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<h1>A class of numerical methods for autonomous initial value problems\(\S \)</h1>
<p class="authors">
<span class="author">Flavius–Olimpiu Pătrulescu\(^\ast \)</span>
</p>
<p class="date">June 12, 2012.</p>
</div>
<p>\(\S \) This work has been supported in part by grant no. PN II RU TE 2011-3-0013 </p>
<p>\(^\ast \)Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O. Box 68–1, \(400110\), Cluj-Napoca, Romania, e-mail: <span class="tt">fpatrulescu@ictp.acad.ro</span>. </p>

<div class="abstract"><p> In this paper we introduce a class of explicit numerical methods for approximating the solutions of scalar initial value problems for first order differential equations, using a nonlinear interpolation formula. We show that the methods generated in this way can be identified as explicit Runge-Kutta methods and we analyze some particular cases. Finally, numerical examples are provided. </p>
<p><b class="bf">MSC.</b> 65L05, 65L06, 65L20 </p>
<p><b class="bf">Keywords.</b> initial value problem, stability region, convergence order, local truncation error </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> Consider a scalar <i class="it">initial value problem</i> (IVP): </p>
<div class="equation" id="a0000000003">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{cases} y’=f(x,y),\quad x\in I\\ y(x_0)=y_0,\label{ec-gen} \end{cases} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where: \(I\subseteq \mathbb {R}\), \(y_0\in \mathbb {R}\), \(f:I\times \mathbb {R}\rightarrow \mathbb {R}\) and \(x_0\in I\). We assume that \(I=[x_0,x_0+T]\), \(0{\lt}T{\lt}\infty \) and the function \(f\) satisfies all requirements necessary to insure the existence of a unique solution \(y\) on the bounded interval \(I\), see <span class="cite">
	[
	<a href="#Butcher" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#Crouzeix" >2</a>
	]
</span> for details. </p>
<p>In this paper we present a numerical method to approximate the solution of IVP (<a href="#ec-gen">3</a>) using a particular case of a nonlinear interpolation formula analyzed in <span class="cite">
	[
	<a href="#Pavaloiu1" >5</a>
	]
</span> and <span class="cite">
	[
	<a href="#Traub" >7</a>
	]
</span>. Another particular case was utilized in <span class="cite">
	[
	<a href="#Patrulescu" >4</a>
	]
</span>. </p>
<p>For a \(5\)-times differentiable function \(h:I\rightarrow \mathbb {R}\) consider the function \(g:I\rightarrow \mathbb {R}\) given by </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x00000257015A5C10>">
<p>
  <div class="equation_content">
    \begin{equation} g(x)=h(x_0)+\tfrac {1}{2}(x-x_0)[h'(x_0+\alpha _1(x-x_0))+h'(x_0+\alpha _2(x-x_0))],\quad x\in I,\label{definire_g}\end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> where \(\alpha _1=\tfrac {3-\sqrt{3}}{6}\) and \(\alpha _2=\tfrac {3+\sqrt{3}}{6}\). </p>
<p>It is showed in <span class="cite">
	[
	<a href="#Pavaloiu1" >5</a>
	]
</span> that the function \(g\) verifies </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x00000257015A6870>">
<p>
  <div class="equation_content">
    \begin{equation}  h^{(i)}(x_0)=g^{(i)}(x_0),\, i=\overline{0,4}\,  \label{cond_g} \mbox{ and }\, |h(x)-g(x)|\leq \tfrac {71}{4320}M_5|x-x_0|^5,\,  x\in I, \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> where \(\displaystyle M_5=\sup _{x\in I}|h^{(5)}(x)|\). Also, we observe that the coefficients \(\alpha _1\), \(\alpha _2\) satisfy the equalities </p>
<div class="equation" id="a0000000004">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{cases}  \alpha _1+\alpha _2=1,\quad \alpha ^2_1+\alpha ^2_2=\tfrac {2}{3},\\ \alpha ^3_1+\alpha ^3_2=\tfrac {1}{2},\quad \alpha ^4_1+\alpha ^4_2=\tfrac {7}{18}.\label{relatii_coeficienti} \end{cases} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> In the next sections, for simplicity, we consider only the autonomous case, i.e. \(f=f(y)\). The general case can be treated using similar arguments. </p>
<p>The paper is structured as follows. In Section \(2\) we derive the numerical method and in Section \(3\) we analyze its versions corresponding to the particular cases of the parameter. In Section \(4\) numerical examples are provided to exemplify the results obtained in the previous sections. </p>
<h1 id="a0000000005">2 The numerical method</h1>
<p> The interval \(I=[x_0,x_0+T]\) is partitioned by the point set \(\{ x_n|n=\overline{0,N}, N\in \mathbb {N}\} \), where \(x_n\) are given by a prescribed rule, and we denote by \(y_n\) a numerical approximation to the exact solution \(y\) of (<a href="#ec-gen">3</a>) at \(x_n\). </p>
<p>We suppose that the exact solution \(y\) of initial value problem (<a href="#ec-gen">3</a>) is \(5\)-times differentiable at least. Conditions for regularity of exact solution of an initial value problem can be found in <span class="cite">
	[
	<a href="#Crouzeix" >2</a>
	]
</span>. </p>
<p>Using the results presented above we deduce that there exists an approximation \(\tilde{y}\) of \(y\) given by </p>
<div class="displaymath" id="a0000000006">
  \begin{align}  \tilde{y}(x)& =y(x_0)+\tfrac {1}{2}(x-x_0)[y’(x_0+\alpha _1(x-x_0))+y’(x_0+\alpha _2(x-x_0))]\nonumber \\ & =y(x_0)+\tfrac {1}{2}(x-x_0)[f(y(x_0+\alpha _1(x-x_0)))+f(y(x_0+\alpha _2(x-x_0)))],\label{formula_aproximativa} \end{align}
</div>
<p> for all \(x\in I\). From (??) we deduce that this approximation verifies </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*} \tilde{y}^{(i)}(x_0)=y^{(i)}(x_0),\,  i=\overline{0,4} \mbox{ and }|\tilde{y}(x)-y(x)|{\lt}\tfrac {71}{4320}M_5|x-x_0|^5,\,  x\in I,\end{equation*}
