# Construction of Two-Derivative Runge-Kutta Methods of Order Six
## 3 Sixth-Order Methods
\begin{table}
\begin{tabular}{l l} \hline
**Tree (\(t\))** & **Elementary Weight Vector \(\boldsymbol{\eta(t)}\)** \\ \hline \(t_{51}\) & \(A\,c^{4}\,+\,4\,\AA\,c^{3}\) \\ \(t_{52}\) & \(A\,c^{2}\,A\,c\,+\,A\,c^{2}\,\hat{c}\,+\,2\,\AA\,c\,c\,+\,\AA\,c^{3}\) \\ \(t_{53}\) & \(A\,c\,A\,c^{2}\,+\,2\,A\,c\,\AA\,c\,+\,\AA\,c^{2}\,+\,2\,\AA^{2}\,c\,+\,\AA\,c^{3}\) \\ \(t_{54}\) & \(A\,c\,A^{2}\,c\,+\,A\,c\,A\,\hat{c}\,+\,A\,c\,\hat{A}\,c\,+\,\AA^{2}\,c\,+\, \AA\,c\,+\) \\ \(\hat{A}^{2}\,c\,+\,\AA\,c\,A\,c\,+\,\AA\,c\,\hat{c}\) \\ \(t_{55}\) & \(A\,((A\,c\,+\,\hat{c})\ast(A\,c\,+\,\hat{c}))\,+\,2\,\AA\,c\,c\,+\,2\,\AA\,c\,\hat {c}\) \\ \(t_{56}\) & \(A^{2}\,c^{3}\,+\,3\,A\,\AA\,c^{2}\,+\,\AA\,c^{3}\) \\ \(t_{57}\) & \(A^{2}\,c\,A\,c\,+\,A^{2}\,c\,\hat{c}\,+\,A\,\AA\,c\,+\,A\,\AA\,\hat{c}\,+\,A\, \AA\,c^{2}\,+\,\AA\,c\,c\,+\,\AA\,c\,\hat{c}\) \\ \(t_{58}\) & \(A^{3}\,c^{2}\,+\,2\,A^{2}\,\AA\,c\,+\,A\,\AA\,c^{2}\,+\,\AA\,c^{2}\,+\,2\AA^{2}\,c\) \\ \(t_{59}\) & \(A^{4}\,c\,+\,A^{3}\,\hat{c}\,+\,A^{2}\,\AA\,c\,+\,A\,\AA\,c\,+\,A\,\AA\,\hat {c}\,+\,\AA\,^{2}\,c\,+\,\AA\,A\,\hat{c}\,+\,\AA^{2}\,c\) \\ \hline \end{tabular}
\end{table} 
Table 2: Elementary weight vectors of order 5. As an application, we shall demonstrate the construction of a specific method. Since the \(c_{i}\)s are free parameters, we chose the equidistant grid \(c_{2}=\frac{1}{4}\), \(c_{3}=\frac{1}{2}\), \(c_{4}=\frac{3}{4}\). From the quadrature conditions (3), we derive  
\[b_{1}=\tfrac{1}{405}\Big{(}101-345\,b_{5}-4440\,\hat{b}_{5}\Big{)},\qquad b_{3}=\frac{16}{15}-12\,b_{5}-136\,\hat{b}_{5},\] \[b_{4}=\tfrac{64}{405}\Big{(}-2+75\,b_{5}+930\,\hat{b}_{5}\Big{)},\qquad\hat{b}_{1}=\frac{1}{54}\Big{(}1-6\,b_{5}+78\,\hat{b}_{5}\Big{)},\] \[\hat{b}_{3}=\tfrac{1}{15}-2\,b_{5}-24\,\hat{b}_{5},\qquad\qquad \hat{b}_{4}=-\frac{16}{135}\Big{(}-1+15\,b_{5}+150\,\hat{b}_{5}\Big{)}.\]  
Assumptions (9) and the first of (10) are satisfied by setting \(a_{32}=a_{42}=a_{52}=0\). From the choice of \(c_{i}\), we have \(\hat{a}_{41}=\hat{a}_{43}\). Let \(\hat{a}_{41}=0\). From the second of (10)  
\[\hat{a}_{42}=\frac{-9(4-45b_{5}+45\hat{a}_{52}b_{5}-510\hat{b}_{5})}{64(-2+75b _{5}+930\hat{b}_{5})}.\]  
From (11)-(14)  
\[\hat{a}_{51} = \frac{1}{180b_{5}}\Big{(}-8+5(26+a_{53})b_{5}+1560\hat{b}_{5}\Big{)}\] \[\hat{a}_{52} = -\frac{-2}{15b_{5}}\Big{(}-1+20b_{5}+240\hat{b}_{5}\Big{)}\] \[\hat{a}_{53} = \frac{1}{6-b_{5}}\Big{(}-8+15(8+a_{53})b_{5}+1560\hat{b}_{5}\Big{)}.\]  
At this point, all order conditions are satisfied. Still, coefficients \(a_{53}\), \(b_{5}\) and \(\hat{b}_{5}\) are not determined. We choose these coefficients so that the next two terms of the phase-lag error and one term of the amplification error are eliminated  
\[a_{53}=\frac{1}{315b_{5}^{2}}\Big{(}-2520b_{5}^{2}+b_{5}(110-35,280\hat{b}_{5} )+224(1-195b_{5})\hat{b}_{5}\Big{)}\]  
and  
\[\hat{b}_{5}=\frac{28,721+31\sqrt{723,121}}{642,600},\quad b_{5}=\frac{-1,396, 559-1669\sqrt{723,121}}{3,901,500}\]  
For this method, the stability function is  
\[b_{1}=\tfrac{1}{405}\Big{(}101-345\,b_{5}-4440\,\hat{b}_{5}\Big{)},\qquad b_{3}=\frac{16}{15}-12\,b_{5}-136\,\hat{b}_{5},\]0and the phase-lag and amplification errors are  
\[b_{1}=\tfrac{1}{405}\Big{(}101-345\,b_{5}-4440\,\hat{b}_{5}\Big{)},\qquad b_{3}=\frac{16}{15}-12\,b_{5}-136\,\hat{b}_{5},\]1  
In Figure 1, we have plotted the stability region both for this method (_new_) and for the seventh-order special TDRK method (_CT_) in [10].