# Thread: 2nd PDE for Exam

1. ## 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

2. 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.