Hello

I have example to resolve:

$\displaystyle yu_{x}(x,y)+xu_{y}(x,y)=u(x,y)$

where $\displaystyle u(1,y)=y+1$

Anyone to help me?

Printable View

- Aug 27th 2012, 09:09 AManka0501Partial differential equation
Hello

I have example to resolve:

$\displaystyle yu_{x}(x,y)+xu_{y}(x,y)=u(x,y)$

where $\displaystyle u(1,y)=y+1$

Anyone to help me? - Aug 27th 2012, 09:59 AMMaxJasperRe: Partial differential equation
You can use these two beauties:

$\displaystyle u[x,y]\to -\frac{\left(-1+\sqrt{1-x^2+y^2}\right) \left(1+\sqrt{1-x^2+y^2}\right)}{x+\sqrt{y^2}}$ = $\displaystyle \frac{x^2-y^2}{x+\sqrt{y^2}}$

$\displaystyle u[x,y]\to -\frac{\left(x+\sqrt{y^2}\right) \left(-1+\sqrt{1-x^2+y^2}\right)}{1+\sqrt{1-x^2+y^2}}$ - Aug 28th 2012, 12:38 AMJJacquelinRe: Partial differential equation
Hi !

Changing the cartesian coordinates system to hyperbolic system leads to a very simple PDE. So, the general solution is obtained (in attachment).

Then the condition u(1,y)=1+y will be easy to apply in order to express the particular solution. - Aug 28th 2012, 12:40 AManka0501Re: Partial differential equation
I don't understand...Maybe this is not true.

- Aug 28th 2012, 12:46 AManka0501Re: Partial differential equation
JJacquelin how you get x and y?

- Aug 28th 2012, 01:04 AMJJacquelinRe: Partial differential equation
I quite not understand your question

x = rho*cosh(theta) and y = rho*sinh(theta)

The inverse is : rho = sqrt(x²-y²) and theta = argtanh(y/x)

This is the usual relationship between cartesian coordinates and hyperbolic coordinates. - Aug 28th 2012, 02:26 AMJJacquelinRe: Partial differential equation
By the way, a much simpler method consists in :

Let u(x,y) = (x+y)*v(x,y)

Binging back into the PDE leads to y*(dv/dx)+x*(dv/dy) = 0

which is very easy to solve. - Aug 28th 2012, 03:36 AManka0501Re: Partial differential equation
$\displaystyle yu_{x}(x,y)+xu_{y}(x,y)=0$

$\displaystyle \frac{dx}{y}=\frac{dy}{x}$

$\displaystyle C=\frac{1}{2}( x^{2}-y^{2})$- shooting fish in a barrel

I case study new method, which seems easier. I've done this example, but I got the another answer.

$\displaystyle \frac{dx}{dt}=y\\ \frac{dy}{dt}=x\\ \frac{dz}{dt}=z\\$

$\displaystyle x(0)=1\\y(0)=L\\z(0)=L+1\\$

$\displaystyle x(t)=yt+1\\y(t)=xt+L\\z(t)Le^{t}+1$

After that I talled t and L

$\displaystyle \\t=(x-1)/y\\L=y-\frac{x(x-1)}{y}$

$\displaystyle \\u(x,y)=e^{\frac{x-1}{y}}(y-\frac{x(x-1)}{y})+1\\$

$\displaystyle u(1,y)=y+1$

If you understand this method, plese give me the answer it is correct? - Aug 28th 2012, 03:50 AMJJacquelinRe: Partial differential equation
Sorry, in fact, I didn't read your method.

I just bing back your result u(x,y) into the PDE. It doesn't aggree.

Clue : The final solution of your problem is very simple : u(x,y) = x+y - Aug 28th 2012, 04:26 AManka0501Re: Partial differential equation
Maybe I give in....or could you give me advice in which book I find this method or where (link)?

- Aug 28th 2012, 05:55 AManka0501Re: Partial differential equation
I understand your solution, thanks for your help, but I have one question, when I have to make a parametrization?

- Aug 28th 2012, 06:37 AMJJacquelinRe: Partial differential equation
I couldn't give a general answer.

As many methods : try and see if it allows simplifications or not. - Aug 28th 2012, 09:09 AManka0501Re: Partial differential equation
Ok, thank you