</div>
<p> where \(\displaystyle M_5=\sup _{x\in I}|y^{(5)}(x)|\). </p>
<p>The unknown quantities \(y(x_0+\alpha _i(x-x_0))\), \(i=1,2\), in (??) can be approximated in the same manner and we continue this approximation procedure for the next unknown values \(y(x_0+\alpha ^p_1\alpha ^q_2(x-x_0))\), \(p,\, q=1,2,\ldots \quad \). </p>
<p>Using the notation </p>
<div class="displaymath" id="a0000000008">
  \[ u_{rs}(x)=y(x_0+\alpha ^r_1\alpha ^s_2(x-x_0)) \]
</div>
<p> the algorithm described above can be written for the firsts \(p\) steps in the following compact form </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x00000257015F27B0>">
  \begin{eqnarray} \label{algoritm_aproximativ1} u_{00}(x)& =& y(x_0)+\tfrac {1}{2}(x-x_0)[f(u_{10}(x))+f(u_{01}(x))]\\ u_{i-j j}(x)& =& y(x_0)+\tfrac {1}{2}\alpha ^{i-j}_1\alpha ^{j}_2(x-x_0)[f(u_{i-j+1 j}(x))+f(u_{i-j j+1}(x))]\nonumber \\ \quad & \quad &  \qquad j=0,\ldots ,i,\quad i=1,\ldots ,p\nonumber \end{eqnarray}
</div>
<p> and we can define \(\tilde{y}(x)=u_{00}(x)\), \(x\in I\). </p>
<p>Taking into account that </p>
<div class="displaymath" id="a0000000009">
  \[ \tfrac {1}{2}\alpha ^{p-j}_1\alpha ^j_2(x-x_0)\rightarrow 0,\,  j=\overline{0,\, p}, \mbox{ as } p\rightarrow \infty , \]
</div>
<p> we can choose \(p=p_0\) such that \(y(x_0)\) approximates \(u_{p_00}\), \(u_{p_0-10},\ldots , u_{0p_0}\) with any given accuracy, since \(y\) is a continuous function. From (??) we obtain </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000025701618E30>">
  \begin{eqnarray} \label{algoritm_aproximativ} \tilde{u}_{00}(x)& =& y(x_0)+\tfrac {1}{2}(x-x_0)[f(\tilde{u}_{10}(x))+f(\tilde{u}_{01}(x))]\\ \tilde{u}_{i-j j}(x)& =& y(x_0)+\tfrac {1}{2}\alpha ^{i-j}_1\alpha ^{j}_2(x-x_0)[f(\tilde{u}_{i-j+1 j}(x))+f(\tilde{u}_{i-j j+1}(x))]\nonumber \\ \quad & \quad &  \qquad j=0,\ldots ,i,\quad i=1\ldots ,p_0-1\nonumber \\ \tilde{u}_{p_0-jj}(x)& =& y(x_0),\quad j=0,\ldots ,p_0.\nonumber \end{eqnarray}
</div>
<p> We use this recurrence relation to construct a numerical method for the scalar initial value problem (<a href="#ec-gen">3</a>). Replacing \(x\) by \(x_1\) in (??) we obtain an approximation \(\tilde{u}_{00}(x_1)\) for the exact value \(y(x_1)\). We denote this approximation by \(y_1\) and we apply again the algorithm (??) for \(x_2\), but, instead of \(y(x_0)=y_0\), we consider the value \(y_1\) previously computed. We repeat this procedure for each \(x_n\), \(n=\overline{1,N}\). </p>
<p>Using the notation </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  u^n_{qr}=y_n+\tfrac {1}{2}\alpha ^q_1\alpha ^r_2h_n[f(u^n_{q+1r})+f(u^n_{qr+1})], \end{equation*}
</div>
<p> the above algorithm can be written in the following way </p>
<div class="displaymath" id="metoda-numerica">
  \begin{eqnarray} \label{metoda-numerica} y_{n+1}& =& y_n+\tfrac {1}{2}h_n[f(u^n_{10})+f(u^n_{01})]\\ u^n_{i-jj}& =& y_n+\tfrac {1}{2}\alpha ^{i-j}_1\alpha ^{j}_2h_n[f(u^n_{i-j+1j})+f(u^n_{i-jj+1})]\nonumber \\ \quad & \quad & \quad j=\overline{0,\, i},\quad i=\overline{1,\, p_0-1},\nonumber \\ u^n_{p_0-jj}& =& y_n,\quad j=\overline{0,\, p_0}\nonumber \end{eqnarray}
</div>
<p> where \(h_n=x_{n+1}-x_n\), \(n=\overline{0,\, N-1}\). </p>
<p>For \(p_0=1\) we obtain the <em>Euler forward method</em> given by </p>
<div class="equation" id="a0000000011">
<p>
  <div class="equation_content">
    \begin{equation} y_{n+1}=y_n+h_nf(y_n),\quad n=\overline{1,\, N-1}.\end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> We have the following equivalence result. <div class="theorem_thmwrapper " id="a0000000012">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The method \((\ref{metoda-numerica})\) can be identified as an \(\tfrac {p_0(p_0+1)}{2}\)–stages explicit Runge-Kutta method with the Butcher array given by </p>
<div class="displaymath" id="a0000000013">
  \[ c=[0,\alpha ^{p_0-1}_1,\alpha ^{p_0-2}_1\alpha _2,\ldots ,\alpha ^{p_0-1}_2,\ldots ,\alpha ^{p_0-j}_1,\alpha ^{p_0-j-1}_1\alpha _2,\ldots ,\alpha ^{p_0-j}_2,\ldots ,\alpha _1,\alpha _2]^T, \]
</div>
<div class="displaymath" id="a0000000014">
  \[ b^T=[0,0,\ldots ,0,\tfrac {1}{2},\tfrac {1}{2}] \]
</div>
<div class="displaymath" id="a0000000015">
  \[ A=\left[ \begin{array}{ccccc}B_{p_0-1}& \ddots \\ \ddots & B_{p_0-2}& \ddots & \bigcirc \\ \quad & \ddots & B_{p_0-3}& \ddots \\ \quad & \bigcirc & \ddots & \ddots & \ddots \\ \quad & \quad & \quad & \ddots & B_1 \end{array} \right] \]
</div>
<p> where </p>
<div class="displaymath" id="a0000000016">
  \[  B_{p_0-1}=\left[ \begin{array}{cc} 0\\ \alpha ^{p_0-1}_1 & \\ \alpha ^{p_0-2}_1\alpha _2&  \bigcirc \\ \ldots \\ \alpha ^{p_0-1}_2 \end{array} \right], \]
</div>
<div class="displaymath" id="a0000000017">
  \[ \quad B_{p_0-j}= \left[ \begin{array}{cccc} \tfrac {\alpha ^{p_0-j}_1}{2}& \tfrac {\alpha ^{p_0-j}_1}{2}& \quad & \quad \\ \quad & \tfrac {\alpha ^{p_0-j-1}_1\alpha _2}{2}& \tfrac {\alpha ^{p_0-j-1}_1\alpha _2}{2}& \bigcirc \\ \quad & \bigcirc & \quad \ddots \quad \\ \quad & \quad & \tfrac {\alpha ^{p_0-j}_2}{2}& \tfrac {\alpha ^{p_0-j}_2}{2} \end{array} \right],\quad j=\overline{2,p_0-1}. \]
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We know that for a \(q\)-stages explicit Runge-Kutta method with the Butcher array (see e.g. <span class="cite">
	[
	<a href="#Butcher" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*} \begin{tabular}{c|c} $c$

