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

[Krizalid1]

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 $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$

 

[chisigma]

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 $\chi$ $\sigma$