Proof on Linear 1st Order IVP solution being bounded

In summary: Here is what I have tried:If I take the absolute value of both sides of the equation then I can overestimate the right hand side by some constant.In summary, the solution to the IVP is bounded for t≥0.
  • #1
marvalos
3
0
Proof on Linear 1st Order IVP solution being "bounded"

A function h(t) is called "bounded" for t≥t0 if there is a constant M>0 such that

|h(t)|≤M for all t≥0

The constant M is called a bound for h(t). Consider the IVP

x'=-x+q(t), x(0)=x0

where the nonhomogeneous term q(t) is bounded for t≥0. Show the solution of this IVP is bounded for t≥0. (Hint: Use the Variation of Constants Formula.)

Any help on where to go for this problem would be great. Thanks
 
Physics news on Phys.org
  • #2
You are given the "Hint: use the Variation of Constants Formula". Okay, what is that formula?
 
  • #3
The Variation of Constants formula is a generalized formula for First Order Linear DE's that can be solved with the Integrating Factor Method.

I would put the exact formula down but I am not too familiar with this equation editor.
 
  • #4
You have to show some effort. Show us what you have tried. I would give a second hint: It is a linear equation.
 
  • #5
Here is what I have tried:

The Variation of Constants formula gave me this

x=(x0+∫e^u q(u) du) e^-t

the integral is definite and goes from 0 to t.

Since q(t) is bounded, would that remain true if the integral is taken from it?
 
  • #6
marvalos said:
Here is what I have tried:

The Variation of Constants formula gave me this

x=(x0+∫e^u q(u) du) e^-t

the integral is definite and goes from 0 to t.

Since q(t) is bounded, would that remain true if the integral is taken from it?

To rephrase your question, if you take the absolute value of both sides of that equation, can you overestimate the right hand side by some constant. So try it and show us what happens.
 

1. What is a Linear 1st Order IVP?

A Linear 1st Order IVP (Initial Value Problem) is a type of differential equation that involves a first-order derivative and an initial condition. It can be represented in the form of dy/dx = f(x,y) with an initial condition y(x0) = y0, where x0 and y0 are given values.

2. What does it mean for a Linear 1st Order IVP solution to be bounded?

A bounded solution means that the values of the solution remain within a certain range or do not approach infinity as x approaches infinity. In other words, the solution does not grow or shrink without bound.

3. Why is it important to prove that a Linear 1st Order IVP solution is bounded?

It is important to prove that a solution is bounded because it ensures the stability and validity of the solution. A bounded solution can be reliably used for further analysis and predictions.

4. How can one prove that a Linear 1st Order IVP solution is bounded?

There are various methods for proving the boundedness of a solution, such as the Mean Value Theorem, the Comparison Theorem, and the use of inequalities. These methods involve analyzing the behavior of the solution and its derivatives at different points.

5. What are the implications of a Linear 1st Order IVP solution not being bounded?

If a solution is not bounded, it means that the values of the solution grow or shrink without bound, which can lead to unstable and unreliable predictions. It may also indicate that the initial conditions or the differential equation itself are invalid.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
4
Views
245
  • Calculus and Beyond Homework Help
Replies
4
Views
602
  • Calculus and Beyond Homework Help
Replies
24
Views
797
  • Calculus and Beyond Homework Help
Replies
8
Views
622
  • Calculus and Beyond Homework Help
Replies
8
Views
235
  • Calculus and Beyond Homework Help
Replies
1
Views
283
  • Calculus and Beyond Homework Help
Replies
6
Views
300
  • Calculus and Beyond Homework Help
Replies
5
Views
914
  • Calculus and Beyond Homework Help
Replies
26
Views
2K
Back
Top