[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 $\chi$ $\sigma$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 $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$
Continuity of IVP refers to the property of a function in which a small change in the initial value results in a small change in the solution. In other words, if the initial value of a function is changed by a small amount, the resulting solution should also change by a small amount.
Continuity of IVP is important because it ensures that the solution to a differential equation is stable and reliable. It also allows us to use numerical methods to approximate the solution, as small changes in the initial value will not significantly affect the accuracy of the solution.
The existence and uniqueness theorem states that a differential equation has a unique solution if the initial value is specified. Continuity of IVP is a necessary condition for this theorem to hold. If a function is not continuous at the initial value, there may be multiple solutions to the differential equation.
Yes, it is possible for continuity of IVP to be violated. This can happen if the function is not continuous at the initial value, or if the function is not differentiable at the initial value. In these cases, the solution to the differential equation may not be unique and may not exhibit the expected behavior.
Continuity of IVP can be verified by checking if the function is continuous and differentiable at the initial value. This can be done analytically or numerically by using numerical methods to approximate the solution and checking for small changes in the initial value.