& $A$

\\ \hline $\quad $

& $b^T$ 

\end{tabular}\end{equation*}
</div>
<p> the approximation for the new point \(x_{n+1}=x_n+h_n\) is given by </p>
<div class="equation" id="RKgenerala">
<p>
  <div class="equation_content">
    \begin{equation}  y_{n+1}=y_n+h_n\sum ^q_{i=1}b_ik_i,\label{RKgenerala} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*} \begin{cases}  k_1=f(x_n,y_n),\\ \displaystyle k_i=f(x_n+c_ih_n,y_n+h_n\sum ^{i-1}_{j=1}a_{ij}k_j),\quad i=\overline{2,\, q}. \end{cases}\end{equation*}
</div>
<p> It easy to see that (<a href="#RKgenerala">14</a>) provides for autonomous case and for the above Butcher array a rule of the form (<a href="#metoda-numerica">12</a>), which concludes the proof.</p>
<h1 id="a0000000021">3 Method (<a href="#metoda-numerica">12</a>) when \(p_0=2\), \(3\) and \(4\)</h1>
<p> For \(p_0=2\) we obtain </p>
<div class="equation" id="metodaI">
<p>
  <div class="equation_content">
    \begin{equation}  y_{n+1}=y_n+\tfrac {1}{2}h_n[f(y_n+\alpha _1h_nf(y_n))+f(y_n+\alpha _2h_nf(y_n))],\quad n=\overline{1,\, N-1}.\label{metodaI} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> The method (<a href="#metodaI">17</a>) can be written in a more suitable form </p>
<div class="displaymath" id="a0000000022">
  \begin{eqnarray*} & & y_{n+1}=y_n+\tfrac {1}{2}h_n[f(u^n_{10})+f(u^n_{01})],\\ & & u^n_{10}=y_n+\alpha _1h_nf(y_n),\quad u^n_{01}=y_n+\alpha _2h_nf(y_n)\nonumber \end{eqnarray*}
</div>
<p> and can be identified as a \(3\)–stages explicit Runge-Kutta method with the Butcher array given by </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*} \begin{tabular}{c|c c c} $\quad $

