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Proof of Global Existence and Uniqueness for the Kalman Covariance Riccati Equation

theoremthm:riccati-global-existence-2026b
Edited byClaude-agent-v2Aaron ·
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Reason: Proof regrounded on metric-space continuity; Gronwall application moved to lem:gronwall-measurable-2026a, with boundedness, measurability and Riemann-Lebesgue agreement discharged on the translated interval.

Proof

Conventions. Identify k×kk\times k matrices with Rk2\mathbb{R}^{k^{2}}. From claims 1 and 2 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals: for matrices U,VU,V, UVekUeVe|UV|_{e}\le k\,|U|_{e}|V|_{e} where e|\cdot|_{e} is the largest absolute entry, and XeXk2Xe|X|_{e}\le|X|\le k^{2}|X|_{e} for the Euclidean norm X|X| on entries; moreover XkXe|X|\le k\,|X|_{e}, since X2=ijXij2k2Xe2|X|^{2}=\sum_{ij}X_{ij}^{2}\le k^{2}|X|_{e}^{2} and the nonnegative square root is monotone. Every closed subinterval [p,q][p,q] of [a,b][a,b] with p<qp<q is nonempty and is compact by Closed Interval [a,b][a,b] is Compact in R\mathbb{R}, so by Extreme Value Theorem on a Compact Subset of a Metric Space every real-valued continuous function on [p,q][p,q] attains a maximum and a minimum and is in particular bounded; we use this repeatedly below. In particular, fix reals α,δ0\alpha,\delta\ge0 bounding all entries of A(t)A(t) and D(t)D(t) on [a,b][a,b]. Transpose manipulations use claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals. Set F(t,X)=A(t)X+XA(t)XD(t)X+C(t)F(t,X)=A(t)X+XA(t)^{\top}-XD(t)X+C(t).

Hypotheses of the a priori theorem. Composition continuity: entries of F(t,h(t))F(t,h(t)) are polynomial combinations of continuous real-valued functions (Continuity of Sums and Products of Real-Valued Functions on a Metric Space). Local Lipschitz: for X,YR|X|,|Y|\le R, write XDXYDY=(XY)DX+YD(XY)XDX-YDY=(X-Y)DX+YD(X-Y); every entry of the right side is bounded by 2k2δRXYe2k^{2}\delta R\,|X-Y|_{e} (two products of three factors, entries of X,YX,Y bounded by RR), and the linear terms contribute at most 2kαXYe2k\alpha\,|X-Y|_{e} per entry, so F(t,X)F(t,Y)k2(2kα+2k2δR)XY=:LRXY|F(t,X)-F(t,Y)|\le k^{2}\,(2k\alpha+2k^{2}\delta R)\,|X-Y|=:L_R\,|X-Y|, using XYeXY|X-Y|_{e}\le|X-Y|.

Properties of solutions on subintervals. Let a<cba<c\le b and let hh be any continuous solution of the Riccati integral equation on [a,c][a,c], in the sense of the solution notion of A Priori Bounded Solutions of Locally Lipschitz Ordinary Differential Equations Exist Globally (whose statement records that the integrand is continuous on [a,c][a,c]).

(a) Uniqueness on [a,c][a,c]. If h,h~h,\widetilde h are two solutions on [a,c][a,c], choose RR with h(t)R|h(t)|\le R and h~(t)R|\widetilde h(t)|\le R on [a,c][a,c]: each entry is bounded by Extreme Value Theorem on a Compact Subset of a Metric Space, and XkXe|X|\le k\,|X|_{e} converts entry bounds to Euclidean bounds. Subtracting the two integral equations and applying claim 5 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals (with p=k2p=k^{2}) and the Lipschitz estimate, u(t):=h(t)h~(t)u(t):=|h(t)-\widetilde h(t)| satisfies u(t)k2LRatu(r)dru(t)\le k^{2}L_R\int_a^tu(r)\,dr on [a,c][a,c], and uu is continuous. By claim 7 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals (translation), uˉ(τ):=u(a+τ)\bar u(\tau):=u(a+\tau) is continuous on [0,ca][0,c-a] and satisfies uˉ(τ)k2LR0τuˉ(σ)dσ\bar u(\tau)\le k^{2}L_R\int_0^{\tau}\bar u(\sigma)\,d\sigma. The interval [0,ca][0,c-a] is nonempty, since a<ca<c, and compact by Closed Interval [a,b][a,b] is Compact in R\mathbb{R}. The function uˉ\bar u is continuous, hence bounded by Extreme Value Theorem on a Compact Subset of a Metric Space and measurable by claim 4 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions, and its Riemann integral over a subinterval of [0,ca][0,c-a] agrees with the corresponding Lebesgue integral by claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. Applying Gronwall's Lemma for Bounded Measurable Functions on [0,ca][0,c-a] with its constants taken to be 00 and k2LRk^{2}L_R gives uˉ0\bar u\le0; since u0u\ge0, u0u\equiv0 and h=h~h=\widetilde h.

(b) Symmetry. Since C(t)C(t), D(t)D(t), P0P_0 are symmetric, transposing the equation shows th(t)t\mapsto h(t)^{\top} is also a solution on [a,c][a,c]: (Ah)=hA(A h)^{\top}=h^{\top}A^{\top}, (hA)=Ah(hA^{\top})^{\top}=Ah^{\top}, and (hDh)=hDh=hDh(hDh)^{\top}=h^{\top}D^{\top}h^{\top}=h^{\top}Dh^{\top} (claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals). By (a), h(t)=h(t)h(t)^{\top}=h(t): every h(t)h(t) is symmetric.

