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I am trying to learn to apply the uniqueness and existence theorem to concrete problems.

I have attached the statements of theorems at the end of post, in case to avoid any confusion.

Let $$y'=\sqrt{x^2+y^2}, y(0)=0$$ be the given differential equation.

I have the following question?

Does uniqueness and existence theorem tell(imply) that this ODE has solution or not?

To answer this question, let's look at the existence theorem. Note that $f(x,y)$ is nothing but norm of $(x,y)$ and it is standard result that norms are continuous.

So $f(x,y)$ is continuous. Since it is continuous over whole of $\mathbb{R}$, we can take any region around $(0,0)$ say $|x|\leq a, |y|\leq b$ and there will be atleast one solution due to the existence theorem.

Now I want to describe all points where IVP has a solution.

Since we know that function $f(x,y)$ is continuous on whole of $\mathbb{R}$, solution will exist for all values of $x$.

Last question is I want to use uniqueness theorem to comment on " what are all points where IVP has a unique solution"

How to solve it. I am not able to proceed?

Also please look at my work on question $1$ and $2$?


Below theorem concentrate on following problem, $$y'=f(x,y), y(x_0)=y_0$$

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1 Answer 1

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Existence Your existence theorem applies on any domain since $f$ is continuous everywhere.

Uniqueness Away from the origin $(x,y)=(0,0)$ it is clear that $f$ has continuous partial derivative with respect to $y$,

$$\frac{\partial f}{\partial y}(x,y) = \frac{y}{\sqrt{x^2 + y^2}},$$

however this is not continuous at the origin, and so your uniqueness theorem does not hold in any neighbourhood of the origin.

Note that uniqueness does hold on any domain not containing the origin. Hence a unique solution does exist on any domain not containing the origin.

See also the Picard-Lindelof theorem, where the condition on $f$ is usually that it is just Lipschitz, and continuous differentiability in $y$ is not required.

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  • $\begingroup$ What about my work on question $1$ and $2$. Have I done it correctly? $\endgroup$ Feb 8, 2019 at 15:06
  • $\begingroup$ yep, I have just added some more detail to the answer to cover those two points. $\endgroup$ Feb 8, 2019 at 15:10
  • $\begingroup$ The function $f$ is Lipschitz (similar to $|x|$ in one dimension), so there is uniqueness of solutions. $\endgroup$ Feb 8, 2019 at 16:09

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