Energy, existence and algebraic constraints

Put $z=(e,b)\in {\mathbb{R}}^{n_e+n_b}$ and define $A:=\left(\begin{array}{cc}-M_e^{-1}S& M_e^{-1}{C}^T{M}_b\\ -C& 0\end{array}\right)$ and $f(t):=\left(\begin{array}{c}-M_e^{-1}j(t)\\ 0\end{array}\right)$. The inverses exist by positive definiteness in [F1]: if $M_{e}x=0$, then $x^T{M}_{e}x=0$ forces $x=0$, and an injective square map in finite dimensions is invertible; the same holds for $M_b$. Thus the system is exactly $z^{\mathrm{\prime }}=Az+f(t)$.

Let $g(s):=e^{-(s-t_0)A}f(s)$ and $H(t):={\int }_{t_0}^{t}g(s)ds$, interpreting each coordinate as a Riemann integral and using the oriented convention ${\int }_a^b=-{\int }_b^a$ when $b<a$ and ${\int }_a^a=0$. If $n_e+n_b=0$, then $g(s)$ and $H(t)$ are the unique empty vector for all $s,t$, so $H^{\mathrm{\prime }}=g$ vacuously. Otherwise, on every nondegenerate compact segment in $I$, each coordinate of $g$ is continuous, so [F3] gives integrability and [F4] defines $H$. For each coordinate $k$, the cut-additivity in [F6], together with the oriented convention, gives $H_k(t+h)-H_k(t)={\int }_t^{t+h}{g}_k(s)ds$ whenever $t,t+h\in I$: positive-oriented additivity follows by splitting at an interior point, and reversing endpoints or using a zero-length interval gives the other endpoint orders. Thus, for $h\mathrm{\ne }0$, $\frac{H_k(t+h)-H_k(t)}{h}-g_k(t)=\frac{1}{h}{\int }_t^{t+h}(g_k(s)-g_k(t))ds$, where [F7] evaluates the integral of the constant $g_k(t)$. Given $\epsilon >0$, continuity of $g_k$ at $t$ gives $\delta >0$ such that $\mathrm{\mid }{g}_k(s)-g_k(t)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }s-t\mathrm{\mid }<\delta$; since $I$ is open, reduce $\delta$ so $[t-\delta ,t+\delta ]\subset I$. If $0<\mathrm{\mid }h\mathrm{\mid }<\delta$, then on $J=[\min (t,t+h),\max (t,t+h)]$ the absolute-value estimate and monotonicity in [F6], followed by [F7], give $\mid \frac{1}{h}{\int }_t^{t+h}(g_k(s)-g_k(t))ds\mid \le \frac{1}{\mathrm{\mid }h\mathrm{\mid }}{\int }_J\mathrm{\mid }{g}_k(s)-g_k(t)\mathrm{\mid }ds\le \epsilon$. Therefore each coordinate has derivative $g_k(t)$ and $H^{\mathrm{\prime }}(t)=g(t)$ throughout $I$.

Define $z(t):=e^{(t-t_0)A}(z_0+H(t))$, where $z_0=(e_0,b_0)$. By [F2] and step 1.2, the product rule gives $z^{\mathrm{\prime }}=Az+f(t)$, since $e^{(t-t_0)A}{e}^{-(t-t_0)A}=I$ by the commuting-product identity in [F2]; also $z(t_0)=z_0$. Thus a $C^1$ solution exists on all of $I$. If $n_e+n_b=0$, every state is the unique empty vector, so the solution is unique. Otherwise, if $\tilde{z}$ is another solution with the same initial value, then $\xi (t):=e^{-(t-t_0)A}(\tilde{z}(t)-z(t))$ has derivative zero by [F2] and the two differential equations. Each coordinate of $\xi$ is constant by [F5] and vanishes at $t_0$, so $\tilde{z}=z$.

Differentiate $W=\frac{1}{2}(e^T{M}_{e}e+b^T{M}_{b}b)$ and substitute the system equations. Symmetry of $M_e,M_b$ gives $\dot{W}=e^T(C^T{M}_{b}b-Se-j)+b^T{M}_b(-Ce)=-e^{T}Se-e^{T}j$, because the real scalar cross terms $e^T{C}^T{M}_{b}b$ and $b^T{M}_{b}Ce$ are equal and cancel.

The magnetic quantity satisfies $(Db{)}^{\mathrm{\prime }}=D\dot{b}=-DCe=0$ by [F1]; each coordinate is constant by [F5], and its value at $t_0$ is $Db(t_0)$.

Since $q=-G^T{M}_{e}e$, differentiation and the first system equation give $\dot{q}=-G^T(C^T{M}_{b}b-Se-j)=-(CG{)}^T{M}_{b}b+G^T(Se+j)=G^T(Se+j)$, using $CG=0$ from [F1]. This proves all three identities.