 1 ) ! a k t n − t k 2 m + 2 p − q − 1 = β ( q ) (11)  (11)  ∑ i = 0 m − 1   ( 2 p + 1 ) ! i ! A 2 p + i t j i = ∑ k = 0 n   ( 2 m + 2 p − 1 ) ! ( 2 m − 1 ) ! a k t j − t k 2 m − 1 = γ j , j = 0 , n ¯ ∑ k = 0 n   a k t k i = 0 , i = 0 , m − 1 ¯ {:[sum_(i=0)^(m+2p-q-1)((q+1)!)/(i!)A_(q+i)t_(0)^(i)=alpha^((q))","q= bar(0,p-1)],[sum_(i=0)^(m+2p-q-1)((q+1)!)/(i!)A_(q+i)t_(n)+sum_(k=0)^(n)((2m+2p-1)!)/((2m+2p-q-1)!)a_(k)(t_(n)-t_(k))^(2m+2p-q-1)=beta^((q))],[(11)" (11) "quadsum_(i=0)^(m-1)((2p+1)!)/(i!)A_(2p+i)t_(j)^(i)=sum_(k=0)^(n)((2m+2p-1)!)/((2m-1)!)a_(k)(t_(j)-t_(k))^(2m-1)=gamma_(j)","quad j= bar(0,n)],[sum_(k=0)^(n)a_(k)t_(k)^(i)=0","i= bar(0,m-1)]:}\begin{align*} & \sum_{i=0}^{m+2 p-q-1} \frac{(q+1)!}{i!} A_{q+i} t_{0}^{i}=\alpha^{(q)}, q=\overline{0, p-1} \\ & \sum_{i=0}^{m+2 p-q-1} \frac{(q+1)!}{i!} A_{q+i} t_{n}+\sum_{k=0}^{n} \frac{(2 m+2 p-1)!}{(2 m+2 p-q-1)!} a_{k}\left(t_{n}-t_{k}\right)^{2 m+2 p-q-1}=\beta^{(q)} \\ & \text { (11) } \quad \sum_{i=0}^{m-1} \frac{(2 p+1)!}{i!} A_{2 p+i} t_{j}^{i}=\sum_{k=0}^{n} \frac{(2 m+2 p-1)!}{(2 m-1)!} a_{k}\left(t_{j}-t_{k}\right)^{2 m-1}=\gamma_{j}, \quad j=\overline{0, n} \tag{11}\\ & \sum_{k=0}^{n} a_{k} t_{k}^{i}=0, i=\overline{0, m-1} \end{align*}∑i=0m+2p−q−1(q+1)!i!Aq+it0i=α(q),q=0,p−1―∑i=0m+2p−q−1(q+1)!i!Aq+itn+∑k=0n(2m+2p−1)!(2m+2p−q−1)!ak(tn−tk)2m+2p−q−1=β(q)(11) (11) ∑i=0m−1(2p+1)!i!A2p+itji=∑k=0n(2m+2p−1)!(2m−1)!ak(tj−tk)2m−1=γj,j=0,n―∑k=0naktki=0,i=0,m−1―
having 2 p + n + 1 + m 2 p + n + 1 + m 2p+n+1+m2 p+n+1+m2p+n+1+m equations and the same number of unknowns: A 0 , A 1 , … A m + 2 p − 1 , a 0 , a 1 , … , a n A 0 , A 1 , … A m + 2 p − 1 , a 0 , a 1 , … , a n A_(0),A_(1),dotsA_(m+2p-1),a_(0),a_(1),dots,a_(n)A_{0}, A_{1}, \ldots A_{m+2 p-1}, a_{0}, a_{1}, \ldots, a_{n}A0,A1,…Am+2p−1,a0,a1,…,an. qquad\qquad ,:
(iii)
This system has a unique solution if and only if the associated homogeneous system (obtained for α ( q ) = 0 = β ( q ) , q = 0 , p − 1 ― , γ k = 0 , k = 0 , n ― α ( q ) = 0 = β ( q ) , q = 0 , p − 1 ¯ , γ k = 0 , k = 0 , n ¯ alpha^((q))=0=beta^((q)),q= bar(0,p-1),gamma_(k)=0,k= bar(0,n)\alpha^{(q)}=0=\beta^{(q)}, q=\overline{0, p-1}, \gamma_{k}=0, k=\overline{0, n}α(q)=0=β(q),q=0,p−1―,γk=0,k=0,n― ) has only the null solution.
Let's show that, if s ∈ S 2 m + 2 p − 1 s ∈ S 2 m + 2 p − 1 s inS_(2m+2p-1)s \in S_{2 m+2 p-1}s∈S2m+2p−1 verifies s ( q ) ( a ) = s ( q ) ( b ) = 0 , q = 0 , p − 1 ― s ( q ) ( a ) = s ( q ) ( b ) = 0 , q = 0 , p − 1 ¯ s^((q))(a)=s^((q))(b)=0,quad q= bar(0,p-1)s^{(q)}(a)=s^{(q)}(b)=0, \quad q=\overline{0, p-1}s(q)(a)=s(q)(b)=0,q=0,p−1―; s ( 2 p ) ( t k ) = 0 , k = 0 , n ― s ( 2 p ) t k = 0 , k = 0 , n ¯ s^((2p))(t_(k))=0,k= bar(0,n)s^{(2 p)}\left(t_{k}\right)=0, k=\overline{0, n}s(2p)(tk)=0,k=0,n― then s ≡ 0 s ≡ 0 s-=0s \equiv 0s≡0 or R R R\mathbb{R}R.
Integrating by parts we obtain
∫ l 10 l n [ s ( m + 2 p ) ( t ) ] 2 d t = ∑ j = 0 m − 2 ( − 1 ) j s ( m + 2 p + j ) ( t ) ⋅ s ( m + 2 p − j − 1 ) ( t ) | l 11 t n + + ( − 1 ) m − 1 ∫ t 10 t n s ( 2 m + 2 p − 1 ) ( t ) ⋅ s ( 2 p + 1 ) ( t ) d t ∫ l 10 l n   s ( m + 2 p ) ( t ) 2 d t = ∑ j = 0 m − 2   ( − 1 ) j s ( m + 2 p + j ) ( t ) ⋅ s ( m + 2 p − j − 1 ) ( t ) l 11 t n + + ( − 1 ) m − 1 ∫ t 10 t n   s ( 2 m + 2 p − 1 ) ( t ) ⋅ s ( 2 p + 1 ) ( t ) d t {:[int_(l_(10))^(l_(n))[s^((m+2p))(t)]^(2)dt=sum_(j=0)^(m-2)(-1)^(j)s^((m+2p+j))(t)*s^((m+2p-j-1))(t)|_(l_(11))^(t_(n))+],[+(-1)^(m-1)int_(t_(10))^(t_(n))s^((2m+2p-1))(t)*s^((2p+1))(t)dt]:}\begin{aligned} \int_{l_{10}}^{l_{n}}\left[s^{(m+2 p)}(t)\right]^{2} \mathrm{~d} t & =\left.\sum_{j=0}^{m-2}(-1)^{j} s^{(m+2 p+j)}(t) \cdot s^{(m+2 p-j-1)}(t)\right|_{l_{11}} ^{t_{n}}+ \\ & +(-1)^{m-1} \int_{t_{10}}^{t_{n}} s^{(2 m+2 p-1)}(t) \cdot s^{(2 p+1)}(t) \mathrm{d} t \end{aligned}∫l10ln[s(m+2p)(t)]2 dt=∑j=0m−2(−1)js(m+2p+j)(t)⋅s(m+2p−j−1)(t)|l11tn++(−1)m−1∫t10tns(2m+2p−1)(t)⋅s(2p+1)(t)dt
But s ( m + 2 p + j ) ( t 0 ) = s ( m + 2 p + j ) ( t n ) = 0 , j = 0 , m − 2 ― s ( m + 2 p + j ) t 0 = s ( m + 2 p + j ) t n = 0 , j = 0 , m − 2 ¯ s^((m+2p+j))(t_(0))=s^((m+2p+j))(t_(n))=0,quad j= bar(0,m-2)s^{(m+2 p+j)}\left(t_{0}\right)=s^{(m+2 p+j)}\left(t_{n}\right)=0, \quad j=\overline{0, m-2}s(m+2p+j)(t0)=s(m+2p+j)(tn)=0,j=0,m−2― (by Condition 3 0 3 0 3^(0)3^{0}30 from Definition 1) so that
∫ t 1 t h [ s ( m + 2 p ) ( t ) ] 2 d t = ∫ a b [ s ( m + 2 p ) ( t ) ] 2 d t = = ( − 1 ) m − 1 ∫ a b s ( 2 m + 2 p − 1 ) ( t ) ⋅ s ( 2 p − 1 ) ( t ) d t = (1024) = ( − 1 ) m − 1 ∑ k = 1 n C k ∫ t k i t k s ( 2 p + 1 ) ( t ) d t = = ( − 1 ) m − 1 ∑ k = 1 n C k ( s ( 2 p ) ( t k ) − s ( 2 p ) ( t k − 1 ) ) = 0 ∫ t 1 t h   s ( m + 2 p ) ( t ) 2 d t = ∫ a b   s ( m + 2 p ) ( t ) 2 d t = = ( − 1 ) m − 1 ∫ a b   s ( 2 m + 2 p − 1 ) ( t ) ⋅ s ( 2 p − 1 ) ( t ) d t = (1024) = ( − 1 ) m − 1 ∑ k = 1 n   C k ∫ t k i t k   s ( 2 p + 1 ) ( t ) d t = = ( − 1 ) m − 1 ∑ k = 1 n   C k s ( 2 p ) t k − s ( 2 p ) t k − 1 = 0 {:[int_(t_(1))^(t_(h))[s^((m+2p))(t)]^(2)dt=int_(a)^(b)[s^((m+2p))(t)]^(2)dt=],[=(-1)^(m-1)int_(a)^(b)s^((2m+2p-1))(t)*s^((2p-1))(t)dt=],[(1024)=(-1)^(m-1)sum_(k=1)^(n)C_(k)int_(t_(ki))^(t_(k))s^((2p+1))(t)dt=],[=(-1)^(m-1)sum_(k=1)^(n)C_(k)(s^((2p))(t_(k))-s^((2p))(t_(k-1)))=0]:}\begin{align*} & \int_{t_{1}}^{t_{h}}\left[s^{(m+2 p)}(t)\right]^{2} \mathrm{~d} t=\int_{a}^{b}\left[s^{(m+2 p)}(t)\right]^{2} \mathrm{~d} t= \\ & =(-1)^{m-1} \int_{a}^{b} s^{(2 m+2 p-1)}(t) \cdot s^{(2 p-1)}(t) \mathrm{d} t= \\ & =(-1)^{m-1} \sum_{k=1}^{n} C_{k} \int_{t_{k i}}^{t_{k}} s^{(2 p+1)}(t) \mathrm{d} t= \tag{1024}\\ & =(-1)^{m-1} \sum_{k=1}^{n} C_{k}\left(s^{(2 p)}\left(t_{k}\right)-s^{(2 p)}\left(t_{k-1}\right)\right)=0 \end{align*}∫t1th[s(m+2p)(t)]2 dt=∫ab[s(m+2p)(t)]2 dt==(−1)m−1∫abs(2m+2p−1)(t)⋅s(2p−1)(t)dt=(1024)=(−1)m−1∑k=1nCk∫tkitks(2p+1)(t)dt==(−1)m−1∑k=1nCk(s(2p)(tk)−s(2p)(tk−1))=0
where C k = s ( 2 m + 2 p − 1 ) ( t ) | l k k = l , n ― C k = s ( 2 m + 2 p − 1 ) ( t ) l k k = l , n ¯ C_(k)=s^((2m+2p-1))(t)|_(l_(k))quad k= bar(l,n)C_{k}=\left.s^{(2 m+2 p-1)}(t)\right|_{l_{k}} \quad k=\overline{l, n}Ck=s(2m+2p−1)(t)|lkk=l,n― (by Condition 2 ( 1 ) 2 ( 1 ) 2^((1))2^{(1)}2(1) from Definition 1).
