Proof of The Control and Filter Riccati Families of the Ising Equilibrium in Closed Form

Throughout write

$$y_{+}=\frac{\Delta-1}{4\chi},\qquad y_{-}=-\frac{\Delta+1}{4\chi},\qquad p_{+}=\frac{\Gamma-2}{2\tilde{d}},\qquad p_{-}=-\frac{\Gamma+2}{2\tilde{d}} ,$$

and $N=\mathsf{n}\mathsf{n}^{\top}$. Since $\chi>0$ and $\psi>0$ we have $1+\chi\psi>1$, so $\Delta>1$, $y_{+}>0>y_{-}$ and $\rho=(\Delta-1)/(\Delta+1)\in(0,1)$; since $\tilde{d}>0$ we have $\Gamma>2$, $p_{+}>0>p_{-}$ and $\rho_{\Pi}\in(0,1)$. Note $y_{+}=-\rho\,y_{-}$ and $p_{+}=-\rho_{\Pi}\,p_{-}$, and, from $\Gamma^{2}-4=2\tilde{d}$, that $p_{+}=(\Gamma-2)/\bigl((\Gamma-2)(\Gamma+2)\bigr)=1/(\Gamma+2)$. Two factorisations will be used: by the quadratic formula the roots of $8\chi y^{2}+4y-\tfrac12\psi$ are $(-1\pm\Delta)/(4\chi)$, that is $y_{\pm}$, so

$$-4y-8\chi y^{2}+\tfrac12\psi=-8\chi\,(y-y_{+})(y-y_{-})\qquad(y\in\mathbb{R}),$$

and likewise the roots of $2\tilde{d}\,p^{2}+4p-1$ are $(-2\pm\Gamma)/(2\tilde{d})$, that is $p_{\pm}$, so

$$-4p-2\tilde{d}\,p^{2}+1=-2\tilde{d}\,(p-p_{+})(p-p_{-})\qquad(p\in\mathbb{R}).$$

For a continuous integrand on a compact interval the Riemann and Lebesgue integrals agree, by Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, so the integral equations below may be read in either sense.

In the two instantiations below, the symbols $l$ and $k$ are the dimension parameters of the cited Riccati results and are set to $1$; they are unrelated to the number $l=2$ of states of the present model; that model parameter is left as $l$ rather than renamed. Claim 1. Existence and uniqueness. Apply The Control Riccati Equation with Cross Term: Global Existence and Uniqueness, Positive Semidefiniteness, and the Lyapunov Bound with $l=k=1$, horizon $T$, and the constant data $A(t)=-2$, $B(t)=\sqrt{2}$, $Q(t)=\tfrac12\psi$, $V(t)=0$, $R(t)=(4\chi)^{-1}$ and $F=0$. All entries are continuous; $Q(t)=\tfrac12\psi\ge0$ is symmetric, $R(t)>0$ is symmetric positive definite, $F=0$ is symmetric positive semidefinite, and $Q(t)-V(t)R(t)^{-1}V(t)^{\top}=\tfrac12\psi\ge0$. The corollary's equation reads, in this one-dimensional case,

$$Z(t)=0+\int_{t}^{T}\Bigl(-2Z(r)-2Z(r)-\bigl(\sqrt{2}\,Z(r)\bigr)\cdot4\chi\cdot\bigl(\sqrt{2}\,Z(r)\bigr)+\tfrac12\psi\Bigr)dr=\int_{t}^{T}\Bigl(-4Z(r)-8\chi Z(r)^{2}+\tfrac12\psi\Bigr)dr ,$$

which is the displayed equation. The corollary provides exactly one continuous solution on $[0,T]$; call it $z$.

The closed form. Put $g_{t}=\exp\bigl(-4\Delta(T-t)\bigr)=\exp(-4\Delta T)\exp(4\Delta t)$ and $\hat{z}_{t}=y_{+}(1-g_{t})/(1+\rho g_{t})$. The exponential is positive, so $g_{t}>0$; and $g_{t}\le1$ with equality exactly at $t=T$, because $-4\Delta(T-t)\le0$ there and the exponential is increasing. In particular $1+\rho g_{t}\ge1$, so $\hat{z}$ is well defined and continuous. By Derivative and Continuity of the Scaled Exponential Function, $g$ is differentiable with $g'_{t}=4\Delta\,g_{t}$. By the reciprocal rule Reciprocal Rule for One-Dimensional Derivatives and the sum and product rules of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers,

