Proof of Adjoint Energy Identity for the Kalman Covariance Riccati Equation

Claim 1. The assignment $\tilde{A}(r)=-\bigl(A(r)-P(r)D(r)\bigr)^{\top}$ has continuous entries: each entry of $P(r)D(r)$ is a finite sum of products of entries of $P(r)$ and $D(r)$, each entry of a transpose is an entry of the original matrix, and sums and products of continuous real-valued functions are continuous by the continuity of sums and products. Let $\Phi$ be the fundamental solution of $\tilde{A}$ on $[a,b]$. By claims 1 and 2 of that theorem $\Phi$ has continuous entries, $\Phi(a)=I_{k}$ is the identity matrix, and every $\Phi(t)$ is invertible. By claim 3 of that theorem, applied with the inhomogeneity $g$ there equal to $0$, for each $\xi\in\mathbb{R}^{k}$ the assignment $u\mapsto\Phi(u)\xi$ has continuous components and is the unique assignment $y$ with continuous components satisfying $y(u)=\xi+\int_{a}^{u}\tilde{A}(r)\,y(r)\,dr$ for every $u\in[a,b]$.

For existence, put $\xi=\Phi(s)^{-1}x$ and $\lambda(u)=\Phi(u)\xi$. Then $\lambda(a)=\Phi(a)\xi=\xi$, so the equation just quoted is precisely the integral equation of claim 1, and $\lambda(s)=\Phi(s)\Phi(s)^{-1}x=x$. For uniqueness, let $\lambda$ be any assignment as in claim 1. It satisfies that equation with $\xi=\lambda(a)$, so $\lambda(u)=\Phi(u)\lambda(a)$ for every $u\in[a,b]$; evaluating at $s$ gives $\Phi(s)\lambda(a)=x$, hence $\lambda(a)=\Phi(s)^{-1}x$ and $\lambda$ is the assignment just constructed.

Differentiability of $P$ and $\lambda$. Every entry of $r\mapsto A(r)P(r)+P(r)A(r)^{\top}-P(r)D(r)P(r)+C(r)$ and every component of $r\mapsto\tilde{A}(r)\lambda(r)$ is a finite sum of products of continuous functions, hence continuous. Applying claim 3 of the first fundamental theorem of calculus to each entry and each component of the two integral equations, each entry of $P$ and each component of $\lambda$ is differentiable on $[a,b]$, with entrywise derivatives

$$P'(r)=A(r)P(r)+P(r)A(r)^{\top}-P(r)D(r)P(r)+C(r),\qquad\lambda'(r)=-\bigl(A(r)-P(r)D(r)\bigr)^{\top}\lambda(r),$$

both continuous in $r$.

Claim 2. Each component of $\psi=P\lambda$ is a finite sum of products of differentiable real functions, so by claims 2 and 3 of the derivative arithmetic lemma it is differentiable with $\psi'(r)=P'(r)\lambda(r)+P(r)\lambda'(r)$, the products being read entrywise. Since $P(r)$ and $D(r)$ are symmetric and $(MN)^{\top}=N^{\top}M^{\top}$ for the matrix product,

$$P(r)\bigl(A(r)-P(r)D(r)\bigr)^{\top}=P(r)A(r)^{\top}-P(r)D(r)P(r),$$

so that, suppressing the argument $r$,

$$\psi'=\bigl(AP+PA^{\top}-PDP+C\bigr)\lambda-\bigl(PA^{\top}-PDP\bigr)\lambda=AP\lambda+C\lambda=A\psi+C\lambda .$$

This is continuous in $r$, so by the second fundamental theorem of calculus, applied componentwise on $[a,u]$,

$$\psi(u)-\psi(a)=\int_{a}^{u}\bigl(A(r)\psi(r)+C(r)\lambda(r)\bigr)\,dr\qquad(a\le u\le b).$$

As $\psi(a)=P(a)\lambda(a)=P_{0}\lambda(a)$ and $\psi$ has continuous components, being an entrywise sum of products of continuous functions, claim 2 follows.