Therefore, s ( m + 2 p ) ( t ) = 0 s ( m + 2 p ) ( t ) = 0 s^((m+2p))(t)=0s^{(m+2 p)}(t)=0s(m+2p)(t)=0, for all t ∈ [ a , b ] t ∈ [ a , b ] t in[a,b]t \in[a, b]t∈[a,b].
Since s ∈ P m + 2 p − 1 s ∈ P m + 2 p − 1 s inP_(m+2p-1)s \in \mathscr{P}_{m+2 p-1}s∈Pm+2p−1 on I 0 ∪ I n + 1 I 0 ∪ I n + 1 I_(0)uuI_(n+1)I_{0} \cup I_{n+1}I0∪In+1 it follows s ( m + 2 p ) ( t ) = 0 s ( m + 2 p ) ( t ) = 0 s^((m+2p))(t)=0s^{(m+2 p)}(t)=0s(m+2p)(t)=0 for any t ∈ I 0 ∪ I n + 1 t ∈ I 0 ∪ I n + 1 t inI_(0)uuI_(n+1)t \in I_{0} \cup I_{n+1}t∈I0∪In+1. By continuity of s ( m + 2 p ) s ( m + 2 p ) s^((m+2p))s^{(m+2 p)}s(m+2p) on R R R\mathbb{R}R it follows s ( m + 2 p ) ( t ) = 0 s ( m + 2 p ) ( t ) = 0 s^((m+2p))(t)=0s^{(m+2 p)}(t)=0s(m+2p)(t)=0 for all t ∈ R t ∈ R t in Rt \in Rt∈R (see the Condition 1 0 1 0 1^(0)1^{0}10 from Definition 1). Then s ∈ P m + 2 p − 1 s ∈ P m + 2 p − 1 s inP_(m+2p-1)s \in \mathscr{P}_{m+2 p-1}s∈Pm+2p−1 on R R R\mathbb{R}R, implies s ( 2 p ) ∈ m m − 1 s ( 2 p ) ∈ m m − 1 s^((2p))inm_(m-1)s^{(2 p)} \in m_{m-1}s(2p)∈mm−1 on R R R\mathbb{R}R. But s ( 2 p ) ( t k ) = 0 , k = 0 , n ― ( n > m ) s ( 2 p ) t k = 0 , k = 0 , n ¯ ( n > m ) s^((2p))(t_(k))=0,k= bar(0,n)(n > m)s^{(2 p)}\left(t_{k}\right)=0, k=\overline{0, n}(n>m)s(2p)(tk)=0,k=0,n―(n>m) implies s ( 2 p ) ( t ) = 0 s ( 2 p ) ( t ) = 0 s^((2p))(t)=0s^{(2 p)}(t)=0s(2p)(t)=0 for all t ∈ R t ∈ R t inRt \in \mathbb{R}t∈R and, consequently, s ∈ P p − 1 s ∈ P p − 1 s inP_(p-1)s \in \mathscr{P}_{p-1}s∈Pp−1 on R R R\mathbb{R}R.
As s ( q ) ( a ) = s ( q ) ( b ) = 0 , q = 0 , 1 , … , p − 1 s ( q ) ( a ) = s ( q ) ( b ) = 0 , q = 0 , 1 , … , p − 1 s^((q))(a)=s^((q))(b)=0,q=0,1,dots,p-1s^{(q)}(a)=s^{(q)}(b)=0, q=0,1, \ldots, p-1s(q)(a)=s(q)(b)=0,q=0,1,…,p−1 we infer that s ≡ 0 s ≡ 0 s-=0s \equiv 0s≡0 or R R R\mathbb{R}R. But then all the coefficients of s s sss are null, so that the homogeneous system associated to (11) has only the null solution.
Remark 1. By. Theorem 2, if y y yyy is the exact solution of the differental tions (1) with condition (2), then there is only one function s x ∈ S 2 m + 2 p − 1 ( Δ n ) s x ∈ S 2 m + 2 p − 1 Δ n s_(x)inS_(2m+2p-1)(Delta_(n))s_{x} \in S_{2 m+2 p-1}\left(\Delta_{n}\right)sx∈S2m+2p−1(Δn) verifying the conditions (2).
  • Hus Let
(12) H 2 m + 2 p ( [ a , b ] ) := { g : [ a , b ] → R , g ( m + 2 p − 1 ) absolutely continuous on [ a , b ] and g ( m + 2 p ) ∈ L 2 [ a , b ] } (12) H 2 m + 2 p ( [ a , b ] ) := g : [ a , b ] → R , g ( m + 2 p − 1 )  absolutely continuous   on  [ a , b ]  and  g ( m + 2 p ) ∈ L 2 [ a , b ] {:[(12)H_(2)^(m+2p)([a","b]):={g:[a,b]rarrR,g^((m+2p-1)):}" absolutely continuous "],[" on "{:[a,b]" and "g^((m+2p))inL_(2)[a,b]}]:}\begin{align*} H_{2}^{m+2 p}([a, b]):= & \left\{g:[a, b] \rightarrow \mathbb{R}, g^{(m+2 p-1)}\right. \text { absolutely continuous } \tag{12}\\ & \text { on } \left.[a, b] \text { and } g^{(m+2 p)} \in L_{2}[a, b]\right\} \end{align*}(12)H2m+2p([a,b]):={g:[a,b]→R,g(m+2p−1) absolutely continuous  on [a,b] and g(m+2p)∈L2[a,b]}
and let Y = ( α ( 0 ) , α ( 1 ) , … , α ( p − 1 ) , β ( 0 ) , β ( 1 ) , … , β ( p − 1 ) , γ 0 , γ 1 , … , γ n ) ∈ R n + 2 p + 1 Y = α ( 0 ) , α ( 1 ) , … , α ( p − 1 ) , β ( 0 ) , β ( 1 ) , … , β ( p − 1 ) , γ 0 , γ 1 , … , γ n ∈ R n + 2 p + 1 Y=(alpha^((0)),alpha^((1)),dots,alpha^((p-1)),beta^((0)),beta^((1)),dots,beta^((p-1)),gamma_(0),gamma_(1),dots,gamma_(n))inR^(n+2p+1)Y=\left(\alpha^{(0)}, \alpha^{(1)}, \ldots, \alpha^{(p-1)}, \beta^{(0)}, \beta^{(1)}, \ldots, \beta^{(p-1)}, \gamma_{0}, \gamma_{1}, \ldots, \gamma_{n}\right) \in \mathbb{R}^{n+2 p+1}Y=(α(0),α(1),…,α(p−1),β(0),β(1),…,β(p−1),γ0,γ1,…,γn)∈Rn+2p+1 be a fixed vector.
Denote
(13) H 2 m + 2 p ( Δ n , Y ) := { g ∈ H 2 m + 2 p ( [ a , b ] ) : g ( 2 p ) ( t k ) = γ k k = ( 0 , 1 , 2 , … , n ; g ( q ) ( a ) = ( x ( q ) , g ( q ) ( b ) = β ( q ) ( b ) = β ( q ) , q = 0 , p − 1 ― } (13) H 2 m + 2 p Δ n , Y := g ∈ H 2 m + 2 p ( [ a , b ] ) : g ( 2 p ) t k = γ k k = 0 , 1 , 2 , … , n ; g ( q ) ( a ) = x ( q ) , g ( q ) ( b ) = β ( q ) ( b ) = β ( q ) , q = 0 , p − 1 ¯ {:[(13)H_(2)^(m+2p)(Delta_(n),Y):={g inH_(2)^(m+2p)([a,b]):g^((2p))(t_(k))=gamma_(k):}],[quad k=(0,1,2,dots,n;g^((q))(a)=(x^((q)),g^((q))(b)=beta^((q))(b)=beta^((q)),q= bar(0,p-1)}:}]:}\begin{align*} & H_{2}^{m+2 p}\left(\Delta_{n}, Y\right):=\left\{g \in H_{2}^{m+2 p}([a, b]): g^{(2 p)}\left(t_{k}\right)=\gamma_{k}\right. \tag{13}\\ & \quad k=\left(0,1,2, \ldots, n ; g^{(q)}(a)=\left(x^{(q)}, g^{(q)}(b)=\beta^{(q)}(b)=\beta^{(q)}, q=\overline{0, p-1}\right\}\right. \end{align*}(13)H2m+2p(Δn,Y):={g∈H2m+2p([a,b]):g(2p)(tk)=γkk=(0,1,2,…,n;g(q)(a)=(x(q),g(q)(b)=β(q)(b)=β(q),q=0,p−1―}
By Theorem 2, there is only one spline function s y ∈ S 2 m + 2 p − 1 ( Δ n ) s y ∈ S 2 m + 2 p − 1 Δ n s_(y)inS_(2m+2p-1)(Delta_(n))s_{y} \in S_{2 m+2 p-1}\left(\Delta_{n}\right)sy∈S2m+2p−1(Δn) such that s y ∈ H 2 m + 2 p ( Δ n , Y ) s y ∈ H 2 m + 2 p Δ n , Y s_(y)inH_(2)^(m+2p)(Delta_(n),Y)s_{y} \in H_{2}^{m+2 p}\left(\Delta_{n}, Y\right)sy∈H2m+2p(Δn,Y).
