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Pietro Majer
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The answer to (2) is yes, and it's very standard: the corresponding $F$ satisfies the Cauchy-Lipschitz-Picard–LindelöfPicard-Lindelöf hypotheses: being continuous $A$ is locally bounded, so $F$ is locally uniformly Lipschitz in the variable $x$. In fact, for a continuous $A:[a,b]\to L(E)$ you can directly solve the Cauchy problem for the ODE with parameter: $$\begin{cases}G(s,s)=I \\ \partial_1 G(t,s)=A(t)G(t,s) \end{cases}$$ solving the equivalent integral equation $$ G(t,s)- \int_s^t A(\tau)G(\tau,s)d\tau=I \ , $$ which consists in inverting a quasinilpotent perturbation of the identity on the Banach space $C^0([a,b]\times [a,b],L(E))$. This is easily done in terms of the Neumann series: $$ G(t,s)=\sum_{k=0}^\infty W_k(t,s) , $$ $$\begin{cases}W_0(t,s)=I \\ W_{k+1}(t,s)=\int_s^t A(\tau)W_k(\tau ,s)d\tau \end{cases}\ .$$ The variation of constant formula holds too, and produces the unique solution of the non-homogeneous problem $$\begin{cases}u(s)=u_0 \\ \dot u(t)=A(t)u(t)+b(t) \end{cases}$$ as $$u(t)=G(t,s)u_0+\int_s^tG(t,\tau)b(\tau)d\tau \ .$$

The answer to (2) is yes, and it's very standard: the corresponding $F$ satisfies the Cauchy-Lipschitz-Picard–Lindelöf hypotheses: being continuous $A$ is locally bounded, so $F$ is locally uniformly Lipschitz in the variable $x$. In fact, for a continuous $A:[a,b]\to L(E)$ you can directly solve the Cauchy problem for the ODE with parameter: $$\begin{cases}G(s,s)=I \\ \partial_1 G(t,s)=A(t)G(t,s) \end{cases}$$ solving the equivalent integral equation $$ G(t,s)- \int_s^t A(\tau)G(\tau,s)d\tau=I \ , $$ which consists in inverting a quasinilpotent perturbation of the identity on the Banach space $C^0([a,b]\times [a,b],L(E))$. This is easily done in terms of the Neumann series: $$ G(t,s)=\sum_{k=0}^\infty W_k(t,s) , $$ $$\begin{cases}W_0(t,s)=I \\ W_{k+1}(t,s)=\int_s^t A(\tau)W_k(\tau ,s)d\tau \end{cases}\ .$$ The variation of constant formula holds too, and produces the unique solution of the non-homogeneous problem $$\begin{cases}u(s)=u_0 \\ \dot u(t)=A(t)u(t)+b(t) \end{cases}$$ as $$u(t)=G(t,s)u_0+\int_s^tG(t,\tau)b(\tau)d\tau \ .$$

The answer to (2) is yes, and it's very standard: the corresponding $F$ satisfies the Cauchy-Lipschitz-Picard-Lindelöf hypotheses: being continuous $A$ is locally bounded, so $F$ is locally uniformly Lipschitz in the variable $x$. In fact, for a continuous $A:[a,b]\to L(E)$ you can directly solve the Cauchy problem for the ODE with parameter: $$\begin{cases}G(s,s)=I \\ \partial_1 G(t,s)=A(t)G(t,s) \end{cases}$$ solving the equivalent integral equation $$ G(t,s)- \int_s^t A(\tau)G(\tau,s)d\tau=I \ , $$ which consists in inverting a quasinilpotent perturbation of the identity on the Banach space $C^0([a,b]\times [a,b],L(E))$. This is easily done in terms of the Neumann series: $$ G(t,s)=\sum_{k=0}^\infty W_k(t,s) , $$ $$\begin{cases}W_0(t,s)=I \\ W_{k+1}(t,s)=\int_s^t A(\tau)W_k(\tau ,s)d\tau \end{cases}\ .$$ The variation of constant formula holds too, and produces the unique solution of the non-homogeneous problem $$\begin{cases}u(s)=u_0 \\ \dot u(t)=A(t)u(t)+b(t) \end{cases}$$ as $$u(t)=G(t,s)u_0+\int_s^tG(t,\tau)b(\tau)d\tau \ .$$

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Pietro Majer
  • 60.5k
  • 4
  • 122
  • 269

The answer to (2) is yes, and it's very standard: the corresponding $F$ satisfies the Cauchy-Lipschitz-Picard–Lindelöf hypotheses: being continuous $A$ is locally bounded, so $F$ is locally uniformly Lipschitz in the variable $x$. In fact, for a continuous $A:[a,b]\to L(E)$ you can directly solve the Cauchy problem for the ODE with parameter: $$\begin{cases}G(s,s)=I \\ \partial_1 G(t,s)=A(t)G(t,s) \end{cases}$$ solving the equivalent integral equation $$ G(t,s)- \int_s^t A(\tau)G(\tau,s)d\tau=I \ , $$ which consists in inverting a quasinilpotent perturbation of the identity on the Banach space $C^0([a,b]\times [a,b],L(E))$. This is easily done in terms of the Neumann series: $$ G(t,s)=\sum_{k=0}^\infty W_k(t,s) , $$ $$\begin{cases}W_0(t,s)=I \\ W_{k+1}(t,s)=\int_s^t A(\tau)W_k(\tau ,s)d\tau \end{cases}\ .$$ The variation of constant formula holds too, and produces the unique solution of the non-homogeneous problem $$\begin{cases}u(s)=u_0 \\ \dot u(t)=A(t)u(t)+b(t) \end{cases}$$ as $$u(t)=G(t,s)u_0+\int_s^tG(t,\tau)b(\tau)d\tau \ .$$