Understanding the Cauchy Problem with the Picard Theorem

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In summary, the Cauchy problem is an initial value problem where the goal is to solve for x'(t) = f(t,x) with the condition x(t0) = x0. The Picard Theorem states that given a point (x0, y0) and a function f(x,y) that is continuous and Lipschitz in y on a neighborhood of (x0, y0), there exists a unique function y(x) that satisfies y' = f(x,y) and y(x0) = y0. Picard Iterations is a method for solving initial value problems by finding a sequence of functions that converge to the solution. However, if f(x,y) is only continuous and not Lipschitz on
  • #1
JasonRox
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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.
 
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  • #2
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.
 
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  • #3
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.
 
  • #4
Another question...

Picard Iterations: general formula and running a couple of iterations.

What is that?
 
  • #5
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.
 
  • #6
its a limit process for finding successively better approximatiions to a solkution by iterating.
 
  • #7
HallsofIvy said:
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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1. What is Picard's and Peano's Theorem?

Picard's and Peano's Theorem is a fundamental theorem in mathematics that deals with the existence and uniqueness of solutions to differential equations. It was first studied by mathematicians Charles Émile Picard and Giuseppe Peano in the late 19th century.

2. How does Picard's and Peano's Theorem work?

The theorem states that if a differential equation has a continuous and well-behaved function, then there exists a unique solution that passes through a given point. This solution can be approximated by a sequence of functions that converges to the true solution.

3. What is the significance of Picard's and Peano's Theorem?

Picard's and Peano's Theorem is a powerful tool in the study of differential equations. It allows mathematicians to prove the existence and uniqueness of solutions to various types of differential equations, which are essential in many areas of science and engineering.

4. Can Picard's and Peano's Theorem be applied to all differential equations?

No, Picard's and Peano's Theorem can only be applied to differential equations with certain properties, such as being continuous and well-behaved. There are other theorems and techniques that can be used for more complex types of differential equations.

5. What are some real-world applications of Picard's and Peano's Theorem?

Picard's and Peano's Theorem has numerous applications in physics, engineering, and other scientific fields. It is used to model and analyze various physical phenomena, such as heat transfer, fluid dynamics, and population growth. It is also used in the design and optimization of complex systems, such as aircraft and chemical processes.

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