Furthermore, we have:
Theorem 3. ([9], Th. 5 and Th. 6).
a) If g ∈ H 2 m + 2 p ( Δ n , Y ) g ∈ H 2 m + 2 p Δ n , Y g inH_(2)^(m+2p)(Delta_(n),Y)g \in H_{2}^{m+2 p}\left(\Delta_{n}, Y\right)g∈H2m+2p(Δn,Y) then

(14)
( ∂ ∣ ) ‖ s Y ( m + 2 p ) ‖ 2 ≤ ‖ g ( m + 2 p ) ‖ 2 ; ( ∂ ∣ ) s Y ( m + 2 p ) 2 ≤ g ( m + 2 p ) 2 ; {:((del∣)")"||s_(Y)^((m+2p))||_(2) <= ||g^((m+2p))||_(2);:}\begin{equation*} \left\|s_{Y}^{(m+2 p)}\right\|_{2} \leq\left\|g^{(m+2 p)}\right\|_{2} ; \tag{$\partial\mid$} \end{equation*}(∂∣)‖sY(m+2p)‖2≤‖g(m+2p)‖2;
b) If f ∈ H 2 m + 2 p ( Δ n ) f ∈ H 2 m + 2 p Δ n f inH_(2)^(m+2p)(Delta_(n))f \in H_{2}^{m+2 p}\left(\Delta_{n}\right)f∈H2m+2p(Δn) then
(c|) vagui doiriw
(15) ‖ f ( m + 2 p ) − s f ( m + 2 p ) ‖ 2 ≤ ‖ f ( m + 2 p ) − s ( m + 2 p ) ‖ 2 , (15) f ( m + 2 p ) − s f ( m + 2 p ) 2 ≤ f ( m + 2 p ) − s ( m + 2 p ) 2 , {:(15)||f^((m+2p))-s_(f)^((m+2p))||_(2) <= ||f^((m+2p))-s^((m+2p))||_(2)",":}\begin{equation*} \left\|f^{(m+2 p)}-s_{f}^{(m+2 p)}\right\|_{2} \leq\left\|f^{(m+2 p)}-s^{(m+2 p)}\right\|_{2}, \tag{15} \end{equation*}(15)‖f(m+2p)−sf(m+2p)‖2≤‖f(m+2p)−s(m+2p)‖2,
for any s ∈ S 2 m + 2 p − 1 ( Δ n ) s ∈ S 2 m + 2 p − 1 Δ n s inS_(2m+2p-1)(Delta_(n))s \in S_{2 m+2 p-1}\left(\Delta_{n}\right)s∈S2m+2p−1(Δn) (Here s f s f s_(f)s_{f}sf is given by Theorem 2).
Proof. To prove (14) we shall use the identity
(6,91) ‖ g ( m + 2 p ) − s y ( m + 2 p ) ‖ 2 2 = ∫ a b [ g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) ] 2 d t = = ‖ g ( m + 2 p ) ‖ 2 2 − ‖ s y ( m + 2 p ) ‖ 2 2 − 2 ∫ a b s y ( m + 2 p ) ( t ) [ g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) ] d t (6,91) g ( m + 2 p ) − s y ( m + 2 p ) 2 2 = ∫ a b   g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) 2 d t = = g ( m + 2 p ) 2 2 − s y ( m + 2 p ) 2 2 − 2 ∫ a b   s y ( m + 2 p ) ( t ) g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) d t {:[(6,91)||g^((m+2p))-s_(y)^((m+2p))||_(2)^(2)=int_(a)^(b)[g^((m+2p))(t)-s_(y)^((m+2p))(t)]^(2)dt=],[=||g^((m+2p))||_(2)^(2)-||s_(y)^((m+2p))||_(2)^(2)-2int_(a)^(b)s_(y)^((m+2p))(t)[g^((m+2p))(t)-s_(y)^((m+2p))(t)]dt]:}\begin{align*} & \left\|g^{(m+2 p)}-s_{y}^{(m+2 p)}\right\|_{2}^{2}=\int_{a}^{b}\left[g^{(m+2 p)}(t)-s_{y}^{(m+2 p)}(t)\right]^{2} \mathrm{~d} t= \tag{6,91}\\ & =\left\|g^{(m+2 p)}\right\|_{2}^{2}-\left\|s_{y}^{(m+2 p)}\right\|_{2}^{2}-2 \int_{a}^{b} s_{y}^{(m+2 p)}(t)\left[g^{(m+2 p)}(t)-s_{y}^{(m+2 p)}(t)\right] \mathrm{d} t \end{align*}(6,91)‖g(m+2p)−sy(m+2p)‖22=∫ab[g(m+2p)(t)−sy(m+2p)(t)]2 dt==‖g(m+2p)‖22−‖sy(m+2p)‖22−2∫absy(m+2p)(t)[g(m+2p)(t)−sy(m+2p)(t)]dt
where the last term is null. Indeed, integrating by parts, we find

( 5 ) + ( 1 ) ( 5 ) + ( 1 ) (5)+(1)(5)+(1)(5)+(1)
∫ a b s ( m + 2 p ) ( t ) [ g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) ] d t = = ( − 1 ) m − 1 ∑ k = 1 n C k [ ( g ( 2 p ) − a ( 2 p ) ) ( t k ) − ( g ( 2 p ) − s y ( 2 p ) ) ( t k − 1 ) ] = 0 ∫ a b   s ( m + 2 p ) ( t ) g ( m + 2 p ) ( t ) − s y ( m + 2 p ) ( t ) d t = = ( − 1 ) m − 1 ∑ k = 1 n   C k g ( 2 p ) − a ( 2 p ) t k − g ( 2 p ) − s y ( 2 p ) t k − 1 = 0 {:[int_(a)^(b)s^((m+2p))(t)[g^((m+2p))(t)-s_(y)^((m+2p))(t)]dt=],[=(-1)^(m-1)sum_(k=1)^(n)C_(k)[(g^((2p))-a^((2p)))(t_(k))-(g^((2p))-s_(y)^((2p)))(t_(k-1))]=0]:}\begin{aligned} & \int_{a}^{b} s^{(m+2 p)}(t)\left[g^{(m+2 p)}(t)-s_{y}^{(m+2 p)}(t)\right] \mathrm{d} t= \\ & =(-1)^{m-1} \sum_{k=1}^{n} C_{k}\left[\left(g^{(2 p)}-a^{(2 p)}\right)\left(t_{k}\right)-\left(g^{(2 p)}-s_{y}^{(2 p)}\right)\left(t_{k-1}\right)\right]=0 \end{aligned}∫abs(m+2p)(t)[g(m+2p)(t)−sy(m+2p)(t)]dt==(−1)m−1∑k=1nCk[(g(2p)−a(2p))(tk)−(g(2p)−sy(2p))(tk−1)]=0
aroh किसी
where C k = s ( 2 m + 2 p − 1 ) | I k k = 1 , 2 , … , n C k = s ( 2 m + 2 p − 1 ) I k k = 1 , 2 , … , n C_(k)=s^((2m+2p-1))|_(I_(k))k=1,2,dots,nC_{k}=\left.s^{(2 m+2 p-1)}\right|_{I_{k}} k=1,2, \ldots, nCk=s(2m+2p−1)|Ikk=1,2,…,n and s y ( m + 2 p + j ) ( a ) = s y ( m + 2 p + j ) ( b ) = 0 s y ( m + 2 p + j ) ( a ) = s y ( m + 2 p + j ) ( b ) = 0 s_(y)^((m+2p+j))(a)=s_(y)^((m+2p+j))(b)=0s_{y}^{(m+2 p+j)}(a)=s_{y}^{(m+2 p+j)}(b)=0sy(m+2p+j)(a)=sy(m+2p+j)(b)=0 for j = 0 , 1 , … , m − 2 j = 0 , 1 , … , m − 2 j=0,1,dots,m-2j=0,1, \ldots, m-2j=0,1,…,m−2 (by Condition 3 0 3 0 3^(0)3^{0}30 from Definition 1).

It follows

implying the relation (14).
"जनहीं en All Aows
To prove (15) we shall use the identity
‖ s ( m + 2 p ) − f ( m + 2 p ) ‖ 2 2 = ‖ s ( m + 2 p ) − s f ( m + 2 p ) ‖ 2 2 + ‖ s f ( m + 2 p ) − f ( m + 2 p ) ‖ 2 2 + + 2 ∫ a b [ s ( m + 2 p ) ( t ) − s f ( m + 2 p ) ( t ) ] [ s f ( m + 2 p ) ( t ) − f ( m + 2 p ) ( t ) ] d t s ( m + 2 p ) − f ( m + 2 p ) 2 2 = s ( m + 2 p ) − s f ( m + 2 p ) 2 2 + s f ( m + 2 p ) − f ( m + 2 p ) 2 2 + + 2 ∫ a b   s ( m + 2 p ) ( t ) − s f ( m + 2 p ) ( t ) s f ( m + 2 p ) ( t ) − f ( m + 2 p ) ( t ) d t {:[||s^((m+2p))-f^((m+2p))||_(2)^(2)=||s^((m+2p))-s_(f)^((m+2p))||_(2)^(2)+||s_(f)^((m+2p))-f^((m+2p))||_(2)^(2)+],[+2int_(a)^(b)[s^((m+2p))(t)-s_(f)^((m+2p))(t)][s_(f)^((m+2p))(t)-f^((m+2p))(t)]dt]:}\begin{aligned} & \left\|s^{(m+2 p)}-f^{(m+2 p)}\right\|_{2}^{2}=\left\|s^{(m+2 p)}-s_{f}^{(m+2 p)}\right\|_{2}^{2}+\left\|s_{f}^{(m+2 p)}-f^{(m+2 p)}\right\|_{2}^{2}+ \\ & +2 \int_{a}^{b}\left[s^{(m+2 p)}(t)-s_{f}^{(m+2 p)}(t)\right]\left[s_{f}^{(m+2 p)}(t)-f^{(m+2 p)}(t)\right] \mathrm{d} t \end{aligned}‖s(m+2p)−f(m+2p)‖22=‖s(m+2p)−sf(m+2p)‖22+‖sf(m+2p)−f(m+2p)‖22++2∫ab[s(m+2p)(t)−sf(m+2p)(t)][sf(m+2p)(t)−f(m+2p)(t)]dt
where, by integrating by parts, the last term is again null. To show this one uses the equalities
graí ow sampration
(이이2 ( h ) ( s ( m + 2 p + i ) − s f ( m + 2 p + j ) ) ( a ) = ( s ( m + 2 p + i ) − s f ( m + 2 p + j ) ) ( b ) = 0 ( h ) s ( m + 2 p + i ) − s f ( m + 2 p + j ) ( a ) = s ( m + 2 p + i ) − s f ( m + 2 p + j ) ( b ) = 0 (h)quad(s^((m+2p+i))-s_(f)^((m+2p+j)))(a)=(s^((m+2p+i))-s_(f)^((m+2p+j)))(b)=0(h) \quad\left(s^{(m+2 p+i)}-s_{f}^{(m+2 p+j)}\right)(a)=\left(s^{(m+2 p+i)}-s_{f}^{(m+2 p+j)}\right)(b)=0(h)(s(m+2p+i)−sf(m+2p+j))(a)=(s(m+2p+i)−sf(m+2p+j))(b)=0
for j = 0 , 1 , 2 , … , m − 2 j = 0 , 1 , 2 , … , m − 2 j=0,1,2,dots,m-2j=0,1,2, \ldots, m-2j=0,1,2,…,m−2, and
( s ( m + 2 p − 1 ) ( t ) − s f ( m + 2 p − 1 ) ( t ) ) | A k = c k ( s ) , k = 1 , n ― s ( m + 2 p − 1 ) ( t ) − s f ( m + 2 p − 1 ) ( t ) A k = c k ( s ) , k = 1 , n ¯ (s^((m+2p-1))(t)-s_(f)^((m+2p-1))(t))|_(A_(k))=c_(k)(s),quad k= bar(1,n)\left.\left(s^{(m+2 p-1)}(t)-s_{f}^{(m+2 p-1)}(t)\right)\right|_{A_{k}}=c_{k}(s), \quad k=\overline{1, n}(s(m+2p−1)(t)−sf(m+2p−1)(t))|Ak=ck(s),k=1,n―
(constants depending on s s sss ). In conclusion
(16) ‖ s ( m + 2 p ) − f ( m + 2 p ) ‖ 2 2 = ‖ s ( m + 2 p ) − s f ( m + 2 p ) ‖ 2 2 + ‖ s f ( m + 2 p ) − f ( m + 2 p ) ‖ 2 2 (16) s ( m + 2 p ) − f ( m + 2 p ) 2 2 = s ( m + 2 p ) − s f ( m + 2 p ) 2 2 + s f ( m + 2 p ) − f ( m + 2 p ) 2 2 {:(16)||s^((m+2p))-f^((m+2p))||_(2)^(2)=||s^((m+2p))-s_(f)^((m+2p))||_(2)^(2)+||s_(f)^((m+2p))-f^((m+2p))||_(2)^(2):}\begin{equation*} \left\|s^{(m+2 p)}-f^{(m+2 p)}\right\|_{2}^{2}=\left\|s^{(m+2 p)}-s_{f}^{(m+2 p)}\right\|_{2}^{2}+\left\|s_{f}^{(m+2 p)}-f^{(m+2 p)}\right\|_{2}^{2} \tag{16} \end{equation*}(16)‖s(m+2p)−f(m+2p)‖22=‖s(m+2p)−sf(m+2p)‖22+‖sf(m+2p)−f(m+2p)‖22
which imply (15).