Two elementary identities. For a real $k\times k$ matrix $M$ and $u,v\in\mathbb{R}^{k}$ the index formulas for the matrix-vector product and the dot product give $u\cdot(Mv)=\sum_{\gamma,\delta}M_{\gamma\delta}u^{\gamma}v^{\delta}$. Hence the expression is linear in $u$ for fixed $v$ and in $v$ for fixed $u$; moreover $(M^{\top}u)\cdot v=\sum_{\gamma,\delta}M_{\delta\gamma}u^{\delta}v^{\gamma}=u\cdot(Mv)$ after exchanging the names of the indices, and if $M$ is symmetric then $u\cdot(Mv)=v\cdot(Mu)$.

Claim 3. First, $\psi(s)=P(s)\lambda(s)=P(s)x$, so $x\cdot\psi(s)=x\cdot\bigl(P(s)x\bigr)$.

Define $q(r)=\lambda(r)\cdot\bigl(P(r)\lambda(r)\bigr)$ for $r\in[a,b]$. Written in indices, $q$ is a finite sum of products of differentiable real functions, so by the derivative arithmetic lemma it is differentiable with

$$q'(r)=\lambda'(r)\cdot\bigl(P(r)\lambda(r)\bigr)+\lambda(r)\cdot\bigl(P'(r)\lambda(r)\bigr)+\lambda(r)\cdot\bigl(P(r)\lambda'(r)\bigr).$$

Suppressing $r$ and using the identities just recorded together with the symmetry of $P$ and $D$,

$$\lambda'\cdot(P\lambda)=-\bigl((A-PD)^{\top}\lambda\bigr)\cdot(P\lambda)=-\lambda\cdot\bigl((A-PD)P\lambda\bigr)=-\lambda\cdot(AP\lambda)+\lambda\cdot(PDP\lambda),$$

next, by the same two identities,

$$\lambda\cdot(P\lambda')=-\lambda\cdot\bigl(P(A-PD)^{\top}\lambda\bigr)=-\lambda\cdot(PA^{\top}\lambda)+\lambda\cdot(PDP\lambda),$$

and, from the Riccati equation itself,

$$\lambda\cdot(P'\lambda)=\lambda\cdot(AP\lambda)+\lambda\cdot(PA^{\top}\lambda)-\lambda\cdot(PDP\lambda)+\lambda\cdot(C\lambda).$$

Adding the three lines, every term cancels except

$$q'(r)=\lambda(r)\cdot\bigl(C(r)\lambda(r)\bigr)+\lambda(r)\cdot\bigl(P(r)D(r)P(r)\lambda(r)\bigr),$$

and by the symmetry of $P$ the second summand equals $\bigl(P\lambda\bigr)\cdot\bigl(D(P\lambda)\bigr)=\psi(r)\cdot\bigl(D(r)\psi(r)\bigr)$. Thus $q'$ is the function displayed in claim 3. It is continuous, being a finite sum of products of continuous functions, and nonnegative because $C(r)$ and $D(r)$ are positive semidefinite. By the second fundamental theorem of calculus applied on $[a,s]$,

$$q(s)-q(a)=\int_{a}^{s}\Bigl(\lambda(r)\cdot\bigl(C(r)\lambda(r)\bigr)+\psi(r)\cdot\bigl(D(r)\psi(r)\bigr)\Bigr)\,dr .$$

Finally $q(s)=\lambda(s)\cdot\bigl(P(s)\lambda(s)\bigr)=x\cdot\bigl(P(s)x\bigr)$ and $q(a)=\lambda(a)\cdot\bigl(P_{0}\lambda(a)\bigr)$, the latter nonnegative because $P_{0}$ is positive semidefinite. Rearranging the last display gives the identity of claim 3.