# Picard's and Peano's Theorem

by JasonRox
Tags: peano, picard, theorem
HW Helper
PF Gold
P: 2,327
What are they? I can't find it on the internet or in my two different textbooks. It's not anywhere in the index.

Also, what is the Cauchy problem?

 Statement of the Picard Theorem on the existence and uniqueness of solution for the Cauchy problem x’ = f(t,x), x(t0) = x0 in case f(t,x) is continuous and Lipschitz in x.
That's from our review. I don't know what the Picard Theorem is (I think I know what it is just by reading the above.), but most importantly what is the Cauchy Problem?

Thanks.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 The 'Cauchy problem' is exactly what you give: the "initial value" problem: Solve x'(t)= f(t,x) with the condition x(t0)= x0. Picard's theorem is what you probably guessed: Given any point in the plane, (x0, y0), and a function f(x,y), continuous on some neighborhood of (x0, y0) and Lipschitz in y on that neighborhood, then there exist a unique function y(x) satisfying y'= f(x,y) and y(x0)= y0. A "neighborhood" of a point is an open set containing that point. A function, f(x), is "Lipschitz" on a set if and only if there exist a positive number C such that for any x, y in that set, |f(x)-f(y)|< C|x-y|. It's easy to see that if f(x) is Lipschitz on a set then it is continuous at every point of that set. You can use the mean value theorem to show that if a function is differentiable at every point of a set, then it is Lipschitz on the set (notice that, while "continuous" and "differentiable" are defined at points, Lipschitz is defined on a set). Many introductory differential equations texts say "differentiable with respect to y at every point of the set" which is "sufficient" (since that function must be Lipschitz) but not necessary since there exist functions that are Lipschitz on a set but not differentiable. If f(x,y) is continuous but not Lipschitz on a set, then there may be many functions satisfying the differential equation and "initial condition". For example, f(x,y)= y1/3 is continuous on any neighborhood of (0,0) but is not Lipschitz on such a neighborhood. dy/dx= y1/3 can be integrated as y-1/3dy= dx so (3/2)y2/3= x+ C so y(x)= (2/3)(x+ C)3/2. Taking C= 0, y(x)= (2/3)x3/2 satisfies that equation as well as y(0)= 0. But y(x)= 0 for all x also obviously satisfies that equation as well as y(0)= 0. In fact, given any a, y(x)= (2/3)(x-a)3/2 satisfies that equation as well as y(a)= 0. Taking any negative number b, the function defined as y(x)= (2/3)(x-b)3/2 if x< b, y(x), y(x)= 0 if b< x< a, y(x)= (2/3)(x-a)3/2 satisfies the differential equation (even at x=b and x= a) and y(0)= 0.
 HW Helper PF Gold P: 2,327 Thanks a lot. I appreciate it. It's what I guessed for the Picard's theorem. Now, I got the Cauchy problem out of the way too.
HW Helper
PF Gold
P: 2,327
Picard's and Peano's Theorem

Another question...

 Picard Iterations: general formula and running a couple of iterations.
What is that?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 Picard's method for solving an initial value problem really formed the basis for his proof. Given dx/dt= f(x,t), x(t0)= X0 for a sequence as follows. x0(t)= X0 (i.e. is a constant function) Solve dx/dt= f(X0,t) by integrating to get x1(t) (choosing the constant of integration so that x1(t0)= X0) Now solve dx/dt= f(x1(t),t) by integrating. Iterate, getting a sequence of functions {xn(t)} that, if it converges, converges to the solution to the initial value problem. For example, suppose the differential equation is dx/dt= x, x(0)= 1. 1) Solve dx/dt= 1, x(0)= 1: Integrating, x(t)= t+ C and x(0)= C= 1 so x(t)= t+ 1. 2) Solve dx/dt= t+ 1, x(0)= 1: Integrating, x(t)= (1/2)t2+ t+ C and x(0)= C= 1 so x(t)= (1/2)t^2+ t+ 1. 3) Solve dx/dt= (1/2)t^2+ t+ 1, x(0)= 1: Integrating, x(t)= (1/6)t^3+ (1/2)t+ t+ C and x(0)= C= 1 so x(t)= (1/6)t^3+ (1/2)t^2+ t+ 1. You see what's happening: with each iteration we get one more term in the Taylor's series for ex which is, of course, the solution to this problem. This mimics the sequence method used in Picard's proof of his theorem.
 Sci Advisor HW Helper P: 9,488 its a limit process for finding successively better approximatiions to a solkution by iterating.
P: 1
 Quote by HallsofIvy If f(x,y) is continuous but not Lipschitz on a set, then there may be many functions satisfying the differential equation and "initial condition". For example, f(x,y)= y1/3 is continuous on any neighborhood of (0,0) but is not Lipschitz on such a neighborhood. dy/dx= y1/3 can be integrated as y-1/3dy= dx so (3/2)y2/3= x+ C so y(x)= (2/3)(x+ C)3/2. Taking C= 0, y(x)= (2/3)x3/2 satisfies that equation as well as y(0)= 0. But y(x)= 0 for all x also obviously satisfies that equation as well as y(0)= 0. In fact, given any a, y(x)= (2/3)(x-a)3/2 satisfies that equation as well as y(a)= 0. Taking any negative number b, the function defined as y(x)= (2/3)(x-b)3/2 if x< b, y(x), y(x)= 0 if b< x< a, y(x)= (2/3)(x-a)3/2 satisfies the differential equation (even at x=b and x= a) and y(0)= 0.
Hi,
sorry about bringing an old post up, but i need help about this example. I've an exam this monday and I looked at last years', there were some questions about Cauchy-Lipschitz-Picard theorem with this example. Question 1 is:

Using "Existence and Uniqueness" theorem, is it possible to show that the problem y' = y1/5 , y(0) = 0 has only one solution?

Question 2 is the same but the function is:
f(x,y) is defined,

(4x2y)/(x4 + y2), (x,y) is not (0,0)
0, (x,y) = (0,0)

y' = f(x,y)
y(0) = 0

Since f(x,y) is continuous and differentiable, and it's derivative is continuous, I couldn't figure out how to show ||f(x, y1) - f(x, y2)|| <= N ||y1-y2|| (as it seems from the quote, first one is cannot be written like this?)

I'm sorry if i did sth wrong, but I don't have much time, so I just asked it here.

Thanks

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