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Some clarification needed about basic concepts

  1. Feb 16, 2009 #1
    Dear All,

    I have learned the Uniqueness and Existence theorem in last lecture, however, the instructor told us that the proof is omitted because it is beyond scope of the course.

    I am more concerned with the derivation of the Uniqueness theorem now.

    I need some clarification here.

    * As far as I understood, if f(x,y) satisfies the lipschitz condition, then the uniqueness theorem is valid. If fy is continuous on the rectangle, then there is unique solution to the equation; because continuity of partial y implies lipschitz condition to be valid.

    Moreover, in order to find a solution, instead of a diff. eq. we could write an integral equation and define an operator. (I'm not sure I know what an operator is, though. ) It turns out that we have an operator such as F(y) = y and F(y) - y = 0 => g(y) = F(y) - y and the values of y which make g(y) = 0 is also a solution to the diff eq.

    I have learned that this is called a fixed point and could be found by fixed point iteration (picard iteration).

    For this, I learned that we could write F(yn) = yn + 1 . This part is totally unclear to me; because I don't see why F(yn) = yn+1. We know that F(yn) = yn and saying F(yn) = yn+1 means we are making an error, aren't we ? I tend to think that this error is so small that we are neglecting it, but why is it so ?

    My priority now is to understand the part above.

    Moreover, I also wonder how do we know that those fixed points are in fact unique ?

    Regarding to uniqueness of fixed points, I don't need the proof in a fully mathematically described in cooperation with other theorems, because I have just started learning differential equations. I have been searching and found some theorems such as Banach's but was unable to understand those. I have some knowledge of calculus; and what I need is a sketch of the proof.
  2. jcsd
  3. Feb 19, 2009 #2


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    Yes, that is true. Strictly speaking the Lispschitz condition applies to functions of one variable (a function is said to be "Lipschitz" on set A if and only if there exist a number C such that [itex]|f(x_1)- f(x_0)|\le C|x_1- x_0|[/itex] for any x1 and x0 in that set.) and for differential equation dy/dx= f(x,y) we require "Lipschitz" in the y variable only. It is easy to show that "Lipschitz" on a set lies between "continuous on the set" and "differentiable" on the set: If a function is differentiable on a set then it must be Lipschitz there and if it is Lipschitz on a set it must be continuous at every point of that set.

    Yes, dy/dx= f(x,y), with condition [itex]y(x_0)= y_0[/itex] is equivalent to the integral equation [itex]y(x)= \int_{x_0}^x f(t, y(t))dt+ y_0[/itex] in the sense that any function y(x) that satisfies one must satisfy the other.

    An operator is just a "function" that works on functions rather than numbers: it changes one function into another. In the differential equation dy/dx= f(x,y), it is derivative that is the operator and in the integral equation [itex]y= \int_{x_0}^x f(t, y(t))dt+ y_0[/itex] it is the integral that is the operator.

    You may be mis-reading this- you are certainly mis-writing it. It is NOT F(yn)= yn+ 1 in the sense that we add 1 to yn. It is F(yn)= yn+1. In any case, it is NOT correct that "F(yn)= yn"; we get each y by applying F to the previous y. What is true is that, under certain conditions, If we have an iterative sequence in which we define yn+1= F(yn), in other words, each term in the sequence is F applied to the previous term, then F has a fixed point: there is a function y, NOT necessarily in the sequence and so NOT yn, such that F(y)= y. It is true that each of the yn in the sequence is NOT equal to the solution but, hopefully the sequence converges to it: each term is closer than the previous one.

    It might be helpful to look at an example. The differential equation dy/dx= y, with condition y(0)= 1, is equivalent to the integral equation [itex]y= \int_0^x y(t)dt+ 1[/itex] which I got just by "integrating" both sides of the differential equation, and then choosing the constant of integration to fit the condition.

    Now define the "iteration" [itex]y_{n+1}= \int_0^x y_n(t)dt+ 1[/itex]: each term in the sequence is that integration applied to the previous one. Of course, the first term in the sequence does not have "previous one" so that must be given. If we were lucky enough to choose the solution to the equation as the first term, applying that integral would just give the same thing again so we would have a "constant" sequence. But we don't know the solution. The one thing we really know about y is that y(0)= 1. Okay, the simplest function we could take then, is the constant function y(x)= 1 for all x. Putting that into the integral gives
    [tex]y_1= \int_0^x y_0(t)dt+ 1= \int_0^x 1dt+ 1= x+ 1[/tex]
    so now we have x+ 1 as y1(x). Repeating
    [tex]y_2= \int_0^x y_1(t)dt+ 1= \int_0^x (t+ 1)dt+ 1= \frac{1}{2}x^2+ x+ 1[/tex]
    so now we have y2(x)= (1/2)x2+ x+ 1. Repeating
    [tex]y_3= \int_0^x y_1(t)dt+ 1= \int_0^x(\frac{1}{2}t^2+ t+ 1)dt= \frac{1}{6}x^3+ \frac{1}{2}x^2+ x+ 1[/itex]
    So now we have y3(x)= (1/6)x3+ (1/2)x2+ x+ 1.

    Now, this is NOT a very good way of actually solving a differential equation- but if you have a good eye, you might notice that what is happening here is that we are getting more and more terms in a power series. Specifically, the power series for ex and you might guess that y(x)= ex is the limit of that sequence. It's not difficult to show that it is, and further to show that it is a "fixed point" for the operator: if y(x)= ex, then
    [tex]\int_0^x y(t)dt+ 1= \int_0^x e^t dt+ 1= e^x- e^0+ 1= e^x[/itex]
    Further, it is very easy to see that y= ex satisfies the differential equation, d(ex)/dx= ex= y, and the condition, y(0)= e0= 1.

    Again, y(x)= ex is NOT any "yn" in the sequence but is the limit of the sequence.

    Yes, Banach's fixed point theorem says "if S is a subset of a complete metric space and f is a contraction map from S to itself then there exist a unique point, x, in S such that f(x)= x (x is a "fixed point" of f)."

    I recently posted a proof of that, with commentary, on this board:

    A "contraction" map on a set is one such that there exist a constant c< 1 such that,for any points x and y on the set, [itex]|f(x)- f(y)|\le c|x- y|[/itex] (the absolute value |p- q| represents the distance between p and q. That says that the distance between f(x) and f(y) is less than the distance between x and y: it "contracts" distances.)

    Suppose f is a contraction map and x and y are fixed points: f(x)= x and f(y)= y.
    By definition of "contraction map", we must have [itex]|f(x)- f(y)|\le c|x- y|[/itex] for some c< 1. But f(x)= x and f(y)= y so that is [itex]|x-y|\le c|x-y|[/itex] or [itex](1-c)|x-y|\le 0[/itex]. c is, by definition, less than 1 so 1- c is positive and |x-y| cannot be negative. For that to be true we must have |x-y|= 0 or x= y.
    Last edited by a moderator: Feb 19, 2009
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