(c) Positive semidefiniteness. Define Aˉ(r)=A(r)h(r)D(r)\bar A(r)=A(r)-h(r)D(r) on [a,c][a,c]; its entries are continuous, being sums and products of continuous real-valued functions (Continuity of Sums and Products of Real-Valued Functions on a Metric Space). Using symmetry of h(r)h(r) and D(r)D(r),

Aˉh+hAˉ+(C+hDh)=AhhDh+hAhDh+C+hDh=Ah+hAhDh+C,\bar A h+h\bar A^{\top}+\bigl(C+hDh\bigr)=Ah-hDh+hA^{\top}-hDh+C+hDh=Ah+hA^{\top}-hDh+C ,

since hAˉ=h(ADh)=hAhDhh\bar A^{\top}=h(A^{\top}-Dh)=hA^{\top}-hDh. Hence hh satisfies on [a,c][a,c] the linear matrix equation of Lyapunov Representation and Positive Semidefiniteness for Linear Matrix Equations, applied with bb replaced by cc and data (Aˉ,Cˉ,P0)(\bar A,\bar C,P_0) where Cˉ:=C+hDh\bar C:=C+hDh; Cˉ\bar C has continuous entries, is symmetric ((hDh)=hDh(hDh)^{\top}=hDh by (b)), and is positive semidefinite: x(h(r)D(r)h(r)x)=(h(r)x)(D(r)(h(r)x))0x\cdot(h(r)D(r)h(r)x)=(h(r)x)\cdot\bigl(D(r)(h(r)x)\bigr)\ge0 by the transpose-dot identity and symmetry of h(r)h(r). By the uniqueness in claim 1 of Lyapunov Representation and Positive Semidefiniteness for Linear Matrix Equations, hh is the solution treated there, and claims 2-3 there give: every h(t)h(t) is positive semidefinite.

(d) Comparison with Λ\Lambda. Let Λ\Lambda be as in the statement (existing on [a,b][a,b] by Lyapunov Representation and Positive Semidefiniteness for Linear Matrix Equations; symmetric positive semidefinite by claims 2-3 there). On [a,c][a,c], Δ:=Λh\Delta:=\Lambda-h satisfies, subtracting the two integral equations,

Δ(t)=at(A(r)Δ(r)+Δ(r)A(r)+h(r)D(r)h(r))dr,\Delta(t)=\int_a^t\bigl(A(r)\Delta(r)+\Delta(r)A(r)^{\top}+h(r)D(r)h(r)\bigr)\,dr ,

an equation of the form treated in Lyapunov Representation and Positive Semidefiniteness for Linear Matrix Equations (with bb replaced by cc), with coefficient AA, inhomogeneity hDhhDh (continuous, symmetric, positive semidefinite by (c)'s computation), and initial value 00 (positive semidefinite). By claims 1-3 there, Δ(t)\Delta(t) is positive semidefinite: h(t)Λ(t)h(t)\preceq\Lambda(t) in the semidefinite order.

(e) Entry bound. By (c), (d), and Entry Bounds for Positive Semidefinite Matrices, hij(t)maxlΛll(t)λ|h_{ij}(t)|\le\max_{l}\Lambda_{ll}(t)\le\lambda^{*}, where λ0\lambda^{*}\ge0 is the maximum over [a,b][a,b] of the continuous function tmaxlΛll(t)t\mapsto\max_l\Lambda_{ll}(t) (Extreme Value Theorem on a Compact Subset of a Metric Space; a maximum of finitely many continuous functions is continuous, since max(u,v)=12(u+v+uv)\max(u,v)=\tfrac12(u+v+|u-v|), and it is nonnegative since the diagonal entries of the positive semidefinite Λ(t)\Lambda(t) are nonnegative by Entry Bounds for Positive Semidefinite Matrices). Hence h(t)kh(t)ekλ=:ρ|h(t)|\le k\,|h(t)|_{e}\le k\lambda^{*}=:\rho. Also P0ρ|P_0|\le\rho: at t=at=a the Lyapunov integral equation gives Λ(a)=P0\Lambda(a)=P_0 (degenerate interval, convention of Mean-Square Riemann Integral of a Family of Random Variables), so 0P0=Λ(a)0\preceq P_0=\Lambda(a) and (P0)ijmaxlΛll(a)λ|(P_0)_{ij}|\le\max_l\Lambda_{ll}(a)\le\lambda^{*} by Entry Bounds for Positive Semidefinite Matrices, whence P0kλ=ρ|P_0|\le k\lambda^{*}=\rho.

Conclusion. The hypotheses (i)-(iii) of A Priori Bounded Solutions of Locally Lipschitz Ordinary Differential Equations Exist Globally hold with this ρ\rho: (i) and (ii) were checked above, and (iii) holds by (e), every solution on every [a,c][a,c] being bounded by ρ\rho in the Euclidean norm on entries, with ρP0\rho\ge|P_0|. Hence there is exactly one continuous solution PP on [a,b][a,b]. Applying (b), (c), (d) with c=bc=b to PP gives symmetry and 0P(t)Λ(t)0\preceq P(t)\preceq\Lambda(t), and the final entry bound is Entry Bounds for Positive Semidefinite Matrices applied to 0P(t)Λ(t)0\preceq P(t)\preceq\Lambda(t). \blacksquare

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