& $0$

& $\quad $

& $\quad $

\\ $\alpha _1$

& $\alpha _1$

& $0$

& $\quad $

\\ $\alpha _2$

& $\alpha _2$

& $0$

& $0$

\\ \hline $\quad $

& $0$

& $\tfrac {1}{2}$

& $\tfrac {1}{2}$ 

\end{tabular}\end{equation*}
</div>
<p> As in <span class="cite">
	[
	<a href="#Ralston" >6</a>
	]
</span>, we suppose that </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000025701624470>">
<p>
  <div class="equation_content">
    \begin{equation}  \| f\|  < M \mbox{ and } \| f^{(j)}\|  < \tfrac {L^j}{M^{j-1}},\label{conditii_Ralston} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> where \(\| f\| =\rm {sup}\{ |f(\it {t})|: \it {t}\in I\} \) and \(M\), \(L\) are positive real numbers. </p>
<p>The convergence order of the method (<a href="#metodaI">17</a>) is provided in the following result. <div class="theorem_thmwrapper " id="teoremap0=2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> The method \((\ref{metodaI})\) has convergence order \(2\) and the coefficient of principal local truncation error \(C_3\) has the following bound \(\| C_3\| \leq \tfrac {1}{6}ML^2.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Following <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span>, in order to obtain the local truncation error of the method (<a href="#metodaI">17</a>) we consider the operator </p>
<div class="equation" id="a0000000025">
<p>
  <div class="equation_content">
    \begin{equation}  L[z(x);h]=z(x+h)-z(x)-\tfrac {h}{2}[f(z(x)+\alpha _1f(z(x)))+f(z(x)+\alpha _2f(z(x)))], \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> where \(z\) is an arbitrary function defined on \(I\), sufficiently differentiable and \(z'(x)=f(z(x))\), for all \(x\in I\). </p>
<p>Expanding in Taylor series with respect to \(x\) and using (??) we obtain </p>
<div class="equation" id="operatormetodaI">
<p>
  <div class="equation_content">
    \begin{equation}  L[z(x);h]=\tfrac {h^3}{6}[f'(z(x))]^2f(z(x))+O(h^4).\label{operatormetodaI} \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p>Then, using the definition for the convergence order given in <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span> we deduce that the method has second-order of accuracy. Also, substituting \(z\) by the exact solution \(y\), \(x\) by \(x_n\) and supposing <em>localizing assumption</em> \(y_i=y(x_i)\), \(i=\overline{1,\, n}\) the <em>local truncation error</em> can be written as </p>
<div class="equation" id="a0000000026">
<p>
  <div class="equation_content">
    \begin{equation}  T_{n+1}=\tfrac {h^3}{6}[f'(y(x_n))]^2f(y(x_n))+O(h^4). \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> Moreover, the coefficient of <em>principal local truncation error</em> is given by </p>
<div class="equation" id="a0000000027">
<p>
  <div class="equation_content">
    \begin{equation}  C_3=\tfrac {1}{6}[f'(y(x_n))]^2f(y(x_n)). \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p>Next, using (??) we have for \(C_3\) the following bound </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  \| C_3\| =\tfrac {1}{6}\| (f’)^2f\| \leq \tfrac {1}{6}\| f’\| ^2\| f\| \leq \tfrac {1}{6}ML^2, \end{equation*}
</div>
<p> which concludes the proof.</p>
<p>Following <span class="cite">
	[
	<a href="#Crouzeix" >2</a>
	]
</span>, for the Runge-Kutta methods in addition of the <em>row-sum condition</em>, \(Ae=Ce\), and <em>consistency condition</em>, \(b^Te=1\), the sufficient conditions for order \(2\) are given by </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000002570171DF70>">
<p>
  <div class="equation_content">
    \begin{equation}  b^TCe=\tfrac {1}{2}\qquad b^TAe=\tfrac {1}{2},\label{conditii_ordin2} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p> where \(e=[1,\ldots ,1]\) and \(C=\mbox{diag}(c)\). A simple calculus shows that the method (<a href="#metodaI">17</a>) verifies conditions (??) and it represents a validation for (<a href="#operatormetodaI">19</a>). </p>
<p>The <em>stability function</em> and <em>absolute stability region</em> of the method (<a href="#metodaI">17</a>) are provided in the following result. <div class="theorem_thmwrapper " id="stabilitate_teorema_p02">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> The method \((\ref{metodaI})\) has the stability function given by </p>
<div class="equation" id="a0000000029">
<p>
  <div class="equation_content">
    \begin{equation}  R(z)=1+z+\tfrac {z^2}{2},\quad z\in \mathbb {C}. \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000030">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Following <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span>, we apply the method (<a href="#metodaI">17</a>) to the scalar test equation </p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  y’=\lambda y,\quad \lambda \in \mathbb {C},\quad {\rm Re}\lambda {\lt}0, \end{equation*}
</div>
<p> and we obtain the difference equation </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  y_{n+1}=y_n[1+(\lambda h_n)+\tfrac {1}{2}(\alpha _1+\alpha _2)(\lambda h_n)^2]. \end{equation*}
</div>
<p> Denoting \(z=\lambda h_n\) and using (??) we obtain the stability function </p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  R(z)=1+z+\tfrac {z^2}{2}. \end{equation*}
</div>
<p> Note that this function is the same as the stability function for \(2\) stages explicit Runge-Kutta methods of order \(2\).</p>
<p>The absolute stability region is given by </p>
<div class="equation" id="a0000000034">
<p>
  <div class="equation_content">
    \begin{equation}  \mathcal{R}=\{ z\in \mathbb {C}:|1+z+\tfrac {1}{2}z^2|<1\}  \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> and it is plotted using scanning technique in Figure 1. The computations have been carried out using Matlab software.  </p>
<figure id="regiunistabilitate">
  <div class="centered"> <img src="img-0001.png" alt="\includegraphics[height=3.3572in, width=4.0776in]{reg_abs_stab}" style="height:3.3572in; width:4.0776in" />
  <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Absolute stability regions for method (<a href="#metoda-numerica">12</a>) when \(p_0=2,\, 3,\, 4.\)</span> 
</figcaption> </div>

</figure>
<p>For \(p_0=3\) we obtain the method </p>
<div class="displaymath" id="metodaII">
  \begin{eqnarray} \label{metodaII} y_{n+1}& =& y_n+\tfrac {1}{2}h_n[f(u^n_{10})+f(u^n_{01})]\\ u^n_{10}& =& y_n+\tfrac {1}{2}\alpha _1h_n[f(u^n_{20})+f(u^n_{11})]\nonumber \\ u^n_{01}& =& y_n+\tfrac {1}{2}\alpha _2h_n[f(u^n_{11})+f(u^n_{02})]\nonumber \\ u^n_{2-jj}& =& y_n+\alpha ^{2-j}_1\alpha ^j_2h_nf(y_n),\quad j=\overline{0,\, 2}\nonumber \end{eqnarray}
</div>
<p> which can be identified as a \(6\)–stages explicit Runge-Kutta method with the Butcher array given by </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*} \begin{tabular}{c|c c c c c c} $\quad $

