# Implicit A-Stable Peer Triplets for ODE Constrained Optimal Control Problems
## 2 Materials and Methods
### Existence of Boundary Methods Imposes Restrictions on the Standard Method
#### 2.3.1 Combined Conditions for the End Method
The corresponding matrix \(\mathcal{L}_{\mu,s}(\hat{V}_{q}^{\mathsf{T}}K_{N}V_{s})\) with appropriate Vandermonde matrices \(\hat{V}\) still has Hankel form._  
However, even with a full end matrix \(K_{N}\), Lemma 3 and the Fredholm alternative enforce restrictions on the standard method \((A,B,K)\) due to the singularity of \(\mathcal{L}_{\mu,s}\). This is discussed for the present situation with \(s=4,q=3\), only. The matrix belonging to the map \(\mathcal{L}_{q,s}\) is \(I_{s}\otimes\tilde{E}_{q}^{\mathsf{T}}+\tilde{E}_{s}^{\mathsf{T}}\otimes I_{q}\) and its transpose is \(I_{s}\otimes\tilde{E}_{q}+\tilde{E}_{s}\otimes I_{q}\). Hence, the adjoint of the map \(\mathcal{L}_{q,s}\) is given by  
\[\mathcal{L}_{q,s}^{\mathsf{T}}:\,\mathbb{R}^{q\times s}\to\mathbb{R}^{q\times s },\quad X\mapsto\tilde{E}_{q}X+X\tilde{E}_{s}^{\mathsf{T}}.\]  
Component-wise the map acts as  
\[\mathcal{L}_{q,s}^{\mathsf{T}}:\,(x_{ij})\mapsto\left(ix_{i+1,j}+jx_{i,j+1}\right)\]  
with elements having indices \(i>q\) or \(j>s\) being zero. For \(q=3\), \(s=4\), one gets  
\[\mathcal{L}_{3,4}(X)^{\mathsf{T}}=\left( \begin{array}{ccc}x_{12}+x_{21}&2x_{ 13}+x_{22}&3x_{14}+x_{23}&x_{24}\\ x_{22}+2x_{31}&2x_{23}+2x_{32}&3x_{24}+2x_{33}&2x_{34}\\ x_{32}&2x_{33}&3x_{34}&0\end{array} \right).\]  
It is seen that the kernel of \(\mathcal{L}_{3,4}^{\mathsf{T}}\) has dimension \(3\) and is given by  
\[X=\left( \begin{array}{ccc}\tilde{\xi}_{1}&\tilde{\xi}_{2}&\tilde{\xi}_{3}&0 \\ -\tilde{\xi}_{2}&-2\tilde{\xi}_{3}&0&0\\ \tilde{\xi}_{3}&0&0&0\end{array} \right). \tag{66}\]  
In (65), the Fredholm condition leads to restrictions on the matrix \(B\) from the standard scheme. However, since the matrix \(A\) should have triangular form, it is the more natural variable in the search for good methods and an equivalent reformulation of these conditions for \(A\) is of practical interest. **Lemma 5**. : _Assume that the standard method \((A,B,K)\) has local order \((s,q)=(4,3)\). Then, end methods \((A_{N},B_{N},K_{N})\) of order \((s,q)=(4,3)\) only exist if the standard method \((A,B,K)\) satisfies the following set of three conditions, either for \(B\) or for \(A\),_  
\[ \begin{array}{c}\mathbf{1}^{\mathsf{T}}B\mathbf{1}=1\\ \mathbf{1}^{\mathsf{T}}B\mathbf{c}-\mathbf{c}^{\mathsf{T}}B\mathbf{1}=1\\ \mathbf{1}^{\mathsf{T}}B\mathbf{c}^{2}-2\mathbf{c}^{\mathsf{T}}B\mathbf{1}= 1\end{array} \left\{ \begin{array}{c}1=\mathbf{1}^{\mathsf{T}}A\mathbf{1}\\ 1=\mathbf{1}^{\mathsf{T}}A\mathbf{c}-\mathbf{c}^{\mathsf{T}}A\mathbf{1}\\ 0=\mathbf{1}^{\mathsf{T}}A\mathbf{c}^{2}-2\mathbf{c}^{\mathsf{T}}A\mathbf{c}+( \mathbf{c}^{2})^{\mathsf{T}}A\mathbf{1}.\end{array} \right.\end{array} \tag{67}\]  
Proof. : Multiplying equation (65) by \(\mathcal{P}_{s}\) from the right and using \(\tilde{E}_{s}\mathcal{P}_{s}=\mathcal{P}_{s}\tilde{E}_{s}\), an equivalent form is\[\mathcal{L}_{q,s}(V_{q}^{\mathsf{T}}K_{N}V_{s}\mathcal{P}_{s})=\mathds{1}_{q} \mathbb{I}_{s}^{\mathsf{T}}\mathcal{P}_{s}-V_{q}^{\mathsf{T}}BV_{s}=:R,\]  
with \(\mathds{1}_{s}^{\mathsf{T}}\mathcal{P}_{s}=(1,2,4,8,\ldots)\) by the binomial formula. The Fredholm alternative requires that \(tr(X^{\mathsf{T}}R)=0\) for all \(X\) from (66). We now frequently use the identities \(tr\big{(}(\
u u^{\mathsf{T}})R\big{)}=u^{\mathsf{T}}Rv\) and \(V_{\varepsilon_{i}}=\mathbf{c}^{i-1}\) with \(\varepsilon_{i}\in\mathbb{R}^{s}\). The kernel in (66) is spanned by three basis elements. The first, \(X_{1}=\bar{\varepsilon}_{1}\varepsilon_{1}^{\mathsf{T}}\) (with the convention \(\bar{\varepsilon}_{i}\in\mathbb{R}^{q}\)) leads to  
\[0\stackrel{{!}}{{=}}tr(X_{1}^{\mathsf{T}}R)=\bar{\varepsilon}_{1 }^{\mathsf{T}}\mathds{1}_{q}\mathbb{I}_{s}^{\mathsf{T}}e_{1}-\bar{\varepsilon }_{1}^{\mathsf{T}}V_{q}^{\mathsf{T}}BV_{s}e_{1}=1-\mathds{1}_{s}^{\mathsf{T}} B\mathds{1}_{s}.\]  
The second basis element is \(X_{2}=\bar{\varepsilon}_{1}\varepsilon_{2}^{\mathsf{T}}-\bar{\varepsilon}_{2 }\varepsilon_{1}^{\mathsf{T}}\).