Solution to Linear Time-invariant Matrix ODE $\frac{d\mathbf{P}}{dt} = \mathbf{A}\mathbf{P} + \mathbf{B}$

Question

Do we have a solution for the following matrix ODE: $$\frac{d\mathbf{P}}{dt} = \mathbf{A}\mathbf{P} + \mathbf{B}$$ where all matrices are square.

My guess is: $$ \mathbf{P}(t) = \exp{(\mathbf{A}t)}\mathbf{P}(0) + \mathbf{A}^{-1}[\exp{(\mathbf{A}t)}\mathbf{B} - \mathbf{B}] $$ when $\mathbf{A}$ is invertible. I am not sure it is correct and not clear when $\mathbf{A}$ is not full rank.

Answer 1

Assuming $A$ is constant (i.e. independent of $t$), you can use $\exp(-At)$ as an integrating factor; left-multiplying by it gives $\frac{d}{dt} \bigl( e^{-At} P(t) \bigr) = e^{-At} B(t)$, so that $$ e^{-At} P(t) - P(0) = \int_0^t e^{-As} B(s) \, ds , $$ and hence $$ P(t) = e^{At} P(0) + \int_0^t e^{A(t-s)} B(s) \, ds , $$ which holds in general, and reduces to your formula if $A$ is invertible and $B$ is constant.

Answer 2

Guessing the solution is a time-honored way of solving differential equations.

Prove that your ansatz is correct by calculating its time derivative $$\eqalign{ AP &= Ae^{At}P_0 + e^{At}B-B \quad\implies\quad &AP+B = Ae^{At}P_0 + e^{At}B \\ A\dot P &= A^2e^{At}P_0 + Ae^{At}B \\ &= A(AP+B) \qquad &\big({\rm Eq\;}{\tt\#1}\big) \\ \dot P &= AP+B \qquad &\big({\rm QED}\big) \\\\ }$$

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If $A$ is singular, try substituting the pseudoinverse $A^+$ $$\eqalign{ P &= e^{At}P_0 + A^+(e^{At}-I)\,B \qquad&\big({\rm ansatz}\big) \\ AP &= Ae^{At}P_0 + (e^{At}-I)\,B \qquad&\big({\rm recall}\!:\,AA^+A=A\big) \\ A\dot P &= A^2e^{At}P_0 + Ae^{At}B \qquad&\big({\rm time\;derivative}\big) \\ &= A(AP+B) \qquad&\big({\rm simplify}\big) \\ }$$ So this assumption yielded a solution of $\big({\rm Eq\;}{\tt\#1}\big)$ $$A\dot P = A(AP+B)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ if not the original ODE.