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Consider the initial value problem y = f(t, y), y(to) = y0 Under what conditions does an initial value problem of the form (3) have at least one solution and it is unique? Theorem 1. Suppose the right hand side function f(t, y) is continuous at all points (t, y) in some rectangle R: |t – tol < a, ly – yol < b (4) and bounded in R; i.e. there is a number K such that |f(t, y)| < K, for all (t, y) E R. (5) Then the initial value problem (3) has at least one solution y(t). This solution erists at least for all t in the subinterval |t – tol < a of the interval t – tol < a; here, a = min a, -. Note 1. Remind the definition of a function being continuous? of be continuous for all (t, y) in the rectangle dy Theorem 2. Let f and its partial derivative fy and bounded; \S(t, y)| < K, ISyl < M for all (t, y) E R. Then the problem has at most one solution.

Consider the initial value problem y = f(t, y), y(to) = y0 Under what conditions does an initial value problem of the form (3) have at least one solution and it is unique? Theorem 1. Suppose the right hand side function f(t, y) is continuous at all points (t, y) in some rectangle R: |t – tol < a, ly – yol < b (4) and bounded in R; i.e. there is a number K such that |f(t, y)| < K, for all (t, y) E R. (5) Then the initial value problem (3) has at least one solution y(t). This solution erists at least for all t in the subinterval |t – tol < a of the interval t – tol < a; here, a = min a, -. Note 1. Remind the definition of a function being continuous? of be continuous for all (t, y) in the rectangle dy Theorem 2. Let f and its partial derivative fy and bounded; \S(t, y)| < K, ISyl < M for all (t, y) E R. Then the problem has at most one solution.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can someone explain the concept in the image in an easy to understand manner?

I don't understand why there is at least one solution in one case and why there is at most one solution in one case.

Kindly buidl your explanation covering all the key points in the image.

Thank you! 

Consider the initial value problem y = f(t, y), y(to) = y0 Under what conditions does an initial value problem of the form (3) have at least one solution and it is unique? Theorem 1. Suppose the right hand side function f(t, y) is continuous at all points (t, y) in some rectangle R: |t – tol < a, |y – yo| < b (4) and bounded in R; i.e. there is a number K such that |f(t, y)| < K, for all (t, y) E R. (5) Then the initial value problem (3) has at least one solution y(t). This solution exists at least for all t in the subinterval |t – tol <a of the interval |t – tol < a; here, a = min a, - K Note 1. Remind the definition of a function being continuous? se dy Theorem 2. Let f and its partial derivative fy be continuous for all (t, y) in the rectangle and bounded; |f(t, y)| < K, |Syl < M for all (t, y) E R. Then the problem has at most one solution.
Transcribed Image Text:Consider the initial value problem y = f(t, y), y(to) = y0 Under what conditions does an initial value problem of the form (3) have at least one solution and it is unique? Theorem 1. Suppose the right hand side function f(t, y) is continuous at all points (t, y) in some rectangle R: |t – tol < a, |y – yo| < b (4) and bounded in R; i.e. there is a number K such that |f(t, y)| < K, for all (t, y) E R. (5) Then the initial value problem (3) has at least one solution y(t). This solution exists at least for all t in the subinterval |t – tol <a of the interval |t – tol < a; here, a = min a, - K Note 1. Remind the definition of a function being continuous? se dy Theorem 2. Let f and its partial derivative fy be continuous for all (t, y) in the rectangle and bounded; |f(t, y)| < K, |Syl < M for all (t, y) E R. Then the problem has at most one solution.
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