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<title>Optimal properties for deficient quartic splines <br />of Marsden type: Optimal properties for deficient quartic splines <br />of Marsden type</title>
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<h1>Optimal properties for deficient quartic splines <br />of Marsden type</h1>
<p class="authors">
<span class="author">A.M. Bica\(^\ast \), D. Curilă-Popescu\(^{\ast \ast }\) M. Curilă\(^{\ast \ast \ast }\)</span>
</p>
<p class="date">November 14, 2020; accepted: December 15, 2020; published online: February 20, 2021.</p>
</div>
<div class="abstract"><p> In this work, we obtain an improved error estimate in the interpolation with the Hermite C\(^{2}\)-smooth deficient complete quartic spline that has the distribution of nodes following the Marsden type scheme and investigate the possibilities to compute the derivatives on the knots such that the obtained spline \(S\in C^{1}[a,b]\) has minimal curvature and minimal \(L^{2}\)-norm of \(S^{\prime }\) and \(S^{\prime \prime \prime }\). In each case, the interpolation error estimate is performed in terms of the modulus of continuity. </p>
<p><b class="bf">MSC.</b> 65D07, 65D10 </p>
<p><b class="bf">Keywords.</b> Marsden type deficient quartic splines, error estimates, optimal properties, minimal curvature, minimal average slope of the graph </p>
</div>
<p>\(^\ast \)Department of Mathematics and Computer Science, University of Oradea, str. Universităţii no. 1, 410087 Oradea, Romania, e-mail: <span class="ttfamily">abica@uoradea.ro, curila_diana@yahoo.com</span>. </p>
<p>\(^{\ast \ast }\)Department of Mathematics and Computer Science, University of Oradea, str. Universităţii no. 1, 410087 Oradea, Romania, e-mail: <span class="ttfamily">curila_diana@yahoo.com</span> </p>
<p>\(^{\ast \ast \ast }\)Department of Environmental Engineering, str. G. Magheru, no. 26, Oradea, Romania, e-mail: <span class="ttfamily">mirceacurila@yahoo.com</span> </p>
<h1 id="a0000000002">1 Introduction</h1>
<p>Motivated by the nice properties of complete cubic splines, and trying to increase the degree of \(C^{2}\)-smooth complete spline interpolant, but preserving the bandwidth of the diagonally dominant system, Howell introduced in <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span> the deficient complete quartic spline that match the interpolated function \(f\in C[a,b]\) on the interpolation nodes and on midpoints of a grid following the Marsden scheme (see <span class="cite">
	[
	<a href="#marsden" >17</a>
	]
</span> and <span class="cite">
	[
	<a href="#volkov2014" >25</a>
	]
</span>). More precisely, considering a grid </p>
<div class="displaymath" id="a0000000003">
  \begin{align*}  a=x_{0}{\lt}x_{1}{\lt}\ldots {\lt}x_{n-1}{\lt}x_{n}=b \end{align*}
</div>
<p>and the midpoints \(z_{i}=x_{i-1/2}=\tfrac {x_{i-1}+x_{i}}{2},\) \(i=\overline{1,n},\) Howell proved the existence and uniqueness of the deficient quartic spline \(S\in C^{2}[a,b]\) taking prescribed values on \(x_{i},\) \(i=\overline{0,n}\) and on \(x_{i-1/2},\) \(i=\overline{1,n},\) with the endpoint interpolation conditions \(S^{\prime }\left( a\right) =f^{\prime }\left(a\right) ,\) \(S^{\prime }\left( b\right) =f^{\prime }\left( b\right) .\) Moreover, by the continuity condition \(S\in C^{2}[a,b]\) the spline must satisfy the tridiagonal system of equations </p>
<div class="displaymath" id="eq1">
  \begin{align}  \label{eq1} & -\tfrac {1}{h_{i}}\cdot m_{i-1}+\left( \tfrac {4}{h_{i}}+\tfrac {4}{h_{i+1}}\right) \cdot m_{i}-\tfrac {1}{h_{i+1}}\cdot m_{i+1} = \\ & =\tfrac {5}{h_{i}^{2}} f\left( x_{i-1}\right) -\tfrac {5}{h_{i+1}^{2}} f\left( x_{i+1}\right) \! \! +\! \! \big( \tfrac {11}{h_{i}^{2}}-\tfrac {11}{h_{i+1}^{2}}\big) f\left(x_{i}\right) \! +\! \tfrac {16}{h_{i+1}^{2}} f( x_{i+1/2})\!  -\! \tfrac {16 }{h_{i}^{2}} f( x_{i-1/2}) \nonumber \\ & =d_{i},\quad i=\overline{1,n-1} \nonumber \end{align}
</div>
<p>with known \(m_{0}=f^{\prime }\left( a\right) ,\) \(m_{n}=f^{\prime }\left(b\right) \), where \(h_{i}=x_{i}-x_{i-1},\) \(i=\overline{1,n},\) and \(m_{i}=S^{\prime }\left( x_{i}\right) ,\) \(i=\overline{0,n}.\) With the notation \(y_{i}=f\left( x_{i}\right) ,\) \(i=\overline{0,n},\) and \(t=\tfrac {x-x_{i-1}}{h_{i}}\in \lbrack 0,1]\) for \(x\in \lbrack x_{i-1},x_{i}],\) this quartic spline has the following expression on each interval \([x_{i-1},x_{i}],\) \(i=\overline{1,n}:\) </p>
<div class="displaymath" id="eq2">
  \begin{align}  \label{eq2} S\left( x\right) & =\left( 1-t\right) ^{2}\left( 1-2t\right) \left(4t+1\right) \cdot y_{i-1}+16t^{2}\left( 1-t\right) ^{2}\cdot y_{i-1/2}+ \\ & \quad +t^{2}\left( 2t\! -\! 1\right) \left( 5\! -\! 4t\right) y_{i}+h_{i}t\left(1\! -\! t\right) ^{2}\left( 1\! -\! 2t\right) m_{i-1}+h_{i}t^{2}\left( 1\! -\! t\right) \left( 1\! -\! 2t\right) m_{i} \nonumber \\ & =A_{i}\left( x\right) \cdot y_{i-1}+B_{i}\left( x\right) \cdot y_{i-1/2}+C_{i}\left( x\right) \cdot y_{i}+D_{i}\left( x\right) \cdot m_{i-1}+E_{i}\left( x\right) \cdot m_{i}. \nonumber \end{align}
</div>
<p>Solving this system, the local derivatives \(m_{i},\) \(i=\overline{0,n}\) are uniquely determined. Concerning the interpolation error estimate, in <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span> was established the following result. </p>
<p><div class="theorem_thmwrapper " id="th1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="cite">
	[
	<a href="#howell" >10</a>
	, 
	Th.2
	]
</span>  Let \(f\in C^{5}\left[ 0,1\right]\). Then we have: </p>
<div class="displaymath" id="a0000000004">
  \begin{align*}  \left\vert f\left( x\right) -S\left( x\right) \right\vert \leq \tfrac {C_{0}h^{5}}{5!}\cdot \max \limits _{x\in \lbrack 0,1]}\left\vert f^{V}\left(x\right) \right\vert ,\quad x\in \lbrack 0,1] \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="eq3">
  \begin{align} &  C_{0}=\left( \tfrac {1}{30}+\tfrac {\sqrt{30}}{3}\right) \cdot \sqrt{\left(\tfrac {1}{4}-\tfrac {1}{\sqrt{30}}\right) }=\max \limits _{x\in \lbrack 0,1]}\left\vert c\left( t\right) \right\vert \label{eq3} \\ &  c\left( t\right) =\tfrac {3t^{2}\left( 1-2t\right) \left( 1-t\right)^{2}+t\left( 1-t\right) \left( 1-2t\right) }{6}.\nonumber \end{align}
</div>
<p>Also we have</p>
<div class="displaymath" id="eq4">
  \begin{align}  \label{eq4} \left\vert f^{\prime }\left( x_{i}\right) -S^{\prime }\left( x_{i}\right)\right\vert \leq \tfrac {h^{4}}{6!}\cdot \max \limits _{x\in \lbrack 0,1]}\left\vert f^{V}\left( x\right) \right\vert ,\quad i=\overline{1,n-1}. \end{align}
</div>
<p>Furthermore, the constant \(C_{0}\) in <a href="#eq3" class="eqref">3</a> cannot be improved for an equally spaced partition. Inequality <a href="#eq4" class="eqref">4</a> is also best possible. Also we have</p>
<div class="displaymath" id="eq5">
  \begin{align}  \label{eq5} \left\vert f^{\prime }\left( x\right) -S^{\prime }\left( x\right)\right\vert \leq C_{1}\tfrac {h^{4}}{6!}\cdot \max \limits _{x\in \lbrack 0,1]}\left\vert f^{V}\left( x\right) \right\vert . \end{align}
</div>

  </div>
</div> </p>
<p>In <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span> Howell conjectured that \(C_{1}=1\). Volkov <span class="cite">
	[
	<a href="#volkov1998" >24</a>
	]
</span> proved that this conjecture holds true. For much less smooth class of functions \(f\in C[a,b]\) and in the case of uniform partition, considering a simplified endpoint condition \(S^{\prime }\left(a\right)=S^{\prime }\left( b\right) =0\), the corresponding error estimate is obtained in terms of the modulus of continuity as follows: </p>
<p><div class="theorem_thmwrapper " id="th2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
    <span class="theorem_thmtitle"><span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span></span>
  </div>
  <div class="theorem_thmcontent">
  <p>  Let \(f\in C[a,b]\). If \(\{ x_{i}\} _{i=0}^{n}\) is the partition of equally spaced knots, then for \(x_{i-1}\leq x\leq z_{i}=\tfrac {x_{i-1}+x_{i}}{2}\) and \(t=\tfrac {x-x_{i-1}}{h_{i}},\) \(i=\overline{1,n}\), we have</p>
<div class="displaymath" id="eq6">
  \begin{align}  \label{eq6} \left\vert f\left( x\right) -S\left( x\right) \right\vert \leq c\left(t\right) \omega \left( f,h\right) \leq c_{2} \omega \left( f,h\right) ,\quad t\in \lbrack 0,\tfrac {1}{2}] \end{align}
</div>
<p>and for \(x_{i}\leq x\leq z_{i+1}\), or \(\frac{1}{2}\leq t\leq 1\)</p>
<div class="displaymath" id="eq7">
  \begin{align}  \label{eq7} \left\vert f\left( x\right) -S\left( x\right) \right\vert \leq c\left(1-t\right) \omega \left( f,h\right) \end{align}
</div>
<p>where \(c\left( t\right) =1+\tfrac {13}{3}t-3t^{2}-\tfrac {58}{3}t^{3}+16t^{4}\) and \(c_{2}=\max \limits _{t\in \lbrack 0,\tfrac {1}{2}]}\left\vert c\left(t\right) \right\vert \cong 1.6572.\) </p>

  </div>
</div> </p>
<p>In <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span> the author had extended this result for nonuniform partitions considering the constant \(\beta =\max \{ h_{i}:i=\overline{1,n}\} / \min \{ h_{i}:i=\overline{1,n}\} \) and \(h=\max \{ h_{i}:i=\overline{1,n}\} \), obtaining: </p>