$$\hat{z}'(t)=y_{+}\,\frac{-g'_{t}\,(1+\rho g_{t})-(1-g_{t})\,\rho g'_{t}}{(1+\rho g_{t})^{2}}=-\,y_{+}\,\frac{g'_{t}\,(1+\rho)}{(1+\rho g_{t})^{2}}=-\,\frac{4\Delta\,(1+\rho)\,y_{+}\,g_{t}}{(1+\rho g_{t})^{2}} ,$$

the middle step because $(1+\rho g)+\rho(1-g)=1+\rho$. On the other hand, using $y_{-}=-y_{+}/\rho$,

$$\hat{z}_{t}-y_{+}=y_{+}\frac{(1-g_{t})-(1+\rho g_{t})}{1+\rho g_{t}}=-\,\frac{y_{+}(1+\rho)g_{t}}{1+\rho g_{t}},\qquad \hat{z}_{t}-y_{-}=y_{+}\frac{\rho(1-g_{t})+1+\rho g_{t}}{\rho\,(1+\rho g_{t})}=\frac{y_{+}(1+\rho)}{\rho\,(1+\rho g_{t})} ,$$

so that

$$-8\chi\,(\hat{z}_{t}-y_{+})(\hat{z}_{t}-y_{-})=\frac{8\chi\,y_{+}^{2}(1+\rho)^{2}g_{t}}{\rho\,(1+\rho g_{t})^{2}} .$$

Since $8\chi y_{+}=2(\Delta-1)$ and $(1+\rho)/\rho=2\Delta/(\Delta-1)$, we have $8\chi y_{+}(1+\rho)/\rho=4\Delta$, and the last display equals $4\Delta(1+\rho)y_{+}g_{t}/(1+\rho g_{t})^{2}=-\hat{z}'(t)$. By the factorisation recorded above this says

$$\hat{z}'(t)=-\Bigl(-4\hat{z}_{t}-8\chi\hat{z}_{t}^{2}+\tfrac12\psi\Bigr)\qquad(t\in[0,T]).$$

The right-hand side is continuous in $t$, so by Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative applied on $[t,T]$, $\hat{z}_{T}-\hat{z}_{t}=\int_{[t,T]}\hat{z}'(r)\,dr$; and $\hat{z}_{T}=0$ because $g_{T}=1$. Rearranging gives exactly the integral equation of the claim, so $\hat{z}=z$ by the uniqueness above.

Bounds and monotonicity. From $0<g_{t}\le1$ we get $0\le1-g_{t}<1\le1+\rho g_{t}$, hence $0\le z_{t}<y_{+}$, with $z_{T}=0$. The displayed formula for $\hat{z}'$ is negative at every $t$, so $z$ is nonincreasing. Finally, from the first displayed factorisation of $\hat{z}_{t}-y_{+}$, $\rho<1$ and $1+\rho g_{t}\ge1$,

$$0\le y_{+}-z_{t}=\frac{y_{+}(1+\rho)g_{t}}{1+\rho g_{t}}\le2y_{+}g_{t}=\frac{\Delta-1}{2\chi}\exp\bigl(-4\Delta(T-t)\bigr).$$

Claim 2. Apply The Filter Riccati Equation: Global Existence and Uniqueness, Symmetry, and the Lyapunov Bound on $[0,T]$ with $k=1$ and the constant data $A(t)=-2$, $C(t)=1$, $D(t)=2\tilde{d}$ and $P_{0}=0$: the entries are continuous, and $C(t)$, $D(t)$, $P_{0}$ are positive semidefinite because $1>0$, $2\tilde{d}>0$ and $0\ge0$. Its equation reads

$$P(t)=0+\int_{0}^{t}\bigl(-2P(r)-2P(r)-2\tilde{d}\,P(r)^{2}+1\bigr)dr=\int_{0}^{t}\bigl(-4P(r)-2\tilde{d}P(r)^{2}+1\bigr)dr ,$$

which is the displayed equation, and the theorem provides exactly one continuous solution; call it $p$.