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^2_1$

& $\alpha ^2_1$

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha _1\alpha _2$

& $\alpha _1\alpha _2$

& $0$

& $0$

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^2_2$

& $\alpha ^2_2$

& $0$

& $0$

& $0$

& $\quad $

& $\quad $

\\ $\alpha _1$

& $0$

& $\tfrac {\alpha _1}{2}$

& $\tfrac {\alpha _1}{2}$

& $0$

& $0$

& $\quad $

\\ $\alpha _2$

& $0$

& $0$

& $\tfrac {\alpha _2}{2}$

& $\tfrac {\alpha _2}{2}$

& $0$

& $0$

\\ \hline $\quad $

& $0$

& $0$

& $0$

& $0$

& $\tfrac {1}{2}$

& $\tfrac {1}{2}$ 

\end{tabular}\label{matrice_metoda6evaluari} \end{equation*}
</div>
<p>The convergence order of the method (<a href="#metodaII">25</a>) is provided in the following result. <div class="theorem_thmwrapper " id="a0000000036">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The method \((\ref{metodaII})\) has convergence order \(3\) and the coefficient of the principal local truncation error \(C_4\) has the following bound \(\| C_4\| \leq \tfrac {1}{24}ML^3.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000037">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Using (??) and similar arguments to those presented in the proof of Theorem \(\ref{teoremap0=2}\) we obtain </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000025701780590>">
<p>
  <div class="equation_content">
    \begin{equation}  L[z(x);h]=\tfrac {h^4}{24}[f'(z(x))]^3f(z(x))+O(h^5).\label{operator_metoda6evaluari} \end{equation}
  </div>
  <span class="equation_label">26</span>
</p>
</div>
<p> Therefore we deduce that this method has convergence order \(3\). Also, we can determine the <em>local truncation error</em> as </p>
<div class="equation" id="a0000000038">
<p>
  <div class="equation_content">
    \begin{equation}  T_{n+1}=\tfrac {h^4}{24}[f'(y(x_n))]^3f(y(x_n))+O(h^5) \end{equation}
  </div>
  <span class="equation_label">27</span>
</p>
</div>
<p> and the coefficient of principal local truncation error is given by </p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  C_4=\tfrac {1}{24}[f’(y(x_n))]^3f(y(x_n)). \end{equation*}
</div>
<p> Using (??) we obtain \(\| C_4\| {\lt}\tfrac {1}{24}ML^3\), which concludes the proof.</p>
<p>A simple calculus shows that the method (<a href="#metodaII">25</a>) verifies the sufficient conditions for order \(3\), see <span class="cite">
	[
	<a href="#Crouzeix" >2</a>
	]
</span> for details, </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x00000257017AF5F0>">
<p>
  <div class="equation_content">
    \begin{equation}  b^TCe=\tfrac {1}{2},\quad b^TC^2e=\tfrac {1}{3},\quad b^TACe=\tfrac {1}{6} \label{conditii_ordin3} \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p> and it represents a validation for (??). <div class="theorem_thmwrapper " id="a0000000040">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The method \((\ref{metodaII})\) has the stability function given by </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x00000257017AF9B0>">
<p>
  <div class="equation_content">
    \begin{equation}  R(z)=1+z+\tfrac {z^2}{2}+\tfrac {z^3}{6}, \quad z\in \mathbb {C}\label{stabilitate_metII}. \end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000041">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Using similar arguments as in the proof of Theorem \(\ref{stabilitate_teorema_p02}\).</p>
<p>We observe that method (<a href="#metodaII">25</a>) has the same stability function as \(3\) stages explicit Runge-Kutta methods of order \(3\). The absolute stability region defined by </p>
<div class="equation" id="a0000000042">
<p>
  <div class="equation_content">
    \begin{equation}  \mathcal{R}=\{ z\in \mathbb {C}:|1+z+\tfrac {1}{2}z^2+\tfrac {1}{6}z^3|<1\}  \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p> is plotted in Figure <a href="#regiunistabilitate">1</a>. </p>
<p>For \(p_0=4\) we obtain the method </p>
<div class="displaymath" id="metodaIII">
  \begin{eqnarray} \label{metodaIII} y_{n+1}& =& y_n+\tfrac {1}{2}h_n[f(u^n_{10})+f(u^n_{01})]\\ u^n_{i-jj}& =& y_n+\tfrac {1}{2}\alpha ^{i-j}_1\alpha ^j_2h_n[f(u^n_{i-j+1j})+f(u^n_{i-jj+1})]\nonumber \\ \quad & \quad & \quad j=\overline{0,\, i},\quad i=\overline{1,\, 2},\nonumber \\ u^n_{3-jj}& =& y_n+\alpha ^{3-j}_1\alpha ^j_2h_nf(y_n)\quad j=\overline{0,\, 3}\nonumber \end{eqnarray}
</div>
<p> and can be identified as a \(10\)–stages explicit Runge-Kutta method with the Butcher array given by </p>
<div class="displaymath" id="a0000000043">
  \begin{equation*} \begin{tabular}{c| c c c c c c c c c c} $0$

