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I have a few functions which i have got from solving differential equations, and i am a bit confused about the maximal interval of existence and …

09.09.2007 · Suppose f (t,x) is a continuous vector valued function on . If f is locally Lipschitz with respect to x with the property that for some positive constant C > 0, then prove that the maximum interval of the existence of the initial value problem x' = f …

maximal interval of existence of ordinary differential equations Let E ⊂ W where W is a normed vector space , f ∈ C 1 ⁢ ( E ) is a continuous differentiable map f : E → W . Furthermore consider the ordinary differential equation

This is a Ricatti's equation Riccati equation (Riccati equation - Wikipedia). Exploring the solution of this equation numerically I've found 1.

The interval (α,β) in Theorem 1 is called the maximal interval of existence of the solution x(t) of the initial value problem (1) or simply the maximal interval of existence of the initial value problem (1). Corollary 1. Let E be an open subset of Rn and assume that f ∈ C1(E) and let (α,β) be the maximal interval of existence of the ...

So By Picard's theorem the solution exists and unique in an interval around. −1. To find the maximal interval, We notice that the nonlinearity is Lipschitz in ...

21.11.2014 · $\begingroup$ Note also that maximal intervals of existence in general are not of the form $[t_0 - a, t_0 + a]$. $\endgroup$ – Travis Willse Nov 21 '14 at 1:53

+) is the maximal interval of existence. An extension argument similar to the one in Step 1 shows that every interval of existence is contained in an open interval of existence. Every open interval of existence is, in turn, a subset of (!−;! +). Step 4: xis the only solution of (1) on (!−;! +). This is a special case of Step 1.

Intervals of Existence. Lecture 5. Math 634. 9/10/99. Maximal Interval of Existence. We begin our discussion with some definitions and an important theorem ...

16.05.2015 · The domain may not be all of $\mathbb{R}$, for example, but a small interval. It may be possible to 'extend' the solution to a larger domain. If there is a domain that cannot be extended, then it is a maximal interval of existence/validity for the solution. $\endgroup$ –

Theorem (Maximal Interval of Existence). An IVP has a maximal interval of existence, and it is of the form ( t −, t +), with t − ∈ [ − ∞, ∞) and t + ∈ ( − ∞, ∞]. There is a unique solution x ( t) on ( t −, t +), and ( t, x ( t)) leaves every compact subset K of D as t ↓ t − and as t ↑ t +. Proof See ODE Notes.

Since the initial condition is at x = 0, the interval of existence should be ... Thus the solution y(x) of the initial value problem attains its maximum ...

Thus the existence and uniqueness theorem can only guarantee the existence of a solution in a small interval [t 0 −h,t 0 +h] whereas in practice the solution will exist in a much larger interval. To find the largest interval of existence we apply the …

The domain of a particular solution to a differential equation is the largest open interval containing the initial value on which the solution satisfies the differential equation. Theorem (Maximal Interval of Existence).

An easy example of solving a differential equation, and then considering the interval of existence for the ...

The domain of a particular solution to a differential equation is the largest open interval containing the initial value on which the solution satisfies the ...

Sep 08, 2007 · Homework Statement Suppose f(t,x) is a continuous vector valued function on \\mathbb{R} \\times \\mathbb{R}^n. If f is locally Lipschitz with respect to x with the property that \\|f(t,x)\\| \\le C \\|x\\| for some positive constant C > 0, then prove that the maximum interval of the existence of the...

Next, if the interval in the theorem is the largest possible ... So, we will know that a unique solution exists if the conditions of the ...

We refer to (−∞,x−10) maximal interval of existence for the differential equation. Note that there is also solutions with x0<0 tend to ...

Dec 24, 2021 · Please solve the following ODE problem: For every nonzero xo belonging to R, find the maximal interval of existence of the following initial problem: x' = f (x) , x (0) = xo , where f: R {0} into R and f (x) = 1/x^2 . Guive proofs for your result. Skectch the region.