23 Example 1
Consider the DAE
\left\{ \begin{aligned} x'(t) &= u(t)\\ y'(t) &= v(t)\\ 2u'(t)+v'(t)+2\,t\,x(t)\,\lambda(t) &= t\\ u'(t)+2\,v'(t)-\lambda(t) &= 1-t \\ t\,x(t)^2-y(t) &= 1 \end{aligned} \right.
define
\bm{Z} = \begin{pmatrix} x \\ y \\ u \\ v \\ \lambda \end{pmatrix}, \quad \bm{E}(\bm{Z},t) = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, \qquad \bm{G}(\bm{Z},t) = \begin{pmatrix} u \\ v \\ t-2\,t\,x\,\lambda \\ 1-t + \lambda \\ 1+y-t\,x^2 \end{pmatrix}
so that the DAE can be written as
\bm{E}(\bm{Z},t)\,\bm{Z}^\prime(t) = \bm{G}(\bm{Z},t)
23.1 Index reduction
The table
\bm{T} = \left[\begin{array}{c|c} \bm{E}(\bm{Z},t) & \bm{G}(\bm{Z},t) \end{array} \right] = \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 2 & 1 & 0 & t-2\,t\,x\,\lambda \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 0 & 0 & 0 & 1+y-t\,x^2 \end{array} \right]
is used to memorize the operation done on the DAE. The reduction is done by using linear algebra and avoid explicit construction of LU decomposition.
First reduction of the index by one
Exchange row 3 and 4 by using
\bm{P}_1 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}
so that
\bm{T}_1 = \bm{P}_1\bm{T} = \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 2 & 1 & 0 & t-2\,t\,x\,\lambda \\ 0 & 0 & 0 & 0 & 0 & 1+y-t\,x^2 \end{array} \right]
Subtract 2 times row 3 to row 4
\bm{L}_2 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}
so that
\bm{T}_2 = \bm{L}_2\bm{T}_1 = \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 0 & -3 & 0 & 3t-2(\lambda(1+t\,x)+1) \\ \hline 0 & 0 & 0 & 0 & 0 & 1+y-t\,x^2 \end{array} \right]
Scale 4 the row by -1/3
\bm{T}_2 = \bm{L}_2\bm{T}_1 = \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 0 & 1+y-t\,x^2 \end{array} \right]
Now algebraic constraint is separated and is
\color{blue} \textrm{[alg1]} = 1+y-t\,x^2=0
\implies\quad \textrm{[alg2]}=\dfrac{\mathrm{d}\textrm{[alg1]}}{\mathrm{d} t}=\color{blue}y'-x^2-2txx'=0
apply derivation to the algebraic constraint (reduce the index by one)
\bm{T}_3 = \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline -2tx & 1 & 0 & 0 & 0 & x^2 \end{array} \right]
Second reduction of the index by one
Eliminate -2tx and 1 of the last rpw by multiplying
\bm{L}_3 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 2tx & -1 & 0 & 0 & 1 \end{pmatrix}
so that
\bm{T}_4 = \bm{L}_3\bm{T}_3= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 2 & 0 & 1-t + \lambda \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 0 & x^2 + 2t\,x\,u-v \end{array} \right]
Eliminate also 2 my multiplying
\bm{U}_4 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}
so that
\bm{T}_5 = \bm{U}_4\bm{T}_4= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 0 & x^2 + 2t\,x\,u-v \end{array} \right]
The algebraic constraint is
\color{blue} \textrm{[alg2]}= x^2 + 2t\,x\,u-v=0
\implies\quad \textrm{[alg3]}= \dfrac{\mathrm{d}\textrm{[alg2]}}{\mathrm{d} t} = \color{blue}2(x+tu)x^\prime+2x\,u + 2tx\,u^\prime - v^\prime=0
apply derivation to the algebraic constraint (reduce the index by one)