Put $h_{t}=\exp(-2\Gamma t)$ and $\hat{p}_{t}=p_{+}(1-h_{t})/(1+\rho_{\Pi}h_{t})$, so that $0<h_{t}\le1$ with equality exactly at $t=0$, and $h'_{t}=-2\Gamma h_{t}$ by Derivative and Continuity of the Scaled Exponential Function. Exactly as in claim 1,

$$\hat{p}'(t)=-\,\frac{h'_{t}(1+\rho_{\Pi})p_{+}}{(1+\rho_{\Pi}h_{t})^{2}}=\frac{2\Gamma(1+\rho_{\Pi})p_{+}h_{t}}{(1+\rho_{\Pi}h_{t})^{2}},\qquad -2\tilde{d}(\hat{p}_{t}-p_{+})(\hat{p}_{t}-p_{-})=\frac{2\tilde{d}\,p_{+}^{2}(1+\rho_{\Pi})^{2}h_{t}}{\rho_{\Pi}(1+\rho_{\Pi}h_{t})^{2}} ,$$

using $p_{-}=-p_{+}/\rho_{\Pi}$. Since $2\tilde{d}\,p_{+}=\Gamma-2$ and $(1+\rho_{\Pi})/\rho_{\Pi}=2\Gamma/(\Gamma-2)$, we get $2\tilde{d}p_{+}(1+\rho_{\Pi})/\rho_{\Pi}=2\Gamma$, so the two displayed quantities are equal; by the second factorisation, $\hat{p}'(t)=-4\hat{p}_{t}-2\tilde{d}\hat{p}_{t}^{2}+1$, a continuous function of $t$. By Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative on $[0,t]$ and $\hat{p}_{0}=0$ we obtain the integral equation, so $\hat{p}=p$. The bounds $0\le p_{t}<p_{+}=(\Gamma+2)^{-1}$, the monotonicity ($\hat{p}'>0$) and the estimate $0\le p_{+}-p_{t}=p_{+}(1+\rho_{\Pi})h_{t}/(1+\rho_{\Pi}h_{t})\le2p_{+}h_{t}$ follow as in claim 1.

Claim 3. We use the rank-one algebra $v^{\top}v=2$, $\mathsf{n}^{\top}v=0$, hence

$$(vv^{\top})(vv^{\top})=2\,vv^{\top},\qquad (vv^{\top})N=N(vv^{\top})=0 .$$

Each $Z_{t}$ is symmetric because $vv^{\top}$ and $N$ are, and its entries are continuous in $t$ because $z$ is continuous and $t\mapsto\mu(T-t)$ is. At $t=T$ both terms vanish, so $Z_{T}=0=\hat{F}$ by claim The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity (ii).

By claim The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity (ii), $E_{r}=-vv^{\top}$ is symmetric, $\mathsf{B}_{r}=-\tfrac12vv^{\top}$, $V_{r}=0$, $R_{r}^{-1}=4\chi I$ and $Q_{r}=\tfrac12\psi\,vv^{\top}+\mu N$. Hence

$$E_{r}^{\top}Z_{r}+Z_{r}E_{r}=-\bigl(vv^{\top}\bigr)\bigl(z_{r}vv^{\top}+\mu(T-r)N\bigr)-\bigl(z_{r}vv^{\top}+\mu(T-r)N\bigr)\bigl(vv^{\top}\bigr)=-4z_{r}\,vv^{\top},$$ $$W_{r}=Z_{r}\mathsf{B}_{r}+\tfrac12V_{r}=-\tfrac12\bigl(z_{r}vv^{\top}+\mu(T-r)N\bigr)\bigl(vv^{\top}\bigr)=-z_{r}\,vv^{\top},$$ $$W_{r}R_{r}^{-1}W_{r}^{\top}=4\chi\,z_{r}^{2}\,(vv^{\top})(vv^{\top})=8\chi\,z_{r}^{2}\,vv^{\top}=\Xi_{r} .$$

Therefore

$$E_{r}^{\top}Z_{r}+Z_{r}E_{r}-W_{r}R_{r}^{-1}W_{r}^{\top}+Q_{r}=\Bigl(-4z_{r}-8\chi z_{r}^{2}+\tfrac12\psi\Bigr)vv^{\top}+\mu N .$$

Integrating entrywise over $[t,T]$ and using claim 1 for the first coefficient and $\int_{[t,T]}\mu\,dr=\mu(T-t)$ for the second gives

$$\int_{[t,T]}\bigl(E_{r}^{\top}Z_{r}+Z_{r}E_{r}-W_{r}R_{r}^{-1}W_{r}^{\top}+Q_{r}\bigr)dr=z_{t}\,vv^{\top}+\mu(T-t)N=Z_{t}=\hat{F}+Z_{t}-\hat{F} ,$$

which, since $\hat{F}=0$, is the asserted integral equation entry by entry. The integrand is continuous in $r$, so subtracting the equations at $t$ and at $0$ gives the equivalent forward form $Z^{\gamma\delta}_{t}=Z^{\gamma\delta}_{0}+\int_{[0,t]}\dot{z}^{\gamma\delta}(r)\,dr$ with the stated continuous densities. Symmetry, continuity of the entries, the density identity and the terminal condition $Z_{T}=\hat{F}$ are precisely what hypothesis (H2) asks for.