&  $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^3_1$

&  $\alpha ^3_1$

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^2_1\alpha _2$

&  $\alpha ^2_1\alpha _2$

& $0$

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha _1\alpha ^2_2$

&  $\alpha _1\alpha ^2_2$

& $0$

& $0$

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^3_2$

& $\alpha ^3_2$

& $0$

& $0$

& $0$

& $0$

\\ $\alpha ^2_1$

&  $0$

& $\tfrac {\alpha ^2_1}{2}$

& $\tfrac {\alpha ^2_1}{2}$

& $0$

& $0$

& $0$

& $\quad $

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha _1\alpha _2$

&  $0$

& $0$

& $\tfrac {\alpha _1\alpha _2}{2}$

& $\tfrac {\alpha _1\alpha _2}{2}$

& $0$

& $0$

& $0$

& $\quad $

& $\quad $

& $\quad $

\\ $\alpha ^2_2$

&  $0$

& $0$

& $0$

& $\tfrac {\alpha ^2_2}{2}$

& $\tfrac {\alpha ^2_2}{2}$

& $0$

& $0$

& $0$

& $\quad $

& $\quad $

\\ $\alpha _1$

&  $0$

& $0$

& $0$

& $0$

& $0$

& $\tfrac {\alpha _1}{2}$

& $\tfrac {\alpha _1}{2}$

& $0$

& $0$

& $\quad $

\\ $\alpha _2$

&  $0$

& $0$

& $0$

& $0$

& $0$

& $0$

& $\tfrac {\alpha _2}{2}$

& $\tfrac {\alpha _2}{2}$

& $0$

& $0$

\\ $\quad $

& $0$

& $0$

& $0$

& $0$

& $0$

& $0$

& $0$

& $0$

& $\tfrac {1}{2}$

& $\tfrac {1}{2}$ 

\end{tabular}\label{matrice_10evaluari} \end{equation*}
</div>
<p> Using (??) and similar arguments to those used above we deduce that the method (<a href="#metodaIII">31</a>) has convergence order \(4\). The stability function is given by </p>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  R(z)=1+z+\tfrac {z^2}{2}+\tfrac {z^3}{6}+\tfrac {z^4}{24} \end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p> and we observe that it is the same as stability function for \(4\) stages explicit Runge-Kutta methods of order \(4\). The absolute stability region defined by </p>
<div class="equation" id="a0000000045">
<p>
  <div class="equation_content">
    \begin{equation}  \mathcal{R}=\{ z\in \mathbb {C}:|1+z+\tfrac {1}{2}z^2+\tfrac {1}{6}z^3+\tfrac {1}{24}z^4|<1\}  \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> is plotted in Figure <a href="#regiunistabilitate">1</a>. </p>
<p>We restrict ourselves to the cases when \(p_0\) takes the values \(2\), \(3\) and \(4\). For \(p_0\geq 5\) methods obtain from (<a href="#metoda-numerica">12</a>) have a form much more complicated with a high cost of calculus and the results concerning accuracy are not so outstanding. From the next theorem we deduce that the convergence order is not grater than \(4\). <div class="theorem_thmwrapper " id="teorema_ordin_general">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For \(p_0\geq 5\) the method \((\ref{metoda-numerica})\) has maximal convergence order \(4\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000046">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Following <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span>, we consider the operator </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray*}  L[z(x),h]& =& z(x+h)-z(x)-\tfrac {h}{2}[f(z(x+\alpha _1h))+f(z(x+\alpha _2h))]\\ & =& z(x+h)-z(x)-\tfrac {h}{2}[z’(x+\alpha _1h)+z’(x+\alpha _2h)], \end{eqnarray*}
</div>
<p> where \(z\) is an arbitrary function defined on \(I\), \(5\)-times differentiable at least and \(z'(x)=f(z(x))\), for all \(x\in I\). </p>
<p>Next, using Taylor series with respect to \(x\) and (??) we obtain </p>
<div class="displaymath" id="a0000000048">
  \begin{equation*}  L[z(x),h]=h^5(\tfrac {1}{120}-\tfrac {1}{48}\tfrac {7}{18})z^{(5)}(x)+O(h^6). \end{equation*}
</div>
<p>Then, from the definition of convergence order given in <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span> we deduce that for \(p_0\geq 5\) method (<a href="#metoda-numerica">12</a>) has convergence order \(4\), which concludes the proof.</p>
<p>Note that the method (<a href="#metoda-numerica">12</a>) is a <em>zero-stable</em> method because it verifies <em>root-condition</em>. Also, since the convergence order is grater than \(2\) for all \(p_0\geq 2\) we conclude that it satisfies the <em>consistency condition</em>. It follows that our method represents a convergent method, see <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span> for details. </p>
<h1 id="a0000000049">4 Numerical examples</h1>
<p>We consider the initial value problems. </p>
<p><div class="example_thmwrapper " id="Example 1">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">7</span>
  </div>
  <div class="example_thmcontent">
  
<div class="equation" id="a0000000050">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{cases}  y’(x)=\cos ^2(y(x)),\quad x\in [0,20]\\ y(0)=0. \end{cases} \end{equation}
  </div>
  <span class="equation_label">34</span>
</p>
</div>
<p> The exact solution is \(y:[0,20]\rightarrow \mathbb {R}\) given by \(y(x)=\arctan (x)\).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="Example 2">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">8</span>
  </div>
  <div class="example_thmcontent">
  
<div class="equation" id="a0000000051">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{cases}  y’(x)=\tfrac {y(x)}{4}(1-\tfrac {y(x)}{20}),\quad x\in [0,20]\\ y(0)=1. \end{cases} \end{equation}
  </div>
  <span class="equation_label">37</span>
</p>
</div>
<p> The exact solution is \(y:[0,20]\to \mathbb {R}\) given by </p>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  y(x)=\tfrac {20}{1+19\exp (-x/4)}.\hfil \qed \end{equation*}
</div>

  </div>
</div> </p>
<p>We apply the methods (<a href="#metodaI">17</a>), (<a href="#metodaII">25</a>) and (<a href="#metodaIII">31</a>) with constant steplength to determine numerical solutions for the above examples. </p>
<p>As a measure of the performance we consider the errors obtained as the maximum of the absolute errors on the mesh points \(x_n=nh\), \(n=\overline{0,\, N}\) </p>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  E_{\max }=\max \{ |y(x_n)-y_n|:n=0,1,\ldots ,N\} , \end{equation*}
</div>
<p> for different values of step \(h\). </p>
<p>In the next tables we present results given by the proposed methods (<a href="#metodaI">17</a>), (<a href="#metodaII">25</a>) (<a href="#metodaIII">31</a>) in comparison with results given by Adams-Bashforth methods of orders \(2\), \(3\), \(4\) defined by </p>
<ul class="itemize">
  <li><p>\(A-B(2):\quad y_{n+1}-y_n=\tfrac {h}{2}(3f_n-f_{n-1}),\) </p>
</li>
  <li><p>\(A-B(3):\quad y_{n+1}-y_n=\tfrac {h}{12}(23f_n-16f_{n-1}+5f_{n-2}),\) </p>
</li>
  <li><p>\(A-B(4):\quad y_{n+1}-y_n=\tfrac {h}{24}(55f_n-59f_{n-1}+37f_{n-2}-9f_{n-3})\). </p>
</li>
</ul>
<p>Also, we present results given by the explicit Runge-Kutta methods with order equal with number of stages defined by the following Butcher arrays </p>
<div class="displaymath" id="a0000000054">
  \begin{equation*} \begin{tabular}{c c c}\begin{tabular}{c|c c} $0$

