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

  • Context: MHB 
  • Thread starter Thread starter Krizalid1
  • Start date Start date
  • Tags Tags
    Continuity Ivp
Click For Summary
SUMMARY

The discussion focuses on the continuity of the solution to the initial value problem (IVP) defined by the differential equation $\overset{\cdot }{\mathop{x}}\,=Ax$ with initial condition $x(0)=x_0$. It is established that for a matrix $A_{n\times n}$ with real and distinct eigenvalues, the limit $\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0)$ holds for each fixed $t\in\mathbb{R}$. The general solution is given by $u(t,x_{0}) = c_{0}\ A\ e^{\lambda\ t}$, where $c_{0}= x_{0}\ A^{-1}$, confirming that $u(t,x_{0})$ is continuous in $x_{0}$.

PREREQUISITES
  • Understanding of linear differential equations
  • Knowledge of matrix theory, specifically eigenvalues and eigenvectors
  • Familiarity with the concept of initial value problems (IVPs)
  • Basic proficiency in calculus and limits
NEXT STEPS
  • Study the properties of matrices with distinct eigenvalues
  • Learn about the stability of solutions to linear differential equations
  • Explore the application of the exponential matrix $e^{At}$ in solving IVPs
  • Investigate continuity and differentiability in the context of functional analysis
USEFUL FOR

Mathematicians, students studying differential equations, and researchers in applied mathematics focusing on the continuity of solutions in initial value problems.

Krizalid1
Messages
106
Reaction score
0
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 $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$
 
Physics news on Phys.org
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 $\chi$ $\sigma$
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
493
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
7
Views
4K