Remark 2. By (15), we obtain for s ≡ 0 s ≡ 0 s-=0s \equiv 0s≡0
(17) ‖ f ( m + 2 p ) − s f ( m + 2 p ) ‖ 2 ≤ ‖ f ( m + 2 p ) ‖ 2 . (17) f ( m + 2 p ) − s f ( m + 2 p ) 2 ≤ f ( m + 2 p ) 2 . {:(17)||f^((m+2p))-s_(f)^((m+2p))||_(2) <= ||f^((m+2p))||_(2).:}\begin{equation*} \left\|f^{(m+2 p)}-s_{f}^{(m+2 p)}\right\|_{2} \leq\left\|f^{(m+2 p)}\right\|_{2} . \tag{17} \end{equation*}(17)‖f(m+2p)−sf(m+2p)‖2≤‖f(m+2p)‖2.
Returning to the problem (1)-(2) we deduce
COROLLARY 4. If the exact solution y y yyy of the problem (1)-(2) is in H ( m + 2 p ) ( [ a , b ] ) H ( m + 2 p ) ( [ a , b ] ) H^((m+2p))([a,b])H^{(m+2 p)}([a, b])H(m+2p)([a,b]) and s Y ∈ S 2 m + p − 1 ( Δ n ) s Y ∈ S 2 m + p − 1 Δ n s_(Y)inS_(2m+p-1)(Delta_(n))s_{Y} \in S_{2 m+p-1}\left(\Delta_{n}\right)sY∈S2m+p−1(Δn) is the spline function associated to y y yyy, verifing the same boundary conditions as y y yyy, then the following evaluation
holds.
THEOREM 5. If y y yyy is the exact solution of the problem (1)-(2), y ∈ H ( m + 2 p ) ( [ a , b ] ) y ∈ H ( m + 2 p ) ( [ a , b ] ) y inH^((m+2p))([a,b])y \in H^{(m+2 p)}([a, b])y∈H(m+2p)([a,b]) and s y ∈ S 2 m + p − 1 ( Δ n ) s y ∈ S 2 m + p − 1 Δ n s_(y)inS_(2m+p-1)(Delta_(n))s_{y} \in S_{2 m+p-1}\left(\Delta_{n}\right)sy∈S2m+p−1(Δn) is the approximant spline function, then the following inequalities:
(19) ‖ v ( m + 2 p − 1 ) − s ( m + 2 p − 1 ) ‖ ∞ ≤ ≤ m ( m − 1 ) ⋯ ( m − 1 + 1 ) ‖ Δ n ‖ l − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 (19) v ( m + 2 p − 1 ) − s ( m + 2 p − 1 ) ∞ ≤ ≤ m ( m − 1 ) ⋯ ( m − 1 + 1 ) Δ n l − 1 2 ⋅ y ( m + 2 p ) 2 {:[(19)||v^((m+2p-1))-s^((m+2p-1))||_(oo) <= ],[ <= sqrtm(m-1)cdots(m-1+1)||Delta_(n)||^(l-(1)/(2))*||y^((m+2p))||_(2)]:}\begin{align*} & \left\|v^{(m+2 p-1)}-s^{(m+2 p-1)}\right\|_{\infty} \leq \tag{19}\\ & \leq \sqrt{m}(m-1) \cdots(m-1+1)\left\|\Delta_{n}\right\|^{l-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2} \end{align*}(19)‖v(m+2p−1)−s(m+2p−1)‖∞≤≤m(m−1)⋯(m−1+1)‖Δn‖l−12⋅‖y(m+2p)‖2
holds, for l = { 2 , 3 , … , m } l = { 2 , 3 , … , m } l={2,3,dots,m}l=\{2,3, \ldots, m\}l={2,3,…,m} and ‖ Δ n ‖ = max { t i − t i − 1 , i = 1 , n ― } Δ n = max t i − t i − 1 , i = 1 , n ¯ ||Delta_(n)||=max{t_(i)-t_(i-1),i= bar(1,n)}\left\|\Delta_{n}\right\|=\max \left\{t_{i}-t_{i-1}, i=\overline{1, n}\right\}‖Δn‖=max{ti−ti−1,i=1,n―}.
Proof. We have
y ( 2 p ) ( t i ) − s y ( 2 p ) ( t i ) = 0 , i = 0 , 1 , 2 , … , n . y ( 2 p ) t i − s y ( 2 p ) t i = 0 , i = 0 , 1 , 2 , … , n . y^((2p))(t_(i))-s_(y)^((2p))(t_(i))=0,quad i=0,1,2,dots,n.y^{(2 p)}\left(t_{i}\right)-s_{y}^{(2 p)}\left(t_{i}\right)=0, \quad i=0,1,2, \ldots, n .y(2p)(ti)−sy(2p)(ti)=0,i=0,1,2,…,n.
By Rôlle's Theorem it follows the existence of the points t i ( 1 ) ∈ ( t i , t i + 1 ) t i ( 1 ) ∈ t i , t i + 1 t_(i)^((1))in(t_(i),t_(i+1))t_{i}^{(1)} \in\left(t_{i}, t_{i+1}\right)ti(1)∈(ti,ti+1), i = 0 , 1 , 2 , … , n − 1 i = 0 , 1 , 2 , … , n − 1 i=0,1,2,dots,n-1i=0,1,2, \ldots, n-1i=0,1,2,…,n−1 such that
y ( 2 p + 1 ) ( t i ( 1 ) ) − s y ( 2 p + 1 ) ( t i ( 1 ) ) = 0 , i = 0 , 1 , 2 , … , n − 1 y ( 2 p + 1 ) t i ( 1 ) − s y ( 2 p + 1 ) t i ( 1 ) = 0 , i = 0 , 1 , 2 , … , n − 1 y^((2p+1))(t_(i)^((1)))-s_(y)^((2p+1))(t_(i)^((1)))=0,quad i=0,1,2,dots,n-1y^{(2 p+1)}\left(t_{i}^{(1)}\right)-s_{y}^{(2 p+1)}\left(t_{i}^{(1)}\right)=0, \quad i=0,1,2, \ldots, n-1y(2p+1)(ti(1))−sy(2p+1)(ti(1))=0,i=0,1,2,…,n−1
Furthermore, we have
201tiliupy 20it
| t i ( 1 ) − t i + 1 ( 1 ) | ≤ 2 ‖ Δ n ‖ , i = 0 , 1 , 2 , … , n − 2 t i ( 1 ) − t i + 1 ( 1 ) ≤ 2 Δ n , i = 0 , 1 , 2 , … , n − 2 |t_(i)^((1))-t_(i+1)^((1))| <= 2||Delta_(n)||,quad i=0,1,2,dots,n-2\left|t_{i}^{(1)}-t_{i+1}^{(1)}\right| \leq 2\left\|\Delta_{n}\right\|, \quad i=0,1,2, \ldots, n-2|ti(1)−ti+1(1)|≤2‖Δn‖,i=0,1,2,…,n−2
Applying again Rôlle's Theorem for y ( 2 p + 1 ) y ( 2 p + 1 ) y^((2p+1))y^{(2 p+1)}y(2p+1) one obtains the existence of the points t i ( 2 ) ∈ ( t i ( 1 ) , t i + 1 ( 1 ) ) , i = 0 , 1 , 2 , … , n − 2 t i ( 2 ) ∈ t i ( 1 ) , t i + 1 ( 1 ) , i = 0 , 1 , 2 , … , n − 2 t_(i)^((2))in(t_(i)^((1)),t_(i+1)^((1))),i=0,1,2,dots,n-2t_{i}^{(2)} \in\left(t_{i}^{(1)}, t_{i+1}^{(1)}\right), i=0,1,2, \ldots, n-2ti(2)∈(ti(1),ti+1(1)),i=0,1,2,…,n−2 such that
y ( 2 p + 2 ) ( t i ( 2 ) ) − s y ( 2 p + 2 ) ( t i ( 2 ) ) = 0 , i = 0 , 1 , 2 , … , n − 2 y ( 2 p + 2 ) t i ( 2 ) − s y ( 2 p + 2 ) t i ( 2 ) = 0 , i = 0 , 1 , 2 , … , n − 2 y^((2p+2))(t_(i)^((2)))-s_(y)^((2p+2))(t_(i)^((2)))=0,i=0,1,2,dots,n-2y^{(2 p+2)}\left(t_{i}^{(2)}\right)-s_{y}^{(2 p+2)}\left(t_{i}^{(2)}\right)=0, i=0,1,2, \ldots, n-2y(2p+2)(ti(2))−sy(2p+2)(ti(2))=0,i=0,1,2,…,n−2
and
| t i ( 2 ) − t i + 1 ( 2 ) | ≤ 3 ‖ Δ n ‖ , i = 0 , 1 , 2 , … , n − 3 t i ( 2 ) − t i + 1 ( 2 ) ≤ 3 Δ n , i = 0 , 1 , 2 , … , n − 3 |t_(i)^((2))-t_(i+1)^((2))| <= 3||Delta_(n)||,i=0,1,2,dots,n-3\left|t_{i}^{(2)}-t_{i+1}^{(2)}\right| \leq 3\left\|\Delta_{n}\right\|, i=0,1,2, \ldots, n-3|ti(2)−ti+1(2)|≤3‖Δn‖,i=0,1,2,…,n−3