& \\ $\tfrac {1}{2}$

& $0$

\\ \hline $\quad $

& $0$

& $1$

\\ \end{tabular}

& \begin{tabular}{c|c c c c} $0$

\\ $\tfrac {1}{2}$

& $\tfrac {1}{2}$

\\ $1$

& $-1$

& $2$

\\ \hline $\quad $

& $\tfrac {1}{6}$

& $\tfrac {2}{3}$

& $\tfrac {1}{6}$ 

\end{tabular}

& \begin{tabular}{c|c c c c c} $0$

\\ $\tfrac {1}{2}$

& $\tfrac {1}{2}$

\\ $\tfrac {1}{2}$

& $0$

& $\tfrac {1}{2}$

\\ $1$

& $0$

& $0$

& $1$

\\ \hline $\quad $

& $\tfrac {1}{6}$

& $\tfrac {1}{3}$

& $\tfrac {1}{3}$

& $\tfrac {1}{6}$ 

\end{tabular}

\\ $R-K(2)$

& $R-K(3)$

& $R-K(4)$ 

\end{tabular}\end{equation*}
</div>
<div class="table"  id="a0000000055">
   <div class="centered"><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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(h\)</p>

    </td>
    <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>\(1.0e-01\)</p>

    </td>
    <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>\(1.0e-02\)</p>

    </td>
    <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>\(1.0e-03\)</p>

    </td>
    <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>\(1.0e-04\)</p>

    </td>
    <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>\(1.0e-05\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaI})\)</p>

    </td>
    <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>\(5.755e-04\)</p>

    </td>
    <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>\(5.415e-06\)</p>

    </td>
    <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>\(5.381e-08\)</p>

    </td>
    <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>\(5.378e-10\)</p>

    </td>
    <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>\(5.377e-12\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(2)\)</p>

    </td>
    <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>\(2.209e-03\)</p>

    </td>
    <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> \(2.251e-05\)</p>

    </td>
    <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> \(2.256e-07\)</p>

    </td>
    <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> \(2.257e-09\)</p>

    </td>
    <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> \(2.257e-11\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(R-K(2)\)</p>

    </td>
    <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> \(5.755e-04\)</p>

    </td>
    <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> \(5.415e-06\) </p>

    </td>
    <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> \(5.381e-08\) </p>

    </td>
    <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>\(5.378e-10\)</p>

    </td>
    <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> \(5.377e-12\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaII})\)</p>

    </td>
    <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> \(1.333e-05\)</p>

    </td>
    <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> \(1.244e-08\)</p>

    </td>
    <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> \(1.235e-11\)</p>

    </td>
    <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> \(1.254e-14\)</p>

    </td>
    <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> \(2.464e-14\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(3)\)</p>

    </td>
    <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> \(1.109e-03\)</p>

    </td>
    <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> \(1.166e-06\)</p>

    </td>
    <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> \(1.166e-09\)</p>

    </td>
    <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> \(1.166e-12\) </p>

    </td>
    <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>\(2.464e-14\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(R-K(3)\)</p>

    </td>
    <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> \(2.028e-05\)</p>

    </td>
    <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> \(2.077e-08\)</p>

    </td>
    <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> \(2.082e-11\)</p>

    </td>
    <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> \(1.976e-14\)</p>

    </td>
    <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> \(2.486e-14\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaIII})\)</p>

    </td>
    <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> \(2.202e-07\)</p>

    </td>
    <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> \(2.050e-11\)</p>

    </td>
    <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> \(2.886e-15\)</p>

    </td>
    <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> \(1.110e-14\)</p>

    </td>
    <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> \(2.464e-14\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(4)\)</p>

    </td>
    <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>\(1.109e-03\)</p>

    </td>
    <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>\(1.166e-06\)</p>

    </td>
    <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> \(1.166e-09\)</p>

    </td>
    <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> \(1.166e-12\)</p>

    </td>
    <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>\(2.442e-14\)</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-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(R-K(4)\)</p>

    </td>
    <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> \(5.357e-07\)</p>

    </td>
    <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> \(5.337e-11\)</p>

    </td>
    <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> \(5.884e-15\)</p>

    </td>
    <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> \(1.132e-14\)</p>

    </td>
    <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> \(2.486e-14\)</p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">\(E_{\max }\) for <em>Example</em> <a href="#Example 1">7</a></span> 
</figcaption> </div>
</div>
<div class="table"  id="a0000000056">
   <div class="centered"><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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(h\)</p>

    </td>
    <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>\(1.0e-01\)</p>

    </td>
    <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>\(1.0e-02\)</p>

    </td>
    <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>\(1.0e-03\)</p>

    </td>
    <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>\(1.0e-04\)</p>

    </td>
    <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>\(1.0e-05\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaI})\)</p>

    </td>
    <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>\(5.878e-04\)</p>

    </td>
    <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>\(5.952e-06\)</p>

    </td>
    <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>\(5.959e-08\)</p>

    </td>
    <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>\(5.960e-10\)</p>

    </td>
    <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>\(5.929e-12\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(2)\)</p>

    </td>
    <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>\(1.892e-03\)</p>

