• February 3rd 2010, 03:46 AM
ryanhorne
Solve:

ut + ((x-1)u)x=0, x E(0,1) & t > 0

I want to try the solution u(x,t) = f(x + ct), but think this might be incorrect, any ideas?
• February 3rd 2010, 03:48 AM
ryanhorne
By the way ut, means differnatiate u with t, and ((x-1)u) is differentiated by x
• February 3rd 2010, 07:15 AM
Jester
Quote:

Originally Posted by ryanhorne
Solve:

ut + ((x-1)u)x=0, x E(0,1) & t > 0

I want to try the solution u(x,t) = f(x + ct), but think this might be incorrect, any ideas?

If you directly substitute your solution form into the PDE you'll get

$
cf' +(x-1)f' + f = 0.
$

If you let $x = r - ct$ this becomes

$cf'(r) + (r - ct - 1)f'(r) + f(r) = 0$

Now differentiate wrt $t$ gives $f'(r) = 0$. So the only form is $f(r) =$constant.

The actual solution of the PDE is $u = e^{-t}f\left((x-1)e^{-t}\right)$.
• February 5th 2010, 10:58 AM
ryanhorne
You a Genius! Cheers