## 3 Computation of Inertial Manifolds for DDEs
### Elimination of the Fastest Relaxation Mode
This ansatz leads to the invariance equation  
\[\dot{r}=\frac{\partial\mathcal{R}}{\partial e}\dot{e}+\frac{\partial\mathcal{R }}{\partial s}\dot{s}+\frac{\partial\mathcal{R}}{\partial c}\dot{c}. \tag{30}\]  
The general plan is as before: we will use the expansion (24) to replace the delayed term in Equation (21c) prior to substitution into the functional equation, and then iteratively solve the invariance equation for \(\mathcal{R}(e,s,c)\) from some suitable initial guess. If we expand \(\rho\) to first order in \(\delta\), the resulting iteration is\[\mathcal{R}_{k+1}^{(1)}=\left\{\rho(s)+\varepsilon_{2}e\frac{\partial \mathcal{R}_{k}^{(1)}}{\partial e}-\left[\kappa-s(e-c)+\alpha c\right]\left( \frac{\delta}{\varepsilon_{1}}\frac{d\rho}{ds}+\frac{\varepsilon_{2}}{ \varepsilon_{1}}\frac{\partial\mathcal{R}_{k}^{(1)}}{\partial s}\right)\right.\\ -\frac{\varepsilon_{2}}{\beta\varepsilon_{1}}\frac{\partial \mathcal{R}_{k}^{(1)}}{\partial c}\left[s(e-c)-(\gamma+\beta\varepsilon_{1})c \right]\Bigg{\}}\left(1+\varepsilon_{2}\frac{\partial\mathcal{R}_{k}^{(1)}}{ \partial e}\right)^{-1}. \tag{31}\]  
To go beyond first order, we need  
\[\ddot{s}=\frac{1}{\varepsilon_{1}}\left[-\dot{s}(e-c)-s\dot{e}+\dot{c}(s+ \alpha)\right] \tag{32}\]  
and  
\[\ddot{s}=(\beta\varepsilon_{1}^{2})^{-1}\bigg{\{}-\beta(\dot{e}- \dot{c})(\kappa+\alpha c)-\beta(e-c)\left[\alpha\dot{c}-\dot{s}(e-c)\right]+2 \beta s(\dot{e}-\dot{c})(e-c)\\ -\beta\varepsilon_{1}\left[\dot{s}(r-e)+\frac{s}{\varepsilon_{2} }(\rho(s)-r)-s\dot{e}\right]+s\left[s(e-c)-(\gamma+\beta\varepsilon_{1})c \right]\\ +(s+\alpha)\left[\dot{s}(e-c)+s(\dot{e}-\dot{c})-(\gamma+\beta \varepsilon_{1})c\right]\bigg{\}}. \tag{33}\]  
Up to third order in \(\delta\), it is possible to analytically isolate \(\mathcal{R}\) from Equation (30). The functional equations for \(\mathcal{R}^{(m)}\) become more and more complicated, but the work is easily carried out in a symbolic algebra system. Iteration of these functional equations can thus be carried out analytically, also in a symbolic algebra system. The size of the algebraic expressions generated by the iteration process grows rapidly, limiting the number of iterates that can practically be computed. Maple, the symbolic algebra system used in this study, lacks a function that gives a useful measure of the size of an expression. A simple workaround is to have Maple generate (say) Matlab code, and to count the number of bytes in these files. The optimize option was _not_ used in order to measure the size of the expressions as stored by Maple. Also note that it is not feasible to use simplify() on the expression generated during iteration as the computation time required to execute this function grows very rapidly with the size of the expression. Table 1 shows the growth of the size of the expressions for the iterates with both \(k\) and \(m\). Note particularly the enormous sizes of the expressions generated for \(m=2\) and \(3\) after a modest number of iterations. This is another reason to prefer the lowest-\(m\) expansion that suffices to reproduce the behavior of the original system. If we take as an initial guess \(\mathcal{R}_{0}^{(1)}=0\), then from Equation (31) we get the following \(O(\delta)\) first approximation to the inertial manifold:  
\[\mathcal{R}_{1}^{(1)}(e,s,c)=\rho(s)-\frac{\delta}{\varepsilon_{1}}\left[ \kappa-s(e-c)+\alpha c\right]\frac{d\rho}{ds}. \tag{34}\]  
This is the simplest approximation to the inertial manifold that takes account of the delay. For \(m=1\), there is no significant change in the shape of the manifold for \(k>2\). This is difficult to show directly since we cannot plot a function of three variables. We could take cuts through \begin{table}
\begin{tabular}{c c c c} \hline \hline \(k\) & \(m=1\) & 2 & 3 \\ \hline
1 & 152 & 628 & 4.07 k \\
2 & 4.15 k & 26.9 k & 190 k \\
3 & 252 k & 2.09 M & 14.3 M \\ \hline \hline \end{tabular}
\end{table}