# Picard existence theorem and IVP

1. Apr 15, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Consider the initial value problem
\begin{align*} \left\{ \begin{array}{l} \displaystyle \frac{dy}{dx} = \exp(xy) \\ y(0) = 1 \end{array} \right. \end{align*}

1. Verify that this IVP has a unique solution in a neighborhood
of $$x = 0$$.

2. Following the notation of the lectures, find the values of $$K$$,
$$M$$, and $$\delta$$ that will work for this case.

2. Relevant equations

3. The attempt at a solution

1. Let $$\displaystyle f(x,y) = \frac{dy}{dx} = \exp(xy)$$. Then
$$\displaystyle \frac{df}{dy} = x \exp(xy)$$ which is continuous and
hence has upper bound $$K$$. Hence according to Picard's theorem the
IVP has a unique solution in $$\abs{x - x_0} \leq \delta$$.

2. How can I find the values of $$M$$, $$K$$ and $$\delta$$? Does $$M$$ mean the upper bound of function $$f(x,y)$$?

2. Apr 15, 2010

### ystael

We cannot possibly help you with this unless you explain "the notation of the lectures".

3. Apr 16, 2010

### complexnumber

I think $$\delta$$ relates to the neighborhood $$|x - x_0| < \delta$$ where the differential equation has a unique solution. $$K$$ is the constant in Lipschitz condition. $$M$$ is the upper bound of function $$f(x,y)$$.