2nd PDE for Exam

**woody198403** · September 13th 2010, 06:43 AM

Hi again,

$x^2u_x - y^2u_y = 0$ subject to $u(1,y) = F(y)$

Solution

$x = 1, y = s, u(s) = F(s)$

$\frac{dx}{dt} = x^2$

$\frac{dx}{x^2} = dt$

$\frac{-1}{x} = t +c$ . Sub in $x = 1$ we have $- 1 = 0 + c$ so

$\frac{-1}{x} = t - 1$

$x = \frac{1}{1 - t}$

Now $\frac{dy}{dt} = -y^2$

$\frac{dy}{y^2} = -dt$

$\frac{-1}{y^2} = -t + c$

but I don't understand this next line....

$\frac{-1}{y} = -t - \frac{1}{s}$ I don't know how $c = \frac{-1}{s}$

From there, $y = \frac{1}{t+ \frac{1}{s}}$

From $x$ and $y$ it can be shown that

$s = \frac{1}{\frac{1}{x} + \frac{1}{y} - 1}$

Then
$\frac{du}{dt} = 0, \ \ u = F(s)$ Why is $u = F(s)$ ?

so

$u(x,y) = F \Big( \frac{1}{\frac{1}{x} + \frac{1}{y} - 1} \Big) <br />$
$= F \Big(\frac{xy}{x+y - xy} \Big)$ 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

**Danny** · September 13th 2010, 03:43 PM

Let me put in some details that might help. In your problem you are actually transforming coordinates from $(x,y) \to (t,s)$ . So by the chain rule

$u_t = u_x x_t + u_y y_t$

Choosing $x_t = x^2$ , and $y_t = -y^2$ gives

$u_t = u_x x_t + u_y y_t = x^2 u_x - y^2 u_y = 0$ (from your PDE).

Thus, we are required to solve

$x_t = x^2,\;\;y_t = -y^2,\;\;\;u_t = 0$ (these are PDEs but independent of $s$ so really ODEs).

What about the IC $x = 1$ where $u = F(y).$ We need some kind of IC to go with this in the $(t,s)$ plane. Here we choose simple and let $t = 0$ and connect the two boundary curves via $y = s$ (as you've done).

Thus, solve

$x_t = x^2,\;\;y_t = -y^2,\;\;\;u_t = 0,\;\;(*)$

subject to when $t = 0$ then

$x = 1,\;\;y = s\;\;\;u = F(y) = F(s),\;\;(**)$ .

Remember when interating (*) we get functions of integration and not constants of integration. Use (**) to find these.

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