MHB Continuity of IVP: Show $\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0)$

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For the initial value problem defined by the matrix equation $\overset{\cdot }{\mathop{x}}\,=Ax$ with distinct real eigenvalues, the solution $u(t,x_0)$ is expressed as $u(t,x_{0}) = c_{0}\ A\ e^{\lambda\ t}$, where $c_{0}= x_{0}\ A^{-1}$. This formulation demonstrates that $u(t,x_0)$ is linear in $x_0$, ensuring its continuity. Consequently, it follows that for any fixed time $t \in \mathbb{R}$, the limit $\lim_{y_0\to x_0}u(t,y_0)$ equals $u(t,x_0)$. Thus, the continuity of the solution with respect to initial conditions is established.
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Let $A_{n\times n}$ be a matrix with real and distincts eigenvalues. Let $u(t,x_0)$ be a solution for the initial value problem $\overset{\cdot }{\mathop{x}}\,=Ax$ with $x(0)=x_0,$ then show that for each fixed $t\in\mathbb R,$ we have $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$
 
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Krizalid said:
Let $A_{n\times n}$ be a matrix with real and distincts eigenvalues. Let $u(t,x_0)$ be a solution for the initial value problem $\overset{\cdot }{\mathop{x}}\,=Ax$ with $x(0)=x_0,$ then show that for each fixed $t\in\mathbb R,$ we have $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$
[sp]The general solution of the IVP is $\displaystyle u(t,x_{0}) = c_{0}\ A\ e^{\lambda\ t}$, where $\displaystyle c_{0}= x_{0}\ A^{-1}$, so that $u(t,x_{0})$ is linear in $x_{0}$ and therefore continous...[/sp]Kind regards $\chi$ $\sigma$
 
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