\bm{T}_6= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 2(x+tu) & 0 & 2tx & -1 & 0 & -2x\,u \end{array} \right]
Third reduction of the index by one
Apply elimination for the last row
\bm{L}_6 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ -2(x+tu) & 0 & -2tx & 1 & 1 \end{pmatrix}
so that
\bm{T}_7=\bm{L}_6\bm{T}_6= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 0 & \frac{\left(8 \lambda \,x^{2}-6 x \right) t^{2}}{3}+\frac{\left(\left(4 \lambda +2\right) x -6 u^{2}-3\right) t}{3}-4 x\,u +\frac{2 \lambda}{3}+\frac{2}{3} \end{array} \right]
Scale last row by 3/2
\bm{S}_7 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \frac{3}{2} \end{pmatrix}
so that
\bm{T}_8=\bm{S}_7\bm{T}_7= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 0 & \textrm{[alg3]} \end{array} \right]
where \color{blue} \textrm{[alg3]}= \left(4 x^{2} \lambda-3 x\right) t^{2}+ \left(\left(2 \lambda+1\right) x-3 u^{2}-\frac{3}{2}\right) t -6 xu+\lambda+1
The algebraic constraint \textrm{[alg3]} is derived and become
\dfrac{\mathrm{d}\textrm{[alg3]}}{\mathrm{d} t} = (A) x' + (B) u' + (C) \lambda' + (D), \qquad \left\{ \begin{aligned} (A) &= (8x\lambda-3)t^2 + (2\lambda+1)t - 6u \\ (B) &= -6(tu+x) \\ (C) &= (4xt+2)xt + 1 \\ (D) &= 8tx^2\lambda + (1 - 6t + 2\lambda)x - 3u^2 -\dfrac{3}{2} \end{aligned} \right.
reducing the index by one
\bm{T}_9= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline (A) & 0 & (B) & 0 & (C) & (D) \end{array} \right]
DAE reduced to an ODE
(D)\neq 0 so that DAE is reduced to an ODE.
Eliminating in the last row and scaling
\bm{T}_9= \left[ \begin{array}{rrrrr|c} 1 & 0 & 0 & 0 & 0 & u \\ 0 & 1 & 0 & 0 & 0 & v \\ 0 & 0 & 1 & 0 & 0 & t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ 0 & 0 & 0 & 1 & 0 & \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \hline 0 & 0 & 0 & 0 & 1 & (E)/(C) \end{array} \right]
where
\begin{aligned} (E) &= (D) -x\cdot(A)- u\cdot(B) \\ &= ( (16x\lambda - 9)t + 4\lambda+3)tu + 16x^2\lambda t + (4\lambda + 3-12t)x -\frac{3}{2} - 9u^2 \end{aligned}
The DAE reduced to ODE become:
\left\{ \begin{aligned} x' &= u \\ y' &= v \\ u' &= t - \frac{1}{3}\left(1+\lambda(1+4\,t\,x)\right) \\ v' &= \frac{2}{3}(\lambda(1+t\,x)+1)-t \\ \lambda' &= \dfrac{ ( (16x\lambda - 9)t + 4\lambda+3)tu + 16x^2\lambda t + (4\lambda + 3-12t)x -\frac{3}{2} - 9u^2}{(4xt+2)xt + 1 } \end{aligned} \right.
The differential index is 3 and the invariants are
\begin{aligned} \textrm{[alg1]} & = 1+y-t\,x^2 = 0\\ \textrm{[alg2]} & = x^2 + 2t\,x\,u-v = 0\\ \textrm{[alg3]} & = \left(4 x^{2} \lambda-3 x\right) t^{2}+ \left(\left(2 \lambda+1\right) x-3 u^{2}-\frac{3}{2}\right) t -6 xu+\lambda+1 = 0 \end{aligned}
notice that
\dfrac{\mathrm{d}\textrm{[alg1]}}{\mathrm{d}t}=\textrm{[alg2]}
but
\dfrac{\mathrm{d}\textrm{[alg2]}}{\mathrm{d}t}\neq\textrm{[alg3]}
and Baumgarte stabilization cannot be used.
23.2 Index reduction using dummy variables
This DAE system exhibits mixed differential and algebraic characteristics. To analyze its structure, we first convert it to semi-explicit form.