<p><div class="theorem_thmwrapper " id="th3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
    <span class="theorem_thmtitle"><span class="cite">
	[
	<a href="#howell1986" >9</a>
	, 
	Th.&#160;4.1.3
	]
</span></span>
  </div>
  <div class="theorem_thmcontent">
  <p>  Let \(f\in C[a,b]\) and let \(S\in C^{2}[a,b]\) be the quartic spline <a href="#eq2" class="eqref">2</a>. Then for \(x_{i-1}\leq x\leq z_{i}=\tfrac {x_{i-1}+x_{i}}{2}\) (i.e. for \(0\leq t\leq \tfrac {1}{2}\) with \(t=\tfrac {x-x_{i-1}}{h_{i}}\))</p>
<div class="displaymath" id="eq8">
  \begin{align}  \label{eq8} \left\vert f\left( x\right) -S\left( x\right) \right\vert \leq c_{1}\left(t\right) \omega \left( f,h\right) \end{align}
</div>
<p>and for \(x_{i}\leq x\leq z_{i+1}\) (i. e. for \(\tfrac {1}{2}\leq t\leq 1\))</p>
<div class="displaymath" id="a0000000005">
  \begin{align*}  \left\vert f\left( x\right) -S\left( x\right) \right\vert \leq c_{1}\left(1-t\right) \omega \left( f,h\right) \end{align*}
</div>
<p>where \(c_{1}\left( t\right) =1+10t^{2}-28t^{3}+16t^{4}+\tfrac {8}{3}\left(\beta ^{2}+\beta \right) t\left( 1-t\right) \left( 1-2t\right) .\) </p>

  </div>
</div> </p>
<p>Error estimates for C\(^{3}\)-smooth quartic splines and for quartic splines that on midpoints matches with the first derivative were established in <span class="cite">
	[
	<a href="#dubey2017" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#eltarazi" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#karaballi" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#micula2003" >18</a>
	]
</span>, <span class="cite">
	[
	<a href="#ranadubey" >20</a>
	]
</span> and <span class="cite">
	[
	<a href="#rana" >21</a>
	]
</span>. In <span class="cite">
	[
	<a href="#ranadubey" >20</a>
	]
</span>, <a href="#th1">theorem 1</a> is generalized by replacing the position of midpoints to a general type of interior points of the form \(z_{i}=x_{i-1}+\theta h_{i},\) \(i=\overline{1,n}\) with \(\theta \in \lbrack \frac{1}{4},\frac{3}{4}]\). In this paper we improve the error estimates (<a href="#eq6">6</a>)–(<a href="#eq7">7</a>) obtaining a smaller constant in terms of the modulus of continuity in the case \(f\in C[a,b]\), and investigate the corresponding error estimates for the situations when the endpoint condition \(S^{\prime }\left( a\right) =S^{\prime }\left( b\right)=0 \) is replaced by other classical ones such as \(S^{\prime \prime }\left(a\right) =S^{\prime \prime }\left( b\right) =0\), \(S^{\prime }\left(a\right)=f^{\prime }\left( a\right) \) and \(S^{\prime }\left( b\right) =f^{\prime } \left( b\right) \), or \(S^{\prime \prime }\left( a\right)=f^{\prime \prime }\left( a\right) \) and \(S^{\prime \prime }\left( b\right) =f^{\prime \prime }\left( b\right)\) . </p>
<p>On the other hand, in the present work we investigate the posibilities to determine the local derivatives \(m_{i},\) \(i=\overline{0,n}\), on the nodes of the deficient quartic spline \(S\in C^{1}[a,b]\) given in (<a href="#eq2">2</a>), such that the mean curvature </p>
<div class="displaymath" id="a0000000006">
  \[  \sqrt{\sum \limits _{i=1}^{n}\int _{x_{i-1}}^{x_{i}}\left( S^{\prime \prime }\left( x\right) \right) ^{2}dx}  \]
</div>
<p> and the functionals </p>
<div class="displaymath" id="a0000000007">
  \[ J_{k}\left( S\right) =\sum \limits _{i=1}^{n}\int _{x_{i-1}}^{x_{i}}\left( S^{\left( k\right)}\left( x\right) \right) ^{2}dx  \]
</div>
<p> for \(k\in \{ 1,2,3\} ,\) are minimized. </p>
<p>Shape preserving properties for quartic splines were investigated in <span class="cite">
	[
	<a href="#grandine" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#sallam" >22</a>
	]
</span> and <span class="cite">
	[
	<a href="#zhu" >26</a>
	]
</span>. Optimal properties for quartic splines were obtained in <span class="cite">
	[
	<a href="#ghizzetti" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#kobza1997" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#kobza2001" >15</a>
	]
</span>, and <span class="cite">
	[
	<a href="#micula2003" >18</a>
	]
</span>. In <span class="cite">
	[
	<a href="#kobza1997" >14</a>
	]
</span> and <span class="cite">
	[
	<a href="#kobza2001" >15</a>
	]
</span> the properties of \(C^{3}\)-smooth quartic splines having interpolation points \(t_{i},\) \(i=\overline{1,n}\) with \(x_{i-1}{\lt}t_{i}{\lt}x_{i}\), different by the grid nodes, were investigated. Natural, complete and periodic \(C^{3}\)-smooth quartic splines in connection with the minimization of the curvature are considered in <span class="cite">
	[
	<a href="#kobza1997" >14</a>
	]
</span>, while the parameters of spline are determined in <span class="cite">
	[
	<a href="#kobza2001" >15</a>
	]
</span> under the minimization of the functionals \(J_{k}\left( S\right) \), \(k\in \{ 0,1,2,3\} \), by using quadratic programming and the technique of pseudoinverse solution of linear systems in a similar way as was performed in <span class="cite">
	[
	<a href="#kobza2002" >13</a>
	]
</span> for cubic splines. </p>
<p>The paper is organized as follows. In <a href="#sec: 2 error est">section 2</a> we obtain some improvements of the error estimates presented in <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span> and in <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span>, regarding the interpolation of a continuous function \(f\in C[a,b]\) by the deficient complete quartic spline \(S\in C^{2}[a,b]\) given in (<a href="#eq2">2</a>). <a href="#sec: 3 optimal prop">section 3</a> is devoted to the optimal properties of the deficient quartic spline \(S\in C^{1}[a,b]\) in connection with the minimization of the functionals \(J_{k}\left( S\right) \) above presented, of a special interest being the minimal curvature and the minimal slope of the graph of \(S\). In order to illustrate the obtained theoretical results, a numerical example is presented in <a href="#tab:1">section 4</a> and some concluding remarks are pointed out in the last section. </p>
<h1 id="sec: 2 error est">2 Error estimates for deficient quartic splines</h1>

<p>Consider the quartic spline \(S\in C^{2}[a,b]\) given in (<a href="#eq2">2</a>) under the endpoint condition \(S^{\prime }\left( a\right) =S^{\prime }\left( b\right)=0 \) interpolating a continuous function \(f\in C[a,b]\) on a uniform parition \(\{ x_{i}\} _{i=0}^{n}\) and matching \(f\) on the midpoints \(x_{i-1/2},\) \(i= \overline{1,n}.\) In that follows, we get an improvement of the estimates (<a href="#eq6">6</a>)–(<a href="#eq7">7</a>) considering both the cases \(f\in C[a,b]\) and \(f\in \operatorname {Lip}[a,b],\) where \(\operatorname {Lip}[a,b]=\{ f\in C[a,b]:\exists L{\gt}0\) with \(\left\vert f\left( x\right) -f\left( y\right) \right\vert \leq L\left\vert x-y\right\vert ,\) \(\forall x,y\in \lbrack a,b]\} \). </p>
<p><div class="theorem_thmwrapper " id="th4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If the quartic spline \(S\) presented in <a href="#eq2" class="eqref">2</a> interpolates \(f\in C[a,b]\) on the uniform partition \(\{ x_{i}\} _{i=0}^{n}\) with the endpoint condition \(S^{\prime }\left( a\right) =S^{\prime }\left( b\right) =0\), then</p>
<div class="displaymath" id="eq9">
  \begin{align}  \label{eq9} & \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \\ \nonumber & \leq \left\{  \begin{array}{c} \max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert P\left( t\right) \right\vert \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192} \omega \left( f,h\right) ,\quad x\in \lbrack x_{i-1},x_{i-1/2}] \\ \max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert P\left( 1\! -\! t\right) \right\vert \omega \left( f,\tfrac {h}{2}\right) \! +\! \tfrac {1125}{8192} \omega \left( f,h\right) ,\  x\in \lbrack x_{i-1/2},x_{i}]\end{array}\right. \  i=\overline{1,n} \end{align}
</div>
<p>where \(P\left( t\right) =8t^{4}-\tfrac {10}{3}t^{3}-11t^{2}+\tfrac {16}{3}t+1\), and \(h=\tfrac {b-a}{n}.\) If \(f\in \operatorname {Lip}[a,b]\) with the Lipschitz constant \(L\), then the error estimate becomes</p>
<div class="displaymath" id="eq10">
  \begin{align}  \label{eq10} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq 0.95084\cdot Lh,\quad \forall x\in \lbrack a,b]. \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Firstly, we observe that \(A_{i}\left( x\right) \geq 0,\) \(B_{i}\left( x\right) \geq 0,\) \(C_{i}\left( x\right) \leq 0\), \(D_{i}\left( x\right) \geq 0,\) \(E_{i}\left( x\right) \geq 0\) for \(x\in \lbrack x_{i-1},x_{i-1/2}]\) and \(A_{i}\left( x\right) \leq 0,\) \(B_{i}\left( x\right) \geq 0,\) \(C_{i}\left( x\right) \geq 0\), \(D_{i}\left( x\right) \leq 0,\) \(E_{i}\left( x\right) \leq 0 \) for \(x\in \lbrack x_{i-1/2},x_{i}]\). Since \(A_{i}\left( x\right) +B_{i}\left( x\right) +C_{i}\left( x\right) =1,\) \(\forall x\in \lbrack x_{i-1},x_{i}]\) we infer that on the interval \([x_{i-1},x_{i-1/2}]\) we have</p>
<div class="displaymath" id="eq11">
