# Thread: Uniqeness and Existence in ODE

1. ## Uniqeness and Existence in ODE

Hello,

I would like to know why is uniqeness defined as partial derivative of f(x,y) with respect to y? Here we are testing the condition if the derivative of f(x,y) is a continous function. I don't understand how this is connected.

Thank you.

2. ## Re: Uniqeness and Existence in ODE

Sorry, do I need to reformulate the question? Because this is not a complex question.

3. ## Re: Uniqeness and Existence in ODE

I have no clue what you are asking.

4. ## Re: Uniqeness and Existence in ODE

I think Nforce is referring to the basic "existence and uniqueness" theorem for differential equations as found in most introductory differential equations texts:

If f(x, y) is continuous in some neighborhood of $(x_0, y_0)$ then there exist a function, y(x), satisfying $\frac{dy}{dx}= f(x,y)$ with initial condition $y(x_0)= y_0$. If, further, f is differentiable with respect to y, then that solution is unique.

Actually, you don't need "differentiable"- Lipschitz (There exist an neighborhood of $(x_0, y_0)$ such that if $(x, y_1)$ and $(x, y_2)$ is in that neighborhood then $|f(x, y_1)- f(x, y_2)|\le c|y_1- y_2|$ for some c< 1.) is sufficient. Every differentiable function is Lipshitz but not the other way around.

To see why that is true, that requires a detailed look at its use in the proof of the "existence and uniqueness" theorem: look in a differentiable equation textbook. As an example look at $y'= y^{1/2}$ with initial condition y(0)= 0. Note that, here, $f(x, y)= y^{1/2}$ is continuous at x= 0 but $f_y(x, y)= (1/2)y^{-1/2}$ is not. If y is not identically 0, we can write that equation as $y^{-1/2}dy= dx$. Integrating both sides of the equation, $2y^{1/2}= x+ C$ or $y= \frac{(x+ C)^2}{4}$. When x= 0, we have $0= \frac{C^2}{4}$ so C= 0. That is, $y= \left(\frac{x}{4}\right)^2= \frac{x^2}{4}$ is a solution.

But it is obvious that the function y(x)= 0 for all x, that is, y identically 0, also satisfies the differential equation $y'= y^{1/2}$ as well as the condition that y(0)= 0. In fact. of we define y(x) as 0 for all $x\le x_0$, $y(x)= \frac{(x- x_0)^2}{16}$, that also satisfies the differential equation and the initial condition- there are, in fact, an infinite number of functions satisfying this differential equation and initial condition.