Introduce dummy variables:
- u'(t) = u_{\text{dot}}(t)
- v'(t) = v_{\text{dot}}(t)
The system decomposes into differential (blue) and algebraic (red) components:
\left\{ \color{blue} \begin{aligned} x' &= u \\ y' &= v \\ u' &= u_{\text{dot}} \\ v' &= v_{\text{dot}} \end{aligned} \right. \quad \bigoplus \quad \left\{ \color{DarkRed} \begin{aligned} 0 &= 2u_{\text{dot}} + v_{\text{dot}} + 2tx\lambda - t & \text{(Constraint 1)}\\ 0 &= u_{\text{dot}} + 2 v_{\text{dot}} - \lambda - 1 + t& \text{(Constraint 2)} \\ 0 &= t\,x^2 - y - 1& \text{(Constraint 3)} \end{aligned} \right.
where blue equations are the ODE part explicited and the red part are the algebraic constraints. If the algebraic constraints determine a solution for u_{\text{dot}}(t), v_{\text{dot}}(t) and \lambda the DAE is of index 1 and we can stop here because we can write (formally) the solution of the algebraic part as
- u_{\text{dot}}=u_{\text{dot}}(x,y,u,v)
- v_{\text{dot}}=v_{\text{dot}}(x,y,u,v)
- \lambda=\lambda(x,y,u,v)
Variable classification
The variable are classified, in practice the variable that appears with a derivative are differential variable, the other are algebraic:
- Differential variables: x, y, u, v
- Algebraic variables: \lambda, u_{\text{dot}}, v_{\text{dot}}
Notice that algebraic variables appears linearly in the DAE.
Index Reduction Procedure
We attempt to solve for u_{\text{dot}} and v_{\text{dot}}, but due to the presence of \lambda, the system is not yet solvable explicitly. We proceed by differentiating.
Step 1: Differentiate algebraic constraints
Now we must derivate the constraints in order to reduce the index. Here we want to differentiate only the algebraic equation that can reduce the index and do not introduce new differential equation. Here we want to avoid to introduce derivative of the algebraic variable unless is strictly necessary.
We construct a dependency table to identify which constraints to differentiate:
\begin{array}{c|ccc} & u_{\text{dot}} & v_{\text{dot}} & \;\lambda\;\; \\[1em] \hline \\[0.1em] \text{(Constraint 1)} & 1 & 1 & 1 \\[1em] \text{(Constraint 2)} & 1 & 1 & 1 \\[1em] \text{(Constraint 3)} & 0 & 0 & 0 \end{array}
a 0 means the the function do not depend on the variable on the column. A 1 means dependence. \text{(Constraint 3)} is the optimal candidate for differentiation as it doesn’t introduce new algebraic variables. Differentiate thus \text{(Constraint 3)} and substitute known variable in differetial form using the ODE part
\text{(Constraint 3)}' = x^2 + 2txx' - y' = x^2 + 2txu - v
the new equation \text{(Constraint 3)}' is algebraic and thus is used to replace \text{(Constraint 3)}. The DAE thus become:
\left\{ \color{blue} \begin{aligned} x' &= u \\ y' &= v \\ u' &= u_{\text{dot}} \\ v' &= v_{\text{dot}} \end{aligned} \right. \quad \bigoplus \quad \left\{ \color{DarkRed} \begin{aligned} 0 &= 2u_{\text{dot}} + v_{\text{dot}} + 2tx\lambda - t && \text{(Constraint 1)}\\ 0 &= u_{\text{dot}} + 2v_{\text{dot}} - \lambda - 1 + t && \text{(Constraint 2)} \\ 0 &= x^2 + 2txu - v && \text{(Constraint 3)}'\\[1em] \color{DarkGreen}0 & \color{DarkGreen} = t\,x^2 - y - 1 && \color{DarkGreen} \text{(Constraint 3)} \end{aligned} \right.
The red algebraic equations are the new algebraic equation after the derivation. The green algebraic equation is the hidden constraints that is removed from the DAE system but that is satisfied by choosing the right initial conditions.
The red part cannot determine the algebraic variable and must be differentiated again.