  \begin{align}  \label{eq11} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq &  \left\vert A_{i}\left( x\right) +B_{i}\left( x\right) \right\vert \cdot \max \big\{ \left\vert y_{i-1}-f\left( x\right) \right\vert ,\vert y_{i-1/2}-f\left( x\right) \vert \big\}  \\ &  +\left\vert C_{i}\left( x\right) \right\vert \cdot \left\vert y_{i}-f\left( x\right) \right\vert +\left\vert D_{i}\left( x\right) +E_{i}\left( x\right) \right\vert \cdot \max \{ \left\vert m_{i-1}\right\vert ,\left\vert m_{i}\right\vert \}  \nonumber \end{align}
</div>
<p>and on \([x_{i-1/2},x_{i}]\) we get</p>
<div class="displaymath" id="eq12">
  \begin{align}  \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq & \left\vert A_{i}\left( x\right) \right\vert \cdot \left\vert y_{i-1}-f\left( x\right) \right\vert \label{eq12} \\ & +\left\vert B_{i}\left( x\right) +C_{i}\left( x\right) \right\vert \cdot \max \{ \left\vert y_{i}-f\left( x\right) \right\vert ,\vert y_{i-1/2}-f\left( x\right) \vert \}  \nonumber \\ & +\left\vert D_{i}\left( x\right) +E_{i}\left( x\right) \right\vert \cdot \max \{ \left\vert m_{i-1}\right\vert ,\left\vert m_{i}\right\vert \}  \nonumber \end{align}
</div>
<p> with \(\left\vert D_{i}\left( x\right) +E_{i}\left( x\right) \right\vert =t\left( 1-t\right) \left\vert 1-2t\right\vert \cdot h\), where \(t=\frac{x-x_{i-1}}{h}\in \lbrack 0,1].\) For estimating \(\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\} \) we see that the tridiagonal system (<a href="#eq1">1</a>) becomes in the case of equally spaced knots:</p>
<div class="displaymath" id="a0000000009">
  \begin{align*}  -\tfrac {1}{8}\cdot m_{i-1}+m_{i}-\tfrac {1}{8}\cdot m_{i+1}=\tfrac {5\left( y_{i-1}-y_{i+1}\right) }{8h}+\tfrac {2\left( y_{i+1/2}-y_{i-1/2}\right) }{h}=d_{i} \end{align*}
</div>
<p>for \(i=\overline{1,n-1},\) with \(m_{0}=m_{n}=0.\) Intending to estimate \(\left\vert d_{i}\right\vert \), \(i=\overline{1,n-1}\), we get</p>
<div class="displaymath" id="a0000000010">
  \begin{align*}  \left\vert d_{i}\right\vert \leq & \tfrac {5\left\vert y_{i-1}-y_{i-1/2}\right\vert }{8h}+\tfrac {11\left\vert y_{i}-y_{i-1/2}\right\vert }{8h}+\tfrac {11\left\vert y_{i+1/2}-y_{i}\right\vert }{8h} +\tfrac {5\left\vert y_{i+1/2}-y_{i+1}\right\vert }{8h} \\ \leq &  \tfrac {4}{h}\cdot \omega \left( f,\tfrac {h}{2}\right) ,\quad \forall i=\overline{1,n-1} \end{align*}
</div>
<p>and since the matrix \(A\) of the tridiagonal system is diagonally dominant, we infer that \(\left\Vert A^{-1}\right\Vert \leq \frac{4}{3}\) and then,</p>
<div class="displaymath" id="a0000000011">
  \begin{align*}  \left\Vert m\right\Vert _{\infty }=\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\} \leq \Vert A^{-1}\Vert \cdot \max \{ \left\vert d_{i}\right\vert :i=\overline{1,n-1}\} \leq \tfrac {16\omega \left( f,\frac{h}{2}\right) }{3h}. \end{align*}
</div>
<p>Consequently, for the last term in (<a href="#eq11">11</a>) and (<a href="#eq12">12</a>) we get </p>
<div class="displaymath" id="a0000000012">
  \begin{align*}  \left\vert D_{i}\left( x\right) +E_{i}\left( x\right) \right\vert \cdot \max \{ \left\vert m_{i-1}\right\vert ,\left\vert m_{i}\right\vert \} \leq t\left( 1-t\right) \left\vert 1-2t\right\vert \cdot \tfrac {16}{3}\cdot \omega \left( f,\tfrac {h}{2}\right) \end{align*}
</div>
<p>for all \(x\in \lbrack x_{i-1},x_{i}],\) \(i=\overline{1,n}\). Now, by (<a href="#eq11">11</a>) we obtain</p>
<div class="displaymath" id="a0000000013">
  \begin{align*} & \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \\ &  \leq \left( \left( 1\! -\! t\right) ^{2}\left( 1\! -\! 2t\right) \left( 4t\! +\! 1\right) +16t^{2}\left( 1\! -\! t\right) ^{2}+\tfrac {16}{3}t\left( 1\! -\! t\right) \left( 1\! -\! 2t\right) \right) \cdot \omega \left( f,\tfrac {h}{2}\right) \\ &  \quad +\left\vert t^{2}\left( 2t-1\right) \left( 5-4t\right) \right\vert \omega \left( f,h\right) \\ &  \leq \max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert 8t^{4}-\tfrac {10}{3}t^{3}-11t^{2}+\tfrac {16}{3}t+1\right\vert \cdot \omega \left( f,\tfrac {h}{2}\right) +\max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert C_{i}\left( x\right) \right\vert \cdot \omega \left( f,h\right) \\ &  \leq \max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert P\left( t\right) \right\vert \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\omega \left( f,h\right) \end{align*}
</div>
<p>for \(x\in \lbrack x_{i-1},x_{i-1/2}]\). Analogously, by (<a href="#eq12">12</a>) it obtains</p>
<div class="displaymath" id="a0000000014">
  \begin{align*}  \left\vert s\left( x\right) -f\left( x\right) \right\vert \leq & \max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert 8t^{4}\! -\! \tfrac {86}{3}t^{3}\! +\! 27t^{2}\! -\! \tfrac {16}{3}t\right\vert \cdot \omega \left( f,\tfrac {h}{2}\right) +\max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert A_{i}\left( x\right) \right\vert \cdot \omega \left( f,h\right) \\ \leq &  \max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert P\left( 1-t\right) \right\vert \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\omega \left( f,h\right) \end{align*}
</div>
<p>when \(x\in \lbrack x_{i-1/2},x_{i}],\) and the estimate (<a href="#eq9">9</a>) follows. After elementary calculus, we see that </p>
<div class="displaymath" id="a0000000015">
  \[  \max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert P\left( t\right) \right\vert =\max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert P\left( 1-t\right) \right\vert \simeq 1.627,  \]
</div>
<p> and when \(f\in \operatorname {Lip}[a,b]\), the estimate (<a href="#eq9">9</a>) becomes</p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq 1.627\cdot \tfrac {Lh}{2}+\tfrac {1125}{8192}\cdot Lh\simeq 0.95084\cdot Lh,\quad \forall x\in \lbrack x_{i-1},x_{i}],i=\overline{1,n} \end{equation*}
</div>
<p>that is the estimate (<a href="#eq10">10</a>). <div class="proof_wrapper" id="a0000000017">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="rem1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5</span>
  </div>
  <div class="remark_thmcontent">
  <p>By <a href="#th2">theorem 2</a>, the estimate (<a href="#eq6">6</a>) is \(\left\vert s\left( x\right) -f\left( x\right) \right\vert \leq 1.6572\cdot \omega \left( f,h\right) ,\) \(\forall x\in \lbrack a,b]\) and when \(f\in \operatorname {Lip}[a,b]\) it becomes \(\left\vert s\left( x\right) -f\left( x\right) \right\vert \leq 1.6572\cdot Lh.\) So, the estimate (<a href="#eq10">10</a>) is better because gives a smaller constant in the case of Lischitzian functions. We can assert that the estimate (<a href="#eq9">9</a>) is better than (<a href="#eq6">6</a>) because in (<a href="#eq9">9</a>) it appears \(\omega ( f,\tfrac {h}{2}) \) near \(\max _{t\in \lbrack 0,\frac{1}{2}]}\left\vert P\left( t\right) \right\vert \simeq 1.627\) and the factor \(\omega ( f,\tfrac {h}{2}) \) considerable reduces the error in comparison with \(\omega \left( f,h\right) .\) Therefore the estimates obtained in <a href="#th4">theorem 4</a> represent an improvement of <a href="#th2">theorem 2</a>, especially in the case of Lipschitzian functions. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000018">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6</span>
  </div>
  <div class="remark_thmcontent">
  <p>An interesting property of the deficient quartic spline \(S\in C^{2}[a,b]\) given in (<a href="#eq2">2</a>) can be observed in the case of uniform partition when integrate this spline over the interval \([a,b]\). It obtains the corrected Simpson composite quadrature formula which is exact for polynomials of degree 5 or less. This corrected quadrature formula is usually obtained by applying the Richardson extrapolation and Grüss type inequalities (see <span class="cite">
	[
	<a href="#liu" >16</a>
	]
</span>), or by using a finite difference technique (see <span class="cite">
	[
	<a href="#ujevic" >23</a>
	]
</span>). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>Concerning the estimate from <a href="#th3">theorem 3</a>, in the case of nonuniform partition, we can state the following. </p>
<p><div class="corollary_thmwrapper " id="th5">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">7</span>
  </div>
  <div class="corollary_thmcontent">
  <p>If the quartic spline \(S\) presented in <a href="#eq2" class="eqref">2</a> interpolates \(f\in C[a,b]\) on a nonuniform partition \(\{ x_{i}\} _{i=0}^{n}\) with the endpoint condition \(S^{\prime }\left( a\right) =S^{\prime }\left( b\right)=0 \) and \(\beta =\max \{ h_{i}:i=\overline{1,n}\} /\min \{ h_{i}:i=\overline{1,n}\}  \), \(h=\max \{ h_{i}:i=\overline{1,n}\} \), \(\underline{h}=\min \{ h_{i}:i=\overline{1,n}\} \), then</p>
<div class="equation" id="eq13">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq13} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\tfrac {4\sqrt{3}\left( \beta ^{2}+1\right)} {27}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left(f,h\right) \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>for all \(x\in \lbrack a,b]\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The matrix \(A\) of the tridiagonal, diagonally dominant system (<a href="#eq1">1</a>) has \(\left\Vert A^{-1}\right\Vert \leq \frac{4}{3}\) and when estimate \(\left\vert d_{i}\right\vert ,\) \(i=\overline{1,n-1}\), we get</p>
