# Thread: initial or boundary value problem

1. ## initial or boundary value problem

hello,

..............(y/5)-1,... y>=0
dy/dx = ........................,y(0)=2, y(x) in C_0[0,inf)
..............(-y/5)-1,... y<0

i am inclined to say this is an initial value problem becuase its to first order and the single condition is (obviously) defined at one (and the same) point, however the initilal condition is only valid for the first equation y'-y/5 -1 so that makes me think twice...

thanks

2. Originally Posted by pepsi
hello,

..............(y/5)-1,... y>=0
dy/dx = ........................,y(0)=2, y(x) in C_0[0,inf)
..............(-y/5)-1,... y<0

i am inclined to say this is an initial value problem becuase its to first order and the single condition is (obviously) defined at one (and the same) point, however the initilal condition is only valid for the first equation y'-y/5 -1 so that makes me think twice...

thanks
I presume you mean:

$\displaystyle \frac{dy}{dx}=\begin{cases} \frac{y}{5}-1,& y\ge 0\\ -\frac{y}{5}-1&y<0 \end{cases},\ \ y(0)=0,\ y(x) \in C_0([0,\infty))$

This is an initial value problem (where $\displaystyle y(x)$ changes sign the continuity of $\displaystyle y$ provides a new initial value for the solution to continue)

You are given the value of $\displaystyle y$ at $\displaystyle x=0$ and required to find the solution for all positive $\displaystyle x$.

CB

3. Originally Posted by pepsi
hello,

..............(y/5)-1,... y>=0
dy/dx = ........................,y(0)=2, y(x) in C_0[0,inf)
..............(-y/5)-1,... y<0

i am inclined to say this is an initial value problem becuase its to first order and the single condition is (obviously) defined at one (and the same) point, however the initilal condition is only valid for the first equation y'-y/5 -1 so that makes me think twice...

thanks
There is only one equation: y= f(x,y) where x is piecewise defined. The initial condition applies to both pieces. Solve y'= y/5- 1 with y(0)= 2 and then solve y'= -y/5- 1 with y(0)= 2. Because the initial condition occurs at the "break point", x= 0, between the two pieces, the solution will be continuous there.

4. Originally Posted by HallsofIvy
There is only one equation: y= f(x,y) where x is piecewise defined. The initial condition applies to both pieces. Solve y'= y/5- 1 with y(0)= 2 and then solve y'= -y/5- 1 with y(0)= 2. Because the initial condition occurs at the "break point", x= 0, between the two pieces, the solution will be continuous there.
The piecewise derivative definition gives different functional forms for the derivative when y is positive or negative. The function y(x) only takes values for x in [0,\infty) so there would be no point in defining the derivative for negative x.

CB