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?
You can use these two beauties:
$\displaystyle u[x,y]\to -\frac{\left(-1+\sqrt{1-x^2+y^2}\right) \left(1+\sqrt{1-x^2+y^2}\right)}{x+\sqrt{y^2}}$ = $\displaystyle \frac{x^2-y^2}{x+\sqrt{y^2}}$
$\displaystyle u[x,y]\to -\frac{\left(x+\sqrt{y^2}\right) \left(-1+\sqrt{1-x^2+y^2}\right)}{1+\sqrt{1-x^2+y^2}}$
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 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.
$\displaystyle yu_{x}(x,y)+xu_{y}(x,y)=0$
$\displaystyle \frac{dx}{y}=\frac{dy}{x}$
$\displaystyle C=\frac{1}{2}( x^{2}-y^{2})$- 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?
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