<div class="displaymath" id="a0000000020">
  \begin{align*}  \left\vert d_{i}\right\vert \leq &  \tfrac {h_{i+1}}{4h_{i}\left( h_{i}+h_{i+1}\right) }\cdot (5\vert y_{i-1}-y_{i-1/2}\vert +11\vert y_{i}-y_{i-1/2}\vert )+ \\ & +\tfrac {h_{i}}{4h_{i+1}\left( h_{i}+h_{i+1}\right) }\left( 5\vert y_{i+1/2}-y_{i+1}\vert +11\vert y_{i+1/2}-y_{i}\vert \right) \\ \leq & \tfrac {4}{\left( h_{i}+h_{i+1}\right) }\cdot \lbrack \tfrac {h_{i+1}}{h_{i}}\cdot \omega \left( f,\tfrac {h_{i}}{2}\right) +\tfrac {h_{i}}{h_{i+1}}\omega \left( f,\tfrac {h_{i+1}}{2}\right) ] \\ \leq &  \tfrac {4\omega \left( f,\frac{h}{2}\right) }{\left( h_{i}+h_{i+1}\right) }\left( \beta +\tfrac {1}{\beta }\right) \leq \tfrac {2\left( \beta ^{2}+1\right)}{\beta \underline{h}} \omega \left( f, \tfrac {h}{2}\right) \end{align*}
</div>
<p>for all \(i=\overline{1,n-1}\). Then, \(\max \{ \left\vert m_{i-1}\right\vert ,\left\vert m_{i}\right\vert \} \leq \frac{8}{3\underline{h}}\left( \beta +\frac{1}{\beta }\right) \cdot \omega \left( f,\frac{h}{2}\right) \) and since</p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert A_{i}\left( x\right) +B_{i}\left( x\right) \right\vert =\max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert B_{i}\left( x\right) +C_{i}\left( x\right) \right\vert =\tfrac {9317}{8192} \end{equation*}
</div>
<p>and \(\max \limits _{t\in \lbrack 0,\frac{1}{2}]}\left\vert C_{i}\left( x\right) \right\vert =\max \limits _{t\in \lbrack \frac{1}{2},1]}\left\vert A_{i}\left( x\right) \right\vert =\tfrac {1125}{8192}\), the maximum values being attained in \(t=\tfrac {5}{16}\) and \(t=\tfrac {11}{16}\), respectively, by (<a href="#eq11">11</a>) and (<a href="#eq12">12</a>) we obtain,</p>
<div class="displaymath" id="a0000000022">
  \begin{align*} & \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \\ & \leq \tfrac {9317}{8192}\cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left( f,h\right) + \max \limits _{t\in \lbrack 0,1]}\left\vert D_{i}\left( x\right) +E_{i}\left(x\right) \right\vert \tfrac {8\left( \beta ^{2}+1\right)} {3\beta \underline{h}}\omega \left( f,\tfrac {h}{2}\right) \\ & =\left( \tfrac {9317}{8192}+\tfrac {h\sqrt{3}}{18}\cdot \tfrac {8\left( \beta ^{2}+1\right) }{3\beta \underline{h}}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\omega \left( f,h\right) \\ & =\left( \tfrac {9317}{8192}+\tfrac {4\sqrt{3}\left( \beta ^{2}+1\right) }{27}\right) \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\omega \left( f,h\right) \end{align*}
</div>
<p>for all \(x\in \lbrack a,b].\) <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>If we compare the estimate (<a href="#eq13">13</a>) with the result from Theorem 4.1.3 in <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span>, since \(\max _{t\in \lbrack 0,\frac{1}{2}]}\left\vert t\left( 1-2t\right) \left( 1-t\right) \right\vert =\tfrac {\sqrt{3}}{18}\) and \(\max _{t\in \lbrack 0,\frac{1}{2}]}\left\vert 16t^{4}-28t^{3}+10t^{2}+1\right\vert \) \(=\tfrac {9317}{8192}+\tfrac {1125}{8192}=1.2747\), we see that the presence in (<a href="#eq13">13</a>) of the factor \(\omega ( f,\tfrac {h}{2}) \) instead of \(\omega \left( f,h\right) \) represents an improvement of the result from Theorem 4.1.3 in <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span>. </p>
<p>The endpoint condition \(S^{\prime }\left( a\right) =S^{\prime }\left(b\right) =0\) imposed in <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span> in order to simplify the study of the error estimate in the case \(f\in C[a,b]\) can be replaced by other classical ones. For instance, in that follows we investigate the modification of the error estimate when other two supplementary endpoint conditions are included. In the first case, mentioned in <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span>, we can consider \(S^{\prime }\left( a\right) =f^{\prime }\left( a\right) ,\) \(S^{\prime }\left(b\right) =f^{\prime }\left( b\right) \) and the linear system (<a href="#eq1">1</a>) has the central lines</p>
<div class="displaymath" id="a0000000024">
  \begin{align*}  -\tfrac {h_{i+1}\cdot m_{i-1}+h_{i}\cdot m_{i+1}}{4\left( h_{i}+h_{i+1}\right)}+m_{i} = & \tfrac {h_{i+1}\cdot \left( 5y_{i-1}-16y_{i-1/2}+11y_{i}\right) } {4h_{i}\left( h_{i}+h_{i+1}\right) } -\tfrac {h_{i}\cdot \left( 5y_{i+1}-16y_{i+1/2}+11y_{i}\right) }{4h_{i+1}\left( h_{i}+h_{i+1}\right) } \\ = &  d_{i},\quad i=\overline{2,n-2} \end{align*}
</div>
<p>and the first and the last equations becomes</p>
<div class="displaymath" id="a0000000025">
  \begin{align*}  m_{1}& -\tfrac {h_{1}\cdot m_{2}}{4\left( h_{1}+h_{2}\right) }=\tfrac {h_{2}^{2}\left( 5y_{0}-16y_{1/2}+11y_{1}-f^{\prime }\left( a\right) \right) }{4h_{1}h_{2}\left( h_{1}+h_{2}\right) }- \tfrac {h_{1}^{2}\left( 5y_{2}-16y_{2-1/2}+11y_{1}\right) }{4h_{1}h_{2}\left(h_{1}+h_{2}\right) } \end{align*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000026">
  \begin{align*} & -\tfrac {h_{n}\cdot m_{n-2}}{4\left( h_{n-1}+h_{n}\right) }+m_{n-1}= \\ & =\tfrac {h_{n}\cdot \left( 5y_{n-2}-16y_{n-1-1/2}+11y_{n-1}\right) -h_{n}\cdot f^{\prime }\left( b\right) }{4h_{n-1}\left( h_{n-1}+h_{n}\right) }- \tfrac {h_{n-1}\cdot \left( 5y_{n}-16y_{n-1/2}+11y_{n-1}\right) }{4h_{i+1}\left( h_{i}+h_{i+1}\right) }=d_{n-1} \end{align*}
</div>
<p>obtaining the same inequality \(\left\vert d_{i}\right\vert \leq \tfrac {4h}{ \underline{h}^{2}}\cdot \omega \left( f,\tfrac {h}{2}\right) ,\) \(i=\overline{2,n-2}\) and</p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  \left\vert d_{1}\right\vert \leq \tfrac {4h}{\underline{h}^{2}}\cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {h}{8\underline{h}^{2}}\left\vert f^{\prime }\left( a\right) \right\vert ,\quad \left\vert d_{n-1}\right\vert \leq \tfrac {4h}{\underline{h}^{2}}\cdot \omega \left( f,\tfrac {h}{2}\right) + \tfrac {h}{8\underline{h}^{2}}\left\vert f^{\prime }\left( b\right) \right\vert . \end{equation*}
</div>
<p>Then, \(\left\Vert m\right\Vert _{\infty }=\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\} \leq \tfrac {4}{3}\cdot \left( \tfrac {4h}{\underline{h}^{2}} \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {h\cdot m^{\prime }}{8 \underline{h}^{2}}\right) \) and the error estimate becomes</p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\left( \tfrac {4\sqrt{3}}{27}+\tfrac {\sqrt{3}\cdot m^{\prime }}{108} \right) \left( 1+\beta ^{2}\right) \right) \cdot \omega \left( f,\tfrac {h}{2} \right) +\tfrac {1125}{8192}\cdot \omega \left( f,h\right) \end{equation*}
</div>
<p>for all \(x\in \lbrack a,b]\), where \(m^{\prime }=\max \{ \left\vert f^{\prime }\left( a\right) \right\vert ,\left\vert f^{\prime }\left( b\right)\right\vert \} .\) </p>
<p><div class="remark_thmwrapper " id="rem2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p> If \(f\in C^{1}[a,b],\) then in the case of equally spaced knots this error estimate is,</p>
<div class="equation" id="estder">
<p>
  <div class="equation_content">
    \begin{equation}  \label{estder} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {11\, 567}{16\, 384}+\tfrac {11\sqrt{3}}{72}\right) \cdot M^{\prime }h\simeq 0.97061\cdot M^{\prime }h,\quad \forall x\in \lbrack a,b] \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p>where \(M^{\prime }=\max \{ \left\vert f^{\prime }\left( x\right) \right\vert :x\in \lbrack a,b]\} .\) <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>Taking the natural type endpoint condition \(S^{\prime \prime }\left(a\right) =S^{\prime \prime }\left( b\right) =0\) the linear system (<a href="#eq1">1</a>) receives two supplementary equations</p>
<div class="equation" id="nat">
<p>
  <div class="equation_content">
    \begin{equation}  \label{nat} \left\{  \begin{array}{c} m_{0}-\tfrac {1}{4}m_{1}=\tfrac {-11y_{0}+16y_{1-1/2}-5y_{1}}{4h_{1}}=d_{0} \\[2mm] -\tfrac {1}{4}m_{n-1}+m_{n}=\tfrac {5y_{n-1}-16y_{n-1/2}+11y_{n}}{4h_{n}}=d_{n}\end{array}\right. \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p>obtaining the estimates \(\left\vert d_{0}\right\vert \leq \tfrac {4}{ \underline{h}}\cdot \omega \left( f,\tfrac {h}{2}\right) \), \(\left\vert d_{n}\right\vert \leq \tfrac {4}{\underline{h}}\cdot \omega \left( f,\frac{h}{2}\right) \) and the interpolation error estimate is the same as in (<a href="#eq13">13</a>). </p>
<p>We can consider now, the second type of complete endpoint conditions \(S^{\prime \prime }\left( a\right) =f^{\prime \prime }\left( a\right) ,\) \(S^{\prime \prime }\left( b\right) =f^{\prime \prime }\left( b\right) \) when the values \(f^{\prime \prime }\left( a\right) \) and \(f^{\prime \prime }\left( b\right) \) are given, the supplementary equations becoming </p>
<div class="displaymath" id="a0000000029">