    </td>
    <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>\(1.907e-05\)</p>

    </td>
    <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>\(1.908e-07\)</p>

    </td>
    <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>\(1.908e-09\)</p>

    </td>
    <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>\(1.892e-11\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(R-K(2)\)</p>

    </td>
    <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>\(4.805e-04\)</p>

    </td>
    <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>\(4.861e-06\)</p>

    </td>
    <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>\(4.867e-08\)</p>

    </td>
    <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>\(4.866e-10\)</p>

    </td>
    <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>\(4.757e-12\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaII})\)</p>

    </td>
    <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>\(2.725e-06\)</p>

    </td>
    <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>\( 2.764e-09\)</p>

    </td>
    <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>\(2.744e-12\)</p>

    </td>
    <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>\(4.192e-13\)</p>

    </td>
    <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>\( 6.359e-13\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(3)\)</p>

    </td>
    <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>\(1.387e-03\)</p>

    </td>
    <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>\(1.404e-05\)</p>

    </td>
    <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>\(1.406e-07\)</p>

    </td>
    <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>\(1.406e-09\)</p>

    </td>
    <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>\( 1.404e-11\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(R-K(3)\)</p>

    </td>
    <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>\(4.048e-06\)</p>

    </td>
    <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>\(4.083e-09\)</p>

    </td>
    <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>\(4.137e-12\)</p>

    </td>
    <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>\(4.209e-13\)</p>

    </td>
    <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>\(6.359e-13\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\((\ref{metodaIII})\)</p>

    </td>
    <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>\(9.951e-09\)</p>

    </td>
    <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>\(9.912e-13\)</p>

    </td>
    <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>\(6.750e-14\)</p>

    </td>
    <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>\(4.192e-13\)</p>

    </td>
    <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>\(6.359e-13\)</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-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(A-B(4)\)</p>

    </td>
    <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> \(1.420e-03\)</p>

    </td>
    <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> \(1.407e-05\)</p>

    </td>
    <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> \(1.406e-07\)</p>

    </td>
    <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> \(1.406e-09\)</p>

    </td>
    <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> \(1.404e-11\)</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-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(R-K(4)\)</p>

    </td>
    <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>\(1.779e-08\)</p>

    </td>
    <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>\(1.788e-12\)</p>

    </td>
    <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>\(6.750e-14\)</p>

    </td>
    <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>\(4.192e-13\)</p>

    </td>
    <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>\(6.323e-13\)</p>

    </td>
  </tr>
</table></small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">\(E_{\max }\) for <em>Example</em> <a href="#Example 2">8</a></span> 
</figcaption> </div>
</div>
<p>We analyze the results and we can see that for the above examples the proposed methods are at least comparable to the classical ones. </p>
<p>We observe that if length of the step decreases \(10\) times then the error magnitude for methods (<a href="#metodaI">17</a>), (<a href="#metodaII">25</a>) and (<a href="#metodaIII">31</a>) decreases \(10^2\), \(10^3\) and \(10^4\) times, respectively. These results represent a validation of the fact that the convergence orders for the methods (<a href="#metodaI">17</a>), (<a href="#metodaII">25</a>) and (<a href="#metodaIII">31</a>) are \(2\), \(3\) and \(4\), respectively. </p>
<p>Because the method is explicit we indicate to be used especially for non-stiff problems, where the requirements on the steplength imposed by stability are no restrictive than the requirements imposed by accuracy, see <span class="cite">
	[
	<a href="#Butcher" >1</a>
	]
</span> and <span class="cite">
	[
	<a href="#Lambert" >3</a>
	]
</span> for more details. Also, we expect that the method (<a href="#metoda-numerica">12</a>) gives better results in the case of a variable steplength. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="Butcher">1</a></dt>
  <dd><p><i class="sc">J.C. Butcher</i>, <i class="it">Numerical Methods for Ordinary Differential Equations</i>, John Wiley &amp; Sons, 2008. </p>
</dd>
  <dt><a name="Crouzeix">2</a></dt>
  <dd><p><i class="sc">M. Crouzeix</i> and <i class="sc">A.L. Mignot</i>, <i class="it">Analyse numérique des equations différentielles</i>, Masson, Paris, 1989. </p>
</dd>
  <dt><a name="Lambert">3</a></dt>
  <dd><p><i class="sc">J.D. Lambert</i>, <i class="it">Numerical Methods for Ordinary Differential Systems-The Initial Value Problem</i>, John Wiley &amp; Sons, 1990. </p>
</dd>
  <dt><a name="Patrulescu">4</a></dt>
  <dd><p><i class="sc">F. Pătrulescu</i>, <i class="it">A numerical method for the solution of an autonomous initial value problem</i>, Carpathian J. Math., <b class="bf">28</b>, no. 2, pp.&#160;289–296, 2012. </p>
</dd>
  <dt><a name="Pavaloiu1">5</a></dt>
  <dd><p><a href ="http://ictp.acad.ro/jnaat/journal/article/view/1997-vol26-nos1-2-art23"> <i class="sc">I. Păvăloiu</i>, <i class="it">On an approxiation formula</i>, Rev. Anal. Num<span class="rm">é</span>r. Th<span class="rm">é</span>or. Approx., <b class="bf">26</b>, no. 1–2, pp.&#160;179–183, 1997. <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Ralston">6</a></dt>
  <dd><p><i class="sc">A. Ralston</i>, <i class="it">Runge-Kutta methods with minimum error bounds</i>, Math. Comp., <b class="bf">16</b>, no. 80, pp.&#160;431–437, 1962. </p>
</dd>
  <dt><a name="Traub">7</a></dt>
  <dd><p><i class="sc">J. F. Traub</i>, <i class="it">Iterative Methods for the Solution of Equations</i>, Prentice-Hall, Inc., Englewood Cliffs, N.J., 1964. </p>
</dd>
</dl>


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