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The following constitute the Existence and Uniqueness Theorems from the text: Existence Theorem: If f(t;x) is de ned and continuous on a rectangle R in the tx-plane, then given any point (t 0;x 0) 2R, the initial value problem x0= f(t;x)andx(t 0)=x 0 has a solution x(t) de ned in an interval containing t 0. Uniqueness Theorem: If f(t;x) and @f @x

Existence Uniqueness Theorem We will now see that rather mild conditions on the right hand side of an ordi-nary di erential equation give us local existence and uniqueness of solutions. De nition. Let f: D!Rnbe a coninuous function de ned in the open set D Rn+1. We say that f is locally Lipschitz in the Rn variable if for each (t 0;x

These notes on the proof of Picard's Theorem follow the text Fundamentals of Differential. Equations and Boundary Value Problems, 3rd edition, by Nagle, ...

Picard’s Theorem so important? One reason is it can be generalized to establish existence and uniqueness results for higher-order ordinary di↵erential equations and for systems of di↵erential equations. Another is that it is a good introduction to the broad class of existence and uniqueness theorems that are based on fixed points.

Theorem 1 (Existence). Suppose that F(x;y) is a continuous function de ned in some region R = f(x;y) : x0 < x < x0 + ;y0 < y < y0 + g containing the point (x0;y0). Then there exists a number 1 (possibly smaller than ) so that a solution y = f(x) to (*) is de ned for x0 1 < x < x0 + 1. Theorem 2 (Uniqueness). Suppose that both F(x;y) and @F

Proof of Existence via Picard Iterates. This is the same proof as found on pages 734-739 of “Ordinary. Differential Equations” by M. Tenenbaum and H. Pollard.

Theorem (Existence and Uniqueness Theorem for ODE). Suppose f(t;x) is continuous in the open set D Rn+1 and is locally Lipschitz in xin D. Let (t 0;x 0) 2D. Then, the initial value problem x_ = f(t;x);x(t 0) = x 0 (1) has a unique solution de ned in a small interval Iabout t 0 in R. Proof. Let Ube an open neighborhood about (t 0;x 0) in Dso that

Picard’s Theorem so important? One reason is it can be generalized to establish existence and uniqueness results for higher-order ordinary di↵erential equations and for systems of di↵erential equations. Another is that it is a good introduction to the broad class of existence and uniqueness theorems that are based on fixed points.

Existence and Uniqueness. Picard Iteration. Uniqueness. Examples. Existence and Uniqueness Theorem. 2. Sketch of Proof of Existence and Uniqueness Theorem.

theorem on existence and uniqueness of first order ODE (with initial value), basically, under a very simple (easily verified) condition (which is a strong ...

existence and uniqueness theorem for (1.1) we just have to establish that the equation (3.1) has a unique solution in [x0 −h,x0 +h]. IV. Proof of the uniqueness part of the theorem. Here we show that the problem (3.1) (and thus (1,1)) has at most one solution (we have not yet proved that it has a solution at all).

The initial value problem (1.1) is equivalent to an integral equation. For the proof of existence and uniqueness one first shows the equivalence of the problem ...

uniqueness part of the PLT. Repeat the process on [t2,t3], and after a finite number of such steps we proved the existence of a unique solution on [t0,a]. For the very last step if t n > a, then we can only guarantee a solution as far as t = a by the theorem. Similarly we can extend to the left so that we have a solution on t0 −a ≤ t ≤ ...

Theorem 2 (Uniqueness). Suppose that both F(x;y) and @F @y (x;y) are continuous functions de ned on a re-gion R as in Theorem 1. Then there exists a number 2 (possibly smaller than 1) so that the solution y = f(x) to (*), whose existence was guaranteed by Theorem 1, is the unique solution to (*) for x0 2 < x < x0 + 2. x − 0 δ 2 x + 0 δ 2 0 ...

So by the Mean Value Theorem, given any x;y 2 J there is some z between x and y such that jf(x)¡f(y)j jx¡yj = jfy(z)j • K and therefore f is Lipschitz on J with constant K. Hence Theorem 1.1 implies the existence of a unique solution of dy dt = y2=3 y(0) = y 0 on some time interval. 3 To prove the existence and uniqueness theorem, we need ...

NOTES ON THE EXISTENCE AND UNIQUENESS THEOREM FOR FIRST ORDER DIFFERENTIAL EQUATIONS I. Statement of the theorem. We consider the initial value problem (1.1) ˆ y′(x) = F(x,y(x)) y(x0) = y0. Here we assume that F is a function of the two variables (x,y), defined in a rectangle R = {(x,y) :x0 − a ≤ x ≤ x0 +a, (1.2) y0 −b ≤ y ≤ y0 +b}

We highlight the proof of Theorem 2.8.1, the existence. / uniqueness theorem for first order differential equations. In par- ticular, we review the needed ...

PDF | It has been proved that uncertain differential equation (UDE) has a unique solution, under the condi-tions that the coefficients are global.

(ii) Under what conditions can we be sure that there is a unique solution to (*)? Here are the answers. Theorem 1 (Existence). Suppose that F(x, y) is a.

Comment: The existence and uniqueness theorem are also valid for certain system of rst order equations. These theorems are also applicable to a certain higher order ODE since a higher order ODE can be reduced to a system of rst order ODE. Example 1. Consider the ODE y0= xy siny; y(0) = 2: Here fand @f=@yare continuous in a closed rectangle about x

is continuous is used in the proof of uniqueness. 3 Equivalent integral equation. The key to both the existence and uniqueness proofs is the equivalent integral ...

2.7: Existence and Uniqueness of Solutions Basic Existence and Uniqueness Theorem (EUT): Suppose f(t,x) is defined and continuous, and has a continuous partial derivative ∂f(t,x)/∂x on a rectangle R in the tx–plane. Then, given any initial point (t0,x0) in R, the initial value problem x′ = f(t,x), x(t0) = x0

where K = max{|A(t)| : t ∈ I}. This is the content of our previous existence and uniqueness theorem for first order linear equations. Similarly the non-linear equation y′ = A(t)siny +B(t), with y(t0) = a0 has a unique solution on any closed interval containing t0 on which A and B are continuous, because |sin(y)−sin(z)| ≤ y Z z −costdt