  \begin{equation*}  \left\{  \begin{array}{c} m_{0}-\frac{1}{4}m_{1}=\frac{-11y_{0}+16y_{1-1/2}-5y_{1}}{4h_{1}}+\frac{h_{1}}{8}f^{\prime \prime }\left( a\right) =d_{0} \\[2mm] -\frac{1}{4}m_{n-1}+m_{n}=\frac{5y_{n-1}-16y_{n-1/2}+11y_{n}}{4h_{n}}+\frac{h_{n}}{8}f^{\prime \prime }\left( b\right) =d_{n}\end{array}\right. \end{equation*}
</div>
<p>with \(\left\vert d_{0}\right\vert \leq \frac{4}{\underline{h}}\cdot \omega \left( f,\frac{h}{2}\right) +\frac{h}{8}\left\vert f^{\prime \prime }\left( a\right) \right\vert ,\) and \(\left\vert d_{n}\right\vert \leq \frac{4}{ \underline{h}}\cdot \omega \left(f,\frac{h}{2}\right) +\frac{h}{8} \left\vert f^{\prime \prime }\left( b\right) \right\vert .\) In this case, the error estimate is </p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\tfrac {4\sqrt{3}\left( 1+\beta ^{2}\right) }{27}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left(f,h\right) +\tfrac {m^{\prime \prime } \sqrt{3}h^{2}}{108} \end{equation*}
</div>
<p>where \(m^{\prime \prime }=\max \{ \left\vert f^{\prime \prime }\left(a\right) \right\vert ,\left\vert f^{\prime \prime }\left( b\right) \right\vert \} .\) </p>
<p><div class="remark_thmwrapper " id="a0000000031">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>When the values \(f^{\prime \prime }\left( a\right) \) and \(f^{\prime \prime }\left( b\right) \) are not available, in order to preserve \(\mathcal{O}\left(h^{5}\right) \) accuracy, we can consider the endpoint conditions \(S^{\prime \prime }\left( x_{0}\right) =p_{0}^{\prime \prime } \left( x_{0}\right) \), \(S^{\prime \prime }\left( x_{n}\right) =p_{n}^{\prime \prime }\left( x_{n}\right) \), where \(p_{0}\) is the quartic Lagrange polynomial interpolating the points \(\left( x_{0},y_{0}\right) ,\) \(( x_{1-1/2},y_{1-1/2}) ,\) \(\left( x_{1},y_{1}\right) ,\) \((x_{2-1/2},y_{2-1/2}) ,\) \(\left( x_{2},y_{2}\right) \) and \(p_{n}\) is the quartic Lagrange polynomial interpolating the points \(\left(x_{n-2},y_{n-2}\right) ,\) \(( x_{n-1-1/2},y_{n-1-1/2}) ,\) \(\left(x_{n-1},y_{n-1}\right) ,\) \((x_{n-1/2}, y_{n-1/2}) ,\) \(\left(x_{n}.y_{n}\right) \). Based on <span class="cite">
	[
	<a href="#howell1991" >11</a>
	]
</span>, we have \(f^{\prime \prime } \left( x_{0}\right) =p_{0}^{\prime \prime }\left( x_{0}\right) +\tfrac {u_{0}^{\prime \prime }\left( x_{0}\right) }{5!}\cdot f^{\left( 5\right)} \left( \xi _{1}\right) \) and \(f^{\prime \prime }\left( x_{n}\right) =p_{n}^{\prime \prime }\left( x_{n}\right) +\tfrac {u_{n}^{\prime \prime } \left( x_{n}\right)} {5!} \cdot f^{\left( 5\right) }\left( \xi _{n}\right) \), where \(\xi _{1}\in \left( x_{0},x_{2}\right) \), \(\xi _{n}\in \left(x_{n-2},x_{n}\right) \), and</p>
<div class="displaymath" id="a0000000032">
  \begin{align*}  u_{0}\left( x\right) = &  \left( x-x_{0}\right) ( x-x_{1-1/2}) \left(x-x_{1}\right) ( x-x_{2-1/2}) \left( x-x_{2}\right) \\ u_{n}\left( x\right) = &  \left( x-x_{n-2}\right) ( x-x_{n-1-1/2}) \left( x-x_{n-1}\right) ( x-x_{n-1/2}) \left( x-x_{n}\right) .\end{align*}
</div>
<p>Similar treatment at endpoints can be realized when the values \(f^{\prime } \left( a\right) \) and \(f^{\prime }\left( b\right) \) are not available, by considering the conditions \(S^{\prime }\left( x_{0}\right) =p_{0}^{\prime } \left( x_{0}\right) \) and \(S^{\prime }\left( x_{n}\right) =p_{n}^{\prime }\left( x_{n}\right) .\) <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="sec: 3 optimal prop">3 Optimal properties for deficient quartic splines</h1>

<h2 id="a0000000033">3.1 Minimal mean curvature</h2>
<p>In this section we consider the deficient quartic spline (<a href="#eq2">2</a>) \(S\in C^{1}[a,b]\) and in this case the spline local derivatives \(m_{i},\) \(i=\overline{0,n}\) remain free. We will determine \(m_{i},\) \(i=\overline{0,n},\) in order to minimize the functionals </p>
<div class="displaymath" id="a0000000034">
  \[  J_{k}\left( S\right) =\sum \limits _{i=1}^{n}\int _{x_{i-1}}^{x_{i}}\left( S^{\left( k\right) }\left( x\right) \right) ^{2}dx, \quad k\in \{ 1,2,3\} .  \]
</div>
<p> According to <span class="cite">
	[
	<a href="#bini" >2</a>
	]
</span>, \(\sqrt{J_{1}\left( S\right) }\) is an average of the slope of the graph of \(S\) and \(\sqrt{J_{2}\left( S\right) }\) represents the mean curvature of the graph of \(S\), while \(\sqrt{J_{3}\left( S\right) }\) is related to the mean curvature of the graph of \(S^{\prime }\). Since \(S^{\left( 4\right) }\) is piecewise constant discontinuous function, the minimization of \(J_{4}\left( S\right) \) is without of interest. </p>
<p>By (<a href="#eq2">2</a>) we see that in each interval \([x_{i-1},x_{i}],\) \(i=\overline{1,n}\) we have </p>
<div class="displaymath" id="a0000000035">
  \begin{align*}  S^{\prime \prime }\left( x\right) =A_{i}^{\prime \prime }\left( x\right) y_{i-1}+B_{i}^{\prime \prime }\left( x\right) y_{i-1/2}+C_{i}^{\prime \prime }\left( x\right) y_{i}+D_{i}^{\prime \prime }\left( x\right) m_{i-1}+E_{i}^{\prime \prime }\left( x\right) m_{i} \end{align*}
</div>
<p>and therefore </p>
<div class="displaymath" id="a0000000036">
  \begin{align*} & J_{2}\left( S\right) \left( m_{0},m_{1},\ldots ,m_{n}\right)= \\ & =\sum \limits _{i=1}^{n}\int \limits _{x_{i-1}}^{x_{i}}\left( S^{\prime \prime }\left( x\right) \right) ^{2}dx \\ & =\sum \limits _{i=1}^{n}\int \limits _{x_{i-1}}^{x_{i}}\! [A_{i}^{\prime \prime }\left( x\right) y_{i-1}\! +\!  B_{i}^{\prime \prime }\left( x\right) y_{i-1/2}\! +\! C_{i}^{\prime \prime }\left( x\right) y_{i}\! +\! D_{i}^{\prime \prime }\left( x\right) m_{i-1}\! +\! E_{i}^{\prime \prime }\left( x\right) m_{i}]^{2}dx. \end{align*}
</div>
<p>In order to minimize \(J_{2}\left( S\right) \) the system of normal equations \(\frac{\partial J_{2}}{\partial m_{i}}=0,\) \(i=\overline{0,n},\) is </p>
<div class="displaymath" id="a0000000037">
  \begin{align*} &  m_{0}\int \limits _{x_{0}}^{x_{1}}\left( D_{1}^{\prime \prime }\left( x\right) \right) ^{2}dx+m_{1}\int \limits _{x_{0}}^{x_{1}}D_{1}^{\prime \prime }\left( x\right) E_{1}^{\prime \prime }\left( x\right) dx= \\ & =-\! \! \int \limits _{x_{0}}^{x_{1}}\! A_{1}^{\prime \prime }\left( x\right) D_{1}^{\prime \prime }\left( x\right) dx\cdot y_{0}- \! \!  \int \limits _{x_{0}}^{x_{1}}\! B_{1}^{\prime \prime }\left( x\right) D_{1}^{\prime \prime }\left( x\right) dx\cdot y_{1-1/2}-\! \! \int \limits _{x_{0}}^{x_{1}}\! C_{1}^{\prime \prime }\left( x\right) D_{1}^{\prime \prime }\left( x\right) dx\cdot y_{1} , \\ &  m_{i-1}\int \limits _{x_{i-1}}^{x_{i}}E_{i}^{\prime \prime }\left( x\right) D_{i}^{\prime \prime }\left( x\right) dx+m_{i}\left( \int \limits _{x_{i-1}}^{x_{i}}\left( E_{i}^{\prime \prime }\left( x\right) \right) ^{2}dx+\int \limits _{x_{i}}^{x_{i+1}}\left( D_{i+1}^{\prime \prime }\left( x\right) \right) ^{2}dx\right) + \\ &  \quad +m_{i+1}\int \limits _{x_{i}}^{x_{i+1}}D_{i+1}^{\prime \prime }\left( x\right) E_{i+1}^{\prime \prime }\left( x\right) dx= \\ & =-\int \limits _{x_{i-1}}^{x_{i}}E_{i}^{\prime \prime }\left( x\right) [A_{i}^{\prime \prime }\left( x\right) y_{i-1}+ B_{i}^{\prime \prime }\left( x\right) y_{i-1/2}+C_{i}^{\prime \prime }\left( x\right) \cdot y_{i}]dx \\ &  \quad -\int \limits _{x_{i}}^{x_{i+1}}D_{i+1}^{\prime \prime }\left( x\right) [A_{i+1}^{\prime \prime }\left( x\right) y_{i}+ B_{i+1}^{\prime \prime }\left( x\right) y_{i+1/2}+C_{i+1}^{\prime \prime }\left( x\right) y_{i+1}]dx,\quad i=\overline{1,n-1} \\ &  m_{n-1}\int \limits _{x_{n-1}}^{x_{n}}D_{n}^{\prime \prime }\left( x\right) E_{n}^{\prime \prime }\left( x\right) dx+m_{n}\int \limits _{x_{n-1}}^{x_{n}}\left( E_{n}^{\prime \prime }\left( x\right) \right) ^{2}dx= \\ &  =-\int \limits _{x_{n-1}}^{x_{n}}E_{n}^{\prime \prime }\left( x\right) [A_{n}^{\prime \prime }\left( x\right) \cdot y_{n-1}+B_{n}^{\prime \prime }\left( x\right) \cdot y_{n-1/2}+C_{n}^{\prime \prime }\left( x\right) \cdot y_{n}]dx\end{align*}
</div>
<p>and after elementary calculus becomes</p>
<div class="equation" id="eq16">
<p>
  <div class="equation_content">
    \begin{equation}  \left\{  \begin{array}{l} m_{0}-\tfrac {1}{6}m_{1}=\tfrac {-47y_{0}+64y_{1-1/2}-17y_{1}}{18h_{1}}=d_{0}^{\prime \prime } \\[2mm] -\tfrac {h_{i+1}\cdot m_{i-1}}{6\left( h_{i}+h_{i+1}\right) }+m_{i}-\tfrac {h_{i}\cdot m_{i+1}}{6\left( h_{i}+h_{i+1}\right) }= \\[2mm] =\tfrac {h_{i+1}\cdot \left( 17y_{i-1}-64y_{i-1/2}\right) }{18h_{i}\left( h_{i}+h_{i+1}\right) } +\tfrac {47y_{i}}{18}\cdot \left( \tfrac {1}{h_{i}}-\tfrac {1}{h_{i+1}}\right) +\tfrac {h_{i}\cdot \left( 64y_{i+1/2}-17y_{i+1}\right) }{18h_{i+1}\left( h_{i}+h_{i+1}\right) } \\[2mm] =d_{i}^{\prime \prime },\quad i=\overline{1,n-1} \\[2mm] -\tfrac {1}{6}m_{n-1}+m_{n}=\tfrac {17y_{n-1}-64y_{n-1/2}+47y_{n}}{18h_{n}}=d_{n}^{\prime \prime }.\end{array}\right. \label{eq16} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p>We see that the matrix \(A^{\prime \prime }\) of this system is strictly diagonally dominant with the index of diagonally dominance 1/6, better than for the system (<a href="#eq1">1</a>), and \(\left\Vert \left( A^{\prime \prime }\right) ^{-1}\right\Vert \leq \frac{6}{5}\)(see <span class="cite">
	[
	<a href="#plato" >19</a>
	]
