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I am stuck on the derivation of the identity in the title which is used in step three and four in the proof of

The Unique Solutions to Linear Nonhomogeneous Systems of First Order ODEs.

It is used in step three/four. Correct me if I am wrong.

Theorem 1

Let $\mathbf{x}^{\prime}=A(t) \mathbf{x}+g(t)$ with $\mathbf{x}(\tau)=\xi$ be a linear nonhomogeneous system of first order ODEs where $(\tau, \xi) \in D$, and let $\Phi(t, \tau)$ be the state transition matrix of the corresponding linear homogeneous system of first order ODEs $\mathbf{x}^{\prime}=A(t) \mathbf{x}$ with $\mathbf{x}(\tau)=\xi$. Then the unique solution to the linear nonhomogeneous system is given by $\varphi(t)=\Phi(t, \tau) \xi+\int_\tau^t \Phi(t, \tau) g(\eta) d \eta$.

Proof:

Let $\varphi(t)=\Phi(t, \tau) \xi+\int_\tau^t \Phi(t, \tau) g(\eta) d \eta$. Then we have that: $$ \begin{aligned} \varphi^{\prime}(t) & =\left[\Phi(t, \tau) \xi+\int_\tau^t \Phi(t, \tau) g(\eta) d \eta\right]^{\prime} \\ & =\frac{d}{d t} \Phi(t, \tau) \xi+\frac{d}{d t} \int_\tau^t \Phi(t, \tau) g(\eta) d \eta \\ & =\Phi^{\prime}(t, \tau) \xi+I g(t)+\int_\tau^t \Phi^{\prime}(t, \eta) g(\eta) d \eta \\ & =A(t) \Phi(t, \tau) \xi+g(t)+\int_\tau^t A(t) \Phi(t, \eta) g(\eta) d \eta \\ & =A(t)\left[\Phi(t, \tau) \xi+\int_\tau^t \Phi(t, \eta) g(\eta) d \eta\right]+g(t) \quad=A(t) \varphi(t)+g(t) \end{aligned} $$

. I found $\frac{\partial \boldsymbol{\Phi}\left(t, t_0\right)}{\partial t}=\mathbf{A}(t) \boldsymbol{\Phi}\left(t, t_0\right)$ with initial conditions $\boldsymbol{\Phi}\left(t_0, t_0\right)=I$ but I am not sure why.

  • What is the definition of $\Phi$? – arkeet Jun 30 '24 at 22:14
  • State transition matrix –  Jun 30 '24 at 23:14
  • Yes, you have said that, but how do you define the state transition matrix? Essentially I am looking for something that would indicate that $\mathbf{x}(t) = \Phi(t,t_0)$ is a solution of the homogeneous system. – arkeet Jun 30 '24 at 23:19
  • Then, the state transition matrix Ф(t) satisfies the following equation ẋ(t) = A * x(t) + B * u(t) , where Ф(t) is a matrix such that x(t)= Ф(t)*x(0). In the above equation , x(t) represents the state of the system at time t, and x(0) is the initial state at time t=0. –  Jul 01 '24 at 00:01
  • What does satisfies mean. Plug that state transition matrix as A or B? –  Jul 01 '24 at 00:02
  • State transition matrix is the matrix exponential of the companion matrix as a function of t –  Jul 01 '24 at 00:19
  • That the derivative of matrix exponential is the matrix exponential times the companion matrix. –  Jul 01 '24 at 00:57

1 Answers1

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Since the transition matrix $\Phi(t,t_0)$ is defined by the relation $x(t) = \Phi(t,t_0)x(t_0)$, where $x(t)$ is the solution to the homogeneous problem $\dot{x} = Ax$, one has : $$ \dot{\Phi}(t,t_0)x(t_0) = \frac{\mathrm{d}}{\mathrm{d}t}(\Phi(t,t_0)x(t_0)) = \dot{x}(t) = A(t)x(t) = A(t)\Phi(t,t_0)x(t_0), $$ hence $\dot{\Phi}(t,t_0) = A(t)\Phi(t,t_0)$, as desired.

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