A k k kkk-times applications of Rôlle's Theorem yields the points
t i ( k ) ∈ ( t i ( k − 1 ) , t i + 1 ( k − 1 ) ) , i = 0 , n − k ― , k = 1 , 2 , … , m − 1 t i ( k ) ∈ t i ( k − 1 ) , t i + 1 ( k − 1 ) , i = 0 , n − k ¯ , k = 1 , 2 , … , m − 1 t_(i)^((k))in(t_(i)^((k-1)),t_(i+1)^((k-1))),quad i= bar(0,n-k),k=1,2,dots,m-1t_{i}^{(k)} \in\left(t_{i}^{(k-1)}, t_{i+1}^{(k-1)}\right), \quad i=\overline{0, n-k}, k=1,2, \ldots, m-1ti(k)∈(ti(k−1),ti+1(k−1)),i=0,n−k―,k=1,2,…,m−1
such that
y ( 2 p + k ) ( t i ( k ) ) − s y ( 2 p + k ) ( t i ( k ) ) = 0 y ( 2 p + k ) t i ( k ) − s y ( 2 p + k ) t i ( k ) = 0 y^((2p+k))(t_(i)^((k)))-s_(y)^((2p+k))(t_(i)^((k)))=0y^{(2 p+k)}\left(t_{i}^{(k)}\right)-s_{y}^{(2 p+k)}\left(t_{i}^{(k)}\right)=0y(2p+k)(ti(k))−sy(2p+k)(ti(k))=0
and
| t i ( k ) − t i + 1 ( k ) | ≤ ( k + 1 ) ‖ Δ n ‖ , i = 0 , n − k ― and k = 1 , 2 , … , n − 1 t i ( k ) − t i + 1 ( k ) ≤ ( k + 1 ) Δ n , i = 0 , n − k ¯  and  k = 1 , 2 , … , n − 1 |t_(i)^((k))-t_(i+1)^((k))| <= (k+1)||Delta_(n)||,quad i= bar(0,n-k)" and "k=1,2,dots,n-1\left|t_{i}^{(k)}-t_{i+1}^{(k)}\right| \leq(k+1)\left\|\Delta_{n}\right\|, \quad i=\overline{0, n-k} \text { and } k=1,2, \ldots, n-1|ti(k)−ti+1(k)|≤(k+1)‖Δn‖,i=0,n−k― and k=1,2,…,n−1
For k = m − 1 k = m − 1 k=m-1k=m-1k=m−1 we obtain
and
y ( m + 2 p − 1 ) ( t i ( m − 1 ) ) − s y ( m + 2 p − 1 ) ( t i ( m − 1 ) ) = 0 , i = 0 , n − m + 1 ― y ( m + 2 p − 1 ) t i ( m − 1 ) − s y ( m + 2 p − 1 ) t i ( m − 1 ) = 0 , i = 0 , n − m + 1 ¯ y^((m+2p-1))(t_(i)^((m-1)))-s_(y)^((m+2p-1))(t_(i)^((m-1)))=0,quad i= bar(0,n-m+1)y^{(m+2 p-1)}\left(t_{i}^{(m-1)}\right)-s_{y}^{(m+2 p-1)}\left(t_{i}^{(m-1)}\right)=0, \quad i=\overline{0, n-m+1}y(m+2p−1)(ti(m−1))−sy(m+2p−1)(ti(m−1))=0,i=0,n−m+1―
Indise
| t i ( m − 1 ) − t i + 1 ( m − 1 ) | ≤ m ‖ Δ n ‖ , i = 0 , n − m + 1 ― t i ( m − 1 ) − t i + 1 ( m − 1 ) ≤ m Δ n , i = 0 , n − m + 1 ¯ |t_(i)^((m-1))-t_(i+1)^((m-1))| <= m||Delta_(n)||,quad i= bar(0,n-m+1)\left|t_{i}^{(m-1)}-t_{i+1}^{(m-1)}\right| \leq m\left\|\Delta_{n}\right\|, \quad i=\overline{0, n-m+1}|ti(m−1)−ti+1(m−1)|≤m‖Δn‖,i=0,n−m+1―
Since | a − t 0 ( m − 1 ) | < m | | Δ n ‖ a − t 0 ( m − 1 ) < m | | Δ n ‖ |a-t_(0)^((m-1))| < m||Delta_(n)||\left|a-t_{0}^{(m-1)}\right|<m| | \Delta_{n} \||a−t0(m−1)|<m||Δn‖ and | b − t n − m + 1 ( m − 1 ) | < ‖ Δ n ‖ b − t n − m + 1 ( m − 1 ) < Δ n |b-t_(n-m+1)^((m-1))| < ||Delta_(n)||\left|b-t_{n-m+1}^{(m-1)}\right|<\left\|\Delta_{n}\right\||b−tn−m+1(m−1)|<‖Δn‖ it follows that for every t ∈ [ a , b ] t ∈ [ a , b ] t in[a,b]t \in[a, b]t∈[a,b] there is i 0 ∈ { 0 , 1 , … , n − m + 1 } i 0 ∈ { 0 , 1 , … , n − m + 1 } i_(0)in{0,1,dots,n-m+1}i_{0} \in\{0,1, \ldots, n-m+1\}i0∈{0,1,…,n−m+1} such that
| t − t i 1 ( m − 1 ) | ≤ m ‖ Δ n ‖ t − t i 1 ( m − 1 ) ≤ m Δ n |t-t_(i_(1))^((m-1))| <= m||Delta_(n)||\left|t-t_{i_{1}}^{(m-1)}\right| \leq m\left\|\Delta_{n}\right\||t−ti1(m−1)|≤m‖Δn‖
and
| y ( m + 2 p − 1 ) ( t ) − s y ( m + 2 p − 1 ) ( t ) | = | ∫ t i 1 ( m − 1 ) t [ y ( m + 2 p ) ( u ) − s y ( m + 2 p ) ( u ) ] | ≤ y ( m + 2 p − 1 ) ( t ) − s y ( m + 2 p − 1 ) ( t ) = ∫ t i 1 ( m − 1 ) t   y ( m + 2 p ) ( u ) − s y ( m + 2 p ) ( u ) ≤ |y^((m+2p-1))(t)-s_(y)^((m+2p-1))(t)|=|int_(t_(i1)^((m-1)))^(t)[y^((m+2p))(u)-s_(y)^((m+2p))(u)]| <=\left|y^{(m+2 p-1)}(t)-s_{y}^{(m+2 p-1)}(t)\right|=\left|\int_{t_{i 1}^{(m-1)}}^{t}\left[y^{(m+2 p)}(u)-s_{y}^{(m+2 p)}(u)\right]\right| \leq|y(m+2p−1)(t)−sy(m+2p−1)(t)|=|∫ti1(m−1)t[y(m+2p)(u)−sy(m+2p)(u)]|≤
≤ ( ∫ t i , l ( m − 1 ) t d u ) 1 2 ⋅ ( ∫ t i , l ( m − 1 ) t ( y ( m + 2 p ) ( u ) − s y ( m + 2 p ) ( u ) ) 2 d u ) 1 2 ≤ ≤ m ‖ Δ n ‖ ⋅ ‖ y ( m + 2 p ) − s y ( m + 2 p ) ‖ 2 ≤ m ‖ Δ n ‖ 1 2 ‖ y ( m + 2 p ) ‖ 2 ≤ ∫ t i , l ( m − 1 ) t   d u 1 2 ⋅ ∫ t i , l ( m − 1 ) t   y ( m + 2 p ) ( u ) − s y ( m + 2 p ) ( u ) 2 d u 1 2 ≤ ≤ m Δ n ⋅ y ( m + 2 p ) − s y ( m + 2 p ) 2 ≤ m Δ n 1 2 y ( m + 2 p ) 2 {:[ <= (int_(t_(i,l)^((m-1)))^(t)(d)u)^((1)/(2))*(int_(t_(i,l)^((m-1)))^(t)(y^((m+2p))(u)-s_(y)^((m+2p))(u))^(2)(d)u)^((1)/(2)) <= ],[ <= sqrt(m||Delta_(n)||)*||y^((m+2p))-s_(y)^((m+2p))||_(2) <= sqrtm||Delta_(n)||(1)/(2)||y^((m+2p))||_(2)]:}\begin{aligned} & \leq\left(\int_{t_{i, l}^{(m-1)}}^{t} \mathrm{~d} u\right)^{\frac{1}{2}} \cdot\left(\int_{t_{i, l}^{(m-1)}}^{t}\left(y^{(m+2 p)}(u)-s_{y}^{(m+2 p)}(u)\right)^{2} \mathrm{~d} u\right)^{\frac{1}{2}} \leq \\ & \leq \sqrt{m\left\|\Delta_{n}\right\|} \cdot\left\|y^{(m+2 p)}-s_{y}^{(m+2 p)}\right\|_{2} \leq \sqrt{m}\left\|\Delta_{n}\right\| \frac{1}{2}\left\|y^{(m+2 p)}\right\|_{2} \end{aligned}≤(∫ti,l(m−1)t du)12⋅(∫ti,l(m−1)t(y(m+2p)(u)−sy(m+2p)(u))2 du)12≤≤m‖Δn‖⋅‖y(m+2p)−sy(m+2p)‖2≤m‖Δn‖12‖y(m+2p)‖2
(according to (17)).