</span>). Moreover, since \(a_{ii}^{\prime \prime }{\gt}0\) and \(a_{ij}^{\prime \prime }{\lt}0\) for \(i\neq j,\) \(i,j=\overline{0,n}\), we infer that all elements of the matrix \(\left( A^{\prime \prime }\right) ^{-1} \) are nonnegative. The strictly diagonally dominance ensures the existence and uniqueness of the solution of system (<a href="#eq16">16</a>) and the numerical stability of the LU factorization method for solving this system. In this way we obtain the following result. </p>
<p><div class="theorem_thmwrapper " id="th6">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">10</span>
  </div>
  <div class="theorem_thmcontent">
  <p>There is a unique solution \(\left(m_{0},m_{1},\ldots ,m_{n}\right) \) which minimize the functional \(J_{2}\left( S\right) \) and the mean curvature of the graph of \(S\), too. The interpolation error estimate of the obtained quartic spline \(S\in C^{1}[a,b]\) interpolating a function \(f\in C[a,b]\) is,</p>
<div class="equation" id="eq17">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq17} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\tfrac {32\sqrt{3}\beta ^{2}}{135}\right) \cdot \omega \left( f, \tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left( f,h\right) \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p>for all \(x\in \lbrack a,b].\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000038">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> As above, the system of normal equations associated to the functional \(J_{2}\left( S\right) \) has unique solution due to the strictly diagonally dominance of its matrix and this solution can be obtained by applying the iterative algorithm presented in <span class="cite">
	[
	<a href="#ahlberg" >1</a>
	, 
	pp.
	
	14–15
	]
</span>, for tridiagonal linear systems. Moreover, by the strictly diagonally dominance of the matrix \(A^{\prime \prime }\) we infer that all the diagonal minors of the Hessian matrix \(\left( \tfrac {\partial ^{2}J_{2}}{\partial m_{i}\partial m_{j}}\right) _{i,j=\overline{0,n}}\) are strictly positive and thus \(\left(m_{0},m_{1},\ldots ,m_{n}\right) \) is a real minimum point of \(J_{2} \left( S\right)\). For obtaining the estimate (<a href="#eq17">17</a>) we observe by (<a href="#eq16">16</a>) that </p>
<div class="displaymath" id="a0000000039">
  \begin{align*}  \left\vert d_{0}^{\prime \prime } \right\vert \leq \tfrac {47\left\vert y_{1-1/2}-y_{0}\right\vert +17\left\vert y_{1-1/2}-y_{1}\right\vert }{18h_{1}} \leq \tfrac {32}{9\underline{h}}\cdot \omega \left(f,\tfrac {h}{2}\right), \end{align*}
</div>
<div class="displaymath" id="a0000000040">
  \begin{align*}  \left\vert d_{n}^{\prime \prime } \right\vert \leq \tfrac {17\left\vert y_{n-1}-y_{n-1/2}\right\vert +47\left\vert y_{n}-y_{n-1/2}\right\vert }{18h_{n}}\leq \tfrac {32}{9\underline{h}}\cdot \omega \left(f,\tfrac {h}{2}\right) \end{align*}
</div>
<p> and</p>
<div class="displaymath" id="a0000000041">
  \begin{align*}  \left\vert d_{i}^{\prime \prime }\right\vert & \leq \tfrac {h_{i+1}(17\left\vert y_{i-1}-y_{i-1/2}\right\vert +47\left\vert y_{i}-y_{i-1/2}\right\vert )}{18h_{i}\left( h_{i}+h_{i+1}\right) }+\tfrac {47h_{i}\left\vert y_{i+1/2}-y_{i}\right\vert }{18h_{i+1}\left( h_{i}+h_{i+1}\right)}+\\ & \quad +\tfrac {17h_{i}\left\vert y_{i+1/2}-y_{i+1}\right\vert }{18h_{i+1}\left(h_{i}+h_{i+1}\right) }\leq \tfrac {32h_{i+1}\cdot \omega \left( f, \frac{h}{2} \right) }{9h_{i}\left( h_{i}+h_{i+1}\right) }+\tfrac {32h_{i}\cdot \omega \left( f,\frac{h}{2}\right) }{9h_{i+1}\left( h_{i}+h_{i+1}\right) } \leq \tfrac {32h}{9\underline{h}^{2}}\cdot \omega \left( f,\tfrac {h}{2}\right) \end{align*}
</div>
<p>for all \(i=\overline{1,n-1},\) and consequently, \(\left\Vert m\right\Vert _{\infty }=\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\}  \leq \tfrac {64h}{15\underline{h}^{2}}\cdot \omega \left( f,\tfrac {h}{2}\right)\). Similarly as in the proof of <a href="#th5">corollary 7</a> we obtain</p>
<div class="displaymath" id="a0000000042">
  \begin{align*}  \left\vert S\left( x\right) -f\left( x\right) \right\vert & \leq \tfrac {9317}{8192}\cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192} \cdot \omega \left( f,h\right) +\tfrac {h\sqrt{3}}{18}\cdot \tfrac {64h}{15\underline{h}^{2}}\omega \left( f,\tfrac {h}{2}\right) \\ & \leq \left( \tfrac {9317}{8192}+\tfrac {32\sqrt{3}\beta ^{2}}{135}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192} \cdot \omega \left(f,h\right) ,\quad \forall x\in \lbrack a,b]. \end{align*}
</div>
<p>Since the solution of (<a href="#eq16">16</a>) minimize \(J_{2}\left( S\right) \) we infer that it minimize the mean curvature \(\sqrt{J_{2}\left( S\right) }\) of the graph of \(S\), too. <div class="proof_wrapper" id="a0000000043">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="rem3">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">11</span>
  </div>
  <div class="remark_thmcontent">
  <p>The solution of (<a href="#eq16">16</a>) provides a less smooth quartic spline \(S\in C^{1}[a,b]\) with increased deficiency, and for the Marsden’s type quartic spline \(S\in C^{2}[a,b]\) given in (<a href="#eq2">2</a>) we cannot obtain neither a natural kind spline, nor the continuity property \(S\in C^{3}[a,b]\). Altough, natural quartic spline can be obtained with high degree of continuity \(S\in C^{3}[a,b]\), and a way that leads to natural quartic spline were obtained in <span class="cite">
	[
	<a href="#blaga" >3</a>
	]
</span>, for the class of interpolating-derivative splines with minimal \(J_{3}\left( S\right)\). For quartic splines \(S\in C^{3}[a,b]\) with different type interpolation points \(t_{i},\) \(i=\overline{1,n}\) with \(x_{i-1}{\lt}t_{i}{\lt}x_{i}\), the property of minimal curvature (minimal \(J_{2}\)) was obtained in <span class="cite">
	[
	<a href="#kobza1997" >14</a>
	]
</span> in connection with the endpoint type conditions \(S^{\prime \prime }\left(a\right) =S^{\prime \prime }\left( b\right) =S^{\prime \prime \prime } \left(a\right) =S^{\prime \prime \prime }\left( b\right) =0\) (natural quartic spline) and \(S^{\left( j\right) }\left( a\right) =f^{\left( j\right) }\left(a\right) ,\) \(S^{\left( j\right) }\left( b\right) =f^{\left( j\right) }\left(b\right) ,\) \(j\in \{ 0,1\} \) (complete quartic spline). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>Concerning the minimal mean curvature \(\sqrt{J_{3}\left( S\right) }\) of the graph of \(S^{\prime }\), since </p>
<div class="displaymath" id="a0000000044">
  \begin{equation*}  J_{3}\left( S\right) \left( m_{0},m_{1},\ldots ,m_{n}\right) =\sum \limits _{i=1}^{n}\int _{x_{i-1}}^{x_{i}}[S^{\prime \prime \prime }\left( x\right) ]^{2}dx \end{equation*}
</div>
<p>the minimization of \(\sqrt{J_{3}\left( S\right) }\) can be obtained by solving the system of normal equations \(\frac{\partial J_{3}}{\partial m_{i}},\) \(i=\overline{0,n}\). After elementary calculus, this system becomes </p>
<div class="equation" id="eq18">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq18} \left\{  \begin{array}{l} m_{0}-\tfrac {13}{19}m_{1}=\tfrac {-70y_{0}+157y_{1-1/2}-87y_{1}}{19h_{1}}=d_{0}^{\prime \prime \prime } \\[2mm] -\tfrac {13h_{i+1}^{3}}{19\left( h_{i}^{3}+h_{i+1}^{3}\right) }\cdot m_{i-1}+m_{i}-\tfrac {13h_{i}^{3}}{19\left( h_{i}^{3}+h_{i+1}^{3}\right) }\cdot m_{i+1}=d_{i}^{\prime \prime \prime }= \\[2mm] =\tfrac {h_{i+1}^{3}\cdot \left( 87y_{i-1}-157y_{i-1/2}+70y_{i}\right) }{19h_{i}\left( h_{i}^{3}+h_{i+1}^{3}\right) }+\tfrac {h_{i}^{3}\cdot \left( -70y_{i}+157y_{i+1/2}-87y_{i+1}\right) }{19h_{i+1}\left( h_{i}^{3}+h_{i+1}^{3}\right) },\quad i=\overline{1,n-1} \\[2mm] -\tfrac {13}{19}m_{n-1}+m_{n}=\tfrac {87y_{n-1}-157y_{n-1/2}+70y_{n}}{19h_{n}}=d_{n}^{\prime \prime \prime }\end{array}\right. \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p>being diagonally dominant and thus has unique solution that minimizes \(J_{3}\left( S\right) \). Since </p>
<div class="displaymath" id="a0000000045">
  \begin{align*}  \left\vert d_{0}^{\prime \prime \prime }\right\vert \leq & \tfrac {70\left\vert y_{0}-y_{1-1/2}\right\vert +87\left\vert y_{1-1/2}-y_{1}\right\vert }{19h_{1}}\leq \tfrac {157}{19\underline{h}}\cdot \omega \left( f,\tfrac {h}{2}\right) , \\ \left\vert d_{n}^{\prime \prime \prime }\right\vert \leq &  \tfrac {157}{19\underline{h}}\cdot \omega \left( f,\tfrac {h}{2}\right) \\ \left\vert d_{i}^{\prime \prime \prime }\right\vert \leq & \tfrac {157h_{i+1}^{3}}{19h_{i}\left( h_{i}^{3}+h_{i+1}^{3}\right) } \omega \left( f,\tfrac {h_{i}}{2}\right) +\tfrac {157h_{i}^{3}}{19h_{i+1}\left( h_{i}^{3}+h_{i+1}^{3}\right) } \omega \left( f,\tfrac {h_{i+1}}{2}\right) \leq \tfrac {157h^{3}}{19\underline{h}^{4}}\cdot \omega \left( f,\tfrac {h}{2}\right) , \end{align*}
</div>
<p> \(\forall i=\overline{1,n-1}\) and since the matrix \(A^{\prime \prime \prime }\) of the linear system (<a href="#eq18">18</a>) has the inverse with \(\Vert \left( A^{\prime \prime \prime }\right) ^{-1}\Vert \leq \frac{19}{6}\), for the solution \(m=\left( m_{0},m_{1},\ldots ,m_{n}\right) \) of (<a href="#eq18">18</a>) we obtain the estimate</p>
