1. ## Existence of solution

Thm: Suppose that both the function f(x,y) and its partial Derivative fy(x,y) are continuous on some rectangle R in the xy plane that contains the point (a,b) in its interior. Then, for some open interval I containing the point a, the initial value problem has one and only one solution that is defined on the interval I.

I have to determine if the above Thm does or does not guarantee existence of a solution of:

dy/dx = x lny ; y(1) = 1

Can someone show how this is done? Thanks a lot!!

2. Originally Posted by jzellt
Thm: Suppose that both the function f(x,y) and its partial Derivative fy(x,y) are continuous on some rectangle R in the xy plane that contains the point (a,b) in its interior. Then, for some open interval I containing the point a, the initial value problem has one and only one solution that is defined on the interval I.

I have to determine if the above Thm does or does not guarantee existence of a solution of:

dy/dx = x lny ; y(1) = 1

Can someone show how this is done? Thanks a lot!!
This is a special case of the Picard–Lindelöf Theorem

You just need to verify the hypothesis of the theorem...

$\displaystyle \frac{dy}{dx}=f(x,y)=x\ln(y)$ and

$\displaystyle \frac{\partial f}{\partial y}=\frac{x}{y}$

The question is can you find a rectangle around the point $(1,1)$

Where both $f(x,y)$ and $f_y(x,y)$ are continuous?

3. I guess I should've mentioned that Ive gotten this far. I don;t know what to do from here...

4. Ask yourself this question what is the domain of $f(x,y)$ and what is the domain of $f_y(x,y)$?

The domain for $f(x,y)$ is
$\displaystyle \left\{ (x,y) \in \mathbb{R}^2\bigg| y > 0\right\}$

The domain for $f_y(x,y)$ is
$\displaystyle \left\{ (x,y) \in \mathbb{R}^2\bigg| y \ne 0\right\}$

Now the question is can you find a rectangle around the point $(1,1)$

That is in both of these domains? I bet you can.

5. The proposed DE has something 'not fully clear' because if we change the roles of x and y it becomes...

$\displaystyle \frac{dx}{dy}= \frac{1}{x\ \ln y}$ , $x(1)=1$ (1)

In that case the conditions of existence of a solution are not satisfied because $f(x,y)= \frac{1}{x\ \ln y}$ has a singularity in y=1. That 'anomaly' is confirmed by the fact that is we try to solve (1) we obtain as general solution...

$\displaystyle \frac{x^{2}}{2} = Li(y) + c$ (2)

... and it is well known that the function $Li(z)$ has a 'pole' in z=1... May be that all that requires further examination...

Kind regards

$\chi$ $\sigma$

6. The 'thriller' about the DE...

$y^{'} = x\ \ln y$ , $y(1)=1$ (1)

... has been 'solved': the solution is [simply...] $y=1$. The solution cannot be derived from the 'general solution' with an appropriate value of the 'arbitrary constant' and it is called 'singular solution'...

Kind regards

$\chi$ $\sigma$