ODE initial values and continuity

In summary: It should address the uniqueness aspect of the question. In summary, the conversation discusses finding a continuous solution for t > 0 to the initial value problem y'(t)+p(t)y(t)=0, y(0)=1, where p(t)=2 for t [0,1] and p(t)=1 for t > 1. One proposed solution is y = e-2t for t [0,1], but the question remains about finding a continuous solution for t > 0. The mention of a theorem on existence and uniqueness of solutions may provide insight into this issue.
  • #1
mathman44
207
0

Homework Statement



Find a continuous y(t) for t > 0 to the initial value prob:

[tex]y'(t)+p(t)y(t)=0, y(0)=1[/tex]
where
[tex]p(t)=2[/tex] for 0 < t < 1
[tex]p(t)=1[/tex] for t > 1

and determine if the soln is unique.

The Attempt at a Solution



By standard ODE techniques I arrive at

[tex]y=\exp(-2t)[/tex] for 0 < t < 1
[tex]y=\exp(-t)[/tex] for t > 1

The problem is that this soln y(t) isn't continuous.. what's wrong here? As far as I know the only way to do this is to solve for y(t) in both intervals of t.
 
Last edited:
Physics news on Phys.org
  • #2
Anyone please?
 
  • #3
One problem with this is that p(t) isn't defined at 0, yet your initial condition is y(0). So your differential equation isn't defined at 0, but you are supposed to find a solution y(t) that is defined at 0.
 
  • #4
Oops... I should have said that p(t) is 2 for t [0,1].
 
  • #5
Then y = e-2t is a solution that is continuous on [0, 1], the interval that contains the initial value t = 0.

I think that's what we're looking for, but your text should have a theorem about existence and uniqueness of solutions of DEs. See what that theorem has to say about this situtation.
 
  • #6
But the question is asking for a continuous soln for t > 0, not just t belonging to [0,1].
 
  • #7
Take a look at the theorem I mentioned.
 

Related to ODE initial values and continuity

1. What is an initial value in ODE?

An initial value in ODE (Ordinary Differential Equations) is a starting point or condition for the solution of a differential equation. It is a known value of the dependent variable at a specific point in the independent variable's domain.

2. How is continuity defined in ODE?

In ODE, continuity refers to the smoothness of a function. It means that the function is defined and has no abrupt changes or discontinuities. In other words, the function is continuous if its value changes gradually as the independent variable changes.

3. What is the significance of initial values in solving ODEs?

The initial values are essential in solving ODEs as they provide a starting point for the solution. They help determine the unique solution to the differential equation and also ensure that the solution is continuous. Without initial values, the solution would not be well-defined.

4. How are initial values and continuity related in ODEs?

Initial values and continuity are closely related in ODEs. The initial values are used to determine the unique solution to the ODE, and this solution must be continuous. If the initial values are not chosen carefully, the resulting solution may not be continuous, and the solution will not be valid.

5. What are some methods for ensuring continuity in ODE solutions?

There are several methods for ensuring continuity in ODE solutions, such as using the Euler method or higher-order numerical methods, which involve approximating the solution at smaller intervals. Another approach is to use adaptive step sizes, where the step size is adjusted based on the smoothness of the solution, ensuring continuity. Additionally, using a more accurate initial value can also help ensure continuity in the solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
465
  • Calculus and Beyond Homework Help
Replies
3
Views
641
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
770
  • Calculus and Beyond Homework Help
Replies
6
Views
387
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
969
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
647
Back
Top