<div class="displaymath" id="a0000000046">
  \begin{equation*}  \left\Vert m\right\Vert _{\infty }=\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\} \leq \tfrac {157h^{3}}{6\underline{h}^{4}}\cdot \omega \left( f,\tfrac {h}{2}\right) . \end{equation*}
</div>
<p>In this way it obtains the following result. </p>
<p><div class="corollary_thmwrapper " id="cor2">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">12</span>
  </div>
  <div class="corollary_thmcontent">
  <p>The functional \(J_{3}\left( S\right) \) has unique minimum point which minimize the mean curvature of the graph of the derivative \(S^{\prime } \). The interpolation error estimate of the obtained deficient quartic spline with minimal curvature of the graph of \(S^{\prime }\) is, </p>
<div class="equation" id="eq19">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq19} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\tfrac {157\sqrt{3}\beta ^{4}}{108}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left(f,h\right) ,\quad \forall x\in \lbrack a,b]. \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>

  </div>
</div> </p>
<h2 id="a0000000047">3.2 Minimal average slope of the graph</h2>
<p>In order to establish the paramaters \(m_{i}\), \(i=\overline{0,n}\), of the quartic spline with minimal average slope of the graph \(\sqrt{J_{1}\left( S\right) }\), we minimize the functional \(J_{1}\left( S\right) \) which has the expression </p>
<div class="displaymath" id="a0000000048">
  \begin{equation*}  J_{1}\left( S\right) \left( m_{0},m_{1},\ldots ,m_{n}\right) =\sum \limits _{i=1}^{n}\int \limits _{x_{i-1}}^{x_{i}}[S^{\prime }\left( x\right) ]^{2}dx \end{equation*}
</div>
<p>The corresponding system of normal equations \(\frac{\partial J_{1}}{\partial m_{i}}\), \(i=\overline{0,n}\), will be after some computation,</p>
<div class="equation" id="eq20">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq20} \left\{  \begin{array}{l} m_{0}+\frac{5}{16}m_{1}=\frac{-29y_{0}+16y_{1-1/2}+13y_{1}}{16h_{1}}=d_{0}^{\prime } \\[2mm] \frac{5h_{i}}{16\left( h_{i}+h_{i+1}\right) }\cdot m_{i-1}+m_{i}+\frac{5h_{i+1}}{16\left( h_{i}+h_{i+1}\right) }\cdot m_{i+1}= \\[2mm] =\frac{-13y_{i-1}-16y_{i-1/2}+16y_{i+1/2}+13y_{i+1}}{16\left( h_{i}+h_{i+1}\right) }=d_{i}^{\prime },\quad i=\overline{1,n-1} \\[2mm] \frac{5}{16}m_{n-1}+m_{n}=\frac{-13y_{n-1}-16y_{n-1/2}+29y_{n}}{16h_{n}}=d_{n}^{\prime }\end{array}\right. \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p>and since the matrix of this system is diagonally dominant, the system (<a href="#eq20">20</a>) has unique solution. In this way we obtain the following result. </p>
<p><div class="theorem_thmwrapper " id="th7">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">13</span>
  </div>
  <div class="theorem_thmcontent">
  <p> The functional \(J_{1}\left( S\right)\) has unique minimum point \(\left(m_{0},m_{1},\ldots ,m_{n}\right)\) and for the corresponding quartic spline \(S\in C^{1}[a,b]\) interpolating a function \(f\in C[a,b]\), we have the following error estimate: </p>
<div class="equation" id="eq21">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq21} \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \left( \tfrac {9317}{8192}+\tfrac {7\sqrt{3}\beta }{33}\right) \cdot \omega \left(f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left(f,h\right) ,\quad \forall x\in \lbrack a,b]. \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000049">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Based on the strictly diagonal dominance of the matrix \(A^{\prime }\) of the system (<a href="#eq20">20</a>) we infer that \(\Vert \left(A^{\prime } \right)^{-1}\Vert _{\infty } \leq \tfrac {16}{11}\), this system has unique solution and all the diagonal minors of the Hessian matrix \(\left( \tfrac {\partial ^{2}J_{1}}{\partial m_{i}\partial m_{j}}\right) _{i,j=\overline{0,n}} \) are strictly positive. Then, this solution is the unique extremal point of the functional \(J_{1}\left( S\right) \) and it is a minimum point. In order to obtain the error estimate, firstly we see that</p>
<div class="displaymath" id="a0000000050">
  \begin{align*}  \left\vert d_{0}^{\prime }\right\vert \leq &  \tfrac {29\left\vert y_{1-1/2}-y_{0}\right\vert +13\left\vert y_{1}-y_{1-1/2}\right\vert } {16h_{1}} \leq \tfrac {21}{8\underline{h}}\cdot \omega \left(f,\tfrac {h}{2} \right) , \\ \left\vert d_{n}^{\prime }\right\vert \leq &  \tfrac {21}{8\underline{h}} \cdot \omega \left( f,\tfrac {h}{2}\right),\\ d_{i}^{\prime }= &  \tfrac {-13y_{i-1}-16y_{i-1/2}}{16\left( h_{i}+h_{i+1} \right) } +\tfrac {29y_{i}}{16\left(h_{i}+h_{i+1}\right)} +\tfrac {16y_{i+1/2}+13y_{i+1}}{16\left( h_{i}+h_{i+1}\right)} -\tfrac {29y_{i}}{16\left( h_{i}+h_{i+1}\right)} \\ =& \tfrac {13 \left( y_{i-1/2}-y_{i-1} \right) }{16 \left(h_{i}+h_{i+1} \right)}+ \tfrac {29\left( y_{i}-y_{i-1/2}\right) }{16 \left(h_{i} +h_{i+1}\right) }+ \tfrac {29\left( y_{i+1/2}-y_{i}\right) }{16 \left(h_{i}+h_{i+1}\right) }+ \tfrac {13\left(y_{i+1}-y_{i+1/2}\right)} {16\left( h_{i}+h_{i+1}\right)} \end{align*}
</div>
<p>and consequently,</p>
<div class="displaymath" id="a0000000051">
  \begin{equation*}  \left\vert d_{i}^{\prime }\right\vert \leq \tfrac {42\cdot \omega (f, \frac{h_{i}}{2}) }{16\left( h_{i}+h_{i+1}\right)} +\tfrac {42 \cdot \omega ( f,\frac{h_{i+1}}{2}) }{16\left( h_{i}+h_{i+1}\right) } \leq \tfrac {21}{8\underline{h}}\cdot \omega ( f,\tfrac {h}{2}) , \quad i=\overline{1,n-1}. \end{equation*}
</div>
<p>It follows that \(\left\Vert m\right\Vert _{\infty }=\max \{ \left\vert m_{i}\right\vert :i=\overline{0,n}\} \leq \tfrac {42} {11\underline{h}} \cdot \omega ( f,\tfrac {h}{2}) \) obtaining,</p>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  \left\vert S\left( x\right) -f\left( x\right) \right\vert \leq \tfrac {9317}{8192} \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left( f,h\right) +\tfrac {h\sqrt{3}}{18}\cdot \tfrac {42}{11\underline{h}} \omega \left(f,\tfrac {h}{2}\right) \leq \end{equation*}
</div>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  \leq \left( \tfrac {9317}{8192}+\tfrac {7\sqrt{3}\beta }{33}\right) \cdot \omega \left( f,\tfrac {h}{2}\right) +\tfrac {1125}{8192}\cdot \omega \left(f,h\right),\quad \forall x\in \lbrack a,b]. \end{equation*}
</div>
<p> <div class="proof_wrapper" id="a0000000054">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="rem4">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">14</span>
  </div>
  <div class="remark_thmcontent">
  <p>Considering the index of the diagonally dominant property introduced in <span class="cite">
	[
	<a href="#han2018" >8</a>
	]
</span> for a matrix \(A=\left( a_{ij}\right) _{i,j= \overline{0,n}},\) be the constant</p>
<div class="displaymath" id="a0000000055">
  \begin{equation*}  D\left( A\right) =\max \limits _{i=\overline{0,n}} \left(\tfrac {1} {\left\vert a_{ii} \right\vert } \cdot \sum \limits _{j=0,j \neq i} ^{n} \left\vert a_{ij}\right\vert \right) \end{equation*}
</div>
<p>and denoting by \(D,\) \(D^{\prime \prime },\) \(D^{\prime \prime \prime },\) \(D^{\prime },\) the index of the diagonally dominant property of the matrices of the linear systems (<a href="#eq1">1</a>), (<a href="#eq16">16</a>), (<a href="#eq18">18</a>), (<a href="#eq20">20</a>), respectively, we see that \(D=\frac{1}{4},\) \(D^{\prime \prime }= \frac{1}{6},\) \(D^{\prime \prime \prime }=\frac{13}{19},\) \(D^{\prime }=\frac{5}{16},\) and \(D^{\prime \prime }{\lt}D{\lt}D^{\prime }{\lt}D^{\prime \prime \prime }\). So, the matrix of the system of normal equations associated to the minimal curvature of the graph has stronger diagonally dominant property than the others. Investigating the error estimates obtained in (<a href="#eq13">13</a>), (<a href="#eq17">17</a>), (<a href="#eq19">19</a>), and (<a href="#eq21">21</a>), and considering even the case of equally spaced knots, when \(\beta =1,\) we see that</p>
<div class="displaymath" id="a0000000056">
  \begin{equation*}  \tfrac {7\sqrt{3}}{33}\simeq 0.3674{\lt}\tfrac {32\sqrt{3}}{135}\simeq 0.41056{\lt}\tfrac {8\sqrt{3}}{27}\simeq 0.5132{\lt}\tfrac {157\sqrt{3}}{108}\simeq 2.\, \allowbreak 5179. \end{equation*}
</div>
<p>Consequently, the quartic spline with minimal average slope of the graph has the best error estimate both for uniform and nonuniform partitions. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="tab:1">4 Numerical experiment</h1>

<p>In order to illustrate the theoretical results consider \(n=5\) and the following data presented in <a href="#tab:1">section 4</a>. </p>
<div class="table"  id="tab:2">
   <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(i\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(x_{i}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right" 
        rowspan=""
        colspan="">
      <p> 10 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(y_{i}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 16 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 20 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 28 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 21 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 24 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right" 