Step 2: Differentiate (again) the algebraic constraints
To find the right cadidate(s) here we write a table of the dependencies
\begin{array}{c|ccc} & u_{\text{dot}} & v_{\text{dot}} & \;\lambda\;\;\\[1em] \hline \\[0.1em] \text{(Constraint 1)} & 1 & 1 & 1 \\[1em] \text{(Constraint 2)} & 1 & 1 & 1 \\[1em] \text{(Constraint 3)}' & 0 & 0 & 0 \end{array}
\text{(Constraint 3)}' is the right candidate because it do not depend on algebraic variable and its derivative do not introduce extra differential equation. Differentiate thus \text{(Constraint 3)}' and substitute known variable in differential form using the ODE part
\begin{aligned} \text{(Constraint 3)}'' & = 2xx' + 2xu + 2tx'u + 2txu' - v' \\ & = 4xu + 2tu^2 + 2txu_{\text{dot}}-v_{\text{dot}} \end{aligned}
the new equation \text{(Constraint 3)}'' is algebraic and thus is used to replace \text{(Constraint 3)}'. The DAE thus become:
\left\{ \color{blue} \begin{aligned} x' &= u \\ y' &= v \\ u' &= u_{\text{dot}} \\ v' &= v_{\text{dot}} \end{aligned} \right. \quad \bigoplus \quad \left\{ \color{DarkRed} \begin{aligned} 0 &= 2u_{\text{dot}} + v_{\text{dot}} + 2tx\lambda - t && \text{(Constraint 1)}\\ 0 &= u_{\text{dot}} + 2v_{\text{dot}} - \lambda - 1 + t && \text{(Constraint 2)}\\ 0 &= 4xu + 2tu^2 + 2txu_{\text{dot}}-v_{\text{dot}} && \text{(Constraint 3)}'' \\[1em] \color{DarkGreen}0 & \color{DarkGreen} = t\,x^2 - y - 1 && \color{DarkGreen} \text{(Constraint 3)} \\ \color{DarkGreen}0 & \color{DarkGreen} = x^2 + 2txu - v && \color{DarkGreen}\text{(Constraint 3)}' \end{aligned} \right.
Te red part is a linear system in the algebraic variable:
\begin{pmatrix} 2 & 1 & 2 t x \\ 1 & 2 & -1 \\ 2tx & -1 & 0 \end{pmatrix} \begin{pmatrix} u_{\text{dot}} \\ v_{\text{dot}} \\ \lambda \end{pmatrix} = \begin{pmatrix} t \\ 1-t \\ -2u(2x+tu) \end{pmatrix}
Check if solvable
\begin{aligned} \begin{vmatrix} 2 & 1 & 2 t x \\ 1 & 2 & -1 \\ 2tx & -1 & 0 \end{vmatrix} & = 2tx \begin{vmatrix} 1 & 2 t x \\ 2 & -1 \end{vmatrix} + \begin{vmatrix} 2 & 2 t x \\ 1 & -1 \end{vmatrix} \\ & = 2tx (-1-4tx) -2 -2tx \\ & = -4tx -8(tx)^2 - 2 \end{aligned}
Let z=tx then the roots of the polynomial -2-4z-8z^2 are
z_{1,2} = \dfrac{-1-\imath \sqrt{3}}{4}
that are complex, thus the linear system is always solvable for tx real. Thus blue part plus red part constitute a DAE of index one. We have done two differentiation and this implies that the original DAE is of index 3.
DAE reduced to ODE
The index-1 DAE reduced to ODE become
\left\{ \color{blue} \begin{aligned} x' &= u \\ y' &= v \\ u' &= u_{\text{dot}}(t,x,y,u,v) \\ v' &= v_{\text{dot}}(t,x,y,u,v) \end{aligned} \right.
