Partial differential equation

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August 27th 2012, 09:09 AM
anka0501

Partial differential equation

Hello
I have example to resolve:
$yu_{x}(x,y)+xu_{y}(x,y)=u(x,y)$
where $u(1,y)=y+1$
Anyone to help me?
August 27th 2012, 09:59 AM
MaxJasper

Re: Partial differential equation

You can use these two beauties:

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

$u[x,y]\to -\frac{\left(x+\sqrt{y^2}\right) \left(-1+\sqrt{1-x^2+y^2}\right)}{1+\sqrt{1-x^2+y^2}}$
August 28th 2012, 12:38 AM
JJacquelin

1 Attachment(s)

Re: 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.
August 28th 2012, 12:40 AM
anka0501

Re: Partial differential equation

I don't understand...Maybe this is not true.
August 28th 2012, 12:46 AM
anka0501

Re: Partial differential equation

JJacquelin how you get x and y?
August 28th 2012, 01:04 AM
JJacquelin

Re: 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.
August 28th 2012, 02:26 AM
JJacquelin

Re: 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.
August 28th 2012, 03:36 AM
anka0501

Re: Partial differential equation

$yu_{x}(x,y)+xu_{y}(x,y)=0$
$\frac{dx}{y}=\frac{dy}{x}$
$C=\frac{1}{2}( x^{2}-y^{2})$ - shooting fish in a barrel
I case study new method, which seems easier. I've done this example, but I got the another answer.
$\frac{dx}{dt}=y\\ \frac{dy}{dt}=x\\ \frac{dz}{dt}=z\\$
$x(0)=1\\y(0)=L\\z(0)=L+1\\$
$x(t)=yt+1\\y(t)=xt+L\\z(t)Le^{t}+1$
After that I talled t and L
$\\t=(x-1)/y\\L=y-\frac{x(x-1)}{y}$
$\\u(x,y)=e^{\frac{x-1}{y}}(y-\frac{x(x-1)}{y})+1\\$
$u(1,y)=y+1$
If you understand this method, plese give me the answer it is correct?
August 28th 2012, 03:50 AM
JJacquelin

Re: 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
August 28th 2012, 04:26 AM
anka0501

Re: Partial differential equation

Maybe I give in....or could you give me advice in which book I find this method or where (link)?
August 28th 2012, 05:55 AM
anka0501

Re: Partial differential equation

I understand your solution, thanks for your help, but I have one question, when I have to make a parametrization?
August 28th 2012, 06:37 AM
JJacquelin

Re: Partial differential equation

I couldn't give a general answer.
As many methods : try and see if it allows simplifications or not.
August 28th 2012, 09:09 AM
anka0501

Re: Partial differential equation

Ok, thank you