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I am investigating solutions of the initial value problem $y^{\prime}=f(x,y)=\frac{2y}{x}$, $y(x_0)=y_0$ using the Existence and Uniqueness Theorem (the one involving Picard-Lindelof).

Clearly, if $x_0 = 0$ then the discontinuity means that there is no solution to the problem. However, if $x_0 \neq 0$, then we could solve the ODE to obtain $y = Cx^2$ for some constant $C$. Now, this is the equation of a parabola, and there is clearly only one value of $C$ which will satisfy the initial conditions $y(x_0)=y_0$, that being $C_0 = \frac{y_0}{x_0^2}$.

My question is this: can we guarantee a unique solution to this initial value problem for all real $x$?

I would say this isn't true, but is this because the function $f(x,y)$ is not continuous at $x=0$? I know this theorem only provides a local statement.

Also, is there a reason why certain texts prefer to define the intervals in the theorem as open, for example $0 < |x - x_0| < a$, and other texts define them as closed, for example $|x - x_0| \leqslant a$? I have seen both of these being used, and have seen a proof of the theorem using the second closed interval definition, but would this proof be invalid, since the first definition seems to imply that $x$ cannot equal $x_0$?

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  • $\begingroup$ You're intuition is correct. Suppose we consider the initial value $y(0)=0$. Then $y=Cx^2$ is a solution for any $C$ you choose. Worse, if you consider the initial value $y(0)=1$ then there is no continuous solution to the ODE. $\endgroup$ – Matt Apr 11 '17 at 14:02
  • $\begingroup$ Sorry but what would be the meaning of an initial condition at $x_0=0$ for a differential equation $y'=f(x,y)$ such that $f(0,v)$ exists for no value of $v$? Even if one could extend solutions defined on $(0,x_1)$, for some $x_1>0$, to $[0,x_1)$ or even to $(x_2,x_1)$ for some $x_2<0$, these extended functions would not be solutions of the differential equation $y'=f(x,y)$. $\endgroup$ – Did Apr 11 '17 at 14:55
  • $\begingroup$ There wouldn't be any point in setting initial conditions like that, but I think Matt was just giving some examples to show that in these situations there exists no unique solution (in this case there can be infinitely many solutions, or no solution at all!) $\endgroup$ – wrb98 Apr 11 '17 at 15:37

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