Existence and uniqueness for the finite-dimensional Maxwell ODE

Put $x=(e,b)\in {\mathbb{R}}^N$, $N=N_e+N_f$, and $A=\left(\begin{array}{cc}0& M_e^{-1}{C}^T{M}_b\\ -C& 0\end{array}\right)$, $f(t)=(-M_e^{-1}j(t),0)$. By [F2], $x^{\mathrm{\prime }}=Ax+f(t)$. Form the continuous $(N+1)\times (N+1)$ matrix $\mathcal{A}(t)=\left(\begin{array}{cc}A& f(t)\\ 0& 0\end{array}\right)$ and the initial matrix $Y_0$ whose last column is $(x_0,1{)}^T$ and whose other columns are zero.

If $T=0$, the prescribed initial coefficients determine the unique singleton map. Hold $j$ constant at $j(0)$ and use the matrix $\mathcal{A}$ of step 1.1 on $[-1,1]$. By [F1] it has a unique matrix solution there through $Y_0$ at zero. Its last column has last entry one and its top block solves the constant-load coefficient ODE, giving the asserted continuation. Uniqueness on any common interval containing zero follows by applying [F1] to each compact subinterval joining zero to a point of overlap. No singleton derivative is taken.

Now assume $T>0$. By [F1], $Y^{\mathrm{\prime }}=\mathcal{A}(t)Y$, $Y(0)=Y_0$ has a unique global solution. Its last column $(x(t),s(t){)}^T$ satisfies $s^{\mathrm{\prime }}(t)=0$ and $s(0)=1$, so $s(t)=1$, while the top block satisfies $x^{\mathrm{\prime }}=Ax+f(t)$ and $x(0)=x_0$. Thus it gives a global solution of the Maxwell coefficient system. Any two such coefficient solutions yield two matrix solutions with the same last-column initial data; uniqueness in [F1] forces their last columns to agree.