# Thread: Existence of Piecewise Function

1. ## Existence of Piecewise Function

(1) Consider the initial value problem given by

y'=
1 if y>0
0 if y=0
1/y if y<0

y(0)=0

(a) According to the existence theorem, do solutions to this initial value problem exist?
(b) Can you construct a solution to the initial value problem? If yes, do so. If not, explain why not.
(c) Explain briefly why your answers in (1a) and (1b) don’t contradict anything.

For part a, I thought at first that since both F(t,y)=1,0,1/y and the partial derivative of each F with respect to y are continuous along each of the given intervals, then a solution must exist. But I know a piecewise function cannot be continuous if it has an asymptote, which 1/y is. I have no idea about piecewise functions for differential equations.

For part b, you can definitely find solutions to each of the functions, but I don't know if constitutes a solution of the whole initial value problem, considering the problems I encountered in part a.

Help please!

2. Originally Posted by veritaserum2002
(1) Consider the initial value problem given by

y'=
1 if y>0
0 if y=0
1/y if y<0

y(0)=0

(a) According to the existence theorem, do solutions to this initial value problem exist?
(b) Can you construct a solution to the initial value problem? If yes, do so. If not, explain why not.
(c) Explain briefly why your answers in (1a) and (1b) don’t contradict anything.

For part a, I thought at first that since both F(t,y)=1,0,1/y and the partial derivative of each F with respect to y are continuous along each of the given intervals, then a solution must exist. But I know a piecewise function cannot be continuous if it has an asymptote, which 1/y is. I have no idea about piecewise functions for differential equations.

For part b, you can definitely find solutions to each of the functions, but I don't know if constitutes a solution of the whole initial value problem, considering the problems I encountered in part a.

Help please!
It exists one and only one solution to the DE of the type $y^{'} = f(y)$ with the condition $y(0)=0$ You have proposed and it is $y=0$. There is no contradiction with the 'Cauchy criterion' that requires $f(x,y)$ to be continous with its partial derivatives of first order in $(x_{0},y_{0})$ because this criterion isn't necessary condition. In other words if the 'Cauchy criterion' is not satisfied the solutuion may exist or not and it may be only one or not... in this case You have one and only one solution...

Kind regards

$\chi$ $\sigma$