Thread: Partial differential equation

Hello
I have example to resolve:

where
Anyone to help me?

You can use these two beauties:

=

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.

I don't understand...Maybe this is not true.

JJacquelin how you get x and y?

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.

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.

- shooting fish in a barrel
I case study new method, which seems easier. I've done this example, but I got the another answer.



After that I talled t and L



If you understand this method, plese give me the answer it is correct?

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

Maybe I give in....or could you give me advice in which book I find this method or where (link)?

I understand your solution, thanks for your help, but I have one question, when I have to make a parametrization?

I couldn't give a general answer.
As many methods : try and see if it allows simplifications or not.

Ok, thank you