Solve:

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

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

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- Feb 3rd 2010, 02:46 AMryanhorneAdvection Equation
Solve:

*u*t + ((*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? - Feb 3rd 2010, 02:48 AMryanhorne
By the way ut, means differnatiate u with t, and ((x-1)u) is differentiated by x

- Feb 3rd 2010, 06:15 AMJester
If you directly substitute your solution form into the PDE you'll get

$\displaystyle

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

$

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

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

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

The actual solution of the PDE is $\displaystyle u = e^{-t}f\left((x-1)e^{-t}\right)$. - Feb 5th 2010, 09:58 AMryanhorne
You a Genius! Cheers