This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 15 ExistenceUniqueness of Initial Value Problems: The Picard Method of Successive Substitution In this lecture, we shall address some theoretical matters, namely the question of existence and uniqueness of solutions to initial value problems for systems of first order differential euqations. These IVPs take the form x ′ ( t ) = f ( x , t ) , x ( t ) = x . (1) For the moment, we shall restrict our discussion to the onedimensional case. And to complicate matters further, we shall revert back to notation used in earlier lectures, i.e., letting x denote the independent variable and y the dependent variable. The initial value problem associated with a single differential equation then takes the form dy dx = f ( y, x ) , y ( x ) = y . (2) If f is “sufficiently nice,” then the IVP in (2) exists and is unique. Of course, the question remains, “What is ‘sufficiently nice’ in this case?” First of all, we shall show that it is not sufficient for f to be continuous in its variables. For example, the IVPs dy dx = y, y ( x ) = y . (3) has a unique solution for any values x , y ∈ R . On the other hand, the following IVPs dy dx = y 1 / 2 , y ( x ) = y . (4) have unique solutions for all x ∈ R and y > 0 but not when y = 0. To see this, consider the problem dy dx = y 1 / 2 , y (0) = 0 . (5) It is not too hard to see that y ( x ) = 0 is a solution. But if we solve the DE as a separable DE: 1 y 1 / 2 dy dx = 1 , (6) 1 so that 2 y 1 / 2 = x + C, (7) so that the solution to the IVP y (0) = 0 is y ( x ) = 1 4 x 2 . (8) We have therefore produced two solutions to the IVP in (4). Obviously, there is existence but not uniqueness. The problem lies in the behaviour of y 1 / 2 at y = 0. Certainly the function g ( y ) = y 1 / 2 is continuous at y = 0 but it is not differentiable at y = 0. It turns out that differentiability is not necessarily required for existenceuniqueness but, in this case, the fact that the derivative g ′ ( y ) = y − 1 / 2 tends to infinity as y → + is responsible for the problem. More on this later. Let us now go back to the general IVP in (2) and integrate both sides from s = 0 to s = x : y ( x ) − y (0) = integraldisplay x x f ( y ( s ) , s ) ds, (9) which we’ll rewrite as y ( x ) = y + integraldisplay x x f ( y ( s ) , s ) ds, (10) Differentiating both sides with respect to x will give us Eq. (2), as it should. Eq. (10) is known as an integral equation , for obvious reasons. A solution of the IVP in (2) must satisfy Eq. (10). So why bother to rewrite our original IVP as an integral equation? The answer is that integral equations are much nicer to deal with in theoretical analysis – because integration is basically a summation procedure, one can derive inequalities involving absolute values. These will then lead to important theorems....
View
Full
Document
This note was uploaded on 09/20/2010 for the course AMATH 351 taught by Professor Sivabalsivaloganathan during the Spring '08 term at Waterloo.
 Spring '08
 SivabalSivaloganathan

Click to edit the document details