        rowspan=""
        colspan="">
      <p> 28 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(y_{i-1/2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 12 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 23 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 32 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 18 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 30 </p>

    </td>
  </tr>
</table></small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">The input data.</span> 
</figcaption> 
</div>
<p>According to (<a href="#eq1">1</a>), (<a href="#nat">15</a>) and (<a href="#eq16">16</a>), let \(SC\in C^{2}[0,10]\) be the deficient quartic spline with the endpoint conditions as in (<a href="#nat">15</a>) and \(SMC\in C^{1}[0,10]\) be the deficient quartic spline with minimal mean curvature of the graph \(\sqrt{J_{2}}\) obtained according to <a href="#th6">theorem 10</a>. With \(SD\) we denote the deficient quartic spline with minimal average slope of the graph \(\sqrt{J_{1}}\) obtained in <a href="#th7">theorem 13</a> after solving the linear system (<a href="#eq18">18</a>), and with \(SDC\) we denote the quartic spline with minimal mean curvature of the graph of the first derivative \(\sqrt{J_{3}}\), according to (<a href="#eq20">20</a>). By solving the linear systems (<a href="#eq1">1</a>)+(<a href="#nat">15</a>), (<a href="#eq16">16</a>), (<a href="#eq18">18</a>) and (<a href="#eq20">20</a>) we obtain the corresponding local derivatives \(m_{i},\) \(i=\overline{0,5}\), for each of the above mentioned quartic splines, and the results are presented in <a href="#tab:2">table 1</a>. </p>
<div class="table"  id="a0000000057">
   <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>splines </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(m_{0}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(m_{1}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(m_{2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(m_{3}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(m_{4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(m_{5}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(SC\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -8.7018 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.1929 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8.2452 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -10.731 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.9057 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> -4.5236 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(SMC\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -7.8476 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6.9145 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.488 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -10.225 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.8167 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> -4.1417 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(SD\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -1.9689 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5.1006 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.5249 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -5.5601 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5.4596 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> -1.0811 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(SDC\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> -21.343 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6.3453 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 13.621 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> -17.530 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6.2924 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> -13.116 </p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">The local derivatives \(m_{i},\) \(i=\overline{0,5}\).</span> 
</figcaption> 
</div>
<figure id="fig:1">
  <div class="centered"> <img src="img-0001.png" alt="\includegraphics[height=6.5cm, width=10cm]{Figure1.png}" style="height:6.5cm; width:10cm" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">The graphs of the \(C^{2}\)-smooth quartic spline \(SC\) ( .... ), and of \(SD\) ( — ) with minimal \(\protect \sqrt{J_{1}\left( S\right) }\).</span> 
</figcaption>


</figure>
<figure id="tab:3">
  <div class="centered"> <img src="img-0002.png" alt="\includegraphics[height=6.5cm, width=10cm]{Figure2.png}" style="height:6.5cm; width:10cm" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Graphs of \(SMC\) (—) with minimal \(\protect \sqrt{J_{2}\left(S\right)}\) and \(SDC\) (....) with minimal \(\protect \sqrt{J_{3}\left( S\right)}\).</span> 
</figcaption>


</figure>
<p>These quartic splines are represented in <a href="#fig:1">figure 1</a> and <a href="#tab:3">figure 2</a> illustrating their interpolation properties. In <a href="#fig:1">figure 1</a> we represent with solid line the quartic spline \(SD\) having minimal average slope of the graph and in dots is plotted the quartic spline \(SC\in C^{2}[0,10]\) obtained by solving the system (<a href="#eq1">1</a>)+(<a href="#nat">15</a>). The quartic spline \(SMC\) with minimal mean curvature of the graph is represented under solid line in <a href="#tab:3">figure 2</a>, while the quartic spline \(SDC\) with minimal mean curvature of the graph of the first derivative is plotted in dots. Investigating ?? we see that smaller oscillation can be observed at the quartic spline \(SD\) with minimal average slope of the graph and at the quartic spline \(SMC\) with minimal mean curvature of the graph, respectively. In order to illustrate this geometric property observed in ?? we compute the length \(L\left( S\right) \) of the graph for this four quartic splines and the results are sumarized in <a href="#tab:3">figure 2</a>. As was expected, the deficient quartic spline \(SD\) with minimal average slope of the graph has the smallest graph length. </p>
<p>The graphs and figures were obtained by using the Matlab application. </p>
<div class="table"  id="a0000000058">
   <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> splines: </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(SC\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(SD\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(SMC\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(SDC\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(L\left( S\right) :\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 68.676 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 63.15 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 68.237 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 72.735 </p>

    </td>
  </tr>
</table></small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">The length of graph.</span> 
</figcaption> 
</div>
<h1 id="a0000000059">5 Conclusions</h1>
<p>Observing the possibility to express the error estimate of the deficient quartic spline interpolant \(S\in C^{2}[a,b]\) in terms of both \(\omega \left( f,\frac{h}{2}\right) \) and \(\omega \left( f,h\right) \), in the case of interpolated functions \(f\in C[a,b]\), in this work we improve the error estimates from <span class="cite">
	[
	<a href="#howell" >10</a>
	]
</span> and <span class="cite">
	[
	<a href="#howell1986" >9</a>
	]
</span>, the results being obtained in <a href="#th4">theorem 4</a> and <a href="#th5">corollary 7</a>. This fact is revealed in <a href="#rem1">remark 5</a> by observing a smaller constant for equally spaced knots in the case of Lipschitzian interpolated functions. The possibility to determine the local derivatives \(m_{i}\), \(i=\overline{0,n}\), for obtaining certain optimal properties of the deficient quartic spline \(S\in C^{1}[a,b]\), is investigated. In this context, the deficient quartic spline with minimal mean curvature of the graph of \(S\) and \(S^{\prime }\), respectively, are obtained in <a href="#th6">theorem 10</a> and <a href="#cor2">corollary 12</a>. Related to the error estimate in terms of the modulus of continuity, a better bound is observed at the deficient quartic spline with minimal average slope of the graph, which is obtained in <a href="#th7">theorem 13</a>. </p>
<p>The numerical example shows the quality of the interpolation properties of the above presented four quartic splines. The technique of minimizing the mean curvature and the average slope of the graph, presented in this work, can be extended to parametric quartic spline curves, too, and this is the subject of a future work, investigating the use of both the chordal and centripetal parametrization. </p>
<p><small class="footnotesize">  </small></p>
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</a> </p>
</dd>
</dl>


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