Claim 4. The matrices $\mathcal{E}_{t}$, $\Theta^{\star}_{t}$ and $\tilde{D}_{t}$ are constant in $t$ by claims The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity (iii) and The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity (iv), hence have continuous entries. For $z\in\mathbb{R}^{2}$ we have $z\cdot\Theta^{\star}_{t}z=(v\cdot z)^{2}\ge0$, so $\Theta^{\star}_{t}$ is symmetric positive semidefinite; and, writing $c=\tfrac{q}{2}+q_{0}>0$, $\tilde{D}_{t}=c^{-1}\tilde{\mathcal{E}}_{t}^{\top}\tilde{\mathcal{E}}_{t}$ is symmetric with $z\cdot\tilde{D}_{t}z=c^{-1}|\tilde{\mathcal{E}}_{t}z|^{2}\ge0$, hence positive semidefinite; the zero matrix is positive semidefinite. So The Filter Riccati Equation: Global Existence and Uniqueness, Symmetry, and the Lyapunov Bound applies and furnishes a unique $\Pi$.

Put $\hat{\Pi}_{t}=p_{t}vv^{\top}$, which has continuous entries. Using $\mathcal{E}_{t}=-vv^{\top}$, the rank-one algebra of claim 3 and the identity $vv^{\top}\tilde{D}_{t}=\tilde{d}\,vv^{\top}$ of claim The Fluctuation LQG Data of the Ising Equilibrium and Its Joint Coercivity (iv),

$$\mathcal{E}_{r}\hat{\Pi}_{r}+\hat{\Pi}_{r}\mathcal{E}_{r}^{\top}=-4p_{r}\,vv^{\top},\qquad \hat{\Pi}_{r}\tilde{D}_{r}\hat{\Pi}_{r}=p_{r}^{2}\,\tilde{d}\,(vv^{\top})(vv^{\top})=2\tilde{d}\,p_{r}^{2}\,vv^{\top},\qquad \Theta^{\star}_{r}=vv^{\top},$$

so the integrand of the Kalman equation is $\bigl(-4p_{r}-2\tilde{d}p_{r}^{2}+1\bigr)vv^{\top}$. Integrating entrywise over $[0,t]$ and using claim 2 gives $\int_{0}^{t}(\cdots)dr=p_{t}vv^{\top}=\hat{\Pi}_{t}$, and $P_{0}=0$; so $\hat{\Pi}$ solves the equation. By the uniqueness in The Filter Riccati Equation: Global Existence and Uniqueness, Symmetry, and the Lyapunov Bound, $\Pi_{t}=p_{t}vv^{\top}$.

Claim 5. For real $2\times2$ matrices of the form $\lambda vv^{\top}$ and $\kappa vv^{\top}$ the entrywise pairing is

$$\sum_{\gamma,\delta}\lambda v^{\gamma}v^{\delta}\,\kappa v^{\gamma}v^{\delta}=\lambda\kappa\Bigl(\sum_{\gamma}(v^{\gamma})^{2}\Bigr)^{2}=4\lambda\kappa ,$$

while

$$\sum_{\gamma,\delta}\mathsf{n}^{\gamma}\mathsf{n}^{\delta}\,v^{\gamma}v^{\delta}=\Bigl(\sum_{\gamma}\mathsf{n}^{\gamma}v^{\gamma}\Bigr)^{2}=0 .$$

With $Z_{t}=z_{t}vv^{\top}+\mu(T-t)N$ and $\Theta^{\star}_{t}=vv^{\top}$ the first pairing is $4z_{t}\cdot1+\mu(T-t)\cdot0=4z_{t}$; with $\Xi_{t}=8\chi z_{t}^{2}vv^{\top}$ and $\Pi_{t}=p_{t}vv^{\top}$ the second is $4\cdot8\chi z_{t}^{2}\cdot p_{t}=32\chi z_{t}^{2}p_{t}$. Neither expression contains $\mu$. Both $z$ and $p$ are continuous by claims 1 and 2, so $t\mapsto4z_{t}+32\chi z_{t}^{2}p_{t}$ is continuous by Continuity of Sums and Products of Real-Valued Functions on a Metric Space.