# Existence of solution

• Sep 13th 2010, 08:10 PM
jzellt
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!!
• Sep 13th 2010, 08:52 PM
TheEmptySet
Quote:

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...

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

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

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

Where both $\displaystyle f(x,y)$ and $\displaystyle f_y(x,y)$ are continuous?
• Sep 13th 2010, 08:58 PM
jzellt
I guess I should've mentioned that Ive gotten this far. I don;t know what to do from here...
• Sep 13th 2010, 09:20 PM
TheEmptySet
Ask yourself this question what is the domain of $\displaystyle f(x,y)$ and what is the domain of $\displaystyle f_y(x,y)$?

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

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

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

That is in both of these domains? I bet you can.
• Sep 13th 2010, 11:11 PM
chisigma
The proposed DE has something 'not fully clear' because if we change the roles of x and y it becomes...

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

In that case the conditions of existence of a solution are not satisfied because $\displaystyle f(x,y)= \frac{1}{x\ \ln y}$ 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...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 14th 2010, 04:12 AM
chisigma
$\displaystyle y^{'} = x\ \ln y$ , $\displaystyle y(1)=1$ (1)
... has been 'solved': the solution is [simply...] $\displaystyle y=1$. The solution cannot be derived from the 'general solution' with an appropriate value of the 'arbitrary constant' and it is called 'singular solution'...
$\displaystyle \chi$ $\displaystyle \sigma$