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?

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?

The 'Cauchy problem' is exactly what you give: the "initial value" problem: Solve x'(t)= f(t,x) with the condition x(t_{0})= x_{0}.

Picard's theorem is what you probably guessed:

Given any point in the plane, (x_{0}, y_{0}), and a function f(x,y), continuous on some neighborhood of (x_{0}, y_{0}) and Lipschitz in y on that neighborhood, then there exist a unique function y(x) satisfying y'= f(x,y) and y(x_{0})= y_{0}.

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)= y^{1/3} is continuous on any neighborhood of (0,0) but is not Lipschitz on such a neighborhood. dy/dx= y^{1/3} can be integrated as y^{-1/3}dy= dx so
(3/2)y^{2/3}= x+ C so y(x)= (2/3)(x+ C)^{3/2}. Taking C= 0, y(x)= (2/3)x^{3/2} 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.

Picard's method for solving an initial value problem really formed the basis for his proof.

Given dx/dt= f(x,t), x(t_{0})= X_{0}
for a sequence as follows.
x_{0}(t)= X_{0} (i.e. is a constant function)
Solve dx/dt= f(X_{0},t) by integrating to get x_{1}(t) (choosing the constant of integration so that x_{1}(t_{0})= X_{0})

Now solve dx/dt= f(x_{1}(t),t) by integrating. Iterate, getting a sequence of functions {x_{n}(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)t^{2}+ 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 e^{x} which is, of course, the solution to this problem.

This mimics the sequence method used in Picard's proof of his theorem.

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' = y^{1/5} , y(0) = 0 has only one solution?

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

(4x^{2}y)/(x^{4} + y^{2}), (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, y_{1}) - f(x, y_{2})|| <= N ||y_{1}-y_{2}|| (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.