Here from we deduce
‖ y ( m + 2 p − 1 ) − s x ( m + 2 p − 1 ) ‖ ∞ ≤ m ⋅ ‖ Δ n ⋅ ‖ 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( m + 2 p − 1 ) − s x ( m + 2 p − 1 ) ∞ ≤ m ⋅ Δ n ⋅ 1 2 ⋅ y ( m + 2 p ) 2 ||y^((m+2p-1))-s_(x)^((m+2p-1))||_(oo) <= sqrtm*||Delta_(n)*||(1)/(2)*||y^((m+2p))||_(2)\left\|y^{(m+2 p-1)}-s_{x}^{(m+2 p-1)}\right\|_{\infty} \leq \sqrt{m} \cdot\left\|\Delta_{n} \cdot\right\| \frac{1}{2} \cdot\left\|y^{(m+2 p)}\right\|_{2}‖y(m+2p−1)−sx(m+2p−1)‖∞≤m⋅‖Δn⋅‖12⋅‖y(m+2p)‖2
Similarly, for every t ∈ [ a , b ] t ∈ [ a , b ] t in[a,b]t \in[a, b]t∈[a,b] there exists i 0 ∈ { 0 , 1 , … , n − m + 2 } i 0 ∈ { 0 , 1 , … , n − m + 2 } i_(0)in{0,1,dots,n-m+2}i_{0} \in\{0,1, \ldots, n-m+2\}i0∈{0,1,…,n−m+2} such that
| t − t i 1 ( m − 2 ) | ≤ ( m − 1 ) ‖ Δ n ‖ t − t i 1 ( m − 2 ) ≤ ( m − 1 ) Δ n |t-t_(i_(1))^((m-2))| <= (m-1)||Delta_(n)||\left|t-t_{i_{1}}^{(m-2)}\right| \leq(m-1)\left\|\Delta_{n}\right\||t−ti1(m−2)|≤(m−1)‖Δn‖
so that
| y ( m + 2 p − 2 ) ( t ) − s y ( m + 2 p − 2 ) ( t ) | = | ∫ t i ′ ( m − 2 , 1 [ y ( m + 2 p − 1 ) ( u ) − s y ( m + 2 p − 1 ) ( u ) ] | ≤ ≤ | t − t i 0 ( m − 2 ) | ⋅ ‖ y ( m + 2 p + 1 ) − s y ( m + 2 p − 1 ) ‖ ∞ ≤ ≤ m ( m − 1 ) ‖ Δ n ‖ 1 + 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( m + 2 p − 2 ) ( t ) − s y ( m + 2 p − 2 ) ( t ) = ∫ t i ′ ( m − 2 , 1   y ( m + 2 p − 1 ) ( u ) − s y ( m + 2 p − 1 ) ( u ) ≤ ≤ t − t i 0 ( m − 2 ) ⋅ y ( m + 2 p + 1 ) − s y ( m + 2 p − 1 ) ∞ ≤ ≤ m ( m − 1 ) Δ n 1 + 1 2 ⋅ y ( m + 2 p ) 2 {:[|y^((m+2p-2))(t)-s_(y)^((m+2p-2))(t)|=|int_(t_(i)^(')(m-2,)^(1)[y^((m+2p-1))(u)-s_(y)^((m+2p-1))(u)]| <= ],[ <= |t-t_(i_(0))^((m-2))|*||y^((m+2p+1))-s_(y)^((m+2p-1))||_(oo) <= ],[ <= sqrt(m(m-1))||Delta_(n)||^(1+(1)/(2))*||y^((m+2p))||_(2)]:}\begin{aligned} & \left|y^{(m+2 p-2)}(t)-s_{y}^{(m+2 p-2)}(t)\right|=\left|\int_{t_{i}^{\prime}(m-2,}^{1}\left[y^{(m+2 p-1)}(u)-s_{y}^{(m+2 p-1)}(u)\right]\right| \leq \\ & \leq\left|t-t_{i_{0}}^{(m-2)}\right| \cdot\left\|y^{(m+2 p+1)}-s_{y}^{(m+2 p-1)}\right\|_{\infty} \leq \\ & \leq \sqrt{m(m-1)}\left\|\Delta_{n}\right\|^{1+\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2} \end{aligned}|y(m+2p−2)(t)−sy(m+2p−2)(t)|=|∫ti′(m−2,1[y(m+2p−1)(u)−sy(m+2p−1)(u)]|≤≤|t−ti0(m−2)|⋅‖y(m+2p+1)−sy(m+2p−1)‖∞≤≤m(m−1)‖Δn‖1+12⋅‖y(m+2p)‖2
It follows
‖ y ( m + 2 p − 2 ) − s y ( m + 2 p − 2 ) ‖ ∞ ≤ m ( m − 1 ) ‖ Δ n ‖ 1 + 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( m + 2 p − 2 ) − s y ( m + 2 p − 2 ) ∞ ≤ m ( m − 1 ) Δ n 1 + 1 2 ⋅ y ( m + 2 p ) 2 ||y^((m+2p-2))-s_(y)^((m+2p-2))||_(oo) <= sqrtm(m-1)||Delta_(n)||^(1+(1)/(2))*||y^((m+2p))||_(2)\left\|y^{(m+2 p-2)}-s_{y}^{(m+2 p-2)}\right\|_{\infty} \leq \sqrt{m}(m-1)\left\|\Delta_{n}\right\|^{1+\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2}‖y(m+2p−2)−sy(m+2p−2)‖∞≤m(m−1)‖Δn‖1+12⋅‖y(m+2p)‖2
In general we find
‖ y ( m + 2 p − 1 ) − s y ( m + 2 p − 1 ) ‖ ∞ ≤ m ( m − 1 ) ⋯ ( m − l + 1 ) ‖ Δ m ‖ | 1 l − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( m + 2 p − 1 ) − s y ( m + 2 p − 1 ) ∞ ≤ m ( m − 1 ) ⋯ ( m − l + 1 ) Δ m 1 l − 1 2 ⋅ y ( m + 2 p ) 2 ||y^((m+2p-1))-s_(y)^((m+2p-1))||_(oo) <= sqrtm(m-1)cdots(m-l+1)||Delta_(m)|||_(1)^(l-(1)/(2))*||y^((m+2p))||_(2)\left\|y^{(m+2 p-1)}-s_{y}^{(m+2 p-1)}\right\|_{\infty} \leq\left.\sqrt{m}(m-1) \cdots(m-l+1)\left\|\Delta_{m}\right\|\right|_{1} ^{l-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2}‖y(m+2p−1)−sy(m+2p−1)‖∞≤m(m−1)⋯(m−l+1)‖Δm‖|1l−12⋅‖y(m+2p)‖2
for all l = 2 , 3 , 4 … , m l = 2 , 3 , 4 … , m l=2,3,4dots,ml=2,3,4 \ldots, ml=2,3,4…,m.
Remark 3. For 1 = m 1 = m 1=m1=m1=m we find
(20) ‖ y ( 2 p ) − 5 y ( 2 p ) ‖ ∞ ≤ m ( m − 1 ) ! ‖ Δ n ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 . (20) ‖ y ( 2 p ) − 5 y ( 2 p ) ∞ ≤ m ( m − 1 ) ! Δ n m − 1 2 ⋅ y ( m + 2 p ) ‖ 2 . {:(20)||y^((2p)-5y^((2p))||_(oo) <= sqrtm(m-1)!||Delta_(n)||^(m-(1)/(2))*||y^((m+2p))||_(2).):}\begin{equation*} \| y^{(2 p)-5 y^{(2 p)}\left\|_{\infty} \leq \sqrt{m}(m-1)!\right\| \Delta_{n}\left\|^{m-\frac{1}{2}} \cdot\right\| y^{(m+2 p)} \|_{2} .} \tag{20} \end{equation*}(20)‖y(2p)−5y(2p)‖∞≤m(m−1)!‖Δn‖m−12⋅‖y(m+2p)‖2.
In the following we shall give estimations for the norms
(2I) ‖ y ( q ) − s y ( q ) ‖ ∞ , q = 0 , 1 , 2 , … , 2 p − 1 , (2I) y ( q ) − s y ( q ) ∞ , q = 0 , 1 , 2 , … , 2 p − 1 , {:(2I)||y^((q))-s_(y)^((q))||_(oo)","q=0","1","2","dots","2p-1",":}\begin{equation*} \left\|y^{(q)}-s_{y}^{(q)}\right\|_{\infty}, q=0,1,2, \ldots, 2 p-1, \tag{2I} \end{equation*}(2I)‖y(q)−sy(q)‖∞,q=0,1,2,…,2p−1,
necessary for the numerical treatment of the problem (1)-(2).
COROLLARY 6. If the exact solution y y yyy of the problem (1)-(2) belongs to H 2 m + 2 p ( [ a , b ] ) H 2 m + 2 p ( [ a , b ] ) H_(2)^(m+2p)([a,b])H_{2}^{m+2 p}([a, b])H2m+2p([a,b]) and s y ∈ S 2 m + 2 p − 1 ( Δ n ) s y ∈ S 2 m + 2 p − 1 Δ n s_(y)inS_(2m+2p-1)(Delta_(n))s_{y} \in S_{2 m+2 p-1}\left(\Delta_{n}\right)sy∈S2m+2p−1(Δn) is the associated spline solution, then the following estimation hold:
‖ y ( q ) − s ( q ) ‖ ∞ ≤ ( b − a ) 2 p − q m ( m − 1 ) ! ⋅ ‖ Δ a ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( q ) − s ( q ) ∞ ≤ ( b − a ) 2 p − q m ( m − 1 ) ! ⋅ Δ a m − 1 2 ⋅ y ( m + 2 p ) 2 ||y^((q))-s^((q))||_(oo) <= (b-a)^(2p-q)sqrtm(m-1)!*||Delta_(a)||^(m-(1)/(2))*||y^((m+2p))||_(2)\left\|y^{(q)}-s^{(q)}\right\|_{\infty} \leq(b-a)^{2 p-q} \sqrt{m}(m-1)!\cdot\left\|\Delta_{a}\right\|^{m-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2}‖y(q)−s(q)‖∞≤(b−a)2p−qm(m−1)!⋅‖Δa‖m−12⋅‖y(m+2p)‖2
for q = 0 , 1 , 2 … .2 p − 1 q = 0 , 1 , 2 … .2 p − 1 q=0,1,2dots.2 p-1q=0,1,2 \ldots .2 p-1q=0,1,2….2p−1.
