Existence and Uniqueness For ODE

In summary, the ODE is a equation that describes the change in y over time. It is continuous for all x∈R and all y∈R, and there is a solution for all (x0, y0) according to the theorem of existence. If f is continuous on an open rectangle that contains (x0, y0) then the IVP has at least one solution on some open subinterval of (a,b) that contains x0. If fy is continuous on the rectangle then there is a unique solution for all (x0, y0) except when y=1.
  • #1
ver_mathstats
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Homework Statement
1. For which initial points exists a solution in some interval containing x-naught?
2. For which initial points exists a unique solution in some interval containing x-naught?
Relevant Equations
Existence and Uniqueness
I'm new to learning about ODE's and I just want to make sure I am on the right track and understanding everything properly.

We have our ODE which is y' = 6x3(y-1)1/6 with y(x0)=y0.

I know that existence means that if f is continuous on an open rectangle that contains (x0, y0) then the IVP has at least one solution on some open subinterval of (a,b) that contains x0. Uniqueness is when f and fy are continuous on the rectangle then we will have a unique solution on some open subinterval of (a,b) that contains x0.

Here is my attempt at a solution for the questions:

1. y' = 6x3(y-1)1/6 is continuous for all x∈R and all y∈R, thus there is a solution for all (x0, y0) according to the theorem of existence.

2. Since fy = x3/(y-1)5/6 is continuous for all x∈R and all y∈R except for y = 1 we see there is a unique solution on some open interval containing x0 for all (x0, y0) except when y = 1.

Could someone please check this over to see if I have the right idea and if this is correct? Thank you.
 
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  • #2
ver_mathstats said:
if f is continuous
What is f in this context?
Some of your post reads as though you are considering some function f of x and y (continuous on a rectangle, etc.) but in the question there is only that y is a function of x.
 
  • #3
Sorry I should've been more clear y' = 6x3(y-1)1/6 and f(x,y) = 6x3(y-1)1/6 and then fy is the partial derivative. So "if f is continuous" pertains to f(x,y).
 
  • #4
ver_mathstats said:
Sorry I should've been more clear y' = 6x3(y-1)1/6 and f(x,y) = 6x3(y-1)1/6 and then fy is the partial derivative. So "if f is continuous" pertains to f(x,y).
Then all looks ok except that the ODE cannot apply for y<1.
 
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  • #5
haruspex said:
Then all looks ok except that the ODE cannot apply for y<1.
Sorry this is for uniqueness right? Thank you though.
 
  • #6
ver_mathstats said:
Sorry this is for uniqueness right? Thank you though.
No, for existence.
The ODE is only defined for y≥1, so you cannot really say that a solution exists for the point (0,0).
 
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  • #7
haruspex said:
No, for existence.
The ODE is only defined for y≥1, so you cannot really say that a solution exists for the point (0,0).
Oh okay, sorry I got a bit confused, I understand now. Thank you for the help.
 

FAQ: Existence and Uniqueness For ODE

1. What is the concept of existence and uniqueness for ODE?

The existence and uniqueness theorem for ordinary differential equations (ODEs) states that for a given initial value problem, there exists a unique solution that satisfies the differential equation and initial conditions. This means that there is only one possible solution for a specific set of initial conditions.

2. Why is the existence and uniqueness theorem important in the study of ODEs?

The existence and uniqueness theorem is important because it guarantees the existence of a solution to a given ODE and ensures that this solution is unique. This is crucial in many applications, as it allows us to confidently use ODEs to model real-world phenomena and make predictions based on the solutions.

3. What are the conditions for the existence and uniqueness of solutions to ODEs?

The existence and uniqueness theorem requires that the function defining the ODE is continuous and satisfies a Lipschitz condition. Additionally, the initial conditions must be well-defined and within the domain of the function.

4. Can the existence and uniqueness theorem be extended to systems of ODEs?

Yes, the existence and uniqueness theorem can be extended to systems of ODEs. In this case, the same conditions apply, but the solution is a vector function instead of a scalar function.

5. Are there any cases where the existence and uniqueness theorem does not hold for ODEs?

Yes, there are some cases where the existence and uniqueness theorem does not hold. For example, if the function defining the ODE is not continuous or does not satisfy the Lipschitz condition, the theorem may not hold. Additionally, if the initial conditions are not well-defined or are outside the domain of the function, the theorem may not hold. In these cases, alternative methods must be used to find solutions to the ODE.

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