Existence and Uniqueness For ODE

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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.
 

Answers and Replies

  • #2
haruspex
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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
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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).
 
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haruspex
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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
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Then all looks ok except that the ODE cannot apply for y<1.
Sorry this is for uniqueness right? Thank you though.
 
  • #6
haruspex
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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
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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.
 

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