Proof. Since y ( t 0 ) − s v ( t 0 ) = y ( t n ) − s y ( t n ) = 0 y t 0 − s v t 0 = y t n − s y t n = 0 y(t_(0))-s_(v)(t_(0))=y(t_(n))-s_(y)(t_(n))=0y\left(t_{0}\right)-s_{v}\left(t_{0}\right)=y\left(t_{n}\right)-s_{y}\left(t_{n}\right)=0y(t0)−sv(t0)=y(tn)−sy(tn)=0, it follows that there exists at least one point t 0 ( 1 ) ∈ ( t 0 , t i j ) t 0 ( 1 ) ∈ t 0 , t i j t_(0)^((1))in(t_(0),t_(ij))t_{0}^{(1)} \in\left(t_{0}, t_{i j}\right)t0(1)∈(t0,tij) such that
y ′ ( t 0 ( 1 ) ) − s y ′ ( t 0 ( 1 ) ) = 0 y ′ t 0 ( 1 ) − s y ′ t 0 ( 1 ) = 0 y^(')(t_(0)^((1)))-s_(y)^(')(t_(0)^((1)))=0y^{\prime}\left(t_{0}^{(1)}\right)-s_{y}^{\prime}\left(t_{0}^{(1)}\right)=0y′(t0(1))−sy′(t0(1))=0
implying
y ′ ( t 0 ) − s y ′ ( t 0 ) = y ′ ( t 0 ( 1 ) ) − s y ′ ( t 0 ( 1 ) ) = y ′ ( t n ) − s y ′ ( t n ) = 0 y ′ t 0 − s y ′ t 0 = y ′ t 0 ( 1 ) − s y ′ t 0 ( 1 ) = y ′ t n − s y ′ t n = 0 y^(')(t_(0))-s_(y)^(')(t_(0))=y^(')(t_(0)^((1)))-s_(y)^(')(t_(0)^((1)))=y^(')(t_(n))-s_(y)^(')(t_(n))=0y^{\prime}\left(t_{0}\right)-s_{y}^{\prime}\left(t_{0}\right)=y^{\prime}\left(t_{0}^{(1)}\right)-s_{y}^{\prime}\left(t_{0}^{(1)}\right)=y^{\prime}\left(t_{n}\right)-s_{y}^{\prime}\left(t_{n}\right)=0y′(t0)−sy′(t0)=y′(t0(1))−sy′(t0(1))=y′(tn)−sy′(tn)=0
Then it will exist the points t 0 ( 2 ) ∈ ( t 0 , t 0 ( 1 ) ) , t 1 ( 2 ) ∈ ( t 0 ( 1 ) , t n ) , t 0 < t 0 ( 2 ) << t 1 ( 2 ) < t n t 0 ( 2 ) ∈ t 0 , t 0 ( 1 ) , t 1 ( 2 ) ∈ t 0 ( 1 ) , t n , t 0 < t 0 ( 2 ) << t 1 ( 2 ) < t n t_(0)^((2))in(t_(0),t_(0)^((1))),t_(1)^((2))in(t_(0)^((1)),t_(n)),quadt_(0) < t_(0)^((2))<<t_(1)^((2)) < t_(n)t_{0}^{(2)} \in\left(t_{0}, t_{0}^{(1)}\right), t_{1}^{(2)} \in\left(t_{0}^{(1)}, t_{n}\right), \quad t_{0}<t_{0}^{(2)}< <t_{1}^{(2)}<t_{n}t0(2)∈(t0,t0(1)),t1(2)∈(t0(1),tn),t0<t0(2)<<t1(2)<tn such that
y ′ ′ ( t 0 ) − s y ′ ′ ( t 0 ) = y ′ ′ ( t 0 ( 2 ) ) − s y ′ ′ ( t 0 ( 2 ) ) = y ′ ′ ( t 1 ( 2 ) ) − s y ′ ′ ( t 1 ( 2 ) ) = y ′ ′ ( t n ) − s y ′ ′ ( t n ) = 0 . y ′ ′ t 0 − s y ′ ′ t 0 = y ′ ′ t 0 ( 2 ) − s y ′ ′ t 0 ( 2 ) = y ′ ′ t 1 ( 2 ) − s y ′ ′ t 1 ( 2 ) = y ′ ′ t n − s y ′ ′ t n = 0 . {:[y^('')(t_(0))-s_(y)^('')(t_(0))=y^('')(t_(0)^((2)))-s_(y)^('')(t_(0)^((2)))=y^('')(t_(1)^((2)))-s_(y)^('')(t_(1)^((2)))=],[y^('')(t_(n))-s_(y)^('')(t_(n))=0.]:}\begin{gathered} y^{\prime \prime}\left(t_{0}\right)-s_{y}^{\prime \prime}\left(t_{0}\right)=y^{\prime \prime}\left(t_{0}^{(2)}\right)-s_{y}^{\prime \prime}\left(t_{0}^{(2)}\right)=y^{\prime \prime}\left(t_{1}^{(2)}\right)-s_{y}^{\prime \prime}\left(t_{1}^{(2)}\right)= \\ y^{\prime \prime}\left(t_{n}\right)-s_{y}^{\prime \prime}\left(t_{n}\right)=0 . \end{gathered}y′′(t0)−sy′′(t0)=y′′(t0(2))−sy′′(t0(2))=y′′(t1(2))−sy′′(t1(2))=y′′(tn)−sy′′(tn)=0.
In general, for every q ∈ { 2 , … , p − 1 } q ∈ { 2 , … , p − 1 } q in{2,dots,p-1}q \in\{2, \ldots, p-1\}q∈{2,…,p−1} there are the points t 0 ( q ) ∈ ( t 0 , t 0 ( q − 1 ) ) , t 1 ( q ) ∈ ( t 0 ( q − 1 ) , t 1 ( q − 1 ) ) , ⋯ , t q − 1 ( q ) ∈ ( t q − 2 ( q − 1 ) , t n ) , Δ t 0 < t 0 ( q ) < t 1 ( q ) < ⋯ < t q − 1 ( q ) < t n t 0 ( q ) ∈ t 0 , t 0 ( q − 1 ) , t 1 ( q ) ∈ t 0 ( q − 1 ) , t 1 ( q − 1 ) , ⋯ , t q − 1 ( q ) ∈ t q − 2 ( q − 1 ) , t n , Δ t 0 < t 0 ( q ) < t 1 ( q ) < ⋯ < t q − 1 ( q ) < t n t_(0)^((q))in(t_(0),t_(0)^((q-1))),quadt_(1)^((q))in(t_(0)^((q-1)),t_(1)^((q-1))),cdots,t_(q-1)^((q))in(t_(q-2)^((q-1)),t_(n)),Deltat_(0) < t_(0)^((q)) < t_(1)^((q)) < cdots < t_(q-1)^((q)) < t_(n)t_{0}^{(q)} \in\left(t_{0}, t_{0}^{(q-1)}\right), \quad t_{1}^{(q)} \in\left(t_{0}^{(q-1)}, t_{1}^{(q-1)}\right), \cdots, t_{q-1}^{(q)} \in\left(t_{q-2}^{(q-1)}, t_{n}\right), \Delta t_{0}<t_{0}^{(q)}<t_{1}^{(q)}<\cdots <t_{q-1}^{(q)}<t_{n}t0(q)∈(t0,t0(q−1)),t1(q)∈(t0(q−1),t1(q−1)),⋯,tq−1(q)∈(tq−2(q−1),tn),Δt0<t0(q)<t1(q)<⋯<tq−1(q)<tn such that
y ( q ) ( t 0 ) − s y ( q ) ( t 0 ) = y ( q ) ( t 0 ( q ) ) − s y ( q ) ( t 0 ( q ) ) = ⋯ ⋯ = y ( q ) ( t q − 1 ( q ) ) − s y ( q ) ( t q − 1 ( q ) ) = y ( q ) ( t n ) − s ( q ) ( t n ) = 0 . y ( q ) t 0 − s y ( q ) t 0 = y ( q ) t 0 ( q ) − s y ( q ) t 0 ( q ) = ⋯ ⋯ = y ( q ) t q − 1 ( q ) − s y ( q ) t q − 1 ( q ) = y ( q ) t n − s ( q ) t n = 0 . {:[y^((q))(t_(0))-s_(y)^((q))(t_(0))=y^((q))(t_(0)^((q)))-s_(y)^((q))(t_(0)^((q)))=cdots],[ cdots=y^((q))(t_(q-1)^((q)))-s_(y)^((q))(t_(q-1)^((q)))=y^((q))(t_(n))-s^((q))(t_(n))=0.]:}\begin{aligned} & y^{(q)}\left(t_{0}\right)-s_{y}^{(q)}\left(t_{0}\right)=y^{(q)}\left(t_{0}^{(q)}\right)-s_{y}^{(q)}\left(t_{0}^{(q)}\right)=\cdots \\ & \cdots=y^{(q)}\left(t_{q-1}^{(q)}\right)-s_{y}^{(q)}\left(t_{q-1}^{(q)}\right)=y^{(q)}\left(t_{n}\right)-s^{(q)}\left(t_{n}\right)=0 . \end{aligned}y(q)(t0)−sy(q)(t0)=y(q)(t0(q))−sy(q)(t0(q))=⋯⋯=y(q)(tq−1(q))−sy(q)(tq−1(q))=y(q)(tn)−s(q)(tn)=0.
For q = p − 1 q = p − 1 q=p-1q=p-1q=p−1 we deduce the existence of p + 1 p + 1 p+1p+1p+1 distinct points
t 0 < t 0 ( p − 1 ) < t 1 ( p − 1 ) < ⋯ < t p − 2 ( p − 1 ) < t n t 0 < t 0 ( p − 1 ) < t 1 ( p − 1 ) < ⋯ < t p − 2 ( p − 1 ) < t n t_(0) < t_(0)^((p-1)) < t_(1)^((p-1)) < cdots < t_(p-2)^((p-1)) < t_(n)t_{0}<t_{0}^{(p-1)}<t_{1}^{(p-1)}<\cdots<t_{p-2}^{(p-1)}<t_{n}t0<t0(p−1)<t1(p−1)<⋯<tp−2(p−1)<tn
at which the ( p − 1 ) ( p − 1 ) (p-1)(p-1)(p−1) - derivative of the difference y ( t ) − s y ( t ) y ( t ) − s y ( t ) y(t)-s_(y)(t)y(t)-s_{y}(t)y(t)−sy(t) vanishes
Finally, we deduce the existence of a point t ¯ 1 ∈ ( a , b ) t ¯ 1 ∈ ( a , b ) bar(t)_(1)in(a,b)\bar{t}_{1} \in(a, b)t¯1∈(a,b) such that
y ( 2 p − 1 ) ( t ¯ 1 ) − s y ( 2 p − 1 ) ( t ¯ 1 ) = 0 y ( 2 p − 1 ) t ¯ 1 − s y ( 2 p − 1 ) t ¯ 1 = 0 y^((2p-1))( bar(t)_(1))-s_(y)^((2p-1))( bar(t)_(1))=0y^{(2 p-1)}\left(\bar{t}_{1}\right)-s_{y}^{(2 p-1)}\left(\bar{t}_{1}\right)=0y(2p−1)(t¯1)−sy(2p−1)(t¯1)=0
But then, for all t ∈ [ a , b ] t ∈ [ a , b ] t in[a,b]t \in[a, b]t∈[a,b] we have
-|yothic who (2p-1) ( t ) − s y ( 2 p − 1 ) ( t ) | = | ∫ t 1 t [ y ( 2 p ) ( h ) − s y ( 2 p ) ( h ) ] d h | ≤ ( t ) − s y ( 2 p − 1 ) ( t ) = ∫ t 1 t   y ( 2 p ) ( h ) − s y ( 2 p ) ( h ) d h ≤ (t)-s_(y)^((2p-1))(t)|=|int_(t_(1))^(t)[y^((2p))(h)-s_(y)^((2p))(h)]dh| <= :}(t)-s_{y}^{(2 p-1)}(t)\left|=\left|\int_{t_{1}}^{t}\left[y^{(2 p)}(h)-s_{y}^{(2 p)}(h)\right] d h\right| \leq\right.(t)−sy(2p−1)(t)|=|∫t1t[y(2p)(h)−sy(2p)(h)]dh|≤ intoline sith to zehon and an ≤ | t − t 1 | ⋅ ‖ y ( 2 n ) − s 2 ( 2 p ) ‖ ∞ ≤ t − t 1 ⋅ y ( 2 n ) − s 2 ( 2 p ) ∞ <= |t-t_(1)|*||y^((2n))-s_(2)^((2p))||_(oo)\leq\left|t-t_{1}\right| \cdot\left\|y^{(2 n)}-s_{2}^{(2 p)}\right\|_{\infty}≤|t−t1|⋅‖y(2n)−s2(2p)‖∞
so that
‖ v ( 2 p − 1 ) − s y ( 2 p − 1 ) ‖ ∞ ≤ ( a − b ) m ( m − 1 ) ! ‖ Δ n ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 v ( 2 p − 1 ) − s y ( 2 p − 1 ) ∞ ≤ ( a − b ) m ( m − 1 ) ! Δ n m − 1 2 ⋅ y ( m + 2 p ) 2 ||v^((2p-1))-s_(y)^((2p-1))||_(oo) <= (a-b)sqrtm(m-1)!||Delta_(n)||^(m-(1)/(2))*||y^((m+2p))||_(2)\left\|v^{(2 p-1)}-s_{y}^{(2 p-1)}\right\|_{\infty} \leq(a-b) \sqrt{m}(m-1)!\left\|\Delta_{n}\right\|^{m-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2}‖v(2p−1)−sy(2p−1)‖∞≤(a−b)m(m−1)!‖Δn‖m−12⋅‖y(m+2p)‖2
Similarly, there is t ¯ 2 ∈ ( a , b ) t ¯ 2 ∈ ( a , b ) bar(t)_(2)in(a,b)\bar{t}_{2} \in(a, b)t¯2∈(a,b) such that
| y ( 2 p − 2 ) ( t ) − s y ( 2 p − 2 ) ( t ) | = | ∫ 1 t t [ y ( 2 p − 1 ) ( h ) − s y ( 2 p − 1 ) ( h ) ] d h | ≤ ≤ ( h − a ) ‖ y ( 2 p − 1 ) − s y ( 2 p − 1 ) ‖ ∞ y ( 2 p − 2 ) ( t ) − s y ( 2 p − 2 ) ( t ) = ∫ 1 t t   y ( 2 p − 1 ) ( h ) − s y ( 2 p − 1 ) ( h ) d h ≤ ≤ ( h − a ) y ( 2 p − 1 ) − s y ( 2 p − 1 ) ∞ {:[|y^((2p-2))(t)-s_(y)^((2p-2))(t)|=|int_((1)/(t))^(t)[y^((2p-1))(h)-s_(y)^((2p-1))(h)]dh| <= ],[ <= (h-a)||y^((2p-1))-s_(y)^((2p-1))||_(oo)]:}\begin{gathered} \left|y^{(2 p-2)}(t)-s_{y}^{(2 p-2)}(t)\right|=\left|\int_{\frac{1}{t}}^{t}\left[y^{(2 p-1)}(h)-s_{y}^{(2 p-1)}(h)\right] \mathrm{d} h\right| \leq \\ \leq(h-a)\left\|y^{(2 p-1)}-s_{y}^{(2 p-1)}\right\|_{\infty} \end{gathered}|y(2p−2)(t)−sy(2p−2)(t)|=|∫1tt[y(2p−1)(h)−sy(2p−1)(h)]dh|≤≤(h−a)‖y(2p−1)−sy(2p−1)‖∞
where from
‖ y ( 2 p − 2 ) − s y ( 2 p − 2 ) ‖ ∞ ≤ ( b − a ) 2 m ( m − 1 ) ! ‖ Δ n ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 . y ( 2 p − 2 ) − s y ( 2 p − 2 ) ∞ ≤ ( b − a ) 2 m ( m − 1 ) ! Δ n m − 1 2 ⋅ y ( m + 2 p ) 2 . ||y^((2p-2))-s_(y)^((2p-2))||_(oo) <= (b-a)^(2)sqrtm(m-1)!||Delta_(n)||^(m-(1)/(2))*||y^((m+2p))||_(2).\left\|y^{(2 p-2)}-s_{y}^{(2 p-2)}\right\|_{\infty} \leq(b-a)^{2} \sqrt{m}(m-1)!\left\|\Delta_{n}\right\|^{m-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2} .‖y(2p−2)−sy(2p−2)‖∞≤(b−a)2m(m−1)!‖Δn‖m−12⋅‖y(m+2p)‖2.
