Theorem of existence and unicity.

In summary, the theorem of existence and unicity guarantees a unique solution for the given initial value problem of x'(t) = log (3t (x(t) - 2)) with x(3) = 5 if the initial value lies within the common intervals of the domain of f(x,t) and the domain of ∂f(x,t)/∂x. The function is continuous and differentiable for all x and t except when t = 0 and x = 2. Therefore, the common intervals are x cannot equal 2 and t cannot equal 0. The initial values of t_0 = 3 and x_0 = 5 are within these common intervals, which means there is a unique solution
  • #1
Lengalicious
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Homework Statement


In the following case: x'(t) = log (3t (x(t) - 2)) does the theorem of existence and unicity guarantee a unique solution for the initial value problem x(3) = 5, justify your answer?

Homework Equations


x'(t) = log (3t (x(t) - 2))

The Attempt at a Solution


Ok what I would do is take the domain of f(x,t) and the domain of ∂f(x,t)/∂x and then find the common interval and if the initial value problem is within this interval then there is a unique solution. However there is x(t) and t in the function so I don't understand how to find the domain?
 
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  • #2
For what x and t is that function, log(3t(x- 2), continuous and differentiable? What is the largest interval containing (3, 5) on which the function is continuous and differentiable?
 
  • #3
Yeh, how would I find that out? Thats what I'm confused about.

EDIT: Would I just plug numbers into the x and t? Until I got an x and t that came up with 'Math error'? Because if that's the case then when t = 0 the function is discontinuous and when x = 2? Am I right?
and then the partial derivative is 3t/(3t(x-2)) = 1/(x-2) in which case x cannot equal 2 and t is eqal to all real numbers? So the common intervals are x cannot = 2 and t cannot = 0, with intial values t_0 = 3 and x_0 = 5 which are within the common intervals, therefore there is a unique solution? On a side note, should i have taken the partial derivative with respect to t or x? Or do I take partial derivative with respect to both and then compare all 3 intervals.
 
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Related to Theorem of existence and unicity.

1. What is the Theorem of Existence and Unicity?

The Theorem of Existence and Unicity is a mathematical principle that states that for a given set of conditions, there exists a unique solution or entity that satisfies those conditions. In other words, there can only be one solution that meets the specified criteria.

2. How is the Theorem of Existence and Unicity used in mathematics?

The Theorem of Existence and Unicity is used to prove the existence and uniqueness of solutions in various mathematical problems. It is often used in calculus, differential equations, and optimization problems to show that there is only one solution that satisfies the given conditions.

3. What are the key assumptions of the Theorem of Existence and Unicity?

The Theorem of Existence and Unicity relies on two key assumptions: existence and uniqueness. Existence assumes that there is at least one solution that satisfies the given conditions, while uniqueness assumes that there is only one solution that meets those conditions.

4. Can the Theorem of Existence and Unicity be proven?

The Theorem of Existence and Unicity is considered a fundamental principle in mathematics and is often taken as an axiom rather than proven. However, it can be proven in certain cases using mathematical proofs and logic.

5. How is the Theorem of Existence and Unicity related to other mathematical concepts?

The Theorem of Existence and Unicity is closely related to other mathematical concepts such as continuity, differentiability, and convexity. These concepts are often used to prove the existence and uniqueness of solutions in various mathematical problems.

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