
2nd PDE for Exam
Hi again,
$\displaystyle x^2u_x  y^2u_y = 0$ subject to $\displaystyle u(1,y) = F(y)$
Solution
$\displaystyle x = 1, y = s, u(s) = F(s)$
$\displaystyle \frac{dx}{dt} = x^2$
$\displaystyle \frac{dx}{x^2} = dt$
$\displaystyle \frac{1}{x} = t +c$. Sub in $\displaystyle x = 1$ we have $\displaystyle  1 = 0 + c$ so
$\displaystyle \frac{1}{x} = t  1$
$\displaystyle x = \frac{1}{1  t}$
Now $\displaystyle \frac{dy}{dt} = y^2$
$\displaystyle \frac{dy}{y^2} = dt$
$\displaystyle \frac{1}{y^2} = t + c$
but I don't understand this next line....
$\displaystyle \frac{1}{y} = t  \frac{1}{s}$ I don't know how $\displaystyle c = \frac{1}{s}$
From there, $\displaystyle y = \frac{1}{t+ \frac{1}{s}}$
From $\displaystyle x$ and $\displaystyle y$ it can be shown that
$\displaystyle s = \frac{1}{\frac{1}{x} + \frac{1}{y}  1}$
Then
$\displaystyle \frac{du}{dt} = 0, \ \ u = F(s)$ Why is $\displaystyle u = F(s)$?
so
$\displaystyle u(x,y) = F \Big( \frac{1}{\frac{1}{x} + \frac{1}{y}  1} \Big)
$
$\displaystyle = F \Big(\frac{xy}{x+y  xy} \Big)$ Again I'm not sure where these values are coming from.
Is it just a simple substitution somewhere?
Any help would be really appreciated.
Thanks

Let me put in some details that might help. In your problem you are actually transforming coordinates from $\displaystyle (x,y) \to (t,s)$. So by the chain rule
$\displaystyle u_t = u_x x_t + u_y y_t$
Choosing $\displaystyle x_t = x^2$, and $\displaystyle y_t = y^2$ gives
$\displaystyle u_t = u_x x_t + u_y y_t = x^2 u_x  y^2 u_y = 0 $ (from your PDE).
Thus, we are required to solve
$\displaystyle x_t = x^2,\;\;y_t = y^2,\;\;\;u_t = 0$ (these are PDEs but independent of $\displaystyle s$ so really ODEs).
What about the IC $\displaystyle x = 1$ where $\displaystyle u = F(y). $ We need some kind of IC to go with this in the $\displaystyle (t,s)$ plane. Here we choose simple and let $\displaystyle t = 0$ and connect the two boundary curves via $\displaystyle y = s$ (as you've done).
Thus, solve
$\displaystyle x_t = x^2,\;\;y_t = y^2,\;\;\;u_t = 0,\;\;(*)$
subject to when $\displaystyle t = 0$ then
$\displaystyle x = 1,\;\;y = s\;\;\;u = F(y) = F(s),\;\;(**)$.
Remember when interating (*) we get functions of integration and not constants of integration. Use (**) to find these.