where u_{\text{dot}}(t,x,y,u,v) and v_{\text{dot}}(t,x,y,u,v) are the solution of the linear system
\color{DarkRed} \begin{pmatrix} 2 & 1 & 2 t x \\ 1 & 2 & -1 \\ 2tx & -1 & 0 \end{pmatrix} \begin{pmatrix} u_{\text{dot}}(t,x,y,u,v) \\ v_{\text{dot}}(t,x,y,u,v) \\ \lambda(t,x,y,u,v) \end{pmatrix} = \begin{pmatrix} t \\ 1-t \\ -2u(2x+tu) \end{pmatrix} \tag{23.1}
there are also two hidden constraints
\color{DarkGreen} \begin{aligned} \text{(Constraint 3)} &= t\,x^2 - y - 1 = 0, \\ \text{(Constraint 3)}' &= x^2 + 2txu - v = 0. \end{aligned}
Baumgarte stabilization
For this DAE the hidden constraint can be used to stabilize the algebric constraint: \text{(Constraint 3)}:
\text{(Constraint 3)}'' + {\color{DarkOrange}2\eta\omega} \text{(Constraint 3)}' + {\color{DarkOrange}\omega^2} \text{(Constraint 3)} = 0
that become
4xu + 2tu^2 + 2txu_{\text{dot}}-v_{\text{dot}} + {\color{DarkOrange}2\eta\omega} (x^2 + 2txu - v) + {\color{DarkOrange}\omega^2} (t\,x^2 - y - 1) = 0
using this stabilized constraint in the linear system Equation 23.1 become
\begin{aligned} \color{DarkRed} \begin{pmatrix} 2 & 1 & 2 t x \\ 1 & 2 & -1 \\ 2tx & -1 & 0 \end{pmatrix} \begin{pmatrix} u_{\text{dot}}(t,x,y,u,v) \\ v_{\text{dot}}(t,x,y,u,v) \\ \lambda(t,x,y,u,v) \end{pmatrix} & = \begin{pmatrix} t \\ 1-t \\ -2u(2x+tu) - {\color{DarkOrange}(\star)} \end{pmatrix} \\[2em] {\color{DarkOrange}(\star)} &= {\color{DarkOrange}2\eta\omega} (x^2 + 2txu - v) + {\color{DarkOrange}\omega^2} (t\,x^2 - y - 1) \end{aligned} \tag{23.2}
Jacobians
For implicit solvers we require derivative of algebraic variables. Let
\begin{aligned} \bm{A}(t,x) &= \begin{pmatrix} 2 & 1 & 2 t x \\ 1 & 2 & -1 \\ 2tx & -1 & 0 \end{pmatrix}, \\ \bm{X}(t,x,y,u,v)&= \begin{pmatrix} u_{\text{dot}}(t,x,y,u,v) \\ v_{\text{dot}}(t,x,y,u,v) \\ \lambda(t,x,y,u,v) \end{pmatrix}, \\ \bm{B}(t,x,y,u,v) &= \begin{pmatrix} t \\ 1-t \\ -2u(2x+tu) -{\color{DarkOrange}2\eta\omega} (x^2 + 2txu - v) - {\color{DarkOrange}\omega^2} (t\,x^2 - y - 1) \end{pmatrix} \end{aligned}
Deriving \bm{A}(t,x)\bm{X}(t,x,y,u,v)=\bm{B}(t,x,y,u,v)
\begin{aligned} \bm{A}(t,x) \partial_t \bm{X}(t,x,y,u,v) & = \partial_t \bm{B}(t,x,y,u,v) - \partial_t\bm{A}(t,x)\bm{X}(t,x,y,u,v) \\ \bm{A}(t,x) \partial_t \bm{X}(t,x,y,u,v) & = \begin{pmatrix} 1 \\ -1 \\ -2u^2 -4xu{\color{DarkOrange}\eta\omega} - {\color{DarkOrange}\omega^2} x^2 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 2 x \\ 0 & 0 & 0 \\ 2x & 0 & 0 \end{pmatrix} \bm{X}(t,x,y,u,v) \\ \partial_t \bm{X}(t,x,y,u,v) &= \bm{A}(t,x)^{-1} \left( \begin{pmatrix} 1 \\ -1 \\ -2u^2 -4xu{\color{DarkOrange}\eta\omega} - {\color{DarkOrange}\omega^2} x^2 \end{pmatrix} - 2x \begin{pmatrix} \lambda(t,x,y,u,v) \\ 0 \\ u_{\text{dot}}(t,x,y,u,v) \end{pmatrix} \right) \end{aligned}
analogously all the other derivatives can be obtained.