Prove Uniqueness Theorem: |\phi(t) - \psi(t)| ≤ ∫0t

In summary, the uniqueness theorem requires a proof to justify the steps involved. One step involves showing why the inequality follows from the initial equation. In the given interval, the value of s is always positive, but the other term may vary in sign. Taking the absolute value of the integrand can only increase the result, leading to the inequality shown.
  • #1
thedude36
30
0
I am having to justify the steps in a proof of the uniqueness theorem. I am supposed to show why the inequality follows from the initial equation.

http://i.imgur.com/AxApogj.png

[itex]\phi[/itex](t) - [itex]\psi[/itex](t) =∫0t 2s[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)] ds


|[itex]\phi[/itex](t) - [itex]\psi[/itex](t)| =|∫0t 2s[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)] ds| [itex]\leq[/itex] ∫0t 2s|[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)]| ds , with t>0

I have no idea where to start. Mostly, I am unsure as to why pulling the 2s out will make it larger than the initial absolute value. Could anyone help?
 
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  • #2
In the interval, s is always positive, but the other term may vary in sign. That would lead to some cancellation during the integration process. If we take the absolute value of the integrand, it can only increase the result:
|∫0t 2s[ϕ(t) - ψ(t)] ds| ≤ ∫0t |2s[ϕ(t) - ψ(t)]| ds = ∫0t 2s|[ϕ(t) - ψ(t)]| ds
 

Related to Prove Uniqueness Theorem: |\phi(t) - \psi(t)| ≤ ∫0t

1. What is the uniqueness theorem?

The uniqueness theorem states that if two solutions to a differential equation have the same initial conditions, then they must be equal for all values of t.

2. How is the uniqueness theorem proven?

The uniqueness theorem is proven using the method of contradiction. We assume that there are two solutions, \phi(t) and \psi(t), that satisfy the given differential equation and initial conditions. Then, we show that the absolute value of their difference, |\phi(t) - \psi(t)|, is equal to zero for all values of t, proving that they are in fact the same solution.

3. What is the importance of the uniqueness theorem?

The uniqueness theorem is important because it guarantees that there can only be one solution to a given differential equation with a specific set of initial conditions. This allows us to confidently use differential equations to model real-world phenomena and make accurate predictions.

4. Can the uniqueness theorem be applied to all differential equations?

No, the uniqueness theorem only applies to certain types of differential equations, specifically those that are linear and have continuous coefficients. Nonlinear or discontinuous differential equations may not have unique solutions.

5. How is the uniqueness theorem related to the existence theorem?

The uniqueness theorem and the existence theorem are often used together to prove the existence and uniqueness of solutions to a given differential equation. The existence theorem states that a solution exists for a given set of initial conditions, while the uniqueness theorem ensures that this solution is the only possible one.

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