Continuing in this manner, we obtain
‖ y ( 2 p − 1 ) − s y ( 2 ) ( 2 p − 1 ) ‖ ∞ ≤ ( b − a ) l m ( m − 1 ) ! ‖ Δ l ‖ ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 y ( 2 p − 1 ) − s y ( 2 ) ( 2 p − 1 ) ∞ ≤ ( b − a ) l m ( m − 1 ) ! Δ l m − 1 2 ⋅ y ( m + 2 p ) ‖ 2 ||y^((2p-1))-s_(y^((2)))^((2p-1))||_(oo) <= (b-a)^(l)sqrtm(m-1)!||Delta_(l)||||^(m-(1)/(2))*||y^((m+2p))||_(2)\left\|y^{(2 p-1)}-s_{y^{(2)}}^{(2 p-1)}\right\|_{\infty} \leq(b-a)^{l} \sqrt{m}(m-1)!\left\|\Delta_{l}\right\|\left\|^{m-\frac{1}{2}} \cdot\right\| y^{(m+2 p)} \|_{2}‖y(2p−1)−sy(2)(2p−1)‖∞≤(b−a)lm(m−1)!‖Δl‖‖m−12⋅‖y(m+2p)‖2
for l = 0 , 1 , 2 , … , 2 p − 1 l = 0 , 1 , 2 , … , 2 p − 1 l=0,1,2,dots,2p-1l=0,1,2, \ldots, 2 p-1l=0,1,2,…,2p−1.
Therefore
‖ y ( q ) − s y ( q ) ( q ) ‖ ∞ ≤ ( b − a ) 2 p − q m ( m − 1 ) ! ‖ Δ n ‖ m − 1 2 ⋅ ‖ y ( m + 2 p ) ‖ 2 , y ( q ) − s y ( q ) ( q ) ∞ ≤ ( b − a ) 2 p − q m ( m − 1 ) ! Δ n m − 1 2 ⋅ y ( m + 2 p ) 2 , ||y^((q))-s_(y^((q)))^((q))||_(oo) <= (b-a)^(2p-q)sqrtm(m-1)!||Delta_(n)||^(m-(1)/(2))*||y^((m+2p))||_(2),\left\|y^{(q)}-s_{y^{(q)}}^{(q)}\right\|_{\infty} \leq(b-a)^{2 p-q} \sqrt{m}(m-1)!\left\|\Delta_{n}\right\|^{m-\frac{1}{2}} \cdot\left\|y^{(m+2 p)}\right\|_{2},‖y(q)−sy(q)(q)‖∞≤(b−a)2p−qm(m−1)!‖Δn‖m−12⋅‖y(m+2p)‖2,
which ends the proof.
Remark 4. For q = 0 q = 0 q=0q=0q=0 one obtains
‖ y − s y ‖ ∞ ≤ ( b − a ) 2 p m ( m − 1 ) ! ‖ Δ n ‖ m − 1 2 ‖ y ( m + 2 p ) ‖ 2 . y − s y ∞ ≤ ( b − a ) 2 p m ( m − 1 ) ! Δ n m − 1 2 y ( m + 2 p ) 2 . ||y-s_(y)||_(oo) <= (b-a)^(2p)sqrtm(m-1)!||Delta_(n)||^(m-(1)/(2))||y^((m+2p))||_(2).\left\|y-s_{y}\right\|_{\infty} \leq(b-a)^{2 p} \sqrt{m}(m-1)!\left\|\Delta_{n}\right\|^{m-\frac{1}{2}}\left\|y^{(m+2 p)}\right\|_{2} .‖y−sy‖∞≤(b−a)2pm(m−1)!‖Δn‖m−12‖y(m+2p)‖2.
By Definition 1, m ≥ 2 m ≥ 2 m >= 2m \geq 2m≥2 and then ‖ y − s y ‖ ∞ y − s y ∞ ||y-s_(y)||_(oo)\left\|y-s_{y}\right\|_{\infty}‖y−sy‖∞ is at least of order
O ( ‖ Δ n ‖ 3 2 ) O Δ n 3 2 O(||Delta_(n)||^((3)/(2)))O\left(\left\|\Delta_{n}\right\|^{\frac{3}{2}}\right)O(‖Δn‖32)
Example.
Consider the problem
( P 1 ) { y ( 4 ) ( t ) = ( t 4 + 14 t 3 + 49 t 2 + 32 t − 12 ) e t , t ∈ [ 0 , 1 ] y ( 0 ) = y ′ ( 0 ) = y ( 1 ) = y ′ ( 1 ) = 0 ( P 1 ) y ( 4 ) ( t ) = t 4 + 14 t 3 + 49 t 2 + 32 t − 12 e t , t ∈ [ 0 , 1 ] y ( 0 ) = y ′ ( 0 ) = y ( 1 ) = y ′ ( 1 ) = 0 (P1){[y^((4))(t)=(t^(4)+14t^(3)+49t^(2)+32 t-12)e^(t)","t in[0","1]],[y(0)=y^(')(0)=y(1)=y^(')(1)=0]:}(P 1)\left\{\begin{array}{l} y^{(4)}(t)=\left(t^{4}+14 t^{3}+49 t^{2}+32 t-12\right) e^{t}, t \in[0,1] \\ y(0)=y^{\prime}(0)=y(1)=y^{\prime}(1)=0 \end{array}\right.(P1){y(4)(t)=(t4+14t3+49t2+32t−12)et,t∈[0,1]y(0)=y′(0)=y(1)=y′(1)=0
Problem (P1) is a problem of Karpilovskaja type for a fourth order differential equation which is studied also in [12].
In Table 1 the maximum values of the error at the nodes of the uniform partition Δ n : n = 5 , 10 , 20 , 30 , 40 Δ n : n = 5 , 10 , 20 , 30 , 40 Delta_(n):n=5,10,20,30,40\Delta_{n}: n=5,10,20,30,40Δn:n=5,10,20,30,40 are presented

Table 1

n maximum values of the error at the nodes of Δ ‖ Δ ‖ Delta_(||)\Delta_{\|}Δ‖
5 0.0000380786
10 0.0000023077
20 0.0000001162
30 0.0000000198
40 0.0000000055
n maximum values of the error at the nodes of Delta_(||) 5 0.0000380786 10 0.0000023077 20 0.0000001162 30 0.0000000198 40 0.0000000055| n | maximum values of the error at the nodes of $\Delta_{\\|}$ | | ---: | :---: | | 5 | 0.0000380786 | | 10 | 0.0000023077 | | 20 | 0.0000001162 | | 30 | 0.0000000198 | | 40 | 0.0000000055 |

REFERENCES

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Received Mars 03, 1998
"T. Popoviciu" Institut of Numerical Analysis
P.O. Box 68
3400 Cluj-Napoca 1
Romania
  1. THEOREM 2. Suppose that f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R verifies the conditions:
    f ( q ) ( a ) = α ( q ) , q = 0 , 1 , 2 , … , p − 1 f ( q ) ( a ) = α ( q ) , q = 0 , 1 , 2 , … , p − 1 f^((q))(a)=alpha^((q)),quad q=0,1,2,dots,p-1f^{(q)}(a)=\alpha^{(q)}, \quad q=0,1,2, \ldots, p-1f(q)(a)=α(q),q=0,1,2,…,p−1
    (9) f ( q ) ( b ) = β ( q ) , q = 0 , 1 , 2 , … , p − 1 (9) f ( q ) ( b ) = β ( q ) , q = 0 , 1 , 2 , … , p − 1 {:(9)f^((q))(b)=beta^((q))","quad q=0","1","2","dots","p-1:}\begin{equation*} f^{(q)}(b)=\beta^{(q)}, \quad q=0,1,2, \ldots, p-1 \tag{9} \end{equation*}(9)f(q)(b)=β(q),q=0,1,2,…,p−1
1999

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